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Feynman V2 OCR

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Textbook by Richard Feynman, Robert Leighton and Matthew Sands (Addison-Wesley, 1964), based on the second-year Caltech lectures of 1962-63. The front matter shown includes Feynman's preface and the foreword, which describe a treatment of electricity and magnetism with vector field calculus, followed by chapters on elasticity and fluid flow. This is a downloaded copy of a published book, not Phil's own work.

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-U;IN-<2c/23%I;Z MAINLY ELECTROMAGNETISM AND MATTER RICHARD P.FEYNMAN Richard Chacc Tolman Professor ofTheoreticiil Physics California Institute ofTechnology ROBERT B.LEIGHTON Professor ofPhysics California Institute ofTechnology MATTHEW SAN DS Professor Stanford University OXNARD PUBLEC LIBRARY251soum AsmearOXNARD, CALIFORNIA 93030 v‘v ADDlSON—WE5l.EY PUBLISHING COMPANY, INC. READING, MASSACHUSETTS 'PALO ALTO 'LONDON Copyright ©1964 CALIFORNIA INSTITUTE OFTECHNOLOGY Printed intheUnited States ofAmerica ALL RIGHTS RESERVED. THIS BOOK, OR PARTS THEREOF MAY NOT BEREPRODUCED INANY FORM WITHOUT WRITTEN PERMISSION OFTHE PUBLISHER Library ofCongress Catalog Card No.63-20717 Second pmntzng—N01'ember, 1.964 Feynman ’sPreface These arethelectures inphysics thatIgave lastyear andtheyear before tothe freshman and sophomore classes atCaltech. The lectures are, ofcourse, not verbatim——they have been edited, sometimes extensively andsometimes lessso. The lectures form only part ofthecomplete course. The whole group of180 students gathered inabiglecture room twice aweek tohear these lectures and then they broke upinto small groups of15to20students inrecitation sections under theguidance ofateaching assistant. Inaddition, there wasalaboratory session onceaweek. Thespecial problem wetried togetatwith these lectures wastomaintain the interest ofthevery enthusiastic andrather smart students coming outofthehigh schools andintoCaltech. They have heard alotabout how interesting andexcit- ingphysics lS—~—Ih€ theory ofrelativity, quantum mechanics, and other modern ideas. Bytheendoftwoyears ofourprevious course, many would bevery dis- couraged because there were really very fewgrand, new, modern ideas presented tothem. They were made tostudy inclined planes, electrostatics. andsoforth, andafter twoyears itwasquite stultifying. Theproblem waswhether ornotwe could make :1course which would save themore advanced andexcited student by maintaining hisenthusiasm. Thelectures here arenotinanywaymeant tobeasurvey course, butarevery serious. 1thought toaddress them tothemost intelligent intheclass andtomake sure, ifpossible, that even themost intelligent student wasunable tocompletely encompass everything thatwasinthelectures—by putting insuggestions ofappli- cations oftheideas andconcepts invarious directions outside themain lineof attack. Forthisreason, though, Itried very hard tomake allthestatements as accurate aspossible, topoint outinevery case where theequations andideas ﬁtted intothebody ofphysics, and how—when they learned more~things would be modiﬁed. lalso feltthat forsuch students itisimportant toindicate what itis thattheyshould—if they aresiifﬁciently c1ever—be able tounderstand bydeduc- tionfrom what hasbeen said before, andwhat isbeing putinassomething new. When newideas came in,Iwould tryeither todeduce them ifthey were deducible, ortoexplain that itwasanewideawhich hadn’t anybasis interms ofthings they hadalready learned andwhich wasnotsupposed tobeprovable—but wasJust added in Atthestartofthese lectures. lassumed thatthestudents knew something when theycame outofhigh schoolAsuch things asgeometrical optics, simple chemistry ideas, andsoon.lalsodidn’t seethat there wasanyreason tomake thelectures 3 inadeﬁnite order, inthesense thatIwould notbeallowed tomention something untilIwasready todiscuss itindetail. There wasagreat dealofmention ofthings tocome, without complete discussions. These more complete discussions would come later when thepreparation became more advanced. Examples arethedis- cussions ofinductance, and ofenergy levels, which areatﬁrst brought inina veryqualitative wayandarelater developed more completely. Atthesame time that Iwasaiming atthemore active student, Ialso wanted totake care ofthefellow forwhom theextra ﬁreworks andsideapplications are merely disquieting andwho cannot beexpected tolearn most ofthematerial in thelecture atall.Forsuch students Iwanted there tobeatleast acentral core or backbone ofmaterial which hecould get. Even ifhedidn’t understand everything inalecture, Ihoped hewouldn’t getnervous. Ididn’t expect himtounderstand everything, butonly thecentral andmost direct features. Ittakes, ofcourse, a certain intelligence onhispart tosecwhich arethecentral theorems andcentral ideas, andwhich arethemore advanced sideissues andapplications which hemay understand onlyinlateryears. Ingiving these lectures there wasoneserious difficulty: inthewaythecourse wasgiven, there wasn’t anyfeedback from thestudents tothelecturer toindicate howwellthelectures were going over. This isindeed avery serious difﬁculty, andIdon’t know howgood thelectures really are.Thewhole thing wasessentially anexperiment. And ifIdiditagain Iwouldn’t doitthesame way—I hope I don’t have todoitagain! Ithink, though, that things worked out——so farasthe physics isconcerned—quite satisfactorily intheﬁrstyear. Inthesecond year Iwasnotsosatisﬁed. Intheﬁrstpart ofthecourse, dealing with electricity andmagnetism, Icouldn’t think ofanyreally unique ordifferent way ofdoing it—of anyway that would beparticularly more exciting than the usual way ofpresenting it.SoIdon‘t think Ididvery much inthelectures on electricity andmagnetism. Attheendofthesecond year Ihadoriginally intended togoon,after theelectricity andmagnetism, bygiving some more lectures onthe properties ofmaterials, butmainly totake upthings like fundamental modes, solutions ofthediffusion equation, vibrating systems, orthogonal functions, ... developing theﬁrststages ofwhat areusually called “the mathematical methods of physics.” Inretrospect, Ithink that ifIwere doing itagain Iwould goback to that original idea. Butsince itwasnotplanned that Iwould begiving these lec- tures again, itwassuggested thatitmight beagood ideatotrytogiveanintroduc- tiontothequantum mechanics—what youwillﬁndinVolume III. Itisperfectly clear that students who willmajor inphysics canwait until their third year forquantum mechanics. Ontheother hand, theargument wasmade that many ofthestudents inourcourse study physics asabackground fortheir primary interest inother ﬁelds. And theusual way ofdealing with quantum mechanics makes thatSLlII)_|6Ci almost unavailable forthegreat majority ofstudents because they have totake solong tolearn it.Yet, initsrealapplications—espe- cially initsmore complex applications, such asinelectrical engineering andchem- istry-—the fullmachinery ofthediﬁerential equation approach isnotactually used. SoItried todescribe theprinciples ofquantum mechanics inawaywhich wouldn’t require that oneﬁrstknow themathematics ofpartial diﬁerential equa- tions. Even foraphysicist Ithink that isaninteresting thing totrytodo—t0 present quantum mechanics inthis reverse fashion——for several reasons which maybeapparent inthelectures themselves. However, Ithink thattheexperiment inthequantum mechanics part was notcompletely successful—in large part because Ireally didnothave enough time attheend(Ishould, forinstance, have hadthree orfourmore lectures inorder todealmore completely withsuchmatters asenergy bands andthespatial dependence ofamplitudes). Also, Ihadnever presented thesubject thisway before, sothelack offeedback was particularly serious. Inow believe thequantum mechanics should begiven atalater time. Maybe I’llhave achance todoitagain someday. Then I’lIdoitright. Thereason there arenolectures onhow tosolve problems isbecause there were recitation sections. Although Ididputinthree lectures intheﬁrstyear onhow to solve problems, they arenotincluded here. Also there wasalecture oninertial 4 guidance which certainly belongs after thelecture onrotating systems, butwhich was, unfortunately, omitted. The ﬁfth and sixth lectures areactually due to Matthew Sands, asIwasoutoftown. ' Thequestion, ofcourse, ishow well thisexperiment hassucceeded. Myown point ofVlCW—~WI'1lCII, however, does notseem tobeshared bymost ofthepeople whoworked with thestudents~is pessimistic. Idon't think Ididvery wellbythe students. When llook atthewaythemajority ofthestudents handled theproblems ontheexaminations, Ithink that thesystem isafailure Ofcourse, myfriends point outtomethatthere were oneortwodozen students who——very surprisingly —understood almost everything inallofthelectures, andwho were quite active inworking with thematerial andworrying about themany points inanexcited andinterested way. These people have now, Ibelieve, aﬁrst-rate background in physics—and they are,after all,theones Iwastrying togetat.Butthen, “The power ofinstruction isseldom ofmuch efﬁcacy except inthose happy dispositions where itisalmost superﬂuous "(Gibbons) Still, Ididn‘t want toleave anystudent completely behind, asperhaps Idid. Ithink onewaywecould help thestudents more would bebyputting more hard work intodeveloping asetofproblems which would elucidate some oftheideas inthelectures. Problems give agood opportunity toﬁlloutthematerial ofthe lectures andmake more realistic, more complete, andmore settled inthemind theideas thathave beenexposed. Ithink, however, that there isn’t anysolution tothisproblem ofeducation other than torealize thatthebestteaching canbedone only when there isadirect individual relationship between astudent andagood teacher——a situation inwhich thestudent discusses theideas, thinks about thethings. andtalks about thethings. It’simpossible tolearn very much bysimply sitting inalecture, oreven bysimply doing problems that areassigned. Butinourmodern times wehave somany students toteach thatwehave totrytoﬁndsome substitute fortheideal. Perhaps mylectures canmake some contribution. Perhaps insome small place where there areindividual teachers andstudents, they may getsome inspiration orsome ideas from thelectures. Perhaps they willhave funthinking them through-—or going ontodevelop some oftheideas further. RICHARD P.FEYNMAN June, I963 5 Foreword Forsome forty years Richard P.Feynman focussed hiscuriosity onthemys- terious workings ofthephysical world, andbent hisintellect tosearching outthe order initschaos. Now, hehasgiven twoyears ofhisability andhisenergy to hisLectures onPhysics forbeginning students. For them hehasdistilled the essence ofhisknowledge, andhascreated interms they canhope tograsp a picture ofthephysicist’s universe. Tohislectures hehasbrought thebrilliance andclarity ofhisthought, theoriginality andvitality ofhisapproach, andthe contagious enthusiasm ofhisdelivery. Itwasajoytobehold. The ﬁrst year‘s lectures formed thebasis fortheﬁrst volume ofthis setof books. Wehave tried inthisthesecond volume tomake some kind ofarecord ofapart ofthesecond year’s lectures—which were given tothesophomore class during the1962-1963 academic year. Therestofthesecond year’s lec- tureswillmake upVolume III. Ofthesecond year oflectures, theﬁrst two-thirds were devoted toafairly complete treatment ofthephysics ofelectricity andmagnetism. Itspresentation wasintended toserve adual purpose. Wehoped, ﬁrst, togivethestudents a complete view ofoneofthegreat chapters ofphysics—from theearly gropings ofFranklin, through thegreat synthesis ofMaxwell, ontotheLorentz electron theory ofmaterial properties, andending with thestillunsolved dilemmas of theelectromagnetic self-energy. And wehoped, second, byintroducing atthe outset thecalculus ofvector ﬁelds, togive asolid introduction tothemathe- matics ofﬁeld theories Toemphasize thegeneral utility ofthemathematical methods, related subjects from other parts ofphysics were sometimes analyzed together with their electric counterparts. Wecontinually tried todrive home thegenerality ofthemathematics. (“The same equations have thesame solu- tions.”) And weemphasized thispoint bythekinds ofexercises andexamina- tionswegavewiththecourse. Following theelectromagnetism there aretwochapters each onelasticity and ﬂuid ﬂow. Intheﬁrstchapter ofeach pair, theelementary andpractical aspects aretreated. Thesecond chapter oneach subject attempts togiveanoverview of thewhole complex range ofphenomena which thesubject canleadto.These fourchapters canwellbeomitted without serious loss,since theyarenotatalla necessary preparation forVolume III. Thelastquarter, approximately, ofthesecond year wasdedicated toanintro- duction toquantum mechanics. Thismaterial hasbeen putintothethird volume. Inthisrecord oftheFeynman Lectures wewished todomore than provide a transcription ofwhat wassaid. Wehoped tomake thewritten version asclear anexposition aspossible oftheideas onwhich theoriginal lectures were based Forsome ofthelectures thiscould bedone bymaking only minor adjustments ofthewording intheoriginal transcript. Forothers ofthelectures amajor re- working and rearrangement ofthematerial was required. Sometimes wefelt weshould addsome new material toimprove theclarity orbalance ofthepres- entation. Throughout theprocess webeneﬁtted from thecontinual help and advice ofProfessor Feynman The translation ofover 1,000,000 spoken words into acoherent text ona tightschedule isaformidable task, particularly when itisaccompanied bythe 7 other onerous burdens which come with theintroduction ofanew course— preparing forrecitation sections, andmeeting students, designing exercises and examinations, and grading them, and soon. Many hands—and heads—-were involved. Insome instances wehave, Ibelieve, been able torender afaithful image—or atenderly retouched portrait—-of theoriginal Feynman. Inother instances wehave fallen farshort ofthisideal. Oursuccesses areowed toall those whohelped. Thefailures, weregret. Asexplained indetail intheForeword toVolume I,these lectures were but oneaspect ofaprogram initiated andsupervised bythePhysics Course Revision Committee (R.B.Leighton, Chairman, H.V.Neher, andM.Sands) atthe California Institute ofTechnology, andsupported ﬁnancially bytheFord Foun- dation. Inaddition, thefollowing people helped with oneaspect oranother of thepreparation oftextual material forthis second volume: T.K.Caughey, M.L.Clayton, J.B.Curcio, J.B.Hartle, T.W.H.Harvey, M.H.Israel, W.J.Karzas, R.W.Kavanagh, R.B.Leighton, J.Mathews, M.S.Plesset, F.L.Warren, W.Whaling, C.H.Wilts, and B.Zimmerman. Others con- tributed indirectly through their work onthecourse: J.Blue, G.F.Chapline, M.I.Clauser, R.Dolen, H.H.Hill, andA.M.Title. Professor Gerry Neuge- bauer contributed inallaspects ofourtask with adiligence and devotion far beyond thedictates ofduty. The story ofphysics youﬁndhere would, however, nothave been, except for theextraordinary ability andindustry ofRichard P.Feynman. MATTHEW S/mos March, I964 8 Contents CHAPTER 1.ELECTROMAGNETISM CHAPTER 6.THEEi.Ec'rRic FiEu> INVARious b-lb-lt—l ,.. CHAPTER 2.DIFFERENTIAL CALCULUS OFVEcToR FIELDS IQYOIQNNIO(AL NM7-‘bié\lJl-LUll\Jl—* '~»l\)r—~ ®\lO\Electrical forces l-1 Electric andmagnetic ﬁelds 1-3 Characteristics ofvector ﬁelds 1-4 Thelaws ofelectromagnetism 1-5 What aretheﬁelds? 1-9 Understanding physics 2-1 Scalar andvector ﬁelds—T andh2-2 Derivatives ofﬁelds—the gradient 2-4 Theoperator V2-6 Operations withV2-7 Thedilferential equation ofheatﬂow2-8 Second derivatives ofvector ﬁelds 2-9 CHAPTER 3.VEcToR INTEGRAL CALCULUS UQLBUQLRU-3UILb-I U3G\ Lb) U)OO~lUJIQVector integrals; thelineintegral ofV\1/3-1 Theﬁuxofavector ﬁeld3-2 Theﬂuxfrom acube; Gauss’ theorem 3-4 Heat conduction; thediffusion equation 3-6 Thecirculation ofavector ﬁeld3-8Electromagnetism inscience andtechnology 1-106-1 6-2 6-3 6-4 6-5 6-6 6-7 6-8 6-9CIRCUMSTANCES Equations oftheelectrostatic potential 6-l Theelectric dipole 6-2 Remarks onvector equations 6-4 Thedipole potential asagradient 6-4 Thedipole approximation foranarbitrary distribution 6-6 Theﬁelds ofcharged conductors 6-8 Themethod ofimages 6-8 Apoint charge near aconducting plane 6-9 Apoint charge near aconducting sphere 6-10 6-10 Condensers; parallel plates 6-11 6-11 High-voltage breakdown 6-13 6-12 Theﬁeld-emission microscope 6-14 \I\I[\)>—* 7-3 7-4 7-5Pitfalls 2_l1 CHAPTER 7.THEEi.EcTR1c FIELD INVARIovs CIRCUMSTANCES (Continued) Methods forﬁnding theelectrostatic ﬁeld7-1 Two-dimensional ﬁelds; functions ofthecomplex variable 7-2 Plasma oscillations 7-5 Colloidal particles inanelectrolyte 7-8 Theelectrostatic ﬁeldofagrid7-10 Thecirculat'o d -1naroun asquare’ CHAPTER 8.Ei.EcTRosTAT1c ENERGYStokes‘ theorem 3-9 Curl freeanddiver ence-free ﬁelds 3-10 8-1 Theelectrostatic energy ofcharges. Auniform — - 2Summary 3-11 CHAPTER 4.ELEcTRos'rATics 4-1 4-2 4-3 4-4 4-5 4-6 4-7 4-8Statics 4-1 Coulomb's law;superposition 4-2 Electric potential 4-4 E=-V4,4-68-2 G>O0®(»O\U\-Lusphere 8-1 Theenergy ofacondenser. Forces oncharged conductors 8-2 Theelectrostatic energy ofanionic crystal 8-4 Electrostatic energy innuclei 8-6 Energy intheelectrostatic ﬁeld8-9 Theenergy ofapoint charge 8-12 Theﬂuxof 4'7 CHAPTER 9.ELECTRICITY INTHEATMOSPHEREGauss’ law;divergence ofE4-9 Field ofasphere ofcharge 4-10 Field lines; equipotential surfaces 4-11 CHAPTER 5.APPLICATION OFGAuss’ LAw 'JlL!l'~Jl'~ItLIl\I|'~Jt\IlUlU|<4»Lt.»[Qt-IElectrostatics isGauss’s lawplus ...5-1 Equilibrium inanelectrostatic ﬁeld5-1 —Equilibrium withconductors 5-2 LII '—'\OUJStability ofatoms 5-39-1 \O\O\O\O@O\U\-Lb-lb)Theelectric potential gradient ofthe atmosphere 9-1 Electric currents intheatmosphere 9-2 Origin oftheatmospheric currents 9-4 Thunderstorms 9-5 Themechanism ofcharge separation 9-7 Lightning 9-10 ThﬁIdf1.h _ CHAPTER 10.DIELECTRICSee0ainec arge5 3 Asheet ofcharge; twosheets 5-4 Asphere ofcharge; aspherical shell 5-4 Istheﬁeldofapoint charge exactly 1/r2?5-5 Theﬁelds ofaconductor 5-7 0Theﬁeldinacavity ofaconductor 5-810-1 10-2 10-3 10-4 10-5Thedielectric constant 10-1 Thepolarization vector P10-2 Polarization charges 10-3 Theelectrostatic equations withdielectrics 10-6 Fields andforces withdielectrics 10-7 CHAPTER ll. INSIDE DIELECTRICS CHAPTER 17. THE LAws OFINr>ucTi0N ll-1 Molecular dipoles ll-1 17-1 11-2 Electronic polarization 11-1 17-2 11-3 Polar molecules; orientation polarization 11-3 17-3 11-4 Electric ﬁelds incavities ofadielectric 11-5 11-5 Thedielectric constant ofliquids; theClausius- 17-4 Mossotti equation 11-6 17-5 11-6 Solid dielectrics ll-8 17-6 11-7 Ferroelectricity; BaTiO3 11-8 17-7 17-8Thephysics ofinduction 17-1 Exceptions tothe“ﬁux rule” 17-2 Particle acceleration byaninduced electric ﬁeld; thebetatron 17-3 Aparadox 17-5 Alternating-current generator 17-6 Mutual inductance 17-9 Self-inductance 17-11 Inductance andmagnetic energy 17-12 CHAPTER 12. Ei.EcTRosTATic ANALOGS 12-1 Thesame equations have thesame solutions 12-1 12-2 Theﬂowofheat; apoint source nearaninﬁnite plane boundary 12-2 12-3 Thestretched membrane 12-5 I2-4 Thediffusion ofneutrons; auniform spherical source inahomogeneous medium 12-6 12-5 Irrotational ﬁuid ﬂow; theﬂowpastasphere 12-8 12-6 Illumination; theuniform lighting ofaplane 12-10 12-7 The“underlying unity” ofnature 12-12 CHAPTER 13.MAoNETosTAT1cs 13-1 Themagnetic ﬁeld 13-1 13-2 Electric current; theconservation ofcharge 13-1 13-3 Themagnetic force onacurrent 13-2 13-4 Themagnetic ﬁeldofsteady currents; Ampere’s law13-3 13-5 The magnetic ﬁeld ofastraight wire and ofa solenoid; atomic currents 13-5 13-6 Therelativity ofmagnetic andelectric ﬁelds 13-6 13-7 Thetransformation ofcurrents andcharges 13-11 13-8 Superposition; theright-hand rule13-1 1 CHAPTER 14. THE‘ MAoNETic FIELD INVARIoos SiTuATioNs 14-1 Thevector potential 14-1 14-2 Thevector potential ofknown currents 14-3 14-3 Astraight wire 14-4 14-4 Alongsolenoid 14-5 14-5 Theﬁeld ofasmall loop; themagnetic dipole 14-7 14-6 Thevector potential ofacircuit 14-8 14-7 ThelawofBiotandSavart 14-9 CHAPTER 15. THE VEcToR POTENTIAL 15-1 Theforces onacurrent loop; energy of adipole 15-1 15-2 Mechanical andelectrical energies 15-3 15-3 Theenergy ofsteady currents 15-6 15-4 Bversus A15-7 15-5 Thevector potential andquantum mechanics 15-8 15-6 What istrueforstatics isfalse fordynamics 15-14 CHAPTER 16. INDUCED CuRRENTs 16-1 Motors andgenerators 16-1 16-2 Transformers andinductances 16-4 16-3 Forces oninduced currents 16-5 16-4 Electrical technology 16-8 10CHAPTER 18. THE MAxwELi. EQuATioNs 18-1 Maxwell’s equations 18-1 18-2 How thenewterm works 18-3 18-3 Allofclassical physics 18-5 18-4 Atravelling ﬁeld 18-5 18-5 Thespeed oflight 18-8 18-6 Solving Maxwell's equations; thepotentials andthe wave equation 18-9 CHAPTER 19.THEPRiNciPLE orLEAsT AcT1oN Aspecial lecture—almost verbatim 19-1 Anoteadded after thelecture 19-14 CHAPTER 20. Soi.uTioNs orMAxwELi.‘s EQuATioNs INFREE SPAcE 20-1 Waves infreespace; plane waves 20-1 20-2 Three-dimensional waves 20-8 20-3 Scientiﬁc imagination 20-9 20-4 Spherical waves 20-12 CHAPTER 21. SoLuTioNs OFMAxwELL’s EQuATioNs WITH CuRRENTs ANDCHARoEs 21-1 Light andelectromagnetic waves 21-1 21-2 Spherical waves from apoint.source 21-2 21-3 Thegeneral solution ofMaxwell’s equations 21-4 21-4 Theﬁelds ofanoscillating dipole 21-5 21-5 The potentials ofamoving charge; thegeneral solution ofLiénard andWiechert 21-9 21-6 The potentials foracharge moving with constant velocity; theLorentz formula 21-12 CHAPTER 22.ACCiRcutTs 22-1 Impedances 22-1 22-2 Generators 22-5 22-3 Networks ofideal elements; Kirchhoff’s rules 22-7 22-4 Equivalent circuits 22-10 22-5 Energy 22-1 1 22-6 Aladder network 22-12 22-7 Filters 22-14 22-8 Other circuit elements 22-16 CHAPTER 23. CAviTY REsoNAToRs 23-1 Real circuit elements 23-1 23-2 Acapacitor athigh frequencies 23-2 23-3 Aresonant cavity 23-6 23-4 Cavity modes 23-9 23-5 Cavities andresonant circuits 23-10 CHAPTER 24.WAVEGUIDES CHAPTER 30.THE INTERNAL GEoMETRY orCRYsTAi.s 24-1 Thetransmission line24-1 24-2 Therectangular waveguide 24-4 24-3 Thecutoff frequency 24-6 24-4 Thespeed oftheguided waves 24-7 24-5 Observing guided waves 24-7 24-6 Waveguide plumbing 24-8 24-7 Waveguide modes 24-10 24-8 Another wayoflooking attheguided waves 24-10 CHAPTER 25.ELECTRODYNAMICS INRELATivisT1c30-1 3O-2 30-3 30-4 30-5 30-6 30-7 30-8 30-9Theinternal geometry ofcrystals 30-1 Chemical bonds incrystals 30-2 Thegrowth ofcrystals 30-3 Crystal lattices 30-3 Symmetries intwodimensions 30-4 Symmetries inthree dimensions 30-7 Thestrength ofmetals 30-8 Dislocations andcrystal growth 30-9 TheBragg-Nye crystal model 30-10 CHAPTER 31.TENsoRsNoTATioN 31-1 Thetensor ofolarizab'1it 1-25-1 Four-vectors 25-1 .P 1y3131-2 Transforming thetensor components 31-325-2 Thescalar product 25-3 ... . . 31-3 Theenergy ellipsoid 31-3 25% Thefoupdlmenslonal gradient 25-6 31-4 Other tensors‘ thetensor ofinertia 31625-4 Electrodynamics infour-dimensional notation 25-8 ' 25-5 Thefour-potential ofamoving charge 25-931-5 Thecross product 31-8 _5 _ 25-6 Theinvariance oftheequations of 31 Thetensor ofstress 319. 31-7 Tensors ofhigher rank 31-11electrodynamics 25-10 CHAPTER 26. LoRENTz TRANSFORMATIONS OFTHE FiELDs 26-1 Thefour-potential ofamoving charge 26-1 26-2 Theﬁelds ofapoint charge with aconstant velocity 26-2 26-3 Relativistic transformation oftheﬁelds 26-5 26-4 Theequations ofmotion inrelativistic notation 26-1 1 CHAPTER 27. FIELD ENERGY AND FiELD MOMENTUM 27-1 Local conservation 27-1 27-2 Energy conservation andelectromagnetism 27-2 27-3 Energy density andenergy ﬂow inthe electromagnetic ﬁeld27-3 27-4 Theambiguity oftheﬁeldenergy 27-6 27-5 Examples ofenergy ﬂow27-6 27-6 Field momentum 27-9 CHAPTER 28.ELEcTRoMAGNETic MAss 28-1 Theﬁeldenergy ofapoint charge 28-1 28-2 Theﬁeldmomentum ofamoving charge 28-2 28-3 Electromagnetic mass 28-3 28-4 The force ofanelectron onitself 28-4 28-5 Attempts tomodify theMaxwell theory 28-6 28-6 Thenuclear force ﬁeld28-12 CHAPTER 29. THE MoTioN o1=CHARGEs INELEcTRic AND MAGNETIC FiELDs 29-1 Motion inauniform electric ormagnetic ﬁeld 29-1 C 29-2 Momentum analysis 29-1 HAP-FER 29-3 Anelectrostatic lens29-2 29-4 Amagnetic lens29-3 29-5 Theelectron microscope 29-3 29-6 Accelerator guide ﬁelds 29-4 29-7 Alternating-gradient focusing 29-631-8 Thefour-tensor ofelectromagnetic momentum 31-12 CHAPTER 32. REi=RAcTivE INDEX oi=DENsE MATERiALs 32-1 32-2 32-3 32-4 32-5 32-6 32-7Polarization ofmatter 32-1 Maxwell’s equations inadielectric 32-3 Waves inadielectric 32-5 Thecomplex index ofrefraction 32-8 Theindex ofamixture 32-8 Waves inmetals 32-10 Low-frequency andhigh-frequency approximations; theskin depth andtheplasma frequency 32-11 CHAPTER 33.REFLECTION FROM SuREAcEs 33-1 33-2 33-3 33-4 33-5 33-6Reﬂection andrefraction oflight 33-1 Waves indense materials 33-2 Theboundary conditions 33-4 Thereﬂected andtransmitted waves 33-7 Reﬂection from metals 33-11 Total internal reﬂection 33-12 CHAPTER 34.THE MAGNETisivi OFMATTER 34-1 34-2 34-3 34-4 34-5 34-6 34-7 34-8 35-1 35-2 35-3 35-4 35-5 29-8 Motion incrossed electric andmagnetic ﬁelds 29-8 35-6Diamagnetism andparamagnetism 34-1 Magnetic moments andangular momentum 34-3 Theprecession ofatomic magnets 34-4 Diamagnetism 34-5 Larmor’s theorem 34-6 Classical physics gives neither diamagnetism nor paramagnetism 34-8 Angular momentum inquantum mechanics 34-8 Themagnetic energy ofatoms 34-11 35. PARAMAGNETisM ANDMAGNETic REsoNANcE Quantized magnetic states 35-1 TheStern-Gerlach experiment 35-3 TheRabi molecular-beam method 35-4 Theparamagnetism ofbulk materials 35-6 Cooling byadiabatic demagnetization 35-9 Nuclear magnetic resonance 35-10 11 CHAPTER 36.FERRoMAGNETisM 36-1 Magnetization currents 36-1 36-2 TheﬁeldH 36-5 36-3 Themagnetization curve 36-6 36-4 Iron-core inductances 36-8 36-5 Electromagnets 36-9 36-6 Spontaneous magnetization 36-11 CHAPTER 37. MAGNETic MATERiALs 37-1 Understanding ferromagnetism 37-1 37-2 Thermodynamic properties 37-4 37-3 Thehysteresis curve 37-5 37-4 Ferromagnetic materials 37-10 37-5 Extraordinary magnetic materials 37-11 CHAPTER 38.ELAsTiciTY 38-1 Hooke’s law38-1 38-2 Uniform strains 38-2 38-3 Thetorsion bar;shear waves 38-5 38-4 Thebentbeam 38-9 38-5 Buckling 38-11CHAPTER 39.ELAsTic MATERiALs 39-l Thetensor ofstrain 39-1 39-2 Thetensor ofelasticity 39-4 39-3 Themotions inanelastic body 39-6 39-4 Nonelastic behavior 39-8 39-5 Calculating theelastic constants 39-10 CHAPTER 40. THE FLow OFDRY WATER 40-1 Hydrostatics 40-1 40-2 Theequations ofmotion 40-2 40-3 Steady ﬂow-—Bernoulli’s theorem 40-6 40-4 Circulation 40-9 40-5 Vortex lines 40-10 CHAPTER 41. THE FLow oi=WET WATER 41-1 Viscosity 41-1 41-2 Viscous ﬂow41-4 41-3 The Reynolds number 41-5 41-4 Flow pastacircular cylinder 41-7 41-5 Thelimit ofzeroviscosity 41-9 41-6 Couette ﬂow41-10 INDEX I Electromagnetism 1-1Electrical forces Consider aforce likegravitation which varies predominantly inversely asthe square ofthedistance, butwhich isabout abillion-billion-billion-billion times stronger. Andwithanother difference. There aretwokinds of“matter,” which wecancallpositive andnegative. Like kinds repel andunlike kinds attract— unlike gravity where there isonlyattraction. What would happen? Abunch ofpositives would repel withanenormous force andspread outin alldirections. Abunch ofnegatives would dothesame. Butanevenly mixed bunch ofpositives andnegatives would dosomething completely different. The opposite pieces would bepulled together bytheenormous attractions. Thenet result would bethattheterriﬁc forces would balance themselves outalmost per- fectly, byforming tight, ﬁnemixtures ofthepositive andthenegative, andbetween twoseparate bunches ofsuchmixtures there would bepractically noattraction or repulsion atall. There issuchaforce: theelectrical force. Andallmatter isamixture ofposi- tiveprotons andnegative electrons which areattracting andrepelling with this great force. Soperfect isthebalance, however, thatwhen youstand nearsomeone elseyoudon’t feelanyforce atall.Ifthere were even alittlebitofunbalance you would know it.Ifyouwere standing atarm’s length from someone andeach of youhadonepercent more electrons thanprotons, therepelling force would bein- credible. How great? Enough tolifttheEmpire State Building? No! Tolift Mount Everest? No! Therepulsion would beenough tolifta“weight” equal to thatoftheentire earth! IWith such enormous forces soperfectly balanced inthisintimate mixture, it Qnothard tounderstand thatmatter, trying tokeep itspositive andnegative charges intheﬁnest balance, canhave agreat stiffness andstrength. TheEmpire State Building, forexample, swings onlyeight feetinthewind because theelectrical forces hold every electron andproton more orlessinitsproper place. Ontheother hand, ifwelook atmatter onascale small enough thatweseeonlyafewatoms, anysmall piece willnot, usually, have anequal number ofpositive andnegative charges, andsothere willbestrong residual electrical forces. Even when there are equal numbers ofboth charges intwoneighboring small pieces, there maystillbe large netelectrical forces because theforces between individual charges vary inversely asthesquare ofthedistance. Anetforce canariseifanegative charge of piece iscloser tothepositive than tothenegative charges oftheother piece. Theattractive forces canthenbelarger thantherepulsive onesandthere canbea netattraction between twosmall pieces withnoexcess charges. Theforce thatholds theatoms together, andthechemical forces thathold molecules together, are really electrical forces acting inregions where thebalance ofcharge isnotperfect, orwhere thedistances areverysmall. You know, ofcourse, thatatoms aremade with positive protons inthe nucleus andwithelectrons outside. You mayask: “Ifthiselectrical force isso terriﬁc, whydon‘t theprotons andelectrons justgetontopofeachother? Ifthey want tobeinanintimate mixture, whyisn’titstillmore intimate?” Theanswer hastodowiththequantum effects. Ifwetrytoconﬁne ourelectrons inaregion thatisveryclose totheprotons, then according totheuncertainty principle they must havesome mean square momentum which islarger themore wetrytocon- ﬁnethem. Itisthismotion, required bythelawsofquantum mechanics, thatkeeps lheelectrical attraction from bringing thecharges anycloser together. 1-11-1Electrical forces 1-2Electric andmagnetic ﬁelds 1-3Characteristics ofvector ﬁelds 1-4Thelawsofelectromagnetism 1-5What aretheﬁelds? 1-6Electromagnetism inscience andtechnology Review: Chapter l2,Vol.I,Character- istics ofForce Lower case Greek letters andcommonly used capitals £1 Q-$><9.<:~\q"o=q=m!‘:=>¢a~<s=x-rmm~<1:sl>'ﬁ ® > I-4 :1 ‘Cl M '6"-3 5'6alpha beta gamma delta epsilon zeta eta theta iota kappa lambda mu nu xi(ksi) omicron pi rho sigma tau upsilon phi chi(khi) psi omegaThere isanother question: “What holds thenucleus together”? Inanucleus there areseveral protons, allofwhich arepositive. Why don’t theypush them- selves apart? Itturns outthatinnuclei there are,inaddition toelectrical forces, nonelectrical forces, called nuclear forces, which aregreater than theelectrical forces andwhich areabletohold theprotons together inspite oftheelectrical repulsion. Thenuclear forces, however, have ashort range—their force fallsoil’ much more rapidly than l/r2. And thishasanimportant consequence. Ifa nucleus hastoomany protons init,itgetstoobig,anditwillnotstaytogether. An example isuranium, with92protons. Thenuclear forces actmainly between each proton (orneutron) anditsnearest neighbor, while theelectrical forces actover larger distances, giving arepulsion between each proton andalloftheothers in thenucleus. Themore protons inanucleus, thestronger istheelectrical repulsion, until, asinthecaseofuranium, thebalance issodelicate thatthenucleus isalmost ready toﬂyapart from therepulsive electrical force. Ifsuch anucleus isjust “tapped” lightly (ascanbedone bysending inaslowneutron), itbreaks intotwo pieces, eachwithpositive charge, andthese pieces ﬂyapart byelectrical repulsion. Theenergy which isliberated istheenergy oftheatomic bomb. This energy is usually called “nuclear” energy, butitisreally “electrical” energy released when electrical forces have overcome theattractive nuclear forces. Wemayask,ﬁnally, what holds anegatively charged electron together (since ithasnonuclear forces). Ifanelectron isallmade ofonekindofsubstance, each partshould repel theother parts. Why, then, doesn’t itﬂyapart? Butdoes the electron have “parts”? Perhaps weshould saythattheelectron isjustapoint and thatelectrical forces onlyactbetween diﬁerent point charges, sothattheelectron does notactupon itself. Perhaps. Allwecansayisthatthequestion ofwhat holds theelectron together hasproduced many diﬂiculties intheattempts toform acomplete theory ofelectromagnetism. Thequestion hasnever been answered. Wewillentertain overselves bydiscussing thissubject some more inlaterchapters. Aswehave seen, weshould expect thatitisacombination ofelectrical forces andquantum-mechanical effects that willdetermine thedetailed structure of materials inbulk, and,therefore, their properties. Some materials arehard, some aresoft. Some areelectrical “conductors”—because their electrons arefreeto move about; others are“insulators”—because their electrons areheldtightly to individual atoms. Weshallconsider laterhowsome ofthese properties come about, butthatisaverycomplicated subject, sowewillbegin bylooking attheelectrical forces onlyinsimple situations. Webegin bytreating onlythelawsofelectricity- including magnetism, which isreally apartofthesame subject. Wehave saidthattheelectrical force, likeagravitational force, decreases inversely asthesquare ofthedistance between charges. This relationship iscalled Coulomb’s law. Butitisnotprecisely truewhen charges aremoving—the elec- trical forces depend alsoonthemotions ofthecharges inacomplicated way. One partoftheforce between moving charges wecallthemagnetic force. Itisreally oneaspect ofanelectrical effect. That iswhywecallthesubject “electromag- netism.” There isanimportant general principle thatmakes itpossible totreat elec- tromagnetic forces inarelatively simple way. Weﬁnd, from experiment, thatthe force thatactsonaparticular charge—no matter howmany other charges there areorhow they aremoving—depends only ontheposition ofthat particular charge, onthevelocity ofthecharge, andontheamount ofcharge. Wecanwrite theforce Fonacharge qmoving withavelocity vas F=q(E+v><B). (1.1) WecallEtheelectric ﬁeld andBthemagnetic ﬁeld atthelocation ofthecharge. Theimportant thing isthattheelectrical forces from alltheother charges inthe universe canbesummarized bygiving justthese twovectors. Their values will depend onwhere thecharge is,andmaychange with time. Furthermore, ifwe replace thatcharge withanother charge, theforce onthenewcharge willbejust inproportion totheamount ofcharge solongasalltherestofthecharges inthe 1-2 world donotchange theirpositions ormotions. (Inrealsituations, ofcourse, each charge produces forces onallother charges intheneighborhood andmaycause these other charges tomove, andsoinsome cases theﬁelds canchange ifwereplace ourparticular charge byanother.) Weknow from Vol.Ihowtoﬁndthemotion ofaparticle ifweknow theforce onit.Equation (1.1) canbecombined withtheequation ofmotion togive %[—————(l_;’;‘;c,),,,] =F=q(E+v><B). (1.2) SoifEandBaregiven, wecanﬁndthemotions. Now weneed toknow howthe E’sandB’sareproduced. Oneofthemost important simplifying principles about thewaytheﬁelds are produced isthis: Suppose anumber ofcharges moving insome manner would produce aﬁeldE1,andanother setofcharges would produce E2.Ifboth setsof charges areinplace atthesame time (keeping thesame locations andmotions theyhadwhen considered separately), thentheﬁeldproduced isjustthesum E=E1+E2. (l.3) Thisfactiscalled theprinciple ofsuperposition ofﬁelds. Itholds alsoformagnetic ﬁelds. This principle means thatifweknow thelawfortheelectric andmagnetic ﬁelds produced byasingle charge moving inanarbitrary way, thenallthelawsof electrodynamics arecomplete. Ifwewant toknow theforce oncharge Aweneed onlycalculate theEandBproduced byeachofthecharges B,C,D,etc.,andthen addtheE’sandB’sfrom allthecharges toﬁndtheﬁelds, andfrom them the forces acting oncharge A.Ifithadonlyturned outthattheﬁeldproduced bya single charge wassimple, thiswould betheneatest waytodescribe thelaws of electrodynamics. Wehave already given adescription ofthislaw(Chapter 28, Vol.I)anditis,unfortunately, rather complicated. Itturns outthattheform inwhich thelaws ofelectrodynamics aresimplest arenotwhat youmight expect. Itisnotsimplest togiveaformula fortheforce that onecharge produces onanother. Itistruethatwhen charges arestanding stillthe Coulomb force lawissimple, butwhen charges aremoving about therelations are complicated bydelays intime andbytheeffects ofacceleration, among others. Asaresult, wedonotwish topresent electrodynamics only through theforce lawsbetween charges; weﬁnditmore convenient toconsider another point of view—a point ofviewinwhich thelawsofelectrodynamics appear tobethemost easily manageable. 1-2Electric andmagnetic ﬁelds First, wemust extend, somewhat, ourideas oftheelectric andmagnetic vectors, EandB.Wehave deﬁned them interms oftheforces thatarefeltbya charge. Wewishnowtospeak ofelectric andmagnetic ﬁelds atapoint evenwhen there isnocharge present. Wearesaying, ineffect, thatsince there areforces “acting on”thecharge, there isstill“something” there when thecharge isremoved. Ifacharge located atthepoint (x,y,z)atthetime tfeels theforce Fgiven by Eq.(1.1)weassociate thevectors EandBwiththepoint inspace (x,y,z).Wemay think ofE(x,y,z,t)andB(x,y,z,t)asgiving theforces thatwould beexperienced atthetimetbyacharge located at(x,y,z),withthecondition thatplacing thecharge there didnotdisturb thepositions ormotions ofalltheother charges responsible fortheﬁelds. Following thisidea, weassociate withevery point (x,y,z)inspace twovectors EandB,which maybechanging withtime. Theelectric andmagnetic ﬁelds are, then, viewed asvector functions ofx,y,z,andt.Since avector isspeciﬁed byits components, each oftheﬁelds E(x,y,z,t)andB(x,y,z,t)represent three mathe- matical functions ofx,y,z,and2. 1-3 / O/v ‘-4-I7 >/"-I 0-0» .___’¢_._> ‘QOi} \\ Fig. l-1. Avector ﬁeld moy be represented bydrowing osetofarrows whose mognitudes anddirections indicate thevolues ofthevector ﬁeld ofthepoints from which theorrows oredrown. / _-/$4-xi Fig.l—-2. Avector ﬁeld can be represented bydrowing lines which ore tangent tothedirection oftheﬁeld vector ofeoch point, andbydrawing thedensity oflines proportional tothemagnitude of theﬁeld vector. I\/Vector '\c:n1'|:=ne'1:r pxpondieu lor / / / Fig.l—3. Theﬂuxofovector ﬁeld through 0surface isdeﬁned usthe overoge volue ofthenormal component ofthevector times theoreo ofthesurface._,¢-@-IIItisprecisely because E(orB)canbespeciﬁed atevery point inspace thatitis called a“ﬁeld.” A“ﬁeld” isanyphysical quantity which takes ondifferent values atdifferent points inspace. Temperature, forexample, isaﬁeld—in thiscasea scalar ﬁeld, which wewrite asT(x,y,z).Thetemperature could alsovaryintime, andwewould saythetemperature ﬁeldistime-dependent, andwrite T(x,y,z,t). Another example isthe“velocity ﬁeld” ofaﬂowing liquid. Wewrite v(x,y,2,t) forthevelocity oftheliquid ateachpoint inspace atthetimet.Itisavector ﬁeld. Returning totheelectromagnetic ﬁelds——although they areproduced by charges according tocomplicated formulas, they have thefollowing important characteristic: therelationships between thevalues oftheﬁelds atonepoint and thevalues atanearby point areverysimple. With onlyafewsuchrelationships in theform ofdifferential equations wecandescribe theﬁelds completely. Itisin terms ofsuchequations thatthelawsofelectrodynamics aremost simply written. There have been various inventions tohelp themind visualize thebehavior of ﬁelds. Themost correct isalsothemost abstract: wesimply consider theﬁelds as mathematical functions ofposition andtime. Wecanalsoattempt togetamental picture oftheﬁeldbydrawing vectors atmany points inspace, eachofwhich gives theﬁeldstrength anddirection atthatpoint. Such arepresentation isshown in Fig. 1-1. Wecangofurther, however, anddraw lines which areeverywhere tangent tothevectors-—which, sotospeak, follow thearrows andkeep track of thedirection oftheﬁeld. When wedothiswelosetrack ofthelengths ofthe vectors, butwecankeep track ofthestrength oftheﬁeldbydrawing thelines far apart when theﬁeldisweak andclose together when itisstrong. Weadopt the convention thatthenumber oflinesperunitareaatright angles tothelines ispro- portional totheﬁeld strength. Thisis,ofcourse, onlyanapproximation, andit willrequire, ingeneral, thatnewlines sometimes start upinorder tokeep the number uptothestrength oftheﬁeld. Theﬁeld ofFig.l—lisrepresented by ﬁeldlines inFig.1-2. 1-3Characteristics ofvector ﬁelds There aretwomathematically important properties ofavector ﬁeld which wewilluseinourdescription ofthelawsofelectricity from theﬁeldpoint ofview. Suppose weimagine aclosed surface ofsome kind andaskwhether wearelosing “something” from theinside; thatis,does theﬁeldhave aquality of“outﬂow”? Forinstance, foravelocity ﬁeldwemight askwhether thevelocity isalways out- ward onthesurface or,more generally, whether more ﬂuid ﬂows out(perunit time) thancomes in.Wecallthenetamount ofﬂuidgoing outthrough thesurface perunittime the“ﬂux ofvelocity” through thesurface. Theﬂow through an element ofasurface isjustequal tothecomponent ofthevelocity perpendicular tothesurface times theareaofthesurface. Foranarbitrary closed surface, the netoutward ﬂow—or ﬂux—is theaverage outward normal component ofthe velocity, times theareaofthesurface: Flux =(average normal component)-(surface area). (1.4) Inthecase ofanelectric ﬁeld, wecanmathematically deﬁne something analogous toanoutﬂow, andweagain callittheﬂux, butofcourse itisnotthe ﬂowofanysubstance, because theelectric ﬁeldisnotthevelocity ofanything. It turns out,however, thatthemathematical quantity which istheaverage normal component oftheﬁeld stillhasauseful signiﬁcance. Wespeak, then, ofthe electric ﬂux-—also deﬁned byEq.(1.4). Finally, itisalsouseful tospeak ofthe ﬂuxnotonlythrough acompletely closed surface, butthrough anybounded sur- face. Asbefore, theﬂuxthrough suchasurface isdeﬁned astheaverage normal component ofavector times theareaofthesurface. These ideas areillustrated in Fig.1-3. . There isasecond property ofavector ﬁeldthathastodowithaline,rather than asurface. Suppose again thatwethink ofavelocity ﬁeldthatdescribes the ﬂowofaliquid. Wemight askthisinteresting question: Istheliquid circulating? 1-4 Bythatwemean: Isthere anetrotational motion around some loop? Suppose thatweinstantaneously freeze theliquid everywhere except inside ofatube which isofuniform bore, andwhich goes inaloop thatcloses back onitself asin Fig.l-4. Outside ofthetube theliquid stops moving, butinside thetubeitmay keep onmoving because ofthemomentum inthetrapped 1iquid—that is,ifthere is more momentum heading onewayaround thetubethan theother. Wedeﬁne a quantity called thecirculation astheresulting speed oftheliquid inthetubetimes its circumference. Wecanagain extend ourideas anddeﬁne the“circulation” forany vector ﬁeld (even when there isn’t anything moving). Foranyvector ﬁeld the circulation around anyimagined closed curve isdeﬁned astheaverage tangential component ofthevector (inaconsistent sense) multiplied bythecircumference oftheloop (Fig. 1-5). Circulation =(average tangential component)-(distance around). (1.5) Youwillseethatthisdeﬁnition does indeed giveanumber which isproportional tothecirculation velocity inthequickly frozen tubedescribed above. With justthese twoideas—ﬁux andcirculation—we candescribe allthelaws ofelectricity andmagnetism atonce. Youmaynotunderstand thesigniﬁcance of thelaws right away, buttheywillgiveyousome ideaofthewaythephysics of electromagnetism willbeultimately described. 1-4Thelawsofelectromagnetism Theﬁrstlawofelectromagnetism describes theﬂuxoftheelectric ﬁeld: TheﬂuxofEthrough anyclosed surface =@ ,(1.6)0 where eoisaconvenient constant. (The constant eoisusually read as“epsilon- zero” or“epsilon-naught”.) Ifthere arenocharges inside thesurface, eventhough there arecharges nearby outside thesurface, theaverage normal component ofE iszero, sothere isnonetﬂuxthrough thesurface. Toshow thepower ofthis typeofstatement, wecanshow thatEq.(1.6) isthesame asCoulomb’s law,pro- vided onlythatwealsoaddtheideathattheﬁeldfrom asingle charge isspherically symmetric. Forapoint charge, wedraw asphere around thecharge. Then the average normal component isjustthevalue ofthemagnitude ofEatanypoint, since theﬁeldmust bedirected radially andhave thesame strength forallpoints on thesphere. Ourrulenowsaysthattheﬁeldatthesurface ofthesphere, times the areaofthesphere—that is,theoutgoing ﬂux—is proportional tothecharge inside. Ifwewere tomake theradius ofthesphere bigger, thearea would increase as thesquare oftheradius. Theaverage normal component oftheelectric ﬁeldtimes thatareamust stillbeequal tothesame charge inside, andsotheﬁeldmust decrease asthesquare ofthedistance—we getan“inverse square” ﬁeld. Ifwehave anarbitrary curve inspace andmeasure thecirculation ofthe electric ﬁeldaround thecurve, wewillﬁndthatitisnot,ingeneral, zero(although itisfortheCoulomb ﬁeld). Rather, forelectricity there isasecond lawthatstates: foranysurface S(notclosed) whose edge isthecurve C, Circulation ofEaround C=gt(ﬂux ofBthrough S). (1.7) Wecancomplete thelaws oftheelectromagnetic ﬁeldbywriting twocorre- sponding equations forthemagnetic ﬁeldB. Flux ofBthrough anyclosed surface =0. (1.8) Forasurface Sbounded bythecurve C, cz(circulation ofBaround C)=%(ﬂux ofEthrough S) .1.9) 1-5_|_ﬂuxofelectric current through S ( E0(0) ‘ﬁﬁlb) _________ __“~\\ _.-_-_ \\ \\_.\\\ \\\ \\ _——’ _ ‘—_‘ \ \\\\ \X.’\\I\\I \-CI\/\~.i \\\\‘_‘/ \Tube »~' \ \\\\ \\\ \ \ (cl \\ ' tI ,/¢ I . _, _-, . /¢ 1 I\ .» _ , _|-a. .. . 4. . .-' /‘ v ‘“\\ -solid / ‘_\ ' \ . II ,’ .pI , ./. ' ~n-o ''_ _ liquid ‘ a_ _. Fig. 1-4. la)Thevelocity ﬁeld ina liquid. Imagine atube ofuniform cross section that follows anarbitrary closed curve asin(bl. Iftheliquid were suddenly frozen everywhere except inside the tube, theliquid inthetube would circulate asshown in(c). 1-direction __,_'._ /’ ,+ + 17 Arbitrary \ CIGOCIICIIVO \/ _> Fig. l-5. Thecirculation ofavector ﬁeld istheaverage tangential compo- nent ofthevector (inaconsistent sensel times thecircumference oftheloop. B(otmagnet) _ Q‘Q +1T'§m|uAi. F sit(onwire) TO N“TERMINAL S BAR MAGNET Fig.1-6. Abar magnet gives a ﬁeld Batawire. When there isacurrent along thewire, thewire moves because oftheforce F=qvXB. Theconstant cgthatappears inEq.(1.9)isthesquare ofthevelocity oflight. Itappears because magnetism isinreality arelativistic eﬁ'ect ofelectricity. The constant cohasbeen stuck intomake theunits ofelectric current come outina convenient way. Equations (1.6) through (1.9), together with Eq.(1.1), areallthelaws of electrodynamics*. Asyouremember, thelaws ofNewton were verysimple to write down, buttheyhadalotofcomplicated consequences andittook usalong timetolearn about them all.These lawsarenotnearly assimple towrite down, which means thattheconsequences aregoing tobemore elaborate anditwilltake usquite alotoftimetoﬁgure them allout. ' Wecanillustrate some ofthelaws ofelectrodynamics byaseries ofsmall experiments which show qualitatively theinterrelationships ofelectric and magnetic ﬁelds. Youhave experienced theﬁrstterm ofEq.(l.l) when combing your hair, sowewon’t show thatone. Thesecond partofEq.(1.1)canbedemon- strated bypassing acurrent through awirewhich hangs above abarmagnet, as shown inFig.1-6.Thewirewillmove when acurrent isturned onbecause ofthe force F=qvXB.When acurrent exists, thecharges inside thewirearemoving, sotheyhaveavelocity v,andthemagnetic ﬁeldfrom themagnet exerts aforce on them, which results inpushing thewiresideways. When thewireispushed totheleft,wewould expect thatthemagnet must feelapush totheright. (Otherwise wecould putthewhole thing onawagon and have apropulsion system thatdidn‘t conserve momentum!) Although theforce is toosmall tomake movement ofthebarmagnet visible, amore sensitively sup- ported magnet, likeacompass needle, willshow themovement. Howdoesthewirepushonthemagnet? Thecurrent inthewireproduces a magnetic ﬁeldofitsownthatexerts forces onthemagnet. According tothelast \- Linu ot8 from wire .¢\° +‘ll’-gRMlNAL /\‘ en‘. \N _.f§;%|||A|_ F(enmagnet) 5anmower Fig.l—7. Themagnetic ﬁeld ofthe wire exerts aforce onthemagnet. 'Weneed onlytoaddaremark about some conventions forthesignofthecirculation. 1-6 IV-' |/ .AC‘IQ: Fig.1-8. Two wires, carrying cur- \/ rent, exert forces oneach other. terminEq.(1.9), acurrent must have acirculation ofB—in thiscase, thelines of Bareloops around thewire, asshown inFig.1-7. ThisB-ﬁeld isresponsible for theforce onthemagnet. Equation (1.9)tellsusthatforaﬁxed current through thewirethecirculation ofBisthesame foranycurve thatsurrounds thewire. Forcurves—say circles— thatarefarther away from thewire, thecircumference islarger, sothetangential component ofBmust decrease. Youcanseethatwewould, infact,expect Bto decrease linearly withthedistance from alongstraight wire. Now, wehave saidthatacurrent through awireproduces amagnetic ﬁeld, andthatwhen there isamagnetic ﬁeldpresent there isaforce onawirecarrying a current. Then weshould alsoexpect thatifwemake amagnetic ﬁeldwithacurrent inonewire, itshould exert aforce onanother wirewhich alsocarries acurrent. Thiscanbeshown byusing twohanging wires asshown inFig.1-8. When the currents areinthesame direction, thetwowires attract, butwhen thecurrents are opposite, theyrepel. Inshort, electrical currents, aswellasmagnets, make magnetic ﬁelds. Butwait, what isamagnet, anyway? Ifmagnetic ﬁelds areproduced bymoving charges, is itnotpossible thatthemagnetic ﬁeldfrom apiece ofironisreally theresult of currents? Itappears tobeso.Wecanreplace thebarmagnet ofourexperiment withacoilofwire, asshown inFig.1-9. When acurrent ispassed through the coil-—as wellasthrough thestraight wireabove it——we observe amotion ofthe wireexact1y'as before, when wehadamagnet instead ofacoil. Inother words, thecurrent inthecoilimitates amagnet. Itappears, then, thatapiece ofironacts asthough itcontains aperpetual circulating current. Wecan,infact,understand magnets interms ofpermanent currents intheatoms oftheiron. Theforce onthe magnet inFig.1-7isduetothesecond term inEq.(1.1). B (from coil) ‘é +Tr%RmNAL 9,6 F Ionwire) -rdgm|uot con.orWIRE i l:l\"::l‘l' Fig. l-9. Thebarmagnet ofFig.1-6 canbereplaced byacoilcarrying an electrical current. Asimilar force acts onthewire. 1-7 Where dothecurrents come from? Onepossibility would befromthemotion oftheelectrons inatomic orbits. Actually, thatisnotthecaseforiron, although itisforsome materials. Inaddition tomoving around inanatom, anelectron alsospins about onitsownaxis-—something likethespinoftheearth—and itis thecurrent from thisspinthatgives themagnetic ﬁeldiniron. (Wesay“some- thing likethespinoftheearth” because thequestion issodeep inquantum me- chanics thattheclassical ideas donotreally describe things toowell.) Inmost substances, some electrons spinonewayandsome spintheother, sothemag- netism cancels out,butiniron——for amysterious reason which wewilldiscuss later—many oftheelectrons arespinning withtheir axeslined up,andthatisthe source ofthemagnetism. Since theﬁelds ofmagnets arefrom currents, wedonothave toaddanyextra term toEqs. (1.8) or(1.9) totakecareofmagnets. Wejusttakeallcurrents, including thecirculating currents ofthespinning electrons, andthen thelawis right. You should alsonotice thatEq.(1.8) saysthatthere arenomagnetic “charges” analogous totheelectrical charges appearing ontheright sideof Eq.(1.6). None hasbeen found. efz e\ Current /Z cumm .__> __ _______ i Fig. l—lO.The circulation of B / / around thecurve Cisgiven either bythe s\//B current passing through thesurface $1, / / " orbytherateofchange oftheﬂuxofE Curve O through thesurface S1. Surface 5| Surface S2 Theﬁrstterm ontheright-hand sideofEq.(1.9)wasdiscovered theoretically byMaxwell andisofgreat importance. Itsaysthatchanging electric ﬁelds produce magnetic effects. Infact, without thisterm theequation would notmake sense, because without itthere could benocurrents incircuits thatarenotcomplete loops. Butsuchcurrents doexist, aswecanseeinthefollowing example. Imagine acapacitor made oftwoﬁatplates. Itisbeing charged byacurrent thatﬂows toward oneplate andaway from theother, asshown inFig.1-10. Wedraw a curve Caround oneofthewires andﬁllitinwithasurface which crosses thewire, asshown bythesurface S1intheﬁgure. According toEq.(1.9),thecirculation of Baround Cisgiven bythecurrent inthewire(times c2).Butwhat ifweﬁllinthe curve withadzﬂerent surface S2,which isshaped likeabowl andpasses between theplates ofthecapacitor, staying always away fromthewire? There iscertainly nocurrent through thissurface. But,surely, justchanging thelocation ofan imaginary surface isnotgoing tochange arealmagnetic ﬁeld! Thecirculation of Bmust bewhat itwasbefore. Theﬁrstterm ontheright-hand sideofEq.(1.9) does, indeed, combine with thesecond term togivethesame result forthetwo surfaces S1andS2.ForS2thecirculation ofBisgiven interms oftherateof change oftheﬂuxofEbetween theplates ofthecapacitor. Anditworks outthat thechanging Eisrelated tothecurrent injustthewayrequired forEq.(1.9)tobe correct. Maxwell sawthatitwasneeded, andhewastheﬁrsttowrite thecomplete equation. “ With thesetup shown inFig.1-6wecandemonstrate another ofthelawsof electromagnetism. Wedisconnect theends ofthehanging wirefrom thebattery andconnect them toagalvanometer which tellsuswhen there isacurrent through thewire. When wepush thewire sideways through themagnetic ﬁeld ofthe magnet, weobserve acurrent. Such aneffect isagain justanother consequence of Eq.(1.l)—the electrons inthewire feeltheforce F=qvXB.Theelectrons have asidewise velocity because theymove withthewire. Thisvwithavertical B from themagnet results inaforce ontheelectrons directed along thewire, which starts theelectrons moving toward thegalvanometer. 1-8 Suppose, however, thatweleave thewirealone andmove themagnet. We guess from relativity thatitshould make nodiﬁerence, andindeed, weobserve a similar current inthegalvanometer. How doesthemagnetic ﬁeldproduce forces on charges atrest? According toEq.(1.1)there must beanelectric ﬁeld. Amoving magnet must make anelectric ﬁeld. How thathappens issaidquantitatively by Eq.(1.7). This equation describes many phenomena ofgreat practical interest, suchasthose thatoccur inelectric generators andtransformers. Themost remarkable consequence ofourequations isthatthecombination of Eq.(1.7) andEq.(1.9) contains theexplanation oftheradiation ofelectromag- netic effects over large distances. Thereason isroughly something likethis: suppose thatsomewhere wehave amagnetic ﬁeld which isincreasing because, say,acurrent isturned onsuddenly inawire. Then byEq.(1.7) there must bea circulation ofanelectric ﬁeld. Astheelectric ﬁeldbuilds uptoproduce itscircula- tion,thenaccording toEq.(1.9)amagnetic circulation willbegenerated. Butthe building upofthismagnetic ﬁeld willproduce anewcirculation oftheelectric ﬁeld, andsoon.Inthiswayﬁelds work their waythrough space without theneed ofcharges orcurrents except attheir source. That isthewayweseeeach other! Itisallintheequations oftheelectromagnetic ﬁelds. 1-5What aretheﬁelds? Wenowmake afewremarks onourwayoflooking atthissubject. Youmay besaying: “Allthisbusiness ofﬂuxes andcirculations ispretty abstract. There are electric ﬁelds atevery point inspace; then there arethese ‘laws.’ Butwhat is actually happening? Why can’t youexplain it,forinstance, bywhatever itisthat goesbetween thecharges.” Well, itdepends onyour prejudices. Many physicists usedtosaythatdirect action withnothing inbetween wasinconceivable. (How could theyﬁndanideainconceivable when ithadalready been conceived?) They would say:“Look, theonlyforces weknow arethedirect action ofonepiece of matter onanother. Itisimpossible thatthere canbeaforce withnothing totrans- mitit.”Butwhat really happens when westudy the“direct action” ofonepiece of matter right against another? Wediscover thatitisnotonepiece right against theother; theyareslightly separated, andthere areelectrical forces acting ona tinyscale. Thus weﬁndthatwearegoing toexplain so-called direct-contact action interms ofthepicture forelectrical forces. Itiscertainly notsensible totryto insist thatanelectrical force hastolook liketheold,familiar, muscular push or pull,when itwillturnoutthatthemuscular pushes andpulls aregoing tobeinter- preted aselectrical forces! Theonly sensible question iswhat isthemost con- venient waytolook atelectrical eﬁ'ects. Some people prefer torepresent them as theinteraction atadistance ofcharges, andtouseacomplicated law. Others love theﬁeldlines. They draw ﬁeldlines allthetime, andfeelthatwriting E’sandB’s istooabstract. Theﬁeldlines, however, areonlyacrude wayofdescribing aﬁeld, anditisverydifﬁcult togivethecorrect, quantitative lawsdirectly interms ofﬁeld lines. Also, theideas oftheﬁeld lines donotcontain thedeepest principle of electrodynamics, which isthesuperposition principle. Even though weknow how theﬁeldlineslookforonesetofcharges andwhat theﬁeldlines looklikeforan- other setofcharges, wedon’t getanyideaabout what theﬁeldlinepatterns will look likewhen both setsarepresent together. From themathematical stand- point, ontheother hand, superposition iseasy—we simply addthetwovectors. Theﬁeldlines have some advantage ingiving avivid picture, buttheyalsohave some disadvantages. Thedirect interaction wayofthinking hasgreat advantages when thinking ofelectrical charges atrest,buthasgreat disadvantages when dealing withcharges inrapid motion. Thebestwayistousetheabstract ﬁeldidea. That itisabstract isunfortunate, butnecessary. Theattempts totrytorepresent theelectric ﬁeldasthemotion of some kindofgearwheels, orinterms oflines, orofstresses insome kindofmate- rialhaveused upmore eﬁort ofphysicists thanitwould have taken simply toget therightanswers about electrodynamics. Itisinteresting thatthecorrect equations forthebehavior oflight incrystals were worked outbyMcCullough in1843. But 1-9 people saidtohim: “Yes, butthere isnorealmaterial whose mechanical properties could possibly satisfy those equations, andsince light isanoscillation thatmust vibrate insomething, wecannot believe thisabstract equation business.” Ifpeople hadbeen more open-minded, theymight have believed intheright equations for thebehavior oflight alotearlier thantheydid. Inthecaseofthemagnetic ﬁeldwecanmake thefollowing point: Suppose thatyouﬁnally succeeded inmaking upapicture ofthemagnetic ﬁeldinterms of some kind oflines orofgear wheels running through space. Then youtryto explain what happens totwocharges moving inspace, both atthesame speed and parallel toeachother. Because theyaremoving, theywillbehave liketwocurrents andwillhave amagnetic ﬁeldassociated withthem (likethecurrents inthewires ofFig.1-8). Anobserver whowasriding along withthetwocharges, however, would seebothcharges asstationary, andwould saythatthere isnomagnetic ﬁeld. The“gear wheels” or“lines” disappear when youridealong withtheobject! All wehave done istoinvent anewproblem. How canthegearwheels disappear?! Thepeople whodraw ﬁeldlines areinasimilar difﬁculty. Notonlyisitnotpos- sible tosaywhether theﬁeldlines move ordonotmove withcharges-—they may disappear completely incertain coordinate frames What wearesaying, then, isthatmagnetism isreally arelativistic eﬂect. In thecaseofthetwocharges wejustconsidered, travelling parallel toeach other, we would expect tohavetomake relativistic corrections totheirmotion, withterms of order v2/c2. These corrections must correspond tothemagnetic force. Butwhat about theforce between thetwowires inourexperiment (Fig. 1-8). There the magnetic force isthewhole force. Itdidn’t look likea“relativistic correction.” Also, ifweestimate thevelocities oftheelectrons inthewire (you candothis yourself), weﬁndthattheir average speed along thewireisabout 0.01centimeter persecond. Sov2/c2 isabout 10"“. Surely anegligible “correction.” Butno! Although themagnetic forceis,inthiscase,10*“ ofthe“normal” electrical force between themoving electrons, remember thatthe“normal” electrical forces have disappeared because ofthealmost perfect balancing out——because thewires have thesame number ofprotons aselectrons. Thebalance ismuch more precise than onepartin1025, andthesmall relativistic termwhich wecallthemagnetic force is theonlyterm left. Itbecomes thedominant term. Itisthenear-perfect cancellation ofelectrical effects which allowed relativity effects (that is,magnetism) tobestudied andthecorrect equations—to order v’/c2——to bediscovered, even though physicists didn’t know that’s what was happening. Andthatiswhy, when relativity wasdiscovered, theelectromagnetic laws didn’t need tobechanged. They——unlikc mechanics—were already correct toaprecision ofv2/c2. 1-6Electromagnetism inscience andtechnology Letusendthischapter bypointing outthatamong themany phenomena studied bytheGreeks there were twoverystrange ones: thatifyourubbed apiece ofamber youcould liftuplittle pieces ofpapyrus, andthatthere wasastrange rockfrom theisland ofMagnesia which attracted iron. Itisamazing tothink that these were theonlyphenomena known totheGreeks inwhich theeffects ofelec- tricity ormagnetism were apparent. Thereason thatthese were theonly phe- nomena thatappeared isdueprimarily tothefantastic precision ofthebalancing ofcharges thatwementioned earlier. Study byscientists whocame aftertheGreeks uncovered onenewphenomena after another thatwere really some aspect ofthese amber and/orlodestone etfects. Now werealize thatthephenomena ofchemical interaction and,ultimately, oflifeitself aretobeunderstood interms ofelectro- magnetism. Atthesame time thatanunderstanding ofthesubject ofelectromagnetism wasbeing developed, technical possibilities thatdeﬁed theimagination ofthepeople thatcame before were appearing: itbecame possible tosignal bytelegraph over longdistances, andtotalktoanother person miles away without anyconnections between, andtorunhuge power systems——a great water wheel, connected by 1-10 ﬁlaments overhundreds ofmiles toanother engine thatturns inresponse tothe master wheel-—many thousands ofbranching ﬁlaments—ten thousand engines in tenthousand places running themachines ofindustries andhomes—a1l turning because oftheknowledge ofthelawsofelectromagnetism. Today weareapplying even more subtle effects. Theelectrical forces, enor- mous astheyare,canalsobeverytiny,andwecancontrol them andusethem in verymany ways. Sodelicate areourinstruments thatwecantellwhat amanis doing bythewayheaffects theelectrons inathinmetal rodhundreds ofmiles away. Allweneed todoistousetherodasanantenna foratelevision receiver! From alongview ofthehistory ofmankind——seen from, say,tenthousand years from now——there canbelittle doubt thatthemost signiﬁcant event ofthe 19thcentury willbejudged asMaxwell’s discovery ofthelawsofelectrodynamics. TheAmerican Civil Warwillpaleintoprovincial insigniﬁcance incomparison with thisimportant scientiﬁc event ofthesame decade. 1-ll 2 Differential Calculus ofVector Fields 2-1Understanding physics Thephysicist needs afacility inlooking atproblems from several points of view. Theexact analysis ofrealphysical problems isusually quite complicated, andanyparticular physical situation maybetoocomplicated toanalyze directly bysolving thedifferential equation. Butonecanstillgetaverygood ideaofthe behavior ofasystem ifonehassome feelforthecharacter ofthesolution indiffer- entcircumstances. Ideas such astheﬁeldlines, capacitance, resistance, andin- ductance are,forsuch purposes, veryuseful. Sowewillspend much ofourtime analyzing them. Inthiswaywewillgetafeelastowhat should happen indifferent electromagnetic situations. Ontheother hand, none oftheheuristic models, such asﬁeldlines, isreally adequate andaccurate forallsituations. There isonlyone precise wayofpresenting thelaws, andthatisbymeans ofdifferential equations. They have theadvantage ofbeing fundamental and, sofarasweknow, precise. Ifyouhave learned thedifferential equations youcanalways goback tothem. There isnothing tounlearn. Itwilltake yousome time tounderstand what should happen indifferent circumstances. You willhave tosolve theequations. Each time yousolve the equations, youwilllearn something about thecharacter ofthesolutions. Tokeep these solutions inmind, itwillbeuseful alsotostudy theirmeaning interms ofﬁeld linesandofother concepts. Thisisthewayyouwillreally “understand” theequa- tions. That isthedifference between mathematics andphysics. Mathematicians, orpeople whohaveverymathematical minds, areoften ledastray when “studying” physics because theylosesight ofthephysics. They say:“Look, these differential equations—-the Maxwell equations——are allthere istoelectrodynamics; itis admitted bythephysicists thatthere isnothing which isnotcontained intheequa- tions. Theequations arecomplicated, butafter allthey areonlymathematical equations andifIunderstand them mathematically inside out,Iwillunderstand thephysics inside out.” Only itdoesn’t work thatway. Mathematicians whostudy physics with thatpoint ofview——and there have been many ofthem—usually make little contribution tophysics and,infact, little tomathematics. They fail because theactual physical situations intherealworld aresocomplicated thatitis necessary tohave amuch broader understanding oftheequations. What itmeans really tounderstand anequation—that is,inmore than a strictly mathematical sense—was described byDirac. Hesaid: “Iunderstand what anequation means ifIhave awayofﬁguring outthecharacteristics ofitssolution without actually solving it.”Soifwehave awayofknowing what should happen ingiven circumstances without actually solving theequations, then we“under- stand” theequations, asapplied tothese circumstances. Aphysical understanding isacompletely unmathematical, imprecise, andinexact thing, butabsolutely neces- saryforaphysicist. 'Ordinarily, acourse likethisisgiven bydeveloping gradually thephysical ideas—by starting withsimple situations andgoing ontomore andmore compli- cated situations. Thisrequires thatyoucontinuously forget things youpreviously learned—-things thataretrueincertain situations, butwhich arenottrueingeneral. Forexample, the“law” thattheelectrical force depends onthesquare ofthe distance isnotalways true. Weprefer theopposite approach. Weprefer totake ﬁrstthecomplete laws, andthen tostepback andapply them tosimple situa- tions, developing thephysical ideas aswegoalong. Andthatiswhat wearegoing todo. 2-12-1Understanding physics 2-2Scalar andvector ﬁelds-—T andh 2-3Derivatives ofﬁelds—the gradient 2-4Theoperator V 2-5Operations withV 2-6Thediﬁerential equation of heatﬂow 2-7Second derivatives ofvector ﬁelds 2-8Pitfalls Review: Chapter ll,Vol.I,Vectors \§au~ar.a.n-1~'-I»-4-1 “I _> _ Ewe E ntud E. QQMA.-r:u+\n.Ourapproach iscompletely opposite tothehistorical approach inwhich one develops thesubject interms oftheexperiments bywhich theinformation was obtained. Butthesubject ofphysics hasbeen developed overthepast200years bysome veryingenious people, andaswehave onlyalimited timetoacquire our knowledge, wecannot possibly cover everything theydid. Unfortunately oneof thethings thatweshall have atendency toloseinthese lectures isthehistorical, experimental development. Itishoped thatinthelaboratory some ofthislackcan becorrected. You canalsoﬁllinwhat wemust leave outbyreading theEncy- clopedia Brittanica, which hasexcellent historical articles onelectricity andon other parts ofphysics. Youwillalsoﬁndhistorical information inmany textbooks onelectricity andmagnetism. 2-2Scalar andvector ﬁelds—T andIt Webegin nowwiththeabstract, mathematical view ofthetheory ofelectricity andmagnetism. Theultimate ideaistoexplain themeaning ofthelawsgiven in Chapter l.Buttodothiswemust ﬁrstexplain anewandpeculiar notation that wewant touse. Soletusforget electromagnetism forthemoment anddiscuss the mathematics ofvector ﬁelds. Itisofverygreat importance, notonlyforelectro- magnetism, butforallkinds ofphysical circumstances. Justasordinary differential andintegral calculus issoimportant toallbranches ofphysics, soalsoisthe differential calculus ofvectors. Weturntothatsubject. Listed below areafewfacts from thealgebra ofvectors. Itisassumed that youalready know them. A-B=scalar =/4,8,, +A,,B,, +A,B, (2.1) AXB=vector (2.2) E XB): =A::B1l '_A1181:an ' (AXB),=A,,B, —A,B,, (A><B)u=AzB:z__AxBz G AB co an HIJlKlLMN 0\>QRs1r\u VWXVZA><A=0 (2.3) A-(A><B)=0 (2.4) A-(B,><c)=(A><B)-C (2.5) A><(B><c)=B(A~C) —C(A-B) (2.6) swdl buy“ “N ¢ |I, Also wewillwant tousethetwofollowing equalities from thecalculus: R.A- 8~ii)3 0 fa t, 1:-Ar xlzllwvt ILA-I.0~ 1-I1- J G I 1, mwm=%m+gw+gM on 02f_a’f mrnn Q” Theﬁrst equation (2.7) is,ofcourse, true only inthelimit that Ax,Ay,andAz gotoward zero. Thesimplest possible physical ﬁeldisascalar ﬁeld. Byaﬁeld, youremember, wemean aquantity which depends upon position inspace. Byascalar ﬁeld we merely mean aﬁeld which ischaracterized ateach point byasingle number—a scalar. Ofcourse thenumber maychange intime, butweneed notworry about thatforthemoment. Wewilltalkabout what theﬁeldlooks likeatagiven instant. Asanexample ofascalar ﬁeld, consider asolid block ofmaterial which hasbeen heated atsome places andcooled atothers, sothatthetemperature ofthebody varies from point topoint inacomplicated way. Then thetemperature willbea function ofx,y,andz,theposition inspace measured inarectangular coordinate system. Temperature isascalar ﬁeld. 2-2 vi Hot // -40° '2'4T/ T-30°T(X,y,1) ) Cold I “=20. '+iT... 0 |God x Onewayofthinking about scalar ﬁelds istoimagine “contours” which are imaginary surfaces drawn through allpoints forwhich theﬁeldhasthesame value, justascontour linesonamapconnect points withthesame height. Foratempera- tureﬁeldthecontours arecalled “isothermal surfaces” orisotherms. Figure 2-1 illustrates atemperature ﬁeld andshows thedependence ofTonxandywhen z=0.Several isotherms aredrawn. There arealsovector ﬁelds. Theideaisverysimple. Avector isgiven foreach point inspace. Thevector varies from point topoint. Asanexample, consider a rotating body. Thevelocity ofthematerial ofthebody atanypoint isavector which isafunction ofposition (Fig. 2—2). Asasecond example, consider theﬁow ofheatinablock ofmaterial. Ifthetemperature intheblock ishighatoneplace andlowatanother, there willbeaflowofheatfrom thehotter places tothecolder. Theheatwillbeﬂowing indifferent directions indifferent parts oftheblock. The heatflowisadirectional quantity which wecallh.Itsmagnitude isameasure of how much heat isﬂowing. Examples oftheheat ﬁow vector arealsoshown inFig.2-1. Y T2 h A0 Tnheotflow 7 Let’s make amore precise deﬁnition ofh:Themagnitude ofthevector heat ﬂowatapoint istheamount ofthermal energy thatpasses, perunittimeandper unitarea, through aninﬁnitesimal surface element, atright angles tothedirection ofﬂow. Thevector points inthedirection offlow(seeFig.2-3). Insymbols: IfAJ isthethermal energy thatpasses perunittimethrough thesurface element Aa,thenZ AJh=E1¢,, (2.9) where e;isaunitvector inthedirection ofﬂow. Thevector Ircanbedeﬁned inanother way——in terms ofitscomponents. We askhowmuch heatﬁows through asmall surface atanyangle withrespect tothe ﬂow. InFig.2-4weshow asmall surface A02inclined withrespect toA01,which isperpendicular totheﬂow. Theunitvector nisnormal tothesurface Aaz. The 2-3Fig.2-1. Temperature Tisonexample ofo scalar ﬁeld. With each point lx,y,z)inspace there isassociated 0number T(x,y,z).Allpoints on thesurface marked T=20°(shown asocurve at = z=O)areofthesome temperature. Thearrows > aresamples oftheheat flow vector ll. _’_,'A’, .w\- . .-\' .'/.".' §’ ,7 \ ~. - ~ 2"IO-OIiTION Fig.2-2. Thevelocity oftheatoms inurotating object isanexample ofo vector ﬁeld. Fig.2-3. Heot ﬂow isovector ﬁeld. Thevector hpoints along thedirection oftheflow. Itsmagni- tude istheenergy transported perunittimeacross o surface element oriented perpendicular totheﬂow, divided bythearea ofthesurface element. I1 \\9 ,/'l&\ hf at A02 Fig.2-4. Thehoof ﬂow through A03 isthesome asthrough Acn. angle 6between nandhisthesame astheangle between thesurfaces (since hisnor- maltoAal). Now what istheheatﬂowperunitareathrough Aa2? Theﬂow through Aa2isthesame asthrough Aal; only theareas aredifferent. Infact, Aal=Aa2cos6.Theheatﬂowthrough Aa2is —A~'£= £cos6= h-n. (2.10)A02 Aal Weinterpret thisequation: theheatﬂow(perunittimeandperunitarea) through anysurface element whose unitnormal..is n,isgiven byh-n.Equally, wecould say:thecomponent oftheheatﬂowperpendicular tothesurface element Aa2is h-n.Wecan,ifwewish, consider thatthese statements deﬁne h.Wewillbeapply- ingthesame ideas toother vector ﬁelds. 2-3Derivatives ofﬁelds-—the gradient When ﬁelds varyintime, wecandescribe thevariation bygiving their deriva- tiveswithrespect tot.Wewant todescribe thevariations withposition inasimilar way, because weareinterested intherelationship between, say,thetemperature in oneplace andthetemperature atanearby place. How shallwetakethederivative ofthetemperature with respect toposition? Dowedifferentiate thetemperature withrespect tox?Orwithrespect toy,or2? Useful physical laws donotdepend upon theorientation ofthecoordinate system. They should, therefore, bewritten inaform inwhich either both sides are scalars orboth sides arevectors. What isthederivative ofascalar ﬁeld, say OT/6x? Isitascalar, oravector, orwhat? Itisneither ascalar noravector, as youcaneasily appreciate, because ifwetook adifferent x-axis, 8T/6x would cer- tainly bedifferent. Butnotice: Wehave three possible derivatives: 8T/6x, 6T/6y, and6T/62. Since there arethree kinds ofderivatives andweknow thatittakes three numbers toform avector, perhaps these three derivatives arethecomponents ofavector: 6T8T6T l5,5595) —avector. Ofcourse itisnotgenerally truethatanythree numbers form avector. Itis trueonlyif,when werotate thecoordinate system, thecomponents ofthevector transform among themselves inthecorrect way. Soitisnecessary toanalyze how these derivatives arechanged byarotation ofthecoordinate system. Weshall show that(2.11) isindeed avector. Thederivatives dotransform inthecorrect waywhen thecoordinate system isrotated. Wecanseethisinseveral ways. Onewayistoaskaquestion whose answer is independent ofthecoordinate system, andtrytoexpress theanswer inan“in- variant” form. Forinstance, ifS=A'B,andifAandBarevectors, wekn0w— because weproved itinChapter llofVol.I—that Sisascalar. Weknow thatS isascalar without investigating whether itchanges with changes incoordinate systems. Itcan’t, because it’sadotproduct oftwovectors. Similarly, ifweknow thatAisavector, andwehavethree numbers B1,B2,andB3,andweﬁndoutthat A,B1 +A,,B2 —l-A,B3 =S, (2.12) where Sisthesame foranycoordinate system, then itmust bethatthethree numbers B1,B2,B3arethecomponents B2,Bu,B,ofsome vector B. Now let’sthink ofthetemperature ﬁeld. Suppose wetaketwopoints P1and P2,separated bythesmall interval AR. Thetemperature atP1isT1andatP2is T2,andthedifference AT=T2—T1.Thetemperatures atthese real, physical points certainly donotdepend onwhat axiswechoose formeasuring thecoordi- nates. Inparticular, ATisanumber independent ofthecoordinate system. Itisa scalar. 2-4 Ifwechoose some convenient setofaxes, wecould write T1=T(x,y,z)and T2=T(x+Ax,y+Ay,z+Az),where Ax,Ay,andAzarethecomponents of thevector AR(Fig. 2-5). Remembering Eq.(2.7), wecanwrite 8T 6T 8T TheleftsideofEq.(2.13) isascalar. Theright sideisthesumofthree products withAx,Ay,andAz,which arethecomponents ofavector. Itfollows thatthe three numbers ﬁilﬂ6x0y62 arealsothex-,y-,andz-components ofavector. Wewrite thisnewvector with thesymbol VT.Thesymbol V(called “del”) isanupside-down A,andissupposed toremind usofdifferentiation. People read VTinvarious ways: “del-T,” or “gradient ofT,”or“grad T;” er6T6T*gradT= VT= (2.14) Using thisnotation, wecanrewrite Eq.(2.13) inthemore compact form AT=VT-AR. (2.15) Inwords, thisequation saysthatthedifference intemperature between twonearby points isthedotproduct ofthegradient ofTandthevector displacement between thepoints. Theform ofEq.(2.15) alsoillustrates clearly ourproof above that VTisindeed avector. Perhaps youarestillnotconvinced? Let’s prove itinadifferent way. (Al- though ifyoulookcarefully, youmaybeabletoseethatit’sreally thesame proof inalonger-winded form!) Weshallshow thatthecomponents ofVTtransform in justthesame waythatcomponents ofRdo.Iftheydo,VTis avector according to ouroriginal deﬁnition ofavector inChapter llofVol.I.Wetakeanewcoordi- natesystem x’,y’,z’,andinthisnewsystem wecalculate 6T/6x’, 6T/By’, and 6T/62’. Tomake things alittlesimpler, weletz=z’,sothatwecanforget about thez-coordinate. (You cancheck outthemore general caseforyourself.) Wetakeanx'y’-system rotated anangle 0withrespect tothexy-system, as inFig.2-6(a). Forapoint (x,y)thecoordinates intheprime system are x’=xcos0 +ysin 0, (2.16) y’=——xsin0+ycos 0. (2.17) Or,solving forxandy, x=x’cos0—y’sin0, (2.18) y=x’sin0+y’cos9. (2.19) Ifanypairofnumbers transforms withthese equations inthesame waythatx andydo,theyarethecomponents ofavector. Now let’slook atthedifference intemperature between thetwonearby points P1andP2,chosen asinFig.2—6(b). Ifwecalculate with thex-andy- coordinates, wewould write BTAT-5;Ax (2.20) —-since Ayiszero. "‘Inournotation, theexpression (a,b,c)represents avector with components a,b, andc.Ifyouliketousetheunitvectors i,j,andk,youmaywrite ,aT .ar ar vT"ox+'ay+"az' 2-5Y \\,1\/5%-9->~< F___l\\_-D\\1'-‘T-aIl|I>||Ix|l1Il_i___l>l ‘<\\,,-viii~L___J~'<’f_ X AX AZ/Q 'l_\_ _‘u’ /K \@ Z Fig.2-5. Thevector AR,whose com- ponents areAx,Ay,andAz. yl H tn) -——L-/ PI --;»’ ,YYXI \e X vly’ (bl A//x'j9’\\BY' -< —— —$ Pl 4* P2 xl X Fig.2-6. lo)Transformation toa rotated coordinate system. lb)Special case ofaninterval ARparallel tothe x-axis. Ifwechoose some convenient setofaxes, wecould write T1=T(x,y,z)and T2=T(x+Ax,y+Ay,z+Az),where Ax,Ay,andAzarethecomponents of thevector AR(Fig. 2-5). Remembering Eq.(2.7), wecanwrite 6T 6T 8T TheleftsideofEq.(2.13) isascalar. Theright sideisthesumofthree products withAx,Ay,andAz,which arethecomponents ofavector. Itfollows thatthe three numbers EEEBx6y62 arealsothex-,y-,andz-components ofavector. Wewrite thisnewvector with thesymbol VT.Thesymbol V(called “de1”) isanupside-down A,andissupposed toremind usofdifferentiation. People read VTinvarious ways: “del-T,” or “gradient ofT,”or“grad T;” erara:r*gI'adT= VT: <33:-’Fy-’3;)' Using thisnotation, wecanrewrite Eq.(2.13) inthemore compact form AT=VT-AR. (2.15) Inwords, thisequation saysthatthedifference intemperature between twonearby points isthedotproduct ofthegradient ofTandthevector displacement between thepoints. Theform of‘Eq.(2.15) alsoillustrates clearly ourproof above that VTisindeed avector. Perhaps youarestillnotconvinced? Let’s prove itinadifferent way. (Al- though ifyoulookcarefully, youmaybeabletoseethatit’sreally thesame proof inalonger-winded form!) Weshallshow thatthecomponents ofVTtransform in justthesame waythatcomponents ofRdo.Iftheydo,VTis avector according to ouroriginal deﬁnition ofavector inChapter llofVol.I.Wetakeanewcoordi- natesystem x’,y’,2',andinthisnewsystem wecalculate 6T/6x’, 6T/6y’, and 6T/62’. Tomake things alittlesimpler, weletz=z’,sothatwecanforget about thez-coordinate. (You cancheck outthemore general caseforyourself.) Wetakeanx'y’-system rotated anangle 0withrespect tothexy-system, as inFig.2-6(a). Forapoint (x,y)thecoordinates intheprime system are x’=xcos0 +ysin 0, (2.16) y’=-—xsin0+ycos 0. (2.17) Or,solving forxandy, x=x’cos0—y’sin0, (2.18) y=x’sin0+y’cos9. (2.19) Ifanypairofnumbers transforms with these equations inthesame waythatx andydo,theyarethecomponents ofavector. Now let’slook atthedifference intemperature between thetwonearby points P1andP2,chosen asinFig.2—6(b). Ifwecalculate with thex-andy- coordinates, wewould write BTAT--6;Ax (2.20) —since Ayiszero. *Inournotation, theexpression (a,b,c)represents avector with components a,b, andc.Ifyouliketousetheunitvectors i,j,andk,youmaywrite ,8T .8T BTVT-r0x+1ay+kaz. 2-5Y \\\/5%-3-I»< F___l\\_-D\\1—-*-alll l>||ixIl1Il_p__bl expg,~L___J~'<’ \ F‘ X AXAz/" 11“ _""/K \; z Fig.2-5. Thevector AR,whose com- ponents areAx,Ay,andAz. y’ yf tn) -——L/'> PI -'? ' YrXI \e X rtY’ (bl A ,’x’3p’\\4Y'I<—— -1 PI N‘ P2 xl X Fig.2-6. lo)Transformation toa rotated coordinate system. lb)Special case ofaninterval ARparallel tothe x-axis. What would acomputation intheprime system give? Wewould havewritten _6T ,6T ,AT-WAx+by,Ay. (2.21) Looking atFig.2—6(b), weseethat Ax’=Axcos0 (2.22) and Ay'=—~Ax sin6, (2.23) since Ayisnegative when Axispositive. Substituting these inEq.(2.21), weﬁnd that AT= Axcos0—5%:Axsin0 (2.24) = cos0—ST?sin0)Ax. (2.25) Comparing Eq.(2.25) with (2.20), weseethat 6T 6T 6T.5 —Ix-; CQS 0—5-J7 S111 0. This equation saysthat8T/6x isobtained from 6T/6x’ and6T/6y’, justasxis obtained from x’andy’inEq.(2.18). So6T/6x isthex-component ofavector. Thesame kindofarguments would show that6T/6y and8T/62 arey-andz-com- ponents. SoVTisdeﬁnitely avector. Itisavector ﬁeldderived from thescalar ﬁeldT. 2-4Theoperator V Now wecandosomething thatisextremely amusing andingenious-—and characteristic ofthethings that make mathematics beautiful. The argument that grad T,orVT,isavector didnotdepend upon what scalar ﬁeld wewere differ- entiating. Allthearguments would gothesame ifTwere replaced byanyscalar ﬁeld. Since thetransformation equations arethesame nomatter what wediffer- entiate, wecould justaswell omit theTandreplace Eq.(2.26) bytheoperator equation 6 O 6.5;-5;;cos0-3-y-,Sll'16. (2.27) Weleave theoperators, asJeans said, “hungry forsomething todifferentiate.” Since thedifferential operators themselves transform asthecomponents ofa vector should, wecancallthem components ofavector operator. Wecanwrite 6 66V-(6;-swaa) 1 (2.28) which means, ofcourse, 6 6 6= -1 9 = -- q = 1 - 2' V’ 6x V” 6y V‘ 62 (29) Wehave abstracted thegradient away from theT—-that isthewonderful idea. You must always remember, ofcourse, that Visanoperator. Alone, it means nothing. IfVbyitself means nothing, what does itmean ifwemultiply itbyascalar-—say T—to gettheproduct TV? (One canalways multiply avector byascalar.) Itstilldoes notmean anything. Itsx-component is aT5. (2.30) which isnotunumber, butisstillsome kind ofoperator. However. according to thealgebra ofvectors wewould stillcallTVavector. 2-6 Now let’smultiply Vbyascalar ontheother side,sothatwehave theproduct (VT). Inordinary algebra TA=AT, (2.31) butwehave toremember thatoperator algebra isalittle different from ordinary vector algebra. With operators wemust always keep thesequence right, sothat theoperations make proper sense. Youwillhavenodifficulty ifyoujustremember thattheoperator Vobeys thesame convention asthederivative notation. What is tobedifferentiated must beplaced ontheright oftheV.Theorder isimportant. Keeping inmind thisproblem oforder, weunderstand thatTVisanoperator, buttheproduct VTisnolonger ahungry operator; theoperator iscompletely satisﬁed. Itisindeed aphysical vector having ameaning. Itrepresents thespatial rateofchange ofT.Thex-component ofVTishowfastTchanges inthex-direc- tion. What isthedirection ofthevector VT? Weknow thattherateofchange of Tinanydirection isthecomponent ofVTinthatdirection (seeEq.2.15). lt follows thatthedirection ofVTisthatinwhich ithasthelargest possible com- ponent—in other words, thedirection inwhich Tchanges thefastest. Thegradient ofThasthedirection ofthesteepest uphill slope (inT). 2-5Operations withV Canwedoanyother algebra withthevector operator V?Letustrycombining itwith avector. Wecancombine twovectors bymaking adotproduct. Wecould make theproducts (avector) -V, or V-(avector). Theﬁrst onedoesn’t mean anything yet,because itisstillanoperator. What it might ultimately mean would depend onwhat itismade tooperate on. The second product issome scalar ﬁeld. (A'Bisalways ascalar.) Let’s trythedotproduct ofVwith avector ﬁeld weknow, sayh.Wewrite outthecomponents: V-It=V,h,, +Vyhu +Vzh; (2.32) or ahahah.v-1»=7x“1+T;'+-5- (2.33) Thesumisinvariant under acoordinate transformation. Ifwewere tochoose a different system (indicated byprimes), wewould have* __(iii 811,,’ 6h,’v’It_ax,+W+3?. (2.34) which isthesame number aswould begotten from Eq.(2.33), even though it looks different. That is. V’-h =V-h (2.35) {orevery point inspace. SoV-Iiisascalar ﬁeld, which must represent some physical quantity. You should realize thatthecombination ofderivatives in V'hisrather special. There areallsorts ofother combinations likeOh,/6x, which areneither scalars norcomponents ofvectors. Thescalar quantity V-(avector) isextremely useful inphysics. Ithasbeen given thename thedivergence. Forexample, V-h=divh=“divergence ofIi.” (2.36) MwedidforVT,wecanascribe aphysical signiﬁcance toV~h.Weshall, how- ever,postpone thatuntil later. 'Wethink ofhasaphysical quantity thatdepends onposition inspace, andnot strictly asamathematical function ofthree variables. When Iiis“differentiated” with respect tox,y,andz,orwithrespect tox’,y’,andz’,themathematical expression forIi must ﬁrstbeexpressed asafunction oftheappropriate variables. 2-7 5'->- ‘(G09 FIQI + -> I la-*1->l (0) >2» h Area ISOTHERMALArea A AA T|+AT 1', (bl Fig. 2-7. (0) Heat ﬂow through a slab. (blAninﬁnitesimal slab parallel to onisothermal surface inalarge block.First, wewish toseewhat elsewecancook upwith thevector operator V. What about across product? Wemust expect that VXIi=avector. (2.37) Itisavector whose components wecanwrite bytheusual ruleforcross products seeE.2.2: ‘q’ ea_an (V X,7): =Vrhy T"Vi/hr =ax ay~ (2.38) Similarly, tv><1|).=v,/1.-v./1,,=-96’? (2.39) and 9%_92Fix(v></1),,=v.h.—v,h,=62 (2.40) Thecombination VXhiscalled “thecurlofh.”Thereason forthename andthephysical meaning ofthecombination willbediscussed later. Summarizing, wehave three kinds ofcombinations withV: VT =gradT =avector, V'h =divh=ascalar, VXh=curlh =avector. Using these combinations, wecanwrite about thespatial variations ofﬁelds ina convenient way——in awaythatisgeneral, inthatitd0esn‘t depend onanyparticular setofaxes. Asanexample oftheuseofourvector differential operator V,wewrite aset ofvector equations which contain thesame lawsofelectromagnetism thatwegave inwords inChapter 1.They arecalled Maxwell's equations. Maxwell’s Equations 1 -=11() VE 60 6B (2)VXE=-‘at <2-41)(3) v~B=0 2 _§£ .1. (4)Cv><B_at+E0 wliere p(rho). the“electric charge density,” istheamount ofcharge perunit volume, andj,the“electric current density," istherateatwhich charge ﬂows through aunitarea persecond. These four equations contain thecomplete classical theory oftheelectromagnetic ﬁeld. You seewhat anelegantly simple form wecangetwithournewnotation! 2-6Thedifferential equation ofheat ﬂow Letusgiveanother example ofalawofphysics written invector notation. Thelawisnotaprecise one, butformany metals andanumber ofother sub- stances thatconduct heatitisquite accurate. Youknow thatifyoutakeaslabof material andheat onefacetotemperature T2andcooltheother toadifferent temperature T1,theheatwillﬂowthrough thematerial from T2toT1[Fig. 2-7(a)]. Theheatflowisproportional totheareaAofthefaces, andtothetemperature difference. Itisalsoinversely proportional tod,thedistance between theplates. (Foragiven temperature difference, thethinner theslabthegreater theheatﬂow.) Letting Jbethethermal energy thatpasses perunittimethrough theslab,wewrite J=,<(r2-T1)-‘:7 (2-42) Theconstant ofproportionality K(kappa) iscalled thethermal conductivity. 2-8 What willhappen inamore complicated case? Sayinanodd-shaped block of material inwhich thetemperature varies inpeculiar ways? Suppose welook ata tinypiece oftheblock andimagine aslablikethatofFig.2-7(a) onaminiature scale. Weorient thefaces parallel totheisothermal surfaces, asinFig.2—7(b), so thatEq.(2.42) iscorrect forthesmall slab. Iftheareaofthesmall slabisAA,theheatflowperunittimeis AAAJ_KATE. (2.43) where Asisthethickness oftheslab. Now AJ/AA wehave deﬁned earlier asthe magnitude ofh,whose direction istheheat flow. Theheat ﬂow willbefrom T1+ATtoward T1,andsoitwillbeperpendicular totheisotherms, asdrawn in Fig.2—7(b). Also, AT/As isjusttherateofchange ofTwithposition. Andsince theposition change isperpendicular totheisotherms, ourAT/As isthemaximum rateofchange. Itis,therefore, justthemagnitude ofVT.New since thedirection ofVTisopposite tothatofh,wecanwrite (2.43) asavector equation: h=—KVT. (2.44) (The minus sign isnecessary because heat ﬂows “downhill” intemperature.) Equation (2.44) isthedifferential equation ofheatconduction inbulk materials. Youseethatitisaproper vector equation. Each sideisavector ifxisjustanum- ber. Itisthegeneralization toarbitrary cases ofthespecial relation (2.42) for rectangular slabs. Later weshould learn towrite allsorts ofelementary physics relations like(2.42) inthemore sophisticated vector notation. This notation is useful notonlybecause itmakes theequations looksimpler. Italsoshows most clearly thephysical content oftheequations without reference toanyarbitrarily chosen coordinate system. 2-7Second derivatives ofvector fields Sofarwehave hadonlyﬁrstderivatives. Why notsecond derivatives? We could have several combinations: (=1)V'(VT) (b)VX(VT) (c) V(V-h) (2.45) (d)v-(vXIi) (e)VX(VXh) Youcancheck thatthese areallthepossible combinations. Let’s lookﬁrstatthesecond one,(b).Ithasthesame form as A><(AT)= (AXA)T=0, since AXAisalways zero. Soweshould have curl(gradT) =VX(VT) =0. (2.46) Wecanseehowthisequation comes about ifwegothrough once withthecom- ponents: [VX(vT)]= =Vw(vT)i/ "VU(VT)Z , 86T 06T =5(ail"a <2”) which iszero(byEq.2.8). Itgoesthesame fortheother components. SoVX (VT) =O,foranytemperature distribution-in fact,foranyscalar function. 2-9 Now letustake another example. Letusseewhether wecanﬁndanother zero. Thedotproduct ofavector withacross product which contains thatvector iszero: A-(AXB)=0. (2.48) because AXBisperpendicular toA,andsohasnocomponents inthedirection A. Thesame combination appears in(d)of(2.45), sowehave v-(v><II)=div(curlll)=0. (2.49) Again, itiseasytoshow thatitiszerobycarrying through theoperations with components. Now wearegoing tostate twomathematical theorems thatwewillnotprove. They areveryinteresting anduseful theorems forphysicists toknow. Inaphysical problem _wefrequently ﬁndthatthecurlofsome quantity-say ofthevector ﬁeld A—is zero. Now wehave seen (Eq. 2.46) thatthe‘curl ofa gradient iszero, which iseasytoremember because ofthewaythevectors work. Itcould certainly be.then. thatAisthegradient ofsome quantity. because then itscurlwould necessarily bezero. Theinteresting theorem isthatifthecurlAis zero, thenAisalways thegradient ofsomething-—there issome scalar ﬁeldtlr(psi) such thatAisequal togradlb.Inother words, wehave the THEOREMI If VXA=0 there isa i// such that A=Vih. (2.50) There isasimilar theorem ifthedivergence ofAiszero. Wehave seenin Eq.(2.49) thatthedivergence ofacurlofsomething isalways zero. Ifyoucome across avector ﬁeldDforwhich divDiszero, thenyoucanconclude thatDis thecurlofsome vector ﬁeld C. THEOREM: If V-D=O there isa C such that D=VXC. (2.51) Inlooking atthepossible combinations oftwoVoperators, wehave found thattwoofthem always givezero. Now welook attheones thatarenotzero. Take thecombination V-(VT), which wasﬁrstonourlist. ltisnot,ingeneral, zero. Wewrite outthecomponents: vT=v,T+v,,T+v,T.Then VI =Vr(VrT) "l'Vi;(Vi/T) +Vz(V:T) .327" azr a=’T _€;§+5F-1-452-2, (2.52) which would, ingeneral, come outtobesome number. Itisascalar ﬁeld. Youseethatwedonotneed tokeep theparentheses, butcanwrite, without anychance ofconfusion, v-(VT)=v-VT=(v-v)T=V22". (2.53) WelookatV2asanewoperator. Itisascalar operator. Because itappears often inphysics, ithasbeen given aspecial name-—the Lap/acian. . 0* a’ a‘Laplacian =V2=Z)-J-(-5-l-5}-2+(-3?~ (2.54) 2-10 Since theLaplacian isascalar operator, wemayoperate withitonavector- bywhich wemean thesame operation oneach component inrectangular coor- dinates: V21:=(V2h,, V2h,,,V2h,). Let’s look atonemore possibility: VX(VXh),which was(e)inthelist (2.45). Now thecurlofthecurlcanbewritten differently ifweusethevector equality (2.6): AX(BXC)=B(A -C)—C(A -B). (2.55) Inorder tousethisformula, weshould replace AandBbytheoperator Vand putC=ll.Ifwedothat, weget v><(V><h)=v(v-h)-h(v-v)...??'! Wait aminute! Something iswrong. The ﬁrst twoterms arevectors allright (theoperators aresatisﬁed), butthelastterm doesn’t come outtoanything. It’s stillanoperator. Thetrouble isthatwehaven’t beencareful enough about keeping theorder ofourterms straight. Ifyoulook again atEq.(2.55), however, yousee thatwecould equally wellhave written itas AX(BXC)=B(A-C)—(A-B)C. (2.56) Theorder ofterms looks better. Now let’smake oursubstitution in(2.56). Weget VX(VXh)=V(V~Ia)—(V-V)h. (2.57) Thisform looks allright. Itis,infact,correct, asyoucanverify bycomputing the components. Thelastterm istheLaplacian, sowecanequally wellwrite vX(v><h)=V(V'h) —V211. (2.58) Wehave hadsomething tosayabout allofthecombinations inourlistof double V’s,except for(c),V(V-h).Itisapossible vector ﬁeld, butthere isnothing special tosayabout it.It’sjustsome vector ﬁeldwhich mayoccasionally come up. Itwillbeconvenient tohave atable ofourconclusions: (a) V-(VT) =V2T=ascalar ﬁeld (b) VX(VT) =0 (c) V(V-I1)=avector ﬁeld (d) V'(VXh)=0 (e)VX(VXh)=V(V'h)—-V21: (f) (V-V)h=V2]:=avector ﬁeld(2.59) Youmaynotice thatwehaven’t tried toinvent anewvector operator (VXV). Doyouseewhy? 2-8Pitfalls Wehave been applying ourknowledge ofordinary vector algebra tothealge- braoftheoperator V.Wehave tobecareful, though, because itispossible togo astray. There aretwopitfalls which wewillmention, although theywillnotcome upinthiscourse. What would yousayabout thefollowing expression, thatin- volves thetwoscalar functions 1,0and¢(Phi): (V1//)><(V¢)? Youmight want tosay:itmust bezerobecause it’sjustlike (Aa)><(Ab), 2-11 which iszerobecause thecrossproduct oftwoequalvectors AXAisalways zero. Butinourexample thetwooperators Varenotequal! Theﬁrstoneoperates on onefunction, ti/;theother operates onadifferent function, ¢.Soalthough werep- resent them bythesame symbol V,theymust beconsidered asdiﬂ'erent operators. Clearly, thedirection ofV¢depends onthefunction ¢,soitisnotlikely tobe parallel toV¢. (V¢) X(V¢) ¢0(generally). Fortunately, wewon’t have tousesuch expressions. (What wehave saiddoesn't change thefactthatVXVill=0foranyscalar ﬁeld, because here both V’s operate onthesame function.) Pitfall number two(which, again, weneed notgetintoinourcourse) isthe following: Therules thatwehave outlined herearesimple andnicewhen weuse rectangular coordinates. Forexample, ifwehave V2]:andwewant thex-com- ponent, itis 2 2 2 (vet),= +5%+ /1,=v’/1,. (2.60) Thesame expression would notwork ifwewere toaskfortheradial component ofV"h. Theradial component ofV2];isnotequal toV2/1,. Thereason isthat when wearedealing withthealgebra ofvectors, thedirections ofthevectors are allquite deﬁnite. Butwhen wearedealing withvector ﬁelds, their directions are different atdifferent places. lfwetrytodescribe avector ﬁeldin,say,polar coordi- nates, what wecallthe“radial” direction varies from point topoint. Sowecan getintoalotoftrouble when westart todifferentiate thecomponents. Forex- ample, even foracom-tan! vector ﬁeld. theradial component changes from point topoint. Itisusually safest andsimplest justtostick torectangular coordinates and avoid trouble. butthere isoneexception worth mentioning: Since theLaplacian V2,isascalar, wecanwrite itinanycoordinate system wewant to(forexample, inpolar coordinates). Butsince itisadifferential operator, weshould useitonly onvectors whose components areinaﬁxed direction—that means rectangular coordinates. Soweshall express allofourvector ﬁelds interms oftheir x-,y-, andz-components when wewrite ourvector differential equatlons outincom- ponents. 2—I2 3 Vector Integral Calculus 3-1Vector integrals; thelineintegral ofVlll‘ Wefound inChapter 2that there were various ways oftaking derivatives of ﬁelds. Some gave vector ﬁelds; some gave scalar ﬁelds. Although wedeveloped many different formulas, everything inChapter 2could besummarized inonerule: theoperators 6/6x, 6/dy, and6/dz arethethree components ofavector operator V.Wewould nowliketogetsome understanding ofthesigniﬁcance ofthederiva- tivesofﬁelds. Wewillthenhave abetter feeling forwhat avector ﬁeldequation means. Wehave already discussed themeaning ofthegradient operation (Vona scalar). Now weturn tothemeanings ofthedivergence andcurloperations. Theinterpretation ofthese quantities isbestdone interms ofcertain vector integrals andequations relating such integrals. These equations cannot, unfor- tunately, beobtained from vector algebra bysome easysubstitution, soyouwill justhave tolearn them assomething new. Ofthese integral formulas, oneis practically trivial, buttheother twoarenot. Wewillderive them andexplain their implications. Theequations weshall study arereally mathematical theorems. They willbeuseful notonlyforinterpreting themeaning andthecontent ofthe divergence andthecurl, butalsoinworking outgeneral physical theories. These mathematical theorems are,forthetheory ofﬁelds, what thetheorem ofthecon- servation ofenergy istothemechanics ofparticles. General theorems likethese areimportant foradeeper understanding ofphysics. Youwillﬁnd, though, that theyarenotveryuseful forsolving problems——except inthesimplest cases. Itis delightful, however, thatinthebeginning ofoursubject there willbemany simple problems which canbesolved with thethree integral formulas wearegoing to treat. Wewillsee,however, astheproblems getharder, thatwecannolonger use thesesimple methods. Wetake upﬁrstanintegral formula involving thegradient. Therelation contains averysimple idea: Since thegradient represents therateofchange ofa ﬁeldquantity, ifweintegrate thatrateofchange, weshould getthetotal change. Suppose wehave thescalar ﬁeld ¢(x,y,z).Atanytwopoints (l)and(2),the function ll!willhave thevalues ¢(l)and¢(2), respectively. [Weuseaconvenient notation, inwhich (2)represents thepoint (x2,yz,22)and(b(2)means thesame thing as\//(X2, yg,22).] IfI‘(gamma) isanycurve joining (1)and(2),asinFig.3-1, thefollowing relation istrue: Tmaonm 1. <1’) ¢(2)—¢(1)=1“) (Vii/)'d& (3-1) along I‘ Theintegral isalineintegral, from (1)to(2)along thecurve I‘,ofthedotproduct ofVi]/—a vector——with ds—another vector which isaninﬁnitesimal lineelement ofthecurve I‘(directed away from (1)andtoward (2)). First, weshould review what wemean byalineintegral. Consider ascalar function f(x,y,z),andthecurve I‘joining twopoints (1)and(2).Wemark olf thecurve atanumber ofpoints andjointhese points bystraight-line segments, as Il1OWn inFig.3-2. Each segment hasthelength As,,where iisanindex thatruns l,2,3,...Bythelineintegral <2> f fds(1)along I‘ 3-13-1Vector integrals; theline integral ofV\I/' 3-2Theﬂuxofavector ﬁeld 3-3Theﬂuxfrom acube; Gauss’ theorem 3-4Heat conduction; thediffusion equation 3-5Thecirculation ofavector ﬁeld 3-6Thecirculation around asquare; Stokes’ theorem 3-7Curl-free anddivergence-free ﬁelds 3-8Summary V\l'(2) Curve I‘ ds [ll Fig.3—l. Theterms used inEq.(3.1). Thevector V¢isevaluated attheline element di. W.tvvi.K/\ (2) /1 Curve I‘I Ass Ast A$| C lll Ob A52 Fig.3-2. The line integral isthe limitofctsum. .::::::. Z h ........///)\\)d “ ///‘;-3 ri- \§r§*i" Fig. 3-3. The closed surface S deﬁnes thevolume V.Theunitvector n istheoutward normal tothesurface element do,andIIistheheat-ﬂow vector atthesurface element.wemean thelimit ofthesum Asia where f,isthevalue ofthefunction attheithsegment. Thelimiting value iswhat thesumapproaches asweaddmore andmore segments (inasensible way, sothat thelargest As,—>0). Theintegral inourtheorem, Eq.(3.1), means thesame thing, although it looks alittle different. Instead off,wehave another scalar—the component of Vrbinthedirection ofAs.Ifwewrite (Vt;/), forthistangential component, itis clear that (V11), As=(Vtk) -As. (3.2) Theintegral inEq.(3.1)means thesumofsuch terms. Now let’sseewhyEq.(3.1) istrue. InChapter 1,weshowed thatthecom- ponent ofVrpalong asmall displacement ARwastherateofchange of(0inthe direction ofAR. Consider thelinesegment Asfrom (1)topoint ainFig.3-2. According toourdeﬁnition, AW1 =Ma) —WU) =(Vt/‘)1 ‘A-t'1- (3-3) Also, wehave if/(b)—¢(¢1)=(W/)2 'A82. (3-4) where, ofcourse, (Vi!/)1 means thegradient evaluated atthesegment Asl,and (V¢)2, thegradient evaluated atAs-2. IfweaddEqs.(3.3) and(3.4), weget v(b)—1!/(1)=(W/)1'As1 +(Vll’)2'A$2- (3-5) Youcanseethatifwekeep adding suchterms, wegettheresult 1!/(2)-((1)=Z(W/)r 'A-rt (3-6) Theleft-hand sidedoesn’t depend onhowwechoose ourintervals—if (1)and(2) arekeptalways thesame—so wecantakethelimit oftheright-hand side. Wehave therefore proved Eq.(3.1). Youcanseefrom ourproof thatjustastheequality doesn’t depend onhow thepoints a,b,c,...,arechosen, similarly itdoesn’t depend onwhat wechoose forthecurve Ftojoin(1)and(2).Ourtheorem iscorrect foranycurve from (1) to(2). Oneremark onnotation: Youwillseethatthere isnoconfusion ifwewrite, forconvenience, (Val/) -ds=Val-ds. (3.7) With thisnotation, ourtheorem is THEOREM l. <2) 1!/(2)—¢(1)=fa) Val'dc (3-3) any curve from (1)to(2) 3-2Theﬂuxofavector ﬁeld Before weconsider ournextintegral theorem—a theorem about thedivergence —we would liketostudy acertain ideawhich hasaneasily understood physical signiﬁcance inthecaseofheatﬂow. Wehavedeﬁned thevector It,which represents theheatthatﬂows through aunitareainaunittime. Suppose thatinside ablock ofmaterial wehavesome closed surface Swhich encloses thevolume V(Fig. 3-3). Wewould liketoﬁndouthowmuch heatisﬂowing outofthisvolume. Wecan, ofcourse, ﬁnditbycalculating thetotal heatﬂowoutofthesurface S. Wewrite dafortheareaofanelement ofthesurface. Thesymbol stands for atwo-dimensional differential. If,forinstance, theareahappened tobeinthe xy-plane wewould have da=dxdy. 3-2 Later weshall have integrals over volume andforthese itisconvenient tocon- sider adifferential volume thatisalittle cube. Sowhen wewrite dVwemean dV=dxdydz. Some people liketowrite dzainstead ofdatoremind themselves thatitis kind ofasecond-order quantity. They would alsowrite d3Vinstead ofdV. We willusethesimpler notation, andassume thatyoucanremember thatanarea hastwodimensions andavolume hasthree. Theheatﬂowoutthrough thesurface element daistheareatimes thecom- ponent ofhperpendicular toda.Wehavealready deﬁned nasaunitvector pointing outward atright angles tothesurface (Fig. 3-3). Thecomponent ofhthatwe want is h,,=h-n. (3.9) Theheatﬂowoutthrough daisthen h-nda. (3.10) Togetthetotal heatﬂowthrough anysurface wesumthecontributions from all theelements ofthesurface. Inother words, weintegrate (3.10) over thewhole surface: Total heatﬂowoutward through S=ISh-nda. (3.11) Wearealsogoing tocallthissurface integral “theﬂuxofhthrough thesur- face.” Originally theword ﬂuxmeant ﬂow, sothatthesurface integral justmeans theﬂowofhthrough thesurface. Wemaythink: histhe“current density” of heatﬂowandthesurface integral ofitisthetotal heatcurrent directed outofthe surface; thatis,thethermal energy perunittime(joules persecond). Wewould liketogeneralize thisideatothecasewhere thevector does not represent theﬂowofanything; forinstance, itmight betheelectric ﬁeld. Wecan certainly stillintegrate thenormal component oftheelectric ﬁeldoveranareaif wewish. Although itisnottheflowofanything, westillcallitthe“ﬂux.” Wesay Flux ofE through thesurface S=IE-nda. (3.12)s Wegeneralize theword “ﬂux” tomean the“surface integral ofthenormal com- ponent” ofavector. Wewillalsousethesame deﬁnition even when thesurface considered isnotaclosed one,asitishere. Returning tothespecial caseofheatﬂow, letustake asituation inwhich heatisconserved. Forexample, imagine some material inwhich after aninitial heating nofurther heatenergy isgenerated orabsorbed. Then, ifthere isanet heatﬂow outofaclosed surface, theheat content ofthevolume inside must decrease. So,incircumstances inwhich heatwould beconserved, wesaythat .__£’_Q. /shnda- dt, (3.13) where Qistheheatinside thesurface. TheheatﬂuxoutofSisequal tominus the rateofchange withrespect totimeofthetotal heatQinside ofS.Thisinterpreta- tionispossible because wearespeaking ofheatﬂowandalsobecause wesupposed thattheheat wasconserved. Wecould not,ofcourse, speak ofthetotal heat inside thevolume ifheatwere being generated there. Now weshall point outaninteresting factabout theﬁuxofanyvector. You maythink oftheheatﬂowvector ifyouwish, butwhat wesaywillbetrueforany vector ﬁeldC.Imagine thatwehave aclosed surface Sthatencloses thevolume V. Wenowseparate thevolume intotwoparts bysome kind ofa“cut,” asinFig. 3-4. Now wehave twoclosed surfaces andvolumes. Thevolume V1isenclosed inthesurface S1,which ismade upofpartoftheoriginal surface S,andofthe surface ofthecut,Sub. Thevolume V2isenclosed byS2,which ismade upof therestoftheoriginal surface Si,andclosed oﬂ'bythecutSa_b-Now consider the 3-3 Fig.3-4. Avolume Vcontained inside thesurface Sisdivided intotwopieces byu"cut" atthesurface $111,. Wenow have thevolume V1enclosed inthe surface $1=Sq+subandthevolume V;enclosed inthesurface S2=Si,-l—Sub. (X,HM. 1-) if s c c.-1/‘("1 l n l"| - -4—-—b' I ;i=-1-u_ __, ats ,MI I(14-A1. 1.1) (1.1.1-wt s Fig.3-5. Computation oftheﬁuxof Coutofosmall cube.S» f\\\_\?:"\\\§>.3l -I nl %<’<f cut following question: Suppose wecalculate thefluxoutthrough surface S1and addtoittheﬂuxthrough surface S2.Does thesumequal theﬂuxthrough the whole surface thatwestarted with? Theanswer isyes.Theﬁuxthrough thepart ofthesurfaces Subcommon toboth S1andS2justexactly cancels out. Forthe ﬂuxofthevector CoutofV1,wecanwrite Fluxthrough S1=/SC-nda +[SC-n1 da, (3.14) a ab andfortheﬂuxoutofV2, mmm@m=Lc1m+LcM@. cu)b ab Note thatinthesecond integral wehave written n1fortheoutward normal for S,,1,when itbelongs toS1,andn2when itbelongs toS2,asshown inFig.3-4. Clearly, n1=—n2, sothat IC-n1da=—/ C-n2da. (3.l6) Salt Sat lfwenowaddEqs. (3.14) and(3.15), weseethatthesumoftheﬂuxes through S1andS2isjustthesumoftwointegrals which, taken together, givetheﬂux through theoriginal surface S=S1,+S1,. Weseethattheﬂuxthrough thecomplete outer surface Scanbeconsidered asthesumoftheﬂuxes from thetwopieces intowhich thevolume wasbroken. Wecansimilarly subdivide again—say bycutting V1intotwopieces. You see thatthesame arguments apply. SoforanyWayofdividing theoriginal volume, it must begenerally truethattheﬂuxthrough theouter surface, which istheoriginal integral, isequal toasumoftheﬂuxes outofallthelittle interior pieces. 3-3Theﬂuxfrom acube; Gauss’ theorem Wenow take thespecial case ofasmall cube* andﬁndaninteresting formula fortheﬂuxoutofit.Consider acube whose edges arelined upwith theaxes asin Fig. 3-5. Letussuppose that thecoordinates ofthecorner nearest theorigin arex,y,2.LetAxbethelength ofthecube inthex-direction, Aybethelength inthey-direction, andAzbethelength inthez-direction. Wewish toﬁnd the ﬁuxofavector ﬁeld Cthrough thesurface ofthecube. Weshall dothisbymaking asum oftheﬁuxes through each ofthesixfaces. First, consider thefacemarked lintheﬁgure. Theﬂuxoutward onthisface isthenegative ofthex-component ofC,integrated overtheareaoftheface. Thisﬁuxis -[gnu Since weareconsidering asmall cube, wecanapproximate thisintegral bythe *Thefollowing development applies equally welltoanyrectangular parallelepiped. 3-4 value ofC,atthecenter ofthefaccwwhich wecallthepoint (l)——multiplied by theareaoftheface, AyAz: Flux outof1=—C,(l) AyAz. Similarly, fortheﬂuxoutofface2,wewrite Flux outof2=C,(2) AyAz. Now C,(l) andC,(2) are,ingeneral, slightly different. IfAxissmall enough, we canwrite ac C,(2) =C,(l) +Tc‘Ax. There are,ofcourse, more terms, buttheywillinvolve (A,)2 andhigher powers, andsowillbenegligible ifweconsider only thelimit ofsmall Ax. Sotheﬂux through face2is Flux outof2=[C,,(1) +879%Ax]AyAz. Summing theﬂuxes forfaces 1and2,weget Flux outofland2=éagiAxAyAz. Thederivative should really beevaluated atthecenter offace 1;thatis,at [x,y+(Ay/2), z+(Az/2)]. Butinthelimit ofaninﬁnitesimal cube, wemake anegligible error ifweevaluate itatthecorner (x,y,z). Applying thesame reasoning toeach oftheother pairs offaces, wehave Flux outof3and4=9%AxAyAz and C Flux outof5and6=Q37‘AxAyAz. Thetotal ﬂuxthrough allthefaces isthesumofthese terms. Weﬁndthat 6C, 6C 6C,1;,’ C'ndt1 = i +—a';)AXAyAZ, 011 B andthesumofthederivatives isjustV'C.Also, AxAyAz=AV,thevolume of thecube. Sowecansaythatforaninﬁnitesimal cube /C-nda =(V-C)AV. (3.17) surface Wehave shown thattheoutward ﬂuxfrom thesurface ofaninﬁnitesimal cube is equal tothedivergence ofthevector multiplied bythevolume ofthecube. We nowseethe“meaning” ofthedivergence ofavector. Thedivergence ofavector atthepoint Pistheﬂux—the outgoing “ﬂow” ofC—-per unitvolume, intheneigh- borhood ofP. Wehaveconnected thedivergence ofCtotheﬂuxofCoutofeachinﬁnitesimal volume. Foranyﬁnite volume wecanusethefactweproved above—that the totalﬂuxfrom avolume isthesumoftheﬂuxes outofeachpart. Wecan,thatis, integrate thedivergence overtheentire volume. Thisgives usthetheorem thatthe integral ofthenormal component ofanyvector overanyclosed surface canalsobe written astheintegral ofthedivergence ofthevector overthevolume enclosed bythesurface. Thistheorem isnamed after Gauss. GAUSS’ Trmomsu. /C'nda =/V~CdV, (3.18)S V where Sisanyclosed surface andVisthevolume inside it. 3-5 ll ,LI //’;;:1_,a*/:> f’, Source \I/\\ 0!hilt Block atnetll Fig.3-6. Intheregion near apoint source ofheat, theheat ﬂow isradially outward.3-4Heat conduction; thediffusion equation Let’s consider anexample oftheuseofthistheorem, justtogetfamiliar withit.Suppose wetakeagain thecaseofheatﬂowin,say,ametal. Suppose we have asimple situation inwhich alltheheathasbeen previously putinandthe body isjustcooling oﬁ".There arenosources ofheat, sothatheatisconserved. Then howmuch heatisthere inside some chosen volume atanytime’? Itmust be decreasing byjusttheamount thatﬂows outofthesurface ofthevolume. Ifour volume isalittle cube, wewould write, following Eq.(3.17), I-Ieatout =fh-nda =V‘/IAV. (3.19) cube Butthismust equal therateoflossoftheheatinside thecube. Ifqistheheatper unitvolume, theheatinthecube isqAV,andtherateoflossis d _ a'q Comparing (3.19) and(3.20), weseethat dq_ _—E-Vh. (3.21) Take careful noteoftheform ofthisequation; theform appears often inphys- ics. Itexpresses aconservation law——here theconservation ofheat. Wehave expressed thesame physical factinanother wayinEq.(3.13). Here wehave the dzﬂerential form ofaconservation equatioﬁ, while Eq.(3.13) istheintegral form. Wehave obtained Eq.(3.21) byapplying Eq.(3.13) toaninﬁnitesimal cube. Wecanalsogotheother way. Forabigvolume Vbounded byS,Gauss’ law saysthat [Sh-nda=-/‘V-hdV. (3.22) Using (3.21), theintegral ontheright-hand sideisfound tobejust—dQ/dt, andagain wehave Eq.(3.13). Now let’sconsider adifferent case. Imagine thatwehave ablock ofmaterial andthatinside itthere isavery tinyhole inwhich some chemical reaction is taking place andgenerating heat. Orwecould imagine thatthere aresome wires running intoatinyresistor thatisbeing heated byanelectric current. Weshall suppose thattheheatisgenerated practically atapoint, andletWrepresent the energy liberated persecond atthatpoint. Weshall suppose thatintherestofthe volume heatisconserved, andthattheheatgeneration hasbeen going onfora long time——so thatnow thetemperature isnolonger changing anywhere. The problem is:What does theheatvector hlooklikeatvarious places inthemetal? How much heatﬂowisthere ateach point? Weknow thatifweintegrate thenormal component ofItoveraclosed surface thatencloses thesource, wewillalways getW.Alltheheatthatisbeing generated atthepoint source must ﬂow outthrough thesurface, since wehave supposed thattheﬂow issteady. Wehave thedifficult problem ofﬁnding avector ﬁeld which, when integrated overanysurface, always gives W.Wecan,however, ﬁnd theﬁeldrather easily bytaking asomewhat special surface. Wetakeasphere of radius R,centered atthesource, andassume thattheheatﬂowisradial (Fig. 3-6). Ourintuition tellsusthathshould beradial iftheblock ofmaterial islarge and Wedon’t gettooclose totheedges, anditshould also have thesame magnitude atallpoints onthesphere. Youseethatweareadding acertain amount ofguess- work—usually called “physical intuition”~—to ourmathematics inorder toﬁnd theanswer. When hisradial andspherically symmetric, theintegral ofthenormal com- ponent ofhover thearea isvery simple, because thenormal component isjust 3-6 themagnitude ofhandisconstant. Theareaoverwhich weintegrate is41rR2. Wehave thenthat [Sn-nda =h'41rR2 (3.23) (where histhemagnitude ofh).Thisintegral should equal W,therateatwhich heatisproduced atthesource. Weget W h_41rR2’ or 1.=5%,e., (3.24) where, asusual, e,represents aunitvector intheradial direction. Ourresult saysthathisproportional toWandvaries inversely asthesquare ofthedistance from thesource. Theresult wehavejustobtained applies totheheatﬁowinthevicinity ofa point source ofheat. Let’s nowtrytoﬁndtheequations thathold inthemost general kind ofheat ﬂow, keeping only thecondition thatheat isconserved. Wewillbedealing only with what happens atplaces outside ofanysources or absorbers ofheat. Thedifferential equation fortheconduction ofheatwasderived inChapter 2. According toEq.(2.44), h=—-xVT. (3.25) (Remember thatthisrelationship isanapproximate one,butfairly good forsome materials likemetals.) Itisapplicable, ofcourse, onlyinregions ofthematerial where there isnogeneration orabsorption ofheat. Wederived above another relation, Eq.(3.21), thatholds when heatisconserved. Ifwecombine thatequation with (3.25), weget dq__ ____ _—E—Vh— V(xVT), or . g=Kv-VT=KV2T, (3.26) ifKisaconstant. You remember that qistheamount ofheat inaunitvolume andV-V=V2istheLaplacian operator 2 2 22_9 L *9V_6x2+6y?+622 Ifwenowmake onemore assumption wecanobtain averyinteresting equa- tion. Weassume thatthetemperature ofthematerial isproportional totheheat content perunitvolume—that is,thatthematerial hasadeﬁnite speciﬁc heat. When thisassumption isvalid (asitoften is),wecanwrite Aq=0,,AT or d J1‘7‘;=c,,75- (3.27) Therateofchange ofheatisproportional totherateofchange oftemperature. Theconstant orproportionality c,,is,here, thespeciﬁc heat perunitvolume of thematerial. Using Eq.(3.27) with(3.26), weget dT'_ L 2 F;_cuvT. (3.28) Weﬁndthatthetimerateofchange ofT—atevery point—is proportional tothe Laplacian ofT,which isthesecond derivative ofitsspatial dependence. Wehave adifferential equation—in x,y,z,andt——for thetemperature T. 3-7 Loop!‘ C 1Ict ~ \ dc d8.\ C CtI 65 11 c Fig.3-7. Thecirculation ofCaround thecurve Pisthelineintegral ofC1,the tangential component ofC. (1) To rib rb T1 dsl\ (5 (2) Fig.3-8. Thecirculation around the whole loop isthesumofthecirculations around thetwoloops: I‘,=I‘,+I‘¢b dudF2=Pb+Fab.Thedifferential equation (3.28) iscalled theheatdzﬂusion equation. Itis often written as J1"_ 2-‘,7_nvT, (3.29) where Discalled thediffusion constant, andishereequal tox/c,,. Thediﬂusion equation appears inmany physical problems—in thediffusion ofgases, inthediffusion ofneutrons, andinothers. Wehave already discussed thephysics ofsome ofthese phenomena inChapter 43ofVol.I.Now youhave thecomplete equation thatdescribes diffusion inthemost general possible situa- tion. Atsome later time wewilltakeupways ofsolving thediﬂusion equation toﬁndhow thetemperature varies inparticular cases. Weturn back now to consider other theorems about vector ﬁelds. 3-5Thecirculation ofavector ﬁeld Wewish nowtolook atthecurlinsomewhat thesame waywelooked atthe divergence. Weobtained Gauss’ theorem byconsidering theintegral over a surface, although itwasnotobvious atthebeginning thatwewere going tobe dealing withthedivergence. How didweknow thatwewere supposed tointegrate overasurface inorder togetthedivergence? Itwasnotatallclear thatthiswould betheresult. Andsowithanapparent equal lackofjustiﬁcation, weshallcalculate something elseabout avector andshow thatitisrelated tothecurl. Thistimewe calculate what iscalled thecirculation ofavector ﬁeld. IfCisanyvector ﬁeld, wetakeitscomponent along acurved lineandtaketheintegral ofthiscomponent allthewayaround acomplete loop. Theintegral iscalled thecirculation ofthe vector ﬁeldaround theloop. Wehave already considered alineintegral ofV\// earlier inthischapter. Now wedothesame kind ofthing foranyvector ﬁeldC. LetI‘beanyclosed loopinspace-—imaginary, ofcourse. Anexample isgiven inFig.3-7. Thelineintegral ofthetangential component ofCaround theloop iswritten as frc,ds=frC-ds. (3.30) Youshould notethattheintegral istaken allthewayaround, notfrom onepoint toanother aswedidbefore. Thelittle circle ontheintegral signistoremind us thattheintegral istobetaken allthewayaround. This integral iscalled the circulation ofthevector ﬁeldaround thecurve P.Thename came originally from considering thecirculation ofaliquid. Butthename-—like ﬁux—has beenextended toapply toanyﬁeldevenwhen there isnomaterial “circulating.” Playing thesame kind ofgame wedidwith theﬂux, wecanshow thatthe circulation around aloopisthesumofthecirculations around twopartial loops. Suppose webreak upourcurve ofFig.3-7intotwoloops, byjoining twopoints (1)and(2)ontheoriginal curve bysome linethatcutsacross asshown inFig. 3-8. There arenowtwoloops, F1andF2.F1ismade upof1",,which isthatpart oftheoriginal curve totheleftof(1)and(2),plusFab,the“short cut." T2ismade upoftherestoftheoriginal curve plustheshort cut. Thecirculation around F1isthesumofanintegral along l‘,,andalong Fab. Similarly, thecirculation around F2isthesumoftwoparts, onealong 1",,andthe other along l‘,,;,. Theintegral along I‘,,1,willhave, forthecurve F2,theopposite signfrom what ithasforP1,because thedirection oftravel isopposite-—we must takeboth ourlineintegrals withthesame “sense” ofrotation. Following thesame kind ofargument weused before, youcanseethatthe sumofthetwocirculations willgivejustthelineintegral around theoriginal curve F.Theparts duetoI‘,,;,cancel. Thecirculation around theonepartplusthecir- culation around thesecond part equals thecirculation about theouter line. Wecancontinue theprocess ofcutting theoriginal loopintoanynumber ofsmaller loops. When weaddthecirculations ofthesmaller loops, there isalways acan- cellation oftheparts ontheir adjacent portions, sothatthesumisequivalent tothe circulation around theoriginal single loop. 3-8 Nowletussuppose thattheoriginal loopistheboundary ofsome surface. There are,ofcourse, aninﬁnite number ofsurfaces which allhave theoriginal loops astheboundary. Ourresults willnot,however, depend onwhich surface wechoose. First, webreak ouroriginal loopintoanumber ofsmall loops thatall lieonthesurface wehave chosen, asinFig.3-9. Nomatter what theshape of thesurface, ifwechoose oursmall loops small enough, wecanassume thateach ofthesmall loops willenclose anareawhich isessentially ﬂat.Also, wecanchoose oursmall loops sothateach isverynearly asquare. Now wecancalculate the circulation around thebigloop I‘byﬁnding thecirculations around allofthe little squares andthentaking their sum. 3-6Thecirculation around asquare; Stokes’ theorem How shall weﬁndthecirculation foreach little square? Onequestion is, howisthesquare oriented inspace? Wecould easily make thecalculation ifit hadaspecial orientation. Forexample, ifitwere inoneofthecoordinate planes. Since wehave notassumed anything asyetabout theorientation ofthecoordinate axes, wecanjustaswellchoose theaxessothattheonelittle square wearecon- centrating onatthemoment liesinthexy-plane, asinFig.3-10. Ifourresult is expressed invector notation, wecansaythatitwillbethesame nomatter what the particular orientation oftheplane. Wewant nowtoﬁndthecirculation oftheﬁeldCaround ourlittle square. Itwillbeeasytodothelineintegral ifwemake thesquare small enough thatthe vector Cdoesn’t change much along anyonesideofthesquare. (The assumption isbetter thesmaller thesquare, sowearereally talking about inﬁnitesimal squares.) Starting atthepoint (x,y)-—the lower leftcorner oftheﬁgure—we goaround in thedirection indicated bythearrows. Along theﬁrstside—marked (l)—the tangential component isC,,(l) andthedistance isAx.Theﬁrstpartoftheintegral isC,,(l) Ax. Along thesecond leg,wegetC,,(2) Ay. Along thethird, weget -C,(3) Ax,andalong thefourth, -C,,(4) Ay. Theminus signs arerequired because wewant thetangential component inthedirection oftravel. Thewhole lineintegral isthen fC-ds=-C,(l)Ax +C,,(2) Ay—C,,(3) Ax-C,,(4) Ay. (3.31) Now let’slook attheﬁrstandthird pieces. Together theyare [C,,(l) —C,,(3)] Ax. (3.32) Youmight think thattoourapproximation thedifference iszero. That istrueto theﬁrstapproximation. Wecanbemore accurate, however, andtakeintoaccount therateofchange ofC,,.Ifwedo,wemaywrite . 6C,CA3) =CA1) +WAy- (3-33) Ifweincluded thenextapproximation, itwould involve terms in(Ay)2, butsince wewillultimately think ofthelimit asAy->0,such terms canbeneglected. Putting (3.33) together with(3.32), weﬁndthat 6C,[C,,(l) —C,(3)]Ay =—-5AxAy. (3.34) Thederivative can,toourapproximation, beevaluated at(x,y). Similarly, fortheother twoterms inthecirculation, wemaywrite 6CC,,(2) Ay-C,,(4) Ay=iAxAy. (3.35) Thecirculation around oursquare isthen ac, ac,(-5 '-' AX Ay, 3-9ﬂnlllll4|cEhI'l/"lama!ileum/I‘mm5-1 Fig.3-9. Some surface bounded by theloop I‘ischosen. Thesurface is divided into anumber ofsmall areas, each approximately asquare. The circulation around I‘isthesumofthe circulations around thelittle loops. c__ cT 5 Av‘t ’ 2 C 1 l (1,?) C,|-em?--lH Fig.3-10. Computing thecirculation ofCaround asmall square..,p- X C Loop!‘ Surface S Q > ‘c I I I I VXC Fig.3-11. The circulation ofC around Fisthesurface integral ofthe normal component ofVXC. (2) C,4? b (1) Fig.3-12. IfVXCiszero, the circulation around theclosed curve I‘is zero. Thelineintegral ofC-dzfrom (1) to(2)along amust bethesame asthe lineintegral along b.which isinteresting, because thetwoterms intheparentheses arejustthez-com- ponent ofthecurl. Also, wenotethatAxAyistheareaofoursquare. Sowe canwrite ourcirculation (3.36) as (VXC),Aa. Butthez-component really means thecomponent normal tothesurface element. Wecan,therefore, write thecirculation around adifferential square inaninvariant vector form: C-ds=(v><C),,Aa =(v><C)-nAa. (3.37) Ourresult is:thecirculation ofanyvector Caround aninﬁnitesimal square isthecomponent ofthecurlofCnormal tothesurface, times theareaofthesquare. Thecirculation around anyloop I‘cannowbeeasily related tothecurlof thevector ﬁeld. Weﬁllintheloopwithanyconvenient surface S,asinFig.3-11, andaddthecirculations around asetofinﬁnitesimal squares inthissurface. The sumcanbewritten asanintegral. Ourresult isaveryuseful theorem called Stokes’ theorem (after Mr.Stokes). STOKES’ THEOREM. 3C~ds=f(v><C).,da, (3.38)I‘ S where Sisanysurface bounded byI‘. Wemust now speak about aconvention ofsigns. InFig.3-10 thez-axis would point toward youina“usual”—that is,“right-handed”—system ofaxes. When wetook ourlineintegral witha“positive” sense ofrotation, wefound that thecirculation wasequal tothez-component ofVXC.Ifwehadgone around theother way, wewould have gotten theopposite sign. Now howshall weknow, ingeneral, what direction tochoose forthepositive direction ofthe“normal” component ofVXC?The“positive” normal must always berelated tothe sense ofrotation, asinFig.3-10. Itisindicated forthegeneral caseinFig.3-11. Onewayofremembering therelationship isbythe“right-hand rule.” Ifyou make theﬁngers ofyour right hand goaround thecurve I‘,with theﬁngertips pointed inthedirection ofthepositive sense ofds,thenyour thumb points inthe direction ofthepositive normal tothesurface S. 3-7Curl-free anddivergence-free ﬁelds Wewould like,now, toconsider some consequences ofournewtheorems. Take ﬁrstthecaseofavector whose curliseverywhere zero. Then Stokes’ theorem says that thecirculation around anyloop iszero. Now ifwechoose twopoints (1)and(2)onaclosed curve (Fig. 3-12), itfollows thatthelineintegral ofthe tangential component from (1)to(2)isindependent ofwhich ofthetwopossible paths istaken. Wecanconclude thattheintegral from (1)to(2)candepend only onthelocation ofthese points-that istosay,itissome function ofposition only. Thesame logic wasusedinChapter 14ofVol.I,where weproved thatiftheintegral around aclosed loop ofsome quantity isalways zero, then thatintegral canbe represented asthedifference ofafunction oftheposition ofthetwoends. This factallowed ustoinvent theideaofapotential. Weproved, furthermore, thatthe vector ﬁeldwasthegradient ofthispotential function (seeEq.14.13 ofVol.I). Itfollows thatanyvector ﬁeldwhose curliszeroisequal tothegradient of some scalar function. That is,ifVXC=0,everywhere, there issomeqb (psi)for which C=V¢—a useful idea. Wecan,ifwewish, describe thisspecial kind of vector ﬁeldbymeans ofascalar ﬁeld. Let’s show something else. Suppose wehave anyscalar ﬁeld4>(phi). Ifwe takeitsgradient, V¢,theintegral ofthisvector around anyclosed loop must_be zero. Itslineintegral from point (1)topoint (2)is[4>(2) —¢(l)]. If(1)and(2) 3-10 arethesame points, ourTheorem 1,Eq.(3.8), tellsusthatthelineintegral iszero: fV¢-ds=0. loop Using Stokes’ theorem, wecanconclude that fv><(V¢)da =0 overanysurface. Butiftheintegral iszerooveranysurface, theintegrand must bezero. So VX(V¢) =0,always. Weproved thesame result inSection 2-7byvector algebra. Let’s looknowataspecial caseinwhich weﬁllinasmall loop I‘withalarge surface S,asindicated inFig.3-13. Wewould like,infact,toseewhat happens when theloopshrinks down toapoint, sothatthesurface boundary disappears- thesurface becomes closed. Now ifthevector Ciseverywhere ﬁnite, theline integral around I‘must gotozeroasweshrink theloop—the integral isroughly proportional tothecircumference ofP,which goestozero. According toStokes’ theorem, thesurface integral of(VXC),must alsovanish. Somehow, aswe close thesurface weaddincontributions thatcancel outwhat wasthere before. Sowehave anewtheorem: I(v><C),,da =0. (3.39) any closed surface Now thisisinteresting, because wealready have atheorem about thesurface integral ofavector ﬁeld. Such asurface integral isequal tothevolume integral ofthedivergence ofthevector, according toGauss’ theorem (Eq.3.18). Gauss’ theorem, applied toVXC,says I(v><C),,da =fv-(v ><C)dV. (3.40) closed v_olumesurface inside Soweconclude thatthesecond integral must alsobezero: fv-(v ><c)av=0, (3.41) any volume andthisistrueforanyvector ﬁeldCwhatever. Since Eq.(3.41) istrueforany volume, itmust betruethatatevery point inspace theintegrand iszero. Wehave V'(VXC)=0,always. Butthisisthesame result wegotfrom vector algebra inSection 2-7. Now we begin toseehoweverything ﬁtstogether. 3-8Summary Letussummarize what wehave found about thevector calculus. These are really thesalient points ofChapters 2and3: 1.The operators 6/6x, 6/0y, and 6/dz can beconsidered asthethree components ofavector operator V,andtheformulas which result from vector algebra bytreating thisoperator asavector arecorrect: 066 "'2.Thedifference ofthevalues ofascalar ﬁeldattwopoints isequal tothe lineintegral ofthetangential component ofthegradient ofthatscalar along 3-lln. ta.’ Loop I‘ Surface S V" Fig.3-l3. Going tothelimit ofa closed surface, wefindthatthesurface integral of(VXC),must vanish. anycurve atallbetween theﬁrstandsecond points: tom-an=](:’w-as. <3-42>any curve 3.Thesurface integral ofthenormal component ofanarbitrary vector overaclosed surface isequal totheintegral ofthedivergence ofthevector over thevolume interior tothesurface: C'nJa = v-cw. (-3.43)f I closed volumesurface llilﬂlde 4.Thelineintegral ofthetangential component ofanarbitrary vector around aclosed loop isequal tothesurface integral ofthenormal component ofthecurlofthatvector overanysurface which isbounded bytheloop. fC-ds= f(VXC)-nda. (3.44) boundary surface 3-12 4 Electrostutics 4-1Statics Webegin nowourdetailed study ofthetheory ofelectromagnetism. Allof electromagnetism iscontained intheMaxwell equations. Maxwell’s equations: V-E= 5. (4.1)60 asVXE--5?, (4.2) 2 _Q€ L cVXB-at+e0, (4.3) V-B=0. (4.4) Thesituations thataredescribed bythese equations canbeverycomplicated. Wewillconsider ﬁrst relatively simple situations, andlearn how tohandle them before wetake upmore complicated ones. Theeasiest circumstance totreat isone inwhich nothing depends onthetime-—called thestatic case. Allcharges are permanently ﬁxed inspace, oriftheydomove, theymove asasteady ﬂowina circuit (sopandjareconstant intime). Inthese circumstances, alloftheterms in theMaxwell equations which aretime derivatives oftheﬁeld arezero. Inthis case, theMaxwell equations become: Electrostatics: v-E=ll. (4.5)50 VXE=0. (4.6) Magnetostatics : V><B=$, (4.7) V-B=0. (4.8) You willnotice aninteresting thing about thissetoffourequations. Itcan beseparated intotwopairs. Theelectric ﬁeldEappears onlyintheﬁrsttwo,and themagnetic ﬁeldBappears onlyinthesecond two. Thetwoﬁelds arenotinter- connected. This means thatelectricity andmagnetism aredistinct phenomena so longascharges andcurrents arestatic. Theinterdependence ofEandBdoes not appear until there arechanges incharges orcurrents, aswhen acondensor is charged, oramagnet moved. Only when there aresuﬁiciently rapid changes, so thatthetime derivatives inMaxwell’s equations become signiﬁcant, willEandB depend oneach other. Now ifyoulook attheequations ofstatics youwillseethatthestudy ofthe twosubjects wecallelectrostatics andmagnetostatics isideal from thepoint of view oflearning about themathematical properties ofvector ﬁelds. Electrostatics isaneatexample ofavector ﬁeldwithzerocurlandagiven divergence. Magnet- ostatics isaneatexample ofaﬁeldwithzerodivergence andagiven curl. Themore conventional—and youmay bethinking, more satisfactory——way ofpresenting 4-14-1Statics 4-2 Coulomb’s law; superposition 4-3Electric potential 4-4E=—V¢ 4-5TheﬂuxofE 4-6Gauss’ law;thedivergence ofE 4-7Field ofasphere ofcharge 4-8Field lines; equipotential surfaces Review: Chapters 13and14,Vol.I, Work andPotential Energy 107 2=__ £06 41r 1i z9 10941l'€° X [so]=coulomb’/newton-meter’ thetheory ofelectromagnetism istostartﬁrstwithelectrostatics andthustolearn about thedivergence. Magnetostatics andthecurlaretaken uplater. Finally, electricity andmagnetism areputtogether. Wehave chosen tostart with the complete theory ofvector calculus. Now weshall apply ittothespecial caseof electrostatics, theﬁeldofEgiven bytheﬁrstpairofequations. Wewillbegin withthesimplest situations—ones inwhich thepositions ofall charges arespeciﬁed. Ifwehadonlytostudy electrostatics atthislevel (aswe shall dointhenext twochapters), lifewould bevery simple—in fact, almost trivial. Everything canbeobtained from Coulomb’s lawandsome integration, asyouwillsee. Inmany realelectrostatic problems, however, wedonotknow, initially, where thecharges are. Weknow onlythattheyhave distributed them- selves inways thatdepend ontheproperties ofmatter. Thepositions thatthe charges takeupdepend ontheEﬁeld, which inturndepends onthepositions of thecharges. Then things cangetquite complicated. If,forinstance, acharged body isbrought near aconductor orinsulator, theelectrons andprotons inthe conductor orinsulator willmove around. Thecharge density pinEq.(4.5) may have onepartthatweknow about, from thecharge thatwebrought up;butthere willbeother parts from charges thathave moved around intheconductor. And allofthecharges must betaken intoaccount. Onecangetintosome rather subtle andinteresting problems. Soalthough thischapter istobeonelectrostatics, itwill notcover themore beautiful andsubtle parts ofthesubject. Itwilltreat onlythe situation where wecanassume thatthepositions ofallthecharges areknown. Naturally, youshould beabletodothatcasebefore youtrytohandle theother ones. 4-2Coulomb’s law;superposition Itwould belogical touseEqs. (4.5) and(4.6) asourstarting points. Itwill beeasier, however, ifwestart somewhere elseandcome back tothese equations. Theresults willbeequivalent. Wewillstart withalawthatwehave talked about before, called Coulomb’s law,which saysthatbetween twocharges atrestthere is aforce directly proportional totheproduct ofthecharges andinversely propor- tional tothesquare ofthedistance between. Theforce isalong thestraight line from onecharge totheother. Coulomb’s law: 1qlqz F1 —Hg 75¢}; ——F2. F,istheforce oncharge ql,e12istheunitvector inthedirection toqlfrom qz, andr12isthedistance between qlandq2.Theforce F2onqzisequal andopposite toF1. Theconstant ofproportionality, forhistorical reasons, iswritten as1/41re0. Inthesystem ofunits which weuse—the mkssystem—it isdeﬁned asexactly l0_7 times thespeed oflight squared. Now since thespeed oflight isapproxi- mately 3X108meters persecond, theconstant isapproximately 9X109,and theunitturns outtobenewton-meterz percoulombz orvolt-meter percoulomb. 1 _ _7 2 .. Ga) -10c (bydeﬁnition) =9.0X10°(byexperiment). (4.10) Unit: newton-meterz/coulombz, orvoltmeter/coulomb. When there aremore than twocharges present—~the only really interesting times——we must supplement Coulomb’s lawwith oneother factofnature: the force onanycharge isthevector sumoftheCoulomb forces from eachoftheother charges. Thisfactiscalled “theprinciple ofsuperposition.” That’s allthere isto electrostatics. Ifwecombine theCoulomb lawandtheprinciple ofsuperposition, there isnothing else. Equations (4.5) and(4.6)—the electrostatic equations—say nomore andnoless. 4-2 When applying Coulomb's law,itisconvenient tointroduce theideaofan electric ﬁeld. WesaythattheﬁeldE(l) istheforce perunitcharge onql(dueto allother charges). Dividing Eq.(4.9) byql,wehave, foroneother charge besides qlr E1 _ 1 qz ()— F0 ,€¢12- (4-11) Also, weconsider thatE(l) describes something about thepoint (1)even ifql were notthere—assuming thatallother charges keep their same positions. We say:E(l)istheelectric ﬁeldatthepoint (1). Theelectric ﬁeldEisavector, sobyEq.(4.11) wereally mean three equations ——one foreach component. Writing outexplicitly thex-component, Eq.(4.11) means E“(""y"’1) =43;.[(x1-av+of1-viii+(:1-Z9213/2’ ‘"12’ andsimilarly fortheother components. Ifthere aremany charges present, theﬁeldEatanypoint (1)isasumofthe contributions from each oftheother charges. Each term ofthesumwilllooklike (4.11) or(4.12). Letting q,~bethemagnitude ofthejthcharge, andr1,thedis- placement from q,tothepoint (1),wewrite E(l)-Z 142%. (4.13) __ ____ __1]J41re() r1] Which means, ofcourse, _ 1 q.1'(x1 _xl) E”"‘*""’1’ "41re(>[(x1 -an+o.-me+(Z.-z.>2r/2’ “'14) andsoon. Often itisconvenient toignore thefactthat charges come inpackages like electrons andprotons, andthink ofthem asbeing spread outinacontinuous smear ——orina“distribution,” asitiscalled. ThisisO.K. solongaswearenotinterested inwhat ishappening ontoosmall ascale. Wedescribe acharge distribution by the“charge density,” p(x,y,z).Iftheamount ofcharge inasmall volume AV2 located atthepoint (2)isAqz,thenpisdeﬁned by Aqz=p(2)AV2. (4.15) TouseCoulomb’s lawwithsuch adescription, wereplace thesums ofEqs. (4.13) or(4.14) byintegrals overallvolumes containing charges. Then wehave 15(1)=1%” I . (4_15) aH space Some people prefer towriteV 812 ==—l2: 1'12 where r12isthevector displacement to(1)from (2),asshown inFig.4-1. The integral forEisthenwritten as 1 2 V5(1)=1;; I . (4_17) aﬂ space When wewant tocalculate something with these integrals, weusually have to write them outinexplicit detail. Forthex-component ofeither Eq.(4.16) or (4.17), wewould have _ (X1-X2)P(x2, P2,Z2)dxzdyzI122 _ EM’""0‘41re0[(x1 -we+o.~we+(2.-z2>21='*/2 “'18)B11803 4-3'|2 P(3u’vI) 1; 4' ~ (2)i(xz Y:Z2) Fig.4-1. The electric ﬁeld Eat point (l),from ucharge distribution, is obtained from onintegral over the distribution. Point (llcould alsobeinside thedistribution. F b Q onepath‘ ‘another path O Fig.4-2. Thework done incarrying 0charge from atobisthenegative of theintegral ofF-dzalong thepath taken.Wearenotgoing tousethisformula much. Wewrite ithereonlytoempha- sizethefactthatwehave completely solved alltheelectrostatic problems inwhich weknow thelocations ofallofthecharges. Given thecharges, what aretheﬁelds ? Answer: Dothisintegral. Sothere isnothing tothesubject; itisjustacaseof doing complicated integrals overthree dimensions—strictly ajobforacomputing machine! With ourintegrals wecanﬁndtheﬁelds produced byasheet ofcharge, from alineofcharge, from aspherical shellofcharge, orfrom anyspeciﬁed distribution. Itisimportant torealize, aswegoontodraw ﬁeldlines, totalkabout potentials, ortocalculate divergences, thatwealready have theanswer here. Itismerely a matter ofitbeing sometimes easier todoanintegral bysome clever guesswork than byactually carrying itout. Theguesswork requires learning allkinds of strange things. Inpractice, itmight beeasier toforget trying tobeclever andal- ways todotheintegral directly instead ofbeing sosmart. Weare,however, going totrytobesmart about it.Weshall goontodiscuss some other features ofthe electric ﬁeld. 4-3Electric potential Firstwetakeuptheideaofelectric potential, which isrelated tothework done incarrying acharge from onepoint toanother. There issome distribution of charge, which produces anelectric ﬁeld. Weaskabout howmuch work itwould taketocarry asmall charge from oneplace toanother. Thework done against theelectrical forces incarrying acharge along some pathisthenegative ofthecom- ponent oftheelectrical force inthedirection ofthemotion, integrated along the path. Ifwecarry acharge from point atopoint b, b W=—/ F~d.1, where Fistheelectrical force onthecharge ateach point, anddsisthediﬁerential vector displacement along thepath. (See Fig.4-2.) Itismore interesting forourpurposes toconsider thework that would be done incarrying oneunitofcharge. Then theforce onthecharge isnumerically thesame astheelectric ﬁeld. Calling thework done against electrical forces inthis caseW(unit), wewrite b W(unit) =—/ E~ds. (4.19)G Now, ingeneral, what wegetwiththiskind ofanintegral depends onthepathwe take. Butiftheintegral of(4.19) depended onthepathfrom atob,wecould get work outoftheﬁeldbycarrying thecharge tobalong onepathandthenback toa ontheother. Wewould gotobalong thepath forwhich Wissmaller andback along theother, getting outmore work than weputin. There isnothing impossible, inprinciple, about getting energy outofaﬁeld. Weshall, infact,encounter ﬁelds where itispossible. Itcould bethatasyoumove acharge youproduce forces ontheother part ofthe“machinery.” Ifthe“ma- chinery” moved against theforce itwould loseenergy, thereby keeping thetotal energy intheworld constant. Forelectrostatics, however, there isnosuch “ma- chinery.” Weknow what theforces back onthesources oftheﬁeld are. They are theCoulomb forces onthecharges responsible fortheﬁeld. Iftheother charges areﬁxed inposition——as weassume inelectrostatics only—these back forces can donowork onthem. There isnowaytogetenergy from them—provided, of course, thattheprinciple ofenergy conservation works forelectrostatic situations. Webelieve thatitwillwork, butlet’sjustshow thatitmust follow from Coulomb’s lawofforce. Weconsider ﬁrstwhat happens intheﬁeld duetoasingle charge q.Let point abeatthedistance r1from q,andpoint batr2.Now wecarry adifferent charge, which wewillcallthe“test” charge, andwhose magnitude wechoose to 4-4 beoneunit, from atob.Let’s startwiththeeasiest possible pathtocalculate. We carry ourtestcharge ﬁrstalong thearcofacircle, thenalong aradius, asshown in part(a)ofFig.4-3. Now onthatparticular pathitischild’s playtoﬁndthework done (otherwise wewouldn’t have picked it).First, there isnowork done atall onthepathfrom ato11’.Theﬁeldisradial (from Coulomb’s law), soitisatright angles tothedirection ofmotion. Next, onthepathfrom a’tob,theﬁeldisinthe direction ofmotion andvaries as1/r2. Thus thework done onthetestcharge incarrying itfrom atobwould be b b _ .__q Q__4L_L)./,,Eds_ 41reQ r2_ 41reo (ra r), (420) Now let’stakeanother easypath. Forinstance, theoneshown inpart(b)of Fig.4-3. Itgoesforawhile along anarcofacircle, thenradially forawhile, then along anarcagain, thenradially, andsoon.Every timewegoalong thecircular parts, wedonowork. Every time wegoalong theradial parts, wemust just integrate 1/r”. Along theﬁrstradial stretch, weintegrate from r,,torat,then along thenextradial stretch from rattor,,~,andsoon.Thesumofallthese in- tegrals isthesame asasingle integral directly from r.,torb.Wegetthesame answer forthispath thatwedidfortheﬁrstpath wetried. Itisclear thatwewould get thesame answer foranypathwhich ismade upofanarbitrary number ofthesame kinds ofpieces. What about smooth paths? Would wegetthesame answer? Wediscussed thispoint previously inChapter 13ofVol.I.Applying thesame arguments used there, wecanconclude thatwork done incarrying aunitcharge from atobis independent ofthepath. W(unit)} =_'/‘bE.ds Gb a——> P351 _ Since thework done depends onlyontheendpoints, itcanberepresented as thediiierence between twonumbers. Wecanseethisinthefollowing way. Let’s choose areference point P0andagree toevaluate ourintegral byusing apaththat always goesbywayofpoint P0.Let¢(a)stand forthework done against theﬁeld ingoing from P0topoint a,andlet¢(b)bethework done ingoing from P0to point b(Fig. 4-4). Thework ingoing toPofrom a(onthewaytob)isthenegative of¢(a), sowehave that b -IE-ds=¢(1>)-¢(a). (4.21) Since onlythedifference inthefunction ¢attwopoints iseverinvolved, we donotreally have tospecify thelocation ofP0. Once wehave chosen some reference point, however, anumber ¢isdetermined foranypoint inspace; ¢is thenascalar ﬁeld. Itisafunction ofx,y,z.Wecallthisscalar function theelec- trostatic potential atanypoint. Electrostatic potential: P ¢(P) =—f E-ds. (4.22)Po Forconvenience, wewilloften take thereference point atinﬁnity. Then, forasingle charge attheorigin, thepotential ¢isgiven foranypoint (x,y,z)— using Eq.(4.20): __41. ¢(X,}’, Z)_47r60 r Theelectric ﬁeldfrom several charges canbewritten asthesumoftheelectric ﬁeldfrom theﬁrst, from thesecond, from thethird, etc.When weintegrate the sumtoﬁndthepotential wegetasumofintegrals. Each oftheintegrals isthe 4-5b 1°) OI 0 b (bl O q U Fig.4-3. Incarrying utestcharge from atobthesome work isdone along either path. v(=-bi-to)-¢<=) b w(1=°~n)-¢(t=) B "<13,-=>-¢(=> P, Fig.4-4. The work done ingoing along anypath from atobisthenegative ofthework from some point P0touplus thework from Potob. potential from oneofthecharges. Weconclude thatthepotential ¢from alotof charges isthesumofthepotentials from alltheindividual charges. There isa superposition principle alsoforpotentials. Using thesame kind ofarguments by which wefound theelectric ﬁeldfrom agroup ofcharges andforadistribution of charges, wecangetthecomplete formulas forthepotential ¢atapoint wecall(1): ¢<1>—Z)1"5, (4.24) _ j4110; ¢(1)=Z-1%/'_’_(27)1;_"§. (4.25) Remember thatthepotential ¢hasaphysical signiﬁcance: itisthepotential energy which aunitcharge would have ifbrought tothespeciﬁed point inspace from some reference point. 4-4E=-v¢ Who cares about 4»?Forces oncharges aregiven byE,theelectric ﬁeld. The point isthatEcanbeobtained easily from ¢—it isaseasy, infact, astaking a derivative. Consider twopoints, oneatxandoneat(x+dx),butboth atthe same yandz,andaskhowmuch work isdone incarrying aunitcharge from one point totheother. Thepathisalong thehorizontal linefrom xtox+dx.The work done isthedifference inthepotential atthetwopoints: 8AW =¢(x +A-x,J’,Z) T’¢(-xaysz) =£Ax' Butthework done against theﬁeldforthesame pathis AW= -[12-as =-2,“. Weseethat E,=_%. (4.26) Similarly, E,=-64»/6y, E,=—64>/82, or,summarizing with thenotation of vector analysis, E=—V¢. (4.27) Thisequation isthediﬂ'erential form ofEq.(4.22). Anyproblem withspeciﬁed charges canbesolved bycomputing thepotential from (4.24) or(4.25) andusing (4.27) togettheﬁeld. Equation (4.27) alsoagrees withwhat wefound from vector calculus: thatforanyscalar ﬁeld4» b IV¢'J8=4>(b)—4>(¢1)- (4-23) According toEq.(4.25) thescalar potential 4»isgiven byathree-dimensional integral similar totheonewehadforE.Isthere anyadvantage tocomputing ¢ rather thanE?Yes. There isonlyoneintegral for4»,while there arethree integrals forE—because itisavector. Furthermore, l/risusually alittleeasier tointegrate than x/r3. Itturns outinmany practical cases thatitiseasier tocalculate 4:and then take thegradient toﬁndtheelectric ﬁeld, than itistoevaluate thethree integrals forE.Itismerely apractical matter. There isalsoadeeper physical signiﬁcance tothepotential 4».Wehaveshown thatEofCoulomb’s lawisobtained from E=—-grad4»,when ¢isgiven by (4.22). ButifEisequal tothegradient ofascalar ﬁeld, thenweknow from the vector calculus thatthecurlofEmust vanish: VXE=0. (4.29) 4-6 Butthatisjustoursecond fundamental equation ofelectrostatics, Eq.(4.6). We have shown thatCoulomb’s lawgives anEﬁeldthatstaisﬁes thatcondition. So far,everything isallright. Wehadreally proved thatVXEwaszerobefore wedeﬁned thepotential. Wehadshown thatthework done around aclosed pathiszero. That is,that fr:-¢¢=o foranypath. WesawinChapter 3thatforanysuch ﬁeldVXEmust bezero everywhere. Theelectric ﬁeldinelectrostatics isanexample ofacurl-free ﬁeld. Youcanpractice your vector calculus byproving thatVXEiszeroinadif- ferent way—by computing thecomponents ofVXEfortheﬁeldofapoint charge, asgiven byEq.(4.11). Ifyougetzero, thesuperposition principle saysyouwould getzerofortheﬁeldofanycharge distribution. Weshould point outanimportant fact. Foranyradial force thework done is independent ofthepath, andthere exists apotential. Ifyouthink about it,the entire argument wemade above toshow thatthework integral wasindependent ofthepath depended only onthefactthattheforce from asingle charge was radial andspherically symmetric. Itdidnotdepend onthefactthatthedependence ondistance wasasl/r2——there could have been anyrdependence. Theexistence ofapotential, andthefactthatthecurlofEiszero, comes really onlyfrom the symmetry anddirection oftheelectrostatic forces. Because ofthis,Eq.(4—28)—— or(4.29)—can contain onlypartofthelawsofelectricity. 4-5TheﬂuxofE Wewillnowderive aﬁeldequation thatdepends speciﬁcally anddirectly on thefactthattheforce lawisinverse square. That theﬁeldvaries inversely asthe square ofthedistance seems, forsome people, tobe“only natural,” because “that’s thewaythings spread out.” Take alight source with light streaming out:the amount oflight that passes through asurface cutoutbyacone with itsapex at thesource isthesame nomatter atwhat radius thesurface isplaced. Itmust beso ifthere istobeconservation oflight energy. Theamount oflight perunitarea- theintensity——must varyinversely astheareacutbythecone, i.e.,inversely asthe square ofthedistance from thesource. Certainly theelectric ﬁeldshould vary inversely asthesquare ofthedistance forthesame reason! Butthere isnosuch thing asthe“same reason” here. Nobody cansaythattheelectric ﬁeldmeasures theﬂow ofsomething likelight which must beconserved. Ifwehad21“model” oftheelectric ﬁeldinwhich theelectric ﬁeldvector represented thedirection and speed—say thecurrent—of some kind oflittle “bullets” which were ﬂying out, andifourmodel required thatthese bullets were conserved, thatnone could ever disappear once itwasshotoutofacharge, thenwemight saythatwecan“see” thattheinverse square lawisnecessary. Ontheother hand, there would necessarily besome mathematical waytoexpress thisphysical idea. Iftheelectric ﬁeldwere likeconserved bullets going out,thenitwould varyinversely asthesquare ofthe distance andwewould beabletodescribe thatbehavior byanequation—which ispurely mathematical. Now there isnoharm inthinking thisway, solongaswe donotsaythattheelectric ﬁeld ismade outofbullets, butrealize thatweare using amodel tohelpusﬁndtheright mathematics. Suppose, indeed, thatweimagine foramoment thattheelectric ﬁeld did represent theﬂow ofsomething thatwasconserved——everywhere, thatis,except atcharges. (Ithastostartsomewhere!) Weimagine thatwhatever itisﬂows out ofacharge intothespace around. IfEwere thevector ofsuch aﬂow(ashisfor heatﬂow), itwould have a1/r2dependence nearapoint source. Now wewish to usethismodel toﬁndouthowtostate theinverse square lawinadeeper ormore abstract way, rather than simply saying “inverse square.” (You may wonder whyweshould want toavoid thedirect statement ofsuch asimple law,andwant instead toimply thesame thing sneakily inadiﬂ"erent way. Patience! Itwillturn outtobeuseful.) 4-7 / // E// b En/\‘// Closed Surface S / / /’/ // // // ’// // ////@/ Fig.4-5. Theﬂux of P95" CMFOQ surface Siszero. % 5...... v‘ E Fig.4-7. Anyvolume canbethought ofascompletely mode upofinﬁnitesimal truncated cones. ThefluxofEfrom one endofeach conical segment isequal and opposite totheﬂuxfrom theother end. The totol flux from thesurface Sis therefore zero.\‘| \\\\\\ \\\ \\\\\\!\*E" 5- b Q E 0/, Fig.4-8. Ifacharge isinside q surface, theﬂuxoutisnotzero.,3, / E ’ Surface S w / / // A,/ / » /,"' Eoutofthe Q; ’’ Fig.4-6. Theﬂux ofEoutofthe Point Charge surface Siszero. Weask:What isthe“ﬂow” ofEoutofanarbitrary closed surface inthe neighborhood ofapoint charge? First let’stakeaneasysurface——the oneshown inFig.4-5. IftheEﬁeld islikeaﬂow, thenetﬂow outofthisboxshould bezero. That iswhat wegetifbythe“ﬂow” from thissurface wemean thesurface integral ofthenormal component ofE——that is,theﬂuxofE.Ontheradial faces, thenor- malcomponent iszero. Onthespherical faces, thenormal component Enisjust themagnitude ofE-—minus forthesmaller faceandplusforthelarger face. The magnitude ofEdecreases asl/r2, butthesurface areaisproportional tor2,so theproduct isindependent ofr.TheﬂuxofEintofaceaisjustcancelled bythe ﬂuxoutoffaceb.Thetotal ﬂow outofSiszero, which istosaythatforthis surface Lmm=a mm Next weshow thatthetwoendsurfaces may betilted with respect tothe radial linewithout changing theintegral (4.30). Although itistrueingeneral, for ourpurposes itisonlynecessary toshow thatthisistruewhen theendsurfaces are small, sothattheysubtend asmall angle from thesource——in fact,aninﬁnitesimal angle. InFig.4—6weshow asurface Swhose “sides” areradial, butwhose “ends” aretilted. Theendsurfaces arenotsmall intheﬁgure, butyouaretoimagine the situation forverysmall endsurfaces. Then theﬁeldEwillbesuﬂiciently uniform overthesurface thatwecanusejustitsvalue atthecenter. When wetiltthesur- facebyanangle 0,theareaisincreased bythefactor 1/cos0.ButE,.,thecompo- nent ofEnormal tothesurface, isdecreased bythefactor cos0.Theproduct EnAaisunchanged. Theﬂuxoutofthewhole surface Sisstillzero. Now itiseasytoseethattheﬂuxoutofavolume enclosed byanysurface S must bezero. Anyvolume canbethought ofasmade upofpieces, likethatin Fig.4-6. Thesurface willbesubdivided completely intopairs ofendsurfaces, andsince theﬂuxes inandoutofthese endsurfaces cancel bypairs, thetotal ﬂux outofthesurface willbezero. Theideaisillustrated inFig.4-7. Wehave the completely general result thatthetotal ﬂuxofEoutofanysurface Sintheﬁeld ofapoint charge iszero. Butnotice! Ourproof works onlyifthesurface Sdoesnotsurround thecharge. What would happen ifthepoint charge were inside thesurface? Wecould still divide oursurface intopairs ofareas thatarematched byradial lines through the charge, asshown inFig.4~8. Theﬂuxes through thetwosurfaces arestillequal- bythesame arguments asbefore—only nowtheyhave thesame sign. Theflux outofasurface thatsurrounds acharge isnotzero. Then what isit?Wecanﬁnd outbyalittle trick. Suppose we“remove” thecharge from the“inside” bysur- rounding thecharge byalittle surface S’totally inside theoriginal surface S,as shown inFig.4-9. Now thevolume enclosed between thetwosurfaces SandS’ hasnocharge init.Thetotal ﬂuxoutofthisvolume (including thatthrough S’) iszero, bythearguments wehave given above. Thearguments tellus,infact,that theﬂuxintothevoltune through S’isthesame astheﬂuxoutward through S. 4-8 Wecanchoose anyshape wewishforS’,solet’smake itasphere centered on thecharge, asinFig.4-10. Then wecaneasily calculate theﬂuxthrough it.Ifthe radius ofthelittle sphere isr,thevalue ofEeverywhere onitssurface is "La4-zreo r2’ andisdirected always normal tothesurface. Weﬁndthetotal ﬂuxthrough S’if wemultiply thisnormal component ofEbythesurface area: __la 2-1 Flux through thesuface S’"(41r¢0 ,2)(41rr )—60s (4.31) anumber independent oftheradius ofthesphere! Weknow then thattheflux outward through Sisalsoq/e0—a value independent oftheshape ofSsolongas thecharge qisinside. Wecanwrite ourconclusions asfollows: 0;qoutside S Ed= ./ "a 2-;qinside S (432)anysurface S 50 Let’s return toour“bullet” analogy andseeifitmakes sense. Ourtheorem saysthatthenetﬂowofbullets through asurface iszeroifthesurface does not enclose thegunthatshoots thebullets. Ifthegunisenclosed inasurface, whatever sizeandshape itis,thenumber ofbullets passing through isthesame——it isgiven bytherateatwhich bullets aregenerated atthegun. Itallseems quite reasonable forconserved bullets. Butdoes themodel tellusanything more than weget simply bywriting Eq.(4.32)? Noonehassucceeded inmaking these “bullets” do anything elsebutproduce thisonelaw. After that, they produce nothing but errors. That iswhytoday weprefer torepresent theelectromagnetic ﬁeldpurely abstractly. 4-6Gauss’ law;thedivergence ofE Ourniceresult, Eq.(4.32), wasproved forasingle point charge. Now suppose thatthere aretwocharges, acharge qlatonepoint andacharge Q2atanother. Theproblem looks more diﬂicult. Theelectric ﬁeldwhose normal component we integrate fortheﬂuxistheﬁeldduetobothcharges. That is,ifE1represents the electric ﬁeldthatwould have been produced byqlalone, andE2represents the electric ﬁeldproduced byqzalone, thetotal electric ﬁeldisE=E1+E2. The ﬂuxthrough anyclosed surface Sis A(E...+122,.)da=[SE...da+/S22.4.1. (4.33) Theﬂuxwithboth charges present istheﬂuxduetoasingle charge plustheflux duetotheother charge. Ifboth charges areoutside S,theﬂuxthrough Siszero. Ifqlisinside Sbutq2isoutside, thentheﬁrstintegral gives q1/soandthesecond integral gives zero. Ifthesurface encloses bothcharges, eachwillgiveitscontribu- tionandwehave thattheﬂuxis(q1+q2)/e0. Thegeneral ruleisclearly thatthe total ﬁuxoutofaclosed surface isequal tothetotal charge inside, divided byso. Ourresult isanimportant general lawoftheelectrostatic ﬁeld, called Gauss’ law. “MW” fg@=mmQmm@2, @%0 anyclosed surface S 01' I E~nda = , (4.35)E0anyclosed h surface S W6115 m~=Zq. mminside S 4-9/6;...na/§$\Z\7Surface S Surface SI \ Fig.4-9. Theﬂux through Sisthe some asthefluxthrough S’. E sl Fig.4-lO.Theﬂuxthrough ospheri- culsurface containing opoint charge qisq/so. \ E P»/' 4\( Charge 'R/v\6aussiun Dl¢rlbution\ Suﬂqgq 5 P \ //,_ / § Fig.4-ll. Using Gauss‘ lowtoﬁnd theﬁeld of0uniform sphere ofcharge.Ifwedescribe thelocation ofcharges interms ofacharge density p,wecancon- sider thateach inﬁnitesimal volume dVcontains a“point” charge pdV. The sum overallcharges isthentheintegral Qt...=fpdV. (4.31) volume inside S From ourderivation youseethatGauss’ lawfollows from thefactthatthe exponent inCoulomb’s lawisexactly two. Al/r3ﬁeld, oranyl/r"ﬁeld with naé2,would notgiveGauss’ law. SoGauss’ lawisjustanexpression, inadif- ferent form, oftheCoulomb lawofforces between twocharges. Infact,working back from Gauss’ law,youcanderive Coulomb’s law. Thetwoarequite equiva- lentsolongaswekeep inmind therulethattheforces between charges isradial. Wewould nowliketowrite Gauss’ lawinterms ofderivatives. Todothis, weapply Gauss’ lawtoaninﬁnitesimal cubical surface. Weshowed inChapter 3 thattheﬂuxofEoutofsuchacube isV-Etimes thevolume dVofthecube. The charge inside ofdV,bythedeﬁnition ofp,isequal topdV,soGauss’ lawgives 60 O1‘ v-E=£- (4.38)60 Thedifferential form ofGauss’ lawistheﬁrstofourfundamental ﬁeldequations of electrostatics, Eq.(4.5). Wehave nowshown thatthetwoequations ofelectro- statics, Eqs. (4.5) and(4.6), areequivalent toCoulomb’s lawofforce. Wewill nowconsider oneexample oftheuseofGauss’ law. OM:willcome later tomany more examples.) 4-7Field ofasphere ofcharge Oneofthediﬁicult problems wehadwhen westudied thetheory ofgravita- tional attractions wastoprove thattheforce produced byasolid sphere ofmatter wasthesame atthesurface ofthesphere asitwould beifallthematter were concentrated atthecenter. Formany years Newton didn’t make public his theory ofgravitation, because hecouldn’t besure thistheorem wastrue. We proved thetheorem inChapter 13ofVol. Ibydoing theintegral forthe potential andthenﬁnding thegravitational force byusing thegradient. Now we canprove thetheorem inamost simple fashion. Only thistimewewillprove the corresponding theorem forauniform sphere ofelectrical charge. (Since thelaws ofelectrostatics arethesame asthose ofgravitation, thesame proof could be done forthegravitational ﬁeld.) Weask:What istheelectric ﬁeldEatapoint Panywhere outside thesurface ofasphere ﬁlled withauniform distribution ofcharge? Since there isno“special” direction, wecanassume thatEiseverywhere directed away from thecenter ofthe sphere. Weconsider animaginary surface thatisspherical andconcentric with thesphere ofcharge, andthatpasses through thepoint P(Fig. 4-11). Forthis surface, theﬂuxoutward is fa.da=E-4112*. Gauss’ lawtellsusthatthisﬂuxisequal tothetotalcharge Qofthesphere (over so): E-4-rrR2 =Q,60 01' 1 4-10 \ ' , ,|"”\\ * \ \ / - / \ \Lines ofEl ,2"'$ / /-El \ \ I.It.Dl.l \ \ ‘I, / I¢=Constant &_ Z 1 \ y / \ X X‘it -1} 1 1 \ Fig.4-12. Field lines andequipotential surfaces forapositive point charge. which isthesame formula wewould have forapoint charge Q.Wehave proved Newton’s problem more easily than bydoing theintegral. Itis,ofcourse, afalse kindofeasiness—it hastaken yousome timetobeabletounderstand Gauss’ law, soyoumaythink thatnotimehasreally been saved. Butafter youhave usedthe theorem more andmore, itbegins topay. Itisaquestion ofeﬂiciency. 4-8Field lines; equipotential surfaces Wewould likenowtogiveageometrical description oftheelectrostatic ﬁeld. Thetwolawsofelectrostatics, onethattheﬂuxisproportional tothecharge inside andtheother thattheelectric ﬁeldisthegradient ofapotential, canalsoberepre- sented geometrically. Weillustrate thiswithtwoexamples. First, wetaketheﬁeldofapoint charge. Wedraw linesinthedirection ofthe ﬁeld—lines which arealways tangent totheﬁeld, asinFig.4-12. These arecalled ﬁeldlines. Thelines show everywhere thedirection oftheelectric vector. Butwe alsowish torepresent themagnitude ofthevector. Wecanmake therulethatthe strength oftheelectric ﬁeldwillberepresented bythe“density” ofthelines. By thedensity ofthelines wemean thenumber oflines perunitareathrough asur- faceperpendicular tothelines. With these tworules wecanhave apicture ofthe electric ﬁeld. Forapoint charge, thedensity ofthelines must decrease asl/r2. Buttheareaofaspherical surface perpendicular tothelinesatanyradius rincreases asr2,soifwealways keep thesame number oflines foralldistances from the charge, thedensity willremain inproportion tothemagnitude oftheﬁeld. Wecan guarantee thatthere arethesame number oflines atevery distance ifweinsist thatthelines becontinuous-——that once alineisstarted from thecharge, itnever stops. Interms oftheﬁeldlines, Gauss’ lawsaysthatlines should start only at pluscharges andstopatminus charges. Thenumber which leave acharge qmust beequal toq/so. Now, wecanﬁndasimilar geometrical picture forthepotential ¢.Theeasiest waytorepresent thepotential istodraw surfaces onwhich ¢isaconstant. Wecall them equipotential surfaces—surfaces ofequal potential. Now what isthegeometri- 4-ll §t/\-/ / \ // // § \ \ / / ' // I 01‘! ANote about Units Quantity Unit NQQ71newton coulomb meter W joule p~Q/L3 coulomb/metera 1/60~FL2/ Q2newton-meter2/coulomb E~F/Q newton/ coulomb 4>~W/Q joule/coulomb =volt E~4:/L volt/meter 1/co~EL2/ Qvolt-meter/coulomb\ \ ‘ / / ‘ii) QT r ‘ Q \\ . / 1 \ \\\\r / / / \\+ // $Z IZ g i Fig.4-13. Field lines andequipotentials fortwoequal andopposite point charges. calrelationship oftheequipotential surfaces totheﬁeldlines? Theelectric ﬁeldis thegradient ofthepotential. Thegradient isinthedirection ofthemost rapid change ofthepotential, andistherefore perpendicular toanequipotential surface. IfEwere notperpendicular tothesurface, itwould have acomponent inthe surface. Thepotential would bechanging inthesurface, butthenitwouldn’t be anequipotential. Theequipotential surfaces must then beeverywhere atright angles totheelectric ﬁeldlines. Forapoint charge allbyitself, theequipotential surfaces arespheres centered atthecharge. Wehave shown inFig.4-12 theintersection ofthese spheres witha plane through thecharge. Asasecond example, weconsider theﬁeldneartwoequal charges, apositive oneandanegative one. Togettheﬁeldiseasy. Theﬁeldisthesuperposition of theﬁelds from each ofthetwocharges. So,wecantaketwopictures likeFig.4-12 andsuperimpose them—impossible! Then wewould have ﬁeldlinescrossing each other, andthat’s notpossible, because Ecan’t have twodirections atthesame point. Thedisadvantage oftheﬁeld-line picture isnowevident. Bygeometrical argu- ments itisimpossible toanalyze inavery simple waywhere thenewlines go. From thetwo independent pictures, wecan’t getthecombined picture. The principle ofsuperposition, asimple anddeep principle about electric ﬁelds, does nothave, intheﬁeld-line picture, aneasyrepresentation. Theﬁeld-line picture hasitsuses, however, sowemight stillliketodraw the picture forapairofequal (and opposite) charges. Ifwecalculate theﬁelds from Eq.(4.13) andthepotentials from (4.23), wecandraw theﬁeld lines andequi- potentials. Figure 4-13 shows theresult. Butweﬁrst hadtosolve theproblem mathematically! 4-l2 5 Application ofGauss’ Law 5-1Electrostatics isGauss’ lawplus... There aretwolaws ofelectrostatics: thatthefluxoftheelectric ﬁeldfrom a volume isproportional tothecharge inside—Gauss’ law,andthatthecirculation oftheelectric ﬁeldiszero—E isagradient. From these twolaws, allthepredictions ofelectrostatics follow. Buttosaythese things mathematically isonething; to usethem easily, andwithacertain amount ofingenuity, isanother. Inthischapter wewillwork through anumber ofcalculations which canbemade withGauss’ law directly. Wewillprove theorems anddescribe some effects, particularly incon- ductors, thatcanbeunderstood veryeasily from Gauss’ law. Gauss’ lawbyitself cannot givethesolution ofanyproblem because theother lawmust beobeyed to'o. Sowhen weuseGauss’ lawforthesolution ofparticular problems, wewillhave to addsomething toit.Wewillhave topresuppose, forinstance, some ideaofhow theﬁeldlooks—based, forexample, onarguments ofsymmetry. Orwemayhave tointroduce speciﬁcally theidea thattheﬁeld isthegradient ofapotential. 5-2Equilibrium inanelectrostatic ﬁeld Consider ﬁrstthefollowing question: When canapoint charge beinstable mechanical equilibrium intheelectric ﬁeld ofother charges? Asanexample, imagine three negative charges atthecorners ofanequilateral triangle inahori- zontal plane. Would apositive charge placed atthecenter ofthetriangle remain there? (Itwillbesimpler ifweignore gravity forthemoment, although including itwould notchange theresults.) Theforce onthepositive charge iszero, but istheequilibrium stable? Would thecharge return totheequilibrium position if displaced slightly? Theanswer isno. There arenopoints 'ofstable equilibrium inanyelectrostatic ﬁeld—except right ontopofanother charge. Using Gauss’ law,itiseasytoseewhy. First, fora charge tobeinequilibrium atanyparticular point P0,theﬁeld must bezero. Second, iftheequilibrium istobeastable one,werequire thatifwemove the charge away from Poinanydirection, there should bearestoring force directed opposite tothedisplacement. Theelectric ﬁeld atallnearby points must be pointing inward—toward thepoint P0.Butthatisinviolation ofGauss’ lawif there isnocharge atP0,aswecaneasily see. Consider atinyimaginary surface thatencloses P0,asinFig.5—l. Ifthe electric ﬁeldeverywhere inthevicinity ispointed toward P0,thesurface integral ofthenormal component iscertainly notzero. Forthecaseshown intheﬁgure, theﬁuxthrough thesurface must beanegative number. ButGauss’ lawsaysthat theﬂuxofelectric ﬁeld through anysurface isproportional tothetotal charge inside. Ifthere isnocharge atPo,theﬁeldwehave imagined violates Gauss’ law. Itisimpossible tobalance apositive charge inempty space—at apoint where there isnotsome negative charge. Apositive charge canbeinequilibrium ifitis inthemiddle ofadistributed negative charge. Ofcourse, thenegative charge distribution would have tobeheldinplace byother than electrical forces! Ourresult hasbeen obtained forapoint charge. Does thesame conclusion hold foracomplicated arrangement ofcharges held together inﬁxed relative positions——with rods, forexample? Weconsider thequestion fortwoequal charges ﬁxed onarod. Isitpossible thatthiscombination canbeinequilibrium insome electrostatic ﬁeld? Theanswer isagain no.Thetotal force ontherod cannot berestoring fordisplacements inevery direction. 5-15-1 Electrostatics isGauss’ law plus... 5-2 Equilibrium inanelectrostatic ﬁeld 5-3 Equilibrium withconductors 5-4 Stability ofatoms 5—5 Theﬁeldofalinecharge 5-6 Asheet ofcharge; twosheets 5-7 Asphere ofcharge; aspherical shell 5-8 Istheﬁeldofapoint charge exactly 1/r2? 5-9 Theﬁelds ofaconductor 5-10 Theﬁeldinacavity ofa conductor /”/I \ A \/' '/ P00 ' .l 4,./Imaginary \{ 1\, surtocc_ I surrounding Po Fig.5-l. IfPowere aposition of stable equilibrium forapositive charge, the electric ﬁeld everywhere inthe neighborhood would point toward Po. CallFthetotalforce ontherodinanyposition—F isthenavector ﬁeld. Following theargument used above, weconclude thatataposition ofstable equi- librium, thedivergence ofFmust beanegative number. Butthetotalforce onthe rodistheﬁrstcharge times theﬁeldatitsposition, plusthesecond charge times theﬁeldatitsposition: F=q1E1 ‘l’q2E2- (5-1) Thedivergence ofFisgiven by V'F =q1(V'E1) +q2(V'E2)- Ifeach ofthetwocharges qlandqgisinfreespace, both V~E1andV~E2are zero, andV-Fiszero—not negative, aswould berequired forequilibrium. You canseethatanextension oftheargument shows thatnorigid combination ofany number ofcharges canhave aposition ofstable equilibrium inanelectrostatic ﬁeld infreespace. * ,............ _-..- _: ——- W0 - 7-___________ 11¢-‘I VFig5-2 Acharge canbeinequili- l Houow bnum ifthere aremechanical constraints. Tube Now wehave notshown thatequilibrium isforbidden ifthere arepivots or other mechanical constraints. Asanexample, consider ahollow tubeinwhich a charge canmove back andforth freely, butnotsideways. Now itisveryeasyto devise anelectric ﬁeldthatpoints inward atboth ends ofthetubeifitisallowed thattheﬁeldmaypoint laterally outward nearthecenter ofthetube. Wesimply place positive charges ateachendofthetube, asinFig.5-2. There cannowbean equilibrium point even though thedivergence ofEiszero. Thecharge, ofcourse, would notbeinstable equilibrium forsideways motion were itnotfor“non- electrical” forces from thetubewalls. 5-3Equilibrium withconductors There isnostable spotintheﬁeldofasystem ofﬁxed charges. What about asystem ofcharged conductors? Canasystem ofcharged conductors produce a ﬁeldthatwillhave astable equilibrium point forapoint charge? (Wemean ata point other than onaconductor, ofcourse.) Youknow thatconductors have the property thatcharges canmove freely around inthem. Perhaps when thepoint charge isdisplaced slightly, theother charges ontheconductors willmove inaway thatwillgivearestoring force tothepoint charge? Theanswer isstillno—-al- though theproof wehave justgiven doesn’t show it.Theproof forthiscaseis more dilﬁcult, andwewillonlyindicate howitgoes. First, wenote thatwhen charges redistribute themselves ontheconductors, theycanonlydosoiftheir motion decreases their total potential energy. (Some energy islosttoheatastheymove intheconductor.) Now wehavealready shown thatifthecharges producing aﬁeldarestationary, there is,nearanyzeropoint P0 intheﬁeld, some direction forwhich moving apoint charge away from P0will decrease theenergy ofthesystem (since theforce isaway from P0). Anyreadjust- ment ofthecharges ontheconductors canonly lower thepotential energy still more, so(bytheprinciple ofvirtual work) their motion willonlyincrease theforce inthat particular direction away from P0,andnotreverse it. Ourconclusions donotmean thatitisnotpossible tobalance acharge by electrical forces. Itispossible ifoneiswilling tocontrol thelocations orthesizes ofthesupporting charges with suitable devices. You know thatarodstanding on itspoint inagravitational ﬁeld isunstable, butthisdoes notprove thatitcannot bebalanced ontheendofaﬁnger. Similarly, acharge canbeheld inonespot by electric ﬁelds iftheyarevariable. Butnotwithapassive—that is,astatic—system. 5-2 5-4Stability ofatoms Ifcharges cannot beheldstably inposition, itissurely notproper toimagine matter tobemade upofstatic point charges (electrons andprotons) governed only bythelaws ofelectrostatics. Such astatic conﬁguration isimpossible; itwould collapse! , Itwasonce suggested thatthepositive charge ofanatom could bedistributed uniformly inasphere, andthenegative charges, theelectrons, could beatrest inside thepositive charge, asshown inFig.5-3. Thiswastheﬁrstatomic model, proposed byThompson. ButRutherford concluded from theexperiment ofGeiger andMarsden thatthepositive charges were verymuch concentrated, inwhat he called thenucleus. Thompson’s static model hadtobeabandoned. Rutherford andBohr thensuggested thattheequilibrium might bedynamic, withtheelectrons revolving inorbits, asshown inFig.5-4.Theelectrons would bekeptfrom falling intoward thenucleus bytheir orbital motion. Wealready know atleast one diﬂiculty withthispicture. With suchmotion, theelectrons would beaccelerating (because ofthecircular motion) andwould, therefore, beradiating energy. They would losethekinetic energy required tostayinorbit, andwould spiral intoward thenucleus. Again unstable! Thestability oftheatoms isnowexplained interms ofquantum mechanics. Theelectrostatic forces pulltheelectron asclose tothenucleus aspossible, butthe electron iscompelled tostayspread outinspace over adistance given bythe uncertainty principle. Ifitwere conﬁned intoosmall aspace, itwould have a great uncertainty inmomentum. Butthatmeans thatitwould have ahigh ex- pected energy-—which itwould usetoescape from theelectrical attraction. The netresult isanelectrical equilibrium nottoodifferent from theideaofThompson —only itisthenegative charge thatisspread out(because themass oftheelectron issomuch smaller thanthemass oftheproton). 5-5Theﬁeldofalinecharge Gauss’ lawcanbeused tosolve anumber ofelectrostatic ﬁeldproblems in- volving aspecial symmetry—usually spherical, cylindrical, orplanar symmetry. Intheremainder ofthischapter wewillapply Gauss’ lawtoafewsuchproblems. Theeasewith which these problems canbesolved maygivethemisleading impres- sionthatthemethod isverypowerful, andthatoneshould beabletogoonto many other problems. Itisunfortunately notso.Onesoon exhausts thelistof problems thatcanbesolved easily with Gauss’ law. Inlater chapters wewill develop more powerful methods forinvestigating electrostatic ﬁelds. Asourﬁrstexample, weconsider asystem withcylindrical symmetry. Suppose thatwehave averylong, uniformly charged rod. Bythiswemean thatelectric charges aredistributed uniformly along anindeﬁnitely longstraight line,withthe charge Aperunitlength. Wewishtoknow theelectric ﬁeld. Theproblem can,of course, besolved byintegrating thecontribution totheﬁeldfrom every partof theline. Wearegoing todoitwithout integrating, byusing Gauss’ lawandsome guesswork. First, wesurmise thattheelectric ﬁeldwillbedirected radially outward from theline. Anyaxial component from charges ononesidewould beaccom- panied byanequal axial component from charges ontheother side. Theresult could onlybearadial ﬁeld. Italsoseems reasonable thattheﬁeldshould have the same magnitude atallpoints equidistant from theline. Thisisobvious. (Itmay notbeeasytoprove, butitistrueifspace issymmetric-—as webelieve itis.) WecanuseGauss’ lawinthefollowing way. Weconsider animaginary surface intheshape ofacylinder coaxial with theline, asshown inFig.5-5. According toGauss’ law,thetotalﬂuxofEfrom thissurface isequal tothecharge inside divided byen.Since theﬁeldisassumed tobenormal tothesurface, the normal component isthemagnitude oftheﬁeld. Let’s callitE.Also, lettheradius ofthecylinder ber,anditslength betaken asoneunit, forconvenience. Theﬂux through thecylindrical surface isequal toEtimes theareaofthesurface, which is 21rr. Theﬂuxthrough thetwoendfaces iszerobecause theelectric ﬁeldistan- 5-3umroau sm-ten:9orPOSITIVEcanes: IIIIII IIIIIIII IIIIIIIII IIIIIIIIII IIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIINIIIIIIIIIIIIITIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIlIIIIIaIIII =NEGATIVE OIMRGE CGJCENTRATED ATTHE CENTER Fig.5-3. TheThompson model ofan atom. POSITIVE NUCLEUS ATTHE CENTER NEGITIVE ELECTRQPB IN PLANETARY ORII T8 Fig.5-4. TheRutherford-Bohr model ofonatom. ._ 5 49§ﬁ‘i?=i'€é‘ vf‘\LINE CHARGE Fig.5-5. Acylindrical gaussian sur- face coaxial withalinecharge. as.GAUSSIAN SURFAC E Fig.5-6. The electric ﬁeld near 0 uniformly charged sheet canbefound by applying Gauss’ lawtoanimaginary box. I |1 | + - + lei E-0 ‘E 5-0+ — + _ (bl .,-e + 4. - (cl *4-E " .,. _. +_ - ' II I Fig.5-7. The ﬁeld between two charged sheets isa/eo.gcntial tothem. Thetotal charge inside oursurface isjustA,because thelength of thelineinside isoneunit. Gauss‘ lawthengives E-21rr =A/co, AE—ﬁ ' (5.2) Theelectric ﬁeld ofalinecharge depends inversely ontheﬁrst power ofthe distance from theline. 5-6Asheet ofcharge; twosheets Asanother example, wewillcalculate theﬁeldfrom auniform plane sheet of charge. Suppose thatthesheet isinﬁnite inextent andthatthecharge perunit areaistr.Wearegoing totakeanother guess. Considerations ofsymmetry lead ustobelieve thattheﬁelddirection iseverywhere normal totheplane, andifwe have noﬁeldfrom anyother charges intheworld, theﬁelds must bethesame (in magnitude) oneach side. This time wechoose forourGaussian surface arec- tangular boxthatcutsthrough thesheet, asshown inFig.5-6. Thetwofaces parallel tothesheet willhave equal areas, sayA.Theﬁeldisnormal tothese two faces, andparallel totheother four. Thetotal ﬂuxisEtimes theareaoftheﬁrst face, plusEtimes theareaoftheopposite face——with nocontribution from the other fourfaces. Thetotal charge enclosed intheboxiso'A.Equating theﬂuxto thecharge inside, wehave '< EA+EA=‘-'5. from which 0-E-E1 (5.3) asimple butimportant result. Youmayremember thatthesame result wasobtained inanearlier chapter byanintegration overtheentire surface. Gauss’ lawgives ustheanswer, inthis instance, much more quickly (although itisnotasgenerally applicable asthe earlier method). Weemphasize thatthisresult applies onlytotheﬁeldduetothecharges on thesheet. Ifthere areother charges intheneighborhood, thetotalﬁeldnearthe sheet would bethesumof(5.3) andtheﬁeld oftheother charges. Gauss’ law would thentellusonlythat E,+E,= (5.4) where E1andE2aretheﬁelds directed outward oneachsideofthesheet. Theproblem oftwoparallel sheets withequal andopposite charge densities, +crand—a,isequally simple ifweassume again thattheoutside world isquite symmetric. Either bysuperposing twosolutions forasingle sheet orbyconstruct- ingagaussian boxthatincludes both sheets, itiseasily seenthattheﬁeldiszero outside ofthetwosheets (Fig. 5—7a). Byconsidering aboxthatincludes onlyone surface ortheother, asin(b)or(c)oftheﬁgure, itcanbeseen thattheﬁeld between thesheets must betwice what itisforasingle sheet. Theresult is E(between thesheets) =tr/co, (5.5) E(outside) =O. (5.6) 5-7Asphere ofcharge; aspherical shell Wehave already (inChapter 4)used Gauss’ lawtoﬁndtheﬁeld outside a uniformly charged spherical region. Thesame method canalsogiveustheﬁeld atpoints inside thesphere. Forexample, thecomputation canbeused toobtain agood approximation totheﬁeldinside anatomic nucleus. lnspite ofthefact thattheprotons inanucleus repel each other, theyare,because ofthestrong nu- clear foroes, spread nearly uniformly throughout thebody ofthenucleus. 5-4I Suppose thatwehaveasphere ofradius Rﬁlleduniformly withcharge. Let pbethecharge perunitvolume. Again using arguments ofsymmetry, weassume theﬁeldtoberadial andequal inmagnitude atallpoints atthesame distance from thecenter. Toﬁndtheﬁeld atthedistance rfrom thecenter, wetake a spherical gaussian surface ofradius r(r<R),asshown inFig.5-8. Theﬂuxout ofthissurface is 41rr2E. Thecharge inside ourgaussian surface isthevolume inside times p,or %1rr2p. Using Gauss’ law,itfollows thatthemagnitude oftheﬁeldisgiven by 1-:=Bl (r<R). (5.7)3C0 Youcanseethatthisformula gives theproper result forr=R.Theelectric ﬁeld isproportional totheradius andisdirected radially outward. Thearguments wehave justgiven forauniformly charged sphere canbe applied alsotoathinspherical shell ofcharge. Assuming thattheﬁeldisevery- where radial andisspherically symmetric, onegetsimmediately from Gauss’ lawthattheﬁeld outside theshell islikethatofapoint charge, while theﬁeld everywhere inside theshell iszero. (Agaussian surface inside theshell willcon- tainnocharge.) S-8Istheﬁeld ofapoint charge exactly 1/1'2? Ifwelookinalittlemore detail athowtheﬁeldinside theshellgetstobezero, wecanseemore clearly whyitisthatGauss’ lawistrueonlybecause thecoulomb force depends exactly onthesquare ofthedistance. Consider anypoint Pinside auniform spherical shell ofcharge. Imagine asmall cone whose apex isatPand which extends tothesurface ofthesphere, where itcutsoutasmall surface area Aa1,asinFig.5-9. Anexactly symmetric cone diverging from theopposite side ofPwould cutoutthesurface areaAC2. Ifthedistances from Ptothese twoele- ments ofareaarer1andr2,theareas areintheratio 2 ‘L=2.Aal rf (You canshow thisbygeometry foranypoint Pinside thesphere.) Ifthesurface ofthesphere isuniformly charged, thecharge Aqoneachofthe elements ofareaisproportional tothearea, so E=A2.A91 A411 Coulomb’s lawthensaysthatthemagnitudes oftheﬁelds produced atPbythese twosurface elements areintheratio Q=_‘l2_/"5 =1_ E1 111/Ff Theﬁelds cancel exactly. Since allparts ofthesurface canbepaired offinthesame way, thetotal ﬁeldatPiszero. Butyoucanseethatitwould notbesoifthe exponent ofrinCoulomb’s lawwere notexactly two. Thevalidity ofGauss’ lawdepends upon theinverse square lawofCoulomb. Iftheforce lawwere notexactly theinverse square, itwould notbetruethatthe ﬁeldinside auniformly charged sphere would beexactly zero. Forinstance, ifthe force varied more rapidly, like,say,theinverse cube ofr,thatportion ofthesur- facewhich isnearer toaninterior point would produce aﬁeldwhich islarger than thatwhich isfarther away, resulting inaradial inward ﬁeldforapositive surface 5-5UNIFORM CHARGE DENSITY eel’ ‘i"I I‘ Fig.5-8. Gauss‘ lawcanbeused to ﬁndtheﬁeld inside 0uniformly charged sphere. Mi '1 P I’ A03 Fig.5-9. Theﬁeld iszero atany point Pinside aspherical shell ofcharge. Q ‘I O V(ol »%‘J‘.§5° .——- »amen:O 0 0 O msuuron etscvnousrsn (bl Q _. Fig.5-l0.Theelectric ﬁeld iszero inside aclosed conducting shell. lcharge. These conclusions suggest anelegant wayofﬁnding outwhether thein- verse square lawisprecisely correct. Weneed onlydetermine whether ornotthe ﬁeldinside ofauniformly charged spherical shellisprecisely zero. Itislucky thatsuchamethod exists. Itisusually dilﬁcult tomeasure aphysical quantity tohighprecisi0n—-a onepercent result maynotbetoodiﬁicult, buthow would onegoabout measuring, say,Coulomb’s lawtoanaccuracy ofonepartin abillion? Itisalmost certainly notpossible with thebestavailable techniques to measure theforce between twocharged objects with such anaccuracy. Butby determining only that theelectric ﬁelds inside acharged sphere aresmaller than some value wecanmake ahighly accurate measurement ofthecorrectness of Gauss’ law,andhence oftheinverse square dependence ofCoulomb’s law. What onedoes, ineffect, iscompare theforce lawtoanideal inverse square. Such com- parisons ofthings thatareequal, ornearly so,areusually thebases ofthemost precise physical measurements. How shall weobserve theﬁeldinside acharged sphere? Onewayistotry tocharge anobject bytouching ittotheinside ofaspherical conductor. You know thatifwetouch asmall metal balltoacharged object andthentouch itto anelectrometer themeter willbecome charged andthepointer willmove from zero(Fig. 5—l0a). Theballpicks upcharge because there areelectric ﬁelds outside thecharged sphere thatcause charges torunonto (oroﬁ’)thelittleball. Ifyoudo thesame experiment bytouching thelittleballtotheinside ofthecharged sphere, youﬁndthatnocharge iscarried totheelectrometer. With such anexperiment youcaneasily show thattheﬁeldinside is,atmost, afewpercent oftheﬁeldout- side, andthatGauss’ lawisatleast approximately correct. Itappears thatBenjamin Franklin wastheﬁrsttonotice thattheﬁeldinside a conducting shell iszero. Theresult seemed strange tohim. When hereported his observation toPriestley, thelatter suggested thatitmight beconnected with an inverse square law,since itwasknown thataspherical shell ofmatter produced nogravitational ﬁeld inside. ButCoulomb didn’t measure theinverse square dependence until 18years later, andGauss’ lawcame evenlater still. Gauss’ lawhasbeen checked carefully byputting anelectrometer inside a large sphere andobserving whether anydeﬂections occur when thesphere is charged toahighvoltage. Anullresult isalways obtained. Knowing thegeometry oftheapparatus andthesensitivity ofthemeter, itispossible tocompute the minimum ﬁeldthatwould beobserved. From thisnumber itispossible toplace an upper limit onthedeviation oftheexponent from two. Ifwewrite thattheelec- trostatic force depends onr"2+‘, wecanplace anupper bound one.Bythismethod Maxwell determined thatewaslessthan 1/10,000. Theexperiment wasrepeated andimproved upon in1936byPlimpton andLaughton. They found thatCoulomb’s exponent differs from twobylessthan onepartinabillion. Now thatbrings upaninteresting question: How accurate doweknow this Coulomb lawtobeinvarious circumstances? Theexperiments wejustdescribed measure thedependence oftheﬁeld ondistance fordistances ofsome tensof centimeters. Butwhat about thedistances inside anatom-—in thehydrogen atom, forinstance, where webelieve theelectron isattracted tothenucleus by thesame inverse square law? Itistruethatquantum mechanics must beused for themechanical part ofthebehavior oftheelectron, buttheforce istheusual electrostatic one. Intheformulation oftheproblem, thepotential energy ofan electron must beknown asafunction ofdistance from thenucleus, andCoulomb’s lawgives apotential which varies inversely withtheﬁrstpower ofthedistance. How accurately istheexponent known forsuch small distances? Asaresult of very careful measurements in1947 byLamb andRetherford ontherelative positions oftheenergy levels ofhydrogen, weknow thattheexponent iscorrect again toonepartinabillion ontheatomic scale-—that is,atdistances oftheorder ofoneangstrom (l0“8 centimeter). The accuracy oftheLamb-Retherford measurement waspossible again because ofaphysical “accident.” Two ofthestates ofahydrogen atom are expected tohave almost indentical energies onlyifthepotential varies exactly as l/r.Ameasurement wasmade oftheveryslight dzﬂerence inenergies byﬁnding 5-6 thefrequency wofthephotons thatareemitted orabsorbed inthetransition from onestate totheother, using fortheenergy difference AE=hw. Computations showed thatAEwould have been noticeably different from what wasobserved if theexponent intheforce lawl/r2differed from 2byasmuch asonepartinabillion. Isthesame exponent correct atstillshorter distances? From measurements in nuclear physics itisfound thatthere areelectrostatic forces attypical nuclear diStances—at about IOTI3 centimeter——and that they stillvary approximately as theinverse square. Weshall look atsome oftheevidence inalater chapter. Coulomb’s lawis,weknow, stillvalid, atleast tosome extent, atdistances ofthe order of10"” centimeter. How about l0““ centimeter? Thisrange canbeinvestigated bybombarding protons with veryenergetic electrons andobserving howtheyarescattered. Re- sults todateseem toindicate thatthelawfailsatthese distances. Theelectrical force seems tobeabout l0times tooweak atdistances lessthan l0'14centimeter. Now there aretwopossible explanations. OneisthattheCoulomb lawdoesnot work atsuch small distances; theother isthat ourobjects, theelectrons and protons, arenotpoint charges. Perhaps either theelectron orproton, orboth, is some kindofasmear. Most physicists prefer tothink thatthecharge oftheproton issmeared. Weknow thatprotons interact strongly with mesons. This implies thataproton will,from timetotime, exist asaneutron witha1r"'meson around it.Such aconﬁguration would act—on theaverage—like alittle sphere ofpositive charge. Weknow thattheﬁeldfrom asphere ofcharge doesnotvaryas1/r2all thewayintothecenter. Itisquite likely thattheproton charge issmeared, but thetheory ofpions isstillquite incomplete, soitmayalsobethatCoulomb’s law failsatverysmall distances. Thequestion isstillopen. Onemore point: Theinverse square lawisvalid atdistances likeonemeter andalsoatl0"1°m; butisthecoefﬁcient l/41re0 thesame? Theanswer isyes; atleasttoanaccuracy ofl5parts inamillion. Wegoback nowtoanimportant matter thatweslighted when wespoke of theexperimental veriﬁcation ofGauss’ law. You may have wondered how the experiment ofMaxwell orofPlimpton andLaughton could givesuchanaccuracy unless thespherical conductor theyused wasaperfect sphere. Anaccuracy of onepartinabillion isreally something toachieve, andyoumight wellaskwhether theycould make asphere which wasthatprecise. There arecertain tobeslight irregularities inanyrealsphere andifthere areirregularities, willtheynotproduce ﬁelds inside? Wewishtoshow nowthatitisnotnecessary tohaveaperfect sphere. Itispossible, infact,toshow thatthere isnoﬁeldinside aclosed conducting shell ofanyshape. Inother words, theexperiments depended onl/r2, buthadnothing todowiththesurface being asphere (except thatwithasphere itiseasier tocal- culate what theﬁelds would beifCoulomb hadbeen wrong), sowetakeupthat subject now. Toshow this, itisnecessary toknow some oftheproperties of electrical conductors. 5—9Theﬁelds ofaconductor Anelectrical conductor isasolid thatcontains many “free” electrons. The electrons canmove around freely inthematerial, butcannot leave thesurface. Inametal there aresomany freeelectrons thatanyelectric ﬁeldwillsetlarge numbers ofthem intomotion. Either thecurrent ofelectrons sosetupmust be continually kept moving byexternal sources ofenergy, orthemotion ofthe electrons willcease astheydischarge thesources producing theinitial ﬁeld. In “electrostatic” situations, wedonotconsider continuous sources ofcurrent (they willbeconsidered later when westudy magnetostatics), sotheelectrons move only until they have arranged themselves toproduce zero electric ﬁeld everywhere inside theconductor. (This usually happens inasmall fraction ofasecond.) If there were anyﬁeldleft,thisﬁeld would urge stillmore electrons tomove; the onlyelectrostatic solution isthattheﬁeldiseverywhere zeroinside. Now consider theinterior ofacharged conducting object. (By“interior” we mean inthemetal itself.) Since themetal isaconductor, theinterior ﬁeld must 5-7 ‘P conoucron " eaussum’_ 1.SURFACE /I 52'‘E ‘P ’ cm.5URFxE CNQRGE 9' 4- DENSITY 0‘ Fig.5-1l.Theelectric ﬁeld iustout- side thesurface ofaconductor ispro- portional tothelocal surface density of charge. if4;-FFig.5-12. What istheﬁeld inan empty cavity ofaconductor, forany shape?bezero, andsothegradient ofthepotential ¢iszero. That means that¢does not vary from point topoint. Every conductor isanequipotential region, andits surface isanequipotential surface. Since inaconducting material theelectric ﬁeldiseverywhere zero, thedivergence ofEiszero, andbyGauss’ lawthecharge density intheinterior oftheconductor must bezero. Ifthere canbenocharges inaconductor, howcaniteverbecharged ?What dowemean when wesayaconductor is“charged”? Where arethecharges? Theanswer isthattheyreside atthesurface oftheconductor, where there are strong forces tokeep them from leaving—they arenotcompletely “free.” When westudy solid-state physics, weshall ﬁndthattheexcess charge ofanyconductor isontheaverage within oneortwoatomic layers ofthesurface. Forourpresent purposes, itisaccurate enough tosaythatifanycharge isputon,orin,aconductor itallaccumulates onthesurface; there isnocharge intheinterior ofaconductor. Wenotealsothattheelectric ﬁeldjustoutside thesurface ofaconductor must benormal tothesurface. There canbenotangential component. Ifthere were a tangential component, theelectrons would move along thesurface; there areno forces preventing that. Saying itanother way: weknow thattheelectric ﬁeldlines must always goatright angles toanequipotential surface. Wecanalso,using Gauss’ law,relate theﬁeldstrength justoutside aconductor tothelocal density ofthecharge atthesurface. Foragaussian surface, wetakea small cylindrical boxhalfinside andhalfoutside thesurface, liketheoneshown inFig.5-1l.There isacontribution tothetotalﬂuxofEonlyfrom thesideofthe boxoutside theconductor. Theﬁeldjustoutside thesurface ofaconductor isthen Outside aconductor: E=5, (5.3)60 where tristhelocal surface charge density. Why does asheet ofcharge onaconductor produce adifferent ﬁeldthanjust asheet ofcharge? Inother words, whyis(5.8)twice aslarge as(5.3)? Thereason, ofcourse, isthatwehave notsaidfortheconductor thatthere areno“other” charges around. There must, infact,besome tomake E=0intheconductor. Thecharges intheimmediate neighborhood ofa-point Ponthesurface do,infact, giveaﬁeld E1,,c,,1 =o1,,,,,,1/2e0 both inside andoutside thesurface. Butallthe restofthecharges ontheconductor “conspire” toproduce anadditional ﬁeld at thepoint Pequal inmagnitude toE1,,,,,1. Thetotal ﬁeldinside goestozeroand theﬁeldoutside to2E1,,,,,,; =a/co. 5-10 Theﬁeldinacavity ofaconductor Wereturn nowtotheproblem ofthehollow container—a conductor witha cavity. There isnoﬁeldinthemetal, butwhat about inthecavity? Weshallshow thatifthecavity isempty thenthere arenoﬁelds init,nomatter what theshape of theconductor orthecavity—say fortheoneinFig.5-12. Consider agaussian surface, likeSinFig.5—l2, thatencloses thecavity butstays everywhere inthe conducting material. Everywhere onStheﬁeldiszero, sothere isnoﬂuxthrough Sandthetotalcharge inside Siszero. Foraspherical shell, onecould thenargue from symmetry thatthere could benocharge inside. But,ingeneral, wecanonly saythatthere areequal amounts ofpositive andnegative charge ontheinner surface oftheconductor. There could beapositive surface charge ononepart andanegative onesomewhere else,asindicated inFig.5-12. Such athing cannot beruled outbyGauss’ law. What really happens, ofcourse, isthatanyequal andopposite charges on theinner surface would slidearound tomeet eachother, cancelling outcompletely. Wecanshow thattheymust cancel completely byusing thelawthatthecirculation ofEisalways zero(electrostatics). Suppose there were charges onsome parts of theinner surface. Weknow thatthere would have tobeanequal number ofop- posite charges somewhere else. Now anylines ofEwould have tostart onthe 5-8 7 positive charges andendonthenegative charges (since weareconsidering onlythe casethatthere arenofreecharges inthecavity). Now imagine aloopPthatcrosses thecavity along alineofforce from some positive charge tosome negative charge, andreturns toitsstarting point viatheconductor (asinFig.5-12). Theintegral along such alineofforce from thepositive tothenegative charges would notbe zero. Theintegral through themetal iszero, since E=0.Sowewould have fr-as #02?? Butthelineintegral ofEaround anyclosed loopinanelectrostatic ﬁeldisalways zero. Sothere canbenoﬁelds inside theempty cavity, noranycharges onthe inside surface. You should notice carefully oneimportant qualiﬁcation wehave made. Wehave always said“inside anempty” cavity. Ifsome charges areplaced atsome ﬁxed locations inthecavity-—as onaninsulator oronasmall conductor insulated from themain one——then there canbeﬁelds inthecavity. Butthenthatisnotan “empty” cavity. Wehave shown that ifacavity iscompletely enclosed byaconductor, no static distribution ofcharges outside canever produce anyﬁelds inside. This explains theprinciple of“shielding” electrical equipment byplacing itinametal can.Thesan-lg ?:‘§Ll‘Ia%?tS canbeusedtoshow thatnostatic distribution ofcharges inside aclose conduc orcanproduce anyﬁelds outside. Shielding works both ways! Inelectrostatics——but notinvarying ﬁelds—the ﬁelds onthetwosides ofa closed conducting shell arecompletely independent. Now youseewhy itwaspossible tocheck Coulomb’s lawtosuch agreat precision. Theshape ofthehollow shell used doesn’t matter. Itdoesn’t need to bespherical; itcould besquare! IfGauss’ lawisexact, theﬁeldinside isalways zero. Now youalsounderstand whyitissafetositinside thehigh-voltage terminal ofamillion-volt vandeGraaff generator, without worrying about getting a shock-—because ofGauss’ law. S-9 6 The Electric Field inVarious Circumstances 6-1Equations oftheelectrostatic potential Thischapter willdescribe thebehavior oftheelectric ﬁeldinanumber of diﬁerent circumstances. Itwillprovide some experience with thewaytheelectric ﬁeld behaves, and willdescribe some ofthemathematical methods which are usedtoﬁndthisﬁeld. Webegin bypointing outthatthewhole mathematical problem isthesolution oftwoequations, theMaxwell equations forelectrostatics: v-E=ll. (61)E0 VXE=O. (6.2) Infact,thetwocanbecombined intoasingle equation. From thesecond equation, weknow atonce that wecandescribe theﬁeld asthegradient ofascalar (see Section 3-7):E=—V¢. (6.3) Wemay, ifwewish, completely describe anyparticular electric ﬁeld interms ofitspotential ¢>.Weobtain thedifferential equation that ¢must obey bysub- stituting Eq.(6.3) into (6.1), toget vv¢=—%- mo Thedivergence ofthegradient of¢isthesame asV2operating on<15: 02 02 a2v-v¢=v2¢=§+,y‘Z+;,z—‘§» <6-5) sowewrite Eq.(6.4) as 2 pV¢~-5- (6.6) Theoperator V2lScalled theLaplacian, andEq(66)IScalled thePoisson equa- tion. Theentire subject ofelectrostatics, from amathematical point ofview, is merely astudy ofthesolutions ofthesingle equation (6.6). Once ¢isobtained by solving Eq.(6.6)wecanﬁndEimmediately from Eq.(6.3). Wetake upﬁrstthespecial class ofproblems inwhich pisgiven asafunction ofx,y,z.Inthat case theproblem isalmost trivial, forwealready know the solution ofEq.(6.6) forthegeneral case. Wehave shown that ifpisknown at every point, thepotential atpoint (1)is =i>(l)—/”‘—2l‘1@. <61) _ 47r€Or12 where p(2)isthecharge density, dV2 isthevolume element atpoint (2),andr12 isthedistance between points (l)and(2).Thesolution ofthedzﬂerential equation (6.6)isreduced toanintegration over space. Thesolution (6.7) should beespecially noted, because there aremany situations inphysics that lead toequations like V2(something) =(something else), andEq.(6.7) isaprototype ofthesolution foranyofthese problems. Thesolution ofelectrostatic ﬁeld problems isthus completely straightforward when thepositions ofallthecharges areknown. Let’s seehow itworks inafew examples. 6-16-1 Equations oftheelectrostatic potential 6-2 Theelectric dipole 6-3 Remarks onvector equations 6-4 Thedipole potential asa gradient 6-5 Thedipole approximation for anarbitrary distribution 6-6 Theﬁelds ofcharged conductors 6-7 Themethod ofimages 6-8 Apoint charge near a conducting plane 6-9 Apoint charge near a conducting sphere 6-10 Condensers; parallel plates 6-11 High-voltage breakdown 6-12 Theﬁeld emission microscope Review. Chapter 23,Vol. I,Resonance XZ P(==,y,1)0 Ml " r-—q Fig. 6-1. Adipole: two charges +qand —qthedistance dopcirt. ..Fig. 6—2. The water molecule H20. The hydrogen atoms hove slightly less than their shore oftheelectron cloud; the oxygen, slightly more.6—2Theelectric dipole First, take twopoint charges, +qand—q,separated bythedistance d.Let thez-axis gothrough thecharges, andpicktheorigin halfway between, asshown inFig. 6—l. Then, using (4.24), thepotential from thetwocharges isgiven by ¢(x.y.Z) l q “q =41%\/[Z-(d/2)]2+x2+yz+\/[Z+(d/2)]2+x2+yd‘(63) Wearenotgoing towrite outtheformula fortheelectric ﬁeld, butwecanalways calculate itonce wehave thepotential. Sowehave solved theproblem oftwo charges. There isanimportant special case inwhich thetwocharges arevery close together—which istosaythatweareinterested intheﬁelds onlyatdistances from thecharges large incomparison with their separation. Wecallsuch aclose pair ofcharges adipole. Dipoles arevery common. A“dipole” antenna canoften beapproximated bytwocharges separated bya small distance—if wedon’t askabout theﬁeld tooclose totheantenna. (Weare usually interested inantennas with moving charges; then theequations ofstatics donotreally apply, butforsome purposes they areanadequate approximation.) More important perhaps, areatomic dipoles. Ifthere isanelectric ﬁeld in anymaterial, theelectrons andprotons feelopposite forces andaredisplaced relative toeach other. Inaconductor, youremember, some oftheelectrons move tothesurfaces, sothattheﬁeld inside becomes zero. Inaninsulator the electrons cannot move very far;they arepulled back bytheattraction ofthenu- cleus. They do,however, shift alittle bit. Soalthough anatom, ormolecule, remains neutral inanexternal electric ﬁeld, there isavery tinyseparation ofits positive andnegative charges anditbecomes amicroscopic dipole. Ifweare interested intheﬁelds ofthese atomic dipoles intheneighborhood ofordinary- sized objects, wearenormally dealing with distances large compared with the separations ofthepairs ofcharges. Insome molecules thecharges aresomewhat separated even intheabsence ofexternal ﬁelds, because oftheform ofthemolecule. Inawater molecule, for example, there isanetnegative charge ontheoxygen atom andanetpositive charge oneach ofthetwohydrogen atoms, which arenotplaced symmetrically butasinFig.6—2. Although thecharge ofthewhole molecule iszero, there isa charge distribution with alittle more negative charge ononeside andalittle more positive charge ontheother. This arrangement iscertainly notassimple astwopoint charges, butwhen seen from faraway thesystem actslikeadipole. Asweshall seealittle later, theﬁeld atlarge distances isnotsensitive tothe ﬁnedetails. Let’s look, then, attheﬁeld oftwoopposite charges with asmall separation d.Ifa’becomes zero, thetwocharges areontopofeach other, thetwopotentials cancel, andthere isnoﬁeld. Butifthey arenotexactly ontopofeach other, we cangetagood approximation tothepotential byexpanding theterms of(6.8) in apower series inthesmall quantity d(using thebinomial expansion). Keeping terms only toﬁrstorder ind,wecanwrite 2 (z—-31)»-=z2—zd. x2+y2+z2=r2. 2 (2-55) +x2+y2==r2—zd=r2(l—€§-g),Itisconvenient towrite Then and 1 l 1 zd)_”2 \/[Z—(d/2)]2if+yeI~/en—(Z11/r2)]z7(1_'7'6—2 Using thebinomial expansion again for[1—(zd/r2)]_”2—and throwing away terms with higher powers than thesquare ofd—-we get l lzd 7(1+ta)" _~__1__-__ cl(1_121).\/[Z+(d/2)]2+'T+‘yar1'2 Thediﬁerence ofthese twoterms gives forthepotentialSimilarly, lz¢>(x,y Z)=——-5qd- (6-9)’ 41r60 r Thepotential, andhence theﬁeld, which isitsderivative, isproportional toqd, theproduct ofthecharge andtheseparation. This product isdeﬁned asthe dipole moment ofthetwocharges, forwhich wewillusethesymbol p(donot confuse with momentum!): p=qd. (6.10) Equation (6.9) canalsobewritten as l 0 <y<x.y.z> =;4,,—60’i9§,i-y (6.11) since z/r=cos0,where 0istheangle between theaxis ofthedipole andthe radius vector tothepoint (x,y,2)-see Fig.6-l. Thepotential ofadipole decreases asl/r2 foragiven direction from theaxis(whereas forapoint charge itgoes as 1/r). Theelectric ﬁeld Eofthedipole willthen decrease asl/r3. Wecanputourformula intoavector form ifwedeﬁne pasavector whose magnitude ispandwhose direction isalong theaxisofthedipole, pointing from q_toward q+.Then cos0=p-e,, (6.12) where e,istheunit radial vector (Fig. 6-3). Wecanalso represent thepoint (x,y,z)byr.Then D'poI tfl: 1-T 1~ l 617067110 = 1;eZ (6.13) 41re0 r2 41re0 r3 This formula isvalid foradipole with anyorientation andposition ifrrepresents thevector from thedipole tothepoint ofinterest. Ifwewant theelectric ﬁeld ofthedipole wecangetitbytaking thegradient of¢.Forexample, thez-component oftheﬁeldis—6¢/dz. Foradipole oriented along thez-axis wecanuse(6.9): _%= __P_i(£)= _.L i_L”),62 41re0 62 r3 41re0 r3 r5 or 320 1 COS — E,=ﬁ___rT__.. (6.14) Thex-andy-components are 32x p3zyEx=L __,E=_._ _.41re0 r5 1' 41re0 r5 These twocanbecombined togiveonecomponent directed perpendicular tothe z-axis, which wewillcallthetransverse component EL: Ei-\/E2+E2— 1’3z\/x2+y2— "7 7/_ 4-7l'€()‘7E or__p3cos0sin0_E_L——-4M0 —-—-irs (6.15) 6-3P Fig. 6- dipole.l I I 3.P 9 r er Vector notation for l-l (0)Y Fig. 6-4. The electric ﬁeld ofq dipole.Q\\“..Thetransverse component Ejisinthex-yplane andpoints directly away from theaxisofthedipole. Thetotal ﬁeld, ofcourse, is E=\/Ef+Ei. Thedipole ﬁeld varies inversely asthecube ofthedistance from thedipole. Ontheaxis, at0=0,itistwice asstrong asat0=90°. Atboth ofthese special angles theelectric ﬁeld hasonly az-component, butofopposite sign atthetwo places (Fig. 6-4). 6-3Remarks onvector equations This isagood place tomake ageneral remark about vector analysis. The fundamental proofs canbeexpressed byelegant equations inageneral form, but inmaking various calculations andanalyses itisalways agood idea tochoose theaxes insome convenient way. Notice thatwhen wewere ﬁnding thepotential ofadipole wechose thez-axis along thedirection ofthedipole, rather than atsome arbitrary angle. This made thework much easier. Butthen wewrote theequations invector form sothatthey would nolonger depend onanyparticular coordinate system. After that, weareallowed tochoose anycoordinate system wewish, knowing thattherelation 1S,ingeneral, true. Itclearly doesn’t make anysense to bother with anarbitrary coordinate system atsome complicated angle when you canchoose aneat system fortheparticular problem—provided thattheresult can ﬁnally beexpressed asavector equation. Sobyallmeans take advantage ofthe factthatvector equations areindependent ofanycoordinate system. Ontheother hand, ifyouaretrying tocalculate thedivergence ofavector, instead ofjustlooking atV-Eandwondering what itis,don’t forget thatitcan always bespread outas BE, 6E, 6E, 6x+dy+W Ifyoucanthen work outthex-,y-,andz-components oftheelectric ﬁeld and differentiate them, youwillhave thedivergence. There often seems tobeafeeling that there issomething inelegant-—some kind ofdefeat involved—in writing out thecomponents; thatsomehow there ought always tobeawaytodoeverything with thevector operators. There isoften noadvantage toit.Theﬁrsttime we encounter aparticular kind ofproblem, itusually helps towrite outthecomponents tobesureweunderstand what isgoing on.There isnothing inelegant about put- tingnumbers intoequations, andnothing inelegant about substituting thederiva- tives forthefancy symbols. Infact, there isoften acertain cleverness indoing justthat. Ofcourse when youpublish apaper inaprofessional journal itwilllook better—and bemore easily understood—if youcanwrite everything invector form. Besides, itsaves print. 6-4Thedipole potential asagradient Wewould liketopoint outarather amusing thing about thedipole formula, Eq.(6.13). Thepotential canalsobewritten as ¢=-Ly-v(§) (6.16)47l'€Q Ifyoucalculate thegradient of1/r,youget vi =__'L= _ﬁ,r r3 r2 andEq.(6.16) isthesame asEq.(6.13). How didwethink ofthat? Wejustremembered thate./r2 appeared inthe formula fortheﬁeld ofapoint charge, andthattheﬁeld wasthegradient ofa potential which hasal/rdependence. 6-4 There isaphysical reason forbeing able towrite thedipole potential inthe form ofEq.(6.16). Suppose wehave apoint charge qattheorigin. Thepotential atthepoint Pat(x,y,z)is ¢0 = -£1- (Let’s leave offthel/4-rreo while wemake these arguments; wecanstick itinat theend.) Now ifwemove thecharge +qupadistance Az,thepotential atPwill change alittle, by,say,A¢+. How much isA¢+? Well, itisjusttheamount that thepotential would change ifwewere toleave thecharge attheorigin andmove Pdownward bythesame distance Az(Fig. 6-5). That is, 19¢A¢+ = —'EQ AZ, where byAzwemean thesame asd/2. So,using ¢=q/r,wehave thatthepo- tential from thepositive charge is d¢+=g_363(3) 5- (6.17)P‘ Applying thesame reasoning forthepotential from thenegative charge, wecanwrite _1i:1é.¢__ r+8z<r>2 (618) Thetotal potential isthesum of(6.17) and(6.18): N’-“I>§¢=¢++¢_=—~%(~)d (6-19) =(—)Forother orientation ofthedipole, wecould represent thedisplacement of thepositive charge bythevector Ar+. Weshould then write Eq.(6.17) as A¢+ = —V¢0 'Al'+, where Aristhen tobereplaced byd/2. Completing thederivation asbefore, Eq.(6.19) would then become 1. This isthesame asEq.(6.16), ifwereplace qd=p,andputback the1/41re0. Looking atitanother way, weseethat thedipole potential, Eq.(6.13), canbe interpreted as ¢=—p-V0, (6.20) where 0=l/41r60r isthepotential ofaunitpoint charge. Although wecanalways ﬁndthepotential ofaknown charge distribution by anintegration, itissometimes possible tosave time bygetting theanswer with a clever trick. Forexample, onecanoften make useofthesuperposition principle. Ifwearegiven acharge distribution thatcanbemade upofthesum oftwodis- tributions forwhich thepotentials arealready known, itiseasy toﬁndthede- sired potential byjustadding thetwoknown ones. One example ofthisisour derivation of(6.20), another isthefollowing. Suppose wehave aspherical surface with adistribution ofsurface charge thatvaries asthecosine ofthepolar angle. Theintegration forthisdistribution is fairly messy. But, surprisingly, such adistribution canbeanalyzed bysuper- position. Forimagine asphere with auniform volume density ofpositive charge, andanother sphere with anequal uniform volume density ofnegative charge, 6-5AZ P -~Az/ I //)P ////// // // // Az _ 0 Y x Fig. 6-5. The potential citPfrom 0 point charge citAzabove theorigin isthe some asthepotential atP’(Az below P) from thesome charge attheorigin. + + + _'_ + + Fig. 6-6. Two uniformly charged + 4. spheres, superposed withaslight disp|ace- + ‘ _ ment, are equivalent toanonuniform e — — _ distribution ofsurface Fig.6—7. Computation ofthe p tential atc|point Patalarge distance from asetofcharges.0 charge. (0) '1' (b) = (C) originally superposed tomake aneutral—that is,uncharged—s;;here. Ifthe positive sphere isthen displaced slightly with respect tothenegative sphere. the body oftheuncharged sphere would remain neutral, butalittle positive charge will appear ononeside, andsome negative chargt willappear ontheopposite side, asillustrated inFig.6-6. Iftherelative displacement ofthetwospheres issmall, thenetcharge isequivalent toasurface charge (onaspherical surface), andthe surface charge density willbeproportional tothecosine ofthepolar angle. Now ifwewant thepotential from thisdistribution. wedonotneed todoan integral. Weknow thatthepotential from each ofthespheres ofcharge is——for points outside thesphere—the same asfrom apoint charge. The two displaced spheres areliketwo point charges; thepotential isjust that ofadipole. Inthisway you canshow that acharge distribution onasphere ofradius a with asurface charge density 0=0'0cos9 produces aﬁeld outside thesphere which isjustthatofadipole whose moment is 41ra0a3 P=*"3—" Itcanalsobeshown thatinside thesphere theﬁeld isconstant, with thevalue E=E.3G0 If6istheangle from thepositive z-axis, theelectric ﬁeldinside thesphere isinthe negative z-direction. Theexample wehave justconsidered isnotasartiﬁcial as itmay appear; wewillencounter itagain inthetheory ofdielectrics. 6-5Thedipole approximation foranarbitrary distribution The dipole ﬁeld appears inanother circumstance both interesting andim- portant. Suppose thatwehave anobject thathasacomplicated distribution of charge——like thewater molecule (Fig. 6—2)—and weareinterested only inthe ﬁelds faraway. Wewillshow thatitispossible toﬁndarelatively simple expression fortheﬁelds which isappropriate fordistances large compared with thesizeof theobject. Wecanthink ofourobject asanassembly ofpoint charges q,inacertain limited region, asshown inFig. 6-7. (We can, later, replace q,bypdVifwewish.) Let each charge q,belocated atthedisplacement d,from anorigin chosen somewhere A P r. °+6 qt R__ O +dl+0o 0+ O_ ' > O- 6- Q+ 6—6 inthemiddle ofthegroup ofcharges. What isthepotential atthepoint P,located atR,where Rismuch larger than themaximum d,? Thepotential from the whole collection isgiven by _1 ql 4._4% rt. (6.21) where r,isthedistance from Ptothecharge q,(thelength ofthevector R—d,). Now ifthedistance from thecharges toP,thepoint ofobservation, isenormous, each ofther,’scanbeapproximated byR.Each term becomes q,/R, andwe cantake 1/Routasafactor infront ofthesummation. This gives usthesimple resultl1 Q¢=%'ﬁZqi= # where Qisjustthetotal charge ofthewhole object. Thus weﬁndthatforpoints farenough from anylump ofcharge, thelump looks likeapoint charge. The result isnottoosurprising. Butwhat ifthere areequal numbers ofpositive andnegative charges? Then thetotal charge Qoftheobject iszero. This isnotanunusual case; infact, aswe know, objects areusually neutral. Thewater molecule isneutral, butthecharges arenotallatonepoint, soifweareclose enough weshould beabletoseesome effects oftheseparate charges. Weneed abetter approximation than (6.22) for thepotential from anarbitrary distribution ofcharge inaneutral object. Equation (6.21) isstillprecise, butwecannolonger justsetr,=R.Weneed amore accu- rateexpression forr,.Ifthepoint Pisatalarge distance, r,willdiffer from Rto anexcellent approximation bytheprojection ofdonR,ascanbeseen from Fig.6-7. (You should imagine thatPisreally farther away than isshown inthe ﬁgure.) Inother words, ife,istheunitvector inthedirection ofR,then ournext approximation tor,is r,zR—d,-e,. (6.23) What wereally want isl/r,,which, since d,<<R,canbewritten toourapproxima- tionas1 1 d,-e,71~i(1+-T) (6.24) Substituting thisin(6.21), wegetthatthepotential is _l Q dye, ¢"2iT.,(ﬁ+;q@TeT+"'>' “'25) Thethree dots indicate theterms ofhigher order ind/Rthatwehave neglected. These, aswellastheones wehave already obtained, aresuccessive terms inaTaylor expansion ofl/r,about l/Rinpowers ofd,/R. Theﬁrstterm in(6.25) iswhat wegotbefore; itdrops outiftheobject is neutral. Thesecond term depends on1/R2, justasforadipole. Infact,ifwedeﬁne asaproperty ofthecharge distribution, thesecond term ofthepotential (6.25) is .1.=ﬁ"7,»2"’l. (6.27) precisely adipole potential. Thequantity piscalled thedipole moment ofthe distribution. Itisageneralization ofourearlier deﬁnition, andreduces toitfor thespecial caseoftwopoint charges. Our result isthat, farenough away from anymess ofcharges thatisasa whole neutral, thepotential isadipole potential. Itdecreases as1/R2 andvaries ascos0—and itsstrength depends onthedipole moment ofthedistribution of charge. Itisforthese reasons thatdipole ﬁelds areimportant, since thesimple caseofapairofpoint charges isquite rare. 6—7 \_ is 1‘ 1///\ / \ ,, \m" __/ / / \ 1’ \\4-’ I \ Fig. 6-8. Theﬁeld lines and equipo- tentials fortwopoint charges. I / +q I CONDUCTOR Fig. 6-9. The ﬁeld outside acon- ductor shaped like theequipotential A ofFig.6-8.Thewater molecule, forexample, hasarather strong dipole moment. The electric ﬁelds thatresult from thismoment areresponsible forsome oftheim- portant properties ofwater. Formany molecules, forexample CO2, thedipole moment vanishes because ofthesymmetry ofthemolecule. Forthem weshould expand stillmore accurately, obtaining another term inthepotential which de- creases asl/R3, andwhich iscalled aquadrupole potential. Wewilldiscuss such cases later. 6-6Theﬁelds ofcharged conductors Wehave now ﬁnished with theexamples wewish tocover ofsituations in which thecharge distributions isknown from thestart. Ithasbeen aproblem Without serious complications, involving atmost some integrations. Weturn now toanentirely new kind ofproblem, thedetermination oftheﬁelds near charged conductors. Suppose thatwehave asituation inwhich atotal charge Qisplaced onan arbitrary conductor. Now wewillnotbeable tosayexactly where thecharges are. They willspread outinsome way onthesurface. How canweknow how thecharges have distributed themselves onthesurface? They must distribute themselves sothatthepotential ofthesurface isconstant. Ifthesurface were not anequipotential, there would beanelectric ﬁeld inside theconductor, andthe charges would keep moving until itbecame zero. The general problem ofthis kind canbesolved inthefollowing way. Weguess atadistribution ofcharge and calculate thepotential. Ifthepotential turns outtobeconstant everywhere on thesurface, theproblem isﬁnished. Ifthesurface isnotanequipotential, we have guessed thewrong distribution ofcharges, andshould guess again—hopefully with animproved guess! This cangoonforever, unless wearejudicious about thesuccessive guesses. Thequestion ofhowtoguess atthedistribution ismathematically diﬁicult. Nature, ofcourse, hastimetodoit;thecharges push andpulluntil theyallbalance themselves. When wetrytosolve theproblem, however, ittakes ussolong to make each trial that that method isvery tedious With anarbitrary group of conductors andcharges theproblem canbevery complicated, andingeneral it cannot besolved without rather elaborate numerical methods. Such numerical computations, these days, aresetuponacomputing machine that willdothe work forus,once wehave toldithow toproceed. Ontheother hand, there arealotoflittle practical cases where itwould benicetobeabletoﬁndtheanswer bysome more direct method—without having towrite aprogram foracomputer. Fortunately, there areanumber ofcases where theanswer canbeobtained bysqueezing itoutofNature bysome trick orother. Theﬁrsttrick wewilldescribe involves making useofsolutions wehave already obtained forsituations inwhich charges have speciﬁed locations. 6-7Themethod ofimages Wehave solved, forexample, theﬁeld oftwopoint charges. Figure 6-8 shows some oftheﬁeld lines andequipotential surfaces weobtained bythecom- putations inChapter 5.Now consider theequipotential surface marked A.Sup- pose wewere toshape athinsheet ofmetal sothatitjustﬁtsthissurface. Ifwe place itright atthesurface andadjust itspotential totheproper value, noone would ever know itwasthere, because nothing would bechanged. Butnotice! Wehave really solved anewproblem. Wehave asituation in which thesurface ofacurved conductor with agiven potential isplaced near a point charge. Ifthemetal sheet weplaced attheequipotential surface eventually closes onitself (or,inpractice, ifitgoes farenough) wehave thekind ofsituation considered inSection 5-10, inwhich ourspace isdivided intotworegions, one inside andoneoutside aclosed conducting shell. Wefound there thattheﬁelds in thetworegions arequite independent ofeach other. Sowewould have thesame ﬁelds outside ourcurved conductor nomatter what isinside. Wecaneven ﬁllup 6—8 thewhole inside with conducting material. Wehave found, therefore, theﬁelds forthearrangement ofFig. 6-9. Inthespace outside theconductor theﬁeld is justlikethatoftwopoint charges, asinFig.6-8. Inside theconductor, itiszero Also—as itmust be-the electric ﬁeld just outside theconductor isnormal to thesurface. Thus wecancompute theﬁelds inFig.6-9bycomputing theﬁeld duetoq andtoanimaginary point charge —qatasuitable point. Thepoint charge we “imagine” existing behind theconducting surface iscalled animage charge. Inbooks youcanﬁndlong listsofsolutions forhyperbolic-shaped conductors andother complicated looking things, andyouwonder how anyone ever solved these terrible shapes. They were solved backwards! Someone solved asimple problem with given charges. Hethen sawthatsome equipotential surface showed upinanewshape, andhewrote apaper inwhich hepointed outthat theﬁeld outside thatparticular shape canbedescribed inacertain way. 6-8Apoint charge near aconducting plane Asthesimplest application oftheuseofthismethod, let's make useofthe plane equipotential surface BofFig.6-8. With it,wecansolve theproblem ofa charge infront ofaconducting sheet. Wejustcross outtheleft-hand halfofthe picture. Theﬁeld lines foroursolution areshown inFig.6-10. Notice that the plane, since itwashalfway between thetwocharges, haszero potential. Wehave solved theproblem ofapositive charge next toagrounded conducting sheet. Wehave now solved forthetotal ﬁeld, butwhat about therealcharges that areresponsible forit?There are,inaddition toourpositive point charge, some induced negative charges ontheconducting sheet thathave been attracted bythe positive charge (from large distances away). Now suppose thatforsome technical reason—or outofcuriosity-you would liketoknow how thenegative charges aredistributed onthesurface. You canﬁndthesurface charge density byusing theresult weworked outinSection 5-6with Gauss’ theorem. Thenormal com- 4;,4,41»;/0//‘// _/V . / / \ \. lcououcrmis /\ \\PLATE — \ \ \\lll//// // - _P \\ \ \ ll,/////1 \\ \ \\\ '/ / \\s\‘."//c\ \ / _ ° 4 — -— - —lMAGE CHARGE —& . Illl/II/04' Q > / / //'l\\\\/ / ///l\\\\ / l\ \/ / I \ \ // //ll\\\\\ / jl\\ - /IIIII/]\\\ _ l\\ _ V Fig. 6-lO. Theﬁeld ofacharge near aplane conducting surface, found bythe method ofimages. 6-9 \ P Ix ., \ “Qq'=-%q Fig. 6-ll. The point charge qin- duces charges onagrounded conducting sphere whose ﬁelds are those ofan image charge q’placed atthepoint shown.ponent oftheelectric ﬁeldjustoutside aconductor isequal tothedensity ofsurface charge 0divided byen.Wecanobtain thedensity ofcharge atanypoint onthe surface byworking backwards from thenormal component oftheelectric ﬁeld at thesurface. Weknow that, because weknow theﬁeldeverywhere. Consider apoint onthesurface atthedistance pfrom thepoint directly be- neath thepositive charge (Fig. 6-10). Theelectric ﬁeld atthispoint isnormal to thesurface andisdirected intoit.Thecomponent normal tothesurface ofthe ﬁeld from thepositive point charge is _ 1 aqEn-i_ — Zhrso (a2+pg)’;/~_) (6.28) Tothiswemust addtheelectric ﬁeldproduced bythenegative image charge. That justdoubles thenormal component (and cancels allothers), sothecharge density 0atanypoint onthesurface is 2 I <r(/>)=60E(P)=-;,-,@-g‘ﬂ‘i;,-),,,-,- 1,(629) Aninteresting check onourwork istointegrated overthewhole surface. We ﬁndthatthetotal induced charge 1S-q.asitshould be. * Onefurther question: Isthere aforce onthepoint charge? Yes,because there isanattraction from theinduced negative surface charge ontheplate. Now that weknow what thesurface charges are(from Eq.(6.29)), wecould compute the force onourpositive point charge byanintegral. Butwealsoknow thattheforce acting onthepositive charge isexactly thesame asitwould bewith thenegative image charge instead oftheplate, because theﬁelds intheneighborhood arethe same inboth cases. Thepoint charge feels aforce toward theplate whose magni- tude is F=~-]— —-‘f-- - (630)47T€() (Z0)! Wehave found theforce much more easily than byintegrating over allthenega- tivecharges. 6-9Apoint charge near aconducting sphere What other surfaces besides aplane have asimple solution” The next most simple shape isasphere. Let's ﬁnd theﬁelds around ametal sphere which hasa point charge qnear it,asshown inFig. 6-ll.Now wemust look forasimple physical situation which gives asphere foranequipotential surface. Ifwelook around atproblems people have already solved, weﬁndthatsomeone hasnoticed that thefield oftwo unequal point charges hasanequipotential that isasphere Aha‘ Ifwechoose thelocation ofanimage charge-and pick theright amount ofcharge——maybe wecanmake theequipotential surface ﬁtoursphere. Indeed, itcanbedone with thefollowing prescription. Assume thatyouwant theequipotential surface tobeasphere ofradius a with itscenter atthedistance bfrom thecharge q.Putanimage charge ofstrength q’=—q(a/b) onthelinefrom thecharge tothecenter ofthesphere, and ata distance a2/b from thecenter. Thesphere Wlllbeatzeropotential. Themathematical reason stems from thefactthatasphere isthelocus ofall points forwhich thedistances from twopoints areinaconstant ratio Referring toFig.6-11,thepotential atPfrom qandq’isproportional to rt+£1..Vi F2 Thepotential willthus bezero atallpoints forwhich 97:-2 0,Q=_iT.F2 Vi "i q 6-10 Ifweplace q’atthedistance a2/b from thecenter, theratio r2/r1 hastheconstant value a/b. Then if 92-=-g. (6.31) thesphere isanequipotential. Itspotential is,infact, zero. What happens ifweareinterested inasphere thatisnotatzero potential ? That would besoonlyifitstotal charge happens accidentally tobeq’Ofcourse ifit isgrounded, thecharges induced onitwould have tobejustthat. Butwhat ifit isinsulated, andwehave putnocharge onit”Orifweknow thatthetotal charge Qhasbeen putonit?Orjustthatithasagiven potential notequal tozero?All these questions areeasily answered. Wecanalways addapoint charge q"atthe center ofthesphere Thesphere stillremains anequipotential bysuperposition: only themagnitude ofthepotential willbechanged. Ifwehave, forexample, aconducting sphere which isinitially uncharged andinsulated from everything else, andwebring near toitthepositive point charge q,thetotal charge ofthesphere Wlllremain zero. Thesolution isfound byusing animage charge q’asbefore, but,inaddition. adding acharge q”atthe center ofthesphere, choosing . q”=-4’=-in (6-32) The ﬁelds everywhere outside thesphere aregiven bythesuperposition ofthe ﬁelds ofq,q’,andq”.Theproblem issolved. Wecanseenow thatthere willbeaforce ofattraction between thesphere andthepoint charge q.Itisnotzeroeven though there isnocharge ontheneutral sphere. Where does theattraction come from? When youbring apositive charge uptoaconducting sphere, thepositive charge attracts negative charges tothe sidecloser toitself andleaves positive charges onthesurface ofthefarside. The attraction bythenegative charges exceeds therepulsion from thepositive charges. there isanetattraction. Wecanﬁndouthowlarge theattraction isbycomputing theforce onqintheﬁeld produced byq’andq”.Thetotal force isthesumofthe attractive force between qandacharge q’=—(a/li)q, atthedistance b-((12//7), andtherepulsive force between qandacharge q”:—l-(a/b)q atthedistance b. Those who were entertained inchildhood bythebaking powder boxwhich hasonitslabel apicture ofabaking powder boxwhich hasonitslabel apicture ofabaking powder boxwhich has.may beinterested inthefollowing problem. Two equal spheres, onewith atotal charge of+Qandtheother withatotal charge of—Q,areplaced atsome distance from each other. What IStheforce between them‘? Theproblem canbesolved with aninﬁnite number ofimages. Oneﬁrst approximates each sphere byacharge atitscenter. These charges willhave image charges intheother sphere. Theimage charges Wlllhave images, etc, etc, etc The solution islike thepicture onthebox ofbaking powder—and itconverges pretty fast. 6-10 Condensers; parallel plates Wetake upnow another kind ofaproblem involving conductors. Consider twolarge metal plates which areparallel toeach other andseparated byadistance small compared with their width. Let’s suppose thatequal andopposite charges have been putontheplates. Thecharges oneach plate willbeattracted bythe charges ontheother plate, andthecharges willspread outuniformly ontheinner surfaces oftheplates. Theplates willhave surface charge densities +0and-a, respectively. asinFig.6-l2. From Chapter 5weknow thattheﬁeld between the plates is0/cu, andthattheﬁeld outside theplates iszero. Theplates willhave Llll:l_€l"CIl[ potentials 4;,and4>2. Forconvenience wewillcallthedifference V;it isoften called the“voltage”: Q51—¢2=V- (You willﬁndthatsometimes people useVforthepotential, butwehave chosen touse¢>.) 6-ll+Q‘\Area =A -+\ l"\\\ '*\V/i/+/4 /T/{ } / /*/ { d I//_///— //_//_/ //_/ /i;//;/| -0‘ Fig. denser.6-12 Aparallel-plate con Fig.6-13. Theelectric ﬁeld near the edge oftwo parallel plates.The potential difference Visthework perunitcharge required tocarry a small charge from oneplate totheother, sothat a d where ¢Qisthetotal charge oneach plate, Aisthearea oftheplates, anda’is theseparation. Weﬁndthatthevoltage isproportional tothecharge. Such aproportionality between VandQisfound foranytwoconductors inspace ifthere isapluscharge ononeandanequal minus charge ontheother. Thepotential difference between them—-that is,thevoltage-will beproportional tothecharge. (Weareassuming thatthere arenoother charges around.) Why thisproportionality? Just thesuperposition principle. Suppose we know thesolution foronesetofcharges, andthen wesuperimpose two such solutions. Thecharges aredoubled, theﬁelds aredoubled, andthework done in carrying aunitcharge from onepoint totheother isalsodoubled. Therefore the potential difference between anytwopoints isproportional tothecharges. In particular, thepotential difference between thetwoconductors isproportional tothecharges onthem. Someone originally wrote theequation ofproportionality theother way. That is,they wrote Q=CV, where Cisaconstant This coeﬁicient ofproportionality iscalled thecapacity. andsuch asystem oftwoconductors iscalled ac'nna'enser.* Forourparallel-plate condenser c:F154(parallel plates). (6.34) This formula isnotexact, because theﬁeld isnotreally uniform everywhere between theplates, asweassumed. Theﬁeld does notjustsuddenly quitatthe edges, butreally ismore asshown inFig6-13 Thetotal charge ISnotcr/4,aswe have assumed—-there isalittle correction fortheeffects attheedges. Toﬁndout what thecorrection is,wewillhave tocalculate theﬁeld more exactly andﬁnd outjust what does happen attheedges. That isacomplicated mathematical problem which can, however, besolved bytechniques which wewillnotdescribe now. Theresult ofsuch calculations isthat thecharge density rises somewhat near theedges oftheplates This means thatthecapacity oftheplates isalittle higher than wecomputed. [Avery good approximation forthecapacity isob- tained ifweuseEq.(6.34) buttakeforAthearea onewould getiftheplates were extended artiﬁcially byadistance 3/8oftheseparation between theplates.] Wehave talked about thecapacity fortwoconductors only. Sometimes people talk about thecapacity ofasingle object. They say, forinstance, that the capacity ofasphere ofradius ais41re(,a. What they imagine isthattheother terminal isanother sphere ofinﬁnite radius-that when there isacharge +Q on thesphere. theopposite charge, —Q,isonaninﬁnite sphere. Onecanalsospeak ofcapacities when there arethree ormore conductors, adiscussion weshall, however, defer. Suppose that wewish tohave acondenser with avery large capacity We could getalarge capacity bytaking avery bigarea andavery small separation Wecould putwaxed paper between sheets ofaluminum foilandrollitup. (If wesealitinplastic, wehave atypical radio-type condenser.) What good isit" ltisgood forstoring charge. Ifwetrytostore charge onaball, forexample, its potential rises rapidly aswecharge itup.ltmayeven getsohigh thatthecharge begins toescape intotheairbywayofsparks Butifweputthesame charge ona condenser whose capacity isvery large, thevoltage developed across thecon- denser willbesmall. *Some people think thewords “capacitance” and“capacitor" should beused, instead of“capacity” and“condensor "Wehave decided tousetheolder terminology, because itisstillmore commonly heard inthephysics laboratory—even ifnotintextbooks! 6-I2 Inmany applications inelectronic circuits, itisuseful tohave something which canabsorb ordeliver large quantities ofcharge without changing itspo- tential much. Acondenser (or“capacit0r”) does justthat. There arealsomany applications inelectronic instruments and incomputers where acondenser is used togetaspeciﬁed change involtage inresponse toaparticular change in charge. Wehave seen asimilar application inChapter 23,Vol. I,where wede- scribed theproperties ofresonant circuits. From thedeﬁnition ofC,weseethatitsunitisonecoul/volt. This unitis alsocalled afarad. Looking atEq.(6.34), weseethatonecanexpress theunits ofe0asfarad/meter, which istheunit most commonly used. Typical sizes of condensers runfrom onemicro-microfarad (=1picofarad) tomillifarads. Small condensers ofafewpicofarads areused inhigh-frequency tuned circuits, and capacities uptohundreds orthousands ofmicrofarads arefound inpower-supply ﬁlters. Apairofplates onesquare centimeter inareawith aonemillimeter separa- tionhave acapacity ofroughly onemicro-microfarad. 6-11 High-voltage breakdown Wewould likenow todiscuss qualitatively some ofthecharacteristics ofthe ﬁelds around conductors. Ifwecharge aconductor thatisnotasphere, butone thathasonitapoint oravery sharp end, as,forexample, theobject sketched inFig.6-14, theﬁeld around thepoint ismuch higher than theﬁeld intheother regions. Thereason is,qualitatively, thatcharges trytospread outasmuch as possible onthesurface ofaconductor, andthetipofasharp point isasfaraway asitispossible tobefrom most ofthesurface. Some ofthecharges ontheplate getpushed allthewaytothetip. Arelatively small amount ofcharge onthetip canstillprovide alarge surface density; ahigh charge density means ahigh ﬁeld justoutside. Onewaytoseethattheﬁeld ishighest atthose places onaconductor where theradius ofcurvature issmallest istoconsider thecombination ofabigsphere andalittlesphere connected byawire, asshown inFig.6-15. Itisasomewhat idealized version oftheconductor ofFig.6-14. Thewire willhave little inﬂuence ontheﬁelds outside; itisthere tokeep thespheres atthesame potential. Now, which ballhasthebiggest ﬁeld atitssurface? Iftheballonthelefthastheradius aandcarries acharge Q,itspotential isabout _1Q4"r1;;2" (Ofcourse thepresence ofoneballchanges thecharge distribution ontheother, sothatthecharges arenotreally spherically symmetric oneither. Butifweare interested only inanestimate oftheﬁelds, wecanusethepotential ofaspherical charge.) Ifthesmaller ball, whose radius isb,carries thecharge q,itspotential isabout I =___ £1.¢2 47T€0 b But¢i=¢2,$0 Qzg‘ a b Ontheother hand, theﬁeld atthesurface (seeEq.5.8)isproportional tothe surface charge density, which islikethetotal charge over theradius squared. Wegetthat Fl=Q_/1'3 =Q. (635) Eb q/172 ll i Therefore theﬁeld ishigher atthesurface ofthesmall sphere. Theﬁelds areinthe inverse proportion oftheradii. This result istechnically very important, because airwillbreak down ifthe electric ﬁeld istoogreat. What happens isthataloose charge (electron, orion) somewhere intheairisaccelerated bytheﬁeld, andiftheﬁeld isvery great, the charge canpickupenough speed before ithitsanother atom tobeabletoknock an 6-131 I 50z , farad/meter ‘ * \\ _ \-i -“"-1 W )- // / / conoucron / , / // / / / / / / / / Fig. 6-14. Theelectric ﬁeld near a sharp point onaconductor isvery high. WIRE Viz / Fig. 6—l5. The ﬁeld ofapointed object canbeapproximated bythat of twospheres atthesame potential. FLUORESCENT ’,..-__ COATING/ \\ /.k&V%*y//,_--_ / /i\____,///,umii..POINT cnouuo st.-ssauua toVACUUMPUMP ———L 1HIGH VOLTAGE Fig. 6-16. Field-emission microscope. Fig. 6-17. Image produced bya ﬁeld-emission microscope. [Courtesy of Erwin W. Mueller, Research Prof. of Physics, Pennsylvania State University]electron oﬁ"that atom. Asaresult, more andmore ions areproduced. Their motion constitutes adischarge, orspark. Ifyouwant tocharge anobject toa highpotential andnothave itdischarge itself bysparks intheair,youmust be sure thatthesurface issmooth, sothatthere isnoplace where theﬁeld isab- normally large. 6-12 Theﬁeld-emission microscope There isaninteresting application oftheextremely high electric ﬁeld which surrounds anysharp protuberance onacharged conductor. Theﬁeld-emission microscope depends foritsoperation onthehigh ﬁelds produced atasharp metal point.* Itisbuilt inthefollowing way. Averyﬁneneedle, withatipwhose diameter isabout 1000 angstroms, isplaced atthecenter ofanevacuated glass sphere (Fig. 6-16.) Theinner surface ofthesphere iscoated with athinconducting layer of ﬂuorescent material, andavery high potential diﬂerence isapplied between the ﬂuorescent coating andtheneedle. Let’s ﬁrstconsider what happens when theneedle isnegative with respect to theﬂuorescent coating. Theﬁeld lines arehighly concentrated atthesharp point. The electric ﬁeld canbeashigh as40million volts percentimeter. Insuch intense ﬁelds, electrons arepulled outofthesurface oftheneedle andaccelerated across thepotential diﬂerence between theneedle andtheﬂuorescent layer. When they arrive there they cause light tobeemitted, _]LlStasinatelevision picture tube. Theelectrons which arrive atagiven point ontheﬂuorescent surface are,to anexcellent approximation, those which leave theother endoftheradial ﬁeld line, because theelectrons willtravel along theﬁeld linepassing from thepoint tothe surface. Thus weseeonthesurface some kind ofanimage ofthetipoftheneedle. More precisely, weseeapicture oftheemissivity ofthesurface oftheneedle——that IStheeasewith which electrons canleave thesurface ofthemetal tip.Iftheresolu- tionwere high enough, onecould hope toresolve thepositions oftheindividual atoms onthetipoftheneedle. With electrons, thisresolution isnotpossible for thefollowing reasons. First, there isquantum-mechanical diffraction ofthe electron waves which blurs theimage. Second, duetotheinternal motions ofthe electrons inthemetal they have asmall sideways initial velocity when theyleave theneedle, andthisrandom transverse component ofthevelocity causes some smearing oftheimage. Thecombination ofthese twoeﬂects limits theresolution to25Aorso. If,however, wereverse thepolarity andintroduce asmall amount ofhelium gasintothebulb, much higher resolutions arepossible. When ahelium atom col- lides with thetipoftheneedle, theintense ﬁeld there strips anelectron offthe helium atom, leaving itpositively charged. The helium ionisthen accelerated outward along aﬁeldlinetotheﬂuorescent screen. Since thehelium ionissomuch heavier than anelectron, thequantum-mechanical wavelengths aremuch smaller. Ifthetemperature isnottoohigh, theeﬂect ofthethermal velocities isalsosmaller than intheelectron case. With lesssmearing oftheimage amuch sharper picture ofthepoint isobtained. Ithasbeen possible toobtain magniﬁcations upto 2,000,000 times with thepositive ionﬁeld-emission microscope-—a magniﬁcation tentimes better than isobtained with thebestelectron microscope. Figure 6-17 isanexample oftheresults which were obtained with aﬁeld- ionmicroscope, using atungsten needle. Thecenter ofatungsten atom ionizes ahelium atom ataslightly diﬂerent ratethan thespaces between thetungsten atoms. Thepattern ofspots ontheﬂuorescent screen shows thearrangement of theindividual atoms onthetungsten tip.Thereason thespots appear inrings can beunderstood byvisualizing alarge boxofballs packed inarectangular array, representing theatoms inthemetal. Ifyoucutanapproximately spherical section outofthisbox, youwillseetheringpattern characteristic oftheatomic structure. Theﬁeld-ion microscope provided human beings with themeans ofseeing atoms fortheﬁrsttime. This isaremarkable achievement, considering thesimplicity of theinstrument. *SeeE.W.Mueller: “The ﬁeld-ion microscope,” Advances II1Electronics andElectron Physics, 13,83-179 (1960). Academic Press, New York 6-14 7 The Electric Field inVurious Cireumstunees (Continued) 7-1Methods forﬁnding theelectrostatic ﬁeld This chapter isacontinuation ofourconsideration ofthecharacteristics of electric ﬁelds invarious particular situations. Weshall ﬁrstdescribe some ofthe more elaborate methods forsolving problems with conductors. Itisnotexpected thatthese more advanced methods canbemastered atthistime. Yetitmay beof interest tohave some idea about thekinds ofproblems thatcanbesolved, using techniques thatmay belearned inmore advanced courses. Then wetake uptwo examples inwhich thecharge distribution isneither ﬁxed noriscarried byacon- duct_or, butinstead isdetermined bysome other lawofphysics. Aswefound inChapter 6,theproblem oftheelectrostatic ﬁeld isfundamen- tallysimple when thedistribution ofcharges isspeciﬁed; itrequires onlytheevalua- tionofanintegral. When there areconductors present, however, complications arise because thecharge distribution ontheconductors isnotinitially known; thecharge must distribute itself onthesurface oftheconductor insuch awaythat theconductor isanequipotential. Thesolution ofsuch problems isneither direct norsimple. Wehave looked atanindirect method ofsolving such problems, inwhich we ﬁndtheequipotentials forsome speciﬁed charge distribution andreplace oneof them byaconducting surface. Inthiswaywecanbuild upacatalog ofspecial solutions forconductors intheshapes ofspheres, planes, etc. Theuseofimages, described inChapter 6,isanexample ofanindirect method. Weshall describe another inthischapter. Iftheproblem tobesolved does notbelong totheclass ofproblems forwhich wecanconstruct solutions bytheindirect method, weareforced tosolve theprob- lembyamore direct method. Themathematical problem ofthedirect method is thesolution ofLaplace’s equation, v2¢=0, (7.1) subject tothecondition that¢isasuitable constant oncertain boundaries—the surfaces oftheconductors. Problems which involve thesolution ofadiﬂerential ﬁeld equation subject tocertain boundary conditions arecalled boundary-value problems. They have been theobject ofconsiderable mathematical study. In thecase ofconductors having complicated shapes, there arenogeneral analytical methods. Even such asimple problem asthat ofacharged cylindrical metal can closed atboth ends——a beer can—presents formidable mathematical difﬁculties. Itcanbesolved only approximately, using numerical methods. Theonly general methods ofsolution arenumerical. There areafewproblems forwhich Eq.(7.1) canbesolved directly. For example, theproblem ofacharged conductor having theshape ofanellipsoid of revolution canbesolved exactly interms ofknown special functions. Thesolution forathindisccanbeobtained byletting theellipsoid become inﬁnitely oblate. Inasimilar manner, thesolution foraneedle canbeobtained byletting theellipsoid become inﬁnitely prolate. However, itmust bestressed thattheonlydirect methods ofgeneral applicability arethenumerical techniques. Boundary-value problems canalsobesolved bymeasurements ofaphysical analog. Laplace’s equation arises inmany diﬂerent physical situations: insteady- state heat ﬁow, inirrotational ﬂuid ﬂow, incurrent ﬂow inanextended medium, 7—l7_ 7-2 7-3 7-4 7-5Methods forﬁnding the electrostatic ﬁeld Two-dimensional ﬁelds; functions ofthecomplex variable Plasma oscillations Colloidal particles inan electrolyte Theelectrostatic ﬁeld ofagrid andinthedeﬂection ofanelastic membrane. Itisfrequently possible tosetupa physical model which isanalogous toanelectrical problem which wewishtosolve. Bythemeasurement ofasuitable analogous quantity onthemodel, thesolution totheproblem ofinterest canbedetermined. Anexample oftheanalog technique istheuseoftheelectrolytic tankforthesolution oftwo-dimensional problems in electrostatics. Thisworks because thedifferential equation forthepotential ina uniform conducting medium isthesame asitisforavacuum. There aremany physical situations inwhich thevariations ofthephysical ﬁelds inonedirection arezero, orcanbeneglected incomparison with thevaria- tions intheother twodirections. Such problems arecalled two-dimensional; the ﬁelddepends ontwocoordinates only. Forexample, ifweplace alongcharged wirealong thez-axis, thenforpoints nottoofarfrom thewiretheelectric ﬁeld depends onxandy,butnotonz;theproblem istwo-dimensional. Since inatwo- dimensional problem 6/82 =0,theequation for¢>infreespace is a2 a2%+5y-‘Q=0. (7.2) Because thetwo-dimensional equation iscomparatively simple, there isawide range ofconditions under which itcanbesolved analytically. There is,infact, avery powerful indirect mathematical technique which depends onatheorem from themathematics offunctions ofacomplex variable, andwhich wewillnow describe. 7-2Two-dimensional ﬁelds; functions ofthecomplex variable Thecomplex variable 3isdeﬁned as a=x+iy. (Donotconfuse 3withthez-coordinate, which weignore inthefollowing dis- cussion because weassume there isnoz-dependence oftheﬁelds.) Every point in xandythen corresponds toacomplex number 3.Wecanuseatasasingle (complex) variable, andwith itwrite theusual kinds ofmathematical functions F(a). Forexample, F(a)=2’, or F(3)=1/33, or F(&) =alog 3, andsoforth. Given anyparticular F(a)wecansubstitute Z»=x+iy,andwehave a function ofxandy—-with realandimaginary parts. Forexample, a2=(x+iy)2=x2-yz+2ixy. (1.3) Anyfunction F(a)canbewritten asasumofapurerealpartandapure imaginary part, each partafunction ofxandy: F(3) =U(X,y) +iV(X,y). (7-4) where U(x,y)andV(x,y)arerealfunctions. Thus from anycomplex function F(3)twonewfunctions U(x,y)andV(x,y)canbederived. Forexample, F(3)=32 gives usthetwofunctions , U(xa =x2 —yza and V(x,y)=2xy. (7.6) Now wecome toamiraculous mathematical theorem which issodelightful that weshall leave aproof ofitforoneofyour courses inmathematics. (We should notreveal allthemysteries ofmathematics, orthatsubject matter would 7-2 become toodull.) Itisthis. Forany“ordinary function” (mathematicians will deﬁne itbetter) thefunctions UandVautomatically satisfy therelations 6U 6V-5;_-9;, (7.7) 8V 6U-9;---67- (7.8) Itfollows immediately thateachofthefunctions UandVsatisfy Laplace’s equation: a2U 6x2 9.2!6x2a2U 62V These equations areclearly trueforthefunctions of(7.5) and(7.6). Thus, starting with anyordinary function, wecanarrive attwo functions U(x,y)andV(x,y),which areboth solutions ofLaplace’s equation intwodimen- sions. Each function represents apossible electrostatic potential. Wecanpickany function F(a) anditshould represent some electric ﬁeld problem—in fact, two problems, because UandVeach represent solutions. Wecanwrite down asmany solutions aswewish-by justmaking upfunctions-—then wejusthave toﬁndthe problem thatgoes with each solution. Itmaysound backwards, butit’sapossible approach. II Ay - ‘_ - / BI-I B-II \ \\,/ / \4 // \\ ___- A A=o \ _ 432 I A=l I4 a-0 _7 _ -a-i mo a--i /’-\ \ / , \\ \\ 2 A-0 A=O// _ / \\ A-—| ’/ T\ \ 3 y / _3/ //I_\\\ \ 4 \ \ -2 I "4_ / // \ 5 \ \ I _ / 6 \ \ \ -3/ I / \\\ \-4' I / \ t II/ \ ' , \\\ \\\ ,/I / I Fig. 7-l. Two sets oforthogonal curves which can represent equipotentials inatwo-dimensional electrostatic ﬁeld.\\\ ‘<5 *at\\ \u\\\ro\\\\ \\\I. "bus111;/ /// / N/I/ // / I0»Ill/1"llI Asanexample, let’sseewhat physics thefunction F(a) =32gives us.From itwegetthetwopotential functions of(7.5) and(7.6). Toseewhat problem the function Ubelongs to,wesolve fortheequipotential surfaces bysetting U=A, aconstant: x2—y2=A. Thisistheequation ofarectangular hyperbola. Forvarious values ofA,weget thehyperbolas shown inFig.7-1. When A=0,wegetthespecial caseofdiagonal straight linesthrough theorigin. Such asetofequipotentials corresponds toseveral possible physical situations. First, itrepresents theﬁnedetails oftheﬁeld near thepoint halfway between two 7-3 CONDUCTOR + "IIIIIIIIﬂllﬁlllllilllllbii “ etc. etc. '5__IIJIIJIJQ IIIIIIIIII __ O . . "9Fig. 7-2. Theﬁeld near thepoint C O. CONDUCTOR _ isthesame asthat inFig.7-l. ¢=+V ¢=-v ¢=-v CONDUCTOR ¢-+v Fig. 7-3. Theﬁeld inaquadrupole lens.equal point charges. Second, itrepresents theﬁeld ataninside right-angle corner ofaconductor. Ifwehave twoelectrodes shaped likethose inFig.7-2,which are held atdifferent potentials, theﬁeld near thecorner marked Cwilllook justlike theﬁeldabove theorigin inFig.7-l. Thesolid linesaretheequipotentials, and thebroken lines atright angles correspond tolines ofE.Whereas atpoints or protuberances theelectric ﬁeld tends tobehigh, ittends tobelowindents or hollows. Thesolution wehave found alsocorresponds tothatforahyperbola-shaped electrode near aright-angle corner, orfortwohyperbolas atsuitable potentials. You willnotice thattheﬁeld ofFig.7-lhasaninteresting property. Thex-com- ponent oftheelectric ﬁeld, E,,,isgiven by __§2-_._E,- ax- 2x. Theelectric ﬁeld isproportional tothedistance from theaxis. This factisused to make devices (called quadrupole lenses) thatareuseful forfocusing particle beams (seeSection 29-9). Thedesired ﬁeld isusually obtained byusing four hyperbola- shaped electrodes, asshown inFig.7-3. Fortheelectric ﬁeldlinesinFig.7-3, wehave simply copied from Fig.7-1thesetofbroken-line curves thatrepresent V=constant. Wehave abonus! Thecurves forV=constant areorthogonal totheones forU=constant because oftheequations (7.7) and(7.8). Whenever wechoose afunction F(a),wegetfrom UandVboththeequipotentials andﬁeld lines. Andyouwillremember thatwehavesolved either oftwoproblems. depend- ingonwhich setofcurves wecalltheequipotentials. Asasecond example, consider thefunction F(a) =\/5. (7.11) Ifwewrite 3=x+iy=pe’°, where P=v53+-yi and tan0=y/x, then I;-(3) =pl/2et0/2 =pl/2(cosg+isin . from which 2 21/2 1/2 2 21/2 l/2 F(,,=[.<.>:_t__r22___ic1] + .(7,1)) 7-4 B=4/ y/ A=4 / / //a=3, A=3 / / / // / =2a=2/ / // / /,-r/= =,-’I II/ B/I 1 I ’, lA=o' ’_aio ______;_ I \ \\ Thecurves forU(x,y)=AandV(x,y)=B,using UandVfrom Eq.(7.12), areplotted inFig.7-4. Again, there aremany possible situations thatcould be described bythese ﬁelds. Oneofthemost interesting istheﬁeldneartheedgeofa thinplate. IfthelineB=0-to theright ofthey-axis—-represents athincharged plate, theﬁeld lines near itaregiven bythecurves forvarious values ofA.The physical situation isshown inFig.7-5. Further examples are F(3) =23/2,, (7.13) which yields theﬁeld outside arectangular corner F(3) =log3, (7.14) which yields theﬁeldforalinecharge, and F(3) =l/3, (7.15) which gives theﬁeld forthetwo-dimensional analog ofanelectric dipole, i.e., twoparallel linecharges with opposite polarities, very close together. Wewillnotpursue thissubject further inthiscourse, butshould emphasize that although thecomplex variable technique isoften powerful, itislimited to two-dimensional problems; andalso,itisanindirect method. 7-3Plasma oscillations Weconsider nowsome physical situations inwhich theﬁeldisdetermined neither byﬁxed charges norbycharges onconducting surfaces, butbyacom- bination oftwophysical phenomena. Inother words, theﬁeld willbegoverned simultaneously bytwosetsofequations: (1)theequations from electrostatics relating electric ﬁelds tocharge distribution, and(2)anequation from another partofphysics thatdetermines thepositions ormotions ofthecharges inthe presence oftheﬁeld. Theﬁrstexample thatwewilldiscuss isadynamic oneinwhich themotion ofthecharges isgoverned byNewton’s laws. Asimple example ofsuch asituation occurs inaplasma, which isanionized gasconsisting ofionsandfreeelectrons distributed overaregion inspace. Theionosphere—an upper layer oftheatmos- phere—is anexample ofsuchaplasma. Theultraviolet raysfrom thesunknock 7-5l \ l\ \\ \ \\\ \ \ \\\ \ \ \\ \ \ \\ Fig.7-4. Curves ofconstant U(x,y) \ \ \\ and Vlx,y)from Eq.(7.12). \ § \ T~ \ \ 7 I enouuoeo T'Trum: 5 Fig. 7-5. Theelectric ﬁeld near the edge ofathingrounded plate. .,_.._, ---__..._+_.>...s.___.| Fig.7-6. Motion inaplasma wave. Theelectrons attheplane amove toa‘, andthose atbmove tob‘.____1___electrons offthemolecules oftheair,creating freeelectrons andions. lnsuch a plasma thepositive ions arevery much heavier than theelectrons, sowemay neglect theionic motion, incomparison tothatoftheelectrons. Letnobethedensity ofelectrons intheundisturbed, equilibrium state. This must also bethedensity ofpositive ions, since theplasma iselectrically neutral (when undisturbed). Now wesuppose that theelectrons aresomehow moved from equilibrium andaskwhat happens. lfthedensity oftheelectrons in oneregion isincreased, they willrepel each other andtend toreturn totheir equilibrium positions. Astheelectrons move toward their original positions they pickupkinetic energy, andinstead ofcoming torestintheir equilibrium conﬁgura- tion, they overshoot themark. They willoscillate back andforth. Thesituation issimilar towhat occurs insound waves, inwhich therestoring force isthegas pressure. Inaplasma, therestoring force istheelectrical force ontheelectrons. Tosimplify thediscussion, wewillworry only about asituation inwhich the motions areallinonedimension, sayx.Letussuppose thattheelectrons origi- nally atxare,attheinstant t,displaced from their equilibrium positions byasmall amount s(x,t).Since theelectrons havebeen displaced, their density will,ingeneral, bechanged. Thechange indensity iseasily calculated. Referring toFig.7-6. theelectrons initially contained between thetwoplanes aandbhave moved and arenow contained between theplanes a’andb’.Thenumber ofelectrons that were between aandbisproportional ton0Ax; thesame number arenowcontained inthespace whose width isAx+As.Thedensity haschanged to _ n@Ax__ :__ no _ ”TAx+As 1+(As/Ax) (H6) lfthechange indensity issmall, wecanwrite [using thebinomial expansion for (l+e)“1] A.H=n.,(1_ (7.17) Weassume thatthepositive ionsdonotmove appreciably (because ofthemuch larger inertia), sotheir density remains no.Each electron carries thecharge —q,, sotheaverage charge density atanypoint isgiven by P=—(~-m>)q..O1‘ dP=Mn§ (Mo (where wehave written thedifferential form forAs/Ax). Thecharge density isrelated totheelectric ﬁeld byMaxwell's equations, in particular, v-E=-3- (7.19)60 Iftheproblem isindeed one-dimensional (and ifthere arenoother ﬁelds butthe oneduetothedisplacements oftheelectrons), theelectric ﬁeld Ehasasingle component E,.Equation (7.19), together with (7.18), gives 6E,_noq, 0s 'n"§a' mm Integrating Eq.(7.20) gives E,=5’-2%1-+K. (7.21)0 Since E,=Owhen s==0.theintegration constant Kiszero. Theforce onanelectron inthedisplaced position is 2 1-",=-%1-, (7.22) 7-6 arestoring force proportional tothedisplacement softheelectron. Thisleads to aharmonic oscillation oftheelectrons. Theequation ofmotion ofadisplaced electron is d2s_ noqf Weﬁndthatswillvaryharmonically. Itstimevariation willbeascoswt,or—- using theexponential notation ofVol.I——as e“°i=‘. (7.24) Thefrequency ofoscillation w,,isdetermined from (7.23): 2 of.= (7.25) andiscalled theplasma frequency. Itisacharacteristic number oftheplasma. When dealing withelectron charges many people prefer toexpress their an- swers interms ofaquantity e2deﬁned by 2 e2=1%? =2.3068 X10"” newton-meterz. (7.26)0 Using thisconvention, Eq.(7.25) becomes 2 1»;=%. (7.27) which istheform youwillﬁndinmost books. Thus wehavefound thatadisturbance ofaplasma wiHsetupfreeoscillations oftheelectrons about their equilibrium positions atthenatural frequency w,,, which isproportional tothesquare rootofthedensity oftheelectrons. Theplasma electrons behave likearesonant system, such asthose wedescribed inChapter 23ofVol.I. Thisnatural resonance ofaplasma hassome interesting effects. Forexample, ifonetriestopropagate aradiowave through theionosphere, oneﬁnds thatit canpenetrate onlyifitsfrequency ishigher thantheplasma frequency. Otherwise thesignal isreﬂected back. Wemust usehighfrequencies ifwewishtocommuni- catewithasatellite inspace. Ontheother hand, ifwewishtocommunicate with aradio station beyond thehorizon, wemust usefrequencies lower thantheplasma frequency, sothatthesignal willbereﬂected back totheearth. Another interesting example ofplasma oscillations occurs inmetals. Ina metal wehaveacontained plasma ofpositive ions, andfreeelectrons. Thedensity noisveryhigh, sowpisalso. Butitshould stillbepossible toobserve theelectron oscillations. Now, according toquantum mechanics, aharmonic oscillator with anatural frequency oi,hasenergy levels which areseparated bythetheenergy increment hwp. If,then, oneshoots electrons through, say,analuminum foil,and makes verycareful measurements oftheelectron energies ontheother side,one might expect toﬁndthattheelectrons sometimes losetheenergy ha,totheplasma oscillations. This does indeed happen. Itwasﬁrstobserved experimentally in 1936thatelectrons withenergies ofafewhundred toafewthousand electron volts lostenergy injumps when scattering from orgoing through athinmetal foil.The eﬂect wasnotunderstood until 1953 when Bohm andPines* showed thatthe observations could beexplained interms ofquantum excitations oftheplasma oscillations inthemetal. “Forsome recent work andabibliography seeC.J.Powell andJ.B.Swann, Phys. Rev.115,869(1959). 7-7 7-4Colloidal particles inanelectrolyte Weturntoanother phenomenon inwhich thelocations ofcharges isgoverned byapotential that arises inpart from thesame charges. Theresulting effects inﬂuence inanimportant waythebehavior ofcolloids. Acolloid consists ofa suspension inwater ofsmall charged particles which, though microscopic, from anatomic point ofview arestillvery large. Ifthecolloidal particles were not charged, they would tend tocoagulate into large lumps: butbecause oftheir charge, they repel each other andremain insuspension. Now ifthere isalsosome saltdissolved inthewater, itwillbedissociated into positive andnegative ions. (Such asolution ofionsiscalled anelectrolyte.) The negative ionsareattracted tothecolloid particles (assuming their charge ispositive) andthepositive ionsarerepelled. Wewilldetermine howtheionswhich surround such acolloidal particle aredistributed inspace. Tokeep theideas simple, wewillagain solve only aone-dimensional case. Ifwethink ofacolloidal particle asasphere having avery large radius—on an atomic scale!——we canthen treat asmall partofitssurface asaplane. (Whenever oneistrying tounderstand anewphenomenon itisagood ideatotakeasomewhat oversimpliﬁed model; then, having understood theproblem with thatmodel, one isbetter abletoproceed totackle themore exact calculation.) Wesuppose thatthedistribution ofionsgenerates acharge density p(x), and anelectrical potential ¢,related bytheelectrostatic lawV24; =—p/er, or,for ﬁelds thatvary inonly onedimension, by d2¢_PF_-Q (7.28) Now supposing there were such apotential ¢(x), howwould theionsdis- tribute themselves init?Thiswecandetermine bytheprinciples ofstatistical mechanics. Ourproblem thenistodetermine ¢sothattheresulting charge density from statistical mechanics alsosatisﬁes (7.28). According tostatistical mechanics (seeChapter 40,Vol.l),particles inthermal equilibrium inaforce ﬁeldaredistributed insuch awaythatthedensity nof particles attheposition xisgiven by no)=nne""">’“'. (1.29) where U(x) isthepotential energy, kisBoltzmann’s constant, andTistheabsolute temperature. Weassume that theions carry oneelectronic charge, positive ornegative. Atthedistance xfrom thesurface ofacolloidal particle, apositive ionwillhave potential energy q,¢(x), sothat I/(X)=qt-¢(X)- Thedensity ofpositive ions, n+,isthen n+(x) :noe-aewm/kT_ Similarly, thedensity ofnegative ionsis n_(x) =n0e+<ii¢<z>/kT_ Thetotal charge density is P=‘]e”+ _qv”—, or p=qen0(e—11r»¢/IrT_e+q¢¢/kT)_ (730) Combining thiswith Eq.(7.28), weﬁndthatthepotential ¢must satisfy 2 1 II gxs’:=_%)l0 (e—qe4>//¢1_e+q¢¢/kl)_ (731) 7-8 Thisequation isreadily solved ingeneral [multiply both sides by2(d¢/dx), and integrate withrespect tox],buttokeeptheproblem assimple aspossible, wewill consider hereonlythelimiting easeinwhich thepotentials aresmall orthetem- perature Tishigh. Thecasewhere tpissmall corresponds toadilute solution. For these cases theexponent issmall, andwecanapproximate e*"~t"”' =1¢%'Z- (7.32) Equation (7.31) thengives d2¢ 2nqf Notice thatthistimethesignontheright ispositive. Thesolutions for¢>arenot oscillatory, butexponential. Thegeneral solution ofEq.(7.33) is ¢=Ae-'/1’ +Be+*”’, (7.34) with D2=51'-‘L 7.352n0q2 ( ) Theconstants AandBmust bedetermined from theconditions oftheproblem. Inourcase, Bmust bezero; otherwise thepotential would gotoinﬁnity forlarge x.Sowehavethat ¢=Ae””'D, (7.36) inwhich Aisthepotential atx=0,thesurface ofthecolloidal particle. A ¢ Fig.7~7. Thevariation ofthepo- tential near thesurface ofacolloidal particle. DistheDebye length. L Ob I!) QID 3'0 ﬁx’ Thepotential decreases byafactor l/eeachtimethedistance increases byD, asshown inthegraph ofFig.7-7. Thenumber Discalled theDebye length, and isameasure ofthethickness oftheionsheath thatsurrounds alarge charged particle inanelectrolyte. Equation (7.36) saysthatthesheath getsthinner with increasing concentration oftheions(no)orwithdecreasing temperature. Theconstant AinEq.(7.36) iseasily obtained ifweknow thesurface charge 0 onthecolloid particle. Weknow that E,=E,,(0)= (7.37) ButEisalsothegradient of4»: AE,(O)=-g0=+5, (7.38) from which weget A=‘L9 (7.39)E0 7-9 Using thisresult in(7.36), weﬁnd (bytaking x=0)that thepotential ofthe colloidal particle is ¢(0)=%- (7.40) You willnotice thatthispotential isthesame asthepotential difference across a condenser with aplate spacing Dandasurface charge density <1. Wehave said that thecolloidal particles arekept apart bytheir electrical repulsion. Butnow weseethattheﬁeld alittle wayfrom thesurface ofaparticle isreduced bytheionsheath thatcollects around it.Ifthesheaths getthinenough, theparticles have agood chance ofknocking against each other. They willthen stick, andthecolloid willcoagulate andprecipitate outoftheliquid. From our analysis, weunderstand why adding enough salttoacolloid should cause itto precipitate out. Theprocess iscalled “salting outacolloid.” Another interesting example istheeffect thatasaltsolution hasonprotein molecules. Aprotein molecule isalong, complicated, andﬂexible chain ofamino acids. The molecule hasvarious charges onit,anditsometimes happens that there isanetcharge, saynegative, which isdistributed along thechain. Because ofmutual repulsion ofthenegative charges, theprotein chain iskept stretched out. Also, ifthere areother similar chain molecules present inthesolution, they will bekept apart bythesame repulsive eﬂects. Wecan,therefore, have asuspension ofchain molecules inaliquid. Butifweaddsalttotheliquid wechange theproper- tiesofthesuspension. Assaltisadded tothesolution, decreasing theDebye distance, thechain molecules canapproach oneanother, andcanalso coilup. Ifenough saltisadded tothesolution, thechain molecules willprecipitate outof thesolution. There aremany chemical effects ofthiskind thatcanbeunderstood interms ofelectrical forces. 7-5Theelectrostatic ﬁeldofagrid Asourlastexample, wewould liketodescribe another interesting property ofelectric ﬁelds. Itisonewhich ismade useofinthedesign ofelectrical instru- ments, intheconstruction ofvacuum tubes, andforother purposes. This isthe character oftheelectric ﬁeld near agridofcharged wires. Tomake theproblem assimple aspossible, letusconsider anarray ofparallel wires lying inaplane, thewires being inﬁnitely long andwith auniform spacing between them. Ifwelook attheﬁeld alarge distance above theplane ofthewires, weseea constant electric ﬁeld, just asthough thecharge were uniformly spread over a plane. Asweapproach thegrid ofwires, theﬁeld begins todeviate from the uniform ﬁeld wefound atlarge distances from thegrid. Wewould liketoestimate how close tothegridwehave tobeinorder toseeappreciable variations inthe potential. Figure 7-8shows arough sketch oftheequipotentials atvarious distances from thegrid. Thecloser wegettothegrid. thelarger thevariations. Aswetravel parallel tothegrid, weobserve thattheﬁeld ﬂuctuates inaperiodic manner. _-¢_-—-—_¢—__---_—_ fz--____,-___¢—~_ _—'~_ _,-~__-— - — - 4-'~ /_\ —T T\,' _~" \~/ ~' ~4/ T *\ r~\ /~ ’— ~ —\\ \\\ Q".__/\ +0 1*-+I -~\\\ -+.I'I.\\+0_z,- ‘-‘I -\ _-' ,_‘\~_,I ’\ \v \_¢ 'O\ lOl l<—O—~lTFig. 7-8. Equipotential surfaces above aunﬁonn gﬁd ofcharged whee 7-10 Now wehave seen (Chapter 50,Vol. I)that anyperiodic quantity canbe expressed asasum ofsinewaves (Fourier’s theorem). Let’s seeifwecanﬁnda suitable harmonic function thatsatisﬁes ourﬁeldequations. Ifthewires lieinthexy-plane andrunparallel tothey-axis, then wemight tryterms like ¢(x,Z)=F,,(z)cos@, (7.41) where aisthespacing ofthewires andnistheharmonic number. (Wehave as- sumed longwires, sothere should benovariation withy.)Acomplete solution would bemade upofasumofsuchterms forn=1,2,3,.... Ifthisistobeavalid potential, itmust satisfy Laplace’s equation inthe region above thewires (where there arenocharges). That is, a2¢ a2¢_ m+5?"°- Trying thisequation onthe¢in(7.41), weﬁndthat 41r2n2 21rnx d2F,, 21rnx—-25- F,,(z) cos-7- +72?cos? =0, (7.42) orthatF,,(z) must satisfy d2F,, 412712F2? =7- F". Sowemust have F,,=A,,e"'/’°, (7.44)where Z0= (7.45) Wehave found thatifthere isaFourier component oftheﬁeldofharmonic n, thatcomponent willdecrease exponentially with acharacteristic distance zo= a/21m. Fortheﬁrstharmonic (n=1),theamplitude falls bythefactor e_2" (alarge decrease) eachtimeweincrease zbyonegridspacing a.Theother har- monics falloﬂeven more rapidly aswemove a'way from thegrid. Weseethatif weareonly afewtimes thedistance aaway from thegrid, theﬁeld isvery nearly uniform, i.e.,theoscillating terms aresmall. There would, ofcourse, always remain the“zero harmonic” ﬁeld ¢0=“E02 togivetheuniform ﬁeld atlarge z.Foracomplete solution, wewould combine thisterm with asumofterms like(7.41) with F,,from (7.44). Thecoefﬁcients A,, would beadjusted sothatthetotal sumwould, when differentiated, giveanelectric ﬁeld thatwould ﬁtthecharge density Aofthegridwires. Themethod wehave justdeveloped canbeused toexplain whyelectrostatic shielding bymeans ofascreen isoften justasgood aswith asolid metal sheet. Except within adistance from thescreen afewtimes thespacing ofthescreen wires, theﬁelds inside aclosed screen arezero. Weseewhy copper screen— lighter andcheaper than copper sheet—is often used toshield sensitive electrical equipment from external disturbing ﬁelds. '7-ll 8 Electrostatic Energy 8-1Theelectrostatic energy ofcharges. Auniform sphere Inthestudy ofmechanics, oneofthemost interesting anduseful discoveries wasthelawoftheconservation ofenergy. Theexpressions forthekinetic and potential energies ofamechanical system helped ustodiscover connections between thestates ofasystem attwodifferent times without having tolookintothedetails ofwhat wasoccurring inbetween. Wewishnowtoconsider theenergy ofelectro- static systems. Inelectricity alsotheprinciple oftheconservation ofenergy will beuseful fordiscovering anumber ofinteresting things. Thelawoftheenergy ofinteraction inelectrostatics isverysimple; wehave, infact,already discussed it.Suppose wehavetwocharges qland:12separated by thedistance rm.There issome energy inthesystem, because acertain amount of work wasrequired tobring thecharges together. Wehave already calculated the work done inbringing twocharges together from alarge distance. Itis ‘I142 _ 41reor12 (8.1) Wealsoknow, from theprinciple ofsuperposition, thatifwehavemany charges present, thetotalforce onanycharge isthesumoftheforces from theothers. It follows, therefore, thatthetotalenergy ofasystem ofanumber ofcharges isthe sumofterms duetothemutual interaction ofeachpairofcharges. Ifq,andq,- areanytwoofthecharges andr,,-isthedistance between them (Fig. 8-1), the energy ofthatparticular pairis qt-qt_41l'€0)','j Thetotalelectrostatic energy Uisthesumoftheenergies ofallpossible pairs of charges: U= _i='_qL_ . g_3 all?1irs 47'-eorij ( ) Ifwehaveadistribution ofcharge speciﬁed byacharge density p,thesumofEq. (8.3)is,ofcourse, tobereplaced byanintegral. Weshallconcern ourselves withtwoaspects ofthisenergy. Oneistheapplica- tionoftheconcept ofenergy toelectrostatic problems; theother istheevaluation oftheenergy indifferent ways. Sometimes itiseasier tocompute thework done forsome special casethantoevaluate thesuminEq.(8.3), orthecorresponding integral. Asanexample, letuscalculate theenergy required toassemble asphere ofcharge withauniform charge density. Theenergy isjustthework done in gathering thecharges together from inﬁnity. Imagine thatweassemble thesphere bybuilding upasuccessibn ofthin spherical layers ofinﬁnitesimal thickness. Ateachstage oftheprocess, wegather asmall amount ofcharge andputitinathinlayer from rtor+dr.Wecontinue theprocess untilwearrive attheﬁnalradius a(Fig.8-2). IfQ,isthecharge ofthe sphere when ithasbeenbuiltuptotheradius r,thework done inbringing acharge dQtoitis _2-_d.2.av-41“or (8.4) 8-18-1Theelectrostatic energy of charges. Auniform sphere 8-2Theenergy ofacondenser. Forces oncharged conductors 8-3Theelectrostatic energy ofan ionic crystal 8-4Electrostatic energy innuclei 8-5Energy intheelectrostatic ﬁeld 8-6Theenergy ofapoint charge Review: Chapter 4,Vol.I,Conservation ofEnergy Chapters 13and14,Vol. I, Work andPotential Energy O O O O O oqi O ° \\-- O O 0 Q\[ll 0\ \\ 0 ° \\0cl] O O O Fig. 8-l. Theelectrostatic energy of asystem ofparticles isthesum ofthe electrostatic energy ofeach pair. eaFig. 8-2. The energy ofauniform sphere ofcharge can becomputed by imagining that itisassembled from successive spherical shells.Ifthedensity ofcharge inthesphere isp,thecharge Q,is 4Qt=p-§1rr3, andthecharge dQis dQ=p-41rr2 dr. Equation (8.4)becomes 24 av=?-4"’; d'- (8.5)O Thetotal energy required toassemble thesphere istheintegral ofdUfrom r= 0tor=a,or 4 25 u=_’{";_:-;-- (8.6) Orifwewishtoexpress theresult interms ofthetotal charge Qofthesphere, _3Q2U_§Z@- (8.7) Theenergy isproportional tothesquare ofthetotal charge andinversely pro- portional totheradius. Wecanalsointerpret Eq.(8.7)assaying thattheaverage of(1/r,,)forallpairs ofpoints inthesphere is3/5a. 8-2Theenergy ofacondenser. Forces oncharged conductors Weconsider nowtheenergy required tocharge acondenser. Ifthecharge Q hasbeentaken from oneoftheconductors ofacondenser andplaced ontheother, thepotential difference between them is V= (8.8) where Cisthecapacity ofthecondenser. How much work isdone incharging thecondenser? Proceeding asforthesphere, weimagine thatthecondenser has been charged bytransferring charge from oneplate totheother insmall increments dQ.Thework required totransfer thecharge dQis dU=VdQ. Taking Vfrom Eq.(8.8), wewrite _219..dU-C Orintegrating from zerocharge totheﬁnalcharge Q,wehave |\)v-AoQv=_-- (8.9) This energy canalsobewritten as U=-L-CV2. (8.10) Recalling thatthecapacity ofaconducting sphere (relative toinﬁnity) is Cspherc =47500: wecanimmediately getfrom Eq.(8.9)theenergy ofacharged sphere, _1Q’ u_5z;55- (8.11) 8-2 This, ofcourse, isalsotheenergy ofathinspherical shelloftotalcharge Qandis just5/6oftheenergy ofauniformly charged sphere, Eq.(8.7). Wenowconsider applications oftheideaofelectrostatic energy. Consider thefollowing questions: What istheforce between theplates ofacondenser? Or what isthetorque about some axisofacharged conductor inthepresence ofan- other with opposite charge? Such questions areeasily answered byusing our result Eq.(8.9)forelectrostatic energy ofacondenser, together withtheprinciple ofvirtual work (Chapters 4,13,andl4ofVol.I). Let’s usethismethod fordetermining theforce between theplates ofa parallel-plate condenser. Ifweimagine thatthespacing oftheplates isincreased bythesmall amount Az,then themechanical work done from theoutside in moving theplates would be AW=FAz, (8.12) where Fistheforce between theplates. Thiswork must beequal tothechange intheelectrostatic energy ofthecondenser. ' ByEq.(8.9), theenergy ofthecondenser wasoriginally l\)r—~<19.U=—-—- Thechange inenergy (ifwedonotletthecharge change) is _121AU—5QA(z,) - (8.13) Equating (8.12) and(8.13), wehave _Q(1) FAz- 2AC- (8.14) Thiscanalsobewritten as 2 FA:=-5%AC. (8.15) Theforce, ofcourse, results from theattraction ofthecharges ontheplates, but weseethatwedonothave toworry indetail about howtheyaredistributed; everything weneed istaken careofinthecapacity C. Itiseasytoseehowtheideaisextended toconductors ofanyshape, andfor other components oftheforce. InEq.(8.14), wereplace Fbythecomponent we arelooking for,andwereplace Azbyasmall displacement inthecorresponding direction. Orifwehaveanelectrode withapivot andwewant toknow thetorque 1',wewrite thevirtual work as AW=-rA0, where A0isasmall angular displacement. Ofcourse, A(l/C)must bethechange in l/Cwhich corresponds toA0.Wecould, inthisway,ﬁndthetorque onthemov- ableplates inavariable condenser ofthetypeshown inFig.8-3. Returning tothespecial caseofaparallel-plate condenser, wecanusethe formula wederived inChapter 6forthecapacity: l d-C,-E57, (8.16) where Aistheareaofeachplate. Ifweincrease theseparation byAz, 1 Az “(El"From Eq.(8.14) wegetthattheforce between theplates is 2 _Q_ F-Z07 (8.17) 8-34> 4% 2 Fig.8-3. What isthetorque ona variable capacitor? Q CONDUCTING LAYER OF M75 sunrncscums: a- E0—-> lEl E0 Fig.8-4. Theﬁeld atthesurface of aconductor varies from zero toE0= a/co, asonepasses through thelayer of surface charge.Let’slookatEq.(8.17) alittlemore closely andseeifwecantellhowtheforce arises. Ifforthecharge ononeplatewewrite Q=<1/1. Eq.(8.17) canberewritten as _l 0'F-iQg- Or,since theelectric ﬁeldbetween theplates is j E0=G10, then F=-i}QEo. (8.18) Onewould immediately guess thattheforce acting ononeplate isthecharge Qontheplate times theﬁeldacting onthecharge. Butwehaveasurprising factor ofone-half. Thereason isthatE0isnottheﬁeldatthecharges. Ifweimagine that thecharge atthesurface oftheplate occupies athinlayer, asindicated inFig.8-4, theﬁeldwillvaryfrom zeroattheinner boundary ofthelayer toE0inthespace outside oftheplate. Theaverage ﬁeldacting onthesurface charges isE0/2. That iswhythefactor one-half isinEq.(8.18). Youshould notice thatincomputing thevirtual work wehaveassumed that thecharge onthecondenser wasconstant——-that itwasnotelectrically connected toother objects, andsothetotalcharge could notchange. Suppose wehadimagined thatthecondenser washeldataconstant potential difference aswemade thevirtual displacement. Then weshould havetaken U=i1rCV2 andinplace ofEq.(8.15) wewould havehad FAZ=11-V2AC, which givesaforceequal inmagnitude totheoneinEq.(8.15) (because V=Q/C), butwith theopposite sign! Surely theforce between thecondenser plates doesn’t reverse insignaswedisconnect itfrom itscharging source. Also, weknow that twoplates withopposite electrical charges must attract. Theprinciple ofvirtual work hasbeen incorrectly applied inthesecond case—we have nottaken into account thevirtual work done onthecharging source. That is,tokeepthepo- tential constant atVasthecapacity changes, acharge VAC must besupplied by asource ofcharge. Butthischarge issupplied atapotential V,sothework done bytheelectrical system which keeps thepotential constant isV2AC.Themechan- icalwork FAzplusthiselectrical work V2ACtogether make upthechange inthe totalenergy -QV”ACofthecondenser. Therefore FAzis—§V2 AC,asbefore. 8-3Theelectrostatic energy ofanionic crystal Wenowconsider anapplication oftheconcept ofelectrostatic energy inatomic physics. Wecannot easily measure theforces between atoms, butweareoften interested intheenergy differences between oneatomic arrangement andanother, as,forexample, theenergy ofachemical change. Since atomic forces arebasically electrical, chemical energies areinlarge partjustelectrostatic energies. Let’s consider, forexample, theelectrostatic energy ofanionic lattice. An ionic crystal likeNaCl consists ofpositive andnegative ionswhich canbethought ofasrigidspheres‘. They attract electrically untiltheybegin totouch; thenthere is arepulsive force which goesupveryrapidly ifwetrytopushthem closer together. Forourﬁrstapproximation, therefore, weimagine asetofrigid spheres thatrepresent theatoms inasaltcrystal. Thestructure ofthelattice hasbeen determined byx-ray diﬂraction. Itisacubic lattice-like athree-dimensional 8-4 checkerboard. Figure 8-5shows across-sectional view. Thespacing oftheionsis 2.81A(=2.8l X10‘8cm). Ifourpicture ofthissystem iscorrect, weshould beabletocheck itbyasking thefollowing question: How much energy willittaketopullallthese ionsapart- thatis,toseparate thecrystal completely intoions? Thisenergy should beequal totheheatofvaporization ofNaCl plustheenergy required todissociate the molecules intoions. Thistotalenergy toseparate NaCl toionsisdetermined experi- mentally tobe7.92electron volts permolecule. Using theconversion lev=1.602 Xl0“°joule, andAvogadro’s number forthenumber ofmolecules inamole, N0=6.02X1023, theenergy ofvaporization canalsobegiven as W=7.64X105joules/mole. Physical chemists prefer foranenergy unitthekilocalorie, which is4190joules; sothat1evpermolecule is23kilocalories permole. Achemist would thensay thatthedissociation energy ofNaCl is W=183kcal/mole. Can weobtain thischemical energy theoretically bycomputing how much work itwould take topullapart thecrystal? According toourtheory, thiswork is thesumofthepotential energies ofallthepairs ofions. Theeasiest waytoﬁgure outthissumistopickoutaparticular ionandcompute itspotential energy with eachoftheother ions. Thatwillgiveustwice theenergy perion,because theenergy belongs tothepairs ofcharges. Ifwewant theenergy tobeassociated withone particular ion,weshould takehalfthesum. Butwereally want theenergy per molecule, which contains twoions, sothatthesumwecompute willgivedirectly theenergy permolecule. Theenergy ofanionwithoneofitsnearest neighbors ise2/a, where e2= qf/41reo andaisthecenter-to-center spacing between ions. (Weareconsidering monovalent ions.) Thisenergy is5.12ev,which wealready seeisgoing togiveus aresult ofthecorrect order ofmagnitude. Butitisstillalongwayfrom theinﬁnite sumofterms weneed. Let’s begin bysumming alltheterms from theionsalong astraight line. Considering thattheionmarked NainFig.8-5isourspecial ion,weshallconsider ﬁrstthose ionsonahorizontal linewithit.There aretwonearest Clionswith negative charges, eachatthedistance a.Then there aretwopositive ionsatthe distance 2a,etc.Calling theenergy ofthissumU1,wewrite U1=7(—-+.i__§+_+...) 28” 111=_.?.(1_.i_|_§_.z_|_...). (3_19)NM v—t->IO t-I -BM Theseries converges slowly, soitisdiﬂicult toevaluate numerically, butitisknown tobeequal tohi2.So 2 2 U,=-3;-1112 =-1.3865’; (8.20) Now consider thenextadjacent lineofionsabove. Thenearest isnegative andatthedistance a.Then there aretwopositives atthedistance \/2a.Thenext pairareatthedistance \/5a,thenextat\/10a,andsoon.Soforthewhole line wegettheseries a<le2 12 2 2 T"+~/ft/5+./1?) (821)8-5T3,><>,<><><>.<..>.<><.><>.<.>:.<>< 4-i> uni Fig.8-5. Cross section ofasalt crystal onanatomic scale. Thechecker- board arrangement ofNaandClionsis thesame inthetwocross sections per- pendicular totheoneshown. (See Vol.l, Fig.1-7.) There arefoursuchlines: above, below, infront, andinback. Then there arethe fourlineswhich arethenearest linesondiagonals, andonandon. Ifyouwork patiently through forallthelines, andthentakethesum, you ﬁndthatthegrand totalis 2 U-1.747?-,G which isjustsomewhat more than what weobtained in(8.20) fortheﬁrstline. Using e2/a =5.12ev,weget U=8.94ev. Ouranswer isabout 10%above theexperimentally observed energy. Itshows that ourideathatthewhole lattice isheldtogether byelectrical Coulomb forces is fundamentally correct. This istheﬁrsttime thatwehave obtained aspeciﬁc property ofamacroscopic substance from aknowledge ofatomic physics. We willdomuch more later. Thesubject thattriestounderstand thebehavior of bulkmatter interms ofthelawsofatomic behavior iscalled solid-state physics. Now what about theerror inourcalculation? Why isitnotexactly right? Itisbecause oftherepulsion between theionsatclose distances. They arenot perfectly rigid spheres, sowhen theyareclose together theyarepartly squashed. They arenotverysoft,sotheysquash onlyalittlebit.Some energy, however, is usedindeforming them, andwhen theionsarepulled apart thisenergy isreleased. Theactual energy needed topulltheionsapart isalittlelessthantheenergy that wecalculated; therepulsion helps inovercoming theelectrostatic attraction. Isthere anywaywecanmake anallowance forthiscontribution? Wecould ifweknew thelawoftherepulsive force. Wearenotready toanalyze thedetails ofthisrepulsive mechanism, butwecangetsome ideaofitscharacteristics from some large-scale measurements. From ameasurement ofthecompressibility of thewhole crystal, itispossible toobtain aquantitative ideaofthelawofrepulsion between theionsandtherefore ofitscontribution totheenergy. Inthiswayit hasbeen found thatthiscontribution must bel/9.4 ofthecontribution from the electrostatic attraction and,ofcourse, ofopposite sign. Ifwesubtract thiscontribu- tionfromthepureelectrostatic energy, weobtain 7.99evforthedissociation energy permolecule. Itismuch closer totheobserved result of7.92ev,butstillnotin perfect agreement. There isonemore thing wehaven’t taken intoaccount: we havemade noallowance forthekinetic energy ofthecrystal vibrations. Ifacor- rection ismade forthiseffect, verygood agreement withtheexperimental number isobtained. Theideas arethencorrect; themajor contribution totheenergy ofa crystal likeNaCl iselectrostatic. 8-4Electrostatic energy innuclei Wewillnow take upanother example ofelectrostatic energy inatomic physics, theelectrical energy ofatomic nuclei. Before wedothiswewillhave to discuss some ,properties ofthemain forces (called nuclear forces) thathold the protons andneutrons together inanucleus. Intheearly daysofthediscovery of nuclei—and oftheneutrons andprotons thatmake them up—it washoped that thelawofthestrong, nonelectrical partoftheforce between, say,aproton and another proton would have some simple law,liketheinverse square lawofelec- tricity. Foronceonehaddetermined thislawofforce, andthecorresponding ones between aproton andaneutron, andaneutron andaneutron, itwould bepossible todescribe theoretically thecomplete behavior ofthese particles innuclei. There- foreabigprogram wasstarted forthestudy ofthescattering ofprotons, inthe hope ofﬁnding thelawofforce between them; butafter thirty years ofeffort, nothing simple hasemerged. Aconsiderable knowledge oftheforcebetween proton andproton hasbeen accumulated, butweﬁndthattheforce isascomplicated as itcanpossibly be. What wemean by“ascomplicated asitcanbe”isthattheforce depends on asmany things asitpossibly can. 8-6 First, theforce isnotasimple function ofthedistance between thetwoprotons. Atlarge distances there isanattraction, butatcloser distances there isarepulsion. Thedistance dependence isacomplicated function, stillimperfectly known. Second, theforce depends ontheorientation oftheprotons’ spin. Theprotons haveaspin, andanytwointeracting protons maybespinning withtheir angular momenta inthesame direction orinopposite directions. Andtheforce isdifferent when thespins areparallel from what itiswhen theyareantiparallel, asin(a) and(b)ofFig.8-6. Thedifference isquite large; itisnotasmall effect. Third, theforce isconsiderably different when theseparation ofthetwo protons isinthedirection parallel totheirspins, asin(c)and(d)ofFig.8-6,than itiswhen theseparation isinadirection perpendicular tothespins, asin(a)and(b). Fourth, theforce depends, asitdoes inmagnetism, onthevelocity ofthe protons, onlymuch more strongly thaninmagnetism. Andthisvelocity-dependent force isnotarelativistic effect; itisstrong evenatspeeds much lessthanthespeed oflight. Furthermore, thispart oftheforce depends onother things besides the magnitude ofthevelocity. Forinstance, when aproton ismoving nearanother proton, theforce isdifferent when theorbital motion hasthesame direction of rotation asthespin, asin(e)ofFig.8-6,thanwhen ithastheopposite direction ofrotation, asin(f).Thisiscalled the“spin orbit” partoftheforce. Theforce between aproton andaneutron andbetween aneutron anda neutron arealsoequally complicated. Tothisdaywedonotknow themachinery behind these forces—that istosay,anysimple wayofunderstanding them. There is,however, oneimportant wayinwhich thenucleon forces aresimpler thantheycould be.Thatisthatthenuclear force between twoneutrons isthesame astheforce between aproton andaneutron, which isthesame astheforce between twoprotons! If,inanynuclear situation, wereplace aproton byaneutron (orvice versa),- thenuclear interactions arenotchanged. The “fundamental reason” for thisequality isnotknown, butitisanexample ofanimportant principle thatcan beextended alsototheinteraction laws ofother strongly interacting particles- suchasthe1r-mesons andthe“strange” particles. Thisfactisnicely illustrated bythelocations oftheenergy levels insimilar nuclei. Consider anucleus likeB“(boron-eleven), which iscomposed ofﬁve protons andsixneutrons. Inthenucleus theeleven particles interact with one another inamost complicated dance. Now, there isoneconﬁguration ofallthe possible interactions which hasthelowest possible energy; thisisthenormal state ofthenucleus, andiscalled theground state. Ifthenucleus isdisturbed (forexam- ple,bybeing struck byahigh-energy proton orother particle) itcanbeputinto anynumber ofother conﬁgurations, called excited states, each ofwhich willhave acharacteristic energy that ishigher than that oftheground state. Innuclear physics research, suchasiscarried onwithVandeGraaff generator (forexample, inCaltech’s Kellogg andSloan Laboratories), theenergies andother properties ofthese excited states aredetermined byexperiment. Theenergies oftheﬁfteen lowest known excited states ofB11areshown inaone-dimensional graph onthe lefthalfofFig.8-7. Thelowest horizontal linerepresents theground state. Theﬁrstexcited state hasanenergy 2.14Mev higher than theground state, thenext anenergy 4.46 Mev higher than theground state, andsoon.Thestudy ofnuclear physics attempts toﬁnd anexplanation forthisrather complicated pattern ofenergies; there isasyet,however, nocomplete general theory of such nuclear energy levels. Ifwereplace oneoftheneutrons inB11withaproton, wehavethenucleus ofanisotope ofcarbon, C‘1.Theenergies ofthelowest sixteen excited states of C11have alsobeen measured; they areshown intheright halfofFig.8-7. (The broken lines indicate levels forwhich theexperimental information is questionable.) Looking atFig.8-7,weseeastriking similarity between thepattern ofthe energy levels inthetwonuclei. Theﬁrstexcited states areabout 2Mev above the ground states. There isalarge gapofabout 2.3Mevtothesecond excited state, then asmall jump ofonly 0.5Mev tothethird level. Again, between thefourth andﬁfthlevels, abigjump; butbetween theﬁfthandsixth atinyseparation ofthe 8-7a b 8‘->¢ 58¢ °¢ "¢ <> <> \-5e §- -§ P“ §~ K Fig. 8-6. The force between two protons depends onevery possible parameter. , I089 ‘I06! 5 1 I052 992 I| ' 2 892 =28. __.as799A 730 . " 503 1-6b\|8.<7"‘IIIII 4Bl .1i6_._i 4;; i"‘i--- 200 B" ll.982/ C" Fig. 8-7. The energy levels ofB“ and C“(energies inMev). Theground state ofC"isL982 Mev higher than thatofB". order of0.1Mev. Andsoon.After about thetenth level, thecorrespondence seems tobecome lost,butcanstillbeseenifthelevels arelabeled withtheirother deﬁning characteristics—for instance, their angular momentum andwhat theydo tolosetheirextra energy. Thestriking similarity ofthepattern oftheenergy levels ofB11andC11is surely notjustacoincidence. Itmust reveal some physical law. Itshows, infact, thateveninthecomplicated situation inanucleus, replacing aneutron byaproton makes verylittlechange. Thiscanmean onlythattheneutron-neutron andproton- proton forces must benearly identical. Only thenwould weexpect thenuclear conﬁgurations withﬁveprotons andsixneutrons tobethesame aswithsixprotons andﬁveneutrons. Notice thattheproperties ofthese twonuclei tellusnothing about theneutron- proton force; there arethesame number ofneutron-proton combinations inboth nuclei. Butifwecompare twoother nuclei, suchasC1‘,which hassixprotons and eightneutrons, withN14,which hasseven ofeach, weﬁndasimilar correspondence ofenergy levels. Sowecanconclude thatthep-p,n-n,andp-nforces areidentical inalltheir complexities. There isanunexpected principle inthelawsofnuclear forces. Even though theforce between eachpairofnuclear particles isverycompli- cated, theforce between thethree possible different pairs isthesame. ' Butthere aresome small diﬂerences. Thelevels donotcorrespond exactly; also,theground stateofC11hasanabsolute energy (itsmass) which ishigher than theground state ofB11by1.982 Mev. Alltheother levels arealsohigher in absolute energy bythissame amount. Sotheforces arenotexactly equal. But weknow verywellthatthecomplete forces arenotexactly equal; there isanelec- trical force between twoprotons because eachhasapositive charge, while between twoneutrons there isnosuchelectrical force. Canweperhaps explain thediffer- ences between B11andC11bythefactthattheelectrical interaction oftheprotons isdifferent inthetwocases? Perhaps eventheremaining minor differences inthe levels arecaused byelectrical effects? Since thenuclear forces aresomuch stronger thantheelectrical force, electrical eﬂ'ects would haveonlyasmall perturbing effect ontheenergies ofthelevels. Inorder tocheck thisidea, orrather toﬁndoutwhat theconsequences ofthis ideaare,weﬁrstconsider thedifference intheground-state energies ofthetwo nuclei. Totakeaverysimple model, wesuppose thatthenuclei arespheres of radius r(tobedetermined), containing Zprotons. Ifweconsider thatanucleus islikeasphere withuniform charge density, wewould expect theelectrostatic energy (from Eq.8.7)tobe _3(Zq.)”U-3-Z-5:07 . (8.22) where q,istheelementary charge oftheproton. Since ZisﬁveforB11andsixfor C11,their electrostatic energies would bedifferent. With suchasmall number ofprotons, however, Eq.(8.22) isnotquite correct. Ifwecompute theelectrical energy between allpairsofprotons, considered aspoints which weassume tobenearly uniformly distributed throughout thesphere, we ﬁndthat inEq.(8.22) thequantity Z2should bereplaced byZ(Z —1),sothe energy is _3z(z-1)qf_ 3z(z-1);U_3 41re0r _3 r ' (823) Ifweknew thenuclear radius r,wecould use(8.23) toﬁndtheelectrostatic energy difference between B11andC11. Butlet’sdotheopposite; let’sinstead usethe observed energy difference tocompute theradius, assuming thattheenergy differ- enceisallelectrostatic inorigin. Thatis,however, notquite right. Theenergy difference of1.982 Mevbetween theground states ofB11andC11includes therestenergies—that is,theenergy mc2—of alltheparticles. Ingoing from B11toC11,wereplace aneutron bya proton, which haslessmass. Sopartoftheenergy difference isthedifference in therestenergies ofaneutron andaproton, which is0.784 Mev. Thedifference, 8-8 tobeaccounted forbyelectrostatic energy, isthusmore than 1.982 Mev; itis 1.982 +0.784 =2.786 Mev. Using thisenergy inEq.(8.23), fortheradius ofeither B11orC11weﬁnd r=3.12><10-“cm. (8.24) Does thisnumber have anymeaning? Toseewhether itdoes, weshould compare itwith some other determination oftheradius ofthese nuclei. For example, wecanmake another measurement oftheradius ofanucleus byseeing howitscatters fastparticles. From suchmeasurements ithasbeenfound, infact, thatthedensity ofmatter inallnuclei isnearly thesame, i.e.,their volumes are proportional tothenumber ofparticles theycontain. IfweletAbethenumber of protons andneutrons inanucleus (anumber verynearly proportional toitsmass), itisfound thatitsradius isgiven by r=A1/aro, (8.25) where ro=1.2Xl0"111cm. (8.26) From these measurements weﬁndthattheradius ofaB11(oraC11)nucleus isexpected tobe 1-=(1.2><1o~1=*)(11)1"* =2.7x10-"cm. Comparing thisresult with (8.24), weseethatourassumptions thatthe energy difference between B11andC11iselectrostatic isfairly good; thedis- crepancy isonlyabout 15%(notbadforourﬁrstnuclear computationl). Thereason forthediscrepancy isprobably thefollowing. According tothe current understanding ofnuclei, anevennumber ofnuclear particles—in thecase ofB11,ﬁveneutrons together withﬁveprotons—makes akindofcore; when one more particle isadded tothiscore, itrevolves around ontheoutside tomake anew spherical nucleus, rather thanbeing absorbed. Ifthisisso,weshould havetaken adifferent electrostatic energy fortheadditional proton. Weshould have taken theexcess energy ofC11overB11tobejust Z184: 417600 ’ which istheenergy needed toaddonemore proton totheoutside ofthecore. Thisnumber isjust5/6ofwhat Eq.(8.23) predicts, sothenewprediction forthe radius is5/6of(8.24), which isinmuch closer agreement withwhat isdirectly measured. Wecandraw twoconclusions from thisagreement. Oneisthattheelectrical lawsappear tobeworking atdimensions assmall asl0"13 cm.Theother isthat wehaveveriﬁed theremarkable coincidence thatthenonelectrical partoftheforces between proton andproton, neutron andneutron, andproton andneutron are allequal. 8-5Energy intheelectrostatic ﬁeld Wenowconsider other methods ofcalculating electrostatic energy. They can allbederived from thebasic relation Eq.(8.3), thesum, overallpairs ofcharges, ofthemutual energies ofeachcharge-pair. First wewishtowrite anexpression fortheenergy ofacharge distribution. Asusual, weconsider thateachvolume element dVcontains theelement ofcharge pdV.Then Eq.(8.3)should bewritten _1p(1)p(2) U_5[1-4?’; dV,dV2. (8.27) space 8-9 Notice thefactor 1},which isintroduced because inthedouble integral overdV1 anddV2wehavecounted allpairsofcharge elements twice. (There isnoconvenient wayofwriting anintegral thatkeeps track ofthepairs sothateachpairiscounted onlyonce.) Next wenotice thattheintegral overdV2in(8.27) isjustthepotential at(1).That is, / p(2) =4mm dV2 ¢(l). sothat(8.27) canbewritten as 1U=5/P(1)¢(1)dVr Or,since thepoint (2)nolonger appears, wecansimply write U=-1-Ip¢av. (8.28) Thisequation canbeinterpreted asfollows. Thepotential energy ofthecharge pdVistheproduct ofthischarge andthepotential atthesame point. Thetotal energy istherefore theintegral over¢pdV.Butthere isagain thefactor 1}.Itis stillrequired because wearecounting energies twice. Themutual energies oftwo charges isthecharge ofonetimes thepotential atitduetotheother. Or,itcanbe taken asthesecond charge times thepotential atitfrom theﬁrst. Thus fortwo point charges wewould write _ __ 42U-q1¢(1) —111ZR; or U=l12¢(2) =(12 Notice thatwecould alsowrite U=‘Hq1¢(1) +q2¢(2)l- (8-29) Theintegral in(8.28) corresponds tothesumofboth terms inthebrackets of (8.29). That iswhyweneedthefactor §. Aninteresting question is:Where istheelectrostatic energy located? One might alsoask:Who cares? What isthemeaning ofsuchaquestion? Ifthere is apairofinteracting charges, thecombination hasacertain energy. Doweneed tosaythattheenergy islocated atoneofthecharges ortheother, oratboth, orin between? These questions maynotmake sense because wereally know onlythat thetotal energy isconserved. Theideathattheenergy islocated somewhere is notnecessary. Yetsuppose thatitdidmake sense tosay,ingeneral, thatenergy islocated atacertain place, asitdoesforheatenergy. Wemight thenextend ourprinciple oftheconservation ofenergy withtheideathatiftheenergy inagiven volume changes, weshould beabletoaccount forthechange bytheﬂowofenergy into oroutofthatvolume. Yourealize thatourearly statement oftheprinciple ofthe conservation ofenergy isstillperfectly allright ifsome energy disappears atone place andappears somewhere elsefaraway without anything passing (thatis,with- outanyspecial phenomena occurring) inthespace between. Weare,therefore, nowdiscussing anextension oftheideaoftheconservation ofenergy. Wemight callitaprinciple ofthelocalconservation ofenergy. Such aprinciple would say thattheenergy inanygiven volume changes onlybytheamount thatﬂows intoor outofthevolume. Itisindeed possible thatenergy isconserved locally insucha way. Ifitis,wewould haveamuch more detailed lawthanthesimple statement oftheconservation oftotal energy. Itdoes turnoutthatinnature energy is conserved locally. Wecanﬁndformulas forwhere theenergy islocated andhowit travels from place toplace. There isalsoaphysical reason whyitisimperative thatwebeabletosay where energy islocated. According tothetheory ofgravitation, allmass isasource 8-10 ofgravitational attraction. Wealsoknow, byE=mc2,thatmass andenergy are equivalent. Allenergy is,therefore, asource ofgravitational force. Ifwecould not locate theenergy, wecould notlocate allthemass. Wewould notbeabletosay where thesources ofthegravitational ﬁeldarelocated. Thetheory ofgravitation would beincomplete. Ifwerestrict ourselves toelectrostatics there isreally nowaytotellwhere the energy islocated. Thecomplete Maxwell equations ofelectrodynamics giveus much more information (although eventhentheanswer is,strictly speaking, not unique.) Wewilltherefore discuss thisquestion indetail again inalaterchapter. Wewillgiveyounow only theresult fortheparticular caseofelectrostatics. Theenergy islocated inspace, where theelectric ﬁeldis.Thisseems reasonable because weknow thatwhen charges areaccelerated theyradiate electric ﬁelds. Wewould liketosaythatwhen light orradiowaves travel from onepoint toanother, they carry their energy with them. Butthere arenocharges inthewaves. Sowe would liketolocate theenergy where theelectromagnetic ﬁeldisandnotatthe charges from which itcame. Wethusdescribe theenergy, notinterms ofthe charges, butinterms oftheﬁelds theyproduce. Wecan,infact,show thatEq. (8.28) isnumerically equal to U=%1/E-EdV. (8.30) Wecantheninterpret thisformula assaying thatwhen anelectric ﬁeldispresent, there islocated inspace anenergy whose density (energy perunitvolume) is 2 8= =-if- (8.31) Thisideaisillustrated inFig.8-8. Toshow thatEq.(8.30) isconsistent withourlawsofelectrostatics, webegin byintroducing intoEq.(8.28) therelation between pand¢thatweobtained in Chapter 6: p=—e0V2¢. Weget u=-%I8v28dV. (8.32) Writing outthecomponents oftheintegrand, weseethat ,_ 8% 82¢ 82¢¢V¢—¢<5-J-C-5+5-}7§+E5> _888_882888)_<88)’ 8<88>_<8¢)* _6x(¢(ix) (6x) +6y<4’6y 6y +6z¢62 62 =V'(¢W5)"(V05) '(V¢)- (3-33) Ourenergy integral isthen U=%4/(vs)-(v¢)dV -%/Iv-(¢ V¢)dV. WecanuseGauss’ theorem tochange thesecond integral intoasurface integral: Iv-(8v¢)av=I(8v¢)-II<18. (8.34) vol. surface Weevaluate thesurface integral inthecasethatthesurface goestoinﬁnity (sothevolume integrals become integrals overallspace), supposing thatallthe charges arelocated within some ﬁnite distance. Thesimple waytoproceed isto takeaspherical surface ofenormous radius Rwhose center isattheorigin of coordinates. Weknow thatwhen weareveryfaraway from allcharges, ¢varies asl/RandV¢asl/R2. (Both willdecrease even faster withRifthere thenet ' 8-11/T ’ /%//% dxdydzinanelectric ﬁeld contains the energy (co/2)E2 dV. charge inthedistribution iszero.) Since thesurface areaofthelarge sphere in- creases asR2,weseethatthesurface integral fallsoffas(1/R)(l /R2)R2 =(l/R) astheradius ofthesphere increases. Soifweinclude allspace inourintegration (R->co),thesurface integral goestozeroandwehavethat U=%I(v¢)-(v¢)dV =%/E-EdV. (8.35) all all 8138.00 SPEOO Weseethatitispossible forustorepresent theenergy ofanycharge distribution asbeing theintegral overanenergy density located intheﬁeld. 8-6Theenergy ofapoint charge Ournewrelation, Eq.(8.35), saysthatevenasingle point charge qwillhave some electrostatic energy. Inthiscase, theelectric ﬁeldisgiven by =__‘1__ .E 41l'€()!'2 Sotheenergy density atthedistance rfrom thecharge is GOE2 _ q2 l 2—321r2e0r4 Wecantakeforanelement ofvolume aspherical shellofthickness drandarea 41rr2. Thetotalenergy is -_ 0° q2 -_ ‘ q2 1 9'=o0 ‘ U_f81re0r2 dr_ 81re0 r,=0 (8.36)1‘=0 Now thelimit atr=sogives nodifficulty. Butforapoint charge weare supposed tointegrate down tor=0,which gives aninﬁnite integral. Equation (8.35) saysthatthere isaninﬁnite amount ofenergy intheﬁeldofapoint charge, although webegan withtheideathatthere wasenergy onlybetween point charges. Inouroriginal energy formula foracollection ofpoint charges (Eq. 8.3), wedid notinclude anyinteraction energy ofacharge withitself. What hashappened is thatwhen wewent overtoacontinuous distribution ofcharge inEq.(8.27), we counted theenergy ofinteraction ofevery inﬁnitesimal charge with allother inﬁnitesimal charges. Thesame account isincluded inEq.(8.35), sowhen we apply ittoaﬁnite point charge, weareincluding theenergy itwould taketo assemble thatcharge from inﬁnitesimal parts. Youwillnotice, infact,thatwe would alsogettheresult inEq.(8.36) ifweusedourexpression (8.11) fortheenergy ofacharged sphere andlettheradius tendtoward zero. Wemust conclude thattheideaoflocating theenergy intheﬁeldisincon- sistent withtheassumption oftheexistence ofpoint charges. Onewayoutofthe difficulty would betosaythatelementary charges, such asanelectron, arenot points butarereally small distributions ofcharge. Alternatively, wecould say thatthere issomething wrong inourtheory ofelectricity atverysmall distances, orwiththeideaofthelocal conservation ofenergy. There aredifficulties with either point ofview. These difﬁculties havenever beenovercome; theyexisttothis day. Sometime later, when wehave discussed some additional ideas, suchasthe momentum inanelectromagnetic ﬁeld, wewillgiveamore complete account of these fundamental difﬁculties inourunderstanding ofnature. 8-12 9 Electricity inthe Atmosphere 9-1Theelectric potential gradient oftheamosphere Onanordinary dayoverﬂatdesert country, oroverthesea,asonegoesup- ward from thesurface oftheground theelectric potential increases byabout 100 voltspermeter. Thus there isavertical electric ﬁeldEof100volts/m intheair.The signoftheﬁeldcorresponds toanegative charge ontheearth’s surface. This means thatoutdoors thepotential attheheight ofyour noseis200volts higher thanthepotential atyourfeet! Youmight ask:“Why don’t wejuststickapairof electrodes outintheaironemeter apart andusethe100voltstopower ourelectric lights?” Oryoumight wonder: “Ifthere isreally apotential difference of200 voltsbetween mynoseandmyfeet,whyisitIdon’t getashock when Igooutinto thestreet?” Wewillanswer thesecond question ﬁrst. Your body isarelatively good conductor. Ifyouareincontact withtheground, youandtheground willtendto make oneequipotential surface. Ordinarily, theequipotentials areparallel tothe surface, asshown inFig.9-1(a), butwhen youarethere, theequipotentials are distorted, andtheﬁeldlooks somewhat asshown inFig.9—1(b). Soyoustillhave verynearly zeropotential difference between your head andyour feet. There are charges thatcome from theearth toyourhead, changing theﬁeld. Some ofthem maybedischarged byionscollected from theair,butthecurrent ofthese isvery small because airisapoor conductor.9-1Theelectric potential gradient oftheatmosphere 9-2Electric currents inthe atmosphere 9-3Origin oftheatmospheric currents 9-4Thunderstorms 9-5Themechanism ofcharge separation 9-6Lightning Reference: Chalmers, J.Alan, Atmos- pheric Electricity, Pergamon Press, London (1957). §"__-L‘-Z A +3OOV / \\—J s +3901 ____________ \’~‘8\‘i\ \+2oov _ _ _ // -‘#1-_§Z i \ °O‘1/ _ " is=IOOV/I'll /“’/‘ / _ _;i-IOOV _ , _ _ _i L§'_ W Z Z \ Z\ \ XX \ \\ \X \ 9_--- —- —-- ———.~ —-----//////‘//////////‘ '////// ’///////T GROUND GROUND (<1) . lb) Fig. 9-1. (a)Thepotential distribution above theearth. (b)Thepotential distribution near amarﬁﬁinfopen ﬂatplace. How canwemeasure suchaﬁeldiftheﬁeldischanged byputting something there? There areseveral ways. Onewayistoplace aninsulated conductor atsome distance above theground andleave itthere tmtilitisatthesame potential asthe air.Ifweleave itlongenough, theverysmall conductivity intheairwillletthe charges leakoff(oronto) theconductor untilitcomes tothepotential atitslevel. Then wecanbring itbacktotheground, andmeasure theshiftofitspotential as wedoso.Afaster wayistolettheconductor beabucket ofwater withasmall leak. Asthewater drops out,itcarries away anyexcess charges andthebucket willapproach thesame potential astheair.(Thecharges, asyouknow, reside on thesurface, andasthedrops come off“pieces ofsurface” break off.)Wecanmeas- urethepotential ofthebucket withanelectrometer. 9-l ll*lCONNECTION TOGROUND KTAL PLATE ////:}-K/sR/ouiso//// /// /to) lgJ1r77P14/1=/..;....;.' ///7/7?lb) Fig.9-2. (a)Agrounded metal plate willhave thesame surface charge asthe earth. (b)Iftheplate iscovered witha grounded conductor itwill have no surface charge. ll-AIR + ;-- IONS - EV *'*- aousrsn Z MI"MO-l Fig.9-3. Measuring theconductivity ofairduetothemotion ofions.There isanother waytodirectly measure thepotential gradient. Since there isanelectric ﬁeld, there isasurface charge ontheearth (a'=eoE). Ifweplace aﬂatmetal plate attheearth’s surface andground it,negative charges appear on it(Fig.9—2a). Ifthisplateisnowcovered byanother grounded conducting cover B, thecharges willappear onthecover, andthere willbenocharges ontheoriginal plate A.Ifwemeasure thecharge thatﬂows from plate Atotheground (by,say, agalvanometer inthegrounding wire) aswecover it,wecanﬁndthesurface charge density thatwasthere, andtherefore alsoﬁndtheelectric ﬁeld. Having suggested howwecanmeasure theelectric ﬁeldintheatmosphere, wenowcontinue ourdescription ofit.Measurements show, ﬁrstofall,thatthe ﬁeldcontinues toexist, butgetsweaker, asonegoesuptohighaltitudes. Byabout 50kilometers, theﬁeldisverysmall, somost ofthepotential change (theintegral ofE)isatlower altitudes. Thetotalpotential difference from thesurface ofthe earth tothetopoftheatmosphere isabout 400,000 volts. 9-2Electric currents intheatmosphere Another thing thatcanbemeasured, inaddition tothepotential gradient, is thecurrent intheatmosphere. Thecurrent density issmall—about 10micromicro- amperes crosses eachsquare meter parallel totheearth. Theairisevidently nota perfect insulator, andbecause ofthisconductivity, asmall current—caused bythe electric ﬁeldwehavejustbeendescribing——passes from theskydown totheearth. Why doestheatmosphere haveconductivity? Here andthere among theair molecules there isanion—a molecule ofoxygen, say,which hasacquired an extra electron, orperhaps lostone. These ionsdonotstayassingle molecules; because oftheirelectric ﬁeldtheyusually accumulate afewother molecules around them. Each ionthenbecomes alittlelump which, along withother lumps, drifts intheﬁe1d—moving slowly upward ordownward-—-making theobserved current. Where dotheionscome from? Itwasﬁrstguessed thattheionswereproduced by theradioactivity oftheearth. (Itwasknown thattheradiation from radioactive materials would make airconducting byionizing theairmolecules.) Particles likeB-rays coming outoftheatomic nuclei aremoving sofastthattheytearelec- trons from theatoms, leaving ionsbehind. Thiswould imply, ofcourse, thatif weweretogotohigher altitudes, weshould ﬁndlessionization, because theradio- activity isallinthedirtontheground—in thetraces ofradium, uranium, po- tassium, etc. Totestthistheory, some physicists carried anexperiment upinballoons to measure theionization oftheair(Hess, in1912) anddiscovered thattheopposite wastrue—the ionization perunitvolume increased withaltitude! (Theapparatus waslikethatofFig.9-3.Thetwoplates werecharged periodically tothepotential V.Duetotheconductivity oftheair,theplates slowly discharged; therateof discharge wasmeasured with theelectrometer.) This wasamost mysterious result—the most dramatic ﬁnding intheentire history ofatmospheric electricity. Itwassodramatic, infact,thatitrequired abranching offofanentirely new subject——cosmic rays. Atmospheric electricity itself remained lessdramatic. Ionization wasevidently being produced bysomething from outside theearth; theinvestigation ofthissource ledtothediscovery ofthecosmic rays. Wewill notdiscuss thesubject ofcosmic raysnow, except tosaythattheymaintain the supply ofions. Although theionsarebeing swept away allthetime, newonesare being created bythecosmic-ray particles coming from theoutside. Tobeprecise, wemust saythatbesides theionsmade ofmolecules, there are alsoother kinds ofions. Tinypieces ofdirt,likeextremely ﬁnebitsofdust, ﬂoat intheairandbecome charged. They aresometimes called “nuclei.” Forexample, when awave breaks inthesea,littlebitsofspray arethrown intotheair.When oneofthese drops evaporates, itleaves aninﬁnitesimal crystal ofNaCl ﬂoating in theair.These tinycrystals canthen pick upcharges andbecome ions; they arecalled “large ions.” Thesmall ions—those formed bycosmic rays—-are themost mobile. Because theyaresosmall, theymove rapidly through theair—with aspeed ofabout l 9-2 cm/sec inaﬁeldof100volts/meter, or1volt/cm. Themuch bigger andheavier ionsmove much more slowly. Itturns outthatifthere aremany “nuclei,” theywill pickupthecharges from thesmall ions. Then, since the“large ions” move so slowly inaﬁeld, thetotalconductivity isreduced. Theconductivity ofair,there- fore,isquite variable, sinceitisverysensitive totheamount of“dirt” there isinit. There ismuch more ofsuchdirtoverland—where thewinds canblow updust orwhere manthrows allkinds ofpollution intotheair——than there isoverwater. Itisnotsurprising thatfrom daytoday,from moment tomoment, from place toplace, theconductivity neartheearth’s surface varies enormously. Thevoltage gradient observed atanyparticular place ontheearth’s surface alsovaries greatly because roughly thesame current ﬂows down fromhighaltitudes indifferent places, andthevarying conductivity neartheearth results inavarying voltage gradient. Theconductivity oftheairduetothedrifting ofionsalsoincreases rapidly withaltitude——for tworeasons. First ofall,theionization from cosmic raysin- creases withaltitude. Secondly, asthedensity ofairgoesdown, themean freepath oftheionsincreases, sothattheycantravel farther intheelectric ﬁeldbefore they haveacollision—resulting inarapid increase ofconductivity asonegoesup. Although theelectric current-density intheairisonly afewmicromicro- amperes persquare meter, there areverymany square meters ontheearth’s surface. Thetotal electric current reaching theearth’s surface atanytimeisverynearly constant at1800amperes. Thiscurrent, ofcourse, is“positive”—it carries plus charges totheearth. Sowehaveavoltage supply of400,000 volts withacurrent of1800amperes-—a power of700megawatts! With such alarge current coming down, thenegative charge ontheearth should soon bedischarged. Infact,itshould takeonlyabout halfanhour todis- charge theentire earth. Buttheatmospheric electric ﬁeldhasalready lasted more thanahalf-hour since itsdiscovery. How isitmaintained? What maintains the voltage? Andbetween what andtheearth? There aremany questions. Theearth isnegative, andthepotential intheairispositive. Ifyougohigh enough, theconductivity issogreat thathorizontally there isnomore chance for voltage variations. Theair,forthescale oftimes thatwearetalking about, be- comes effectively aconductor. Thisoccurs ataheight intheneighborhood of50 kilometers. Thisisnotashighaswhat iscalled the“ionosphere,” inwhich there areverylarge numbers ofionsproduced byphotoelectricity from thesun. Never- theless, forourdiscussions ofatmospheric electricity, theairbecomes suﬂiciently conductive atabout 50kilometers thatwecanimagine thatthere ispractically a perfect conducting surface atthisheight, from which thecurrents come down. Ourpicture ofthesituation isshown inFig.9—4. Theproblem is:I-Iow isthe positive charge maintained there? How isitpumped back? Because ifitcomes down totheearth, ithastobepumped back somehow. That wasoneofthe greatest puzzles ofatmospheric electricity forquite awhile. Each piece ofinformation wecangetshould giveaclueor,atleast, tellyou something about it.Hereisaninteresting phenomenon: Ifwemeasure thecurrent (which ismore stable thanthepotential gradient) overthesea,forinstance, orin careful conditions, andaverage verycarefully sothatwegetridoftheirregularities, wediscover thatthere isstilladaily variation. Theaverage ofmany measurements overtheoceans hasavariation withtimeroughly asshown inFig.9-5. The current varies byabout :15percent, anditislargest at7:00P.M.inLondon. The strange partofthething isthatnomatter where youmeasure thecurrent—in the Atlantic Ocean, thePaciﬁc Ocean, ortheArctic Ocean—it isatitspeak value when theclocks inLondon say7:00P.M.! Allovertheworld thecurrent isatits maximum at7:00P.M.London timeanditisataminimum at4:00A.M.London time. Inother words, itdepends upon theabsolute timeontheearth, notupon thelocal timeattheplace ofobservation. Inonerespect thisisnotmysterious; itchecks withourideathatthere isaveryhighconductivity laterally atthetop, because thatmakes itimpossible forthevoltage difference from theground to thetoptovarylocally. Anypotential variations should beworldwide, asindeed theyare.What wenowknow, therefore, isthatthevoltage atthe“top” surface isdropping andrising by15percent withtheabsolute timeontheearth. 9-3men+ coi4_oug_|v|rv50,000m—— ——-—— ——-—— cunnsur 00° aslO.|2I votrs Ampll in‘ 5“LEv Eanrws sunncz Fig.9-4. Typical electrical condi- tions inaclear atmosphere. E(V/m) I II I so J t it is 2'4FOURS GMT Fig.9-5. Theaverage daily varia- tionoftheatmospheric potential gradient onaclear dayover theoceans; referred toGreenwich time. 9-3Origin oftheatmospheric currents Wemust next talk about thesource ofthelarge negative currents which must beﬂowing from the“top” tothesurface oftheearth tokeep charging itup negatively. Where arethebatteries thatdothis? The“battery” isshown inFig. 9-6. Itisthethunderstorm anditslightning. Itturns outthatthebolts oflightning donot“discharge” thepotential wehave been talking about (asyoumight at ﬁrstguess). Lightning storms carry negative charges totheearth. When alightning boltstrikes, ten-to—one itbrings down negative charges totheearth inlarge amounts. ltisthethunderstorms throughout theworld thatarecharging theearth with an average of1800 amperes, which isthen being discharged through regions of fairweather. There areabout 300thunderstorms perdayallover theearth, andwecan think ofthem asbatteries pumping theelectricity totheupper layer andmaintain- ingthevoltage difference. Then take intoaccount thegeography oftheearth»- there arethunderstorms intheafternoon inBrazil. tropical thunderstorms in Africa, andsoforth. People have made estimates ofhowmuch lightning isstriking world-wide atanytime. andperhaps needless tosay,their estimates more orless agree with thevoltage difference measurements: thetotal amount ofthunderstorm activity ishighest onthewhole earth atabout 7:00 P.M. inLondon. However, thethunderstorm estimates arevery difﬁcult tomake and were made only after itwas known that thevariation should have occurred. These things arevery difﬁcult because wedon’t have enough observations ontheseasandover allparts oftheworld toknow thenumber ofthunderstorms accurately. Butthose people who think they “doitright” obtain theresult thatthere isapeak intheactivity at7:00 P.M.Greenwich Mean Time. Fig 9—6 Themechanism that generates theatmospheric electric ﬁeld. [Photo byWilliam L.Widmayer.] 9-4 Inorder tounderstand howthese batteries work, wewilllookatathunder- storm indetail. What isgoing oninside athunderstorm? Wewilldescribe this insofar asitisknown. Aswegetintothismarvelous phenomenon ofrealnature—- instead oftheidealized spheres ofperfect conductors inside ofother spheres that wecansolve soneatly—we discover thatwedon’t know verymuch. Yetitisreally quite exciting. Anyone whohasbeeninathunderstorm hasenjoyed it,orhasbeen frightened, oratleasthashadsome emotion. Andinthose places innature where wegetanemotion, weﬁndthatthere isgenerally acorresponding complexity and mystery about it.Itisnotgoing tobepossible todescribe exactly howathunder- storm works, because wedonotyetknow verymuch. Butwewilltrytodescribe alittlebitabout what happens. 9-4Thunderstorms Intheﬁrstplace, anordinary thunderstorm ismade upofanumber of“cells” fairly close together, butalmost independent ofeachother. Soitisbesttoanalyze onecellatatime. Bya“cell” wemean aregion withalimitareainthehorizontal direction inwhich allofthebasic processes occur. Usually there areseveral cells sidebyside,andineachoneabout thesame thing ishappening, although perhaps withadifferent timing. Figure 9-7indicates inanidealized fashion what sucha celllooks likeintheearly stage ofthethunderstorm. Itturns outthatinacertain place intheair,under certain conditions which weshalldescribe, there isageneral rising oftheair,with higher andhigher velocities near thetop. Asthewarm, moist airatthebottom rises, itcools andcondenses. Intheﬁgure thelittlecrosses indicate snow andthedotsindicate rain,butbecause theupdraft currents aregreat enough andthedrops aresmall enough, thesnow andraindonotcome down at thisstage. This isthebeginning stage, andnottherealthunderstorm yet—in the sense thatwedon’t haveanything happening attheground. Atthesame timethat thewarm airrises, there isanentrainment ofairfrom thesides—an important point which wasneglected formany years. Thus itisnotjusttheairfrom below which isrising, butalsoacertain amount ofother airfrom thesides. Whydoestheairriselikethis? Asyouknow, when yougoupinaltitude the airiscolder. Theground isheated bythesun,andthere-radiation ofheattothe skycomes from water vapor highintheatmosphere; soathighaltitudes theair iscold——very cold—whereas lower down itiswarm. Youmaysay,“Then it’s verysimple. Warm airislighter thancold; therefore thecombination ismechan- ically unstable andthewarm airrises.” Ofcourse, ifthetemperature isdiﬂerent atdiﬂerent heights, theairisunstable thermodynamically. Lefttoitself inﬁnitely long, theairwould allcome tothesame temperature. Butitisnotlefttoitself; thesunisalways shining (during theday). Sotheproblem isindeed notoneof thermodynamic equilibrium, butofmechanical equilibrium. Suppose weplot—as inFig.9—8—the temperature oftheairagainst height above theground. In ordinary circumstances wewould getadecrease along acurve liketheonelabeled (a);astheheight goesup,thetemperature goesdown. How cantheatmosphere bestable? Why doesn’t thehotairbelow simply riseupintothecoldair? The answer isthis: iftheairwere togoup,itspressure would godown, andifwe consider aparticular parcel ofairgoing up,itwould beexpanding adiabatically. (There would benoheatcoming inoroutbecause inthelarge dimensions con- sidered here, there isn’ttimeformuch heatﬂow.) Thus theparcel ofairwould coolasitrises. Suchanadiabatic process would giveatemperature-height relation- shiplikecurve (b)inFig.9-8. Anyairwhich rosefrom below would becolder thantheenvironment itgoesinto. Thus there isnoreason forthehotairbelow torise;ifitwere torise,itwould cooltoalower temperature thantheairalready there, would beheavier thantheairthere, andwould justwanttocome down again. Onagood, bright daywithverylittlehumidity there isacertain rateatwhich the temperature intheatmosphere falls, andthisrateis,ingeneral, lower than the “maximum stable gradient,” which isrepresented bycurve (b). Theairisin stable mechanical equilibrium. 9-5FEET I I -...,/i:'ttt‘t*‘rri.. . -.;*;~;r;;-rm,.-'-.-~tf""._'._T:“"“_~-L or.ZSHX) E1100 _l5.lXD ,,¢rf3iiiil\-\\\-._ ._-;—' """"" __'""“~1.-_.__.; can. . . _,// 11II1t\\u\\\\.. -------- "_"‘""~--_-_ nu ‘\\\“é\\\“i\\\¥-\\‘<"-'l§\€\\§§\1\\\3\\5\\§\\\=’a Horizontal Scale ?_._.ﬁ_.imi .Rain_l.Q.QZ] _ﬁ.QQ0 _Siir.lm: Draft Vector ScaIe?_'i:‘_3f,tt/lei: -Snow Fig.9-7. Athunderstorm cellinthe early stages ofdevelopment. [From U.S. Department ofCommerce Weather Bureau Report, June 1949.] A TEMPERATURE/ / / /O /Q.O\ \ \ b\ O \D ALTITUDE Fig.9-8. Atmospheric temperature. (a)Static atmosphere; lb)adiabatic cooling ofdryair;(cladiabatic cooling ofwetair;(d)wetairwith some mixing ofambient air. FEE‘ QED n -1|: f‘ _ _ _ _ _e-s-> /____._- "-.:~><\>\"\.\\ We/1'-.~. .".;;<<;\*"\;‘R‘;k‘?l.*j/‘7}*2{;i§,:;. ."“ weNW/J, ,"“mm 4.-1' ._ .-__._. __.___. - a1%f;E.»-_I.;.;;.nw:.511-r ....»1.11?/~,.;;-,;Y;>;,~; -~ _____ __.____:\. .....I.J; .?-III/I <_“____.______ '...'____ ''' .tt\\\t\.\%\\\\\ n'/1I"b"Hcnzelldiwb L_4._Jm1 'Snuu/\A/—P-_-L-¢\ DwWhc|wScun ?_'.’_§°m»= —lccOvyI\d| Fig. 9-9. Amature thunderstorm cell. [From U.S. Department ofCommerce Weather Bureau Report, June 1949.]Ontheother hand, ifwethink ofaparcel ofairthatcontains alotofwater vapor being carried upintotheair,itsadiabatic cooling curve willbedifferent. As itexpands andcools, thewater vapor initwillcondense, andthecondensing water willliberate heat. Moist air,therefore, doesnotcoolnearly asmuch asdryair does. Soifairthatiswetter thantheaverage starts torise,itstemperature will follow acurve like(c)inFig.9-8. Itwillcooloffsomewhat, butwillstillbewarmer thanthesurrounding airatthesame level. Ifwehave aregion ofwarm moist airandsomething starts itrising, itwillalways ﬁnditself lighter andwarmer than theairaround itandwillcontinue toriseuntilitgetstoenormous heights. This isthemachinery thatmakes theairinthethunderstorm cellrise. Formany years thethunderstorm cellwasexplained simply inthismanner. Butthen measurements showed thatthetemperature ofthecloud atdifferent heights wasnotnearly ashighasindicated bycurve (c).Thereason isthatasthe moist air“bubble” goesup,itentrains airfrom theenvironment andiscooled oﬁ'byit.Thetemperature-versus-height curve looks more likecurve (d),which ismuch closer totheoriginal curve (a)thantocurve (c). After theconvection justdescribed getsunder way, thecross section ofa thunderstorm celllooks likeFig.9—9. Wehave what iscalled a“mature” thunder- storm. There isaveryrapid updraft which, inthisstage, goesuptoabout 10,000 to15,000 meters—sometimes even much higher. Thethunderheads, withtheir condensation, climb wayupoutofthegeneral cloud bank, carried byanupdraft thatisusually about 60miles anhour. Asthewater vapor iscarried upand condenses, itforms tinydrops which arerapidly cooled totemperatures below zerodegrees. They should freeze, butdonotfreeze immediately—they are“super- cooled.” Water andother liquids willusually cool wellbelow their freezing points before crystallizing ifthere areno“nuclei” present tostart thecrystallization process. Only ifthere issome small piece ofmaterial present, likeatinycrystal of NaCl, willthewater drop freeze intoalittlepiece ofice.Then theequilibrium is suchthatthewater drops evaporate andtheicecrystals grow. Thus atacertain point there isarapid disappearance ofthewater andarapid buildup ofice.Also, there maybedirect collisions between thewater drops andtheice—col1isions in which thesupercooled water becomes attached totheicecrystals, which causes it tosuddenly crystallize. Soatacertain point inthecloud expansion there isarapid accumulation oflarge iceparticles. When theiceparticles areheavy enough, theybegin tofallthrough therising air—they gettooheavy tobesupported anylonger intheupdraft. Astheycome down, theydraw alittle airwiththem andstartadowndraft. Andsurprisingly enough, itiseasytoseethatoncethedowndraft isstarted, itwillmaintain itself. Theairnowdrives itself down! Notice thatthecurve (d)inFig.9-8fortheactual distribution oftemperature inthecloud isnotassteep ascurve (c),which applies towetair.Soifwehavewet airfalling, itstemperature willdrop withtheslope ofcurve (c)andwillgobelow thetemperature oftheenvironment ifitgetsdown farenough, asindicated by curve (e)intheﬁgure. Themoment itdoesthat,itisdenser thantheenvironment andcontinues tofallrapidly. Yousay,“That isperpetual motion. First, youargue thattheairshould rise,andwhen youhaveitupthere, youargue equally wellthat theairshould fall.” Butitisn’tperpetual motion. When thesituation isunstable andthewarm airshould rise,thenclearly something hastoreplace thewarm air. Itisequally truethatcoldaircoming down would energetically replace thewarm air,butyourealize thatwhat iscoming down isnottheoriginal air.Theearly arguments, thathadaparticular cloud without entrainment going upandthen coming down, hadsome kindofapuzzle. They needed theraintomaintain the downdraft——an argument which ishardtobelieve. Assoonasyourealize thatthere isalotoforiginal airmixed inwiththerising air,thethermodynamic argument shows thatthere canbeadescent ofthecoldairwhich wasoriginally atsome great height. Thisexplains thepicture oftheactive thunderstorm sketched inFig.9-9. Astheaircomes down, rainbegins tocome outofthebottom ofthethunder- storm. Inaddition, therelatively coldairspreads outwhen itarrives attheearth’s surface. Sojustbefore theraincomes there isacertain littlecoldwind thatgives 9-6 usaforewarning ofthecoming storm. Inthestorm itself there arerapid andir- regular gusts ofair,there isanenormous turbulence inthecloud, andsoon.But basically wehave anupdraft, thenadowndraft—in general, averycomplicated process. Themoment atwhich precipitation starts isthesame moment thatthelarge downdraft begins andisthesame moment, infact,when theelectrical phenomena arise. Before wedescribe lightning, however, wecanﬁnish thestory bylooking atwhat happens tothethunderstorm cellafterabout one-half anhourtoanhour. Thecelllooks asshown inFig.9-10. Theupdraft stops because there isnolonger enough warm airtomaintain it.Thedownward precipitation continues forawhile, thelastlittlebitsofwater come out,andthings getquieter andquieter—although there aresmall icecrystals leftwayupintheair.Because thewinds atverygreat altitude areindifferent directions, thetopofthecloud usually spreads intoan anvil shape. Thecellcomes totheendofitslife. A—‘; *;’I>,T-_ nu _ —,f’ I .- - /7'7’ --. - - - - - - - .\ \0%\\\‘\\\ll ﬂy . - » 9 - _ .. .. _ uurts IIInustum /.. .. - - .. .. ..an mnnunuo rutrenswao _ an ' '- ' - -'11 + ' ' Posmv:~ - - - ems: ezurzn + mm ------- --.---.---.----.----.~ ii“‘IL + + + .. - - .. - 'l' + 4' + 4- 4- ..:...'_\(.1fvr/r~.-.- . - - - .. .min; __ ___ ___ ‘i_‘___ _____i.gg IIMYIVE _""“‘“"‘ ‘ _ enncsezurzn -.~.\\\\\n|::4/ ... -+" + _ _,, +-+ + + § +-|0C +_ onecnm orinmost OG ~.<.\\, \\ 'J_,,‘,_. - _ .W _________ '1 M -_‘___\\.\\\.H_/.1.//V/1:,’ _ -- :mn_._ _.,'____ ;' '_ _,_ ______um\. _....,t- -14‘Ir/ryﬁrI'1|'"“\\\ 1\~..\\..\_‘. L.._._. “U or ,__mm __ ________________________ ___. 01 _. _ -++ _ _ _ *- _4 Fill WIAYOII FUTEIIIIOL QDOQT-0-+ +_' +++I /1/I'11 \uA‘\\\\\ umnv: mu‘ //I/kl""'lL' ‘fl roiufliieaeunemn 2 " /.a-as.» -- \ _'l '++1! | M“ 'a\\\\\—\\€%x=&\{§\wé\<4§\%l$e§\¥i\§a‘*&émQ\\.§_\\&\\a\\e-‘*"au‘ / /1/IIn“ ffl\\ AREAornuvv um lﬂ ' ' , ' II L,‘ ‘ __ _ _ :,'_,‘:",;:,'*%',.;_ ;§;-,,,,__ |:—/37/JWL %_'%/1 /r“7)L /_-W/-*7’=»<=ﬂ=//7] Fig.9-l0. Thelatephase ofathunderstorm Fig.9-1l.Thedistribution ofelectrical charges ina cell.[From U.S.Department ofCommerce Weather mature thunderstorm cell. [From U.S.Department ofCom Bureau Report, June l949.] merce Weather Bureau Report, June l949.] 9-5Themechanism ofcharge separation Wewant nowtodiscuss themost important aspect forourpurposes—the development oftheelectrical charges. Experiments ofvarious kinds—including ﬂying airplanes through thunderstorms (thepilots whodothisarebrave men!)- tellusthatthecharge distribution inathunderstorm cellissomething likethat shown inFig.9-11. Thetopofthethunderstorm hasapositive charge, andthe bottom anegative one—except forasmall local region ofpositive charge inthe bottom ofthecloud, which hascaused everybody alotofworry. Nooneseems to know whyitisthere, how important itis——whether itisasecondary effect ofthe positive rain coming down, orwhether itisanessential part ofthemachinery. Things would bemuch simpler ifitweren’t there. Anyway, thepredominantly negative charge atthebottom andthepositive charge atthetophave thecorrect signforthebattery needed todrive theearth negative. Thepositive charges are 6or7kilometers upintheair,where thetemperature isabout —20°C, whereas thenegative charges are3or4kilometers high, where thetemperature isbetween zeroand—l0°C. Thecharge atthebottom ofthecloud islarge enough toproduce potential differences of20,or30,oreven100million voltsbetween thecloud andtheearth- much bigger thanthe0.4million volts from the“sky” totheground inaclear 9-7 .§___§~,»-.<\\~-/ // kl /'71/ \~\ \\\ 1"!‘>1 x "' " x :1 TOWATER SUPPLY Fig. 9-l2. Aietofwater with an electric ﬁeld near thenozzle.atmosphere. These large voltages break down theairandcreate giant arcdis- charges. When thebreakdown occurs thenegative charges atthebottom ofthe thunderstorm arecarried down totheearth inthelightning strokes. Now wewilldescribe insome detail thecharacter ofthelightning. First of all,there arelarge voltage differences around, sothattheairbreaks down. There arelightning strokes between onepiece ofacloud andanother piece ofacloud, orbetween onecloud andanother cloud, orbetween acloud andtheearth. In eachoftheindependent discharge ﬁashes—the kindoflightning strokes yousee—— there areapproximately 20or30coulombs ofcharge brought down. Onequestion is:How longdoesittakeforthecloud toregenerate the20or30coulombs which aretaken away bythelightning bolt? This canbeseen bymeasuring, farfrom a cloud, theelectric ﬁeldproduced bythecloud’s dipole moment. Insuchmeasure- ments youseeasudden decrease intheﬁeldwhen thelightning strikes, andthen anexponential return totheprevious value with atime constant which isslightly different fordifferent cases butwhich isintheneighborhood of5seconds. Ittakes athunderstorm only5seconds after eachlightning stroke tobuild itscharge up again. That doesn’t necessarily mean thatanother stroke isgoing tooccur in exactly 5seconds every time, because, ofcourse, thegeometry ischanged, andsoon. Thestrokes occur more orlessirregularly, buttheimportant point isthatittakes about 5seconds torecreate theoriginal condition. Thus there areapproximately 4amperes ofcurrent inthegenerating machine ofthethunderstorm. This means thatanymodel made toexplain howthisstorm generates itselectricity must beone withplenty ofjuice——it must beabig,rapidly operating device. Before wegofurther weshall consider something which isalmost certainly completely irrelevant, butnevertheless interesting, because itdoesshow theeffect ofanelectric ﬁeldonwater drops. Wesaythatitmaybeirrelevant because it relates toanexperiment onecandointhelaboratory withastream ofwater to show therather strong effects oftheelectric ﬁeldondrops ofwater. Inathunder- storm there isnostream ofwater; there isacloud ofcondensing iceanddrops of water. Sothequestion ofthemechanisms atwork inathunderstorm isprobably notatallrelated towhat youcanseeinthesimple experiment wewilldescribe. Ifyoutakeasmall nozzle connected toawater faucet anddirect itupward ata steep angle, asinFig.9-12, thewater willcome outinaﬁnestream thateventually breaks upintoaspray ofﬁnedrops. Ifyounowputanelectric ﬁeldacross the stream atthenozzle (bybringing upacharged rod,forexample), theform ofthe stream willchange. With aweak electric ﬁeldyouwillﬁndthatthestream breaks upintoasmaller number oflarge-sized drops. Butifyouapply astronger ﬁeld, thestream breaks upintomany, many ﬁnedrops—smaller than before.* With a weak electric ﬁeldthere isatendency toinhibit thebreakup ofthestream into drops. With astronger ﬁeld, however, there isanincrease inthetendency tosepa- rateintodrops. Theexplanation ofthese effects isprobably thefollowing. Ifwehave the stream ofwater coming outofthenozzle andweputaSmall electric ﬁeldacross it onesideofthewater getsslightly positive andtheother sidegetsslightly negative. Then, when thestream breaks, thedrops ononesidemaybepositive, andthose on theother sidemaybenegative. They willattract eachother andwillhaveatend- ency tostick together more than they would have before——the stream doesn’t break upasmuch. Ontheother hand, iftheﬁeldisstronger, thecharge ineach oneofthedrops getsmuch larger, andthere isatendency forthecharge itself to helpbreak upthedrops through their ownrepulsion. Each drop willbreak into many smaller ones, each carrying acharge, sothatthey areallrepelled, and spread outsorapidly. Soasweincrease theﬁeld, thestream becomes more ﬁnely separated. Theonlypoint wewishtomake isthatincertain circumstances electric ﬁelds canhave considerable inﬂuence onthedrops. Theexact machinery by which something happens inathunderstorm isnotatallknown, andisnotatall necessarily related towhat wehavejustdescribed. Wehaveincluded itjustsothat *Ahandy waytoobserve thesizesofthedrops istoletthestream fallonalarge thin metal plate. Thelarger drops make alouder noise. 9-8 youwillappreciate thecomplexities thatcould come intoplay. Infact,nobody hasatheory applicable toclouds based onthatidea. Wewould liketodescribe twotheories which have beeninvented toaccount fortheseparation ofthecharges inathunderstorm. Allthetheories involve the ideathatthere should besome charge ontheprecipitation particles andadifferent charge intheair.Then bythemovement oftheprecipitation particles—the water ortheice--through theairthere isaseparation ofelectric charge. Theonlyques- tionis:How does thecharging ofthedrops begin? Oneoftheolder theories is called the“breaking-drop” theory. Somebody discovered thatifyouhaveadrop ofwater thatbreaks intotwopieces inawindstream, there ispositive charge onthe water andnegative charge intheair. This breaking-drop theory hasseveral disadvantages, among which themost serious isthatthesigniswrong. Second, inthelarge number oftemperate-zone thunderstorms which doexhibit lightning, theprecipitation effects athighaltitudes areinice,notinwater. From what wehavejustsaid,wenotethatifwecould imagine some wayfor thecharge tobedifferent atthetopandbottom ofadrop andifwecould alsosee some reason whydrops inahigh-speed airstream would break upintounequal pieces-—a large oneinthefront andasmaller oneinthebackbecause ofthemotion through theairorsomething-we would haveatheory. (Different from anyknown theory!) Then thesmall drops would notfallthrough theairasfastasthebig ones, because oftheairresistance, andwewould getacharge separation. You see,itispossible toconcoct allkinds ofpossibilities. Oneofthemore ingenious theories, which ismore satisfactory inmany re- spects thanthebreaking-drop theory, isduetoC.T.R.Wilson. Wewilldescribe it,asWilson did,withreference towater drops, although thesame phenomenon would alsowork withice.Suppose wehaveawater dropthatisfalling intheelec- tricﬁeldofabout 100volts permeter toward thenegatively charged earth. The drop willhave aninduced dipole moment—-with thebottom ofthedrop positive andthetopofthedrop negative, asdrawn inFig.9-13. Now there areintheair the“nuclei” thatwementioned earlier—the large slow-moving ions. (The fast ionsdonothaveanimportant effect here.) Suppose thatasadrop comes down, itapproaches alarge ion.Iftheionispositive, itisrepelled bythepositive bottom ofthedrop andispushed away. Soitdoes notbecome attached tothedrop. Iftheionweretoapproach from thetop,however, itmight attach tothenegative, topside. Butsince thedrop isfalling through theair,there isanairdriftrelative toit,going upwards, which carries theionsaway iftheir motion through theair isslow enough. Thus thepositive ionscannot attach atthetopeither. This would apply, yousee,onlytothelarge, slow-moving ions. Thepositive ionsof thistypewillnotattach themselves either tothefront orthebackofafalling drop. Ontheother hand, asthelarge, slow, negative ionsareapproached byadrop, theywillbeattracted andwillbecaught. Thedrop willacquire negative charge- thesignofthecharge having been determined bytheoriginal potential difference ontheentire earth—and wegettheright sign. Negative charge willbebrought down tothebottom partofthecloud bythedrops, andthepositively charged ions which areleftbehind willbeblown tothetopofthecloud bythevarious updraft currents. Thetheory looks pretty good, anditatleastgives theright sign. Alsoit doesn’t depend onhaving liquid drops. Wewillsee,when welearn about polariza- tioninadielectric, thatpieces oficewilldothesame thing. They alsowilldevelop positive andnegative charges ontheirextremities when theyareinanelectric ﬁeld. There are,however, some problems evenwiththistheory. First ofall,the totalcharge involved inathunderstorm isveryhigh. After ashort time, thesupply oflarge ionswould getusedup.SoWilson andothers havehadtopropose that there areadditional sources ofthelarge ions. Once thecharge separation starts, verylarge electric ﬁelds aredeveloped, andinthese large ﬁelds there maybeplaces where theairwillbecome ionized. Ifthere isahighly charged point, oranysmall object likeadrop, itmayconcentrate theﬁeldenough tomake a“brush discharge.” When there isastrong enough electric ﬁeld—let ussayitispositive—electrons willfallintotheﬁeldandwillpickupalotofspeed between collisions. Their speed willbesuchthatinhitting another atom theywilltearelectrons offatthat 9-9FALLING DROP E \ / G)G9V LARGE IUVS Fig.9-l3.C.T.R.Wilson's theory of charge separation inathundercloud. Fig.9-14. Photograph ofalightning ﬂash taken witha"Boys" camera. [From Schonland, Malan, andCollens, Proc. Roy. Soc.London, Vol.152(l935).] c?5?// _Z 1 I 7+//+ /+/$717 +/7+/=l~717T EARTH Fig.9-15. Theformation ofthe"step leader."atom, leaving positive charges behind. These newelectrons alsopickupspeed andcollide withmore electrons. Soakindofchain reaction oravalanche occurs, andthere isarapid accumulation ofions. Thepositive charges areleftneartheir original positions, sotheneteffect istodistribute thepositive charge onthepoint intoaregion around thepoint. Then, ofcourse, there isnolonger astrong ﬁeld, andtheprocess stops. Thisisthecharacter ofabrush discharge. Itispossible that theﬁelds maybecome strong enough inthecloud toproduce alittlebitofbrush discharge; there mayalsobeother mechanisms, oncethething isstarted, topro- duce alarge amount ofionization. Butnobody knows exactly howitworks. So thefundamental origin oflightning isreally notthoroughly understood. Weknow itcomes from thethunderstorms. (And weknow, ofcourse, thatthunder comes from thelightning—from thethermal energy released bythebolt.) Atleastwecanunderstand, inpart,theorigin ofatmospheric electricity. Due totheaircurrents, ions,andwater drops oniceparticles inathunderstorm, positive andnegative charges areseparated. Thepositive charges arecarried upward to thetopofthecloud (seeFig.9-11), andthenegative charges aredumped intothe ground inlightning strokes. Thepositive charges leave thetopofthecloud, enter thehigh-altitude layers ofmore highly conducting air,andspread throughout the earth. Inregions ofclear weather, thepositive charges inthislayer areslowly conducted totheearth bytheionsintheair—ions formed bycosmic rays, bythe sea,andbyman’s activities. Theatmosphere isabusyelectrical machine! 9-6Lightning Theﬁrstevidence ofwhat happens inalightning stroke wasobtained in photographs taken withacamera heldbyhand andmoved back andforth with theshutter open—while pointed toward aplace where lightning wasexpected. Theﬁrstphotographs obtained thiswayshowed clearly thatlightning strokes are usually multiple discharges along thesame path. Later, the“Boys” camera, which hastwolenses mounted 180°apart onarapidly rotating disc,wasdeveloped. Theimage made byeachlensmoves across theﬁlm—the picture isspread outin time. If,forinstance, thestroke repeats, there willbetwoimages idebyside. Bycomparing theimages ofthetwolenses, itispossible towork outthedetails ofthetimesequence oftheﬂashes. Figure 9-14shows aphotograph taken witha “Boys” camera. Wewillnowdescribe thelightning. Again, wedon’t understand exactly how itworks. Wewillgiveaqualitative description ofwhat itlooks like,butwewon’t gointoanydetails ofwhyitdoeswhat itappears todo.Wewilldescribe onlythe ordinary caseofthecloud withanegative bottom overﬂatcountry. Itspotential ismuch more negative than theearth underneath, sonegative electrons willbe accelerated toward theearth. What happens isthefollowing. Itallstarts witha thing called a“step leader,” which isnotasbright asthestroke oflightning. On thephotographs onecanseealittlebright spotatthebeginning thatstarts from the cloud andmoves downward veryrapidly—-at asixth ofthespeed oflight! Itgoes onlyabout S0meters andstops. Itpauses forabout 50microseconds, andthen takes another step. Itpauses again andthen goesanother step, andsoon.It moves inaseries ofsteps toward theground, along apathlikethatshown inFig. 9-15. Intheleader there arenegative charges from thecloud; thewhole column isfullofnegative charge. Also, theairisbecoming ionized bytherapidly moving charges thatproduce theleader, sotheairbecomes aconductor along thepath traced out. Themoment theleader touches theground, wehave aconducting “wire” thatrunsallthewayuptothecloud andisfullofnegative charge. Now, atlast,thenegative charge ofthecloud cansimply escape andrunout. The electrons atthebottom oftheleader aretheﬁrstonestorealize this;theydump out,leaving positive charge behind thatattracts more negative charge from higher upintheleader, which initsturnpours out,etc.Soﬁnally allthenegative charge inapartofthecloud runsoutalong thecolumn inarapid andenergetic way. Sothelightning stroke youseerunsupwards from theground, asindicated inFig. 9-16. Infact,thismain stroke—-by farthebrightest part—is called thereturn 940 stroke. Itiswhat produces theverybright light, andtheheat, which bycausing arapid expansion oftheairmakes thethunder clap. Thecurrent inalightning stroke isabout 10,000 amperes atitspeak, andit carries down about 20coulombs. Butwearestillnotﬁnished. After atimeof,perhaps, afewhundredths ofa second, when thereturn stroke hasdisappeared, another leader comes down. Butthistimethere arenopauses. Itiscalled a“dark leader” thistime, andit goesallthewaydown——from toptobottom inoneswoop. Itgoesfullsteam on exactly theoldtrack, because there isenough debris there tomake ittheeasiest route. Thenewleader isagain fullofnegative charge. Themoment ittouches the ground—zing!—there isareturn stroke going straight upalong thepath. Soyou seethelightning strike again, andagain, andagain. Sometimes itstrikes only once ortwice, sometimes ﬁveortentimes—once asmany as42times onthesame track wasseen-but always inrapid succession. Sometimes things geteven more complicated. Forinstance, after oneofits pauses theleader maydevelop abranch bysending outtwosteps—both toward the ground butinsomewhat different directions, asshown inFig.9-15. What happens thendepends onwhether onebranch reaches theground deﬁnitely before theother. Ifthatdoeshappen, thebright return stroke (ofnegative charge dumping intothe ground) works itswayupalong thebranch thattouches theground, andwhen it reaches andpasses thebranching point onitswayuptothecloud, abright stroke appears togodown theother branch. Why? Because negative charge isdumping outandthatiswhat lights upthebolt. Thischarge begins tomove atthetopof thesecondary branch, emptying successive, longer pieces ofthebranch, sothe bright lightning boltappears towork itswaydown thatbranch, atthesame time asitworks uptoward thecloud. If,however, oneofthese extra leader branches happens tohavereached theground almost simultaneously withtheoriginal leader, itcansometimes happen thatthedark leader ofthesecond stroke willtakethe second branch. Then youwillseetheﬁrstmain ﬂashinoneplace andthesecond ﬂashinanother place. Itisavariant oftheoriginal idea. Also, ourdescription isoversimpliﬁed fortheregion veryneartheground. When thestepleader getstowithin ahundred meters orsofrom theground, there isevidence thatadischarge rises from theground tomeet it.Presumably, the ﬁeldgetsbigenough forabrush-type discharge tooccur. If,forinstance, there is asharp object, likeabuilding withapoint atthetop,thenastheleader comes down nearby theﬁelds aresolarge thatadischarge starts from thesharp point andreaches uptotheleader. Thelightning tends tostrike suchapoint. Ithasapparently been known foralong time thathigh objects arestruck by lightning. There isaquotation ofArtabanis, theadvisor toXerxes, giving his master advice onacontemplated attack ontheGreeks—during Xerxes’ campaign tobring theentire known world under thecontrol ofthePersians. Artabanis said, “See howGodwithhislightning always smites thebigger animals andwillnot suffer them towaxinsolent, while these ofalesser bulkchafe himnot. How like- wisehisboltsfalleveronthehighest houses andtallest trees.” Andthenheexplains thereason: “So,plainly, dothhelovetobring down everything thatexalts itself.” Doyouthink-—now thatyouknow atrueaccount oflightning striking tall trees—that youhave agreater wisdom inadvising kings onmilitary matters than didArtabanis 2300years ago? Donotexalt yourself. Youcould onlydoitless poetically. 9-111 - 7+/+/ +/+/+/-y+//+//+’ Fig.9-16. Thereturn lightning stroke runsback upthepathmade bytheleader. I0 Dielectrics 10-1 Thedielectric constant Here webegin todiscuss another ofthepeculiar properties ofmatter under theinﬂuence oftheelectric ﬁeld. Inanearlier chapter weconsidered thebehavior ofconductors, inwhich thecharges move freely inresponse toanelectric ﬁeld to such points thatthere isnoﬁeldleftinside aconductor. Now wewilldiscuss insulators, materials which donotconduct electricity. Onemight atﬁrstbelieve thatthere should benoeffect whatsoever. However, using asimple electroscope andaparallel-plate capacitor, Faraday discovered thatthiswasnotso.Hisexperi- ments showed thatthecapacitance ofsuch acapacitor isincreased when anin- sulator isputbetween theplates. Iftheinsulator completely ﬁllsthespace between theplates, thecapacitance isincreased byafactor xwhich depends onlyonthe nature oftheinsulating material. Insulating materials arealsocalled dielectrics; thefactor Kisthenaproperty ofthedielectric, andiscalled thedielectric constant. Thedielectric constant ofavacuum is,ofcourse, unity. Ourproblem nowistoexplain whythere isanyelectrical effect iftheinsulators areindeed insulators anddonotconduct electricity. Webegin withtheexperi- mental factthatthecapacitance isincreased andtrytoreason outwhat might begoing on.Consider aparallel-plate capacitor withsome charges onthesurfaces oftheconductors, letussaynegative charge onthetopplate andpositive charge on thebottom plate. Suppose thatthespacing between theplates isdandtheareaof eachplate isA.Aswehaveproved earlier, thecapacitance is _MC_- d. (10.1) andthecharge andvoltage onthecapacitor arerelated by Q=CV. (10.2) Now theexperimental factisthatifweputapiece ofinsulating material like lucite orglassbetween theplates, weﬁndthatthecapacitance islarger. Thatmeans, ofcourse, thatthevoltage islower forthesame charge. Butthevoltage difference istheintegral oftheelectric ﬁeldacross thecapacitor; sowemust conclude that inside thecapacitor, theelectric ﬁeldisreduced even though thecharges onthe plates remain unchanged. °i=a:: cououcron _ .‘II ‘III ‘IIIII+ w w a ‘‘Q10-1 Thedielectric constant 10-2 Thepolarization vector P 10-3 Polarization charges 10-4 Theelectrostatic equations withdielectrics 10-5 Fields andforces with dielectrics \§i\\ Fig. 10-1. Aparallel-plate capaci- -If ‘l’ If ' A7 I1-‘J torwithadielectric. Thelines ofEareUFREE CONDUCTO R shown Now howcanthatbe?WehavealawduetoGauss thattellsusthattheflux oftheelectric ﬁeldisdirectly related totheenclosed charge. Consider thegaussian surface Sshown bybroken linesinFig.10-1. Since theelectric ﬁeldisreduced withthedielectric present, weconclude thatthenetcharge inside thesurface must 10-1 7%,!!!’ --W-_ _ ;-II,-_I-I-III. _T_ I Fig.10-2. Ifweputaconducting ‘plate inthegapofaparallel-plate con- I+- Abelower thanitwould bewithout thematerial. There isonlyonepossible conclu- sion,andthatisthatthere must bepositive charges onthesurface ofthedielectric. Since theﬁeldisreduced butisnotzero, wewould expect thispositive charge to besmaller thanthenegative charge ontheconductor. Sothephenomena canbe explained ifwecould understand insome waythatwhen adielectric material is placed inanelectric ﬁeldthere ispositive charge induced ononesurface andnega- tivecharge induced ontheother. OONDUCTOR kl._..Pi-O.denser, theinduced charges reduce the '‘ ’ '" ﬁeld intheconductor tozero. c0N|)ucTQR +91“//‘*1.$94F‘§~‘4/.e.e.”e.e.<eIe Fig. 10-3. Amodel ofadielectric: small conducting spheres embedded in anidealized insulator.Wewould expect thattohappen foraconductor. Forexample, suppose that wehadacapacitor withaplate spacing d,andweputbetween theplates aneutral conductor whose thickness isb,asinFig.10-2. Theelectric ﬁeldinduces apositive charge ontheupper surface andanegative charge onthelower surface, sothere is noﬁeldinside theconductor. Theﬁeldintherestofthespace isthesame asit waswithout theconductor, because itisthesurface density ofcharge divided by eo;butthedistance overwhich wehavetointegrate togetthevoltage (thepotential difference) isreduced. Thevoltage is V=-'-(d—b).60 Theresulting equation forthecapacitance islikeEq.(10.1), with (d—b)sub- stituted ford: 60A C- . (10.3) Thecapacitance isincreased byafactor which depends upon (b/d), theproportion ofthevolume which isoccupied bytheconductor. Thisgives usanobvious model forwhat happens withdielectrics—that inside thematerial there aremany littlesheets ofconducting material. Thetrouble with suchamodel isthatithasaspeciﬁc axis,thenormal tothesheets, whereas most dielectrics have nosuch axis. However, thisdifficulty canbeeliminated ifwe assume thatallinsulating materials contain small conducting spheres separated from each other byinsulation, asshown inFig.10-3. Thephenomenon ofthe dielectric constant isexplained bytheeffect ofthecharges which would beinduced oneachsphere. Thisisoneoftheearliest physical models ofdielectrics usedto explain thephenomenon thatFaraday observed. More speciﬁcally, itwasassumed thateachoftheatoms ofamaterial wasaperfect conductor, butinsulated from theothers. Thedielectric constant xwould depend ontheproportion ofspace which wasoccupied bytheconducting spheres. Thisisnot,however, themodel thatisusedtoday. 10-2 Thepolarization vector P Ifwefollow theabove analysis further, wediscover thattheideaofregions ofperfect conductivity andinsulation isnotessential. Each ofthesmall spheres actslikeadipole, themoment ofwhich isinduced bytheexternal ﬁeld. Theonly thing thatisessential totheunderstanding ofdielectrics isthatthere aremany little dipoles induced inthematerial. Whether thedipoles areinduced because there aretinyconducting spheres orforanyother reason isirrelevant. 10-2 Why should aﬁeldinduce adipole moment inanatom iftheatom isnota conducting sphere? Thissubject willbediscussed inmuch greater detail inthe next chapter, which willbeabout theinner workings ofdielectric materials. However, wegivehereoneexample toillustrate apossible mechanism. Anatom hasapositive charge onthenucleus, which issurrounded bynegative electrons. Inanelectric ﬁeld,thenucleus willbeattracted inonedirection andtheelectrons in theother. Theorbits orwave patterns oftheelectrons (orwhatever picture is usedinquantum mechanics) willbedistorted tosome extent, asshown inFig.10-4; thecenter ofgravity ofthenegative charge willbedisplaced andwillnolonger coincide withthepositive charge ofthenucleus. Wehave already discussed such distributions ofcharge. Ifwelookfrom adistance, suchaneutral conﬁguration isequivalent, toaﬁrstapproximation, toalittledipole. Itseems reasonable thatiftheﬁeldisnottooenormous, theamount ofinduced dipole moment willbeproportional totheﬁeld. That is,asmall ﬁeldwilldisplace thecharges alittlebitandalarger ﬁeldwilldisplace them further-and inpropor- tiontotheﬁeld-unless thedisplacement getstoolarge. Fortheremainder ofthis chapter, itwillbesupposed thatthedipole moment isexactly proportional tothe ﬁeld. Wewillnowassume thatineach atom there arecharges qseparated bya distance 5,sothatq5isthedipole moment peratom. (Weuse5because weare already using dfortheplate separation.) Ifthere areNatoms perunitvolume, there willbeadipole moment perunitvolume equal toNqﬁ. Thisdipole moment perunitvolume willberepresented byavector, P.Needless tosay,itisinthe direction oftheindividual dipole moments, i.e.,inthedirection ofthecharge separation 8: P=Nqs. (10.4) Ingeneral, Pwillvaryfrom place toplace inthedielectric. However, atany point inthematerial, Pisproportional totheelectric ﬁeldE.Theconstant of proportionality, which depends ontheeasewithwhich theelectron aredisplaced, willdepend onthekinds ofatoms inthematerial. What actually determines howthisconstant ofproportionality behaves, how accurately itisconstant forverylarge ﬁelds, andwhat isgoing oninside different materials, wewilldiscuss atalatertime. Forthepresent, wewillsimply suppose thatthere exists amechanism bywhich adipole moment isinduced which is proportional totheelectric ﬁeld. 10-3 Polarization charges Now letusseewhat thismodel gives forthetheory ofacondenser withadi- electric. Firstconsider asheet ofmaterial inwhich there isacertain dipole moment perunitvolume. Willthere beontheaverage anycharge density produced bythis‘? NotifPisuniform. Ifthepositive andnegative charges being displaced relative toeachother havethesame average density, thefactthattheyaredisplaced does notproduce anynetcharge inside thevolume. Ontheother hand, ifPwerelarger atoneplace andsmaller atanother, thatwould mean thatmore charge would be moved intosome region thanaway from it;wewould thenexpect togetavolume density ofcharge. Fortheparallel-plate condenser, wesuppose thatPisuniform, soweneedtolookonlyatwhat happens atthesurfaces. Atonesurface thenega- tivecharges, theelectrons, have effectively moved outadistance 6;attheother surface theyhavemoved in,leaving some positive charge effectively outadistance 6.Asshown inFig.10-5, wewillhaveasurface density ofcharge, which willbe called thesurface polarization charge. '.'l'___+_... _4'___;l'_..;|'_.._.'!__.1'__'l'_ _._+_ 1'itt1TP'!ii _||:;.,_ _ _ _ - — - _ - L-___ l_ 10-3ELECTRON DISTRIBUTION E1 Fig.10-4. Anatom inanelectric ﬁeld hasitsdistribution ofelectrons dis- placed with respect tothenucleus. I ' Fig.10-5. Adielectric slab ina uniform ﬁeld. Thepositive charges dis- -— placed thedistance 6with respect to ' thenegatives. Thischarge canbecalculated asfollows. IfAistheareaoftheplate, the number ofelectrons thatappear atthesurface istheproduct ofAandN,the number perunitvolume, andthedisplacement 6,which weassume hereisper- pendicular tothesurface. Thetotal charge isobtained bymultiplying bythe electronic charge qe.Togetthesurface density ofthepolarization charge induced onthesurface, wedivide byA.Themagnitude ofthesurface charge density is UFO] =Nq, 5. Butthisisjustequal tothemagnitude Pofthepolarization vector P,Eq.(10.4): ape;=P. (10.5) Thesurface density ofcharge isequal tothepolarization inside thematerial. The surface charge is,ofcourse, positive ononesurface andnegative ontheother. Now letusassume thatourslabisthedielectric ofaparallel-plate capacitor. Theplates ofthecapacitor alsohave asurface charge, which wewillcallo';,,,,,, because theycanmove “freely” anywhere ontheconductor. Thisis,ofcourse, thecharge thatweputonwhen wecharged thecapacitor. Itshould beemphasized thatapolexists onlybecause oftrim. Ifafm,isremoved bydischarging thecapacitor, then ape;willdisappear, notbygoing outonthedischarging wire, butbymoving back intothematerial-by therelaxation ofthepolarization inside thematerial. Wecannowapply Gauss’ lawtothegaussian surface SinFig.10-1. The electric ﬁeldEinthedielectric isequal tothetotalsurface charge density divided byco.Itisclear thatam;ando,,,,,,have opposite signs, so E=%Z1>_<>1 . (105) Note thattheﬁeldE0between themetal plate andthesurface ofthedielectric ishigher thantheﬁeldE;itcorresponds to03",,alone. Buthereweareconcerned withtheﬁeldinside thedielectric which, ifthedielectric nearly ﬁllsthegap,isthe ﬁeldovernearly thewhole volume. Using Eq.(10.5), wecanwrite E=ﬁﬂl£- (mp60 Thisequation doesn’t telluswhat theelectric ﬁeldisunless weknow what Pis. Here, however, weareassuming thatPdepends onE—in fact,thatitisproportional toE.Thisproportionality isusually written as P=mm (ma Theconstant X(Greek “khi”) iscalled theelectric susceptibility ofthedielectric. Then Eq.(10.7) becomes E=ﬁw_l_, rm60<1+><> ‘) which gives usthefactor 1/(1+x)bywhich theﬁeldisreduced. Thevoltage between theplates istheintegral oftheelectric ﬁeld. Since the ﬁeldisuniform, theintegral isjusttheproduct ofEandtheplate separation d. Wehavethat = = “freed _ VEde.<1"+x) Thetotal charge onthecapacitor isa;,e,,A, sothatthecapacitance deﬁned by(10.2) becomes _e0A(l +x)_xe0A_C-—i—d -id (10.10) Wehave explained theobserved facts. When aparallel-plate capacitor is ﬁlled withadielectric, thecapacitance isincreased bythefactor K=1+X, (10.11) 10-4 which isaproperty ofthematerial. Ourexplanation, ofcourse, isnotcomplete until wehaveexplained—as wewilldolater-how theatomic polarization comes about. Let’s nowconsider something alittlebitmore complicated—the situation in which thepolarization Pisnoteverywhere thesame. Asmentioned earlier, ifthe polarization isnotconstant, wewould expect ingeneral toﬁndacharge density inthevolume, because more charge might come intoonesideofasmall volume elementthanleaves itontheother. Howcanweﬁndouthowmuch charge isgained orlostfrom asmall volume? First let’scompute howmuch charge moves across anyimaginary surface when thematerial ispolarized. Theamount ofcharge thatgoesacross asurface isjustPtimes thesurface area ifthepolarization isnormal tothesurface. Ofcourse, ifthepolarization istangential tothesurface, nocharge moves across it. Following thesame arguments wehavealready used, itiseasytoseethatthe charge moved across anysurface element isproportional tothecomponent ofP perpendicular tothesurface. Compare Fig.10-6 withFig.10-5. Weseethat Eq.(10.5) should, inthegeneral case, bewritten am; =P-n. (10.12) Ifwearethinking ofanimagined surface element inside thedielectric, Eq. (10.12) gives thecharge moved across thesurface butdoesn’t result inanet surface charge, because there areequal andopposite contributions from thedi- electric onthetwosides ofthesurface. Thedisplacements ofthecharges can,however, result inavolume charge density. Thetotal charge displaced outofanyvolume Vbythepolarization‘ isthe integral oftheoutward normal component ofPoverthesurface Sthatbounds the volume (seeFig.10-7). Anequal excess charge oftheopposite signisleftbehind. Denoting thenetcharge inside VbyAQPOI wewrite AQ,,,,,=-/SP-llda. (10.13) Wecanattribute AQDO1 toavolume distribution ofcharge withthedensity pm), andso AQ,,,,,=/Vpp°1dV. (10.14) Combining thetwoequations yields fVp,,,,dV =—[SP-nda. (10.15) WehaveakindofGauss’ theorem thatrelates thecharge density from polarized materials tothepolarization vector P.Wecanseethatitagrees withtheresult wegotforthesurface polarization charge orthedielectric inaparallel-plate capaci- tor. Using Eq.(10.15) with thegaussian surface ofFig. 10-l. thesurface integral gives PAA,andthecharge inside isam;AA,sowegetagain thata=P. JustaswedidforGauss’ lawofelectrostatics, wecanconvert Eq.(10.15) to adifferential form-using Gauss’ mathematical theorem: P-nda = V'PdV./. /VWeget p,,,,,=-V-P. (10.16) Ifthere isanonuniform polarization, itsdivergence gives thenetdensity ofcharge appearing inthematerial. Weemphasize thatthisisaperfectly realcharge density; wecallit“polarization charge” onlytoremind ourselves howitgotthere. 10-5Fig.10-6. Thecharge moved across anelement ofanimaginary surface ina dielectric isproportional tothecom- ponent ofPnormal tothesurface. \\\\\\\ A0 \\ Volume V SUSS: \_. s\ \Fig.10-7. Anonuniform polariza- tionPcanresult inanetcharge inthe body ofadielectric. 1 10-4 Theelectrostatic equations withdielectrics Now let’scombine theabove result with ourtheory ofelectrostatics. The fundamental equation is v-E=11- (10.17)60 Thephereisthedensity ofallelectric charges. Since itisnoteasytokeeptrack of thepolarization charges, itisconvenient toseparate pintotwoparts. Again we callpm)thecharges duetononuniform polarizations, andcallpm,alltherest. Usually pm, isthecharge weputonconductors, oratknown places inspace. Equation (10.17) then becomes v_E= Pfi-ee‘l'Ppol =Pfree _V'P, 69 £0 or v-(E+Z)=E52. (10.18)G0 G0 Ofcourse, theequation forthecurlofEisunchanged: VXE=0. (10.19) Taking Pfrom Eq.(10.8), wegetthesimpler equation v-[(1+x)E]=v-(KE)= (10.20) These aretheequations ofelectrostatics when there aredielectrics. They don’t, ofcourse, sayanything new,buttheyareinaform which ismore convenient for computation incases where pmeisknown andthepolarization Pisproportional toE. Notice thatwehave nottaken thedielectric “constant,” x,outofthediver- gence. Thatisbecause itmaynotbethesame everywhere. Ifithaseverywhere the same value, itcanbefactored outandtheequations arejustthose ofelectrostatics withthecharge density pf,“divided byK.Intheform wehavegiven, theequations apply tothegeneral casewhere different dielectrics maybeindifferent places in theﬁeld. Then theequations maybequite difficult tosolve. There isamatter ofsome historical importance which should bementioned here. Intheearly days ofelectricity, theatomic mechanism ofpolarization was notknown andtheexistence ofppolwasnotappreciated. Thecharge pm,was considered tobetheentire charge density. Inorder towrite Maxwell’s equations inasimple form, anewvector Dwasdeﬁned tobeequal toalinear combination ofEandP: D=e(,E+P. (10.21) Asaresult, Eqs.(10.18) and(10.19) werewritten inanapparently verysimple form: V'D=pfreea VXE= Canonesolve these? Only ifathird equation isgiven fortherelationship between DandE.When Eq.(10.8) holds, thisrelationship is D=eQ(l +X)E=Ke0E. (10.23) Thisequation wasusually written D=eE, (10.24) where eisstillanother constant fordescribjng thedielectric property ofmaterials. Itiscalled the“permittivity.” (Now youseewhywehave soinourequations, itis the“permittivity ofempty space.”) Evidently, c=xeo=(1+X)e0. (10.25) 10-6 Today welookupon these matters from another point ofview, namely, that wehave simpler equations inavacuum, andifweexhibit inevery caseallthe charges, whatever their origin, theequations arealways correct. Ifweseparate some ofthecharges away forconvenience, orbecause wedonotwant todiscuss what isgoing onindetail, thenwecan,ifwewish, write ourequations inanyother form thatmaybeconvenient. Onemore point should beemphasized. Anequation likeD=eEisanattempt todescribe aproperty ofmatter. Butmatter isextremely complicated, andsuch anequation isinfactnotcorrect. Forinstance, ifEgetstoolarge, thenDisno longer proportional toE.Forsome substances, theproportionality breaks down evenwithrelatively small ﬁelds. Also, the“constant” ofproportionality mayde- pend onhowfastEchanges withtime. Therefore thiskindofequation isakind ofapproximation, likeHooke’s law.Itcannot beadeepandfundamental equation. Ontheother hand, ourfundamental equations forE,(10.17) and(10.19),represent ourdeepest andmost complete understanding ofelectrostatics. 10-5 Fields andforces withdielectrics Wewillnowprove some rather general theorems forelectrostatics insituations where dielectrics arepresent. Wehaveseenthatthecapacitance ofaparallel-plate capacitor isincreased byadeﬁnite factor ifitisﬁlled withadielectric. Wecan show thatthisistrueforacapacitor ofanyshape, provided theentire region in theneighborhood ofthetwoconductors isﬁlled withauniform linear dielectric. Without thedielectric, theequations tobesolved are v-E0=l’E and VXE0=0.60 With thedielectric present, theﬁrstofthese equations ismodiﬁed; wehaveinstead theequations v-(KE)= and v><E=0. (10.26) Now since wearetaking Ktobeeverywhere thesame, thelasttwoequations can bewritten as v-(KE)= and v><(ICE)=0. (10.27) Wetherefore havethesame equations forKEasforE0,sotheyhavethesolu- tionKE=E0.Inother words, theﬁeldiseverywhere smaller, 'bythefactor 1/K, thaninthecasewithout thedielectric. Since thevoltage difference isalineintegral oftheﬁeld, thevoltage isreduced bythissame factor. Since thecharge onthe electrodes ofthecapacitor hasbeentaken thesame inbothcases, Eq.(10.2) tells usthatthecapacitance, inthecaseofaneverywhere uniform dielectric, isin- creased bythefactor x. ' Letusnowaskwhat theforce would bebetween twocharged conductors ina dielectric. Weconsider aliquid dielectric thatishomogeneous everywhere. We haveseenearlier thatonewaytoobtain theforce istodifferentiate theenergy with respect totheappropriate distance. Iftheconductors have equal andopposite charges, theenergy U=Q2/2C, where Cistheir capacitance. Using theprinciple ofvirtual work, anycomponent isgiven byadifferentiation; forexample, ___6U____Q26 1)F,_ 73}-_ 75(C- (10.22) Since thedielectric increases thecapacity byafactor K,allforces willbereduced bythissame factor. Onepoint should beemphasized. What wehave saidistrueonlyifthedi- electric isaliquid. Anymotion ofconductors thatareembedded insoliddielectric changes themechanical stress conditions ofthedielectric andalters itselectrical 10-7 \ E F otztscrmcOBJECT \ Fig. 10-8. Adielectric obiect in0 nonuniform ﬁeld feels aforce toward regions ofhigher ﬁeld strength.properties, aswellascausing some mechanical energy change inthedielectric. Moving theconductors inaliquid doesnotchange theliquid. Theliquid moves toanewplace butitselectrical characteristics arenotchanged. Many older books onelectricity start withthe“fundamental” lawthatthe force between twocharges is F=Z5-G1-%. (10.29) apoint ofviewwhich isthoroughly unsatisfactory. Foronething, itisnottrue ingeneral; itistrueonlyforaworld ﬁlled withaliquid. Secondly, itdepends on thefactthatKisaconstant, which isonlyapproximately trueformost realmaterials. Itismuch better tostart with Coulomb’s lawforcharges inavacuum, which is always right (forstationary charges). What doeshappen inasolid? Thisisaveryditﬁcult problem which hasnot been solved, because itis,inasense, indeterminate. Ifyouputcharges inside a dielectric solid, there aremany kinds ofpressures andstrains. Youcannot deal withvirtual work without including alsothemechanical energy required tocom- press thesolid, anditisadifficult matter, generally speaking, tomake aunique distinction between theelectrical forces andthemechanical forces duetothesolid material itself. Fortunately, nooneeverreally needs toknow theanswer tothe question proposed. Hemaysometimes want toknow howmuch strain there is going tobeinasolid, andthatcanbeworked out.Butitismuch more complicated thanthesimple result wegotforliquids. Asurprisingly complicated problem inthetheory ofdielectrics isthefollow- ing:Whydoesacharged object pickuplittlepieces ofdielectric? Ifyoucomb your haironadryday,thecomb readily picks upsmall scraps ofpaper. Ifyouthought casually about it,youprobably assumed thecomb hadonecharge onitandthe paper hadtheopposite charge onit.Butthepaper isinitially electrically neutral. Ithasn’t anynetcharge, butitisattracted anyway. Itistruethatsometimes the paper willcome uptothecomb andthenﬂyaway, repelled immediately afterit touches thecomb. Thereason is,ofcourse, thatwhen thepaper touches thecomb, itpicks upsome negative charges andthenthelikecharges repel. Butthatdoesn’t answer theoriginal question. Why didthepaper come toward thecomb inthe ﬁrstplace? Theanswer hastodowiththepolarization ofadielectric when itisplaced in anelectric ﬁeld. There arepolarization charges ofbothsigns, which areattracted andrepelled bythecomb. There isanetattraction, however, because theﬁeld nearer thecomb isstronger thantheﬁeldfarther away—the comb isnotaninﬁnite sheet. Itscharge islocalized. Aneutral piece ofpaper willnotbeattracted to either plate inside theparallel plates ofacapacitor. Thevariation oftheﬁeldis anessential partoftheattraction mechanism. Asillustrated inFig.10-8, adielectric isalways drawn from aregion ofweak ﬁeldtoward aregion ofstronger ﬁeld. Infact,onecanprove thatforsmall objects theforce isproportional tothegradient ofthesquare oftheelectric ﬁeld. Why doesitdepend onthesquare oftheﬁeld?Because theinduced polarization charges areproportional totheﬁelds, andforgiven charges theforces areproportional to theﬁeld. However, aswehavejustindicated, there willbeanetforce onlyifthe square oftheﬁeldischanging from point topoint. Sotheforce isproportional to thegradient ofthesquare oftheﬁeld. Theconstant ofproportionality involves, among other things, thedielectric constant oftheobject, anditalsodepends upon thesizeandshape oftheobject. There isarelated problem inwhich theforce onadielectric canbeworked out quite accurately. Ifwehave aparallel-plate capacitor withadielectric slabonly partially inserted, asshown inFig.l0—9, there willbeaforce driving thesheet in. Adetailed examination oftheforce isquite complicated; itisrelated tononuni- formities intheﬁeldneartheedges ofthedielectric andtheplates. However, if wedonotlookatthedetails, butmerely usetheprinciple ofconservation ofenergy, wecaneasily calculate theforce. Wecanﬁndtheforce from theformula wede- 10-8 Z/_//////1 1??$221+ +DIELECTRIC F _ \\ \ computed byapplying theprinciple of7 7 7 l Fig. lO—9 Theforce onadielectric ' ' d sheet inaparallel plate capacitor canbe X L . "II rived earlier. Equation (10.28) isequivalent to av V26CF,_-Tx_+75; (10.30) Weneedonlyﬁndouthowthecapacitance varies withtheposition ofthedielectric slab. Let’s suppose thatthetotal length oftheplates isL,thatthewidth oftheplates isW,thattheplate separation anddielectric thickness ared,andthatthedistance towhich thedielectric hasbeen inserted isx.Thecapacitance istheratio ofthe total freecharge ontheplates tothevoltage between theplates. Wehave seen above thatforagiven voltage Vthesurface charge density offreecharge is/<e0V/d. Sothetotalcharge ontheplates is Q=5‘;j;l’xW+i‘-;,i’<L—x>W. from which wegetthecapacitance: c=59;-V(1<x +L-x). (10.31) Using (10.30), wehave 2 F,=-lg-%’ (K-1). (10.32) Now thisequation isnotparticularly useful foranything unless youhappen to need toknow theforce insuch circumstances. Weonly wished toshow thatthe theory ofenergy canoften beusedtoavoid enormous complications indetermining theforces ondielectric materials—as there would beinthepresent case. Ourdiscussion ofthetheory ofdielectrics hasdealt only with electrical phe- nomena, accepting thefactthatthematerial hasapolarization which isproportional totheelectric ﬁeld. Why there issuch aproportionality isperhaps ofgreater interest tophysics. Once weunderstand theorigin ofthedielectric constants fromanatomic point ofview, wecanuseelectrical measurements ofthedielectric constants in varying circumstances toobtain detailed information about atomic ormolecular structure. Thisaspect willbetreated inpartinthenextchapter. 10-9energy conservation ll Inside Dielectrics v ll-1 Molecular dipoles Inthischapter wearegoing todiscuss whyitisthatmaterials aredielectric. Wesaidinthelastchapter thatwecould understand theproperties ofelectrical systems withdielectrics onceweappreciated thatwhen anelectric ﬁeldisapplied toadielectric itinduces adipole moment intheatoms. Speciﬁcally, iftheelectric ﬁeldEinduces anaverage dipole moment perunitvolume P,then 1<,thedielectric constant, isgiven by P_]=__. ' K EOE (ill) Wehave already discussed how thisequation isapplied; now wehave todis- cussthemechanism bywhich polarization arises when there isanelectric ﬁeld inside amaterial. Webegin with thesimplest possible example——the polarization ofgases. Buteven gases already have complications: there aretwotypes. The molecules ofsome gases, likeoxygen, which hasasymmetric pairofatoms ineach molecule, havenoinherent dipole moment. Butthemolecules ofothers, likewater vapor (which hasanonsymmetric arrangement ofhydrogen andoxygen atoms) carry apermanent electric dipole moment. Aswepointed outinChapters 6and7, there isinthewater vapor molecule anaverage pluscharge onthehydrogen atoms andanegative charge ontheoxygen. Since thecenter ofgravity ofthenega- tivecharge andthecenter ofgravity ofthepositive charge donotcoincide, the totalcharge distribution ofthemolecule hasadipole moment. Such amolecule is called apolar molecule. Inoxygen, because ofthesymmetry ofthemolecule, the centers ofgravity ofthepositive andnegative charges arethesame, soitisa nonpolar molecule. Itdoes, however, become adipole when placed inanelectric ﬁeld. Theforms ofthetwotypes ofmolecules aresketched inFig.11-1. 11-2 Electronic polarization Wewillﬁrstdiscuss thepolarization ofnonpolar molecules. Wecanstart with thesimplest case ofamonatomic gas(forinstance, helium). When anatom of suchagasisinanelectric ﬁeld, theelectrons arepulled onewaybytheﬁeldwhile thenucleus ispulled theother way,asshown inFig.10—4. Although theatoms are verystiffwith respect totheelectrical forces wecanapply experimentally, there isa slight netdisplacement ofthecenters ofcharge, andadipole moment isinduced. Forsmall ﬁelds, theamount ofdisplacement, andsoalsothedipole moment, is proportional totheelectric ﬁeld. Thedisplacement oftheelectron distribution which produces thiskind ofinduced dipole moment iscalled electronic polarization. Wehave already discussed theinﬂuence ofanelectric ﬁeld onanatom in Chapter 31ofVol.I,when weweredealing withthetheory oftheindex ofrefrac- tion. Ifyouthink about itforamoment, youwillseethatwhat wemust donowis exactly thesame aswedidthen. Butnowweneed worry onlyabout ﬁelds thatdo notvary with time, while theindex ofrefraction depended ontime-varying ﬁelds. InChapter 31ofVol.Iwesupposed thatwhen anatom isplaced inanoscilla- tingelectric ﬁeldthecenter ofcharge oftheelectrons obeys theequation d2xmW+mwﬁx =q,E. (11.2) ll-111-1 Molecular dipoles 11—2 Electronic polarization 11-3 Polar molecules; orientation polarization ll-4 Electric ﬁelds incavities ofa dielectric 11-5 Thedielectric constant of liquids; theClausius-Mossotti equation 11-6 Solid dielectrics 11-7 Ferroelectricity; BaTiO3 Review: Chapter 31,Vol.I,TheOrigin oftheRefractive Index Chapter 40,Vol. I,ThePrin- ciples ofStatistical Mechanics _ +8 _ _ — I — _ - csursn or_ +mo-CHARGE (0) CENTER OF -'CHARGE CENTER W +CHARGE lb) Fig. ll-1. la)Anoxygen molecule withzero dipole moment. lb)Thewater molecule hasapermanent dipole moment Po- Theﬁrsttermistheelectron mass times itsacceleration andthesecond isarestoring force, while theright-hand sideistheforce from theoutside electric ﬁeld. Ifthe electric ﬁeldvaries withthefrequency w,Eq.(11.2) hasthesolution _ q.Ex_--?m(w?)__0),). (11.3) which hasaresonance atw=0:0.When wepreviously found thissolution, we interpreted itassaying thatwowasthefrequency atwhich light (intheoptical region orintheultraviolet, depending ontheatom) wasabsorbed. Forour purposes, however, weareinterested onlyinthecaseofconstant ﬁelds, i.e.,for 0:=0,sowecandisregard theacceleration term in(11.2), andweﬁndthatthe displacement is .12x= (11.4) From thisweseethatthedipole moment pofasingle atom is q§EP=qex= (11-5) Inthistheory thedipole moment pisindeed proportional totheelectric ﬁeld. People usually write p=ot€0E. (11.6) (Again thesoisputinforhistorical reasons.) Theconstant aiscalled thepolariz- ability oftheatom, andhasthedimensions L3.Itisameasure ofhoweasyitisto induce amoment inanatom withanelectric ﬁeld. Comparing (11.5) and(11.6), oursimple theory saysthat 2 2 0.=--4‘ =-4“- (11.7)eomwg mwg Ifthere areNatoms inaunitvolume, thepolarization P—the dipole moment perunitvolume—is given by P=Np=Ntxe0E. (11.8) Putting (11.1) and(11.8) together, weget PK——l=€i—Na (11.9) or,using (11.7), 41rNe2K-1=7173- (11.10) From Eq.(11.9) wewould predict thatthedielectric constant Kofdifferent gases should depend onthedensity ofthegasandonthefrequency woofitsoptical absorption. Ourformula is,ofcourse, onlyaveryrough approximation, because inEq. (11.2) wehave taken amodel which ignores thecomplications ofquantum me- chanics. Forexample, wehave assumed that anatom hasonly oneresonant frequency, when itreally hasmany. Tocalculate properly thepolarizability aof atoms wemust usethecomplete quantum-mechanical theory, buttheclassical ideas above giveusareasonable estimate. Let’s seeifwecangettheright order ofmagnitude forthedielectric constant ofsome substance. Suppose wetryhydrogen. Wehave onceestimated (Chapter 38,Vol.I)thattheenergy needed toionize thehydrogen atom should beapproxi- mately 1me4E~iz2—- (11.11) 11-2 Foranestimate ofthenatural frequency coo,wecansetthisenergy equal tohw0— theenergy ofanatomic oscillator whose natural frequency is0:0.Weget ~lme4 “Orr?-5" Ifwenowusethisvalue ofweinEq.(11.7), weﬁndfortheelectronic polarizability h2 3 Thequantity (hz/me2) istheradius oftheground-state orbit ofaBohr atom (see Chapter 38,Vol. I)andequals 0.528 angstroms. Inagasatstandard pressure and temperature (1atmosphere, 0°C) there are2.69 X1019atoms/cma, soEq.(11.9) gives us K=1+(2.69 Xl0‘9)l611-(0.528 X10“8)3 =1.00020. (11.13) Thedielectric constant forhydrogen gasismeasured tobe Kexp =1.00026. Weseethatourtheory isabout right. Weshould notexpect anybetter, because themeasurements were, ofcourse, made withnormal hydrogen gas,which has diatomic molecules, notsingle atoms. Weshould notbesurprised ifthepolariza- tionoftheatoms inamolecule isnotquite thesame asthatoftheseparate atoms. Themolecular effect, however, isnotreally thatlarge. Anexact quantum- mechanical calculation ofoiforhydrogen atoms gives aresult about 12% higher than(11.12) (the161ris changed to181r), andtherefore predicts adielectric constant somewhat closer totheobserved one. Inanycase, itisclear thatourmodel ofa dielectric isfairly good. Another check onourtheory istotryEq.(11.12) onatoms which have a higher frequency ofexcitation. Forinstance, ittakes about 24.5volts topullthe electron oﬁ"helium, compared withthe13.5volts required toionize hydrogen. Wewould, therefore, expect thattheabsorption frequency woforhelium would be about twice asbigasforhydrogen andthatozwould beone-quarter aslarge. We expect that Khelmm z1.000050. Experimentally, Khelium = soyouseethatourrough estimates arecoming outontheright track. Sowehave understood thedielectric constant ofnonpolar gas,butonly qualitatively, because wehave notyetused acorrect atomic theory ofthemotions oftheatomic electrons. 11-3 Polar molecules; orientation polarization Next wewillconsider amolecule which carries apermanent dipole moment p0——such asawater molecule. With noelectric ﬁeld, theindividual dipoles point inrandom directions, sothenetmoment perunitvolume iszero. Butwhen an electric ﬁeld isapplied, twothings happen: First, there isanextra dipole moment induced because oftheforces ontheelectrons; thispartgives justthesame kind of electronic polarizability wefound foranonpolar molecule. Forvery accurate work, thiseffect should, ofcourse, beincluded, butwewillneglect itforthe moment. (Itcanalways beadded inattheend.) Second, theelectric ﬁeldtends to lineuptheindividual dipoles toproduce anetmoment perunitvolume. Ifallthe dipoles inagaswere tolineup,there would beavery large polarization, butthat doesnothappen. Atordinary temperatures andelectric ﬁelds thecollisions ofthe molecules intheirthermal motion keepthem from lining upverymuch. Butthere issome netalignment, andsosome polarization (seeFig.ll—2). Thepolarization thatdoes occur canbecomputed bythemethods ofstatistical mechanics we described inChapter 40ofVol.I. 11-3\\)‘ ‘,1 it I \jt *1. t-*~\16’‘m#‘K Quiz8 (0) '°"\\ it..il\/*9» ¢:'fl\ 1, 1’ii’? »°‘ lb) Fig. 11-2. (alInagas ofpolar molecules, the individual moments are oriented atrandom; theaverage moment inasmall volume iszero. lb)When there isanelectric ﬁeld, there issome average alignment ofthemolecules. E(ll+11 d -<12) Fig. 11-3. The energy ofadipole pointheﬁeld Eis-—p°-E.Tousethismethod weneedtoknow theenergy ofadipole inanelectric ﬁeld. Consider adipole ofmoment p0inanelectric ﬁeld, asshown inFig.11-3. The energy ofthepositive charge isq¢(l), andtheenergy ofthenegative charge is —q¢(2). Thus theenergy ofthedipole is U=q¢(1)—q¢(2)=114'W.or U=—p0-E =—p0Ecos 0, (11.14) where 0istheangle between poandE.Aswewould expect, theenergy islower when thedipoles arelined upwith theﬁeld. Wenowﬁndouthowmuch lining upoccurs byusing themethods ofstatis- ticalmechanics. Wefound inChapter 40ofVol.Ithatinastateofthermal equili- brium, therelative number ofmolecules with thepotential energy Uisproportional to e'U"‘T, (11.15) where U(x,y,z)isthepotential energy asafunction ofposition. Thesame argu- ments would saythatusing Eq.(l1.l4)' forthepotential energy asafunction of angle, thenumber ofmolecules at0perunitsolidangle isproportional toe_U”°T. Letting n(0)bethenumber ofmolecules perunit solid angle at0,wehave n(6)=n0e+'”°E°°°’/H. (11.16) Fornormal temperatures andﬁelds, theexponent issmall, sowecanapproximate byexpanding theexponential: n(0)=no<1+ (11.17) Wecanﬁndnoifweintegrate (11.17) overallangles; theresult should bejust N,thetotalnumber ofmolecules perunitvolume. Theaverage value ofcos0over allangles iszero, sotheintegral isjustnotimes thetotal solid angle 41r.Weget no=Z? (11.18) Weseefrom (11.17) thatthere willbemore molecules oriented along theﬁeld (cos0=1)than against theﬁeld (cos0=—l).Soinanysmall volume contain- ingmany molecules there willbeanetdipole moment perunitvolume-—that is, apolarization P.Tocalculate P,wewant thevector sum ofallthemolecular moments inaunitvolume. Since weknow thattheresult isgoing tobeinthe direction ofE,wewilljustsumthecomponents inthatdirection (thecomponents atright angles toEwillsumtozero): P= 2p0cos0,. unit volume Wecanevaluate thesum byintegrating over theangular distribution. The solid angle at0is21rsin0d0,so 7|’ P=/n(0)p0 cos021rsin0d0. (11.19)0 Substituting forn(0)from (11.17), wehave 1|‘ P=—g/0 (1+%9T£cos0)p0cos0d(cos0), which iseasily integrated togive N2EP=-3~;%- (11.20) ll-4 Thepolarization isproportional totheﬁeldE,sothere willbenormal dielectric behavior. Also, asweexpect, thepolarization depends inversely onthetempera- ture,because athigher temperatures there ismore disalignment bycollisions. This 1/Tdependence iscalled Curie’s law.Thepermanent moment pt)appears squared forthefollowing reason: Inagiven electric ﬁeld, thealigning force depends upon po,andthemean moment thatisproduced bythelining upisagain proportional topo.Theaverage induced moment isproportional topg. Weshould nowtrytoseehowwellEq.(11.20) agrees withexperiment. Let’s lookatthecaseofsteam. Since wedon’t know whatpgis,wecannot compute P directly, butEq.(11.20) doespredict thatK-1should varyinversely asthetem- perature, andthisweshould check. From (11.20) weget 2 K-1=;:lE=31:%T. (11.21) soK-1should varyindirect proportion tothedensity N,andinversely asthe absolute temperature. Thedielectric constant hasbeen measured atseveral different pressures andtemperatures, chosen suchthatthenumber ofmolecules in aunitvolume remained ﬁxed.* [Notice thatifthemeasurements hadallbeen taken atconstant pressure, thenumber ofmolecules perunitvolume would decrease linearly with increasing temperature andK—1would vary asT“? instead ofasT“‘.] InFig.11-4weplottheexperimental observations forK—l asafunction of1/T. Thedependence predicted by(11.21) isfollowed quite well. There isanother characteristic ofthedielectric constant ofpolar molecules- itsvariation withthefrequency oftheapplied ﬁeld. Duetothemoment ofinertia ofthemolecules, ittakes acertain amount oftimefortheheavy molecules toturn toward thedirection oftheﬁeld. Soifweapply frequencies inthehighmicrowave region orabove, thepolar contribution tothedielectric constant begins tofall away because themolecules cannot follow. Incontrast tothis,theelectronic polarizability stillremains thesame uptooptical frequencies, because ofthe smaller inertia intheelectrons. 11-4 Electric ﬁelds incavities ofadielectric Wenow turn toaninteresting butcomplicated question—-the problem ofthe dielectric constant indense materials. Suppose that wetake liquid helium or liquid argon orsome other nonpolar material. Westillexpect electronic polari- zation. Butinadense material, Pcanbelarge, sotheﬁeldonanindividual atom willbeinﬂuenced bythepolarization oftheatoms initsclose neighborhood. The question is,what electric ﬁeldactsontheindividual atom? Imagine thattheliquid isputbetween theplates ofacondenser. Iftheplates arecharged theywillproduce anelectric ﬁeldintheliquid. Butthere arealso charges intheindividual atoms, andthetotalﬁeldEisthesumofboth ofthese effects. Thistrueelectric ﬁeldvaries very, veryrapidly from point topoint inthe liquid. Itisveryhigh inside theatoms——particularly right nexttothenucleus-—and relatively small between theatoms. Thepotential difference between theplates is thelineintegral ofthistotalﬁeld. Ifweignore alltheﬁne-grained variations, we canthink ofanaverage electric ﬁeldE,which isjustV/d. (This istheﬁeldwewere using inthelastchapter.) Weshould think ofthisﬁeldastheaverage over aspace containing many atoms. Now youmight think thatan“average” atom inan“average” location would feelthisaverage ﬁeld. Butitisnotthatsimple, aswecanshow byconsidering what happens ifweimagine different-shaped holes inadielectric. Forinstance, suppose thatwecutaslotinapolarized dielectric, with theslotoriented parallel tothe ﬁeld, asshown inpart (a)ofFig.11-5. Since weknow thatVXE=0,theline integral ofEaround thecurve, I‘,which goesasshown in(b)oftheﬁgure, should *Sanger, Steiger, andGachter, Helvetica Physica Acta5,200(1932). 11-5K4’ l l /T 0004- _'/+/ - 4'1‘ 000s- / - / / ' /0002 / / / 0001- / — / /O Ol I0.001 0.002 01003 1/T(°K") Fig.ll-4. Experimental measure- ments ofthedielectric constant ofwater vapor atvarious temperatures. 4%2\\__\\\\ ‘§:_‘'__'l 4-++ ++ 11/1‘ lb) id) Fig. 11-5. Theﬁeld inaslotcutina dielectric depends ontheshape and orientation oftheslot. é\§i.\‘.&\\_NSHRRQX.‘\K\‘wgyi @“‘RO“\_‘Q\\\\_‘“‘Y A“\\ l DIPOLE FIELD OUTSIDE ‘pil'|'I --=-am‘U Fig. 11-7. The electric ﬁeld ofa uniformly polarized sphere.bezero. Theﬁeldinside theslotmust giveacontribution which justcancels the partfrom theﬁeldoutside. Therefore theﬁeldE0actually found inthecenter of alongthinslotisequal toE,theaverage electric ﬁeldfound inthedielectric. Now consider another slotwhose large sides areperpendicular toE,asshown inpart(c)ofFig.ll-5. Inthiscase, theﬁeldE0intheslotisnotthesame asE because polarization charges appear onthesurfaces. lfweapply Gauss’ lawto asurface Sdrawn asin(d)oftheﬁgure, weﬁndthat theﬁeld E0intheslotis given by E0=E+2. (11.22) where Eisagain theelectric ﬁeldinthedielectric. (Thegaussian surface contains thesurface polarization charge 03,01 =P.) Wementioned inChapter 10that e0E+Pisoften called D,soe0E0 =D0isequal toDinthedielectric. Earlier inthehistory ofphysics, when itwassupposed tobeveryimportant todeﬁne every quantity bydirect experiment, people were delighted todiscover thattheycould deﬁne what theymeant byEandDinadielectric without having tocrawl around between theatoms. Theaverage ﬁeldEisnumerically equal to theﬁeldE0thatwould bemeasured inaslotcutparallel totheﬁeld. And theﬁeld Dcould bemeasured byﬁnding E0inaslotcutnormal totheﬁeld. Butnobody evermeasures them thatwayanyway, soitwasjustoneofthose philosophical things. Fig. ll-6. Theﬁeld atany point A inadielectric canbeconsidered asthe sumoftheﬁeld inaspherical hole plus theﬁeld duetoaspherical plug.+ Q17 Formost liquids which arenottoocomplicated instructure, wecould expect thatanatom ﬁnds itself, ontheaverage, surrounded bytheother atoms inwhat would beagood approximation toaspherical hole. And soweshould ask: “What would betheﬁeld inaspherical hole?” Wecanﬁndoutbynoticing thatifwe imagine carving outaspherical holeinauniformly polarized material, wearejust removing asphere ofpolarized material. (Wemust imagine thatthepolarization is“frozen in”before wecutoutthehole.) Bysuperposition, however, theﬁelds inside thedielectric, before thesphere wasremoved, isthesum oftheﬁelds from allcharges outside thespherical volume plustheﬁelds from thecharges within the polarized sphere. That is,ifwecallEtheﬁeld intheuniform dielectric, wecan write E=Ehole +Eplugs where E001, istheﬁeld inthehole andE0103 istheﬁeld inside asphere which is uniformly polarized (seeFig.11-6). Theﬁelds duetoauniformly polarized sphere areshown inFig. ll-7. Theelectric ﬁeld inside thesphere isuniform, andits value is PE0100 =—§-6- (11.24) Using (11.23), weget PE0010 =E+55- (11.25) The ﬁeld inaspherical cavity isgreater than theaverage ﬁeld bytheamount P/3e0. (The spherical holegives aﬁeld1/3ofthewaybetween aslotparallel to theﬁeld andaslotperpendicular totheﬁeld.) 11-5 Thedielectric constant ofliquids; theClausius-Mossotti equation Inaliquid weexpect thattheﬁeld which willpolarize anindividual atom is more likeE001, thanjustE.IfweusetheE),010 of(11.25) forthepolarizing ﬁeldin ll-6 Eq.(11.6), thenEq.(11.8) becomes P=1va¢0(E + (11.26)D OI‘ P=1-_-lz'§,a-/,3 e0E. (11.27) Remembering thatK—1isjustP/e0E, wehave K-1= (11.28) which gives usthedielectric constant ofaliquid interms ofoz,theatomic polar- izability. This iscalled theClausius-Mossotti equation. Whenever Notisverysmall, asitisforagas(because thedensity Nissmall), then theterm Na/3 canbeneglected compared with 1,andwegetouroldresult, Eq.(11.9), that K—1=Na. (11.29) Let’s compare Eq.(11.28) withsome experimental results. Itisﬁrstnecessary tolook atgases forwhich, using themeasurement ofK,wecanﬁndafrom Eq. (11.29). Forinstance, forcarbon disulﬁde atzerodegrees centigrade thedielectric constant is1.0029, soNais0.0029. Now thedensity ofthegasiseasily worked out andthedensity oftheliquid canbefound inhandbooks. At20°C, thedensity of liquid CS2is381times higher thanthedensity ofthegasat0°C. Thismeans that Nis381times higher intheliquid thanitisinthegasso,that—if wemake the approximation thatthebasic atomic polarizability ofthecarbon disulﬁde doesn’t change when itiscondensed intoa1iquid—Na intheliquid isequal to381times 0.0029, or1.11. Notice thattheNa/3 term amounts toalmost 0.4,soitisquite signiﬁcant. With these numbers wepredict adielectric constant of2.76, which agrees reasonably wellwiththeobserved value of2.64. InTable 11-1wegivesome experimental data onvarious materials (taken from theHandbook ofChemistry andPhysics), together withthedielectric constants calculated from Eq.(11.28) inthewayjustdescribed. Theagreement between observation andtheory iseven better forargon andoxygen than forCS2—and notsogood forcarbon tetrachloride. Onthewhole, theresults show thatEq. (11.28) works verywell. Table 11-1 Computation ofthedielectric constants ofliquids from thedielectric constant ofthegas. Gas Liquid x(exp) Na K(predict) K(exp) CS2 1.0029 0.0029 0.00339 1.293 381 O2 1.000523 0.000523 0.00143 1.19 832 CCI4 1.0030 0.0030 0.00489 1.59 325Substance A 1.000545 0.000545 0.00178 1.44 810i Na Density Density Ratio* 1.11 0.435 0.977 0.4412.76 1.509 2.45 1.5172.64 1.507 2.24 1.54 “Ratio =density ofliquid/density ofgas. Ourderivation ofEq.(11.28) isvalid onlyforelectronic polarization inliquids. Itisnotright forapolar molecule likeH20. Ifwegothrough thesame calcu- lations forwater, weget13.2forNa,which means thatthedielectric constant for theliquid isnegative, while theobserved value ofKis80.Theproblem hastodo withthecorrect treatment ofthepermanent dipoles, andOnsager haspointed out therightwaytogo.Wedonothavethetimetotreatthecasenow, butifyouare interested itisdiscussed inKittel’s book, Introduction toSolid State Physics. ll-7 Q)©®@G)6)®©® ®G)®(D@®®®69$ G)(DG)G)@®©®®® ®®®(9G)G)69G9®6)®@(D®@®C)G)C) G)G) CD(D (D C)(DG) CDCD G)CD___ ‘ I ___ | : I I : l I | I | | r Fig. 11-8. Acomplex crystal lattice canhave apermanent intrinsic polariza- tionP. _/_-_.®Wé /—@ § @""'j\*\ ’// .\/ 40> era" Oea" @0'2 Fig. 11-9. The unit cell ofBaTiO3. Theatoms really ﬁllupmost ofthespace; forclarity, only the positions oftheir centers areshown.11-6 Solid dielectrics Now weturntothesolids. Theﬁrstinteresting factabout solids isthatthere canbeapermanent polarization builtin—which exists evenwithout applying an electric ﬁeld. Anexample occurs withamaterial likewax, which contains long molecules having apermanent dipole moment. Ifyoumeltsome waxandputa strong electric ﬁeld onitwhen itisaliquid, sothatthedipole moments getpartly lined up,theywillstaythatwaywhen theliquid freezes. Thesolid material will have apermanent polarization which remains when theﬁeld isremoved. Such a solid iscalled anelectret. Anelectret haspermanent polarization charges onitssurface. Itistheelectrical analog ofamagnet. Itisnotasuseful, though, because freecharges from theair areattracted toitssurfaces, eventually cancelling thepolarization charges. The electret is“discharged” andthere arenovisible external ﬁelds. l Apermanent internal polarization Pisalsofound occurring naturally insome crystalline substances. Insuchcrystals, eachunitcellofthelattice hasanidentical permanent dipole moment, asdrawn inFig.11-8. Allthedipoles point inthesame direction, even with noapplied electric ﬁeld. Many complicated crystals have, in fact, such apolarization; wedonotnormally notice itbecause theexternal ﬁelds aredischarged, justasfortheelectrets. Ifthese internal dipole moments ofacrystal arechanged, however, external ﬁelds appear because there isnottime forstray charges togather andcancel the polarization charges. Ifthedielectric isinacondenser, freecharges willbeinduced ontheelectrodes. Forexample, themoments canchange when adielectric is heated, because ofthermal expansion. Theeﬂ'ect iscalled pyroelectricity. Similarly, ifwechange thestresses inacrysta1——for instance, ifwebend it—again themo- ment may change alittle bit,andasmall electrical effect, called piezoelectricity, canbedetected. Forcrystals thatdonothaveapermanent moment, onecanwork outatheory ofthedielectric constant thatinvolves theelectronic polarizability oftheatoms. Itgoesmuch thesame asforliquids. Some crystals alsohave rotatable dipoles inside, andtherotation ofthese dipoles willalsocontribute toK.Inionic crystals such asNaCl there isalsoionicpolarizability. Thecrystal consists ofacheckerboard ofpositive andnegative ions, andinanelectric ﬁeldthepositive ionsarepulled onewayandthenegatives theother; there isanetrelative motion oftheplusand minus charges, andsoavolume polarization. Wecould estimate themagnitude oftheionic polarizability from ourknowledge ofthestiffness ofsaltcrystals, but wewillnotgointothatsubject here. 11-7 Ferroelectricity; BaTi03 Wewant todescribe nowonespecial class ofcrystals which have, justby accident almost, abuilt-in permanent moment. Thesituation issomarginal that ifweincrease thetemperature alittle bitthey losethepermanent moment com- pletely. Ontheother hand, ifthey arenearly cubic crystals, sothattheir moments canbeturned indifferent directions, wecandetect alarge change inthemoment when anapplied electric ﬁeldischanged. Allthemoments ﬂipoverandwegeta large effect. Substances which have thiskind ofpermanent moment arecalled ferroelectric, after thecorresponding ferromagnetic effects which were ﬁrstdis- covered iniron. Wewould liketoexplain howferroelectricity works bydescribing aparticular example ofaferroelectric material. There areseveral ways inwhich theferro- electric property canoriginate; butwewilltakeuponlyonemysterious case——that ofbarium titanate, BaTiO3. Thismaterial hasacrystal lattice whose basic cellis sketched inFig. ll—9. Itturns outthatabove acertain temperature, speciﬁcally 118°C, barium titanate isanordinary dielectric with anenormous dielectric con- stant. Below thistemperature, however, itsuddenly takes onapermanent moment. Inworking outthepolarization ofsolid material, wemust ﬁrstﬁndwhat are thelocal ﬁelds ineach unitcell. Wemust include theﬁelds from thepolarization ll-8 itself, justaswedidforthecaseofaliquid. Butacrystal isnotahomogeneous liquid, sowecannot useforthelocal ﬁeldwhat wewould getinaspherical hole. Ifyouwork itoutforacrystal, youﬁndthatthefactor 1/3inEq.(11.24) becomes slightly different, butnotfarfrom 1/3.(Forasimple cubic crystal, itisjust1/3.) Wewill,therefore, assume forourpreliminary discussion thatthefactor is1/3 forBaTiO3. Now when wewrote Eq.(11.28) youmayhavewondered what would happen ifNabecame greater than3.Itappears asthough Kwould become negative. But thatsurely cannot beright. Let’s seewhat should happen ifweweregradually to increase ainaparticular crystal. Asagetslarger, thepolarization getsbigger, making abigger local ﬁeld. Butabigger local ﬁeld willpolarize each atom more, raising thelocal ﬁelds stillmore. Ifthe“give” oftheatoms isenough, theprocess keeps going; there isakind offeedback thatcauses thepolarization toincrease without limit—-assuming thatthepolarization ofeach atom increases inproportion totheﬁeld. The“runaway” condition occurs when Na=3.Thepolarization doesnotbecome inﬁnite, ofcourse, because theproportionality between thein- duced moment andtheelectric ﬁeldbreaks down athighﬁelds, sothatourformulas arenolonger correct. What happens isthatthelattice gets“locked in”withahigh, self-generated, internal polarization. InthecaseofBaTiO3, there is,inaddition toanelectronic polarization, also arather large ionic polarization, presumed tobeduetotitanium ionswhich can move alittlewithin thecubic lattice. Thelattice resists large motions, soafterthe titanium hasgone alittle way, itjams upandstops. Butthecrystal cellisthen left with apermanent dipole moment. Inmost crystals, thisisreally thesituation foralltemperatures thatcanbe reached. Theveryinteresting thing about barium titanate isthatthere issucha delicate condition thatifNaisdecreased justalittlebititcomes unstuck. Since Ndecreases withincreasing temperature—because ofthermal expansion—we can vary Nabyvarying thetemperature. Below thecritical temperature itisjust barely stuck, soitiseasy—-by applying anexternal ﬁeld—to shiftthepolarization andhaveitlockinadifferent direction. Let’s seeifwecananalyze what happens inmore detail. WecallT,thecritical temperature atwhich Naisexactly 3.Asthetemperature increases, Ngoes down a littlebitbecause oftheexpansion ofthelattice. Since theexpansion issmall, we can’saythatnearthecritical temperature Na=3_/3(T-T,), (11.30) where Hisasmall constant, ofthesame order ofmagnitude asthethermal expansion coefﬁcient, orabout 10_5 tol0_6 perdegree C.Now ifwesubstitute thisrelation intoEq.(11.28), wegetthat K_.1= . 5(T—T¢)/3 Since wehave assumed thatB(T—Tc)issmall compared withone,wecanap- proximate thisformula by 9 This relation isright, ofcourse, only forT>Tc.Weseethatjustabove the critical temperature Kisenormous. Because Naissoclose to3,there isatremen- dousmagniﬁcation effect, andthedielectric constant caneasily beashighas50,000 to100,000. Itisalsoverysensitive totemperature. Forincreases intemperature, thedielectric constant goesdown inversely asthetemperature, but,unlike thecase ofadipolar gas,forwhich K—1goes inversely astheabsolute temperature, for ferroelectrics itvaries inversely asthedifference between theabsolute temperature andthecritical temperature (thislawiscalled theCurie-Weiss law). When welower thetemperature tothecritical temperature, what happens? Ifweimagine alattice ofunitcellslikethatinFig.11-9, weseethatitispossible 11-9 .t*’°—”!“Ti ‘l’ Lt t i’'t iAt t-t t+t 4» i l>+l> t l l+t(bl Fig. 11-10. Models ofaferroelec- tric: la)corresponds toanantiferro- electric, and(b)toanormal ferroelectric.topickoutchains ofionsalong vertical lines. Oneofthem consists ofalternating oxygen andtitanium ions. There areother lines made upofeither barium or oxygen ions,butthespacing along these linesisgreater. Wemake asimple model toimitate thissituation byimagining, asshown inFig.ll~l0(a), aseries ofchains ofions. Along what wecallthemain chain, theseparation oftheionsisa,which ishalfthelattice constant; thelateral distance between identical chains is2a. There areless-dense chains inbetween which wewillignore forthemoment. To make theanalysis alittleeasier, wewillalsosuppose thatalltheionsonthemain chain areidentical. (Itisnotaserious simpliﬁcation because alltheimportant effects willstillappear. This isoneofthetricks oftheoretical physics. Onedoes adifferent problem because itiseasier toﬁgure outtheﬁrsttime——then when one understands how thething works, itistime toputinallthecomplications.) Now let’strytoﬁndoutwhat would happen with ourmodel. Wesuppose that thedipole moment ofeach atom ispandwewish tocalculate theﬁeld atoneof theatoms ofthechain. Wemustﬁndthesumoftheﬁelds from alltheother atoms. Wewillﬁrstcalculate theﬁeldfrom thedipoles inonlyonevertical chain; wewill talkabout theother chains later. Theﬁeldatthedistance rfrom adipole ina direction along itsaxisisgiven by _121> E-4M0 —r-§- (11.32) Atanygiven atom, thedipoles atequal distances above andbelow itgiveﬁelds in thesame direction, soforthewhole chain weget Ea...= +§+§+ ={i09"j¥- <11-33> Itisnottoohardtoshow thatifourmodel were likeacompletely cubic crystal- thatis,ifthenextidentical lineswereonlythedistance aaway—the number 0.383 would bechanged to1/3.Inother words, ifthenextlineswereatthedistance a theywould contribute only -0.050 unittooursum. However, thenextmain chain weareconsidering isatthedistance 2aand,asyouremember fromChapter 7, theﬁeld from aperiodic structure diesoffexponentially with distance. Therefore these linescontribute much lessthan -0.050 andwecanjustignore alltheother chains. Itisnecessary nowtoﬁndoutwhat polarizability ozisneeded tomake the runaway process work. Suppose thattheinduced moment pofeach atom ofthe chain isproportional totheﬁeld onit,asinEq.(11.6). Wegetthepolarizing ﬁeld ontheatom from Ectwm, using Eq.(11.32). Sowehavethetwoequations P=a€0Echa.in and . 0.383pEchain =T :0' There aretwosolutions: Eandpboth zero, or as a=———,0.383 withEandpbothﬁnite. Thus ifozisaslarge asa3/0.383, apermanent polarization sustained byitsownﬁeldwillsetin.Thiscritical equality must bereached for barium titanate atjustthetemperature Tc.(Notice thatifawerelarger thanthe critical value forsmall ﬁelds, itwould decrease atlarger ﬁelds andatequilibrium thesame equality wehave found would hold.) ForBaTiO3, thespacing ais2X10-8 cm,sowemust expect thatat= 21.8 X10*“ cm3. Wecancompare thiswith theknown polarizabilities ofthe individual atoms. Foroxygen, oz=30.2 X10*“ cma; we’re ontheright track! Butfortitanium, a=2.4Xl0_2‘ cm3;rather small. Touseourmodel weshould probably taketheaverage. (Wecould work outthechain again foralternating 11-10 atoms, buttheresult would beabout thesame.) Soa(average) =16.3X10*“, which isnothigh enough togiveapermanent polarization. Butwait amoment! Wehave sofaronly added uptheelectronic polariz- abilities. There isalsosome ionic polarization duetothemotion ofthetitanium ion.Allweneedisanionic polarizability of9.2X10*“ cm3. (Amore precise computation using alternating atoms shows thatactually 11.9X10"“ isneeded.) Tounderstand theproperties ofBaTiO3, wehave toassume that such anionic polarizability exists. Why thetitanium ioninbarium titanate should have thatmuch ionic polar- izability isnotknown. Furthermore, why,atalower temperature, itpolarizes along thecube diagonal andthefacediagonal equally wellisnotclear. Ifweﬁgure out theactual sizeofthespheres inFig.11-9, andaskwhether thetitanium isalittle bitloose intheboxformed byitsneighboring oxygen atoms—which iswhat you would hope, sothatitcould beeasily shifted-—you ﬁndquite thecontrary. Itﬁts very tightly. Thebarium atoms areslightly loose, butifyouletthem betheones thatmove, itdoesn’t work out.Soyouseethatthesubject isreally notone-hundred percent clear; there arestillmysteries wewould liketounderstand. Returning tooursimple model ofFig.ll-l0(a), weseethattheﬁeldfrom one chain would tendtopolarize theneighboring chain intheopposite direction, which means thatalthough eachchain would belocked, there would benonetpermanent moment perunitvolume! (Although there would benoexternal electric effects, there arestillcertain thermodynamic effects onecould observe.) Such systems exist, andarecalled antiferroelectric. Sowhat wehave explained isreally ananti- ferroelectric. Barium titanate, however, isreally likethearrangement inFig. 1l—10(b). The oxygen-titanium chains areallpolarized inthesame direction because there areintermediate chains ofatoms inbetween. Although theatoms inthese chains arcnotvery polarizable, orvery dense, they willbesomewhat polarized, inthedirection antiparallel totheoxygen-titanium chains. Thesmall ﬁelds produced atthenextoxygen-titanium chain willgetitstarted parallel tothe ﬁrst. SoBaTiO3 isreally ferroelectric, anditisbecause oftheatoms inbetween. Youmaybewondering: “But what about thedirect effect between thetwoO-Ti chains?” Remember, though, thedirect effect diesoffexponentially with the separation; theeffect ofthechain ofstrong dipoles at2acanbelessthantheeffect ofachain ofweak ones atthedistance a. This completes ourrather detailed report onourpresent understanding ofthe dielectric constants ofgases, ofliquids, andofsolids. ll-ll I2 Electrostatic Analogs 12-1 Thesame equations havethesame solutions Thetotal amount ofinformation which hasbeen acquired about thephysical world since thebeginning ofscientiﬁc progress isenormous, anditseems almost impossible thatanyoneperson could know areasonable fraction ofit.Butitis actually quite possible foraphysicist toretain abroad knowledge ofthephysical world rather than tobecome aspecialist insome narrow area. Thereasons for thisarethreefold: First, there aregreat principles which apply toallthedifferent kinds ofphenomena—such astheprinciples oftheconservation ofenergy andof angular momentum. Athorough understanding ofsuch principles gives anunder- standing ofagreat dealallatonce. Second, there isthefactthatmany compli- cated phenomena, such asthebehavior ofsolids under compression, really basically depend onelectrical and quantum-mechanical forces, sothat ifone understands thefundamental laws ofelectricity andquantum mechanics, there is atleast some possibility ofunderstanding many ofthephenomena thatoccur incomplex situations. Finally, there isamost remarkable coincidence: The equations formany diﬂerent physical situations have exactly thesame appearance. Ofcourse, thesymbols may bedifferent—0ne letter issubstituted foranother— butthemathematical form oftheequations isthesame. This means thathaving studied onesubject, weimmediately have agreat deal ofdirect andprecise knowledge about thesolutions oftheequations ofanother. Wearenowﬁnished withthesubject ofelectrostatics, andwillsoon goonto study magnetism andelectrodynamics. Butbefore doing so,wewould liketo show that while learning electrostatics wehave simultaneously learned about a large number ofother subjects. Wewillﬁndthattheequations ofelectrostatics appear inseveral other places inphysics. Byadirect translation ofthesolutions (ofcourse thesame mathematical equations must have thesame solutions) itis possible tosolve problems inother ﬁelds withthesame ease——or withthesame difﬁculty—as inelectrostatics. Theequations ofelectrostatics, weknow, are v-(KE)= (12.1) VXE=0. (12.2) (Wetaketheequations ofelectrostatics withdielectrics soastohave themost general situation.) Thesame physics canbeexpressed inanother mathematical form: E=—V¢, (12.3) v-(Kv¢)=- (12.4) Now thepoint isthat there aremany physics problems whose mathematical equations havethesame form. There isapotential (¢)whose gradient multiplied byascalar function (K)hasadivergence equal toanother scalar function (—p/60). Whatever weknow about electrostatics canimmediately becarried over into thatother subject, andviceversa. (Itworks both ways, ofcourse—if theother subject hassome particular characteristics thatareknown, then wecanapply thatknowledge tothecorresponding electrostatic problem.) Wewant toconsider aseries ofexamples from different subjects thatproduce equations ofthisform. 12-112-1 Thesame equations have the same solutions 12-2 Theﬂowofheat; apoint source near aninﬁnite plane boundary 12-3 Thestretched membrane 12-4 Thediﬂ'usion ofneutrons; a uniform spherical source ina homogeneous medium 12-5 Irrotational ﬂuid ﬂow; the ﬂowpast asphere 12—6 Illumination; thetmiform lighting ofaplane 12-7 The“underlying unity” of nature 12-2 Theflowofheat; apoint source nearaninﬁnite plane boundary Wehave discussed oneexample earlier (Section 3—4)—the flow ofheat. Imagine ablock ofmaterial, which need notbehomogeneous butmay consist of different materials atdifferent places, inwhich thetemperature varies from point topoint. Asaconsequence ofthese temperature variations there isaﬂow ofheat, which canberepresented bythevector h.Itrepresents theamount ofheat energy which ﬂows perunittime through aunitarea perpendicular totheﬂow. Thedi- vergence ofhrepresents therateperunitvolume atwhich heatisleaving aregion: V-h=rateofheat outperunitvolume. (Wecould, ofcourse, write theequation inintegral form—just aswedidinelectro- statics with Gauss’ law—which would saythattheﬂuxthrough asurface isequal totherateofchange ofheat energy inside thematerial. Wewillnotbother to translate theequations back andforth between thedifferential andtheintegral forms, because itgoes exactly thesame asinelectrostatics.) Therateatwhich heat isgenerated orabsorbed atvarious places depends, of course, ontheproblem. Suppose, forexample, thatthere isasource ofheat inside thematerial (perhaps aradioactive source, oraresistor heated byanelectrical current). Letuscallstheheat energy produced perunitvolume persecond by thissource. There mayalsobelosses (orgains) ofthermal energy toother internal energies inthevolume. Ifuistheinternal energy perunitvolume, —du/dz will alsobea“source” ofheat energy. Wehave, then, v-h=s-f,_‘:- (12.5) Wearenotgoing todiscuss justnow thecomplete equation inwhich things change with time, because wearemaking ananalogy toelectrostatics, where no- thing depends onthetime. Wewillconsider only steady heat-ﬂow problems, in which constant sources haveproduced anequilibrium state. Inthese cases, V-h =s. (12.6) Itis,ofcourse, necessary tohave another equation, which describes how the heat ﬂows atvarious places. Inmany materials theheat current isapproximately proportional totherateofchange ofthetemperature with position: thelarger the temperature difference, themore theheat current. Aswehave seen, thevector heat current isproportional tothetemperature gradient. Theconstant ofpro- portionality K,aproperty ofthematerial, iscalled thethermal conductivity. h=——KVT. (12.7) Iftheproperties ofthematerial vary from place toplace, then K=K(x, y,z),a function ofposition. [Equation (12.7) isnotasfundamental as(12.5), which expresses theconservation ofheatenergy, since theformer depends upon aspecial property ofthesubstance.] Ifnowwesubstitute Eq.(12.7) intoEq.(12.6) wehave V-(KVT) =—s, (12.8) which hasexactly thesame form as(12.4). Steady heat-ﬂow problems andelectro- static problems arethesame. Theheat ﬂow vector hcorresponds toE,andthe temperature Tcorresponds to¢.Wehave already noticed thatapoint heat source produces atemperature ﬁeld which varies asl/randaheat ﬂow which varies as 1/r2. Thisisnothing more thanatranslation ofthestatements from electrostatics thatapoint charge generates apotential which varies asl/randanelectric ﬁeld which varies as1/r2. Wecan, ingeneral, solve static heat problems aseasily as wecansolve electrostatic problems. Consider asimple example. Suppose thatwehave acylinder ofradius aatthe temperature T1,maintained bythegeneration ofheatinthecylinder. (Itcould be, forexample, awire carrying acurrent, orapipe with steam condensing inside.) 12-2 Thecylinder iscovered withaconcentric sheath ofinsulating material which hasa conductivity K.Saytheoutside radius oftheinsulation isbandtheoutside is kept attemperature T2(Fig. l2—la). Wewant toﬁndoutatwhat rateheat will belostbythewire, orsteampipe, orwhatever itisinthecenter. Letthetotal amount ofheatlostfrom alength Lofthepipebecalled G——which iswhat weare trying toﬁnd. How canwesolve thisproblem‘? Wehavethedifferential equations, butsince these arethesame asthose ofelectrostatics, wehave really already solved the mathematical problem. Theanalogous problem isthatofaconductor ofradius a atthepotential ¢>1,separated from another conductor ofradius batthepotential 4:2,with aconcentric layer ofdielectric material inbetween, asdrawn inFig. 12-1(b).Now since theheatﬂowhcorresponds totheelectric ﬁeldE,thequantity Gthatwewant toﬁndcorresponds totheﬂuxoftheelectric ﬁeldfrom aunit length (inother words, totheelectric charge perunitlength over so). Wehave solved theelectrostatic problem byusing Gauss’ law. Wefollow thesame pro- cedure forourheat-ﬂow problem. From thesymmetry ofthesituation, weknow that hdepends only onthe distance from thecenter. Soweenclose thepipeinagaussian cylinder oflength Landradius r.From Gauss’ law, weknow that theheat ﬂow hmultiplied by thearea 21rrL ofthesurface must beequal tothetotal amount ofheat generated lnSldC, which iswhat wearecalling G: 21rrLh =G or h= (12.9) Theheatﬂowisproportional tothetemperature gradient: h=—KVT, or,inthiscase, themagnitude ofhis dTh= —K E" ' This, together with(12.9), gives dT G Integrating from r=ator=b,weget G bTg—T1=— lHz' Solving forG,weﬁnd Thisresult corresponds exactly totheresult forthecharge onacylindrical conden- ser: Q=27l'60L(<l>1 —¢2)_ ln(b/a) Theproblems arethesame, andtheyhavethesame solutions. From ourknowledge ofelectrostatics, wealsoknow howmuch heatislostbyaninsulated pipe. Let’s consider another example ofheat ﬂow. Suppose wewish toknow the heatflowintheneighborhood ofapoint source ofheatlocated alittlewaybeneath thesurface oftheearth, ornearthesurface ofalarge metal block. Thelocalized heatsource might beanatomic bomb thatwassetoffunderground, leaving an intense source ofheat, oritmight correspond toasmall radioactive source inside ablock ofiron—-—there arenumerous possibilities. Wewilltreattheidealized problem ofapoint heatsource ofstrength Gatthe distance abeneath thesurface ofaninﬁnite block ofuniform material whose thermal conductivity isK.Andwewillneglect thethermal conductivity ofthe 12-3III"" '0 Q0 ""0011 /?/\(0) III:0,.’ o .."'IIl ‘l’ \\.KQ;(bl Fig.12-1. la)Heat ﬂow inacylin- drical geometry. (b)Thecorresponding electrical problem. \O *-\___/// I/ /I, / / / /a’,, \§k//’»-———->- //\‘\/\\\\\\\\ \\\\ \\‘\\\\\ t-1\‘\u\O\ \\ \ \ \>-_ __» \\ / __, \_,¢ \ ___:/ \\ ‘ ,/ .II\///7/ / // ///7 .. -‘ -Eta:0;‘:TIConstan - / . T SURFACE TEMPERATURE 0 a a p Fig. 12-2. The heat ﬂow and iso- thermals near apoint heat source atthe distance abelow thesurface ofagood thermal conductor. Animage source is shown outside thematerial.airoutside thematerial. Wewant todetermine thedistribution ofthetemperature onthesurface oftheblock. How hotisitright above thesource andatvarious places onthesurface oftheblock? How shallwesolve it?Itislikeanelectrostatic problem withtwomaterials with different dielectric coefficients tconopposite sides ofaplane boundary. Aha! Perhaps itistheanalog ofapoint charge neartheboundary between adielectric andaconductor, orsomething similar. Let’s seewhat thesituation isnearthe surface. Thephysical condition isthatthenormal component ofhonthesurface iszero, since wehave assumed there isnoheat ﬂow outoftheblock. Weshould ask: Inwhat electrostatic problem dowehave thecondition that thenormal component oftheelectric ﬁeld E(which istheanalog ofh)iszero atasurface‘? There isnone! Thatisoneofthethings thatwehavetowatch outfor.Forphysical reasons, there may becertain restrictions inthekinds ofmathematical conditions which arise inanyonesubject. Soifwehave analyzed thedifferential equation only for certain limited cases, wemayhavemissed some kinds ofsolutions thatcanoccur inother physical situations. Forexample, there isnomaterial withadielectric constant ofzero, whereas avacuum does have zerothermal conductivity. Sothere isnoelectrostatic analogy foraperfect heat insulator. Wecan,however, stilluse thesame methods. Wecantrytoimagine what would happen ifthedielectric constant were zero. (Ofcourse, thedielectric constant isnever zero inanyreal situation. Butwemight have acaseinwhich there isamaterial with avery high dielectric constant, sothatwecould neglect thedielectric constant oftheairout- side.) How shall weﬁndanelectric ﬁeld thathasnocomponent perpendicular to thesurface? That is,onewhich isalways tangent atthesurface? You willnotice thatourproblem isopposite totheoneofapoint charge near aplane conductor. There wewanted theﬁeldtobeperpendicular tothesurface, because theconductor wasallatthesame potential. Intheelectrical problem, weinvented asolution byimagining apoint charge behind theconducting plate. Wecanusethesame ideaagain. Wetrytopickan“image source” thatwillautomatically make the normal component oftheﬁeld zero atthesurface. The solution isshown in Fig.l2—2. Animage source ofthesame signandthesame strength placed atthe distance aabove thesurface willcause theﬁeld tobealways horizontal atthesur- face. Thenormal components ofthetwosources cancel out. Thus ourheatﬂowproblem issolved. Thetemperature everywhere isthe same, bydirect analogy, asthepotential duetotwoequal point charges! The temperature Tatthedistance rfrom asingle point source Ginaninﬁnite medium is GT-g (12.13) (This, ofcourse, isjusttheanalog of¢=q/41re0r.) Thetemperature forapoint source, together with itsimage source, is T=%+ %2- (12.14) This formula gives usthetemperature everywhere intheblock. Several isothermal surfaces areshown inFig.12-2. Also shown arelines ofIt,which canbeobtained from h=—KVT. Weoriginally asked forthetemperature distribution onthesurface. Fora point onthesurface atthedistance pfrom theaxis,r1=r2=\/p2 +a2,so T(surface) =Z;-K (12.15)p a This function isalsoshown intheﬁgure. Thetemperature is,naturally, higher right above thesource than itisfarther away. This isthekind ofproblem that geophysicists often need tosolve. Wenow seethatitisthesame kind ofthing we have already been solving forelectricity. 12-4 12-3 Thestretched membrane Now letusconsider acompletely different physical situation which, never- theless, gives thesame equations again. Consider athinrubber sheet—a membrane ——which hasbeen stretched over alarge horizontal frame (like adrumhead). Suppose now thatthemembrane ispushed upinoneplace anddown inanother; asshown inFig.12-3. Canwedescribe theshape ofthesurface? Wewillshow how theproblem canbesolved when thedeﬂections ofthemembrane arenottoo large. There areforces inthesheet because itisstretched. Ifwewere tomake a small cutanywhere, thetwosides ofthecutwould pullapart (seeFig.12-4). So there isasurface tension inthesheet, analogous totheone-dimensional tension inastretched string. Wedeﬁne themagnitude ofthesurface tension 1-astheforce perunitlength which willjusthold together thetwosides ofacutsuch asoneof those shown inFig. 12-4. Suppose now that welook atavertical cross section ofthemembrane. It willappear asacurve, liketheoneinFig.l2—5. Letubethevertical displacement ofthemembrane from itsnormal position, andxandythecoordinates inthe horizontal plane. (The cross section shown isparallel tothex-axis.) Consider alittle piece ofthesurface oflength Axandwidth Ay.There willbe forces onthepiece from thesurface tension along each edge. The force along edge loftheﬁgure willbe11Ay,directed tangent tothesurface—-that is,atthe angle 01from thehorizontal. Along edge 2,theforce willbe12Ayattheangle 02. (There willbesimilar forces ontheother twoedges ofthepiece, butwewillforget them forthemoment.) Thenetupward force onthepiece from edges land2is AF=1'2Aysin02—1'1Aysin61. Wewilllimit ourconsiderations tosmall distortions ofthemembrane, i.e.,to small slopes: wecanthenreplace sin0bytan0,which canbewritten asBu/6x. The force isthen 6u 6u AF- “'1“Thequantity inbrackets canbeequally wellwritten (forsmall Ax)as iT9!Ax‘6x 6x ’ 6 6uAF-5(T5)AxAy. There willbeanother contribution toAFfrom theforces ontheother two edges; thetotalisevidently AF= <1" —l—%<1’ AxAy. (12.16) Thedistortions ofthediaphragm arecaused byexternal forces. Let’s let frepresent theupward force perunitarea onthesheet (akind of“pressure”) from theexternal forces. When themembrane isinequilibrium (thestatic case), thisforce must bebalanced bytheinternal force wehave justcomputed, Eq (l2.l6). That isthen AF f-"Ty Equation (12.16) canthen bewritten f= —V-(1-Vu), (l2.l7) where byVwenow mean, ofcourse, thetwo-dimensional gradient operator (6/6x, 6/By). Wehavethedifferential equation thatrelates u(x,y)totheapplied 12-5.111? \--I-2-"fe*=‘.-.a‘§§\\‘p‘¢\,if-‘.ir1i'i'wi§\;=—-‘e-‘.‘$r\& 7 Z -/ Fig. 12-3. Athin rubber sheet stretched over 0cylindrical frame (like cldrumhead). Ifthesheet ispushed up atAand down otB,what istheshope ofthesurface? \\\\\\Fig. l2-4. Thesurface tension 1'of ostretched rubber sheet istheforce per unitlength across oline. 92 T2 IAx2 A 9 SHEET 1.’ u __.> X Fig.12-5. Cross section ofthede- ﬂected sheet. forces f(x,y)andthesurface tension -r(x,y),which may, ingeneral, varyfrom place toplace inthesheet. (The distortions ofathree-dimensional elastic body are alsogoverned bysimilar equations, butwewillstick totwo-dimensions.) We willworry only about thecase inwhich thetension 1'isconstant throughout the sheet. Wecanthen write forEq.(12.17), Vzu=- (12.18) Wehave another equation that isthesame asforelectrostatics!—only this time, limited totwo-dimensions. Thedisplacement ucorresponds to¢,andf/1' corresponds top/co.Soallthework wehave done forinﬁnite plane charged sheets, orlong parallel wires, orcharged cylinders isdirectly applicable tothestretched membrane. Suppose wepush themembrane atsome points uptoadeﬁnite height—that is, weﬁxthevalue ofuatsome places. Thatistheanalog ofhaving adeﬁnite potential atthecorresponding places inanelectrical situation. So,forinstance, wemay make apositive “potential” bypushing uponthemembrane withanobject having thecross-sectional shape ofthecorresponding cylindrical conductor. Forexample, ifwepush thesheet upwith around rod,thesurface willtake ontheshape shown inFig.12-6. Theheight uisthesame astheelectrostatic potential abofacharged cylindrical rod. Itfalls offasln(1/r). (The slope, which corresponds tothe electric ﬁeld E,drops offasl/r.) LlFig. 12-6. Cross section of a stretched rubber sheet pushed upbya round rod. Thefunction u(x,y)isthesame astheelectric potential ¢lx,y) near a very longcharged rod. Thestretched rubber sheet hasoften been used asawayofsolving complicated electrical problems experimentally. Theanalogy isused backwards! Various rods andbars arepushed against thesheet toheights thatcorrespond tothepo- tentials ofasetofelectrodes. Measurements oftheheight then givetheelectrical potential fortheelectrical situation. Theanalogy hasbeen carried even further. Iflittle balls areplaced onthemembrane, their motion corresponds approximately tothemotion ofelectrons inthecorresponding electric ﬁeld. Onecanactually watch the“electrons” move ontheirtrajectories. Thismethod wasusedtodesign thecomplicated geometry ofmany photomultiplier tubes (such astheones used forscintillation counters, andtheoneused forcontrolling theheadlight beams on Cadillacs). Themethod isstillused, buttheaccuracy islimited. Forthemost accurate work, itisbetter todetermine theﬁelds bynumerical methods, using the large electronic computing machines. 12-4 Thediffusion ofneutrons; auniform spherical source inahomogeneous medium Wetake another example that gives thesame kind ofequation, thistime having todowith diffusion. InChapter 43ofVol. Iweconsidered thediffusion ofions inasingle gas,andofonegasthrough another. This time, let’s take a different example—the diffusion ofneutrons inamaterial likegraphite. Wechoose tospeak ofgraphite (apure form ofcarbon) because carbon doesn’t absorb slow neutrons. Inittheneutrons arefreetowander around. They travel inastraight lineforseveral centimeters, ontheaverage, before being scattered byanucleus anddeﬂected intoanewdirection. Soifwehave alarge block—many meters on aside—the neutrons initially atoneplace willdiffuse toother places. Wewant to ﬁndadescription oftheir average behavior—that is,their average ﬂow. 12-6 LetN(x, y,z)AVbethenumber ofneutrons intheelement ofvolume AV atthepoint (x,y,z).Because oftheir motion, some neutrons willbeleaving AV, andothers willbecoming in.Ifthere aremore neutrons inoneregion thanina nearby region, more neutrons willgofrom theﬁrstregion tothesecond thancome back; there willbeanetﬂow. Following thearguments ofChapter 43inVol. I, wedescribe theﬂow byaﬂow vector J.Itsx-component J,isthenetnumber of neutrons thatpassinunittimeaunitareaperpendicular tothex-direction. We found that azvJ,_-p5;. (12.19) where thediffusion constant Disgiven interms ofthemean velocity v,andthe mean-free-path lbetween scatterings isgiven by lD—3-ll}. Thevector equation forJis J=—DVN. (12.20) Therateatwhich neutrons ﬂow across anysurface element daisJ-nda (where, asusual, nistheunitnormal). Thenetﬂowoutofavolume element isthen (following theusual gaussian argument) V-JdV. This flow would result in adecrease withtimeofthenumber inAVunless neutrons arebeing created in AV(bysome nuclear process). Ifthere aresources inthevolume thatgenerate S neutrons perunittime inaunitvolume, then thenetﬂow outofAVwillbeequal to(S—6N/61) AV. Wehave then that 6NVJ-S-3? (12.21) Combining (12.21) with(12.20), wegettheneutron diﬂusion equation v~(-0vzv)=s- (12.22) Inthestatic case-—where 6N/6t =0—we have Eq.(12.4) allover again! Wecanuseourknowledge ofelectrostatics tosolve problems about thediffusion ofneutrons. Solet’ssolve aproblem. (You may wonder: Why doaproblem if wehavealready done alltheproblems inelectrostatics? Wecandoitfaster this timebecause wehavedone theelectrostatic problems!) Suppose wehaveablock ofmaterial inwhich neutrons arebeing generated- saybyuranium ﬁssion—uniformly throughout aspherical region ofradius a (Fig. 12-7). Wewould liketoknow: What isthedensity ofneutrons everywhere? How uniform isthedensity ofneutrons intheregion where they arebeing gen- erated ‘?What istheratio oftheneutron density atthecenter totheneutron density atthesurface ofthesource region? Finding theanswers iseasy. Thesource density S0replaces thecharge density p,soourproblem isthesame astheproblem ofasphere ofuniform charge density. Finding Nisjustlikeﬁnding thepotential ¢.Wehave already worked outtheﬁelds inside andoutside ofauniformly charged sphere; wecanintegrate them togetthepotential. Outside, thepotential is Q/41re0r, with thetotal charge Qgiven by41ra3p/3.So ._£21. ¢outs1de "360', Forpoints inside, theﬁeldisdueonlytothecharge Q(r)inside thesphere ofradius r,Q(r) =41rr3p/3, so _ll’.. E_360 (12.24) l2-7//\ \siuu>:11rs\\ \ §s:tj.%2.../ \t\'“$%%j"jj\ \\-.\_\_;\I -Ls__ Z QQ____ -\ .1.ELECTRIC /',\FIELD /--. r.- '/"E-\. /./~/,\..\"/‘‘____\__l / \ 4* I l l O U I’ lb) Fig. l2-7. la)Neutrons areproduced uniformly throughout asphere ofradius a inalarge graphite block and diffuse outward. Theneutron density Nisfound asafunction ofr,thedistance from the center ofthesource. lb)Theanalogous electrostatic situation: auniform sphere of charge, where Ncorresponds to¢and J corresponds toE. Theﬁeldincreases linearly withr.Integrating Etoget¢,wehave 9'2 ¢i,,,(de =—6?+aconstant.0 Attheradius a,¢,,,,,de must bethesame as¢.,ut,ide, sotheconstant must be pa2/2&0. (Weareassuming that¢iszeroatlarge distances from thesource, which willcorrespond toNbeing zero fortheneutrons.) Therefore, 32 2 qsinside =£ _ ' Weknow immediately theneutron density inourother problem. Theanswer is S3 Noutside =% ’ and s3;? r2Ninside _E _ ' Nisshown asafunction ofrinFig. 12-7. Nowwhatistheratioofdensity atthecenter tothatattheedge? Atthecenter (r=0),itisproportional to3a2/2. Attheedge (r=a)itisproportional to 2a2/2,sotheratioofdensities is3/2.Auniform source doesn’t produce auniform density ofneutrons. Yousee,ourknowledge ofelectrostatics gives usagood start onthephysics ofnuclear reactors. There aremany physical circumstances inwhich diffusion plays abigpart. Themotion ofions through aliquid, orofelectrons through asemiconductor, obeys thesame equation. Weﬁndagain andagain thesame equations. 12-5 Irrotational ﬂuid ﬂow; theﬂowpast asphere Let’s nowconsider anexample which isnotreally averygood one,because theequations wewillusewillnotreally represent thesubject with complete generality butonlyinanartiﬁcial idealized situation. Wetakeuptheproblem ofwater ﬂow. Inthecaseofthestretched sheet, ourequations were anapproxima- tionwhich wascorrect only forsmall deﬂections. Forourconsideration ofwater ﬂow, wewillnotmake thatkindofanapproximation; wemust make restrictions thatdonotapply atalltorealwater. Wetreatonlythecaseofthesteady ﬂowof anincompressible, nonviscous, circulation-free liquid. Then werepresent theﬂow bygiving thevelocity v(r)asafunction ofposition r.Ifthemotion issteady (theonlycaseforwhich there isanelectrostatic analog) visindependent oftime. Ifpisthedensity oftheﬂuid, thenpvistheamount ofmass which passes perunit timethrough aunitarea. Bytheconservation ofmatter, thedivergence ofpvwill be,ingeneral, thetime rateofchange ofthemass ofthematerial perunitvolume. Wewillassume thatthere arenoprocesses forthecontinuous creation ordestruc- tion ofmatter. Theconservation ofmatter then requires that V-pv=0.(It should, ingeneral, beequal to—6p/6t, butsince ourﬂuid isincompressible, p cannot change.) Since piseverywhere thesame, wecanfactor itout,andourequa- tionissimply V-v=0. Good! Wehaveelectrostatics again (with nocharges); it’sjustlikeV'E=0. Notso!Electrostatics isnotsimply V-E=0.Itisapairofequations. One equation doesnottellusenough; weneedstillanadditional equation. Tomatch electrostatics, weshould havealsothatthecurlofviszero. Butthatisnotgenerally trueforrealliquids. Most liquids willordinarily develop some circulation. So wearerestricted tothesituation inwhich there isnocirculation oftheﬂuid. Such ﬂowisoften called irrotational. Anyway, ifwemake allourassumptions, wecan 12-8 imagine acaseofﬂuidﬂowthatisanalogous toelectrostatics. Sowetake V-v=0 (12.28) and VXv=0. (12.29) Wewant toemphasize thatthenumber ofcircumstances inwhich liquid ﬂowfollows these equations isfarfrom thegreat majority, butthere areafew. They must becases inwhich wecanneglect surface tension, compressibility, and viscosity, andll'1which wecanassume thattheﬂow isirrotational. These assump- tions arevalid sorarely forrealwater thatthemathematician John vonNeumann saidthatpeople whoanalyze Eqs.(12.28) and(12.29) arestudying “dry water”! (Wetakeuptheproblem ofﬂuidﬂowinmore detail inChapters 40and41.) Because VXv=0,thevelocity of“dry water” canbewritten asthe gradient ofsome potential: v=—V1l. (12.30) What isthephysical meaning of1//?There isn’tanyveryuseful meaning. The velocity canbewritten asthegradient ofapotential simply because theﬂowis irrotational. Andbyanalogy withelectrostatics, 1!»iscalled thevelocity potential, butitisnotrelated toapotential energy inthewaythat¢is.Since thedivergence ofviszero, wehave v-(v1p) =vzp=0. (12.31) Thevelocity potential itobeys thesame differential equation astheelectrostatic potential infreespace (p=0). Let’s pick aproblem inirrotational ﬂow andseewhether wecansolve itby themethods wehave learned. Consider theproblem ofaspherical ballfalling through aliquid. Ifitisgoing tooslowly, theviscous forces, which wearedis- regarding, willbeimportant. Ifitisgoing toofast,littlewhirlpools (turbulence) willappear initswake andthere willbesome circulation ofthewater. Butifthe ballisgoing neither toofastnortooslow, itismore orlesstruethatthewater ﬂow willﬁtourassumptions, andwecandescribe themotion ofthewater byour simple equations. Itisconvenient todescribe what happens inaframe ofreference ﬁxed inthe sphere. Inthisframe weareasking thequestion: Howdoeswater ﬂowpastasphere atrestwhen theﬂowatlarge distances isuniform? That is,when, farfrom the sphere, theﬂowiseverywhere thesame. Theﬂownearthesphere willbeasshown bythestreamlines drawn inFig.12-8. These lines, always parallel tov,correspond tolinesofelectric ﬁeld. Wewant togetaquantative description forthevelocity ﬁeld, i.e.,anexpression forthevelocity atanypoint P. Wecanﬁndthevelocity from thegradient ofiy,soweﬁrstwork outthepo- tential. Wewant apotential thatsatisﬁes Eq.(12.31) everywhere, andwhich alsosatisﬁes tworestrictions: (1)there isnoﬂowinthespherical region inside thesurface oftheball,and(2)theﬂowisconstant atlarge distances. Tosatisfy (1),thecomponent ofvnormal tothesurface ofthesphere must bezero. That means that61,1//6r iszeroatr=a.Tosatisfy (2),wemust have 13¢/62 =voat allpoints where r>>a.Strictly speaking, there isnoelectrostatic casewhich corresponds exactly toourproblem. Itreally corresponds toputting asphere of dielectric constant zeroinauniform electric ﬁeld. Ifwehadworked outthe solution totheproblem ofasphere ofadielectric constant 1<inauniform ﬁeld, thenbyputting K=0wewould immediately have thesolution tothisproblem. Wehave notactually worked outthisparticular electrostatic problem inde- tail,butlet’sdoitnow. (Wecould work directly ontheﬂuidproblem withvand 1/,butwewilluseEand¢1because wearesoused tothem.) Theproblem is:Find asolution ofV2¢=0suchthatE=~V¢ isacon- stant, sayEQ,forlarge r,andsuchthattheradial component ofEisequal tozero atr=a.That is, %? =0. (12.32)7'r=a 12-91lt111 111111 Ply 1 111411 uR l.t1 Fig.12-8. Thevelocity ﬁeld ofir rotational ﬂuid ﬂow past asphere. Ourproblem involves anewkindofboundary condition, notoneforwhich ¢isaconstant onasurface, butforwhich 64>/6r isaconstant. That isa littledifferent. Itisnoteasytogettheanswer immediately. First ofall,without thesphere, ¢would be—E0z. Then Ewould beinthez-direction andhave thecon- stant magnitude E0,everywhere. Now wehave analyzed thecase ofadielectric sphere which hasauniform polarization inside it,andwefound thattheﬁeld inside suchapolarized sphere isauniform ﬁeld, andthatoutside itisthesame as theﬁeldofapoint dipole located atthecenter. So1et’s guess thatthesolution we want isasuperposition ofauniform ﬁeldplustheﬁeldofadipole. Thepotential ofadipole (Chapter 6)ispz/41re0r3. Thus weassume that ¢=-E02+;4?1ZF- (12.33) Since thedipole ﬁeldfallsoffasl/r“, atlarge distances wehavejusttheﬁeldE0. Ourguess willautomatically satisfy condition (2)above. Butwhat dowetakefor thedipole strength p?Toﬁndout,wemayusetheother condition on¢,Eq.(12.32). Wemust diﬂcrentiate ¢withrespect tor,butofcourse wemust dosoataconstant angle 9,soitismore convenient ifweﬁrstexpress ¢interms ofrand6,rather than of2andr.Since z=rcos0,weget pcos0¢=~—E0I' COS 0+HF ' Theradial component ofEis 19¢_ pcos6—5 — -l-E0 COS 0+ € ' Thismust bezeroatr=aforall0.Thiswillbetrueif p= —27re0a3E0. Note carefully thatifbothterms inEq.(12.35) hadnothadthesame 0-depen- dence, itwould nothavebeenpossible tochoose psothat(12.35) turned outtobe zeroatr=aforallangles. Thefactthatitworks outmeans thatwehaveguessed wisely inwriting Eq.(12.33). Ofcourse, when wemade theguess wewerelooking ahead; weknew thatwewould need another term that(a)satisﬁed V245 =0(any realﬁeldwould dothat), (b)dependent oncos0,and(c)felltozeroatlarge r. Thedipole ﬁeldistheonlyonethatdoesallthree. Using (12.36), ourpotential is a3 ¢= —E0 COS 0(T + ' I‘ Thesolution oftheﬂuidﬂowproblem canbewritten simply as 3 ¢=—v0cos0(r+2iﬂ) - (12.38) Itisstraightforward toﬁndvfrom thispotential. Wewillnotpursue thematter further. 12-6 Illumination; theuniform lighting ofaplane Inthissection weturntoacompletely different physical problem—we want toillustrate thegreat variety ofpossibilities. Thistimewewilldosomething that leads tothesame kind ofintegral thatwefound inelectrostatics. (Ifwehave a mathematical problem which gives usacertain integral, thenweknow something about theproperties ofthatintegral ifitisthesame integral thatwehadtodofor another problem.) Wetakeourexample from illumination engineering. Suppose there isalightsource atthedistance aabove aplane surface. What istheillumina- tionofthesurface? That is,what istheradiant energy perunittime arriving ata unitareaofthesurface? (SeeFig.12-9.) Wesuppose thatthesource isspherically I2-10 S ____.__Y___'1____ symmetric, sothatlight isradiated equally inalldirections. Then theamount of radiant energy which passes through aunitareaatrightangles toalightﬂowvaries inversely asthesquare ofthedistance. Itisevident thattheintensity ofthelightin thedirection normal totheﬂowisgiven bythesame kindofformula asforthe electric ﬁeldfrom apoint source. Ifthelightraysmeet thesurface atanangle 0to thenormal, thenI,theenergy arriving perunitareaofthesurface, isonlycos19as great, because thesame energy goesontoanarealarger byl/cos 0.Ifwecallthe strength ofourlightsource S,thenI,,,theillumination ofastuface, is S1,,=Fer-ri, (12.39) where e,istheunitvector from thesource andnistheunitnormal tothesurface. Theillumination 1,,corresponds tothenormal component oftheelectric ﬁeldfrom apoint charge ofstrength 41re0S. Knowing that, weseethatforanydistribution of light sources, wecanﬁndtheanswer bysolving thecorresponding electrostatic problem. Wecalculate thevertical component ofelectric ﬁeld ontheplane dueto adistribution ofcharge inthesame wayasforthatofthelightsources.* Consider thefollowing example. Wewish forsome special experimental situation toarrange thatthetopsurface ofatablewillhaveaveryuniform illumina- tion. Wehave available long tubular ﬂuorescent lights which radiate uniformly along their lengths. Wecanilluminate thetable byplacing theﬂuorescent tubes inaregular array ontheceiling, which isattheheight zabove thetable. What is thewidest spacing bfrom tubetotubethatweshould useifwewant thesurface illumination tobeuniform to,say,within onepartinathousand? Answer; (1) Find theelectric ﬁeld from agridofwires with thespacing b,each charged uni- formly; (2)compute thevertical component oftheelectric ﬁeld; (3)ﬁndoutwhat bmust besothattheripples oftheﬁeldarenotmore than onepartinathousand. InChapter 7wesawthattheelectric ﬁeld ofagridofcharged wires could be represented asasumofterms, each oneofwhich gaveasinusoidal variation of theﬁeldwithaperiod ofb/n,where nisaninteger. Theamplitude ofanyoneof these terms isgiven byEq.(7.44): Fn =Ane—211-nz/bl Weneed consider only n=1,solong asweonly want theﬁeld atpoints nottoo close tothegrid. Foracomplete solution, wewould stillneed todetermine the coefficients A.,,which wehave notyetdone (although itisastraightforward calculation). Since weneed only A1,wecanestimate thatitsmagnitude isroughly thesame asthatoftheaverage ﬁeld. Theexponential factor would then giveus directly therelative amplitude ofthevariations. Ifwewant thisfactor tobel0“3, weﬁndthatbmust be0.912. Ifwemake thespacing oftheﬂuorescent tubes 3/4 *Since wearetalking about incoherent sources whose intensities always addlinearly, theanalogous electric charges willalways have thesame sign. Also, ouranalogy applies onlytothelight energy arriving atthetopofanopaque surface, sowemust include in ourintegral only thesources which shine onthesurface (and, naturally, notsources located below thesurface !). 12-11\ S\ I=ff\ \ T,- t,,=$5-cosa 9 Fig. 12-9 The illumination I,,ofa surface istheradiant energy perunit timearriving ataunitarea ofthesurface ofthedistance totheceiling, theexponential factor isthenl/4000, andwehavea safety factor of4,sowearefairly surethatwewillhavetheillumination constant toonepartinathousand. (Anexact calculation shows thatA1isreally twice the average ﬁeld, sotheexact answer isb=0.82.) Itissomewhat surprising thatfor suchauniform illumination theallowed separation ofthetubes comes outsolarge. 12-7 The“underlying imity” ofnature Inthischapter, wewished toshow that inlearning electrostatics youhave learned atthesame time how tohandle many subjects inphysics, andthat by keeping thisinmind, itispossible tolearn almost allofphysics inalimited number ofyears. However, aquestion surely suggests itself attheendofsuch adiscussion: Why aretheequations from diﬁferent phenomena sosimilar? Wemight say: “Itis theunderlying unity ofnature.” Butwhat does thatmean? What could such a statement mean? Itcould mean simply thattheequations aresimilar fordiﬂ'erent phenomena; butthen, ofcourse, wehave given noexplanation. The“underlying unity” might mean thateverything ismade outofthesame stuff, andtherefore obeys thesame equations. That sounds likeagood explanation, butletus think. Theelectrostatic potential, thediﬂusion ofneutrons, heat ﬂow—are we really dealing with thesame stuff? Canwereally imagine thattheelectrostatic po- tential isphysically identical tothetemperature, ortothedensity ofparticles? Certainly 4>isnotexactly thesame asthethermal energy ofparticles. Thedisplace- ment ofamembrane iscertainly notlikeatemperature. Why, then, isthere “an underlying unity” ? Acloser look atthephysics ofthevarious subjects shows, infact, thatthe equations arenotreally identical. Theequation wefound forneutron diffusion is only anapproximation thatisgood when thedistance over which wearelooking islarge compared withthemean freepath. Ifwelookmore closely, wewould see theindividual neutrons running around. Certainly themotion ofanindividual neutron isacompletely different thing from thesmooth variation wegetfrom solving thedifferential equation. Thedifferential equation isanapproximation, because weassume thattheneutrons aresmoothly distributed inspace. Isitpossible thatthisistheclue? That thething which iscommon toallthe phenomena isthespace, theframework intowhich thephysics isput? Aslong as things arereasonably smooth inspace, then theimportant things that willbe involved willbetherates ofchange ofquantities with position inspace. That is why wealways getanequation with agradient. Thederivatives must appear in theform ofagradient oradivergence; because thelaws ofphysics areindependent ofdirection, they must beexpressible invector form. Theequations ofelectro- statics arethesimplest vector equations thatonecangetwhich involve onlythe spatial derivatives ofquantities. Anyother simple problem—or simpliﬁcation ofa complicated problem—must look likeelectrostatics. What iscommon toallour problems isthatthey involve space andthatwehave imitated what isactually a complicated phenomenon byasimple diﬂerential equation. That leads ustoanother interesting question. Isthesame statement perhaps alsotruefortheelectrostatic equations’? Aretheyalsocorrect onlyasasmoothed- outimitation ofareally much more complicated microscopic world? Could itbe thattherealworld consists oflittle X-ons which canbeseen only atverytinydis- tances? And thatinourmeasurements wearealways observing onsuch alarge scale thatwecan’t seethese little X-ons, andthatiswhy wegetthedifferential equations? Ourcurrently most complete theory ofelectrodynamics does indeed have its diﬁiculties atveryshort distances. Soitispossible, inprinciple, thatthese equations aresmoothed-out versions ofsomething. They appear tobecorrect atdistances down toabout 10*“ cm,butthen they begin tolook wrong. Itispossible that there issome asyetundiscovered underlying “machinery,” andthatthedetails of anunderlying complexity arehidden inthesmooth-looking equations—as isso 12-l2 inthe“smooth” diffusion ofneutrons. Butnoonehasyetformulated asuccessful theory thatworks thatway. Strangely enough, itturns out(forreasons thatwedonotatallunderstand) thatthecombination ofrelativity andquantum mechanics asweknow them seems toforbid theinvention ofanequation that isfundamentally diﬂerent from Eq. (12.4), andwhich does notatthesame time lead tosome kind ofcontradiction. Notsimply adisagreement withexperiment, butaninternal contradiction. As,for example, theprediction thatthesumoftheprobabilities ofallpossible occurrences isnotequal tounity, orthatenergies maysometimes come outascomplex numbers, orsome other suchidiocy. Noonehasyetmade upatheory ofelectricity forwhich V2¢=—p/e0 isunderstood asasmoothed-out approximation toamechanism underneath, andwhich does notleadultimately tosome kind ofanabsurdity. But,itmust beadded, itisalsotruethattheassumption thatV2¢=—p/e0 is valid foralldistances, nomatter howsmall, leads toabsurdities ofitsown(the electrical energy ofanelectron isinﬁnite)—absurdities from which nooneyet knows anescape. 12-13 13 Magnetostatics 13-1 Themagnetic ﬁeld Theforce onanelectric charge depends notonlyonwhere itis,butalsoon howfastitismoving. Every point inspace ischaracterized bytwovector quantities which determine theforce onanycharge. First, there istheelectric force, which gives aforce component independent ofthemotion ofthecharge.‘ Wedescribe it bytheelectric ﬁeld, E.Second, there isanadditional force component, called the magnetic force, which depends onthevelocity ofthecharge. This magnetic force hasastrange directional character: Atanyparticular point inspace, both the direction oftheforce anditsmagnitude depend onthedirection ofmotion ofthe particle: atevery instant theforce isalways atright angles tothevelocity vector; also,atanyparticular point, theforce isalways atright angles toaﬁxed direction inspace (seeFig.13-l); andﬁnally, themagnitude oftheforce isproportional to thecomponent ofthevelocity atrightangles tothisunique direction. Itispossible todescribe allofthisbehavior bydeﬁning themagnetic ﬁeldvector B,which speci- ﬁesboththeunique direction inspace andtheconstant ofproportionality withthe velocity, andtowrite themagnetic force asqvXB.Thetotal electromagnetic force onacharge can,then, bewritten as F=q(E+v><B). (13.1) Thisiscalled theLorentz force. Themagnetic force iseasily demonstrated bybringing abarmagnet close toa cathode-ray tube. Thedeﬂection oftheelectron beam shows thatthepresence of themagnet results inforces ontheelectrons transverse totheirdirection ofmotion, aswedescribed inChapter 12ofVol.I. The unit ofmagnetic ﬁeld Bisevidently one newton-second per coulomb-meter. The same unit isalso onevolt-second permeterz. Itisalso called oneweber persquare meter. 13-2 Electric current; theconservation ofcharge Weconsider ﬁrsthowwecanunderstand themagnetic forces onwires carrying electric currents. Inorder todothis,wedeﬁne what ismeant bythecurrent density. Electric currents areelectrons orother charges inmotion withanetdriftorﬂow. Wecanrepresent thecharge ﬂowbyavector which gives theamount ofcharge passing perunitareaandperunittime through asurface element atright angles to theﬂow (just aswedidforthecase ofheat ﬂow). Wecallthisthecurrent density andrepresent itbythevector j.Itisdirected along themotion ofthecharges. Ifwetakeasmall areaASatagiven place inthematerial, theamount ofcharge ﬂowing across thatareainaunittimeis j-nAS, (13.2) where nistheunitvector normal toAS. The current density isrelated totheaverage ﬂow velocity ofthecharges. Suppose thatwehave adistribution ofcharges whose average motion isadrift withthevelocity v.Asthisdistribution passes overasurface element AS,thecharge Aqpassing through thesurface element inatimeAtisequal tothecharge contained inaparallelepiped whose baseisASandwhose height isvAt,asshown inFig.13-2. Thevolume oftheparallelepiped istheprojection ofASatright angles tovtimes 13-113-1 Themagnetic ﬁeld 13-2 Electric current; the conservation ofcharge 13-3 Themagnetic force ona current 13-4 Themagnetic ﬁeldofsteady currents; Ampere’s law 13-5 Themagnetic ﬁeldofa straight wireandofasolenoid; atomic currents 13-6 Therelativity ofmagnetic and electric ﬁelds 13-7 Thetransformation ofcurrents andcharges 13-8 Superposition; theright-hand rule Review: Chapter 15,Vol. I:TheSpecial Theory ofRelativity 90° 9V q 90° F Fig. 13-1. The velocity-dependent component oftheforce onamoving charge isatright angles tovandtothe direction ofB.Itisalsoproportional to thecomponent ofvatright angles toB, that is,tovsin0. // ,//’, \ / /VA! _,/’/ Fig. 13-2. Ifacharge distribution of dénsity pmoves with thevelocity v,the charge per unit time through AS is pv-nAS. .1"1'\<‘£%Z;§l SURFACE S Fig.13-3. Thecurrent lthrough the surface Sisfj-nd$. \// .1‘Z\\We 4?. Cl.0$D \/ 1/\ sun;/ace Fig.13-4. Theintegral ofj-riover aclosed surface istherate ofchange of thetotal charge Qinside.vAt,which when multiplied bythecharge density pwillgiveAq.Thus Aq=pv-nASAt. Thecharge perunittimeisthenpv-nAS,from which weget j=pv. (13.3) Ifthecharge distribution consists ofindividual charges, sayelectrons, each withthecharge qandmoving withthemean velocity v,thenthecurrent density is j=Nqv, (13.4) where Nisthenumber ofcharges perunitvolume. Thetotal charge passing perunit time through anysurface Siscalled the electric current, I.Itisequal totheintegral ofthenormal component oftheﬂow through alloftheelements ofthesurface: 1=/sj-has (135) (seeFig. 13-3). Thecurrent Ioutofaclosed surface Srepresents therateatwhich charge leaves thevolume Venclosed byS.One ofthebasic laws ofphysics isthat electric charge isindestructible; itisnever lostorcreated. Electric charges can move from place toplace butnever appear from nowhere. Wesaythatcharge is conserved. Ifthere isanetcurrent outofaclosed surface, theamount ofcharge inside must decrease bythecorresponding amount (Fig. 13-4). Wecan,therefore, write thelawoftheconservation ofcharge as /1'-nds=-§;<Q...1..). (13.6)anyclosedsurface Thecharge inside canbewritten asavolume integral ofthecharge density: Qinside = [ V inside S Ifweapply (13.6) toasmall volume AV,weknow thattheleft-hand integral isV-jAV.Thecharge inside ispAV,sotheconservation ofcharge canalsobe written as . 6v-1=-;f (13.8) (Gauss’ mathematics onceagainl). 13-3 Themagnetic force onacurrent Now weareready toﬁndtheforce onacurrent-carrying wireinamagnetic ﬁeld. Thecurrent consists ofcharged particles moving withthevelocity valong thewire. Each charge feelsatransverse force F=qvXB (Fig. 13-5a). Ifthere areNsuchcharges perunitvolume, thenumber inasmall volume AVofthewireisNAV.Thetotal magnetic force AFonthevolume AV isthesumoftheforces ontheindividual charges, thatis, AF=(NAV)(qv XB). ButNqvisjustj,so AF=jXBAV (13.9) (Fig. 13-5b). Theforce perunitvolume isjXB. 13-2 Ifthecurrent isuniform across awirewhose cross-sectional areaisA,we maytakeasthevolume element acylinder withthebaseareaAandthelength AL.Then AF=jXBAAL. (13.10) Now wecancalljAthevector current Iinthewire. (Itsmagnitude istheelectric current inthewire, anditsdirection isalong thewire.) Then AF=IXBAL. (13.11) Theforce perunitlength onawireisIXB. Thisequation gives theimportant result thatthemagnetic force onawire, duetothemovement ofcharges init,depends only onthetotal current, andnoton theamount ofcharge carried byeach particle—or even itssign! Themagnetic force onawirenearamagnet iseasily shown byobserving itsdeﬂection when a current isturned on,aswasdescribed inChapter 1(seeFig.1-6). 13-4 Themagnetic ﬁeldofsteady currents; Ampere’s law Wehaveseenthatthere isaforce onawireinthepresence ofamagnetic ﬁeld, produced, say,byamagnet. From theprinciple thataction equals reaction we might expect thatthere should beaforce onthesource ofthemagnetic ﬁeld, i.e., onthemagnet, when there isacurrent through thewire.* There areindeed such forces, asisseenbythedeﬂection ofacompass needle nearacurrent-carrying wire. Now weknow thatmagnets feelforces from other magnets, sothatmeans thatwhen there isacurrent inawire, thewire itself generates amagnetic ﬁeld. Moving charges, then, produce amagnetic ﬁeld. Wewould likenow totryto discover thelawsthatdetermine howsuch magnetic ﬁelds arecreated. Thequestion is:Given acurrent, what magnetic ﬁelddoesitmake? Theanswer tothisquestion wasdetermined experimentally bythree critical experiments andabrilliant theoretical argument given byAmpere. Wewillpassoverthisinteresting historical development andsimply saythatalarge number ofexperiments havedemonstrated thevalidity ofMaxwell’s equations. Wetakethem asourstarting point. Ifwe droptheterms involving timederivatives inthese equations wegettheequations of magnetostatics: V~B=0 (13.12) and c2V><B= (13.13)0 These equations arevalid only ifallelectric charge densities areconstant andall currents aresteady, sothattheelectric andmagnetic ﬁelds arenotchanging with time—all oftheﬁelds are“static.” Wemayremark thatitisrather dangerous tothink thatthere issuchathing asastatic magnetic situation, because there must becurrents inorder togeta magnetic ﬁeldatall—and currents cancome onlyfrom moving charges. “Mag- netostatics” is,therefore, anapproximation. Itrefers toaspecial kind ofdynamic situation withlarge numbers ofcharges inmotion, which wecanapproximate by asteady ﬂowofcharge. Only then canwespeak ofacurrent density jwhich does notchange with time. Thesubject should more accurately becalled thestudy of steady currents. Assuming thatallﬁelds aresteady, wedrop allterms in6E/61 and6B/8t from thecomplete Maxwell equations, Eqs. (2.41), and obtain the twoequations (13.12) and(13.13) above. Also notice thatsince thedivergence of thecurlofanyvector isnecessarily zero, Eq.(13.13) requires thatV-j=0.This istrue, byEq.(13.8), only if6p/6t iszero. Butthatmust besoifEisnotchanging withtime, soourassumptions areconsistent. *Wewillseelater, however, thatsuchassumptions arenotgenerally correct forelectro- magnetic forces! 13-3B \I u->_ up I |I __ I -1» ll V 11 ,1»-> lo-> F (0) i>“~l‘TlCD D in‘P O-"' I—-> -—-> (bl Fig. 13-5. Themagnetic force ona current-carrying wire isthesum ofthe forces ontheindividual moving charges. Mil);Fig. l3-6. The line integral ofthe tangential component ofBisequal tothe surface integral ofthenormal component ofVXB.Therequirement thatV-j=0means thatwemayonlyhavecharges which ﬂowinpaths thatclose back onthemselves. They may, forinstance, ﬂowinwires that form complete loops—called circuits. Thecircuits may, ofcourse, contain generators orbatteries thatkeep thecharges ﬂowing. Butthey may notinclude condensers which arecharging ordischarging. (We will, ofcourse, extend the theory later toinclude dynamic ﬁelds, butwewant ﬁrsttotakethesimpler caseo_f' steady currents.) Now letuslook atEqs. (13.12) and(13.13) toseewhat they mean. Theﬁrst onesaysthatthedivergence ofBiszero. Comparing ittotheanalogous equation inelectrostatics, which says thatV-E=p/so, wecanconclude thatthere isno magnetic analog ofanelectric charge. There arenomagnetic charges from which lines ofBcanemerge. Ifwethink interms of“lines” ofthevector ﬁeldB,theycan never start andthey never stop. Then where dotheycome from? Magnetic ﬁelds “appear” inthepresence ofcurrents; they have acurlproportional tothecurrent density. Wherever there arecurrents, there arelines ofmagnetic ﬁeld making loops around thecurrents. Since lines ofBdonotbegin orend, they willoften close back onthemselves, making closed loops. Butthere canalsobecomplicated situations inwhich thelines arenotsimple closed loops. Butwhatever they do, they never diverge from points. Nomagnetic charges have ever been discovered, soV-B=0.This much istruenotonly formagnetostatics, itisalways true— even fordynamic ﬁelds. Theconnection between theBﬁeldandcurrents iscontained inEq.(13.13). Here wehave anewkind ofsituation which isquite different from electrostatics, where wehadVXE=O.That equation meant thatthelineintegral ofEaround anyclosed path iszero: )£E~ds =O. loop Wegotthatresult from Stokes’ theorem, which saysthattheintegral around any closed pathofanyvector ﬁeldisequal tothesurface integral ofthenormal com- ponent ofthecurlofthevector (taken over anysurface which hastheclosed loop asitsperiphery). Applying thesame theorem tothemagnetic ﬁeld vector and using thesymbols shown inFig.13-6, weget 7;}?-ds =f(VXB)-ndS. (13.14)r s Taking thecurlofBfrom Eq.(13.13), wehave l . Theintegral overj,according to(13.5), isthetotal current Ithrough thesurface S. Since forsteady currents thecurrent through Sisindependent oftheshape ofS, solong asitisbounded bythecurve I‘,oneusually speaks of“thecurrent through theloop I‘.”Wehave, then, ageneral law: thecirculation ofBaround anyclosed curve isequal tothecurrent Ithrough theloop, divided bye0c2: £3.43 =@3521. (13_16)soc? Thislaw—called Ampere’s Iaw—plays thesame roleinmagnetostatics thatGauss’ lawplayed inelectrostatics. Ampere’s lawalone doesnotdetermine Bfrom cur- rents; wemust, ingeneral, also useV-B=0.But, aswewillseeinthenext section, itcanbeused toﬁndtheﬁeld inspecial circumstances which have certain simple symmetries. - 13—4 13-5 Themagnetic ﬁeldofastraight wireandofasolenoid; atomic currents Wecanillustrate theuseofAmpere’s lawbyﬁnding themagnetic ﬁeldnear awire. Weask: What istheﬁeld outside along straight wire with acylindrical cross section‘? Wewillassume something which maynotbeatallevident, butwhich isnevertheless true: thattheﬁeldlines ofBgoaround thewireinclosed circles. Ifwemake thisassumption, then Ampere’s law,Eq.(13.16), tellsushowstrong the ﬁeldis.From thesymmetry oftheproblem, Bhasthesame magnitude atall points onacircle concentric withthewire(seeFig.13-7). Wecanthendotheline integral ofB-dsquite easily; itisjustthemagnitude ofBtimes thecircumference. Ifristheradius ofthecircle, then fB-ds =B-21rr. Thetotal current through theloop ismerely thecurrent Iinthewire, so IB-21rr =—-2,EQC or 12I Thestrength ofthemagnetic ﬁeld drops oﬁinversely asr,thedistance from the axisofthewire. Wecan,ifwewish, write Eq.(13.17) invector form. Remembering thatBisatright angles both toIandtor,wehave _ 121Xe,B-mg); ir—-- (13.18) Wehave separated outthefactor l/41re0c2, because itappears often. Itis worth remembering thatitisexactly 10"’ (inthemkssystem), since anequation like(13.17) isused todeﬁne theunitofcurrent, theampere. Atonemeter from a current ofoneampere themagnetic ﬁeld is2Xl0‘7 webers persquare meter. Since acurrent produces amagnetic ﬁeld, itwillexert aforce onanearby wire which isalsocarrying acurrent. InChapter 1wedescribed asimple demonstration oftheforces between twocurrent-carrying wires. Ifthewires areparallel, each is atright angles totheBﬁeld oftheother; thewires should then bepushed either toward oraway from each other. When currents areinthesame direction, the wires attract; when thecurrents aremoving inopposite directions, thewires repel. Kill go\\\\\\‘ '~IIIIIIIIIIIIEHIIIIIIII '‘l /Fig. 13-7. Themagnetic ﬁeld outside ofalong wire carrying thecurrent l. ."#1)! Fig. l3—8. The magnetic ﬁeld ofa |_|E5 long solenoid.I r+'=:aaaarrm:§::aa::a:::°"ii //-"#»..,~-lllllliilili-iiliilllll -1Illlllllllllllllllllll OFB Let’s take another example thatcanbeanalyzed byAmpere’s lawifweadd some knowledge about theﬁeld. Suppose wehave along coilofwire wound ina tight spiral, asshown bythecross sections inFig. l3—8. Such acoiliscalled a solenoid. Weobserve experimentally thatwhen asolenoid isvery long compared with itsdiameter, theﬁeld outside isvery small compared with theﬁeld inside. Using justthatfact,together withAmpere’s law,wecanﬁndthesizeoftheﬁeld inside. Since theﬁeldstays inside (andhaszerodivergence), itslinesmust goalong parallel totheaxis, asshown inFig.l3—8. That being thecase, wecanuseAmpere’s lawwiththerectangular “curve” I‘shown intheﬁgure. Thisloopgoesthedistance 13-5 Fig.13-9. Themagnetic ﬁeldoutside ofasolenoid.Linside thesolenoid, where theﬁeldis,say,B0,thengoesatright angles tothe ﬁeld, andreturns along theoutside, where theﬁeldisnegligible. Thelineintegral ofBforthiscurve isjustBOL, anditmust be1/eocz times thetotal current through I‘,which isNIifthere areNturns ofthesolenoid inthelength L.Wehave NIBQL = E0?’ Or,letting nbethenumber ofturns perunitlength ofthesolenoid (that is,n= N/L), weget 1B0= (13.19) What happens tothelines ofBwhen they gettotheendofthesolenoid? Presumably, theyspread outinsome wayandreturn toenter thesolenoid atthe other end,assketched inFig.13-9. Such aﬁeldisjustwhat isobserved outside of abarmagnet. Butwhat isamagnet anyway ?Ourequations saythatBcomes from thepresence ofcurrents. Yetweknow thatordinary barsofiron(nobatteries or generators) alsoproduce magnetic ﬁelds. You might expect thatthere should be some other terms ontheright-hand sideof(13.12) or(13.13) torepresent “the density ofmagnetic iron” orsome such quantity. Butthere isnosuch term. Our theory saysthatthemagnetic effects ofironcome from some internal currents which arealready taken careofbythejterm. Matter isverycomplex when looked atfrom afundamental point ofview—as wesawwhen wetried tounderstand dielectrics. Inorder nottointerrupt ourpres- entdiscussion, wewillwaituntil later todealindetail with theinterior mechanisms ofmagnetic materials likeiron. You willhave toaccept, forthemoment, thatall magnetism isproduced from currents, andthatinapermanent magnet there are permanent internal currents. Inthecaseofiron, these currents come from electrons spinning around theirownaxes. Every electron hassuchaspin,which corresponds toatinycirculating current. Ofcourse, oneelectron doesn’t produce much mag- netic ﬁeld, butinanordinary piece ofmatter there arebillions andbillions ofelec- trons. Normally these spin andpoint every which way, sothat there isnonet eﬂ"ect. Themiracle isthatinavery fewsubstances, likeiron, alarge fraction of theelectrons spinwiththeiraxesinthesame direction—for iron,twoelectrons from each atom takes partinthiscooperative motion. Inabarmagnet there arelarge numbers ofelectrons allspinning inthesame direction and, aswewillsee,their total eﬁect isequivalent toacurrent circulating onthesurface ofthebar. (This is quite analogous towhat wefound fordielectrics—that auniformly polarized di- electric isequivalent toadistribution ofcharges onitssurface.) Itis,therefore, no accident thatabarmagnet isequivalent toasolenoid. 13-6 Therelativity ofmagnetic andelectric ﬁelds When wesaidthat themagnetic force onacharge wasproportional toits velocity, youmay have wondered: “What velocity? With respect towhich refer- enceframe?” Itis,infact,clear from thedeﬁnition ofBgiven atthebeginning of thischapter thatwhat thisvector iswilldepend onwhat wechoose asareference frame forourspeciﬁcation ofthevelocity ofcharges. Butwehave saidnothing about which istheproper frame forspecifying themagnetic ﬁeld. Itturns outthatanyinertial frame willdo.Wewillalsoseethatmagnetism andelectricity arenotindependent things—that theyshould always betaken to- gether asonecomplete electromagnetic ﬁeld. Although inthestatic caseMaxwell’s equations separate intotwodistinct pairs, onepairforelectricity andonepairfor magnetism, with noapparent connection between thetwoﬁelds, nevertheless, in nature itself there isaveryintimate relationship between them thatarises from the principle ofrelativity. Historically, theprinciple ofrelativity wasdiscovered after Maxwell’s equations. Itwas, infact, thestudy ofelectricity andmagnetism which ledultimately toEinstein’s discovery ofhisprinciple ofrelativity. Butlet’s see 13-6 l-l"°Q q r S S’ \ \0-» \ ' -=—-¢-:..- -c-.—.— ~~-¢--Cr.:- ta).' " ' 1 ‘‘fr “” ‘"' "" Fig. l3—lO. Theinteraction ofacurrent-carrying wire andaparticle with the charge qasseen intwoframes. lnframe S(part a),thewire isatrest, inframe S’(part bl,thecharge isatrest. whatourknowledge ofrelativity would tellusabout magnetic forces ifweassume thattherelativity principle isapplicable—as itis—to electromagnetism. Suppose wethink about what happens when anegative charge moves with velocity v0parallel toacurrent-carrying wire, asinFig.l3~l0. Wewilltrytounder- stand what goesonintworeference frames: oneﬁxed withrespect tothewire, asinpart(a)oftheﬁgure, andoneﬁxed withrespect totheparticle, asinpart(b). Wewillcalltheﬁrstframe Sandthesecond S’. IntheS-frame, there isclearly amagnetic force ontheparticle. Theforce is directed toward thewire, soifthecharge ismoving freely wewould seeitcurve in toward thewire. ButintheS’-frame there canbenomagnetic force ontheparticle, because itsvelocity iszero. Does it,therefore, staywhere itis?Would wesee diﬁerent things happening inthetwosystems? Theprinciple ofrelativity would saythatinS’weshould alsoseetheparticle move closer tothewire. Wemust trytounderstand whythatwould happen. Wereturn toouratomic description ofawirecarrying acurrent. Inanormal conductor, likecopper, theelectric currents come from themotion ofsome ofthe negative electrons—ca1led theconduction electrons—while thepositive nuclear charges andtheremainder oftheelectrons stayﬁxed inthebody ofthematerial. Weletthedensity oftheconduction electrons bep_andtheirvelocity inSbev. Thedensity ofthecharges atrestinSisp+,which must beequal tothenegative ofp_,since weareconsidering anuncharged wire. There isthus noelectric ﬁeld outside thewire, andtheforce onthemoving particle isjust F=qU()><B. Using theresult wefound inEq.(13.18) forthemagnetic ﬁeld atthedistance rfrom theaxisofawire, weconclude thattheforce ontheparticle isdirected toward thewireandhasthemagnitude __ 1 2IqU[) F_4111002. r Using Eqs. (13.4) and(13.5), thecurrent Icanbewritten asp_vA, where Ais theareaofacross section ofthewire. Then l 2qp_Am10F=__ .___. _4'rre0c2 r (1320) Wecould continue totreat thegeneral case ofarbitrary velocities forvand110, butitwillbejustasgood tolook atthespecial case inwhich thevelocity 00of theparticle isthesame asthevelocity voftheconduction electrons. Sowewrite 00=v,andEq.(13.20) becomes _qP-Af_ F-27% rC2 (13.21) Now weturnourattention towhat happens inS’,inwhich theparticle isat restandthewireisrunning past(toward theleftintheﬁgure) withthespeed v. Thepositive charges moving with thewirewillmake some magnetic ﬁeldB’at theparticle. Buttheparticle isnowatrest,sothere isnomagnetic force onit! Ifthere isanyforce ontheparticle, itmust come from anelectric ﬁeld. Itmust l3-7 0)bethatthemoving wire hasproduced anelectric ﬁeld. Butitcandothatonlyifit appears charged—it must bethataneutral wirewithacurrent appears tobecharged when setinmotion. ' Wemust lookintothis. Wemust trytocompute thecharge density inthe wireinS’from what weknow about itinS.Onemight, atﬁrst, think theyarethe same; butweknow thatlengths arechanged between SandS’(seeChapter 15, Vol. I),sovolumes willchange also. Since thecharge densities depend onthe volume occupied bycharges, thedensities willchange, too. Before wecandecide about thecharge densities inS’,wemust know what happens totheelectric charge ofabunch ofelectrons when thecharges aremoving. Weknow thattheapparent mass ofaparticle changes by1/\/1 —v2/c2. Does itscharge dosomething similar? No! Charges arealways thesame, moving or not. Otherwise wewould notalways observe thatthetotal charge isconserved. Suppose thatwetake ablock ofmaterial, sayaconductor, which isinitially uncharged. Now weheat itup.Because theelectrons have adiﬁ"erent mass than theprotons, thevelocities oftheelectrons andoftheprotons willchange bydiﬁer- entamounts. Ifthecharge ofaparticle depended onthespeed oftheparticle carry- ingit,intheheated block thecharge oftheelectrons andprotons would nolonger balance. Ablock would become charged when heated. Aswehave seenearlier, a verysmall fractional change inthecharge ofalltheelectrons inablock would give risetoenormous electric ﬁelds. Nosucheﬁ"ect haseverbeenobserved. Also, wecanpoint outthatthemean speed oftheelectrons inmatter depends onitschemical composition. Ifthecharge onanelectron changed with speed, the netcharge inapiece ofmaterial would bechanged inachemical reaction. Again, astraightforward calculation shows thateven avery small dependence ofcharge onspeed would giveenormous ﬁelds from thesimplest chemical reactions. No such effect isobserved, andweconclude thattheelectric charge ofasingle particle isindependent ofitsstateofmotion. Sothecharge qonaparticle isaninvariant scalar quantity, independent of theframe ofreference. That means thatinanyframe thecharge density ofa distribution ofelectrons isjustproportional tothenumber ofelectrons perunit volume. Weneed only worry about thefactthatthevolume canchange because oftherelativistic contraction ofdistances. Wenow apply these ideas toourmoving wire. Ifwetake alength L0ofthe wire, inwhich there isacharge density poofstationary charges, itwillcontain thetotal charge Q=p0L0A 0.Ifthesame charges areobserved inadifferent frame tobemoving with velocity v,theywillallbefound inapiece ofthematerial with theshorter length L=Lox/l —v2/c2, (13.22) butwiththesame areaA0(since dimensions transverse tothemotion areun- changed). SeeFig.13-11. Ifwecallpthedensity ofcharges intheframe inwhich they aremoving, the total charge QwillbepLA0.This must alsobeequal top0LOA,because charge is thesame inanysystem, sothatpL=p0L0 or,from (13.22), Pp=Wm (13.23) Loid 5 Pf L’-———-"ll S, I_- . \_-/,p4 A‘r4 __',- 1.'_ _ . '‘--I I A A V=0 AreaA Q! V ‘1'90 Fig. 13-1 l.Ifadistribution ofcharged particles atresthasthecharge density pg,thesame charges willhave thedensity p=pg/\/l -—vi/c1 when seen from a frame with therelative velocity v. 13-8 Thecharge density ofamoving distribution ofcharges varies inthesame wayasthe relativistic mass ofaparticle. Wenow usethisgeneral result forthepositive charge density p+ofourwire. These charges areatrestinframe S.InS’,however, where thewire moves with thespeed v,thepositive charge density becomes P+-p’=————— - 13.24+ \/1—122/c2 ( ) Thenegative charges areatrestinS’.Sotheyhavetheir “rest density” p0in thisframe. InEq.(13.23) p0=p’_,because theyhave thedensity p’_when the wireisatrest,i.e.,inframe S,where thespeed ofthenegative charges isv.For theconduction electrons, wethenhavethat or p’_=p_\/1 —v2/c2. (13.26) Now wecanseewhythere areelectric ﬁelds inS’—-because inthisframe the wirehasthenetcharge density p’given by P’=P5.+p’_- Using (13.24) and(13.26), wehave pd-'=i—+ _\/1- 22.”\/I'T527F ” "’° Since thestationary wire isneutral, p__=—p+, andwehave 22 P’=9+7TL ' (13.27) Ourmoving wireispositively charged andwillproduce anelectric ﬁeldE’atthe external stationary particle. Wehave already solved theelectrostatic problem ofa uniformly charged cylinder. Theelectric ﬁeld atthedistance rfrom theaxisofthe cylinder is 13'=LA=__P+_A_"2i/C2 . (13.28)2750' 21re0r\/l —v2/c2 Theforce onthenegatively charged particle istoward thewire. Wehave, atleast, aforce inthesame direction from thetwopoints ofview; theelectric force inS’ hasthesame direction asthemagnetic force inS. Themagnitude oftheforce inS’is F’=-‘LPl’!_-?_"2/‘2 - (13.29)21re0 r\/Tip? Comparing thisresult forF’with ourresult forFinEq.(13.21), weseethatthe magnitudes oftheforces arealmost identical from thetwopoints ofview. Infact, F, = s —vc soforthesmall velocities wehave been considering, thetwoforces areequal. Wecansaythat forlowvelocities, atleast, weunderstand thatmagnetism and electricity arejust“two ways oflooking atthesame thing.” Butthings areeven better than that. Ifwetake into account thefactthat forces alsotransform when wegofrom onesystem totheother, weﬁndthatthe twoways oflooking atwhat happens doindeed givethesame physical result for anyvelocity. 13—9 S lb) Fig. 13-12. Inframe Sthecharge density iszero and thecurrent density is i.There isonly amagnetic ﬁeld. InS’, there isacharge density p',andadiffer- entcurrent density i’.Themagnetic ﬁeld B’isdifferent and there isanelectric ﬁeld E’.Onewayofseeing thisistoaskaquestion like:What transverse momentum willtheparticle have aftertheforce hasacted foralittlewhile? Weknow from Chapter 16ofVol.Ithatthetransverse momentum ofaparticle should bethesame inboth theS-andS’-frames. Calling thetransverse coordinate y,wewant to compare Ap,,andAp§,. Using therelativistically correct equation ofmotion, F=dp/dt, weexpect thatafter thetimeAtourparticle willhave atransverse momentum Ap,intheS-system given by Ap,,=FAt. (13.31) IntheS’-system, thetransverse momentum willbe Ap;=F’At’. (13.32) Wemust, ofcourse, compare ApyandApf,forcorresponding time intervals Atand At’. Wehave seeninChapter 15ofVol.Ithatthetimeintervals referred toa moving particle appear tobelonger than those intherestsystem oftheparticle. Since ourparticle isinitially atrestinS’,weexpect, forsmall At,that IA —vc andeverything comes outO.K. From (13.31) and(13.32), AL;=F’At' Ap, FAt’ which isjust =1ifwecombine (13.30) and(13.33). Wehave found thatwegetthesame physical result whether weanalyze the motion ofaparticle moving along awireinacoordinate system atrestwithrespect tothewire, orinasystem atrestwithrespect totheparticle. Intheﬁrstinstance, theforce waspurely “magnetic,” inthesecond, itwaspurely “electric.” Thetwo points ofviewareillustrated inFig.13-12 (although there isstillamagnetic ﬁeld B’inthesecond frame, itproduces noforces onthestationary particle). Ifwehadchosen stillanother coordinate system, wewould have found a different mixture ofEandBﬁelds. Electric andmagnetic forces arepart ofone physical phenomenon—the electromagnetic interactions ofparticles. Thesepara- tionofthisinteraction intoelectric andmagnetic parts depends very much onthe reference frame chosen forthedescription. Butacomplete electromagnetic de- scription isinvariant; electricity andmagnetism taken together areconsistent withEinstein’s relativity. Since electric andmagnetic ﬁelds appear indifferent mixtures ifwechange our frame ofreference, wemust becareful about howwelookattheﬁelds EandB. Forinstance, ifwethink of“lines” ofEorB,wemust notattach toomuch reality tothem. Thelinesmaydisappear ifwetrytoobserve them from adilferent co- ordinate system. Forexample, insystem S’there areelectric ﬁeld lines, which we donotﬁnd“moving pastuswith velocity vinsystem S.”Insystem Sthere areno electric ﬁeldlines atall!Therefore itmakes nosense tosaysomething like: When Imove amagnet, ittakes itsﬁeld with it,sothelines ofBarealsomoved. There isnowaytomake sense, ingr1€1'ﬂl, outoftheideaof“the speed ofamoving ﬁeld line.” Theﬁelds areourwayofdescribing what goes onatapoint inspace. In particular, EandBtellusabout theforces thatwillactonamoving particle. The question “What istheforce onacharge from amoving magnetic ﬁeld?”doesn’t mean anything precise. Theforce isgiven bythevalues ofEandBatthecharge, andtheformula (13.1) isnottobealtered ifthesource ofEorBismoving (itis thevalues ofEandBthatwillbealtered bythemotion). Ourmathematical de- scription deals only with theﬁelds asafunction ofx,y,z,andtwith respect to some inertial frame. Wewilllater bespeaking of“awave ofelectric andmagnetic ﬁelds travelling through space,” as,forinstance, alight wave. Butthatislikespeaking ofawave travelling onastring. Wedon’t then mean thatsome partofthestring ismoving 13-10 inthedirection ofthewave, wemean thatthedisplacement ofthestring appears ﬁrstatoneplace andlater atanother. Similarly, inanelectromagnetic wave, the wave travels; butthemagnitude oftheﬁelds change. Sointhefuture when we-or someone else—speaks ofa“moving” ﬁeld, youshould think ofitasjustahandy, short wayofdescribing achanging ﬁeld insome circumstances. 13-7 Thetransformation ofcurrents andcharges You may have worried about thesimpliﬁcation wemade above when we took thesame velocity vfortheparticle andfortheconduction electrons inthe wire. Wecould goback andcarry through theanalysis again fortwodifferent velocities, butitiseasier tosimply notice thatcharge andcurrent density arethe components ofafour-vector (seeChapter 17,Vol. I). Wehaveseenthatifpoisthedensity ofthecharges intheirrestframe, then inaframe inwhich theyhavethevelocity v,thedensity is p=_iL_.\/1—222/c2 Inthatframe theircurrent density is j=pv=—1”°—"2/;- (13.34)\/-1) Now weknow thattheenergy Uandmomentum pofaparticle moving with velocity varegiven by U= "1002 P= mov V1-02/c2 \/1-112/c2 where moisitsrestmass. Wealsoknow thatUandpformarelativistic four-vector. Since pandjdepend onthevelocity vexactly asdoUandp,wecanconclude thatp andjarealsothecomponents ofarelativistic four-vector. Thisproperty isthekey toageneral analysis oftheﬁeldofawiremoving withanyvelocity, which we would needifwewant todotheproblem again withthevelocity v0oftheparticle different from thevelocity oftheconduction electrons. Ifwewish totransform pandjtoacoordinate system moving with avelocity uinthex-direction, weknow thatthey transform justliketand(x,y,z),sothat wehave(seeChapter 15,Vol.I) X— . 1'1—up I Ix ='-i”'—-is J =-i——i*is \/1—-u2/c2 x \/1—uz/c2 -1 - J/"=J’: J11=Jib #=a n=n 1'=-——-’T“"’°21p’=———”_"j"/C2- (13.35)\/l—u2/c2 \/1—uz/c2 With these equations wecanrelate charges andcurrents inoneframe tothose inanother. Taking thecharges andcurrents ineither frame, wecansolve the electromagnetic problem inthat frame byusing ourMaxwell equations. The result weobtain forthemotions ofparticles willbethesame nomatter which frame wechoose. Wewillreturn atalater time totherelativistic transformations ofthe electromagnetic ﬁelds. 13-8 Superposition; theright-hand rule Wewillconclude thischapter bymaking twofurther points regarding the subject ofmagnetostatics. First, ourbasic equations forthemagnetic ﬁeld, V-B=0, V><B=j/czeo, 13-ll arelinear inBandj.That means thattheprinciple ofsuperposition alsoapplies tomagnetic ﬁelds. Theﬁeld produced bytwodifferent steady currents isthesum oftheindividual ﬁelds from eachcurrent acting alone. Oursecond remark con- cerns theright-hand rules which wehave encountered (such astheright-hand ruleforthemagnetic ﬁeld produced byacurrent). Wehave alsoobserved thatthe magnetization ofanironmagnet istobeunderstood from thespinoftheelectrons inthematerial. Thedirection ofthemagnetic ﬁeldofaspinning electron isrelated toitsspinaxisbythesame right-hand rule. Because Bisdetermined bya“handed” rule—involving either across product oracurl-it iscalled anaxial vector. (Vec- torswhose direction inspace does notdepend onareference toaright orlefthand arecalled polar vectors. Displacement, velocity, force, andE,forexample, are’ polar vectors.) Physically observable quantities inelectromagnetism arenot,however, right- (orleft-) handed. Electromagnetic interactions aresymmetrical under reﬂection (seeChapter 52,Vol.I).Whenever magnetic forces between twosetsofcurrents are computed, theresult isinvariant withrespect toachange inthehand convention. Ourequations lead, independently oftheright-hand convention, totheendresult thatparallel currents attract, orthatcurrents inopposite directions repel. (Try working outtheforce using “left-hand rules.”) Anattraction orrepulsion isa polar vector. Thishappens because indescribing anycomplete interaction, we usetheright-hand ruletwice--once toﬁndBfrom currents, again toﬁndtheforce thisBproduces onasecond current. Using theright-hand ruletwice isthesame asusing theleft-hand ruletwice. Ifweweretochange ourconventions toaleft- hand system allourBﬁelds would bereversed, butallforces—or, what isperhaps more relevant, theobserved accelerations ofobjects—would beunchanged. Although physicists have recently found totheir surprise thatallthelaws of nature arenotalways invariant formirror reﬂections, thelaws ofelectromagnetism dohave such abasic symmetry. 13-12 I4 The Magnetic Field inVarious Situations 14-1 Thevector potential Inthischapter wecontinue ourdiscussion ofmagnetic ﬁelds associated with steady currents—the subject ofmagnetostatics. Themagnetic ﬁeldisrelated to electric currents byourbasic equations v-B=0, (14.1) ¢’v><B= (14.2) Wewant nowtosolve these equations mathematically inageneral way, thatis, without requiring anyspecial symmetry orintuitive guessing. Inelectrostatics, wefound thatthere wasastraightforward procedure forﬁnding theﬁeldwhen the positions ofallelectric charges areknown: Onesimply works outthescalar potential ¢bytaking anintegral overthecharges—as inEq.(4.25). Then ifone wants theelectric ﬁeld, itisobtained from thederivatives of¢.Wewillnowshow thatthere isacorresponding procedure forﬁnding themagnetic ﬁeldBifweknow thecurrent density jofallmoving charges. Inelectrostatics wesawthat(because thecurlofEwasalways zero) itwas possible torepresent Easthegradient ofascalar ﬁeldqs.Now thecurlofBisnot always zero, soitisnotpossible, ingeneral, torepresent itasagradient. However, thedivergence ofBisalways zero, andthismeans thatwecanalways represent Bas thecurlofanother vector ﬁeld. For,aswesawinSection 2-8,thedivergence ofa curlisalways zero. Thus wecanalways relate BtoaﬁeldwewillcallAby B=VXA. (14.3) Or,bywriting outthecomponents, £'2_?_/11oz’B:c=(vXA):i:= ay 6A,, 8A 8A 8AB,=(VXA),=-5;"—-5J-;3- Writing B=VXAguarantees thatEq.(14.1) issatisﬁed, since, necessarily, V-B =V-(V XA) =0. TheﬁeldAiscalled thevector potential. Youwillremember thatthescalar potential ¢wasnotcompletely speciﬁed byitsdeﬁnition. Ifwehavefound ¢forsome problem, wecanalways ﬁndanother potential ¢'thatisequally good byadding aconstant: ¢’=¢+C- Thenewpotential ¢’gives thesame electric ﬁelds, since thegradient VCiszero; ¢’and¢represent thesame physics. Similarly, wecanhave diﬁerent vector potentials Awhich givethesame magnetic ﬁelds. Again, because Bisobtained from Abydifferentiation, adding a 14--114-1 Thevector potential 14-2 Thevector potential ofknown currents 14-3 Astraight wire 14-4 Along solenoid 14-5 Theﬁeldofasmall loop; the magnetic dipole 14-6 Thevector potential ofa circuit 14-7 ThelawofBiotandSavart constant toAdoesn’t change anything physical. Butthere isevenmore latitude forA.WecanaddtoAanyﬁeldwhich isthegradient ofsome scalar ﬁeld, without changing thephysics. Wecanshow thisasfollows. Suppose wehave anAthat gives correctly themagnetic ﬁeld Bforsome realsituation, andaskinwhat cir- cumstances some other newvector potential A’willgivethesame ﬁeldBifsub- stituted into(14.3). Then AandA’must have thesame curl: B=VXA’=VXA. Therefore VXA’—VXA=VX(A’—A)=0. Butifthecurlofavector iszero itmust bethegradient ofsome scalar ﬁeld, say 1,0,soA’—A=Vii. That means thatifAisasatisfactory vector potential fora problem then, forany1/1atall, .4’=A+vi (14.5) willbeanequally satisfactory vector potential, leading tothesame ﬁeld B. Itisusually convenient totake some ofthe“latitude” outofAbyarbitrarily placing some other condition onit(inmuch thesame waythatwefound itcon- venient—often—to choose tomake thepotential ¢zero atlarge distances). We can,forinstance, restrict Abychoosing arbitrarily what thedivergence ofAmust be.Wecanalways dothatwithout affecting B.This isbecause although A’and Ahave thesame curl, andgivethesame B,they donotneed tohave thesame divergence. Infact,V-A’=V-A+V2://,andbyasuitable choice ofipwecan make V-A’anything wewish. What should wechoose forV-A?Thechoice should bemade togetthe greatest mathematical convenience andwilldepend ontheproblem wearedoing. Formagnetostatics, wewillmake thesimple choice V-A =0. (14.6) (Later, when wetakeupelectrodynamics, wewillchange ourchoice.) Ourcomplete deﬁnition* ofAisthen, forthemoment, VXA=BandV-A=O. Togetsome experience with thevector potential, let’slook ﬁrstatwhat itis forauniform magnetic ﬁeldB0.Taking ourz-axis inthedirection ofB0,wemust have _6A, 6A,,__ B3-TyW"°' BA 6A _6A,, 6A,, _ B3"W-F;"B"- Byinspection, weseethatonepossible solution ofthese equations is Au=xB0, A,=O, A,=0. Orwecould equally welltake A,=—yB0, A,=0, A,=0. Stillanother solution isalinear combination ofthetwo: A,=—%yB0, Au=%xB0, A,=0. (14.8) *Ourdeﬁnition stilldoes notuniquely determine A.Foraunique speciﬁcation we would alsohave tosaysomething about howtheﬁeldAbehaves onsome boundary, or atlarge distances. Itissometimes convenient, forexample, tochoose aﬁeld which goestozeroatlarge distances. 14-2 Itisclear thatforanyparticular ﬁeldB,thevector potential Aisnotunique; there aremany possibilities. The third solution, Eq.(14.8), hassome interesting properties. Since the x-component isproportional to—yandthey-component isproportional to+x, Amust beatright angles tothevector from thez-axis, which wewillcallr’(the “prime” istoremind usthatitisnotthevector displacement from theorigin). Also, themagnitude ofAisproportional to\/x2 +y2and,hence, tor’.SoA canbesimply written (forouruniform ﬁeld) as .4=in><I". (14.9) Thevector potential Ahasthemagnitude Br’/2 androtates about thez-axis as shown inFig.14-1. If,forexample, theBﬁeldistheaxialﬁeldinside asolenoid, thenthevector potential circulates inthesame sense asdothecurrents ofthe solenoid. Thevector potential forauniform ﬁeldcanbeobtained inanother way. Thecirculation ofAonanyclosed loopPcanberelated tothesurface integral of VXAbyStokes’ theorem, Eq.(3.38): §I_A-ds= (VXA)-nda. (mo) inside I‘ Buttheintegral ontheright isequal totheﬂuxofBthrough theloop, so 95.4-ds= fB-nda. (14.11) 1‘ inside I‘ Sothecirculation ofAaround anyloopisequal totheﬂuxofBthrough theloop. Ifwetakeacircular loop, ofradius r’inaplane perpendicular toauniform ﬁeld B,theﬂuxisjust 1rr'2B. Ifwechoose ourorigin onanaxisofsymmetry, sothatwecantakeAascircum- ferential andafunction onlyofr’,thecirculation willbe fA-ds=21rr’A =1rr’2B. Weget,asbefore, Br’A-3-- Intheexample wehavejustgiven, wehavecalculated thevector potential from themagnetic ﬁeld, which isopposite towhat onenormally does. Incomplicated problems itisusually easier tosolve forthevector potential, andthen determine themagnetic ﬁeldfrom it.Wewillnowshow howthiscanbedone. 14-2 Thevector potential ofknown currents Since Bisdetermined bycurrents, soalsoisA.Wewant nowtoﬁndA in terms ofthecurrents. Westart with ourbasic equation (14.2): c2V><B=i,E0 which means, ofcourse, that GavX(VX,4)= (14.12)0 Thisequation isformagnetostatics what theequation V-V¢=__£1 (14.13)O wasforelectrostatics. 14-3yl e»~A Fig.14-1. Auniform magnetic ﬁeld Binthez-direction corresponds toa vector potential Athatrotates about the z-axis, with the magnitude A=Br’/2 (r'isthedisplacement from thez-axis). Fig. 14-2. Thevector potential Aat point 1isgiven byanintegral over the current elements |'dVatallpoints 2.E1tn!Ourequation (14.12) forthevector potential looks even more likethat for ¢ifwerewrite VX(VXA)using thevector identity Eq.(2.58): VX(VXA)=V(V 'A)—V2A. (14.14) Since wehave chosen tomake V-A=0(and now youseewhy), Eq.(14.12) becomes v2.4=- (14.15) This vector equation means, ofcourse, three equations: 2___1-_, 2=_iv_, 2=_L. vA,,_ 6°C, v.4, G06, v.4, £06, (14.16) And each ofthese equations ismathematically identical to v2¢=-§_ (14.17)0 Allwehave learned about solving forpotentials when p1Sknown canbeused for solving foreach component ofAwhen jisknown! Wehave seeninChapter 4thatageneral solution fortheelectrostatic equation (14.17) is ,0,=_1_/15215.471'€(] 7'12 Soweknow immediately thatageneral solution forA,is .4,,(1)= (14.18)47l'€()C2 7'12 andsimilarly forA,andA,.(Figure 14-2willremind youofourconventions for r12anddV2.) Wecancombine thethree solutions inthevector form 4(1)=_1— /Kl (14.19)47l"€QC2 T12 (You canverify ifyouwish, bydirect differentiation ofcomponents, thatthisinte- gralforAsatisﬁes V-A=0solong asV-j=0,which, aswesaw,must happen forsteady currents.) Wehave, then, ageneral method forﬁnding themagnetic ﬁeld ofsteady cur- rents. Theprinciple is:thex-component ofvector potential arising from acurrent density jisthesame astheelectric potential ¢thatwould beproduced byacharge density pequal toj,,/c2—and similarly forthey-andz-components. (This principle works only with components inﬁxed directions. The“radial” component ofA does notcome inthesame wayfrom the“radial” component ofj,forexample.) Sofrom thevector current density j,wecanﬁndAusing Eq.(14.l9)—that is,we ﬁndeach component ofAbysolving three imaginary electrostatic problems for thecharge distributions pl=j,/c2, p2=j,,/c2, andp3=j,/c2. Then weget Bbytaking various derivatives ofAtoobtain VXA.It’salittle more compli- cated than electrostatics, butthesame idea. Wewillnow illustrate thetheory by solving forthevector potential inafewspecial cases. 14-3 Astraight wire Forourﬁrstexample, wewillagain ﬁndtheﬁeldofastraight wire-—which we solved inthelastchapter byusing Eq.(14.2) andsome arguments ofsymmetry. Wetake along straight wire ofradius a,carrying thesteady current I.Unlike the charge onaconductor intheelectrostatic case, asteady current inawire isuni- formly distributed throughout thecross section ofthewire. Ifwechoose our 14-4 coordinates asshown inFig.14-3, thecurrent density vector jhasonlyaz-com- ponent. Itsmagnitude is . I,lz=W (14.20) inside thewire, andzerooutside. Sincej,andj,,arebothzero, wehaveimmediately .4,=o, .4,=o. TogetA,wecanuseoursolution fortheelectrostatic potential ¢ofawirewitha uniform charge density p=j,/c2. Forpoints outside aninﬁnite charged cylinder, theelectrostatic potential is X¢—'-T1-r?5lI1I", where r’=\/xi +y2and>1isthecharge perunitlength, 7l'£12p. SoA,must be _ 1ra21', A’_ 21re0c2 In’, forpoints outside alongwirecarrying auniform current. Since 1ra2j, =I,we canalsowrite IA, ='-Q-7?;-665 111T’. Now wecanﬁndBfrom (14.4). There areonlytwoofthesixderivatives that arenotzero. Weget _ I 6 _ I yBx — 5?‘)? '5};11']I‘,— 21re0c2 75> I 6 I xBI, —W5 -gin?’ —5;? 75’ B,=0. Wegetthesame result asbefore: Bcircles around thewire, andhasthemagnitude 121 14-4 Alongsolenoid Next, weconsider again theinﬁnitely long solenoid with acircumferential current onthesurface ofnIperunitlength. (Weimagine there arenturns ofwire perunitlength, carrying thecurrent I,andweneglect theslight pitch ofthewinding.) Justaswehavedeﬁned a“surface charge density” 0',wedeﬁne herea“sur- facecurrent density” Jequal tothecurrent perunitlength onthesurface ofthe solenoid (which is,ofcourse, justtheaverage jtimes thethickness ofthethin winding). Themagnitude ofJis,here, nI.This surface current (seeFig.14-4) has thecomponents. J,=—Jsin ¢, J,=Jcos ¢, J,=0. Now wemust ﬁndAforsuchacurrent distribution. First, wewish toﬁndA,forpoints outside thesolenoid. Theresult isthesame astheelectrostatic potential outside acylinder withasurface charge a=0'0sin¢, with0'0=J/c2.Wehavenotsolved suchacharge distribution, butwehavedone something similar. Thischarge distribution isequivalent totwosolidcylinders of charge, onepositive andonenegative, withaslight relative displacement oftheir 14-5zl i. \i\\\\<=7. P , Y// \\\\\_' ____ P / Fig.14-3. Along cylindrical wire along thez-axis with auniform current density i. 111 1 ' 1 I I I r ‘\ y // J-Jr . / \ '3*\\\% Fig. 14-4. Along solenoid with a surface current density J. lwQ)l ’__ ___ I \\ \J=crv (£11¢ _/ \ / \ ‘Bl--w~ .4 I» _\ L37 J-0' /”‘ \ I \ 1 Fig. 14-5. Arotating charged cylin- derproduces amagnetic ﬁeld inside. A short radial wire rotating withthecylinder hascharges induced onitsends.axesinthey-direction. Thepotential ofsuchapairofcylinders isproportional tothederivative with respect toyofthepotential ofasingle uniformly charged cylinder. Wecould work outtheconstant ofproportionality, butlet’snotworry about itforthemoment. Thepotential ofacylinder ofcharge isproportional tolnr’;thepotential ofthepairisthen éllnr’ y ¢°‘W" W‘ Soweknow that A,=—K . (14.25) where Kissome constant. Following thesame argument, wewould ﬁnd xA,=Kr,—2- (14.26) Although wesaidbefore thatthere wasnomagnetic ﬁeldoutside asolenoid, we ﬁndnowthatthere isanA-ﬁeld which circulates around thez-axis, asinFig.14-4. Thequestion is:Isitscurlzero? Clearly, B,andB,arezero, and _6 x__§_ _1 B:-a("P§l of1 2x2 1 2y2 =K<ﬁ_'ﬁ+7§_?? =°- Sothemagnetic ﬁeldoutside averylongsolenoid isindeed zero, eventhough the vector potential isnot. Wecancheck ourresult against something elseweknow: Thecirculation of thevector potential around thesolenoid should beequal tothefluxofBinside the coil(Eq.14.11). Thecirculation isA-2-irr’or,since A=K/r’, thecirculation is 21rK. Notice thatitisindependent ofr’.That isjustasitshould beifthere isno Boutside, because theﬂuxisjustthemagnitude ofBinside thesolenoid times 11-a2. Itisthesame forallcircles ofradius r’>a.Wehave found inthelastchapter thattheﬁeldinside isnI/eocz, sowecandetermine theconstant K: 21rK=M2ii.soc? or nIa2K Z -42 0 2606' Sothevector potential outside hasthemagnitude 2 A=2l€’0‘5c-5 (14.27) andisalways perpendicular tothevector r’. Wehave been thinking ofasolenoidal coilofwire, butwewould produce thesame ﬁelds ifwerotated along cylinder with anelectrostatic charge onthe surface. Ifwehave athincylindrical shell ofradius awith asurface charge a, rotating thecylinder makes asurface current J=av,where v=awisthevelocity ofthesurface charge. There willthen beamagnetic ﬁeld B=aaw/eocz inside thecylinder. Now wecanraise aninteresting question. Suppose weputashort piece of wire Wperpendicular totheaxisofthecylinder, extending from theaxisoutto thesurface, andfastened tothecylinder sothatitrotates with it,asinFig.14-5. This wire ismoving inamagnetic ﬁeld, sothevXBforces willcause theends of thewire tobecharged (they willcharge upuntil theE-ﬁeld from thecharges just balances thevXBforce). Ifthecylinder hasapositive charge, theendofthewire attheaxiswillhave anegative charge. Bymeasuring thecharge 0T1theendofthe 14-6 wire, wecould measure thespeed ofrotation ofthesystem. Wewould have an “angular-velocity meter”! Butareyouwondering: “What ifIputmyself intheframe ofreference ofthe rotating cylinder? Then there isjustacharged cylinder atrest,andIknow thatthe electrostatic equations saythere willbenoelectric ﬁelds inside, sothere willbeno force pushing charges tothecenter. Sosomething must bewrong.” Butthere is nothing wrong. There isno“relativity ofrotation.” Arotating system isnotan inertial frame, andthelawsofphysics aredifferent. Wemust besuretouseequa- tionsofelectromagnetism onlywithrespect toinertial coordinate systems. Itwould beniceifwecould measure theabsolute rotation oftheearth with suchacharged cylinder, butunfortunately theeffect ismuch toosmall toobserve evenwiththemost delicate instruments nowavailable. 14-5 Theﬁeld ofasmall loop; themagnetic dipole Let’s usethevector-potential method toﬁndthemagnetic ﬁeld ofasmall loopofcurrent. Asusual, by“small” wemean simply thatweareinterested in theﬁelds onlya_tdistances large compared withthesizeoftheloop. Itwillturn outthatanysmall loopisa“magnetic dipole.” That is,itproduces amagnetic ﬁeldliketheelectric ﬁeldfrom anelectric dipole. P z z Ry Y I Z-_-11-4 it Fig.14-6. Arectangular loop ofwire with the Fig.14-7. Thedistribution of1,,in current I.What isthemagnetic ﬁeld atP?(R>>ci,orb.) thecurrent loopofFig.14-6 Wetakeﬁrstarectangular loop, andchoose ourcoordinates asshown in Fig.14-6. There arenocurrents inthez-direction, soA,iszero. There arecurrents inthex-direction onthetwosides oflength a.Ineach leg,thecurrent density (and current) isuniform. Sothesolution forA,isjustliketheelectrostatic po- tential fromtwocharged rods(seeFig.14-7). Since therodshaveopposite charges, theirelectric potential atlarge distances would bejustthedipole potential (Section 6-5). Atthepoint PinFig.14-6, thepotential would be __ 1P'¢Ie,¢-4_n_€0 ——R2 (14.28) where pisthedipole moment ofthecharge distribution. Thedipole moment, in thiscase, isthetotalcharge ononerodtimes theseparation between them: 1;=)(ab_ (14.29) Thedipole moment points inthenegative y-direction, sothecosine oftheangle between Randpis-y/R (where yisthecoordinate ofP).Sowehave __1E2L."’“41.6,,R2R WegetA,simply byreplacing )\by1/c2: _ Iab yAx -’ _' ' “RT, ' 14-7<- b '1 X I, I X II I I -L ’ 4-++-1-++-1-+i zl A R /y X 0-ml‘t Fig. 14-8. Thevector potential ofa small current loop attheorigin (inthe xy-plane); amagnetic dipole ﬁeld.Bythesame reasoning, Iab xA,-IE0? ﬁ- (14.31) Again, A,isproportional toxandA,isproportional to—y,sothevector potential (atlarge distances) goes incircles around thez-axis, circulating inthesame sense asIintheloop, asshown inFig.14-8. Thestrength ofAisproportional toIab,which isthecurrent times thearea oftheloop. This product iscalled themagnetic dipole moment (or,often, just “magnetic moment”) oftheloop. Werepresent itby,uI ,.=Iab. (14.32) Thevector potential ofasmall plane loop ofanyshape (circle, triangle, etc.) is alsogiven byEqs. (14.30) and(14.31) provided wereplace Iabby p=I~(area ofloop). (14.33) Weleave theproof ofthistoyou. Wecanputourequation invector form ifwedeﬁne thedirection ofthevector ittobethenormal totheplane oftheloop, with apositive sense given bytheright- hand rule(Fig. 14-8). Then wecanwrite _ 1MXR_ 1MX¢'1c_ A_41re0c2 R3 _41re0c*-’ R2If (1434) Wehave stilltoﬁndB.Using (14.33) and(14.34), together with (14.4), weget _6#><_Z>5£&'_5E£EF_”'m ““” (where by...wemean /.1/41reoc2), 6 y_ 3yz _<9...i‘_ _i __..._1L ..>.1.2.)__m@_£)T rs ,-.5 Thecomponents oftheB-ﬁeld behave exactly likethose oftheE-ﬁeld fora dipole oriented along thez-axis. (See Eqs. (6.14) and (6.15); also Fig. 6-5.) That’s why wecalltheloop amagnetic dipole. Theword “dipole” isslightly misleading when applied toamagnetic ﬁeldbecause there arenomagnetic “poles” thatcorrespond toelectric charges. Themagnetic “dipole ﬁeld” isnotproduced bytwo“charges,” butbyanelementary current loop. Itiscurious, though, thatstarting with completely different laws, V-E=p/so andVXB=j/15002, wecanendupwith thesame kind ofaﬁeld. Why should thatbe? Itisbecause thedipole ﬁelds appear only when wearefaraway from allcharges orcurrents. Sothrough most oftherelevant space theequations for EandBareidentical: both have zero divergence andzero curl. Sothey give the same solutions. However, thesources whose conﬁguration wesummarize bythe dipole moments arephysically quite different-in onecase, it’sacirculating cur- rent; intheother, apairofcharges, oneabove andonebelow theplane oftheloop forthecorresponding ﬁeld. 14-6 Thevector potential ofacircuit Weareoften interested inthemagnetic ﬁelds produced bycircuits ofwire in which thediameter ofthewire isvery small compared with thedimensions ofthe whole system. Insuch cases, wecansimplify theequations forthemagnetic ﬁeld. 14-8 Forathinwire wecanwrite ourvolume element as dV=Sds, where Sisthecross-sectional area ofthewire anddsistheelement ofdistance along thewire. Infact, since thevector dsisinthesame direction asj,asshown in Fig.14-9 (and wecanassume thatjisconstant across anygiven cross section), wecanwrite avector equation: jdV =jSds. (14.37) ButjSisjustwhat wecallthecurrent Iinawire, soourintegral forthevector potential (14.19) becomes ,4(1)- 1/Id” (14.38)_4776062 7'12 (seeFig.14-10). (Weassume thatIisthesame throughout thecircuit. Ifthere are several branches with different currents, weshould, ofcourse, usetheappropriate Iforeachbranch.) Again, wecanﬁndtheﬁelds from (14.38) either byintegrating directly orby solving thecorresponding electrostatic problems. _ 14-7 ThelawofBiot andSavart Instudying electrostatics wefound thattheelectric ﬁeld ofaknown charge distribution could beobtained directly with anintegral (Eq. 4-16): E(1)=G%[ . Aswehaveseen, itisusually more work toevaluate thisintegral—there arereally three integrals, oneforeachcomponent—than todotheintegral forthepotential andtakeitsgradient. There isasimilar integral which relates themagnetic ﬁeld tothecurrents. Wealready have anintegral forA,Eq.(14.19); wecangetanintegral forBby taking thecurlofbothsides: _ _ 1 J'(2)dV2B(1) -VXA(l) -VX (14.39) Now wemust becareful: The curl operator means taking thederivatives of A(1),thatis,itoperates only onthecoordinates (x1,yl,21). Wecanmove the VXoperator inside theintegral sign ifweremember that itoperates only on variables with thesubscript 1,which ofcourse, appear only in T12=[(x1 _X2)2 +(Y1—y2)2 +(Z1—Z2)2]1/2- (14-40) Wehave, forthex-component ofB, B,,=i4_e_%6y1 621 1 .a1 .a1 r it '1”52; "V2 (‘4-4’) __ 1 -J’1—J/2_-Z1—Z2] _ 47r€0c2/|:Jz ‘-‘i Ju—i_ri;2 dV2. Thequantity inbrackets isjustthex-component of J'XI‘12 =jX¢12_ ":i2 "i2 14-9I/_, Fig.14-9. Foraﬁnewire|'dVisthe same asIds. '|2 ' 2 Fig.‘I4-IO. Themagnetic ﬁeld ofa wire canbeobtained from anintegral around thecircuit. Corresponding results willbefound fortheother components, sowehave 1'212(1)=Hr? ['J%flZ at/2. (14.42) Theintegral gives Bdirectly interms oftheknown currents. Thegeometry in- volved isthesame asthatshown inFig.14-2. Ifthecurrents existonlyincircuits ofsmall wires wecan,asinthelastsection, immediately dotheintegral across thewire, replacing jdVbyIds, where dsisan element oflength ofthewire. Then, using thesymbols inFig. 14-10, 11 d3(1)=_2ﬁr$f_£12%s_2. (14.43) (Theminus signappears because wehavereversed theorder ofthecross product.) Thisequation forBiscalled theBiot-Savart law,after itsdiscoverers. Itgives a formula forobtaining directly themagnetic ﬁeld produced bywires carrying currents. Youmaywonder: “What istheadvantage ofthevector potential ifwecan ﬁndBdirectly withavector integral? After all,Aalsoinvolves three integrals!” Because ofthecross product, theintegrals forBareusually more complicated, as isevident from Eq.(14.41). Also, since theintegrals forAarelikethose ofelectro- statics, wemayalready know them. Finally, wewillseethatinmore advanced theoretical matters (inrelativity, inadvanced formulations ofthelaws ofme- chanics, liketheprinciple ofleast action tobediscussed later, andinquantum mechanics) thevector potential plays animportant role. 14-10 I5 The Vector Potential 1S—1 Theforces onacurrent loop; energy ofadipole Inthelastchapter westudied themagnetic ﬁeld produced byasmall rec- tangular current loop. Wefound thatitisadipole ﬁeld, with thedipole moment given by p.=IA, (15.1) where Iisthecurrent andAisthearea oftheloop. Thedirection ofthemoment isnormal totheplane oftheloop, sowecanalsowrite y.=I/Ill, where nistheunitnormal totheareaA. Acurrent loop—-or magnetic dipole—-not only produces magnetic ﬁelds, but willalsoexperience forces when placed inthemagnetic ﬁeldofother currents. Wewilllookﬁrstattheforces onarectangular loopinauniform magnetic ﬁeld. Letthez-axis bealong thedirection oftheﬁeld, andtheplane oftheloop be placed through they-axis, making theangle 0with thexy~plane asinFig. l5—l. Then themagnetic moment oftheloop—which isnormal toitsplane——wi1l make theangle 6withthemagnetic ﬁeld. Since thecurrents areopposite onopposite sides oftheloop, theforces are alsoopposite, sothere isnonetforce ontheloop (when theﬁeldisuniform). Because offorces onthetwosidesmarked 1and2intheﬁgure, however, there isa torque which tends torotate theloopabout they-axis. Themagnitude ofthese forces F1andF2is F1=F2=IBb. Their moment armis asin0, sothetorque is -r=labBsin6, or,since Iabisthemagnetic moment oftheloop, 7'=/4BSin9. Thetorque canbewritten invector notation: -r=p.XB. (15.2) Although-we have only shown thatthetorque isgiven byEq.(15.2) inonerather special case, theresult isright forasmall loop ofanyshape, aswewillsee.Youwill remember thatwefound thesame kind ofrelation forthetorque onanelectric dipole: 1=pXE. Wenow askabout themechanical energy ofourcurrent loop. Since there is atorque, theenergy evidently depends ontheorientation. Theprinciple ofvirtual work saysthatthetorque istherateofchange ofenergy with angle, sowecanwrite dU=—'rd0. 15-115-1 Theforces onacurrent loop; energy ofadipole 15-2 Mechanical andelectrical energies 15-3 Theenergy ofsteady currents 15-4 Bversus A 15-5 Thevector potential and quantum mechanics 15-6 What istrueforstatics is false fordynamics Z Y B .A F. /4 .1“ 1\ X F3 ‘\ 4' >1‘ u 1 1 o\/\/b Fig.l5—l. Arectangular loopcarry- ingthecurrent !sitsinauniform ﬁeld B (inthez-direction). The torque onthe loop is-r=itXB,where themagnetic moment u=lab. Fig.15-2. Aloop iscarried along thex-direction through theﬁeld B,at right angles tox.Setting 1'=—-/.tB sin0,andintegrating, wecanwrite fortheenergy U=—p.Bcos0+aconstant. (15.3) (Thesignisnegative because thetorque triestolineupthemoment withtheﬁeld; theenergy islowest when itandBareparallel.) Forreasons which wewilldiscuss later, thisenergy isnotthetotal energy ofa current loop. (Wehave, foronething, nottaken intoaccount theenergy required tomaintain thecurrent intheloop.) Wewill, therefore, callthisenergy Umech, toremind usthatitisonly partoftheenergy. Also, since weareleaving outsome oftheenergy anyway, wecansettheconstant ofintegration equal tozero inEq. (15.3). Sowerewrite theequation: Umech =_I" ' Again, thiscorresponds toourresult foranelectric dipole: U=—p-E. (15.5) Now theelectrostatic energy UinEq.(15.5) isthetrue energy, butUmech in (15.4) isnottherealenergy. Itcan,however, beused incomputing forces, bythe principle ofvirtual work, supposing thatthecurrent intheloop—-or atleast ;.t—is keptconstant. Wecanshow forourrectangular loop that Umech alsocorresponds tothe mechanical work done inbringing theloop intotheﬁeld. Thetotal force onthe loop iszero only inauniform ﬁeld; inanonuniform ﬁeld there arenetforces ona current loop. Inputting theloop intoaregion with aﬁeld, wemust have gone through places where theﬁeld wasnotuniform, andsowork wasdone. Tomake thecalculation simple, weshall imagine thattheloop isbrought intotheﬁeldwith itsmoment pointing along theﬁeld. (Itcanberotated toitsﬁnal position after it isinplace.) Imagine thatwewant tomove theloopinthex-direction—toward aregion of stronger ﬁeld—and that theloop isoriented asshown inFig. 15-2. Westart somewhere where theﬁeld iszero andintegrate theforce times thedistance aswe bring theloopintotheﬁeld. B F' F2/. _ /, A /X| /X2 First, let’scompute thework done oneachsideseparately andthentakethe sum(rather than adding theforces before integrating). Theforces onsides 3and4 areatright angles tothedirection ofmotion, sonowork isdone onthem. The force onside2isIbB(x) inthex-direction, andtogetthework done against the magnetic forces wemust integrate thisfrom some xwhere theﬁeld iszero, sayat x=—oo,tox2,itspresent position: W2=-F2dx=-11>/"B(x)dx. (15.6) Similarly, thework done against theforces onside1is W,=-/”‘F1dx=11>/”‘B(x)dx. (15.7) 15-2 Toﬁndeach integral, weneed toknow how B(x) depends onx.Butnotice that side lfollows along right behind side2,sothatitsintegral includes most ofthe work done onside2.Infact, thesumof(15.6) and(15.7) isjust W=—Ib/“’B(x)dx. (15.8) Butifweareinaregion where Bisnearly thesame onboth sides 1and2,wecan write theintegral as '/I2B(x) dx=(x2--x1)B =aB, $1 where Bistheﬁeld atthecenter oftheloop. Thetotal mechanical energy wehave putinis Um... =W=—IabB=—;.tB. (15.9) Theresult agrees with theenergy wetook forEq.(15.4). WeWould, ofcourse, have gotten thesame result ifwehadadded theforces ontheloop before integrating toﬁndthework. IfweletB1betheﬁeld atside1 andB2betheﬁeldatside2,thenthetotalforce inthex-direction is F,=Ib(B2 -B1). Iftheloopis“small,” thatis,ifB2andB1arenottoodifferent, wecanwrite 6B 8BB2= B1+5AX= B1-I-'5}-(1. Sotheforce is F,=Iab-3% (15.10) Thetotalwork done ontheloopbyexternal forces is Z —/ F,dx=-—Iab/Qdx =—IabB,_,,, dx which isagain just—;tB. Only nowweseewhyitisthattheforce onasmall current loopisproportional tothederivative ofthemagnetic ﬁeld, aswewould expect from F,Ax=—AU,,,,,,), ==-A(—p.'B). (15.11) Ourresult, then, isthateven though Umech =—;.t-Bmaynotinclude allthe energy ofasystem——it isafakekind ofenergy—it canstillbeused withtheprinciple ofvirtual work toﬁndtheforces onsteady current loops. 15-2 Mechanical andelectrical energies Wewant nowtoshow whytheenergy Umech discussed intheprevious section isnotthecorrect energy associated withsteady currents-—that itdoesnotkeep track ofthetotal energy intheworld. Wehave, indeed, emphasized thatitcan beusedliketheenergy, forcomputing forces from theprinciple ofvirtual work, provided thatthecurrent intheloop (and allother currents) donotchange. Let’s seewhyallthisworks. Imagine thattheloopinFig.15-2ismoving inthe+x-direction andtakethe z-axis inthedirection ofB.Theconduction electrons inside2willexperience a force along thewire, inthey-direction. Butbecause oftheirﬂow—-as anelectric current-—there isacomponent oftheirmotion inthesame direction astheforce. Each electron is,therefore, having work done onitattherateF,,v,,, where 0,,isthe component oftheelectron velocity along thewire. Wewillcallthiswork done on theelectrons electrical work. Now itturns outthat iftheloop ismoving ina umform ﬁeld, thetotalelectrical work iszero, since positive work isdone onsome parts oftheloopandanequal amount ofnegative work isdone onother parts. 15-3 Butthisisnottrueifthecircuit ismoving inanonuniform ﬁeld—then there will beanetamount ofwork done ontheelectrons. Ingeneral, thiswork would tend tochange theﬂowoftheelectrons, butifthecurrent isbeing heldconstant, energy must beabsorbed ordelivered bythebattery orother source thatiskeeping the current steady. This energy wasnotincluded when wecomputed Um“), inEq. (15.9), because ourcomputations included only themechanical forces onthebody ofthewire. Youmaybethinking: Buttheforce ontheelectrons depends onhowfast thewireismoved; perhaps ifthewire ismoved slowly enough thiselectrical energy canbeneglected. Itistruethattherateatwhich theelectrical energy isdelivered isproportional tothespeed ofthewire, butthetotal energy delivered ispropor- tional alsotothetimethatthisrategoeson.Sothetotalelectrical energy ispro- portional tothevelocity times thetime, which isjustthedistance moved. Fora given distance moved inaﬁeld thesame amount ofelectrical work isdone. Let’s consider asegment ofwireofunitlength carrying thecurrent Iandmov- inginadirection perpendicular toitself andtoamagnetic ﬁeldBwiththespeed vwire. Because ofthecurrent theelectrons willhave adrift velocity vdrm along the wire. Thecomponent ofthemagnetic force oneach electron inthedirection ofthe drift isq,vw,,.,B. Sotherateatwhich electrical work isbeing done isFvdm, = (q,vwi,eB)vd,m. Ifthere areNconduction electrons intheunitlength ofthewire, thetotal rateatwhich electrical work isbeing done is dU =NqevwireBvdrift~ ButNqevdrm =I,thecurrent inthewire, so dU O ﬁ%i =I77wireB- Now since thecurrent isheldconstant, theforces ontheconduction electrons donotcause them toaccelerate; theelectrical energy isnotgoing intotheelectrons butintothesource thatiskeeping thecurrent constant. Butnotice thattheforce onthewireisIB,soIBvw,,,, isalsotherateofme- chanical work done onthewire, dUm,,,,1,/dt =IBvw,,,,. Weconclude thatthe mechanical work done onthewire isjustequal totheelectrical work done onthe current source, sotheenergy oftheloop isaconstant! This isnotacoincidence, butaconsequence ofthelawwealready know. Thetotal force oneach charge inthewire is F=q(E+vXB). Therateatwhich work isdone is v-F=q[v-E+v-(vXB)]. (15.12) Ifthere arenoelectric ﬁelds wehave only thesecond term, which isalways zero. Weshall seelater that changing magnetic ﬁelds produce electric ﬁelds, soour reasoning applies only tomoving wires insteady magnetic ﬁelds. How isitthen that theprinciple ofvirtual work gives theright answer? Because westillhave nottaken intoaccount thetotalenergy oftheworld. Wehave notincluded theenergy ofthecurrents thatareproducing themagnetic ﬁeld we start outwith. Suppose weimagine acomplete system suchasthatdrawn inFig.15—3(a), in which wearemoving ourloop with thecurrent I1intothemagnetic ﬁeld B1pro- duced bythecurrent I2inacoil. Now thecurrent I1intheloop willalsobepro- ducing some magnetic ﬁeld B2atthecoil. Iftheloop ismoving, theﬁeld B2will bechanging. Asweshall seeinthenext chapter, achanging magnetic ﬁeld gen- erates anE-ﬁeld; andthisE-ﬁeld willdowork onthecharges inthecoil. This energy must alsobeincluded inourbalance sheet ofthetotal energy. 15-4 I2 l./ ___/'B. it”lF1&1 iB2 +52 I, Loop Y I, Z 2 la) lb) Fig. 15-3. Finding theenergy ofasmall loop inamagnetic ﬁeld. Wecould wait until thenext chapter toﬁndoutabout thisnewenergy term, butwecanalsoseewhat itwillbeifweusetheprinciple ofrelativity inthefollowing way. When wearemoving thelooptoward thestationary coilweknow thatits electrical energy isjustequal andopposite tothemechanical work done. So Umech +Uelect(loOp) = Suppose nowwelook atwhat ishappening from adifferent point ofview, inwhich theloop isatrest,andthecoilismoved toward it.Thecoilisthenmoving intotheﬁeld produced bytheloop. Thesame arguments would givethat Umeoh +Uelect(cOi1) = Themechanical energy isthesame inthetwocases because itcomes from theforce between thetwocircuits. Thesumofthetwoequations gives 2Umech +Ue1ect(l°°P) +Uelect(coil) = Thetotal energy ofthewhole system is,ofcourse, thesumofthetwoelectrical energies plusthemechanical energy taken onlyonce. Sowehave Utotal =Uelect(loOp) +Uelect(coil) +Umech =_Umech- Thetotal energy oftheworld isreally thenegative ofUmech. Ifwewant the trueenergy ofamagnetic dipole, forexample, weshould write Utotal =+I-‘ 'B- Itisonly ifwemake thecondition thatallcurrents areconstant thatwecanuse only apartoftheenergy, Umech (which isalways thenegative ofthetrueenergy), toﬁndthemechanical forces. Inamore general problem, wemust becareful to include allenergies. Wehave seenananalogous situation inelectrostatics. Weshowed thatthe energy ofacapacitor isequal toQ2/2C. When weusetheprinciple ofvirtual work toﬁndtheforce between theplates ofthecapacitor, thechange inenergy isequal toQ2/2 times thechange in1/C.That is, 2 2 AU=—Q2—A(%.)= -92-%% (15.14) Now suppose thatwewere tocalculate thework done inmoving twocon- ductors subject tothedifferent condition thatthevoltage between them isheld constant. Then wecangettheright answers forforce from theprinciple ofvirtual work ifwedosomething artiﬁcial. Since Q=CV,therealenergy is%CV2.But ifwedeﬁne anartiﬁcial energy equal to—%CV2, then theprinciple ofvirtual work canbeusedtogetforces bysetting thechange intheartiﬁcial energy equal tothe 15-5T B mi ,4nn_.unn\ _..Annunnnn\n_ ”nnnannnnnnnunnnnnnnn‘IEIIII~rr1Innn:..‘_'"—_!I I Surface SInI‘:55’ Fig.15-4. Theenergy ofalarge loop inamagnetic ﬁeld canbeconsidered asthesumofenergies ofsmaller loops.mechanical work, provided thatweinsist thatthevoltage Vbeheldconstant. Then 2 2 110...... =A(- =-22-AC, (15.15) which isthesame asEq.(15.14). Wegetthecorrect result eventhough weare neglecting thework done bytheelectrical system tokeep thevoltage constant. Again, thiselectrical energy isjusttwice asbigasthemechanical energy andof theopposite sign. Thus ifwecalculate artiﬁcially, disregarding thefactthatthesource ofthe potential hastodowork tomaintain thevoltages constant, wegettherightanswer. Itisexactly analogous tothesituation inmagnetostatics. 15-3 Theenergy ofsteady currents Wecannowuseourknowledge thatU,,,t,,1 =—Umech toﬁndthetrueenergy ofsteady currents inmagnetic ﬁelds. Wecanbegin with thetrueenergy ofasmall current loop. Calling U,,,,,,1 justU,wewrite U=pt-B. (15.16) Although wecalculated thisenergy foraplane rectangular loop, thesame result holds forasmall plane loop ofanyshape. Wecanﬁndtheenergy ofacircuit ofanyshape byimagining thatitismade upofsmall current loops. Saywehaveawireintheshape oftheloopI‘ofFig. 15-4. Weﬁllinthiscurve withthesurface S,andonthesurface mark outalarge number ofsmall loops, eachofwhich canbeconsidered plane. Ifweletthecurrent Icirculate around each ofthelittle loops, thenetresult willbethesame asacurrent around I‘,since thecurrents willcancel onalllinesinternal toP.Physically, the system oflittle currents isindistinguishable from theoriginal circuit. The energy mustalsobethesame, andsoisjustthesumoftheenergies ofthelittleloops. IftheareaofeachlittleloopisAa,itsenergy isIAaB,,, where B,,isthecom- ponent normal toAa.Thetotalenergy is U=2IB,,Aa. Going tothelimit ofinﬁnitesimal loops, thesumbecomes anintegral, and U=1/B,da=IIB-nda, (15.17) where nistheunitnormal toda. IfwesetB=VXA,wecanconnect thesurface integral toalineintegral, using Stokes’ theorem, 1/S(v><A)-nda =rylrn-ds, (15.18) where dsisthelineelement along I‘.Sowehave theenergy foracircuit ofany shape: U=156A-.11. (15.19) circuit Inthisexpression Arefers, ofcourse, tothevector potential duetothose currents (other thantheIinthewire) which produce theﬁeldBatthewire. Now anydistribution ofsteady currents canbeimagined tobemade upof ﬁlaments thatrunparallel tothelinesofcurrent flow. Foreachpairofsuchcircuits, theenergy isgiven by(15.19), where theintegral istaken around onecircuit, using thevector potential Afrom theother circuit. Forthetotal energy wewant the sumofallsuch pairs. If,instead ofkeeping track ofthepairs, wetakethecomplete sum over alltheﬁlaments, wewould becounting theenergy twice (wesawa similar effect inelectrostatics), sothetotal energy canbewritten U=sf;-.4 dV. (15.20) 15-6 Thisformula corresponds totheresult wefound fortheelectrostatic energy: U=s/pi.av. (15.21) Sowemayifwewish think ofAasakind ofpotential energy forcurrents in magnetostatics. Unfortunately, thisideaisnottoouseful, because itistrueonly forstatic ﬁelds. Infact,neither oftheequations (15.20) and(15.21) gives thecor- rectenergy when theﬁelds change withtime. 15-4 Bversus A _ Inthissection wewould liketodiscuss thefollowing questions: Isthevector potential merely adevice which isuseful inmaking calculations—as thescalar potential isuseful inelectrostatics—or isthevector potential a“real” ﬁeld? Isn’t themagnetic ﬁeldthe“rea1” ﬁeld, because itisresponsible fortheforce ona moving particle? First weshould saythatthephrase “arealﬁeld” isnotvery meaningful. Foronething, youprobably don’t feelthatthemagnetic ﬁeldis very“real” anyway, because even thewhole ideaofaﬁeldisarather abstract thing. Youcannot putoutyourhand andfeelthemagnetic ﬁeld. Furthermore, thevalue ofthemagnetic ﬁeldisnotverydeﬁnite; bychoosing asuitable moving coordinate system, forinstance, youcanmake amagnetic ﬁeldatagiven point disappear. What wemean herebya“real” ﬁeldisthis: arealﬁeldisamathematical function weuseforavoiding theideaofaction atadistance. Ifwehaveacharged particle attheposition P,itisaffected byother charges located atsome distance fromP.Onewaytodescribe theinteraction istosaythattheother charges make some “condition”—whatever itmay be—-in theenvironment atP.Ifweknow thatcondition, which wedescribe bygiving theelectric andmagnetic ﬁelds, then wecandetermine completely thebehavior oftheparticle—with nofurther reference tohowthose conditions came about. Inother words, ifthose other charges were altered insome way, butthe conditions atPthataredescribed bytheelectric andmagnetic ﬁeldatPremain thesame, then themotion ofthecharge willalsobethesame. A“real” ﬁeld is thenasetofnumbers wespecify insuchawaythatwhat happens atapoint depends onlyonthenumbers atthatpoint. Wedonotneedtoknow anymore about what’s going onatother places. Itisinthissense thatwewilldiscuss whether thevector potential isa“real” ﬁeld. You maybewondering about thefactthatthevector potential isnotunique- thatitcanbechanged byadding thegradient ofanyscalar with nochange atall intheforces onparticles. Thathasnot,however, anything todowiththequestion ofreality inthesense thatwearetalking about. Forinstance, themagnetic ﬁeld isinasense altered byarelativity change (asarealsoEandA).Butwearenot worried about what happens iftheﬁeld canbechanged inthisway. That doesn’t really make anydifference; thathasnothing todowiththequestion ofwhether thevector potential isaproper “real”' ﬁeldfordescribing magnetic effects, or whether itisjustauseful mathematical tool. Weshould alsomake some remarks ontheusefulness ofthevector potential A.Wehaveseenthatitcanbeusedinaformal procedure forcalculating themag- netic ﬁelds ofknown currents, justas¢canbeused toﬁndelectric ﬁelds. In electrostatics wesawthat¢wasgiven bythescalar integral ¢(1)=-1-I'12)at/2. (15.22)41l'€() 7'12 From this¢,wegetthethree components ofEbythree differential operations. This procedure isusually easier tohandle than evaluating thethree integrals in thevector formula1 212(1)=Z;-raj’-’(_r-1,)? at/2. (15.23) First, there arethree integrals; andsecond, eachintegral isingeneral somewhat more diﬁicult. 15-7 Theadvantages aremuch lessclear formagnetostatics. Theintegral forAis already avector integral: 4(1)=_1_fj(l)i‘l-I53. (15.24)4-7l'€()C2 T12 which is,ofcourse, three integrals. Also, when wetake thecurlofAtogetB,we have sixderivatives todoandcombine bypairs. Itisnotimmediately obvious whether inmost problems thisprocedure isreally anyeasier than computing B directly from 1 '2><3(1)=WM/'i% at/2. (15.25) Using thevector potential isoften more difﬁcult forsimple problems forthe following reason. Suppose weareinterested only inthemagnetic ﬁeld Batone point, andthattheproblem hassome nicesymmetry——say wewant theﬁeld ata point ontheaxisofaringofcurrent. Because ofthesymmetry, wecaneasily get Bbydoing theintegral ofEq.(15.25). If,however, wewere toﬁndAﬁrst, wewould have tocompute Bfrom derivatives ofA,sowemust know what Aisatallpoints intheneighborhood ofthepoint ofinterest. And most ofthese points areoffthe axisofsymmetry, sotheintegral forAgetscomplicated. Intheringproblem, for example, wewould need touseelliptic integrals. Insuch problems, Aisclearly notvery useful. Itistruethatinmany complex problems itiseasier towork with A,butitwould behard toargue thatthisease oftechnique would justify making youlearn about onemore vector ﬁeld. Wehave introduced Abecause itdoeshave animportant physical signiﬁcance. Notonly isitrelated totheenergies ofcurrents, aswesawinthelastsection, but itisalsoa“real” physical ﬁeld inthesense thatwedescribed above. Inclassical mechanics itisclear thatwecanwrite theforce onaparticle as F=q(E+v><3). (15-26) sothat, given theforces, everything about themotion isdetermined. Inanyregion where B=0even ifAisnotzero, such asoutside asolenoid, there isnodis- cernible effect ofA.Therefore foralong time itwasbelieved thatAwasnota “real” ﬁeld. Itturns out,however, thatthere arephenomena involving quantum mechanics which show thattheﬁeld Aisinfacta“real” ﬁeld inthesense wehave deﬁned it.Inthenext section wewillshow youhow thatworks. 15-5 Thevector potential andquantum mechanics There aremany changes inwhat concepts areimportant when wegofrom classical toquantum mechanics. Wehave already discussed some ofthem in Vol. I.Inparticular, theforce concept gradually fades away, while theconcepts ofenergy andmomentum become ofparamount importance. You remember that instead ofparticle motions, onedeals with probability amplitudes which vary in space andtime. Inthese amplitudes there arewavelengths related tomomenta, andfrequencies related toenergies. Themomenta andenergies, which determine thephases ofwave functions, aretherefore theimportant quantities inquantum mechanics. Instead offorces, wedealwiththewayinteractions change thewave- length ofthewaves. Theideaofaforce becomes quite secondary—if itisthere at all.When people talkabout nuclear forces, forexample, what theyusually analyze andwork witharetheenergies ofinteraction oftwonucleons, andnottheforce between them. Nobody everdifferentiates theenergy toﬁndoutwhat theforce looks like. Inthissection wewant todescribe howthevector andscalar poten- tials enter into quantum mechanics. Itis,infact, justbecause momentum and energy playacentral roleinquantum mechanics thatAandd>provide themost direct wayofintroducing electromagnetic effects intoquantum descriptions. Wemust review alittle how quantum mechanics works. Wewillconsider again theimaginary experiment described inChapter 37ofVol. I,inwhich elec- 15-8 x 1 ‘\ *l\‘FIIIIIIIIIIIILDETECTO SOURCE __________ Irl~f Z’; Z ,:_, T ‘K/’ ‘ I 5: “ii” ) \\ \ -l-o. \\\ 6._? ‘o WALL L Fig.15-5. Aninterference experiment with electrons lseealsoChapter 37ofVol.I). trons arediffracted bytwoslits. Thearrangement isshown again inFig.15-5. Electrons, allofnearly thesame energy, leave thesource andtravel toward awall withtwonarrow slits. Beyond thewallisa“backstop” withamovable detector. Thedetector measures therate,which wecallI,atwhich electrons arrive atasmall region ofthebackstop atthedistance xfrom theaxisofsymmetry. Therateis proportional totheprobability thatanindividual electron thatleaves thesource willreach thatregion ofthebackstop. Thisprobability hasthecomplicated-looking distribution shown intheﬁgure, which weunderstand asduetotheinterference of twoamplitudes, onefrom each slit. Theinterference ofthetwoamplitudes depends ontheirphase difference. Thatis,iftheamplitudes areC1e“’1andC2e“"=, thephase difference 6=11>;—2determines their interference pattern [seeEq. (29.12) inVol.I].Ifthedistance between thescreen andtheslitsisL,andifthe difference inthepath lengths forelectrons going through thetwoslitsisa,as shown intheﬁgure, thenthephase difference ofthetwowaves isgiven by U 6-X- (15.27) Asusual, weletit=A/21r, where Aisthewavelength ofthespace variation ofthe probability amplitude. Forsimplicity, wewillconsider only values ofxmuch lessthanL;thenwecanset a=5dL and xd6-ZX- (15.28) When xiszero, 5iszero; thewaves areinphase, andtheprobability hasamaxi- mum. When 6is1r,thewaves areoutofphase, theyinterfere destructively, andthe probability isaminimum. Sowegetthewavy function fortheelectron intensity. Now wewould liketostate thelawthatforquantum mechanics replaces the force lawF=qvXB.Itwillbethelawthatdetermines thebehavior ofquantum- mechanical particles inanelectromagnetic ﬁeld. Since what happens isdetermined byamplitudes, thelawmust tellushow themagnetic influences affect theampli- tudes; wearenolonger dealing with theacceleration ofaparticle. Thelawisthe following: thephase oftheamplitude toarrive viaanytrajectory ischanged by thepresence ofamagnetic ﬁeldbyanamount equal totheintegral ofthevector potential along thewhole trajectory times thecharge oftheparticle overPlanck’s constant. That is, Magnetic change inphase=gX.4-ds. (15.29) trajectory 15-9 Ifthere were nomagnetic ﬁeldthere would beacertain phase ofarrival. Ifthere is amagnetic ﬁeldanywhere, thephase ofthearriving wave isincreased bytheintegral inEq.(15.29). Although wewillnotneed touseitforourpresent discussion, wemention thattheeffect ofanelectrostatic ﬁeldistoproduce aphase change given bythe negative ofthetimeintegral ofthescalar potential ¢: Electric change inphase =—%/¢dt. These twoexpressions arecorrect notonly forstatic ﬁelds, buttogether givethe correct result foranyelectromagnetic ﬁeld, static ordynamic. This isthelawthat replaces F=q(E+vXB).Wewant now, however, toconsider only astatic magnetic ﬁeld. Suppose thatthere isamagnetic ﬁeld present inthetwo-slit experiment. We want toaskforthephase ofarrival atthescreen ofthetwowaves whose paths pass through thetwoslits. Their interference determines where themaxima inthe probability willbe.Wemay call1thephase ofthewave along trajectory (1). If1(B =0)isthephase without themagnetic ﬁeld, then when theﬁeld isturned onthephase willbe <1>,=q>,(3=0)+gfA'ds. (15.30)<1) Similarly, thephase fortrajectory (2)is <1»,=<1>,(3=0)+51/A-ds. (15.31)71<2) Theinterference ofthewaves atthedetector depends onthephase difference 5=¢,(3= 0)-¢>,(3=0)-riff A-ds— ifA-ds. (15.32)fl(1) ll<2) Theno-ﬁeld difference wewillcall6(B=0);itisjust thephase difference we have calculated above inEq.(15.28). Also, wenotice thatthetwointegrals can bewritten asoneintegral thatgoes forward along (1)andback along (2);wecall thistheclosed path (1-2). Sowehave 8=a(3=0)+,5-gj€1_2).4 'ds. (15.33) This equation tellsushow theelectron motion ischanged bythemagnetic ﬁeld; withitwecanﬁndthenewpositions oftheintensity maxima andminima atthe backstop. Before wedothat, however, wewant toraise thefollowing interesting and important point. You remember that thevector potential function hassome arbitrariness. Two different vector potential functions AandA’whose difference isthegradient ofsome scalar function V1//,both represent thesame magnetic ﬁeld, since thecurlofagradient iszero. They give, therefore, thesame classical force qvXB.Ifinquantum mechanics theeffects depend onthevector potential, which ofthemany possible A-functions iscorrect? Theanswer isthatthesame arbitrariness inAcontinues toexist forquantum mechanics. IfinEq.(15.33) wechange AtoA’=A+V¢,theintegral on Abecomes f A’-ds=f A'ds+f V1//-ds.(1-2) (1-2) (1-2) Theintegral ofV10isaround theclosed path (1-2), buttheintegral ofthetangential component ofagradient onaclosed path isalways zero, byStokes’ theorem. Therefore both AandA’givethesame phase differences andthesame quantum- mechanical interference effects. Inboth classical andquantum theory itisonly the curlofAthatmatters; anychoice ofthefunction ofAwhich hasthecorrect curl gives thecorrect physics. 15-10 Thesame conclusion isevident ifweusetheresults ofSection 14-1. There wefound thatthelineintegral ofAaround aclosed path istheﬂuxofBthrough thepath, which here istheﬂuxbetween paths (1)and(2). Equation (15.33) can, ifwewish, bewritten as 8=.s(3=0)+%[ﬂuxofBbetween (1)and (2)1, (15.34) where bytheﬂuxofBwemean, asusual, thesurface integral ofthenormal com- ponent ofB.Theresult depends onlyonB,andtherefore onlyonthecurlofA. Now because wecanwrite theresult interms ofBaswellasinterms ofA, youmight beinclined tothink thattheBholds itsownasa“real” ﬁeldandthat theAcanstillbethought ofasanartiﬁcial construction. Butthedeﬁnition of “real” ﬁeldthatweoriginally proposed wasbased ontheideathata“real” ﬁeld would notactonaparticle from adistance. Wecan, however, giveanexample inwhich Biszero—or atleastarbitrarily small—at anyplace where there issome chance toﬁndtheparticles, sothatitisnotpossible tothink ofitacting directly onthem. Youremember thatforalongsolenoid carrying anelectric current there is aB-ﬁeld inside butnone outside, while there islotsofAcirculating around outside, asshown inFig.15-6. Ifwearrange asituation inwhich electrons aretobefound onlyoutside ofthesolenoid-—only where there isA—there willstillbeaninﬂuence onthemotion, according toEq.(15.33). Classically, thatisimpossible. Classically, theforce depends onlyonB;inorder toknow thatthesolenoid iscarrying current, theparticle must gothrough it.Butquantum-mechanically youcanﬁndoutthat there isamagnetic ﬁeldinside thesolenoid bygoing around it—without evergoing close toit! Suppose thatweputaverylongsolenoid ofsmall diameter justbehind the wallandbetween thetwoslits, asshown inFig.15-7. Thediameter ofthesolenoid istobemuch smaller thanthedistance dbetween thetwoslits. Inthese circum- stances, thediffraction oftheelectrons attheslitgives noappreciable probability thattheelectrons willgetnearthesolenoid. What willbetheeffect onourinter- ference experiment? ‘ /4''III/IIIIIIII1. \\\\ xiSOURCE —--‘""lq')"_'_1-ifit?-(Fig.15-6. Themagnetic ﬁeld and vector potential ofalongsolenoid. I\\\\0.1‘I'“” 1 /,1O. VIAV.43,’ :_ ,_*_ T T “~.® n|____, SOLENOID LINES OFB L Fig.15-7. Amagnetic ﬁeldcaninﬂuence themotion ofelectrons even though itexists onlyinregions where there isanarbitrarily small probability ofﬁnding the electrons. Wecompare thesituation withandwithout acurrent through thesolenoid. Ifwehavenocurrent, wehavenoBorAandwegettheoriginal pattern ofelec- tronintensity atthebackstop. Ifweturnthecurrent oninthesolenoid andbuild upamagnetic ﬁeldBinside, thenthere isanAoutside. There isashiftinthe phase difference proportional tothecirculation ofAoutside thesolenoid, which will mean thatthepattern ofmaxima andminima isshifted toanewposition. Infact, sincetheﬂuxofBinside isaconstant foranypairofpaths, soalsoisthecircula- tionofA.Forevery arrival point there isthesame phase change; thiscorresponds 15-ll toshifting theentire pattern inxbyaconstant amount, sayxo,thatwecaneasily calculate. Themaximum intensity willoccur where thephase diﬁ'erence between thetwowaves iszero. Using Eq.(15.32) orEq.(15.33) for6andEq.(15.28) for 6(B=0),wehave __5qf , x0- (111,; u_2)A ds, (15.35) Or x,=-511,‘-ll[ﬂuxofBbetween (1)and(2)1. (15.36) Thepattern with thesolenoid inplace should appear* asshown inFig. 15-7. At least, thatistheprediction ofquantum mechanics. Precisely thisexperiment hasrecently been done. Itisavery, very difficult experiment. Because thewavelength oftheelectrons issosmall, theapparatus must beonatinyscaletoobserve theinterference. Theslitsmust beveryclose together, andthatmeans thatoneneeds anexceedingly small solenoid. Itturns outthatin certain circumstances, iron crystals willgrow intheform ofvery long, microsco- pically thinﬁlaments called whiskers. When these iron whiskers aremagnetized they arelikeatinysolenoid, andthere isnoﬁeld outside except near theends. Theelectron interference experiment wasdone with such awhisker between two slits, andthepredicted displacement inthepattern ofelectrons wasobserved. Inoursense then, theA-ﬁeld is“real.” You maysay:“But there wasamag netic ﬁeld.” There was, butremember ouroriginal idea—that aﬁeldis“real” ifitis what must bespeciﬁed attheposition oftheparticle inorder togetthemotion. TheB-ﬁeld inthewhisker actsatadistance. Ifwewant todescribe itsinﬂuence notasaction-at-a-distance, wemust usethevector potential. This subject hasaninteresting history. Thetheory wehave described was known from thebeginning ofquantum mechanics in1926. Thefactthatthevector potential appears inthewave equation ofquantum mechanics (called theSchrod- inger equation) wasobvious fromthedayitwaswritten. Thatitcannot bereplaced bythemagnetic ﬁeldinanyeasywaywasobserved byonemanafter theother whotried todoso.Thisisalsoclear from ourexample ofelectrons moving ina region where there isnoﬁeld andbeing affected nevertheless. Butbecause in classical mechanics Adidnotappear tohave anydirect importance and, further- more, because itcould bechanged byadding agradient, people repeatedly said thatthevector potential hadnodirect physical signiﬁcance—that onlythemagnetic andelectric ﬁelds are“right” even inquantum mechanics. Itseems strange in retrospect that noonethought ofdiscussing thisexperiment until 1956, when Bohm andAharanov ﬁrstsuggested 1tandmade thewhole question crystal clear. Theimplication wasthere allthetime, butnoonepaidattention toit.Thus many people were rather shocked when thematter wasbrought up.That’s why someone thought itwould beworth while todotheexperiment toseethatitreally wasright, even though quantum mechanics, which hadbeen believed forsomany years, gave anunequivocal answer. Itisinteresting thatsomething likethiscan bearound forthirty years but,because ofcertain prejudices ofwhat isandisnot signiﬁcant, continues tobeignored. Now wewish tocontinue inouranalysis alittle further. Wewillshow the connection between thequantum-mechanical formula andtheclassical formula— toshow why itturns outthatifwelook atthings onalarge enough scale itwill look asthough theparticles areacted onbyaforce equal toqvXthecurlofA. Togetclassical mechanics from quantum mechanics, weneedtoconsider cases in which allthewavelengths arevery small compared with distances over which ex- ternal conditions, likeﬁelds, vary appreciably. Weshall notprove theresult in great generality, butonly inavery simple example, toshow how itworks. Again weconsider thesame slitexperiment. Butinstead ofputting allthemagnetic ﬁeld inavery tinyregion between theslits, weimagine amagnetic ﬁeld thatextends *IftheﬁeldBcomes outoftheplane oftheﬁgure, theﬂuxaswehave deﬁned itis negative andx0ispositive. 15-12/ 1 1 \\1 l \ I ,: s.. \'- "\ _a IQ) Ax ‘*~ .., souncs ,_-_ _rl.-_.-_'_T._”__ e”'1I1‘\\ FIZ.-_,_,1-- --1'._; l ‘Q-Q.1!//I‘i..,_..r ,~ T ,’_,v’-=11”\ ‘\_\\ *~ _. t .|'-- - ,. ' /If L *.unssore *\ 12 ___- .,, ,_,--_ o ~42‘I§ ag-I Fig.15-8. Theshiftoftheinterference pattern duetoostrip ofmagnetic ﬁeld. overalarger region behind theslits,asshown inFig.15-8. Wewilltaketheideal- izedcasewhere wehave amagnetic ﬁeldwhich isuniform inanarrow strip of width w,considered small ascompared withL.(That caneasily bearranged; the backstop canbeputasfaroutaswewant.) Inorder tocalculate theshiftinphase, wemust takethetwointegrals ofAalong thetwotrajectories (1)and(2).They differ, aswehaveseen, merely bythefluxofBbetween thepaths. Toourapproxi- mation, theﬂuxisBwd. Thephase difference forthetwopaths isthen 8=8(3=0)+gBwd. (15.37) Wenotethat, toourapproximation, thephase shiftisindependent oftheangle. Soagain theeffect willbetoshiftthewhole pattern upward byanamount Ax. Using Eq.(15.28), Lx LxAx-?/18 =-d-[8-8(3-0)]. Using(15.37) for8-8(3=0), Ax=L8gtBw. (15.38) Such ashift isequivalent todeflecting allthetrajectories bythesmall angle oz (seeFig.15-8), where Ax8.=f=5qBw. (15.39) Now classically wewould alsoexpect athinstripofmagnetic ﬁeldtodeﬂect alltrajectories through some small angle, saya’,asshown inFig.15-9(a). Asthe electrons gothrough themagnetic ﬁeld, theyfeelatransverse force qvXBwhich lastsfor'atimew/v. Thechange intheir transverse momentum isjustequal to thisimpulse, so Ap,=qwB. (15.40) Theangular deﬂection [Fig. 15-9(b)] isequal totheratio ofthistransverse mo- mentum tothetotalmomentum p.Wegetthat 8/=5-5=5114- (15.41)3 3 Wecancompare thisresult withEq.(15.39), which gives thesame quantity computed quantum-mechanically. Buttheconnection between classical mechanics andquantum mechanics isthis:Aparticle ofmomentum pcorresponds toaquan- 15-13-8 I ,. . -.‘ -_i L -7 \ p ,—nr—L -T 7 '1',‘ 8 _ “I '..n.|""-' ' LINESorE 1'-1"--1(0) “I 4-_/Q13 Pver.P (bl Fig.15-9. Deﬂection ofaparticle due topassage through astrip of magnetic ﬁeld. tumamplitude varying with thewavelength ll=h/p. With thisequality, atand0/ areidentical; theclassical andquantum calculations givethesame result. From theanalysis weseehow itisthatthevector potential which appears in quantum mechanics inanexplicit form produces aclassical force which depends only onitsderivatives. Inquantum mechanics what matters istheinterference between nearby paths; italways turns outthattheeffects depend onlyonhowmuch theﬁeld Achanges from point topoint, andtherefore only onthederivatives of Aandnotonthevalue itself. Nevertheless, thevector potential A(together with thescalar potential atthatgoes with it)appears togivethemost direct description ofthephysics. This becomes more andmore apparent themore deeply wego intothequantum theory. Inthegeneral theory ofquantum electrodynamics, one takes thevector and scalar potentials asthefundamental quantities inaset ofequations that replace theMaxwell equations: EandBareslowly disappear- ingfrom themodern expression ofphysical laws; they arebeing replaced byA and¢. 15-6 What istrueforstatics isfalsefordynamics Wearenowattheendofourexploration ofthesubject ofstatic ﬁelds. Already inthischapter wehave come perilously close tohaving toworry about what happens when ﬁelds change with time. Wewere barely able toavoid itinour treatment ofmagnetic energy bytaking refuge inarelativistic argument. Even so, ourtreatment oftheenergy problem wassomewhat artiﬁcial andperhaps even mysterious, because weignored thefactthatmoving coils must, infact, produce changing ﬁelds. Itisnowtime totake upthetreatment oftime-varying ﬁelds-the subject ofelectrodynamics. Wewilldosointhenext chapter. First, however, we would liketoemphasize afewpoints. Although webegan thiscourse with apresentation ofthecomplete andcorrect equations ofelectromagnetism, weimmediately began tostudy some incomplete pieces-—because thatwaseasier. There isagreat advantage instarting withthe simpler theory ofstatic ﬁelds, andproceeding onlylatertothemore complicated theory which includes dynamic ﬁelds. There islessnewmaterial tolearn allat once, andthere istime foryoutodevelop your intellectual muscles inpreparation forthebigger task. Butthere isthedanger inthisprocess thatbefore wegettoseethecomplete story, theincomplete truths learned onthewaymay become ingrained andtaken asthewhole truth—that what istrueandwhat isonly sometimes truewillbecome confused. SowegiveinTable 15-1 asummary oftheimportant formulas wehave covered, separating those which aretrueingeneral from those which aretruefor statics, butfalse fordynamics. This summary alsoshows, inpart, where weare going, since aswetreat dynamics wewillbedeveloping indetail what wemust just state here without proof. Itmay beuseful tomake afewremarks about thetable. First, youshould notice that theequations westarted with arethetrueequations—we have not misled you there. The electromagnetic force (often called theLorentz force) F=q(E+vXB)istrue. Itisonly Coulomb’s lawthatisfalse, tobeused only forstatics. Thefour Maxwell equations forEandBarealsotrue. Theequations wetook forstatics arefalse, ofcourse, because weleftoffallterms with time derivatives. Gauss’ law, V'E=p/e0, remains, butthecurlofEisnotzero ingeneral. SoEcannot always beequated tothegradient ofascalar—the electrostatic po- tential. Wewillseethatascalar potential stillremains, butitisatime-varying quantity thatmust beusedtogether withvector potentials foracomplete descrip- tionoftheelectric ﬁeld. Theequations governing thisnewscalar potential are, necessarily, alsonew. Wemust alsogiveuptheideathatEiszeroinconductors. When theﬁelds are changing, thecharges inconductors donot, ingeneral, have time torearrange themselves tomake theﬁeld zero. They aresetinmotion, butnever reach equili- brium. Theonly general statement is:electric ﬁelds inconductors produce cur- 15-14 Table 15-1 FALSE INGENERAL (trueonlyforstatics) TRUE ALWAYS _ 1 41112 9F-4H0 '2 (Coulombslaw) F=q(E+v><B) (Lorentz force) ->V-E=£78 (Gauss’1aw) VXE=0 E=—V¢ 1 2 E(l) =IT; £7? dV2 Forconductors, E=0,¢=constant. Q=CV->V><E=— E=—V¢-(-9; (Faraday’s law) ‘llat Inaconductor, Emakes currents. c2V XB=é (Ampere’s law)0 1 23(1)=Mf dV2->V-B=0 (Nomagnetic charges) B=VXA ->¢2v><3=f0+‘-25 V24: =—g (Poisson’s equation) V2,,=_Lencz with V-A=01a’¢ 2 =___/L V¢ C2612 60 and2 . v2A_léi‘_=__l_ withc2612 e()c2 ezv-A+%‘?=0 _1Ki)A0) _41l'6()C2 ./27'12 dV2and with__i_ t>(2.t’)¢(11t) _47r€o-/ r12 dV2 1 i(2.1’)A(1,t)=1= M062 Irmat/2 rt'=t__£ ( C 2 U=§fp¢dV+%/j-AdV U=/<%E-3+3’2i3-3)dV Theequations marked byanarrow (—>)areMaxwell’s equations. 15-15 rents. Soinvarying ﬁelds aconductor isnotanequipotential. Italsofollows that theideaofacapacitance isnolonger precise. Since there arenomagnetic charges, thedivergence ofBisalways zero. S0 Bcanalways beequated toVXA.(Everything doesn’t change!) Butthegenera- tionofBisnotonly from currents: VXBisproportional tothecurrent density plusanewterm 6E/61. This means thatAisrelated tocurrents byanewequation. Itisalsorelated to¢>.lfwemake useofourfreedom tochoose V-Aforourown convenience, theequations forAor¢canbearranged totake onasimple andele- gant form. Wetherefore make thecondition that c2V-A=—8¢>/6t, andthe differential equations forAor¢appear asshown inthetable. Thepotentials Aand¢canstillbefound byintegrals over thecurrents and charges, butnotthesame integrals asforstatics. Most wonderfully, though, the trueintegrals arelikethestatic ones, with only asmall andphysically appealing modiﬁcation. When wedotheintegrals toﬁndthepotentials atsome point, say point (1)inFig. l5—l0, wemust usethevalues ofjandpatthepoint (2)atan earlier time t’=t—r12/c. Asyouwould expect, theinﬂuences propagate from point (2)topoint (1)atthespeed c.With thissmall change, onecansolve forthe ﬁelds ofvarying currents andcharges, because once wehave Aand¢,wegetB from VXA,asbefore, andEfrom —V¢ —-6A/6t. (|,t) "12 Fig.I5—l0. Thepotentials atpoint (1)and citthetime toregiven bysum- Kai ming thecontributions from ecich element ofthesource attheroving point (2), using thecurrents andchores which were present attheearlier timet—H2/C. Finally, youwillnotice thatsome results—for example, thattheenergy density inanelectric ﬁeld ise0E2/2—are true forelectrodynamics aswell asforstatics. You should notbemisled intothinking thatthisisatall“natural.” Thevalidity ofanyformula derived inthestatic casemust bedemonstrated over again forthe dynamic case. Acontrary example istheexpression fortheelectrostatic energy in terms ofavolume integral ofp¢.Thisresult istrueonlyforstatics. Wewillconsider allthese matters inmore detail induetime, butitwillperhaps beuseful tokeep inmind thissummary, soyouwillknow what youcanforget, andwhat youshould remember asalways true. 15-16 I6 Induced Currents 16-1 Motors andgenerators Thediscovery in1820thatthere wasaclose connection between electricity andmagnetism wasveryexciting—until then, thetwosubjects hadbeenconsidered asquiteindependent. Theﬁrstdiscovery wasthatcurrents inwires make magnetic ﬁelds; then, inthesame year, itwasfound thatwires carrying current inamagnetic ﬁeldhaveforces onthem. Oneoftheexcitements whenever there isamechanical force isthepossibility ofusing itinanengine todowork. Almost immediately after their discovery, people started todesign electric motors using theforces oncurrent-carrying wires. Theprinciple oftheelectromagnetic motor isshown inbareoutline inFig.16-1. Apermanent magnet—usually withsome pieces ofsoftiron—is usedtoproduce amagnetic ﬁeldintwoslots. Across each slotthere isanorth andsouth pole, asshown. Arectangular coilofcopper isplaced withonesideineachslot. When acurrent passes through thecoil,itﬂows inopposite directions inthetwoslots, sotheforces arealsoopposite, producing atorque onthecoilabout theaxis shown. Ifthecoilismounted onashaft sothatitcanturn, itcanbecoupled to pulleys orgears andcandowork. Thesame ideacanbeused formaking asensitive instrument forelectrical measurements. Thus themoment theforce lawwasdiscovered theprecision of electrical measurements wasgreatly increased. First, thetorque ofsuchamotor canbemade much greater foragiven current bymaking thecurrent goaround many turns instead ofjustone. Then thecoilcanbemounted sothatitturns with verylittletorque—either bysupporting itsshaft onverydelicate jewel bearings or byhanging thecoilonaveryﬁnewireoraquartz ﬁber. Then anexceedingly small current willmake thecoilturn, andforsmall angles theamount ofrotation will beproportional tothecurrent. Therotation canbemeasured bygluing apointer tothecoilor,forthemost delicate instruments, byattaching asmall mirror tothe coilandlooking attheshift oftheimage ofascale. Such instruments arecalled galvanometers. Voltmeters andammeters work onthesame principle. Thesame ideas canbeapplied onalarge scale tomake large motors forpro- viding mechanical power. Thecoilcanbemade togoaround andaround byar- ranging thattheconnections tothecoilarereversed each half-turn bycontacts mounted ontheshaft. Then thetorque isalways inthesame direction. Small dcmotors aremade justthisway. Larger motors, dcorac,areoften made by replacing thepermanent magnet byanelectromagnet, energized from theelectrical power source. With therealization thatelectric currents make magnetic ﬁelds, people im- mediately suggested that, somehow orother, magnets might alsomake electric ﬁelds. Various experiments weretried. Forexample, twowires wereplaced parallel toeach other andacurrent waspassed through oneofthem inthehope ofﬁnding acurrent intheother. Thethought wasthatthemagnetic ﬁeldmight insome way drag theelectrons along inthesecond wire, giving some such lawas“likes prefer tomove alike.” With thelargest available current andthemost sensitive gal- vanometer todetect anycurrent, theresult wasnegative. Large magnets nextto wires alsoproduced noobserved effects. Finally, Faraday discovered in1840the essential feature thathadbeenmissed—that electric effects exist onlywhen there issomething changing. Ifoneofapairofwires hasachanging current, acurrent isinduced intheother, orifamagnet ismoved nearanelectric circuit, there isa current. Wesaythatcurrents areinduced. Thiswastheinduction effect discovered 16-116-1 Motors andgenerators 16-2 Transformers andinductances 16-3 Forces oninduced currents 16-4 Electrical technology N.‘. '0 \ Q ~‘Q/,IQ/\ o-‘OOPPER’ sort ; WIRE Q... IRON 60'. .Q~.‘ 'Q§‘§: § \' 4Q~ . ._ Iz ‘ ,‘Q E"' Fig.l6—l. Schematic outline of0 simple electromagnetic motor. byFaraday. Ittransformed therather dullsubject ofstatic ﬁelds intoaveryex- citing dynamic subject withanenormous range ofwonderful phenomena. This chapter isdevoted toaqualitative description ofsome ofthem. Aswewillsee, onecanquickly getintofairly complicated situations thatarehard toanalyze quantitatively inalltheirdetails. Butnever mind, ourmain purpose inthischapter isﬁrsttoacquaint youwiththephenomena involved. Wewilltakeupthedetailed analysis later. Wecaneasily understand onefeature ofmagnetic induction from what we already know, although itwasnotknown inFaraday’s time. Itcomes from the vXBforce onamoving charge thatisproportional toitsvelocity inamagnetic ﬁeld. Suppose thatwehaveawirewhich passes nearamagnet, asshown inFig. 16-2, andthatweconnect theendsofthewiretoagalvanometer. Ifwemove the wireacross theendofthemagnet thegalvanometer pointer moves. Themagnet produces some vertical magnetic ﬁeld, andwhen wepush the wireacross theﬁeld, theelectrons inthewirefeelasideways force——at right angles totheﬁeldandtothemotion. Theforce pushes theelectrons along thewire. Butwhydoesthismove thegalvanometer, which issofarfromtheforce? Because when theelectrons which feelthemagnetic force trytomove, theypush——by electric repulsion-—the electrons alittle farther down thewire; they, inturn, repel the electrons alittle farther on,andsoonforalong distance. Anamazing thing. Itwassoamazing toGauss andWeber—who ﬁrstbuiltagalvanometer—that theytriedtoseehowfartheforces inthewirewould go.They strung awireallthe wayacross their city. Mr.Gauss, atoneend,connected thewires toabattery (batteries wereknown before generators) andMr.Weber watched thegalvanometer move. They hadawayofsignaling longdistances—it wasthebeginning ofthe telegraph! Ofcourse, thishasnothing directly todowithinduction—it hastodo withthewaywires carry currents, whether thecurrents arepushed byinduction ornot. Now suppose inthesetup ofFig.16-2weleave thewirealone andmove the magnet. Westillseeaneffect onthegalvanometer. AsFaraday discovered, moving themagnet under thewire—one way—has thesame effect asmoving thewireover themagnet~—the other way. Butwhen themagnet ismoved, wenolonger have anyvXBforce ontheelectrons inthewire. This istheneweffect thatFaraday found. Today, wemight hope tounderstand itfrom arelativity argument. Wealready understand thatthemagnetic ﬁeldofamagnet comes from its internal currents. Soweexpect toobserve thesame effect ifinstead ofamagnet inFig.l6-2weuseacoilofwireinwhich there isacurrent. Ifwemove thewire pastthecoilthere willbeacurrent through thegalvanometer, oralsoifwemove thecoilpastthewire. Butthere isnowamore exciting thing: Ifwechange the magnetic ﬁeldofthecoilnotbymoving it,butbychanging itscurrent, there is again aneffect inthegalvanometer. Forexample, ifwehavealoopofwirenear acoil,asshown inFig.l6—3, andifwekeep both ofthem stationary butswitch offthecurrent, there isapulse ofcurrent through thegalvanometer. When we switch thecoilonagain, thegalvanometer kicks intheother direction. Whenever thegalvanometer inasituation such astheoneshown inFig.16-2, orinFig.16-3, hasacurrent, there isanetpush ontheelectrons intheWireinone direction along thewire. There maybepushes indifferent directions atdifferent places, butthere ismore push inonedirection thananother. What counts isthe push integrated around thecomplete circuit. Wecallthisnetintegrated push the electromotive force (abbreviated emf) inthecircuit. More precisely, theemfis deﬁned asthetangential force perunitcharge inthewire integrated over length, once around thecomplete circuit. Faraday’s complete discovery wasthat emf’s canbegenerated inawireinthree different ways: bymoving thewire, bymoving amagnet near thewire, orbychanging acurrent inanearby wire. Let’s consider thesimple machine ofFig. l6-1 again, only now, instead of putting acurrent through thewiretomake itturn, let’sturntheloop byanexternal force, forexample byhand orbyawaterwheel. When thecoilrotates, itswires are moving inthemagnetic ﬁeld andwewillﬁndanemfinthecircuit ofthecoil. Themotor becomes agenerator. 16-2 i /// / / / 5/ _ ,/ A ' \/ I I W GALVANOMETER Fig. l6—2. Moving 0wire through 0magnetic ﬁeld produces ocurrent, usshown bythegalvanometer.\-s\§____- C3 GALVANOME TERW.\=s(.\\BATTERY orifitscurrent ischanged. Thecoilofthegenerator hasaninduced emffrom itsmotion. Theamount of theemfisgiven byasimple rulediscovered byFaraday. (Wewilljuststate the rulenowandwaituntillatertoexamine itindetail.) Theruleisthatwhen themag- netic ﬂuxthatpasses through theloop (thisﬂuxisthenormal component ofB integrated overtheareaoftheloop) ischanging withtime, theemfisequal to therateofchange oftheﬂux. Wewillrefertothisas“theﬂuxrule.” Youseethat when thecoilofFig.16-lisrotated, theﬂuxthrough itchanges. Atthestart some ﬂuxgoesthrough oneway; thenwhen thecoilhasrotated 180°thesame ﬂuxgoesthrough theother way. Ifwecontinuously rotate thecoiltheﬂuxis ﬁrstpositive, thennegative, thenpositive, andsoon.Therateofchange ofthe ﬂuxmust alternate also. Sothere isanalternating emfinthecoil. Ifweconnect thetwoends ofthecoiltooutside wires through some sliding contacts——called slip-rings—(just sothewires won’t gettwisted) wehave analternating-current generator. Orwecanalsoarrange, bymeans ofsome sliding contacts, thatafter every one-half rotation, theconnection between thecoilends andtheoutside wires is reversed, sothatwhen theemfreverses, sodotheconnections. Then thepulses of emfwillalways push currents inthesame direction through theexternal circuit. Wehavewhat iscalled adirect-current generator. Themachine ofFig.16-1iseither amotor oragenerator. Thereciprocity between motors andgenerators isnicely shown byusing twoidentical dc“motors” ofthepermanent magnet kind, withtheir coils connected bytwocopper wires. When theshaft ofoneisturned mechanically, itbecomes agenerator anddrives theother asamotor. Iftheshaft ofthesecond isturned, itbecomes thegenerator anddrives theﬁrstasamotor. Sohereisaninteresting example ofanewkindof equivalence ofnature: motor andgenerator areequivalent. Thequantitative equivalence is,infact, notcompletely accidental. Itisrelated tothelawofcon- servation ofenergy. Another example ofadevice thatcanoperate either togenerate emf’s orto respond toemf’s isthereceiver ofastandard telephone—that is,an“earphone.” Theoriginal telephone ofBellconsisted oftwosuch “earphones” connected by twolongwires. Thebasic principle isshown inFig.16-4. Apermanent magnet produces amagnetic ﬁeldintwo“yokes” ofsoftironandinathindiaphragm that ismoved bysound pressure. When thediaphragm moves, itchanges theamount ofmagnetic ﬁeldintheyokes. Therefore acoilofwire wound around oneofthe yokes willhavetheﬂuxthrough itchanged when asound wave hitsthediaphragm. 16-3DISCFig.l6—3. Acoilwithcurrent produces o current inosecond coiliftheﬁrstcoilismoved mmmom Lsouuo PRESSURE S ’/l/AllhSOFT IRGI -\\.§ PERMANENT BDR IMGNET Fig.l6—4. Atelephone orreceiver.Q oovren con. transmitter I .lb \4ueurgentsB ‘E-' '‘\ ltlllllllksA.C. GENERATOR Fig. l6-5. Two coils, wrapped around bundles ofironsheets, allow a generator tolight abulb with nodirect connection.Sothere isanemfinthecoil. Iftheends ofthecoilareconnected toacircuit, a current which isanelectrical representation ofthesound issetup. Iftheends ofthecoilofFig.16-4 areconnected bytwowires toanother identical gadget, varying currents willﬂow inthesecond coil. These currents will produce avarying magnetic ﬁeldandwillmake avarying attraction ontheiron diaphragm. The diaphragm willwiggle andmake sound waves approximately similar totheones thatmoved theoriginal diaphragm. With afewbitsofironand copper thehuman voice istransmitted over wires! (The modern home telephone uses areceiver liketheonedescribed butuses animproved invention togetamore powerful transmitter. Itisthe“carbon- button microphone,” thatusessound pressure tovarytheelectric current from abattery.) 16-2 Transformers andinductances One ofthemost interesting features ofFaraday’s discoveries isnotthat an emfexists inamoving coil-which wecanunderstand interms ofthemagnetic force qvXB—but thatachanging current inonecoilmakes anemfinasecond coil. And quite surprisingly theamount ofemfinduced inthesecond coilisgiven bythesame “ﬂux rule”: thattheemfisequal totherateofchange ofthemagnetic ﬂuxthrough thecoil. Suppose thatwetaketwocoils, each wound around separate bundles ofiron sheets (these help tomake stronger magnetic ﬁelds), asshown in Fig. l6-5. Now weconnect oneofthecoils—coil (a)—to analternating-current generator. The continually changing current produces acontinuously varying magnetic ﬁeld. This varying ﬁeld generates analternating emfinthesecond coil— coil(b).This emfcan,forexample, produce enough power tolight anelectric bulb. Theemfalternates incoil(b)atafrequency which is,ofcourse, thesame asthe frequency oftheoriginal generator. Butthecurrent incoil(b)canbelarger or smaller thanthecurrent incoil(a).Thecurrent incoil(b)depends ontheemf induced initandontheresistance andinductance oftherestofitscircuit. The emfcanbelessthanthatofthegenerator if,say,there islittleﬂuxchange. Orthe emfincoil(b)canbemade much larger than thatinthegenerator bywinding coil (b)with many turns, since inagiven magnetic ﬁeld theﬂuxthrough thecoilis then greater. (Orifyouprefer tolook atitanother way, theemfisthesame ineach turn, andsince thetotalemfisthesumoftheemf’s oftheseparate turns, many turns inseries produce alarge emf.) Such acombination oftwocoils—usually with anarrangement ofironsheets toguide themagnetic ﬁelds—is called atransformer. Itcan“transform” oneemf (also called a“voltage”) toanother. There arealsoinduction effects inasingle coil. Forinstance, inthesetup in Fig.16-5 there isachanging ﬂuxnotonly through coil(b),which lights thebulb, butalso through coil(a). The varying current incoil(a)produces avarying magnetic ﬁeldinside itself andtheﬂuxofthisﬁeldiscontinually changing, sothere isaself-induced emfincoil(a). There isanemfacting onanycurrent when itis building upamagnetic ﬁeld—or, ingeneral, when itsﬁeld ischanging inanyway. Theeffect iscalled self-inductance. When wegave “theﬂuxrule” thattheemfisequal totherateofchange ofthe ﬂuxlinkage, wedidn’t specify thedirection oftheemf. There isasimple rule, called Lenz’s rule, forﬁguring outwhich waytheemfgoes: theemftries tooppose anyﬂuxchange. That is,thedirection ofaninduced emfisalways such thatifa current were toflow inthedirection oftheemf, itwould produce aﬂuxofBthat opposes thechange inBthatproduces theemf. Lenz’s rulecanbeused toﬁnd thedirection oftheemfinthegenerator ofFig.l6-l,orinthetransformer winding ofFig. 16-3. Inparticular, ifthere isachanging current inasingle coil(orinanywire) there isa“back” emfinthecircuit. This emfactsonthecharges ﬂowing incoil (a)ofFig.16-5tooppose thechange inmagnetic ﬁeld, andsointhedirection to oppose thechange incurrent. Ittriestokeep thecurrent constant; itisopposite to thecurrent when thecurrent isincreasing, anditisinthedirection ofthecurrent 16-4 — SWITCH i L BATTERY iiii—l;¥}’Fig.16-6. Circuit connections foran electromagnet. The lamp allows the passage ofcurrent when theswitch is opened, preventing theappearance of excessive emf’s. when itisdecreasing. Acurrent inaself-inductance has“inertia,” because the inductive eﬂects trytokeep theﬂow constant, justasmechanical inertia tries to keep thevelocity ofanobject constant. Anylarge electromagnet willhave alarge self-inductance. Suppose thata battery isconnected tothecoilofalarge electromagnet, asinFig.16-6, andthata strong magnetic ﬁeld hasbeen built up.(The current reaches asteady value deter- mined bythebattery voltage andtheresistance ofthewireinthecoil.) Butnow suppose thatwetrytodisconnect thebattery byopening theswitch. Ifwereally opened thecircuit, thecurrent would gotozerorapidly, andindoing soitwould generate anenormous emf. Inmost cases thisemfwould belarge enough tode- velop anarcacross theopening contacts oftheswitch. Thehigh voltage thatap- pears might alsodamage theinsulation ofthecoil—or you, ifyouaretheperson whoopens theswitch! Forthese reasons, electromagnets areusually connected in acircuit liketheoneshown inFig. 16-6. When theswitch isopened, thecurrent does notchange rapidly butremains steady, ﬂowing instead through thelamp, being driven bytheemffrom theself~inductance ofthecoil. 16-3 Forces oninduced currents Youhaveprobably seenthedramatic demonstration ofLenz’s rulemade with thegadget shown inFig.16-7. Itisanelectromagnet, justlikecoil(a)ofFig. l6-5. Analuminum ringisplaced ontheendofthemagnet. When thecoilis connected toanalternating-current generator byclosing theswitch, theringﬂies intotheair. Theforce comes, ofcourse, from theinduced currents inthering. Thefactthattheringﬂiesaway shows thatthecurrents initoppose thechange of theﬁeldthrough it.When themagnet ismaking anorth poleatitstop,theinduced current intheringismaking adownward-point north pole. Theringandthecoil arerepelled justliketwomagnets with likepoles opposite. Ifathinradial cutis made intheringtheforce disappears, showing thatitdoes indeed come from the currents inthering. \‘\, F T CONDUCTING RING ....CDl>o% \/\i—-> roANA.C.°°'L czuennon\ >-l i ss/' I\ $WlTCH 3s\\ @((((Fig. l6—7. Aconducting ring isstrongly repelled byanelectromagnet with avarying current. conducting plate. 16-5_///4j7/7////PERFECTLY CONDUCTING PLATE Fig.l6—8. Anelectromagnet near aperfectly ii] Fig. 16-9. Abar magnet issus- pended above asuperconducting bowl, bytherepulsion ofeddy currents. 'IVOT\ 2°.i'¥El SWITCH BATTERYK» it Fig. I6-10. Thebraking ofthepen- dulum shows theforces duetoeddy cur- rents. V_‘/ EDDY CURRENT S \©/W B Fig. 16-1 l.Theeddy currents inthe copper pendulum.If,instead ofthering,weplace adiscofaluminum orcopper across theend oftheelectromagnet ofFig.16-7, itisalsorepelled; induced currents circulate in thematerial ofthedisc,andagain produce arepulsion. Aninteresting effect, similar inorigin, occurs withasheet ofaperfect con- ductor. Ina“perfect conductor” there isnoresistance whatever tothecurrent. So ifcurrents aregenerated init,theycankeep going forever. Infact,theslightest emfwould generate anarbitrarily large current—which really means that there canbenoemf’s atall.Any attempt tomake amagnetic ﬂuxgothrough such a sheet generates currents thatcreate opposite Bﬁelds—all with inﬁnitesimal emf’s, sowith noﬂuxentering. Ifwehaveasheet ofaperfect conductor andputanelectromagnet nexttoit, when weturn onthecurrent inthemagnet, currents called eddy currents appear in thesheet, sothatnomagnetic ﬂuxenters. Theﬁeld lines would look asshown in Fig.16-8. Thesame thing happens, ofcourse, ifwebring abarmagnet neara perfect conductor. Since theeddy currents arecreating opposing ﬁelds, the magnets arerepelled from theconductor. Thismakes itpossible tosuspend abar magnet inairabove asheet ofperfect conductor shaped likeadish, asshown in Fig.16-9. Themagnet issuspended bytherepulsion oftheinduced eddy currents intheperfect conductor. There arenoperfect conductors atordinary tempera- tures, butsome materials become perfect conductors atlowenough temperatures. Forinstance, below 3.8°K tinconducts perfectly. Itiscalled asuperconductor. Iftheconductor inFig.16-8isnotquite perfect there willbesome resistance toﬂow oftheeddy currents. Thecurrents willtend todieoutandthemagnet will slowly settle down. Theeddy currents inanimperfect conductor need anemfto keep them going, andtohave anemftheﬂuxmust keep changing. Theﬂuxof themagnetic ﬁeld gradually penetrates theconductor. Inanormal conductor, there arenotonlyrepulsive forces from eddycurrents, butthere canalsobesidewise forces. Forinstance, ifwemove amagnet sideways along aconducting surface theeddy currents produce aforce ofdrag, because the induced currents areopposing thechanging ofthelocation ofﬂux. Suchforces are proportional tothevelocity andarelikeakindofviscous force. These effects show upnicely intheapparatus shown inFig. 16-10. Asquare sheet ofcopper issuspended ontheendofarodtomake apendulum. Thecopper swings back andforth between thepoles ofanelectromagnet. When themagnet isturned on,thependulum motion issuddenly arrested. Asthemetal plate enters thegapofthemagnet, there isacurrent induced intheplate which actstooppose thechange inﬂuxthrough theplate. Ifthesheet were aperfect conductor, the currents would besogreat that they would push theplate outagain—-it would bounce back. With acopper plate there issome resistance intheplate, so thecurrents atﬁrstbring theplate almost toadead stopasitstarts toenter the ﬁeld. Then, asthecurrents diedown, theplate slowly settles torestinthemagnetic ﬁeld. Thenature oftheeddy currents inthecopper pendulum isshown inFig. 16-l1.Thestrength andgeometry ofthecurrents arequite sensitive totheshape oftheplate. If,forinstance, thecopper plate isreplaced byonewhich hasseveral narrow slots cutinit,asshown inFig.16-12, theeddy-current eﬂects aredrastically reduced. The pendulum swings through themagnetic ﬁeld with only asmall retarding force. Thereason isthatthecurrents ineach section ofthecopper have lessﬂuxtodrive them, sotheeffects oftheresistance ofeach loop aregreater. Thecurrents aresmaller andthedrag isless. Theviscous character oftheforce isseen even more clearly ifasheet ofcopper isplaced between thepoles ofthe magnet ofFig.16-10 andthenreleased. Itdoesn’t fall;itjustsinks slowly down- ward. Theeddy currents exert astrong resistance tothemotion—just likethe viscous drag inhoney. If,instead ofdragging aconductor past amagnet, wetrytorotate itina magnetic ﬁeld, there willbearesistive torque from thesame eﬂects. Alternatively, ifwerotate amagnet—end over end—near aconducting plate orring, thering is dragged around; currents inthering willcreate atorque that tends torotate theringwith themagnet. 16-6 V4" m’||§,,,§|’°'\ -b\\l/ /¢||§/ §|Ixu21/ 2;/ 5 I e-\//if \//t“,5 6 5 5 5 to) (bl (¢) 2 3 2 3 2 3 IQ.//\ I = ' = i 4| 4 \//1*’ \//if6 5 6 5 6 5 ldl (9) (ll Fig. 16-12. Eddy-current effects are drasti- Fig. l6-13. Making arotating magnetic ﬁeld. cally reduced bycutting slots intheplate. Aﬁeldjustlikethatofarotating magnet canbemade withanarrangement ofcoils such asisshown inFig.16-13. Wetake atorus ofiron (that is,aringof ironlikeadoughnut) andwind sixcoils onit.Ifweputacurrent, asshown in part(a),through windings (1)and(4),there willbeamagnetic ﬁeldinthedirection shown intheﬁgure. Ifwenowswitch thecurrent towindings (2)and(5),the magnetic ﬁeldwillbeinanewdirection, asshown inpart(b)oftheﬁgure. Con- tinuing theprocess, wegetthesequence ofﬁelds shown intherestoftheﬁgure. Iftheprocess isdone smoothly, wehave a“rotating” magnetic ﬁeld. Wecaneasily gettherequired sequence ofcurrents byconnecting thecoils toathree-phase power line,which provides justsuchasequence ofcurrents. “Three-phase power” ismade inagenerator using theprinciple ofFig.16-l, except thatthere arethree loops fastened together onthesame shaft inasymmetrical way-—that is,withan angle of120°from oneloop tothenext. When thecoils arerotated asaunit, the emfisamaximum inone,theninthenext, andsooninaregular sequence. There aremany practical advantages ofthree-phase power. Oneofthem isthepossibility ofmaking arotating magnetic ﬁeld. Thetorque produced onaconductor bysuch arotating ﬁeldiseasily shown bystanding ametal ringonaninsulating tablejust above thetorus, asshown inFig.16-14. Therotating ﬁeldcauses theringtospin about avertical axis. Thebasic elements seen here arequite thesame asthose at playinalarge commercial three-phase induction motor. Another form ofinduction motor isshown inFig. 16-15. Thearrangement shown isnotsuitable forapractical high-efficiency motor butwillillustrate the principle. Theelectromagnet M,consisting ofabundle oflaminated ironsheets wound with asolenoidal coil,ispowered with alternating current from agenerator. Themagnet produces avarying ﬂuxofBthrough thealuminum disc. Ifwehave justthese twocomponents, asshown inpart (a)oftheﬁgure, wedonotyethave amotor. There areeddy currents inthedisc,buttheyaresymmetric andthere is notorque. (There willbesome heating ofthediscduetotheinduced currents.) If wenowcover onlyone-half ofthemagnet polewithanaluminum plate, asshown inpart(b)oftheﬁgure, thediscbegins torotate, andwehave amotor. The operation depends ontwoeddy-current effects. First, theeddy currents inthe aluminum plate oppose thechange ofﬂuxthrough it,sothemagnetic ﬁeldabove theplatealways lagstheﬁeldabove thathalfofthepolewhich isnotcovered. This so-called “shaded-pole” effect produces aﬁeldwhich inthe“shaded” region varies 16-7-l--LFig. 16-I4. The rotating ﬁeld of Fig.16-l3canbeusedtoprovide torque onaconducting ring.l ALUMINUMPLATE ALUMINUM DISC 3 ToAc:lllllllllllllllllSOURCE .|||m||m||||||.~lllllllllllllllllllllllllllllll. lllllllllllllroA.C._llll|ll|ll|||sous\|||||||||||||~°l _||||||||||||| » 1 Slllllllllllllllll MAGNET \|||||m|m| “” Fig.16-15. Asimple example ofashaded-pole induction motor. much likethatinthe“unshaded” region except thatitisdelayed aconstant amount intime. Thewhole effect isasifthere were amagnet onlyhalfaswidewhich is continually being moved from theunshaded region toward theshaded one. Then thevarying ﬁelds interact with theeddy currents inthedisctoproduce thetorque onit. 16-4 Electrical technology When Faraday ﬁrstmade public hisremarkable discovery thatachanging magnetic ﬂuxproduces anemf, hewasasked (asanyone isasked when hedis- covers anewfactofnature), “What istheuseofit?” Allhehadfound wasthe oddity that atinycurrent wasproduced when hemoved awire near amagnet. Ofwhat possible “use” could thatbe?Hisanswer was:“What istheuseofanew- born baby?” Yetthink ofthetremendous practical applications hisdiscovery hasledto. What wehavebeendescribing arenotjusttoysbutexamples chosen inmost cases torepresent theprinciple ofsome practical machine. Forinstance, therotating ring intheturning ﬁeldisaninduction motor. There are,ofcourse, some differences between itandapractical induction motor. Theringhasavery small torque; it canbestopped withyourhand. Foragood motor, things havetobeputtogether more intimately: there shouldn’t besomuch “wasted” magnetic ﬁeld outinthe air.First, theﬁeldisconcentrated byusing iron. Wehave notdiscussed how iron doesthat,butironcanmake themagnetic ﬁeldtensofthousands oftimes stronger thancopper coilsalone could do.Second, thegapsbetween thepieces ofironare made small; todothat,some ironisevenbuiltintotherotating ring. Everything isarranged soastogetthegreatest forces andthegreatest eﬁ'iciency—that is, conversion ofelectrical power tomechanical power—until the“ring” canno longer beheld stillbyyour hand. This problem ofclosing thegaps andmaking thething work inthemost practical wayisengineering. Itrequires serious study ofdesign problems, although there arenonewbasic principles from which theforces areobtained. Butthere isalong waytogofrom thebasic principles toapractical andeconomic design. Yetitisjustsuchcareful engineering design thathasmade possible suchatre- mendous thing asBoulder Dam andallthatgoeswithit. What isBoulder Dam? Ahugeriverisstopped byaconcrete wall. Butwhat awall itis!Shaped with aperfect curve thatisvery carefully worked outsothat theleastpossible amount ofconcrete willholdback awhole river. Itthickens at thebottom inthatwonderful shape thattheartists likebutthattheengineers can appreciate because they know that such thickening isrelated totheincrease of pressure withthedepth ofthewater. Butwearegetting away from electricity. Then thewater oftheriverisdiverted intoahugepipe. That’s aniceengineer- ingaccomplishment initself. Thepipefeeds thewater intoa“waterwheel”—a huge turbine—and makes wheels turn. (Another engineering feat.) Butwhyturn wheels? They arecoupled toanexquisitely intricate mess ofcopper andiron, all 16-8 twisted andinterwoven. With twoparts—one thatturns andonethatdoesn’t. Allacomplex intermixture ofafewmaterials, mostly iron andcopper butalso some paper andshellac forinsulation. Arevolving monster thing. Agenerator. Somewhere outofthemess ofcopper andironcome afewspecial pieces ofcopper. Thedam, theturbine, theiron, thecopper, allputthere tomake something special happen toafewbarsofcopper—-an emf. Then thecopper barsgoalittlewayand circle forseveral times around another piece ofiron inatransformer; then their jobisdone. Butaround thatsame piece ofironcurls another cable ofcopper which has nodirect connection whatsoever tothebars from thegenerator; they have just been inﬂuenced because theypassed nearit—to gettheir emf. Thetransformer converts thepower from therelatively lowvoltages required fortheefficient design ofthegenerator totheveryhighvoltages thatarebestforefficient transmission of electrical energy overlongcables. And everything must beenormously efﬁcient—there canbenowaste, noloss. Why? Thepower forametropolis isgoing through. Ifasmall fraction werelost— oneortwopercent—think oftheenergy leftbehind! Ifonepercent ofthepower wereleftinthetransformer, thatenergy would needtobetaken outsomehow. If itappeared asheat, itwould quickly melt thewhole thing. There is,ofcourse, some small inefficiency, butallthatisrequired areafewpumps which circulate some oil through aradiator tokeepthetransformer from heating up. OutoftheBoulder Dam come afewdozen rodsofcopper—long, long, long rodsofcopper perhaps thethickness ofyourwrist thatgoforhundreds ofmiles in alldirections. Small rods ofcopper carrying thepower ofagiant river. Then the rodsaresplittomake more rods...thentomore transformers ...sometimes to great generators which recreate thecurrent inanother form ...sometimes to engines turning forbigindustrial purposes ...tomore transformers ...then more splitting andspreading. ..until ﬁnally theriver isspread throughout the whole city—turning motors, making heat, making light, working gadgetry. The miracle ofhotlights from coldwater over600miles away—all done withspecially arranged pieces ofcopper andiron. Large motors forrolling steel, ortinymotors foradentist’s drill. Thousands oflittlewheels, turning inresponse totheturning ofthebigwheel atBoulder Dam. Stopthebigwheel, andallthewheels stop; the lights goout.They really areconnected. Yetthere ismore. Thesame phenomena thattakethetremendous power of theriverandspread itthrough thecountryside, until afewdrops oftheriver are running thedentist’s drill, come again intothebuilding ofextremely ﬁneinstru- ments ...for thedetection ofincredibly small amounts ofcurrent. ..forthe transmission ofvoices, music, andpictures. ..for computers. ..forautomatic machines offantastic precision. Allthisispossible because ofcarefully designed arrangements ofcopper and iron—eﬂiciently created magnetic ﬁelds ...blocks ofrotating iron sixfeetin diameter whirling with clearances of1/16 ofaninch. ..careful proportions of copper fortheoptimum eﬂiciency ...strange shapes allserving apurpose, like thecurve ofthedam. Ifsome future archaeologist uncovers Boulder Dam, wemay guess that he would admire thebeauty ofitscurves. Butalsotheexplorers from some great future civilizations willlook atthegenerators andtransformers andsay: “Notice thatevery iron piece hasabeautifully eﬂicient shape. Think ofthethought that hasgoneintoevery piece ofcopper!” This isthepower ofengineering andthecareful design ofourelectrical tech- nology. There hasbeencreated inthegenerator something which exists nowhere elseinnature. Itistruethatthere areforces ofinduction inother places. Certainly insome places around thesunandstars there areeffects ofelectromagnetic induc- tion. Perhaps also(though it’snotcertain) themagnetic ﬁeld oftheearth ismain- tained byananalog ofanelectric generator thatoperates oncirculating currents intheinterior oftheearth. Butnowhere havethere beenpieces puttogether with moving parts togenerate electrical power asisdone inthegenerator—with great efficiency andregularity. 16-9 Youmaythink thatdesigning electric generators isnolonger aninteresting subject, thatitisadead subject because theyarealldesigned. Almost perfect generators ormotors canbetaken from ashelf. Even ifthiswere true, wecan admire thewonderful accomplishment ofaproblem solved tonearperfection. Butthere remain asmany unﬁnished problems. Even generators andtransformers arereturning asproblems. Itislikely thatthewhole ﬁeldoflowtemperatures and superconductors willsoonbeapplied totheproblem ofelectric power distribution. With aradically newfactor intheproblem, newoptimum designs willhave tobe created. Power networks ofthefuture may have little resemblance tothose of today. Youcanseethatthere isanendless number ofapplications andproblems that onecould takeupwhile studying thelawsofinduction. Thestudy ofthedesign of electrical machinery isalifework initself. Wecannot goveryfarinthatdirection, butweshould beaware ofthefactthatwhen wehave discovered thelawofinduc- tion, wehave suddenly connected ourtheory toanenormous practical develop- ment. Wemust, however, leave thatsubject totheengineers andapplied scientists whoareinterested inworking outthedetails ofparticular applications. Physics onlysupplies thebase—the basic principles thatapply, nomatter what. (Wehave notyetcompleted thebase, because wehaveyettoconsider indetail theproperties ofironandofcopper. Physics hassomething tosayabout these aswewillseea littlelater.) Modern electrical technology began withFaraday’s discoveries. Theuseless baby developed intoaprodigy andchanged thefaceoftheearth inways itsproud father could never haveimagined. 16-10 I7 The Laws ofInduction 17-1 Thephysics ofinduction Inthelastchapter wedescribed many phenomena which show thattheeffects ofinduction arequite complicated andinteresting. Now wewant todiscuss the fundamental principles which govern these effects. Wehave already deﬁned theemf inaconducting circuit asthetotal accumulated force onthecharges throughout thelength oftheloop. More speciﬁcally, itisthetangential component oftheforce perunitcharge, integrated along thewire once around thecircuit. This quantity isequal, therefore, tothetotal work done onasingle charge that travels once around thecircuit. Wehave alsogiven the“ﬂux rule,” which saysthattheemfisequal totherate atwhich themagnetic ﬂuxthrough such aconducting circuit ischanging. Let’s seeifwecanunderstand whythatmight be.First, we’ll consider acase inwhich theﬂuxchanges because acircuit ismoved inasteady ﬁeld. InFig.17-1weshow asimple loop ofwirewhose dimensions canbechanged. Theloop hastwoparts, aﬁxed U-shaped part (a)andamovable crossbar (b) thatcanslide along thetwolegsoftheU.There isalways acomplete circuit, but itsareaisvariable. Suppose wenowplace theloop inauniform magnetic ﬁeldwith theplane oftheUperpendicular totheﬁeld. According totherule, when thecross- barismoved there should beintheloop anemfthatisproportional totherateof change oftheﬂuxthrough theloop. This emfwillcause acurrent intheloop. Wewillassume that there isenough resistance inthewire thatthecurrents are small. Then wecanneglect anymagnetic ﬁeld from thiscurrent. Theﬂuxthrough theloop iswLB, sothe“ﬂux rule” would givefortheemf—— which wewrite as8— 8=WBPIA =wBv,dt where 1!isthespeed oftranslation ofthecrossbar. Now weshould beable tounderstand thisresult from themagnetic vXB forces onthecharges inthemoving crossbar. These charges willfeelaforce, tangential tothewire. equal tovBperunitcharge. Itisconstant along thelength wofthecrossbar andzero elsewhere, sotheintegral is 8=wvB, which isthesame result wegotfrom therateofchange oftheﬂux. Theargument _]llSIgiven canbeextended toanycase where there isaﬁxed magnetic ﬁeld andthewires aremoved. Onecanprove, ingeneral, thatforany circuit whose parts move inaﬁxed magnetic ﬁeld theemfisthetime derivative oftheﬂux, regardless oftheshape ofthecircuit. Ontheother hand, what happens iftheloop isstationary andthemagnetic ﬁeldischanged? Wecannot deduce theanswer tothisquestion from thesame argument. ItwasFaraday’s discovery—from experiment—that the“ﬂux rule” isstillcorrect nomatter why theﬂuxchanges. Theforce onelectric charges is given incomplete generality byF=q(E+v><B);there arenonew special “forces duetochanging magnetic ﬁelds.” Any forces oncharges atrestina stationary wirecome from theEterm. Faraday’s observations ledtothediscovery thatelectric andmagnetic ﬁelds arerelated byanewlaw: inaregion where the magnetic ﬁeld ischanging with time, electric ﬁelds aregenerated. Itisthiselectric 17-117-1 Thephysics ofinduction 17-2 Exceptions tothe“ﬂux rule” 17-3 Particle acceleration byan induced electric ﬁeld; the betatron 17-4 Aparadox 17-5 Alternating-current generator 17-6 Mutual inductance 17—7 Self-inductance 17-8 Inductance andmagnetic energy f . (O) . L~¥ *l~t~ —-I LINESOFB Fig. l7—l. Anemf isinduced inci loop iftheﬂuxischanged byvarying the area ofthecircuit. ﬁeld which drives theelectrons around thewire—and soisresponsible fortheemf inastationary circuit when there isachanging magnetic ﬂux. Thegeneral lawfortheelectric ﬁeldassociated with achanging magnetic ﬁeld is asv><E--57 (17.1) WewillcallthisFaraday’s law. Itwasdiscovered byFaraday butwasﬁrstwritten indifferential form byMaxwell, asoneofhisequations. Let’s seehowthisequation gives the“ﬂux rule” forcircuits. Using Stokes’ theorem, thislawcanbewritten inintegral form as j€E~ds=/S(V><E)'nda= -I95-nda (17.2)Sai ’ where, asusual, I‘isanyclosed curve andSisanysurface bounded byit.Here, remember, I‘isamathematical curve ﬁxed inspace, andSisaﬁxed surface. Then thetime derivative canbetaken outside theintegral andwehave 6y€E-ds= -57/QB nda =-56;(ﬂuxthrough s). (17.3) Applying thisrelation toacurve Pthatfollows aﬁxed circuit ofconductor, we getthe“ﬂux rule” once again. Theintegral ontheleftistheemf. andthatonthe right isthenegative rateofchange oftheﬂuxlinked bythecircuit. SoEq.(17.1) applied toaﬁxed circuit isequivalent tothe“ﬂux rule.” Sothe“ﬂux rule”——that theemfinacircuit isequal totherateofchange of themagnetic ﬂuxthrough thecircuit—applies whether theﬂuxchanges because the ﬁeldchanges orbecause thecircuit moves (orboth). Thetwopossibilities- “circuit moves” or“ﬁeld changes”——are notdistinguished inthestatement ofthe rule. Yetinourexplanation oftherulewehave used twocompletely distinct laws forthetwocases—v XBfor“circuit moves” andVXE=—6B/61 for“ﬁeld changes.” Weknow ofnoother place inphysics where such asimple andaccurate general principle requires foritsrealunderstanding ananalysis interms oftwo diﬂerent phenomena. Usually such abeautiful generalization isfound tostem from asingle deep underlying principle. Nevertheless, inthiscasethere does notappear tobeanysuch profound implication. Wehave tounderstand the“rule” asthe combined effects oftwoquite separate phenomena. Wemust look atthe“flux rule" inthefollowing way. Ingeneral, theforce per unitcharge isF/q=E+v><B.Inmoving wires there istheforce from the second term. Also, there isanE-ﬁeld ifthere issomewhere achanging magnetic ﬁeld. They areindependent effects, buttheemfaround theloop ofwire isalways equal totherateofchange ofmagnetic ﬂuxthrough it. 17-2 Exceptions tothe“ﬁux rule” Wewillnow give some examples, dueinpart toFaraday, which show the importance ofkeeping clearly inmind thedistinction between thetwoeffects re- sponsible forinduced emf's. Ourexamples involve situations towhich the“ﬁux rule” cannot beapplied—either because there isnowire atallorbecause thepath taken byinduced currents moves about within anextended volume ofaconductor. Webegin bymaking animportant point: Thepartoftheemfthatcomes from theE-ﬁeld does notdepend ontheexistence ofaphysical wire (asdoes thevXB part). TheE-ﬁeld canexist infreespace, anditslineintegral around anyimaginary lineﬁxed inspace istherateofchange oftheﬂuxofBthrough thatline. (Note thatthisisquite unlike theE-ﬁeld produced bystatic charges, forinthatcasethe lineintegral ofEaround aclosed loop isalways zero.) 17—2 change, butthere isnevertheless anemf. Figure l7—2 shows aconducting disc which canberotated onaﬁxed axisinthepresence ofamagnetic ﬁeld. One contact ismade totheshaft andanother rubs ontheouter periphery ofthedisc. Acircuit iscompleted through agalvanometer. Asthediscrotates, the“circuit,” inthesense oftheplace inspace where thecurrents are,isalways thesame. But thepartofthe“circuit” inthediscisinmaterial which ismoving. Although the ﬂuxthrough the“circuit” isconstant, there isstillanemf, ascanbeobserved by thedeﬂection ofthe galvanometer. Clearly, hereisacasewhere thevXBforce in themoving discgives risetoanemfwhich cannot beequated toachange ofﬂux. Now weconsider. asanopposite example, asomewhat unusual situation in which theﬁuxthrough a“circuit” (again inthesense ofthe place where thecurrent is)changes butwhere there isnoemf. Imagine twometal plates withslightly curved edges, asshown inFig. l7—3, placed inauniform magnetic ﬁeld perpendicular to their surfaces. Each plate isconnected tooneoftheterminals ofagalvanometer, asshown. Theplates make contact atonepoint P.sothere isacomplete circuit Iftheplates arenowrocked through asmall angle, thepoint ofcontact willmove toP’.Ifweimagine the“circuit” tobecompleted through theplates onthedotted lineshown intheﬁgure, themagnetic ﬂuxthrough thiscircuit changes byalarge amount astheplates arerocked back andforth. Yettherocking canbedone with small motions, sothat vXBisvery small andthere ispractically noemf. The “flux rule” does notwork inthiscase. Itmust beapplied tocircuits inwhich the material ofthecircuit remains thesame. When thematerial ofthecircuit ischang- ing,wemust return tothebasic laws. Thecorrect physics isalways given bythe twobasic laws F=q(E+v><B), 17-3 Particle acceleration byaninduced electric ﬁeld; thebetatron Wehave saidthattheelectromotive force generated byachanging magnetic ﬁeldcanexist even without conductors; thatis,there canbemagnetic induction without wires. Wemay stillimagine anelectromotive force around anarbitrary mathematical curve inspace. Itisdeﬁned asthetangential component ofE integrated around thecurve. Faraday’s lawsaysthatthislineintegral isequal to therateofchange ofthemagnetic ﬂuxthrough theclosed curve, Eq.(17.3). Asanexample oftheeffect ofsuch aninduced electric ﬁeld, wewant now to consider themotion ofanelectron inachanging magnetic ﬁeld. Weimagine a magnetic ﬁeldwhich, everywhere onaplane, points inavertical direction, asshown inFig.17-4. Themagnetic ﬁeld isproduced byanelectromagnet, butwewillnot worry about thedetails Forourexample wewillimagine thatthemagnetic ﬁeld issymmetric about some axis, ie..that thestrength ofthemagnetic ﬁeld will depend only onthedistance from theaxis. Themagnetic ﬁeld isalsovarying with time Wenow imagine anelectron thatismoving inthisﬁeld onapath thatisa circle ofconstant radius with itscenter attheaxisoftheﬁeld. (We willseelater 17~3'mmMAGNET V/ ‘ \ » ‘W”\._ ERsc L "'4 COPP DI H Fig. l7-2. When the disc rotates there isanemf from vXB,butwith GALVANOMETER nochange inthelinked flux. Now wewilldescribe asituation inwhich theﬁuxthrough acircuit does not OPPER PLATES _ 1*-—— —-.i,P'\ \ 1| \\ l l \l ._ \a\ , @B \ \ . _\ ' e, /, I I GALVANOMETER Fig. l7—3. When the plates are rocked inauniform magnetic ﬁeld, there can bealarge change inthe flux linkage without the generation ofan emf. .SE/' “E.. Q ? B '. ' qE\ ‘E ''LINES ora Fg l7—4. An electron accelerating inanaxially symmetric, time-varying magnetic ﬁeld. how thismotion canbearranged.) Because ofthechanging magnetic ﬁeld, there willbeanelectric ﬁeldEtangential totheelectron’s orbit which willdrive itaround thecircle. Because ofthesymmetry, thiselectric ﬁeld willhave thesame value everywhere onthecircle. Iftheelectron’s orbit hastheradius r,thelineintegral ofEaround theorbit isequal totherateofchange ofthemagnetic ﬂuxthrough thecircle. Thelineintegral ofEis]LlStitsmagnitude times thecircumference of thecircle, 27rr. Themagnetic fluxmust, ingeneral, beobtained from anintegral. Forthemoment, weletBM,represent theaverage magnetic ﬁeld intheinterior of thecircle; then theﬂuxisthisaverage magnetic ﬁeld times thearea ofthecircle. Wewillhave i 6 0 2 27rrE -at(BM. 7rr). Since weareassuming risconstant, Eisproportional tothetime derivative of theaverage ﬁeld: dBE=5_e-Y- 17.42dt ( ) Theelectron willfeeltheelectric force qEandwillbeaccelerated byit.Remember- ingthattherelativistically correct equation ofmotion isthattherateofchange of themomentum isproportional totheforce, wehave qE=dt (17.5) Forthecircular orbit wehave assumed, theelectric force ontheelectron is always inthedirection ofitsmotion, soitstotal momentum willbeincreasing at therategiven byEq.(17.5). Combining Eqs. (17.5) and(17.4), wemay relate the rateofchange ofmomentum tothechange oftheaverage magnetic ﬁeld: dp qrdB,,___=__ ". 7_ dz 2dt (16) Integrating with respect tot,weﬁndfortheelectron’s momentum 1»=Po+§AB... <17-7) where p0isthemomentum with which theelectrons start out,andABM, isthesub- sequent change inBM. Theoperation ofabetatr0n—a machine foraccelerating electrons tohigh energies—is based onthisidea. Toseehow thebetatron operates indetail, wemust now examine how the electron canbeconstrained tomove onacircle. Wehave discussed inChapter ll ofVol. Itheprinciple involved. Ifwearrange thatthere isamagnetic ﬁeld Bat theorbit oftheelectron, there willbeatransverse force qvXBwhich, forasuit- l7-4 ablychosen B,cancause theelectron tokeep moving onitsassumed orbit. Inthe betatron thistransverse force causes theelectron tomove inacircular orbit of constant radius. Wecanﬁndoutwhat themagnetic ﬁeld attheorbit must beby using again therelativistic equation ofmotion, butthistime, forthetransverse component oftheforce. Inthebetatron (seeFigl7—4), Bisatright angles tov,so thetransverse force isqvB. Thus theforce isequal totherateofchange ofthetrans- verse component p,ofthemomentum: qvB =dz (17.8) When aparticle ismoving inacircle, therateofchange ofitstransverse momentum isequal tothemagnitude ofthetotal momentum times to,theangular velocity of rotation (following thearguments ofChapter ll,Vol. I): dpg _ dt—cup, (17.9) where, since themotion iscircular, 1..=9- (17.10)r Setting themagnetic force equal tothetransverse acceleration, wehave qUBorbit =Pg’ where B,,,b,, istheﬁeld attheradius r. Asthebetatron operates, themomentum oftheelectron grows inproportion toB,,,,,according toEq.(17.7), andiftheelectron istocontinue tomove inits proper circle, Eq.(17.11) must continue tohold asthemomentum oftheelectron increases. Thevalue ofB,,,b,, must increase inproportion tothemomentum p. Comparing Eq.(17.11) with Eq.(17.7), which determines p,weseethatthefollow- ingrelation must hold between B.,,,_ theaverage magnetic ﬁeld inside theorbit attheradius r,andthemagnetic ﬁeld BMW attheorbit: A3,,=2AB,,,b,,. (17.12) Thecorrect operation ofabetatron requires thattheaverage magnetic ﬁeld inside theorbit increase attwice therateofthemagnetic ﬁeld attheorbit itself. Inthese circumstances, astheenergy oftheparticle isincreased bytheinduced electric ﬁeldthemagnetic ﬁeld attheorbit increases atjusttheraterequired tokeep the particle moving inacircle. Thebetatron isused toaccelerate electrons toenergies oftensofmillions of volts, oreven tohundreds ofmillions ofvolts. However, itbecomes impractical for theacceleration ofelectrons toenergies much higher than afewhundred million volts forseveral reasons. One ofthem isthepractical difficulty ofattaining the required highaverage value forthemagnetic ﬁeldinside theorbit. Another isthat Eq.(17.6) isnolonger correct atvery high energies because itdoes notinclude the lossofenergy from theparticle duetoitsradiation ofelectromagnetic energy (theso-called synchrotron radiation discussed inChapter 36,Vol. I).Forthese reasons, theacceleration ofelectrons tothehighest energies—to many billions of electron volts—is accomplished bymeans ofadifferent kind ofmachine, called a synchrotron. 17-4 Aparadox Wewould now liketodescribe foryouanapparent paradox. Aparadox isa situation which gives oneanswer when analyzed oneway, andadifferent answer when analyzed another way, sothatweareleftinsomewhat ofaquandary asto actually what should happen. Ofcourse, inphysics there arenever anyrealpara- doxes because there isonly onecorrect answer; atleast webelieve thatnature will l7-5 CHARGED METAL SPHERES COIL OFWIRE V],//5 0.. .0\-/6 I‘.. BATTERY.‘ \|‘.__ a II _$$ PLASTIC DISC ‘ Fig. l7—5. Will thedisc rotate ifthe current Iisstopped? dl> ,1].___>__ I B LOAD 7%___ /// E If Fig. l7—6. Acoilofwire rotating ina uniform magnetic field—-the basic idea oftheacgenerator.actinonly oneway(and thatistheright way, naturally). Soinphysics aparadox isonly aconfusion inourownunderstanding. Here isourparadox. Imagine thatweconstruct adevice likethatshown inFig. 17-5. There isa thin, circular plastic discsupported onaconcentric shaft with excellent bearings, sothatitisquite freetorotate. Onthediscisacoilofwire intheform ofashort solenoid concentric with theaxisofrotation. This solenoid carries asteady current Iprovided byasmall battery, alsomounted onthedisc. Near theedge ofthedisc andspaced uniformly around itscircumference areanumber ofsmall metal spheres insulated from each other andfrom thesolenoid bytheplastic material ofthedisc. Each ofthese small conducting spheres ischarged with thesame electrostatic charge Q.Everything isquite stationary, andthediscisatrest. Suppose nowthat bysome accident—or byprearrangement—the current inthesolenoid isinter- rupted, without, however, anyintervention from theoutside. Solongasthecurrent continued, there wasamagnetic fluxthrough thesolenoid more orlessparallel totheaxisofthedisc. When thecurrent isinterrupted, thisﬁuxmust gotozero. There will, therefore, beanelectric ﬁeld induced which willcirculate around in circles centered attheaxis. Thecharged spheres ontheperimeter ofthediscwill allexperience anelectric ﬁeld tangential totheperimeter ofthedisc. This electric force isinthesame sense forallthecharges andsowillresult inanettorque onthe disc. From these arguments wewould expect thatasthecurrent inthesolenoid disappears, thediscwould begin torotate. Ifweknew themoment ofinertia of thedisc, thecurrent inthesolenoid, andthecharges onthesmall spheres, wecould compute theresulting angular velocity. ButWecould alsomake adifferent argument. Using theprinciple ofthecon- servation ofangular momentum, wecould saythattheangular momentum ofthe discwith allitsequipment isinitially zero, andsotheangular momentum ofthe assembly should remain zero. There should benorotation when thecurrent is stopped. Which argument IScorrect "Willthediscrotate orwillitnot‘? Wewill leave thisquestion foryoutothink about. Weshould warn youthatthecorrect answer does notdepend onanynon- essential feature, such astheasymmetric position ofabattery, forexample. In fact, youcanimagine anideal situation such asthefollowing‘ Thesolenoid is made ofsuperconducting Wire through which there isacurrent. After thedischas been carefully placed atrest,thetemperature ofthe solenoid isallowed toriseslowly When thetemperature ofthewire reaches thetransition temperature between superconductivity andnormal conductivity, thecurrent inthesolenoid willbe brought tozero bytheresistance ofthewire. Theﬂuxwill,asbefore, falltozero, andthere willbeanelectric ﬁeldaround theaxis. Weshould alsowarn youthatthe solution isnoteasy, norisitatrick. When youﬁgure itout,youwillhave dis- covered animportant principle ofelectromagnetism. 17-5 Alternating-current generator Intheremainder ofthischapter weapply theprinciples ofSection 17-1 to analyze anumber ofthe phenomena discussed inChapter 16.Weﬁrstlook inmore detail atthealternating-current generator. Such agenerator consists basically ofa coilofwire rotating inauniform magnetic ﬁeld. Thesame result canalso be achieved byaﬁxed coilinamagnetic ﬁeld whose direction rotates inthemanner described inthelastchapter. Wewillconsider only theformer case. Suppose we have acircular coilofwire which canbeturned onanaxisalong oneofitsdiam- eters. Letthiscoilbelocated inauniform magnetic ﬁeld perpendicular totheaxis of‘rotation, asinFig. 17-6 Wealso imagine that thetwoends ofthecoilare brought toexternal connections through some kind ofsliding contacts. Duetotherotation ofthecoil, themagnetic ﬁuxthrough itwillbechanging. Thecircuit ofthecoilwilltherefore have anemfinit.LetSbetheareaofthecoil and0theangle between themagnetic ﬁeldandthenormal totheplane ofthecoil.* *Now thatweareusing theletter Aforthevector potential, weprefer toletSstand foraSurface area. 17-6 Theﬁuxthrough thecoilisthen BScos0. (17.13) Ifthecoilisrotating attheuniform angular velocity w,0varies with time as 0=wt.Theemf8inthecoilisthen 8=—Z€ (ﬂux) =—dit (BScoswt), or 8=BSw sinwt. (17.14) Ifwebring thewires from thegenerator toapoint some distance from the rotating coil, where themagnetic ﬁeld iszero, oratleast isnotvarying with time, thecurlofEinthisregion willbezero andwecandeﬁne anelectric potential. Infact, ifthere isnocurrent being drawn from thegenerator, thepotential differ- ence Vbetween thetwowires willbeequal totheemfintherotating coil. That is, V=BSw sinwt=V0sinwt. Thepotential difference between thewires varies assinwt.Such avarying potential difference iscalled analternating voltage. Since there isanelectric ﬁeld between thewires, they must beelectrically charged. Itisclear thattheemfofthegenerator haspushed some excess charges outtothewireuntil theelectric ﬁeldfrom them isstrong enough toexactly counter- balance theinduction force. Seen from outside thegenerator, thetwowires appear asthough they hadbeen electrostatically charged tothepotential difference V, andasthough thecharge wasbeing changed with time togiveanalternating po- tential ditference. There isalsoanother difference from anelectrostatic situation. Ifweconnect thegenerator toanexternal circuit thatpermits passage ofacurrent, weﬁndthattheemfdoes notpermit thewires tobedischarged butcontinues to provide charge tothewires ascurrent isdrawn from them, attempting tokeepthe wires always atthesame potential difference. If,infact, thegenerator isconnected inacircuit whose total resistance isR,thecurrent through thecircuit willbepro- portional totheemfofthegenerator andinversely proportional toR.Since the emfhasasinusoidal timevariation, soalsodoes thecurrent. There isanalternating current I= -1%: —I/Igsinwt. Theschematic diagram ofsuch acircuit isshown inFig.17-7. Wecanalsoseethattheemfdetermines how much energy issupplied bythe generator. Each charge inthewire isreceiving energy attherateF-v.where F1S theforce onthecharge andvisitsvelocity. Now letthenumber ofmoving charges perunitlength ofthewire ben;then thepower being delivered intoanyelement dsofthewire is F-l}I1dS. Forawire, visalways along ds,sowecanrewrite thepower as nvF-ds. Thetotal power being delivered tothecomplete circuit istheintegral ofthis expression around thecomplete loop: Power =yfmir ds. (17.15) Now remember thatqnvisthecurrent I,andthattheemfisdeﬁned astheintegral ofF/qaround thecircuit. Wegettheresult Power from agenerator =SI. (17.16) 17-7I—-v A.C. R Generator g=gt5...... Fig. 17-7. Acircuit with an ac generator andaresistance. When there isacurrent inthecoilofthegenerator, there willalsobemechani- calforces onit.Infact, weknow thatthetorque onthecoilisproportional toits magnetic moment, tothemagnetic ﬁeld strength B,andtothesineoftheangle between. Themagnetic moment isthecurrent inthecoiltimes itsarea. Therefore thetorque is 7'=ISBsin6. (17.17) Therateatwhich mechanical work must bedone tokeep thecoilrotating isthe angular velocity wtimes thetorque: 91%’=w7'=wISBsin0. (17.18) Comparing thisequation with Eq.(17.14),weseethattherateofmechanical work required torotate thecoilagainst themagnetic forces isJustequal toE11,therate atwhich electrical energy isdelivered bytheemfofthegenerator. Alloftheme- chanical energy used upinthegenerator appears aselectrical energy inthecircuit. Asanother example ofthecurrents andforces duetoaninduced emf, let’s analyze what happens inthesetup described inSection 12,andshown inFig.17-1. There aretwoparallel wires andasliding crossbar located inauniform magnetic ﬁeld perpendicular totheplane oftheparallel wires. Now let’sassume thatthe “bottom” oftheU(theleftsideintheﬁgure) ismade ofwires ofhigh resistance, while thetwosidewires aremade ofagood conductor likecopper—then wedon’t need toworry about thechange ofthecircuit resistance asthecrossbar ismoved. Asbefore, theemfinthecircuit is 8=t1Bw. (17.19) Thecurrent inthecircuit isproportional tothisemfandinversely proportional totheresistance ofthecircuit: 8 vBwI-7?——R—- (17.20) Because ofthiscurrent there willbeamagnetic force onthecrossbar thatis proportional toitslength, tothecurrent init,andtothemagnetic ﬁeld, such that F=Blw. (17.21) Taking Ifrom Eq.(17.20), wehave fortheforce B2w2F_T v. (17.22) Weseethattheforce isproportional tothevelocity ofthecrossbar. Thedirection oftheforce, asyoucaneasily see,isopposite toitsvelocity. Such a“velocity- proportional” force, which isliketheforce ofviscosity, isfound whenever induced currents areproduced bymoving conductors inamagnetic ﬁeld. Theexamples of eddy currents wegave inthelastchapter alsoproduced forces ontheconductors proportional tothevelocity -oftheconductor, even though such situations, in general, giveacomplicated distribution ofcurrents which isdifﬁcult toanalyze. It1Soften convenient inthedesign ofmechanical systems tohave damping forces which areproportional tothevelocity. Eddy-current forces provide oneof themost convenient ways ofgetting such avelocity-dependent force. Anexample oftheapplication ofsuch aforce isfound intheconventional domestic wattmeter. Inthewattmeter there isathinaluminum discthatrotates between thepoles ofa permanent magnet. This discisdriven byasmall electric motor whose torque is proportional tothepower being consumed intheelectrical circuit ofthehouse. Because oftheeddy-current forces inthedisc, there isaresistive force proportional tothevelocity. Inequilibrium, thevelocity istherefore proportional totherateof consumption ofelectrical energy. Bymeans ofacounter attached totherotating disc, arecord iskept ofthenumber ofrevolutions itmakes. This count isanindi- cation ofthetotal energy consumption, i.e.,thenumber ofwatthours used. 17-8 Wemay alsopoint outthatEq.(17.22) shows thattheforce from induced currents—that is,anyeddy-current force—is inversely proportional tothere- sistance. Theforce willbelarger, thebetter theconductivity ofthematerial. The reason, ofcourse, isthatanemfproduces more current iftheresistance islow,and thestronger currents represent greater mechanical forces. Wecanalsoseefrom ourformulas how mechanical energy isconverted into electrical energy. Asbefore, theelectrical energy supplied totheresistance ofthe circuit istheproduct 81.Therateatwhich work isdone inmoving theconducting crossbar istheforce onthebartimes itsvelocity. Using Eq.(17.21) fortheforce, therateofdoing work is ﬂ_128%dzTR Weseethatthisisindeed equal totheproduct 81wewould getfrom Eqs. (17.19) and(17.20). Again themechanical work appears aselectrical energy. 17-6 Mutual inductance Wenowwant toconsider asituation inwhich there areﬁxed coils ofwirebut changing magnetic ﬁelds. When wedescribed theproduction ofmagnetic ﬁelds by currents, weconsidered onlythecaseofsteady currents. Butsolong asthecurrents arechanged slowly, themagnetic ﬁeldwillateach instant benearly thesame asthe magnetic ﬁeld ofasteady current. Wewillassume inthediscussion ofthissection thatthecurrents arealways varying sufﬁciently slowly thatthisistrue. InFig. 17-8 isshown anarrangement oftwocoils which demonstrates the basic effects responsible fortheoperation ofatransformer. Coil 1consists ofa conducting wire wound intheform ofalong solenoid. Around thiscoil—and insulated from it—is wound coil2,consisting ofafewturns ofwire. Ifnowa current ispassed through coil1,weknow thatamagnetic ﬁeldwillappear inside it. This magnetic ﬁeld alsopasses through coil2.Asthecurrent incoil1isvaried, themagnetic ﬂuxwillalsovary, andthere willbeaninduced emfincoil2.Wewill nowcalculate thisinduced emf. Wehave seen inSection 13-5 thatthemagnetic ﬁeld inside along solenoid is uniform andhasthemagnitude 1N111= —-i i 9 B eocz l (3) where N1isthenumber ofturns incoil1,I1isthecurrent through it,andlisits length. Let’s saythatthecross-sectional area ofcoil1isS;then theﬂuxofBis itsmagnitude times S.Ifcoil2hasN2turns, thisfluxlinks thecoilN2times. Therefore theemfincoil2isgiven by 82=—NgS(j{—€- (17.24) Theonlyquantity inEq.(17.23) which varies with time is11.Theemfistherefore given by NNSdI18,==____l_TZ, 2 eucll dt(17.25) Weseethattheemfincoil2isproportional totherateofchange ofthecurrent incoil1.Theconstant ofproportionality, which isbasically ageometric factor of thetwocoils, iscalled themutual inductance, andisusually designated 31121.Equa- tion(17.25) isthen written 82=am,gal (17.26) Suppose now that wewere topass acurrent through coil2andaskabout theemfincoil1.Wewould compute themagnetic ﬁeld, which iseverywhere 17-9B II 4; "'“.%1% ftCOIL 2 Fig. 17-8. Acurrent incoil lpro duces amagnetic field through coil2. proportional tothecurrent I2.Theﬁuxlinkage through coil1would depend on thegeometry, butwould beproportional tothecurrent I2.Theemfincoil1 would, therefore, again beproportional toall2/dt:Wecanwrite 61=@1112%- (17.27) Thecomputation ofS1112 would bemore difficult than thecomputation wehave justdone forS1121.Wewillnotcarry through thatcomputation now, because we willshow later inthischapter thatS1112 isnecessarily equal to£11121. Since foranycoilitsﬁeld isproportional toitscurrent, thesame kind of result would beobtained foranytwocoils ofwire. Theequations (17.26) and (17.27) would have thesame form; only theconstants iYit21 andN112 would be different. Their values would depend ontheshapes ofthecoils andtheir relative positions. ds, I’ ' ds,, 1,1 Fg 17-9 Any two coils have a l mutual inductance ‘lllproportional tothe integral ofdsi dsgr1; Suppose thatwewish toﬁndthemutual inductance between anytwoarbitrary coils—for example, those shown inFig.17-9. Weknow thatthegeneral expression fortheemfincoil1canbewritten as d81- —Ef(1)B nda, where BISthemagnetic ﬁeldandtheintegral istobetaken over asurface bounded bycircuit 1.Wehave seen inSection 14-1 thatsuch asurface integral ofBcanbe related toalineintegral ofthevector potential. Inparticular, /B-nda =; A~ds1, -(1) (1) where Arepresents thevector potential andds1isanelement ofcircuit 1.Theline integral istobetaken around circuit 1.Theemfincoil1cantherefore bewritten as d81 — —Jt4%(1)A dS1. Now let’s assume thatthevector potential atcircuit 1comes from currents incircuit 2.Then itcanbewritten asalineintegral around circuit 2: A=__l if__I2d‘2, (17.29)47l'€()C2 (2) r12 where I2isthecurrent incircuit 2,andr12isthedistance from theelement ofthe circuit dS2tothepoint oncircuit 1atwhich weareevaluating thevector potential. (See Fig. 17-9.) Combining Eqs. (17.28) and(17.29), wecanexpress theemfin circuit 1asadouble lineintegral: 121%f12.11,t;=___- Q11 .1 4-7l'E()C2 dl (1) (2) V12 S1 Inthisequation theintegrals arealltaken with respect tostationary circuits. The only variable quantity isthecurrent I2,which does notdepend onthevariables of 17-10 integration. Wemay therefore take itoutoftheintegrals. Theemfcanthen bewritten as dl 31=91112 7?’ where thecoefficient STZ12 is =m12= -—~—1 jfyf_—d’2'd’1- (17.30)(2)47T€()C2 (1) V12 Weseefrom thisintegral thatam12depends onlyonthecircuit geometry. Itdepends onakind ofaverage separation ofthetwocircuits, with theaverage weighted most forparallel segments ofthetwocoils. Ourequation canbeused forcalculating themutual inductance ofanytwocircuits ofarbitrary shape. Also, itshows that theintegral for£31112 isidentical totheintegral forSR21.Wehave therefore shown thatthetwocoefﬁcients areidentical. Forasystem with only twocoils, theco- efficients STC12and91121areoften represented bythesymbol mtwithout subscripts, called simply themutual inductance: 571112 -'=37521 =5m- 17—7 Self-inductance Indiscussing theinduced electromotive forces inthetwocoils ofFigs. 17-8 or17-9, wehave considered only thecaseinwhich there wasacurrent inonecoil ortheother. Ifthere arecurrents inthetwocoils simultaneously, themagnetic ﬂuxlinking either coilwillbethesumofthetwoﬂuxes which would exist separately, because thelawofsuperposition applies formagnetic ﬁelds. Theemfineither coilwilltherefore beproportional notonly tothechange ofthecurrent inthe other coil, butalsotothechange inthecurrent ofthecoilitself. Thus thetotal emfincoil2should bewritten* 32=37521 6%;"l-97122 dt' (17-31) Similarly, theemfincoil1willdepend notonly onthechanging current incoil2, butalsoonthechanging current initself: 31=37112 (17?'1‘91311 (17-32) Thecoefficients $11122 and91111 arealways negative numbers. Itisusual towrite STZ11 = —£1, fllfgg = '-£2, where £1and£2arecalled theself-inductances ofthetwocoils. The self-induced emfwill, ofcourse, exist even ifwehave only onecoil. Anycoilbyitself willhave aself-inductance .8.Theemfwillbeproportional tothe rateofchange ofthecurrent init.Forasingle coil, itisusual toadopt thecon- vention thattheemfandthecurrent areconsidered positive iftheyareinthesame direction. With thisconvention, wemay write fortheemfofasingle coil dI8——£ -if (17.34) Thenegative signindicates thattheemfopposes thechange incurrent-it isoften called a“back emf.” Since anycoilhasaself-inductance which opposes thechange incurrent, the current inthecoilhasakind ofinertia. Infact, ifwewish tochange thecurrent in *Thesignofem12andtill;1inEqs. (17.31) and(17.32) depends onthearbitrary choices forthesense ofapositive current inthetwocoils. 17-11 _1- SC’ (<1) V—> F m /////////////7/////// I (bl Fig. l7—l0 (a) Acircuit with a voltage source andaninductance. (b)An analogous mechanical system.acoilwemust overcome thisinertia byconnecting thecoiltosome external voltage source such asabattery oragenerator, asshown intheschematic diagram ofFig. l7—l0(a). Insuchacircuit, thecurrent Idepends onthevoltage ‘Uaccording to therelation d1'0-.8Z;- (17.35) This equation hasthesame form asNewton’s lawofmotion foraparticle in onedimension. Wecantherefore study itbythepI'll'1C1plC that“thesame equations have thesame solutions.” Thus, ifwemake theexternally applied voltage “Ocorre- spond toanexternally applied force F,andthecurrent Iinacoilcorrespond tothe velocity vofaparticle, theinductance .13ofthecoilcorresponds tothemass mofthe particle.* SeeFig. l7—lO(b). Wecanmake thefollowing table ofcorresponding quantities. Particle Coil F(force) v(velocity) x(displacement) F_ Q_mdz*0(potential difference) I(current) q(charge) *0—.0gTdz mv(momentum) £1 %mv2 (kinetic energy) %.£12 (magnetic energy) 17-8 Inductance andmagnetic energy Continuing with theanalogy ofthepreceding section, wewould expect that corresponding tothemechanical momentum p=mv,whose rate ofchange is theapplied force, there should beananalogous quantity equal to£1,whose rateof change is‘O.Wehavenoright, ofcourse, tosaythat£1istherealmomentum ofthe circuit; infact,itisn’t. Thewhole circuit maybestanding stillandhavenomo- mentum. Itisonlythat£1isanalogous tothemomentum mvinthesense ofsatisfy- ingcorresponding equations. Inthesame way, tothekinetic energy %mv2, there corresponds ananalogous quantity @312. Butthere wehave asurprise. This @5312 isreally theenergy intheelectrical casealso. This 1Sbecause therateofdoing work ontheinductance is“OI,andinthemechanical system it1SFI’,thecorre- sponding quantity. Therefore, inthecase oftheenergy, thequantities notonly correspond mathematically, butalsohave thesame physical meaning aswell. Wemay seethisinmore detail asfollows. Aswefound inEq.(17.16), the rateofelectrical work byinduced forces istheproduct oftheelectromotive force andthecurrent: 1;’=8,. Replacing 8byitsexpression interms ofthecurrent from Eq.(17.34), wehave dW»—=— 7.dt dt (136) Integrating thisequation, weﬁndthattheenergy required from anexternal source toovercome theemfintheself-inductance while building upthecurrentt (which must equal theenergy stored, U)is -W=U=@812 (17.37) Therefore theenergy stored inaninductance is5.812. *This is,incidentally, nottheonly way acorrespondence canbesetupbetween me- chanical andelectrical quantities. TWeareneglecting anyenergy losstoheatfrom thecurrent intheresistance ofthecoil. Such losses require additional energy from thesource butdonotchange theenergy which goes intotheinductance. l7—l2 Applying thesame arguments toapairofcoils such asthose inFigs. 17-8 or 17-9, wecanshow thatthetotal electrical energy ofthesystem isgiven by U=5,1211% +aw; +9111112. (17.38) For, starting with I=0inboth coils, wecould ﬁrstturn onthecurrent I1in coil1,with 12=0.Thework done 1Sjust%.,G1I§’. Butnow, onturning upI2, wenotonly dothework $1321 against theemfincircuit 2,butalsoanadditional amount 8111,12, which istheintegral oftheemf[i)TZ(dI2/dt)] incircuit 1times the nowconstant current I1inthatcircuit. Suppose wenow wish toﬁndtheforce between anytwocoils carrying the currents I1andI2.Wemight atﬁrstexpect that wecould usetheprinciple of virtual work, bytaking thechange intheenergy ofEq.(17.38). Wemust remember, ofcourse, thataswechange therelative positions ofthecoils theonly quantity which varies isthemutual inductance 911.Wemight then write theequation of virtual work as —FAx =AU=I1I2AE)1Z (wrong). Butthisequation iswrong because, aswehave seen earlier, itincludes only the change intheenergy ofthetwocoils andnotthechange intheenergy ofthesources which aremaintaining thecurrents I1andI2attheir constant values. Wecannow understand thatthese sources must supply energy against theinduced emf’sinthe coils astheyaremoved. Ifwewish toapply theprinciple ofvirtual work correctly, wemust alsoinclude these energies. AsWehave seen, however, wemay take a short cutandusetheprinciple ofvirtual work byremembering that thetotal energy isthenegative ofwhat wehave called U,,,,.c,,, the“mechanical energy.” We cantherefore write fortheforce —FAx =AU,,,,.,h =—AU. (17.39) Theforce between twocoils isthen given by FAX =I112 Equation (17.38) fortheenergy ofasystem oftwocoils canbeused toshow thataninteresting inequality exists between mutual inductance STZandtheself- inductances £1and£2ofthetwoCO1lS. Itisclear thattheenergy oftwocoils must bepositive. Ifwebegin with zero currents inthecoils andincrease these currents tosome values, wehave been adding energy tothesystem. Ifnot,the currents would spontaneously increase with release ofenergy totherestofthe world-~an unlikely thing tohappen! Now ourenergy equation, Eq.(17.38), can equally wellbewritten inthefollowing form: 1 em 21 5112U=5.21<1,+3112>+§<.c2 --be-1)1;. (17.40) That ISJustanalgebraic transformation. This quantity must always bepositive foranyvalues of11andI2.Inparticular, itmust bepositive ifI2should happen to have thespecial value 12=-%1, (17.41) Butwiththiscurrent forI2,theﬁrstterm inEq.(17.40) iszero. Iftheenergy isto bepositive, thelastterm in(17.40) must begreater than zero. Wehave therequire- ment that .c1.e2 >arc? Wehave thusproved thegeneral result thatthemagnitude ofthemutual inductance ﬁllofanytwocoils isnecessarily lessthan orequal tothegeometric mean ofthe twoself-inductances. (911itself may bepositive ornegative, depending onthesign 17—13 conventions forthecurrents I1andI2.) pm<\/E. (17-42) Therelation between STZandtheself-inductances isusually written as sit=la/53;. (17.43) Theconstant kiscalled thecoefficient ofcoupling. Ifmost oftheﬂuxfrom one coillinks theother coil, thecoelﬁcient ofcoupling isnear one; wesaythecoils are “tightly coupled.” Ifthecoils arefarapart orotherwise arranged sothatthere is very little mutual ﬂuxlinkage, thecoefficient ofcoupling isnear zero and the mutual inductance isvery small. Forcalculating themutual inductance oftwocoils, wehave given inEq. (17.30) aformula which isadouble lineintegral around thetwocircuits. We might think thatthesame formula could beused togettheself-inductance ofa single coilbycarrying outboth lineintegrals around thesame coil. This, however, willnotwork, because inintegrating around thetwocoils, thedenominator r12of theintegrand willgotozero when thetwolineelements areatthesame point. Theself-inductance obtained from thisformula isinﬁnite. Thereason isthatthis formula isanapproximation thatisvalid only when thecross sections ofthewires ofthetwocircuits aresmall compared with thedistance from onecircuit tothe other. Clearly, thisapproximation doesn’t hold forasingle coil. Itis,infact, true thattheinductance ofasingle coiltends logarithmically toinﬁnity asthediameter ofitswire ismade smaller andsmaller. Wemust, then, look foradilferent wayofcalculating theself-inductance ofa single coil. Itisnecessary totake into account thedistribution ofthecurrents within thewires because thesizeofthewireisanimportant parameter. Weshould therefore asknotwhat istheinductance ofa“circuit,” butwhat istheinductance ofadistribution ofconductors. Perhaps theeasiest waytoﬁndthisinductance is tomake useofthemagnetic energy. Wefound earlier, inSection 15-3, anex- pression forthemagnetic energy ofadistribution ofstationary currents: U=%[j~A dV. (17.44) Ifweknow thedistribution ofcurrent density j,wecancompute thevector po- tential Aandthen evaluate theintegral ofEq.(17.44) togettheenergy. This energy isequal tothemagnetic energy oftheself-inductance, 5.812. Equating thetwogives usaformula fortheinductance: .1:=Il2fj~AdV. (17.45) Weexpect, ofcourse, that theinductance isanumber depending only onthe geometry ofthecircuit andnotonthecurrent Iinthecircuit. Theformula ofEq. (17.45) willindeed givesuch aresult, because theintegral inthisequation ispro- portional tothesquare ofthecurrent—the current appears once through jand again through thevector potential A.Theintegral divided byI2willdepend onthe geometry ofthecircuit butnotonthecurrent I. Equation (17.44) fortheenergy ofacurrent distribution canbeputinaquite different form which issometimes more convenient forcalculation. Also, aswe willseelater, itisaform thatisimportant because itismore generally valid. In theenergy equation, Eq.(17.44), both Aandjcanberelated toB,sowecanhope toexpress theenergy interms ofthemagnetic ﬁeld—just aswewere abletorelate theelectrostatic energy totheelectric ﬁeld. Webegin byreplacing jbye(,c2V XB. Wecannot replace Asoeasily, since B=VXAcannot bereversed togiveAin terms ofB.Anyway, wecanwrite EQC2 U=T (VXB)‘A dV. (17.46) 17-14 The interesting thing isthat—with some restrictions—this integral canbe written as 2 U=Bails» (v><A)dV. (17.47) Toseethis, wewrite outindetail atypical term. Suppose thatwetake theterm (VXB);/42 which occurs intheintegral ofEq.(17.46). Writing outthecom- ponents, weget 6B, 6B,, (There are,ofcourse, twomore integrals ofthesame kind.) Wenowintegrate the ﬁrstterm with respect tox—integrating byparts. That is,wecansay aB,, _ I0,4,/>3‘; Azdx —By./12 — By? dx. Now suppose that oursystem-—meaning thesources andﬁelds—is ﬁnite, sothat aswegotolarge distances allﬁelds gotozero. Then iftheintegrals arecarried out over allspace, evaluating theterm B,,A,, atthelimits willgivezero. Wehave left only theterm with B,,(6A,/6x), which isevidently onepart ofB,,(V XA),and, therefore, ofB'(VXA).Ifyouwork outtheother ﬁveterms, youwillseethat Eq.(17.47) isindeed equivalent toEq.(17.46). Butnow wecanreplace (VXA)byB,toget 2 u=%/B-BdV. (17.48) Wehave expressed theenergy ofamagnetostatic situation interms ofthemagnetic ﬁeld only. Theexpression corresponds closely totheformula wefound forthe electrostatic energy: U=%/‘E-EdV. (17.49) Onereason foremphasizing these twoenergy formulas isthatsometimes they aremore convenient touse. More important, itturns outthatfordynamic ﬁelds (when EandBarechanging with time) thetwoexpressions (17.48) and(17.49) remain true, whereas theother formulas wehave given forelectric ormagnetic energies arenolonger correct—they hold only forstatic ﬁelds. Ifweknow themagnetic ﬁeldBofasingle coil,wecanﬁndtheself-inductance byequating theenergy expression (17.48) to55312. Let’s seehow thisworks by ﬁnding theself-inductance ofalong solenoid. Wehave seen earlier thatthemag- netic ﬁeldinside asolenoid isuniform andBoutside iszero. Themagnitude ofthe ﬁeldinside isB=nI/e0c2, where nisthenumber ofturns perunitlength inthe winding andIisthecurrent. Iftheradius ofthecoilisranditslength isL(we takeLverylong, sothatwecanneglect endeffects, i.e.,L>>r),thevolume inside is1rr2L. Themagnetic energy istherefore 2 22 U=992iB2-(v01)=2”T{:,ML, which isequal to%..cI2. Or, 7rr2n2.2=—_2L. (17.50)EQC I7-15 I8 The Maxwell Equations 18-1 Maxwell’s equations Inthischapter wecome backtothecomplete setofthefourMaxwell equations thatwetookasourstarting point inChapter 1.Until now, wehavebeenstudying Maxwell’s equations inbitsandpieces; itistime toaddoneﬁnal piece, andtoput them alltogether. Wewillthen have thecomplete andcorrect story forelectro- magnetic ﬁelds thatmay bechanging with time inanyway. Anything saidinthis chapter thatcontradicts something saidearlier istrueandwhat wassaidearlier is false——because what was said earlier applied tosuch special situations as,for instance, steady currents orﬁxed charges. Although wehavebeenverycareful to point outtherestrictions whenever wewrote anequation, itiseasytoforget allof thequaliﬁcations andtolearn toowellthewrong equations. Now weareready togivethewhole truth, with noqualiﬁcations (oralmost none). Thecomplete Maxwell equations arewritten inTable 18-1, inwords aswell asinmathematical symbols. Thefactthatthewords areequivalent totheequations should bythistimebefamiliar—you should beabletotranslate back andforth from oneform totheother. Theﬁrstequation—that thedivergence ofEisthecharge density over e0——is trueingeneral. Indynamic aswellasinstatic ﬁelds, Gauss’ lawisalways valid. TheﬂuxofEthrough anyclosed surface isproportional tothecharge inside. Thethird equation isthecorresponding general lawformagnetic ﬁelds. Since there arenomagnetic charges, theﬂuxofBthrough anyclosed surface isalways zero. Thesecond equation, thatthecurlofEis—6B/6t, isFaraday’s lawandwas discussed inthelasttwochapters. Italsoisgenerally true. Thelastequation has something new. Wehave seenbefore onlythepartofitwhich holds forsteady currents. InthatcasewesaidthatthecurlofBisj/eocz, butthecorrect general equation hasanewpartthatwasdiscovered byMaxwell. Until Maxwell’s work, theknown laws ofelectricity andmagnetism were those wehave studied inChapters 3through 17.Inparticular, theequation for themagnetic ﬁeld ofsteady currents wasknown only as vXB= (18.1) Maxwell began byconsidering these known lawsandexpressing them asdiffer- ential equations, aswehave done here. (Although theVnotation wasnotyet invented, itismainly duetoMaxwell thattheimportance ofthecombinations of derivatives, which wetoday callthecurlandthedivergence, ﬁrstbecame apparent.) Hethen noticed thatthere wassomething strange about Eq.(18.1). Ifonetakes the divergence ofthisequation, theleft-hand sidewillbezero, because thedivergence ofacurlisalways zero. Sothisequation requires thatthedivergence ofjalsobe zero. Butifthedivergence ofjiszero, then thetotal ﬂuxofcurrent outofany closed surface isalsozero. Theﬂuxofcurrent from aclosed surface isthedecrease ofthecharge inside thesurface. This certainly cannot ingeneral bezero because weknow that the charges canbemoved from oneplace toanother. Theequation V-j= -% (18.2) has,infact,beenalmost ourdeﬁnition ofj.Thisequation expresses theveryfunda- 18-118-1 Maxwell’s equations 18-2 How thenewterm works 18-3 Allofclassical physics 18-4 Atravelling ﬁeld 18-5 Thespeed oflight 18-6 Solving Maxwell’s equations; thepotentials andthewave equation Table 18-1 Classical Physics Maxwell's equations I.v-E=B50 OBII. VXE=—-5 III. V-B=0 j 6E 60+ 6riv.Ev><B= Conservation ofcharge v.,--_<3; Force law F=q(E +vXB) Law ofmotion %(p) =F, where Gravitation m1mg F= 8,- I‘(Flux ofEthrough aclosed surface) =(Charge inSiCle)/e0 (Line integral ofEaround aloop) =—5}(Flux ofBthrough theloop) (Flux ofBthrough aclosed surface) =0 c2(Integralof Baround aloop) =(Current through theloop)/en 8+6?(Flux ofEthrough theloop) (Flux ofcurrent through aclosed surface) =—S;(Charge inside) p=—% (Newton’s law,with Einstein’s modiﬁcation) \/1—v2/c2 mental lawthatelectric charge isconserved—any ﬂow ofcharge must come from some supply. Maxwell appreciated thisdifficulty andproposed that itcould be avoided byadding theterm 8E/6t totheright-hand sideofEq.(18.1); hethen got thefourth equation inTable 18-1: iv. c2VXB=. 95-J EQ 6t Itwasnotyetcustomary inMaxwell’s time tothink interms ofabstract ﬁelds. Maxwell discussed hisideas interms ofamodel inwhich thevacuum waslikean elastic solid. Healsotried toexplain themeaning ofhisnewequation interms of themechanical model. There wasmuch reluctance toaccept histheory, ﬁrstbe- cause ofthemodel, andsecond because there wasatﬁrstnoexperimental justi- ﬁcation. Today, weunderstand better thatwhat counts aretheequations themselves andnotthemodel used togetthem. Wemay only question whether theequations aretrueorfalse. This isanswered bydoing experiments, anduntold numbers of experiments have conﬁrmed Maxwell’s equations. Ifwetake away thescaffolding heusedtobuild it,weﬁndthatMaxwell’s beautiful ediﬁce stands onitsown. He brought together allofthelaws ofelectricity andmagnetism andmade onecomplete andbeautiful theory. Letusshow thattheextra term isjustwhat isrequired tostraighten outthe difficulty Maxwell discovered. Taking thedivergence ofhisequation (IVinTable 18-1), wemust have thatthedivergence oftheright-hand sideiszero: .1' .E_ v60+vat_0. (18.3) 18-2 Inthesecond term, theorder ofthederivatives with respect tocoordinates and time canbereversed, sotheequation canberewritten as . 6V']+e0FtV'E=O. ButtheﬁrstofMaxwell’s equations saysthatthedivergence ofEisp/co. Inserting thisequality inEq.(18.4), wegetback Eq.(18.2), which weknow istrue. Con- versely, ifweaccept Maxwell's equations—and wedobecause noonehasever found anexperiment thatdisagrees withthem—we must conclude thatcharge is always conserved. Thelaws ofphysics have noanswer tothequestion: “What happens ifa charge issuddenly created atthispoint-—what electromagnetic effects arepro- duced?” Noanswer canbegiven because ourequations sayitdoesn’t happen. Ifitwere tohappen, wewould need newlaws, butwecannot saywhat theywould be.Wehave nothadthechance toobserve howaworld without charge con- servation behaves. According toourequations, ifyousuddenly place acharge at some point, youhadtocarry itthere from somewhere else. Inthat case, wecan saywhat would happen. When weadded anewterm totheequation forthecurlofE,wefound thata whole newclass ofphenomena wasdescribed. Weshall seethatMaxwell’s little addition totheequation forVXBalsohasfar-reaching consequences. Wecan touch ononly afewofthem inthischapter. 18-2 How thenewterm works Asourﬁrstexample weconsider what happens with aspherically symmetric radial distribution ofcurrent. Suppose weimagine alittle sphere with radioactive material onit.This radioactive material issquirting outsome charged particles. (Orwecould imagine alarge block ofjello with asmall hole inthecenter into which some charge hadbeen injected withahypodermic needle andfrom which thecharge isslowly leaking out.) Ineither casewewould have acurrent thatis everywhere radially outward. Wewillassume thatithasthesame magnitude in alldirections. Letthetotal charge inside anyradius rbeQ(r). Iftheradial current density atthesame radius isj(r),thenEq.(18.2) requires thatQdecreases attherate 9%’) =—47rr2j(r). (18.5) Wenowaskabout themagnetic ﬁeldproduced bythecurrents inthissituation. Suppose wedraw some loop I‘onasphere ofradius r,asshown inFig. 18-1. There issome current through thisloop, sowemight expect toﬁndamagnetic ﬁeldcirculating inthedirection shown. Butwearealready indifﬁculty. How cantheBhave anyparticular direction onthesphere? Adifferent choice ofI‘would allow ustoconclude thatitsdirection isexactly opposite tothatshown. Sohowcanthere beanycirculation ofBaround thecurrents‘? Wearesaved byMaxwell’s equation. Thecirculation ofBdepends notonly onthetotal current through I‘butalso ontherateofchange with time ofthe electric ﬂuxthrough it.Itmust bethatthese twoparts justcancel. Let’s seeifthat works out. Theelectric ﬁeld attheradius rmust beQ(r)/41re0r2—so long asthecharge issymmetrically distributed, asweassume. Itisradial, anditsrateofchange isthen 6E_1aQ at-Hana" <18-6’ Comparing thiswithEq.(18.5), weseethatatanyradius §§__i. 176!“ 60 18-315 I \ / \ ,’ E \ / \\ F /' )// \\* 7/ \ .' \ EI V \j Fig. 18-1. What isthe magnetic ﬁeld ofaspherically symmetric current? toovr' LOOPr $1 . e--1--------- -_t_. < Il§r> \ ll //I / B \\ ?i// \ / - / as-~>iiiiii» 10> l(b) Fig 18-2. Themagnetic ﬁeld near acharging capacitor. InEq.IVthetwosource terms cancel andthecurlofBisalways zero. There is nomagnetic ﬁeld inourexample. Asoursecond example, weconsider themagnetic ﬁeld ofawire used to charge aparallel-plate condenser (seeFig. 18-2). Ifthecharge Qontheplates is changing with time (butnottoofast), thecurrent inthewires isequal todQ/dt. Wewould expect thatthiscurrent willproduce amagnetic ﬁeldthatencircles the wire. Surely, thecurrent close tothewiremust produce thenormal magnetic ﬁeld—it cannot depend onwhere thecurrent isgoing. Suppose Wetakealoop I‘1which isacircle with radius r,asshown inpart(a) oftheﬁgure. Thelineintegral ofthemagnetic ﬁeld should beequal tothecurrent Idivided byeocz. Wehave I 21rrB =-6-0?; (18.8) Thisiswhat wewould getforasteady current, butitisalsocorrect withMaxwell’s addition, because ifweconsider theplane surface Sinside thecircle, there areno electric ﬁelds onit(assuming thewire tobeavery good conductor). Thesurface integral ofOE/6t iszero. Suppose, however, thatwenow slowly move thecurve Pdownward. Weget always thesame result until wedraw even with theplates ofthecondenser. Then thecurrent Igoes tozero. Does themagnetic ﬁeld disappear? That would be quite strange. Let’s seewhat Maxwell’s equation saysforthecurve F2,which isa circle ofradius rwhose plane passes between thecondenser plates [Fig. l8—2(b)]. Thelineintegral ofBaround I‘2is27rrB. This must equal thetime derivative of theﬂuxofEthrough theplane circular surface S2.This ﬂuxofE,weknow from Gauss’ law,mustbeequal to1/eotimes thecharge Qononeofthecondenser plates. Wehave C221rrB=%- (18.9) That isvery convenient. Itisthesame result wefound inEq.(18.8). Inte- grating over thechanging electric ﬁeld gives thesame magnetic ﬁeld asdoes inte- grating overthecurrent inthewire. Ofcourse, thatisjustwhat Maxwell’s equation says. Itiseasytoseethatthismust always besobyapplying oursame arguments tothetwosurfaces S1andS{that arebounded bythesame circle F1inFig. 18-2(b). Through S1there isthecurrent I,butnoelectric ﬂux. Through S{there isnocurrent, butanelectric ﬂuxchanging attherateI/co. Thesame Bisobtained ifweuseEq.IVwith either surface. From ourdiscussion sofarofMaxwell’s newterm, youmayhave theim- pression thatitdoesn’t addmuch—that itjustﬁxes uptheequations toagree with what wealready expect. Itistruethatifwejustconsider Eq.IVbyitself, nothing particularly new comes out.The words “byitself” are,however, all-important. Maxwell’s small change inEq.IV,when combined with theother equations, does indeed produce much thatisnewandimportant. Before wetakeupthese matters, however, wewant tospeak more about Table 18-1. 18-4 18-3 Allofclassical physics InTable 18-lwehave allthatwasknown offundamental classical physics, thatis,thephysics thatwasknown by1905. Hereitallis,inonetable. With these equations wecanunderstand thecomplete reahri ofclassical physics. First wehavetheMaxwell equations—written inboththeexpanded form and theshort mathematical form. Then there istheconservation ofcharge, which is evenwritten inparentheses, because themoment wehave thecomplete Maxwell equations, wecandeduce from them theconservation ofcharge. Sothetable is evenalittle redundant. Next, wehave written theforce law,because having all theelectric andmagnetic ﬁelds doesn’t tellusanything until weknow what they dotocharges. Knowing EandB,however, wecanﬁndtheforce onanobject with thecharge qmoving withvelocity v.Finally, having theforce doesn’t tellusany- thing until weknow what happens when aforce pushes onsomething; weneed the lawofmotion, which isthattheforce isequal totherateofchange ofthemo- mentum. (Remember? WehadthatinVolume I.)Weeveninclude relativity effects bywriting themomentum asp="1021/\/l —-v2/c2. Ifwereally want tobecomplete, weshould addonemore law-Newton’s lawofgravitation—so weputthatattheend. Therefore inonesmall table wehave allthefundamental laws ofclassical physics—-even withroom towrite them outinwords andwithsome redundancy. Thisisagreat moment. Wehave climbed agreat peak. Weareonthetopof K-2-we arenearly ready forMount Everest, which isquantum mechanics. We haveclimbed thepeak ofa“Great Divide,” andnowwecangodown theother side. Wehavemainly beentrying tolearn howtounderstand theequations. Now thatwehavethewhole thing puttogether, wearegoing tostudy what theequations mean—what newthings theysaythatwehaven’t already seen. We’ve beenworking hardtogetuptothispoint. Ithasbeenagreat effort, butnowwearegoing tohave nicecoasting downhill asweseealltheconsequences ofouraccomplishment. 18-4 Atravelling ﬁeld Now forthenewconsequences. They come from putting together allof Maxwell’s equations. First, let’sseewhat would happen inacircumstance which wepicktobeparticularly simple. Byassuming thatallthequantities varyonlyin onecoordinate, wewillhaveaone-dimensional problem. Thesituation isshown inFig.18-3. Wehaveasheet ofcharge located ontheyz-plane. Thesheet isﬁrst atrest,theninstantaneously given avelocity uinthey-direction, andkeptmoving with thisconstant velocity. You might worry about having such an“inﬁnite” acceleration, butitdoesn’t really matter; justimagine thatthevelocity isbrought to uveryquickly. Sowehavesuddenly asurface current J(Jisthecurrent perunit y MOVING @JNDARY WFIELDS /- / /CHARGE‘: ——\ '1 °\1 _7aze \E§k/ E §\/ .A l1lI 1lm\\l1*"l\\l14-_\\1:<"'\~4.»-8\x‘D Q //No FIELDS 'Fig. 18-3. Ariinﬁnite sheet ofcharge / 5=3=Q issuddenly setintomotion parallel to _ _________._ __._ itself. There aremagnetic and electric71 ’|_ Vt /,l ﬁelds that propagate outfrom thesheet ‘=9 | ;=X0 ataconstant speed. 18-5 BorE" l~—-—-i-vti——-U :- (11) Boril f-—v(t-T)—-——>17 f T v (bl swat V li-—T—-I r U3) Fig.l8-4. lo)Themagnitude ofB (orE)asafunction ofxatthetimetafter thecharge sheet issetinmotion. (b)The ﬁelds foracharge sheet setinmotion, toward negative yatt=T.(c)Thesum of(aland lb).width inthez-direction). Tokeeptheproblem simple, wesuppose thatthere is alsoastationary sheet ofcharge ofopposite signsuperposed ontheyz-plane, so thatthere arenoelectrostatic effects. Also, although intheﬁgure weshow only what ishappening inaﬁnite region, weimagine thatthesheet extends toinﬁnity in=1=yand=*=z.Inother words, wehaveasituation where there isnocurrent, and thensuddenly there isauniform sheet ofcurrent. What willhappen? Well, when there isasheet ofcurrent intheplusy-direction, there is,aswe know, amagnetic ﬁeldgenerated which willbeintheminus z-direction forx>0 andintheopposite direction forx<0.Wecould ﬁndthemagnitude ofBby using thefactthatthelineintegral ofthemagnetic ﬁeldwillbeequal tothecurrent oversoc’. Wewould getthatB=J/2ecc”(since thecurrent Iinastripofwidth wisJwandthelineintegral ofBis2Bw). Thisgives ustheﬁeldnexttothesheet———for small x—-but since weareim- agining aninﬁnite sheet, wewould expect thesame argument togivethemagnetic ﬁeldfarther outforlarger values ofx.However, thatwould mean thatthemoment weturnonthecurrent, themagnetic ﬁeldissuddenly changed from zerotoa ﬁnite value everywhere. Butwait! Ifthemagnetic ﬁeldissuddenly changed, it willproduce tremendous electrical effects. (Ifitchanges inanyway, there are electrical effects.) Sobecause wemoved thesheet ofcharge, wemake achanging magnetic ﬁeld, andtherefore electric ﬁelds must begenerated. Ifthere areelectric ﬁelds generated, theyhadtostartfrom zeroandchange tosomething else. There willbesome 6E/6t thatwillmake acontribution, together withthecurrent'J, tothe production ofthemagnetic ﬁeld. Sothrough thevarious equations there isabig intermixing, andwehavetotrytosolve foralltheﬁelds atonce. Bylooking attheMaxwell equations alone, itisnoteasytoseedirectly how togetthesolution. Sowewillﬁrstshow youwhat theanswer isandthenverify thatitdoesindeed satisfy theequations. Theanswer isthefollowing: TheﬁeldB thatwecomputed is,infact,generated rightnexttothecurrent sheet (forsmall x). Itmust beso,because ifwemake atinylooparound thesheet, there isnoroorn foranyelectric ﬂuxtogothrough it.ButtheﬁeldBoutfarther—for larger x——is, atﬁrst,zero. Itstays zeroforawhile, andthensuddenly turns on.Inshort, we turnonthecurrent andthemagnetic ﬁeldimmediately nexttoitturns ontoa constant value B;thentheturning onofBspreads outfrom thesource region. After acertain time, there isauniform magnetic ﬁeldeverywhere outtosome value x,andthenzerobeyond. Because ofthesymmetry, itspreads inboth the plusandminus x-directions. TheE-ﬁeld doesthesame thing. Before t=0(when weturnonthecurrent), theﬁeldiszeroeverywhere. Then afterthetimet,bothEandBareuniform out tothedistance x=vt,andzerobeyond. Theﬁelds make theirwayforward like atidalwave, withafront moving atauniform velocity which turns outtobec, butforawhile wewilljustcallitv.Agraph ofthemagnitude ofEorBversus x, astheyappear atthetimet,isshown inFig.18-4(a). Looking again atFig.18-3, atthetimet,theregion between x==*=vtis“ﬁlled” withtheﬁelds, buttheyhave notyetreached beyond. Weemphasize again thatweareassuming thatthecurrent sheet and,therefore theﬁelds EandB,extend inﬁnitely farinboththey-andz-di- rections. (Wecannot draw aninﬁnite sheet, sowehaveshown onlywhat happens inaﬁnite area.) Wewant nowtoanalyze quantitatively what ishappening. Todothat, we wanttolookattwocross-sectional views, atopviewlooking down along they-axis, asshown inFig.18-5, andasideviewlooking back along thez-axis, asshown in Fig.18-6. Suppose westartwiththesideview. Weseethecharged sheet moving up;themagnetic ﬁeldpoints intothepage for+x,andoutofthepage for—-x, andtheelectric ﬁeldisdownward everywhere—out tox==I=vt. Let’s seeifthese ﬁelds areconsistent withMaxwell’s equations. Let’s ﬁrst draw oneofthose loops thatweusetocalculate alineintegral, saytherectangle 1",shown inFig.18-6. Younotice thatonesideoftherectangle isintheregion where there areﬁelds, butonesideisintheregion theﬁelds havestillnotreached. There issome magnetic ﬂuxthrough thisloop. Ifitischanging, there should be anemfaround it.Ifthewavefront ismoving, wewillhave achanging magnetic 18-6 TOP VIEW IX III0 0 -a-%SIDE VIEW 831%‘ 61 T‘-___Zma 5—>———- -__T.__\‘\‘-_.y-____-2"""-»ﬁ-G1|: x N4k§Q X=X¢1~ ‘ ($1 +1 :8 ¢ ." if 11 xI tarts“ 1'*I* Igggglm ‘‘|Zx x x - . I X x 1 | vt O x x x x xo-1vAt Fig.18-5. Topview ofFig.18-3. Fig.18-6. Side view ofFig.18-3. ﬂux,because theareainwhich Bexists isprogressively increasing atthevelocity v. Theﬂuxinside F2isBtimes thepartoftheareainside F2which hasamagnetic ﬁeld. Therateofchange oftheﬂux,since themagnitude ofBisconstant, isthe magnitude times therateofchange ofthearea. Therateofchange oftheareais easy. Ifthewidth oftherectangle I‘;isL,theareainwhich Bexists changes by LvAtinthetimeAt.(SeeFig.18-6.) Therateofchange ofﬂuxisthenBLv. According toFaraday’s law,thisshould equal thelineintegral ofEaround F2, which isjustEL.Wehavetheequation E=vB. (18.10) Soiftheratio ofEtoBisv,theﬁelds wehave assumed willsatisfy Faraday’s equation. Butthatisnottheonlyequation ;wehavetheother equation relating EandB: 2 _l EL". cvx3-Eo+at (18.11) Toapply thisequation, welookatthetopviewinFig.18-5. Wehaveseenthat thisequation willgiveusthevalue ofBnexttothecurrent sheet. Also, forany loopdrawn outside thesheet butbehind thewavefront, there isnocurlofBnor anyjorchanging E,sotheequation iscorrect there. Now let’slookatwhat hap- pensforthecurve P1thatintersects thewavefront, asshown inFig.18-5. Here there arenocurrents, soEq.(18.11) canbewritten——in integral form—as Hts-d.<1=-‘Z IE-nda. (18.12)pl df insider, Thelineintegral ofBisjustBtimes L.Therateofchange oftheﬂuxofEisdue onlytotheadvancing wavefront. Theareainside F1,where Eisnotzero, isin- creasing attheratevL.Theright-hand sideofEq.(18.12) isthenvLE. Thatequa- tionbeeomes C23=Ev. (1813) Wehave asolution inwhich wehave aconstant Bandaconstant Ebehind thefront, bothatright angles tothedirection inwhich thefront ismoving andat right angles toeachother. Maxwell’s equations specify theratio ofEtoB.From Eqs.(18.10) and(18.13), 2 E=vB, and E=%B. Butonemoment! Wehavefound twodzﬂerent conditions ontheratio E/B. Can suchaﬁeldaswedescribe really exist? There is,ofcourse, onlyonevelocity vfor which both ofthese equations canhold, namely v=c.Thewavefront must travel withthevelocity c.Wehave anexample inwhich theelectrical inﬂuence from acurrent propagates atacertain ﬁnite velocity c. 18-7 Now let’saskwhat happens ifwesuddenly stopthemotion ofthecharged sheet afterithasbeenonforashort timeT.Wecanseewhat willhappen bythe principle ofsuperposition. Wehadacurrent thatwaszeroandthenwassuddenly turned on.Weknow thesolution forthatcase. Now wearegoing toaddanother setofﬁelds. Wetakeanother charged sheet andsuddenly startitmoving, inthe opposite direction withthesame speed, onlyatthetimeTafterwestarted theﬁrst current. Thetotal current ofthetwoadded together isﬁrstzero, thenonfora time T,then oilagain—because thetwocurrents cancel. Wehave asquare “pulse” ofcurrent. Thenewnegative current produces thesame ﬁelds asthepositive one,only withallthesigns reversed and,ofcourse, delayed intimebyT.Awavefront again travels outatthevelocity c.Atthetime tithasreached thedistance x= ==c(r—T),asshown inFig.l8—4(b). Sowehavetwo“blocks” ofﬁeldmarching outatthespeed c,asinparts (a)and(b)ofFig.18-4. Thecombined ﬁelds areas shown inpart(c)oftheﬁgure. Theﬁelds arezeroforx>ct,theyareconstant (with thevalues wefound above) between x=c(t—T)andx=ct,andagain zeroforx<c(t-—T). Inshort, wehave alittlepiece ofﬁeld—a block ofthickness cT——wl1ich has leftthecurrent sheet andistravelling through space allbyitself. Theﬁelds have “taken off”; theyarepropagating freely through space, nolonger connected inany waywiththesource. Thecaterpillar hasturned intoabutterﬂy! How canthisbundle ofelectric andmagnetic ﬁelds maintain itself? Thean- sweris:bythecombined effects oftheFaraday law,VXE=—6B/6t, andthe newterm ofMaxwell, c2VXB=6E/8!. They cannot helpmaintaining them- selves. Suppose themagnetic ﬁeldweretodisappear. There would beachanging magnetic ﬁeldwhich would produce anelectric ﬁeld. Ifthiselectric ﬁeldtriesto goaway, thechanging electric ﬁeldwould create amagnetic ﬁeldback again. So byaperpetual interplay——by theswishing back andforth from oneﬁeldtothe other—they must goonforever. Itisimpossible forthem todisappear)’ They maintain themselves inakindofadance—one making theother, thesecond making theﬁrst—-propagating onward through space. 18-5 Thespeed oflight Wehave awave which leaves thematerial source andgoesoutward atthe velocity c,which isthespeed oflight. Butlet’sgoback amoment. From ahis- torical point ofview, litwasn’t known thatthecoeﬂicient cinMaxwell’s equations wasalsothespeed oflightpropagation. There wasjustaconstant intheequations. Wehavecalled itcfrom thebeginning, because weknew what itwould turnout tobe.Wedidn’t think itwould besensible tomake youlearn theformulas witha different constant andthengobacktosubstitute cwherever itbelonged. From the point ofview ofelectricity andmagnetism, however, wejuststart outwithtwo constants, soandc2,thatappear intheequations ofelectrostatics andmagneto- statics: v-E=3 (18.14)60 and _JVXB-;()c—2- (18.15) Ifwetakeanyarbitrary deﬁnition ofaunitofcharge, wecandetermine experi- mentally theconstant sorequired inEq.(l8.l4)—-say bymeasuring theforce between twounitcharges atrest,using Coulomb*s law. Wemust alsodetermine experimentally theconstant e002thatappears inEq.(18.15), which wecando,say, bymeasuring theforce between twounitcurrents. (Aunitcurrent means oneunit ofcharge persecond.) Theratio ofthese twoexperimental constants isc2—just another “electromagnetic constant.” *Well, notquite. They canbe“absorbed” iftheygettoaregion where there arecharges. Bywhich wemean thatother ﬁelds canbeproduced somewhere which superpose onthese ﬁelds and“cancel” them bydestructive interference (seeChapter 31,Vol.I). 18-3 Notice nowthatthisconstant c2isthesame nomatter what wechoose for ourunitofcharge. Ifweputtwice asmuch “charge”-—say twice asmany proton charges—-in our“unit” ofcharge, sowould need tobeone-fourth aslarge. When wepasstwoofthese “unit” currents through twowires, there willbetwice asmuch “charge” persecond ineach wire, sotheforce between twowires isfour times larger. Theconstant eoczmust bereduced byone-fourth. Buttheratio eocz/so isunchanged. Sojustbyexperiments withcharges andcurrents weﬁndanumber c2which turns outtobethesquare ofthevelocity ofpropagation ofelectromagnetic in- ﬂuences. From static measurements——by measuring theforces between twounit charges andbetween twounitcurrents——we ﬁndthatc=3.00X103meters/sec. When Maxwell ﬁrstmade thiscalculation withhisequations, hesaidthatbundles ofelectric andmagnetic ﬁelds should bepropagated atthisspeed. Healsore- marked onthemysterious coincidence thatthiswasthesame asthespeed oflight. “Wecanscarcely avoid theinference,” saidMaxwell, “that lightconsists inthe transverse undulations ofthesame medium which isthecause ofelectric and magnetic phenomena.” Maxwell hadmade oneofthegreat uniﬁcations ofphysics. Before histime, there waslight, andthere waselectricity andmagnetism. Thelatter twohadbeen uniﬁed bytheexperimental work ofFaraday, Oersted, andAmpere. Then, all ofasudden, light wasnolonger “something else,” butwasonlyelectricity and magnetism inthisnewform—little pieces ofelectric andmagnetic ﬁelds which propagate through space ontheir own. Wehavecalled yourattention tosome characteristics ofthisspecial solution, which turnouttobetrue, however, foranyelectromagnetic wave: thatthemag- netic ﬁeldisperpendicular tothedirection ofmotion ofthewavefront; thatthe electric ﬁeld islikewise perpendicular tothedirection ofmotion ofthewavefront; andthatthetwovectors EandBareperpendicular toeach other. Furthermore, themagnitude oftheelectric ﬁeldEisequal toctimes themagnitude ofthe magnetic ﬁeldB.These three facts—that thetwoﬁelds aretransverse tothedirec- tionofpropagation, thatBisperpendicular toE,andthatE=cB—-are generally trueforanyelectromagnetic wave. Ourspecial caseisagood one——it shows all themain features ofelectromagnetic waves. 18-6 Solving Maxwell’s equations; thepotentials andthewave equation Now wewould liketodosomething mathematical; wewanttowrite Maxwell’s equations inasimpler form. Youmayconsider thatwearecomplicating them, butifyouwillbepatient alittlebit,theywillsuddenly come outsimpler. Although bythistimeyouarethoroughly usedtoeachoftheMaxwell equations, there are many pieces thatmust allbeputtogether. That’s what wewant todo. Webegin withV-B=0-—the simplest oftheequations. Weknow thatit implies thatBisthecurlofsomething. So,ifwewrite B=vXA, (13-16) wehavealready solved oneofMaxwell’s equations. (Incidentally, youappreciate thatitremains truethatanother vector A’would bejustasgood ifA’=A+Vtp —where itisanyscalar ﬁeld—-because thecurlofV111iszero, andBisstillthesame. Wehavetalked about thatbefore.) WetakenexttheFaraday law,VXE=—6B/ 61,because itdoesn’t involve anycurrents orcharges. Ifwewrite BasVXAanddifferentiate withrespect to t,wecanwrite Faraday’s lawintheform . 6VXE-—;tVXA. Since wecandifferentiate either withrespect totimeortospace ﬁrst,wecanalso write thisequation as v><(E+gt!)=0. (18.17) 13-9 WeseethatE+BA/6t isavector whose curlisequal tozero. Therefore thatvec- toristhegradient ofsomething. When weworked onelectrostatics, wehad VXE=0,andthenwedecided thatEitself wasthegradient ofsomething. Wetook ittobethegradient of—¢(theminus fortechnical convenience). We dothesame thing forE+BA/8t; weset E+%‘;=-v¢. (18.18) Weusethesame symbol ¢sothat, intheelectrostatic casewhere nothing changes withtimeandthe6A/8t termdisappears, Ewillbeourold—V¢. SoFaraday’s equation canbeputintheform E=-v¢-%‘§-- (18.19) Wehavesolved twoofMaxwell’s equations already, andwehavefound that todescribe theelectromagnetic ﬁelds EandB,weneed fourpotential functions: ascalar potential ¢andavector potential A,which is,ofcourse, three functions. Now thatAdetermines partofE,aswellasB,what happens when wechange AtoA’=A+VIII? Ingeneral, Ewould change ifwedidn't takesome special precaution. Wecan,however, stillallow Atobechanged inthiswaywithout affecting theﬁelds EandB—that is,without changing thephysics—if wealways change Aand¢together bytherules 4'=A+w/, ¢'=.1,-%-£4 (18.20) Then neither BnorE,obtained from Eq.(18.19), ischanged. Previously, wechose tomake V-A=0,tomake theequations ofstatics somewhat simpler. Wearenotgoing todothatnow; wearegoing tomake a different choice. Butwe’ll waitabitbefore saying what thechoice is,because lateritwillbeclear whythechoice ismade. Now wereturn tothetworemaining Maxwell equations which willgiveus relations between thepotentials andthesources pandj.Once wecandetermine A and¢from thecurrents andcharges, wecanalways getEandBfrom Eqs. (18.16) and(18.19), sowewillhave another form ofMaxwell's equations. Webegin bysubstituting Eq.(18.19) intoV-E=p/co; weget 6A3 which wecanwrite alsoas a-v’¢-5v-A=5 (18.21) Thisisoneequation relating ¢andAtothesources. Ourﬁnalequation willbethemost complicated. Westart byrewriting the fourth Maxwell equation as 2 _ﬁ_Lat—e09 andthen substitute forBandEinterms ofthepotentials, using Eqs. (18.16) and(18.19): c2VX(VXA)——;%(—V4>—-%)=£~ Theﬁrstterm canberewritten using thealgebraic identity: VX(VXA)= V(V-A) -V2A;weget 2 ~¢2v2.4 +c2V(V -.4)+aitv¢+83?‘!= (18.22) lt’snotverysimple! 18-10 Fortunately, wecannowmake useofourfreedom tochoose arbitrarily the divergence ofA.What wearegoing todoistouseourchoice toﬁxthings sothat theequations forAandfor¢areseparated buthavethesame form. Wecando thisbytaking* .___1_@¢.vA_ C25 (18.23) When wedothat,thetwomiddle terms inAand¢inEq.(18.22) cancel, andthat equation becomes much simpler: 2_L&L_i.v.4 C,at,_W (18.24) Andourequation for¢—Eq. (l8.2l)—takes onthesame form: 2_iﬁ__a.v¢C2at,_Go (18.25) What abeautiful setofequations! They arebeautiful, ﬁrst,because theyare nicely separated-—-with thecharge density, goes¢;withthecurrent, goesA.Further- more, although theleftsidelooks alittle funny-—-a Laplacian together with a (6/6!) 2—when weunfold itwesee a2 a2 a2 12"’+ “’+ ¢- a"’=-”» (18.26)6x2 6y2 6z2 c2812 en Ithasanicesymmetry inx,y,z,t—the -1/c2 isnecessary because, ofcourse, timeandspace aredifferent; theyhavedifferent units. Maxwell’s equations haveledustoanewkindofequation forthepotentials 41andAbuttothesame mathematical form forallfourfunctions ¢,A,,,Ay,and A,.Once welearn howtosolve these equations, wecangetBandEfrom VXAand—V¢ —6A/6t. Wehave another form oftheelectromagnetic laws exactly equivalent toMaxwell’s equations, andinmany situations theyaremuch simpler tohandle. Wehave, infact,already solved anequation much likeEq.(18.26). When westudied sound inChapter 47ofVol.I,wehadanequation oftheform f2_if26x2_c26t2’ andwesawthatitdescribed thepropagation ofwaves inthex-direction atthe speed c.Equation (18.26) isthecorresponding wave equation forthree dimensions. Soinregions where there arenolonger anycharges andcurrents, thesolution of these equations isnotthat11>andAarezero. (Although thatisindeed onepossible solution.) There aresolutions inwhich there issome setof¢andAwhich are changing intimebutalways moving outatthespeed c.Theﬁelds travel onward through freespace, asinourexample atthebeginning ofthechapter. With Maxwell ’snewterminEq.IV,wehavebeenabletowrite theﬁeldequa- tions interms ofAand¢inaform thatissimple andthatmakes immediately apparent thatthere areelectromagnetic waves. Formany practical purposes, it willstillbeconvenient tousetheoriginal equations interms ofEandB.But theyareontheother sideofthemountain wehavealready climbed. Now weare ready tocross overtotheother sideofthepeak. Things willlookdifferent—-we are ready forsome newandbeautiful views. *Choosing theV-Aiscalled “choosing agauge.” Changing Abyadding V(l/iscalled a“gauge transformation.” Equation (18.23) iscalled “theLorentz gauge.” 18-11 I9 The Principle ofLeast Action Aspecial lecture—almost verbatim* “When Iwasinhigh school, myphysics teacher-——whose name wasMr.Bader ——called medown onedayafter physics class andsaid, ‘You look bored; Iwant to tellyousomething interesting.’ Then hetold mesomething which Ifound ab- solutely fascinating, andhave, since then, always found fascinating. Every time thesubject comes up,Iwork onit.Infact, when Ibegan toprepare thislecture Ifound myself making more analyses onthething. Instead ofworrying about the lecture, Igotinvolved inanew problem. The subject isthis—the principle of least action. “Mr. Bader toldmethefollowing: Suppose youhave aparticle (inagravita- tional ﬁeld, forinstance) which starts somewhere andmoves tosome other point byfreemotion—you throw it,anditgoes upandcomes down. Itgoes from theoriginal place totheﬁnal place inacertain amount oftime. Now, youtryadifferent motion. Suppose thattogetfrom heretothere, itwent likethis i butgotthere injustthesame amount oftime. Then hesaidthis: Ifyoucalculate thekinetic energy atevery moment onthepath, take away thepotential energy, andintegrate itover thetime during thewhole path, y0u’ll ﬁndthatthenumber you’ll getisbigger than thatfortheactual motion. *Later chapters donotdepend onthematerial ofthisspecial lecture—which isin- tended tobefor“entertainment.” 19-1i “Inother words, thelawsofNewton could bestated notintheform F=ma butintheform: theaverage kinetic energy lesstheaverage potential energy isas littleaspossible forthepathofanobject going from onepoint toanother. “Let meillustrate alittle bitbetter what itmeans. Ifyoutakethecaseofthe gravitational ﬁeld,theniftheparticle hasthepathx(t)(let’sjusttakeonedimension foramoment; wetake atrajectory that goes upanddown andnotsideways), where xistheheight above theground, thekinetic energy is%m(dx/dt) 2,andthe potential energy atanytime ismgx. Now Itake thekinetic energy minus the potential energy atevery moment along thepathandintegrate thatwithrespect totime from theinitial time totheﬁnal time. Let’s suppose thatattheoriginal timet1westarted atsome height andattheendofthetime12wearedeﬁnitely ending atSOm6 0T.hcl' P1ac¢. “Then theintegral is Hl dx2/M[5m —mgx] dt. Theactual motion issome kindofacurve—it’s aparabola ifweplotagainst the time-—and gives acertain value fortheintegral. Butwecould imagine some other motion thatwent veryhighandcame upanddown insome peculiar way. Iililllllllllllljp Wecancalculate thekinetic energy minus thepotential energy andintegrate for suchapath...orforanyother pathwewant. Themiracle isthatthetruepathis theoneforwhich thatintegral isleast. “Let’s tryitout. First, suppose wetakethecaseofafreeparticle forwhich there isnopotential energy atall.Then therulesaysthatingoing from onepoint toanother inagiven amount oftime, thekinetic energy integral isleast, soitmust goatauniform speed. (Weknow that’s therightanswer—to goatauniform speed.) Whyisthat? Because iftheparticle weretogoanyother way,thevelocities would besometimes higher andsometimes lower than theaverage. Theaverage velocity isthesame forevery casebecause ithastogetfrom ‘here’ to‘there’ inagiven amount oftime. “Asanexample, sayyourjobistostartfromhome andgettoschool inagiven length oftime with thecar. You candoitseveral ways: You canaccelerate like madatthebeginning andslowdown withthebrakes neartheend,oryoucango atauniform speed, oryoucangobackwards forawhile andthen goforward, andsoon.Thething isthattheaverage speed hasgottobe,ofcourse, thetotal distance thatyouhave gone overthetime. Butifyoudoanything butgoatauni- form speed, thensometimes youaregoing toofastandsometimes youaregoing tooslow. Now themean square ofsomething thatdeviates around anaverage, as youknow, isalways greater thanthesquare ofthemean; sothekinetic energy integral would always behigher ifyouwobbled your velocity than ifyouwent ata uniform velocity. Soweseethattheintegral isaminimum ifthevelocity isa constant (when there arenoforces). Thecorrect pathislikethis. —i.¢-.9 “Now, anobject thrown upinagravitational ﬁelddoesrisefaster ﬁrstand thenslowdown. That isbecause there isalsothepotential energy, andwemust havetheleastdiﬂerence ofkinetic andpotential energy ontheaverage. Because thepotential energy risesaswegoupinspace, wewillgetalower dﬂerence ifwe cangetassoon aspossible uptowhere there isahighpotential energy. Then we cantakethatpotential away from thekinetic energy andgetalower average. So itisbetter totake apath which goes upandgetsalotofnegative stuff from the potential energy. 1-ZI—P “Ontheother hand, youcan't gouptoofast,ortoofar,because youwillthen have toomuch kinetic energy involved—you have togovery fasttogetway upandcome down again inthefixed amount oftimeavailable. Soyoudon’t want togotoofarup,butyouwant togoupsome. Soitturns outthatthesolution is some kindofbalance between trying togetmore potential energy withtheleast amount ofextra kinetic energy—trying togetthedifference, kinetic minus the potential, assmall aspossible. 19-2 “That isallmyteacher toldme,because hewasaverygood teacher andknew when tostoptalking. ButIdon’t know when tostoptalking. Soinstead ofleaving itasaninteresting remark, Iamgoing tohorrify anddisgust youwiththecomplexi- tiesoflifebyproving thatitisso.Thekind ofmathematical problem wewill haveisverydiﬂicult andanewkind. Wehave acertain quantity which iscalled theaction, S.Itisthekinetic energy, minus thepotential energy, integrated over time. Action =s=/1”(KE_PE)dt.1 Remember thatthePEandKEareboth functions oftime. Foreach different possible pathyougetadifferent number forthisaction. Ourmathematical problem istoﬁndoutforwhat curve thatnumber istheleast. _ “You say-Oh, that’s justtheordinary calculus ofmaxima andminima. Youcalculate theaction andjustdifferentiate toﬁndtheminimum. “But watch out.Ordinarily wejusthaveafunction ofsome variable, andwe havetoﬁndthevalue ofthatvariable where thefunction isleast ormost. For instance, wehavearodwhich hasbeenheated inthemiddle andtheheatisspread around. Foreachpoint ontherodwehaveatemperature, andwemust ﬁndthe point atwhich thattemperature islargest. Butnowforeachpathinspace wehave anumber—quite adifferent thing—and wehavetoﬁndthepathinspace forwhich thenumber istheminimum. Thatisacompletely different branch ofmathematics. Itisnottheordinary calculus. Infact,itiscalled thecalculus ofvariations. “There aremany problems inthiskind ofmathematics. Forexample, the circle isusually deﬁned asthelocus ofallpoints ataconstant distance from a ﬁxed point, butanother wayofdeﬁning acircle isthis: acircle isthatcurve of given length which encloses thebiggest area. Anyother curve encloses lessareafor agiven perimeter thanthecircle does. Soifwegivetheproblem: ﬁndthatcurve which encloses thegreatest areaforagiven perimeter, wewould have aproblem ofthecalculus ofvariations—a diﬁerent kindofcalculus thanyou’re usedto. “Sowemake thecalculation forthepathofanobject. Here isthewaywe aregoing todoit.Theideaisthatweimagine thatthere isatruepathandthat anyother curve wedraw isafalsepath, sothatifwecalculate theaction forthe falsepathwewillgetavalue thatisbigger thanifwecalculate theaction forthe true Path i “Problem: Find thetruepath. Where isit?Oneway,ofcourse, istocalculate theaction formillions andmillions ofpaths andlook atwhich oneislowest. When youﬁndthelowest one,that’s thetruepath. “That’s apossible way. Butwecandoitbetter thanthat. When wehave a quantity which hasaminimum—for instance, inanordinary function likethe temperature—-one oftheproperties oftheminimum isthatifwegoaway from the minimum intheﬁrstorder, thedeviation ofthefunction from itsminimum value isonlysecond order. Atanyplace elseonthecurve, ifwemove asmall distance thevalue ofthefunction changes alsointheﬁrstorder. Butataminimum, atiny motion away makes, intheﬁrstapproximation, nodifference. .-i-1’ “That iswhat wearegoing tousetocalculate thetruepath. Ifwehavethe truepath, acurve which differs onlyalittlebitfromitwill,intheﬁrstapproxima- tion, make nodifference intheaction. Any difference willbeinthesecond approximation, ifwereally haveaminimum. “That iseasytoprove. Ifthere isachange intheﬁrstorder when Ideviate thecurve acertain way,there isachange intheaction thatisproportional tothe deviation. Thechange presumably makes theaction greater; otherwise wehaven’t gotaminimum. Butthenifthechange isproportional tothedeviation, reversing thesignofthedeviation willmake theaction less. Wewould gettheaction to increase onewayandtodecrease theother way. Theonlywaythatitcould really beaminimum isthatintheﬁrstapproximation itdoesn’t make anychange, that thechanges areproportional tothesquare ofthedeviations from thetruepath. 19-3 “Sowework itthisway: Wecallit)(with anunderline) thetruepath—the onewearetrying toﬁnd. Wetakesome trialpathx(t)thatdiffers from thetrue pathbyasmall amount which wewillcall1;(t)(etaoft). my “Now theideaisthatifwecalculate theaction Sforthepathx(t),thenthe diﬁ'erence between thatSandtheaction thatwecalculated forthepathx(t)—to simplify thewriting wecancallitS—the diﬁerence ofSandSmust bezeroin theﬁrst-order approximation ofsmall 11.Itcandiﬁ"er inthesecond order, but intheﬁrstorder thediﬁerence must bezero. “And thatmust betrueforany1;atall.Well, notquite. Themethod doesn’t mean anything unless youconsider paths which allbegin andendatthesame two points—-each pathbegins atacertain point att1andendsatacertain other point att2,andthose points andtimes arekeptﬁxed. Sothedeviations inour1;haveto bezeroateachend,1;(t1) =0and1;(t2) =0.With thatcondition, wehavespeci- ﬁedourmathematical problem. “Ifyoudidn’t know anycalculus, youmight dothesame kind ofthing to ﬁndtheminimum ofanordinary function f(x). Youcould discuss what happens ifyoutakef(x)andaddasmall amount htoxandargue thatthecorrection tof(x) intheﬁrstorder inhmust bezeroattheminimum. Youwould substitute x+h forxandexpand outtotheﬁrstorder inh...justaswearegoing todowith1;. “The ideaisthenthatwesubstitute x(t)=x(t)+n(t)intheformula for theaction: 1 /s<2-at»miwhere Icallthepotential energy V(x). Thederivative dx/dt is,ofcourse, the derivative ofx(t)plusthederivative ofn(t),sofortheaction Igetthisexpression: in_ mdgg dn2_ ]S—ﬁl[—5(-d7+d—,) V(2r+n) dt- “Now Imust write thisoutinmore detail. Forthesquared termIget da2dxdoday(E)+2Eat+<15' Butwait. l’mnotworrying about higher thantheﬁrstorder, soIwilltakeallthe terms which involve 112andhigher powers andputthem inalittle boxcalled ‘second andhigher order.’ From thistermIgetonlysecond order, butthere will bemore from something else. Sothekinetic energy partis 2 g +mgé(7%+(second andhigher order). “Now weneed thepotential Vatx+1;.Iconsider 11small, soIcanwrite V(x) asaTaylor series. Itisapproximately V(x); inthenextapproximation (from theordinary nature ofderivatives) thecorrection is11times therateofchange ofVwithrespect tox,andsoon: 2 Vt;+in=I/(5)+Wm+§I/"ca+--- Ihavewritten V’forthederivative ofVwithrespect toxinorder tosavewriting. Theterminn2andtheonesbeyond fallintothe‘second andhigher order’ category andwedon’t havetoworry about them. Putting italltogether, $2 _ mdx2 dxdn S‘Ha?) "“<5”“Ia—nV’(x) +(second andhigher order)] dt. 19-4 Now ifwelookcarefully atthething, weseethattheﬁrsttwoterms which Ihave arranged herecorrespond totheaction SthatIwould have calculated withthe truepath_x.Thething Iwant toconcentrate onisthechange inS—the difference between theSandtheSthatwewould getfortheright path. Thisdifference we willwrite as6S,called thevariation inS.Leaving outthe‘second andhigher order’ terms, Ihavefor6S in _ dz;do6S —‘/;1 ["1 -E B7 — 11V,(£)]dt- “Now theproblem isthis:Here isacertain integral. Idon’t know what the xisyet,butIdoknow thatnomatter what 1;is,thisintegral must bezero. Well, youthink, theonlywaythatthatcanhappen isthatwhat multiplies 11must be zero. Butwhat about theﬁrsttermwithd1;/dt? Well, afterall,if1;canbeanything atall,itsderivative isanything also,soyouconclude thatthecoefﬁcient ofd1;/dt must alsobezero. That isn’tquite right. Itisn’tquite right because there isa connection between 11anditsderivative; they arenotabsolutely independent, because 1;(t)must bezeroatboth t1andlg. “The method ofsolving allproblems inthecalculus ofvariations always uses thesame general principle. Youmake theshiftinthething youwant tovary (aswedidbyadding 1;);youlookattheﬁrst-order terms; thenyoualways arrange things insuchaform thatyougetanintegral oftheform ‘some kindofstufftimes theshift(n),’butwithnoother derivatives (nodn/dt). Itmust berearranged soit isalways ‘something’ times 1;.Youwillseethegreat value ofthatinaminute. (There areformulas thattellyouhowtodothisinsome cases without actually calculating, buttheyarenotgeneral enough tobeworth bothering about; thebest wayistocalculate itoutthisway.) “How canIrearrange thetermind1;/dt tomake ithavean1;?Icandothat byintegrating byparts. Itturns outthatthewhole trickofthecalculus ofvariations consists ofwriting down thevariation ofSandthenintegrating byparts sothat thederivatives of1;disappear. Itisalways thesame inevery problem inwhich derivatives appear. “You remember thegeneral principle forintegrating byparts. Ifyouhave anyfunction ftimes d11/dzintegrated withrespect tot,youwrite down thederivative ofnf: %(nf) =ng+fd, Theintegral youwant isoverthelastterm, so [f%;1dt= nf—f1;%dt. “Inourformula for6S,thefunction fismtimes dx/dt; therefore, Ihavethe following formula for6S. _ d " "ddx "6S=m%11(1) ‘I—/Q1E(m 17(t)dt-—/;l V'(gc_)17(t)dt. Theﬁrsttermmust beevaluated atthetwolimits t1andt2.Then Imust havethe integral from therestoftheintegration byparts. Thelasttermisbrought down without change. “Now comes something which always happens—the integrated partdisappears. (Infact,iftheintegrated partdoesnotdisappear, yourestate theprinciple, adding conditions tomake sureitdoes!) Wehavealready saidthat11must bezeroatboth endsofthepath, because theprinciple isthattheaction isaminimum provided thatthevaried curve begins andendsatthechosen points. Thecondition isthat 19-5 1;(t1) =0,and1;(t2) =0.Sotheintegrated term iszero. Wecollect theother terms together andobtain this: H 6S=L1[—m % —V'(§):|11(t)dt. Thevariation inSisnowthewaywewanted it—there isthestuffinbrackets, say F,allmultiplied by1;(t)andintegrated from t1tot2. “Wehavethatanintegral ofsomething orother times 1;(t)isalways zero: IF(t)1(1)at=0. Ihave some function oft;Imultiply itby1;(t); andIintegrate itfrom oneendto theother. And nomatter what the1;is,Igetzero. That means thatthefunction F(t)iszero. That’s obvious, butanyway I’llshow youonekindofproof. “Suppose thatfor1;(t)Itooksomething which waszeroforalltexcept right nearoneparticular value. Itstays zerountilitgetstothist, I--i-up thenitblipsupforamoment andblipsrightbackdown. When wedotheintegral ofthis1;times anyfunction F,theonly place thatyougetanything other than zero waswhere 1;(t)wasblipping, andthenyougetthevalue ofFatthatplace times the integral overtheblip. Theintegral overtheblipalone isn’tzero, butwhen multi- plied byFithastobe;sothefunction Fhastobezerowhere theblipwas. But theblipwasanywhere Iwanted toputit,soFmust bezeroeverywhere. “Weseethatifourintegral iszeroforany1;,thenthecoefﬁcient of1;must be zero. Theaction integral willbeaminimum forthepaththatsatisﬁes thiscompli- cated diﬁerential equation: [-1115%-mp]=0. It’snotreally socomplicated; youhaveseenitbefore. ItisjustF=ma.Theﬁrst termisthemass times acceleration, andthesecond isthederivative ofthepotential energy, which istheforce. “So, foraconservative system atleast, wehave demonstrated thattheprinciple ofleast action gives theright answer; itsaysthatthepath thathastheminimum action istheonesatisfying Newton’s law. “One remark: Ididnotprove itwasaminimum—maybe it’samaximum. In fact,itdoesn’t really havetobeaminimum. Itisquite analogous towhat wefound forthe‘principle ofleasttime’ which wediscussed inoptics. There also,wesaid atﬁrstitwas‘least’ time. Itturned out,however, thatthereweresituations inwhich itwasn’t theleast time. Thefundamental principle wasthatforanyﬁrst-order variation away from theoptical path, thechange intimewaszero; itisthesame story. What wereally mean by‘least’ isthattheﬁrst-order change inthevalue ofS,when youchange thepath, iszero. Itisnotnecessarily a‘minimum.’ “Next, Iremark onsome generalizations. Intheﬁrstplace, thething canbe done inthree dimensions. Instead ofjustx,Iwould havex,y,andzasfunctions oft;theaction ismore complicated. Forthree-dimensional motion, youhave to usethecomplete kinetic energy—(m/2) times thewhole velocity squared. Thatis, m dx2 dy2 dz2 KB"5&2?) +(Ft)+(E) Also, thepotential energy isafunction ofx,y,andz.Andwhat about thepath? Thepathissome general curve inspace, which isnotsoeasily drawn, buttheidea isthesame. Andwhat about the1;?Well, 1;canhave three components. You could shiftthepaths inx,oriny,orinz-—or youcould shiftinallthree directions simultaneously. So1;would beavector. Thisdoesn’t really complicate things too much, though. Since only theﬁrst-order variation hastobezero, wecandothe calculation bythree successive shifts. Wecanshift1;onlyinthex-direction and 19-6 saythatcoeﬂicient must bezero. Wegetoneequation. Then weshiftitinthe y-direction andgetanother. Andinthez-direction andgetanother. Or,ofcourse, inanyorder thatyouwant. Anyway, yougetthree equations. And, ofcourse, Newton’s lawisreally three equations inthethree dimensions—-one foreachcom- ponent. Ithink thatyoucanpractically seethatitisbound towork, butwewill leave youtoshow foryourself thatitwillwork forthree dimensions. Incidentally, youcould useanycoordinate system youwant, polar orotherwise, andgetNewton’s lawsappropriate tothatsystem right oﬁ"byseeing what happens ifyouhavethe shift1;inradius, orinangle, etc. “Similarly, themethod canbegeneralized toanynumber ofparticles. Ifyou have, say,twoparticles with aforce between them, sothatthere isamutual potential energy, thenyoujustaddthekinetic energy ofboth particles andtake thepotential energy ofthemutual interaction. Andwhat doyouvary? You varythepaths ofbothparticles. Then, fortwoparticles moving inthree dimensions, there aresixequations. Youcanvarytheposition ofparticle 1inthex-direction, inthey-direction, andinthez-direction, andsimilarly forparticle 2;sothere are sixequations. Andthat’s asitshould be.There arethethree equations thatdeter- mine theacceleration ofparticle linterms oftheforce onitandthree fortheac- celeration ofparticle 2,from theforce onit.Youfollow thesame game through, andyougetNewton’s lawinthree dimensions foranynumber ofparticles. “Ihavebeensaying thatwegetNewton’s law.That isnotquite true,because Newton’s lawincludes nonconservative forces likefriction. Newton saidthatma isequal toanyF.Buttheprinciple ofleast action onlyworks forconservative systems—where allforces canbegotten from apotential function. Youknow, however, thatonamicroscopic level-—on thedeepest level ofphysics-—there are nononconservative forces. Nonconservative forces, likefriction, appear onlybe- cause weneglect microscopic complications——there arejusttoomany particles to analyze. Butthefundamental lawscanbeputintheform ofaprinciple ofleast action. “Letmegeneralize stillfurther. Suppose weaskwhat happens iftheparticle moves relativistically. Wedidnotgettheright relativistic equation ofmotion; F=maisonlyrightnonrelativistically. Thequestion is:Isthere acorresponding principle ofleastaction fortherelativistic case? There is.Theformula inthecase ofrelativity isthefollowing: S=—moc2 Lax/l -v2/c2 dt—q‘/its [4>(x, y,z,t)—v-A(x, y,2,t)]dt. I I Theﬁrstpartoftheaction integral istherestmass motimes cztimes theintegral ofafunction ofvelocity, \/1—vi/c2. Then instead ofjustthepotential energy, wehaveanintegral overthescalar potential ¢andovervtimes thevector potential A.Ofcourse, wearethenincluding onlyelectromagnetic forces. Allelectric and magnetic ﬁelds aregiven interms of¢andA.Thisaction function gives thecom- plete theory ofrelativistic motion ofasingle particle inanelectromagnetic ﬁeld. “Ofcourse, wherever Ihavewritten v,youunderstand thatbefore youtryto ﬁgure anything out,youmust substitute dx/dt forv,andsoonfortheother com- ponents. Also, youputthepoint along thepathattimet,x(t),y(t), z(t)where Iwrote simply x,y,z.Properly, itisonlyafteryouhavemade those replacements forthe v’sthatyouhavetheformula fortheaction forarelativistic particle. Iwillleave tothemore ingenious ofyoutheproblem todemonstrate thatthisaction formula does, infact,givethecorrect equations ofmotion forrelativity. May Isuggest youdoitﬁrstwithout theA,thatis,fornomagnetic ﬁeld? Then youshould get thecomponents oftheequation ofmotion, dp/dt =-qV¢,where, youremember, p=mu/\/l —02/c3. “Itismuch more difﬁcult toinclude alsothecasewithavector potential. Thevariations getmuch more complicated. Butintheend,theforce term does come outequal toq(E+vXB),asitshould. ButIwillleave thatforyouto playwith. “Iwould liketoemphasize thatinthegeneral case, forinstance intherela- tivistic formula, theaction integrand nolonger hastheform ofthekinetic energy 19-7 minus thepotential energy. That’s onlytrueinthenonrelativistic approximation. Forexample, theterm m0c”\/ l-—112/c2 isnotwhat wehave called thekinetic energy. Thequestion ofwhat theaction should beforanyparticular easemust bedetermined bysome kindoftrialanderror. Itisjustthesame problem asdeter- mining whatarethelawsofmotion intheﬁrstplace. Youjusthavetoﬁddle around withtheequations thatyouknow andseeifyoucangetthem intotheform ofthe principle ofleastaction. “One other point onterminology. Thefunction thatisintegrated overtime togettheaction Siscalled theLagrangian, .8,which isafunction onlyofthe velocities andpositions ofparticles. Sotheprinciple ofleastaction isalsowritten s=f"so.-.1».->dt.it where byx,-andv,aremeant allthecomponents ofthepositions andvelocities. Soifyouhearsomeone talking about the‘Lagrangian,’ youknow theyaretalking about thefunction thatisused toﬁndS.Forrelativistic motion inanelectro- magnetic ﬁeld tc= —m(;C2V — + v'A). “Also, Ishould saythatSisnotreally called the‘action’ bythemost precise andpedantic people. Itiscalled ‘Hamilton’s ﬁrstprincipal function.’ Now Ihate togive alecture on‘the-principle-of-least-Hamilton’s-ﬁrst-principal-function.’ SoIcallit‘theaction.’ Also, more andmore people arecalling ittheaction. You see,historically something elsewhich isnotquite asuseful wascalled theaction, butIthink it’smore sensible tochange toanewer deﬁnition. Sonowyoutoo willcallthenewfunction theaction, andpretty sooneverybody willcallitbythat simple name. “Now Iwant tosaysome things onthissubject which aresimilar tothedis- cussions Igaveabout theprinciple ofleasttime. There isquite adifference inthe characteristic ofalawwhich saysacertain integral from oneplace toanother isa minimum—which tellssomething about thewhole path—and ofalawwhich says thatasyougoalong, there isaforce thatmakes itaccelerate. Thesecond waytells howyouinchyourwayalong thepath, andtheother isagrand statement about the whole path. Inthecaseoflight, wetalked about theconnection ofthese two. Now, Iwould liketoexplain whyitistruethatthere aredifferential lawswhen there isaleastaction principle ofthiskind. Thereason isthefollowing: Consider theactual pathinspace andtime. Asbefore, let’stakeonlyonedimension, so wecanplotthegraph ofxasafunction oft.Along thetruepath, Sisaminimum. Let’s suppose thatwehave thetruepathandthatitgoesthrough some point a inspace andtime, andalsothrough another nearby point b. -1-Z1.) Now iftheentire integral from t1tot2isaminimum, itisalsonecessary thatthe integral along thelittlesection from atobisalsoaminimum. Itcan’t bethatthe partfrom atobisalittlebitmore. Otherwise youcould justﬁddle withjustthat piece ofthepathandmake thewhole integral alittlelower. “Soevery subsection ofthepathmust alsobeaminimum. Andthisistrue nomatter howshort thesubsection. Therefore, theprinciple thatthewhole path gives aminimum canbestated alsobysaying thataninﬁnitesimal section ofpath alsohasacurve suchthatithasaminimum action. Nowifwetakeashort enough section ofpath—between twopoints aandbveryclose together—how thepotential varies from oneplace toanother faraway isnottheimportant thing, because you arestaying almost inthesame place overthewhole littlepiece ofthepath. The onlything thatyouhavetodiscuss istheﬁrst-order change inthepotential. The answer canonlydepend onthederivative ofthepotential andnotonthepotential everywhere. Sothestatement about thegross property ofthewhole pathbecomes astatement ofwhat happens forashort section ofthepath—a differential statement. Andthisdifferential statement onlyinvolves thederivatives ofthepotential, that is,theforce atapoint. That’s thequalitative explanation oftherelation between thegross lawandthediﬁerential law. 19-8 “Inthecaseoflight wealsodiscussed thequestion: How doestheparticle ﬁndtheright path? From thedifferential point ofview, itiseasytounderstand. Every moment itgetsanacceleration andknows onlywhat todoatthatinstant. Butallyourinstincts oncause andeffect gohaywire when yousaythattheparticle decides totakethepaththatisgoing togive'the minimum action. Does it‘smell’ theneighboring paths toﬁndoutwhether ornottheyhavemore action? Inthe caseoflight, when weputblocks inthewaysothatthephotons could nottestall thepaths, wefound thattheycouldn’t ﬁgure outwhich waytogo,andwehadthe phenomenon ofdiffraction. “Isthesame thing trueinmechanics? Isittruethattheparticle doesn’t just ‘take theright path’ butthatitlooks atalltheother possible trajectories? Andif byhaving things intheway, wedon’t letitlook, thatwewillgetananalog of diffraction? Themiracle ofitallis,ofcourse, thatitdoesjustthat. That’s what thelaws ofquantum mechanics say. Soourprinciple ofleast action isincom- pletely stated. Itisn’tthataparticle takes thepath ofleast action butthatit smells allthepaths intheneighborhood andchooses theonethathastheleast action byamethod analogous totheonebywhich lightchose theshortest time. Youremember thatthewaylightchose theshortest timewasthis:Ifitwent ona paththattookadifferent amount oftime, itwould arrive atadiﬁerent phase. And thetotalamplitude atsome point isthesumofcontributions ofamplitude forall thedifferent ways thelight canarrive. Allthepaths thatgivewildly different phases don’t adduptoanything. Butifyoucanﬁndawhole sequence ofpaths which havephases ahnost allthesame, thenthelittlecontributions willaddupand yougetareasonable total amplitude toarrive. Theimportant pathbecomes the oneforwhich there aremany nearby paths which givethesame phase. “Itisjustexactly thesame thing forquantum mechanics. Thecomplete quantum mechanics (forthenonrelativistic caseandneglecting electron spin) works asfollows: Theprobability thataparticle starting atpoint latthetimet1 willarrive atpoint 2atthetimet2isthesquare ofaprobability amplitude. The totalamplitude canbewritten asthesumoftheamplitudes foreachpossible path—- foreach wayofarrival. Forevery x(t)thatwecould have—-for every possible imaginary trajectory—we have tocalculate anamplitude. Then weaddthem all together. What'do wetakefortheamplitude foreachpath? Ouraction integral tellsuswhat theamplitude forasingle pathought tobe.Theamplitude ispro- portional tosome constant times e"S/", where Sistheaction forthatpath. That is,ifwerepresent thephase oftheamplitude byacomplex number, thephase angle isS/h. Theaction Shasdimensions ofenergy times time, andP1anck’s constant h hasthesame dimensions. Itistheconstant thatdetermines when quantum me- chanics isimportant. “Here ishowitworks: Suppose thatforallpaths, Sisverylarge compared to It.Onepathcontributes acertain amplitude. Foranearby path, thephase isquite different, because withanenormous Sevenasmall change inSmeans acompletely different phase—because hissotiny. Sonearby paths willnormally cancel their effects outintaking thesum-—except foroneregion, andthatiswhen apath and anearby pathallgivethesame phase intheﬁrstapproximation (more precisely, thesame action within h).Only those paths willbetheimportant ones. Sointhe limiting caseinwhich Planck’s constant hgoes tozero, thecorrect quantum- mechanical laws.can besummarized bysimply saying: ‘Forget about allthese probability amplitudes. Theparticle doesgoonaspecial path, namely, thatonefor which Sdoes notvary intheﬁrstapproximation.’ That’s therelation between the principle ofleastaction andquantum mechanics. Thefactthatquantum mechanics canbeformulated inthiswaywasdiscovered in1942byastudent ofthatsame teacher, Bader, Ispoke ofatthebeginning ofthislecture. [Quantum mechanics wasoriginally formulated bygiving adiﬁerential equation fortheamplitude (Schrodinger) andalsobysome other matrix mathematics (Heisenberg).] “Now Iwant totalkabout other minimum principles inphysics. There are many veryinteresting ones. Iwillnottrytolistthem allnowbutwillonlydescribe onemore. Later on,when wecome toaphysical phenomenon which hasanice minimum principle, Iwilltellabout itthen. Iwant nowtoshow thatwecande- 19-9 scribe electrostatics, notbygiving adiﬁerential equation fortheﬁeld, butbysaying thatacertain integral isamaximum oraminimum. First, let’stakethecasewhere thecharge density isknown everywhere, andtheproblem istoﬁndthepotential 4; everywhere inspace. Youknow thattheanswer should be V2¢=—P/5o- Butanother wayofstating thesame thing isthis:Calculate theintegral U*,where U*=%f(v¢)2dV— /Mar, which isavolume integral tobetaken overallspace. Thisthing isaminimum forthecorrect potential distribution ¢(x,y,z). “We canshow thatthetwostatements about electrostatics areequivalent. Let’s suppose thatwepickanyfunction ¢.Wewant toshow thatwhen wetake for¢thecorrect potential ¢,plusasmall deviation f,thenintheﬁrstorder, the change inU*iszero. Sowewrite ¢=Q+f- The¢iswhat wearelooking for,butwearemaking avariation ofittoﬁndwhat ithastobesothatthevariation ofU*iszerotoﬁrstorder. Fortheﬁrstpartof U*,weneed (v¢)”=W9)’+2v<g~vf+<vf)”- Theonlyﬁrst-order termthatwillvaryis 2vg-vf. Inthesecond term ofthequantity U"‘,theintegrand is M=@+M whose variable partispf.So,keeping onlythevariable parts, weneed theintegral AU*=f(e0V1S-Vf— pf)dV. “Now, following theoldgeneral rule,wehavetogetthedarn thing allclear ofderivatives off. Let’s lookatwhat thederivatives are.Thedotproduct is 6:28f dtpdf OpOf aa+@5+&a’ which wehavetointegrate withrespect tox,toy,andtoz.Now hereisthetrick: togetridofof/6x weintegrate byparts withrespect tox.That willcarry the derivative overontothe¢.It’sthesame general ideaweusedtogetridofderivatives withrespect tot.Weusetheequality agaf _19¢I029 /aa“-fa- ﬁn“ Theintegrated termiszero, since wehavetomake fzeroatinﬁnity. (That corre- sponds tomaking 1;zeroat:1andt2.Soourprinciple should bemore accurately stated: U*islessforthetrue41thanforanyother ¢(x,y,z)having thesame values atinﬁnity.) Then wedo.the same thing foryandz.Soourintegral AU*is AU*=f(-av”; -p)fdV. 19-10 Inorder forthisvariation tobezeroforanyf,nomatter what, thecoefﬁcient of fmust bezeroand,therefore, V22 =—p/€0. Wegetback ouroldequation. Soour‘minimum’ proposition iscorrect. “Wecangeneralize ourproposition ifwedoouralgebra inalittledifferent way. Let’s goback anddoourintegration by’parts without taking components. Westartbylooking atthefollowing equality: v-(fvg) =vf-vg+fv’<g- IfIdifferentiate outtheleft-hand side,Icanshow thatitisjustequal totheright- hand side. Nowwecanusethisequation tointegrate byparts. Inourintegral AU*, wereplace -VQ 'VfbyfV29—V-(fV9),which getsintegrated overvolume. Thedivergence termintegrated overvolume canbereplaced byasurface integral: fV'(fV_d3)dV=ffVQ-nda. Since weareintegrating overallspace, thesurface overwhich weareintegrating is atinﬁnity. There, fiszeroandwegetthesame answer asbefore. “Only nowweseehowtosolve aproblem when wedon’t know where allthe charges are.Suppose thatwehaveconductors withcharges spread outonthem in some way. Wecanstilluseourminimum principle ifthepotentials ofallthe conductors areﬁxed. Wecarry outtheintegral forU*onlyinthespace outside ofallconductors. Then, since wecan’t vary¢ontheconductor, fiszeroonall those surfaces, andthesurface integral _ /fvg-nda isstillzero. Theremaining volume integral AU*=/(-1., v29-pgfdv isonlytobecarried outinthespaces between conductors. Ofcourse, weget Poisson’s equation again, v22=—p/60. Sowehaveshown thatouroriginal integral U*isalsoaminimum ifweevaluate itoverthespace outside ofconductors allatﬁxed potentials (thatis,suchthatany trial¢(x,y,2)must equal thegiven potential oftheconductors when x,y,zisa point onthesurface ofaconductor). “There isaninteresting casewhen theonlycharges areonconductors. Then U*=-%1[(V¢)2dV. Ourminimum principle saysthatinthecasewhere there areconductors setat certain given potentials, thepotential between them adjusts itself sothatintegral U"'isleast. What isthisintegral? ThetermVqsistheelectric ﬁeld, sotheintegral istheelectrostatic energy. Thetrueﬁeldistheone,ofallthose coming from the gradient ofapotential, withtheminimum totalenergy. “Iwould liketousethisresult tocalculate something particular toshow you thatthese things arereally quite practical. Suppose Itaketwoconductors inthe form ofacylindrical condenser. -1-Z’ Theinside conductor hasthepotential V,andtheoutside isatthepotential zero. Lettheradius oftheinside conductor beaandthatoftheoutside, b.Now wecan suppose anydistribution ofpotential between thetwo. Ifweusethecorrect 12, andcalculate so/2I(Vg)2 dV,itshould betheenergy ofthesystem, %CV2. 19-11 Sowecanalsocalculate Cbyourprinciple. Butifweuseawrong distribution of potential andtrytocalculate thecapacity Cbythismethod, wewillgetacapacity thatistoobig,since Visspeciﬁed. Anyassumed potential ¢thatisnottheexactly correct onewillgiveafakeCthatislarger thanthecorrect value. Butifmyfalse ¢isanyrough approximation, theCwillbeagood approximation, because the error inCissecond order intheerror in¢. “Suppose Idon’t know thecapacity ofacylindrical condenser. Icanusethis principle toﬁndit.Ijustguess atthepotential function 11>untilIgetthelowest C. Suppose, forinstance, Ipickapotential thatcorresponds toaconstant ﬁeld. (You know, ofcourse, thattheﬁeldisn’treally constant here; itvaries asl/r.) Aﬁeld which isconstant means apotential which goeslinearly withdistance. Toﬁtthe conditions atthetwoconductors, itmust be r—a¢-V<l—3-:-7) - Thisfunction isVatr=a,zeroatr=b,andinbetween hasaconstant slope equal to—V/(b-a).Sowhat onedoestoﬁndtheintegral U*ismultiply the square ofthisgradient byso/2andintegrate overallvolume. Let’s dothiscal- culation foracylinder ofunitlength. Avolume element attheradius ris21rrdr. Doing theintegral, Iﬁndthatmyﬁrsttryatthecapacity gives 1CV2(ﬁrst try)=3)‘/'b—L2— 21rrdr2 2,,(b—a)2 ' 1|-V2 . b—a SoIhaveaformula forthecapacity which isnotthetrueonebutisanapproximate job:Theintegral iseasy; itisjust C_b+a_ 21re0_2(b—a) Itis,naturally, different from thecorrect answer C=21re0/ln(b/a), butit’snot toobad. Let’s compare itwiththeright answer forseveral values ofb/a. Ihave computed outtheanswers inthistable: a 21re0 21re0 2 1.4423 1.500 4 0.721 0.833 10 0.434 0.612 100 0.267 0.51 2.4662 2.50ll Cm, C(ﬁrstapprox.) 1.5 1.1 10.492070 10.500000 Even when b/aisasbigas2—which gives apretty bigvariation intheﬁeldcom- pared withalinearly varying ﬁeld—I getapretty fairapproximation. Theanswer is,ofcourse, alittletoohigh, asexpected. Thething getsmuch worse ifyouhave atinywireinside abigcylinder. Then theﬁeldhasenormous variations andifyou represent itbyaconstant, you’re notdoing verywell. With b/a=100,we’re oﬁ' bynearly afactor oftwo. Things aremuch better forsmall b/a. Totaketheop- posite extreme, when theconductors arenotveryfarapart—say b/a=l.l—then theconstant ﬁeldisapretty good approximation, andwegetthecorrect value for Ctowithin atenth ofapercent. “Now Iwould liketotellyouhowtoimprove suchacalculation. (Ofcourse, youknow theright answer forthecylinder, butthemethod isthesame forsome other oddshapes, where youmaynotknow theright answer.) Thenextstepisto tryabetter approximation totheunknown true¢.Forexample, wemight trya 19-12 constant plusanexponential ¢,etc.Buthowdoyouknow when youhaveabetter approximation unless youknow thetrue11>?Answer: Youcalculate C;thelowest Cisthevalue nearest thetruth. Letustrythisideaout.Suppose thatthepotential isnotlinear butsayquadratic inr-—that theelectric ﬁeldisnotconstant butlinear. Themost general quadratic form thatﬁts¢=0atr=band¢=Vatr=ais ¢=1/[1+11(2-{-1) -(1+a)<%)2]1 where aisanyconstant number. Thisformula isalittle more complicated. It involves aquadratic terminthepotential aswellasalinear term. Itisveryeasy togettheﬁeldoutofit.Theﬁeldisjust =__£12=___ aV (r—a)V_ E <11 11-a"'2(l+°‘)(b'-a)2 Now wehavetosquare thisandintegrate overvolume. Butwaitamoment. What should Itakefora?Icantakeaparabola forthe¢;butwhat parabola? Here’s what Ido:Calculate thecapacity withanarbitrary ct.What Igetis C a b0:2 2a 12 1 z....,=t-a[a(6 +3+‘)+s“ +5] Itlooks alittlecomplicated, butitcomes outofintegrating thesquare oftheﬁeld. Now Icanpickmyct.Iknow thatthetruth lieslower thananything thatIam going tocalculate, sowhatever Iputinforctisgoing togivemeananswer toobig. ButifIkeep playing withaandgetthelowest possible value Ican,thatlowest value isnearer tothetruth thananyother value. Sowhat Idonextistopickthe athatgives theminimum value forC.Working itoutbyordinary calculus, Iget thattheminimum Coccurs foroz=—2b/ (b+a).Substituting thatvalue into theformula, Iobtain fortheminimum capacity _C_=Qiiiaiiri.21reQ 3(b2—a2) “I’ve worked outwhat thisformula gives forCforvarious values ofb/a. I callthese numbers C(quadratic). withthetrueC. '3GHere isatable thatcompares C(quadratic) Cm» 21re0C(quadratic) 21re0 2 4 10 100 1.51.4423 0.721 0.434 0.267 2.4662 1.1 10.4920701.444 0.733 0.475 0.346 2.4667 10.492065 “For example, when theratio oftheradii is2tol,Ihave 1.444, which isa verygood approximation tothetrueanswer, 1.4423. Even forlarger b/a,itstays pretty good—it ismuch, much better thantheﬁrstapproximation. Itisevenfairly good—-only offby10percent—when b/ais10to1.Butwhen itgetstobe100to1- well,things begin togowild. IgetthatCis0.346 instead of0.267. Ontheother hand, foraratio ofradii of1.5,theanswer isexcellent; andforab/aof1.1,the answer comes out10.492065 instead of10.492070. Where theanswer should be good, itisvery, verygood. “Ihavegiven these examples, ﬁrst,toshow thetheoretical value oftheprinci- plesofminimum action andminimum principles ingeneral and,second, toshow their practical utility—-not justtocalculate acapacity when wealready know the answer. Foranyother shape, youcanguess anapproximate ﬁeldwith some unknown parameters likeorandadjust them togetaminimum. Youwillgetex- cellent numerical results forotherwise intractable problems.” 19-13 Anoteadded after thelecture “Ishould liketoaddsomething thatIdidn’t have timeforinthelecture. (Ialways seem toprepare more thanIhave timetotellabout.) AsImentioned earlier, Igotinterested inaproblem while working onthislecture. Iwant totell youwhat thatproblem is.Among theminimum principles thatIcould mention, Inoticed thatmost ofthem sprang inonewayoranother from theleast action principle ofmechanics andelectrodynamics. Butthere isalsoaclassthatdoesnot. Asanexample, ifcurrents aremade togothrough apiece ofmaterial obeying Ohrn’s law,thecurrents distribute themselves inside thepiece sothattherateat which heatisgenerated isaslittleaspossible. Alsowecansay(ifthings arekept isothermal) thattherateatwhich energy isgenerated isaminimum. Now, this principle alsoholds, according toclassical theory, indetermining even thedis- tribution ofvelocities oftheelectrons inside ametal which iscarrying acurrent. Thedistribution ofvelocities isnotexactly theequilibrium distribution [Chapter 40,Vol.I;Eq.(40.6)] because theyaredrifting sideways Thenewdistribution canbefound from theprinciple thatitisthedistribution foragiven current for which theentropy developed persecond bycollisions isassmall aspossible. The truedescription oftheelectrons’ behavior ought tobebyquantum mechanics, however. Thequestion is:Does thesame principle ofminimum entropy generation alsoholdwhen thesituation isdescribed quantum-mechanically? Ihaven’t found outyet. “The question isinteresting academically, ofcourse. Such principles are fascinating, anditisalways worth while totrytoseehowgeneral theyare.But alsofrom amore practical point ofview, Iwanttoknow. I,withsome colleagues, have published apaper inwhich wecalculated byquantum mechanics approxi- mately theelectrical resistance feltbyanelectron moving through anionic crystal likeNaCl. [Feynman, Hellworth, Iddings, andPlatzman, “Mobility ofSlow Electrons inaPolar Crystal,” Phys Rev.127,1004 (l962).] Butifaminimum principle existed, wecould useittomake theresults much more accurate, justas theminimum principle forthecapacity ofacondenser permitted ustogetsuch accuracy forthatcapacity eventhough wehadonlyarough knowledge oftheelec- tricﬁeld.” 19-14 20 Solutions ofl!Iaxwell’s Equations in Free Space 20-1 Waves infreespace; plane waves InChapter 18wehadreached thepoint where wehadtheMaxwell equations incomplete form. Allthere istoknow about theclassical theory oftheelectric andmagnetic ﬁelds canbefound inthefour equations: I.vt=£ IL vxE=-Q60 61 . mum.VB=0 1v¢%xB=L+§G0 6t When weputallthese equations together, aremarkable newphenomenon occurs: ﬁelds generated bymoving charges canleave thesources andtravel alone through space. Weconsidered aspecial example inWl'llCh aninﬁnite current sheet is suddenly turned on.After thecurrent hasbeen onforthetime i,there areuniform electric andmagnetic ﬁelds extending outthedistance ctfrom thesource. Suppose that thecurrent sheet liesintheyz-plane with asurface current density Jgoing toward positive y.Theelectric ﬁeld willhave only ay-component, andthemag- netic ﬁeld, only az-component. Themagnitude oftheﬁeldcomponents isgiven by E1,=cB,=—5;-I0?’ (20.2) forpositive values ofxlessthan ct.Forlarger xtheﬁelds arezero. There are, ofcourse, similar ﬁelds extending thesame distance from thecurrent sheet inthe negative x-direction. InFig.20-1 weshow agraph ofthemagnitude oftheﬁelds asafunction ofxattheinstant t.Astime goes on,the“wavefront” atctmoves outward inxattheconstant velocity c. Now consider thefollowing sequence ofevents. Weturn onacurrent ofunit strength forawhile, then suddenly increase thecurrent strength tothree units, andhold itconstant atthisvalue. What dotheﬁelds look likethen? Wecansee what theﬁelds willlook likeinthefollowing way. First, weimagine acurrent of unitstrength thatisturned onatt=0andleftconstant forever. Theﬁelds for positive xarethen given bythegraph inpart (a)ofFig.20-2. Next, weaskwhat would happen ifweturn onasteady current oftwounits atthetime 11. Theﬁelds inthiscase willbetwice ashigh asbefore, butwillextend outin xonly thedistance c(t—t1),asshown inpart (b)oftheﬁgure. When weadd these twosolutions, using theprinciple ofsuperposition, weﬁndthatthesum of thetwosources isacurrent ofoneunitforthetime from zero tot1andacurrent ofthree units fortimes greater than t1.Atthetime ttheﬁelds willvary with x asshown inpart (c)ofFig.20-2. Now let’s take amore complicated problem. Consider acurrent which is turned ontooneunitforawhile, then turned uptothree units, andlater turned offtozero. What aretheﬁelds forsuch acurrent? Wecanﬁndthesolution in thesame way—by adding thesolutions ofthree separate problems. First, weﬁnd theﬁelds forastepcurrent ofunitstrength. (Wehave solved thatproblem already.) Next, weﬁndtheﬁelds produced byastepcurrent oftwounits. Finally, wesolve fortheﬁelds ofastepcurrent ofminus three units. When weaddthethree solutions, wewillhave acurrent which isoneunitstrong from t=Otosome later time, say:1,then three units strong until astilllater time 12,andthen turned off——that 20-l20-1 Waves infreespace; plane waves 20-2 Three-dimensional waves 20-3 Scientiﬁc imagination 20-4 Spherical waves References: Chapter 47,Vol.I:Sound: TheWave Equation Chapter 28,Vol.l:Electro- magnetic Radiation lEl=c|B| -ct ct Ti Fig. 20-1. The electric and mog- netic ﬁeld os0function ofxofthetime t offer thecurrent sheet isturned on. E ll 2_ I O l 1lo) ct x Ell 2 I- 0 _ :(bu Cl | ii E ll 3 2_ ;_ O c(t-i,) ct :x (Cl Fig. 20-2. The electric ﬁeld ofci current sheet. lo)One unit ofcurrent turned onofl= O;(blTwo units of current turned oncitt=t;;(c)Super- position of(cland (bl. _Ey 3 2 l l» O '| 2 1 0)c(t-:2) c(t-t|) ct'1 (bl Fig 20-3. Ifthecurrent source strength varies asshown in(cl,then atthe time tshown bythearrow theelectric ﬁeld asufunction ofxisusshown in(b). is,tozero. Agraph ofthecurrent asafunction oftime isshown inFig.20—3(a). When weaddthethree solutions fortheelectric ﬁeld, weﬁndthat itsvariation with x,atagiven instant t,isasshown inFig. 20—3(b). Theﬁeld isanexact representation ofthecurrent. Theﬁeld distribution inspace isanice graph of thecurrent variation withtime——only drawn backwards. Astime goesonthewhole picture moves outward atthespeed c,sothere isalittle blob ofﬁeld, travelling toward positive x,which contains acompletely detailed memory ofthehistory of allthecurrent variations. Ifwewere tostand miles away, wecould tellfrom the variation oftheelectric ormagnetic ﬁeld exactly how thecurrent hadvaried atthesource. You willalsonotice thatlong after allactivity atthesource hascompletely stopped andallcharges andcurrents arezero, theblock ofﬁeldcontinues totravel through space. Wehave adistribution ofelectric andmagnetic ﬁelds thatexist independently ofanycharges orcurrents. That istheneweffect thatcomes from thecomplete setofMaxwell’s equations. Ifwewant, wecangive acomplete mathematical representation oftheanalysis wehave justdone bywriting thatthe electric ﬁeldatagiven place andagiven timeisproportional tothecurrent atthe source, onlynotatthesame time, butattheearlier timet—x/c. Wecanwrite Ey(t)=- (20.3) Wehave, believe itornot,already derived thissame equation from another point ofview inVol. I,when wewere dealing with thetheory oftheindex ofre- fraction. Then, wehadtoﬁgure outwhat ﬁelds were produced byathinlayer of oscillating dipoles inasheet ofdielectric material with thedipoles setinmotion bytheelectric ﬁeld ofanincoming electromagnetic wave. Ourproblem wasto calculate thecombined ﬁelds oftheoriginal wave andthewaves radiated bythe oscillating dipoles. How could wehave calculated theﬁelds generated bymoving charges when wedidn’t have Maxwell’s equations? Atthattime wetook asour starting point (without anyderivation) aformula fortheradiation ﬁelds produced atlarge distances from anaccelerating point charge. Ifyouwilllook inChapter 31ofVol. I,youwillseethatEq.(31.10) there isjustthesame astheEq.(20.3) thatwehave justwritten down. Although ourearlier derivation wascorrect only atlarge distances from thesource, weseenow thatthesame result continues to becorrect even right uptothesource. Wewant nowtolook inageneral wayatthebehavior ofelectric andmagnetic ﬁelds inempty space faraway from thesources, i.e.,from thecurrents andcharges. Very near thesources—near enough sothatduring thedelay intransmission, the source hasnothadtime tochange much—the ﬁelds arevery much thesame aswe have found inwhat wecalled theelectrostatic ormagnetostatic cases. Ifwegoout todistances large enough sothat thedelays become important, however, the nature oftheﬁelds canberadically different from thesolutions wehave found. Inasense, theﬁelds begin totake onacharacter oftheir own when they have gone along wayfrom allthesources. Sowecanbegin bydiscussing thebehavior oftheﬁelds inaregion where there arenocurrents orcharges. 20-2 Suppose weask: What kind ofﬁelds canthere beinregions where pandjare both zero? InChapter 18wesaw that thephysics ofMaxwell’s equations could alsobeexpressed interms ofdifferential equations forthescalar andvector potentials: 2 2__1_u_ _aV4) c2612_ so’ (204) 2_iﬂ___L.VA_C2at,-606, (20.5) Ifpandjarezero, these equations take onthesimpler form 1 2 v2¢-5%t§=0, (20.6) 2_L221!_ v.4C2at,_0. (20.7) Thus infreespace thescalar potential ¢andeach component ofthevector potential Aallsatisfy thesame mathematical equation. Suppose weletIP(psi) stand for anyoneofthefourquantities ¢,AI,A,,,AZ;then wewant toinvestigate thegeneral solutions ofthefollowing equation: 102Vzlp-C-2Elf=0. (20.8) This equation iscalled thethree-dimensional wave equation-—three-dimensional, because thefunction upmaydepend ingeneral onx,y,andz,andweneed toworry about variations inallthree coordinates. This ismade clear ifwewrite outex- plicitly thethree terms oftheLaplacian operator: aw aw a2-p 162¢ 6x2+By?+822‘Faﬂ=°- 0°’) Infreespace, theelectric ﬁelds EandBalsosatisfy thewave equation. For example, since B=VXA,wecangetadifferential equation forBbytaking thecurlofEq.(20.7). Since theLaplacian isascalar operator, theorder ofthe Laplacian andcurloperations canbeinterchanged: v><(VZA) =v2(v><A)=v2B. Similarly, theorder oftheoperations curland6/62 canbeinterchanged: 1.92.4 162 1628 "><z§'.W- 23:2” XA)" aw‘ Using these results, wegetthefollowing differential equation forB: 2_1911!_ VB gatz-O. (20.10) Soeach component ofthemagnetic ﬁeld Bsatisﬁes thethree-dimensional wave equation. Similarly, using thefactthatE=—V¢ —dA/dt, itfollows thatthe electric ﬁeld Einfreespace alsosatisﬁes thethree-dimensional wave equation: 2_L125_ VE C2atz-0. (20.11) Allofourelectromagnetic ﬁelds satisfy thesame wave equation, Eq.(20.8). Wemight wellask:What isthemost general solution tothisequation? However, rather than tackling thatdiﬂicult question right away, wewilllook ﬁrstatwhat canbesaidingeneral about those solutions inwhich nothing varies inyandz. (Always doaneasy case ﬁrstsothatyoucanseewhat isgoing tohappen, and thenyoucangotothemore complicated cases.) Let’s suppose thatthemagnitudes 20-3 oftheﬁelds depend only upon x-—that there arenovariations oftheﬁelds with yandz.Weare,ofcourse, considering plane waves again. Weshould expect to getresults something likethose intheprevious section. Infact, wewillﬁnd precisely thesame answers. You may ask: “Why doitallover again?” Itisim- portant todoitagain, ﬁrst, because wedidnotshow thatthewaves wefound were themost general solutions forplane waves, andsecond, because wefound theﬁelds only from avery particular kind ofcurrent source. Wewould liketoasknow: What isthemost general kind ofone-dimensional wave there canbeinfreespace? Wecannot ﬁndthatbyseeing what happens forthisorthatparticular source, but must work with greater generality. Also wearegoing towork thistime with differ- ential equations instead ofwith integral forms. Although wewillgetthesame re- sults, itisaway ofpracticing back andforth toshow thatitdoesn’t make any difference which wayyougo.YJJ should know how todothings every which way, because when yougetahard problem, youwilloften ﬁndthatonly oneof thevarious ways istractable. Wecould consider directly thesolution ofthewave equation forsome elec- tromagnetic quantity. Instead, wewant tostart right from thebeginning with Maxwell’s equations infreespace sothatyoucanseetheir close relationship to theelectromagnetic waves. Sowestart with theequations in(20.1), setting the charges andcurrents equal tozero. They become I.V'E=0 II. v><E=-23? (20.12) III.V-B: 0 2 _£€ IV.cVXB_6t Wewrite theﬁrstequation outincomponents: __0E, 05 9%_vE_-3;+by+62_0. (20.13) Weareassuming thatthere arenovariations withyandz,sothelasttwoterms are zero. This equation then tellsusthat 6E, _T’; —0. (20.14) Itssolution isthatE,,,thecomponent oftheelectric ﬁeld inthex-direction. isa constant inspace. Ifyoulook atIVin(20.12), supposing noB-variation inyand zeither, youcanseethatE,isalsoconstant intime. Such aﬁeld could bethe steady DCﬁeld from some charged condenser plates along distance away. Weare notinterested now insuch anuninteresting static ﬁeld; weareatthemoment interested only indynamically varying ﬁelds. Fordynamic ﬁelds, E,=O. Wehave then theimportant result thatforthepropagation ofplane waves inanydirection, theelectric ﬁeld must beatright angles tothedirection ofpropaga- tion. Itcan, ofcourse, stillvary inacomplicated waywith thecoordinate x. Thetransverse E-ﬁeld canalways beresolved intotwocomponents, saythe y-component andthez-component. Solet’sﬁrstwork outacaseinwhich theelec- tricﬁeld hasonly onetransverse component. We’ll take ﬁrstanelectric ﬁeld that isalways inthey-direction, with zero z-component. Evidently, ifwesolve this problem wecanalso solve forthecase where theelectric ﬁeld isalways inthe z-direction. Thegeneral solution canalways beexpressed asthesuperposition of twosuch ﬁelds. How easy ourequations now get. Theonly component oftheelectric ﬁeld thatisnotzeroisEy,andallderivatives—-except those with respect tox—are zero TherestofMaxwell’s equations then become quite simple. 20-4 Let’s look next atthesecond ofMaxwell’s equations [IIofEq.(20.l2)]. Writing outthecomponents ofthecurlE,wehave _aE, aE,,_(v><E),_-5};--$_0, 6E, 6E, (vXE)”=¥_7£=0’ _6E1»_9.€s_%(VXELTIK ayT6x Thex-component ofVXEiszero because thederivatives with respect toyand zarezero. They-component isalsozero; theﬁrstterm iszerobecause thederivative withrespect toziszero, andthesecond term iszero because E,iszero. Theonly components ofthecurlofE thatisnotzero isthez-component, which isequal to 6E,,/6x. Setting thethree components ofVXEequal tothecorresponding components of——6B/6!. wecanconclude thefollowing: ea, B-67=0,%=0. (20.15) as, 6E Since thex-component ofthemagnetic ﬁeld andthey-component ofthemagnetic ﬁeldboth have zerotime derivatives, these twocomponents arejustconstant ﬁelds andcorrespond tothemagnetostatic solutions wefound earlier. Somebody may have leftsome permanent magnets near where thewaves arepropagating. Wewill ignore these constant ﬁelds andsetB,andByequal tozero. Incidentally, wewould already have concluded that thex-component ofB should bezero foradifferent reason. Since thedivergence ofBiszero (from the third Maxwell equation), applying thesame arguments weused above forthe electric ﬁeld, wewould conclude thatthelongitudinal component ofthemagnetic ﬁeld canhave novariation with x.Since weareignoring such uniform ﬁelds in ourwave solutions, wewould have setB,equal tozero. Inplane electromagnetic waves theB-ﬁeld, aswell astheE-ﬁeld, must bedirected atright angles tothe direction ofpropagation. Equation (20.16) gives ustheadditional proposition thatiftheelectric ﬁeld hasonly ay-component, themagnetic ﬁeld willhave only az-component. So EandBareatright angles toeach other. This isexactly what happened inthe special wave wehave already considered. Wearenow ready tousethelastofMaxwell’s equations forfreespace [IV ofEq.(20.l2)]. Writing outthecomponents, wehave 2 BB, 6B 6E,C(vXB)$=C23;—C2-5;1l'=—éT’ ¢2(v><B),,=c21%-C25;?‘= (20.17) c2(VXB),= czgégrf-4- c2§£3==?(%- Ofthesixderivatives ofthecomponents ofB,onlytheterm 6B,/6x isnotequal tozero. Sothethree equations giveussimply _2§&_ %. Cax_at (20.18) Theresult ofallourwork isthatonly onecomponent each oftheelectric and magnetic ﬁelds isnotzero, andthat these components must satisfy Eqs. (20.16) and(20.18). Thetwoequations canbecombined intooneifwedifferentiate the ﬁrstwith respect toxandthesecond with respect tot;theleft-hand sides ofthe 20-5 I f Ci———>| I /+\_ ’_ O /If \ ’ \/ (lb),/ _V‘iC _/ tO _,,~ T': Fig. 20-4. The function f(x—ct) represents aconstant "shape" thattravels toward positive xwith thespeed c.twoequations willthen bethesame (except forthefactor 02). Soweﬁnd that Elysatisﬁes theequation 62E, 10215,,§ —C-2~55 -0. (20.19) Wehave seenthesame differential equation before, when westudied thepropaga- tionofsound. Itisthewave equation forone-dimensional waves. Youshould notethatintheprocess ofourderivation wehave found something more than iscontained inEq.(20ll). Maxwell’s equations have given usthe further information that electromagnetic waves have ﬁeld components only at right angles tothedirection ofthewave propagation. Let’s review what weknow about thesolutions oftheone-dimensional wave equation. Ifanyquantity 11/satisﬁes theone-dimensional wave equation azip 102¢~—--—-—= 20.206x2 c20t2 0’ ( ) then onepossible solution isafunction 1]/(x,t)oftheform ¢(x,t)=f(x—ct), (20.21) that is,some function ofthesingle variable (x—ct). Thefunction f(x——ct) represents a“rigid” pattern inxwhich travels toward positive xatthespeed c (seeFig.20-4). Forexample, ifthefunction fhasamaximum when itsargument iszero, then fort==0themaximum ofipwilloccur atx=0.Atsome later time, sayt=10,(/1willhave itsmaximum atx=10c. Astime goes on,themaximum moves toward positive xatthespeed c. Sometimes itismore convenient tosaythatasolution oftheone-dimensional wave equation isafunction of(t——x/c). However, thisissaying thesame thing, because anyfunction of(t—x/c) isalsoafunction of(x—ct): F(t—x/c) =F[—- =f(x—ct). Let’s show thatf(x—ct)isindeed asolution ofthewave equation. Since itisafunction ofonly onevariable—the variable (x—ct)—we willletf’represent thederivative offwithrespect toitsvariable andf”represent thesecond derivative off.Differentiating Eq.(20.21) with respect tox,wehave §=/'0:-co. since thederivative of(x—ct)with respect toxis1.Thesecond derivative of tpwith respect toxisclearly 2 jg;=f”(x-Cl). (20.22) Taking derivatives ofipwith respect tot,weﬁnd 61,0__5;-/'<xaxC). 82¢ 2/1513=+cf(x—cl) (20.23) Weseethatlldoes indeed satisfy theone-dimensional wave equation. You may bewondering: “IfIhave thewave equation, how doIknow that Ishould takef(x—ct)asasolution? Idon’t likethisbackward method. Isn’t there some forward way toﬁndthesolution?” Well, onegood forward way is toknow thesolution. Itispossible to“cook up”anapparently forward mathe- matical argument, expecially because weknow what thesolution issupposed to be,butwith anequation assimple asthiswedon’t have toplay games. Soon youwillgetsothat when you seeEq.(20.20), you nearly simultaneously see 20-6 up=f(x—ct)asasolution. (Just asnowwhen youseetheintegral ofx2dx,you know right away thattheanswer isx3/3.) Actually youshould alsoseealittle more. Notonlyisanyfunction of(x—ct) asolution, butanyfunction of(x+ct)isalsoasolution. Since thewave equation contains only c2,changing thesign ofcmakes nodifference. Infact, themost general solution oftheone-dimensional wave equation isthesumoftwoarbitrary functions, oneof(x—ct)andtheother of(x+ct): 1,0=f(x—-ct)+g(x+ct). (20.24) Theﬁrstterm represents awave travelling toward positive x,andthesecond term anarbitrary wave travelling toward negative x.Thegeneral solution isthesuper- position oftwosuch waves both existing atthesame time. Wewillleave thefollowing amusing question foryoutothink about. Take afunction (0ofthefollowing form: up=coskxcoskct. Thisequation isn’tintheform ofafunction of(x—ct)orof(x+ct).Yetyoucan easily show thatthisfunction isasolution ofthewave equation bydirect substitution into Eq.(20.20). How canwethen saythatthegeneral solution isoftheform ofEq.(20.24)? Applying ourconclusions about thesolution ofthewave equation tothe y-component oftheelectric ﬁeld, Ey,weconclude thatEucanvary with xinany arbitrary fashion. However, theﬁelds which doexist canalways beconsidered as thesumoftwopatterns. Onewave issailing through space inonedirection with speed c,with anassociated magnetic ﬁeld perpendicular totheelectric ﬁeld; another wave istravelling intheopposite direction withthesame speed. Such waves correspond totheelectromagnetic waves thatweknow about—light, radio- waves, infrared radiation, ultraviolet radiation, x-rays, andsoon.Wehave already discussed theradiation oflight ingreat detail inVol.I.Since everything welearned there applies toanyelectromagnetic wave, wedon’t need toconsider ingreat detail herethebehavior ofthese waves. Weshould perhaps make afewfurther remarks onthequestion ofthepolariza- tionoftheelectromagnetic waves. Inoursolution wechose toconsider thespecial caseinwhich theelectric ﬁeld hasonly ay-component. There isclearly another solution forwaves travelling intheplus orminus x-direction, with anelectric ﬁeld which hasonly az-component. Since Maxwell’s equations arelinear, the general solution forone-dimensional waves propagating inthex-direction isthe sumofwaves ofE,,andwaves ofE,.This general solution issummarized inthe following equations: E=(0,Eu,E,) Eu=f(x—CI)+g(x+CI) E,=F(x—ct)+G(x+ct) B=(0,Bu,B5) cB,=f(x—ct) —g(x +ct) (IB,, =—F(x —ct)+G(x +ct).(20.25) Such electromagnetic waves have anE-vector whose direction isnotconstant but which gyrates around insome arbitrary way intheyz-plane. Atevery point themagnetic ﬁeld isalways perpendicular totheelectric ﬁeld andtothedirection ofpropagation. 20-7 Ifthere areonly waves travelling inonedirection, saythepositive x-direction, there isasimple rulewhich tellstherelative orientation oftheelectric andmag- netic ﬁelds. Therule isthat thecross product EXB-—which is,ofcourse, a vector atright angles toboth EandB—-points inthedirection inwhich thewave is travelling. IfEisrotated intoBbyaright-hand screw, thescrew points inthe direction ofthewave velocity. (We shall seelater thatthevector EXBhasa special physical signiﬁcance: itisavector which describes theﬂow ofenergy inan electromagnetic ﬁeld.) 20-2 Three-dimensional waves Wewant now toturn tothesubject ofthree-dimensional waves. Wehave already seen thatthevector Esatisﬁes thewave equation. Itisalsoeasy toarrive atthesame conclusion byarguing directly from Maxwell’s equations. Suppose we start with theequation 6BVXE-—57 andtake thecurlofboth sides: VX(VXE)=“%(V XB). (20.26) You willremember thatthecurlofthecurlofanyvector canbewritten asthesum oftwoterms, oneinvolving thedivergence andtheother theLaplacian, v><(VXE)= V(V-E)—V2E. Infreespace, however, thedivergence ofEiszero, soonly theLaplacian term remains. Also, from thefourth ofMaxwell’s equations infreespace [Eq. (20.l2)] thetime derivative of02VXBisthesecond derivative ofEwith respect tor: ¢-2£(v X3)=6! dig Equation (20.26) then becomes 2_l<’2_EvE_C2.at2. which isthethree-dimensional wave equation. Written outinallitsglory, this equation is,ofcourse, 62E 62E 62E 162EF _ ____ = 76x2+6y2+622 c28t? 0' (202 ) How shall weﬁndthegeneral wave solution” Theanswer isthatallthesolu- tions ofthethree-dimensional wave equation canberepresented asasuperposition oftheone-dimensional solutions wehave already found. Weobtained theequation forwaves which move inthex-direction bysupposing thattheﬁelddidnotdepend onyandz.Obviously, there areother solutions inwhich theﬁelds donotdepend onxandz,representing waves going inthey-direction. Then there aresolutions which donotdepend onxandy,representing waves travelling inthez-direction. Oringeneral, since wehave written ourequations invector form, thethree- dimensional wave equation canhave solutions which areplane waves moving in anydirection atall.Again, since theequations arelinear, wemayhave simultane- ously asmany plane waves aswewish, travelling inasmany different directions. Thus themost general solution ofthethree-dimensional wave equation isa superposition ofallsorts ofplane waves moving inallsorts ofdirections. Trytoimagine what theelectric andmagnetic ﬁelds look likeatpresent in thespace inthislecture room. First ofall,there isasteady magnetic ﬁeld; itcomes from thecurrents intheinterior oftheearth—that is,theearth’s steady magnetic ﬁeld. Then there aresome irregular, nearly static electric ﬁelds produced perhaps byelectric charges generated byfriction asvarious people move about intheir Z)-8 chairs andrubtheir coat sleeves against thechair arms. Then there areother magnetic ﬁelds produced byoscillating currents intheelectrical wiring—-ﬁelds which vary atafrequency of60cycles persecond, insynchronism with thegenera- toratBoulder Dam. Butmore interesting aretheelectric andmagnetic ﬁelds vary- ingatmuch higher frequencies. Forinstance, aslight travels from window to ﬁoor andwall towall, there arelittle wiggles oftheelectric andmagnetic ﬁelds moving along at186,000 miles persecond. Then there arealsoinfrared waves travelling from thewarm foreheads tothecoldblackboard. And wehave forgotten theultraviolet light, thex-rays, andtheradiowaves travelling through theroom. Flying across theroom areelectromagnetic waves which carry music ofajazz band. There arewaves modulated byaseries ofimpulses representing pictures of events going oninother parts oftheworld, orofimaginary aspirins dissolving in imaginary stomachs. Todemonstrate thereality ofthese waves itisonly necessary toturnonelectronic equipment thatconverts these waves intopictures andsounds. Ifwegointo further detail toanalyze even thesmallest wiggles, there are tinyelectromagnetic waves thathave come intotheroom from enormous distances. There arenow tinyoscillations oftheelectric ﬁeld, whose crests areseparated by adistance ofonefoot, thathave come from millions ofmiles away, transmitted totheearth from theMariner IIspace craft which hasjustpassed Venus. Its signals carry summaries ofinformation ithaspicked upabout theplanets (infor- mation obtained from electromagnetic waves that travelled from theplanet to thespace craft). There arevery tinywiggles oftheelectric andmagnetic ﬁelds thatarewaves which originated billions oflight years away—from galaxies intheremotest corners oftheuniverse. That thisistruehasbeen found by“ﬁlling theroom with wires”— bybuilding antennas aslarge asthisroom. Such radiowaves have been detected from places inspace beyond therange ofthegreatest optical telescopes. Even they, theoptical telescopes, aresimply gatherers ofelectromagnetic waves. What we callthestars areonly inferences, inferences drawn from theonly physical reality wehave yetgotten from them—from acareful study oftheunendingly complex undulations oftheelectric andmagnetic ﬁelds reaching usonearth. There is,ofcourse, more: theﬁelds produced bylightning miles away, the ﬁelds ofthecharged cosmic rayparticles asthey zipthrough theroom, andmore, andmore. What acomplicated thing istheelectric ﬁeld inthespace around you! Yetitalways satisﬁes thethree-dimensional wave equation. 20-3 Scientiﬁc imagination Ihave asked youtoimagine these electric andmagnetic ﬁelds. What doyou do? Doyouknow how? How doIimagine theelectric andmagnetic ﬁeld? What doIactually see? What arethedemands ofscientiﬁc imagination? Isitany different from trying toimagine thattheroom isfullofinvisible angels? No,itis notlikeimagining invisible angels. Itrequires amuch higher degree ofimagination tounderstand theelectromagnetic ﬁeld than tounderstand invisible angels. Why? Because tomake invisible angels understandable, allIhave todoistoalter their properties alittle bit——I make them slightly visible, andthen Icanseetheshapes oftheir wings, andbodies, andhalos. Once Isucceed inimagining avisible angel, theabstraction required——which istotake almost invisible angels andimagine them completely invisible—is relatively easy. Soyousay,“Professor, please give meanapproximate description oftheelectromagnetic waves, even though itmay beslightly inaccurate, sothatItoocanseethem aswellasIcanseealmost invisible angels. Then IWlllmodify thepicture tothenecessary abstraction.” I‘msorry Ican’t dothatforyou. Idon’t know how. Ihave nopicture ofthis electromagnetic ﬁeld thatisinanysense accurate. Ihave known about theelectro- magnetic ﬁeld along time—I wasinthesame position 25years agothatyouare now, andIhave had25years more ofexperience thinking about these wiggling waves. When Istart describing themagnetic ﬁeld moving through space, Ispeak oftheE-andBﬁelds andwave myarms andyoumayimagine thatIcanseethem. 20-9 I’lltellyouwhat Isee. Iseesome kind ofvague shadowy, wiggling lines—here andthere isanEandBwritten onthem somehow, andperhaps some ofthelines have arrows onthem—an arrow here orthere which disappears when Ilook too closely atit.When Italkabout theﬁelds swishing through space, Ihave aterrible confusion between thesymbols Iusetodescribe theobjects andtheobjects them- selves. Icannot really make apicture thatiseven nearly likethetruewaves. So ifyouhave some difficulty inmaking such apicture, youshould notbeworried thatyour dilﬁculty isunusual. Our science makes terriﬁc demands ontheimagination. The degree of imagination thatisrequired ismuch more extreme than thatrequired forsome of theancient ideas. Themodern ideas aremuch harder toimagine. Weusealot oftools, though. Weusemathematical equations andrules, andmake alotof pictures. What Irealize now isthatwhen Italkabout theelectromagnetic ﬁeldin space, Iseesome kind ofasuperposition ofallofthediagrams which I’veever seen drawn about them. Idon’t seelittle bundles ofﬁeld lines running about be- cause itworries methatifIranatadifferent speed thebundles would disappear. Idon’t even always seetheelectric andmagnetic ﬁelds because sometimes Ithink Ishould have made apicture with thevector potential andthescalar potential, forthose were perhaps themore physically signiﬁcant things thatwere wiggling. Perhaps theonly hope, yousay,istotake amathematical view. Now what is amathematical view? From amathematical view, there isanelectric ﬁeld vector andamagnetic ﬁeld vector atevery point inspace; thatis,there aresixnumbers associated with every point. Can youimagine sixnumbers associated with each point inspace? That’s toohard. Canyouimagine even onenumber associated with every point? Icannot! Icanimagine such athing asthetemperature atevery point inspace. That seems tobeunderstandable. There isahotness andcoldness thatvaries from place toplace. ButIhonestly donotunderstand theidea ofa number atevery point. Soperhaps weshould putthequestion: Canwerepresent theelectric ﬁeldby something more likeatemperature, saylikethedisplacement ofapiece ofjello? Suppose thatwewere tobegin byimagining thattheworld wasﬁlled with thin jello andthattheﬁelds represented some distortion—say astretching ortwisting- ofthejello. Then wecould visualize theﬁeld. After we“see” what itislikewe could abstract thejello away. Formany years that’s what people tried todo. Maxwell, Ampere, Faraday, and others tried tounderstand electromagnetism thisway. (Sometimes theycalled theabstract jello “ether.”) Butitturned outthat theattempt toimagine theelectromagnetic ﬁeld inthatwaywasreally standing in theway ofprogress. Weareunfortunately limited toabstractions, tousing in- struments todetect theﬁeld, tousing mathematical symbols todescribe theﬁeld, etc. Butnevertheless, insome sense theﬁelds arereal, because after weareall ﬁnished ﬁddling around with mathematical equations—with orwithout making pictures anddrawings ortrying tovisualize thething—we canstillmake theinstru- ments detect thesignals from Mariner IIandﬁndoutabout galaxies abillion miles away, andsoon. Thewhole question ofimagination inscience isoften misunderstood bypeople inother disciplines. They trytotestourimagination inthefollowing way. They say,“Here isapicture ofsome people inasituation. What doyouimagine will happen next?” When wesay,“Ican’t imagine,” they may think wehave aweak imagination. They overlook thefactthatwhatever weareallowed toimagine in science must beconsistent witheverything elseweknow: thattheelectric ﬁelds and thewaves wetalkabout arenotjustsome happy thoughts which wearefreeto make aswewish, butideas which must beconsistent with allthelaws ofphysics weknow. Wecan’t allow ourselves toseriously imagine things which areobviously incontradiction totheknown laws ofnature. And soourkind ofimagination is quite adifficult game. Onehastohave theimagination tothink ofsomething that hasnever been seen before, never been heard ofbefore. Atthesame time the thoughts arerestricted inastrait jacket, sotospeak, limited bytheconditions that come from ourknowledge ofthewaynature really is.Theproblem ofcreating 20-10 something which isnew, butwhich isconsistent with everything which hasbeen seenbefore, isoneofextreme diﬁiculty. While l‘monthissubject Iwant totalkabout whether itwilleverbepossible toimagine beauty thatWecan't see Itisaninteresting question. When welook atarainbow, itlooks beautiful tous.Everybody says, “Ooh, arainbow.” (You seehowscientiﬁc Iam. Iamafraid tosaysomething isbeautiful unless Ihave an experimental wayofdeﬁning it.)Buthowwould wedescribe arainbow ifwewere blind? Weareblind when wemeasure theinfrared reﬂection coelﬁcient ofsodium chloride, orwhen wetalkabout thefrequency ofthewaves thatarecoming from some galaxy thatwecanit see-—-we make adiagramfwe make aplot. Forinstance, fortherainbow. such aplot would betheintensity ofradiation vs.wavelength measured with aspectrophotometer foreach direction inthesky. Generally. such measurements would giveacurve thatwasrather ﬂat. Then some day, someone would discover thatforcertain conditions oftheweather, andatcertain angles in thesky, thespectrum ofintensity asafunction ofwavelength would behave strangely; itwould have abump. Astheangle oftheinstrument wasvaried onlya little bit,themaximum ofthebump would move from onewavelength toanother. Then onedaythephysical review oftheblind menmight publish atechnical article with thetitle“The Intensity ofRadiation asaFunction ofAngle under Certain Conditions oftheWeather.” Inthisarticle there might appear agraph such as theoneinFig.20~5 Theauthor would perhaps remark thatatthelarger angles there wasmore radiation atlong wavelengths, whereas forthesmaller angles the maximum intheradiation came atshorter wavelengths. (From ourpoint ofview, wewould saythatthelight at40°ispredominantly green andthelight at42°is predominantly red.) ~ csf ta) - Fig. 20-5 The intensity ofelectro ntenstyas"oi. Wavelength dilions. Now doweﬁndthegraph ofFig.20—5 beautiful? Itcontains much more de- tailthan weapprehend when welook atarainbow, because oureyes cannot see theexact details intheshape ofaspectrum. Theeye,however, ﬁnds therainbow beautiful. Dowehave enough imagination toseeinthespectral curves thesame beauty weseewhen welook directly attherainbow? Idon’t know. Butsuppose Ihave agraph ofthereﬂection coefﬁcient ofasodium chloride crystal asafunction ofwavelength intheinfrared, andalsoasafunction ofangle. Iwould have arepresentation ofhow itwould look tomyeyes ifthey could see intheinfrared——perhaps some glowing, shiny “green,” mixed with reﬂections from thesurface ina“metallic red.” That would beabeautiful thing, butIdon’t know whether Icanever look atagraph ofthereﬂection CO€lI'lCl€Ill ofNaCl measured withsome instrument andsaythatithasthesame beauty. Ontheother hand. even ifwecannot seebeauty inparticular measured results, Wecanalready claim toseeacertain beauty intheequations which describe general physical laws. Forexample, inthewave equation (20.9), there’s something nice about theregularity oftheappearance ofthex,they,thez,andthet.And this nicesymmetry inappearance ofthex.y,z,andIsuggests tothemind stillagreater beauty which hastodowith thefour dimensions, thepossibility thatspace has four-dimensional symmetry, thepossibility ofanalyzing thatandthedevelopments ofthespecial theory ofrelativity. Sothere isplenty ofintellectual beauty asso- ciated with theequations. 20-llmagnetic waves asafunction ofwave length forthree angles (measured from thedirection opposite thesun), observed > only with certain meteorological con 20-4 Spherical waves Wehave seen that there aresolutions ofthewave equation which corre- spond toplane waves, andthatanyelectromagnetic wave canbedescribed asa superposition ofmany plane waves. Incertain special cases, however, itismore convenient todescribe thewave ﬁeld inadifferent mathematical form. Wewould liketodiscuss now thetheory ofspherical waves——waves which correspond to spherical surfaces that arespreading outfrom some center. When youdrop a stone intoalake, theripples spread outincircular waves onthesurface they are two-dimensional waves. Aspherical wave isasimilar thing except thatitspreads outinthree dimensions. Before westart describing spherical waves, weneed alittle mathematics. Suppose wehave afunction that depends only ontheradial distance rfrom a certain origin—in other words, afunction that isspherically symmetric. Let’s callthefunction ¢(r), where byrwemean r=\/x2+y2+Z2. theradial distance from theorigin. Inorder toﬁndoutwhat functions i//(r)satisfy thewave equation, wewillneed anexpression fortheLaplacian ofi//.Sowewant toﬁndthesumofthesecond derivatives of11/with respect tox,y,andz.Wewill usethenotation thati//(r) represents thederivative ofll!with respect torandi//’(r) represents thesecond derivative ofipwith respect tor. First, weﬁndthederivatives with respect tox.Theﬁrstderivative is 6¢(r) _,6r 6x_lb(r)07¢- Thesecond derivative ofifwith respect toxis 02¢ 6r2 62r Tie' +*”'z-i;§' Wecanevaluate thepartial derivatives ofrwith respect toxfrom 9:_5.92_11_L26x_r 6x2 Tr r2 Sothesecond derivative ofitwith respect toxis 62¢ 2 1 6x2 Likewise, QdyzI éiw/1+ 2 Z-242$!/+r 1I‘g)ii’. (20.28) 2 g).i/, (20.29) 02¢ Z2,,1 Z2525- ﬁll! +;(l —;_5 (I/’. (20.30) The Laplacian isthesum ofthese three derivatives. Remembering that x2+yz+22=r2,weget W0)=¢”(r)+§i'<ri- (20.31) Itisoften more convenient towrite thisequation inthefollowing form: W=15wi <2032) rdr2 ' ' Ifyoucarry outthedifferentiation indicated inEq.(20.32), youwillseethatthe right-hand side1Sthesame asinEq.(20.31). Ifwewish toconsider spherically symmetric ﬁelds which canpropagate as spherical waves, ourﬁeld quantity must beafunction ofboth randt.Suppose 20-I2 weask, then, what functions ip(r,t)aresolutions ofthethree-dimensional wave equation vaint)-l-‘iiiint)=0 (2033)’ c2at? ’ ' ' Since i//(r,i)depends onlyonthespatial coordinates through r,wecanusetheequa- tionfortheLaplacian wefound above, Eq.(20.32). Tobeprecise, however, since ilisalsoafunction ofz,weshould write thederivatives with respect toraspartial derivatives. Then thewave equation becomes 1a2 1at“*T(H//) *f ll’=0rdrl cl6t2 ' Wemust nowsolve thisequation, which appears tobemuch more complicated than theplane wave case. Butnotice thatifwemultiply thisequation byr,weget a2 1a2 Thisequation tellsusthatthefunction ripsatisﬁes theone-dimensional wave equa- tioninthevariable r.Using thegeneral principle which wehave emphasized so often, thatthesame equations always have thesame solutions, weknow thatif r¢isafunction only of(r—ct)then itwillbeasolution ofEq.(20.34). Sowe know thatspherical waves must have theform r¢(r,1)=ftr—cr)_ Or,aswehave seen before, wecanequally well saythatripcanhave theform rill=f(t—r/c). Dividing byr,weﬁndthattheﬁeldquantity ip(Whatever itmaybe)hasthefollow- ingform: ,):f(' (20.35) Such afunction represents ageneral spherical wave travelling outward from the origin atthespeed c.Ifweforget about therinthedenominator foramoment, theamplitude ofthewave asafunction ofthedistance from theorigin atagiven time hasacertain shape thattravels outward atthespeed c.Thefactor rinthe denominator, however, saysthattheamplitude ofthewave decreases inproportion tol/rasthewave propagates. Inother words, unlike aplane wave inwhich the amplitude remains constant asthewave runs along, inaspherical wave theampli- tudesteadily decreases, asshown inFig.20—6. This effect iseasy tounderstand from asimple physical argument. \ \ r/C)Z / / '~ ft-r f(t-r/ct. \\\\\\ l/I’ “"\ \ lL§§L. If“~—_____ ’=>/\ ‘(\9 r r r O T| 3 ll 72 l l<————c(t2—t|) —A (a) (bl Fig. 20—6. Aspherical wave ilr=flt—r/cl/r. la)(Ixasofunction ofrfort=l1and the some wave forthelater time fg.[blil/asafunction oftforr=r1andthesome wave seen atr; 20-13 Weknow thattheenergy density inawave depends onthesquare ofthewave amplitude. Asthewave spreads, itsenergy isspread over larger andlarger areas proportional totheradial distance squared. Ifthetotalenergy isconserved, the energy density must fallasI/r2, andtheamplitude ofthewave must decrease as l/r. SoEq.(20.35) isthe“reasonable” form foraspherical wave. Wehave disregarded thesecond possible solution totheone-dimensional wave equation: or=s'(I+r/C),Of _80+r/0)i--—»,——- This alsorepresents aspherical wave, butonewhich travels inward from large r toward theorigin. Wearenowgoing tomake aspecial assumption. Wesay,without anydemon- stration whatever, thatthewaves generated byasource areonly thewaves which gooutward. Since weknow thatwaves arecaused bythemotion ofcharges, we Want tothink that thewaves proceed outward from thecharges. Itwould be rather strange toimagine thatbefore charges were setinmotion, aspherical wave started outfrom inﬁnity andarrived atthecharges justatthetime they began to move. That isapossible solution, butexperience shows thatwhen charges are accelerated thewaves travel outward from thecharges. Although Maxwell’s equations would allow either possibility, wewillputinanadditional fact——based onexperience—that only theoutgoing wave solution makes “physical sense.” Weshould remark, however, thatthere isaninteresting consequence tothis additional assumption: weareremoving thesymmetry with respect totime that exists inMaxwell’s equations. Theoriginal equations forEandB,aridalsothe wave equations wederived from them, have theproperty thatifwechange thesign oft,theequation isunchanged. These equations saythat forevery solution corresponding toawave going inonedirection there isanequally valid solution forawave travelling intheopposite direction. Ourstatement thatwewillconsider only theoutgoing spherical waves isanimportant additional assumption. (A formulation ofelectrodynamics inwhich thisadditional assumption isavoided has been carefully studied. Surprisingly, inmany circumstances itdoes notlead to physically absurd conclusions, butitwould take ustoofarastray todiscuss these ideas justnow. Wewilltalkabout them alittle more inChapter 28.) Wemust mention another important point. Inoursolution foranoutgoing wave, Eq.(20.35), thefunction ipisinﬁnite attheorigin. That issomewhat peculiar. Wewould liketohave awave solution which issmooth everywhere. Oursolution must represent physically asituation inwhich there issome source attheorigin. Inother words, wehave inadvertently made amistake. Wehave notsolved the freewave equation (20.33) everywhere; wehave solved Eq.(20.33) with zero on theright everywhere, except attheorigin. Ourmistake crept inbecause some of thesteps inourderivation arenot“legal” when r=0. Let’s show thatitiseasy tomake thesame kind ofmistake inanelectrostatic problem. Suppose wewant asolution oftheequation foranelectrostatic potential infreespace, V24; =0.TheLaplacian isequal tozero, because weareassuming that there arenocharges anywhere. Butwhat about aspherically symmetric solution tothisequation—that is,some function ¢thatdepends only onr.Using theformula ofEq.(20.32) fortheLaplacian, wehave 2 1i0¢i= 0. rdr2 Multiplying thisequation byr,wehave anequation which isreadily integrated: d2 F("¢) =0- Ifweintegrate once with respect tor,weﬁndthattheﬁrstderivative ofr¢isa 20-14 constant, which wemay calla: %(r¢>)=a. Integrating again, weﬁndthatr¢isoftheform r¢=ar+b, where bisanother constant ofintegration. Sowehave found thatthefollowing ¢ isasolution fortheelectrostatic potential infreespace: b ¢:a'l";' Something isevidently wrong. Intheregion where there arenoelectric charges, weknow thesolution fortheelectrostatic potential: thepotential is everywhere aconstant. That corresponds totheﬁrstterm inoursolution. Butwe alsohave thesecond term, which saysthatthere isacontribution tothepotential thatvaries asoneover thedistance from theorigin. Weknow, however, thatsuch apotential corresponds toapoint charge attheorigin. So,although wethought wewere solving forthepotential infreespace, oursolution also gives theﬁeld forapoint source attheorigin. Doyouseethesimilarity between what happened nowandwhat happened when wesolved foraspherically symmetric solution to thewave equation? Ifthere were really nocharges orcurrents attheorigin, there would notbespherical outgoing waves. Thespherical waves must, ofcourse, be produced bysources attheorigin. Inthenextchapter wewillinvestigate thecon- nection between theoutgoing electromagnetic waves andthecurrents andvoltages which produce them. 20-15 21 Solutions ofMaxwell’s Equations with Currents and Charges 21-1 Light andelectromagnetic waves Wesawinthelastchapter thatamong their solutions, Maxwell’s equations have waves ofelectricity andmagnetism. These waves correspond tothephe- nomena ofradio, light, x-rays, andsoon,depending onthewavelength. Wehave already studied light ingreat detail inVol.I.Inthischapter wewant totietogether thetwosubJects—we want toshow thatMaxwell’s equations canindeed form the baseforourearlier treatment ofthephenomena oflight. When westudied light, webegan bywriting down anequation fortheelectric ﬁeldproduced byacharge which moves inanyarbitrary way. That equation was __q_?;' f_'d"r' lag]E_41reQ[r’2+cE<r’2 +c2Jt5er ’ (211) cB= e,/XE. [SeeEq.(28.3), Vol. I.] Ifacharge moves inanarbitrary way, theelectric ﬁeld wewould ﬁndnowat some point depends only ontheposition andmotion ofthecharge notnow, but atanearlier time——at aninstant which isearlier bythetime itwould take light, going atthespeed c,totravel thedistance r’from thecharge totheﬁeld point. Inother words, ifwewant theelectric ﬁeld atpoint (l)atthetime i,wemust cal- culate thelocation (2')ofthecharge anditsmotion atthetime (t-—r’/c), where r’isthedistance tothepoint (l)from theposition ofthecharge (2')atthetime (t—r’/c). Theprime istoremind youthatr’istheso-called “retarded distance" from thepoint (2')tothepoint (l),andnottheactual distance between point (2),the position ofthecharge atthetime i,andtheﬁeld point (l)(seeFig.21—l). Note thatweareusing adifferent convention now forthedirection oftheunitvector e,.InChapters 28and36ofVol. Iitwasconvenient totake r(and hence e,) pointing toward thesource. Now wearefollowing thedeﬁnition wetook forCou- lomb’s law,inwhich risdirectedfrom thecharge, at(2),toward theﬁeldpoint at(l). Theonly difference, ofcourse, 1Sthatournewr(and e,)arethenegatives ofthe oldones. Wehave alsoseen thatifthevelocity I)ofacharge isalways much lessthan c,andifweconsider only points atlarge distances from thecharge, sothatonlythe lastterm ofEq.(21.1) isimportant, theﬁelds canalsobewritten as EZ_ q [acceleration ofthecharge at(t—r’/c) , (211,) 41re0c2r’ projected atright angles tor’ ' and cB=e,’XE. Let's look atwhat thecomplete equation, Eq.(21.1), says inalittle more detail. The vector e,’istheunit vector topoint (I)from theretarded position (2'). Theﬁrstterm, then, iswhat wewould expect fortheCoulomb ﬁeld ofthecharge atitsretarded position—we may callthis“the retarded Coulomb ﬁeld.” The electric ﬁeld depends inversely onthesquare ofthedistance andisdirected away from theretarded position ofthecharge (that is,inthedirection ofe,-). Butthatisonlytheﬁrstterm. Theother terms tellusthatthelaws ofelectricity donotsaythatalltheﬁelds arethesame asthestatic ones, butjustretarded (which iswhat people sometimes liketosay). Tothe“retarded Coulomb ﬁeld” wemust 21-121-1 Light andelectromagnetic waves 21-2 Spherical waves from apoint source 21-3 Thegeneral solution of Maxwell’s equations 21-4 Theﬁelds ofanoscillating dipole 21-5 Thepotentials ofamoving charge; thegeneral solution ofLiénard andWiechert 21-6 Thepotentials foracharge moving with constant velocity; theLorentz formula Review: Chapter 28,Vol. I,Electro- magnetic Radiation Chapter 31,Vol l,T/ie Origin oftheRefractive Index Chapter 36,Vol l,Relativistic Eﬂects inRadiation (I) r'/ er’ Z7 (2')qVf rPOSITION Oi l-r’/c<1(2)/ Positional? Fig. 2l—l. The ﬁelds of(l) citthe time idepend ontheposition (2')occupied bythechcirge qcitthetime (t—r’/c). addtheother twoterms. Thesecond term saysthatthere isa“correction” tothe retarded Coulomb ﬁeld which istherateofchange oftheretarded Coulomb ﬁeld multiplied byr’/c, theretardation delay. Inawayofspeaking, thisterm tends to compensate fortheretardation intheﬁrstterm. Theﬁrsttwoterms correspond to computing the“retarded Coulomb ﬁeld” andthen extrapolating ittoward the future bytheamount r’/c, thatis,right uptothetimellTheextrapolation islinear, asifwewere toassume thatthe“retarded Coulomb ﬁeld” would continue tochange attheratecomputed forthecharge atthepoint (2'). Iftheﬁeldischanging slowly, theeffect oftheretardation isalmost completely removed bythecorrection term, andthetwoterms together giveusanelectric ﬁeld thatisthe“instantaneous Cou- lomb ﬁeld”——that is,theCoulomb ﬁeld ofthecharge atthepoint (2)—to avery good approximation. Finally, there isathird term inEq.(2l.l) which isthesecond derivative ofthe unitvector e,/.Forourstudy ofthephenomena oflight, wemade useofthefact thatfaraway from thecharge theﬁrsttwoterms went inversely asthesquare of thedistance and, forlarge distances, became very weak incomparison tothelast term, which decreases asl/r.Soweconcentrated entirely onthelastterm, andwe showed thatitis(again, forlarge distances) proportional tothecomponent ofthe acceleration ofthecharge atright angles tothelineofsight. (Also, formost ofour work inVol. l,wetook thecaseinwhich thecharges were moving nonrelativistic- ally. Weconsidered therelativistic effects inonly onechapter, Chapter 36.) Now weshould trytoconnect thetwothings together. Wehave theMaxwell equations, andwehave Eq.(21.1) fortheﬁeld ofapoint charge. Weshould cer- tainly askwhether theyareequivalent. Ifwecandeduce Eq.(21.l)from Maxwell’s equations, wewillreally understand theconnection between light andelectro- magnetism. Tomake thisconnection isthemain purpose ofthischapter. ltturns outthatwewon’t quite make it—that themathematical details get toocomplicated forustocarry through inalltheir gory details. Butwewillcome close enough sothatyoushould easily seehow theconnection could bemade. Themissing pieces willonly beinthemathematical details Some ofyoumay ﬁndthemathematics inthischapter rather complicated, andyoumay notwish to follow theargument very closely. Wethink itisimportant, however, tomake the connection between what youhave learned earlier andwhat youarelearning now, oratleast toindicate how such aconnection canbemade. You willnotice, if youlook over theearlier chapters, thatwhenever wehave taken astatement asa starting point foradiscussion, wehave carefully explained whether itisanew “assumption" thatisa“basic law,” orwhether itcanultimately bededuced from some other laws. Weoweittoyouinthespirit ofthese lectures tomake thecon- nection between light andMaxwell’s equations. Ifitgetsditlicult inplaces, well, that’s life—there isnoother way. 2l—2 Spherical waves from apoint source InChapter 18wefound thatMaxwell’s equations could besolved byletting E2-v¢~ (21.2) and B;VXA, (21.3) where ¢andAmust then besolutions oftheequations 2__l__‘f?__B_ 214V¢ C’) (912 G0 ( I) and 2 l62A j andmust alsosatisfy thecondition that '___Wl6¢>VA- C55, (21.6) 21-2 Now wewillﬁndthesolution ofEqs. (21.4) and(21.5). Todothatwehave toﬁndthesolution tpoftheeqtiation 2 152¢__V11/ C2M2- s, (21.7) where s‘,which wecallthesource, isknown. Ofcourse, Ycorresponds top/e(, and iiito¢forEq(21.4), orsIS],/€0t‘2 ifttis/1,,etc,butwewant tosolve Eq(217) asamathematical problem nomatter what itandsarephysically. Inplaces where pandjarezero—-in what wehave called “free” space——the potentials <1;andA,andtheﬁelds EandB,allsatisfy thethree-dimensional wave equation without sources, whose mathematical form is 21a%_ Vit C2517—0. (21.8) InChapter 20wesawthatsolutions ofthisequation canrepresent waves ofvarious kinds: plane waves inthex-direction, i//=f(t—x/c); plane waves inthey-or z-direction, orinanyother direction; orspherical waves oftheform a»»ao=ﬂl}i9- (mm (The solutions canbewritten instillother ways, forexample cylindrical waves thatspread outfrom anaxis.) Wealso remarked that, physically, Eq.(21.9) does notrepresent awave in freespace—that there must becharges attheorigin togettheoutgoing wave started. Inother words, Eq.(21.9) isasolution ofEq.(21.8) everywhere except right near r=O,where itmust beasolution ofthecomplete equation (21.7), including some sources. Let’s seehow thatworks. What kind ofasource sinEq.(21.7) would giverisetoawave likeEq.(21.9)? Suppose wehave thespherical wave ofEq.(21.9) andlook atwhat ishappen- ingforvery small r.Then theretardation —r/c inf(t —r/c)canbeneglected- provided fisasmooth function——and 5!,becomes ii=1;) (r_>0). (21.10) Sot//is_]LlS[likeaCoulomb ﬁeld foracharge attheorigin thatvaries with time. That is,ifwehadalittle lump ofcharge, limited toavery small region near the origin, with adensity p,weknow that Q/4rre0<t>=mjf i where Q=fpdV.Now weknow thatsuch a¢satisﬁes theequation 2=_B. V¢ 60 Following thesame mathematics, wewould saythat theitofEq.(21.10) satisﬁes Vztb =—-s (r-~>O), (21.11) where sisrelated tofby Sf=17;’ with s=fSdV. Theonlydiﬁerence isthatinthegeneral case, s,andtherefore S,canbeafunction oftime. Now theimportant thing isthatifti»satisﬁes Eq.(21.11) forsmall r,italso satisﬁes Eq.(21.7). Aswegovery close totheorigin, thel/rdependence oftp 21-3 causes thespace derivatives tobecome very large. Butthetime derivatives keep their same values [They arejust thetime derivatives ofj(r).] Soasrgoes tozero, theterm 62¢/612 inEq.(21.7) canbeneglected incomparison with V21//, andEq. (21.7) becomes equivalent toEq.(21.11). Tosummarize, then, ifthesource function s(t)ofEq.(217)islocalized at theorigin andhasthetotal strength 5(1)=/8(1)dV, (21.12) thesolution ofEq.(21.7) is if/(x,y,z, 1)=1%§-(lif/if (21.13) Theonly effect oftheterm ()"’¢/(J12 inEq.(21.7) istointroduce theretardation (t—r/c)intheCoulomb-like potential. 21—3 Thegeneral solution ofMaxwell’s equations Wehave found thesolution ofEq.(21.7) fora“point” source. Thenext question is:What isthesolution foraspread-out source‘? That’s easy; wecan think ofanysource s(x,y,z,t)asmade upofthesum ofmany “point" sources, oneforeach volume element dV,andeach with thesource sticngth s(x.y,z,t)dV. Since Eq(21.7) islinear, theresultant ﬁeld isthesuperposition oftheﬁelds from allofsuch source elements. Using theresults ofthepreceding section [Eq. (21.l3)] weknow that the ﬁeld dipatthepoint (x1,y1, z1)Aor (1)forshort—at thetime t,from asource elements dVatthepoint (x2,yg,22) or(2)forshort——is given by _ \(2,l -'F121/C)O1V2 (f\h(l, I)— 4T”_12 > where r12isthedistance from (2)to(1). Adding thecontributions from allthe pieces ofthesource means. ofcourse, doing anintegral over allregions where s¢O;sowehave ¢(1,i) =I5Q’l4»;r1:12/iavg. (21.14) That is,theﬁeld at(1)atthetime tisthesum ofallthespherical waves which leave thesource elements at(2)atthetimes (t—rm/c). This isthesolution of ourwave equation foranysetofsources. Weseenow how toobtain ageneral solution forMaxwell‘s equations. If for11/wemean thescalar potential 4>,thesource function sbecomes p/en. Orwe canlet1/1represent anyoneofthethree components ofthevector potential A, replacing sbythecorresponding component ofj/soc”. Thus, ifweknow the charge density p(x,y,z,t)andthecurrent density j(x,y,z,t)everywhere, wecan immediately write down thesolutions ofEqs (21.4) and(21.5). They are -ate»- d ¢(1,t)e 4mm dV2 (2115) an A(1,1)= dV2. (21.16) Theﬁelds EandBcanthen befound bydifferentiating thepotentials, using Eqs. (21.2) and(21.3). [Incidentally, itispossible toverify thatthe¢andAobtained from Eqs. (21.15) and(21.16) dosatisfy theequality (21.6) ] Wehave solved Maxwell’s equations. Given thecurrents andcharges inany circumstance, wecanﬁnd thepotentials directly from these integrals andthen differentiate andgettheﬁelds. Sowehave finished with theMaxwell theory Also thispermits ustoclose theringback tootirtheory oflight, because toconnect with ourearlier work onlight, weneed only calctilate theelectric ﬁeld from a 2l—4 moving charge. Allthat remains istotake amoving charge, calculate thepo- tentials from these integrals, andthen differentiate toﬁndEfrom —V¢ —6A/8!. Weshould getEq.(21.1). Itturns outtobelotsofwork, butthat’s theprinciple. Sohereisthecenter oftheuniverse ofelectromagnetism—the complete theory ofelectricity andmagnetism, andoflight; acomplete description oftheﬁelds produced byanymoving charges; andmore. Itisallhere. Here isthestructure built byMaxwell, complete inallitspower andbeauty. Itisprobably oneofthe greatest accomplishments ofphysics. Toremind youofitsimportance, wewill putitalltogether inaniceframe. Maxwell’s equations: v.E=GB v-B=0 _ 6B . __] gv><E--3 cZv><B-E0+at Their solutions: 6AE : '—V¢! '— :97 B=v><A ¢(1,;) =_-[P di/247l'€0 7'12 A(1,t)—[L-—-———(2” TF”/C)av_ 4ire0C2i‘12 2 21-4 Theﬁelds ofanoscillating dipole Wehave stillnotlived uptoourpromise toderive Eq.(21.1) fortheelectric ﬁeld ofapoint charge inmotion. Even with theresults wealready have, it1Sa relatively complicated thing toderive. Wehave notfound Eq.(21.1) anywhere in thepublished literature except inVol. 1ofthese lectures.* Soyoucanseethatitis noteasytoderive. (The ﬁelds ofamoving charge have been written inmany othei forms thatareequivalent, ofcourse.) Wewillhave tolimit ourselves hereJUSIto showing that, inafewexamples, Eqs. (21.15) and(21.16) givethesame results as Eq.(21.1). First, wewillshow thatEq(21.1) gives thecorrect ﬁelds with only the restriction that themotion ofthecharged particle isnonrelativistic. (Just this special casewilltakecare of90percent, ormore, ofwhat wesaidabout light.) Weconsider asituation inwhich wehave ablob ofcharge thatismoving about insome way, inasmall region, andwewillﬁndtheﬁelds faraway. Toputit another way, weareﬁnding theﬁeld atanydistance from apoint charge thatis shaking upanddown invery small motion. Since light isusually emitted from neutral Ol)jCCiS such asatoms. wewillconsider thatourwiggling charge qislocated nearanequal andopposite charge atrest. Iftheseparation between thecenters of thecharges isd,thecharges willhave adipole momentp =qd,which wetake to be.1function oftime. Now weshould expect thatifwelook attheﬁelds close to thechaiges, wewon’t have toworry about thedelay; theelectric ﬁeld willbe exactly thesame astheonewehave calculated earlier foranelectrostatic dipole *Theformula wasworked outbyR.P.Feynman, inabout 1950, andgiven insome lectures asagood wayofthinking about synchrotron radiation 21-5 Z 1l) AV2 |'l V1tX,y,l) y X Fig. 21-2. The potentials cit(1)are given by integrals over the Qhqrge density p.—using, ofcourse, theinstantaneous dipole moment p(t). Butifwegovery far out.weought toﬁndaterm intheﬁeld thatgoes asl/randdepends ontheac- celeration ofthecharge perpendicular tothelineofsight. Let’s seeifwegetsuch aresult. Webegin bycalculating thevector potential A,using Eq.(21.16). Suppose thatourmoving charge isinasmall blob whose charge density isgiven byp(x,y,2), andthewhole thing ismoving atanyinstant with thevelocity v.Then thecurrent density j(x,y, z)willbeequal tovp(x, y,z).Itwillbeconvenient totake our coordinate system sothatthez-axis isinthedirection ofv;then thegeometry of ourproblem isasshown inFig.21-2. Wewant theintegral iii/,. (21.17)12 Now ifthesizeofthecharge-blob isreally very small compared with r12.we cansetthermterm inthedenominator equal tor,thedistance tothecenter ofthe blob, andtake routside theintegral. Next, wearealsogoing tosetr12=rin thenumerator, although thatisnotreally quite right. Itisnotright because we should takej at,say,thetopoftheblob ataslightly different time than weused forj atthebottom oftheblob. When wesetrm=rinj(t—r12/c), weare taking thecurrent density forthewhole blob atthesame time (t—r/c). That is anapproximation thatwillbegood only ifthevelocity 7'ofthecharge ismuch lessthan c.Sowearemaking anonrelativistic calculation. Replacingj bypv, theintegral (21.17) becomes é[vp(Z,t -r/c)dV2. Since allthecharge hasthesame velocity, thisintegral isjustv/rtimes thetotal charge q.Butqvis_]LlS[Op/6t, therateofchange ofthedipole moment—which is, ofcourse, tobeevaluated attheretarded time (t—r/c). Wewillwrite itas p(t—r/c). Sowegetforthevector potential 1'- Ourresult saysthatthecurrent inavarying dipole produces avector potential intheform ofspherical waves whose source strength isp/41re0c2. Wecannowgetthemagnetic ﬁeldfrom B=VXA.Sincep istotally inthe z-direction, Ahasonly az-component; there areonly twononzero derivatives in thecurl SoB,=6A2/8y andBy:—6A,,/6x. Let’s ﬁrstlook atB,: _6A,_ 1 6p(t—-r/c)_~ _ ~ - .9 BI 6}’ 4711062 6)’ " (21 1) Tocarry outthedifferentiation, wemust remember thatr:Vx2+y2-1-z2,so 1 ,_a1 11aB, — —I’L) + -I: I‘/C). Remembering thatOr/Oy =y/r,theﬁrstterm gives _1_yak(/C) R062 ;_,__. (21.21) which drops offasl/r2liketheﬁelds ofastatic dipole (because y/risconstant for agiven direction). Thesecond term inEq.(21.20) gives usthenew effects. Carrying outthe differentiation, weget 1y _C? _. _/ _ 471106“, Cr,p(t r,c), (2122) where pmeans, ofcourse, thesecond derivative ofpwith respect toi.This term, 2l—6 which comes from differentiating thenumerator, isresponsible forradiation. First, itdescribes aﬁeld which decreases with distance only asl/r. Second, it depends ontheacceleration ofthecharge. Youcanbegin toseehowwearegoing togetaresult likeEq.(211’),which describes theradiation oflight. Let’s examine inalittle more detail how thisradiation term comes about-it issuch aninteresting andimportant result. Westart with theexpression (21.18), which hasal/rdependence andistherefore likeaCoulomb potential, except for thedelay term inthenumerator. Why isitthen thatwhen wedifferentiate with respect tospace coordinates togettheﬁelds, wedon’t _]LlS'[getal/r3 ﬁeld-—with, ofcourse, thecorresponding time delays? Wecanseewhyinthefollowing way: Suppose thatweletourdipole oscillate upanddown inasinusoidal motion. Then wewould have P=in=iiosinwland 1wpocos(c(t—r/c)A,=-_, -ii -4rre(,c~ r Ifweplotagraph ofA,asafunction ofratagiven instant, wegetthecurve shown inFig21-3. Thepeak amplitude decreases asl/r,butthere is,inaddition. an oscillation inspace, bounded bythel/renvelope. When wetake thespatial de- rivatives, they willbeproportional totheslope ofthecurve. From theﬁgure we seethatthere areslopes much steeper than theslope ofthel/rcurve itself. ltis. infact, evident thatforagiven frequency thepeak slopes areproportional tothe amplitude ofthewave, which varies as1/r. Sothatexplains thedrop-off rateof theradiation term. Itallcomes about because thevariations withtimeatthesource aretranslated intovariations inspace asthewaves arepropagated outward, andthelT1ﬂgl"lC11L ﬁelds depend onthespatial derivatives ofthepotential. Let's goback andﬁnish ourcalculation ofthemagnetic ﬁeld. Wehave for B,thetwoterms (21.21) and(21.22), so B:1[_vi>(r—r/c)_yp'(r—r/c))” 41reOc~' r~* er? With thesame kind ofmathematics, weget _ 1 xp(t -r/c) xp(t —r/c) ‘ B”—4776062 l ri‘ + cr2 Orwecanptititalltogether inanicevector formula: _ 1 +(7/C)I.7llt—r/0 X" B_41re(,c2 rd I (2123) Now let’slook atthisformula. First ofall,ifwegoveryfaroutinr,only the p‘term counts. Thedirection ofBisgiven byp Xr,which isatright angles tothe radius randalsoatright angles totheacceleration, asinFig.21-4. Everything is coming outright; thatisalsotheresult wegetfrom Eq.(21.1'). Now let’slook atwhat wearenotused to—at what happens closer in.ln Section 14-9 weworked outthelawofBiotandSavart forthemagnetic ﬁeld ofan element ofcurrent. Wefound thatacurrent elementj dVcontributes tothemag- netic ﬁeld theamount _ 1 XrdB-2‘-_,-rTC-2- -I? dV. (21.24) Youseethatthisformula looks very much liketheﬁrstterm ofEq(21.23), ifwe remember thatpisthecurrent. Butthere isonedifference. InEq.(21.23), the current istobeevaluated atthetime(t—r/c), which doesn’t appear inEq.(21.24). Actually, however, Eq.(21.24) isstillvery good forsmall r,because thesecond 21-7Azll \ l/r\ \ i'7\T7”\""/\‘-/L/_,_\/_j;\J / / / / / / / Fig. 21-3. Themcignitdue ofAasci function ofrcitthe instant tfor the sphericcil wove from cmoscillating dipole. B(I) E 1' ii (21 Fig. 21-4. Theradiation fields Bond EOfonoscillciting dipole. term ofEq.(2123)tends tocancel outtheeffect oftheretardation intheﬁrstterm. Thetwotogether givearesult very near toEq.(2124)when rissmall. Wecanseethatthisway. When rissmall, (1-r/c)isnotverydifferent from t,sowecanexpand thebracket inEq.(21.23) inaTaylor series. Fortheﬁrstterm, pa-r/c)=pm—211(1)+ac. andtothesame order inr/c, PU-r/c)=ﬁt!)- When wetake thesum, thetwoterms inpcancel, andweareleftwith theun- retarded currentp: thatis,p(t)—plus terms oforder (r/c)2 orhigher [e.g., ._1§(r/c)2p‘] which willbevery small forrsmall enough thatpdoes notalter markedly inthe time r/c. SoEq.(2123)gives ﬁelds very much liketheinstantaneous theory—much closer than theinstantaneous theory with adelay; theﬁrst-order effects ofthedelay aretaken outbythesecond term. Thestatic formulas arevery accurate, much more accurate than youmight think Ofcourse, thecompensation only works for points close in.Forpoints faroutthecorrection becomes very bad, because the time delays produce avery large effect, andwegettheimportant l/rterm ofthe radiation. Westillhave theproblem ofcomputing theelectric ﬁeld anddemonstrating thatitisthesame asEq.(21.1’). Forlarge distances wecanseethattheanswer isgoing tocome outallright. Weknow thatfarfrom thesources, where wehave apropagating wave. Eisperpendicular toB(and alsotor),asinFig.21-4, and thatcB=ESoEisproportional totheacceleration p‘,asexpected from Eq. (21.1’). Togettheelectric ﬁeld completely foralldistances, weneed tosolve forthe electrostatic potential. When wecomputed thecurrent integral forAtoget Eq.(21.18), wemade anapproximation bydisregarding theslight variation ofr inthedelay terms. This willnotwork fortheelectrostatic potential, because we would then getl/rtimes theintegral ofthecharge density, which isaconstant. This approximation 1Stoorough. Weneed togotoonehigher order. Instead of getting involved inthathigher-order computation directly, wecandosomething else—we candetermine thescalar potential from Eq.(21.6), using thevector po- tential wehave already found. Thedivergence ofA,inourcase, isjust6A,/62 —since A,andAyareidentically zero. Differentiating inthesame waythatwe didabove toﬁndB, I 6 1 16 V‘/1 ZEEO-CT_>[P(1 —"/@)5E<;) -lr75%! —"/Cl] __1___[_W-1/£2_ﬂ;;/C2].—41re0c2 rt cr2 Or,invector notation, V_A 2 ___1 ‘ (7/£)2lt—r/0'7 41re()c1 r~‘ Using Eq.(21.6), wehave anequation for¢: 92:L117 +_@/i)I"]a-_r/ii .61 47l'€() 1f-5 Integrating with respect totjust removes onedotfrom each ofthep’s,so 1 ’/ l—r c'I¢,(,~,I)IZ77?“ [ L__' (Zj25) (The constant ofintegration would correspond tosome superposed static ﬁeld which could, ofcourse, exist. Fortheoscillating dipole wehave taken, there 1S nostatic ﬁeld ) 21-8 Wearenow abletoﬁndtheelectric ﬁeld Efrom 6AE--v¢ -Tr- Since thesteps aretedious butstraightforward [providing youremember that p(t-r/c)anditstime derivatives depend onx,y,andzthrough theretardation r/c], wewilljustgivetheresult: __1 *.Ea,Z)=H607, [-p*-3g +E-1,{1'1'(t-r/c)><i}><F](21.26) with i»*=pt:-r/c)+§i>(i-r/c). (21.21) Although itlooks rather complicated, theresult iseasily interpreted. The vector p*isthedipole moment retarded andthen “corrected” fortheretardation, sothetwoterms with p*givejustthestatic dipole ﬁeld when rissmall. [See Chapter 6,Eq.(6.l4).] When rislarge, theterm inpdominates, andtheelectric ﬁeldisproportional totheacceleration ofthecharges, atright angles tor,and, in fact,directed along theprojection ofp‘inaplane perpendicular tor. This result agrees with what wewould have gotten using Eq.(21.1). Of course, Eq.(21.1) ismore general; itworks with anymotion, while Eq.(21.26) 1S valid only forsmall motions forwhich wecantake theretardation r/casconstant over thesource. Atanyrate, wehave now provided theunderpinnings forour entire previous discussion oflight (excepting some matters discussed inChapter 36ofVol. I),foritallhinged onthelastterm ofEq.(21.26). Wewilldiscuss next how theﬁelds canbeobtained formore rapidly moving charges (leading tothe relativistic effects ofChapter 36ofVol. I). 21-5 Thepotentials ofamoving charge; thegeneral solution ofLiénard and Wiechert Inthelastsection wemade asimpliﬁcation incalculating ourintegral forA byconsidering only lowvelocities. Butindoing sowemissed animportant point andalsoonewhere itiseasytogowrong. Wewilltherefore takeupnowacalcula- tionofthepotentials forapoint charge moving inanywaywhatever—even with arelativistic velocity. Once wehave thisresult, wewillhave thecomplete electro- magnetism ofelectric charges. Even Eq.(21.1) canthen bederived bytaking derivatives. Thestory willbecomplete. Sobear with us. Let’s trytocalculate thescalar potential ¢(1)atthepoint (x1,yl,Z1)produced byapoint charge, such asanelectron, moving inanymanner whatsoever. Bya “point” charge wemean avery small ballofcharge, shrunk down assmall asyou like, with acharge density p(x,y,z).Wecanﬁnd¢>from Eq.(21.15): ¢(1,Z)=24-7&6 dV2. (21.28) Theanswer would seem tobe—and almost everyone would, atﬁrst, think-that theintegral ofpover such a“point” charge isjustthetotal charge q,sothat 1¢(1,i)=mo (wrong)- Byr[2wemean theradius vector from thecharge atpoint (2)topoint (1)atthe retarded time (t-r12/c). Itiswrong. Thecorrect answer is _1q 1 where UT’isthecomponent ofthevelocity ofthecharge parallel tor§2—namely, toward point (1). Wewillnow show youwhy. Tomake theargument easier to 21-9 |4—G ——>1 -- - “POINT”CHARGE T/P/O \\ rm (ll \\AV; 7; ll) to/7>-Q >Q +ll<-w V (0) (bi Fig. 21-5. (a)A"point" charge—considered asasmall cubical distribution of charge—moving with thespeed vtoward point (ll lb)Thevolume element _\V, used forcalculating thepotentials. I ilillllllllllit ii—~>~ Wlllllllllllllllll‘I il lI mi. ii*1t~ F‘ (I) ib) ~U ll >~ti ‘ Ill i llW1! it (2 :8) - l ll TeE ‘\§“‘§,4_4_'I///['IIIIl___ <2Y::.l:L:ii'3>9) 4 e+0! (8)l O TA___ _b ____,l Fig. 2l~6. lntegrating p(t—r//c)dV foramoving charge.follow, wewillmake thecalculation ﬁrstfora“point” charge which isintheform ofalittle cube ofcharge moving toward thepoint (1)with thespeed 1),asshown inFig.2l—5(a). Letthelength ofasideofthecube bea,which wetake tobe much, much lessthan r12, thedistance from thecenter ofthecharge tothe point (1). Now toevaluate theintegral ofEq.(21.28), wewillreturn tobasic principles; wewillwrite itasthesum LAV7. Z'97-. (21.30) where r.isthedistance from point (1)totheithvolume element AV,andp,isthe charge density atAV,atthetime IL:t—r,/c. Since r,>>u,always, itwillbe convenient totake ourAVLintheform ofthin, rectangular slices perpendicular to r12,asshown inFig.2l—5(b). Suppose westart bytaking thevolume elements AV,with some thickness w much lessthan a.Theindividual elements willappear asshown inFig.2l—6(a), where wehave putinmore than enough tocover thecharge. Butwehave not shown thecharge, andforagood reason. Where should wedraw it"Foreach volume element AV,, wearetotake patthetime r,=(t—rl/c), butsince the charge ismoving, itisinadzﬂercnr placefor each volume clement AV,! Let’s saythatwebegin with thevolume element labeled “l”inFig.2l—6(a), chosen sothatatthetime 11=(I—r1/c) the“back” edge ofthecharge occupies AV1, asshown inFig.2l—6(b). Then when weevalute p2AV2, wemust usethe position ofthecharge attheslightly later time r2=(r—r2/c), when thecharge willbeintheposition shown inFig.21—6(c). And soon,forAV3, AV4, etc.Now wecanevaluate thesum. Since thethickness ofeach AV‘isw,itsvolume iswag. Then each volume clement that overlaps thecharge distribution contains theamount ofcharge walp, where pisthedensity ofcharge within thecube-—which wetake tobe uniform. When thedistance from thecharge topoint (1)islarge, wewillmake a negligible error bysetting alltherjsinthedenominators equal tosome average value, saytheretarded position r’ofthe center ofthecharge. Then thesum(21.30) is N 2pwu 2Trl ’(=1 where AVN isthelastAV,thatoverlaps thecharge distributions, asshown inFig. 2l—6(e). Thesumis,clearly, NZmy(ta).r r a Now pa"isJustthetotal charge qandNWisthelength bshown inpart(e)ofthe ﬁgure. Sowehave __;1_- Q.¢—41i'ei,r’ (L1) (2131) 2l—l0 What isb?Itisthelength ofthecube ofcharge increased bythedistance moved bythecharge between I1=(t—r1/c) andrt=(t—ry/c)—which is thedistance thecharge moves inthetime Al=iv—11=(V1—W)/<7 I17/6- Since thespeed ofthecharge 1Sti,thedistance moved isiiAt=vb/c. Butthe length bisthisdistance added toa: b=a+%b. Solving forb,weget 1,: . 1—(v/c) Ofcourse byllwemean thevelocity attheretarded time t’=(I——r’/c), which wecanindicate bywriting [1—1»/c],,.,, andEq.(21.31) forthepotential becomes q 1 ‘*0’’)4mr/ [1-(U/C)]rQt Thisresult agrees with ourassertion, Eq.(21.29). There isacorrection term which comes about because thecharge ismoving asourintegral “sweeps overthecharge.” When thecharge ismoving toward thepoint (1),itscontribution totheintegral is increased bytheratio b/a. Therefore thecorrect integral isq/r’multiplied by b/a, which is1/[1 —it/c]r,.,. Ifthevelocity ofthecharge isnotdirected toward theobservation point (1), youcanseethat what matters isthecomponent ofitsvelocity toward point (1). Calling thisvelocity component 1i,,thecorrection factor 1S1/[1 -1',/c]M. Also, theanalysis wehave made goes exactly thesame wayforacharge distribution of anyshape—it doesn’t have tobeacube. Finally, since the“size” ofthecharge u doesn’t enter into theﬁnal result, thesame result holds when weletthecharge shrink toanysize-—even toapoint. Thegeneral result isthatthescalar potential forapoint charge moving with anyvelocity is _ q , "“’)"41re0r’[1 —<1»./oi... (2132) Thisequation isoften written intheequivalent form where risthevector from thecharge tothepoint (1),where ¢isbeing evaluated, andallthequantities inthebracket aretohave their values attheretarded time t’=t—r’/c. Thesame thing happens when wecompute Aforapoint charge, from Eq. (21.16). Thecurrent density ispvandtheintegral over pisthesame aswefound for4;.Thevector potential is qvAll. 1)~@rgc2[7jTJ_—;/6)]; (21.34) Thepotentials forapoint charge were ﬁrstdeduced inthisform byLiénard andWiechert andarecalled theLzénard-Wiechert potentials. Toclose theringback toEq.(21.1) itisonly necessary tocompute EandB from these potentials (using B=VXAandE~—V<1> —6A/6t). Itisnow onlyarithmetic. Thearithmetic, however, isfairly involved, sowewillnotwrite outthedetails. Perhaps youwilltake ourword foritthatEq.(21.1) isequivalent totheLiénard-Wiechert potentials wehave derived.* *Ifyouhave alotofpaper andtime youcantrytowork itthrough yourself. We would, then, make two suggestions" First, don’t forget that thederivatives ofr’are complicated, since itisafunction of1’Second, don’t trytodeme (211),butcarry out allofthederivatives init,andthen compare what yougetwith theEobtained from the potentials (21.33) and(21.34). 21-11 Fig. 21-7. Finding the potential at (At1) Pofacharge moving with uniform -——————————— velocity along thex-axis.21-6 Thepotentials foracharge moving with constant velocity; theLorentz formula Wewant next tousetheLiénard-Wiechert potentials foraspecial case——to ﬁndtheﬁelds ofacharge moving with uniform velocity inastraight line. Wewill doitagain later, using theprinciple ofrelativity. Wealready know what thepo- tentials arewhen wearestanding intherestframe ofacharge. When thecharge ismoving, wecanﬁgure everything outbyarelativistic transformation from one system totheother. Butrelativity haditsorigin inthetheory ofelectricity and magnetism. The formulas oftheLorentz transformation (Chapter 15,Vol. 1) were discoveries made byLorentz when hewasstudying theequations ofelectricity andmagnetism. Sothat youcanappreciate where things have come from, we would liketoshow thattheMaxwell equations doleadtotheLorentz transforma- tion. Webegin bycalculating thepotentials ofacharge moving with uniform velocity, directly from theelectrodynamics ofMaxwell’s equations. Wehave shown thatMaxwell’s equations leadtothepotentials foramoving charge thatwe gotinthelastsection. Sowhen weusethese potentials, weareusing Maxwell’s theory. Y P lxpyrz) <'1 r—\Ni._i N Q-1 \\‘>1"RETARDElT' POSITION(Att'=t-r’/c) r, %\.__;._“_-i"PRESENT" POSITION Z Suppose wehave acharge moving along thex-axis with thespeed ii.Wewant thepotentials atthepoint P(x,y,z),asshown inFig.21-7. lfi:0isthemoment when thecharge 1Sattheorigin, atthetime tthecharge isatx~Ht,y=z=0 What weneed toknow, however, 1Sitsposition attheretarded time t’=1-T1, (2115) C .- I where risthedistance tothepoint Pfrom thecharge attheretarded time. Atthe earlier time t’,thecharge wasatx=vt’,so r’=\/(x—ut’)3 —l—y‘-T-725. (21.36) Toﬁndr’ort’wehave tocombine thisequation with Eq.(21.35). First, we eliminate r’bysolving Eq.(21.35) forr’andsubstituting inEq.(21.36). Then, squaring both sides, weget c2t1—o2=(x—W+ye+Z2. which isaquadratic equation int’.Expanding thesquared binomials andcollecting liketerms int’,weget (F2—c2)t’2 —2(xv —c2t)t’ —l—x2+y2+22~—(ct)2 =O. Solving fort’, 2 ' 1 l l2 0 T (1- 1'I1- —E(X-tr)’+1— (y“+Z2) (21.37) 21-12 Togetr’wehave tosubstitute thisexpression fort’into r’=c(t—t’). Now weareready toﬁnd¢from Eq.(21.33), which, since visconstant, becomes l ¢(X,Jr’,Z,1)=15:0 ' (21-38) Thecomponent ofvinthedirection ofr’isvX(x—vt)/r’, sov~r' isjust 11X(x~—vt’),andthewhole denominator is 2 ¢(¢-t’)—g(x— vt')=¢[¢- ’-g-(1-§,)1'l~ Substituting for(1—v2/c2)t’ from Eq.(21.37), wegetfor¢ _q I _¢(X,J',Z,l) _47T60 U2 (X—~02+(1—;)(y2+Z2) This equation ismore understandable ifwerewrite itas _<1 1 1.¢(-xa J’,Z:t)_4,n_6O \/j X__Ut 2 2 21/2 l -— —————— + —l— 02 \/1—212/02 y Z Thevector potential Aisthesame expression with anadditional factor ofv/c2" U InEq.(21.39) youcanclearly seethebeginning oftheLorentz transformation. Ifthecharge wereattheorigin initsownrestframe, itspotential would be _q 1 _¢(-X: yaZ) “T4,n_€0 [x2 +yg + Z211/2 Weareseeing itinamoving coordinate system, anditappears thatthecoordinates should betransformed by x—vtx—>————>, \/l—02/c2 J/_’J/, z——>z. That isjusttheLorentz transformation, andwhat wehave done isessentially the wayLorentz discovered it. Butwhat about thatextra factor 1/\/1 —v2/c2 thatappears atthefront of Eq.(21.39)? Also, how does thevector potential Aappear, when itiseverywhere zerointherestframe oftheparticle? Wewillsoon show thatAand¢together constitute afour-vector, likethemomentum pandthetotal energy Uofaparticle. Theextra 1/ inEq.(21.39) isthesame factor thatalways comes in when onetransforms thecomponents ofafour-vector—just asthecharge density p transforms top/\/l —212/c2. Infact, itisalmost apparent from Eqs. (214) and(21.5) thatAand¢>arecomponents ofafour-vector, because wehave already shown inChapter 13thatjandparethecomponents ofafour-vector. Later wewilltakeupinmore detail therelativity ofelectrodynamics; here we only wished toshow how naturally theMaxwell equations lead totheLorentz transformation. You willnot,then, besurprised toﬁndthatthelaws ofelectricity andmagnetism arealready correct forEinstein’s relativity. Wewillnothave to “ﬁxthem up.” aswehadtodoforNewton’s laws ofmechanics. 21-13 22 AC Circuits 22-1 Impedances Most ofourwork inthiscourse hasbeen aimed atreaching thecomplete equations ofMaxwell. Inthelasttwochapters Wehave been discussing thecon- sequences ofthese equations. Wehave found that theequations contain allthe static phenomena wehadworked outearlier, aswellasthephenomena ofelectro- magnetic waves andlight thatwehadgone over insome detail inVolume I.The Maxwell equations giveboth phenomena, depending upon whether onecomputes theﬁelds close tothecurrents andcharges, orvery farfrom them There isnot much interesting tosayabout theintermediate region; nospecial phenomena appear there. There stillremain, however, several subjects inelectromagnetism that we want totake up.Wewant todiscuss thequestion ofrelativity andtheMaxwell equations—what happens when onelooks attheMaxwell equations with respect tomoving coordinate systems. There isalsothequestion oftheconservation of energy inelectromagnetic systems. Then there isthebroad subject oftheelectro- magnetic properties ofmaterials; sofar,except forthestudy oftheproperties ofdielectrics, wehave considered onlytheelectromagnetic ﬁelds infreespace And although wecovered thesubject oflight insome detail inVolume I,there are stillafewthings wewould liketodoagain from thepoint ofview oftheﬁeld equations. Inparticular, wewant totake upagain thesubject oftheindex ofre- fraction, particularly fordense materials. Finally, there arethephenomena associated with waves conﬁned inalimited region ofspace. Wetouched onthis kindofproblem brieﬂy when wewere studying sound waves. Maxwell’s equations leadalsotosolutions which represent conﬁned waves oftheelectric andmagnetic ﬁelds. Wewilltake upthissubject, which hasimportant technical applications, insome ofthefollowing chapters. Inorder tolead uptothatsubject, wewill begin byconsidering theproperties ofelectrical circuits atlowfrequencies. We willthen beable tomake acomparison between those situations inwhich the almost static approximations ofMaxwell’s equations areapplicable and those situations inwhich high-frequency effects aredominant. Sowedescend from thegreat andesoteric heights ofthelastfewchapters andturn totherelatively low-level subject ofelectrical circuits. Wewillsee,how- ever, thateven such amundane subject, when looked atinsufficient detail, can contain great complications Wehave already discussed some oftheproperties ofelectrical circuits in Chapters 23and25ofVol. 1.Now Wewillcover some ofthesame material again, butingreater detail. Again wearegoing todealonly with linear systems andwith voltages andcurrents which allvary sinusoidally; wecanthenrepresent allvoltages andcurrents bycomplex numbers, using theexponential notation described in Chapter 22ofVol. I.Thus atime-varying voltage V(t)willbewritten V(t)=Ve"“", (22.1) where Vrepresents acomplex number thatisindependent of1.Itis,ofcourse, understood thattheactual time-varying voltage V(t)isgiven bytherealpart of thecomplex function ontheright-hand sideoftheequation. 22-122-1 Impedances 22-2 Generators 22-3 Networks ofideal elements; KirchhoiI’s rules 22-4 Equivalent circuits 22-5 Energy 22-6 Aladder network 22-7 Filters 22-8 Other circuit elements Review.‘ Chapter 22,Vol. I,Algebra Chapter 23,Vol. l,Resonance Chapter 25,Vol I,Linear Systems andReview I “-0 i_>b I Fig. 22-1. Aninducfcince.Similarly, allofourother time-varying quantities willbetaken tovary sinusoidally atthesame frequency w.Sowewrite I=Iem (current), s=ée“"'(emf), (22.2) E=Ee‘°" (electric ﬁeld), andsoon. Most ofthetime wewillwrite ourequations interms ofV,I,8,...(instead of interms ofI7,i,§;,...),remembering, though, that thetime variations areas given in(22.2). Inourearlier discussion ofcircuits weassumed thatsuch things asinductances, capacitances, andresistances were familiar toyou. Wewant nowtolook inalittle more detail atwhat ismeant bythese idealized circuit elements. Webegin with theinductance. Aninductance ismade bywinding many turns ofwire intheform ofacoil andbringing thetwoends outtoterminals atsome distance from thecoil,asshown inFig.22-1. Wewant toassume thatthemagnetic ﬁeld produced bycurrents in thecoildoes notspread outstrongly allover space andinteract with other parts of thecircuit. This isusually arranged bywinding thecoilinadoughnut-shaped form, orbyconﬁning themagnetic ﬁeldbywinding thecoilonasuitable ironcore, orbyplacing thecoilinsome suitable metal box, asindicated schematically in Fig.22—l. Inanycase, weassume thatthere isanegligible magnetic ﬁeld inthe external region near theterminals aandb.Wearealsogoing toassume thatwe canneglect anyelectrical resistance inthewire ofthecoil. Finally. wewillassume thatwecanneglect theamount ofelectrical charge thatappears onthesurface of awire inbuilding uptheelectric ﬁelds. With allthese approximations wehave what wecallan“ideal” inductance. (Wewillcome back later anddiscuss what happens inarealinductance.) Foran ideal inductance wesaythatthevoltage across theterminals isequal toL(d1/dt). Let’s seewhythatisso.When there isacurrent through theinductance, amagnetic ﬁeld proportional tothecurrent isbuilt upinside thecoil. Ifthecurrent changes with time, themagnetic ﬁeld alsochanges. Ingeneral, thecurlofEisequal to —dB/dt; or,putdifferently, thelineintegral ofEallthewayaround anyclosed path isequal tothenegative ofthe rateofchange oftheﬂuxofBthrough theloop Now suppose weconsider thefollowing path: Begin atterminal aandgoalong thecoil(staying always inside thewire) toterminal b;then return from terminal b toterminal athrough theairinthespace outside theinductance. Thelineintegral ofEaround thisclosed path canbewritten asthesumoftwoparts: /E~ds= /a"E-ds+ E'ds. (22.3) via outside coil Aswehave seen before, there canbenoelectric ﬁelds inside aperfect conductor. (The smallest ﬁelds would produce inﬁnite currents.) Therefore theintegral from atobviathecoilisZero. Thewhole contribution tothelineintegral ofEcomes from thepath outside theinductance from terminal btoterminal a.Since wehave assumed thatthere arenomagnetic ﬁelds inthespace outside ofthe“box,” this part oftheintegral isindependent ofthepath chosen andwecandeﬁne thepo- tentials ofthetwoterminals. Thedifference ofthese twopotentials iswhat we callthevoltage difference, orsimply thevoltage V,sowehave V=-/:5-as: -9§E-ds. Thecomplete lineintegral iswhat wehave before called theelectromotive force 8andis,ofcourse, equal totherateofchange ofthemagnetic ﬂuxinthe coil. Wehave seen earlier thatthisemfisequal tothenegative rateofchange of 22-2 thecurrent. sowehave dlV — *8 -— L2‘? 9 where Listheinductance ofthecoil. Since dl/dt =iwl,wehave V=iwL1. (22.4) Thewaywehave described theideal inductance illustrates thegeneral approach toother ideal circuit elements—usually called “lumped” elements. Theproperties oftheelement aredescribed completely interms ofcurrents andvoltages that appear attheterminals. Bymaking suitable approximations, itispossible to ignore thegreat complexities oftheﬁelds thatappear inside theobject. Aseparation ismade between what happens inside andwhat happens outside. Forallthecircuit elements wewillﬁndarelation liketheoneinEq.(22.4), in which thevoltage isproportional tothecurrent with aproportionality constant thatis,ingeneral, acomplex number. This complex coefficient ofproportionality iscalled theimpedance andisusually written asz(not tobeconfused with the z-coordinate). Itis,ingeneral, afunction ofthefrequency w.Soforanylumped element wewriteA V V-=‘T=, 22.5 IIZ () Foraninductance, wehave z(inductance) =2,,=iwL. (22.6) Now let’slook atacapacitor from thesame point ofview.* Acapacitor con- sistsofapairofconducting plates from which twowires arebrought outtosuitable terminals. Theplates may beofanyshape whatsoever, andareoften separated bysome dielectric material. Weillustrate such asituation schematically inFig. 22-2. Again wemake several simplifying assumptions. Weassume that the plates andthewires areperfect conductors. Wealsoassume thattheinsulation between theplates isperfect, sothat nocharges canflow across theinsulation from oneplate totheother. Next, weassume thatthetwoconductors areclose toeach other butfarfrom allothers, sothatallﬁeld lines which leave oneplate endupontheother. Then there arealways equal andopposite charges onthetwo plates andthecharges ontheplates aremuch larger than thecharges onthesur- faces ofthelead-in wires. Finally, weassume thatthere arenomagnetic ﬁelds close tothecapacitor. \Suppose now weconsider thelineintegral ofEaround aclosed loop which starts atterminal a,goes along inside thewire tothetopplate ofthecapacitor, jumps across thespace between theplates, passes from thelower plate toterminal bthrough thewire. andreturns toterminal ainthespace outside thecapacitor. Since there isnomagnetic ﬁeld, thelineintegral ofEaround thisclosed path is Zero. Theintegral canbebroken down intothree parts: 9§E-ds=/ E-ds+/ E-ds+ E-ds. along between outside wires plates(22.7) Theintegral along thewires iszero, because there arenoelectric ﬁelds inside per- fectconductors. Theintegral from btoaoutside thecapacitor isequal tothenega- tiveofthepotential difference between theterminals. Since weimagined thatthe twoplates areinsome wayisolated from therestoftheworld, thetotal charge on *There arepeople who sayweshould calltheobjects bythenames “inductor” and “capacitor” andcalltheir properties “inductance” and“capacitance” (byanalogy with “resistor” and“resistance”). Wewould rather usethewords youwillhear inthelabora- tory. Most people stillsay“inductance” forboth thephysical coilanditsinductance L. Theword “capacitor” seems tohave caught on although youwillstillhear “condenser” fairly often—and most people stillprefer thesound of“capacity“ to“capacitance.” ' 22-3_L ° /_>b I Fig.22—2. Acapacitor (or con denser). I ‘-0 V _>b I Fig. 22-3. Aresistor. <0) (b) to (d) tiitR_._L l2- I IUJL TUE R Fig. 22-4. The ideal lumped circuit elements (passive).thetwoplates must bezero; ifthere isacharge Qontheupper plate, there isan equal. opposite charge —Qonthelower plate. Wehave seen earlier thatiftwo conductors have equal andopposite charges, plus andminus Q,thepotential difference between theplates isequal toQ/C,where Ciscalled thecapacity ofthe twoconductors. From Eq.(22.7) thepotential difference between theterminals aandbisequal tothepotential difference between theplates. Wehave, therefore, that V: The electric current Ientering thecapacitor through terminal a(and leaving through terminal b)isequal todQ/dt, therateofchange oftheelectric charge on theplates. Writing dV/dt aszwV, wecanputthevoltage current relationship for acapacitor inthefollowing way: Z -gs OT 1V_23- (22.8) Theimpedance zofacapacitor, isthen z(capacitor) =zg= (22.9) Thethird element wewant toconsider isaresistor. However, since wehave notyetdiscussed theelectrical properties ofrealmaterials, wearenotyetready totalkabout what happens inside arealconductor. Wewilljusthave toaccept asfactthatelectric ﬁelds canexist inside realmaterials, thatthese electric ﬁelds giverisetoaﬂow ofelectric charge—that is,toacurrent—and thatthiscurrent isproportional totheintegral oftheelectric ﬁeld from oneendoftheconductor totheother. Wethen imagine anideal resistor constructed asinthediagram of Fig.22-3. Two wires which wetaketobeperfect conductors gofrom theterminals aandbtothetwoends ofabarofresistive material. Following ourusual lineof argument, thepotential difference between theterminals aandbisequal tothe lineintegral oftheexternal electric ﬁeld, which isalsoequal tothelineintegral of theelectric ﬁeld through thebarofresistive material. Itthen follows thatthecur- rentIthrough theresistor isproportional totheterminal voltage V: VI_F, where Riscalled theresistance. Wewillseelater thattherelation between the current andthevoltage forrealconducting materials isonly approximately linear. Wewillalsoseethatthisapproximate proportionality isexpected tobeindependent ofthefrequency ofvariation ofthecurrent andvoltage only ifthefrequency is nottoohigh. Foralternating currents then, thevoltage across aresistor isinphase with thecurrent, which means thattheimpedance isarealnumber. z(resistance) =21¢=R. (22.10) Ourresults forthethree lumped circuit elements——the inductor, thecapacitor, andtheresistor——are summarized inFig.22-4. lnthisfigure, aswell asinthe preceding ones, wehave indicated thevoltage byanarrow thatisdirected from one terminal toanother. Ifthevoltage is“positive”-—that is,iftheterminal aisata higher potential than theterminal b—~the arrow indicates thedirection ofapositive “voltage drop.” Although wearetalking about alternating currents, wecanofcourse include thespecial caseofcircuits with steady currents bytaking thelimit asthefrequency wgoes tozero. Forzero frequency-—that is,forDC——the impedance ofaninduc- tance goestozero; itbecomes ashort circuit. ForDC,theimpedance ofacondenser 22-4\ goestoinﬁnity; itbecomes anopen circuit. Since theimpedance ofaresistor is independent offrequency, it1Stheonly element leftwhen weanalyze acircuit forDC. Inthecircuit elements wehave described sofar,thecurrent andvoltage are proportional toeach other. Ifoneiszero, soalsoistheother. Weusually think in terms likethese: Anapplied voltage is“responsible” forthecurrent, oracurrent “gives riseto”avoltage across theterminals; soinasense theelements “respond” tothe“applied” external conditions. Forthisreason these elements arecalled passive elements. They canthus becontrasted with theactive elements, such as thegenerators wewillconsider inthenext section, which arethesources ofthe oscillating currents orvoltages inacircuit. 22-2 Generators Now wewant totalkabout anactive circuit element—one thatisasource of thecurrents andvoltages inacircuit—name1y, agenerator. Suppose thatwehave acoillikeaninductance except thatithasvery few turns, sothat wemay neglect themagnetic ﬁeld ofitsown current. This coil, however, sitsinachanging magnetic ﬁeldsuch asmight beproduced byarotating magnet, assketched inFig.22-5. (Wehave seen earlier thatsuch arotating mag- netic ﬁeldcanalsobeproduced byasuitable setofcoils with alternating currents.) Again wemust make several simplifying assumptions. Theassumptions weWlll make arealltheones thatwedescribed forthecaseoftheinductance. Inparticular, weassume thatthevarying magnetic ﬁeld isrestricted toadeﬁnite region inthe vicinity ofthecoilanddoes notappear outside thegenerator inthespace between theterminals. Following closely theanalysis wemade fortheinductance, weconsider the lineintegral ofEaround acomplete loop thatstarts atterminal a,goes through the coiltoterminal bandreturns toitsstarting point inthespace between thetwo terminals. Again weconclude thatthepotential difference between theterminals isequal tothetotal lineintegral ofEaround theloop: V=-955-ds. This lineintegral isequal totheemfinthecircuit, sothepotential difference V across theterminals ofthegenerator isalsoequal totherateofchange ofthemag- netic ﬂuxlinking thecoil: V=-a=%(ﬂux). (22.11) Foranideal generator weassume thatthemagnetic ﬂuxlinking thecoilisdeter- mined byexternal conditions—such astheangular velocity ofarotating magnetic ﬁeld—and isnotinﬂuenced inanyway bythecurrents through thegenerator. Thus agenerator~—at least theideal generator weareconsidering—is notan impedance. The potential difference across itsterminals isdetermined bythe arbitrarily assigned electromotive force 8(1). Such anideal generator isrepresented bythesymbol shown inFig.22-6. Thelittle arrow represents thedirection ofthe emfwhen itispositive. Apositive emfin thegenerator ofFig.22-6 willproduce avoltage V=8,with theterminal aatahigher potential than theterminal b. There isanother way tomake agenerator which isquite different onthe inside bywhich isindistinguishable from theonewehave justdescribed insofar aswhat happens beyond itsterminals. Suppose wehave acoilofwire which isrotated inaﬁxed magnetic ﬁeld, asindicated inFig. 22-7. Weshow abar magnet toindicate thepresence ofamagnetic ﬁeld; itcould, ofcourse, bereplaced byanyother source ofasteady magnetic ﬁeld, such asanadditional coilcarrying asteady current. Asshown intheﬁgure, connections from therotating coilare made totheoutside world bymeans ofsliding contacts or“slip rings.” Again, weareinterested inthepotential difference thatappears across thetwoterminals 22-5I Fig. 22-5. Agenerator consisting of aﬁxed coilandarotating magnetic ﬁeld. \../0 b Fig. 22—6. Symbol foranideal gen- erator.b SQFig 22-7 Agenerator consisting of b ciCOllrotating inciﬁxed magnetic ﬁeld.O AIIII' N Q V aandb,which isofcourse theintegral oftheelectric ﬁeld from terminal atoter- minal balong apath outside thegenerator. Now inthesystem ofFig.22-7 there arenochanging magnetic ﬁelds, sowe might atﬁrstwonder how anyvoltage could appear atthegenerator terminals lnfact, there arenoelectric ﬁelds anywhere inside thegenerator. Weare,asusual, assuming forourideal elements thatthewires inside aremade ofaperfectly con- ducting material, andaswehave saidmany times, theelectric ﬁeld inside aperfect conductor isequal tozero. Butthatisnottrue. Itisnottruewhen aconductor ismoving inamagnetic ﬁeld. Thetruestatement isthatthetotal force onany charge inside aperfect conductor must bezero. Otherwise there would bean inﬁnite flowofthefreecharges. Sowhat isalways trueisthatthesumoftheelectric ﬁeld Eandthecross product ofthevelocity oftheconductor andthemagnetic ﬁeld B—which isthetotal force onaunit charge—-must have thevalue zero inside theconductor: F=E+vXB=0(inaperfect conductor), (22.12) where vrepresents thevelocity oftheconductor. Ourearlier statement thatthere isnoelectric ﬁeld inside aperfect conductor isallright ifthevelocity vofthe conductor iszero; otherwise thecorrect statement isgiven byEq.(22.12). Returning toourgenerator ofFig.22-7, wenow seethatthelineintegral of theelectric ﬁeld Efrom terminal 0toterminal bthrough theconducting path of thegenerator must beequal tothelineintegral ofvXBonthesame path, fl’E-ds=_/b (v><B)-ds. (22.13) 1IISﬁ1(3 insfde conductor conductor Itisstilltrue, however, thatthelineintegral ofEaround acomplete loop, including thereturn from btoaoutside thegenerator, must bezero, because there areno changing magnetic ﬁelds. Sotheﬁrstintegral inEq.(22.13) isalsoequal toV, thevoltage between thetwoterminals. Itturns outthat theright-hand integral ofEq.(2213)isjusttherateofchange oftheﬂuxlinkage through thecoilandis therefore—by theﬂuxrule—-equal totheemfinthecoil. Sowehave again that thepotential difference across theterminals isequal totheelectromotive force in thecircuit, inagreement withEq.(22.11). Sowhether wehave agenerator inwhich amagnetic ﬁeld changes near aﬁxed coil, oroneinwhich acoilmoves inaﬁxed magnetic ﬁeld, theexternal properties ofthegenerators arethesame. There isa voltage difference Vacross theterminals, which isindependent ofthecurrent in thecircuit butdepends only onthearbitrarily assigned conditions inside the generator. Solong aswearetrying tounderstand theoperation ofgenerators from the point ofview ofMaxwell’s equations, wemight alsoaskabout theordinary chemi- calcell,likeaﬂashlight battery It1Salsoagenerator, i.e.,avoltage source, al- though itwillofcourse only appear inDCcircuits. Thesimplest kind ofcellto understand isshown inFig.22-8. Weimagine twometal plates immersed insome 22-6 chemical solution. Wesuppose thatthesolution contains positive andnegative ions. Wesuppose alsothatonekind ofion,saythenegative, ismuch heavier than theoneofopposite polarity, sothatitsmotion through thesolution bytheprocess ofdiffusion ismuch slower. Wesuppose next thatbysome means orother itis arranged thattheconcentration ofthesolution ismade tovary from onepartof theliquid totheother, sothatthenumber ofionsofboth polarities near, say,the lower plate ismuch larger than theconcentration ofions near theupper plate. Because oftheir rapid mobility thepositive ions willdrift more readily into the region oflower concentration, sothatthere willbeaslight excess ofpositive charge arriving attheupper plate. Theupper plate willbecome positively charged and thelower plate willhave anetnegative charge. Asmore andmore charges diffuse totheupper plate. thepotential ofthisplate willriseuntil theresulting electric ﬁeld between theplates produces forces onthe ionswhich justcompensate fortheir excess mobility, sothetwoplates ofthecell quickly reach apotential difference which ischaracteristic oftheinternal con- struction. Arguing justaswedidfortheideal capacitor, weseethatthepotential differ- ence between theterminals aandbisjustequal tothelineintegral oftheelectric ﬁeldbetween thetwoplates when there isnolonger anynetdiffusion oftheions. There is,ofcourse, anessential difference between acapacitor andsuch achemical cell. Ifweshort-circuit theterminals ofacondenser foramoment, thecapacitor isdischarged andthere isnolonger anypotential difference across theterminals. Inthecase ofthechemical cellacurrent canbedrawn from theterminals con- tinuously without anychange intheemf—until, ofcourse, thechemicals inside thecellhave been used up.Inarealcellitisfound thatthepotential difference across theterminals decreases asthecurrent drawn from thecellincreases. In keeping with theabstractions wehave been making, however, wemay imagine an ideal cellinwhich thevoltage across theterminals isindependent ofthecurrent. Arealcellcanthen belooked atasanideal cellinseries with aresistor. 22-3 Networks ofideal elements; Kirchhoff ’srules Aswehave seen inthelastsection, thedescription ofanideal circuit element interms ofwhat happens outside theelement isquite simple. Thecurrent and thevoltage arelinearly related. Butwhat isactually happening inside theelement isquite complicated, anditisquite difficult togiveaprecise description interms of Maxwell’s equations. Imagine trying togiveaprecise description oftheelectric andmagnetic ﬁelds oftheinside ofaradio which contains hundreds ofresistors, capacitors, andinductors. Itwould beanimpossible tasktoanalyze such athing byusing Maxwell’s equations. Butbymaking themany approximations wehave described inSection 22-2 and summarizing theessential features ofthereal circuit elements interms ofidealizations, itbecomes possible toanalyze anelec- trical circuit inarelatively straightforward way. Wewillnow show how that isdone. Suppose wehave acircuit consisting ofagenerator andseveral impedances connected together, asshown inFig.22-9. According toourapproximations there isnomagnetic ﬁeldintheregion outside theindividual circuit elements. Therefore thelineintegral ofEaround anycurve which does notpass through anyofthe elements iszero. Consider then thecurve I‘shown bythebroken linewhich goes allthewayaround thecircuit inFig.22-9. Thelineintegral ofEaround thiscurve ismade upofseveral pieces. Each piece isthelineintegral from oneterminal ofa circuit element totheother. This lineintegral wehave called thevoltage drop across thecircuit element. Thecomplete lineintegral isthenjustthesum ofthe voltage drops across alloftheelements inthecircuit: 9512-ds= EV... Since thelineintegral iszero, wehave thatthesum ofthepotential differences 22-7I—-> l; +t V +- --++—+—+—+* b1 Fig. 22-8. Achemical cell. °"?\// ?N<l\ 2| VI V3f / Z3 / \\ P/1 c X-G tn/\ 1\ {as UI<'1 /'5‘J\ II‘ //“"“‘\\ / \ =dFig. 22-9. The sum ofthevoltage drops around anyclosed path iszero. a b c d / lt, VQ \ lI4 8 f g h Fig. 22-10. Thesum ofthecurrents intoanynode iszero. -- = =YT @fI, Ial23 " ’Q Isl Z5 Z6+‘i-( I1 “" Fig. 22-l l.Analyzing acircuit with Kirchhoff's rules.around acomplete loop ofacircuit isequal tozero: ZV,,=0. (22.14)around any loop This result follows from oneofMaxwell’s equations—that inaregion where there arenomagnetic ﬁelds thelineintegral ofEaround anycomplete loop iszero. Suppose weconsider now acircuit likethatshown inFig.22-10. Thehori- zontal linejoining theterminals a,b,c,anddisintended toshow thatthese ter- minals areallconnected, orthatthey arejoined bywires ofnegligible resistance. Inanycase, thedrawing means thatterminals a,b,c,anda’areallatthesame potential and, similarly, thattheterminals e,f,g,andharealsoatonecommon potential. Then thevoltage drop Vacross each ofthefour elements isthesame. Now oneofouridealizations hasbeen thatnegligible electrical charges ac- cumulate ontheterminals oftheimpedances. Wenow assume further thatany electrical charges onthewires joining terminals canalsobeneglected. Then the conservation ofcharge requires thatanycharge which leaves onecircuit element immediately enters some other circuit element. Or,what isthesame thing, we require thatthealgebraic sumofthecurrents which enter anygiven junction must bezero. Byajunction, ofcourse, wemean anysetofterminals such asu,b,c, anddwhich areconnected. Such asetofconnected terminals isusually called a “node.” Theconservation ofcharge thenrequires thatforthecircuit ofFig.22-10, 1,~12_13-1.,=0. (22.15) The sum ofthecurrents entering thenode which consists ofthefour terminals e,f,g,andhmust alsobezero: -1,+1,+13+1.,=0. (22.16) Thisis,ofcourse, thesame asEq.(22.15). Thetwoequations arenotindependent. Thegeneral ruleisthatthesumofthecurrents intoanynodemustbezero. Z1,,=0. (22.17) i.‘?.l§Zis Ourearlier conclusion thatthesumofthevoltage drops around aclosed loop iszero must apply toanyloop inacomplicated circuit. Also, ourresult thatthe sumofthecurrents intoanode iszeromust betrueforanynode. These twoequa- tions areknown asKirchh0ﬁ"s rules. With these tworules itispossible tosolve for thecurrents andvoltages inanynetwork whatever. Suppose weconsider themore complicated circuit ofFig.22-11. How shall weﬁndthecurrents andvoltages inthiscircuit? Wecanﬁndthem inthefollowing straightforward way. Weconsider separately each ofthefour subsidiary closed loops which appear inthecircuit. (For instance, oneloop goes from terminal ato terminal btoterminal etoterminal dand back toterminal a.)Foreach oftheloops wewrite theequation fortheﬁrstofKirchhoff’s rules——that thesumofthevoltages around each loop isequal tozero. Wemust remember tocount thevoltage drop aspositive ifwearegoing inthedirection ofthecurrent andnegative ifweare going across anelement inthedirection opposite tothecurrent; andwemust remember thatthevoltage drop across agenerator isthenegative oftheemfin thatdirection. Thus ifweconsider thesmall loop thatstarts andends atterminal awehave theequation Z111 + Z313 + Z414 '— 81 :0. Applying thesame ruletotheremaining loops, wewould getthree more equations ofthesame kind. Next, wemust write thecurrent equation foreach ofthenodes inthecircuit. Forexample, summing thecurrents intothenode atterminal bgives theequation I1 _"[3 '—I2 =0. 22-8 Similarly, forthenode labeled ewewould have thecurrent equation I3 -1! "b I8 -I5 ==O. Forthecircuit shown there areﬁvesuch current equations. Itturns out,however, thatanyoneofthese equations canbederived from theother four; there are, therefore, only four independent current equations. Wethus have atotal ofeight independent, linear equations: thefour voltage equations andthefour current equations. With these eight equations wecansolve fortheeight unknown currents. Once thecurrents areknown thecircuit issolved. Thevoltage drop across any element isgiven bythecurrent through thatelement times itsimpedance (or,in thecaseofthevoltage sources, itisalready known). Wehave seen thatwhen wewrite thecurrent equations, wegetoneequation which isnotindependent oftheothers. Generally itisalsopossible towrite down toomany voltage equations. Forexample, inthecircuit ofFig.22-11, although wehave considered only thefour small loops, there arealarge number ofother loops forwhich wecould write thevoltage equation. There is,forexample, the loop along thepath abcfeda. There isanother loop which follows thepath abefehgda. You canseethatthere aremany loops. Inanalyzing complicated cir- cuitsitisveryeasytogettoomany equations. There arerules which tellushowto proceed sothat only theminimum number ofequations iswritten down, but usually with alittle thought itispossible toseehow togettheright number of equations inthesimplest form. Besides, writing anextra equation ortwodoesn’t doanyharm. They willnotlead toanywrong answers, only perhaps alittle unnecessary algebra. InChapter 25ofVol. Iweshowed thatifthetwoimpedances 21and22are inseries, they areequivalent toasingle impedance 2,given by 2,=21—l—22. (22.18) Wealsoshowed thatifthetwoimpedances areconnected inparallel, they are equivalent tothesingle impedance 2,,given by ZIZ2 1 Z"=<1/Z1)+<1/Z2)=Z.+Z2‘ (2219) Ifyoulook back youwillseethatinderiving these results wewere ineffect making useofKirchhoff ’srules. Itisoften possible toanalyze acomplicated circuit by repeated application oftheformulas forseries andparallel impedances. Forin- stance, thecircuit ofFig.22-12 canbeanalyzed thatway. First, theimpedances 24andz_-,canbereplaced bytheir parallel equivalent, andsoalsocan20and27. Then theimpedance 22canbecombined with theparallel equivalent of26and27 bytheseries rule. Proceeding inthisway, thewhole circuit canbereduced toa generator inseries with asingle impedance Z.Thecurrent through thegenerator isthen just8/Z. Then byworking backward onecansolve forthecurrents in each oftheimpedances. There are,however, quite simple circuits which cannot beanalyzed bythis method, asforexample thecircuit ofFig.22-13. Toanalyze thiscircuit wemust 0 b c z,lt, Z2112 Z3lI3=-(I,z, 2, 2, 2,, E Z7 Z8 Fig. 22-12. Acircuit which can be analyzed interms ofseries and parallel combinations. Fig. 22-13. Acircuit that cannot be analyzed interms ofseries and parallel d e f combinations. 22-9 O Z6 eQ04?b Fig. 22-14. Abridge circuit. .2.Obe Any (0) Circuit of Z's b LO (b)@ Zeff. b/‘<- Fig. 22-15. Any two-terminal net- work ofpassive elements isequivalent to aneffective impedance.write down thecurrent andvoltage equations from Kirchhoff ’srules. Let’s doit. There isjustonecurrent equation: I1+I2+I3=0, soweknow immediately that Ia=_(I1 -l“I2)- Wecansave ourselves some algebra ifweimmediately make useofthisresult in writing thevoltage equations. Forthiscircuit there aretwoindependent voltage equations; they are -81 —l—I222 —I121 =0 and 52*(I1-l"12)Za —I222 =0- There aretwoequations andtwounknown currents. Solving these equations for I1andI2,weget __Z232 -(Z2—l—Z3)g1 ’*2 <22-2°’and 1= 2221 2 Zi(Z2 ‘l'Z3)'1‘Z223 ( ) Thethird current isobtained from thesumofthese two. Another example ofacircuit thatcannot beanalyzed byusing therules for series andparallel impedance isshown inFig.22-14. Such acircuit iscalled a “bridge.” Itappears inmany instruments used formeasuring impedances. With such acircuit oneisusually interested inthequestion: How must thevarious impedances berelated ifthecurrent through theimpedance 23istobezero? We leave itforyoutoﬁndtheconditions forwhich thisisso. 22-4 Equivalent circuits Suppose weconnect agenerator 8toacircuit containing some complicated interconnection ofimpedances, asindicated schematically inFig.22-15(a). All oftheequations wegetfrom Kirchhoff ’srules arelinear, sowhen wesolve them forthecurrent Ithrough thegenerator, wewillgetthatIisproportional to8. Wecanwrite 8I= ~"—s Zeff where now 2,.“issome complex number, analgebraic function ofalltheelements inthecircuit. (Ifthecircuit contains nogenerators other than theoneshown, there isnoadditional term independent of8.)Butthisequation isjustwhat wewould write forthecircuit ofFig.22—15(b). Solong asweareinterested only inwhat happens totheleftofthetwoterminals aandb,thetwocircuits ofFig.22-15 are equivalent. Wecan, therefore, make thegeneral statement thatanytwo-terminal network ofpassive elements canbereplaced byasingle impedance 2,.“without changing thecurrents andvoltages intherestofthecircuit. This statement is,of course, justaremark about what comes outofKirchhoff ’srules—and ultimately from thelinearity ofMaxwell’s equations. Theidea canbegeneralized toacircuit thatcontains generators aswell as impedances. Suppose welook atsuch acircuit “from thepoint ofview” ofoneof theimpedances, which wewillcall2,,.asinFig.22-l6(a). Ifwewere tosolve the equation forthewhole circuit, wewould ﬁndthatthevoltage V,,between thetwo terminals aandbisalinear function ofI,which wecanwrite V,,=A—BI,,, (22.22) where AandBdepend onthegenerators andimpedances inthecircuit totheleft 22-10 oftheterminals. Forinstance, forthecircuit ofFig.22-13, weﬁnd V1=I121. Thiscanbewritten (byrearranging Eq.(22.20)] as V= -l .-s --is-1. 22.231 l(Z2 "l"Z3)82 1] Z2+Z31 ( ) Thecomplete solution isthen obtained bycombining thisequation with theone fortheimpedance 21,namely, V1=I121, orinthegeneral case, bycombining Eq.(22.22) with V,,=I,,z,,. lfnowweconsider that2,,isattached toasimple series circuit ofageneratoi andacurrent, asinFig.22-15(b), theequation corresponding toEq.(22.22) is Vn=Em—Inzeffs which isidentical toEq.(22.22) provided weset81.11=Aand2011=B.Soifwe areinterested only inwhat happens totheright oftheterminals aandb.thearbi- trary circuit ofFig.22-16 canalways bereplaced byanequivalent combination of agenerator inseries with animpedance. 22-5 Energy Wehave seen that tobuild upthecurrent Iinaninductance, theenergy U=%LI2 must beprovided bytheexternal circuit. When thecurrent fallsback tozero, thisenergy isdelivered back totheexternal circuit. There isnoenergy-loss mechanism inanideal inductance. When there isanalternating current through aninductance, energy ﬂows back andforth between itandtherestofthecircuit, buttheaverage rateatwhich energy isdelivered tothecircuit iszero. Wesaythat aninductance isanondissipative element; noelectrical energy isdissipated—that is, “lost”-—in it. Similarly, theenergy ofacondenser, U=%CV2, isreturned totheexternal circuit when acondenser 1Sdischarged. When acondenser isinanACcircuit energy ﬂows inandoutofit,butthenetenergy ﬂowineach cycle iszero. Anideal condenser isalsoanondissipative element. Weknow thatanemfisasource ofenergy. When acurrent Iﬂows inthe direction oftheemf, energy isdelivered totheexternal circuit attheratedU/dt = SI.lfcurrent isdriven against theemf—by other generators inthecircuit-the emfwillabsorb energy attherateSI;since Iisnegative, dU/dt willalsobenegative. lfagenerator isconnected toaresistor R,thecurrent through theresistor isI=8/R. Theenergy being supplied bythegenerator attherate£31isbeing absorbed bytheresistor. This energy goes into heat intheresistor andislost from theelectrical energy ofthecircuit. Wesaythatelectrical energy isdissipated inaresistor. Therateatwhich energy isdissipated inaresistor isdU/dt =R12. InanACcircuit theaverage rateofenergy losttoaresistor istheaverage of R12over onecycle. Since I=few‘-—by which wereally mean that Ivaries as coswt—the average ofI2over onecycle isIII2/2, since thepeak current isII]and theaverage ofcosz wtis1/2. What about theenergy losswhen agenerator isconnected toanarbitrary impedance 2?(By“1oss” wemean, ofcourse, conversion ofelectrical energy into thermal energy.) Any impedance 2canbewritten asthesum ofitsrealandini- ginary parts. That is, z=R-1-iX, (22.24) where RandXare realnumbers. From thepoint ofview ofequivalent circuits we cansaythat anyimpedance isequivalent toaresistance inseries with apure imaginary impedance-—called areactance—as shown inFig.22-17. Wehave seen earlier thatanycircuit thatcontains only L’sandC'shasan impedance that1Sapure imaginary number. Since there isnoenergy lossintoany oftheL’sandC’sontheaverage, apure reactance containing only L’sandC’s willhave noenergy loss. Wecanseethatthismust betrueingeneral forareactance. 22-llAny Circu_it ofZls andEs 10) cl»/=<\lai\l la’-' InO —-> Zeff (bl Z" b Fig. 22-16. Any two-terminal net- work canbereplaced byagenerator in series with animpedance. R Z E iX Fig. 22-17. Any impedance isequiv- alent toaseries combination ofapure resistance and apure reactance. O Z Q (0) Z2 b b (bl E:== E? Z3‘: A*Z2 (cl U’oE {UIo 0 O == ldl == (9 atD -%;=Zi+Z'—3 z,=z,+z,, Fig. 22-18. Theeffective impedance ofaladden (O2llai Cl CIfagenerator with theemf8isconnected totheimpedance 2ofFig. 22-17, theemfmust berelated tothecurrent Ifrom thegenerator by s=I(R+iX). (22.25) Toﬁndtheaverage rateatwhich energy isdelivered, wewant theaverage ofthe product 81.Now wemust becareful. When dealing with such products, wemust dealwith therealquantities 8(1)andI(t). (The realparts ofthecomplex functions willrepresent theactual physical quantities only when wehave linear equations; now weareconcerned withproducts, which arecertainly notlinear.) Suppose wechoose ourorigin oftsothattheamplitude Iisarealnumber, let’ssayI0;then theactual time variation Iisgiven by I=I0coswt. TheemfofEq.(22.25) istherealpart of I1-,e“”’(R +iX) Of a=1,12coswt-I0Xsin wt. (22.26) The twoterms inEq.(22.26) represent thevoltage drops across RandX inFig.22-17. Weseethatthevoltage drop across theresistance isinphase with thecurrent, while thevoltage drop across thepurely reactive part isoutofphase with thecurrent. Theaverage rateofenergy loss, (P),,,., from thegenerator istheintegral of theproduct 8Iover onecycle divided bytheperiod T;inother words, T T T (P),,v =;_/0 8Idi = I§Rcos2wldl - I§Xcos wtsinwtdt.O Theﬁrstintegral is%I§R, andthesecond integral iszero. Sotheaverage energy lossinanimpedance z=R+iXdepends only ontherealpart ofz, andis13R/2, which isinagreement with ourearlier result fortheenergy lossina resistor. There isnoenergy lossinthereactive part. 22-6 Aladder network Wewould likenow toconsider aninteresting circuit which canbeanalyzed interms ofseries andparallel combinations. Suppose westart with thecircuit of Fig.22-18(a). Wecanseeright away thattheimpedance from terminal atoter- minal bissimply 21+22.Now let’stake alittle harder circuit, theoneshown in Fig. 22-18(b). Wecould analyze thiscircuit using Kirchhoff's rules, butitis also easy tohandle with series andparallel combinations. Wecanreplace the twoimpedances ontheright-hand endbyasingle impedance 23=21—l—22,as inpart (c)oftheﬁgure. Then thetwoimpedances 22and23canbereplaced by their equivalent parallel impedance 2.1,asshown inpart(d)oftheﬁgure. Finally, 21and2.1areequivalent toasingle impedance 25,asshown inpart (e). Now wemayaskanamusing question: What would happen ifinthenetwork ofFig.22-18(b) wekept onadding more sections forever—as weindicate bythe dashed lines inFig.22—l9(a)? Canwesolve such aninﬁnite network? Well, that’s »b d-1111 11.112.1iii; °° ° ¢ --- b b Fig. 22-19. Theeffective impedance ofaninﬁnite ladder. 22-12 notsohard. First, wenotice thatsuch aninﬁnite network isunchanged ifweadd onemore section atthe“front” end. Surely, ifweaddonemore section toan inﬁnite network itisstillthesame inﬁnite network. Suppose wecalltheimpedance between thetwoterminals aandboftheinﬁnite network 20;then theimpedance of allthestuff totheright ofthetwoterminals canddisalso21,.Therefore, sofaras thefront endisconcerned, wecanrepresent thenetwork asshown inFig.22-19(b). Combining theparallel combinations 2220 andadding theresult inseries with 21, wecanimmediately write down theimpedance ofthiscombination: l Z220 2=21—|—-—--— or 2=21-1---- (1/Z2) "lr(1/Z0) Z2+Z0 Butthisimpedance isalsoequal to20,sowehave theequation Z2Zo ZZZ -is 0 i+Z2+z0 2,,=521+1/(Z;/4) +Z122. (22.27)Wecansolve for2,1toget Sowehave found thesolution fortheimpedance ofaninﬁnite ladder ofrepeated series andparallel impedances. The impedance 20iscalled thecharacteristic impedance ofsuch aninﬁnite network. Let’s now consider aspeciﬁc example inwhich theseries element isanin- ductance Landtheshunt element isacapacitance C,asshown inFig. 22-20(a). Inthiscase weﬁndtheimpedance oftheinﬁnite network bysetting 21=l(.0L and22=1/iwC. Notice thattheﬁrstterm, 21/2, inEq.(22.27) isjustone-half theimpedance oftheﬁrstelement. Itwould therefore seem more natural, orat least somewhat simpler, ifwewere todraw ourinﬁnite network asshown inFig. 22-20(b). Looking attheinﬁnite network from theterminal a’wewould seethe characteristic impedance 20=\/(L/C) —(w5’L2/4). (22.28) Nowthere aretwointeresting cases, depending onthefrequency w.If(.02isless than 4/LC, thesecond term intheradical willbesmaller than theﬁrst, andthe impedance 21,willbearealnumber. Ontheother hand, ifw21Sgreater than 4/LC theimpedance 20willbeapure imaginary number which wecanwrite as 20=ix/(w2L2/4) -(L/C). Wehave saidearlier thatacircuit which contains onlyimaginary impedances, such asinductances andcapacitances, willhave animpedance which ispurely imaginary. How canitbethen thatforthecircuit wearenowstiidying—which has onlyL’sandC’s—the impedance isapure resistance forfrequencies below \/4/LC? Forhigher frequencies theimpedance ispurely imaginary, inagreement with our earlier statement. Forlower frequencies theimpedance isapure resistance and willtherefore absorb energy. Buthowcanthecircuit continuously absorb energy, asaresistance does, ifitismade only ofinductances andcapacitances? Answer: Because there isaninﬁnite number ofinductances andcapacitances, sothatwhen asource isconnected tothecircuit, itsupplies energy totheﬁrstinductance and capacitance, then tothesecond, tothethird, andsoon.Inacircuit ofthiskind, energy iscontinually absorbed from thegenerator ataconstant rateandﬂows constantly outinto thenetwork, supplying energy which isstored intheinduc- tances andcapacitances down theline. This ideasuggests aninteresting point about what ishappening inthecircuit. Wewould expect thatifweconnect asource tothefront end, theeffects ofthis source willbepropagated through thenetwork toward theinﬁnite end. The propagation ofthewaves down thelineismuch liketheradiation from anantenna which absorbs energy from itsdriving source; thatis,weexpect such apropagation tooccur when theimpedance isreal, which occurs ifwislessthan \/4/LC. But when theimpedance ispurely imaginary, which happens forwgreater than V4/LC, wewould notexpect toseeanysuch propagation. 22-1331-L‘ZPorn rinm ,'1lllU\ ill i°T‘1°iiiOL/2°, /L/2\ /1./2,‘ /L/2\ — ,,f'°“”T§MT”'”lf."...1»,,T,1.T-2 Fig. 22-20. AnL-C ladder drawn intwoequivalent ways. 0ll. I2.VI V222-7 Filters Wesawinthelastsection thattheinﬁnite ladder network ofFig.22-20 absorbs energy continuously ifitisdriven atafrequency below acertain critical frequency \/4/LC, which wewillcallthecutoﬂ frequency wo.Wesuggested thatthiseffect could beunderstood interms ofacontinuous transport ofenergy down theline. Ontheother hand, athigh frequencies, forw>wo,there isnocontinuous ab- sorption ofenergy; weshould then expect thatperhaps thecurrents don’t “pene- trate” very fardown theline. Let’s seewhether these ideas areright. Suppose wehave thefront endoftheladder connected tosome ACgenerator andweaskwhat thevoltage looks likeat,say,the754th section oftheladder. Since thenetwork isinﬁnite, whatever happens tothevoltage from onesection to thenext isalways thesame; solet’sjustlook atwhat happens when wegofrom some section, saythenthtothenext. Wewilldeﬁne thecurrents 1,.andvoltages V"asshown inFig.22—2l(a). l='H if 12.v, v ___ _ )“C--. /i/ etc. (D) Vn Vn+l Fig. 22-21. Finding thepropagation factor ofaladder. Wecangetthevoltage V,,+1 from V,,byremembering thatwecanalways replace therestoftheladder afterthenthsection byitscharacteristic impedance 20; thenweneedonlyanalyze thecircuit ofFig.22—2l(b). First, wenotice thatany V,,,since itisacross z(,,must equal Inzo. Also, thedifference between V,,andV,,.i_, isjustInzlr VII _Vn+] =inzl =Vn Z0 Sowegettheratio 5.111.:1_5:?_~_:_i1 .V” L'(\ Zr) Wecancallthisratio thepropagation factor foronesection oftheladder; we’ll callita.Itis,ofcourse, thesame forallsections: Zn_Z101=~7 Z0(22.29) Thevoltage after thenthsection isthen V,,=ct"F,_ (22.30) You cannow ﬁndthevoltage after 754sections: itisJUSI01tothe754th power times 8. Suppose weseewhat atislikefortheL-Cladder ofFig.22—20(a) Using 2‘, from Eq.(22.27), and2,:iwL, weget ct=‘"7’:/P_;‘-"”L_’;I‘?_ “‘(“’L’2) (22.31)Wt/C) —(M1/4) +i<wL/2) lfthedriving frequency isbelow thecutoff frequency wt,=\/I/LC, theradical isarealnumber, andthemagnitudes ofthecomplex numbers inthenumerator anddenominator areequal. Therefore, themagnitude ofozisone; wecanwrite _1t$05-6, which means thatthemagnitude ofthevoltage isthesame atevery section. only 22-14 itsphase changes. Thephase change 6is,infact, anegative number andrepresents the“delay” ofthevoltage asitpasses along thenetwork. Forfrequencies above thecutoff frequency wt,itisbetter tofactor outanl from thenumerator anddenominator ofEq.(223|)andrewrite itas 01=-“FEE//4’ ‘-‘HQ ““"”2)- <2212>\/(w‘1L‘1/4) -(L/C) +(wL/2) Thepropagation factor atisnow arealnumber, andanumber lessthanone. That means thatthevoltage atanysection isalways lessthan thevoltage atthepre- ceding section bythefactor Ot.Foranyfrequency above wit,thevoltage dies away rapidly aswegoalong thenetwork. Aplotoftheabsolute value ofOlasa function offrequency looks likethegraph inFig.22-22. Weseethat thebehavior of01,both above andbelow wo,agrees with our interpretation thatthenetwork propagates energy forw<wi,andblocks itfor w>w.,. Wesaythat thenetwork “passes“ lowfrequencies and “reJects“ oi “ﬁlters out" thehigh frequencies. Any network designed tohave itscharacteristics varyinaprescribed waywithfrequency iscalled a“ﬁlter.” Wehave been analyzing a“low-pass ﬁlter.” You may bewondering why allthisdiscussion ofaninﬁnite network which obviously cannot actually occur. The point isthat thesame characteristics are fotind inaﬁnite network ifweﬁnish itoffattheendwith animpedence equal to thecharacteristic impedence 20. Now inpractice itisnotpossible toexuci/Vi‘ reproduce thecharacteristic impedance with afewsimple elements—like R’s. L’s,andC’s. Butitisoften possible todosowith afairapproximation foracertain range offrequencies. Inthisway onecanmake aﬁnite ﬁlter network whose properties arevery nearly thesame asthose fortheinﬁnite case Forinstance, the L-Cladder behaves much aswehave described itifitisterminated inthepure resistance R=\7U_C. IfinourL-Cladder weinterchange thepositions oftheL’sandC’s,tomake theladder shown inFig.22-23(a). wecanhave aﬁlter thatpropagates highfre- quencies and1‘€]€ClS lowfrequencies. ltiseasy toseewhat happens with thisnet- work byusing theresults wealready have. Youwillnotice thatwhenever wechange anLtoaCandviceverso, wealsochange every iwtol/iw. Sowhatever happened atwbefore willnowhappen atl/w. lnparticular, wecanseehowctwillvary with frequency byusing Fig.22-22 andchanging thelabel ontheaxistol/w, aswe have done inFig.22-23(b). Thelow-pass andhigh-pass ﬁlters wehave described have various technical applications. AnL-Clow-pass ﬁlter isoften used asa“smoothing” ﬁlter inaDC power supply. Ifwewant tomanufacture DCpower from anACsource, webegin with arectiﬁer which permits current toﬂow only inonedirection. From the rectiﬁer wegetaseries ofpulses that look like thefunction V(t) shown in Fig22-24, which islousy DC,because itwobbles upanddown. Suppose wewould likeanicepure DC,such asabattery provides. Wecancome close tothat by putting alow-pass ﬁlter between therectiﬁer andtheload. Weknow from Chapter 50ofVol.Ithatthetimefunction inFig.22-24 canbe represented asasuperposition ofaconstant voltage plusasinewave, plusahigher- frequency sinewave, plus astillhigher-frequency sinewave, etc.—by aFourier series. lfourﬁlter islinear (if,aswehave been assuming, theL’sandC’sdon't varywith thecurrents orvoltages) then what comes outoftheﬁlter isthesuper- position oftheoutputs foreach component attheinput. lfwearrange thatthe culolf frequency wt,ofourﬁlter iswellbelow thelowest frequency inthefunction V(i), theDC(forwhich w=0)goes through ﬁne, buttheamplitude oftheﬁrst harmonic willbecutdown alot.And amplitudes ofthehigher harmonics Wlllbe cutdown even more. Sowecangettheoutput assmooth aswewish, depending onlyonhow many ﬁlter sections wearewilling tobuy. Ahigh-pass ﬁlter isused ifonewants toreyect certain lowfrequencies. For instance, inaphonograph ampliﬁer ahigh-pass ﬁlter may beused toletthemusic 22-15ldl I we cu Fig 22-22. The propagation factor of0section ofanL-Cladder C C C C ti“av#1"#1ti(<1) lﬂl O I/we I/ai (b) Fig. 22-23. (0) Ahigh-pass ﬁlter; lb)itspropagation factor asafunction Ofl 0.‘. (U Fig. 22-24. Theoutput voltage of0 full-wove rectiﬁer. (0) (blll it €V _e__________it | -U12 {U Fig. 22-25. lei) Aband-pass ﬁlter. (b)Asimple resonant ﬁlter. I2 t, ‘” (0) I, 12 L, L2 (bl Fig. 22-26. Equivalent circuit of0 mutual inductance.through, while keeping outthelow-pitched rumbling from themotor ofthe turntable. Itisalso possible tomake “band-pass” ﬁlters thatreject frequencies below some frequency wlandabove another frequency w2(greater than wl),butpass the frequencies between wlandw2. This canbedone simply byputting together a high-pass andalow-pass ﬁlter, butitismore usually done bymaking aladder in which theimpedances 21and22aremore coinplicated——being each acombination ofL’sandC’s. Such aband-pass ﬁlter might have apropagation constant like thatshown inFig.22-25(a). Itmight beused, forexample, inseparating signals thatoccupy onlyaninterval offrequencies, such aseach ofthemany voice channels inahigh-frequency telephone cable, orthemodulated carrier ofaradio trans- mission. Wehave seeninChapter 25ofVol.Ithatsuch ﬁltering canalsobedone using theselectivity ofanordinary resonance curve, which wehave drawn forcomparison inFig.22—25(b). Buttheresonant ﬁlter isnotasgood forsome purposes asthe band-pass ﬁlter. You willremember (Chapter 48,Vol. I)thatwhen acarrier of frequency w,ismodulated with a“signal” frequency ws,thetotal signal contains notonly thecarrier frequency butalso thetwoside-band frequencies w,+w, andw.—w,.With aresonant ﬁlter, these side-bands arealways attentuated some- what, andtheattenuation ismore, thehigher thesignal frequency, asyoucansee from theﬁgure. Sothere isapoor “frequency response.” Thehigher musical tones don’t getthrough. Butiftheﬁltering isdone with aband-pass ﬁlter designed sothatthewidth (1)2—wlisatleast twice thehighest signal frequency, thefre- quency response willbe“ﬂat” forthesignals wanted. Wewant tomake onemore point about theladder ﬁlter: theL-Cladder of Fig. 22-20 isalso anapproximate representation ofatransmission line. Ifwe have along conductor thatrtins parallel toanother conductor—such asawireina coaxial cable, orawiresuspended above theearth—there willbesome capacitance between thetwoconductors andalsosome inductance duetothemagnetic ﬁeld between them. Ifweimagine thelineasbroken upintosmall lengths At,each length willlook likeonesection oftheL-Cladder with aseries inductance ALand ashunt capacitance AC. Wecanthen useourresults fortheladder ﬁlter. lfwe take thelimit asA6goes tozero, wehave agood description ofthetransmission line. Notice thatasA6ismade smaller andsmaller, both ALandACdecrease, but inthesame proportion, sothattheratio AL/AC remains constant. Soifwetake thelimit ofEq.(22.28) asALandACgotozero, weﬁndthatthecharacteristic impedance 20isapure resistance whose magnitude is\/A15/iAC. Wecanalso write theratio AL/AC asL0/C0, where L0andC0aretheinductance andcapaci- tance ofaunitlength oftheline; then wehave _£2. 2,,_\/CO (22.33) You willalso notice that asALandACgotozero, thecutoﬁ" frequency wi,=\/4/LC goes toinﬁnity. There isnocutoff frequency foranideal transmission line. 22-8 Other circuit elements Wehave sofardeﬁned only theideal circuit impedances—the inductance, thecapacitance, andtheresistance—as wellastheideal voltage generator. Wewant now toshow that other elements, such asmutual inductances ortransistors or vacuum tubes, canbedescribed byusing only thesame basic elements. Suppose thatwehave twocoils andthatonpurpose, orotherwise, some ﬂuxfrom oneof thecoils links theother, asshown inFig.22-26(a). Then thetwocoils willhave a mutual inductance Msuch thatwhen thecurrent varies inoneofthecoils, there willbeavoltage generated intheother. Canwetake intoaccount such aneﬁect inourequivalent circuits? Wecaninthefollowing way. Wehave seen thatthe 22-l6 induced emf’sineach oftwointeracting coils canbewritten asthesumoftwoparts: 81 “-= —-L1£lL1- ItMidi d’ d’ (22.34) dr.. at 82:"L2diiMTz1' Theﬁrstterm comes from theself-inductance ofthecoil, andthesecond term comes from itsmutual inductance with theother coil. Thesignofthesecond term canbeplusorminus, depending onthewaytheﬂuxfrom onecoillinks theother. Making thesame approximations weused indescribing anideal inductance, we would saythatthepotential difference across theterminals ofeach coilisequal to theelectromotive force inthecoil. Then thetwoequations of(22.34) arethesame astheones wewould getfrom thecircuit ofFig.22-26(b), provided theelectro- motive force ineach ofthetwocircuits shown depends onthecurrent inthe opposite circuit according totherelations 81=d=iwMI2, 82==I=iwMI1. (22.35) Sowhat wecandoisrepresent theeffect oftheself-inductance inanormal waybut replace theeffect ofthemutual inductance byanauxiliary ideal voltage generator. Wemust inaddition, ofcourse, have theequation that relates thisemftothe current insome other partofthecircuit; butsolong asthisequation islinear, we have justadded more linear equations toourcircuit equations, andallofour earlier conclusions about equivalent circuits andsoforth arestillcorrect. Inaddition tomutual inductances there may also bemutual capacitances. Sofar,when wehave talked about condensers wehave always imagined thatthere were only twoelectrodes, butinmany situations, forexample inavacuum tube, there maybemany electrodes close toeach other. Ifweputanelectric charge on anyoneoftheelectrodes. itselectric ﬁeldwillinduce charges oneachoftheother electrodes andaffect itspotential. Asanexample, consider thearrangement of fourplates shown inFig.22-27(a). Suppose these four plates areconnected to external circuits bymeans ofthewires A,B,C,andD.Solong asweareonly worried about electrostatic effects, theequivalent circuit ofsuch anarrangement ofelectrodes isasshown inpart(b)oftheﬁgure. Theelectrostatic interaction of anyelectrode with each oftheothers isequivalent toacapacity between the twoelectrodes. Finally, let’sconsider how weshould represent such complicated devices as transistors andradio tubes inanACcircuit. Weshould point outatthestart that such devices areoften operated insuch awaythat therelationship between the currents andvoltages isnotatalllinear. Insuch cases, those statements wehave made which depend onthelinearity ofequations are,ofcourse, nolonger correct. Ontheother hand, inmany applications theoperating characteristics aresufﬁciently linear thatwemayconsider thetransistors andtubes tobelinear devices. Bythis wemean thatthealternating currents in,say,theplate ofavacuum tubearelinearly proportional tothevoltages that appear ontheother electrodes, saythegrid voltage andtheplate voltage. When wehave such linear relationships, wecan incorporate thedevice intoourequivalent circuit representation. Asinthecaseofthemutual inductance, ourrepresentation willhave toinclude auxiliary voltage generators which describe theinﬂuence ofthevoltages orcurrents inonepartofthedevice onthecurrents orvoltages inanother part. Forexample, theplate circuit ofatriode canusually berepresented byaresistance inseries with anideal voltage generator whose source strength isproportional tothegridvoltage. Wegettheequivalent circuit shown inFig.22-28.* Similarly, thecollector circuit *Theequivalent circuit shown iscorrect only forlowfrequencies. Forhigh frequencies theequivalent circuit gets much more complicated and willinclude various so-called “parasitic” capacitances andinductances. 22-17A B (<1) t t 1 _ ; / C D A B ledFig. 22-27. Equivalent circuit of mutual capacitance. PLATE P GRID G V9 \ THODE C C=-/_4,Vq Fig. 22-28. Alow-frequency equiv- cilent circuit of0vacuum triode. om IOEM|TTE R OGLECTOR 1-eBA$E 0B Fig 22-29 Alowfrequency equiv- A cilent C|l'CUIl' of0transistor 8_KI: ofatransistor isconveniently represented asaresistor inseries with anideal voltage generator whose source strength isproportional tothecurrent from the emitter tothebase ofthetransistor. Theequivalent circuit isthen likethatinFig. 22-29. Solong astheequations which describe theoperation arelinear, wecan usesuch representations fortubes ortransistors. Then, when theyareincorporated inacomplicated network, ourgeneral conclusions about theequivalent representa- tionofanyarbitrary connection ofelements isstillvalid. There isoneremarkable thing about transistor andradio tube circuits which isdifferent from circuits containing only impedances: therealpartoftheelfective impedance z,.;fcanbecome negative. Wehave seenthattherealpartofzrepresents thelossofenergy. Butitistheimportant characteristic oftransistors andtubes thattheysupply energy tothecircuit. (Ofcourse they don’t just“make” energy; they take energy from theDCcircuits ofthepower supplies andconvert itinto ACenergy.) Soitispossible tohave acircuit with anegative resistance. Such a circuit hastheproperty thatifyouconnect ittoanimpedance with apositive real part, i.e.,apositive resistance, andarrange matters sothat thesum ofthetwo realparts isexactly zero, then there isnodissipation inthecombined circuit. If there isnolossofenergy, anyalternating voltage oncestarted willremain forever. Thisisthebasic ideabehind theoperation ofanoscillator orsignal generator which canbeusedasasource ofalternating voltage atanydesired frequency. 22-18 23 Cavity Resonators 23-1 Real circuit elements When looked atfrom anyonepairofterminals, anyarbitrary circuit made upofideal impedances andgenerators is,atanygiven frequency, equivalent toa generator é‘.inseries with animpedance :.That comes about because ifweputa voltage Vacross theterminals andsolve alltheequations toﬁndthecurrent 1, wemust getalinear relation between thecurrent and thevoltage. Since allthe equations arelinear, theresult forImust also depend only linearly onVThe most general linear form canbeexpressed as 1=E(V~8). (23.1) lngeneral, both zand8maydepend insome complicated wayonthefrequency ta. Equation (231),however, istherelation wewould getifbehind thetwoterminals there wasjust thegenerator <‘§(w) inseries with theimpedance 2(0)). There isalsotheopposite kind ofquestion' lfwehave anyelectromagnetic device atallwith twotermiiials andwemeasure therelation between IandVto detei mine 6'.andzasfunctions offrequency. canweﬁndacombination ofourideal elements thatisequivalent totheinternal impedance 2?Theanswer isthat for anyreasonable—tliat is,physically meaningful——function z(w), itISpossible to uf)/J/‘U)tl/Iltllc’ thesituation toashighanaccuracy asyouwishwithacircuit containing aﬁnite setofideal elements. Wedon’t want toconsider thegeneral problem now butonly look atwhat might beexpected from physical arguments forafewcases ll‘wethink ofarealresistor, weknow thatthecurrent through itwillproduce amagnetic ﬁeld. Soanyrealresistor should alsohave some inductance. Also. when aresistor hasapotential difference across it,there must becharges onthe ends oftheresistoi toproduce thenecessary electric ﬁelds Asthevoltage changes. thecharges willchange inproportion, sotheresistor willalsohave some capaci- tance. Weexpect thatarealresistor might have theequivalent circuit shown in Fig23~l lnawell-designed resistor, theso-called “parasitic” elements LandC aresmall, sothatatthefrequencies forwhich itisintended, wLismuch lessthan R,andl/wC ismuch greater than R.ltmaytherefore bepossible toneglect them Asthefrequency israised, however, they willeventually become important, anda resistor begins tolook likearesonant circuit. Arealinductance isalsonotequal totheidealized inductance, whose impe- dance is1wL. Arealcoilofwire willhave some resistance, soatlowfrequencies the coilisreally equivalent toaninductance inseries with some resistance, asShown in Fig.23—2(a) But,youarethinking, theresistance andinductance aretogether ina realcoil—the resistance isspread allalong thewire, soitismixed inwith the inductance. Weshould probably useacircuit more liketheoneinFig.23-Ztb). which hasseveral little R’sandL’sinseries. Butthetotal impedance ofsuch a circuit isjust ZR—l—ZiwL, which isequivalent tothesimpler diagram ofpart (a) Aswegoupinfrequency with arealcoil,theapproximation ofaninductance plusaresistance isnolonger very good. Thecharges thatmust build uponthe wires tomake thevoltages willbecome important. Itisasifthere were little con- densers across theturns ofthecoil, assketched inFig23—3(a). Wemight tryto approximate therealcoilbythecircuit inFig.23—3(b). Atlowfrequencies, this circuit canbeimitated fairly wellbythesimpler oneinpart(c)oftheﬁgure (which isagain thesame resonant circuit wefound forthehigh-frequency model ofa resistor) For higher frequencies. however, themore complicated circuit of 23-123-1 Real circuit elements 23~2 Acapacitor athigh frequencies 23-3 Aresonant cavity 23—4 Cavity modes 23-5 Cavities andresonant circuits Review: Chapter 23.Vol. I.Resonance Chapter 49,Vol l,Modes L C R Fig. 23—l. Equivalent circuit of 0 recil resistor. (<1) (bl Fig. 23—2. The equivalent circuit of cireal inductcince atlowfrequencies. ' _ tut (bi (C) Fig. 23-3. The equivalent circuit of 0recil inductance qthigher frequencies.if ._____~ i~__ ,I / \ I\ N iii LINES OFE‘llilii4Fig.23—3(b) isbetter. Infact, themore accurately youwish torepresent theactual impedance ofareal, physical inductance, themore ideal elements youwillhave to useintheartiﬁcial model ofit. Let's look alittle more closely atwhat goes oninarealcoil. Theimpedance ofaninductance goes ast.iL,soitbecomes zero atlowfrequencies—it isa“short circuit”: allweseeistheresistance ofthewire. Aswegoupinfrequency, wLsoon becomes much larger than R,andthecoillooks pretty much likeanideal induc- tance. Aswegostillhigher, however, thecapacities become important. Their impedance isproportional tol/wC, which islarge forsmall cu.Forsmall enough frequencies acondenser isan“open circuit," andwhen itisinparallel with some- thing else, itdraws nocurrent. Butathigh frequencies, thecurrent prefers toﬂow intothecapacitance between theturns, rather than through theiiiductance So thecurrent inthecoiljumps from oneturn totheother anddoesn‘t bother togo around andaround where ithastobuck theemf Soalthough wemay have intended thatthecurrent should goaround theloop, itwilltaketheeasier path—the path ofleast impedance. 23—2 Acapacitor athigh frequencies Now wewant todiscuss indetail thebehavior ofacapacitor—a geometrically ideal capacitor—as thefrequency getslarger andlarger, sowecanseethetransition ofitsproperties. (We prefer touseacapacitor instead ofaninductance, because thegeometry ofapairofplates ismuch lesscomplicated than thegeometry ofa coil.) Weconsider thecapacitor shown inFig.23—4(a). which consists oftwopar- allelcircular plates connected toanexternal generator byapairofwires. lfwe charge thecapacitor with DC,there willbeapositive charge ononeplate anda negative charge ontheother; andthere willbeauniform electric ﬁeld between the plates. Now suppose thatinstead ofDC,weputanACoflowfrequency ontheplates (Wewillﬁndoutlater what is“low” andwhat is“high".) Sayweconnect theca- pacitor toalower-frequency generator. Asthevoltage alternates, thepositive charge onthetopplate istaken offandnegative charge isputon.While thatis happening, theelectric ﬁelddisappears andthen builds upintheopposite direction ’fQ SURFACE '4 S _l 7 I lIit+CURVE I" ._ \_l_ TTTT ll o0 o, B ‘n‘Q§§_ ’f i® Q/it_ _ __/4 ‘a" V. LINES OFB r___] (<1) (bl Fig. 23-4. Theelectric andmagnetic ﬁelds between theplates ofciccipocitor. 23—2Ifthesubject hadbeen oneofpopular interest, thiseffect would have been called “the high-frequency barrier,” orsome such name. Thesame kind ofthing happens inallsubjects. Inaerodynamics, ifyoutrytomake things gofaster than thespeed ofsound when they were designed forlower speeds, they don’t woik ltdoesn’t mean thatthere isagreat “barrier” there, itjustmeans thattheobject should beredesigned. Sothiscoilwhich wedesigned asan“inductance” isnot going towork asagood inductance, butassome other kind ofthing atvery high frequencies. Forhigh frequencies, wehave toﬁndanewdesign. cunvs F2 Asthecharge sloshes back andforth slowly, theelectric ﬁeld follows Ateach instant theelectric ﬁeld isuniform. asshown inFig23—4(b), except forsome edge effects which wearegoing todisregard. Wecanwrite themagnitude oftheelectric ﬁeld as E=E0e“"', (23.2) where E0isaconstant. Now willthatcontinue toberight asthefrequency goes up? No,because as theelectric field isgoing upanddown, there isaﬂuxofelectric ﬁeld through any loop likeI‘,inFig23—4(a). And, asyouknow, achanging electric ﬁeld actsto produce amagnetic ﬁeld. One ofMaxwell’s equations says thatwhen there isa varying electric ﬁeld, asthere ishere, there hasgottobealineintegral ofthe magnetic ﬁeld. Theintegral ofthemagnetic ﬁeld around aclosed ring, multiplied byc2,isequal tothetime rate-of-change oftheelectric fluxthrough thearea inside thering(ifthere arenocurrents): c2fB'a's=i fE-nda. (23.3)1" 61 inside I‘ Sohowmuch magnetic ﬁeld isthere? That’s notvery hard. Suppose thatwetake theloop F1.which isacircle ofradius r.Wecanseefrom symmetry that the magnetic ﬁeld goes around asshown intheﬁgure. Then thelineintegral ofBis 21rrB. And, since theelectric ﬁeld isuniform, theﬂuxoftheelectric ﬁeld issimply Emultiplied by'rrr2, thearea ofthecircle: @212-27i'r= 7172. (23.4) Thederivative ofEwith respect totime is,forotiralternating ﬁeld, simply iwE(,e'°". Soweﬁndthatourcapacitor hasthemagnetic ﬁeld B=5:-QE.,@“"‘. (23.5) Inother words, themagnetic ﬁeld alsooscillates andhasastrength proportional tor. What istheeffect ofthat? When there isamagnetic ﬁeldthatisvarying, there willbeinduced electric ﬁelds andthecapacitor willbegin toactalittle bitlikean inductance. Asthefrequency goes up,themagnetic ﬁeld getsstronger; itispro- portional totherateofchange ofE,andsotowTheimpedance ofthecapacitor willnolonger besimply l/iwC. Let’s continue toraise thefrequency andtoanalyze what happens more care- fully. Wehave amagnetic ﬁeld thatgoes sloshing back andforth. Btitthen the electric ﬁeld cannot beuniform, aswehave assumed! When there isavarying magnetic ﬁeld. there must bealineintegral ofthe electric ﬁeld—because ofFaraday’s law Soifthere isanappreciable magnetic ﬁeld, asbegins tohappen athigh fre- quencies, theelectric ﬁeldcannot bethesame atalldistances from thecenter. The electric field must change with rsothat thelineintegral oftheelectric ﬁeld can equal thechanging ﬂuxofthemagnetic ﬁeld. Let’s seeifwecanﬁgure outthecorrect electric ﬁeld. Wecandothat by computing a“correction” totheuniform ﬁeld weoriginally assumed forlow frequencies. Let’s calltheuniform ﬁeld E1,which willstillbeE(,e““‘, andwrite thecorrect ﬁeld as E=E1-l"E2, where E2isthecorrection duetothechanging magnetic ﬁeld. Foranycuwewill write theﬁeld atthecenter ofthecondenser asE0e"“' (thereby deﬁning E0), so thatwehave nocorrection atthecenter; E2=0atr=0. ToﬁndE2wecanusetheintegral form ofFaraday’s law: 6£_E ds-—€t(ﬂuxofB). 23-3 E /-E‘ , ,; ,/2 _ // \ / / QE, / Ei‘E2}\ T\\ if“__4 _ _ _ __ i ___o..__ ___ __?£_J___. Fig. 23~5. The electric ﬁeld between the capacitor plotes cithigh frequency. (Edge effects ore neglected.)Theintegrals aresimple ifwetake them forthecurve I‘2,shown iiiFig23—4(b). which goes upalong theaxis, outradially thedistance 1"along thetopplate, down vertically tothebottom plate, andback totheaxis The lineintegral ofE,around thiscurve IS,ofcourse, zero; soonly E2contribtites, and itsintegral isjust —E2(r) -Ii,where /1isthespacing between theplates. (We callEpositive ifit points upward.) This isequal totherateofchange ofthefluxofB,which wehave togetbyanintegral over theshaded area Sinside I‘2inFig.23—4(b). Theflux through avertical strip ofwidth drisB(r)/1 dr.sothetotal fltixis 11/Bo)6])‘. Setting —6/0! oftheﬂuxequal tothelineintegral ofE2,wehave 8E2(r) =5fB(r) dr. (23.6) Notice thattheIicancels out,theﬁelds don’t depend ontheseparation ofthe plates Using Eq(23.5) forB(r), wehave 6IZUF2 Lwj E20’) I5Z67 E09 - Thetime derivativejust brings down another factor iw;weget 22Q3r Tu:E2(r)=-F50¢‘. (23.7) Asweexpect. theinduced ﬁeld tends toreduce theelectric field farther otit. The corrected ﬁeld E=E,+E2isthen 1wzrz ,2,E=Ej + E2 = —Z"C"-2") EQC’ . The electric ﬁeld inthecapacitor isnolonger uniform; ithastheparabolic shape shown bythebroken lineinFig.23-5 You seethatoursimple capacitor is getting slightly complicated. Wecould now useourresults tocalculate theimpedance ofthecapacitor athigh frequencies. Knowing theelectric ﬁeld, wecould compute thecharges on theplates andﬁnd outhow thecurrent through thecapacitor depends onthe frequency w,butwearenotinterested inthatproblem forthemoment. Weare more interested inseeing what happens aswecontinue togoupwith thefrequency ~to seewhat happens ateven higher frequencies Aren't wealready ﬁnished ‘? No,because wehave corrected theelectric ﬁeld, which means thatthemagnetic ﬁeld wehave calculated isnolonger right. The magnetic field ofEq.(23.5) is approximately right, butitisonly aﬁrstapproximation Solet’scallitB, We should then rewrite Eq.(23.5) as iwr twl B1 : E()€ . You willremember thatthisﬁeld wasproduced bythevariation ofE,.Now the correct magnetic ﬁeld willbethatproduced bythetotal electric ﬁeld E1+E2. Ifwewrite themagnetic field asB=B1—l—B2,thesecond term isjusttheaddi- tional ﬁeld produced byE2 ToﬁndB2wecangothrough thesame arguments wehave used toﬁndB1,thelineintegral ofB2around thecurve I‘,isequal to therateofchange oftheﬂuxofE2through F1.Wewilljust have Eq(234)again with Breplaced byB2andEreplaced byE2: c2B2 ~21rr=%(flux ofE2through I‘1). Since E2varies with radius, toobtain itsﬂuxwemust integrate over thecircular 23-4 surface inside F1.Using 21rrdrastheelement ofarea, thisintegral is [TE2(r) '21rrdr. 0 SowegetforB2(r) B2(r)=i3/E2(r)rdr. (23.10)rc26t Using E2(r) from Eq.(23.7), weneed theintegral ofr3dr,which is,ofcourse, r4/4. Ourcorrection tothemagnetic ﬁeld becomes -33 B20)=_51%E0e“”. (23.11) Butwearestillnotﬁnished! Ifthemagnetic ﬁeld Bisnotthesame asweﬁrst thought, then wehave incorrectly computed E2. Wemust make afurther cor- rection toE,which comes from theextra magnetic ﬁeldB2.Let's callthisadditional correction totheelectric ﬁeld E3.Itisrelated tothemagnetic ﬁeld B2inthesame waythatE2wasrelated toB1.WecanuseEq.(23.6) allover again justbychang- ingthesubscripts: 6E3(I') = dr. Using ourresult, Eq.(23.11), forB2,thenewcorrection totheelectric ﬁeld is w4r4 wtE3(r) =-I»g Eoe . (23.13) Writing ourdoubly corrected electric ﬁeld asE=E1-1-E2+E5,weget _ wt__1gg2_1gg4E— E02 [1 §§(c) +2;—_—Z§(c)]- (23.14) Thevariation oftheelectric ﬁeld with radius isnolonger thesimple parabola we drew inFig.23-5, butatlarge radii liesslightly above thecurve (E1+E2). Wearenotquite through yet.Thenewelectric ﬁeldproduces anewcorrection tothemagnetic ﬁeld, andthenewly corrected magnetic ﬁeldwillproduce afurther correction totheelectric ﬁeld, andonandon.However, wealready have allthe formulas thatweneed ForBawecanuseEq.(23.10), changing thesubscripts of BandEfrom 2to3. Thenextcorrection totheelectric ﬁeld is 1 wr6 W E4=“ (Y)E08t- Sotothisorder wehave thatthecomplete electric ﬁeld isgiven by w 1 2 1 4 1 6 E:E“ +<2ir(%) ‘tar(23.15) where wehave written thenumerical coeﬁicients insuch awaythatitisobvious howtheseries istobecontinued. Ourﬁnal result isthattheelectric ﬁeld between theplates ofthecapacitor, foranyfrequency, isgiven byE0e“"‘ times theinﬁnite series which contains only thevariable wr/C. Ifwewish, wecandeﬁne aspecial function, which wewillcall J(,(x), astheinﬁnite series thatappears inthebrackets ofEq.(23.15): 2 4 6 23-5 lJolll‘ 10" \ \ \ OSF l 2405 _./ 5... ..5./2 \ K6552 8W12 g_o‘WT’ml/ ei -O5 Fig. 23-6. The Bessel function JO(x).Then wecanwrite oursolution asEoei“ times thisfunction, with x=wr/c: E=E.,e““J0 - (23.17) Thereason wehave called ourspecial function J0isthat, naturally, thisisnot theﬁrsttime anyone haseverworked outaproblem with oscillations inacylinder. Thefunction hascome upbefore andisusually called J0. ltalways comes up whenever yousolve aproblem about waves with cylindrical symmetry. Thefunc- tion ./(,istocylindrical waves what thecosine function istowaves onastraight line. Soitisanimportant function, invented along time ago. Then aman named Bessel gothisname attached toit.Thesubscript zero means thatBessel invented awhole lotofdifferent functions andthisisjusttheﬁrstofthem. Theother functions ofBessel—J1,J2, andsoon—have todowith cylindrical waves which have avariation oftheir strength with theangle around theaxisof thecylinder. The completely corrected electric ﬁeld between theplates ofourcircular capacitor, given byEq.(23.17), isplotted asthesolid lineinFig. 23—5. For frequencies that arenottoohigh, oursecond approximation was already quite good. Thethird approximation waseven better—so good, infact. thatifwehad plotted it,youwould nothave been able toseethedl1T€l'CﬂC€ between itandthe solid curve. You willseeinthenext section, however, thatthecomplete series is needed togetanaccurate description forlarge radii, orforhigh frequencies. 23-3 Aresonant cavity Wewant tolook nowatwhat oursolution gives fortheelectric ﬁeld between theplates ofthecapacitor aswecontinue togotohigher andhigher frequencies. Forlarge co,theparameter x=wr/c alsogetslarge, andtheﬁrstfewterms inthe series forJ0ofxwillincrease rapidly. That means that theparabola wehave drawn inFig.23-5 curves downward more steeply athigher frequencies Infact. itlooks asthough theﬁeld would fallallthewaytozero atsome high frequency, perhaps when c/<13isapproximately one-half ofa. Let’s seewhether J,,does indeed gothrough zero andbecome negative. Webegin bytrying x:2' J,,(2)=l——l+%;—;,%:()22 Thefunction isstillnotzero, solet'stryahigher value ofx,say,x=25Putting innumbers, wewrite Jn(2.5l =l——1.56 +0.61 —009 :-0.04. The function .10hasalready gone through zero bythetime wegettoxI2.5. Comparing theresults forx=2andx=25,itlooks asthough J,,goes through zeroatone-fifth ofthewayfrom 2.5to2.Wewould guess thatthezerooccurs for xapproximately equal to2.4. Let’s seewhat thatvalue ofxgives: J()(24) :l—1.44 —l—0.52 —008 =000 Weget/erototheaccuracy ofourtwodecimal places. lfwemake thecalculation more accurate (orsince .10isawell-known function, ifwelook itupiiiabook), we ﬁndthatitgoes through zero atxI2405 Wehave worked itoutbyhand to show youthatyoutoocould have discovered these things iatlier than having to borrow them from abook Aslong aswearelooking upJ0inabook, itisinteresting tonotice how it goes forlarger values ofx,itlooks likethegraph inFig23—6. Asxincreases, J,,(x) oscillates between positive andnegative values with adecreasing amplitude ofoscillation Wehave gotten thefollowing interesting result: lfwegohigh enough infre- quency, theelectric ﬁeld atthecenter ofourcondenser willbeonewayandilie electric ﬁeld near theedge willpoint intheopposite direction. Foiexample, 23-6 suppose thatwetake anwhigh enough sothatx=wr/c attheouter edge ofthe capacitor isequal to4;then theedge ofthecapacitor corresponds totheabscissa x=4inFig.23-6. This means thatourcapacitor isbeing operated atthefre- quency to=4c/u Attheedge oftheplates, theelectric ﬁeld willhave arather high magnitude opposite thedirection wewould expect. That istheterrible thing thatcanhappen toacapacitor athighfrequencies. lfwe gotoveryhighfrequencies, thedirection oftheelectric ﬁeld oscillates back andforth many times aswego outfrom thecenter ofthecapacitor. Also there arethemagnetic ﬁelds associated with these electric ﬁelds. ltisnotsurprising thatourcapacitor doesn’t look like theideal capacitance forhigh frequencies. Wemayeven start towonder whether itlooks more likeacapacitor oraninductance Weshould emphasize thatthere areeven more complicated effects thatwehave neglected which happen attheedges ofthe capacitor. Forinstance, there willbearadiation ofwaves outpasttheedges, sotheﬁelds areeven more complicated than theones wehave computed. butwe willnotworry about those effects now. Wecould trytoﬁgure outanequivalent circuit forthecapacitor, butperhaps itisbetter ifwejustadmit thatthecapacitor wehave designed forlow-frequency ﬁelds isjustnolonger satisfactory when thefrequency istoohigh. lfwewant to treat theoperation ofsuch anobject athigh frequencies, weshould abandon the approximations toMaxwell’s equations that wehave made fortreating circuits andreturn tothecomplete setofequations which describe completely theﬁelds inspace Instead ofdealing with idealized circuit elements, wehave todeal with therealconductors asthey are,taking intoaccount alltheﬁelds inthespaces in between. Forinstance, ifwewant aresonant circuit athigh frequencies wewill nottrytodesign oneusing acoilandaparallel-plate capacitor. Wehave already mentioned that theparallel-plate capacitor wehave been analyzing hassome oftheaspects ofboth acapacitor andaninductance. With the electric ﬁeld there arecharges onthesurfaces oftheplates, andwith themagnetic ﬁelds there areback emf’s. lsitpossible thatwealready have aresonant circuit? Wedoindeed. Suppose wepick afrequency forwhich theelectric ﬁeld pattern fallstozeroatsome radius inside theedge ofthedisc; thatis,wechoose wa/c greater than 2.405 Everywhere onacircle coaxial with theplates theelectric ﬁeld willbezero. Now suppose wetake athin metal sheet andcutastrip justwide enough toﬁtbetween theplates ofthecapacitor. Then webend itintoacylinder thatwillgoaround attheradius where theelectric ﬁeld iszero Since there are noelectric ﬁelds there, when weputthisconducting cylinder inplace, nocurrents willflow init;andthere willbenochanges intheelectric andmagnetic ﬁelds. We have been able toputadirect short circuit across thecapacitor without changing anything. And look what wehave; wehave acomplete cylindrical canwith elec- trical andmagnetic ﬁelds inside andnoconnection atalltotheoutside world Theﬁelds inside won’t change even ifwethrow away theedges oftheplates outside ourcan,andalsothecapacitor leads. Allwehave leftisaclosed canwith electric andmagnetic ﬁelds inside, asshown inFig.23—7(a). Theelectric ﬁelds areos- cillating back andforth atthefrequency w—WhlCh, don’t forget, determined the diameter ofthecan Theamplitude oftheoscillating Eﬁeldvaries withthedistance from theaxisofthecan,asshown inthegraph ofFig.23-"/(b). This curve isjust theﬁrstarch oftheBessel function ofzero order. There isalsoamagnetic ﬁeld which goes incircles around theaxisandoscillates intime 90°outofphase with theelectric ﬁeld Wecanalsowrite outaseries forthemagnetic ﬁeld andplotit.asshown in thegraph ofFig.23—7(c). How isitthatwecanhave anelectric andmagnetic ﬁeldinside acanwith no external connections? Itisbecause theelectric andmagnetic ﬁelds maintain them- selves: thechanging Emakes aBandthechanging Bmakes anE~all according totheequations ofMaxwell. Themagnetic ﬁeld hasaninductive aspect, andthe electric ﬁeld acapacitive aspect; together they make something likearesonant circuit. Notice that theconditions wehave described would only happen ifthe radius ofthecanisexactly 2.405 c/w. Foracanofagiven radius, theoscillating electric andmagnetic ﬁelds willmaintain themselves—in thewaywehave described 23—7LINES OFB ——--—--——--—--1 O 0 -1- ® ,. . O O Q ® ________§_-___mesore(0) 1:,/1“'-___—1 l-o-o8bu8 L_.....;>_-.. ,-U’VID 2.405 c/wI’ CB9‘ 1.0- J?Q§I F Fig. 23—7. Theelectric andmagnetic ﬁelds inonenclosed cylindrical con. /‘T“{:>l*“feaﬂgc/\\ .0T'iTT_1l_ipt“J\7\lNPUT- >ee‘A >OUTPUTLOOP LOOP Fig. 23-8. Coupling into and outof aresonant cavity. R-F SIGNAL GENERATOR I CAVITYDETECTOR€ AMPLIFIER “Q Fig. 23—9. Asetup forobserving the cavity resonance. ENT I I CURR OUTPUT Aw=we/Q §__YFrequency Fig. 23—lO. Thefrequency response curve ofaresonant cavity.—only atthatparticular frequency. Soacylindrical canofradius risresonant at thefrequency C wo=2.405 ;- (23.18) Wehave saidthattheﬁelds continue tooscillate inthesame wayafter thecan iscompletely closed. That isnotexactly right. ltwould bepossible ifthewalls ofthecanwere perfect conductors. Forarealcan, however, theoscillating cur- rents which exist ontheinside walls ofthecanloseenergy because oftheresistance ofthematerial. Theoscillations oftheﬁelds willgradually dieaway. Wecansee from Fig. 23-7 that there must bestrong currents associated with electric and magnetic ﬁelds inside thecavity. Because thevertical electrical ﬁeldstops suddenly atthetopandbottom plates ofthecan, ithasalarge divergence there; sothere must bepositive andnegative electric charges ontheinner surfaces ofthecan,as shown inFig.23—7(a) When theelectric ﬁeld reverses, thecharges must reverse also. sothere must beanalternating current between thetopandbottom plates ofthecan These charges willflow inthesides ofthecan,asshown intheﬁgure Wecanalsoseethatthere must becurrents inthesides ofthecanbyconsidering what happens tothemagnetic ﬁeld Thegraph ofFig.23—7(c) tells usthatthe magnetic ﬁeld suddenly drops tozero attheedge ofthecan Such asudden change inthemagnetic ﬁeldcanhappen only ifthere isacurrent inthewall This current iswhat gives thealternating electric charges onthetopandbottom plates ofthe can. You maybewondering about ourdiscovery ofcurrents inthevertical sides of thecan What about ourearlier statement thatnothing would bechanged when we introduced these vertical sides inaregion where theelectric ﬁeld waszero" Re- member, however, that when weﬁrst putinthesides ofthecan. thetopand bottom plates extended outbeyond them, sothatthere were alsomagnetic ﬁelds ontheoutside ofourcan Itwasonly when wethrew away theparts ofthe capacitor plates beyond theedges ofthecanthatnetcurrents hadtoappear onthe insides ofthevertical walls Although theelectric andmagnetic ﬁelds inthecompletely enclosed canWlll gradually dieaway because oftheenergy losses, wecanstop thisfrom happening ifwemake alittle hole inthecanandputinalittle bitofelectrical energy tomake upthelosses Wetakeasmall wire, poke itthrough theholeinthesideofthe can, andfasten ittotheinside wallsothatitmakes asmall loop, asshown inFig23e8. lfwenow connect thiswire toasource ofhigh-frequency altei nating current, this current Wlllcouple energy into theelectric and magnetic ﬁelds ofthecavity and keep theoscillations going. This willhappen, ofcourse, only iftlie frequency ofthe driving source isattheresonant frequency ofthecan lftlie source isatthewrong frequency, theelectric andmagnetic ﬁelds willnotresonate, andtheﬁelds inthe canwillbevery weak. Theresonant behavior caneasily beseen bymaking another small hole in thecanandhooking inanother coupling loop, aswehave alsodrawn inFig.23~8. The changing magnetic ﬁeld through this loop will generate aninduced electro- motive force intheloop. lfthisloop isnowconnected tosome external measuring circuit. thecurrents willbeproportional tothestrength oftheﬁelds inthecavity Suppose wenow connect theinput loop ofourcavity toanRFsignal generator, asshown inFig.23-9. Thesignal generator contains asource ofalternating current whose frequency canbevaried byvarying theknob onthefront ofthegenerator. Then weconnect theoutput loop ofthecavity toa“detector,” which isaninstru- ment thatmeasures thecurrent froni theoutput loop. ltgives ameter reading pro- portional tothiscurrent. lfwenow measure theoutput current asafunction of thefrequency ofthesignal generator, weﬁndacurve likethatshown inFig23-10. Theoutput current issmall forallfrequencies except those verynear thefrequency w.,,which istheresonant frequency ofthe cavity. Theresonance curve isverymuch likethose wedescribed inChapter 23ofVol. l.Thewidth oftheresonance is, however, much narrower than weusually ﬁndforresonant circuits made ofinduc- tances andcapacitors; thatis,theQofthecavity isvery high. ltisnotunusual toﬁndQ‘sashigh as100,000 ormore iftheinside walls ofthecavity aremade of some material with avery good conductivity, such assilver 23-8 23-4 Cavity modes Suppose wenow trytocheck ourtheory bymaking measurements with an actual can. Wetakeacanwhich isacylinder with adiameter of3.0inches anda height ofabout 2.5inches. Thecanisﬁtted with aninput andoutput loop, as shown inFig.23-8. lfwecalculate theresonant frequency expected forthiscan according toEq(23.18), wegetthat fo=wo/211' =3010 megacycles When wesetthefrequency ofoursignal generator near 3000 megacycles andvary it slightly until weﬁnd theresonance, weobserve that themaximum output current occurs forafrequency of3050 megacycles, which isquite close tothepredicted resonant frequency, butnotexactly thesame. There areseveral possible reasons forthediscrepancy. Perhaps theresonant frequency ischanged alittle bitbecause oftheholes wehave cuttoputinthecoupling loops. Alittle thought, however. shows thattheholes should lower theresonant frequency alittle bit,sothatcannot bethereason. Perhaps there issome slight error inthefrequency calibration ofthe signal generator, orperhaps ourmeasurement ofthediameter ofthecavity isnot accurate enough. Anyway, theagreement isfairly close. Much more important issomething thathappens ifwevary thefrequency of oursignal generator somewhat further from 3000 megacycles. When wedothat wegettheresults shown inFig.23-11.Weﬁndthat, inaddition totheresonance weexpected near 3000 megacycles. there isalsoaresonance near 3300 megacycles andonenear 3820 megacycles. What dothese extra resonances mean? Wemight getacluefrom Fig.23-6. Although wehave been assuming thattheﬁrstzero of theBessel function occurs attheedge ofthecan, itcould alsobethatthesecond zero oftheBessel function corresponds totheedge ofthecan, sothatthere isone complete oscillation oftheelectric ﬁeld aswemove from thecenter ofthecanout totheedge, asshown inFig.23-12. Thisisanother possible mode fortheoscillating ﬁelds. Weshould certainly expect thecantoresonate insuch amode. But notice, thesecond zero oftheBessel function occurs atx=5.52, which isover twice aslarge asthevalue attheﬁrstzero. Theresonant frequency ofthismode should therefore behigher than 6000 megacycles. Wewould, nodoubt, ﬁndit there, butitdoesn’t explain theresonance weobserve at3300. Thetrouble isthatinouranalysis ofthebehavior ofaresonant cavity wehave considered only onepossible geometric arrangement oftheelectric andmagnetic ﬁelds. Wehave assumed thattheelectric ﬁelds arevertical andthatthemagnetic ﬁelds lieinhorizontal circles. Butother ﬁelds arepossible. Theonly requirements arethattheﬁelds should satisfy Maxwell’s equations inside thecanandthatthe electric ﬁeldshould meet thewallatright angles. Wehave considered thecase in which thetopandthebottom ofthecanareﬁat,butthings would notbecompletely different ifthetopandbottom were curved. lnfact, how isthecansupposed to know which isitstopandbottom, andwhich areitssides? 1tis,infact, possible toshow thatthere isamode ofoscillation oftheﬁelds inside thecaninwhich the electric ﬁelds gomore orlessacross thediameter ofthecan,asshown inFig.23-13. ltisnottoohard tounderstand why thenatural frequency ofthismode should benotvery diﬂcrent from thenatural frequency oftheﬁrstmode wehave considered. Suppose thatinstead ofourcylindrical cavity wehadtaken acavity which wasacube 3inches onaside. ltisclear thatthiscavity would have three different modes, butallwith thesame frequency. Amode with theelectric ﬁeld going more orlessupanddown would certainly have thesame frequency asthe mode inwhich theelectric ﬁeld wasdirected right andleft lfwenow distort the cube intoacylinder, wewillchange these frequencies somewhat. Wewould still expect them nottobechanged toomuch, provided wekeep thedimensions ofthe cavity more orlessthesame. Sothefrequency ofthemode ofFig.23-13 should notbetoodifferent from themode ofFig.23-8. Wecould make adetailed cal- culation ofthenatural frequency ofthemode shown inFig.23-13, butwewillnot dothatnow. When thecalculations arecarried through, itisfound that, forthe dimensions wehave assumed, theresonant frequency comes outveryclose tothe observed resonance at3300 megacycles. Bysimilar calculations itispossible toshow that there should bestillanother mode attheother resonant frequency wefound near 3800 megacycles Forthis 23-9CLHRENT3050 5300 A sazo ssbb 4066’ w/21r (ﬁlmnmu periocond)OUTPUT Fig. 23-11. Observed resonant fre- quencies ofacylindrical cavity. (ci) E Eo r=552c/ail 1 l\,\ — > r lb) Fig. 23-12 Ahigher-frequency mode. Fig. 23-13. Atransverse mode of thecylindrical cavity. $15‘ jQ $ Fig. 23-14. Another mode ofacy- lindrical cavity.mode, theelectric andmagnetic ﬁelds areasshown inFig.23-14. Theelectric field does notbother togoallthewayacross thecavity 1tgoes from thesides to theends, asshown. Asyouwillprobably nowbelieve, ifwegohigher andhigher infrequency we should expect toﬁndmore andmore resonances. There aremany diﬁerent modes.‘ each ofwhich willhave adifferent resonant frequency corresponding tosome par- ticular complicated arrangement oftheelectric andmagnetic ﬁelds. Each ofthese ﬁeldarrangements iscalled aresonant mode. Theresonance frequency ofeach mode canbecalculated bysolving Maxwell’s equations fortheelectric and magnetic ﬁelds inthecavity. When wehave aresonance atsome particular frequency, how canweknow which mode isbeing excited‘) One way istopoke alittle wire into thecavity through asmall hole 1ftheelectric ﬁeld isalong thewire, asinFig23-15(11), there willberelatively large currents inthewire, sapping energy from theﬁelds. andtheresonance willbesuppressed. lftheelectric ﬁeld isasshown inFig. 23-l5(b), thewire willhave amuch smaller effect. Wecould ﬁndwhich waythe ﬁeldpoints inthismode bybending theendofthewire, asshown inFig23-l5(c) Then, aswerotate thewire, there will beabigeffect when theend ofthewire is parallel toEand asmall etlect when itisiotated soastoheat90°toE. eFig. 23-15. Ashort metal wire inserted into acavity willdisturb the resonance much more when itisparallel toEthan when itisatright angles. 23-5 Cavities andresonant circuits Although theresonant cavity wehave been desciibiiig seems tobequite different from theordinary resonant circuit consisting ofaninductance anda capacitor, thetworesonant systems are,ofcourse, closely related They areboth members ofthesame family; they arejusttwoextreme cases ofelectromagnetic resonators—and there aremany intermediate cases between these two extremes. Suppose westart byconsidering theresonant circuit ofacapacitor inparallel with aninductance, asshown inFig.23—l6(a). This circuit willresonate atthefrequency ev.,=1/v’ZC. lfwewant toraise theresonant frequency ofthiscircuit. wecan dosobylowering theinductance L.Onewayistodecrease thenumber ofturns in thecoil. Wecan,however, goonly sofarinthisdirection. Eventually wewillget down tothelastturn, andwewillhave justapiece ofwire joining thetopand bottom plates ofthecondenser. Wecould raise theresonant frequency stillfurther bymaking thecapacitance smaller; however, wecanalsocontinue todecrease the inductance byputting several inductances inparallel Two one-turn inductances in parallel willhave only halftheinductance ofeach turn. Sowhen ourinductance hasbeen reduced toasingle turn, wecancontinue toraise theresonant frequency byadding other single loops from thetopplate tothebottom plate ofthecondenser. Forinstance, Fig 23—l6(b) shows thecondenser plates connected bysixsuch “single-turn inductances.” lfwecontinue toaddmany such pieces ofwire, wecan make thetransition tothecompletely enclosed resonant system shown inpart(c) oftheﬁgure, which isadrawing ofthecross section ofacylindrically symmetrical 23-10 tines ora "i_"“_ 111111; fl(___.Q___."},X\G Q) Q) ® a ‘Q ___. '' /\ I3hoo co® l ’1 I 0,, C \ - ,' ll: ft ' (1 \ 9 Q, ESorE ® ® \ I ' | _ 14—I—'—-I-1 lcll (bl (l Fig. 23-16. Resonators ofprogressively higher resonant frequencies object. Ourinductance isnow acylindrical hollow canattached totheedges of thecondenser plates. Theelectric andmagnetic ﬁelds willbeasshown inthe ﬁgure. Such anobject is,ofcourse, aresonant cavity. ltiscalled a“loaded” cavity. Butwecanstillthink ofitasanL-Ccircuit inwhich thecapacity section isthe region where weﬁnd most oftheelectric ﬁeld and theinductance section is that region where weﬁndmost ofthemagnetic ﬁeld. Ifwewant tomake thefrequency oftheresonator inFig.23—l6(c) stillhigher, wecandosobycontinuing todecrease theinductance L.Todothat, wemust decrease thegeometric dimensions oftheinductance section, forexample by decreasing thedimension liinthedrawing. As/1isdecreased, theresonant fre- quency willbeincreased Eventually, ofcourse, wewillgettothesituation in which theheight /1isjustequal totheseparation between thecondenser plates Wethen have justacylindrical can, ourresonant circuit hasbecome thecavity resonator ofFig.23-7. You willnotice that intheoriginal L-Cresonant circuit ofFig. 23-16 the electric andmagnetic ﬁelds arequite separate. Aswehave gradually modiﬁed the resonant system tomake higher andhigher frequencies, themagnetic ﬁeld hasbeen brought closer andcloser totheelectric ﬁeld until inthecavity resonator thetwo arequite intermixed Although thecavity resonators wehave talked about inthischapter have been cylindrical cans, there isnothing magic about thecylindrical shape Acanofany shape willhave resonant frequencies corresponding tovarious possible modes of oscillations oftheelectric andmagnetic ﬁelds Forexample, the“cavity” shown inFig23-17 willhave itsown particular setofresonant frequencies—although they would berather diﬁicult tocalculate. 23-ll‘Q .2) ’O /in-‘‘1.1.‘I1 C i Fig. 23-17. Another resonant cavity 24 Waveguides 24-1 Thetransmission line lnthelastchapter westudied what happened tothelumped elements ofcircuits when they were operated atvery high frequencies, andwewere ledtoseethata resonant circuit could bereplaced byacavity with theﬁelds resonating inside. Another interesting technical problem istheconnection ofoneobject toanother, sothatelectromagnetic energy canbetransmitted between them. Inlow-frequency circuits theconnection ismade with wires. butthismethod doesn't work very well athigh frequencies because thecircuits would radiate energy into allthespace around them, anditishard tocontrol where theenergy willgo.Theﬁelds spread outaround thewires; thecurrents andvoltages arenot“guided” very well by thewires. lnthischapter wewant tolook into theways that objects canbe interconnected athigh frequencies Atleast, that’s oneway ofpresenting our SUb_]€Ci. Another wayistosaythatwehave been discussing thebehavior ofwaves in freespace. Now itistime toseewhat happens when oscillating ﬁelds areconﬁned inoneormore dimensions. Wewilldiscover theinteresting new phenomenon when theﬁelds areconﬁned inonly twodimensions andallowed togofreeinthe third dimension, they propagate inwaves. These are“guided waves”—the subject ofthischapter. Webegin byworking outthegeneral theory ofthetransmission line. The ordinary power transmission linethatruns from tower totower over thecountry- sideradiates away some ofitspower. butthepower frequencies (50-60 cycles/sec) aresolowthatthislossisnotserious. Theradiation could bestopped bysurround- ingthelinewith ametal pipe, butthismethod would notbepractical forpower lines because thevoltages andcurrents used would require averylarge, expensive, andheavy pipe. Sosimple “open lines" areused. Forsomewhat higher frequencies—say afewkilocycles~radiation canal- ready beserious However. itcanbereduced byusing “twisted-pair” transmission lines. asisdone forshort-run telephone connections. Athigher frequencies, how- ever, theradiation soon becomes intolerable, either because ofpower losses or because theenergy appears inother circuits where itisn’twanted Forfrequencies from afewkilocycles tosome hundreds ofmegacycles. electromagnetic signals andpower areusually transmitted viacoaxial lines consisting ofawire inside a cylindrical “outer conductor” or“shield "Although thefollowing treatment will apply toatransmission lineoftwoparallel conductors ofanyshape, wewillcarry itoutreferring toacoaxial line. Wetake thesimplest coaxial linethathasacentral conductor, which wesup- pose isathin hollow cylinder, andanouter conductor which isanother thin cylinder onthesame axisastheinner conductor, asinFig.24-1 Webegin by ﬁguring outapproximately how thelinebehaves atrelatively low frequencies Wehave already described some ofthelow-frequency behavior when wesaid earlier that twosuch conductors hadacertain amount ofinductance perunit length oracertain capacity perunitlength. Wecan, infact, describe thelow- frequency behavior ofanytransmission linebygiving itsinductance perunit length, L1,anditscapacity perunitlength, C0. Then wecananalyze thelineas thelimiting case oftheL-Cﬁlter asdiscussed inSection 22—6. Wecanmake a ﬁlter which imitates thelinebytaking small series elements L(,Ax andsmall shunt capacities C0Ax,where Axisanelement oflength oftheline. Using our results fortheinﬁnite ﬁlter. weseethatthere would beapropagation ofelectric 24-124~l Thetransmission line 24-2 Therectangular waveguide 24-3 The cutoff frequency 24-4 Thespeed oftheguided waves 24-5 Observing guided waves 24-6 Waveguide plumbing 24-7 Waveguide modes 24-8 Another wayoflooking atthe guided waves i\§?%§;111ij; Fig. 24-1. Acocixiciltronsmissionline. WIREI El IE1“)E. /5t\V(x)i IV(x+Ax) WIRE2 \} _ x x+Ax Fig. 24-2. The currents and voltages ofcitransmission line.signals along theline. Rather than following that approach, however, wewould now rather look atthelinefrom thepoint ofview ofadliT€I‘€Il[l21l equation. Suppose that weseewhat happens attwo neighboring points along the transmission line, sayatthedistances xandx+Axfrom thebeginning ofthe line. Let’s callthevoltage difference between thetwoconductors V(x). andthe current along the“hot” conductor I(x)(seeFig.24-2). Ifthecurrent intheline isvarying, theinductance willgiveusavoltage drop across thesmall section of linefrom xtox+Axintheamount AV=V(x+Ax)—V(x) =—-L0Axg€- Or,taking thelimit asAx->0,weget 6V 615=—L0at (24.1) Thechanging current gives agradient ofthevoltage. Referring again totheﬁgure, ifthevoltage atxischanging. there must be some charge supplied tothecapacity inthatregion. Ifwetake thesmall piece of linebetween xandx+Ax,thecharge onitisq—C0AxV.Thetime rate-of- change ofthischarge isC0Axa'V/dt. butthecharge changes only ifthecurrent I(x)intotheelement isdiﬁerent from thecurrent I(x+Ax)out. Calling thediﬁer- ence AI,wehave dVAI = —COAX -dt Taking thelimit asAx—>0,weget 61 6V5}-—C0 -97- (24.2) Sotheconservation ofcharge implies thatthegradient ofthecurrent ispropor- tional tothetime rate-of-change ofthevoltage. Equations (24.1) and(24.2) arethen thebasic equations ofatransmission line. Ifwewish, wecould modify them toinclude theeffects ofresistance inthe conductors orofleakage ofcharge through theinsulation between theconductors, butforourpresent discussion wewilljuststaywith thesimple example. Thetwotransmission lineequations canbecombined by(l1lT€l'6l"ltl2lt1f1g one with respect totandtheother with respect toxandeliminating either VorI. Then wehave either 62V 62V or 2 6-5=COLD 6x2 (244) Once more werecognize thewave equation inx.Forauniform transmission line, thevoltage (and current) propagates along thelineasawave. Thevoltage along thelinemust beoftheform V(x, 1)=f(x —vt)orV(x, r)=g(x+wt). orasum ofboth. Now what isthevelocity It?Weknow thatthecoeﬁicient of the62/612 term isjust1/02, so 1»= (24.5) \/LOCO Wewillleave itforyoutoshow that thevoltage foreach wave inalineis proportional tothecurrent ofthatwave andthattheconstant ofproportionality isjust thecharacteristic impedance 20.Calling V+and1+thevoltage andcurrent forawave going intheplusx-direction, youshould get V+ = 201+. 24-2 Similary, forthewave going toward minus xtherelation is V_. : _'ZOI_. Thecharacteristic impedance-—as wefound outfrom ourﬁlter equations-is given by _ll-0 Z0 — G 1 andis,therefore, apure resistance. Toﬁnd thepropagation speed 1'andthecharacteristic impedance 20ofa transmission line, wehave toknow theinductance andcapacity perunit length. Wecancalculate them easily foracoaxial cable, sowewillseehowthatgoes. For theinductance wefollow theideas ofSection 17-8, andset%LI2 equal tothemag- netic energy which wegetbyintegrating e,,c2B2/2 over thevolume. Suppose thatthecentral conductor carries thecurrent I;then weknow thatB=I/21re0c2r, where risthedistance from theaxis. Taking asavolume element acylindrical shell ofthickness drandoflength I,wehave forthemagnetic energy U=ei-8/b —L)2lZ1rrdr2,,,21re0c2r ’ where aandbaretheradii oftheinner andouter conductors, respectively. Carry- ingouttheintegral, weget 2II b Setting theenergy equal to%LI2, weﬁnd l b Itis.asitshould be,proportional tothelength loftheline,sotheinductance per unitlength L0is _lytb/11)L0_2;-6~0c_; - (24.10) Wehave worked outthecharge onacylindrical condenser (seeSection 12-2) Now, dividing thecharge bythepotential diﬁference, weget 21re0l C ln(b/a) Thecapacity perunitlength C0isC/I. Combining thisresult with Eq.(24.10), weseethat theproduct LOCO isjustequal to1/c2. sov=1/\/LOCO isequal toc.Thewave travels down thelinewith thespeed oflight. Wepoint outthatthis result depends onourassumptions: (a)thatthere arenodielectrics ormagnetic materials inthespace between theconductors, and(b)thatthecurrents areallon thesurfaces oftheconductors (asthey Would beforperfect conductors). Wewill seelater that forgood conductors athigh frequencies, allcurrents distribute themselves onthesurfaces asthey would foraperfect conductor, sothisassump- tionisthen valid. Now itisinteresting thatsolong asassumptions (a)and(b)arecorrect, the product L()C() isequal to1/c2 foranyparallel pairofconductors—even, say,fora hexagonal inner conductor anywhere inside anelliptical outer conductor. Solong asthecross section isconstant andthespace between hasnomaterial, waves are propagated atthevelocity oflight. Nosuch general statement canbemade about thecharacteristic impedance. Forthecoaxial line, itis In(b/a)ZQ = ' 24-3 ll)‘\\ -—-->‘< / \’,,-x\ _, \ \\ \\ \ \ \ \ \ \\ \ \Z Fig. 24-3. Coordinates chosen for therectangular waveguide. °*1 Tl15 l,(0) x 5 m a T Fig. 24-4. The electric ﬁeld inthe waveguide atsome value ofz. ‘lillfltliﬁillitlil Q’! (0) Agy > Z/\-i/\ Fig. 24-5. Thez-dependence ofthe ﬁeld intheWGveguide_Thefactor I/soc hasthedimensions ofaresistance andisequal tol201r ohms. Thegeometric factor ln(b/a) depends only logarithmically onthedimensions, so forthecoaxial line—and most lines—the characteristic impedance hastypical values offrom 50ohms orsotoafewhundred ohms. 24-2 Therectangular waveguide Thenext thing wewant totalkabout seems, atﬁrstsight, tobeastriking phenomenon: ifthecentral conductor isremoved from thecoaxial line, itcanstill carry electromagnetic power. Inother words, athigh enough frequencies ahollow tube willwork justaswellasonewith wires. Itisrelated tothemysterious wayin which aresonant circuit ofacondenser andinductance getsreplaced bynothing butacanathigh frequencies. Although itmay seem tobearemarkable thing when onehasbeen thinking interms ofatransmission lineasadistributed inductance andcapacity, weall know that electromagnetic waves cantravel along inside ahollow metal pipe. Ifthepipe isstraight, wecanseethrough it!Socertainly electromagnetic waves gothrough apipe. Butwealsoknow thatitisnotpossible totransmit low-fre- quency waves (power ortelephone) through theinside ofasingle metal pipe. So itmust bethatelectromagnetic waves willgothrough iftheir wavelength isshort enough. Therefore wewant todiscuss thelimiting caseofthelongest wavelength (orthelowest frequency) thatcangetthrough apipe ofagiven size. Since the pipe isthen being used tocarry waves, itiscalled awaveguide. Wewillbegin with arectangular pipe, because itisthesimplest case to analyze. Wewillﬁrstgiveamathematical treatment andcome back later tolook attheproblem inamuch more elementary Way. Themore elementary approach. however, canbeapplied easily only toarectangular guide. Thebasic phenomena arethesame forageneral guide ofarbitrary shape, sothemathematical argument isfundamentally more sound. Ourproblem, then, istoﬁndwhat kindofwaves canexistinside arectangular pipe. Let’s ﬁrstchoose some convenient coordinates: wetake thez-axis along the length ofthepipe, andthex-andy-axes parallel tothetwosides, asshown in Fig.24-3. Weknow that when light waves godown thepipe, they have atransverse electric ﬁeld; sosuppose Welook ﬁrstforsolutions inwhich Eisperpendicular to z,saywith only ay-component, E,,.This electric ﬁeld willhave some variation across theguide; infact,itmust gotozeroatthesides parallel tothey-axis, because thecurrents andcharges inaconductor always adjust themselves sothatthere is notangential component oftheelectric ﬁeld atthesurface ofaconductor. So E,willvary with xinsome arch, asshown inFig.24-4. Perhaps itistheBessel function wefound foracavity? No,because theBessel function hastodowith cylindrical geometries. Forarectangular geometry, waves areusually simple harmonic functions, soweshould trysomething likesinkzx. Since wewant waves that propagate down theguide, weexpect theﬁeld to alternate between positive andnegative values aswegoalong in2,asinFig.24-5, andthese oscillations willtravel along theguide with some velocity I’.Ifwehave oscillations atsome deﬁnite frequency w,wewould guess thatthewave might vary with zlikeCOS(wt —kzz), ortousethemore convenient mathematical form. likee"(“"""/’>. This z-dependence represents awave travelling with thespeed v=to/kz (seeChapter 29,Vol. I). Sowemight guess that thewave intheguide would have thefollowing mathematical form: @=mmmWH@ mm ‘Let’s seewhether thisguess satisﬁes thecorrect ﬁeld equations. First. the electric ﬁeld should have notangential components attheconductors. Ourﬁeld satisﬁes thisrequirement; itisperpendicular tothetopandbottom faces andis zero atthetwosidefaces. Well, itisifwechoose k,sothatone-half acycle of 24-4 sinl<,xj'ust ﬁtsinthewidth oftheguide—that is,if kxa=1r. (24.13) There areother possibilities, likekxa=21r,31r,...,or,ingeneral, kxa =n1r, (2414) where nisanyinteger. These represent various complicated arrangements ofthe ﬁeld, butfornow let’s take only thesimplest one, where k,=11-/a, where ais thewidth oftheinside oftheguide. Next, thedivergence ofEmust bezero inthefreespace inside theguide, since there arenocharges there. OurEhasonly ay-component, anditdoesn’t change with y,sowedohave thatV-E=0. Finally. ourelectric ﬁeld must agree with therestofMaxwell’s equations in thefreespace inside theguide. That isthesame thing assaying that itmust satisfy thewave equation 0215,, a2E,, @215, 162E _> ___ __% _ i : 6x2 +dyz +622 c2012 0' (2415) Wehave toseewhether ourguess, Eq.(24.12), willwork. Thesecond derivative of E,,with respect toxisjust —k§E,, Thesecond derivative with respect toyis zero. since nothing depends onv.Thesecond derivative withrespect tozis—kfE,,, andthesecond derivative with respect totis-w2E,,. Equation (24.15) then says that 2 /<55,+kiis,-2%E,=0. Unless E,,iszeroeverywhere (which isnotveryinteresting), thisequation iscorrect if 2 2_0->2__k,+k, E;-0. (24.16) Wehave already ﬁxed /<,,sothisequation tellsusthatthere canbewaves ofthe type wehave assumed ifkzisrelated tothefrequency tosothat Eq.(2416)is satisﬁed—in other words, if k,=\/(L02/C2) ~(7l'2/(12). (24.17) Thewaves wehave described arepropagated inthez-direction with thisvalue Ofkz Thewave number kzwegetfrom Eq.(24.17) tellsus,foragiven frequency w. thespeed with which thenodes ofthewave propagate down theguide. The phase velocity is 11= (24.18) You willremember that thewavelength Aofatravelling wave isgiven by A=21rv/w, sok,isalsoequal to211-/)\,,, where >\,,isthewavelength oftheoscilla- tions along thez-direction—the “guide wavelength." Thewavelength intheguide isdifferent ,ofcourse, from thefree-space wavelength ofelectromagnetic waves ofthesame frequency. Ifwecallthefree-space wavelength X0,which isequal to 21rc/w, wecanwrite Eq.(24.17) as M=_-__-. (24.19)A”vi-or./we Besides theelectric ﬁelds there aremagnetic ﬁelds that willtravel with the wave. butwewillnotbother towork outanexpression forthem right now. Since c2V XB=6E/6!, thelines ofBwillcirculate around theregions inwhich 6E/61 islargest, thatis,halfway between themaximum andminimum ofE.The loops ofBwilllieparallel tothexz-plane andbetween thecrests andtroughs of E,asshown inFig.24-6. 24-5Y 11E, 1B\-—-- x \ )’°"~ \.l)ls gs‘ l1I1l,:'z__/ MAX \ Z Fig. 24-6. Themagnetic ﬁeld inthe waveguide. I 1 1*). O O 20 Cl Z F T Fig. 24-7. The variation ofEywith zforto<<w¢.24-3 Thecutoif frequency Insolving Eq.(24.16) forkz,there should really betworoots—one plusand oneminus. Weshould write kz=i\/((112/c2) ——(tr?/a2). (24.20) Thetwosigns simply mean thatthere canbewaves which propagate with anega- tivephase velocity (toward -2), aswellaswaves which propagate inthepositive direction intheguide. Naturally, itshould bepossible forwaves togoineither direction. Since both types ofwaves canbepresent atthesame time, there willbe thepossibility ofstanding-wave solutions. Ourequation forkzalsotellsusthathigher frequencies givelarger values of kl,andtherefore smaller wavelengths, until inthelimit oflarge co,kbecomes equal tow/c, which isthevalue wewould expect forwaves infreespace. The light we“see” through apipestilltravels atthespeed c.Butnownotice thatifwe gotoward lowfrequencies, something strange happens. Atﬁrstthewavelength getslonger andlonger, butifcugetstoosmall thequantity inside thesquare root ofEq.(24.20) suddenly becomes negative. This willhappen assoon aswgetsto belessthan rrc/a——or when A0becomes greater than 2a.Inother words, when thefrequency getssmaller than acertain critical frequency to,=rrc/a, thewave number kg(and also Ag)becomes imaginary andwehaven’t gotasolution any more. Ordowe? Who saidthatk,hastobereal? What ifitdoes come out imaginary? Ourﬁeld equations arestillsatisﬁed. Perhaps animaginary kcalso represents awave. Suppose wZSlessthan wc;then wecanwrite k,==*=1'k’, (24.21) where k’isapositive realnumber: k’=\/(7T2/dz) -(£02/C2). (24.22) Ifwenow goback toourexpression, Eq.(24.12), forE,,,wehave E,=E0sin1<,,><t»“‘“""’°"’, (24.23) which wecanwrite as E,=E0sin/<,xe*"'=e“"‘. (24.24) This expression gives anE-ﬁeld that oscillates with time ase“”‘butwhich varies with zasei"". Itdecreases orincreases with zsmoothly asarealexponent- ialInourderivation wedidn’t worry about thesources thatstarted thewaves. butthere must, ofcourse, beasource someplace intheguide. Thesignthatgoes with k’must betheonethat makes theﬁeld decrease with increasing distance from thesource ofthewaves. Soforfrequencies below we=rrc/a, waves donotpropagate down theguide; theoscillating ﬁelds penetrate intotheguide only adistance oftheorder of1/k’. Forthisreason, thefrequency weiscalled the“cutoff frequency” oftheguide. Looking atEq.(2422),weseethatforfrequencies justalittle below 0.1,,thenum- berk’issmall andtheﬁelds canpenetrate along distance intotheguide. Butif (1)ismuch lessthan wc,theexponential coefﬁcient k’isequal to1r/aandtheﬁeld diesoffextremely rapidly, asshown inFig.24-7. Theﬁelddecreases byl/einthe distance a/1r, orinonly about one-third oftheguide width. Theﬁelds penetrate very little distance from thesource. Wewant toemphasize aninteresting feature ofouranalysis oftheguided waves—the appearance oftheimaginary wave number kz.Normally, ifwesolve anequation inphysics andgetanimaginary number, itdoesn’t mean anything physical. Forwaves, however, animaginary wave number does mean something. Thewave equation isstillsatisﬁed; itonly means that thesolution gives expo- nentially decreasing ﬁelds instead ofpropagating waves. Soinanywave problem where kbecomes imaginary forsome frequency, itmeans thattheform ofthewave changes—the sinewave changes intoanexponential. 24-6 24-4 Thespeed oftheguided waves Thewave velocity wehave used above isthephase velocity, which isthespeed ofanode ofthewave; itisafunction offrequency. Ifwecombine Eqs.(24.17) and(24.18), wecanwrite C vat...= . (24.25) Forfrequencies above cutoff—where travelling waves exist—wc/to islessthan one, andv,,h,,5,. isrealandgreater than thespeed oflight. Wehave already seen in Chapter 48ofVol. Ithatphase velocities greater than light arepossible, because it isjust thenodes ofthewave which aremoving andnotenergy orinformation. In order toknow how fastsignals willtravel, wehave tocalculate thespeed ofpulses ormodulations made bytheinterference ofawave ofonefrequency with oneor more waves ofslightly different frequencies (seeChapter 48.Vol. I).Wehave called thespeed oftheenvelope ofsuch agroup ofwaves thegroup velocity: itis notw/kbutdw/dk: d1i,,,,,,,,,=E‘:. (24.26) Taking thederivative ofEq.(24.17) with respect towandinverting togetdw/dk. weﬁndthat vgmup =c\/l —(we/w)2, (24.27) which islessthan thespeed oflight. Thegeometric mean ofvp;,,,,,,, andi'g,,,,,,, isjustc,thespeed oflight: vphasevgroup =02. (24.28) This iscurious, because wehave seen asimilar relation inquantum mechanics. Foraparticle with anyvelocity—even relativistic—the momentum pandenergy Uarerelated by U2=p2c2 +m2c"‘. (24.29) Butinquantum mechanics theenergy ishw.andthemomentum isfi/7t. which is equal tohk;soEq.(24.29) canbewritten (U2 n»l2c2 2'5 =k2 +7? s or k=\/(L02/C2) -(m2c2/h2), (24.31) which looks very much likeEq.(24.17) ...Interesting! Thegroup velocity ofthewaves isalsothespeed atwhich energy istransported along theguide. Ifwewant toﬁndtheenergy ﬁow down theguide, wecangetit from theenergy density times thegroup velocity. Iftheroot mean square electric ﬁeld isE0,then theaverage density ofelectric energy ise(,E§/2. There isalso some energy associated with themagnetic ﬁeld. Wewillnotprove ithere, butin anycavity orguide themagnetic andelectric energies areequal, sothetotal electromagnetic energy density ise0E§. Thepower dU/dr transmitted bytheguide isthen dU-‘-5=¢,,E§ab1»,,,,,,,. (24.32) (Wewillseelater another. more general wayofgetting theenergy ﬂow.) 24-5 Observing guided waves Energy canbecoupled intoawaveguide bysome kind ofan“antenna.” For example, alittle vertical wire or“stub” willdo.Thepresence oftheguided waves canbeobserved bypicking upsome oftheelectromagnetic energy with alittle receiving “antenna,” which again canbealittle stub ofwire orasmall loop. 24-7 Fig 24-8 Awaveguide with adriv- ingstub andapickup probe _FROMSIGNAL /to DETECTOR GENERATOR ,-f\ /I <—I-p <—i—1\ I 1\\\ \ _ Illl1llll111 ._.__1__\\\'\I\ 11P_ / X / InFig.24-8, weshow aguide with some cutaways toshow adriving stub anda pickup “probe”. Thedriving stub canbeconnected toasignal generator viaa coaxial cable, andthepickup probe canbeconnected byasimilar cable toa detector. Itisusually convenient toinsert thepickup probe viaalong thinslot intheguide, asshown inFig.24-8. Then theprobe canbemoved back andforth along theguide tosample theﬁelds atvarious positions. Ifthesignal generator issetatsome frequency 0.1greater than thecutoff frequency w,,there willbewaves propagated down theguide from thedriving stub. These willbetheonly waves present iftheguide isinﬁnitely long, which caneffectively bearranged byterminating theguide with acarefully designed absorber insuch awaythatthere arenoreﬂections from thefarend. Then, since thedetector measures thetime average oftheﬁelds near theprobe, itwillpick upasignal which isindependent oftheposition along theguide; itsoutput will beproportional tothepower being transmitted. Ifnow thefarendoftheguide isﬁnished offinsome waythatproduces a reflected wave—as anextreme example, ifweclosed itoffwithametal plate—there willbeareflected wave inaddition totheoriginal forward wave. These twowaves willinterfere andproduce astanding wave intheguide similar tothestanding waves onastring which wediscussed inChapter 49ofVol. I.Then, asthepickup probe ismoved along theline, thedetector reading willriseandfallperiodically. showing amaximum intheﬁelds ateach loop ofthe standing wave andaminimum ateach node Thedistance between twosuccessive nodes (orloops) isjust>\,,/2. This gives aconvenient wayofmeasuring theguide wavelength. Ifthefrequency isnow moved closer to0.1,,thedistances between nodes increase. showing thatthe guide wavelength increases aspredicted byEq.(24.19). Suppose now thesignal generator issetatafrequency just alittle below (ac Then thedetector output willdecrease gradually asthepickup probe ismoved down theguide Ifthefrequency issetsomewhat lower, theﬁeld strength will fallrapidly, following thecurve ofFig. 24-7. andshowing that waves arenot propagated. 24-6 Waveguide plumbing Animportant practical useofwaveguides isforthetransmission ofhigh- frequency power, as,forexample, incoupling thehigh-frequency oscillator or output ampliﬁer ofaradar settoanantenna. Infact, theantenna itself usually consists ofaparabolic reﬂector fedatitsfocus byawaveguide ﬂared outatthe endtomake a“horn” thatradiates thewaves coming along theguide. Although high frequencies canbetransmitted along acoaxial cable, awaveguide isbetter fortransmitting large amounts ofpower. First. themaximum power thatcanbe transmitted along alineislimited bythebreakdown oftheinsulation (solid orgas) between theconductors. Foragiven amount ofpower, theﬁeld strengths ina guide areusually lessthan they areinacoaxial cable, sohigher powers canbe transmitted before breakdown occurs. Second, thepower losses inthecoaxial cable areusually greater than inawaveguide. Inacoaxial cable there must beinsulating material tosupport thecentral conductor, and there isanenergy loss inthis material—particularly athigh frequencies. Also, thecurrent densities onthe central conductor arequite high, andsince thelosses goasthe.)(]Li(I/(5 ofthecurrent density, thelower currents that appear onthewalls oftheguide result inlower 24-8 ~51‘: /O RESONANT FLANGE CAVITY GUIDE \b Fig. 24-9. Sections ofwaveguide connected Fig. 24-lO. AlOW-lOss connection between with ﬂanges. twosections ofwaveguide. energy losses. Tokeep these losses toaminimum, theinner surfaces oftheguide areoften plated withamaterial ofhigh conductivity, such assilver. The problem ofconnecting a“circuit” with waveguides isquite diﬂerent from thecorresponding circuit problem atlowfrequencies, andisusually called microwave “plumbing.” Many special devices have been developed forthepur- pose. Forinstance, twosections ofwaveguide areusually connected together by means ofﬂanges, ascanbeseen inFig.24-9. Such connections can, however. cause serious energy losses, because thesurface currents must ﬂowacross thejoint which may have arelatively high resistance. One waytoavoid such losses isto make theﬂanges asshown inthecross section drawn inFig.24-10. Asmall space isleftbetween theadjacent sections oftheguide, andagroove iscutinthefaceof oneoftheﬂanges tomake asmall cavity ofthetype shown inFig.23-l6(c) The dimensions arechosen sothatthiscavity isresonant atthefrequency being used This resonant cavity presents ahigh “impedance” tothecurrents, sorelatively little current ﬂows across themetallic joints (atainFig.24-10). Thehigh guide currents simply charge anddischarge the“capacity” ofthegap(atbintheﬁgure), where there islittledissipation ofenergy. Suppose youwant tostop awaveguide inawaythatwon’t result inreﬂected waves. Then youmust putsomething attheendthatimitates aninﬁnite length of guide. You need a“termination” which actsfortheguide likethecharacteristic impedance does foratransmission line-—something thatabsorbs thearriving waves without making reﬂections. Then theguide willactasthough itwent onforever Such terminations aremade byputting inside theguide some wedges ofresistance material carefully designed toabsorb thewave energy while generating almost noreﬂected waves. Ifyouwant toconnect three things together—for instance, onesource to twodifferent antennas—then youcanusea“T”liketheoneshown inFig24-ll. Power fedinatthecenter section ofthe“T"willbesplit andgooutthetwoside arms (and there mayalsobesome reﬂected waves). Youcanseequalitatively from thesketches inFig.24-12 thattheﬁelds would spread outwhen they gettothe endoftheinput section andmake electric ﬁelds thatwillstart waves going outthe twoarms. Depending onwhether electric ﬁelds intheguide areparallel orper- pendicular tothe“top” ofthe“T,” theﬁelds atthejunction would beroughly asshown in(a)or(b)ofFig.24-12. Finally, wewould liketodescribe adevice called an“unidirectional coupler," which isvery useful fortelling what isgoing onafter youhave connected acompli- cated arrangement ofwaveguides. Suppose youwant toknow which waythe waves aregoing inaparticular section ofguide-you might bewondering, for instance, whether ornotthere isastrong reﬂected wave. The unidirectional coupler takes outasmall fraction ofthepower ofaguide ifthere isawave going oneway, butnone ifthewave isgoing theother way. Byconnecting theoutput ofthecoupler toadetector, youcanmeasure the“one-way” power intheguide 24-9Fig. 24-1 l.Awaveguide "T." (The ﬂanges have plastic endcaps tokeep the inside clean while the"T"isnotbeing used.) E <——-— ————> V V lv (<1) E o oooO6 oo tlzillf Fig. 24-l2. The electric ﬁelds ina waveguide "T" fortwo possible ﬁeld orientations. 0 /%>‘ //// TT90 o F5,/\ %\./ 4. A C\ \ D4:74 ° \_ W\\_ AK :?4-;‘q B. / O/// Fig 24-l 3. Aunidirectional coupler. Ar (0) " Eyt > X (bl Fig. 24-14. Another possible varia tionofE,with x.Figure 24-13 isadrawing ofaunidirectional coupler; apiece ofwaveguide ABhasanother piece ofwaveguide CDsoldered toitalong oneface. Theguide CDiscurved away sothatthere isroom fortheconnecting ﬂanges. Before the guides aresoldered together, two(ormore) holes have been drilled ineach guide (matching each other) sothat some oftheﬁelds inthemain guide ABcanbe coupled intothesecondary guide CD. Each oftheholes actslikealittle antenna thatproduces awave inthesecondary guide. Ifthere were only onehole, waves would besentinboth directions andwould bethesame nomatter which waythe wave wasgoing intheprimary guide. Butwhen there aretwoholes with asepara- tionspace equal toone-quarter oftheguide wavelength, theywillmake twosources 90°outofphase Doyouremember thatweconsidered inChapter 29ofVollthe interference ofthewaves from twoantennas spaced A/4apart andexcited 90° outofphase intime? Wefound thatthewaves subtract inonedirection andadd intheopposite direction Thesame thing willhappen here. Thewave produced intheguide CDwillbegoing inthesame direction asthewave inAB. Ifthewave intheprimary guide istravelling from Atoward B,there willbe awave attheoutput Dofthesecondary guide. Ifthewave intheprimary guide goes from Btoward A,there willbeawave going toward theendCofthe secondary guide. This endisequipped with atermination. sothatthiswave 1Sabsorbed and there isnowave attheoutput ofthecoupler 24-7 Waveguide modes Thewave wehave chosen toanalyze isaspecial solution ofthe ﬁeldequations. There aremany more. Each solution iscalled awaveguide “mode ”Forexample, ourx-dependence oftheﬁeld wasjust one-half acycle ofasinewave. There isan equally good solution with afullcycle, then thevariation ofE,,with xisasshown inFig24-14 Thek,forsuch amode istwice aslarge, sothecutoff frequency is much higher. Also, inthewave westudied Ehasonly ay-component, butthere areother modes with more complicated electric ﬁelds. Iftheelectric ﬁeld has components only inxandy—so that thetotal electric ﬁeld isalways attight angles tothe2-direction—-the mode iscalled a“transverse electric” (orTE)mode. Themagnetic ﬁeld ofsuch modes willalways have az-component. Itturns out thatifEhasacomponent inthez-direction (along thedirection ofpropagation). then themagnetic ﬁeld willalways have only transverse components. Sosuch ﬁelds arecalled transverse magnetic (TM) modes. Forarectangular guide, all theother modes have ahigher cutoff frequency than thesimple TEmode wehave described. Itis,therefore, poss1ble—and usual-—to useaguide with afrequency justabove thecutolf forthislowest mode butbelow thecutoﬂ frequency forall theothers, sothatjust theonemode ispropagated. Otherwise. thebehavior gets complicated anddiﬂicult tocontrol. 24-8 Another wayoflooking attheguided waves Wewant now toshow youanother wayofunderstanding why awaveguide attenuates theﬁelds rapidly forfrequencies below thecutolf frequency (0,.Then you will have amore “physical” idea ofwhy thebehavior changes sodrastically between lowandhigh frequencies Wecandothisfortherectangular guide by analyzing theﬁelds interms ofreﬂections—or images—in thewalls oftheguide. Theapproach only works forrectangular guides, however, that’s whywestarted with themore mathematical analysis which works, inprinciple. forguides ofany shape. Forthemode wehave described. thevertical dimension (iny)hadnoeffect, sowecanignore thetopandbottom oftheguide andimagine thattheguide is extended indeﬁnitely inthevertical direction. Weimagine then that theguide justconsists oftwovertical plates with theseparation u. Let’s saythatthesource oftheﬁelds isavertical wire placed inthemiddle of theguide, with thewire carrying acurrent that oscillates atthefrequency 0:. Intheabsence oftheguide walls such awire would radiate cylindrical waves 24-10 Now weconsider thattheguide walls areperfect conductors. Then, justasin electrostatics, theconditions atthesurface willbecorrect ifweaddtotheﬁeld of thewire theﬁeld ofoneormore suitable image wires. Theimage ideaworks just aswellforelectrodynamics asitdoes forelectrostatics, provided, ofcourse, that wealso include theretardations. Weknow that istrue because wehave often seenamirror producing animage ofalight source. And amirror isjusta“perfect” conductor forelectromagnetic waves with optical frequencies. Now let'stake ahorizontal cross section, asshown inFig.24-15, where W1 andW2arethetwoguide walls andS0isthesource wire. Wecallthedirection of thecurrent inthewirepositive. Now ifthere were only onewall, sayW1,wecould remove itifweplaced animage source (with opposite polarity) attheposition marked S1.Butwith both walls inplace there willalsobeanimage ofS0inthe wall W2,which weshow astheimage S2.This source, too,willhave animage in W1,which wecallS3.Now both S1andS3willhave images inW2atthepositions marked S4andS6,andsoon. Forourtwoplane conductors with thesource halfway between, theﬁelds arethesame asthose produced byaninﬁnite lineof sources, allseparated bythedistance a.(Itis,infactjustwhat youwould seeif youlooked atawire placed halfway between twoparallel mirrors.) Fortheﬁelds tobezero atthewalls, thepolarity ofthecurrents intheimages must alternate from oneimage tothenext. Inother words, they oscillate 180° outofphase. Thewaveguide ﬁeld is,then, justthesuperposition oftheﬁelds ofsuch aninﬁnite setoflinesources. Weknow thatifweareclose tothesources, theﬁeld isvery much likethe static ﬁelds. Weconsidered inSection 7-5thestatic ﬁeld ofagridoflinesources andfound thatitisliketheﬁeld ofacharged plate except forterms thatdecrease exponentially with thedistance from thegrid. Here theaverage source strength iszero, because thesignalternates from onesource tothenext. Any ﬁelds which exist should falloffexponentially with distance. Close tothesource, weseethe ﬁeld mainly ofthenearest source; atlarge distances, many sources contribute and their average effect iszero. Sonowweseewhythewaveguide below cutoff fre- quency gives anexponentially decreasing ﬁeld. Atlowfrequencies, inparticular, thestatic approximation isgood, anditpredicts arapid attenuation oftheﬁelds with distance. Now wearefaced with theopposite question: Why arewaves propagated atall? That isthemysterious part! Thereason isthatathigh frequencies the retardation oftheﬁelds canintroduce additional changes inphase which cancause theﬁelds oftheout-of-phase sources toaddinstead ofcancelling. Infact, in Chapter 29ofVol. Iwehave already studied, just forthisproblem, theﬁelds generated byanarray ofantennas orbyanoptical grating. There wefound that when several radio antennas aresuitably arranged, they cangiveaninterference pattern thathasastrong signal insome direction butnosignal inanother. Suppose wegoback toFig.24-15 andlook attheﬁelds which arrive ata large distance from thearray ofimage sources. Theﬁelds willbestrong only in certain directions which depend onthefrequency—only inthose directions for which theﬁelds from allthesources addinphase. Atareasonable distance from thesources theﬁeldpropagates inthese special directions asplane waves. Wehave sketched such awave inFig.24-16, where thesolid lines represent thewave crests andthedashed lines represent thetroughs. Thewave direction willbetheone forwhich thedifference intheretardation fortwoneighboring sources tothecrest ofawave corresponds toone-half aperiod ofoscillation. Inother words, the difference between r2andr0intheﬁgure isone-half ofthefree-space wavelength: f2 - f0 :: Q?- Theangle 6isthen given by sin0= (24.33) There is,ofcourse, another setofwaves travelling downward atthesymmetric angle with respect tothearray ofsources. Thecomplete waveguide ﬁeld (nottoo 24-11s5.- San \ 's"1'>'ii‘i<%EsSI0-/ W1 INE3,,-/Iéouncs Q \0* WAVEGUIDE/ W2s2._ IMAGE SOURCES 540+ $60- Fig. 24-l5. The line source $0be- tween theconducting plane walls W1 and W2. Thewalls canbereplaced by theinﬁnite sequence ofimage sources. U)OI00+ \\//” //. /\(\// //\\o"//3//.// /./\/\Ny/\///\./ ./\‘/ /$9//e/ >/<55K///// /// /////(\\ /( \, , 530+ \ \\ \ 510- /3’’ )/ ‘\ . s \ =\’§> O \ VC $0 Q / S2to/2 , \ mo \ \ \ \ \\ 3\3.\ 52”‘ \ \\ \ \\ \ \ 55"‘ \ \ \\ \\ \ \ \ \ \ Q \ \ \ Fig. 24-16. One set ofcoherent waves from anarray oflinesources. 330+ / /\ \ \ \/\ /\<,\/ /\\//\s,-- w,\~ .\. \/\ /\ \ So» /\A\8/ C \/ __X\//// /W2 \/i\\ \/\ \ S2“ /\ \\Xa /\\ \ / \\540+ g \ Fig. 24-l7. Thewaveguide ﬁeld can beviewed asthe superposition oftwo trains ofplane waves.close tothesource) isthesuperposition ofthese twosetsofwaves, asshown in Fig.24-17. Theactual ﬁelds arereally likethis, ofcourse, only between thetwo walls ofthewaveguide. Atpoints likeAandC,thecrests ofthetwowave patterns coincide, andthe ﬁeld willhave amaximum, atpoints likeB,both waves have their peak negative value, andtheﬁeld hasitsminimum (largest negative) value. Astime goes on theﬁeld intheguide appears tobetravelling along theguide with awavelength X0,which isthedistance from AtoC.That distance isrelated to6by cos0=2-Q- (2434) 11Using Eq.(24.33) for6,wegetthat )‘° -iii. (2435) ),IA_ I”cos6 WAG//2a)2 which isjustwhat wefound inEq(2419) Now weseewhy there isonly wave propagation above thecutoff frequency 11.1,, lfthe free-space wavelength islonger than 211,there isnoangle where thewaves shown inFig 24-16 canappear. The necessary constructive interference appears suddenly when >\1,drops below 2a,orwhen (.0goes above 0:1,-1r('/cl. Ifthefrequency 1Shigh enough, there canbetwo ormore possible directions inwhich thewaves willappear. Forourcase, thiswillhappen ifA1,<§a In general, however, itcould also happen when >111<(1.These additional waves correspond tothehigher guide modes wehave mentioned. Ithasalsobeen made evident byouranalysis why thephase velocity ofthe guided waves isgreater than candwhythisvelocity depends oncuAscoischanged. theangle ofthefreewaves ofFig.24-16 changes, andtherefore sodoes thevelocity along theguide. Although wehave described theguided wave asthesuperposition ofthe ﬁelds ofaninﬁnite array oflinesources, youcanseethatwewould arrive atthesame result ifweimagined twosetsoffree-space waves being continually reﬂected back and forth between two perfect m1rrors—remen1bering that areﬂection means a reversal ofphase. These setsofreﬂecting waves would allcancel each other unless they were going atjust theangle 0given inEq.(2433) There aremany ways of looking atthesame thing. 24-12 25 Electrodynamics inRelativistic Notation 25-1 Four-vectors Wenow discuss theapplication ofthespecial theory ofrelativity toelectro- dynamics. Since wehave already studied thespecial theory ofrelativity inChapters 15through 17ofVol. I,wewilljustreview quickly thebasic ideas. Itisfound experimentally thatthelaws ofphysics areunchanged ifwemove with uniform velocity. You can’t tellifyouareinside aspaceship moving with uniform velocity inastraight line, unless youlook outside thespaceship, orat least make anobservation having todowith theworld outside. Any truelawof physics wewrite down must bearranged sothatthisfactofnature isbuilt in. Therelationship between thespace andtime oftwosystems ofcoordinates, one,S’,inuniform motion inthex-direction with speed 1'relative totheother, S, isgiven bytheLoren/z transformation." I’= »y’=y,\/1—09 (25.1) x—1'tX'=“:i:* Z'=Z- \[l—Zi2 Thelaws ofphysics must besuch thatafter aLorentz transformation, thenew form ofthelaws looks justliketheoldform. This isjustliketheprinciple that thelaws ofphysics don't depend ontheorientation ofourcoordinate system. In Chapter llofVolI,wesawthatthewaytodescribe mathematically theinvariance ofphysics with respect torotations wastowrite ourequations interms ofvectors. Forexample. ifwehave twovectors A:(Ara A1/1 /4:) and B:(Bn B1/1 B2)» wefound thatthecombination A~B =A,B, +A,,B,, +AZBZ wasnotchanged ifwetransformed toarotated coordinate system. Soweknow thatifwe have ascalar product likeA-Bonboth sides ofan equation, theequation willhave exactly thesame form inallrotated coordinate systems. Wealsodis- covered anoperator (seeChapter 2), V;(gee),OX 6y dZ which, when applied toascalar function, gave three quantities which transform justlikeavector With thisoperator wedeﬁned thegradient, ‘andincombination with other vcctois, thedivergence andtheLaplacian. Finally wediscovered that bytaking sums ofcertain products ofpairs ofthecomponents oftwovectors we could getthree newquantities which behaved likeanewvector. Wecalled itthe cross product oftwovectors Using thecross product with ouroperator Vwethen deﬁned thecurlofavector Since wewillbereferring back towhat wehave done invector analysis. we have putinTable 25-1 asummary ofalltheimportant vector operations in three dimensions thatwehave used inthepast. Thepoint isthatitmust bepossible towrite theequations ofphysics sothatboth sides transform thesame wayunder 25-I25-1 Four-vectors 25-2 Thescalar product 25-3 Thefour-dimensional gradient 25-4 Electrodynamics in four-dimensional notation 25-5 Thefour-potential ofa moving charge 25-6 Theinvariance oftheequations ofelectrodynamics Inthischapter: c=1 | Review" Chapter 15,Vol. I,The Special Theory ofRelatlvit y Chapter I6,Vol. I,Rela- tivistic Energy and M0- mentnm Chapter 17,Vol. I,Space- Time Chapter 13.Vol. ll,Mag- netosluttcs Table 25-1 Theimportant quantities andoperations ofvector analysis inthree dimensions Deﬁnition ofa vector Scalar product DllT€f€ﬂ[lal vector operator Gradient Divergence Laplacian Cross product CurlA A V=(Ax, Au, A2) -B We V V A V-A .V: XB XAV2rotations. Ifonesideisavector, theother sidemust alsobeavector, andboth sides willchange together inexactly thesame wayifwerotate ourcoordinate sys- tem Similarly, ifonesideisascalar, theother sidemust alsobeascalar, sothat neither sidechanges when werotate coordinates, andsoon. Now inthecase ofspecial relativity, time andspace areinextricably mixed, andwemust dotheanalogous things forfourdimensions Wewant ourequations toremain thesame notonly forrotations, butalso foranyinertial frame. That means that ourequations should beinvariant under theLorentz traiisformation ofequations (25.1). Thepurpose ofthischapter istoshow youhow thatcanbe done. Before wegetstarted, however, wewant todosomething thatmakes our work aloteasier (and saves some confusion) And thatistochoose ourunits of length andtime sothatthespeed oflight cisequal tolYou canthink ofitas taking ourunit oftime tobethetime that zttakes‘ //g/1! I0goonemeter (which is about 3><lO_‘"' sec) Wecaneven callthistime unit“one meter." Using this unit, allofourequations willshow more clearly thespace-time syninictry Also, allthec'sWllldisappear from ourrelativistic equations. (Ifthisbothers you, youcanalways putthec'sback intoanyequation byreplacing every Ibycr,or.in general, bysticking inacwherever itISneeded tomake thedimensions ol‘the equations come outright.) With this groundwork weareready tobegin Our program istodointhefourdimensions ofspace-time allofthethings wedidwith vectors forthree dimensions. Itisreally quite asimple game, weJustwork by analogy Theonly realcomplications isthenotation (We’ve already used upthe vector symbol forthree dimensions) andoneslight twist ofsigns First, byanalogy with vectors inthree dimensions, wedeﬁne afour-vucmr as asetofthefourquantities (1,,ur,a,,,and(13,,which transform likeI,X,1".and2when wechange toamoving coordinate system. There areseveral (lllT€I'Cﬂl. notations people useforafour-vector; wewillwrite UM,bywhich wemean thegroup ol'four numbers (11,,u,,u,,,uz)—in other words, thesubscript /.1cantake ontheFour “values” z,x,y,zItwillalsobeconvenient, attimcs toindicate thethree space components byathree-vector, likethis: ll“=((1,,a) Wehave already encountered onefour-vector, which consists oftheenergy andmomentum ofaparticle (Chapter l7,Vol. I).Inournewnotation wewrite In=(E41), (25-2) which means that thetour-vector /1,,ismade upoftheenergy Eand thethree components ofthethree-vector pofaparticle. ltlooks asthough thegame isreally very siniple—for each three-vector in physics allwehave todoisﬁndwhat theremaining component should be,andwe have afour-vector Toseethatthisisnotthecase, consider thevelocity vector with components _dz_8;.P_dx P_ Iat’"di The question is:What isthetime component‘? Instinct should give theright answer. Since four-vectors arelike1,x,y,z,wewould guess thatthetime coin- ponent is (IPgzjgzl. T/iis ISwrong Thereason isthat the1ineach denominator isnotaninvariant when wemake aLorentz transformation Thenuinerators have theright behavior tomake afoul-vector, butthedrinthedenominator spoils things; itisunsyninictric andisnotthesame intwodifferent systems. Itturns outthatthefour “velocity” components which wehave written down willbecome thecomponents ofaFour-vector ifwejustdivide by\/l~Y5.We canseethatthatistruebecause it‘westart with themonientum four-vector /1=(E,p) =——LLi~ » (253M <\/l ~—/\—’ \/l —~n‘-’ ) 25—2 anddivide itbytherestmass ma.which isaninvariant scalar infour diinensionv, wehave _/2:._.___1,_,__v_:_ , (244) "7" \/l -—U2\/1 —I12 which must stillbeafour-vector. (Dividing byaninvariant scalar doesn't change thetransformation properties )Sowecandefine the“velocity foiii-vector” upby I ——'L—;—% Z /l 1/ll; U2 s ll,’ 1* U2 s N U (25.5) uz: ! Llz=—————-s \/1—-02 \/l —212 Thefour-velocity isauseful quantity; wecan, foiinstance, write pl‘ 2 /’l’l()ll#. This isthetypical sortofform anequation which isrelativistically correct must have; each side isafour-vector. (The right-hand side isaninvariant times a four-vector, which isstillafour-vector.) 25-2 Thescalar product ltisanaccident oflife, ifyouwish, that under coordinate rotations the distance ofapoint from theorigin does notchange. This means mathematically that r2=xi+yg+22isaninvariant lnother words, after arotation r’2=r“),or X/2 + V/2 + Z/2 :X2 +VV2 + Z2- Now thequestion is"lsthere asimilar quantity which isinvariant under the Lorentz transformation" There is.From Eq.(25.1) youcanseethat 1) ; tI/.. __xi. =,2___ X1 That ispiclty nice, except thatitdepends onaparticular choice ofthex-direction Wecanﬁxthatupbysubtracting yzand22.Then anyLorentz transformation pliisarotation Wlllleave thequantity unchanged. Sothequantity which l\anal- agous tor“)forfourdimensions, inthree dimensions is Ia__X2_ya_Z2_ ItISaninvariant under what iscalled the“complete Lorentz group’"—which means fortransformation ofboth translations atconstant velocity androtations. Now since thisinvariance isanalgebraic matter depending only onthe transforniation rules ofEq(25.l)~plus rotations—it istrueforanyfour-vector (bydeﬁnition they alltransform thesame). Soforafour-vector [JMwehave that (122—-a',2—a_§,2—aQ2=(1?—af—a5—af. Wewillcallthisquantity thesquare of“the length" ofthefour-vector a,,(Some- times people change thesign ofalltheterms andcallthelength ai—l—af—l- af—af,soyou’ll have towatch out) Now ifwehave twovectors a,,and bu,their corresponding components transform inthesame way, sothecombination a,b, —all), —a,,b,, —-agbz isalsoaninvariant (scalar) quantity. (We have infact already proved thisin Chapter 17ofVol. l.)Clearly thisexpression isquite analogous tothedotproduct forvectors. Wewill, infact, callitthedotproduct orscalar product oftwofour- vectors ltwould seem logical towrite itasKIMhp,soitwould looklikeadotprod- uct But.unhappily. it’snotdone thatway; itisusually written without thedot. 25-3 Fg 25-1 The reaction P+P—> 3P+l3viewed intheloborcitory and 0/ bl C, CMsystems Theincident proton issup- pp pp pp posed tohove rustbarely enough energy ."i‘—’ I '-_* tomake thereaction go Protons ore denoted bysolid circles, ontiprotons, bySowewillfollow theconvention andwrite thedotproduct simply asa,,b,,. So, bydeﬁnition, a,,b,, =aib, —axln —a,,b,, —a,/i, (25.7) Whenever youseetwoidentical subscripts together (wewilloccasionally have touse1/orsome other letter instead oftt)itmeans thatyouaretotake thefour products andsum, remembering theminus sign fortheproducts ofthespace components. With thisconvention theinvariance ofthescalar product under a Lorentz transformation canbewritten as IV__a“b# —a,,b,,. Since thelastthree terms in(25.7) arejustthescalar dotproduct inthree dimensions, itisoften more convenient towrite Ugh” =(lib; “—a‘ ltisalso obvious that thefour-dimensional length wedescribed above canbe written asa,.a,,: _ 2 2 2 2_ 2a,,a,, —at—a,—ay—az—(1,,—a-a. (25.8) Itwillalsobeconvenient tosometimes write thisquantity asaf: affEaua“. Wewillnow give youanillustration oftheusefulness offour-vector dot products. Antiprotons (P)areproduced inlarge accelerators bythereaction P+P-+P+P+P+ P. That is,anenergetic proton collides with aproton atrest(forexample, inahy- drogen target placed inthebeam), andiftheincident proton hasenough energy, aproton-antiproton pairmaybeproduced, inaddition tothetwooriginal protons.* Thequestion is:How much energy must begiven totheincident proton tomake thisreaction energetically possible” Theeasiest waytogettheanswer istoconsider what thereaction looks like inthecenter~of-mass (CM) system (seeFig.25-1). We'll calltheincident proton aanditsfour-momentum pf}Similarly, we'll callthetarget proton I)anditsfour- BEFORE AFTER 0 b cPu FL P -—---CENTER-OF-MASSSYSTEM LABORATORYSYSTEM *You may wellask: Why notconsider thereactions P+P~P+P+E oreven _ P4-P-+P4-P which clearly require lessenergy‘? Theanswer isthat aprinciple called ('UIlS€I‘V(llI()l1 of baryiins tells usthequantity “number ofprotons minus number ofantiprotons" cannot change. This quantity is2ontheleftsideofourreaction. Therefore, ifwewant an antiproton ontheright side, wemust have also three protons (orother baryons). 25-4 momentum pf.Iftheincident proton hasjustbarely enough energy tomake the reaction go.theﬁnal state—the situation after thecollision—will consist ofa glob containing three protons andanantiproton atrestintheCMsystem If theincident energy were slightly higher. theﬁnal state particles would have some kinetic energy andbemoving apart; iftheincident energy were slightly lower, there would notbeenough energy tomake thefour particles Ifwecallpﬂthetotal four-momentum ofthewhole glob intheﬁnal state, conservation ofenergy andmomentum tellsusthat 11"+Pb=11‘.and E“+Eb=E”. Combining these twoequations, wecanwrite that pg+pi=pg. (25.9) Now theimportant thing isthatthisisanequation among four-vectors, and is,therefore, true inanyinertial frame. Wecanusethisfacttosimplify our calculations. Westart bytaking the“length” ofeach sideofEq.(259);they are, ofcourse, alsoequal. Weget 013+pf1><pZ+p{1>=pips. (25.10) Since pﬁpﬂ isinvariant, wecanevaluate itinanycoordinate system. IntheCM system, thetime component ofpﬂistherestenergy offour protons, namely 4M, andthespace partpiszero; sopﬁ=(4M, 0).Wehave used thefactthatthe restmass ofanantiproton equals therestmass ofaproton, andwehave called thiscommon mass M. Thus, Eq.(25.10) becomes 1>;‘p:+2rZ11lZ+ pfipii’=16M? (25.11) Now pf}/)1,‘ andp,',’/2,1’ arevery easy, SIHCC the“length” ofthemomentum four-vector ofanyparticle is_]UStthemass oftheparticle squared: 17.111)/1 :E2_P2 :M2- This canbeshown bydirect calculation or,more cleverly, bynoting that fora particle at!‘(’.\l12,,=(M,0),sopypﬂ =M3 Butsince itisaninvariant, itisequal toM2inanyframe. Using these results inEq.(25.11), wehave Zpﬂpfj =l4M2 or pjpﬁI7M2. (25.12) Now wecanalso evaluate pﬂpfj inthelaboratory system. Thefour-vector pffcanbewritten (E”.p"), while pf]=(M,0),since itdescribes aproton atrest. Thus, pjfpl,’ must also beequal toME“, andsince weknow thescalar product is aninvariant thismust benumerically thesame aswhat wefound in(25.12). So wehave that E"=7M, which istheresult wewere after Thetotal energy oftheinitial proton must be atleast 7M(about 6.6Gevsince M=938Mev) or,subtracting therestmass M, the/\IIl('/l(' energy must beatleast 6M(about 5.6Gev). TheBevatron accelerator atBerkeley wasdesigned togiveabout 62Gevofkinetic energy totheprotons it accelerates, inorder tobeabletomake antiprotons Since scalar products areinvariant. they arealways interesting toevaluate. What about the“length” ofthefour-velocity u,,u,,‘7 1 112 __ 2__ 2____________ _____* =:uﬂuu ~u) u—1_U2 1__U2 l. Thus, L1,,istheLl!11ff0ur‘-veC!0l" Z5—5 25—3 Thefour-dimensional gradient Thenext thing thatwehave todiscuss isthefour-dimensional analog ofthe gradient. Werecall (Chapter 14,Vol. l)that thethree differential operators 6/6x, 6/6y, 6/62 transform likeathree-vector andarecalled thegradient. The same scheme ought towork infour dimensions; thatis,wemight guess thatthe four-dimensional gradient should be(6/6!, 6/6x, 6/6y, 6/82). T/HS‘iswrong. Toseetheerror, consider ascalar function ¢which depends only onxand1. Thechange in¢,ifwemake asmall change Arintwhile holding xconstant, is A¢=59$At. (25.13) Ontheother hand, according toamoving observer, _E IE ,Ad)-ax,Ax—l—at,Al. Wecanexpress Ax’andA1’interms ofAtbyusing Eq(25.1) Remembering thatweareholding xconstant, sothatAx=0,wewrite Ax’:___"~x,. At’=i—-\/1-122 \/1-112 _£41 _ 2' 62 ___At Ad’'6x’<\/j*_hﬁ At)+aw<\/T;?> _(931_.252)___“-.Taw Ox’\/Tip-2Thus, Comparing thisresult with Eq.(25.13), welearn that 6¢____‘ id:_.-s°25?T\/7:32 <61’ L6x’> (25'14) Asimilar calculation gives 8¢_ 1 6¢ 64>W_Vi (M-1»5). (25.15)6x 1172 dx’ ' Now wecanseethatthegradient israther strange. Theformulas forxandI interms ofx’andI’[obtained bysolving Eq.(251)]are: t_t'+vx’ X_x’-l—vt’ \/l—v2 \/l—1>é This isthewayafour-vector must transform. ButEqs. (25.14) and(25l5)have acouple ofsigns wrong‘ Theanswer isthatinstead oftheincorrect (0/81, V).wemust define thefour- dimensional gradient operator, which wewillcallV”,by 6 8 6 6 6V,,-(&,—\">_<Iﬁ,—-5}i—5y,—O;)- (25.16) With thisdeﬁnition, thesign diﬁiculties encountered above goaway, and V) behaves as.1four-vector should. (It'srather awkward tohave those minus signs, butthat's thewaytheworld is.)Ofcourse, what itmeans tosaythatTH“behaves likeafour-vector” issimply thatthefour-gradient ofascalar isafour-vector. lf ¢isatruescalar invariant ﬁeld (Lorentz. invariant) then Vp_¢isafour-vector ﬁeld Allright, now that wehave vectors, gradients, anddotproducts, thenext thing istolook foraninvariant which isanalogous tothedivergence ofthree- dimensional vector analysis. Clearly, theanalog istoform theexpression V,,b,,, where buisafour-vector ﬁeld whose components arefunctions oi‘space andtime. 25-6 Wedeﬁne therliieige/ice ofthefour-vector by=([1,,b)asthedotproduct of V)and1),): ') 6 6 6 Vt”‘it”‘“<‘be“l‘552'"'(“ale 0 —;)’Ibt'l"V by(25.17) where V-bistheordinary three-divergence ofthethree-vector b.Note thatone hastobecareful with thesigns. Some oftheminus signs come from thedeﬁnition ofthescalar product. Eq.(25.7); theothers arerequired because thespace coin- ponents ofV)are-6/6x, etc, asinEq.(25.16) Thedivergence asdeﬁned by (2517)isaninvariant and gives thesame answer inallcoordinate systems which differ byaLorentz transformation. Let’s look ataphysical example inwhich thefour-divergence shows tip Wecanuseittosolve theproblem oftheﬁelds around amoving wire Wehave already seen (Section l3~7) that theelectric charge density pandthecurrent densityj form afour-vectorj,, =(p.j). lfanuncharged wire carries thecurrent j,,then inaframe moving pastitwith velocity (1(along x),thewire willhave the charge and current density [obtained from theLorentz transformation Eqs. (25.l)] asfollows: jr pr *7)/I ’ I-1 \/l_:Ttt3 \/1—U2 These kll‘€JUSl what wefound inChapter 13Wecanthen usethese sources inMaxwelfs equation int/iemoving system toﬁndtheﬁelds. Thecharge conservation law, Section 13-2, also takes onasimple form in thefotir-vector notation. Consider thefourdivergence of/,,: .0 .v,,),,=-5+v-1. (25.18) Thelawoftheconservation ofcharge says thattheoutflow ofcurrent perunit voltiine must equal thenegative rateofincrease ofcharge density. Inother words, that _ 6V-1: Putting thisintoEq.(25.18), thelawofconservation ofcharge takes onthesimple form V)/,, =0. (25.19) Since T)/,,isaninvariant scalar, ifitiszero inoneframe itiszero inallframes. Wehave theresult thatifcharge isconserved inonecoordinate system, itiscon- served inallcoordinate systems moving with uniform velocity. Asourlastexample wewant toconsider thescalar product ofthegradient operator V)with itself. Inthree dimensions, such aproduct gives theLaplacian 2 02 62-3_ _i9 V‘V v_dx2+8y2+6z2' What dowegetinfour dimensions" That’s easy Following ourrules fordot products andgradients, weget v,.v,.=T<‘;i;>(“ T(T T(T5a2><T a” c'V~ This operator, which istheanalog ofthethree-dimensional Laplacian, iscalled 25-7 Vector Scalar pioduct Vector opera tor Gradient Divergence Laplacian and D’AlembertiantheD’Alembertian andhasaspecial notation: 2 32 2El VV V. (25.20) :I1HZET From itsdeﬁnition itisaninvariant scalar operator; ifitoperates onafour-vector ﬁeld. itproduces anewfour-vector ﬁeld. (Some people deﬁne theD’Aleinbertian with theopposite signtoEq.(25.20), soyouwillhave tobecareful when reading theliterature.) Wehave now found four-dimensional equivalents ofmost ofthethree- dimensional quantities wehadlisted inTable 25-1. (We donotyethave the equivalents ofthecross product andthecurloperation; wewon’t gettothem until thenextchapter) Itmay helpyouremember howtheygoifweputalltheimpor- tantdeﬁnitions andresults together inoneplace, sowehave made such asummary inTable 25-2. Table 25-2 Theimportant quantities ofvector analysis inthree andfourdimensions. Three dimensions Four dimensions A: A-B=A,,B,, +A,,B,, +AZBZ u,,l1,, =cub, —(1,1), —u,,/2,, -—azbz =ail), ——a-b V=(6,/’6x,6/6y,6/dz) V,(6/61, —6/6x, —6/6y, —6/6:) =(6/dr, —V) _QL/’‘?l,%WTfa./ay dz) an, a/1,, 6,4, _ _ VA-‘a‘;+:5;F+‘@7 W" 2>,+@.*.>,+‘" I(Ala An, /42) an = (uh Uri at/1 a2) : (ah a) a2 02 <52 2 2V'V=§;§+5;5+;9; ViiVit="-c-"" ‘c —‘-="c—V =13_<'»»2 a@.~@_<*¢_<1¢)V”T<61‘ 6x 6y 62TatW’ dag+duz da., 61¢:Ha) | weir) 62 62 62 62 32 (it? 6x3 6)3 622 (312 25-4 Electrodynamics infour-dimensional notation Wehave already encountered theD'Alembertian operator, without giving it thatname, inSection 18-6. thedifferential equations wefound there forthepo- tentials canbewritten inthenewnotations as: (324,=B, i:;i=’,4=!_- (25.21)60 60 Thefour quantities ontheright-hand side ofthetwoequations in(25.21) are p,1,,jy,/Z,divided byco,which isauniversal constant which willbethesanie inallcoordinate systems ifthe same unitofcharge isused inallframes. Sothefour quantities p/co, /',/en, /,,/e,,, /‘Z/et, alsotransform asafour-vector Wecanwrite them as1,,/e0 The D'Aleinbertian doesn’t change when thecoordinate system ischanged, sothequantities ¢>,A1,A.,,A;must alsotransform likeafour-vector— which means thatthey arethecomponents ofafour-vector. lnshort. All Z ((1)5 isafour-vector. What wecallthescalar andvector potentials arereally different aspects ofthesame physical thing. They belong together And ifthey arekept together therelativistic invariance oftheworld isobvious WecallAMthefour- potential 25-8 Inthefour-vector notation Eqs. (25.21) become simply [)2/1,,=, (25.22) Thephysics ofthisequation isjustthesame asMaxwell’s equations. Butthere is some pleasure inbeing able torewrite them inanelegant form. Thepretty form isalsomeaningful, itshows directly theinvariance ofelectrodynamics under the Lorentz transformation. Remember thatEqs (25.21) could bededuced from Maxwell’s equations only ifweimposed thegauge condition 53$—l—V-A=0, (25.23) which justsays V,,A,, =0;thegauge condition says that thedivergence ofthe four-vector A),iszero. This condition iscalled theLorentz condition. Itisvery convenient because itisaninvariant condition andtherefore Maxwell’s equations stayintheform ofEq.(25.22) forallframes. 25-5 Thefour-potential ofamoving charge Although itisimplicit inwhat wehave already said, letuswrite down the transformation laws which give¢andAinamoving system interms of¢andA inastationary system. Since A,,=(<75,A)isafour-vector, theequations must lookjustlikeEqs. (25.1). except thattisreplaced by¢,andxisreplaced byA Thus, __Ax I¢I= , Ag:/4”, \/l 11 A_W (25.24) A;=—i——-—, A’,=A,. \/1—212 This assumes that theprimed coordinate system ismoving with speed 0inthe positive x-direction. asmeasured intheunprimed coordinate system. Wewillconsider oneexample ofthe usefulness oftheideaofthefour-potential What arethevector andscalar potentials ofacharge qmoving with speed iialong thex-axis‘? Theproblem iseasy inacoordinate system moving with thecharge, since inthissystem thecharge isstanding still. Let’s saythatthecharge isatthe origin oftheS’-frame. asshown inFig.25—2. Thescalar potential inthemoving system isthen given by I=__‘1__. 25.25¢ 41re0r’ ( ) r’being thedistance from qtotheﬁeld point, asmeasured inthemoving system Thevector potential A’is,ofcourse, zero. Now itisstraightforward toﬁnd¢andA,thepotentials asmeasured inthe stationary coordinates. Theinverse relations toEqs. (2524)are ¢'—l-A2 ‘*=viT_%’ ”r="‘1~,, (25.26)A,=A-_~”+”¢. A,=A’z.\/1 —212 Using the¢’given byEq.(25.25), andA’=0,weget =___L_,L__4) 41re() r/,/1 __U2 i —~+q 1 0 47I'€() ,/1_U2\/3<7Zj_ y/2 +Z/2 25-9ys(2, b.-_,\,,/ /r/'_ /<\ // / T\ q/ \N "1/ (___\>< \L' Fig. 25-2. Thetrome S’moves with velocity v(inthex-direction) wifh respect toS.Achcirge citrestoftheorigin ofS’ isotx=vtinS.Thepotentials atPcon becomputed ineither frcime. This gives usthescalar potential ¢>wewould seeinS,but,unfortunately, expressed interms oftheS’coordinates. Wecangetthings interms oft,x,y,zbysubstituting fort’,x’,y’,andz’,using (25.1). Weget q 1 14;=~~-—~ 9 ‘___ (25.27)4‘n'6o \/1 _0- _m)V/\/T__ U212 _|_y2 +Z2 Following thesame procedure forthecomponents ofA.youcanshow that A=v¢. (25.28) These arethesame formulas wederived byadifferent method inChapter 21. 25-6 Theinvariance oftheequations ofelectrodynamics Wehave found thatthepotentials <15andAtaken together form afour-vector which wecallA,,,andthatthewave eqtiations—the fullequations which determine theA,,interms ofthej,,—can bewritten asinEq.(2522). This equation, together with theconservation ofcharge, Eq.(2519),gives usthefundamental lawofthe electromagnetic ﬁeld: 1. .512.4,,=20.1,, v,,),,=0. (25.29) There, inonetinyspace onthepage. arealloftheMaxwell equations—beautiful andsimple. Didwelearn anything from writing theequations thisway, besides thatthey arebeautiful andsimple? Intheﬁrstplace, isitanything different from what wehadbefore when wewrote everything outinallthevarious components? Canwefrom thisequation deduce something thatcould notbededuced from the wave equations forthepotentials interms ofthecharges andcurrents? Theanswer isdeﬁnitely no.Theonlything wehave been doing ischanging thenames ofthings —using anewnotation. Wehave written asquare symbol torepresent thede- rivatives, butitstillmeans nothing more norlessthan thesecond derivative with respect tot,minus thesecond derivative with respect tox,minus thesecond derivative with respect toy,minus thesecond derivative with respect toz.Andthe /.1means thatwehave fourequations, oneeach for/.t=t,x,y,orz.What then is thesigniﬁcance ofthefactthattheequations canbewritten inthissimple form? From thepoint ofview ofdeducing anything directly, itdoesn’t mean anything. Perhaps, though, thesimplicity oftheequations means that nature also hasa certain simplicity. Letusshow yousomething interesting thatwehave recently discovered: All ofthelaws ofphysics canbecontained inoneequation. That equation is U=0. (25.30) What asimple equation! Ofcourse, itisnecessary toknow what thesymbol means. Uisaphysical quantity which wewillcallthe“unworldliness” ofthe situation. And wehave aformula forit.Here ishow youcalculate theunworld- liness. You take alloftheknown physical laws andwrite them inaspecial form. Forexample, suppose youtake thelawofmechanics, F=ma,andrewrite itas F—ma=0Then youcancall(F—ma)~—which should, ofcourse, bezero—- the“mismatch,” ofmechanics. Next, youtake thesquare ofthismismatch and callitU1,which canbecalled the“unworldl1ness ofmechanical effects.” Inother words, youtake U1=(F—ma)2_ (25.31) Now youwrite another physical law,say,V-E=p/er, anddeﬁne U4 ‘= <V'E"B->2: 60 which youmight call“the gaussian unworldliness ofelectricity.” You continue towrite U3,U4,andsoon—one forevery physical lawthere is 25-10 Finally youcallthetotal unworldliness Uoftheworld thesumofthevarious tinworldlinesses U,from allthesubphenoniena that areinvolved; that is,U= ZU, Then thegreat “law ofnature” is lU=0. (2522) This “law” means, ofcourse, that thesum ofthesquares ofalltheindividual mismatches iszero, andtheonly waythesumofalotofsquares canbezero isfor each oneoftheterms tobezero Sothe“beautifully simple" lawinEq.(25.32) isequivalent tothewhole series ofequations thatyouoriginally wrote down Itistherefore absolutely obvious thatasimple notation thatjusthides thecomplexity inthedeﬁnitions ofsymbols isnotrealsimplicity. Itis]llSlatrick. Thebeauty thatappears inEq.(2532)— justfrom thefactthatseveral equations arehidden within it—is nomore than a trick. When youunwrap thewhole thing, yougetback where youwere before However, there ismore tothesimplicity ofthelaws ofelectromagnetism written intheform ofEq.(25.29). Itmeans more, _]LlSt asatheory ofvector analysis means more. Thefactthattheelectromagnetic equations canbewritten inaveryparticular notation which wasdesigned forthefour-dimensional geometry oftheLorentz transforinations—in other words, asavector equation inthefour- space—means thatitisinvariant under theLorentz transformations. ltisbecause theMaxwell equations areinvariant under those transformations that they can bewritten inabeautiful form. Itisnoaccident thattheequations ofelectrodynamics canbewritten inthe beautifully elegant form ofEq.(2529). Thetheory ofrelativity wasdeveloped becaiise itwasfound experimentally that thephenomena predicted byMaxwell’s equations were thesame inallinertial systems. And itwasprecisely bystudying thetransformation properties ofMaxwell’s equations that Lorentz discovered histransformation astheonewhich lefttheequations invariant. There is,however, another reason forwriting ourequations thisway. Ithas been discovered—after Einstein guessed thatitmight beso— thatallofthelaws ofphysics areinvariant under theLorentz transformation. That istheprinciple ofrelativity. Therefore, ifweinvent anotation which shows immediately when a lawiswritten down whether itisinvariant ornot,wecanbesure thatintrying tomake newtheories wewillwrite only equations which areconsistent with the principle ofrelativity. ThefactthattheMaxwell equations aresimple inthisparticular notation is notamiracle, because thenotation wasinvented with them inmind. Butthe interesting physical thing isthatevery lawofphysics-the propagation ofmeson waves orthebehavior ofneutrinos inbeta decay, andsoforth—must have this same invariance under thesame transformation Then when youaremoving ata uniform velocity inaspaceship, allofthelaws ofnature transform together in such awaythatnonewphenomenon willshow up.Itisbecause theprinciple of relativity isafactofnature thatinthenotation offour-dimensional vectors the equations oftheworld willlook simple. 25-11 26 Lorentz Transformations ofthe Fields 26-1 Thefour-potential ofamoving charge Wesawinthelastchapter that thepotential Ap=(¢,A)isafour-vector. Thetime component isthescalar potential ¢,andthethree space components are thevector potential A.Wealsoworked outthepotentials ofaparticle moving with uniform speed onastraight linebyusing theLorentz transformation (We had already found them byanother method inChapter 21.) Forapoint charge whose position atthetime tis(Hf,0,O),thepotentials atthepoint (x.y,z)are Q ¢ Z ‘ _._ A,= H eQ”1 . 26.147,6“/1 _U2 +yz+Z2]i/2 () Ay=A,=O Equations (261)givethepotentials atx.y,andzatthetime t,foracharge whose “present” position (bywhich wemean theposition atthetimet)isatx=1*! Notice thattheequations areinterms of(x—wt),y,andz,which arethecoordi- nates measured from thecurrent positzon Pofthemoving charge (seeFig.26-1) Theactual inﬂuence weknow really travels atthespeed c,soitisthebehavior of thecharge back attheretarded position P’thatreally counts.T Thepoint P’isat x=iii’(where, r’=I—r’/cistheretarded time). Butwesaidthatthecharge was moving with uniform velocity inastraight line,sonaturally thebehavior atP’and thecurrent position aredirectly related. lnfact,ifwemake theadded assumption thatthepotentials depend only upon theposition andthevelocity attheretarded moment, wehave inequations (26.1) acomp/ere formula forthepotentials fora charge moving anyway. Itworks thisway. Suppose that youhave acharge moving insome arbitrary fashion, saywith thetrajectory inFig.26—2, andyou aretrying toﬁndthepotentials atthepoint (x,y,z).First, youﬁndtheretarded position P’andthevelocity It’atthatpoint. Then youimgaine thatthecharge would keep onmoving with thisvelocity during thedelay time (r’—I).sothat itwould then appear atanimaginary position P,,,‘,,, which wecancallthe“pro- jected position,” andwould arrive there with thevelocity ti’.(Ofcourse, itdoesn‘t dothat; itsrealposition atIisatP.)Then thepotentials at(x,y,z)arejustwhat equations (261)would give fortheimaginary charge attheprojected position Pm”. What wearesaying isthat since thepotentials depend only onwhat the charge isdoing attheretarded time, thepotentials willbethesame whether the charge continued moving ataconstant velocity orwhether itchanged itsvelocity after t’—that is,after thepotentials thatwere going toappear at(x,y,z)atthe timeIwere already determined. Youknow, ofcourse, thatthemoment thatwehave theformula forthepo- tentials from acharge moving inanymanner whatsoever, wehave thecomplete electrodynamics; wecangetthepotentials ofanycharge distribution bysuper- 1‘Theprimes used heretoindicate theretarded positions andtimes should notbeconfused withtheprimes referring toaLorentz-transformed frame inthepreceding chapter. 26-126-1 Thefour-potential ofa moving charge 26-2 Theﬁelds ofapoint charge with aconstant velocity 26—3 Relativistic transformation oftheﬁelds 26-4 Theequations ofmotion in relativistic notation I Inthischapter: c=1 ' Review.‘ Chapter 20,Vol. ll,Solution ofMaxwell’s‘ Equations in Free Space yli lX,y,11 \\1RETARDED /POSITION( F ,7 PRE$$NT 1@951 _--Hi,‘ ’1‘___x—vt ___ X Ytm, ‘ _,___ __!_ Fig. 26-1. Finding theﬁelds atPdue toacharge qmoving along thex-axis with theconstant speed v.The ﬁeld "now" atthepoint (x,y,z)can beex- pressed interms ofthe"present" position P,aswell asinterms ofP’,the"retarded" position (att’=t—r’/c). (x,y,z) r, ‘ii §s;.'t:sa ,aree» \ p y' ,/ ll ll\\ ' \PROJECTbfD q POSITIO PRESENTP4" —POSITIONTRAJECTORY V Fig. 26~2. Acharge moves onan arbitrary trajectory. The potentials at (x,y,z) atthetime tare determined by the position P’and velocity v’atthe retarded time t'—r’/c. They are con- veniently expressed interms oftheco- ordinates from the"pro|ected" position Ppml. (The actual position attisP.)position. Therefore wecansummarize allthephenomena ofelectrodynamics either bywriting Maxwell’s equations orbythefollowing series ofremarks. (Remember them incaseyouareeveronadesert island. From them, allcanbe reconstructed. You will, ofcourse, know theLorentz transformation; youwill never forget thatonadesert island oranywhere else) First, A“isafour-vector. Second, theCoulomb potential forastationary charge isq/41rei,r. Third, thepotentials produced byacharge moving inanyway depend only upon thevelocity andposition attheretarded time With those three facts wehave everything From thefactthat/1,,isafour-vector, wetransform theCoulomb potential, which weknow, andgetthepotentials foraconstant velocity. Then. bythelaststatement thatpotentials depend only upon thepast velocity attheretarded time, wecanusetheprojected position game toﬁndthem. Itisnotaparticularly useful wayofdoing things. butitisinteresting toshow that thelaws ofphysics canbeputinsomany diﬁerent ways ltissometimes said, bypeople who arecareless, thatallofelectrodynamics canbededuced solely from theLorentz transformation andCoulomb’s law. Of course, that iscompletely false. First, wehave tosuppose thatthere isascalar potential andavector potential thattogether make afour-vector That tells us how thepotentials transform Then why isitthat theeffects attheretarded time aretheonlythings thatcount? Better yet,whyisitthatthepotentials depend only ontheposition andthevelocity andnot,forinstance, ontheacceleration‘? Thefit?/C/S‘ EandBdodepend ontheacceleration. lfyoutrytomake thesame kind ofanargument with respect tothem, youwould saythatthey depend only upon theposition andvelocity attheretarded time Butthen theﬁelds from an accelerating charge would bethesame astheﬁelds from acharge attheprojected position——which isfalse. Theﬁelds depend notonlyontheposition andthevelocity along thepath butalsoontheacceleration. Sothere areseveral additional tacit assumptions inthisgreat statement that everything canbededuced from the Lorentz transformation (Whenever youseeasweeping statement thatatremen- dous amount cancome from avery small number ofassumptions, you always ﬁndthatitisfalse. There areusually alarge number ofimplied assumptions that arefarfrom obvious ifyouthink about them suﬁiciently carefully.) 26-2 Theﬁelds ofapoint charge with aconstant velocity Now that wehave thepotentials from apoint charge moving atconstant velocity, weought toﬁndtheﬁelds—for practical reasons There aremany cases where wehave uniformly moving partic1es—for instance, cosmic raysgoing through acloud chamber, oreven slow-moving electrons inawire. Solet'satleast see what theﬁelds actually dolook likeforanyspeed—even forspeeds nearly that oflight assuming only thatthere isnoacceleration. ltisaninteresting question. Wegettheﬁelds from thepotentials bytheusual rules" E=—v¢—%%, B=V><A. F1fSt,fOTEz _ 64> 6A, E2-K27-F" ButA,iszero; sodifferentiating ¢inequations (261),weget E2= ‘1 _Z-6- 26.247,60,/1 _U2 +ya+z2:|3/2 ( ) Similarly, forEU, E= " ~~.~ y 26.3 1/ ‘W60 ml_U2 +y2+22:13/2 ( ) Thex-component isalittle more work. Thederivative of¢ismore complicated 26-2 andA,isnotzero. First, __ai=_i‘1__.__._.(x.IL”’)/(1 (264)5) ,,_ ~ . at-X 41re(,\/l —01 —l—yz—l—22] Then, differentiating A,with respect tot,weﬁnd _,2__j _2_agill : q __U‘:/2 ’ 47l'€0\/1 —U2 +'J12 + Z2J Andﬁnally. taking thesum, E,= ‘I -e» —”’ _- (26.6)47,-60,/1 _U2 +ya+22,3/2 We’ll look atthephysics ofEinaminute. let'sﬁrstﬁndBForthez-compo- nent. __6A,, 6A, B1"WTy‘ Since A,,iszero, wehave justonederivative toget. Notice, however, that A, isjust1'45,and6/6y ofzi¢isjust~—1'E,,. So B,=1'E,,. (26.7) Similarly, _aA,, aA,_ 04> B1/"W a?"+'/5’and BU=—-uE,. (26.8) Finally, B,iszero. since A,,andA2areboth zero. Wecanwrite themagnetic ﬁeld simply as B=vXE (26.9) Now let’sseewhat theﬁelds look like. Wewilltrytodraw apicture ofthe ﬁeldatvarious positions around thepresent position ofthecharge. ltistruethat theinfluence ofthecharge comes, inacertain sense. from theretarded position. butbecause themotion isexactly speciﬁed, theretarded position isuniquely given interms ofthepresent position Foruniform velocities, it’snicer torelate the ﬁelds tothecurrent position, because theﬁeld components at(x,y, z)depend only on(x—vt),y,and z-which arethecomponents ofthedisplacements VPfrom thepresent position to(x,y,z)(seeFig.26-3). Consider ﬁrstapoint with z=O.Then Ehasonly x—andy-components. From Eqs. (26.3) and(26.6), theratio ofthese components isjustequal tothe ratio ofthex-andy-components ofthedisplacement. That means thatEisin the.\‘(IH7(’ direction asrp,asshown inFig.26-3. Since E2isalsoproportional to2. itisclear thatthisresult holds inthree dimensions. lnshort, theelectric ﬁeld is radial from thecharge, andtheﬁeld lines radiate directly outofthecharge. just asthey doforastationary charge. Ofcourse, theﬁeld isn't exactly thesame as forthestationary charge. because ofalltheextra factors of(1—~v2) Butwe canshow something rather interesting. Thedifference isjustwhat youwould get ifyouwere todraw theCoulomb ﬁeld with apeculiar setofcoordinates inwhich thescale ofxwassquashed upbythefactor \/l—v2.lfyoudothat, theﬁeld lines willbespread outahead andbehind thecharge andwillbesqueezed together around thesides, asshown inFig.26-4. lfwe relate thestrength ofEtothedensity oftheﬁeldlines intheconventional way, weseeastronger ﬁeld atthesides andaweaker ﬁeld ahead andbehind. which isjustwhat theequations say. First, ifwelook atthestrength oftheﬁeld atright angles tothelineofmotion, thatis,for(x—1)!)=0,thedistance from 26-3ily E’ E <X Ex ___ ______ P xlvl i ﬁx» q\1?\\ vt—rr"*l\PRESENTPOSITION Fig. 26-3. Foracharge moving with constant speed, theelectric ﬁeld points radially from the“present” position of thecharge. \\ E/ \ \ll /2 \\ /‘-\\‘\\ ,2 -—<-l- g ~--->—- to)v=O _\ / \ / V \ \E \l\ll \\\\/\\,/\/ V (b) v=O.9c / .\ ll1 Fig. 26-4. The electric ﬁeld ofo charge moving with theconstant speed v=O.9c, part (b),compared with the ﬁeld ofcicharge atrest, part la). B \ 9>Me.lV / Fig. 26-5. The magnetic ﬁeld near amoving charge isvXE.(Compare with Fig. 26-4.)thecharge is(y2—l—22). Here thetotal ﬁeld strength is\/E3 —l——E—Z, which is 1E=_-‘Ii -__- (26.10)41re0\/1 —112Y2+Z2 Theﬁeld isproportional totheinverse square ofthedistance—just liketheCou- lomb ﬁeld except increased bytheconstant, extra factor 1/\/1 —112,which is always greater than one. Soatthesides ofamoving charge, theelectric ﬁeld is stronger than yougetfrom theCoulomb law. Infact, theﬁeld inthesidewise direction isbigger than theCoulomb potential bytheratio oftheenergy ofthe particle toitsrestmass. Ahead ofthecharge (and behind), yandzarezero and _ _q(l—112)_E_E’_41re0(x —U02 (2611) Theﬁeld again varies astheinverse square ofthedistance from thecharge butis nowreduced bythefactor (1—v2),inagreement with thepicture oftheﬁeldlines. IfI’/Cissmall, 1'2/c2 isstillsmaller, andtheeffect ofthe(1—I12)terms isvery small; wegetback toCoulomb‘s law. Butifaparticle ismoving very close to thespeed oflight, theﬁeld intheforward direction isenormously reduced, and theﬁeld inthesidewise direction isenormously increased. Our results fortheelectric ﬁeld ofacharge canbeputthisway: Suppose youwere todraw onapiece ofpaper theﬁeld lines foracharge atrest, andthen setthepicture totravelling with thespeed v.Then, ofcourse, thewhole picture would becompressed bytheLorentz contraction; that is,thecarbon granules onthepaper would appear indifferent places Themiracle ofitisthatthepicture youwould seeasthepage ﬂiesbywould stillrepresent theﬁeld lines ofthepoint charge. Thecontraction moves them closer together atthesides andspreads them outahead andbehind, justintheright waytogivethecorrect linedensities. We have emphasized before thatﬁeld lines arenotrealbutareonly onewayofrepre- senting theﬁeld. However, here they almost seem tobereal. lnthisparticular case, ifyoumake themistake ofthinking thattheﬁeld lines aresomehow really there inspace, andtransform them, yougetthecorrect ﬁeld. That doesn’t, however, make theﬁeld lines anymore real Allyouneed dotoremind yourself thatthey aren’t realistothink about theelectric ﬁelds produced byacharge together with amagnet; when themagnet moves, newelectric ﬁelds areproduced, anddestroy thebeautiful picture Sotheneat idea ofthecontracting picture doesn't work in general. ltis,however, ahandy way toremember what theﬁelds from afast- moving charge arelike. Themagnetic ﬁeldisvXE[from Eq.(26.9)]. lfyou takethevelocity crossed intoaradial E-ﬁeld, yougetaBwhich circles around thelineofmotion, asshown inFig. 26-5. Ifweputback thec’s,you willseethat it’sthesame result wehad forlow-velocity charges. Agood waytoseewhere thec’smust goistorefer back totheforce law, F=q(E—l—vXB). You seethat avelocity times themagnetic ﬁeld hasthesame dimensions asan electric ﬁeld. Sotheright-hand sideofEq(26.9) must have afactor 1/c2: _v><EB-—c5—- (2612) Foraslow-moving charge (ll<<c),wecantake forEtheCoulomb ﬁeld: then BI__‘1___ "ll 4Tl'€iiC2 I‘; This formula corresponds exactly toequations forthemagnetic ﬁeld ofacurrent thatwefound inSection 14-7. 26-4 Wewould liketopoint out,inpassing, something interesting foryoutothink about. (Wewillcome back todiscuss itagain later.) Imagine twoelectrons with velocities atright angles, sothatonewillcross over thepath oftheother, butin front ofit,sothey don’t collide. Atsome instant, their relative positions willbe asinFig26-6(a). Welook attheforce onqiduetoqgandviceversa. Onqg there isonly theelectric force from ql,since qimakes nomagnetic ﬁeld along its lineofmotion. Onql,however, there isagain theelectric force but,inaddition, amagnetic force, since itismoving inaB-ﬁeld made byqg.Theforces areasdrawn inFig.26—6(b). Theelectric forces onqiandq2areequal andopposite. However, there isasidewise (magnetic) force onqiandnosidewise force onqg.Does action notequal reaction? Weleave itforyoutoworry about. 26-3 Relativistic transformation oftheﬁelds Inthelastsection wecalculated theelectric andmagnetic ﬁelds from the transformed potentials. Theﬁelds areimportant, ofcourse, inspite oftheargu- ments given earlier that there isphysical meaning andreality tothepotentials. Theﬁelds, too,arereal. Itwould beconvenient formany purposes tohave away tocompute theﬁelds inamoving system ifyoualready know theﬁelds insome “rest” system. Wehave thetransformation laws for¢andA,because A,,isa four-vector. Now wewould liketoknow thetransformation laws ofEandB. Given EandBinoneframe, how dothey look inanother frame moving past? Itisaconvenient transformation tohave. Wecould always work back through the potentials, butitisuseful sometimes tobeable totransform theﬁelds directly. Wewillnow seehowthatgoes. How canweﬁndthetransformation laws oftheﬁelds? Weknow thetrans- formation laws ofthe¢andA,andweknow how theﬁelds aregiven interms of ¢>andA-—it should beeasy toﬁndthetransformation fortheBandE.(You might think thatwith every vector there should besomething tomake itafour- vector, sowith Ethere’s gottobesomething elsewecanuseforthefourth com- ponent. And alsoforB.Butit’snotso.It’squite different from what youwould expect.) Tobegin with, let’s take just amagnetic ﬁeld B,which is,ofcourse VXA.Now weknow thatthevector potential with itsx-,y-,andz-components isonly apiece ofsomething; there isalsoat-component. Also weknow thatfor derivatives likeV,besides thex,y,zparts, there isalsoaderivative with respect to t.Solet’strytoﬁgure outwhat happens ifwereplace a“y"bya“t",ora“z" bya“I,”orsomething likethat. First, notice theform oftheterms inVXAwhen wewrite outthecom- ponents: _6A, 6A,, _6A,, 6A, __6A,, 6A,Bx -—' F Bu -- "7; '"‘ *5; 9 B2 — F ‘ Thex-component isequal toacouple ofterms thatinvolve only y-andz-com- ponents. Suppose wecallthiscombination ofderivatives andcomponents a “zy-thing,” andgiveitashorthand name, F,,,. Wesimply mean that =ﬁe_L11/. F,,,_ ay 62 (26.15) Similarly. Byisequal tothesame kind of“thing,” butthistime itisan“xz-thing.” AndB,is,ofcourse, thecorresponding “yx-thing.” Wehave B,=F,,,,, By=F12, B,=F,,,. (26.16) Now what happens ifwesimply trytoconcoct also some “t”-type things likeF”andF),(since nature should beniceandsymmetric inx,y,z,andI)?For instance, what isFM? Itis,ofcourse, 242_£141.62 6t 26-5VIq,'”_’ qz (0) Iva Fl q|v'xBI '5- " q2E2=F2 qlEl Q51 v2 Fig. 26-6. The forces between two moving charges arenotalways equal and opposite. Itappears that "action" isnot equal to"reaction." Table 26-1 Thecomponents ofF,,,, FM :_FV# FPPi0 l l J F111; _Bz Fxt:Ez Fl/z:_‘Bz F://ZE FZ,---12, F,,=EllButremember thatA,=¢,soitisalso at_5’ZA 02 8t You’ve seen that before. ltisthez-component ofE.Well, almost—there isa signwrong Butweforgot thatinthefour-dimensional gradient thet-dertvative comes with theopposite signfrom x,y,and:Soweshould really have taken the more consistent extension ofF,zas 6/1 6/{ZFjz Z "J + *2)!‘ ()2 Then itisexactly equal to—E, Trying alsoF,,andF,,,,weﬁndthatthethree poss1bilities give F”,=—E,, Fm,=—E,,, Fm=-E2. (26.18) What happens ifboth subscripts are1°Or,forthat matter, ifboth arex? Wegetthings like _6A, 8A, F”-WTi’and 6A,, 6A,, F"-W'37’which givenothing butzero. Wehave then sixofthese F-things. There aresixmore which yougetby reversing thesubscripts, butthey givenothing really new, since F,,,=—F,j,,. andsoon.So,otitofsixteen possible combinations ofthefour subscripts taken inpairs. wegetonly sixdifferent physical objects; andthey arethe('()II'l]7()I1(.’I1fS' ofBandE. Torepresent thegeneral term ofF,wewillusethegeneral subscripts /.1and1/. where each canstand forO,1,2,or3—meaning inourusual four-vector notation t,x,y.andzAlso, everything willbeconsistent with ourfour-vector notation if wedeﬁne F”,by F,”=V,,A, —V,,A,,, (2619) remembering that V,,=(6/dt. -6/dx, -6/6y, -6/02) andthat A,,:(¢,A,,A,,. AZ) What wehave found isthatthere aresixquantities that belong together in nature-that aredifferent aspects ofthesame thing. Theelectric andmagnetic ﬁelds which wehave considered assepaiate vectors 1nourslow-moving world (where wedon’t worry about thespeed oflight) arenotvectors infour-space. They areparts ofa new “thing.” Our physical “ﬁeld” isreally thesix-component object F,,,. That isthewaywemust look atitforrelativity Wesummaiize our results onF,”inTable 26-1 You seethat what wehave done here istogeneralize thecross product We began with thecurloperation, andthefactthatthetransformation properties of thecurlarethesame asthetransformation properties oftwovectors—the ordinary three-dimensional vector Aandthegradient operator which weknow alsobehaves likeavector Let’s look foramoment atanordinary cross product inthree di- mensions, forexample, theangular momentum ofaparticle When anobject is moving inaplane. thequantity (xix, —yz~,) isimportant. Formotion inthree dimensions, there arethree such important quantities, which wecalltheangular momentum: L,,,Im(x/1,, —y1',), L,,,:i11(j'r_ -—Il',,), Li,=Hl(Il', -.\/1,) Then (although youmay have forgotten bynow) wediscovered inChapter 20 ofVol Ithemiracle thatthese three quantities could beidentiﬁed with thecoin- 26-6 ponents ofavector. Inorder todoso,wehadtomake anartiﬁcial rulewith a right-hand convention ItwasJust luck. Itwasluck because L,,(with tand1 equal tox,y,orz)wasanantisymmetric object L,,:~L,,. L,,,=0 Ofthenine possible quantities. there areonly three independent numbers. And itjust happens that when you change coordinate systems these three objects transform inexactly thesame wayasthecomponents ofavector. Thesame thing letsusrepresent anelement ofsurface asavector Asurface element hastwoparts—say dxanddy—which wecanrepresent bythevector da normal tothesuiface. Butwecan’t dothat infour dimensions What isthe “normal” todxdy‘? lsitalong zoralong 1? lnshort, forthree dimensions ithappens byluck thatafter you’ve taken a combination oftwovectors likeL,,,youcanrepresent itagain byanother vector because there are]llStthree terms that happen totransform likethecomponents ofavector Butinfourdimensions thatisevidently lmp0SS1bl€, because there are sixindependent terms, and youcan’t represent si\things byfour things. Even inthree dimensions itispossible tohave combinations ofvectors that can't berepresented byvectors. Suppose wetake anytwovectors a:((1,,a,,,(1,) andb=(h,,b,,,1),),andmake thevarious possible combinations ofcomponents. likeall». u,b.,. etc There would benine possible quantities: a,b,. a,b,,, a,bz, (1,,[i,. u,,b,,, a,,b,, azbx, (13/11,, 11,173. Wemight callthese quantities T,,. lfwenow gotoarotated coordinate system (sayrotated about thez-axis). thecomponents ofaandbarechanged. lnthenewsystem, (1,,forexample. gets replaced by a;=axcos0—l—aysin6, andh,,gets replaced by bf,=b,,cos6—bxsin0. And similarly forother components The nine components ofthe product quantity T,,wehave invented areallchanged too, ofcourse Forinstance, T,,,,1a,l>,, getschanged to T1,,:u,l>,,(cos2 0)—a,b,,(cos 0sin6)—l—a,,b,,(siii 6cos6)—a_,,b,,(sin2 0), or T,Q,,=T,,,cosz 0——T”cos0sin0+T,,,,sin0cosB—T,,,,sing0. Each component ofT,’,isalinear combination ofthecomponents ofT,,. Sowcdiscover that itisnotonly possible tohave a“vector product" like a><bwhich hasthree components that transform likeavector, btitwecan— artiﬁcially—also make another kind of“product” oftwovectors T,,with nine components thattransform under arotation byacomplicated setofrules that wecould figure out Such anOIJJBCI which hastwoindices todescribe it,instead ofone, iscalled atensor. Itisatensor ofthe“second rank,” because youcan playthisgame with three vectors tooandgetatensor ofthethird rank,—or with four, togetatensor ofthefourth rank, andsoon.Atensor oftheﬁrstrank isa vector Thepoint ofallthisisthatourelectromagnetic quantity F,,,,isalsoatensor ofthesecond rank, because ithastwoindices init.ltis,however, atensor in fourdimensions. Ittransforms inaspecial waywhich wewillwork outinamo- ment—it is]Ll§l theway aproduct ofvectors transforms. ForF,,,,ithappens thatifyouchange theindices around, F,”changes sign. That’s aspecial case—it is 26-7 anantisymmetric tensor. Sowesay:theelectric andmagnetic ﬁelds areboth part ofanantisymmetric tensor ofthesecond rank infour dimensions. You’ve come along way. Remember way back when wedeﬁned what a velocity meant? Now wearetalking about “anantisymmetric tensor ofthe second rank infour dimensions.” Now wehave toﬁnd thelawofthetransformation ofF,,,,. ltisn’t atall difficult todo;it’sjust1aborious—the brains involved arenil,butthework isnot. What wewant istheLorentz transformation ofV,,A,, —V,,A,,. Since V,,isjusta special case ofavector, wewillwork with thegeneral antisymmetric vector com- bination. which wecancallG,,,,: 0,,=a,,b,,—0,11,, (26.20) (For ourpurposes, a,,willeventually bereplaced byV,,and11,,willbereplaced by thepotential A,,.) Thecomponents of0,,andb,,transform bytheLorentz formulas, which are _a,—va, b,_b,—vb,,_iis W9 \/l—v5 t \/l—v9I at I a,=a—“”I_”"_‘, b5,=—-b‘_fLbi» (26.21)\/1 —v2 \/l —vl a§,=a,,, bf,=by, a'Z=az. b’,=b3- Now let’stransform thecomponents ofGM. Westart with Gm: Gil=Gib;—H252 :at—va, bx—vb, _ax—vat b,——vb, \/W \/TTF \/YT? \/if = (lib; '—Clxbt. ButthatisJLISIG,,,; sowehave thesimple result G22: : GM- Wewilldoonemore ,__u,~va,, bi—_zl7_,, _(a,b,, —a,,b,) —i,£a,b,,_— iI,,b;¢)G13, — -“-“““*‘4 b * a T“ ‘ a 1/ y _ * ____A \/l—v2 \/l—v2 \/l—v3 Sowegetthat GQDI£'JL__f__".(_7f_a. \/1 v’ And, ofcourse, inthesame way, ; G12 —UGxz G52 :“*“h*Z \/1 ——v3 ltisclear how therestwillgo.Let’s make atable ofallsixterms. only now we may aswellwrite them forF,,,,: m=m, m=@;§L\/1 vi F,”: , FLZ=Fyz, (2612) a \/l —v‘ Fgz I F52 * ZiFxz: F, : ﬁg * i,iFZ/_ \/1-U2 \/1-? Ofcourse, westillhave F,f,,=—F;,, andF,j,,=O. 26-8 Sowehave thetransformation oftheelectric andmagnetic ﬁelds. Allwehave todoislook atTable 26-1 toﬁndoutwhat ourgrand notation interms ofF,,,, means interms ofEandB.It’sjustamatter ofsubstitution. Sothatwecansee how itlooks intheordinary symbols, we'll rewrite ourtransformation ofthe ﬁeldcomponents inTable 26-2. Table 26-2 TheLorentz transformation oftheelectric andmagnetic ﬁelds (Note: c=1) E§=E, B§,=B,, E,/:E,,-UB2 B,:B,,+vEz V1-—v2 \/1-—v2 ,%2+vB,, ,_B,—-vE,,E,,— --—- Bz-€——- \/1-—v2 \/1-—v2 J Theequations inTable 26-2 tellushowEandBchange ifwegofrom oneinertial frame toanother. lfweknow EandBinonesystem, wecanﬁndwhat they are inanother thatmoves bywith thespeed I’. Wecanwrite these equations inaform thatiseasier toremember ifwenotice thatsince visinthex-direction, alltheterms with varecomponents ofthecross products vXEandvXB.Sowecanrewrite thetransformations asshown in Table 26-3 Table 26-3 Analternative form fortheﬁeld transformations (Note: c=1) E§=E, B§=B, E; _ + vXB)1/ B, __ T’ UXE)y 11- --- ll_ -—- \/1 —v3 \/1- v2 E,=(E+v><B)z B,:(B_U><E)z \/YT-“V113 \/TI]? Itisnoweasier toremember which components gowhere lnfact, thetransforma- tioncanbewritten even more simply ifwedeﬁne theﬁeld components along x asthe“parallel” components EHandB1,(because they areparallel totherelative velocity ofSandS’),andthetotal transverse components-the vector sums of they-andz-components-as the“perpendicular” components EiandBl Then wegettheequations inTable 26-4. (We have alsoputback thec’s,soitwillbe more convenient when wewant torefer back later ) Table 26-4 Still another form fortheLorentz transformation ofEandB Ei'|=E Bi'i=B v><EB______ B (> E1=(E,1L_”.>.<__)i~ B1= _CZ.;Ly1—1'2/c \/T -v’/c‘~ Theﬁeld transformations give usanother wayofsolving some problems we have done before—-for instance. forﬁnding theﬁelds ofamoving point charge. Wehave worked outtheﬁelds before bydifferentiating thepotentials. Butwe could now doitbytransforming theCoulomb ﬁeld. lfwehave apoint charge atrestintheS-frame, then there isonly thesimple radial E-ﬁeld IntheS’-frame wewillseeapoint charge moving with thevelocity u,iftheS’-franie moves bythe 26-9 4+- + T““T21+ (D\ < l ll/' Fig. 26-7. The coordinate frame moving through 0static electric ﬁeld.S!S-frame with thespeed v=—u. Wewillletyoushow that thetransformations ofTables 26-3 and26-4 givethesame electric andmagnetic ﬁelds wegotinSection 26-2. Thetransformation ofTable 26-2 gives usaninteresting andsimple answer forwhat weseeifwemove pastanysystem ofﬁxed charges. Forexample. suppose wewant toknow theﬁelds inourframe S’ifwearemoving along between the plates ofacondenser, asshown inFig.26-7. (ltis,ofcourse, thesame thing if wesaythatacharged condenser ismoving pastus.)What dowesee" Thetrans- formation iseasy inthiscase because theB-ﬁeld intheoriginal system iszero. Suppose, ﬁrst, thatourmotion isperpendicular toE,then wewillseeanE’= E/\/1 —~1’?/02 which isstillcompletely transverse. Wewillsee,inaddition, a magnetic ﬁeldB’=—vXE’/cg. (The \/l——1'2doesn’t appear inourformula forB’because wewrote itinterms ofE’rather than E;butit'sthesame thing.) Sowhen wemove along perpendicular toastatic electric ﬁeld, weseeareduced Eandanadded transverse B.lfourmotion isnotperpendicular toE,webreak EintoEHandEl.Theparallel partisunchanged, E,’|=E,i,andtheperpendicular component does asjustdescribed. Let’s take theopposite case, andimagine wearemoving through apure static magnetic ﬁeld. This time wewould seeanelectric ﬁeld E’equal tovXB’, andthemagnetic ﬁeldchanged bythefactor 1/\/l —1'2/c2 (assuming itistrans- verse). Solong asvismuch lessthan c,wecanneglect thechange inthemagnetic ﬁeld, andthemain effect isthatanelectric ﬁeld appears. Asoneexample ofthis effect, consider thisonce famous problem ofdetermining thespeed ofanairplane. lt’snolonger famous, since radar cannow beused todetermine theairspeed from ground reﬂections, butformany years itwasvery hard toﬁndthespeed of anairplane inbadweather. You could notseetheground andyoudidn't know which waywasup,andsoon.Yetitwasimportant toknow how fastyouwere moving relative totheearth. How canthisbedone without seeing theearth? Many whoknew thetransformation formulas thought oftheideaofusing thefact thattheairplane moves inthemagnetic ﬁeldoftheearth Suppose thatanairplane isﬂying where there isamagnetic ﬁeld more orlessknown. Let’s justtake the simple casewhere themagnetic ﬁeld isvertical. Ifwewere ﬂying through itwith ahorizontal velocity v,then, according toourformula, weshould seeanelectric ﬁeld which isvXB,i.e,perpendicular tothelineofmotion lfwehang an insulated wireacross theairplane, thiselectric ﬁeldWlllinduce charges ontheends ofthewire. That isnothing new. From thepoint ofview ofsonieone ontheground, wearemoving awirethrough aﬁeld. andthevXBforce causes charges tomove totheends ofthewire Thetransformation equations justsaythesame thing in adiﬂerent way. (The factthatwecansaythething more than onewaydoesn’t mean that oneway isbetter than another Wearegetting somany different methods andtools thatwecanusually getthesame result in65different ways‘) Sotomeasure v,allwehave todoismeasure thevoltage between theends of thewire. Wecan’t doitwith avoltmeter because thesame ﬁelds willactonthe wires inthevoltmeter, butthere areways ofmeasuring such ﬁelds. Wetalked about some ofthem when wediscussed atmospheric electricity inChapter 9So itshould bepossible tomeasure thespeed oftheairplane. This important problem was, however, never solved thisway. Thereason is thattheelectric ﬁeld thatisdeveloped isoftheorder ofmillivolts permeter. lt ispossible tomeasure such ﬁelds, butthetrouble isthatthese ﬁelds are.unfortun- ately, notanydifferent from anyother electric ﬁelds. Theﬁeld thatisproduced bymotion through themagnetic ﬁeld can’t bedistinguished from some electric ﬁeld thatwasalready intheairfrom another cause, sayfrom electrostatic charges intheair,orontheclouds Wedescribed inChapter 9thatthere are,typically, electric ﬁelds above thesurface oftheearth with strengths ofabout 100volts per meter Butthey arequite irregular. Soastheairplane ﬂiesthrough theair,it seesﬂuctuations ofatmospheric electric ﬁelds which areenormous incomparison tothetinyﬁelds produced bythevXBterm, anditturns outforpractical reasons tobeimpossible tomeasure speeds ofanairplane byitsmotion through theearth's magnetic ﬁeld. 26-10 26-4 Theequations ofmotion inrelativistic notation* ltdoesn‘t doniucfi good toﬁndelectric andmagnetic ﬁelds from Maxwell's equations unless weknow what theﬁelds dowhen wehave them. You may re- member thattheﬁelds arerequired toﬁndtheforces oncharges, andthatthose forces determine themotion ofthecharge. So,ofcourse, part ofthetheory of electrodynamics istherelation between themotion ofcharges andtheforces. Forasingle charge intheﬁelds EandB,theforce is F=q(E+v><B). <2623) This force isequal tothemass times theacceleration forlowvelocities, butthe correct lawforanyvelocity isthat theforce isequal todp/dt. Writing p=[_A__ B. muv/V l-1'-'/cl, weﬁndthattherelativistically correct equation ofmotion is ITIUU cl Wewould likenowtodiscuss thisequation from thepoint ofview ofrelativity. Since wehave putourMaxwell equations inrelativistic form, itwould beinteresting toseewhat theequations ofmotion would look likeinrelativistic form Let’s see whether wecanrewrite theequation inafour-vector notation. Weknow that themomentum ispart ofafour-vector 12,,whose time coni- ponent istheenergy /11.,/\/If-‘T5/Z5. Sowemight think toreplace theleft-hand sideofEq.(2624)byd/2,,/dr Then weneed only ﬁndafourth component togo withF.This fourth component must equal therate-of-change oftheenergy, orthe rateofdoing work, which isF-v.Wewould then liketowrite theright-hand sideofliq(26.24) asafour-vector like(F-v,F,,,F,,,F3). Butthisdoes notmake afour-vector. Thetime derivative ofafour-vector isnolonger afour-vector, because the d/d!requires thechoice ofsome special frame formeasuring I.Wegotintothat trouble before when wetried tomake vintoafour-vector. Ourﬁrstguess was thatthetime component would betdr/dz =c.Butthequantities tlxdydz _(C, 9zit" 2 —- (C, U) arenotthecomponents ofafour-vector Wefound thatthey could bemade into onebymultiplying each component by1/\/l —-vi/c2. The “four-velocity" 14,,isthefour-vector C. 11,,=--—-,-ii)-,—( - (26.26) \/l—-vi/02 \/l —v2/cl Soitappears thatthetrick istomultiply a’/di byl/V1 —1'2/c2. ifwewant the derivatives tomake afour-vector. Oursecond guess then isthat -‘—— 5(pa (26.21) ,/1_,,2/C2 dt should beafour-vector. Butwhat isv?ltisthevelocity oftheparticle—not ofa coordinate frame! Then thequantity/'1, deﬁned by F- Ff,,= ,-?—— (26.28) \/l—vi/c2 \/1—v2/c2 istheextension intofour dimensions ofaforce—we cancallitthe“four-force.” ltisindeed afour-vector, anditsspace components arenotthecomponents of Fbut ofF/\/l ——I’-3/C2. *Inthissection wewillputback allofthec‘s. 26-ll Thequestion is—why isf,,afour-vector” ltwould benicetogetalittle under- standing ofthat l/\/l -v2/c2 factor Since ithascome uptwice now, itistime toseewhythed/dt canalways beﬁxed bythesame factor. Theanswer isinthe following: When wetake thetime derivative ofsome function x,wecompute the increment Axinasmall interval Atinthevariable t.Butinanother frame. the interval Atmight correspond toachange inboth t’andx’,soifwevary only t’, thechange inxwillbedifferent. Wehave toﬁndavariable forourdiﬂerentiation thatisameasure ofan“interval" inspace-time, which willthen bethesame in allcoordinate systems. When wetake Axforthatinterval, itwillbethesame for allcoordinate frames. When aparticle “moves” infour-space, there arethechanges At,Ax,Ay,Az.Canwemake aninvariant interval outofthem? Well, they are thecomponents ofthefour-vector x,,=(ct,x,y,z)soifwedeﬁne aquantity Asby (As)2=?12Ax,,Ax,, =C-1-2(Razz -Axz-A)?-A22) (26.29) —which isafour-dimensional dotproduct—we then have agood four-scalar to useasameasure ofafour-dimensional interval. From As—or itslimit dv—we candeﬁne aparameter s=jds. And aderivative with respect tos,d/ds, isa nicefour-dimensional operation, because itisinvariant with respect toaLorentz transformation. Itiseasytorelate dstodtforamoving particle. Foramoving point particle. dx=tr,dt. dy=v,,dt, dz=1'2dt, (Z630) and ds=\,'(d,2/c2)(,.2 _,5_,5_,3)=dtx/1-1'2/C2. (26.31) Sotheoperator _1_i,/1_,,2/C2 dt isaninvariant operator. Ifweoperate onanyfour-vector with it,wegetanother four-vector. Forinstance, ifweoperate on(ct.x,y,z),wegetthefour-velocity u,,: d1%“Weseenow whythefactor ﬁxes things up. Theinvariant variable sisauseful physical quantity. ltiscalled the“proper time” along thepath ofaparticle, because dsisalways aninterval oftime ina frame that ismoving with theparticle atanyparticluar instant. (Then, Ax= Ay=Az=O,andAs=At.) Ifyou canimagine some “clock” whose rate doesn’t depend ontheacceleration, such aclock carried along with theparticle would show thetime s Wecannow goback andwrite Newton’s law(ascorrected byEinstein) in theneat form dpu _jars“—fa, (26-32) wheref,, 1Sgiven inEq.(26.28). Also, themomentum 12,,canbewritten as a’1),,=mOu,,=mo%. (26.33) where thecoordinates x,,=(ct,x,y,z)nowdescribe thetrajectory oftheparticle. Finally, thefour-dimensional notation gives usthisvery simple form oftheequa- tions ofmotion: d2xf,‘ Z I710 -6-is-TH 5 which isreminiscent ofF=ma ltisimportant tonotice thatEq.(26.34) isno! thesame asF=ma,because thefour-vector formula Eq.(26.34) hasinitthe 26-12 relativistic mechanics which aredifferent from Newton’s lawforhigh velocities. Itisunlike thecase ofMaxwell’s equations, where wewere able torewrite the equations intherelativistic form without anychange inthemeaning atall—but with justachange ofnotation. Now let’sreturn toEq.(26.24) andseehow wecanwrite theright-hand side infour-vector notation. Thethree components—when divided by\/l-v2/c2- arethecomponents of)1,so f:q(E —l—vXB), :q E, V+ v,,Bz _ vZB,, _ x \/l —v2/c2 xﬁ —v2/C2 \/1 —v2/02 \/l -v2/c2 (26.35) Now wemust putallquantities intheir relativistic notation. First, c/\/TT— v27/cl andv,,/c/l —v2/02 and1:,/\/ilﬁ-T/t‘? arethei-,y-,andz-components ofthe four-velocity 11,,And thecomponents ofEandBarecomponents ofthesecond- rank tensor oftheﬁelds F,,,,. Looking back inTable 26-1 forthecomponents of F,,,,thatcorrespond toEx,B2,andBU,weget fr :q(utFart T’ui/Fry _uZFIZ)v which begins tolook interesting. Every term hasthesubscript x,which isreason- able, since we’re ﬁnding anx-component. Then alltheothers appear inpairs: tt,yy,22-—except that thexx-term ismissing. Sowejuststick itin.andwrite fr=q(u,F,t —u,,F,,, -—u,,F,,y —u_,F,,,). (2636) Wehaven’t changed anything because F,,,,isantisymmetric, andF”iszero. The reason forwanting toputinthexx-term issothat wecanwrite Eq.(26.36) in theshort-hand form f,,=qu,,F,,,,. (26.37) This equation isthesame asEq(26.36) ifwemake therulethatwhenever any subscript occurs twice (as1/does here), youautomatically sum over terms inthe same wayasforthescalar product, using thesame convention forthesigns. You caneasily believe that(26.37) works equally wellfor/.t=yora=2. butwhat about /.t=t?Let’s see,forfun,what itsays: fr=q(uiFii _‘uzFtx —u!]F[y _llzFtz) Now wehave totranslate back toE’sandB’s. Weget =0+<9”; _-E.+~—“”—-E. +—i-E”fi q( xﬁ—v2/c2 \/T—v2/c2 J \/l—v2/c2 °r (2638)f,: i \/l —v3/c2 Butfrom Eq.(26.28),f, issupposed tobe F-v =q(E+vXB)-v_ \/l —U2/C2 \/l —U2/C2 Thisisthesame thing asEq.(26.38), since (vXB)-viszero. Soeverything comes outallright. Summarizing, ourequation ofmotion canbewritten intheelegant form dzx"--F (2639 /’n()."W'—-f,,—qH;,,,;,. -) Although itisnicetoseethattheequations canbewritten that way, thisform isnotparticularly useful lt’susually more convenient tosolve forparticle motions byusing theoriginal equations (26.24), andthat's what wewillusually do. 26-13 27 Field Energy and Field Momentum 27-1 Local conservation Itisclear thattheenergy ofmatter isnotconserved. When anobject radiates light itloses energy. However, theenergy lostispossibly describable insome other form, sayinthelight. Therefore thetheory oftheconservation ofenergy is incomplete without aconsideration oftheenergy which isassociated with thelight or,ingeneral, with theelectromagnetic ﬁeld. Wetakeupnowthelawofconserva- tionofenergy and, also, ofmomentum fortheﬁelds. Certainly, wecannot treat onewithout theother, because intherelativity theory they aredifferent aspects of thesame four-vector. Very early inVolume I,wediscussed theconservation ofenergy; wesaid thenmerely thatthetotal energy intheworld isconstant. Now wewant toextend theideaoftheenergy conservation lawinanimportant way—in awaythatsays something indetail about howenergy isconserved. Thenewlawwillsaythatif energy goesaway from aregion, itisbecause itﬂows away through theboundaries ofthatregion. Itisasomewhat stronger lawthan theconservation ofenergy without such arestriction. Toseewhat thestatement means, let’slook athowthelawoftheconservation ofcharge works. Wedescribed theconservation ofcharge bysaying thatthere is acurrent density jandacharge density p,andthatwhen thecharge decreases at some place there must beaﬂow ofcharge away from thatplace. Wecallthatthe conservation ofcharge. Themathematical form oftheconservation lawis V-j=-52% (27.1) Thislawhastheconsequence thatthetotal charge intheworld isalways constant— there isnever anynetgain orlossofcharge. However, thetotal charge inthe world could beconstant inanother way. Suppose thatthere issome charge Q1 near some point (1)while there isnocharge near some point (2)some distance away (Fig. 27-1). Now suppose that, astime goes on,thecharge Q1were to gradually fade away andthatsimultaneously with thedecrease ofQ1some charge Q2would appear near point (2),andinsuch awaythatatevery instant thesumof Q1andQ2wasaconstant. Inother words, atanyintermediate state theamount ofcharge lostbyQ1would beadded toQ2. Then thetotal amount ofcharge in theworld would beconserved. That’s a“world-wide” conservation, butnotwhat wewillcalla“local” conservation, because inorder forthecharge togetfrom (l)to(2).itdidn’t have toappear anywhere inthespace between point (l)and point (2).Locally, thecharge wasjust“lost.” There isadifficulty with such a“world-wide” conservation lawinthetheory ofrelativity. Theconcept of“simultaneous moments” atdistant points isonewhich isnotequivalent indifferent systems. Two events thataresimultaneous inone system arenotsimultaneous foranother system moving past. For“world-wide" conservation ofthekind described, itisnecessary that thecharge lostfrom Q1 should appear simultaneously inQ2. Otherwise there would besome moments when thecharge wasnotconserved. There seems tobenoway tomake the lawofcharge conservation relativistically invariant without making ita“local” conservation law. Asamatter offact, therequirement oftheLorentz relativistic invariance seems torestrict thepossible laws ofnature insurprising ways. In modern quantum ﬁeld theory, forexample, people have often wanted toalter the theory byallowing what wecalla“nonlocal” interaction—where something here 27-127-1 Local conservation 27-2 Energy conservation and electromagnetism 27-3 Energy density andenergy ﬂowintheelectromagnetic ﬁeld 27-4 Theambiguity oftheﬁeld energy 27-5 Examples ofenergy ﬂow 27-6 Field momentum ll) (2) /'/-1%, 0,\/ Q2 (bl Fig. 27-1. Two wciys toconserve charge: la)Q1—l—Q2isconstant; (b) dQi/dt =ff-nda =—dQ2/dt. hasadirect effect onsomething there—but wegetintrouble with therelativity principle. “Local” conservation involves another idea. Itsays that acharge canget from oneplace toanother onlyifthere issomething happening inthespace between. Todescribe thelawweneed notonly thedensity ofcharge, p,butalsoanother kind ofquantity, namely j,avector giving therateofﬂow ofcharge across a surface. Then theﬂow isrelated totherateofchange ofthedensity byEq.(27.1). This isthemore extreme kind ofaconservation law. Itsays that charge iscon- served inaspecial way—conserved “locally.” Itturns outthatenergy conservation isalsoalocal process. There isnotonly anenergy density inagiven region ofspace butalsoavector torepresent therate ofﬂowoftheenergy through asurface. Forexample, when alight source radiates, wecanﬁndthelight energy moving outfrom thesource. Ifweimagine some mathe- matical surface surrounding thelight source, theenergy lostfrom inside thesurface isequal totheenergy thatﬂows outthrough thesurface. 27-2 Energy conservation andelectromagnetism Wewant now towrite quantitatively theconservation ofenergy forelectro- magnetism. Todothat, wehave todescribe how much energy there isinany volume element ofspace, andalsotherateofenergy ﬂow. Suppose wethink ﬁrst only oftheelectromagnetic ﬁeldenergy. Wewillleturepresent theenergy density intheﬁeld (that is,theamount ofenergy perunitvolume inspace) andletthe vector Srepresent theenergy ﬂux oftheﬁeld (that is,theﬂow ofenergy perunit time across aunitarea perpendicular totheﬂow). Then, inperfect analogy with theconservation ofcharge, Eq(271),wecanwrite the“local” lawofenergy conservation intheﬁeld as 6u5-1-—V-S. (27.2) Ofcourse, thislawisnottrueingeneral; itisnottruethattheﬁeld energy is conserved. Suppose youareinadark room andthen turnonthelight switch. All ofasudden theroom isfulloflight, sothere isenergy intheﬁeld, although there wasn’t anyenergy there before. Equation (27.2) isnotthecomplete conservation law, because theﬁeld energy alone isnotconserved, only thetotal energy inthe world—there isalsotheenergy ofmatter. Theﬁeld energy willchange ifthere is some work being done bymatter ontheﬁeld orbytheﬁeld onmatter However, ifthere ismatter inside thevolume ofinterest, weknow how much energy ithas: Each particle hastheenergy m1,c2/\/l ——('2/c2. Thetotal energy ofthematter isjust thesumofalltheparticle energies, andtheﬂow ofthisenergy through asurface isjustthesumoftheenergy carried byeach particle thatcrosses thesurface Wewant now totalkonly about theenergy oftheelectromagnetic ﬁeld. Sowemust write anequation which saysthatthetotal fieldenergy inagiven volume decreases either because ﬁeld energy ﬂows outofthevolume orbecause theﬁeld loses energy tomatter (orgains energy, which isjustanegative loss). Theﬁeld energy inside avolume Vis /udV,v anditsrateofdecrease isminus thetime derivative ofthisintegral. Theﬂow of ﬁeldenergy outofthevolume Vistheintegral ofthenormal component ofSover thesurface Zthatencloses V, /S'nda.Z So -g [Viia'V=/S-nda—l—(work done onmatter inside V). (27.3). 2) 27-2 Wehave seen before thattheﬁeld does work oneach unitvolume ofmatter attherateE-j.[The force onaparticle isF=q(E+v><B),andtherateof‘ doing work isF~v=qE-v.Ifthere areNparticles perunitvolume, therateof doing work perunitvolume isNqE- v,butNqv =j]Sothequantity E~jmust beequal tothelossofenergy perunit time andperunit volume bytheﬁeld. Equation (27.3) then becomes -3; udV=/S~nda+/E-jdV. (27.4)<91V 2 V This isourconservation lawforenergy intheﬁeld. Wecanconvert itintoa differential equation likeEq.(27.2) ifwecanchange thesecond term toavolume integral That iseasy todowith Gauss’ theorem. The surface integral ofthe normal component ofSistheintegral ofitsdivergence over thevolume inside. SoEq.(27.3) isequivalent to _/@dV=/ v-sdV+/E-jdV,Vdl V V where wehave putthetime derivative oftheﬁrstterm inside theintegral. Since thisequation istrueforanyvolume, wecantake away theintegrals andwehave theenergy equation fortheelectromagnetic ﬁelds: -3-i:=v-S+E-j. (27.5) Now thisequation doesn’t dousabitofgood unless weknow what uandS are. Perhaps weshould _]USItellyouwhat they areinterms ofEandB,because allwereally want istheresult. However, wewould rather show youthekind of argument thatwasused byPoynting in1884 toobtain formulas forSandu,so youcanseewhere they come from. (You won’t, however, need tolearn thisde- rivation forourlater work.) 27-3 Energy density andenergy ﬂowintheelectromagnetic ﬁeld Theideaistosuppose thatthere isaﬁeld energy density uandafluxSthat depend onlyupon theﬁelds EandB.(For example, weknow thatinelectrostatics, atleast, theenergy density canbewritten %eOE -E.)Ofcourse, theuandSmight depend onthepotentials orsomething else. butlet’sseewhat wecanwork out Wecantrytorewrite thequantity E-jinsuch awaythatitbecomes thesumof twoterms. onethatisthetime derivative ofonequantity andanother thatisthe divergence ofasecond quantity. Theﬁrstquantity would then beuandthesecond would beS(with suitable signs). Both quantities must bewritten interms ofthe ﬁelds only; thatis,wewant towrite ourequality as _ 6E-]=-—FL;—V-S. (27.6) Theleft-hand sidemust ﬁrstbeexpressed interms oftheﬁelds only. How canwedothat" Byusing Maxwell’s equations. ofcourse. From Maxwell’s equation forthecurlofB, . 6E_]= €QC2V><B'-"€()H' Substituting thisin(276)wewillhave only E’sandB’s: E-j= e0c2E-(V><B)— e0E-%- (27.7) Wearealready partly ﬁnished. The last term isatime derivative—it is (6/<9/)(§e0E -E). So§e0E -Eisatleast onepartofu.It’sthesame thing we found inelectrostatics. Now, allwehave todoistomake theother term intothe divergence ofsomething. 27-3 Notice thattheﬁrstterm ontheright-hand sideof(27.7) isthesame as (VXB)'E. (27.8) And, asyouknow from vector algebra, (aXb)cisthesame asa-(bXc); soourterm isalsothesame as v-(B><E) (27.9) andwehave thedivergence of“something,” just aswewanted. Only that's wrong‘ Wewarned youbefore that Vis“like” avector, butnot“exactly” the same. Thereason itisnotisbecause there isanadditional C0!lvL’!1ll0I1 from cal- culus: when aderivative operator isinfront ofaproduct, itworks oneverything totheright. InEq(27.7), theVoperates only onB,notonEButintheform (279),thenormal convention would saythat Voperates onboth BandESo it’snotthesame thing Infact, ifwework outthecomponents ofV-(BXE) wecanseethatitisequal toE~(VXB)plussome other terms. It’slikewhat happens when wetake aderivative ofaproduct iiialgebra Forinstance, §§</g>= gg+/gg Rather than working outallthecomponents ofV-(BXE),wewould like toshow youatrick thatisvery useful forthiskind ofproblem. Itisatrick that allows youtousealltherules ofvector algebra onexpressions withtheVoperator, without getting intotrouble Thetrick istothrow out~for awhile atleast—~the ruleofthecalculus notation about what thederivative operator works on You see,ordinarily, theorder ofterms isused forrimseparate purposes. One isfor calculus: f(d/dx)g isnotthesame asg(d/dx)f; andtheother isforvectors: aXbisdiﬁerent from bXa.Wecan,ifwe want,chooseto abandon momentarily thecalculus rule. Instead ofsaying thataderivative operates oneverything tothe right, wemake anewrulethatdoesn't depend ontheorder inwhich terms arewrit- tendown Then wecanJuggle terms around without worrying Here isournewconvention. weshow, byasubscript. what adifferential op- erator works on;theorder hasnomeaning. Suppose welettheoperator Dstand for0/6x. Then Dfmeans that only thederivative ofthevariable quantity fis taken. Then ifDH’=5;" D/fg=<%)g- Butnotice now thataccording toournewrule,fD,g means thesame thing We canwrite thesame thing anywhich way. D/fg =gD/f=fD¢g =fgD7- You see,theD;caneven come after everything. (It"ssurprising thatsuch ahandy notation isnever taught inbooks onmathematics orphysics.) You may wonder: What ifIwant towrite thederivative of/g? Iwant the derivative ofboth terms. That’s easy, youJustsayso;youwrite D/(jg) +Dg(fg)- That isJUSIg(6f/6x) —l—f(6g/Bx), which iswhat youmean intheoldnotation by <9(fg)/<9X-You willseethatitisnowgoing tobeveryeasytowork outanewexpression forV-(BXE).Westart bychanging tothenewnotation; wewriteButifwehave D/fg, itmeans V-(B><E)=V);-(BXE)+VE-(BXE). (27.10) Themoment wedothatwedon’t have tokeep theorder straight anymore We always know that VEoperates onEonly, andV1,operates onBonly Inthese circumstances, wecanuseVasthough itwere anordinary vector. (Ofcourse, 27—4 when weareﬁnished, wewillwant toreturn tothe“standard” notation that everybody usually uses)Sonow wecandothevarious things likeinterchanging dots andcrosses andmaking other kinds ofrearrangements oftheterms. For instance, themiddle term ofEq.(27.10) canberewritten asE-VBXB.(You remember thata-bXc=b-cXa.)And thelastterm isthesame asB-EX VE.Itlooks freakish, butitisallright. Now ifwetrytogoback totheordinary convention, wehave toarrange thattheVoperates only onits“own” variable. Theﬁrstoneisalready thatway, sowecanjust leave oﬁthesubscript. Thesecond oneneeds some rearranging toputtheVinfront oftheE,which wecandoby reversing thecross product andchanging sign: B-(EX VE)=—B-(VE ><E). Now itisinaconventional order, sowecanreturn totheusual notation. Equation (27.10) isequivalent to V-(BXE)=E-(VXB)—B-(VXE). (27.11) (Aquicker way would have been tousecomponents inthisspecial case, butit wasworth taking thetime toshow youthemathematical trick. You probably won’t seeitanywhere else, anditisvery good forunlocking vector algebra from therules about theorder ofterms with derivatives.) Wenowreturn toourenergy conservation discussion anduseournewresult, Eq.(27.11), totransform theVXBterm ofEq.(27.7). That energy equation becomes . aE-1=@.,¢2v-(B><E)+6083- (v><E)-5(%e0E-E) (27.12) Now youseewe’re almost ﬁnished. Wehave oneterm which isanicederivative withrespect tottouseforuandanother thatisabeautiful divergence torepresent S.Unfortunately, there isthecenter term leftover, which isneither adivergence noraderivative withrespect tot.Sowealmost made it,butnotquite. After some thought, welook back atthedifferential equations ofMaxwell anddiscover thatVXEis,fortunately, equal to—6B/6t, which means thatwecanturn the extra term intosomething thatisapure time derivative: 6B 6B'BB(VXE)- B-(— E) --—5<-—2—)- Now wehave exactly what wewant. Ourenergy equation reads . a 2E-1=v-(@.,¢2B>< E)-5(i§B-B+529E-E), (27.13) which isexactly likeEq.(27.6), ifwemake thedeﬁnitions 2 u=%E-E+‘°T°12-B (27.14) and S=e.,c2E XB. (27.15) (Reversing thecross product makes thesigns come outright.) Ourprogram wassuccessful. Wehave anexpression fortheenergy density thatisthesum ofan“electric” energy density anda“magnetic” energy density, whose forms arejustliketheones wefound instatics when weworked outthe energy interms oftheﬁelds. Also, wehave found aformula fortheenergy ﬂow vector oftheelectromagnetic ﬁeld. This new vector, S=e0c2E XB,iscalled “Poynting’s vector,” after itsdiscoverer. Ittells ustherateatwhich theﬁeld energy moves around inspace. Theenergy which ﬂows through asmall area da persecond isS-nda,where nistheunitvector perpendicular toda.(Now that wehave ourformulas foruandS,youcanforget thederivations ifyouwant.) 27—5 E S —~4*—~——~4—e>V B MRECUON OFWAVE PROPAGNHON Fig. 27-2. The vectors E,B,and S forolight wove.27-4 Theambiguity oftheﬁeld energy Before wetake upsome applications ofthePoynting formulas [Eqs. (27.14) and(27.l5)], wewould liketosaythatwehave notreally “proved” them. All wedidwastoﬁndapossible “ii”andapossible “S.” How doweknow thatby juggling theterms around some more wecouldn't ﬁndanother formula for“u” andanother formula for“S”? ThenewSandthenewuwould bedifferent, but they would stillsatisfy Eq.(27.6). It’spossible. Itcanbedone, buttheforms that have been found always involve various derivatives oftheﬁeld (and always with second-order terms likeasecond derivative orthesquare ofaﬁrstderivative). There are,infact, aninﬁnite number ofdifferent possibilities foruandS,and sofarnoonehasthought ofanexperimental waytotellwhich oneisright! People have guessed thatthesimplest oneisprobably thecorrect one, butwemust say thatwedonotknow forcertain what istheactual location inspace oftheelectro- magnetic ﬁeld energy. Sowetoowilltake theeasy wayoutandsaythattheﬁeld energy isgiven byEq.(27.14). Then theﬁowvector Smust begiven byEq.(27.15). Itisinteresting thatthere seems tobenounique waytoresolve theindeﬁnite- nessinthelocation oftheﬁeld energy. Itissometimes claimed thatthisproblem canberesolved byusing thetheory ofgravitation inthefollowing argument. Inthetheory ofgravity, allenergy isthesource ofgravitational attraction. There- foretheenergy density ofelectricity must belocated properly ifwearetoknow in which direction thegravity force acts. Asyet,however, noonehasdone such a delicate experiment that theprecise location ofthegravitational inﬂuence of electromagnetic ﬁelds could bedetermined. That electromagnetic ﬁelds alone can bethesource ofgravitational force isanideaitishard todowithout. Ithas,in fact, been observed that light isdeﬂected asitpasses near thesun——we could saythatthesunpulls thelight down toward it.Doyounotwant toallow thatthe light pulls equally onthesun? Anyway, everyone always accepts thesimple expressions wehave found forthelocation ofelectromagnetic energy anditsﬂow. And although sometimes theresults obtained from using them seem strange, noboby hasever found anything wrong with them—that is,nodisagreement with experiment. Sowewillfollow therestoftheworld——besides, webelieve thatitis probably perfectly right. Weshould make onefurther remark about theenergy formula. Intheﬁrst place, theenergy perunitvolume intheﬁeld isvery simple: Itistheelectrostatic energy plus themagnetic energy, i'fwe write theelectrostatic energy interms of E2andthemagnetic energy asB2. Wefound twosuch expressions aspossible expressions fortheenergy when wewere doing static problems. Wealsofound a number ofother formulas fortheenergy intheelectrostatic ﬁeld, such asp¢, which isequal totheintegral ofE-Eintheelectrostatic case However, inan electrodynamic ﬁeld theequality failed, andthere wasnoobvious choice asto which wastheright one. Now weknow which istheright one. Similarly, wehave found theformula forthemagnetic energy that iscorrect ingeneral Theright formula fortheenergy density ofdynamic ﬁelds isEq(27.14) 27-5 Examples ofenergy ﬂow Ourformula fortheenergy ﬂow vector Sissomething quite new. Wewant now toseehow itworks insome special cases andalsotoseewhether itchecks outwith anything thatweknew before. Theﬁrstexample wewilltake islight. Inalight wave wehave anEvector andaBvector atright angles toeach other andtothedirection ofthewave propagation. (See Fig27-2.) Inanelectroniag- netic wave, themagnitude ofBisequal to1/ctimes themagnitude ofE,andsince they areatright angles, E2 ll?XB|I—-c Therefore, forlight, theﬁow ofenergy perunitarea persecond is S=e0cE2. (27.16) 27—6 Foralight wave inwhich E=E0cosw(t—x/c), theaverage rate ofenergy ﬂowperunitarea, <S)av —Which iscalled the“intensity” ofthelight—is themean value ofthesquare oftheelectric ﬁeld times soc: Intensity =(S),,,, =e0c(E2).,,,. (27.17) Believe itornot,wehave already derived thisresult inSection 31-3 ofVol. I, when wewere studying light. Wecanbelieve thatitisright because italsochecks against something else When wehave alight beam, there isanenergy density in space given byEq.(27.14). Using cB=Eforalight wave, wegetthat 22__Q2 606 §_ = 2H-— (C2) EOE. ButEvaries inspace, sotheaverage energy density is (ulav =€0<E2>av~ Now thewave travels atthespeed c.soweshould think thattheenergy thatgoes through asquare meter inasecond isctimes theamount ofenergy inonecubic meter. Sowewould saythat (S>;w =6()C<E2>_,v. Andit’sright; itisthesame asEq.(27.17). Now wetake another example. Here isarather curious one. Welook atthe energy ﬂowinacapacitor thatwearecharging slowly. (Wedon’t want frequencies sohighthatthecapacitor isbeginning tolook likearesonant cavity, butwedon't want DCeither.) Suppose weuseacircular parallel plate capacitor ofourusual kind, asshown inFig.27-3. There isanearly uniform electric ﬁeldinside which is changing with time. Atanyinstant thetotal electromagnetic energy inside isll times thevolume. Iftheplates have aradius aandaseparation h,thetotal energy between theplates is U= E2)(1ra2h). (27.19) This energy changes when Echanges. When thecapacitor isbeing charged, the volume between theplates isreceiving energy attherate -‘-g-t]=e07ra2hEE. (27.20) Sothere must beaflow ofenergy intothatvolume from somewhere. Ofcourse youknow thatitmust come inonthecharging wires—not atall! Itcan’t enter thespace between theplates from thatdirection, because EISperpendicular to theplates; EXBmust beparallel totheplates. You remember, ofcourse, that there isamagnetic ﬁeld thatcircles around theaxiswhen thecapacitor ischarging. Wediscussed thatinChapter 23.Using thelastofMaxwell’s equations, wefound thatthemagnetic ﬁeld attheedge ofthe capacitor isgiven by 27rac2B =E'7ra2, ora.B=-2; E. Itsdirection isshown inFig. 27-3. Sothere isanenergy ﬂow proportional toEXBthat comes inallaround theedges, asshown intheﬁgure. The energy isn’t actually coming down thewires, butfrom thespace surrounding the capacitor. Let's check whether ornotthetotal amount ofﬂowthrough thewhole surface between theedges oftheplates checks withtherateofchange oftheenergy inside— ithadbetter; wewent through allthat work proving Eq.(27.15) tomake sure, 27-7‘kg,’ _s,ﬁ’,E‘Dl) + Fig. 27-3. Near cicharging capaci- tor,thePoynting vector Spoints inward toward theaxis. I -\ V Fig.27-4. Theﬁelds outside 0capacitor when itisbeing charged bybringing two charges from 0large distcince. /‘\ 4‘ii llIE EI I S B L} Fig. 27-5 ThePoynting vector$ necir owire carrying ocurrent. ‘___ E N i B / XX S Fig. 27-6. Acharge and omcignet produce 0Poynting vector that circulates inclosed loops.butlet's see. Thearea ofthe surface is21ra/1, andS=c,,c“)E XBisinmagnitude a I 6[)C2E<“iE‘§ s 7ra2he0EE.sothetotal ﬂuxofenergy is Itdoes check with Eq.(27.20). Butittellsusapeculiar thing: thatwhen weare charging acapacitor. theenergy isnotcoming down thewires; itiscoming in through theedges ofthegap. That’s what thistheory says! How canthatbe?That’s notaneasyquestion, buthereisonewayofthinking about it.Suppose thatwehadsome charges above andbelow thecapacitor and faraway. When thecharges arefaraway, there isaweak butenormously spread- outﬁeld that surrounds thecapacitor. (See Fig.27-4.) Then, asthecharges come together, theﬁeld getsstronger nearer tothecapacitor. Sotheﬁeld energy which iswayoutmoves toward thecapacitor andeventually ends upbetween the plates. Asanother example, weaskwhat happens inapiece ofresistance wirewhen it iscarrying acurrent. Since thewirehasresistance, there isanelectric ﬁeldalong it, driving thecurrent. Because there isapotential drop along thewire, there isalso anelectric ﬁeld just outside thewire, parallel tothesurface. (See Fig. 27-5.) There is,inaddition, amagnetic ﬁeld which goes around thewire because ofthe current. TheEandBareatright angles; therefore there isaPoynting vector directed radially inward, asshown intheﬁgure. There isaﬂow ofenergy intothe wire allaround. Itis,ofcourse, equal totheenergy being lostinthewire inthe form ofheat. Soour“crazy” theory says that theelectrons aregetting their energy togenerate heat because oftheenergy ﬂowing intothewire from theﬁeld outside. Intuition would seem totellusthattheelectrons gettheir energy from being pushed along thewire, sotheenergy should beﬂowing down (orup)along thewire. Butthetheory saysthattheeiectrons arereally being pushed byanelectric ﬁeld, which hascome from some charges very faraway. andthattheelectrons get their energy forgenerating heat from these ﬁelds. Theenergy somehow ﬂows from thedistant charges intoawide area ofspace andthen inward tothewire. Finally, inorder toreally convince youthat thistheory isobviously nuts, wewilltake onemore example—-an example inwhich anelectric charge anda magnet areatrestnear each other—both sitting quite still. Suppose wetake the example ofapoint charge sitting near thecenter ofabarmagnet, asshown in Fig.27-6 Everything isatrest, sotheenergy isnotchanging with time. Also, EandBarequite static. ButthePoynting vector saysthatthere isaﬂowofenergy, because there isanEXBthatisnotzero. Ifyoulook attheenergy ﬂow, youﬁnd thatitjustcirculates around andaround. There is1i't anychange intheenergy anywhere—everything which ﬂows into onevolume ﬂows outagain Itislike incompressible water ﬂowing around. Sothere isacirculation ofenergy inthis so-called static condition. How absurd itgets! Perhaps itisn’t soterribly puzzling, though, when youremember thatwhat wecalled a“static” magnet isreally acirculating permanent current. Inaperma- nent magnet theelectrons arespinning permanently inside. Somaybe acirculation oftheenergy outside isn’t soqueer after all. You nodoubt begin togettheimpression thatthePoynting theory atleast partially violates your intuition astowhere energy islocated inanelectromagnetic ﬁeld. You might believe thatyoumust revamp allyour intuitions, and, therefore have alotofthings tostudy here. Butitseems really notnecessary You don’t need tofeelthatyouwillbeingreat trouble ifyouforget once inawhile thatthe energy inawire isﬂowing intothewire from theoutside, rather than along the wire. Itseems tobeonly rarely ofvalue, when using theideaofenergy conserva- tion, tonotice indetail what path theenergy istaking. Thecirculation ofenergy around amagnet andacharge seems, inmost circumstances, tobequite unimpor- tant. Itisnotavital detail, butitisclear thatourordinary intuitions arequite wrong. 27-8 27-6 Field momentum Next wewould liketotalkabout themomentum intheelectromagnetic ﬁeld. Justastheﬁeld hasenergy. itwillhave acertain momentum perunit volume. Letuscallthatmomentum density g.Ofcourse, momentum hasvarious possible directions, sothatgmust beavector. Let’s talkabout onecomponent atatime; ﬁrst, wetakethex-component. Since each component ofmomentum isconserved weshould beabletowrite down alawthatlooks something likethis: _2momentum __dig+momentum 61 ofmatter I76t outﬂow ,' Theleftsideiseasy. Therate-of-change ofthemomentum ofmatter isjustthe force onit.Foraparticle, itisF=q(E+vXB);foradistribution ofcharges, theforce perunit volume is(pE+jXB). The “momentum outﬂow” term, however, isstrange. Itcannot bethedivergence ofavector because itisnota scalar; itis,rather, anx-component ofsome vector. Anyway, itshould probably look something like 6a 6b 6c a+5+e because thex-momentum could beﬂowing inanyoneofthethree directions. Inanycase, whatever a,b,andcare,thecombination issupposed toequal the outﬂow ofthex-momentum. Now thegame would betowrite pE—l—jXBinterms only ofEandB- eliminating panjbyusing Maxwell’s equations—and thentojuggle terms andmake substitutions togetitintoaform thatlooks like ag, 6a 6b 6c w+&+5+e Then, byidentifying terms, wewould have expressions forg,,,a,b,andc.It’sa lotofwork, andwearenotgoing todoit.Instead, weareonly going toﬁndan expression forg,themomentum density—-and byadiﬂ'erent route. There isanimportant theorem inmechanics which isthis: whenever there is aﬂowofenergy inanycircumstance atall(ﬁeld energy oranyother kind ofenergy). theenergy ﬂowing through aunitarea perunittime, when multiplied by1/02, is equal tothemomentum perunit volume inthespace Inthespecial case ofelec- trodynamics, thistheorem gives theresult thatgisI/c2times thePoynting vector g=gs. (27.21) SothePoynting vector gives notonly energy ﬂow but,ifyoudivide byc2,alsothe momentum density. Thesame result would come outoftheother analysis we suggested. butitismore interesting tonotice thismore general result. Wewill nowgiveanumber ofinteresting examples andarguments toconvince youthat thegeneral theorem istrue. First example: Suppose thatwehave alotofparticles inabox——let’s sayN percubic meter—and thatthey aremoving along with some velocity v.Now let’s consider animaginary plane surface perpendicular tov.Theenergy ﬂow through aunitareaofthissurface persecond isequal toN0,thenumber which ﬂowthrough thesurface persecond, times theenergy carried byeach one. Theenergy ineach particle ismocz/\/l ——7/2/c2. Sotheenergy ﬂow persecond is 2mc NU*i0i--\/I—v2/02 Butthemomentum ofeach particle ismgzi/\/l —112/c2, sothedensity ofmo- mentum isWIQU N I \/1-U2/C2 27-9 l‘ L ‘l U\ GD);/@’ 2, Q“, Cs)“ @(bl @C I I Hg. 27-7. The energy Uin moﬁon at thespeed cccirries themomentum U/c.which isjustI/c2 times theenergy ﬂow—-as thetheorem says. Sothetheorem is trueforabunch ofparticles. Itisalsotrueforlight. When westudied light inVolume I,wesawthatwhen theenergy ISabsorbed from alight beam, acertain amount ofmomentum ISde- livered totheabsorber. Wehave, infact, shown inChapter 36ofVol. Ithatthe momentum is1/ctimes theenergy absorbed [Eq. (36.24) ofVol. I].IfweletU0 betheenergy arriving ataunitarea persecond, then themomentum arriving ata unitareapersecond isU0/c. Butthemomentum istravelling atthespeed c,soits density infront oftheabsorber must beU0/c2. Soagain thetheorem isright. Finally wewillgive anargument duetoEinstein which demonstrates the same thing once more. Suppose thatwehave arailroad caronwheels (assumed frictionless) with acertain bigmass M.Atoneendthere isadevice which will shoot outsome particles orlight (oranything, itdoesn’t make anydifference what itis),which arethen stopped attheopposite endofthecar. There wassome energy originally atoneend—say theenergy Uindicated inFig.27—7(a)—-and then later itisattheopposite end, asshown inFig.27-7(c). Theenergy Uhasbeen displaced thedistance L,thelength ofthecar. Now theenergy Uhasthemass U/c2, soifthecarstayed still, thecenter ofgravity ofthecarwould bemoved. Einstein didn’t liketheideathatthecenter ofgravity ofanobject could bemoved byfooling around only ontheinside, soheassumed thatitisimpossible tomove thecenter ofgravity bydoing anything inside. Butifthatisthecase,when we moved theenergy Ufrom oneendtotheother, thewhole carmust have recoiled some distance x,asshown inpart (c)oftheﬁgure. You cansee,infact, thatthe total mass ofthecar,times x,must equal themass oftheenergy moved, U/c2 times L(assuming that U/c2 ismuch lessthan M): UMX=E5L. (27.22) Let’s nowlook atthespecial caseoftheenergy being carried byalight ﬂash. (The argument would work aswellforparticles, butwewillfollow Einstein, who wasinterested intheproblem oflight )What causes thecartobemoved” Einstein argued asfollows: When thelight isemitted there must bearecoil, some unknown recoil with momentum p.Itisthisrecoil which makes thecarrollbackward. Therecoil velocity 0ofthecarwillbethismomentum divided bythemass ofthe car: 2): Thecarmoves with thisvelocity until thelight energy Ugetstotheopposite end. Then, when ithits, itgives back itsmomentum andstops thecar. Ifxissmall, then thetime thecarmoves isnearly equal toL/c; sowehave that __L_pL X—Ul—U;—MC Putting thisxinEq.(27.22), wegetthat UP=‘LT' Again wehave therelation ofenergy andmomentum forlight. Dividing bycto getthemomentum density g=p/c, wegetonce more that ug=C2 (27.23) You may well wonder: What issoimportant about thecenter-of-gravity theorem? Maybe itiswrong. Perhaps, butthen wewould alsolosetheconserva- tionofangular momentum. Suppose thatourboxcar ismoving along atrack at some speed Uandthatweshoot some light energy from thetoptothebottom of thecar—say, from AtoBinFig.27-8. Now welook attheangular momentum of thesystem about thepoint PBefore theenergy Uleaves A,ithasthemass 27-10 m=U2/c andthevelocity I’,soithastheangular momentum Wlllfa When it arrives atB,ithasthesame mass and, ifthelinear momentum ofthewhole boxcar isnottochange, itmust stillhave thevelocity ll.It’sangular momentum about P isthen mvrB. The angular momentum willbechanged unless theright recoil momentum wasgiven tothecarwhen thelight wasemitted—that is,unless the light carries themomentum U/c. Itturns outthattheangular momentum con- servation andthetheorem ofcenter-of-gravity areclosely related intherelativity theory. Sotheconservation ofangular momentum would alsobedestroyed ifour theorem were nottrue Atanyrate, itdoes turn outtobeatruegeneral law,and inthecaseofelectrodynamics wecanuseittogetthemomentum intheﬁeld. Wewillmention twofurther examples ofmomentum intheelectromagnetic ﬁeld. Wepointed outinSection 26-2 thefailure ofthelawofaction andreaction when twocharged particles were moving onorthogonal trajectories. Theforces onthetwoparticles don’t balance out,sotheaction andreaction arenotequal. therefore thenetmomentum ofthematter must bechanging. Itisnotconserved Butthemomentum intheﬁeld isalsochanging insuch asituation. Ifyouwork outtheamount ofmomentum given bythePoynting vector, itisnotconstant. However, thechange oftheparticle momenta isjust made upbytheﬁeld momen- tum, sothetotal momentum ofparticles plusﬁeld isconserved. Finally, another example isthesituation with themagnet andthecharge. shown inFig.27-6. Wewere unhappy toﬁndthatenergy wasﬂowing around in circles, butnow, since weknow thatenergy ﬂow andmomentum areproportional, weknow alsothatthere ismomentum circulating inthespace. Butacirculating momentum means thatthere isangular momentum. Sothere isangular momentum intheﬁeld. Doyouremember theparadox wedescribed inSection 17-4 about a solenoid andsome charges mounted onadisc? Itseemed that when thecurrent turned off,thewhole discshould start toturn Thepuzzle was: Where didthe angular momentum come from ?Theanswer isthatifyouhave amagnetic ﬁeldand some charges, there willbesome angular momentum intheﬁeld. Itmust have been putthere when theﬁeldwasbuilt up.When theﬁeldisturned off,theangular momentum isgiven back. Sothediscintheparadox would start rotating. This mystic circulating ﬂow ofenergy, which atﬁrstseemed soridiculous, isab- solutely necessary. There isreally amomentum ﬂow. Itisneeded tomaintain the conservation ofangular momentum inthewhole world. 27-11A @jli'-@-P ___, _ Fig. 27-8. The energy Umust corry themomentum U/c ifthecingulor mo- mentum about Pistobeconserved. 28 Electromagnetic Mass 28-1 Theﬁeldenergy ofapoint charge Inbringing together relativity andMaxwell’s equations, wehave ﬁnished our main work onthetheory ofelectromagnetism. There are,ofcourse, some details wehaveskipped overandonelarge areathatwewillbeconcerned withinthefuture —-the interaction ofelectromagnetic ﬁelds with matter. Butwewant tostop fora moment toshow you that thistremendous ediﬁce, which issuch abeautiful success inexplaining somany phenomena, ultimately falls onitsface. When youfollow anyofourphysics toofar,youﬁndthatitalways getsintosome kind oftrouble. Now wewant todiscuss aserious trouble—the failure oftheclassical electromagnetic theory. You canappreciate thatthere isafailure ofallclassical physics because ofthequantum-mechanical effects. Classical mechanics isamathe- matically consistent theory; itjustdoesn’t agree with experience. Itisinteresting, though, thattheclassical theory ofelectromagnetism isanunsatisfactory theory allbyitself. There aredifficulties associated with theideas ofMaxwell’s theory which arenotsolved byandnotdirectly associated with quantum mechanics. Youmaysay,“Perhaps there’s nouseworrying about these difﬁculties. Since the quantum mechanics isgoing tochange thelaws ofelectrodynamics, weshould wait toseewhat diﬂiculties there areafter themodiﬁcation.” However, when electromagnetism isjoined toquantum mechanics, thedifficulties remain. Soit willnotbeawaste ofourtimenowtolookatwhat these difﬁculties are. Also, theyareofgreat historical importance. Furthermore, youmaygetsome feeling ofaccomplishment from being abletogofarenough withthetheory toseeevery- thing-including allofitstroubles. Thedifficulty wespeak ofisassociated with theconcepts ofelectromagnetic momentum andenergy, when applied totheelectron oranycharged particle. Theconcepts ofsimple charged particles andtheelectromagnetic ﬁeld areinsome wayinconsistent. Todescribe thedifficulty, webegin bydoing some exercises withourenergy andmomentum concepts. First, wecompute theenergy ofacharged particle. Suppose wetakeasimple model ofanelectron inwhich allofitscharge qisuniformly distributed onthe surface ofasphere ofradius a,which wemaytaketobezeroforthespecial case of apoint charge. Now let’s calculate theenergy intheelectromagnetic ﬁeld. If thecharge isstanding still, there isnomagnetic ﬁeld, andtheenergy perunit volume isproportional tothesquare oftheelectric ﬁeld. Themagnitude ofthe electric ﬁeldisq/41re,,r2, andtheenergy density is _E 2_ Q2 _ ”"2EV327i'2e0r4 Togetthetotal energy, wemust integrate thisdensity over allspace. Using the volume element 41rr2 dr,thetotal energy, which wewillcallU,.1,.,,. is 2 Uelec : /‘L. dr-87T€(]V2 Thisisreadily integrated. Thelower limit isa,andtheupper limit isoo,so _11121Uelcc —E1;?) 5' (28-1) 28-128-1 28-2 28-3 28-4 28-5 28-6Theﬁeld energy ofapoint charge Theﬁeld momentum ofa moving charge Electromagnetic mass Theforce ofanelectron on itself Attempts tomodify the Maxwell theory Thenuclear force ﬁeld F>\ t,__/ g- Q4? 0 1/ \_;, +\ ¢ SPHERICAL ELECTRON (1')7\IO Fig. 28-1. Theﬁelds EcindBandthe momentum density gfor0positive elec- tron. Forcinegative electron, Eand B arereversed butgisnot. *‘ l<—dr O9 ‘ rd8 G/L- MFig. 28-2. The volume element 21rr2 sin0d0drused forcalculating the ﬁeld momentum.QIfweusetheelectronic charge qeforqandthesymbol e2forq?/41re0, then 2 [\)>—n areUelec : ' Itisallﬁneuntil wesetaequal tozero forapoint charge—there’s thegreat difficulty. Because theenergy oftheﬁeld varies inversely asthefourth power of thedistance from thecenter, itsvolume integral isinﬁnite. There isaninﬁnite amount ofenergy intheﬁeld surrounding apoint charge. What’s wrong with aninﬁnite energy‘? Iftheenergy can’t getout,butmust staythere forever, isthere anyrealdifficulty Wl[l'1 aninﬁnite energy? Ofcourse, a quantity thatcomes outinﬁnite may beannoying, butwhat really matters isonly whether there areanyobservable physical effects. Toanswer thatquestion, we must turn tosomething elsebesides theenergy. Suppose weaskhow theenergy changes when wemove thecharge. Then, ifthechanges areinﬁnite, wewillbe introuble. 28-2 Theﬁeld momentum ofamoving charge Suppose anelectron ismoving atauniform velocity through space, assuming foramoment thatthevelocity islowcompared with thespeed oflight. Associated with thismoving electron there isamomentum—even iftheelectron hadnomass before itwascharged—because ofthemomentum intheelectromagnetic ﬁeld. Wecanshow thattheﬁeld momentum isinthedirection ofthevelocity vofthe charge andis,forsmall velocities, proportional to11.Forapoint Patthedistance rfrom thecenter ofthecharge andattheangle 6with respect tothelineofmotion (seeFig.28-1) theelectric ﬁeld isradial and, aswehave seen, themagnetic ﬁeld isvXE/c2. Themomentum density, Eq.(27.21), is g=€QE><B. It1Sdirected obliquely toward thelineofmotion, asshown intheﬁgure, andhas themagnitudeev .g=T‘;E2sin0. Theﬁelds aresymmetric about thelineofmotion, sowhen weintegrate over space, thetransverse components willsumtozero, giving aresultant momentum parallel tov.Thecomponent ofginthisdirection isgsin0.which wemust inte- grate over allspace. Wetake asourvolume element aringwith itsplane per- pendicular tov,asshown inFig.28-2. Itsvolume is21172 sin0d0dr.Thetotal momentum isthen p=/% E2sin2027172 sin0d0dr. Since Eisindependent of0(foru<<c),wecanimmediately integrate over 0;the integral is 3 />sin30d0 =—/‘(I —cos20)d(cos0) =—cos0+EO?%9- Thelimits of0areOand1r,sothe9-integral gives merely afactor of4/3, and p= £37-r%fE2r2dr. Theintegral (forv<<c)istheonewehave justevaluated toﬁndtheenergy; itis q2/l61r2e§a, and _2q2 v 1’-51;;F’or 2 2 p=3%v. (28.3) 28-2 Themomentum intheﬁeld—the electromagnetic momentum—is proportional to v.Itisjustwhat weshould have foraparticle with themass equal tothecoefficient ofv.Wecan,therefore, callthiscoelﬁcient theelectromagnetic mass, mam, and writeitas 22 .=§ (28.4) 28-3 Electromagnetic mass Where does themass come from? Inourlaws ofmechanics wehave supposed thatevery object “carries” athing wecallthemass—which also means that it “carries” amomentum proportional toitsvelocity. Now wediscover thatitis understandable that acharged particle carries amomentum proportional toits velocity. Itmight, infact, bethatthemass isjusttheeﬁect ofelectrodynamics. Theorigin ofmass hasuntil now been unexplained. Wehave atlastinthetheory ofelectrodynamics agrand opportunity tounderstand something that wenever understood before. Itcomes outoftheblue~or rather, from Maxwell and Poynting—that anycharged particle willhave amomentum proportional toits velocity justfrom electromagnetic inﬂuences. Let’s beconservative andsay,foramoment, thatthere aretwokinds ofmass— thatthetotalmomentum ofanobject could bethesumofamechanical momentum andtheelectromagnetic momentum. Themechanical momentum isthe“mechan- ical”mass, m,,,,.,.|,, times v.Inexperiments where wemeasure themass ofaparticle byseeing howmuch momentum ithas, orhow itswings around inanorbit, we aremeasuring thetotal mass. Wesaygenerally thatthemomentum isthetotal mass (m,,,c,,,, +mom.) times thevelocity. Sotheobserved mass canconsist oftwo pieces (orpossibly more ifweinclude other ﬁelds): amechanical piece plus an electromagnetic piece. Weknow thatthere isdeﬁnitely anelectromagnetic piece, andwehave aformula forit.And there isthethrilling possibility thattheme- chanical piece isnotthere atall—that themass isallelectromagnetic. Let’sseewhatsizetheelectron must haveifthere istobenomechanical mass. Wecanﬁndoutbysetting theelectromagnetic mass ofEq.(28.4) equal tothe observed massmuofanelectron. Weﬁnd 2 (1= (23.5) Thequantity r0=-‘ii (28.6)m,,c2 iscalled the“classical electron radius”; ithasthenumerical value 2.82 X10'“ cm,about oneone-hundred-thousandth ofthediameter ofanatom. Why isrocalled theelectron radius, rather than oura?Because wecould equally welldothesame calculation with other assumed distributions ofcharges— thecharge might bespread uniformly through thevolume ofasphere oritmight besmeared outlikeafuzzy ball. Foranyparticular assumption thefactor 2/3 would change tosome other fraction. Forinstance, foracharge uniformly dis- tributed throughout thevolume ofasphere, the2/3getsreplaced by4/5. Rather thantoargue over which distribution iscorrect, itwasdecided todeﬁne r0asthe “nominal” radius. Then different theories could supply their petcoefﬁcients. Let’s pursue ourelectromagnetic theory ofmass. Ourcalculation wasfor v<<c;what happens ifwegotohigh velocities? Early attempts ledtoacertain amount ofconfusion, butLorentz realized thatthecharged sphere would contract intoaellipsoid athigh velocities andthattheﬁelds would change inaccordance withtheformulas (266)and(26.7) wederived fortherelativistic caseinChapter 26. Ifyoucarry through theintegrals forpinthatcase, youﬁndthatforanarbitrary velocity v,themomentum isaltered bythefactor 1/\/l—112/c2: 2e2 v ”r2 ' <28”28-3 Inother words, theelectromagnetic mass rises with velocity inversely as \/l—v2/c2——a discovery thatwasmade before thetheory ofrelativity. Early experiments were proposed tomeasure thechanges with velocity inthe observed mass ofaparticle inorder todetermine how much ofthemass was mechanical andhowmuch waselectrical. Itwasbelieved atthetime thattheelec- trical part would vary with velocity, whereas themechanical partwould not. But while theexperiments were being done, thetheorists were alsoatwork. Soon the theory ofrelativity wasdeveloped, which proposed thatnomatter what theorigin ofthemass, itallshould vary asmo/\/1 —v2/c2. Equation (28.7) wasthe beginning ofthetheory thatmass depended onvelocity. Let’s now goback toourcalculation oftheenergy intheﬁeld, which ledto Eq.(28.2). According tothetheory ofrelativity, theenergy Uwillhave themass U/c2; Eq.(28.2) then says that theﬁeld oftheelectron should have themass Us cc 192 mblec =*5?‘ =52?’ (28-8) which isnotthesame astheelectromagnetic mass, m,.1,.,., ofEq.(28.4). Infact, if wejustcombine Eqs. (28.2) and(28.4), wewould write 3 2Uelec =Zmelecc - This formula wasdiscovered before relativity, andwhen Einstein andothers began torealize thatitmust always bethat U=mcz, there wasgreat confusion. 28-4 Theforce ofanelectron onitself Thediscrepancy between thetwoformulas fortheelectromagnetic mass is especially annoying, because wehave carefully proved thatthetheory ofelectro- dynamics isconsistent with theprinciple ofrelativity. Yetthetheory ofrelativity implies without question that themomentum must bethesame astheenergy times v/c2. Soweareinsome kind oftrouble; wemust have made amistake. Wedidnotmake analgebraic mistake inourcalculations, butwehave leftsome- thing out. Inderiving ourequations forenergy andmomentum, weassumed thecon- servation laws. Weassumed thatallforces were taken intoaccount andthatany work done andanymomentum carried byother “nonelectrical” machinery was included. Now ifwehave asphere ofcharge, theelectrical forces areallrepulsive andanelectron would tend toﬂyapart. Because thesystem hasunbalanced forces, wecangetallkinds oferrors inthelaws relating energy andmomentum. Togeta consistent picture, wemust imagine that something holds theelectron together. Thecharges must beheldtothesphere bysome kind ofrubber bands—something thatkeeps thecharges from ﬂying off.Itwasﬁrstpointed outbyPoincaré thatthe rubber bands—or whatever itisthatholds theelectron together—must beincluded intheenergy andmomentum calculations. Forthisreason theextra nonelectrical forces arealsoknown bythemore elegant name “the Poincaré stresses.” Ifthe extra forces areincluded inthecalculations, themasses obtained intwoways are changed (inawaythatdepends onthedetailed assumptions). And theresults are consistent with relativity; i.e.,themass thatcomes outfrom themomentum cal- culation isthesame astheonethatcomes from theenergy calculation. However, both ofthem contain twocontributions: anelectromagnetic mass andcontribution from thePoincaré stresses. Only when thetwoareadded together dowegeta consistent theory. Itistherefore impossible togetallthemass tobeelectromagnetic intheway wehoped. Itisnotalegal theory ifwehave nothing butelectrodynamics. Some- thing elsehastobeadded. Whatever youcallthem—“rubbcr bands,” or“Poincaré stresses,” orsomething else—there have tobeother forces innature tomake a consistent theory ofthiskind. 28-4 Clearly, assoon aswehave toputforces ontheinside oftheelectron, the beauty ofthewhole ideabegins todisappear. Things getvery complicated. You would want toask: How strong arethestresses? How does theelectron shake? Does itoscillate? What areallitsinternal properties? And soon.Itmight be possible thatanelectron does have some complicated internal properties. Ifwe made atheory oftheelectron along these lines, itwould predict oddproperties, likemodes ofoscillation, which haven’t apparently been observed. Wesay“ap- parently” because weobserve alotofthings innature thatstilldonotmake sense. Wemaysomeday ﬁndoutthatoneofthethings wedon't understand today (for example, themuon) can, infact, beexplained asanoscillation ofthePoincaré stresses. Itdoesn’t seem likely, butnoonecansayforsure. There aresomany things about fundamental particles thatwestilldon’t understand. Anyway, the complex structure implied bythistheory isundesirable, andtheattempt toexplain allmass interms ofelectromagnetism—at least inthewaywehave described-—-has ledtoablind alley. Wewould liketothink alittle more about whywesaywehave amass when themomentum intheﬁeld ISproportional tothevelocity. Easy! Themass isthe coeﬂicient between momentum andvelocity. Butwecanlook atthemass inanother way: aparticle hasmass ifyouhave toexert aforce inorder toaccelerate it.So itmayhelpourunderstanding ifwelook alittle more closely atwhere theforces come from. How doweknow thatthere hastobeaforce? Because wehave proved thelawoftheconservation ofmomentum fortheﬁelds. Ifwehave a charged particle andpush onitforawhile, there willbesome momentum inthe electromagnetic ﬁeld. Momentum must have been poured intotheﬁeldsomehow. Therefore there must have been aforce pushing ontheelectron inorder togetit going—a force inaddition tothatrequired byitsmechanical inertia, aforce due toitselectromagnetic interaction. And there must beacorresponding force back onthe“pusher.” Butwhere does thatforce come from? - -dF / d2F - -dF / 1°' 4_ (O) (b) (C),?\,\ \ Fig. 28-3. Theself-force onanaccelerating electron isnotzero because ofthe retardation. (BydFwemean theforce onasurface element da;byd2Fwemean the force onthesurface element da.,from thecharge onthesurface element dﬂg.) Thepicture issomething likethis. Wecanthink oftheelectron asacharged sphere. When itisatrest,each piece ofcharge repels electrically each other piece, buttheforces allbalance inpairs, sothatthere isnonetforce. [SeeFig.28—3(a).] However, when theelectron isbeing accelerated, theforces willnolonger bein balance because ofthefactthat theelectromagnetic inﬂuences take time togo from onepiece toanother. Forinstance, theforce onthepiece ainFig.28—3(b) from apiece 5ontheopposite sidedepends ontheposition of6atanearlier time, asshown. Both themagnitude anddirection oftheforce depend onthemotion ofthecharge. Ifthecharge isaccelerating, theforces onvarious parts ofthe electron might beasshown inFig.28—3(c). When allthese forces areadded up, theydon’t cancel out. They would cancel forauniform velocity, even though itlooks atﬁrstglance asthough theretardation would giveanunbalanced force even forauniform velocity. Butitturns outthatthere isnonetforce unless the electron isbeing accelerated. With acceleration, ifwelook attheforces between 28-5 thevarious parts oftheelectron, action andreaction arenotexactly equal, and theelectron exerts afdrce onitself thattriestohold back theacceleration. Itholds itself back byitsown bootstraps. Itispossible, butdifficult, tocalculate thisself-reaction force; however, we don’t want togointosuch anelaborate calculation here. Wewilltellyouwhat theresult isforthespecial caseofrelatively uncomplicated motion inonedimension, sayx.Then, theself-force canbewritten inaseries. Theﬁrstterm intheseries depends ontheacceleration X‘,thenext term isproportional tox,andsoon.* Theresult is >’. 21. »’F=¢ta°C_,)<-3-Z.-,»<+viC,»-,:‘)<+~~» (28.9) where orandyarenumerical coefficients oftheorder of1.Thecoefﬁcient <1of theatterm depends onwhat charge distribution isassumed; ifthecharge isdis- tributed uniformly onasphere, then or=2/3. Sothere isaterm. proportional totheacceleration, which varies inversely astheradius aoftheelectron andagrees exactly with thevalue wegotinEq.(28.4) form,,1,,c. Ifthecharge distribution is chosen tobedifferent, sothatorischanged, thefraction 2/3inEq.(28.4) would bechanged inthesame way Theterm inX"isindependent oftheassumed radius a,andalsooftheassumed distribution ofthecharge; itscoefficient isalways 2/3. Thenext term isproportional totheradius a,anditscoefﬁcient 'ydepends onthe charge distribution. You willnotice thatifwelettheelectron radius agotozero, thelastterm (and allhigher terms) willgotozero; thesecond term remains con- stant, buttheﬁrstterm—the electromagnetic mass—goes toinﬁnity. And wecan seethattheinﬁnity arises because oftheforce ofonepartoftheelectron onanother ——because wehave allowed what isperhaps asilly thing, thepossibility ofthe “point” electron acting onitself. 28-5 Attempts tomodify theMaxwell theory Wewould likenow todiscuss how itmight bepossible tomodify Maxwell’s theory ofelectrodynamics sothattheidea ofanelectron asasimple point charge could bemaintained. Many attempts have been made, andsome ofthetheories were even abletoarrange things sothatalltheelectron mass waselectromagnetic. Butallofthese theories have died. Itisstillinteresting todiscuss some ofthe possibilities thathave been suggested—to seethestruggles ofthehuman mind. Westarted outourtheory ofelectricity bytalking about theinteraction of onecharge with another. Then wemade upatheory ofthese interacting charges andended upwith aﬁeld theory. Webelieve itsomuch thatweallow ittotell usabout theforce ofonepartofanelectron onanother. Perhaps theentire diffi- culty isthatelectrons donotactonthemselves; perhaps wearemaking toogreat anextrapolation from theinteraction ofseparate electrons totheidea that an electron interacts withitself. Therefore some theories have been proposed inwhich thepossibility thatanelectron actsonitself isruled out. Then there isnolonger theinﬁnity duetotheself-action. Also, there isnolonger anyelectromagnetic mass associated with theparticle; allthemass isback tobeing mechanical, but there arenewdifﬁculties inthetheory. Wemust sayimmediately that such theories require amodiﬁcation ofthe ideaoftheelectromagnetic ﬁeld. You remember wesaidatthestart thattheforce onaparticle atanypoint wasdetermined byjusttwoquantities—E andB.If weabandon the“self-force” thiscannolonger betrue, because ifthere isanelec- troninacertain place, theforce isn’t given bythetotal EandB,butbyonly those parts duetoother charges. Sowehave tokeep track always ofhow much ofE andBisduetothecharge onwhich youarecalculating theforce andhow much isduetotheother charges. This makes thetheory much more elaborate, butit getsridofthediﬂiculty oftheinﬁnity. *Weareusing thenotation: x=dx/dt, it=d2x/dt2, x=d3x/dt3, etc. 28-6 Sowecan,ifwewant to,saythatthere isnosuch thing astheelectron acting upon itself, andthrow away thewhole setofforces inEq.(28.9). However, we have thenthrown away thebaby with thebath! Because thesecond term inEq. (28.9), theterm in>2‘,isneeded. That force does something very deﬁnite. Ifyou throw itaway, you’re introuble again. When weaccelerate acharge, itradiates electromagnetic waves, soitloses energy. Therefore, toaccelerate acharge, we must require more force than isrequired toaccelerate aneutral object ofthesame mass; otherwise energy wouldn’t beconserved. Therateatwhich wedowork on anaccelerating charge must beequal totherateoflossofenergy persecond by radiation. Wehave talked about thiseffect before—-it iscalled theradiation re- sistance. Westillhave toanswer thequestion: Where does theextra force, against which wemust dothiswork, come from’? When abigantenna isradiating, the forces come from theinﬂuence ofonepart oftheantenna current onanother. Forasingle accelerating electron radiating into otherwise empty space, there would seem tobeonly oneplace theforce could come from—the action ofone partoftheelectron onanother part. Wefound back inChapter 32ofVol. Ithat anoscillating charge radiates energy attherate dW 22"271}-= 3-5£%- (28.10) Let’s seewhat wegetfortherateofdoing work onanelectron against theboot- strap force ofEq.(28.9). Therateofwork istheforce times thevelocity, orFx; 2 2 %/=a%xx-%‘§-5>'ex+--- (28.11) Theﬁrstterm isproportional to11262/dt, andtherefore justcorresponds totherate ofchange ofthekinetic energy %mvzassociated with theelectromagnetic mass. Thesecond term should correspond totheradiated power inEq.(28.10). Butit isdifferent. Thediscrepancy comes from thefactthattheterm inEq.(28.11) is generally true,whereas Eq.(28.10) isright onlyforanoscillating charge. Wecan show thatthetwoareequivalent ifthemotion ofthecharge isperiodic. Todo that,werewrite thesecond term ofEq.(28.11) as 2e2d,,_ 2e2_, 'it-—3a("") +ta(")3 which isjustanalgebraic transformation. Ifthemotion oftheelectron isperiodic, thequantity returns periodically tothesame value, sothat ifwetake the average ofitstime derivative, wegetzero. Thesecond term, however, isalways positive (it‘sasquare), soitsaverage 1Salsopositive This term gives thenetwork done andisjustequal toEq.(28.10). Theterm inJEofthebootstrap force isrequired inorder tohave energy conservation inradiating systems, andwecan’t throw itaway. Itwas, infact, one ofthetriumphs ofLorentz toshow thatthere issuch aforce andthatitcomes from theaction oftheelectron onitself. Wemust believe intheideaoftheaction ofthe electron onitself, andweneed theterm in Theproblem ishowwecangetthat term without getting theﬁrstterm inEq.(28.9), which gives allthetrouble. We don’t know how. You seethat theclassical electron theory haspushed itself intoatight corner. There have been several other attempts tomodify thelawsinorder tostraighten thething out. Oneway, proposed byBorn andInfeld, istochange theMaxwell equations inacomplicated waysothatthey arenolonger linear. Then theelectro- magnetic energy andmomentum canbemade tocome outﬁnite. Butthelaws theysuggest predict phenomena which have never been observed. Their theory alsosuﬂers from another diﬂiculty wewillcome tolater, which iscommon toall theattempts toavoid thetroubles wehave described. Thefollowing peculiar possibility wassuggested byDirac. Hesaid: Let’s admit thatanelectron actsonitself through thesecond term inEq.(28.9) butnot through theﬁrst. Hethen hadaningenious ideaforgetting ridofonebutnotthe 28-7 other. Look, hesaid, wemade aspecial assumption when wetook only the retarded wave solutions ofMaxwell’s equations; ifwewere totake theadvanced waves instead, wewould getsomething different. Theformula fortheself-force would be e2 2e2 02:1F=a—.x+»~.-"x='+v_- X.‘ (28.12)acz 3c-t c4 This equation isJLISIlikeEq.(28.9) except forthesignofthesecond terin—and some higher terms~of theseries [Changing from retarded toadvanced waves isjustchanging thesignofthedelay which, itisnothard tosee,isequivalent to changing thesignofreverywhere. Theonly effect onEq.(28.9) istochange the signofalltheoddtime derivatives.] So,Dirac said, let’smake thenewrulethat anelectron actsonitself byone-half thedzﬁference oftheretarded andadvanced ﬁelds which itproduces. Thedifference ofEqs. (28.9) and(28.12). divided bytwo, isthen 2 F=— x+higher terms. Inallthehigher terms, theradius aappears tosome positive power inthenumera- tor. Therefore. when wegotothelimit ofapoint charge, wegetonly theone term—just what isneeded. Inthisway, Dirac gottheradiation resistance force andnone oftheinertial forces. There isnoelectromagnetic mass, andtheclassical theory issaved—but attheexpense ofanarbitrary assumption about theself-force. Thearbitrariness oftheextra assumption ofDirac wasremoved, tosome ex- tentatleast, byWheeler and Feynman, who proposed astillstranger theory They suggest thatpoint charges interact onlywith other charges, butthattheinter- action ishalfthrough theadvanced andhalfthrough theretarded waves. Itturns out, most surprisingly, that inmost situations youwon’t seeanyeffects ofthe advanced waves, butthey dohave theeffect ofproducing _]LlSltheradiation re- action force. Theradiation resistance isnotduetotheelectron acting onitself, butfrom thefollowing peculiar effect. When anelectron isaccelerated atthe time r,itshakes alltheother charges intheworld atalater time 1'Ir+r/c (where risthedistance totheother charge), because oftheretarded waves. But then these other charges react back ontheoriginal electron through their advanced waves, which willarrive atthetime t”,equal tot’minus r/c,which is,ofcourse, JustI.(They also react back with their retarded waves too, butthatjustcorre- sponds tothenormal “reﬂected” waves.) Thecombination oftheadvanced and retarded waves means that attheinstant itisaccelerated anoscillating charge feels aforce from allthecharges thatare“going to”absorb itsradiated waves. You seewhat tight knots people have gotten intointrying togetatheory ofthe electron! We’ll describe now stillanother kind oftheory, toshow thekind ofthings thatpeople think ofwhen theyarestuck. This isanother modiﬁcation ofthelaws ofelectrodynamics, proposed byBopp. You realize thatonce youdecide tochange theequations ofelectromagnetism youcanstart anywhere youwant. You can change theforce lawforanelectron, oryoucanchange theMaxwell equations (aswesawintheexamples wehave described). oryoucanmake achange some- where else. One possibility istochange theformulas that give thepotentials in terms ofthe charges andcurrents. Oneofourformulas hasbeen thatthepotentials atsome point aregiven bythecurrent density (orcharge) ateach other point atan earlier time Using ourfour-vector notation forthepotentials, wewrite AA], I)2____1_ -l.#(2_iIL!?_/C_) d|/2 (2313) 4Tl'€()C2 _ /'12 Bopp’s beautifully simple idea isthat: Maybe thetrotible isinthel/rfactor in theintegral. Suppose wewere tostart outbyassuming only thatthepotential at onepoint depends onthecharge density atanyother point assome function of thedistance between thepoints, sayasf(r12). Thetotal potential atpoint (l) 28-8 willthenbegiven bytheintegral ofj,,times thisfunction over allspace: Am)=/j#(2)f(r12) dVZ- That’s all.Nodifferential equation, nothing else. Well, onemore thing. Wealso askthattheresult should berelativistically invariant. Soby“distance” weshould take theinvariant “distance” between twopoints inspace-time. This distance squared (within asignwhich doesn’t matter) is Siz=52(t1— l2)2 "Viz =C201 -lzlz_(X1—X2)2 "‘(J/1"J'2)2 _(Z1—Z2)2- (28-14) S0,forarelativistically invariant theory, weshould take some function ofthe magnitude ofs12,orwhat isthesame thing, some function ofsf, SoB0pp’s theory isthat A,.<1,:1)=/1;.<2,:2>F<s122)dV2 do (28.15) (Theintegral must, ofcourse, beoverthefour-dimensional volume dfgd./Y2dygdz;) Allthatremains istochoose asuitable function forF.Weassume only one thing about F-—that itisverysmall except when itsargument isnearzero—so thata graph ofFwould beacurve liketheoneinFig.28-4. Itisanarrow spike with a ﬁnite areacentered ats2=O,andwith awidth which wecansayisroughly a2. Wecansay,crudely, thatwhen wecalculate thepotential atpoint (1),only those points (2)produce anyappreciable effect ifS?2=c2(t2 —t1)2 —~rigiswithin =a2ofzero. Wecanindicate thisbysaying thatFisimportant only for £2=ﬁn,-z2)2-£2~:02. (28.16) Youcanmake itmore mathematical ifyouwant to,butthat’s theidea. Now suppose thataisvery small incomparison with thesizeofordinary objects likemotors, generators, andthelikesothatfornormal problems r12>>a. Then Eq.(28.16) saysthatcharges contribute totheintegral ofEq.(28.15) only when 11—12isinthesmall range a2 C(t1 — I2) Z \/7&2 :1:G2 %V12‘/1 =5 r12 Since a2/rig <<1,thesquare root canbeapproximated byI=ta2/2r%2, so 2 2 ,._t2=’1"=(i 1“)=’12i “_c Zrfz c 2712C What isthesigniﬁcance? This result saysthattheonly times :2thatareim- portant intheintegral ofA,,arethose which differ from thetime t1,atwhich we want thepotential, bythedelay r12/c——with anegligible correction solong as r12>>a.Inother words, thistheory ofBopp approaches theMaxwell theory—so longaswearefaraway from anyparticular charge——in thesense thatitgives the retarded wave effects. Wecan,infact, seeapproximately what theintegral ofEq.(28.15) isgoing togive. Ifweintegrate ﬁrstover t2from —ooto+<>o—keeping r12ﬁxed~then sf;isalsogoing togofrom —ooto+oo.Theintegral willallcome from t2’sin asmall interval ofwidth A12=2><a2/2r12c, centered at:1—r12/c. Say thatthefunction F(s2) hasthevalue Kats2=0;then theintegral over I2gives approximately K]',,At2, or Eit.C7'12 Weshould, ofcourse, take thevalue ofj,,atI2=t1—r12/c, sothatEq.(28.15) becomes 2 - _ A#(1, ,1)= dV2_ 28-9If(s2) O2 6 $2 tot I '12 lE\ \\2//I lb) Fig. 28-4. Thefunction F(s2l used in thenonlocul theory ofBopp. Ifwepick K=q2c/41re0a2, weareright back totheretarded potential solution ofMaxwell’s eqtiations—including automatically thel/rdependence! And it allcame outofthesimple proposition that thepotential atonepoint inspace- time depends onthecurrent density atallother points inspace-time, butwith aweighting factor thatissome narrow function ofthefour-dimensional distance between thetwopoints. This theory again predicts aﬁnite electromagnetic mass fortheelectron, andtheenergy andmass have theright relation fortherelativity theory They must, because thetheory isrelativistically invariant from thestart, andeverything seems tobeallright There is,however, onefundamental objection tothistheory andtoallthe other theories wehave described. Allparticles weknow obey thelaws ofquantum mechanics, soaquantum-mechanical modiﬁcation ofelectrodynamics hastobe made Light behaves likephotons Itisn’t lO0percent liketheMaxwell theory. Sotheelectrodynamic theory hastobechanged. Wehave already mentioned that itmight beawaste oftime towork sohard tostraighten outtheclassical theory, because itcould turn outthat inquantum electrodynamics thedifficulties Wlll disappear ormay beresolved insome other fashion. Butthedifficulties donot disappear inquantum electrodynamics. That isoneofthereasons that people have spent somuch effort trying tostraighten outtheclassical difficulties, hoping thatifthey could straighten outtheclassical difficulty andthenmake thequantum modiﬁcations, everything would bestraightened out The Maxwell theory still hasthed1lTlCLllil€S after thequantum mechanics modiﬁcations aremade. The quantum effects domake some changes—the formula forthemass is modiﬁed, andPlanck’s constant liappears—but theanswer stillcomes outinﬁnite unless youcutoffanintegration somehow—just aswehadtostop theclassical integrals atr=a.And theanswers depend onhow youstop theintegrals. We cannot, unfortunately, demonstrate foryou here that thedifficulties arereally basically thesame, because wehave developed solittle ofthetheory ofquantum mechanics andeven lessofquantum electrodynamics. Soyoumust justtake otir word that thequantized theory ofMaxwell’s electrodynamics gives aninﬁnite mass forapoint electron. Itturns out,however.that nobody haseversucceeded inmaking a90//-C()I1\[\I(3I1f quantum theory outofanyofthemodiﬁed theories. Born andlnfeld's ideas have never been satisfactorily made intoaquantum theory Thetheories with thead- vanced and retarded waves ofDirac, orofWheeler and Feynman, have never been made into asatisfactory quantum theory Thetheoiy ofBopp hasnever been made intoasatisfactory quantum theory. Sotoday, there isnoknown solution tothisproblem. Wedonotknow how tomake aconsistent theory—including thequantum mechanics—which does notproduce aninﬁnity fortheself-energy of anelectron, oranypoint charge. And atthesame time, there isnosatisfactory theory thatdescribes anon—point charge. It’sanunsolved problem Incaseyouaredeciding torushofftomake atheory inwhich theaction ofan electron onitself iscompletely removed, sothat electromagnetic mass isnolonger meaningful, andthen tomake aquantum theory ofit,youshould bewarned that youarecertain tobeintrouble. There isdeﬁnite experimental evidence ofthe existence ofelectromagnetic inertia—there ISevidence that some ofthemass of charged particles iselectromagnetic inorigin ltused tobesaid intheolder books that since Nature Wlllobviously notpre- sentuswith twoparticles~one neutral andtheother charged, butotherwise the same—we willnever beabletotellhow much ofthemass iselectromagnetic and how much ismechanical Butitturns outthat Nature /I(l\been kind enotigh to present uswith justsuch objects, sothatbycomparing theobserved mass ofthe charged onewith theobserved mass oftheneutral one, wecantellwhether there isanyelectromagnetic mass. Forexample, there aretheneutrons andprotons. They interact with tremendous forces—the nuclear forces~whose origin isun- known However, aswehave already described, thenuclear forces have one re- markable property. Sofarasthey areconcerned, theneutron andproton are exactly thesame Thenuclear forces between neutron andneutron, neutron and proton, andproton andproton areallidentical asfaraswecantell Only thelittle 28—l0 electromagnetic forces aredifferent; electrically theproton andneutron areas different asnight andday. This isjustwhat wewanted There aretwoparticles, identical from thepoint ofview ofthestrong interactions, butdifferent electrically. Andtheyhave asmall difference inmass. Themass difference between theproton andtheneutron—expressed asthedifference intherest-energy mc2 inunits of Mev—is about 1.3Mev, which isabout 2.6times theelectron mass. Theclassical theory would then predict aradius ofabout §to%theclassical electron radius, orabout 10*“ cm. Ofcourse, oneshould really usethequantum theory, butby some strange accident, alltheconstants—-21r‘s andh's,etc.—come outsothatthe quantum theory gives roughly thesame radius astheclassical theory. Theonly trouble isthatthesigniswrong! Theneutron isheavier than theproton. Table 28-1 Particle Masses Charge Particle (electronic)Mass (Mev)Am* (Mev) n(neutron) 0 p(proton) +1 1r(1r-meson) O d:1 K(K-meson) 0 =i=1 Z(sigma) 0 +1939.5 938.2 135.0 139.6 497.8 493.9 1191.5 1189.4-1.3 +4.6 —3.9 -2.1 ——1 1196.0 +4.5 *Am=(mass ofcharged) —(mass ofneutral). Nature hasalsogiven usseveral other pairs—-or triplets—of particles which appear tobeexactly thesame except fortheir electrical charge. They interact with protons andneutrons, through theso-called “strong” interactions ofthenuclear forces. Insuch interactions, theparticles ofagiven kind—say the1r-mesons- behave inevery waylikeoneobject except fortheir electrical charge. lnTable 28-1 wegivealistofsuch particles, together with their measured masses. The charged tr-mesons—positive ornegative—have amass of139.6 Mev, butthe neutral 1r-meson is46Mev lighter. Webelieve thatthismass difference iselectro- magnetic; itwould correspond toaparticle radius of3to4Xl0_1“ cm. You will seefrom thetable thatthemass differences oftheother particles areusually ofthe same general size. Now thesizeofthese particles canbedetermined byother methods, forin- stance bythediameters they appear tohave inhigh-energy collisions. Sothe electromagnetic mass seems tobeingeneral agreement with electromagnetic theory, ifwestop ourintegrals oftheﬁeld energy atthesame radius obtained by these other methods. That’s why webelieve that thedifferences dorepresent electromagnetic mass. Youarenodoubt worried about thedifferent signs ofthemass differences in thetable. Itiseasytoseewhythecharged ones should beheavier than theneutral ones. Butwhat about those pairs liketheproton andtheneutron, where themea- sured mass comes outtheother way? Well, itturns outthatthese particles are complicated, andthecomputation oftheelectromagnetic mass must bemore elaborate forthem. Forinstance, although theneutron hasnonetcharge, itdoes haveacharge distribution inside it—-it isonly thenetcharge thatiszero Infact, webelieve thattheneutron 1ooks—at least sometimes—1ike aproton with anega- tive1r-meson ina“cloud" around it,asshown inFig.28-5. Although theneutron is“neutral,” because itstotal charge iszero, there arestillelectromagnetic energies 28-11i_' <-_(— -Negative ‘~ - -ir-meson \ .iI i \ ‘PROTON Fig. 28-5. Aneutron may exist, at times, asciproton surrounded byo negative Tr-meson. (forexample, ithasamagnetic moment), soit’snoteasy totellthesignofthe electromagnetic mass difference without adetailed theory oftheinternal structure. Weonly wish toemphasize herethefollowing points" (1)theelectromagnetic theory predicts theexistence ofanelectromagnetic mass, btititalso falls onits faceindoing so,because itdoes notproduce aconsistent theory~and thesame is true with thequantum modiﬁcations; (2)there isexperimental evidence forthe existence ofelectromagnetic mass; and(3)allthese masses areroughly thesame asthemass ofanelectron. Sowetome back again totheoriginal ideaofLorentz- maybe allthemass ofanelectron ispurely electromagnetic, maybe thewhole 0511Mev isduetoelectrodynamics lsitorisn’t it?Wehaven’t gotatheory. so wecannot say. Wemust mention onemore piece ofinformation, which isthemost annoying There isanother particle intheworld called amuon-—or it-meson—whicli, sofar aswecantell,differs innowaywhatsoever from anelectron except foritsmass. It actsinevery waylikeanelectron: itinteracts with neutrinos andwith theelectro- magnetic ﬁeld, andithasnonuclear forces. Itdoes nothing different from what anelectron does-at least, nothing which cannot beunderstood asmerely acon- sequence ofitshigher mass (20677times theelectron mass). Therefore, whenever someone ﬁnally getstheexplanation ofthemass ofanelectron. hewillthen have thepuzzle ofwhere amuon getsitsmass. Why? Because whatever theelectron does, themuon does thesame—so themass ought tocome outthesame There arethose who believe faithfully intheideathatthemuon andtheelectron arethe same particle and that, intheﬁnal theory ofthemass, theformula forthemass willbeaquadratic equation with tworoots—one foreach particle There arealso those who propose itwillbeatranscendental equation with aninﬁnite number of roots, andwho areengaged inguessing what themasses oftheother particles in theseries must be,andwhythese particles haven‘t been discovered yet 28-6 Thenuclear force ﬁeld Wewould liketomake some further remarks about thepart ofthemass of nuclear particles thatisnotelectromagnetic Where does thisother large fraction come from 2There areother forces besides electrodynamics—like nuclear forces that have their own ﬁeld theories, although nooneknows whether thecurrent theories areright These theories also predict aﬁeld energy which gives the nuclear particles amass term analogous toelectromagnetic mass; wecotild call itthe“Tr-mesic-ﬁeld-mass." ltispresumably very large, because theforces are great, anditisthepossible origin ofthemass oftheheavy particles. Butthe meson ﬁeld theories arestillinamost rudimentary state Even with thewell- developed theory ofelectromagnetism, wefound itimpossible togetbeyond ﬁrst baseinexplaining theelectron mass. With thetheory ofthemesons, westrike out Wemay take amoment tooutline thetheory ofthemesons, because ofits interesting connection with electrodynamics. lnelectrodynamics, thefield canbe described interms ofafour-potential that satisﬁes theequation [12/1,, Isources. Now wehave seen thatpieces oftheﬁeld canberadiated away sothatthey exist separated from thesources. These arethephotons oflight, andtheyaredescribed byadifferential equation without sources: D2/4,, =0. People have argued that theﬁeld ofnuclear forces ought also tohave itsown “photons"—they would presumably betherr-mesons—and that they should be described byananalogous differential equation. (Because oftheweakness ofthe human brain, wecan’t think ofsomething really new, soweargue byanalogy with what weknow.) Sothemeson equation might be m2¢=0.28-12 where d>could beadifferent four-vector orperhaps ascalar. Itturns outthatthe pion hasnopolarization, so¢>should beascalar. With thesimple equation El2¢ =0,themeson ﬁeld would vary with distance from asource as1/r2, just astheelectric ﬁeld does. Butweknow thatnuclear forces have much shorter dis- tances ofaction, sothesimple equation won’t work. There isoneway wecan change things without disrupting therelativistic invariance: wecanaddorsubtract from theD’Alembertian aconstant, times ¢.SoYukawa suggested thatthefree quanta ofthenuclear force_ ﬁeld might obey theequation 52¢-82¢=0. (2817) where ,u2isaconstant—that is,aninvariant scalar. (Since C]2isascalar differ- ential operator infour dimensions, itsinvariance isunchanged ifweaddanother scalar toit.) Let’s seewhat Eq.(28.17) gives forthenuclear force when things arenot changing with time. Wewant aspherically symmetric solution of V2—13¢=0 around some point source at,say,theorigin. If¢depends only onr,weknow that 2 2_liV4) -rar2 (rd). Sowehave theequation 1a2 2;§(r¢) _H¢=0 or air?(rt)=M2(r¢)- Thinking of(r¢>)asourdependent variable, thisisanequation wehave seenmany times. It’ssolution is r¢>=Keif”. Clearly, ¢cannot become inﬁnite forlarge r,sothe+sign intheexponent is ruled out. Thesolution is e“T¢=K-;-- (28.18) Thisfunction iscalled theYukawa potential. Foranattractive force, Kisanegative number whose magnitude must beadjusted toﬁttheexperimentally observed strength oftheforces. TheYukawa potential ofthenuclear forces diesoffmore rapidly than l/r bytheexponential factor. Thepotential—and therefore theforce—falls to'zero much more rapidly than l/rfordistances beyond l/,Lt, asshown inFig. 28-6 The“range” ofnuclear forces ismuch lessthan the“range” ofelectrostatic forces Itisfound experimentally that thenuclear forces donotextend beyond about 10-” cm,soitav10”’ INTI Finally, let'slook atthefree-wave solution ofEq(2817) Ifwesubstitute ¢:¢0et(wi—lcz) intoEq.(2817),wegetthat 2OJ 2 2__'63-‘k _pL -0. Relating frequency toenergy andwave number tomomentum, aswedidatthe endofChapter 36ofVol. I,wegetthat E2 3 _ P2 Z M2,,’/2’ which saysthattheYukawa “photon” hasamass equal touh/c. lfweuseforit 28-13¢l \ l l \ \ \ \\‘/I/r \ \ 8“/‘I ,4 ’// r \ \ \ W \ O I I ) o 1/)2 2/,2 3/); r Fig. 28-6. The Yukowo potential e_'”/r, compared with the Coulomb potential l/r. theestimate 1015m—‘, which gives theobserved range ofthenuclear forces, the mass comes outto3X10'25 gm,or170Mev, which isroughly theobserved mass oftherr-meson. So,byananalogy with electrodynamics, wewould saythatthe 1r-meson isthe“photon” ofthenuclear force ﬁeld Butnow wehave pushed the ideas ofelectrodynamics intoregions where theymay notreally bevalid—we have gone beyond electrodynamics totheproblem ofthenuclear forces. 28-14 29 The Motion ofCharges inElectrio and Magnetic Fields 29-1 Motion inauniform electric ormagnetic ﬁeld Wewant now todescribe-mainly inaqualitative way—the motions of charges invarious circumstances. Most oftheinteresting phenomena inwhich charges aremoving inﬁelds occur invery complicated situations. with many, many charges allinteracting with each other Forinstance, when anelectroinagne- ticwave goes through ablock ofmaterial oraplasma, billions andbillions of charges areinteracting with thewave andwith each other. Wewillcome tosuch problems later, butnow wejust want todiscuss themuch simpler problem ofthe motions ofasingle charge inagiven ﬁeld. Wecanthen disregard allother charges —except, ofcourse, those charges andcurrents which exist somewhere toproduce theﬁelds wewillassume. Weshould probably askﬁrstabout themotion ofaparticle inauniform elec- tricﬁeld Atlowvelocities, themotion isnotparticularly interesting—it isjusta uniform acceleration inthedirection oftheﬁeld. However, iftheparticle picks upenough energy tobecome relativistic, then themotion getsmore complicated. Butwewillleave thesolution forthatcase foryoutoplay with Next, weconsider themotion inauniform magnetic ﬁeld with zero electric ﬁeld. Wehave already solved thisproblem——one solution isthattheparticle goes inacircle Themagnetic force qvXBisalways atright angles tothemotion, sodp/dt isperpendicular topandhasthemagnitude zip/R, where Ristheradius ofthecircle. F=qvB =U1? Theradius ofthecircular orbit isthen R=li- 29.1 qB () That isonly onepossibility. lftheparticle hasacomponent ofitsmotion along theﬁeld direction, thatmotion isconstant, since there canbenocomponent ofthe magnetic force inthedirection oftheﬁeld. Thegeneral motion ofaparticle inauniform magnetic ﬁeldisaconstant velocity parallel toBandacircular motion atright angles toB—the trajectory isacylindrical helix (Fig. 29-1). Theradius ofthehelix isgiven byEq(291)ifwereplace pbypi,thecomponent ofnio- nientuin atright angles totheﬁeld. 29-2 Momentum analysis Auniform magnetic ﬁeld isoften used inmaking a"momentum analyzer," or“momentum spectrometer," forhigh-energy charged particles. Suppose that charged particles areshot into auniform magnetic ﬁeld atthepoint AinFig. 29-2(a), themagnetic ﬁeld being perpendicular totheplane ofthedrawing. Each particle willgointoanorbit which isacircle whose radius isproportional toits momentum. lfalltheparticles enter perpendicular totheedge oftheﬁeld, they willleave theﬁeldatadistance x(from A)which isproportional totheir momentum p.Acounter placed atsome point such asCwilldetect only those particles whose momentum isinaninterval A/2near themomentum p=qBx/2 ltis,ofcourse, notnecessary thattheparticles gothrough 180° before they arecounted. buttheso-called “180° spectrometer" hasaspecial property ltisnot 29-129-1 Motion inauniform electric ormagnetic ﬁeld 29-2 Momentum analysis 29-3 Anelectrostatic lens 29-4 Amagnetic lens 29-5 Theelectron microscope 29-6 Accelerator guide ﬁelds 29-7 Alternating-gradient focusing 29-8 Motion incrossed electric andmagnetic ﬁelds Review Chapter 30,Vol l,Diffraction ls iZ_’_ Y “\ VL X X '\ f 1°) lb) Fig. 29-1. Motion ofopcirticle ino uniform magnetic ﬁeld. V /71/ /'/UNIFORM MAGNETIC FIELD lfl TC ozrecton l-~—~1(~)i——l O / / POINT SOURCE (bl Fig. 29—2. Auniform-ﬁeld, momen- tum spectrometer with l80° focusing: la) different momenta; lb) different angles. (The magnetic ﬁeld isdirected perpendicular totheplane ofthefigure,) A __ Fig. 29-3. Anaxial-ﬁeld spectrom- eter.y”.\\\ 1////1/0 / // ,//// *l ‘U000 .0‘0H ~ED;..0.090H 000055000 I‘. Ax Fig. 29-4. Anellipsoidal coil with equal currents ineach axial interval Ax produces auniform magnetic field inside.necessary thatalltheparticles enter atright angles totheﬁeldedge. Figure 29—2(b) shows thetI‘3_]CCtOI‘l6S ofthree particles, allwith thesame momentum butentering theﬁeld atdifferent angles. You seethatthey take different trajectories, butall leave theﬁeld very close tothepoint C.Wesaythatthere isa“focus.” Such a focusing property hastheadvantage that larger angles canbeaccepted atA—- although some limit isusually imposed, asshown intheﬁgure. Alarger angular acceptance usually means thatmore particles arecounted inagiven time, decreasing thetime required foragiven measurement. Byvarying themagnetic ﬁeld, ormoving thecounter along inx,orbyusing many counters tocover arange ofx,the“spectrum” ofmomenta intheincoming beam canbemeasured. [Bythe“momentum spectrum”f(p), wemean thatthe number ofparticles with momenta between pand(p+dp)isf(p) dp.] Such measurements have been made, forexample, todetermine thedistribution of energies intheB-decay ofvarious nuclei. There aremany other forms ofmomentum spectrometers, butwewilldescribe justonemore, which hasanespecially large solid angle ofacceptance. Itisbased onthehelical orbits inauniform ﬁeld, liketheoneshown inFig.29—l. Let’s think ofacylindrical coordinate system—p. 0,z—set upwith thez-axis along the direction oftheﬁeld. Ifaparticle isemitted from theorigin atsome angle at with respect tothez-axis, itwillmove along aspiral whose equation is p=asinkz, 6=bz, where a,b,andkareparameters youcaneasily work outinterms ofp, 01,andthe magnetic ﬁeld B.Ifweplotthedistance pfrom theaxisasafunction ofzfora given momentum, butforseveral starting angles, wewillgetcurves likethesolid ones drawn inFig.29-3. (Remember thatthisisjustakind ofprojection ofa helical trajectory.) When theangle between theaxisandthestarting direction islarger, thepeak value ofpislarge butthelongitudinal velocity isless, sothe trajectories fordiﬁerent angles tend tocome toakind of“focus” near thepoint Aintheﬁgure. Ifweputanarrow aperture ofA,particles with arange ofinitial angles canstillgetthrough andpass ontotheaxis, where theycanbecounted by thelong detector D. Particles which leave thesource attheorigin with ahigher momentum but atthesame angles, follow thepaths shown bythebroken lines anddonotget through theaperture atA.Sotheapparatus selects asmall interval ofmomenta Theadvantage over theﬁrstspectrometer described isthattheaperture A—and theaperture A’—can beanannulus, sothatparticles which leave thesource ina rather large solid angle areaccepted. Alarge fraction oftheparticles from the source areused—an important advantage forweak sources orforvery precise measurements. One pays aprice forthisadvantage, however, because alarge volume of uniform magnetic ﬁeldisrequired, andthisisusually only practical forlow-energy particles Onewayofmaking auniform ﬁeld, youremember. istowind aCOllon asphere, with asurface current density proportional tothesine oftheangle Youcanalsoshow thatthesame thing istrueforanellipsoid ofrotation. Sosuch spectrometers areoften made bywinding anelliptical coilonawooden (oralumi- num) frame. Allthatisrequired isthatthecurrent ineach interval ofaxial distance Axbethesame, asshown inFig29-4 29-3 Anelectrostatic lens Particle focusing hasmany applications. Forinstance. theelectrons thatleave thecathode inaTVpicture tube arebrought toafocus atthescreen—to make a ﬁnespot. Inthiscase, onewants totake electrons allofthesame energy butwith different initial angles andbring them together inasmall spot. Theproblem is likefocusing light with lllens, anddevices which dothecorresponding jobfor particles arealso called lenses. 29—2 QQQQQQQQQQQQQY LQQQQQQQQQQQQQQ T k .ll)ll)XQQQQQQQQQQQQQY #—>-Z -T ‘—<_-’ I i i fl).LQQQQQQQQQQQQXI __ d n\§Q\\Q\§§§§§§Q"\\§\‘Q§§§§§§§\‘ qE Tv Fig. 29—5. Anelectrostatic lens. Theﬁeld lines shown are“lines of force," thatis,ofqE. Oneexample ofanelectron lensissketched inFig29—5. Itisan“electro- static” lenswhose operation depends ontheelectric ﬁeld between twoadjacent electrodes. Itsoperation canbeunderstood byconsidering what happens toa parallel beam thatenters from theleft. When theelectrons arrive attheregion a, theyfeelaforce withasidewise component andgetacertain impulse thatbends them toward theaxis You might think thatthey would getanequal andopposite im- pulse intheregion b,butthatisnotso.Bythetime theelectrons reach btheyhave gained energy andsospend lesslIH’I€intheregion b.Theforces arethesame, but thetime isshorter, sotheimpulse isless. Ingoing through theregions aandb, there isanetaxial impulse, andtheelectrons arebent toward acommon point Inleaving thehigh-voltage region, theparticles getanother kick toward theaxis Theforce isoutward inregion candinward inregion d,buttheparticles staylonger inthelatter region, sothere isagain anetimpulse Fordistances nottoofarfrom theaxis, thetotal impulse through thelensisproportional tothedistance from the axis(Can youseewhy”), andthisisjust thecondition necessary forlens-type focusing. You canusethesame arguments toshow that there isfocusing ifthe potential ofthemiddle electrode iseither positive ornegative with respect tothe other two. Electrostatic lenses ofthistype arecommonly used incathode-ray tubes andinsome electron microscopes. 29-4 Amagnetic lens Another kind oflens——often found inelectron microscopes—is themagnetic lenssketched schematically inFig.29-6. Acylindrically symmetric electromagnet hasvery sharp circular pole tipswhich produce astrong, nonuniform ﬁeld ina small region. Electrons which travel vertically through thisregion arefocused Youcanunderstand themechanism bylooking atthemagniﬁed view ofthepole-tip region drawn inFig.29-7. Consider twoelectrons aandbthatleave thesource Satsome angle with respect totheaxis. Aselectron areaches thebeginning ofthe ﬁeld, itisdeﬂected (Iw(1yfr0H’l youbythehorizontal component oftheﬁeld But thenitwillhave alateral velocity, sothatwhen itpasses through thestrong vertical ﬁeld, itwillgetanimpulse toward theaxis. Itslateral motion istaken outbythe magnetic force asitleaves theﬁeld, sotheneteffect isanimpulse toward the axis, plus a“rotation" about theaxis. Alltheforces onparticle bareopposite, soitalsoisdeflected toward theaxis. Intheﬁgure, thedivergent electrons are brought intoparallel paths. Theaction islikealenswith anobject atthefocal point. Another similar lensupstream canbeused tofocus theelectrons back toa single point, making animage ofthesource S. 29-5 Theelectron microscope You know thatelectron microscopes can“see” objects toosmall tobeseen byoptical microscopes. Wediscussed inChapter 30ofVol. Ithebasic limitations ofanyoptical system duetodiffraction ofthelensopening lfalensopening sub- 29-331%'/'9 ' IFig. 29-6. Amagnetic lens. 7/ '// /L_.___._./ S )l B \N / b a /--"'7// // 4__Y_/ S Fig. 29—7. Electron motion inthe magnetic lens. LENS OPENING 9 SOURCE Fig. 29-8. Theresolution ofamicro- scope islimited bytheangle subtended from thesource. BLURRED IMAGE l4 .ENS% Vl(_)PENlNG sPOINT souncs Fig. 29—9. Spherical aberration of cilens. HELD STRONGER HERE Fig. 29—lO. Particle motion ina slightly nonuniform ﬁeld.tends theangle 20from asource (seeFig.29-8), twoneighboring spots atthesource cannot beseen asseparate ifthey arecloser than about X5~ , where Aisthewavelength ofthelight. With thebestoptical microscope, 0ap- proaches thetheoretical limit of90°,so6isabout equal toX,orapproximately 5000 angstroms. Thesame limitation would alsoapply toanelectron microscope, butthere thewavelength is—for 50-kilovolt electrons——about 0.05 angstrom. Ifonecould usealensopening ofnear 30°,itwould bepossible toseeobjects only §ofan angstrom apart. Since theatoms inmolecules aretypically 1or2angstroms apart. wecould getphotographs ofmolecules. Biology would beeasy; wewould have aphotograph oftheDNA structure. What atremendous thing that would be! Most ofpresent-day research inmolecular biology isanattempt toﬁgure outthe \l12lp6S ofcomplex organic molecules. Ifwecould only seethem! Unfortunately, thebestresolving power thathasbeen achieved inanelectron microscope ismore like20angstroms. Thereason isthatnoonehasyetdesigned alenswith alarge opening. Alllenses have “spherical aberration,” which means thatraysatlarge angles from theaxishave adifferent point offocus than therays nearer theaxis. asshown inFig.29-9 Byspecial techniques, optical microscope lenses canbemade with anegligible spherical aberration, butnoonehasyet been able tomake anelectron lenswhich avoids spherical aberration. Infact, onecanshow thatanyelectrostatic ormagnetic lensofthetypes we have described must have anirreducible amount ofspherical aberration. This aberration—together with diffraction—limits theresolving power ofelectron microscopes totheir present value. The limitation wehave mentioned does notapply toelectric andmagnetic ﬁelds which arenotaxially symmetric orwhich arenotconstant intime. Perhaps some daysomeone willthink ofanewkind ofelectron lensthatWlllovercome the inherent aberration ofthesimple electron lens. Then wewillbeabletophotograph atoms directly. Perhaps onedaychemical compounds willbeanalyzed bylooking atthepositions oftheatoms rather than bylooking atthecolor ofsome pre- cipitatel 29-6 Accelerator guide ﬁelds Magnetic ﬁelds arealsoused toproduce special particle trajectories inhigh- energy particle accelerators. Machines likethecyclotron andsynchrotron bring particles tohigh energies bypassing theparticles repeatedly through astrong electric ﬁeld. Theparticles areheld intheir cyclic orbits byamagnetic ﬁeld. Wehave seen thataparticle inauniform magnetic ﬁeld willgoinacircular orbit. This, however, istrue only foraperfectly uniform ﬁeld. Imagine aﬁeld Bwhich isnearly uniform over alarge area butwhich isslightly stronger inone region than inanother. Ifweputaparticle ofmomentum pinthisﬁeld, itwillgo inanearly circular orbit with theradius R=p/qB. Theradius ofcurvature will, however, beslightly smaller intheregion where theﬁeld isstronger. Theorbit is notaclosed circle butwill“walk” through theﬁeld, asshown inFig.29-10. Wecan, ifwewish, consider thattheslight “error” intheﬁeld produces anextra angular kickwhich sends theparticle offonanewtrack. Ifthe particles aretomake millions ofrevolutions inanaccelerator, some kind of“radial focusing” isneeded which willtend tokeep thetrajectories close tosome design orbit. Another difficulty with auniform ﬁeld isthattheparticles donotremain ina plane Ifthey start outwith theslightest angle-—or aregiven aslight angle by anysmall error intheﬁeld—they willgoinahelical path thatwilleventually take them into themagnet pole ortheceiling orﬁoor ofthevacuum tank Some arrangement must bemade toinhibit such vertical drifts; theﬁeld must provide “vertical focusing” aswellasradial focusing. 294 MAGNETIC Masiiiigric ./rs”"’\* /F'E'-D M“‘3NFl'3|;'LCD =/FIELD f T ,/T \ / 7, \\ // \ // \ / \ /./me- \ \/ f S / \ // \‘ \ / \l l l i I I \ \ \\ / /' \ l/ \ \\7/r/I /l \k \ & / // \ //l \\ /2 ’__ / // ‘_ // k 0 // l CIRCULAR) I CIRCULAR ORBIT / | OR CIRCULAR’ l i B” B 1/ OPBIT B \'~R\i l l ’1 “I / " r )T’ lrT/l7\\\ --X\/ \l \ as7"‘\l'”Z\ /-_s__-__s_ Fig.29—l l.Radial motion ofapar- Fig. 29-12. Radial motion ofapar- Fig. 29—l3. Radial motion ofapar- ticle inamagnetic ﬁeld with alarge ticle inamagnetic ﬁeld with asmall ticle inamagnetic ﬁeld with cilarge positive slope. negative slope. negative slope. Onewould, atﬁrst, guess thatradial focusing could beprovided bymaking a magnetic ﬁeldwhich increases withincreasing distance from thecenter ofthedesign path Then ifaparticle goes outtoalarge radius, itwillbeinastronger ﬁeldwhich willbend itback toward thecorrect radius. Ifitgoes totoosmall aradius, the bending willbeless,anditwillbereturned toward thedesign radius. lfaparticle isonce started atsome angle with respect totheideal circle, itwilloscillate about theideal circular orbit, asshown inFig.29-11. Theradial focusing would keep the particles near thecircular path. Actually there isstillsome radial focusing even with theopposite ﬁeld slope Thiscanhappen iftheradius ofcurvature ofthetrajectory does notincrease more rapidly than theincrease inthedistance oftheparticle from thecenter oftheﬁeld. Theparticle orbits willbeasdrawn inFig29—l2. Ifthegradient oftheﬁeldistoo large, however. theorbits willnotreturn tothedesign radius butwillspiral inward oroutward, asshown inFig.29-13. Weusually describe theslope oftheﬁeld interms ofthe“relative gradient” orﬁeldindex, n: dB/B \SVI:W ' ll Aradial ﬁeld gradient willalso produce vertical forces ontheparticles /lll 8§'§l{‘“ Suppose wehave aﬁeldthatisstronger nearer tothecenter oftheorbit andweaker i attheoutside. Avertical cross section ofthemagnet atright angles totheorbit might beasshown inFig.29—l4. (For protons theorbits would becoming outof thepage )Iftheﬁeld istobestronger totheleftandweaker totheright, thelines ofthemagnetic ﬁeld must becurved asshown .Wecanseethatthismust beso byusing thelawthatthecirculation ofBiszeroinfreespace. lfwe takecoordinates 59-2944- A\'eI'll¢°l Qulde ﬁeld Q5 asShown mtheﬁgure, then seen inacross section perpendicular to theorbits.toCENTER B X Aguide ﬁeld gives radial focusing ifthisrelative gradient isgreater than -1. oF<fiT—— —~ll—l— j--»-> N / as, as’._(v><B),_-52»--6-;_0, Of aB,_aB, 3Z“m' ‘”” Since weassume thatOB;/ox ISnegative, there must beanequal negative GB,/62. 29—5 quadrupole lens.Ifthe“nominal” plane oftheorbit isaplane ofsymmetry where B,=0,then the radial component B,willbenegative above theplane andpositive below Thelines must becurved asshown. Such aﬁeld willhave vertical focusing properties. Imagine aproton thatis travelling more orlessparallel tothecentral orbit butabove it.Thehorizontal component ofBwillexert adownward force onit.Ifthe proton isbelow thecentral orbit, theforce isreversed. Sothere isaneffective “restoring force” toward the central orbit. From ourarguments there willbevertical focusing, provided that thevertical ﬁelddecreases withincreasing radius; butifthe ﬁeldgradient ispositive, there willbe“vertical defocusing.” Soforvertical focusing, theﬁeld index nmust belessthan zero. Wefound above that forradial focusing nhadtobegreater than -1. Thetwoconditions together givethecondition that —l<n<0 iftheparticles aretobekept instable orbits. Incyclotrons, values very near zero areused; inbetatrons andsynchrotrons, thevalue n=-0.6 istypically used. 29-7 Alternating-gradient focusing Such small values ofngive rather “weak” focusing. Itisclear that much more effective radial focusing would begiven byalarge positive gradient (n>>l),but then thevertical forces would bestrongly defocusing Similarly, large negative slopes (n<<—I)would give stronger vertical forces butwould cause radial de- focusing. Itwasrealized about IOyears ago, however, thataforce thatalternates between strong focusing andstrong defocusing canstillhave anetfocusing force Toexplain how alternating-gradientfocusing works, wewillﬁrst describe the operation ofaquadrupole lens, which ISbased onthesame principle. Imagine that auniform negative magnetic ﬁeld isadded totheﬁeld ofFig 29-14, with the strength adjusted tomake zero ﬁeld attheorbit. Theresulting ﬁeld-for small displacements from theneutral point—would beliketheﬁeldshown inFig29-I5 Such afour-pole magnet iscalled a“quadrupole lens." Apositive particle that enters (from thereader) totheright orleftofthecenter ispushed back toward thecenter. Iftheparticle enters above orbelow, itispushed away from thecenter. This isahorizontal focusing lens Ifthehorizontal gradient isreversed—-as can bedone byreversing allthepolarities—the signs ofalltheforces arereversed andwehave avertical focusing lens, asinFig.29-16 Forsuch lenses, theﬁeld strength—and therefore thefocusing forces—increase linearly with thedistance ofthelensfrom theaxis. \l II \i l))(( ii %/at.\rupole lens. 29-6 VERTICA L DISPLACEMENT FROM AXIS(I) 21%>50Ernzoz;"Ii-Z-'l 2> 0 0 / DISTANCE \ / "°‘Z“L HORIZONTAL VERTICAL VERTIGALDISTANCE F°¢U$'"° F0usms ozrocusims FocusmcFiEi_D EEELE FiEi_D FIELD (0) (bl Fig. 29-17. Horizontal andvertical focusing with apair ofquadrupole lenses. Now imagine thattwosuch lenses areplaced inseries. Ifaparticle enters with some horizontal displacement from theaxis, asshown inFig.29-l7(a), itwillbe deﬂected toward theaxisintheﬁrstlens. When itarrives atthesecond lensitis closer totheaxis, sotheforce outward islessandtheoutward deﬂection isless There isanetbending toward theaxis; theaverage effect ishorizontally focusing Ontheother hand, ifwelook ataparticle which enters offtheaxisinthevertical direction, thepath willbeasshown inFig.29-l7(b). Theparticle isﬁrstdeﬂected away from theaxis, butthen itarrives atthesecond lenswith alarger displacement, feelsastronger force, andsoisbenttoward theaxis. Again theneteffect isfocusing. Thus apair ofquadrupole lenses actsindependently forhorizontal andvertical motion—very much likeanoptical lens. Quadrupole lenses areused toform and control beams ofparticles inmuch thesame waythatoptical lenses areused for light beams. Weshould point outthat analternating-gradient system does notalways produce focusing. Ifthegradients aretoolarge (inrelation totheparticle momen- tumortothespacing between thelenses), theneteffect canbeadefocusing one. Youcanseehow thatcould happen ifyouimagine thatthespacing between the twolenses ofFig.29-17 were increased, say,byafactor ofthree orfour. Let’s return now tothesynchrotron guide magnet. Wecanconsider thatit consists ofanalternating sequence of“positive” and“negative” lenses with a superimposed uniform ﬁeld. Theuniform ﬁeld serves tobend theparticles, onthe average, inahorizontal circle (with noeffect onthevertical motion), andthe alternating lenses actonanyparticles thatmight tend togoastray-pushing them always toward thecentral orbit (ontheaverage). There isanice mechanical analog which demonstrates that aforce which alternates between a“focusing” force anda“defocusing” force canhave anet “focusing” effect. Imagine amechanical “pendulum” which consists ofasolid rodwithaweight ontheend,suspended from apivot which isarranged tobemoved rapidly upanddown byamotor driven crank. Such apendulum hastwoequili- brium positions. Besides thenormal, downward-hanging position, thependulum isalsoinequilibrium “hanging upward”—with its“bob” above thepivot! Such a pendulum isdrawn inFig.29-I8. Bythefollowing argument you canseethat thevertical pivot motion is equivalent toanalternating focusing force. When thepivot isaccelerated down- ward, the“bob” tends tomove inward, asindicated inFig. 29-I9. When the pivot isaccelerated upward, theeffect isreversed. Theforce restoring the“bob” toward theaxisalternates, buttheaverage effect isaforce toward theaxis. Sothe pendulum willswing back andforth about aneutral position which isjust opposite thenormal one. There is,ofcourse, amuch easier wayofkeeping apendulum upside down, andthat isbybalancing itonyour ﬁnger’ Buttrytobalance twoindependent sticks onthesame ﬁnger! Oronestick with your eyesclosed! Balancing involves making acorrection forwhat isgoing wrong. And thisisnotpossible, ingeneral. ifthere areseveral things going wrong atonce. Inasynchrotron there arebillions ofparticles going around together, each oneofwhich maystart outwith adifferent “error.” Thekind offocusing wehave been describing works onthem all. 29-7I\ 1/ s_.I\// II ll// // // jl ll / ’_:5&_-*- f i ~=-Di’‘QP Fig. 29-18. Apendulum with an oscillating pivot canhave astable posi- tionwith thebob above thepivot. /'\\i I qt\ \‘\\\,\ \\Al\/ Fig. 29-I9. Adownward accelera- tionofthepivot causes thependulum to move toward thevertical. vd-i» V0 TE (Da Fig. 29-20. Path ofaparticle in crossed electric and magnetic ﬁelds.29-8 Motion incrossed electric andmagnetic ﬁelds Sofarwehave talked about particles inelectric ﬁelds only orinmagnetic ﬁelds only. There aresome interesting effects when there areboth kinds ofﬁelds atthesame time. Suppose wehave auniform magnetic ﬁeld Bandanelectric ﬁeld Eatright angles. Particles thatstart outperpendicular toBwillmove ina curve liketheoneinFig 29-20 (The ﬁgure isaplane curve, no!ahelix!) Wecan understand thismotion qualitatively. When theparticle (assumed positive) moves inthedirection ofE,itpicks upspeed, andsoitisbent lessbythemagnetic ﬁeld. When itisgoing against theE-ﬁeld, itloses speed andiscontinually bent more by themagnetic ﬁeld. Theneteffect isthatithasanaverage “drift” inthedirection ofE ><B. Wecan. infact, show that themotion isauniform circular motion super- imposed onauniform sidewise motion atthespeed iy,=E/B—the trajectory in Fig.29-20 isacycloid. Imagine anobserver whoismoving totheright atacon- stant speed. Inhisframe ourmagnetic ﬁeld getstransformed toanewmagnetic ﬁeldplus‘anelectric ﬁeld inthedownward direction. Ifhehasjusttheright speed, histotal electric ﬁeldwillbezero, andhewillseetheelectron going inacircle. So themotion weseeisacircular motion, plus atranslation atthedrift speed 1;,=E/B Themotion ofelectrons incrossed electric andmagnetic ﬁelds isthe basis ofthemagnetron tubes, ie.,oscillators used forgenerating microwave energy. There aremany other interesting examples ofparticle motions inelectric and magnetic ﬁelds-such astheorbits oftheelectrons andprotons trapped inthe VanAllen belts—but wedonot,unfortunately, have thetimetodealwiththem here 29-8 30 The Internal Geometry ofCrystals 30-1 Theinternal geometry ofcrystals Wehave ﬁnished thestudy ofthebasic laws ofelectricity andmagnetism, and wearenow going tostudy theelectromagnetic properties ofmatter. Webegin bydescribing solids—that is,crystals. When theatoms ofmatter arenotmoving around very much, they getstuck together andarrange themselves inaconﬁgura- tionwith aslowanenergy aspossible. Iftheatoms inacertain place have found a pattern which seems tobeoflowenergy, then theatoms somewhere elsewill probably make thesame arrangement. Forthese reasons, wehave inasolid ma- terial arepetitive pattern ofatoms Inother words, theconditions inacrystal arethisway: Theenvironment ofa particular atom inacrystal hasacertain arrangement, andifyoulook atthesame kindofanatom atanother place farther along, youwillﬁndonewhose surround- ingsareexactly thesame. Ifyoupickanatom farther along bythesame distance, youwillﬁndtheconditions exactly thesame once more. Thepattern isrepeated overandover aga1n——and, ofcourse, inthree dimensions. lniagine theproblem ofdesigning awallpaper-—or acloth, orsome geometric design foraplane area-—in which youaresupposed tohave adesign element which repeats andrepeats andrepeats, sothatyoucanmake theareaaslarge asyouwant. Thisisthetwo-dimensional analog ofaproblem which acrystal solves inthree dimensions. Forexample, Fig.30-1(a) shows acommon kind ofwallpapei design. There isasingle element repeated inapattern thatcangoonforever. Thegeometric characteristics ofthiswallpaper design, considering only itsrepetition properties andnotworrying about thegeometry oftheﬂower itself oritsartistic merit, are contained inFig.30-l(b). Ifyoustart atanypoint, youcanﬁndthecorresponding point bymoving thedistance iialong thedirection ofarrow 1You canalsoget toacorrespoliding point ifyoumove thedistance binthedirection oftheother arrow. There are,ofcourse, many other directions. You cango,forexample. from point attopoint Bandreach acorresponding position, butsuch astep canbeconsidered asacombination ofastepalong direction 1,followed byastep along direction 2.One ofthebasic properties ofthepattern canbedescribed by thetwoshortest steps tonearby equal positions. By“equal” positions wemean that ifyou were tostand inanyoneofthem andlook around you,youwould seeexactly thesame thing asifyouwere tostand inanother one. That’s thefundamental property ofacrystal. Theonly difference isthatacrystal isathree-dimensional arrangement instead ofatwo-dimensional arrangement; andnaturally, instead of ﬂowers, each element ofthelattice issome kind ofanarrangement ofatoms- perhaps sixhydrogen atoms andtwocarbon atoms~in some kind ofpattern Thepattern ofatoms inacrystal canbefound outexperimentally byx-ray diffrac- tion. Wehave mentioned thismethod brieﬂy before, andwon’t sayanymore now except thattheprecise arrangement oftheatoms inspace hasbeen worked outfor most simple crystals andalsoforsome fairly complex ones. Theinternal pattern ofacrystal shows upinseveral ways. First, thebinding strength oftheatoms incertain directions isusually stronger than inother direc- tions. This means thatthere arecertain planes through thecrystal where itismore easily broken than others. They arecalled thecleavage planes. Ifyoucrack a crystal with aknife blade itwilloften split apart along such aplane. Second, the internal structure often appears atthesurface because ofthewaythecrystal was formed. Imagine acrystal being deposited outofasolution. There aretheatoms ﬂoating around inthesolution andﬁnally settling down when they ﬁndaposition 30-130-1 Theinternal geometry of crystals 30-2 Chemical bonds incrystals 30-3 Thegrowth ofcrystals 30-4 Crystal lattices 30-5 Symmetries intwodimensions 30-6 Symmetries inthree dimensions 30-7 Thestrength ofmetals 30-8 Dislocations andcrystal growth 30-9 TheBragg-Nye crystal model Reference: C.Kittel, Introduction to Solid State Physics, John Wiley andSons, Inc., New York, 2nded.,I956. ﬁﬁ.§;eeiegeee(<1) /*/0- aO~lO— /is. li-l ////ib)/<=- /=- /is ./=- lZ-l Fig. 30—l. Arepeating pattern in twodimensions. ltil -iH‘, [bl lcl Fig. 30-2. Natural crystals: (a) quartz, (blsodium chloride, (clmica. Fig. 30-3. The lattice ofamolecular crystal.oflowest energy. (It’s asifthewallpaper gotmade byﬂowers drifting around until onedrifted accidentally intoplace andgotstuck, andthen thenext, andthe next sothatthepattern gradually grows.) You canappreciate thatthere willbe certain directions inwhich itwillgrow atadifferent speed than inother directions, thereby growing intosome kind ofgeometrical shape. Because ofsuch effects, the outside surfaces ofmany crystals show some ofthecharacter oftheinternal arrangement oftheatoms Forexample, Fig.30—2(a) shows theshape ofatypical quartz crystal whose internal pattern ishexagonal. Ifyoulook closely atsuch acrystal, youwillnotice thattheoutside does notmake avery good hexagon because thesides arenotall ofequal length—they are,infact, often very unequal. Butinonerespect itisa verygood hexagon: theangles between thefaces areexactly 120°. Clearly, thesize ofanyparticular face isanaccident ofthegrowth, buttheangles arearepresenta- tion oftheinternal geometry Soevery crystal ofquartz hasadifferent shape, even though theangles between corresponding faces arealways thesame. Theinternal geometry ofacrystal ofsodium chloride isalsoevident from its external shape Figure 30-2(b) shows theshape ofatypical grain ofsalt. Again thecrystal isnotaperfect cube, butthefaces areexactly atright angles toone another Amore complicated crystal ismica, which hastheshape shown inFig30—2(c) Itisahighly anisotropic crystal, asiseasily seen from thefactthatitisvery tough ifyoutrytopullitapart inonedirection (horizontally intheﬁgure), butveryeasy tosplit bypulling apart intheother direction (vertically) Ithascommonly been used toobtain very tough, thin sheets Mica andquartz aretwoexamples of natural minerals containing silica. Athird example ofamineral with silica is asbestos, which hastheinteresting property that itiseasily pulled apart intwo directions butnotinthethird. Itappears tobemade ofvery strong, linear ﬁbers. 30-2 Chemical bonds incrystals Themechanical properties ofcrystals clearly depend onthekind ofchemical bindings between theatoms. Thestrikingly different strength ofmica along differ- entdirections depends onthekinds ofinteratomic binding inthediﬁerent directions. You have already learned inchemistry, nodoubt, about thedifferent kinds of chemical bonds First, there areionic bonds, aswehave already discussed for sodium chloride Roughly speaking, thesodium atoms have lostanelectron and become positive ions, thechlorine atoms have gained anelectron andbecome negative ions. Thepositive andnegative ionsarearranged inathree-dimensional checkerboard andareheld together byelectrical forces. The covalent bond—in which electrons areshared between twoatoiiis—is more common andisusually very strong. Inadiamond, forexample, thecarbon atoms have covalent bonds inallfour directions tothenearest neighbors. sothe crystal isvery hard indeed. There isalsocovalent bonding between silicon and oxygen inaquartz crystal, butthere thebond isreally only partially covalent. Because there isnotcomplete sharing oftheelectrons, theatoms arepartly charged, and thecrystal issomewhat ionic Nature isnotassimple aswetrytomake it; there arereally allpossible gradations between covalent andionic bonding Asugar crystal hasstillanother kind ofbinding Initthere arelarge molecules inwhich theatoms areheldstrongly together bycovalent bonds, sothattheiiiole- cule isatough structure. Butsince thestrong bonds arecompletely satisﬁed, there areonly relatively weak attractions between theseparate, individual molecules Insuch molecular crystals themolecules keep their individual identity, sotospeak, andtheinternal arrangement might beasshown inFig.30-3. Since themolecules arenotheldstrongly toeach other, thecrystals areeasy tobreak They arequite different from something likediamond, which isreally onegiant molecule that cannot bebroken anywhere without disrupting strong covalent bonds. Parifﬁn isanother example ofamolecular crystal. Anextreme example ofamolecular crystal occurs inasubstance likesolid argon. There isverylittle attraction between theatoms—each atom isacompletely 30-2 saturated monatomic molecule. Butatverylowtemperatures, thethermal motion isverysmall, sotheslight interatomic forces cancause theatoms tosettle down into aregular array likeapileofclosely packed spheres. Themetals form acompletely different class ofsubstances Thebonding is ofanentirely different kind. Inametal thebonding isnotbetween adjacent atoms butisaproperty ofthewhole crystal. Thevalence electrons arenotattached to oneatom ortoapairofatoms butareshared throughout thecrystal. Each atom contributes anelectron toauniversal pool ofelectrons, andtheatomic positive ionsreside intheseaofnegative electrons. Theelectron seaholds theionstogether likesome kind ofglue. Inthemetals, since there arenospecial bonds inanyparticular direction, there isnostrong directionality inthebinding. They arestillcrystalline, however, be- cause thetotal energy islowest when theatomic ionsarearranged insome deﬁnite array—although theenergy ofthepreferred arrangement isnotusually much lower thanother possible ones. Toaﬁrstapproximation, theatoms ofmany metals are likesmall spheres packed inastightly aspossible. 30-3 Thegrowth ofcrystals Trytoimagine thenatural formation ofcrystals intheearth. Intheearth’s surface there ISabigmixture ofallkinds ofatoms. They arebeing continually churned about byvolcanic action, bywind, andbywater—continually being moved about andmixed. Yet. bysome trick, silicon atoms gradually begin toﬁndeach other, andtoﬁndoxygen atoms, tomake silica. One atom atatime isadded to theothers tobuild upacrystal—the mixture gets unmixed. And somewhere nearby, sodium andchlorine atoms areﬁnding each other andbuilding upacrystal ofsalt. How does ithappen thatonce acrystal 1Sstarted, itpermits only aparticular kindofatom tojoinon” Ithappens because thewhole system isworking toward thelowest possible energy. Agrowing crystal willaccept anewatom ifitisgoing tomake theenergy aslowaspossible. Buthow does itknow thatasilicon——or anoxygen—-atom atsome particular spot isgoing toresult inthelowest possible energy‘? Itdoes itbytrialanderror. Intheliquid, alloftheatoms areinperpetual motion. Each atom bounces against itsneighbors about 101" times every second. Ifithitsagainst theright spotofgrowing crystal, ithasasomewhat smaller chance ofjumping oﬁ"again iftheenergy ISlow. Bycontinually testing over periods of millions ofyears atarateofI013 tests persecond. theatoms gradually build up attheplaces where they ﬁndtheir lowest energy. Eventually they grow intobig crystals. 30-4 Crystal lattices Thearrangement oftheatoms inacrystal——the crystal latrice—can take on many geometric forms. Wewould liketodescribe ﬁrstthesimplest lattices, which arecharacteristic ofmost ofthemetals andofthesolid form oftheinert gases. They arethecubic lattices which canoccur intwoforms: thebody-centered cubic. shown inFig.30—4(a), andtheface-centered cubic shown inFig.30—4(b). The drawings show, ofcourse, only onecube ofthelattice; youaretoimagine thatthe pattern isrepeated indeﬁnitely inthree dimensions. Also, tomake thedrawing clearer, only the“centers” oftheatoms areshown. Inanactual crystal, theatoms aremore likespheres incontact with each other. Thedark andlight spheres in thedrawings may, ingeneral, stand fordifferent kinds ofatoms ormay bethe same kind. Forinstance, ironhasabody-centered cubic lattice atlowtemperatures, butaface-centered cubic lattice athigher temperatures. Thephysical properties arequite different inthetwocrystalline forms. How dosuch forms come about? Imagine that youhave theproblem of packing spherical atoms together astightly aspossible. Onewaywould betostart bymaking alayer ina“hexagonal close-packed array,” asshown inFig.30—5(a) Then youcould build upasecond layer liketheﬁrst, butdisplaced horizontally, 30-3(0) IFig. 30-4. The unit cell ofcubic crystals: (0)body-centered, (b)face- centered. §**®\<§ \i<®><®>§><<§s§§i§@$$‘ P\O®§\TL as’ Fig. 30-6. lsthiscihexagon or0cube seen from one corner?~\/\//t/"'/\ / \/ asshown inFig. 30—5(b) Next, you can putonthethird layer. But notice‘ There aretwodistinct ways ofplacing thethird layer Ifyoustart thethird layer byplacing anatom atAinFig30—5(b), each atom inthethird layer isdirectly above anatom ofthebottom layer Ontheother hand, ifyoustart thethird layer byputting anatom attheposition B,theatoms ofthethird layer willbecentered atpoints exactly inthemiddle ofatriangle formed bythree atoms ofthebottom layer. Any other starting place isequivalent toAorB,sothere areonly twoways ofplacing thethird layer. Ifthethird layer hasanatom atpoint B,thecrystal lattice isaface-centered cubic—but seen atanangle Itseems funny thatstarting with liexagons youcan endupwith cubes. Butnotice thatacube looked atfrom acorner hasahexagonal outline Forinstance, Fig.30-6 could represent aplane hexagon oracube seen in perspective! Ifathird layer isadded toFig. 30—5(b) bystarting with anatom atA,there is nocubical structure, andthelattice hasinstead only ahexagonal symmetry. Itis clear that both possibilities wehave described areequally close-packed Some metals—for example, copper andsilver—choose theﬁrst alternative, theface-centered cubic. Others—for example, beryllium andmagnesiuiii—choose theother alternatives; they form hexagonal crystals. Clearly, which crystal lattice appears cannot depend onlyonthepacking oflittle spheres, butmust alsobedeter- mined inpart byother factors Inparticular, itdepends ontheslight remaining angular dependence oftheinteratoniic forces (or,inthecaseofthemetals, onthe energy oftheelectron pool) You will, nodoubt, learn allabout such things in your chemistry courses. 30-5 Symmetries intwodimensions Wewould nowliketodiscuss some oftheproperties ofcrystals from thepoint ofview oftheir internal symmetries. Themain feature ofacrystal isthatifyou start atone atom and move toacorresponding atom onelattice unit away, you areagain inthesame kind ofanenvironment. That’s thefundamental proposition. Butifyouwere anatom, there would beanother kind ofchange thatcould take you again tothesame environnient—that is,another possible “symmetry.” Figure 30—7(a) shows another possible “wallpaper-type” design (though oneyou have probably never seen). Suppose wecompare theenvironments forpoints AandBYou might, atﬁrst, think thatthey arethesame—but notquite Points CandDareequivalent toA,buttheenvironment ofBislikethatofAonly ifthe surroundings arereversed, asinamirror reﬂection. 30-4 IiiQtaX0!‘<-:0 ~11 ‘"2wto~ 50*es°___69m bar’°Q “Q69% 60’ 6O\O’ Q 6 0*Q 6 0*Q o~O:T;tas@ @6.,¢ O~ R——————————————~06 av ~oQty 9°‘ '°‘2 9b,»°‘{ R—~———————--—— ----~-— Q W5s »O (0)~<~6tiQ,.0 9»~oQ9 <659<>em—-__ G0*~o~Qm_.______.--6o~,0 by~06»oQ9o~ ~0653>" Fig. 30-7. Apattern ofhigh symmetry. There areother kinds of“equivalent” points inthepattern. Forinstance, thepoints EandFhave the“same” environments except thatoneisrotated 90° withrespect totheother. Thepattern isquite special. Arotation of90°——or any multiple ofit—about avertex such asAgives thesame pattern allover again. A crystal with such astructure would have square corners ontheoutside, butinside itismore complicated than asimple cube. Now that wehave described some special examples, let’s trytoﬁgure outall thepossible symmetries acrystal canhave. First, weconsider what happens ina plane. Aplane lattice canbedeﬁned bythetwoso-called prtmitive vectors that go from onepoint ofthelattice tothetwonearest equivalent points. Thetwovectors 1and2aretheprimitive vectors ofthelattice ofFig. 30-1. The two vectors aand bofFig30—7(a) aretheprimitive vectors ofthepattern there Wecould, ofcourse, / equally wellreplace aby—a,orbby—b. Since aandbareequal inmagnitude / andatright angles, arotation of90°turns aintob,andbinto—a,giving thesame D // C latticeonceagain.‘ ~———— —x -- ——— ———— Weseethatthere arelattices which have a“four-sided" symmetry. And we b,\ /b havedescribed earlier aclose-packed array based onahexagon which could have \\I60“ asix-sided symmetry. Arotation ofthearray ofcircles inFig.30—5(a) byanangle _.______ ___ of60°about thecenter ofanycircle brings thepattern back toitself. A 0 B What other kinds ofrotational symmetry arethere? Canwehave, forexample, to) aﬁvefold oraneightfold rotational symmetry" Itiseasy toseethatthey are impossible. Theonlysymmetry withmore sides thanfour tsastx-sided symmetry. C First, let’sshow thatmore than sixfold symmetry isimpossible. Suppose wetryto TTTTTTTT TTTTTT imagine alattice with twoequal primitive vectors with anenclosed angle lessthan D D 60°,asinFig.30-8(a). Wearetosuppose that points BandCareequivalent 7 toA,andthataandbarethetwoshortest vectors from Atoitsequivalent neighbors b’,/ Butthatisclearly wrong, because thedistance between BandCisshorter than from /1 Q 2° either onetoA.There must beaneighbor atDequivalent toAwhich iscloser _‘_ __is__ ____ than BorC.Weshould have chosen b’asoneofourprimitive vectors. Sothe E A G B angle between thetwoprimitive vectors must be60°orlarger. Octagonal symmetry (b) isnotpossible. What about ﬁvefold symmetry? Ifweassume that theprimitive vectors a Fig.30-8. la)Rotational symmetries andbhave equal lengths andmake anangle of21r/5 =72°,asinFig.30—8(b), greater than sixfold arenotpossible. thenthere should alsobeanequivalent lattice point atD,at72°from C.Butthe lblFivefold rotational symmetry isnot vector b’from EtoDisthen lessthan b,sobisnotaprimitive vector. There can P°5§ibl°- benoﬁvefold symmetry. Theonly possibilities thatdonotgetusintothiskind ofdifﬁculty are0=60°, 90°, or120°. Zero or180° arealso clearly possible. Onewayofstating ourresult isthatthepattern canbeleftunchanged byarotation ofonefullturn (nochange atall),one-half ofaturn, one-third, one-fourth, or one-sixth ofaturn And those areallthepossible rotational symmetries ina plane——a total ofﬁve. If0=21r/n, wespeak ofan“n-fold” symmetry. Wesay 30-5 /s As /6 A} /,//7 R777/ ,7.6.7/77 ..,/ <0)/ /e A /e As -7./6777 727 777 7%7777 777T(c) (<1) Fig 30—9. Symmetry under inversion. Pattern lb)isunchanged ifR—+—R, but pattern aISchanged. lnthree dimensions pattern (d)ISsymmetric under aninversion but(clisnot. thatapattern with nequal to4orto6hasa“higher syninietry” than onewith nequal tolorto2. Returning toFig.30—7(a), weseethatthepattern hasafourfold rotational symmetry. Wehave drawn inFig.30—7(b) another design which hasthesame symmetry properties aspart (a). Thelittle comiiia-like ﬁgures areasymmetric 0b_]€CllS which serve todeﬁne thesymmetry ofthedesign inside ofeach square Notice thatthecommas arereversed inalternate squares, sothattheunitcellis larger than oneofthesmall squares. Ifthere were nocommas, thepattern would stillhave fourfold symmetry, buttheunitcellwould besmaller. Thepatterns of Fig.30—7 alsohave other symmetry properties. Forinstance, areﬂection about any ofthebroken lines R—R reproduces thesame pattern The patterns ofFig. 30-7 have stillanother kind ofsymmetiy. Ifthepattern isreﬂected about theline Y—Yandshifted onesquare totheright (orleft), weget back theoriginal pattern ThelineY—Y iscalled a"glide" line. These areallthepossible symmetries intwodimensions There isonemore spatial symmetry operation which isequivalent tntwodtmcnsions toal80°rotation, butwhich isaquite distinct operation inthree dimensions. ItislI’lVL’i’.\I()I’!. Byan inversion wemean that anypoint atthevector displacement Rfrom some origin [forinstance, thepoint AinFig.30—9(b)] ismoved tothepoint at—R Aninversion ofpattern (a)ofFig. 30~9 produces anew pattern, butanin- version ofpattern (b)reproduces thesame pattern. Foratwo-dimensional pattern (asyoucanseefrom theﬁgure), aninversion ofthepattern (b)through thepoint Aisequivalent toarotation of180° about thesame point Suppose, however, wemake thepattern inFig. 30~9(b) three dimensional byimagining that thelittle 6'sand 9’seach have an“arrow” potnttng outofthepage. After aninversion in three dimensions allthearrows willbereversed, sothepattern isnotreproduced. Ifweindicate theheads andtailsofthearrows bydots andcrosses, respectively, wecanmake al/II‘€€-dlI’H€/’lS‘IOI’l£ll pattern, asinFig 30—9(c), which isnotsyninietric under aninversion, orwecanmake apattern like theone shown in(d),which does have such asymmetry. Notice that itisnotpossible toimitate athree- dimensional inversion byanycombination ofrotations. Ifwecharacterize the“symnietry“ ofapattern—or lattice—by thekinds of symmetry operations wehave been describing, itturns outthat fortwodimensions l7distinct patterns arepossible Wehave drawn onepattern ofthelowest possible 30-6 symmetry inFig.30-1, andoneofhigh symmetry inFig.30-7. Wewillleave you with thegame oftrying toﬁgure outallofthel7possible patterns. Itispeculiar how fewofthel7possible patterns areused inmaking wall- paper andfabrics. Onealways seesthesame three orfour basic patterns. Isthis because ofalack ofimagination ofdesigners, orbecause many ofthepossible patterns arenotpleasing totheeye? 30-6 Symmetries inthree dimensions Sofarwehave talked only about patterns intwodimensions. What weare really interested in,however, arepatterns ofatoms inthree dimensions. First, itisclear that athree-dimensional crystal willhave three primitive vectors. If wethen askabout thepossible symmetry operations inthree dimensions, weﬁnd thatthere are230different possible symmetries! Forsome purposes, these 230 UmmnnmgmmwhmomwndwwswmmamdmwnmFgJ040'Mehmm withtheleast symmetry iscalled thetrzclinic. Itsunitcellisaparallelepiped. The primitive vectors areofdifferent lengths, andnotwooftheangles between them are equal. There isnopossibility ofanyrotational orreﬂection symmetry. There are, however, stilltwopossible symmetries——the unitcellis,orisnot,changed byan inversion through thevertex. (Byaninversion inthree dimensions, weagain mean thatspatial displacements Rarereplaced by—R—in other words, that (x,y,z) goesinto(—x, ——y,—z) Sothetriclinic lattice hasonlytwopossible symmetries, unless there issome special relation among theprimitive vectors. Forexample, if allthevectors areequal andareseparated byequal angles, onehasthetrtgonal lattice shown intheﬁgure. This ﬁgure canhave anadditional symmetry, itmay beunchanged byarotation about thelong, body diagonal. Ifoneoftheprimitive vectors, sayc,isatright angles totheother two, we getamonoclintc unitcell. Anewsymmetry ispossible—a rotation by180°about c Thehexagonal cellisaspecial caseinwhich thevectors aandbareequal andthe angle between them is60°,sothatarotation of60°, or120°, or180°about thevector crepeats thesame lattice (forcertain internal symmetries). Ifallthree primitive vectors areatright angles, butofdifferent lengths, we gettheorthorhombic cell. Theﬁgure issymmetric forrotations of180°about the three axes. Higher-order symmetries arepossible with theierragonal cell, which hasallright angles andtwoequal primitive vectors Finally, there isthecubic cell,which isthemost symmetric ofall. Thepoint ofallthisdiscussion about symmetries isthattheinternal symmetries ofthecrystals show up—-sometimes insubtle ways—in themacroscopic physical properties ofthecrystal Forinstance, acrystal will, ingeneral, have atensor electric polarizability. Ifwedescribe thetensor interms oftheellipsoid ofpolari- zation, weshould expect that some ofthecrystal symmetries should show up alsointheellipsoid. Forexample, acubic crystal issymmetric with respect to arotation of90°about anyoneofthree orthogonal directions. Clearly, the only ellipsoid with thisproperty isasphere. Acubic crystal must beanisotropic dielectric‘. Ontheother hand, atetragonal crystal hasafourfold rotational symmetry Itsellipsoid must have twoofitsprincipal axes equal, andthethird must be parallel totheaxisofthecrystal. Similarly, since theorthorhombic crystal has twofold rotational symmetry about three orthogonal axes, itsaxes must coincide withtheaxes ofthepolarization ellipsoid. lnalikemanner, oneoftheaxes ofa monoclinic crystal must beparallel tooneoftheprincipal axes oftheellipsoid, though wecan’t sayanything about theother axes. Since atriclinic crystal hasno rotational symmetry, theellipsoid canhave anyorientation atall. Asyoucansee,wecanmake abiggame ofﬁguring outthepossible sym- metries andrelating them tothepossible physical tensors. Wehave considered only thepolarization tensor, butthings getmore complicated forothers—for instance, forthetensor ofelasticity. There isabranch ofmathematics called “group theory” thatdeals with such subjects, butusually youcanﬁgure outwhat youwant with common sense. 30-7//-r"'_’_’/—/7 _._ 7/__--7» /c / / / b ——7—/-,=’ a TR|CLlNlC //7“:77 0 ,// ’4-/"/3 ci TRIGONAL 0I|\I ___|_\il|I ____a'\I\\|\| L_____t/ /// O MONOCL INIC O 0\___-ill—i~——\}\l\| __4J/ __-/4X,ea ° HEXAGONAL /i__TT"7|’_L Q U¢_ ——'“\\u___/ G ORTHORHOMBIC __[__t\\\\t___si//1 // l ’__i_.._ C _G O TETRAGONAL ol\\ __'_\ITl _,___i\\\\\\L____A.|G _ _ O CUBIC Fig. 30—lO. The seven classes of crystal lattices. L234______ /\/'\r“\ QXXXXXECF -7. ) (b) (G Fig. 30—l2. Aphotograph ofasmall crystal ofcopper after stretching. [Cour- tesy of5.S.Brenner, Senior Scientist, United States Steel Research Center, Monroeville, Pal O *Q>‘§DQ/7iO><O“O<-Q><§>$J'Q.OIz<~ JHg-“J,i‘.‘OCr~<Q ">‘SO>//‘f‘ OQG“GQigC)O _>4’.9'49-9511K-.. §4 Fig. 30—l3. Adislocation inacry- stal.Fig. 30—l l.Slippcige ofcrystal planes. 30—7 Thestrength ofmetals Wehave saidthat metals usually have asimple cubic crystal structure; we want now todiscuss their mechanical properties——which depend onthisstructure. Metals are,generally speaking, very “soft,” because itiseasy toslide onelayer ofthecrystal overthenext. You maythink: “That’s ridiculous; metals arestrong.” Notso,asingle crystal ofametal canbedistorted very easily. Suppose welook attwolayers ofacrystal subjected toashear force, asshown inthediagram ofFig.30—l1(a). You might atﬁrstthink thewhole layer would resist motion until theforce wasbigenough topush thewhole layer “over the hump,” sothatitshifted onenotch totheleft. Although slipping does occur along aplane, itdoesn’t happen thatway (Ifitdid,youwould calculate thatthemetal ismuch stronger than itreally is.)What happens ismore likeoneatom going ata time; ﬁrsttheatom ontheleftmakes itsjuinp, thenthenext, andsoon,asindicated inFig.30—ll(b). Ineﬂect itisthevacant space between twoatoms thatquickly travels totheright, with thenetresult that thewhole second layer hasmoved over oneatomic spacing. Theslipping goes thiswaybecause ittakes much less energy toliftoneatom atatime over thehump than toliftawhole row. Once theforce isenough tostart theprocess, itgoes therestofthewayvery fast Itturns outthatinarealcrystal, slipping willoccur repeatedly atoneplane. then willstopthere andstart atsome other plane. Thedetails ofwhyitstarts and stops arequite mysterious. Itis,infact, quite strange thatsuccessive regions of slipareoften fairly evenly spaced. Figure 30-12 shows aphotograph ofatiny, thincopper crystal thathasbeen stretched. You canseethevarious planes where slipping hasoccurred. Thesudden slipping ofindividual crystal planes isquite apparent ifyoutake apiece oftinwire thathaslarge crystals initandstretch itwhile holding itnext toyour ear. You canhear arush of“ticks” astheplanes snap totheir newposi- tions, oneafter theother. Theproblem ofhaving a“missing” atom inonerowissomewhat more ditlicult than itmight appear from Fig.30—1l. When there aremore layers, thesituation must besomething likethatshown inFig.30-13. Such animperfection inacrystal iscalled adislocation. Itispresumed that such dislocations areeither present when thecrystal wasformed oraregenerated atsome notch orcrack atthesurface Once they areproduced, they canmove relatively freely through thecrystal The gross distortions result from themotions ofmany ofsuch dislocations. Dislocations canmove freely-that is,they require little extia enei'gy—so longastherestofthecrystal hasaperfect lattice. Butthey mayget“stuck“ ifthey encounter some other kind ofimperfection inthecrystal. lfittakes alotofenergy forthem topass theimperfection, they willbestopped. This isprecisely the mechanism thatgives strength toimperfect metal crystals. Pure ironcrystals are quite soft, butasmall concentration ofimpurity atoms maycause enough imper- fections toeffectively immobilize thedislocations. Asyouknow, steel, which is primarily iron, isvery hard. Tomake steel, asmall amount ofcarbon isdissolved intheironmelt; ifthemelt iscooled rapidly, thecarbon precipitates outinlittle grains, making many microscopic distortions inthelattice. Thedislocations can nolonger move about, andthemetal ishard. Pure copper isvery soft, butcanbe“work-hardened.“ This isdone byham- mering onitorbending itback andforth. Inthiscase, many newdislocations of various kinds aremade which interfere with oneanother, cutting down their 30—8 mobility. Perhaps you’ve seen thetrick oftaking abarof“dead soft” copper andgently bending itaround someone’s wrist asabracelet. Intheprocess, it becomes work-hardened andcannot easily beunbent again‘ Awork-hardened metal likecopper canbemade softagain byannealing atahigh temperature. Thethermal motion oftheatoms “irons out” thedislocations andmakes large single crystals again. Wehave, sofar,described only theso-called slipdislocation. There aremany other kinds, oneofwhich isthescrew dislocation shown inFig. 30-14. Such dislocations often play animportant partincrystal growth. 30-8 Dislocations andcrystal growth Oneofthegreat puzzles foralong time washowcrystals canpossibly grow. Wehave described how itisthateach atom might, byrepeated testing, determine whether itwasbetter tobeinthecrystal ornot. Butthatmeans thateach atom must ﬁndaplace oflowenergy. However, anatom putonanewsurface isonly bound byoneortwobonds from below, anddoesn’t have thesame energy it would have ifitwere placed inacorner, where itwould have atoms onthree sides Suppose weimagine agrowing crystal asastack ofblocks, asshown inFig.30-15 Ifwetryanewblock at,say,position A,itwillhave only oneofthesixneighbors itshould ultimately get. With somany bonds lacking, itsenergy isnotvery low. Itwould bebetter oﬂatposition B,where italready hasone-half ofitsquota of bonds. Crystals doindeed grow byattaching new atoms atplaces likeB. What happens, though, when thatlineisﬁnished? Tostart anew line, an atom must come torestwith only twosides attached, andthatisagain notvery likely. Even ifitdid,what would happen when thelayer wasﬁnished? How could anewlayer getstarted? Oneanswer isthatthecrystal prefers togrow ata dislocation, forinstance around ascrew dislocation liketheoneshown inFig. 30-14. Asblocks areadded tothiscrystal, there isalways some place where there arethree available bonds. Thecrystal prefers, therefore, togrow with adislocation built in.Such aspiral pattern ofgrowth isshown inFig.30-16, which isaphoto- graph ofasingle crystal ofparaﬂin. KTTTTTTTTTTT TTTTTTTTTT0‘’ 1 }i‘T”.1“T ”/" il Fig. 30-14. Ascrew dislocation. [From Charles Kittel, Introduction to Solid State Physics, John Wiley and Sons, Inc.,New York, 2nded., l956.] do\s;;:+Fig. 30-l5. Crystal growth. Fig. 30-l6. Aparaffin crystal which 7 ;l has grown around ascrew dislocation. up 30—9 TheBragg-Nye crystal model Wecannot, ofcourse, seewhat goes onwith theindividual atoms inacrystal. Also, asyourealize bynow, there aremany complicated phenomena thatarenot easytotreat quantitatively. SirLawrence Bragg andJ.F.Nye have devised a scheme formaking amodel ofametallic crystal which shows inastriking way many ofthephenomena thatarebelieved tooccur inarealmetal. Inthefollowing pages wehave reproduced their original article, which describes their method and shows some oftheresults they obtained with it.(The article isreprinted from the Proceedings oftheRoyal Society ofLondon, Vol. 190,September 1947, pp.474-481 —-with thepermission oftheauthors andoftheRoyal Society.) 30-9[From Charles Kittel, Introduction toSolid State Physics, John Wiley and Sons, Inc., M»m *-"'-""“"' .» ‘ ""3 New York, 2nd ed., l956.] Adynamical model ofacrystal structure BYSmLAWRENCE Basso, FRS.ANDJ.F.NYE Cavendish LG.b0rlll01‘y, Unwersity ofCambridge (Received 9January l947—Read 19June 1947) [Plates sto2i] The crystal structure ofametal isrepiesented byanassemblage ofbubbles, amillunetre or leaindiameter, ﬂoating onthesurface ofasoap solution The bubbles areblown from aﬁne pipette beneath thesurface with aconstant airpressure, and areremarkably uniform insize They areheld together bysurface tension, either masmgle layer onthesurface orinathree- dunensional mass Anassemblage may contain hundreds ofthousands ofbubbles and persists foranhour ormore The assemblages show structures which nave been supposed toexist inmetals, and simulate effects which have been observed, such asgmin boundaries, dl8l008- tions and other types offault, slip, recrystallization, snneahng, and strains due to‘foreign‘ atoms 1THE BUBBLE MODEL Models ofcrystal structure have been described from time totime inwhich the atoms arerepresented bysmall ﬂoating orsuspended magnets, orbycircular disks ﬂoating onawater surface and held together bytheforces ofcapillary attraction These models have certain disadvantages ,forinstance, inthecase ofﬂoating objects incontact, frictional forces impede then‘ free relative movement Amore serious disadvantage isthat thenumber ofcomponents islimited foralarge number of components isrequired inorder toapproach thestate ofaffairs inareal crystal The present paper describes thebehaviour ofamodel inwhich theatoms arerepre- sented bysmall bubbles from 20to0lmm indiameter ﬂoating onthesurface of asoap solution These small bubbles aresufﬁciently persistent forexperiments lasting anhour ormore, they slide past each other without friction, and they can beproduced inlarge numbers Some oftheillustrations inthis paper were taken from assemblages ofbubbles numbering 100,000 ormore The model most nearly represents thebehaviour ofametal stnicture, because thebubbles areofonetype only and areheld together byageneral capillary attraction which represents the binding force ofthefreeelectrons mthemetal Abrief description ofthemodel has been given uitheJ0ll""lGl ofScientiﬁc I1ic¢rumenu(Bragg 1942b) /J; ';=1151%,,m.:31.:,=..::;,.,2:..;:..=:.r 2 5, 5|llll I Ibe //// ///////////Fiouaiz lApparatus forproducing rafts ofbubbles. 2METHOD OFFORMATION Thebubbles areblown from aﬁneoriﬁce, beneath thesurface ofasoap solution. Wehave had thebest results with asolution theformula ofwhich was given tous byMrGreen oftheRoyal Institution l52ccofoleic acid(pure redistilled) iswell shaken in50c cofdistilled water This ismixed thoroughly with 73c cof10% solution oftn-ethanolamine and themixture made upto2000 cTothis isadded 164ccofpure glycennc Itislefttostand and theclear hquid isdrawn offfrom below Insome expenments this was diluted mthree times itsvolume ofwater to reduw viscosity The onﬁce ofthejetisabout 5mm below thesurface Aconstant airpressure of50to200cm ofwater issupplied bymeans oftwo Winchester ﬂasks Normally the bubbles areremarkably uiuform insize Occasionally they issue inanirregular manner, butthis canbecorrected byachange ofjet orofpres- sure Unwanted bubbles caneasily bedestroyed byplaying asmall ﬂame over the surface Figure lshows theapparatus Wehave found itofadvantage toblacken thebotwm ofthevessel, because details ofstructure, such asgram boundanes and dislocations, then show upmore clearly Figure 2,plate 8,shows aportion ofa raft'ortwo-dimensional crystal ofbubbles Itsregularity can bejudged bylooking attheﬁgure inaglancing direction The size ofthebubbles vanes with theaperture, butdoes notappear tovary toany marked degree with thepressure orthedepth oftheorifice beneath thesurface Themam effect ofincreasing thepressure istoincrease therate ofissue ofthe bubbles Asanexample, athick-walled jetof49/ibore with apressure of100cin produced bubbles ofl~2mm indiameter Athin-walled jetof27/4diameter and apressure ofl80cm produced bubbles of06mm diameter Itisconvenient to refer tobubbles of20to10mm diameter as‘large’ bubbles, those from 08to 0-6mm diameter as‘medium’ bubbles, andthose from 0-3to0-1mm diameter assmall‘ bubbles, since their behaviour vanes with their size 30-10"ﬁt " mFIGURE 3Apparatus forproducing bubbles ofsmall size With thisapparatus wehave notfound itpossible toreduce thesizeofthejet andsoproduce bubbles ofsmaller diameter than 06mm Asitwasdesired toexperr ment with very small bubbles, wehadrecourse toplacing thesoap solution ina rotating vessel andintroducing aﬁnejetasnearly aspossible parallel toastream hne Thebubbles areswept away asthey form, andunder steady conditions are reasonably imiform They issue atarateofonethousand ormore persecond, giving ahigh-pitched note The soap solution mounts upinasteep wall around thepen- meter ofthevessel while itisrotating butcarries back most ofthebubbles with it when rotation ceases With thisdevice, illustrated inﬁgure 3,bubbles down to 0-12 mm mdiameter can beobtamed Asanexample, anoriﬁce 38p across ina thin-walled jet,with apressure of190cmofwater, andaspeed oftheﬂuid of 180cm/sec past theoriﬁce, produced bubbles of014mm diameter. Intlus case adish ofdiameter 9~5cmand speed of6rev/sec was used Figure 4,plate 8,isan enlarged picture ofthese ‘small’ bubbles andshows their degree ofregularity, the pattern isnotasperfect with arotating aswith astationary vessel, therows being seen tobeshghtly irregular when viewed i.naglancing direction These two-dimensional crystals show structures which have been supposed to exist inmetals, and simulate effects which have been observed, such asgrain boundaries, dislocations and other types offault, shp, recrystallization, annealing, andstrains dueto‘foreign’ atoms. 3Guam BOUNDARIES Figures 5a,5band50,plates 9and10,show typical grain boundaries forbubbles of1-87, 076and0-30mm diameter respectively Thewidth ofthedisturbed area attheboundary, where the bubbles have anirregular distribution, isingeneral greater the smaller the bubbles Inﬁgure 5a, which shows portions ofseveral adjacent grains, bubbles ataboundary between two gi'ains adhere deﬁnitely toone crystalline arrangement ortheother Inﬁgure 50there isamarked ‘Beilby layer’ between thetwo grains The small bubbles, aswill beseen, have agreater rigidity thanthelarge ones, andthisappears togiverisetomore irregularity attheinterface Separate grains show updistinctly when photographs ofpolycrystalline rafts such asﬁgures 5ato50,plates 9and 10,and ﬁgures 12a tol2e, plates 14to16, areviewed obhquely. With suitable hghtmg, theﬂoating raft ofbubbles itself when viewed obhquely resembles apolished and etched metal inaremarkable way Itoften happens that some ‘impurity atoms’, orbubbles which aremarkedly larger orsmaller than theaverage, arefound inapolycrystalline raft, andwhen this issoalarge proportion ofthem aresituated atthegrain boundaries Itwould be moorrect tosaythat theirregular bubbles make their way totheboundaries, itis adefect ofthemodel that nodiffusion ofbubbles through thestructure cantake place, mutual adjustments ofneighbours alone being possible Itappears that the boundaries tend toreadjust themselves bythegrowth ofonecrystal attheexpense ofanother tillthey pass through theirregular atoms 4DISLOCATIONS When asingle crystal orpolycrystalline raft iscompressed, extended, orother- wisedeformed itexhibits abehaviour very similar tothatwluch hasbeen pictured formetals subjected tostrain Uptoacertain limit themodel iswithin itselastic range Beyond that point ityields byshpalong oneofthethree equally inchned directions ofclosely packed rows Sliptakes place bythebubbles inonerowmoving forward over those inthenext row byanamount equal tothedistance between neighbours. Itisvery mterestmg towatch this process taking place The movement isnot simultaneous along thewhole row but begins atone end with theappearance ofa‘dislocation’, where there islocally onemore bubble inthe rows ononesideofthesliphneascompared with those ontheother This dis- location then runs along thesliplinefrom onesideofthecrystal totheother, the ﬁnal result bemg ashp byone ‘mter-atomic’ distance Such aprocess has been mvcked byOrowan, byPolanyi and byTaylor toexplain thesmall forces required toproduce plastic gliding Illmetal structures The theory put forward byTaylor (1934) toexplain themechanism ofplastic deformation ofcrystals considers the mutual action and equihbrium ofsuch dislocations The bubbles afford avery stnknig picture ofwhat hasbeen supposed totake place inthemetal Sometimes thedislocations runalong quite slowly, taking amatter ofseconds tocross acrystal, ltationary dislocations alsoaretobeseen incrystals which arenothomogeneously strained They appear asshort black hnes, andcanbeseen mtheseries ofphoto- graphs, ﬁgures 12atol2e,plates 14to16.When apolycrystalline raftiscompreﬂﬂedi these dark lmes areseen tobedashing about inalldirections across thecrystals. Figures 6a,6band 6c,plates 10and ll,show examples ofdislocations In ﬁgure 6a,where thediameter ofthebubbles isl9mm ,thedislocation isvery local, extendmg over about sixbubbles Inﬁgure 6b(diameter 076mm)itextends over twelve bubbles, andinﬁgure 60(diameter 030mm) itsinﬂuence canbe traced foralength ofabout ﬁfty bubbles Thegreater rigidity ofthesmall bubbles leads tolonger dislocations. Thestudy ofanymass ofbubbles shows, however, thatthere isnotastandard length ofdislocation foreach size Thelength depends upon thenature ofthestram inthecrystal Aboundary between twocrystals with corresponding axes atapproximately 30°(themaximum angle which canoccur) may beregarded asaseries ofdislocations inaltemate rows, and inthis case the dislocations arevery short Astheangle between theneighbouring crystals decreases, thedislocations occur atwider intervals and atthesame time become longer, till oneﬁnally hassingle dislocations inalarge body ofperfect structure asshown in ﬁgures Ga,6band6c Figure 7,plate ll,shows three parallel dislocations Ifwecallthem positive and negative (following Taylor) they arepositive. negative. positive. reading from left toright Thestrip between thelasttwohasthree bubbles mexcess, ascanbeseen bylookmg along therows inahorizontal direction Figure 8,plate 12,shows a dislocation prolectmg from agram boundary, aneffect often observed. Figure 9,plate l2,shows aplace where twobubbles take theplace ofone This may beregarded asahmiting caseofpositive andnegative dislocations onneigh- bouring rows, with thecompressive sides ofthedislocations facing each other. The contrary casewould leadtoaholeinthestructure, onebubble bemg nussmg atthe pomt where thedislocations met 5.OTHER TYPES orFAULT Figure 10,plate 12,shows anarrow strip between twocrystals ofparallel or1en~ tation, thestrip being crossed byanumber offault hnes where thebubbles arenot inclose packing Itisinsuch places asthese thatrecrystallization may beexpected The boundaries approach and thestrip isabsorbed into awider area ofperfect crystal Figures 11atollg,plates l3and 14areexamples ofarrangements which frequently appear inplaces whcrc there isalocal deﬁciency ofbubbles While adislocation is seenasadark stripe inageneral view, these structures show upintheshape ofthe letter Vorastnangles Atypical Vstructure isseeninﬁgure llaWhen themodel is bemg distorted, aVstructure isformed bytwodislocations meeting ataninclination of60°,itisdestroyed bythedislocations continuing along their paths. Figure lIb shows asmall triangle, which also embodies adislocation, foritwill benoticed that therows below thefault have onemore bubble than those below Ifamild amount of‘thermal movement’ isimposed bygentle agitation ofonesideofthecrystal, suchfaulty places disappear andaperfect structure isformed Here andthere inthecrystals there isablank space where abubble ismissing. showing asablack dotinageneral view Examples occur inﬁgure llg.Such agap cannot beclosed byalocal readiustment, smce ﬁlling thehole causes another to appear Such holes both appear anddisappear when thecrystal is‘cold-worked ’. These structures inthemodel suggest that similar local faults may exist inan actual metal They may play apart inprocesses such asdiffusion ortheorder- disorder change byreducing energy barriers intheir neighbourhood. andactas nuclei forcrystalhzation manallotropic change 6RECRYSTAILIZATION AND ANNEALHVG Figures 12a,tol2e,plates 14to16,show thesame raftofbubbles atsuccessive times Araftcovering thesurface ofthesolution wasgiven avigorous stirring with iiglass rake, andthen lefttoBdjllﬂl; itself Figure 12¢shows itsaspect about lsec. after stirrmg hasceased. Theraftisbroken intoanumber ofsmall ‘crystallites’, these aremahigh state ofnon-homogeneous strain asisshown bythenumerous dislocations andother faults The following photograph (ﬁgure 12b) shows the same raft32seclater. Thesmall grains have coalesced toform larger grams, and much ofthestrain hasdisappeared intheprocess. Recrystallization takes place right through theseries, thelastthree photographs ofwhich show theappearance oftheraft2,14and25min after theinitial stirring. Itisnotpossible tofollow the rearrangement formuch longer times, because thebubbles shnnk after longstanding, apparently duetothediffusion ofairthrough their walls, andthey alsobecome thin andtend toburst Noagitation wasgiven tothemodel during thisprocess An everslower process ofrearrangement goes on,themovement ofthebubbles mone partoftheraftsetting upstrains which activate arearrangement inaneighbouring part, and that initsturn still another Anumber ofinteresting pomts aretobeseen inthisseries. Note thethree small grains atthepoints indicated bytheco-ordinates AA, BB, CC. Apersists, though 30-11changed mform, throughout thewhole series Bisstillpresent after l4min,but hasdisappeared m25mm,leaving behind itfour dislocations marking mternsl strain inthegrain. Grain C’shrinks andﬁnally disappears inﬁgure 12d, leaving a holeandaVwhich hasdisappeared mﬁgure 12¢Atthesame time theill—deﬁned boundary inﬁgure I211atDDhasbecome adeﬁnite oneinﬁgure 12¢ Note also thestraightening outofthegram boundary mtheneighbourhood ofEEmﬁgures l2btol2e. Dislocations ofvarious lengths canbeseen, marking allstages between aslight warping ofthestructure andadeﬁnite boundary Holes where bubbles aremissmg show upasblack dots Some ofthese holes areformed orﬁlled upby movements ofdislocations, butothers represent places where abubble hasburst Many examples ofV’sandsome oftriangles canbeseen Other interesting points willbeapparent ﬁ-om astudy ofthissenes ofphotographs. Figures 13a, 13band 13¢, plate 17,show aportion ofaraft 1sec,4secand 4min. after thestirnng process, andisinteresting asshowing twosuccessive stages mthe relaxation towards amore perfect arrangement Thechanges show upwellwhen one looks maglancing direction across thepage Thearrangement isvery broken in ﬁgure 13a Inﬁgure 13bthebubbles have grouped themselves inrows, butthe curvature ofthese rows indicates ahigh degree ofintemal strain Inﬁgure 13¢‘this strain hasbeen reheved bytheformation ofanewboundary atA—A, therows on either side now being straight Itwould appear that theenergy ofthis strained crystal isgreater than that oftheiittercrystalline boundary Weareindebted to Messrs Kodak forthephotographs ofﬁgure 13,which were taken when thecine- matograph ﬁlmreferred tobelow wasproduced 7EFFECT orIMPUBITY non Figure 14,plate 18,shows thewidespread effect ofa bubble which isofthewrong size Ifthis ﬁgure iscompared with theperfect rafts shown inﬁgures 2and 4, plate 8,itwill beseen that three bubbles, one larger and two smaller than normal, disturb theregularity oftherows over thewhole oftheﬁgure Ashasbeen mentioned above, bubbles ofthewrong size aregenerally found inthegrain boun- daries, where holes ofirregular sizeoccur which canaccommodate them 8MECIIAXICAL PROPERTIES OF‘THE TWO~l)IMENSlONAL MODEL The mechanical properties ofatwo-dimensional perfect raft have been described inthepaper referred toabove (Bragg 1942b) The raft liesbetween two parallel springs dipping horizontally inthesurface ofthesoap solution The pitch ofthe springs isad]usted toﬁtthespacmg oftherows ofbubbles, wluch then adhere ﬁrmly tothem One spring canbetranslated parallel toitself byamicrometer screw, and theother issupported bytwothinvertical glass ﬁbres Theshearing stress canbe measured bynoting thedeﬂexion oftheglass ﬁbres When sub]ected toashearmg stram, theraftobeys Hooke‘s lawofelasticity uptothepoint where theelastic hmit isreached Itthen slips along some mtermediate rowbyanamount equal to thewidth ofonebubble The elastic shear andshpcanberepeated several times The elastic limit isapproximately reached when onesideoftherafthasbeen sheared byanamount equal toabubble width past theother side This feature supports thebasic assumption made byoneofusinthecalculation oftheelastic limit ofa metal (Bragg 194211), inwhich itissupposed that each crystalhte inacold-worked metal only yields when the strain inithasreached such avalue that energy is released bytheshp Acalculation hasbeen made byMMNicolson oftheforces between thebubbles, and will bepublished shortly Itshows two interesting points The curve for thevariation ofpotential energy with distance between centres isvery sirmlar to those which have been plotted foratoms Ithasaminimum foradistance between centres slightly lessthan afree bubble diameter, and rises sharply forsmaller dis~ tances Further, therise isextremely sharp forbubbles of0lmm diameter but much lesssoforbubbles oflmm diameter, thus conﬁrmmg theimpression given bythemodel that thesmall bubbles behave asifthey were much more rigid than thelarge ones. 9THREE-DIMENSIONAL ASSEMBLAGES Ifthebubbles areallowed toaccumulate inmultiple layers onthesurface, they form amass ofthree-dimensional ‘crystals ’with oneofthearrangements ofclosest Packing Figure 15,plate 18,shows anobhque view ofsuch amass, itsresemblance to8-polished andetched metal surface isnoticeable Inﬁgure 16,plate 20.asimilar mass isseen viewed normally Parts ofthestructure aredeﬁnitely incubic closest packing, theouter surface being the(1l1)face or(100) face Figure 17a, plate 19, shows a(Ill) face The outlines ofthethree bubbles onwhich each upper bubble rests canbeclearly seen, and thenext layer ofthese bubbles isfaintly visible ina position notbeneath theuppermost layer, showing that thepacking ofthe(Ill) planes hasthewelbknown cubic succession Figure l7bplate 19,shows a(100) face with each bubble restmg onfour others The cubic axes areofcourse inclined at 45°totheclose-packed rows ofthesurface layer Figure 170, plate 19,shows a twin inthecubic structure across the face (lll) The uppermost faces are(lll) and (100), and they make asmall angle with each other, though this isnotapparent mtheﬁgure, itshows upinanoblique view Figure 17d, plate 19,appears toshow both thecubic andhexagonal succession ofclosely packed planes, butitisdiﬁicult toverify whether theleft-hand sidefollows thetruehexagonal close-packed struc» tum because 1tlsnot eertam that theassemblage had adepth ofmore than two layers atthlspomt Many mstanoes oftwms, andofmtercrystallme boundarles, canbeseen mﬁgure 16,plate 20 Flgure 18,plate 21,shows several dlslocatrons lnathree-drmenslonal structure subjected toabendmg stram 10 DEMONSTRATION OFTHE MODEL Wxth theco~operat1on ofMessrs Kodak, a16mm clnematograph ﬁlmhasbeen made ofthemovements ofthedzslocamons andgram boundanes when smgle crystal and polycrystalhne rafts aresheared compressed, orextended Moreover, 1fthe soap solutlon lsplaced maglass vessel w1th aﬂatbottom, themodel lends ltself to projectlon onalarge scale bytransmltted hght Smoe aeertam depth 1sreqmrecl for producmg thebubbles, and thesolutlon 1srather opaque, It1sdesuable tomake the pro]ect1on through aglass block restmg onthebottom ofthevessel andJustsub- merged beneath thesurface Inconclusnon, wewxsh toexpress ourthanks toMrCEHarrold, ofKmg’s College, Cambndge, whomade forassome oftheprpettes whxch were used toproduce the bubbles Rnrmnsucss Bragg. WL1942a Natura, I49, 511 Bragg, W.L19420 JS01. Irwtrum I9,148. Taylor, GI1934 Prbc Roy Soc A,145, 362. 30-12 1>_ 2COOgOO%OwOCOCOSOOO&CmgmmﬂmwCOSmﬁﬂ‘OOOUlwgp gm....mm[w__..AmJﬂv,,y_v_wfM“p_¥WmWWWn3_MMMMMVk.Hﬁhg A‘OSiOS,1OOOAW%____“__“W”__“__"__“__“M%_“hOOOO§€’OO@fOOOOOQ‘yvOO O_OCOOOOOOSmv S‘OOO‘QCOOOﬁvO O‘OOCOOOOOCOOOA‘$‘COOOOOOOOO O‘jOOOOOOCOOO{,6COOOOOO “DEGCOOCO _DQOOO O “QCOO MEMOOU%U%Upaw OSOOnmﬂﬂﬂﬁuwvW_'__“§g__ OOOOOOOIOCOCOOOm‘ g“OO ﬁﬂvﬁOHwwﬂwwwﬂw_‘ ._jgQ._.@w’.%m_&§‘g....._‘ﬁﬂv‘... my..@Sw..@ﬂP....ﬁﬂv.....’.‘O.mw....’....._O0..__i%__?ﬁ...“§pOO@£wO_____“__“"£"“_ﬁQQQ@_QQU;‘ ........... ._UQUQQUDQUQQOQ Q....’.’. UQUQQDUUQUQ‘ .0...’ £QUQQUDQQQ'_ ....§&_ UQUQUQUQ_ﬁ J‘_... __\‘_’_v___). HA_HAHM@Q)_\m_£U_‘__\J_ gW__‘__\_/_\._’_ _)__ MA_H@UU U%@_@£UUwUmW33\H%*NU)))“)Hkykyvk\»$*J’ kaggAm“_@3_Hv‘’Vkxvwl(Lg arm‘ONO“ﬁgQC‘ipg’\Vg __ gxiD_\)_\} \ v__)_)_\]T\k____.__"1‘I‘l\)\’_‘_.___‘__‘‘IX’\)_“_‘ ‘n_v‘I‘NW“,\'Q_"'v7__'>_,%\_J'_,_'_Prkkk__‘__ T_\'\,|\ ___\ ,5N_’.paiwﬁm’,5.N.’hgﬁﬁwwﬁhyxgx_._\)_>\I_ W)Nkwhwg_Aﬁ*¥% MlHIPMmaDwMuMRmmHmSyW_b MMP2HUmF MmW0PMmdD%WuMRMMMyGMMP 4“MW“M Grain boundaries . ~_ ,-~. . \._. _..1-~._‘~, _~-'_~_ \_.‘_' 5-\ .~_\. '.v_‘- ."'k“1.‘*_~~‘_.‘- l‘>‘\L'-L‘=1>.:.‘,-1~€.,,‘\».‘:‘:_ I1§_\,*;".:' \':_-.},-.‘,- :_-.'~,‘-_._“, .~§'.‘- ~~\‘ \\.\~"s‘~”. '*-1\"§.J“ *3»: '~$L ‘~"'s».‘WM ‘§_.,,"1_'\ :~\"1 >14.,1'-~,,~—~:~’4\\_» -.,\,_2'1,,:1_Y"*1'1»,*\_'~.2'1 ‘.011f3>I f’F'”¢“.4.1 ,r¥‘“i"‘I1'*',55‘ r'.~}f‘w*.-,5F14w‘]A’.F;In-.:4,4‘,1'4'41<.'.;4§[email protected];v1..r‘41*,~~--» u.1,431,,‘*1 :1~»"\1,3111; ~~11;;~.11 sf‘-,"~ .,1.*1;1. fl r*;f"'~1-.31‘. ~_.,Y-_\,*r1~~ »\4 /4"TI;'-\,¢_,";>*~.-T{Y/Q 35;; -§>**‘?¢‘,‘l,lrés?5'5?Jr” ‘Miilxgu §¢~.¢ g {K Kg.‘ ?‘~;"<»?:‘Z"?>i’~" s ‘illX»-, - ~.,17\‘,- 1- KKK -1\§'\‘§,¢~\(€'.‘€_.\<,<v '~ .1. ‘_.X,_>;w ,_-_\-g‘_.,.§${,,_. _;',_~_..»., .,~745%.’?-;,. 2‘.5/,\_',.‘ . x»\ﬁ?1 if W;-24».*~ .,~».;'~',.~,;--.‘_~.;-I ~_*_, “*1 '.__\;,~;,:¢_‘Lj\1>.1‘:’_".,"-~ '\v‘:'“" ‘—’~,‘1_;~,\; '4,-,1-1~\,~'»‘:v ,1It ,‘y;"L,‘.""' '1% ‘1;~iﬁ?f.."w;\,,, ,--.*\.__'».7,-:~.7»*'.»\w 'vsZ7-.ggaa 9 ::\-~'~~1*1%‘s.* 11$; ii ,. § t ‘@5213?‘ ii;“\?r,4“‘~ '\.“’\.»)‘;; '-’»,.:-»_t¢QI ‘ifs?-1 ‘*1,17‘?A,,/~/‘<~-“.1~:;._,’;>~‘,,"'3'-'~.‘,».ﬂljw V~'Z‘§‘1' "41§1§4:\;~".'- >4.*13"’:‘;'‘:7 QR‘; ,‘\~;,;‘;L;;_:,,;'?~a-...;~;( Qfa‘:-‘ .K=.1§e§£@§§¢¢@q§>@$¢t~~v¢¢s*»w<ss>¢wwx%:>¢<<+t a§—“<s*5‘:zs>»*@x¢>®**s* §r<$n;3£¥c<"»“ ‘*%>i*K€Qtt*<”€W%4‘~tiKi*~f§K¥*¥~**$ ..~_' ._4‘, _, »,_{U ,;.v~. 1- ;\~&_$;,» ~Us/._;~\,,, /\ 1'W"’ ;§\?i<***T<.‘§~<.5gm ¢?¢:V““i.¢ ‘fx‘1';>3‘,i~1‘liq;§-1}P‘ ﬂﬁx\.*4¢*.—'Pwﬁ .\H__ 44.'‘#§.1 ‘$4.3&6Vt‘MK,»A 174'.‘ FIGURE 5a. Dxameter I87mm.7“-I-\"eeﬁfﬁfA Q1 s§<it ¢x¥<$=<igsqsg<>~x11*!‘Po“ﬁ~ it§:§<P~, ...&{:f&,' .(_~1.,,»1\,(*_i,‘-_, ~.‘~g‘.‘§\i;_> ks ~2 .1e ;~:€,;5‘~{ 01¢“: <11- §-§',;f<z@?-m*;§?:Qe;1 5-,_, ,_lag “4€{='<l e is ‘*‘f‘f ~=-at"§'*€€€¢ Qs i;Q‘<1_ *‘>{§{f§§1€§ii1*1<$!<»£i</5iliT%# :~é=§s<:§§%§%,@1<s* Frouma 56.Dzameter 0~76 mm. 3044 . ,; . _ .Q -_ _,\\ ‘\-\‘\_‘\‘\\'\. -\_-‘_-,\_-,‘ \\ l’ —_ ‘.~'.; 1; -_; 1;;,__ ._‘ r__,._ ,_‘,, _‘ ., .,¢__»_» _-. _.’, ‘‘“\\\\‘s‘\'\'~,'\~&\"w,"\»j‘\ \»’\'‘\;‘V--'. -... ., ».;4'- \ \ \ I\ \ ‘\ '\ \ \'- - - ~ -’-\\ 'v‘ v‘'v‘ \FIGURE 50. Agrain boundary. Diameter O-30 mm '‘-‘»-'-- '»~'.</.=».~1_1.,_../ , ‘N \‘.,\., -1,» A1.'.‘-’»'»"—', . ',< -_» -_~. 1_ ,_ ,. ,. 4 _ ,_ V ,., ,._ ,_ ,_- ,_=g_-1; L\1‘\\-‘~,'\'\--\-;\-;‘-,\-,‘-;‘-ﬁg,‘ ~\-_.'\, \.\. \; 1, 1/."Ill?“\‘\\e\i~»'~,‘~‘)‘»‘),.'> ‘J‘T ’»‘.'\ '\ —;\ -;‘ -;\“.;‘_»_;‘ ,_.IJ_;\"!“s‘I/";\,'/,_\:¢\,v_L;\;_\f,§).(» .‘»,‘ »\ -_\ »_\ -\ ._‘1,‘ 1‘ 1_\ ; 1' J_ 1’\/_\I,\l, {'. 1 .-:-'.,._,'._ .,\-¢‘J\\J,\J ~1_t,-,\J \‘J,\J\\J »~1.-10-1 \~1,-1\‘J \‘10J\‘1 »~I,»1,I v‘1\1‘\,1‘ -2,~.1\\I vI,\IP1 v‘Ju‘J\‘7 P}u‘J‘xi ,\J‘\Iv‘1_ 91,\J‘0‘1:5 9’\\I\\ \\\‘P), »iv‘?}. \_\¥'{. \\‘"\'1,.4. I\\>41‘. t/'Ifl‘~v‘.‘\),""‘\~\‘~.‘‘$1v“_\\~4 xi2J‘-x1’‘\$\,1_-up1."‘Rx >1.‘-JI\‘/ 4\,JJ-\4’ I\~/ 1_\\‘1 z‘.\J I-\I .1,\\J 1‘t\‘J 1,.\I .4‘‘\\‘{ 1,\,\1J,\\‘1‘- \\{'\ W4‘I'\1""9; W4“7.\\_J‘ I\\‘~‘.',\ \;~‘\ ...e. ‘"'\\l‘ .»-V,.-;-; -;1;2 1“ 3-.->_-. .,,-_,._,.,,..».1,». »_> 1;» _*\ \ \' \ \\ ‘\ ‘\ ‘\"t "\"\ "\ '\-'\- \~ '\ '\- '\~ti)tO‘\{' \_}'-4?'\l'"\--"Nt.» » -, \-\'at --»1.»1.\-i\»I\. ‘EJ » -‘v‘.-'.'~~_A. A..4',. ..-, '1. _=»,-_ ‘\\ \ t ,., .\ , _; _ _. _5 »_. /_>»_ '4_\).\. “I ,‘I \‘ FIGURE 6a. Adislocation. Diameter 1-9mm. 30-15,'\,'\.>/‘»,,/~ .1.'\',‘\'."\/"\ \ji />. _}*._A_ I, ',."'\/"V \- \ \\\x‘\ \‘\\\\‘vi‘\'\ _J. .j»1, _ - ',~ "5 v'\ :-_ 1" ,~_ 4~ ,. . . _ _ , L '5 - ‘. 1' ' ' ‘ \\‘\ ‘~.'-_-\-"~-a'-;'-;'-.’-;'~..’;.=’».,"\,'\:' \»5,\-'"'\-‘ZJ ‘)5._,» L- L -‘: >,‘ \ » ,_\ , \"\-"}-.A. I , Q \'\'\ ‘~I1.\. »'\ >'\/’ I1,./,/\.1 \;’\ '-K-,’—;\,— L.1.1\: ’\ 1.",'\. » -‘\ Dislocations \~\.\._._._. --\\\\\\\\ \\\\\\\\\» \’\'\'\’\"\"\‘\’\;\;\L\’\*\"\"\;\'\'\’\"\’\"'\"\" '‘''" \'\‘\"\"\*\‘\‘\‘\"\‘\"\‘\‘\‘\*\*\*\‘\*\‘\ ‘\'\"'."\"~‘\'"\‘\‘\'\"\‘\‘\'*\‘\*\‘\’\‘\"\‘\*'\ \"\,"\A"\‘\‘\L\'\‘\"\‘§‘\’\*\"§"§L\‘§"\‘§'j'”§; ",'?T"?'?"_\.'\‘»\\'\'\-\'\'\"'§'7"‘?'§'?7‘\'\““\'\\\w"\'\'\"\‘\‘\"\"\‘\“~'\'\'\'\'\ 5aaaa\a<<aa\\aaaa\aa5\\ w#5\H\\\<=Aaa\\aaa\<w\\='\'\'\*\‘\‘\‘\"\‘\‘\‘\‘\H*\‘\‘\‘\‘\‘\"\"\‘§\.\. _"\,'*\'j\1-\R5\R'-\?\,-\3~.'-R"\?\R4\3\9\§") FIGURE 6b.Diameter 0-76 mm.-\.- X4;‘.$_»;1'~L__‘k-_ ’\LT“J.\\- KK1 t»1 _1~»\ 4‘. L\L ﬁx-4 lffﬁﬁw 4:1“..\“\11ru./,1 1,1/‘.I.».1.1~\ L11»v~4,1.1\,¢4-4‘; ~11IvI‘‘JJJJ-3‘j!"h‘J»}”“‘/‘igt 41-;'/JJ".a.¢." .}{"”~ "('1-$U$U1-I //Q //J ‘,4~k'_.(-"17.1‘ "'‘"""'f ‘Pf’-, -{'*~44»I-TU.---' AV’_ '~ .»4"‘.—’H.", -414y%&} QM;§u¢&4£;@&%@¢@;p@{g$g§y /V/"‘}’(fI¢'} '':'~z 1 ' 9 '*,’I"#»,>" '- *‘W /;U1.-.,(I -I._- ‘ J../.__,.v].f,:,./.'_,l_,:,._)'-I); .v\4 , . '-."‘fZ"-"1-)f'*f‘§"'f'-1"?’ ' ‘ 'l’-?"7'¥'/"": +.'f' »'‘‘‘I‘U (7-I,:,.‘,,¢'\{:ﬂ_ _- ‘v ,‘~ (. ‘/. ‘)1 YV~4>‘4 V‘-I-‘ !IJ JVxiak335$;»+»+l ;¢~-my_,-6- .vi‘-.I‘- gs5.0%01-1»1'&'1?£9§i. ;.§z§;$lézI*.§§§l!z ‘TI‘I‘-1?Ek%%H+% P-=;‘\.{\xv1 '::,_ ,0“K FI(.L'RE 6c. Dlameter 0-30 mm. XsXH>551#hkh'Z’~"XXxmﬁ¥¥?$“X191?¢ 5,.$6 ‘..I'~_r»_)‘)‘»r.r_|_‘~r 1‘-~---~' ----" --'----‘- I.P1 ' X. 3xmm1ﬁﬁxmmr$xm$%$$$J1i»1m1»11“ - -»~»w~»_Mxxxx)n.ymxm1J**1%1xmm1w1m)))-‘~xy§Qgmx»»yx»»y§»ﬁ$§§§mx1x>¢»>»ym)r>r>}y))__;yKRQ¢Wﬁm»ymlyi;»>n§ynynylr:y)“vﬁ§i5ﬁ3*~">3mRYX' _ _ ~>)>))ry1>xx»»»»>>>»»y»»wn»rr>r¢mr§¥r>yrm1%)) N V‘ Q.~q»J&hkhj%)QJ.LkkJj)J,k)kl))))MLL)))) ,.“mm§%ﬁ%%§xnnymypyyylyymlﬁymmyymmlrlyarxpllj )(:)C)()U‘();_~i\~)~( ‘),-‘)-v)\._)‘...)~_)~ )1_)\_)~£_)'l_XvX)‘,)\,)&_)<_)\.}‘».)\.)‘.)\_)\.X.)- )~_)~)-)~,)_'_)- )>-,););_)_-__); V4 ;xKyKyRy;§XkKAnn»Mxrxhrymyyymmmml>»yymJy)y)I0“ "yﬁyggﬁyyhﬂilmyxhjﬂlllllhyj)h)m))}})}1);)yyynX;M&§khkXA;" ~l%il)yh)yr%))y)))))f}J)__>)_()(:)'(")‘()‘( XXX,4‘.)-.;_)'~)i)-1 >e _~)-»)=J:.)=.)=.)~_)=_1~ )~_)-._)~ )~,r_)~.)-_)~ ),);.>1_)_~_)-_)_ x1myyX&xxx&;“**»1»_¢r>¥rmJ1mmmr»¢}m11¢m1@1_j%?XyyyKR&kXhixxgxnnxxy§§§§%§%%%§%§3§§x§§%%>,.xxxn‘** '=.=,-q_x»rr="»'- --@¢ ~tap »*““****’ “*1 >>>>>>>r>>r>r>r> _%$yRxxxyxxi¥»x;*P*%¥})fjk )‘-1:E’~.‘L/€é;‘£»f&a+ » ___v_)(_)(:)<_).;)'._),_)")\)_‘§_;‘3_)},.)~)1)-):_))~)~); ))))>))))-)~)-_)~)') )~)-)_)-)) )(_)(_)\'_)~l )-.;~_'> ).>1 .=w~ >>>»p»»y;.;a))qn)¢»)»»»»)¢¢"*“”’>>>y>;,;;§§ﬂ>>>>)>>>>>>)>>>; ).‘7$)\)\‘:)\)-.‘ ‘. "‘7~. 1=-x_»xhr»>1>f§/}f§)}]}];”,”»§], 2xr,»>~>>>_»>.y,>>>114>>>>>Ii»!>1K,)'.)i)\)‘.)‘}zH)4}.).;1),).).;,3.,):}'))‘)")‘))I ));))))))))))))))‘4 U k.)()'.)1 >1)\‘,~\)':»1» . ,...)2 A.)\_)‘..):‘.;_,~ V21.‘ ‘:2. ._;".>“ ; '._;,,:,_~-_“',. _b2)‘.}. FIGURE 7.Parallel dlslocations. Dlameter 0-76 mm 30-16;-)).))).). y,); )‘)-)‘)))))));)I}'))))))))))) IIP!r)‘)’)y)))IJ)))t)'}')‘)')')'}‘)')))')))))))))))wr \»»~,,,.,H,.,,)>>>>>;>>>;;,>>>>>>>>>>)>>r\)):~))::)‘I))r‘)1)))))‘):))))),))))7))))‘))))3I }'-.'"-I-)')'.}1}')‘}-I-}~\-k;})~}-)~)‘)'7')‘)))‘)')‘)')')))'}‘)')')')')")')‘)')')} ‘I‘)'::):>!I))1/)))))l))))))7))))))))))))l)))) r:~>r>,»\>w,>, ,, >>1>>)>r>>>>)>>,,>>>>)>,>)r)))!*:)1)!)))}‘I))))))))))))))))))))))\ ‘:):r)}l*):r\-- ' })))))))))I)))))))) )))) "l )')‘) \ ~)))))))))))))))J))))) ) ) ) """-" "*1)“)-)’;))));)\); 1-) J)1-)))-) )'}‘ )-) 1)') 73') 33)1)I‘2\ \ l \ so,~.. O‘...-‘Qme... "‘_0vn ';f¢¢,4 it_ I -iii. O Cl-.,_ 3"‘'.°¢|\ -'-ax: FIGURE 8.Dislocation pI‘OJ8ClZlI1g from agrain boundary. Diameter 0~30 mm.-> ?.'.5.,’.4,‘vi.’__.’121-3;".__‘-'-_‘?-37._‘-,-.-,-.:,.,_‘?- \-1.\-’-.\-,-___- ‘."K1"Q1“iv“t1'l‘t1"ivisiv‘\?"\?L\?L\?¢\7i\7L\7*\7Q5,7*\1‘P 513kt:'‘if,"59.-r"~1145.7'39,?'-‘Ki,'*\‘¢£\?£\\?L\7L\7a\?L\7£§7Q5\?£\ .g:3,‘-'._/_\,_?-:\,_‘_'<‘,_7 48;.’13:’-Lg74\¢(i\&L\‘7a\?a\7a\7¢\7L\7¢\?'. “-"“'“F'3?"*5.“K“*1?'47?3715'‘~'~"“‘v‘\'r"r\'/’-K‘/"\'¢'-“<'¢'*\'/'""\f»'-\-. I""-'“’f/ =1‘~-i7"“’~f"“’~""‘*_""‘\.—,'5*~~,"?\’ 3"’"‘§?’~‘§%-’~\'¢'"\T"T"\'¢'-\'¢'~\'/-5-,'‘$1"fa" ‘Sf.'\?.',’‘pv,’\-,"\'_1'"§9,7/'\-,'—‘$; ’~§—'3''3”"\'/'\ ,"M3?13?“H\?'\1":1'~\.+‘\?£\'L "\" ‘'‘"3""\__’,-\_/\_‘,\_,\‘~‘»'¢\,;\ .4‘ .;\____ ____ ,»..,ix,»s¢<s,v:\,<\,i<\,.s, -\\v (Q 1‘Qi ‘I IQti‘2?‘I f“»/l\1"f“<.-((1/‘~ e/‘~x‘ex/‘~,l"\/(‘\J"\.¢('\\\jf\~/l"'~//‘X/f‘~/f"\/I'\._/1*_/I“/\‘,JP1\ 104.‘Q \¢I\ J\<I‘G‘Q FIGURE 9.Dislocations inadJacent rows. Diameter 1-9min. FIGURE 10. Series offault lines between two areas ofparallel orientation. Diameter 0-30 mm 30-17 6;; §§§§:=§'¥-'*:=:'<19'\2ig.3"gg:9 glzi I§§§#:1§:§§§:§‘r§*§:¢§$v$2$3$*=2"'3»'*3%-\'5’€‘3'~:-e&§§§§&$*“*AM-'X>t;,~(.l~g;:x~ ;~¢5“;~¢~°~"~~“'*.APCQ-. "'V"""' ‘\""" .¢'\/'\f\l\¢‘y\fy-5,-5 I‘> ‘ 4‘ -.. >,><>@,e>@,o‘e>"1 ...._.,....>m0O090099890.Q.;<>g<>g<><>000-0-r983909999999.)Og)OOOO%OO'OOO(_",%?QQQU_QQQQ.)0,‘-)O0OOOO80O(;%»'Q8Q.O,3‘Q9Q._5060‘O0-OOOO%(9-.-Q-~.>~\.>-.0or0o0--r!Q8Qg(;8QS)QS-7'IOO(OO8O880(*0-0»0°09"-300’<>‘0:’;OQQ»QUQOQk)Q»,g%,,°<>@,>, bl‘~IA‘-£‘§;-:2‘!Diameter 0-68 mm. u O G ,-s, Diameter 0-6mm. c v.‘ Diameter 06mm eDiameter 0-68mm. b -iY¢3v??7s?2?¢Y§?y17¢ v?e7é?v71?v7i?y-2-s»w»w»e-2 2s9-M-y-9-2-M-y 0 byi'i's’i?‘Qa’»'.’~Q,¢’;’p’ ?:’>Ql’Ag yi’ ,21-9;!-9,»e-v»:-2-we-2 Q-N’-1-9 y-2-?,»e»v»*y<:»w'¢r-2-2-1 v»g~g1»y '1,$.99.’ gig.;-¢,g,Q.Q ?'1' yPa 00?? 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Dlameter 0-70 mm 30-26 31 Tensors 31-1 Thetensor ofpolarizability Physicists always have ahabit oftaking thesimplest example ofanyphenome- nonandcalling it“physics,” leaving themore complicated examples tobecome theconcern ofother ﬁelds—say ofapplied mathematics, electrical engineering, chemistry, orcrystallography. Even solid-state physics isalmost only halfphysics because itworries toomuch about special substances. Sointhese lectures weWlll beleaving outmany interesting things. Forinstance, oneoftheimportant proper- tiesofcrystals~or ofmost substances—is that their electric polarizability IS different indifferent directions. Ifyouapply aﬁeld inanydirection, theatomic charges shift alittle andproduce adipole moment, butthemagnitude ofthe moment depends very much onthedirection oftheﬁeld. That is,ofcourse, quite acomplication. Butinphysics weusually start outbytalking about the special case inwhich thepolarizability isthesame inalldirections, tomake life easier. Weleave theother cases tosome other ﬁeld. Therefore, forourlater work, wewillnotneed atallwhat wearegoing totalkabout inthischapter. Themathematics oftensors isparticularly useful fordescribing properties ofsubstances which vary indirection—although that’s only oneexample oftheir use. Since most ofyouarenotgoing tobecome physicists, butaregoing togo intotherealworld, where things depend severely upon direction, sooner orlater youwillneed tousetensors. Inorder nottoleave anything out,wearegoing to describe tensors, although notingreat detail. Wewant thefeeling thatourtreat- ment ofphysics iscomplete. Forexample, ourelectrodynamics iscomplete—as complete asanyelectricity andmagnetism course, even agraduate course. Our mechanics isnotcomplete, because westudied mechanics when youdidn’t have a highlevel ofmathematical sophistication, andwewere notabletodiscuss subjects liketheprinciple ofleast action, orLagrangians, orHamiltonians, andsoon, which aremore elegant ways ofdescribing mechanics. Except forgeneral relativity, however, wedohave thecomplete lawsofmechanics. Ourelectricity andmagnetism iscomplete, andalotofother things arequite complete. Thequantum mechanics, naturally, willnotbe—we have toleave something forthefuture. Butyoushould atleast know what atensor is. Weemphasized inChapter 30thattheproperties ofcrystalline substances are different indifferent directions—We saythey areanisotropic. The variation of theinduced dipole moment with thedirection oftheapplied electric ﬁeld isonly oneexample, theonewewilluseforourexample ofatensor. Let's saythatfora given direction oftheelectric ﬁeld theinduced dipole moment perunitvolume P isproportional tothestrength oftheapplied ﬁeldE.(This isagood approximation formany substances ifEisnottoolarge.) Wewillcalltheproportionality constant oz." Wewant now toconsider substances inwhich ctdepends onthe direction oftheapplied ﬁeld, as,forexample, incrystals likecalcite, which make double images when youlook through them. Suppose, inaparticular crystal, weﬁndthatanelectric ﬁeldE1inthex-direc- tionproduces thepolarization P1ll'1thex-direction. Then weﬁndthatanelectric ﬁeldE2inthey-direction, with thesame strength, asE1produces adifferent polar- *InChapter 10wefolloyved theusual convention andwrote P=er>xE andcalled x(“khi”) the“susceptibility.” Here, itwillbemore convenient touseasingle letter, so wewritea foreqx Forisotropic dielectrics, a=(K-l)eQ, where Kisthedielectric constant (seeSection 10-4). 31-131-l Thetensor ofpolarizability 31-2 Transforming thetensor components 31-3 Theenergy ellipsoid 31-4 Other tensors; thetensor of inertia 31-5 Thecross product 31-6 Thetensor ofstress 31-7 Tensors ofhigher rank 31-8 Thefour-tensor of electromagnetic momentum R€VI€W' Chapter ll,Vol I,Vectors Chapter 20,Vol.l,Rotation tn Space E2 P2 Pl E1(0) ifFig. 3l—l. The vector addition of aolorizations inananisotropic crystal.ization P2inthey-direction. What would happen ifweputanelectric ﬁeld at 45°? Well, that’s asuperposition oftwoﬁelds along xandy,sothepolarization Pwillbethevector sumofP1andP2,asshown inFig.31-1(a). Thepolarization isnolonger inthesame direction astheelectric ﬁeld. You canseehowthatmight come about. There may becharges which canmove easily upanddown, but which arerather stiffforsidewise motions. When aforce isapplied at45°,the charges move farther upthan they dotoward theside. Thedisplacements are notinthedirection oftheexternal force, because there areasymmetric internal elastic forces. There is,ofcourse, nothing special about 45°. Itisgenerally truethatthe induced polarization ofacrystal isnotinthedirection oftheelectric ﬁeld. Inour example above, wehappened tomake a“lucky” choice ofourx-andy-axes, forwhich Pwasalong Eforboth thex-andy-directions. lfthecrystal were rotated with respect tothecoordinate axes, theelectric ﬁeld E2inthey-direction would have produced apolarization Pwith both anx-and ay-component. Similarly, thepolarization duetoanelectric ﬁeld inthex-direction would have produced apolarization with anx-component anday-component. Then the polarizations would beasshown inFig.3l—l(b), instead ofasinpart(a).Things getmore complicated—but foranyﬁeld E,themagnitude ofPisstillproportional tothemagnitude ofE. Wewant now totreat thegeneral caseofanarbitrary orientation ofacrystal with respect tothecoordinate axes. Anelectric ﬁeldinthex-direction willproduce apolarization Pwith x-,y-,andz-components; wecanwrite P,=a,,,,E,,, P,,=a,,,E,, P,=a,,,E,,. (31.1) Allwearesaying here isthat iftheelectric ﬁeld isinthex-direction, the polarization does nothave tobeinthatsame direction, butrather hasanx-,ay-, andaz-component—each proportional toE2. Wearecalling theconstants of proportionality oz“,041,2,andan,respectively (theﬁrstletter totelluswhich com- ponent ofPisinvolved, thelasttorefertothedirection oftheelectric ﬁeld). Similarly, foraﬁeld inthey-direction, wecanwrite P,=an/Ey, Pg=am,Ey, P2=ot2yEZ,; (31.2) andforaﬁeld inthez-direction, P,=a,,2E2, P1,=a,,2E,, P,=ot22E2. (31.3) Now wehave saidthatpolarization depends linearly ontheﬁelds, soifthere isan electric ﬁeld Ethathasboth anx-anday-component, theresulting x-component ofPwillbethesumofthetwoPjsofEqs. (31.1) and(31.2). IfEhascomponents along x,y,andz,theresulting components ofPwillbethesum ofthethree contributions inEqs. (31.1), (31.2), and(31.3). Inother words, Pwillbegiven by Pa: =azzEx + axyEy + azzEz> Pl,=ozy,,E,, +aWEy —l—ozy2E2, (31.4) P2 =azxEz + azyEy + azzEz- Thedielectric behavior ofthecrystal isthen completely described bythenine quantities (am, ax”, 0112,aw,...),which wecanrepresent bythesymbol ct”. (The subscripts iandjeach stand foranyoneofthethree possible letters x,y, andz.)Anyarbitrary electric ﬁeld Ecanberesolved with thecomponents E1,E1, andE2;from these wecanusethe(1,,toﬁndP1,Pg,andP,,which together give thetotal polarization P.Thesetofnine coeﬁicients 01,,iscalled atensor—in this instance, thetensor ofpolarizability. Justaswesaythatthethree numbers (E2, E1,E2)“form thevector E,”wesaythatthenine numbers (012,, aw,...)“form thetensor 01,1.” 31-2 31-2 Transforming thetensor components Youknow thatwhen wechange toadifferent coordinate system x’,y’,andz’, thecomponents E1»,E,/,andE,ofthevector willbequite different—as will alsothecomponents ofP.Soallthecoefficients ev,,willbedifferent foradifferent setofcoordinates. You can,infact, seehowthea'smust bechanged bychanging thecomponents ofEandPintheproper way, because ifwedescribe thesame physical electric ﬁeldinthenewcoordinate system weshould getthesame polariza- tion. Foranynewsetofcoordinates, P,’isalinear combination ofP2,P1,,andP2: P,»=aP,,+bP,,-|—cP,,, andsimilarly fortheother components. Ifyousubstitute forP1,P1,,andP,in terms oftheE‘s,using Eq.(31.4), youget Pr’ =a(azrE:c +ax;/Ely +azzEz) +b(°lziIEr +at/1tEo +'' +L-(QHEI + + Then youwrite E1,E1,andE2interms ofE,/,E1,/,andE2’;forinstance, E,=a'E,,i+b'E,,»+c'E,., where a’,b’,c’arerelated to,butnotequal to,a,b,c.Soyouhave Pzi,expressed interms ofthecomponents Ext,Eu’,andI2,/;thatis,youhave thenewct”. Itis fairly messy, butquite straightforward. When wetalkabout changing theaxes weareassuming thatthecrystal stays putinspace. Ifthecrystal were rotated withtheaxes, thea’swould notchange. Conversely, iftheorientation ofthecrystal were changed with respect totheaxes, wewould have anewsetofas Butifthey areknown foranyoneorientation of thecrystal, they canbefound foranyother orientation bythetransformation we haveJustdescribed. Inother words, thedielectric property ofacrystal isdescribed completely bygiving thecomponents ofthepolarization tensor anwith respect toanyarbitrarily chosen setofaxes. Just aswecanassociate avector velocity v=(15,,v,,,21,)with aparticle, knowing that thethree components willchange inacertain deﬁnite way ifwechange ourcoordinate axes, sowith acrystal we associate itspolarization tensor ct”,whose nine components willtransform ina certain deﬁnite wayifthecoordinate system ischanged. Therelation between PandEwritten inEq.(31.4) canbeputinthemore compact notation: P,=Za,,i2,, (31.5) J where itisunderstood thatirepresents either x,y,orzandthatthesumIStaken on] =x,y,andz.Many special notations have been invented fordealing with tensors, buteach ofthem 1Sconvenient only foralimited class ofproblems. One common convention istoomit thesum sign (Z)inEq.(31.5), leaving itunder- stood thatwhenever thesame subscript occurs twice (here _]),asumistobetaken over that index. Since wewillbeusing tensors solittle, wewillnotbother to adopt anysuch special notations orconventions. 31-3 Theenergy ellipsoid Wewant now togetsome experience with tensors. Suppose weaskthein- teresting question: What energy isrequired topolarize thecrystal (inaddition to theenergy intheelectric ﬁeldwhich weknow ise11E2/2 perunitvolume)? Consider foramoment theatomic charges thatarebeing displaced. Thework done indis- placing thecharge thedistance dxisqE2dx,andifthere areNcharges perunit volume, thework done isqE,,N dx. ButqNdx isthechange dP,,inthedipole 31-3 moment perunitvolume. Sotheenergy required perunitvolume is E1dP,,. Combining thework forthethree components oftheﬁeld, thework perunit volume isfound tobe E-a'P. Since themagnitude ofPisproportional toE,thework done perunitvolume in bringing thepolarization from 0toPistheintegral ofE-dP. Calling thiswork UP,* wewrite ' LIP=115-P =125,15. (31.6) Now wecanexpress Pinterms ofEbyEq.(31.5), andwehave that up=1ZZa11E1E1. (31.7)1 J Theenergy density UPisanumber independent ofthechoice ofaxes, soitisa scalar. Atensor hasthen theproperty thatwhen itissummed over oneindex (with avector), itgives anew vector; andwhen itissummed over both indexes (with twovectors), itgives ascalar. Thetensor a1,should really becalled a“tensor ofsecond rank,” because it hastwoindexes. Avector—with oneindex—is atensor oftheﬁrstrank, anda scalar—with noindex—is atensor ofzero rank. Sowesaythattheelectric ﬁeld Eisatensor oftheﬁrstrank andthattheenergy density upisatensor ofzero rank. Itispossible toextend theideas ofatensor tothree ormore indexes, and sotomake tensors ofranks higher than two. Thesubscripts ofthepolarization tensor range over three possible values— they aretensors inthree dimensions. The mathematicians consider tensors in four, ﬁve,ormore dimensions. Wehave already usedafour-dimensional tensor F1,inourrelativistic description oftheelectromagnetic ﬁeld (Chapter 26). Thepolarization tensor a1,hastheinteresting property thatitissymmetric, that is,that 0111=0111,,andsoonforanypair ofindexes. (This isaphysical property ofarealcrystal andnotnecessary foralltensors.) You canprove for yourself that thismust betrue bycomputing thechange inenergy ofacrystal through thefollowing cycle: (1)Turn onaﬁeld inthex-direction; (2)turn ona ﬁeld inthey-direction; (3)turn oﬂthex-ﬁeld; (4)turn offthey-ﬁeld. Thecrystal isnow back where itstarted, andthenetwork done onthepolarization must be back tozero. You canshow, however, thatforthistobetrue, 1111must beequal toa111,. Thesame kind ofargument can, ofcourse, begiven foram,etc. Sothe polarization tensor issymmetric. This alsomeans thatthepolarization tensor canbemeasured byjustmeasuring theenergy required topolarize thecrystal invarious directions. Suppose weapply anE-ﬁeld with only anx-anday-component; then according toEq.(31.7), up=%[ai1,1,EZ +(01111, +a11,)E,E1 +a11EZ]. (31.8) With anE1,alone, wecandetermine 011,1;with anE1alone, wecandetermine 0111; with both E,andE1,wegetanextra energy duetotheterm with (a,1 +111,). Since the04,1and0111,areequal, thisterm is21111 andcanberelated totheenergy. The energy expression, Eq. (31.8), hasanice geometric interpretation. Suppose weaskwhat ﬁelds E1,andE1correspond tosome given energy density—say uo.That isjustthemathematical problem ofsolving theequation a,,E;’+2a,1E,E1, +0111133=2u1,. This isaquadratic equation, soifweplotE1andE1,thesolutions ofthisequation *This work done inproducing thepolarization byanelectric ﬁeld isnottobeconfused with thepotential energy -pg-E ofapermanent dipole moment 120. 31-4 areallthepoints onanellipse (Fig. 31-2). (Itmust beanellipse, rather than a parabola orahyperbola, because theenergy foranyﬁeld isalways positive and ﬁnite.) Thevector Ewith components E1andE1canbedrawn from theorigin totheellipse. Sosuch an“energy ellipse” isanicewayof“visualizing” thepolar- ization tensor. Ifwenow generalize toinclude allthree components, theelectric vector Ein anydirection required togiveaunitenergy density gives apoint which willbeon thesurface ofanellipsoid, asshown inFig.31-3. Theshape ofthisellipsoid of constant energy uniquely characterizes thetensor polarizability. Now anellipsoid hastheniceproperty thatitcanalways bedescribed simply bygiving thedirections ofthree “principal axes” andthediameters oftheellipse along these axes. The “principal axes” arethedirections ofthelongest and shortest diameters andthedirection atright angles toboth. They areindicated bytheaxes a,b,andcinFig.31-3. With respect tothese axes, theellipsoid has theparticularly simple equation aaaEZ + abbEl% + ace-E3 :2u0- Sowith respect tothese axes, thedielectric tensor hasonly three components thatarenotzero: 01,11,a111,,and0:1, That istosay,nomatter howcomplicated a crystal is,itisalways possible tochoose asetofaxes (not necessarily thecrystal axes) forwhich thepolarization tensor hasonly three components. With such a setofaxes, Eq.(31.4) becomes simply Pa :aaaEas Pb =abbEb> Po :accEr- Anelectric ﬁeld along anyoneoftheprincipal axesproduces apolarization along thesame axis, butthecoefficients forthethree axes may, ofcourse, bedifferent. Often, atensor isdescribed bylisting thenine coefficients inatable inside of apairofbrackets: am: aary an aw aw aw ' (31.10) an azy azz Fortheprincipal axes a,b,andc,only thediagonal terms arenotzero; wesay thenthat“the tensor isdiagonal.” Thecomplete tensor is Olaa O O O 041,1, O' (31.1 l) 0 0 oi“ Theimportant point isthatanypolarization tensor (infact, anysymmetric tensor ofrank twoinanynumber ofdimensions) canbeputinthisform bychoosing a suitable setofcoordinate axes. Ifthethree elements ofthepolarization tensor indiagonal form areallequal, thatis,if aaa :abb =ace =as theenergy ellipsoid becomes asphere, andthepolarizability isthesame inall directions. Thematerial isisotropic. Inthetensor notation, at’; =(1511; where 611istheunittensor 1 O 0 6,1=0 l 0- (31.14) 0 0 1 That means, ofcourse, 51'] =1, i=j; 511=0, if i¢j. (31.15) 31-5A I71‘<[Tl Ex Fig. 31-2. Locus ofthevector E= (E,,E1)that gives aconstant energy of polarization. UFig. 31-3. The energy ellipsoid of thepolarization tensor.15%) Thetensor 61-1isoften called the“Kronecker delta.” You may amuse yourself byproving thatthetensor (31.14) hasexactly thesame form ifyouchange the coordinate system toanyother rectangular one. Thepolarization tensor ofEq. (31.13) gives P,=.125,115,=aE1, J which means thesame asouroldresult forisotropic dielectrics: I”==all The shape andorientation ofthepolarization ellipsoid cansometimes be related tothesymmetry properties ofthecrystal. Wehave saidinChapter 30 that there are230different possible internal symmetries ofathree-dimensional lattice andthatthey can, formany purposes, beconveniently grouped intoseven classes, according totheshape oftheunitcell. Now theellipsoid ofpolarizability must share theinternal geometric symmetries ofthecrystal. For example, a triclinic crystal haslowsymmetry—-the ellipsoid ofpolarizability willhave unequal axes, anditsorientation willnot,ingeneral, bealigned with thecrystal axes. On theother hand. amonoclinic crystal hastheproperty that itsproperties areun- changed ifthecrystal isrotated 180° about oneaxis. Sothepolarization tensor must bethesame after such arotation. Itfollows thattheellipsoid ofthepolariz- ability must return toitself after a180°rotation. That canhappen only ifoneof theaxesoftheellipsoid isinthesame direction asthesymmetry axisofthecrystal. Otherwise, theorientation anddimensions oftheellipsoid areunrestricted Foranorthorhombic crystal, however, theaxes oftheellipsoid must corre- spond tothecrystal axes, because a180° rotation about anyoneofthethree axes repeats thesame lattice. Ifwegotoatetragonal crystal, theellipse must have the same symmetry, soitmust have twoequal diameters. Finally. foracubic crystal, allthree diameters oftheellipsoid must beequal, itbecomes asphere, andthe polarizability ofthecrystal isthesame inalldirections. There isabiggame ofﬁguring outthepossible kinds oftensors forallthe possible symmetries ofacrystal. Itiscalled a“group-theoretical” analysis. But forthesimple case ofthepolarizability tensor, itisrelatively easy toseewhat the relations must be. 31-4 Other tensors; thetensor ofinertia There aremany other examples oftensors appearing inphysics. Forexample, inametal, orinanyconductor, oneoften ﬁnds thatthecurrent density 1'isap- proximately proportional totheelectric ﬁeld E;theproportionality constant is called theconductivity tr: j=0E. Forcrystals, however, therelation between jandEismore complicated; the conductivity isnotthesame inalldirections. Theconductivity isatensor, and wewrite 11=Zo11E1. Another example ofaphysical tensor isthemoment ofinertia. InChapter 18 ofVolume Iwesawthatasolid object rotating about aﬁxed axishasanangular momentum Lproportional totheangular velocity w,andwecalled theproportion- ality factor 1,themoment ofinertia: L=Iw. Foranarbitrarily shaped object, themoment ofinertia depends onitsorientation with respect totheaxisofrotation. Forinstance, arectangular block willhave different moments about each ofitsthree orthogonal axes. Now angular velocity atandangular momentum Lareboth vectors. Forrotations about oneoftheaxes ofsymmetry, they areparallel. Butifthemoment ofinertia isdifferent forthe 31-6 three principal axes, then 0.1andLare,ingeneral, notinthesame direction (seeFig. 31-4). They arerelated inaway analogous totherelation between EandP.Ingeneral, wemust write L: :120103.: +Ixywy 'l‘Ixzwz, L1=111031 —l—I1,1w1 +11,012, (31.16) L2 =Izxwx +[zywy +leewa- Thenine coefficients I11arecalled thetensor ofinertia. Following theanalogy with thepolarization, thekinetic energy foranyangular momentum must be some quadratic form inthecomponents 031,(.01,andwz: KE=22I11w1w1. (31.17) 17 Wecanusetheenergy todeﬁne theellipsoid ofinertia. Also, energy arguments canbeused toshow thatthetensor issymmetric—that 111=111. Thetensor ofinertia forarigid body canbeworked outiftheshape ofthe object isknown. Weneed only towrite down thetotal kinetic energy ofallthe particles inthebody. Aparticle ofmass mandvelocity vhasthekinetic energy %mv2,andthetotal kinetic energy isjustthesum Ztmvi over alloftheparticles ofthebody. Thevelocity vofeach particle isrelated to theangular velocity wofthesolid body. Let’s assume thatthebody isrotating about itscenter ofmass, which wetake tobeatrest. Then ifristhedisplacement ofaparticle from thecenter ofmass, itsvelocity visgiven bywXr.Sothetotal kinetic energy is 1413=Z2m(...><r)2. (31.18) Now allwehave todoiswrite toXroutinterms ofthecomponents w,,,o.>,,.w,, andx,y,z,andcompare theresult with Eq.(31.17); weﬁndI11byidentifying terms. Carrying outthealgebra, wewrite <<»><r>2- <~><i)3+(w><r)§+(<»><r)§ =(w1z —w1y)2 -1-(wzx —w1z)2 +(w,y —w,,x)3 =+0:322 —2w1w,zy -l-o.§§y2 +wfxz —2w2w,xz +@322 -l-wfy2 —2w1,w1yx +wZx2. Multiplying thisequation bym/2, summing over allparticles, and comparing withEq.(31.17), weseethatI11,forinstance, isgiven by Ira: :E "1012 +Z2)- This istheformula wehave hadbefore (Chapter 19,Vol. I)forthemoment of inertia ofabody about thex-axis. Since r2=x2+y2+22,wecanalsowrite thisterm as 2 11,1,=Zm(r —x2). Working outalloftheother terms, thetensor ofinertia canbewritten as Zm(r2 -x2) —Zmxy —Zmxz I1]= —Zmyx Zm(r2—1/2) —Zmyz - (31-19) —Zmzx —Zmzy Zm(r2 -22) Ifyouwish, thismay bewritten in“tensor notation” as 1,]=Em(r2 611—-r1r1). (31.20) 31-7atL‘ L _,> Fig. 31-4. The angular momentum Lofasolid object isnot, ingeneral, parallel toitsangular velocity w. where ther1arethecomponents (x,y,z)oftheposition vector ofaparticle and theZmeans tosumover alltheparticles. Themoment ofinertia, then, isatensor ofthesecond rank whose terms areaproperty ofthebody andrelate Ltowby L,=21,113, (3121) J Forabody ofanyshape whatever, wecanﬁndtheellipsoid ofinertia and, therefore, thethree principal axes. Referred tothese axes, thetensor will be diagonal, soforanyobject there arealways three orthogonal axes forwhich the angular velocity andangular momentum areparallel They arecalled theprincipal axes ofinertia. 31-5 Thecross product Weshould point outthat wehave been using tensors ofthesecond rank since Chapter 20ofVolume I.There, wedeﬁned a“torque inaplane,” such as r11,by 1'11,=xF1 —yF,. Generalized tothree dimensions, wecould write T11=r1F1 —r1F1. (31.22) Thequantity T11isatensor ofthesecond rank. Onewaytoseethatthisissoisby combining T11with some vector, saytheunitvector e,according to E TUGJ. J Ifthisquantity isavector, then111must transform asatensor—this isourdeﬁnition ofatensor. Substituting forT11,wehave gT1181 ==Er,F1e1 ——gr1e1f71 J 1 J =r1(F'e) —(r-e)F,. Since thedotproducts arescalars, thetwoterms ontheright-hand sidearevectors, andlikewise their difference. SoT11isatensor. But1'1,ISaspecial kind oftensor; itisantisymmetric, thatis, Ti] :“Tit: soithasonly three nonzero terii1s—-T11. 1-1,,and1-1,. Wewere able toshow in Chapter 20ofVolume lthatthese three terms, almost “byaccident,” transform likethethree components ofavector, sothatwecould define T:(Tn Ty; T2) = (T3/29 7-21- Try) Wesay“byaccident,” because ithappens only inthree dimensions. Infour dimensions, forinstance, anantisymmetric tensor ofthesecond rank hassix nonezero terms andcertainly cannot bereplaced byavector withfour components. Justastheaxial vector -r=rXFisatensor, soalsoisevery cross product oftwopolar vectors——all thesame arguments apply. Byluck, however. they are alsorepresentable byvectors (really pseudovectors), soourmathematics hasbeen made easier forus. Mathematically, ifaandbareanytwovectors, thenine quantities a,b1form atensor (although itmay have nouseful physical purpose). Thus, fortheposition vector F1,r,r1isatensor, andsince 6,1isalso, weseethat theright sideofEq. (3120)isindeed atensor. Likewise Eq(31.22) isatensor, since thetwoterms on theright-hand sidearetensors. 31-8 31-6 Thetensor ofstress Thesymmetric tensors wehave described sofararose ascoefficients inre- lating onevector toanother. Wewould liketolook now atatensor which hasa different physical signiﬁcance—the tensor ofstress. Suppose wehave asolid object with various forces onit.Wesaythatthere arevarious “stresses” inside, bywhich wemean thatthere areinternal forces between neighboring parts ofthe material. Wehave talked alittle about such stresses inatwo-dimensional case when weconsidered thesurface tension inastretched diaphragm inSection 12-3. Wewillnowseethattheinternal forces inthematerial ofathree-dimensional body canbedescribed interms ofatensor. Consider abody ofsome elastic material—say ablock ofjello. Ifwemake acutthrough theblock, thematerial oneach sideofthecutWlll, ingeneral. get displaced bytheinternal forces. Before thecutwasmade, there must have been forces between thetwoparts oftheblock thatkept thematerial inplace; wecan deﬁne thestresses interms ofthese forces. Suppose welook atanimaginary plane perpendicular totheX-klX1S—-l1l(C theplanea inFig3l—5—and askabout theforce across asmall areaAyAzinthisplane Thematerial ontheleftofthearea exerts theforce AF1 onthematerial totheright, asshown inpart (b)oftheﬁgure There is,ofcourse, theopposite reaction force -AF1 exerted onthematerial to theleftofthesurface. Ifthearea issmall enough, weexpect thatAF1 ispropor- tional tothearea AyAz. You arealready familiar with onekind ofstress—the pressure inastatic liquid. There theforce isequal tothepressure times theareaand1Satright angles tothesurface element. Forsolids—also forviscous liquids inmotion-the force need notbenormal tothesurface; there areshear forces inaddition topressures (positive ornegative) (Bya“shear” force wemean thetangential components oftheforce across asurface.) Allthree components oftheforce must betaken intoaccount. Notice alsothat ifwemake ourcutonaplane with some other orientation, theforces willbedifferent. Acomplete description oftheinternal stress requires atensor. AF).1 5&1\ 150 <7 / \ //1 \ /AF, // W,ll1/,/i ’if,/J/ 1,/v.1//lQ// ////’ /1 \ //////// _AF‘ __ .--X (0) lb) Fig 31-5. The material totheleftof the plane 0‘exerts across the area AyAzthe force AF1 onthe material to theright oftheplane. Fig. 31-6. Theforce AF1 across an , element ofarea AyAzperpendicular to AFZ1 the x-axis isresolved into the three MM? components AFX1, AFY1, andAFI1.Y Wedeﬁne thestress tensor inthefollowing way: First, weimagine acut perpendicular tothex-axis andresolve theforce AF1across thecutintoitsconi- ponents AF11, AF11, AFZ1, asinFig.31-6. Theratio ofthese forces tothearea AyA2,wecallS1,,S11,andS2,. Forexample, AF Se=521 Theﬁrstindex yrefers tothedirection force component; thesecond index xis normal tothearea. Ifyouwish, youcanwrite theareaAyA2asAa,, meaning an element ofarea perpendicular tox.Then AF11 S“:Aa Next, wethink ofanimaginary cutperpendicular tothey-axis. Across asmall 31-9 AFy2 \ \ AF2 AFX2 1\\i3\ F—r*Ax-9/\ \\ XAFR Fig. 3l—7. Theforce across anele- ment ofarea perpendicular toyisre- solved into three rectangular components AFyn F AF“ A//l4/ \ Ay — AFxn //AFzn 9/ Ax Fig. 3l—8. The force F,,across the face N(whose unitnormal isn)isresolved into components.area AxA2there willbeaforce AF2. Again weresolve thisforce intothree com- ponents, asshown inFig 31-7, anddeﬁne thethree components ofthestress, S,,,.SW,SM,astheforce perunitareainthethree directions. Finally, wemake an imaginary cutperpendicular tozanddeﬁne thethree components S”,S,,2,andS22. Sowehave thenine numbers Sm: Sony S12 SW=SW SW SW - (31.23) Szar Say S22 Wewant toshow now thatthese nine numbers aresufficient todescribe com- pletely theinternal state ofstress. andthatS”isindeed atensor Suppose wewant toknow theforce across asurface oriented atsome arbitrary angle Canweﬁnd itfrom SN‘? Yes, inthefollowing way: Weimagine alittle solid ﬁgure which has onefaceNinthenewsurface, andtheother faces parallel tothecoordinate axes. lfthefaceNhappened tobeparallel tothez-axis, wewould have thetriangular piece shown inFig.3l—8. (This isasomewhat special case, butwillillustrate well enough thegeneral method.) Now thestress forces onthelittle solid triangle in Fig 3l~8 areinequilibrium (atleast inthelimit ofinﬁnitesimal dimensions), sothetotal force onitmust bezero. Weknow theforces onthefaces parallel to thecoordinate axes directly from S,, Their vector sum must equal thefoice on thelaceN,sowecanexpress thisforce interms ofSH. Our assumption that thesurface forces onthesmall triangular volume arein equilibrium neglects any other body forces that might bepresent, such asgravity oipseudo forces ifourcoordinate system isnotaninertial frame Notice, however, thatsuch body forces willbeproportional tothevolume ofthelittle triangle and, therefore. toAx,Ay,A2,whereas allthesurface forces areproportional tothe areas such asAxAy,AyA2.etc. Soifwetake thescale ofthelittle wedge small enough, thebody forces canalways beneglected incomparison with thesurface forces. Let’s nowadduptheforces onthelittle wedge. Wetakeﬁrstthex-component, which isthesumofﬁveparts—one from each face However, ifA: issmall enough, theforces onthetriangular faces (perpendicular tothez-axis) willbeequal and opposite, sowecanforget them. Thex-component oftheforce onthebottom rectangle is AF,”=SwAx AZ. Thex-component oftheforce onthevertical rectangle is AF“ =SmAyA2. These twomust beequal tothex-component oftheforce outward across theface N.Let’s callntheunitvector normal tothefaceN,andtheforce onitF,,,then wehave AF,“ =SmAyA2—l—SWAxAz. Thex-component S,"oftheitress across thisplane isequal toAF,,, divided by thearea, which isA\/zAx2 +Ay2, or Sam :Spa; "a" '4LT TT+ S11; TTT’;TA_‘¥_L_‘il; ' \/Ax? +Ayz \/Ax! —l—Ayg Now Ax/\/fAxf2ﬂ—l- Ayiiisthecosine oftheangle 6between nandthey-axis, as shownﬁin Fig.31-8, soitcanalsobewritten asny,they-component ofn.Similarly, Ay/\/Ax? —l—Ay2issin6=nx.Wecanwrite Sm=Sun, +Sfynya Ifwenow generalize toanarbitrary surface element, wewould getthat Szri :Szznic + Sp;/ny —l— Srznz 31-10 or,ingeneral, S...=Zs,,n,. (31.24)J Wecanﬁndtheforce across anysurface element interms oftheS”,soitdoes describe completely thestate ofinternal stress ofthematerial. Equation (3124)saysthatthetensor S”relates theforce Sntotheunitvector n,just asev,,relates PtoE.Since nandSnarevectors, thecomponents ofSHmust transform asatensor with changes incoordinate axes. SoS,,isindeed atensor. Wecanalsoshow thatS”isasymmetric tensor bylooking attheforces ona littlecube ofmaterial. Suppose wetakealittle cube, oriented with itsfaces parallel toourcoordinate axes, andlook atitincross section, asshown inFig3l—9. If welettheedge ofthecube beoneunit, thex-andy-components oftheforces on thefaces normal tothex-andy-axes might beasshown intheﬁgure. Ifthecube issmall, thestresses donotchange appreciably from onesideofthecube tothe opposite side, sotheforce components areequal andopposite asshown Now there must benotorque onthecube, oritwould start spinning. Thetotal torque about thecenter is(S,,,, —SW) (times theunitedge ofthecube), andsince the total isZero, S,,,,isequal toSM,andthestress tensor 1Ssymmetric. Since SHisasymmetric tensor, itcanbedescribed byanellipsoid which will have three principal axes. Forsurfaces normal tothese axes, thestresses are particularly simple—they correspond topushes orpulls perpendicular tothesur- faces There arenoshear forces along these faces. Foranystress, wecanalways choose ouraxessothattheshear components arezero. Iftheellipsoid isasphere, there areonly normal forces inanydirection. This corresponds toahydrostatic pressure (positive ornegative). Soforahydrostatic pressure, thetensor isdiagonal andallthree components areequal; they are,infact,justequal tothepressure p. Wecanwrite S”=p5,,. (31.25) Thestress tensor—and alsoitsellipsoid——will, ingeneral, vary from point to point inablock ofmaterial; todescribe thewhole block weneed togivethevalue ofeach component ofS”asafunction ofposition. Sothestress tensor isaﬁeld. Wehave hadscalar ﬁe/ds, likethetemperature T(x,y,z),which giveonenumber foreach point inspace, andvectorﬁelds likeE(x,y,z),which givethree numbers foreach point. Now wehave atensor ﬁeld which gives nine numbers foreach point inspace—or really sixforthesymmetric tensor S”.Acomplete description oftheinternal forces inanarbitrarily distorted solid requires sixfunctions of x,y,and2. 31-7 Tensors ofhigher rank Thestress tensor S”describes theinternal forces ofmatter. Ifthematerial is elastic, itisconvenient todescribe theinternal dzstortion interms ofanother tensor T,,—called thestrain tensor. Forasimple object likeabarofmetal, youknow thatthechange inlength, AL,isapproximately proportional totheforce, sowe sayitobeys Hooke’s law: AL='YF. Forasolid elastic body with arbitrary distortions, thestrain T”1Srelated tothe stress S”byasetoflinear equations: T,,=Zv,,,,,s,,,. (31.26)kJ Also, youknow thatthepotential energy ofaspring (orbar)is %FAL =%vF2. Thegeneralization fortheelastic energy density inasolid body is U.,,,,,,,, =Z%v.,,,,S,,S,,,. (31.27)tjkl 31-11Syy syx s,y SXX SXX sxy sy, Sty Fig. 31-9. The x-and y-forces four faces ofosmall unitcube. Thecomplete description oftheelastic properties ofacrystal must begiven in terms ofthecoefﬁcients V”k1.This introduces ustoanewbeast. Itisatensor ofthe fourth rank. Since each index cantakeonanyoneofthree values, x,y,orz,there are34=81coefficients. Butthere arereally only 21diﬁerent numbers. First, since S”issymmetric, ithasonly sixdifferent values, andonly 36dzﬂerent co- efficients areneeded inEq.(31.27). Butalso, S,,canbeinterchanged with S1,; without changing theenergy, S0V111,; must besymmetric ifweinterchange ij andkl.This reduces thenumber ofdifferent coeﬁicients to21.Sotodescribe the elastic properties ofacrystal ofthelowest possible symmetry requires 21elastic constants! This number is,ofcourse, reduced forcrystals ofhigher symmetry. Forexample, acubic crystal hasonly three elastic constants, andanisotropic substance hasonly two. That thelatter istruecanbeseen asfollows How canthecomponents of 'Y,,;,; beindependent ofthedirection oftheaxes, asthey must beifthematerial isisotropic? Answer: They canbeindependent onlyiftheyareexpressible interms ofthetensor 6,,There aretwopossible expressions, 5,,6;,; and6,145,; +6,16,,” which have therequired symmetry, so'Y,,;,; must bealinear combination ofthem. Therefore, forisotropic materials, 711/1"! :a(62]6kZ) + b(52k6)l Tl’ 6215170; andthematerial requires twoconstants. aandb.todescribe itselastic properties. Wewillleave itforyoutoshow thatacubic crystal needs only three Asaﬁnal example, thistime ofathird-rank tensor, wehave thepiezoelectric effect. Under stress, acrystal generates anelectric ﬁeld proportional tothestress; hence, ingeneral, thelawis E, IZ P,]kS]l~- ],k where E,istheelectric ﬁeld, andtheP,,;,arethepiezoelectric coefﬁcients———or the piezoelectric tensor Can youshow thatifthecrystal hasacenter ofinversion (invariant under x,y,2—>——x,—y,~2)thepiezoelectric coefficients areallzero‘? 31-8 Thefour-tensor ofelectromagnetic momentum Allthetensors wehave looked atsofarinthischapter relate tothethree dimensions ofspace; they aredeﬁned tohave acertain transformation property under spatial rotations. InChapter 26wehadoccasion touseatensor inthefour dimensions ofrelativistic space-time——the electromagnetic ﬁeld tensor F,,,, The components ofsuch afour-tensor transform under aLorentz transformation of thecoordinates inaspecial waythatweworked out. (Although wedidnotdoit thatway, wecould have considered theLorentz transformation asa“rotation” inafour-dimensional “space” called Minkowski space; then theanalogy withwhat wearedoing here would have been clearer) Asourlastexample, wewant toconsider another tensor inthefourdimensions (t,x,y,z)ofrelativity theory. When wewrote thestress tensor, wedeﬁned S” asacomponent ofaforce across aunitarea. Butaforce isequal tothetime rateofchange ofamomentum. Therefore, instead ofsaying “Sn, isthex-compon- entoftheforce across aunitarea perpendicular toy,”wecould equally wellsay, “S1,, istherateofﬁow ofthex-component ofmomentum through aunitarea perpendicular toy.” Inother words, each term ofS”alsorepresents theﬁow of the1'-component ofmomentum through aunitareaperpendicular tothe/-direction These arepure space components, butthey areparts ofa“larger” tensor S,,,,in four dimensions (itand 1/=t,x,y,z)containing additional components like Sm,SW,S”,etc. Wewillnow trytoﬁnd thephysical meaning ofthese extra components. Weknow thatthespace components represent ﬂow ofmomentum. Wecan getaclueonhowtoextend thistothetime dimension bystudying another kind of “ﬂow”—the ﬂow ofelectric charge. Forthescalar quantity, charge, therateof ﬂow (perunitareaperpendicular totheﬂow) isaspace vect0r—the current density 31-12 vector j.Wehave seen thatthetime component ofthisflow vector isthedensity ofthestuffthatisﬂowing Forinstance, jcanbecombined with atimecomponent, j,=p,thecharge density, tomake thefour-vector /',,=(p,j); that is,theitin j,,takes onthevalues I,x,y,ztomean “density, rateofﬁow inthex-direction, rateofllow iny,rateofﬂow in2”ofthescalar charge. Now byanalogy with ourstatement about thetime component oftheﬁow of ascalar quantity, wemight expect thatwith Sm,S,,,,,andSm,describing theﬁow ofthex-component ofmomentum, there should beatime component S,”which would bethedensity ofwhatever isﬂowing; thatis,SMshould bethedensity of x-momentum. Sowecanextend ourtensor horizontally toinclude at-component Wehave SM=density ofx-momentum, SN=x-ﬂow ofx-momentum, SW=y-ﬂow ofx-momentum, SM=z-ﬂow ofx-momentum. Similarly, forthey-component ofmomentum wehave thethree components of ﬁow——S,,z, SM,S,,,~—to which weshould addafourth term: S,”=density ofy-momentum. And, ofcourse, toSn,Sz,,,S22wewould add S2,=density ofz-momentum. Infour dimensions there isalsoat-component ofmomentum, which is,we know, energy Sothetensor S”should beextended vertically with Sm,S,,,,and S”,where S”,=x-ﬂow ofenergy, SM=y-ﬁow ofenergy, (31.28) Sn=z-ﬂow ofenergy; thatis,St,istheﬂow ofenergy perunitarea andperunittime across asurface perpendicular tothex-axis, andsoon.Finally, tocomplete ourtensor weneed S”,which would bethedensity ofenergy. Wehave extended ourstress tensor S”ofthree dimensions tothefour-dimensional stress-energy tensor S,,,,. The index itcantake onthefourvalues t,x,y,and2,meaning, respectively, “density,” “ﬂow perunit area inthex-direction,” “ﬁow perunitarea inthey-direction," and“ﬁow perunitarea inthez-direction "Inthesame way, 1/takes onthefour values t,x,y,ztotelluswhat ﬁows, namely, “energy,” “momentum inthex-direc- tion,” “momentum inthey-direction,” and“momentum inthez-direction.” Asanexample, wewilldiscuss thistensor notinmatter. butinaregion offree space inwhich there isanelectromagnetic ﬁeld. Weknow thattheﬁowofenergy is thePoynting vector S=e,,c2E ><B.Sothex-,y-,andz-components ofSare, from therelativistic point ofview, thecomponents SM,S,,,,andS”ofourfour- diniensional stress-energy tensor. Thesymmetry ofthetensor SHcarries over into thetime components aswell, sothefour-dimensional tensor S),issymmetric: S,,,,=SW. (31.29) Inother words, thecomponents S,,,S,,,,Szt,which arethedensities ofx,y,and 2momentum, arealsoequal tothex-,y-,andz-components ofthePoynting vector S,theenergy fl0W—~2lS wehave already shown inanearlier chapter byadifferent kindofargument. Theremaining components oftheelectromagnetic stress tensor S,”canalso beexpressed interms oftheelectric andmagnetic ﬁelds EandBThat istosay, wemust admit stress or,toputitlessmysteriously, ﬂow ofmomentum inthe electromagnetic ﬁeld Wediscussed thisinChapter 27inconnection with Eq (27.21), butdidnotwork outthedetails 31-13 Those whowant toexercise their prowess intensors infourdimensions might liketoseetheformula forS,.,,interms oftheﬁelds: SP” I FIHIFWI _7}auvxﬁ Fl9=1F5<1> l where sums onoi,Bareont,x,y,zbut(asusual inrelativity) weadopt aspecial meaning forthesumsignZandforthesymbol 6.Inthesums thex,y,zterms aretobesubtracted and 6,,=+1,while 6,,=6,”,=6,,=-1and 6,,,,=O forit¢1/(c=1).Can you verify that itgives theenergy density S),= (co/2) (E2—l—B2)andthePoynting vector e0EXB?Canyoushow thatinan electrostatic ﬁeld with B=0theprincipal axes ofstress areinthedirection ofthe electric ﬁeld, thatthere isatension (en/2)E2 along thedirection oftheﬁeld, andthat there isanequal pressure indirections perpendicular totheﬁeld direction? 31-14 32 Refractive Index ofDense Materials 32-1 Polarization ofmatter Wewant nowtodiscuss thephenomenon oftherefraction oflight——and also, therefore, theabsorption oflight bydense materials. InChapter 31ofVolume I wediscussed thetheory oftheindex ofrefraction, butbecause ofourlimited mathematical abilities atthat time, wehadtorestrict ourselves toﬁnding theindex onlyformaterials oflowdensity, likegases. Thephysical principles thatproduced theindex were, however, made clear The electric ﬁeld ofthelight wave polarizes themolecules ofthegas,producing oscillating dipole moments. Theacceleration oftheoscillating charges radiates new waves oftheﬁeld. This new ﬁeld, interfering withtheoldﬁeld, produces achanged ﬁeldwhich isequivalent toaphase shift of theoriginal wave. Because thisphase shift isproportional tothethickness ofthe material, theeffect isequivalent tohaving adifferent phase velocity inthematerial. When welooked atthesubject before, weneglected thecomplications thatarise from such effects asthenewwave changing theﬁelds attheoscillating dipoles. Weassumed thattheforces onthecharges intheatoms came justfrom theincoming wave, whereas, infact, their oscillations aredriven notonly bytheincoming wave butalsobytheradiated waves ofalltheother atoms Itwould have been difficult forusatthat time toinclude thiseffect, sowestudied only therareﬁed gas, where such effects arenotimportant. Now, however, wewillﬁndthatitisveryeasytotreat theproblem bytheuse ofdifferential equations. This method obscures thephysical origin oftheindex (ascoming from there-radiated waves interfering with theoriginal waves), but itmakes thetheory fordense materials much simpler. This chapter willbring together alarge number ofpieces from ourearlier work. We’ve taken uppractically everything wewillneed, sothere arerelatively fewreally newideas tobeintroduced. Since youmay need torefresh your memory about what wearegoing toneed, wegiveinTable 32-1 alistoftheequations wearegoing touse,together with a reference totheplace where each canbefound. Inmost instances, wewillnottake thetimetogivethephysical arguments again, butwilljustusetheequations. Table 32-1 Ourwork inthischapter willbebased onthefollowing material, already covered inearlier chapters Subject Reference Equanon Damped oscillations Index ofgases Mobility Electrical conductivity Polarizability Inside dielectricsVol Vol Vol Vol Vol VolI,Chap. 23 I,Chap. 31 I,Chap. 41 I,Chap. 43 II,Chap. 10 II,Chap llm(ic n Z )1; WIX H: ppol E100+Vx+wgx)=F 1NE1+z__2% 2e0(o.>0—w) n’—in" —l—iix=F r_ _Nqgr,0' I71 m =—V-P i.=12I I +36‘ 32-1)32-1 Polarization ofmatter 32-2 Maxwell’s equations ina dielectric 32-3 Waves inadielectric 32-4 Thecomplex index ofrefraction 32-5 The index ofamixture 32-6 Waves inmetals 32-7 Low-frequency and high-frequency approximations; theskindepth andtheplasma frequency Review: SeeTable 32-1. Webegin byrecalling themachinery oftheindex ofrefraction foragas. Wesuppose thatthere areNparticles perunitvolume andthateach particle be- haves asaharmonic oscillator. Weuseamodel ofanatom ormolecule inwhich theelectron isbound with aforce proportional toitsdisplacement (asthough the electron were heldinplace byaspring). Weemphasized thatthiswasnotalegiti- mate classical model ofanatom, butwewillshow later thatthecorrect quantum mechanical theory gives results equivalent tothismodel (insimple cases). Inour earlier treatment, wedidnotinclude thepossibility ofadamping force intheatomic oscillators, butwewilldosonow. Such aforce corresponds toaresistance tothe motion, thatis,toaforce proportional tothevelocity oftheelectron. Then the equation ofmotion is F=q,E=m()'c'+vx+wgix), (32.1) where xisthedisplacement parallel tothedirection ofE.(Weareassuming an isotropic oscillator whose restoring force isthesame inalldirections. Also, we aretaking, forthemoment, alinearly polarized wave, sothatEdoesn’t change direction.) Iftheelectric ﬁeld acting ontheatom varies sinusoidally with time, wewrite E=E0e’°". (32.2) The displacement willthen oscillate with thesame frequency, andwecanlet x=x(,e“”’ Substituting X=iwxandX=—w2x, wecansolve forxinterms ofE: 11¢/'2i_,-__ ' X=23" E , (32.3)—w —l—i'Yw +cog Knowing thedisplacement, wecancalculate theacceleration Xand ﬁnd the radiated wave responsible fortheindex. This wasthewaywecomputed theindex inChapter 31ofVolume I. Now, however, wewant totake adifferent approach. Theinduced dipole moment pofanatom isqpxor,using Eq.(32.3), 2/, p=W2~‘1*“[’7’5~~* 2E. (32.4)—w +1'Yw—l—o.>(, Since pisproportional toE,wewrite p=€0a(w)E, (32.5) where oziscalled theatomic p0larizab1li'ty.* With thisdeﬁnition, wehave ct=all/m‘L (326—w2 —l—l’Yw+0:?) The quantum mechanical solution forthemotions ofelectrons inatoms gives asimilar answer except with thefollowing modiﬁcations. Theatoms have several natural frequencies, each frequency with itsown dissipation constant V.Also theeffective “strength” ofeach mode isdiﬁerent, which wecanrepresent bymultiplying thepolarizability foreach frequency byastrength factorf. which isanumber weexpect tobeoftheorder of1Representing thethree parameters w,7,andj byw/,,Wk,andﬁ-foreach mode ofoscillation, andsumming over the *Throughout thischapter wefollow thenotation ofChapter 31ofVolume l,andlet ctrepresent theum/iiic polarizability asdeﬁned here. Inthelastchapter, weused orto represent thevolume polarizability—the ratio ofPtoEInthenotation ofI/uschapter P=Nae()E (seeEq32.8) 32—2 various modes, wemodify Eq.(32.6) toread > _-51) _ft. _, awTW”2:—<»*+rm»+wﬁk (327) IfNisthenumber ofatoms perunitvolume inthematerial, thepolarization PlS_]llSll Np=6UN0zE, andisproportional toE: P=EU/Vcv(w)E. (32.8) Inother words, when there isasinusoidal electric ﬁeld acting inamaterial, there isaninduced dipole moment perunitvolume which isproportional totheelectric ﬁeld——with aproportionality constant 04that, weemphasize, depends upon the frequency. Atvery high frequencies, 02issmall; there 1Snotmuch response. How- ever, atlowfrequencies there canbeastrong response. Also, theproportionality constant isacomplex ntimber, which means thatthepolarization does notexactly follow theelectric ﬁeld. butmay beshifted inphase tosome extent Atanyrate, there isapolarization perunitvolume whose magnitude isproportional tothe strength oftheelectric ﬁeld. 32-2 l\/IaxweIl’s equations inadielectric The existence ofpolarization inmatter means that there arepolarization charges andcurrents inside ofthe material, andthese must beputintothecomplete Maxwell eqtiations inorder toﬁndtheﬁelds Wearegoing tosolve Maxwell’s equations thisftime inasituation inwhich thecharges andcurrents arenotzero, asinavactiuiii, butaregiven implicitly bythepolarization vector Our ﬁrst stepistoﬁndexplicitly thecharge density pandcurrent density 1',averaged over asmall volume ofthesame sizewehadinmind when wedeﬁned P.Then the pandj weneed canbeobtained from thepolarization. Wehave seen inChapter IOthatwhen thepolarization Pvaries from place toplace, there isacharge density given by ppiil : —V ' Atthattime, wewere dealing with static ﬁelds, butthesame formula isvalid also fortime-varying ﬁelds However, when Pvaries with time, there arecharges in motion, sothere isalso apolarization current. Each oftheoscillating charges contributes acurrent equal toitscharge (1,,times itsvelocity UWith Nsuch charges perunitvolume, thecurrent densityj is i=Nqcv. Since weknow thatll=dx/dr, then /=Nq,.(dx/dr), which 1S_]LlSt dP/dr. There- forethecurrent density from thevarying polarization is . dP [pol : Ourproblem isnow direct andsimple. Wewrite Maxwell’s equations with thecharge density andcurrent density expressed interms ofP,using Eqs. (32.9) and(32l0). (We assume that there arenoother currents andcharges inthe material.) Wethen relate PtoEwith Eq.(32.5), andwesolve theequation for EandB—looking forthewave solutions Before wedothis, wewould liketomake anhistorical note. Maxwell origi- nally wrote hisequations inaform which wasdifferent from theonewehave been using. Because theequations were written inthisdillerent form formany years-—- andarestillwritten thatwaybymany people—we willexplain theClllTCf€HC€ In theearly days. themechanism ofthedielectric constant wasnotfully andclearly appreciated. Thenature ofatoms wasnotunderstood, northatthere wasapolar- ization ofthematerial. Sopeople didnotappreciate thatthere wasacontribution 32-3 tothecharge density pfrom V*P.They thought only interms ofcharges that were notbound toatoms (such asthecharges thatflow inwires orarertibbed offsurfaces). Today, weprefer toletprepresent thetotal charge density, including thepart from thebound atomic charges. Ifwecallthatpartppol, wecanwrite p=ppol +potlivrs where p,,,1,,., isthecharge density considered byMaxwell andrefers tothecharges notbound toindividual atoms. Wewould then write 191'V : +i 50 Substituting pm;from Eq.(32.9), v.E:F_)"_tl‘_'“'__l_V.P 50 60 Of VI(€OE + :p()l;lI('f' Thecurrent density intheMaxwell equations forVXBalsohas,ingeneral, contributions from bound atomic currents. Wecantherefore write i:jpol +j()ll\(‘fJ andtheMaxwell equation becomes 2 _lbw“-r lat 2?. cVXB— en—l—€0+(,” (3212) Using Eq.(32.10), weget GQCZV><B=/"...i....+<@..E+P). (3213) Now youcanseethatifwewere todeﬁne anewvector Dby D=e0E+P, (32.14) thetwoﬁeld equations would become V-D=p,,,1,,., (32.15) and @,¢2v ><B=,',,,,,,.,+961,’- (32.16) These areactually theforms thatMaxwell used fordielectrics. Histworemaining equations were 6B and V-B=0, which arethesame aswehave been using. Maxwell andtheother early workers also hadaproblem with magnetic materials (which wewilltake upsoon) Because they didnotknow about the circulating currents responsible foratomic magnetism, they used acurrent density thatwasmissing stillanother part Instead ofEq.(32.16), they actually wrote v><H=j’+ (32.17) where Hdiffers from e0c2B because itincludes theeffects ofatomic currents. (Thenj’ represents what isleftofthe currents.) SoMaxwell had/our ﬁeldvectors— E,D,B,andH—the DandHwere hidden ways ofnotpaying attention towhat 32-4 wasgoing oninside thematerial You willﬁndtheequations written thiswayin many places. Tosolve theequations, itisnecessary torelate DandHtotheother ﬁelds, andpeople used towrite D=eE and B=pH. (32.18) However, these relations areonly approximately true forsome materials and even then only iftheﬁelds arenotchanging rapidly with time. (For sinusoidally varying ﬁelds oneoften canwrite theequations thiswaybymaking eand/.tcomplex functions ofthefrequency, butnotforanarbitrary time variation oftheﬁelds.) Sothere used tobeallkinds ofcheating insolving theequations. Wethink the right wayistokeep theequations interms ofthefundamental quantities aswe nowunderstand them—and that’s how wehave done it. 32-3 Waves inadielectric Wewant now toﬁndoutwhat kind ofelectromagnetic waves canexist ina dielectric material inwhich there arenoextra charges other than those bound in atoms. Sowetakep =—V-Pandj =6P/6!. Maxwell's equations thenbecome . .2_Y"’ 2 _i!I (a)VE- 60 (b)cV><B-al<€u+E> (32.19) (c)V><E=—9£ (d)v-B=0 Wecansolve these equations aswehave done before. Westart bytaking thecurlofEq.(32.l9c): V><(V><E)=—§-tv><B. Next, wemake useofthevector identity v><(v><E)=V(V-E) -V2E, andalsosubstitute forV><B,using Eq.(3219b); weget V(V-E)—V2E=—— ,_ —-~-51a‘P 162E soc’ 6t2 c261‘ Using Eq.(3219a)forV-E,weget 2_l_825 _ L . 12212 22 VE C2all~ 60V(V P)+E002 M2 (3.0) Soinstead ofthewave equation, wenowgetthattheD’Alembertian ofEisequal totwoterms involving thepolarization P. Since Pdepends onE,however, Eq.(32.20) canstillhave wave solutions. Wewillnowlimit ourselves toisozropzc dielectrics, sothatPisalways inthesame direction asE.Let’s trytoﬁndasolution forawave going inthez-direction Then, theelectric ﬁeld might vary ase"""—'”). Wewillalsosuppose thatthewave ispolarized inthex-direction—that theelectric ﬁeld hasonly anx-component. Wewrite E,=E.,t»“*"'*'”>. (32.21) You know thatanyfunction of(z—vi)represents awave thattravels with thespeed 2».Theexponent ofEq.(32.21) canbewritten as —l/€<Z—€I%I), so,Eq.(3221)represents awave with thephase velocity - Upll :w/k ' 32-5 Theindex ofrefraction nisdeﬁned (seeChapter 31,Vol. I)byletting cUph =Z' Thus Eq.(3221)becomes Ex =E0e1w(t—nz/0)‘ Sowecanﬁndnbyﬁnding what value ofkisrequired ifEq.(32.21) istosatisfy theproper ﬁeld equations, andthen using n= (32.22)(J) Inanisotropic material, there willbeonly anx-component ofthepolarization; then Phasnovariation with thex-coordinate, soV~P=O,andwegetridof theﬁrstterm ontheright-hand sideofEq.(32.20) Also, since weareassuming a linear dielectric, P,willvary ase“"‘,and62P,,/6r2 =—w2P,,. TheLaplacian in Eq.(32.20) becomes simply 62E,/622 =—k2E,,, soweget 2 2 -185, +99;E,=-“’_,P, (32.23)C EQC“ Now letusassume forthemoment that since Eisvarying sinusoidally, we cansetPproportional toE,asinEq.(32.5). (We’ll come back todiscuss this assumption later.) Wewrite PI :€0N(1E,;. Then E,drops outofEq.(32.23), andweﬁnd 2 /<2=%(1+Na). (32.24) Wehave found thatawave likeEq.(32.21). with thewave number kgiven by Eq.(3224),willsatisfy theﬁeldequations. Using Eq.(32.22), theindex nisgiven by n2=1+Na. (32.25) Let’s compare thisformula with what weobtained inourtheory oftheindex ofagas(Chapter 31,Vol. I).There, wegotEq(31.29), which is 2 n=1+$3’-1:'_3j1;» 2. (32.26)-0-? wt) Taking afrom Eq.(32.6), Eq(32.25) would giveus 2 n2=1+ivﬁ (32.27)me‘) —w2 +i'Yo.>+cu?) First, wehave thenewterm inWm, because weareincluding thedissipation of theoscillators. Second, theleft-hand sideisninstead ofn2,andthere isanextra factor of1/2. Butnotice thatifNissmall enough sothatn1Sclose toone(asit isforagas), then Eq.(32.27) saysthatn2isoneplusasmall number: n2=1+e. Wecanthen write n=\/1—l—eQ1—l—e/2,andthetwoexpressions areequiva- lent. Thus ournewmethod gives foragasthesame result wefound earlier. Now youmight think that Eq.(32.27) should give theindex ofrefraction fordense materials also. Itneeds tobemodiﬁed, however, forseveral reasons. First, thederivation ofthisequation assumes that thepolarizing ﬁeld oneach atom istheﬁeld Ex. That assumption isnotright, however, because indense materials there isalsotheﬁeldproduced byother atoms inthevicinity, which may becomparable toEx.Weconsidered asimilar problem when westudied thestatic ﬁelds indielectrics. (See Chapter ll.) You willremember thatweestimated the ﬁeldatasingle atom byimagining thatitsatinaspherical holeinthesurrounding dielectric. Theﬁeld insuch ahole—which wecalled thelocal ﬁeld—is increased 32-6 over theaverage ﬁeld Ebytheamount P/3e0. (Remember, however, that this result isonly strictly true inisotropic materials—including thespecial case ofa cubic crystal.) Thesame arguments willhold fortheelectric ﬁeld inawave, solong asthe wavelength ofthewave ismuch longer than thespacing between atoms. Limiting ourselves tosuch cases, wewrite PEloml =E—l—360 (32.28) Thislocal ﬁeldistheonethatshould beused forEinEq.(32.3); thatis,Eq.(32.8) should berewritten: P=e0NaE1,,c,,1. (32.29) Using Eiomi from Eq.(32.28), weﬁnd 360P= GQNIX + *5) OI‘ lVa Inother words, fordense materials Pisstillproportional toE(forsinusoidal ﬁelds). However, theconstant ofproportionality isnot€0Na, aswewrote below Eq.(32.23), butshould be€0Na/[l —(Na/3)]. Soweshould correct Eq(32.25) toread n2=1+1%. (32.31) Itwillbemore convenient ifwerewrite thisequation as n2—13W2 = Na, which isalgebraically equivalent. This isknown astheClausius-Mosotti equation. There isanother complication indense materials. Because neighboring atoms aresoclose, there arestrong interactions between them. Theinternal modes of oscillation are,therefore, modiﬁed. Thenatural frequencies oftheatomic oscilla- tions arespread outbytheinteractions, andthey areusually quite heavily damped —the resistance coefﬁcient becomes quite large. Sothew0’s andv’softhesolid willbequite different from those ofthefreeatoms. With these reservations, we canstillrepresent a,atleast approximately, byEq.(32.7). Wehave then that 2-1 N3352+2=mgZ 2ft 2- (32.33)9Ic—w -l-l'Y1¢w+w0k Oneﬁnal complication. Ifthedense material isamixture ofseveral compo- nents, each willcontribute tothepolarization. Thetotal L!willbethesumofthe contributions from each component ofthemixture [except fortheinaccuracy of thelocal ﬁeldapproximation, Eq.(32.28), inordered crystals—effects wediscussed when analyzing ferroelectrics]. Writing N,asthenumber ofatoms ofeach com- ponent perunitvolume, weshould replace Eq.(32.32) by 2-13+2)=ZN,a,, (32.34)J where each oz,willbegiven byanexpression likeEq.(32.7). Equation (32.34) completes ourtheory oftheindex ofrefraction. Thequantity 3(n2 —1)/(n2 +2) isgiven bysome complex function offrequency, which isthemean atomic polariz- ability 01(0)). Theprecise evaluation ofa(w) (that is,ﬁnding fk,V),andwok)indense substances isadilﬁcult problem ofquantum mechanics. Ithasbeen done from ﬁrstprinciples only forafewespecially simple substances. 32-7 \ \ \\ e—umIz/c \/ \ \\ \\\\\ \ —l ' -+- -/T />4" T /// \‘\\eIu1(!—nRz/c) / // / / / / / / Fig. 32—l. Agrciph ofEXforsome instant f,ifn1%nR/2Ti'.32-4 Thecomplex index ofrefraction Wewant tolook now attheconsequences ofourresult, Eq(32.33). First. wenotice thatoriscomplex, sotheindex nisgoing tobeacomplex number. What does thatmean" Let’s saythatwewrite nasthesumofarealandanimaginary part: n=nR~in), (3235) where nlfandn;arerealfunctions ofwWewrite in,with aminus sign, sothatn; willbeapositive quantity inallordinary optical materials. (Inordinary inactive materials—that arenot,likelasers, light sources themselves~v isapositive number, andthatmakes theimaginary partofnnegative.) Ourplane wave ofEq.(32.21) iswritten interms ofnas El: :EOe—-tw(t—n2/1) Writing nasinEq.(32.35), wewould have Ex = E0e—mn[z'reim(!AriKz/U. The term e"”(’_"It‘/fl represents awave travelling with thespeed c/np, soii),- represents what wenormally think ofastheindex ofrefraction. Buttheamplitude ofthiswave is Ene—w7lIZ/C 7 which decreases exponentially with zAgraph ofthestrength oftheelectric ﬁeld atsome instant asafunction ofzisshown inFig.32-1, forn,~nk/21r. The imaginary part oftheindex represents theattenuation ofthewave duetothe energy losses intheatomic oscillators. Theintensity ofthewave isproportional tothesquare oftheamplitude, so Intensity =<e_2°’”Iz/“. This isoften written as Intensity cce“"‘, where B=2w/1;/c iscalled theabsorption coeﬂﬁcienr. Thus wehave inEq(32.33) notonly thetheory oftheindex ofrefraction ofmaterials, butthetheory oftheir absorption oflight aswell. Inwhat weusually consider tobetransparent material, thequantity c/wn;— which hasthedimensions ofalength—is quite large incomparison with the thickness ofthematerial. 32-5 Theindex ofamixture There isanother prediction ofourtheory oftheindex ofrefraction thatwe cancheck against experiment. Suppose weconsider amixture oftwomaterials. Theindex ofthemixture isnottheaverage ofthetwoindexes, butshould be given interms ofthesumofthetwopolarizabilities, asinEq.(32.34). Ifweask about theindex of,say,asugar solution, thetotal polarizability isthesumofthe polarizability ofthewater andthatofthesugar. Each must, ofcourse, becal- culated using forNthenumber perunitvolume ofthemolecules oftheparticular kind. Inother words, ifagiven solution hasN1molecules ofwater, whose polariz- ability isa1,andN2molecules ofsucrose (C12H22O11), whose polarizability is <12,weshould have that n“,—l3 = Nidl + Ngdg. Wecanusethisformula totestourtheory against experiment bymeasuring theindex forvarious concentrations ofsucrose inwater. Wearemaking several assumptions here, however. Ourformula assumes thatthere isnochemical action when thesucrose isdissolved andthatthedisturbances totheindividual atomic 32-8 Refractive index ofsucrose solutions, andcomparison withpredictions ofEq.(32.37). Data from Handbook A B C Fraction ofsucrose density byweight (gm/cm3) at20°CHTable 32-2 D E F G Moles of sucrosed perliter,Moles of n2_1 water“ 3("5—— —-Nperliter, n+2 lal N2/N0 Ni/No 0“ 0.9982 0.30 1.1270 0.50 1.2296 085 1.4454 1.00“ 1.5881.333 1.3811 1.4200 1.5033 1.5577 C0 55.5 0.970 43.8 1.798 '3415 3.59 1202 464 00698 0.759 0.886 0.960 “pure water "sugar crystals °average (seetext) ‘Imolecular weight ofsucrose °molecular weight ofwater =18 oscillators arenottoodiﬂ"erent forvarious concentrations. Soourresult iscertainly only approximate. Anyway, let’sseehow good itis. Wehave picked theexample ofasugar solution because there isagood table ofmeasurements oftheindex ofrefraction intheHandbook ofChemistry and Physics andalsobecause sugar isamolecular crystal thatgoes intosolution with- outionizing orotherwise changing itschemical state. Wegiveintheﬁrstthree columns ofTable 32-2 thedata from thehandbook. Column Aisthepercent ofsucrose byweight, column Bisthemeasured density (gm/cm3), andcolumn Cisthemeasured index ofrefraction forlight whose wavelength is589.3 millimicrons. Forpure sugar wehave taken themeasured index ofsugar crystals. Thecrystals arenotisotropic, sothemeasured index is different along different directions. Thehandbook gives three values: n1=1.5376, I12=1.5651, n3=1.5705. Wehave taken theaverage. Now wecould trytocompute nforeach concentration, butwedon’t know what value totake fora1or042.Let’s testthetheory thisway: Wewillassume thatthepolarizability ofwater (011)isthesame atallconcentrations andcompute thepolarizability ofsucrose byusing theexperiment ofvalues fornandsolving Eq.(38.27) for<12. Ifthetheory iscorrect, weshould getthesame 022forall concentrations. First, weneed toknow N1andN2:let’sexpress them interms ofAvogadro’s number, N0.Let’s takeoneliter(1000 ems) forourunitofvolume. Then N,/N0 is theweight perliterdivided bythegram-molecular weight. And theweight per literisthedensity (multiplied by1000 togetgrams perliter) times thefractional weight ofeither thesucrose orthewater. Inthisway, wegetN2/N0 andN1/N0 asincolumns DandEofthetable. Incolumn Fwehave computed 3(n2 —1)/(n2 +2)from theexperimental values ofnincolumn C.Forpure water, 3(n2 —1)/(n2 —l—2)is0.617, which is equal toJustNlal. Wecanthen ﬁllintherestofColumn G,since foreach row rowG/E may beinthesame ratio—namely, 0.6l7:55.5. Subtracting column G from column F,wegetthecontribution N2a2 ofthesucrose, shown incolumn H Dividing these entries bythevalues ofN2/N0 incolumn D,wegetthevalue of Noﬂg shown incolumn J From ourtheory wewould expect allthevalues ofN0012tobethesame They arenotexactly equal, butpretty close. Wecanconclude thatourideas arefairly correct. Even more, weﬁndthatthepolarizability ofthesugar molecule doesn’t seem todepend much onitssurroundings—its polarizability isnearly thesame ina dilute solution asitisinthecrystal. 32-90617 0617 l 0.487 0.379 0.1335 0 =342H J N(il12N.. lag (gm/liter) 0 0.21 1 0.380 0.752 0.960()213 0211 0210 0.207 Vdftfi _" AVE ETWEEN COLﬁe‘£0gm ml‘__Zmm31(1) Fig. 32-2. The motion of ofree electron.32-6 Waves inmetals Thetheory wehave worked outinthischapter forsolid materials canalso beapplied togood conductors, likemetals, with verylittle modiﬁcation. Inmetals some oftheelectrons have nobinding force holding them toanyparticular atom; itisthese “free” electrons which areresponsible fortheconductivity. There are other electrons which arebound, andthetheory above isdirectly applicable to them. Their inﬂuence, however, isusually swamped bytheeffects ofthecon- duction electrons. Wewillconsider now only theellects ofthefreeelectrons Ifthere isnorestoring force onanelectron—but stillsome resistance toits motion—its equation ofmotion differs from Eq.(32.1) only because theterm in wgxislacking. Soallwehave todoisset61%=0intherestofourderivations— except thatthere isonemore difference. Thereason thatwehadtodistinguish between theaverage ﬁeld andthelocal ﬁeld inadielectric isthatinaninsulator each ofthedipoles isﬁxed inposition, sothatithasadeﬁnite relationship tothe position oftheothers. Butbecause theconduction electrons inametal move around allover theplace, theﬁeld onthem ontheaverage isjusttheaverage ﬁeld E.Sothecorrection wemade toEq.(325)byusing Eq.(32.28) should notbe made forconduction electrons Therefore theformula fortheindex ofrefraction formetals should look likeEq.(32.27), except with wesetequal tozero, namely, N2 2= _ﬁ__l_*- 323H I+men -651 +W6: (0I8) This isonly thecontribution from theconduction electrons, which wewillassume isthemajor term formetals Now weeven know how toﬁndwhat value tousefor7,because itisrelated totheconductivity ofthemetal. InChapter 43ofVolume Iwediscussed howthe conductivity ofametal comes from thediffusion ofthefreeelectrons through the crystal. Theelectrons goonajagged path from onescattering tothenext. and between scatterings theymove freely except foranacceleration duetoanyaverage electric ﬁeld (asshown inFig32-2). Wefound inChapter 43ofVolume Ithat theaverage drift velocity isjusttheacceleration times theaverage time 7'between collisions. Theacceleration isq,E/m, so .5i~,,,,,,=‘lgT. (32.39) This formula assumed thatEwasconstant, sothat 11.1,,“ wasasteady velocity. Since there isnoaverage acceleration, thedrag force isequal totheapplied force. Wehave deﬁned Wbysaying thatWm)isthedrag force [seeEq.(32.l)], which is q,E; therefore wehave that 1V-;- (32.40) Although wecannot easily measure 'rdirectly, wecandetermine itbymeasur- ingtheconductivity ofthemetal. Itisfound experimentally thatanelectric ﬁeldE inametal produces acurrent with thedensityj proportional toE(forisotropic materials): j=0E. Theproportionality constant aiscalled theconductivity. This isjust what weexpect from Eq.(32.39) ifweset j:Nqevtlrift- Then 2 _Neetr—~—m 7'. (32.41) SoT——and therefore 't—can berelated totheobserved electrical conductivity. Using Eqs (32.40) and(3241),wecanrewrite ourformula fortheindex, Eq. 32-10 (32.38), inthefollowing form: 2_ __g/66n_1+100 +ilwT), (32.42) where 1 maT=-= (32.43)IN413 This isaconvenient formula fortheindex ofrefraction ofmetals. 32-7 Low-frequency andhigh-frequency approximations; theskindepth andthe plasma frequency Ourresult, Eq.(32.42), fortheindex ofrefraction formetals predicts quite different characteristics forwave propagation atdifferent frequencies. Let’s ﬁrst seewhat happens atveryl0wfrequencies. Ifwissmall enough, wecanapproximate Eq.(32.42) by "2=-12:0) (32.44) Now, asyoucancheck bytaking thesquare,* 1—iy/__.i Z 4? ; \/2 soforlowfrequencies,--- AMPLITU DE n_\/6/2@,,<.3(i -1"). (32.45) , Therealandimaginary parts ofnhave thesame magnitude. With such alarge imaginary part ton,thewave israpidly attenuated inthemetal. Referring to _z/8 Eq.(32.36), theamplitude ofawave going inthez-direction decreases as ° exp[—vGa 26,62Z]. (32.46)Let’swrite thisas e_’/“, (32.47)0 1 i 1 > 9 swhere 6isthen thedistance inwhich thewave amplitude decreases bythefactor LSURFACE 28 38 Z e_1-1/2.72—or roughly one-third. Theamplitude ofsuch awave asafunction ofzisshown inFig. 32-3. Since electromagnetic waves willpenetrate into a Fig.32-3. Theamplitude ofatrans- metal only thisdistance, 6iscalled theskindepth. Itisgiven by Verse @|@¢tr<>m<19"eti<I Wave OS9fv"¢ti°" ofdistance intoametal. 6=\/2e0c2/aw. (32.48) Now what dowemean by“low” frequencies? Looking atEq.(32.42), we seethatitcanbeapproximated byEq.(32.44) only if0.17"ismuch lessthan one andifweo/0 isalsomuch lessthan one—that is,ourlow-frequency approximation applies when 1w<<-T and 63<<55- (32.49)60 Let’s seewhat frequencies these correspond toforatypical metal likecopper. Wecompute 7'byusing Eq.(32.43), and0/eo, byusing themeasured conductivity. Wetake thefollowing data from ahandbook: 0=5.76 X107(ohm-meter)“, atomic weight =63.5grams, density =8.9grams —cm_3, Avogadro’s number =6.02 X102“ (gram atomic weight)_1. *Orwriting —l=e_”'/2; \/:1 =e_"'/4 =cos7r/4 -—lSll17I'/4, which gives the same result. 32-ll Ifweassume thatthere isonefreeelectron peratom, then thenumber ofelectrons percubic meter is N=8.5X1028 meter“. Using q,=1.6X10*” coulomb, so=8.85 X10*” farad-meter”, m=9.11 Xl0—31kgm, weget 'r=2.4Xl0“1“sec, 1 . —=4.1X101"sec_1,T -0;=6.5X1018sec_1. 60 Soforfrequencies lessthan about 1012 cycles persecond, copper willhave the “low-frequency” behavior wedescribe (that means forwaves whose free-space wavelength islonger than 0.3millimeters—-very short radio wavesl). Forthese waves, theskindepth incopper is 5_ 0. m2-sec-T10 \/028 C0 Formicrowaves of10,000 megacycles persecond (3-cm waves) 6=6.7Xl0_4 cm. Thewave penetrates avery small distance. Wecanseefrom thiswhyinstudying cavities (orwaveguides) weneeded to worry only about theﬁelds inside thecavity, andnotinthemetal oroutside the cavity. Also, weseewhy thelosses inacavity arereduced byathinplating of silver orgold. Thelosses come from thecurrent, which areappreciable only ina thinlayer equal totheskindepth. Suppose welook now attheindex ofametal likecopper athigh frequencies. Forvery high frequencies o.>*rismuch greater than one, andEq.(32.42) iswell approximated by 2_ _ 0'_n—1 T0327 (32.50) Forwaves ofhigh frequencies theindex ofametal becomes real—and lessthan one! Thisisalsoevident from Eq.(32.38) ifthe dissipation term with 'Yisneglected, ascanbedone forvery large w.Equation (32.38) gives 2 I12=1-Nq” (32.51)I’l’l€()(.i)2 which is,ofcourse, thesame asEq.(32.50). Wehave seen before thequantity Nqf/men, which wecalled thesquare oftheplasma frequency (Section 7-3): 2 2 Nqcwp i >7 9 €()l’l’l sowecanwrite Eq.(32.50) orEq.(32.51) as "2:1_ E)? (I) Theplasma frequency isakind of“critical” frequency. Forto<wptheindex ofametal hasanimaginary part, andwaves are attenuated; butforw>>6.1,,theindex isreal, andthemetal becomes transparent. You know, ofcourse, that metals arereasonably transparent tox-rays. But some metals areeven transparent intheultraviolet. InTable 32-3 wegive for 32-12 several metals theexperimental observed wavelength atwhich theybegin tobecome transparent. Inthesecond column wegive thecalculated critical wavelength A,,=27rc/mp. Considering that theexperimental wavelength isnottoowell deﬁned, theﬁtofthetheory isfairly good. You may wonder why theplasma frequency wpshould have anything todo withthepropagation ofelectromagnetic waves inmetals. Theplasma frequency came upinChapter 7asthenatural frequency ofdensity oscillations ofthefree electrons. (Aclump ofelectrons isrepelled byelectric forces, andtheinertia ofthe electrons leads toanoscillation ofdensity.) Solongitudinal plasma waves are resonant atmp.Butwearenow talking about transverse electromagnetic waves, andwehave found thattransverse waves areabsorbed forfrequencies below wp. (It’saninteresting andnotaccidental coincidence.) Although wehave been talking about wave propagation inmetals, youap- preciate bythistime theuniversality ofthephenomena ofphysics—that itdoesn’t make anydifference whether thefreeelectrons areinametal orwhether they are intheplasma oftheionosphere oftheearth, orintheatmosphere ofastar. To understand radio propagation intheionosphere, wecanusethesame expressions— using, ofcourse, theproper values forNand7'.Wecanseenow why long radio waves areabsorbed orreﬂected bytheionosphere, whereas short waves goright through. (Short waves must beused forcommunication with satellites.) Wehave talked about thehigh- andlow-frequency extremes forwave propaga- tioninmetals. Forthein-between frequencies thefull-blown formula ofEq. (32.42) must beused. Ingeneral, theindex willhave realandimaginary parts; thewave isattenuated asitpropagates intothemetal. Forverythinlayers, metals aresomewhat transparent even atoptical frequencies. Asanexample, special goggles forpeople who work around high-temperature furnaces aremade by evaporating athinlayer ofgold onglass. Thevisible light istransmitted fairly well—with astrong green tinge—but theinfrared isstrongly absorbed. Finally. itcannot have escaped thereader that many ofthese formulas re- semble insome ways those forthedielectric constant Kdiscussed inChapter 10. Thedielectric constant Kmeasures theresponse ofthematerial toaconstant ﬁeld, thatis,forw=0.Ifyoulook carefully atthedeﬁnition ofnandKyouseethat Kissimply thelimit ofn2as6.»—>0.Indeed, placing w=0andn2=Kinequa- tions ofthischapter willreproduce theequations ofthetheory ofthedielectric constant ofChapter 11. 32-13Table 32-3* Wavelengths below which themetal becomes transparent Metal )\(experimental) A,,=21rc/lib], Li 1550 A 1550 A Na 2100 2090 K 3150 2870 Rb 3400 3220 1 .._._ J *From: C.Kittel, Introduction toSolid State Physics, John Wiley andSons, Inc., New York, 2nded.,1956, p.266. 33 Roflovtion from Surfaces 33-1 Reﬂection andrefraction oflight Thesubject ofthischapter isthereﬂection andrefraction oflight—or electro- magnetic waves ingeneral—at surfaces. Wehave already discussed thelaws of reﬂection andrefraction inChapter 35ofVolume I.Here’s what wefound out there: 1.Theangle ofreﬂection isequal totheangle ofincidence. With theangles deﬁned asshown inFig.33-1, 0,=6,. (33.1) 2.The product nsin0isthesame fortheincident and transmitted beams (Snell’s law). n1sin0,=n2sin0,. (33.2) 3.Theintensity ofthereﬂected light depends ontheangle ofincidence and alsoonthedirection ofpolarization. ForEperpendicular totheplane of incidence, thereﬂection coefﬁcient RLis 1,S11126-0 _ _ (7 ) Re"7;7 <3“) ForEparallel totheplane ofincidence, thereﬂection coefficient R11is 2 RZQ:l§_13_Qk_:E0') H Ii tang (0L + 6!)~ (33.4) 4.Fornormal incidence (any polarization, ofcoursel), 2 4-(4%)3(Earlier, weused ifortheincident angle andrfortherefracted angle Since we can’t userforboth “refracted” and“reﬂected” angles, wearenow using 6,= incident angle, 6,=reﬂected angle, and6,=transmitted angle.) Ourearlier discussion isreally about asfarasanyone would normally need togowith thesubject, butwearegoing todoitallover again adifferent way Why"One reason isthatweassumed before thattheindexes were real(noab- sorption inthematerials) Butanother reason isthatyoushould know how to dealwith what happens towaves atsurfaces from thepoint ofview ofMaxwell's equations. We'll getthesame answers asbefore, butnow from astraightforward solution ofthewave problem, rather than bysome clever arguments. Wewant toemphasize that theamplitude ofasurface reﬂection isnota property ofthematerial, asistheindex ofrefraction Itisa“surface property,” onethatdepends precisely onhowthesurface ismade. Athinlayer ofextraneous junk onthesurface between twomaterials ofindices n1andn2willusually change thereﬂection. (There areallkinds ofpossibilities ofinterference here-like the colors ofoilﬁlms Suitable thickness caneven reduce thereﬂected amplitude to zero foragiven frequency; that’s how coated lenses aremade.) The formulas wewillderive arecorrect only ifthechange ofindex issudden—within adistance very small compared with onewavelength. Forlight, thewavelength isabout 5000 A,sobya“smooth” surface wemean oneinwhich theconditions change in 33-133-1 Reﬂection andrefraction of light 33-2 Waves indense materials 33-3 The boundary conditions 33-4 Thereﬂected andtransmitted waves 33-5 Reﬂection from metals 33-6 Total internal reﬂection Review. Chapter 35,Vol.I,Polarization _r / - _ ~ <0 I., - ,6“.'~\__ ‘_ Q9‘tsqp .~ ‘,6',_C‘; . - °.i . ,_ \ ._ 1, _. »\ . ‘pr '9 t -.9 sit-.'-lg(x“F3"-sui=ii=/.\cE ‘.‘\\s<‘-.‘° ~ ~‘ ._( fl] 1'12 Fig. 33-1. Reflection and refraction oflight waves atasurface. (The wave directions are normal tothewave crests.) ly \\ F \ ///X//I/§>//3A\ Fig. 33-2. Forawave movin'g inthe direction It,thephase atany point Pis (wt—l(-rl.going adistance ofonly afewatoms (orafewangstroms). Our equations will work forlight forhighly polished surfaces. Ingeneral, iftheindex changes grad- ually over adistance ofseveral wavelengths, there isvery little reﬂection atall. 33-2 Waves indense materials First, weremind youabout theconvenient way ofdescribing asinusoidal plane wave weused inChapter 36ofVolume I.Any ﬁeld component inthewave (weuseEasanexample) canbewritten intheform E=E0e“"”_""’, (33.6) where Erepresents theamplitude atthepoint r(from theorigin) atthetime t. The vector kpoints inthedirection thewave istravelling, anditsmagnitude lkl=k=27r/>1 isthewave number. Thephase velocity ofthe wave is13,),=6.1/k, foralight wave inamaterial ofindex n,up),=c/n,so 60!’! k_6-- (33.7) Suppose kisinthez-direction, then k-risjustkz,aswehave often used itFor kinanyother direction, weshould replace zbyrk,thedistance from theorigin inthek-direction; thatis,weshould replace kzbykrk,which isjustkr.(See Fig.33-2.) SoEq.(33.6) isaconvenient representation ofawave inanydirection. Wemust remember, ofcourse, that k-r= k,,x—l—kyy—l—kzz, where k,,k,,,andkgarethecomponents ofkalong thethree axes. Infact,we pointed outonce that (w,kl,/(,1,kz)isafour-vector, andthatitsscalar product with (1,x,y,z)isaninvariant. Sothephase ofawave isaninvariant, andEq. (33.6) could bewrittenE=Eoellculg Butwedon’t need tobethatfancy now. Forasinusoidal E,asinEq.(33.6), 6E/61 isthesame asiwE, andHE/6x is —ik,,E, andsoonfortheother components. Youcanseewhyitisveryconvenient tousetheform inEq.(336)when working with differential equations—differentia- tions arereplaced bymultiplications. One further useful point: The operation V=(6/6x, 6/6y, 8/62) getsreplaced bythethree multiplications (—ik,,, —ik,,, -ikz). Butthese three factors transform asthecomponents ofthevector k,so theoperator Vgetsreplaced bymultiplication with —ik: fl_,,0,at ’ v_,—ik. (33.8) This remains trueforanyVoperation—whether itisthegradient, orthediver- gence, orthecurl. Forinstance, thez-component ofVXEis Q5_9?».6x 6)‘ Ifboth El,andE,vary ase*"‘',then weget —ik,,Ey +ikyE,, which is,yousee,thez-component of—ik XE. Sowehave thevery useful general factthatwhenever youhave totake the gradient ofavector thatvaries asawave inthree dimensions (they areanimportant part ofphysics), youcanalways take thederivations quickly andalmost without thinking byremembering thattheoperation Visequivalent tomultiplication by —ik. 33-2 Forinstance, theFaraday equation 0BVXE——-6; becomes forawave —ik XE=—l(.uB. Thistellsusthat _k><EB»4;“. (33.9) which corresponds totheresult wefound earlier forwaves infreespace—that B, inawave, isatright angles toEandtothewave direction. (Infreespace, co/k = c.)Youcanremember thesigninEq.(339)from thefactthatkisinthedirection ofPoynting's vector S=e()c2E XB. Ifyouusethesame rulewith theother Maxwell equations, yougetagain the results ofthelastchapter and, inparticular, that ‘ C02’/[2 6 i Ll kk-kw C2 (3310) Butsince weknow that, wewon’t doitagain. Ifyouwant toentertain yourself, youcantrythefollowing terrifying problem thatwastheultimate testforgraduate students back in1890: solve Maxwell’s equations forplane waves inananisotropic crystal, thatis,when thepolarization Pisrelated totheelectric ﬁeld Ebyatensor ofpolarizability. You should, of course, choose your axesalong theprincipal axesofthetensor, sothattherelations aresimplest (then P,—a,,E,,, P,,=a),E,,, andP,=at-E2), butletthewaves haveanarbitrary direction andpolarization. You should beabletoﬁndtherela- tions between EandB,andhow kvaries with direction andwave polarization. Then youwillunderstand theoptics ofananisotropic crystal. Itwould bebest tostart with thesimpler case ofabirefringent crystal—like calcite—for which twoofthepolarizabilities areequal (say, 011,=a,),andseeifyoucanunderstand whyyouseedouble when youlook through such acrystal Ifyoucandothat, thentrythehardest case, inwhich allthree a’saredifferent. Then youwillknow whether youareuptothelevel ofagraduate student of1890. Inthischapter, however, wewillconsider only isotropic substances .‘Y Er- _A‘ lg D El I»W' it’, \' i<’; _ ~'_“er atD , _‘e, x 1' ~ '1 5 ~il -_ Fig. 33-3. The propagation vectors El'~. 3—' Ir,k’,and k”fortheincident, reﬂected,. J ‘ - \ n| Weknow from experience that when aplane wave arrives attheboundary between twodifferent materials say,airandglass, orwater andoil—there isa wave reﬂected andawave transmitted Suppose weassume nomore than thatand seewhat wecanwork out. Wechoose ouraxes with theyz-plane inthesurface andthexy-plane perpendicular totheincident wave surfaces, asshown inFig.33-3. 33-3andtransmitted waves -1y iI _-1" IEy,.,15,2 \--‘ ' >H. H2 x Fig. 33-4. Aboundary condition Ey;=E71isobtained from fl.Eds=O.Theelectric vector oftheincident wave canthen bewritten as E,=E()e“"’T"'). (33.11) Since kisperpendicular tothez-axis, k-r=kxx+kyy. (3312) Wewrite thereﬂected wave as E,=E(,e“°"‘_""", (33.13) sothat itsfrequency is65’,itswave number isk’,anditsamplitude isE(,.(We know, ofcourse, thatthefrequency isthesame andthemagnitude ofkisthesame asfortheincident wave, butwearenotgoing toassume even that. Wewillletit come outofthemathematical machinery.) Finally, wewrite forthetransmitted wave, E,=E{)'e“°""_"""). (33.14) Weknow thatoneofMaxwell’s equations gives Eq(33.9), soforeach ofthe waves wehave I N 3,Zﬁll, BrZ 3,Z5‘___>f/ E2. (33_|5)(.0 60 (.0 Also, ifwecalltheindexes ofthetwomedia n1andn2,wehave from Eq.(33.10) 22k2_k2+k2_w"i_—1 ii——c2—' (3316) Since thereﬂected wave isinthesame material, then k/2 _(0,211?__T , (33.17) whereas forthetransmitted wave, /22 kl/2Zf’:’I”__2.C2(33.18) 33-3 Theboundary conditions Allwehave done sofaristodescribe thethree waves; ourproblem nowis towork outtheparameters ofthereﬂected andtransmitted waves interms of those oftheincident wave. How canwedothat? Thethree waves wehave de- scribed satisfy Maxwell’s equations intheuniform material, butMaxwell’s equa- tions must alsobesatisﬁed attheboundary between thetwodifferent materials. Sowemust now look atwhat happens right attheboundary. Wewillﬁndthat Maxwell‘s equations demand thatthethree waves ﬁttogether inacertain way. Asanexample ofwhat wemean, they-component oftheelectric ﬁeldEmust bethesame onboth sides oftheboundary. This isrequired byFaraday’s law, 6BVXE- —E, (33.19) aswecanseeinthefollowing way. Consider alittle rectangular loop I‘which straddles theboundary, asshown inFig33-4. Equation (33.19) saysthattheline integral ofEaround I‘isequal totherateofchange oftheﬂuxofBthrough the loop: _ 6%E-ds =—--fB'nda.1‘ 6! Now imagine thattherectangle isvery narrow, sothattheloop encloses anin- ﬁnitesimal area. IfBremains ﬁnite (and there‘s noreason itshould beinﬁnite attheboundary!) theﬂuxthrough thearea iszero Sothelineintegral ofEmust 33-4 bezero. IfE,,1andEH2arethecomponents oftheﬁeld onthetwosides ofthe boundary andifthelength oftherectangle isl,wehave E,/1, _ E,)2l : 0 Of E,,1 =E,,2, (33.20) aswehave said. This gives iisonerelation among theﬁelds ofthethree waves. The procedtire ofworking outtheconsequences ofMaxwell’s equations at theboundary iscalled “determining theboundary conditions.” Ordinarily, itis done byﬁnding asmany equations likeEq.(3320)asonecan, bymaking argu- ments about little rectangles likeFinFig.33-4, orbyusing little gaussian surfaces thatstraddle theboundary Although thatisaperfectly good wayofproceeding, itgives theimpression that theproblem ofdealing with aboundary isdifferent forevery different physical problem Forexample, inaproblem ofheatﬂow across aboundary, how arethetem- peratures onthetwosides related? Well, youcould argue, foronething, thatthe heatﬂow totheboundary from onesidewould have toequal theﬂow awa_i~ from theother side. Itisusually possible, andgenerally quite useful, towork outthe boundary conditions bymaking such physical arguments. There may betimes, however, when inworking onsome problem youhave only some equations, and youmaynotseeright away what physical arguments touse. Soalthough weare atthemoment interested only inanelectromagnetic problem, where wecanmake thephysical arguments, wewant toshow youamethod thatcanbeused forany problem—-a general wayofﬁnding what happens ataboundary directly from the C111T€l‘€l1[l£ll equations Webegin bywriting alltheMaxwell equations foradielectr1c—and thistime wearevery speciﬁc andwrite outexplicitly allthecomponents: (-EZ__‘_?’5|) 6..++3"?)--(‘lpf+954+W?) (33.21) I.X (ly ii- d.\ 6y 02 4')B “XE"—a OE; ('lE,) _ dB, (fl) ()2 -_ 0t OE OE, GB”--'—---“-=—--- 3.2('12 Ox ('1! (3 2b)(33.22a) 9,5"_GE"I_35% (33226)().\‘ 6}’ at \"‘B-0 951+fl?’+"1":-0 (3323> (l.\ ii)4 (JZ > l6P (YE.~ B: _ ___ CV X F4) + )(JB, 68,, 1OP, OE, 22 .~ ~__g :_, \ ,7 __4. (<0)‘ dz) 6‘)(71+81‘ (3(1) _,0B,_ (iii,_L919, 515,, 2 ((OZ ilk‘) T G1) (CF + (91 .6B,, OB iaP 6E,~~Z--Zr =-ﬁr Z 3.24‘<(ix dy) 6,,atTat (3 C) 33-5 O> ‘U‘Uto (<1) / 0'710/ ________.|,..___"___-__XF1= X_ REGION 3 REGION l REGION 2 AaP, ‘ax ti lb) /,\1 > X (Cl ,\ Fig. 33-5. The ﬁelds inthetransition region (3) between two different ma- terials inregions (lland (2).> XNow these equations must allhold inregion 1(totheleftoftheboundary) andinregion 2(totheright oftheboundary). Wehave already written thesolu- tions inregions 1and2.Finally, theymust alsobesatisﬁed intheboundary, which wecancallregion 3.Although weusually think oftheboundary asbeing sharply discontinuous, inreality itisnot. The physical properties change very rapidly butnotinﬁnitely fast. Inanycase, wecanimagine thatthere isavery rapid, but continuous, transition oftheindex between region 1and2,inashort distance we cancallregion 3.Also, anyﬁeld quantity likeP,,,orE,,,etc.,willmake asimilar kind oftransition inregion 3.Inthisregion, thedifferential equations must still besatisﬁed, anditisbyfollowing thedifferential equations inthisregion thatwe canarrive attheneeded “boundary conditions.” Forinstance, suppose thatwehave aboundary between vacuum (region 1) andglass (region 2).There isnothing topolarize inthevacuum, soP1=0. Let's saythere issome polarization P2intheglass. Between thevacuum andthe glass there isasmooth, butrapid, transition Ifwelook atanycomponent of P,sayPx,itmight vary asdrawn inFig.33—5(a). Suppose now wetake theﬁrst ofourequations, Eq(33.21). Itinvolves derivatives ofthecomponents ofPwith respect tox,y,and2.They-andz-derivatives arenotinteresting; nothing spec- tacular ishappening inthose directions. Butthex-derivative ofP,willhave some verylarge values inregion 3.because ofthetremendous slope ofP,.Thederivative (JP,/6x willhave asharp spike attheboundary, asshown inFig.33-5(b). Ifwe imagine squashing theboundary toaneven thinner layer, thespike would get much higher Iftheboundary isreally sharp forthewaves weareinterested in, themagnitude of6P,/6x inregion 3willbemuch, much greater than anycontribu- tions wemight have from thevariation ofPinthewave away from theboundary-— soweignore anyvariations other than those duetotheboundary. Now howcanEq.(3321)besatisﬁed ifthere isawhopping bigspike onthe right-hand side? Only ifthere isanequally whopping bigspike ontheother side. Something ontheleft-hand sidemust alsobebig. Theonly candidate isGE,/6x, because thevariations withyandzareonly those small eﬂects inthewave wejust mentioned. So—e0(6E/6x) must beasdrawn inFig.33—5(c)—just acopy of OP,/6x. Wehave that 6595-z__‘lPr06xT ox l Ifweintegrate thisequation with respect toxacross region 3,weconclude that 5o(Ex2 _Em) :_(Px2 -'P11)‘ (33-25) Inother words, thejump in(EOE, ingoing from region 1toregion 2must beequal tothejump in—P,,. Wecanrewrite Eq.(33.25) as €0Ez2 +P12 :€()Ex1 Tl’P21, which saysthatthequantity (e()E, +PI)hasequal values inregion 2andregion 1. People say:thequantity (e(,E, +P,)iscontinuous across theboundary. Wehave, inthisway, oneofourboundary conditions. Although wetook asanillustration thecase inwhich P1waszero because region 1wasavacuum, itisclear that thesame argument applies foranytwo materials inthetworegions, soEq.(33.26) istrueingeneral. Let’s now gothrough therestofMaxwell’s equations andseewhat each of them tellsus.Wetake next Eq.(33.22a). There arenox-derivatives, soitdoesn’t tellusanything. (Remember thattheﬁelds themselves donotgetespecially large attheboundary; only thederivatives with respect toxcanbecome sohuge that they dominate theequation.) Next, welook atEq.(3322b). Ah‘ There isan x-derivative! Wehave GE)/6x ontheleft-hand side. Suppose ithasahuge de- rivative Butwait amoment! There isnothing ontheright-hand sidetomatch it with; therefore E,cannot have anyjump ingoing from region 1toregion 2. [Ifitdid,there would beaspike ontheleftofEq.(33.22a) butnone ontheright, 33-6 andtheequation would befalse ]Sowehave anewcondition: E,2=E51. (33.27) Bythesame argument, Eq(33.22c) gives EU;=E,,1. (33.28) Thislastresult isjustwhat wegotinEq.(3320)byalineintegral argument. WegoontoEq.(3323) Theonly term thatcould have aspike isGB,/ox. Butthere’s nothing ontheright tomatch it,soweconclude that BF2 =B,1. (33.29) OntothelastofMaxwell’s equations! Equation (3324a) gives nothing, because there arenox-derivatives Equation (3323b) hasone, —c2 GB,/6x, but again, there isnothing tomatch itwith. Weget B22 =B21. (33.30) Thelastequation isquite similar, andgives B,,2=B!/1| (33.31) Thelastthree equations gives usthat B)=B1. Wewant toemphasize, however, that wegetthisresult only when thematerials onboth sides ofthe boundary arenonmagnetic—or rather, when wecanneglect anymagnetic effects ofthematerials. This canusually bedone formost materials, except ferromagnetic opes (Wewilltreat themagnetic properties ofmaterials insome later chapters.) EOur program hasnetted usthesixrelations between theﬁelds inregion 1and those inregion 2.Wehave putthem alltogether inTable 33-l. Wecannow use them tomatch thewaves inthetworegions. Wewant toemphasize, however, that theideawehave justused willwork inanyphysical situation inwhich youhave differential equations andyou want asolution that crosses asharp boundary between tworegions where some property changes. Forourpresent purposes, wecould have easily derived thesame equations byusing arguments about the ﬂuxes andcirculations attheboundary. (You might seewhether youcangetthe same result thatway.) Butnowyouhave seen amethod thatwillwork incaseyou evergetstuck anddon't seeanyeasyargument about thephysics ofwhat ishappen- ingattheboundary—you canjustwork with theequations. 33-4 Thereﬂected andtransmitted waves Now weareready toapply ourboundary conditions tothewaves wewrote down inSection 33-2. Wehad: E,=E,,i»"“"’""I"-‘W’, (3332) E,=E,',@”‘“"-"¥‘*"4"’, (33.33) E,=E§,’e““’ ""1"-’°r”1 (33.34) B,Z532-5’ . (33.35) B,=_"lQ>i,_5. (33.36)OJ l‘l'>§§i.CollB)= (3337) Wehave onefurther bitofknowledge: Eisperpendicular toitspropagation vector kforeach wave. 33-7Table 33-1 Boundary conditions atthesurface of dielectric (60Ei +Pl): I(60152 -f"P2); (El)l/ =(E2)1i (El): =(E2): B1 =-B3 (The surface isintheyz-plane) _Y - I ‘ kn l,k _ Br.‘ ' El >E~ .r B, .- > x,k . E,.~' ‘\suRFAcE ,B,1‘ . "I'1F12 Fig. 33-6. Polarization ofthe re- ﬂected and transmitted waves when the E-ﬁeld oftheincident wave isperpendicu- lartotheplane ofincidence.Theresults Willdepend onthedirection oftheE-vector (the“polarization”) oftheincoming wave. Theanalysis ismuch simpliﬁed ifwe treat separately thecase ofanincident wave with itsE-vector parallel tothe“plane ofincidence” (that is, thexy-plane) andthecase ofanincident wave with theE-vector perpendicular to theplane ofincidence. Awave ofanyother polarization isjustalinear conibina- tionoftwosuch waves. Inother words, thereﬂected andtransmitted intensities aredifferent fordifferent polarizations, anditiseasiest topick thetwosimplest cases andtreat them separately. Wewillcarry through theanalysis foranincoming wave polarized per- pendicular totheplane ofincidence andthenjust giveyoutheresult fortheother. Wearecheating alittle bytaking thesimplest case, buttheprinciple isthesame forboth. Sowetake thatE,hasonly az-component, andsince alltheE-vectors areinthesame direction wecanleave otlthevector signs. Solong asboth materials areisotropic, theinduced oscillations ofcharges in thematerial willalsobeinthez-direction, andtheE-ﬁeld ofthetransmitted and radiated waves willhave only z-components. Soforallthewaves, ExandE1, andP,andPgarezero. Thewaves willhave their E-andB-vectors asdrawn in Fig.33-6 (Wearecutting acorner here onouroriginal plan ofgetting everything from theequations. This result would alsocome outoftheboundary conditions, butwecansave alotofalgebra byusing thephysical argument When youhave some spare time, seeifyoucangetthesame result from theequations. Itisclear thatwhat wehave saidagrees with theequations; itlS]USl thatwehave notshown thatthere arenoother possibilities.) Now ourboundary conditions, Eqs. (3326)through (33.31), give relations between thecomponents ofEandBinregions land2.Forregion 2wehave only thetransmitted wave, butinregion 1wehave twowaves. Which onedoweuse? Theﬁelds inregion lare,ofcourse, thesuperposition ol‘theﬁelds oftheincident andreﬂected waves. (Since each satisﬁes Maxwell’s equations, sodoes thesum.) Sowhen weusetheboundary conditions, wemust usethat 'E1=Et-l-Er, E2’-=Et, andsimilarly fortheB's. \\ Forthepolarization weareconsidering, Eqs. (33.26) and(33.28) giveusno newinformation; only Eq(33.27) isuseful. ltsaysthat El + Er : E3 attheboundary, thatis,forx=0.Sowehave that E0e1(wt--kl/1/) + E6et(ui'l—lrUt/) :Eti)/et(ui"t—Iti,'_i/)’ which must betrueforalltandforally. Suppose welook ﬁrstaty=O.Then we have Eoetwl + E6eL(4)//1 Ez,/elm’/I This equation says that two oscillating terms areequal toathird oscillation. That canhappen only ifalltheoscillations have thesame frequency. (ltisim- possible forthree—or anynumber—of such terms with diﬂerent frequencies to addtozero foralltimes.) So w”=w’=w. (33.39) Asweknew allalong, thefrequencies ofthereﬂected andtransmitted waves are thesame asthatoftheincident wave. Weshould really have saved ourselves some trouble byputting thatinatthe beginning, butwewanted toshow youthatitcanalsobegotoutoftheequations. When youaredoing arealproblem, itisusually thebestthing toputeverything you know intotheworks right atthestart andsaveyourself alotoftrouble. Bydeﬁnition, themagnitude ofkisgiven byk2=n2¢-:2/(:2, sowehave also thatk//2 k/2 k2 —~=J=< (33.40)ng nf nf 33-8 Now look atEq.(33.38) fort=0.Using again thesame kind ofargument wehave Justmade, butthistime based onthefactthattheequation must hold forallvalues ofy,wegetthat kl,’=kl,1kg: (33.41) From Eq.(33.40), k’2=kz,so k;2+/<32=kg?+/<3. Combining thiswith Eq.(33.41), wehave that /<12I/<2), orthat k’,==*=k,,. The positive sign makes nosense; that would notgive a reﬂected wave, butanother incident wave, andwesaidatthestart thatwewere solving theproblem ofonly oneincident wave. Sowehave /<1.=-1<,. (3342) Thetwoequations (33.41) and(33.42) giveusthattheangle ofreﬂection isequal totheangle ofincidence, asweexpected. (See Fig.33-3 )Thereﬂected wave is E,=E{,e““"_"""+"””). (33.43) Forthetransmitted wave wealready have that kg;=1<,,,and /2 2 5L=5- (33.44)"5 "i’ sowecansolve these toﬁndk§,'.Weget /<5,”=W-kg,"=/<2-k3. (33.45)i Suppose foramoment thatn1and113arerealnumbers (that theimaginary parts oftheindexes arevery small). Then allthek’sarealsorealnumbers, and from Fig.33—3 weﬁndthat k . k” .=sin0,, If=sin0,. (33.46) From (33.44) wegetthat n2sin0,=n1sin0,, (33.47) which isSnell’s lawofrefraction—again, something wealready knew. lfthe indexes arenotreal.thewave numbers arecomplex, andwehave touseEq.(33.45). [Wecould stilldefine theangles 0,and0,byEq.(33.46), andSnell’s law,Eq.(33.47), would betrueingeneral. Butthen the“angles” alsoarecomplex numbers, thereby losing their simple geometrical interpretation asangles. Itisbestthen todescribe thebehavior ofthewaves bytheir complex k,ork’,’values ] Sofar,wehaven’t found anything new. Wehavejust hadthesimple-niinded delight ofgetting some obvious answers from acomplicated mathematical ma- chinery. Now weareready toﬁndtheamplitudes ofthewaves which wehave notyetknown. Using ourresults forthe(v’sandk’s,theexponential factors in Eq.(33.38) canbecancelled, andWeget E0+El,=El)’. (3348) Since both E(,andE{,'areunknown, Weneed onemore relationship. Wemust useanother oftheboundary conditions. Theequations forE,andE,,arenohelp, because alltheE'shave only az-component Sowemust usetheconditions on B.Let’s tryEq.(3329): B12 :Bz1- 33—9 .U - -'H ll .. .Er Kk_ .'Br._ B’ .'E;' ' > I x E ''. I k‘ .'B"I."‘\suRi=AcE ~ r .>n." n2 Fig. 33-7. Polarization ofthewaves when the E-ﬁeld oftheincident wave is parallel totheplane ofincidence.From Eqs. (33.35) through (33.37). kE k§,E l<1’E,Brt :Tl/Tl’ Brr :TT/ll’ Bast :“L/T 'Cl) (.0 (J) Recalling thatca”=w’=wandkj,’=k,',=k,,,wegetthat E0+El)2El)’- Butthisisjust Eq.(3348)alloveragain‘ We'vejust wasted time getting something wealready knew. Wecould tryEq.(33.30), Bzg=B21, butthere areno2-coniponents ofB‘ Sothere’s only oneequation left: Eq.(33.31), B,,2=B,,1. Forthethree waves. k,,E, /<;E, "/<gE3,,=-~07». ByrI-457. By,=-~w,,’- (33.49) Putting forE,,E,.andE,thewave expression forx=0(tobeattheboundary), theboundary condition 1S kx i(wif~lc,i/) 1i(w'i‘—le,"y) kin’ /1t(<ii"l—li"i/)G E06 l+ Z0‘; E(;€ J :(77 EUC U . Again allw’sandkjsareequal, sothisreduces to k,E,, +k}E(, =k§’E(,’. (33.50) This gives usanequation fortheE’sthatisdifferent from Eq.(3348). With the two, wecansolve forE6andE6’. Remembering thatkj:—k,, weget k, /(QEl] I El)» E61=__2ﬁ__ E0_ kl‘+kt’ (3352) These, together with Eq.(33.45) orEq.(3346)forkl’,giveuswhat wewanted to know. Wewilldiscuss theconsequences ofthisresult inthenext section. lfwebegin with awave polarized with itsE-vector parallel totheplane of incidence, Ewillhave both x-andy-components. asshown inFig.33-7. The algebra isstraightforward butmore complicated (The work canbesomewhat reduced byexpressing things inthiscaseinterms ofthemagnetic ﬁelds, which are allinthez-direction.) Oneﬁnds that 2_2 {Ea=’%5”~~~1’~.il‘-5"IEOI (3353)/12/<3+niké’ and lE6’l=—,-2”1””—‘-3 iE..i. (3354)43/<.+nikﬁi’ Let’s seewhether ourresults agree with those wegotearlier Equation (333) istheresult weworked outinChapter 35ofVolume lfortheratio oftheintensity ofthereﬂected wave totheintensity oftheincident wave Then, however, wewere considering only realindexes Forrealindexes (and k’s), wecanwrite wnk,=kcos 0,=-C—1cos0,, wnk§,’=k”cos0,=-C-2cos0, Substituting inEq.(33.51), wehave E6 ll;cos0,—'12cos0, E0 n1cos0,—l—n2@637, ’ (3355) 33-10 which does notlook thesame asEq.(33.3). Itwill, however, ifweuseSnell’s law togetridofthen’s.Setting n2=n1sin0,/sin 0,.andmultiplying thenumerator anddenominator bysin0,,weget El,_cos0,sin0,—sin0.cos0, E0 cos0,sin0,—l—sin0,cos0, The numerator anddenominator arejUSt thesines of(0,—0,)and(0,—l—0,); weget El) _Sin (61 _' 66) E0—sin('t§,>7-l—?,) (3356) Since Er’,andE0areinthesame material, theintensities areproportional tothe squares oftheelectric ﬁelds, andwegetthesame result asbefore. Similarly, Eq. (33.53) isthesame asEq.(33.4). Forwaves which arrive atnormal incidence, 0,=Oand0,=0.Equation (33.56) gives O/0, which isnotvery useful. Wecan, however, goback toEq. (33.55), which gives I7‘ El)>2 (/11 —"2>2-=- =—--— - 33.57It (Eu "1'l'"2 ( ) This result, naturally, applies for“either” polarization, since fornormal incidence there isnospecial “plane ofincidence.” 33-5 Reﬂection from metals Wecannow useourresults tounderstand theinteresting phenomenon of reﬂection from metals. Why isitthatmetals areshiny? Wesawinthelastchapter thatmetals have anindex ofrefraction which, forsome frequencies, hasalarge imaginary part. Let’s seewhat wewould getforthereﬂected intensity when light shines from air(with n=1)onto amaterial with n=—ll’l1. Then Eq.(33.55) gives (fornormal incidence) .§§I:lei?l”1.E0 l* lit] Fortheintensity ofthereﬂected wave, wewant thesquare oftheabsolute values ofE5andEU: QZliar2U+I, lE(,|3 |1—ll’l1l2 or 112Lilli? _1 11 l+n?(33ss) Foramaterial with anindex which isapure imaginary number, there is100per- centreﬂection‘ Metals donotreﬂect 100percent, butmany doreﬂect visible light very well. lriother words, theimaginary partoftheir indexes isverylarge Butwehave seen thatalarge imaginary part oftheindex means astrong absorption. Sothere 1Sa general rulethatifanymaterial getstobeaverygood absorber atanyfrequency. thewaves arestrongly reﬂected atthesurface andvery little getsinside tobeab- sorbed You canseethiseffect with strong dyes Pure crystals ofthestrongest dyeshave a“metallic” shine. Probably youhave noticed thatattheedge ofabottle ofpurple inkthedried dyewillgiveagolden metallic reﬂection, orthatdried red inkwillsometimes give agreenish metallic reﬂection. Redinkabsorbs outthe greens oftransmitted light, soiftheinkisveryconcentrated, itwillexhibit astrong surface reﬂection forthefrequencies ofgreen light. You caneasily show thiseffect bycoating aglass plate with redinkand letting itdry. Ifyoudirect abeam ofwhite light attheback oftheplate, asshown inFig.33-8, there willbeatransmitted beam ofredlight andareﬂected beam of green light. 33-11l/,/ \)\\ \\\ .\ ,RED 2;/7 /’ 'GLASS PLATE DRIED RED INK Fig. 33-8. Amaterial which absorbs light strongly atthe frequency walso reﬂects light ofthat frequency. (~_.'|'.lY |Ey| 4 \ "-\.~ \ .. i s1 Y j 1 X ‘_'I/k~..)\o x i ‘ I \ \ AtI \ I) ,. ‘n| n2 , _ \ / i i v - 1 V I 1‘ I i i i 3 ‘ A \ . .i .1 . ~ A I .‘.n,= n‘.i.n2=0 nsin . Fig. 33-10. Ifthere isasmall gap, internal reﬂection isnot"total"; atrans- mitted wave appears beyond thegap.Fig. 33-9. Total internal reﬂection. 33-6 Total internal reﬂection Iflight goes from amaterial likeglass, with arealindex llgreater than l. toward, say,air.with anindex n2equal to1,Snell's lawsaysthat sin0,=nsin0,. Theangle 0,ofthetransmitted wave becomes 90°when theincident angle 0,is equal tothe“critical angle" 0,given by nsin0,=1. (33.59) What happens for0,greater than thecritical angle‘? You know thatthere istotal internal reﬂection. Buthowdoes thatcome about" Let's goback toEq(33.45) which gives thewave nuinbei kl’forthetrans- mitted wave. Wewould have) -> k 3 /<j"=W—l<,J. Now k,,=ksin 0,andk=om/c, so ~>. Q)“ 1 . kf=F(i-ifSH123,). lfiisin0,1Sgreater than one, l<§’2isnegative andk',’isapure imaginary, say =*=I/(1. You know bynow what thatmeans‘ The“transmitted” wave (Eq. 33.34) willhave theform El2E6,eiii,3ei(ai_i.,,y)_ The wave amplitude either grows ordrops offexponentially with increasing x. Clearly, what wewant here isthenegative sign. Then theamplitude ofthewave totheright oftheboundary willgoasshown inFig.33-9. Notice thatA,isof theorder to/c—which isA...thefree-space wavelength ofthelight. When light is totally reﬂected from theinside ofaglass-air surface, there areﬁelds intheair. butthey extend beyond thesurface only adistance oftheorder ofthewavelength ofthelight Wecannowseehowtoanswer thefollowing question: lfalight wave inglass arrives atthesurface atalarge enough angle, itisreﬂected, ifanother piece of glass isbrought uptothesurface (sothatthe“surface” ineffect disappears) the light istransmitted. Exactly when does thishappen‘? Surely there must becon- tinuous change froni total reﬂection tonoreﬂection‘ Theanswer, ofcourse, is thatiftheairgapissosmall thattheexponential tailofthewave intheairhasan appreciable strength atthesecond piece ofglass, itwillshake theelectrons there andgenerate anewwave, asshown inFig.33-10. Some light willbetransmitted. (Clearly, oursolution isincomplete, weshould solve alltheequations again fora thinlayer ofairbetween tworegions ofglass.) 33-12 lllll(<1) A.’ TRANSMITTER DETECTOR DETECTOR8 __ lC _,_illlll)llLl illlllll B‘ (bl i () TRANSMITTER DETECTOR DETECTOR TRANSMITTER DETECTOR DETECTOR Fig. 33-l l.Ademonstration ofthepenetration ofinternally reﬂected waves. This transmission effect canbeobserved with ordinary light only iftheair gapisvery small (oftheorder ofthewavelength oflight, likel0”5 cm), butitis easily demonstrated with three-centimeter waves. Then theexponentially de- creasing ﬁeld extends several centimeters. Amicrowave apparatus thatshows the effect isdrawn inFig.33-11 Waves from asmall three-centimeter transmitter are directed ata45°prism ofparaffin. Theindex ofrefraction ofparaffin forthese frequencies is1.50, andtherefore thecritical angle is4l.5°. Sothewave istotally reﬂected from the45°face and ispicked upbydetector A,asindicated in Fig.33-l1(a). Ifasecond paraffin prism isplaced incontact with theﬁrst, as shown inpart(b)oftheﬁgure, thewave passes straight through andispicked up atdetector B.Ifagapofafewcentimeters isleftbetween thetwoprisms, asin part (c).there areboth transmitted andreﬂected waves. Theelectric ﬁeld outside the45°faceoftheprism inFig.33-ll(a) canalso beshown bybringing detector Btowithin afewcentimeters ofthesurface. 33-13 34 The Magnetism ofMatter 34-1 Diamagnetism andparamagnetism Inthischapter wearegoing totalkabout themagnetic properties ofmaterials. Thematerial which hasthemost striking magnetic properties is,ofcourse, iron. Similar magnetic properties areshared alsobytheelements nickel, cobalt, and-at suﬂiciently lowtemperatures (below l6°C)—by gadolinium, aswellasbyanumber ofpeculiar alloys. That kind ofmagnetism, called ferromagnetism, issuﬂiciently striking andcomplicated that wewilldiscuss itinaspecial chapter. However, allordinary substances doshow some magnetic eﬂects, although very small ones-a thousand toamillion times lessthan theeffects inferromagnetic materials. Here wearegoing todescribe ordinary magnetism, thatistosay.themagnetism ofsubstances other than theferromagnetic ones. This small magnetism isoftwokinds. Some materials areattracted toward magnetic ﬁelds; others arerepelled. Unlike theelectrical effect inmatter, which always causes dielectrics tobeattracted, there aretwo signs tothemagnetic effect. These twosigns canbeeasily shown with thehelpofastrong electromagnet which hasonesharply pointed pole piece andoneﬂatpole piece, asdrawn in Fig.34-1. Themagnetic ﬁeldismuch stronger near thepointed pole than near the ﬂatpole. Ifasmall piece ofmaterial isfastened toalong string andsuspended between thepoles, there will, ingeneral, beasmall force onit.This small force canbeseenbytheslight displacement ofthehanging material when themagnet isturned on.Thefewferromagnetic materials areattracted very strongly toward thepointed pole; allother materials feelonly averyweak force. Some areweakly attracted tothepointed pole; andsome areweakly repelled. STRING ___ SMALL PIECE OFMATERIAL V/ ¢//44The effect ismost easily seen with asmall cylinder ofbismuth, which is repelled from thehigh-ﬁeld region. Substances which arerepelled inthiswayare called diamagnetic. Bismuth isoneofthestrongest diamagnetic materials, but even with it,theeffect isstillquite weak. Diamagnetism isalways very weak. Ifasmall piece ofaluminum issuspended between thepoles, there isalsoaweak force, buttoward thepointed pole. Substances likealuminum arecalled para- magnetic. (Insuch anexperiment, eddy-current forces arise when themagnet is turned onandoff,andthese cangive offstrong impulses. You must becareful tolook forthenetdisplacement after thehanging object settles down.)s .\\\ 34-134-1 Diamagnetism and paramagnetism 34-2 Magnetic moments andangular momentum 34-3 Theprecession ofatomic magnets 34-4 Diamagnetism 34-5 Larmor’s theorem 34-6 Classical physics gives neither diamagnetism nor paramagnetism 34-7 Angular momentum inquantum mechanics 34-8 Themagnetic energy ofatoms Review: Section 15-l, “The forces on acurrent loop; energy ofa dipole.” Fig. 34-1. Asmall cylinder ofbis- muth isweakly repelled bythesharp pole; apiece ofaluminum isattracted. Wewant now todescribe brieﬂy themechanisms ofthese two effects. First, inmany substances theatoms have nopermanent magnetic moments, orrather, allthemagnets within each atom balance outsothatthenetmoment oftheatom iszero. Theelectron spins andorbital motions allexactly balance out,sothatanyparticular atom hasnoaverage magnetic moment. Inthese cir- cumstances, when youturn onamagnetic ﬁeld little extra currents aregenerated inside theatom byinduction. According toLenz’s law, these currents arein such adirection astooppose theincreasing ﬁeld. Sotheinduced magnetic mo- ments oftheatoms aredirected opposite tothemagnetic ﬁeld. This istheiiiech- anism ofdiamagnetism. Then there aresome substances forwhich theatoms dohave apermanent magnetic moment—in which theelectron spins andorbits have anetcirculating current thatisnotzero Sobesides thediamagnetic effect (which isalways present), there isalsothepossibility oflining uptheindividual atomic magnetic moments Inthiscase, themoments trytolineupwith themagnetic ﬁeld (inthewaythe permanent dipoles ofadielectric arelined upbytheelectric ﬁeld), andtheinduced magnetism tends toenhance themagnetic ﬁeld. These aretheparamagnetic sub- stances. Paramagnetism isgenerally fairly weak because thelining-up forces are relatively small compared with theforces from thethermal motions which tryto derange theorder. Italsofollows thatparamagnetism isusually sensitive tothe temperature (The paramagnetism arising from thespins oftheelectrons re- sponsible forconduction inametal constitutes anexception. Wewillnotbe discussing thisphenomenon here.) Forordinary paramagnetism, thelower the temperature, thestronger theeffect. There ismore lining-up atlowtemperatures when thederanging effects ofthecollisions areless. Diamagnetism, ontheother hand, ismore orlessindependent ofthetemperature. lnanysubstance with built-in magnetic moments there isadiamagnetic aswellasaparamagnetic effect, butthe paramagnetic eﬂiect usually dominates. InChapter llwedescribed aferroelectric material, inwhich alltheelectric dipoles getlined upbytheir ownmutual electric ﬁelds ltisalsopossible toimagine themagnetic analog offerroelectricity, inwhich alltheatomic moments would lineupandlock together. Ifyoumake calculations ofhow thisshould happen, youwillﬁndthatbecause themagnetic forces aresomuch smaller than theelectric forces, thermal motions should knock outthisalignment even attemperatures as lowasafewtenths ofadegree Kelvin. Soitwould beimpossible atroom tempera- turetohave anypermanent lining upofthemagnets. Ontheother hand, thisisexactly what does happen iniron—it does getlined up.There isaneffective force between themagnetic moments ofthedifferent atoms ofironwhich ismuch, much greater than thedirect magnetic interaction Itisan indirect effect which canbeexplained only byquantum mechanics. Itisabout tenthousand times stronger than thedirect magnetic interaction, andiswhat lines upthemoments inferromagnetic materials. Wediscuss thisspecial interaction inalater chapter. Now thatwehave tried togiveyouaqualitative explanation ofdianiagnetism andparamagnetism, wemust correct ourselves andsaythat iiISnotpossible to understand themagnetic effects ofmaterials inanyhonest way from thepoint ofview ofclassical physics. Such magnetic effects areacompletely quantum- mechanical phenomenon. Itis,however, possible tomake some plioney classical arguments andtogetsome idea ofwhat isgoing on. Wemight putitthisway. You canmake some classical arguments andgetguesses astothebehavior ofthe material, butthese arguments arenot“legal” inanysense because itisabsolutely essential that quantum mechanics beinvolved inevery oneofthese magnetic phenomena. Ontheother hand, there aresituations, such asinaplasma ora region ofspace with many freeelectrons, where theelectrons doobey thelaws ofclassical mechanics Andinthose circumstances, some ofthetheorems from clas- sical magnetism areworth while Also, theclassical arguments areofsome value forhistorical reasons. Theﬁrstfewtimes thatpeople were abletoguess attheincan- ingandbehavior ofmagnetic materials, they used classical arguments Finally, aswehave already illustrated, classical mechanics cangiveussome useful guesses 34-2 astowhat might happen—even though thereally honest waytostudy thissubject would betolearn quantum mechanics ﬁrstandthen tounderstand themagnetism interms ofquantum mechanics. Ontheother hand, wedon’t want towaituntilwelearn quantum mechanics inside outtounderstand asimple thing likediamagnetism Wewillhave to leanontheclassical mechanics askind ofhalfshowing what happens, realizing, however, thatthearguments arereally notcorrect. Wetherefore make aseries of theorems about classical magnetism thatwillconfuse youbecause they willprove different things. Except forthelasttheorem, every oneofthem Wlllbewrong Furthermore, theywillallbewrong asadescription ofthephysical world, because quantum mechanics isleftout. 34-2 Magnetic moments andangular momentum Theﬁrsttheorem wewant toprove from classical mechanics isthefollowing: Ifanelectron ismoving inacircular orbit (forexample, revolving around anucleus under theinﬂuence ofacentral force), there isadeﬁnite ratio between themagnetic moment andtheangular momentum. Let’s callJtheangular momentum and p.themagnetic moment oftheelectron intheorbit. Themagnitude oftheangular momentum isthemass oftheelectron times thevelocity times theradius (See Fig.34-2.) Itisdirected perpendicular totheplane oftheorbit. J=mvr. (34.1) (This is.ofcourse, anonrelativistic formula. butitisagood approximation for atoms, because fortheelectrons involved v/cisgenerally oftheorder ofe2/he = l/137, orabout lpercent) Themagnetic moment ofthesame orbit isthecurrent times thearea. (See Section l4—5 )Thecurrent isthecharge perunittime which passes anypoint on theorbit, namely, thecharge qtimes thefrequency ofrotation. Thefrequency isthe velocity divided bythecircumference oftheorbit; so I) I: q21rr TheareaisTrrz,sothemagnetic moment is M=9”’ (342) Itisalsodirected perpendicular totheplane oftheorbit. SoJandptareinthe same direction: M=%J(Orbit). (34.3) Their ratio depends neither onthevelocity norontheradius. Foranyparticle moving inacircular orbit themagnetic moment isequal toq/2m times theangular momentum. Foranelectron, thecharge isnegative——we cancallit—q,; sofor anelectron MI-29;/EJ(electron Ofblll). (34.4) That’s what wewould expect classically and, miraculously enough, itisalso truequantum-mechanically lt’soneofthose things. However, ifyoukeep going with theclassical physics, youﬁndother places where itgives thewrong answers, anditisagreat game totrytoremember which things areright andwhich things arewrong. Wemight aswell give youimmediately what istrue ingeneral in quantum mechanics. First, Eq.(344)istruefororbiial motion, butthat’s notthe only magnetism thatexists. Theelectron alsohasaspin rotation about itsown axis(something liketheearth rotating onitsaxis), andasaresult ofthatspinit hasboth anangular momentum andamagnetic moment Butforreasons thatare purely quantum—mechanical—there isnoclassical explanation—the ratio of;.t 34—3J P- V "liq Fig. 34-2. Foranycirculcir orbit the magnetic moment ;tisq/2m times the ongulor momentum J. toJfortheelectron spinistwice aslarge asitisfororbital motion ofthespinning electron: M=~J(electroii spin). (345) Inanyatom there are,generally speaking, several electrons andsome combina- tionofspinandorbit rotations which builds upatotal angular momentum anda total magnetic moment. Although there isnoclassical reason whyitshould beso, itisalways trueinquantum mechanics that(foranisolated atom) thedirection of themagnetic moment 1Sexactly opposite tothedirection oftheangular momentum. Theratio ofthetwoisnotnecessarily either —q,,/m or—q,/2m, butsomewhere in between, because there isamixture ofthecontributions from theorbits andthe spins. Wecanwrite M=—g(2%)1, (34.6) where gisafactor which ischaracteristic ofthestate oftheatom. Itwould be1 forapure orbital moment, or2forapure spin moment, orsome other number inbetween foracomplicated system likeanatom. This formula does not,ofcourse, tellusverymuch. Itsaysthatthemagnetic moment isparallel totheangular mo- mentum, butcanhave anymagnitude. Theform ofEq.(34.6) isconvenient, how- ever, because g—called the“Landé g-factor"—is adimensionless constant whose magnitude isoftheorder ofone. Itisoneofthe]ObS ofquantum mechanics to predict theg-factor foranyparticular atomic state. You might also beinterested inwhat happens innuclei. Innuclei there are protons andneutrons which may move around insome kind oforbit andatthe same time, likeanelectron, have anintrinsic spin. Again themagnetic moment isparallel totheangular momentum. Only now theorder ofmagnitude ofthe ratio ofthetwoiswhat youwould expect foraproion going around inacircle, with minEq.(34.3) equal totheproton mass Therefore itisusual towrite for nuclei qt» ~Age.>1» wewhere mpisthemass oftheproton, andg—called thenuclear g-factor—-is anumber near one, tobedetermined foreach nucleus. Another important difference foranucleus isthatthespinmagnetic moment oftheproton does nothave ag-factor of2,astheelectron does Foraproton, g=2(279). Surprisingly enough, theneutron alsohasaspinmagnetic moment, anditsmagnetic moment relative toitsangular momentum is2(—l.93). The neutron, inother words, isnotexactly “neutral” inthemagnetic sense. ltislike alittle magnet, andithasthekind ofmagnetic moment thatarotating negative charge would have. 34-3 Theprecession ofatomic magnets Oneoftheconsequences ofhaving themagnetic moment proportional tothe angular momentum isthatanatomic magnet placed inamagnetic ﬁeldwillprecess. First wewillargue classically. Suppose that wehave themagnetic moment ;.t suspended freely inauniform magnetic ﬁeld. Itwillfeelatorque T,equal to /4XB,which tries tobring itinlinewith theﬁeld direction Buttheatomic magnet isagyroscope—it hastheangular momentum J.Therefore thetorque duetothemagnetic ﬁeldwillnotcause themagnet tolineup.Instead, themagnet willprecess, aswesawwhen weanalyzed agyroscope inChapter 20ofVolume I. Theangular momentum—and with itthemagnetic moment——precesses about an axisparallel tothemagnetic ﬁeld. Wecanﬁndtherateofprecession bythesame method weused inChapter 20oftheﬁrstvolume. Suppose thatinasmall time A!theangular momentum changes from JtoJ’, asdrawn inFig.34~3, staying always atthesame angle 6with respect tothedirec- tionofthemagnetic ﬁeld B.Let’s callmptheangular velocity oftheprecession, sothatinthetime Artheangle ofprecession iswpAt.From thegeometry ofthe 34-4 ﬁgure, weseethatthechange ofangular momentum inthetime Atis AJ=(Jsin6)(w,, At). Sotherateofchange oftheangular momentum is 5%=w,,Jsin6, (34.8) which must beequal tothetorque: 1'=,aBsin0. (34.9) Theangular velocity ofprecession isthen wp=B. (34.10) Substituting it/Jfrom Eq.(34.6), weseethatforanatomic system CB(407)=g%;, (34.11) theprecession frequency isproportional toB.Itishandy toremember that foranatom (orelectron) wej,,=Zr=(1.4megacycles/gauss)gB, (34.12) andthatforanucleus wef,,=Z?=(0.76 kilocycles/gauss)gB. (34.13) (The formulas foratoms andnuclei aredifferent only because ofthedifferent conventions forgforthetwocases.) According tothecla.s'sical theory, then, theelectron orbits—and spins——in anatom should precess inamagnetic ﬁeld. Isitalsotruequantum-mechanically? Itisessentially true, butthemeaning ofthe“precession” isdiﬂferent. Inquantum mechanics onecannot talkabout thedirection oftheangular momentum inthe same sense asonedoes classically, nevertheless, there isavery close analogy——so close thatwecontinue tocallit“precession.” Wewilldiscuss itlater when wetalk about thequantum-mechanical point ofview. 34-4 Diamagnetism Next wewant tolook atdiamagnetism from theclassical point ofview. It canbeworked outinseveral ways, butoneofthenice ways isthefollowing. Suppose thatweslowly turn onamagnetic ﬁeld inthevicinity ofanatom. As themagnetic ﬁeld changes anelectric ﬁeld isgenerated bymagnetic induction. From Faraday’s law, thelineintegral ofEaround anyclosed path istherateof change ofthemagnetic ﬂuxthrough thepath. Suppose wepick apath Fwhich 1S acircle ofradius rconcentric with thecenter oftheatom, asshown inFig.34~4. Theaverage tangential electric ﬁeld Earound thispath isgiven by E21rr=—z€(B1rr2), andthere isacirculating electric ﬁeld whose strength is E_ rdB 2dt Theinduced electric ﬁeld acting onanelectron intheatom produces atorque equal to—qeEr, which must equal therateofchange oftheangular momentum dJ/dz: d 2dB J_qer at“TE‘ (3“"“‘) 34-5/Jl$,'n0 J AJ W9C) 9 B Fig. 34-3. Anobiect with angular momentum Jand aparallel magnetic moment ptplaced inamagnetic ﬁeld B precesses with theangular velocity nip. B / Path1" \\\ Fig. 34-4. The induced electric forces ontheelectrons inanatom. Integrating with respect totime from zeroﬁeld, weﬁndthatthechange inangular momentum duetoturning ontheﬁeld is r2 AJ=13-B. (34.15) This istheextra angular momentum from thetwist given totheelectrons asthe ﬁeld isturned on. This added angular momentum makes anextra magnetic moment which, because itisanorbital motion, isjust—q,/2m times theangular momentum. The induced diamagnetic moment is __a __£?5:2. an_2mAJ_4mB. (34.16) Theminus sign(asyoucanseeisright byusing Lenz’s law) means thattheadded moment isopposite tothemagnetic ﬁeld. Wewould liketowrite Eq(34.16) alittle differently. Ther2which appears istheradius from anaxisthrough theatom parallel toB,soifBisalong thez-direc- tion, itisx2+y2.Ifweconsider spherically symmetric atoms (oraverage over atoms with their natural axes inalldirections) theaverage ofxi—l—yzis2/3of theaverage ofthesquare ofthetrueradial distance from thecenter point ofthe atom. Itistherefore usually more convenient towrite Eq.(34.16) as 2 A/3=~5%<r2>...B. (34.17) Inanycase, wehave found aninduced atomic moment proportional tothe magnetic ﬁeldBandopposing it.Thisisdiamagnetism ofmatter. Itisthismagnetic effect thatisresponsible forthesmall force onapiece ofbismuth inanonuniform magnetic ﬁeld. (You could compute theforce byworking outtheenergy ofthe induced moments intheﬁeld andseeing how theenergy changes asthematerial ismoved intooroutofthehigh-ﬁeld region.) Wearestillleftwith theproblem: What isthemean square radius, (r2)_,.? Classical mechanics cannot supply ananswer. Wemust goback andstart over with quantum mechanics. Inanatom wecannot really saywhere anelectron is, butonly know theprobability thatitwillbeatsome place lfweinterpret (r2)_,, tomean theaverage ofthesquare ofthedistance from thecenter fortheprobability distribution, thediamagnetic moment given byquantum mechanics is_]USt the same asformula (34.17). This equation, ofcourse, isthemoment foroneelectron. Thetotal moment isgiven bythesum over alltheelectrons intheatom. The surprising thing isthat theclassical argument andquantum mechanics givethe same answer, although, asweshall see,theclassical argument thatgives Eq.(34.17) isnotreally valid inclassical mechanics. Thesame diamagnetic effect occurs even when anatom already hasaperma- nent moment. Then thesystem willprecess inthemagnetic ﬁeld. Asthewhole atom precesses, ittakes upanadditional small angular velocity, andthat slow turning gives asmall current which represents acorrection tothemagnetic moment. This isjust thediamagnetic effect represented inanother way. Butwedon't really have toworry about that when wetalkabout paramagnetism. Ifthedia- magnetic effect isﬁrstcomputed, aswehave done here, wedon’t have toworry about thefactthatthere isanextra little current from theprecession. That has already been included inthediamagnetic term. 34-5 Larmor’s theorem Wecanalready conclude something from ourresults sofar First ofall,in theclassical theory themoment piwasalways proportional toJ,with agiven con- stant ofproportionality foraparticular atom There wasn’t anyspin ofthe electrons, andtheconstant ofproportionality wasalways —q,,/2m; thatistosay, inEq.(34.6) weshould setg=1.Theratio ofatoJwasindependent ofthein- ternal motion oftheelectrons. Thus, according totheclassical theory, allsystems 34~6 ofelectrons would precess with thesame angular velocity. (This isnottrue in quantum mechanics.) This result isrelated toatheorem inclassical mechanics thatwewould nowliketoprove Suppose wehave agroup ofelectrons which are allheldtogether byattraction toward acentral point—as theelectrons areattracted byanucleus. Theelectrons willalsobeinteracting with each other, andcan, in general, have complicated motions. Suppose you have solved forthemotions withnomagnetic ﬁeld andthen want toknow what themotions would bewitha weak magnetic ﬁeld. The theorem says that themotion with aweak magnetic ﬁeldisalways oneoftheno-ﬁeld solutions with anadded rotation, about theaxis oftheﬁeld, with theangular velocity ail,=qeB/2m. (This isthesame astop, ifg=1.)There are,ofcourse, many possible motions. The point isthat for every motion without themagnetic ﬁeld there isacorresponding motion inthe ﬁeld, which istheoriginal motion plusauniform rotation This iscalled Larmor’s theorem, andcu1,iscalled theLarmor frequency. Wewould liketoshow how thetheorem canbeproved, butwewillletyou work outthedetails. Take, ﬁrst, oneelectron inacentral force ﬁeld. Theforce on itisjustF(r), directed toward thecenter. Ifwenow turn onauniform magnetic ﬁeld, there isanadditional force, qvXB;sothetotal force is F(r)+qv><B. (34.18) Now let’slook atthesame system from acoordinate system rotating with angular velocity toabout anaxisthrough thecenter offorce andparallel toB.This isno longer aninertial system, sowehave toputintheproper pseudoforces——the cen- trifugal andCoriolis forces wetalked about inChapter 19ofVolume I.Wefound there thatinaframe rotating withangular velocity ai,there isanapparent tangential force proportional to1),,theradial component ofvelocity: F,=—2mw1i, (34.19) And there isanapparent radial force which isgiven by F,=mwzr —l—21110311,, (3420) where 11,isthetangential component ofthevelocity, measured intherotating frame. (The radial component B,forrotating andinertial frames isthesame ) Now forsmall enough angular velocities (that is,ifwr<<0,),wecanneglect theﬁrstterm (centrifugal) inEq.(34.20) incomparison with thesecond (Coriolis) Then Eqs. (34.19) and(34.20) canbewritten together as F=—2mw ><v. (34.21) Ifwenow combine arotation andamagnetic ﬁeld, wemust addtheforce in Eq.(34.21) tothatinEq(34.18). Thetotal force is F(t)+qv><B+2mv><w (34.22) [wereverse thecross product andthesign ofEq.(3421)togetthelastterm]. Looking atourresult, weseethatif Zmw =—qB thetwoterms ontheright cancel, andinthemoving frame theonly force isF(r). Themotion oftheelectron isjustthesame aswith nomagnetic ﬁeld—and, of course, norotation. Wehave proved Larmor’s theorem foroneelectron. Since theproof assumes asmall w,italsomeans thatthetheorem istrueonly forweak magnetic ﬁelds. Theonly thing wecould askyoutoimprove onistotake thecase ofmany electrons mutually interacting with each other, butallinthesame central ﬁeld, andprove thesame theorem. Sonomatter howcomplex anatom is,ifithas acentral ﬁeld thetheorem istrue. Butthat’s theendoftheclassical mechanics, because itisn’t trueinfactthatthemotions precess inthatway. Theprecession frequency wpofEq.(34.11) isonly equal to60Lifghappens tobeequal tol. 34-7 34-6 Classical physics gives neither diamagnetism norparamagnetism Now wewould liketodemonstrate that according toclassical mechanics there canbenodiamagnetism andnoparamagnetism atall.Itsounds crazy—ﬁrst, wehave proved that there areparamagnetism, diamagnetism, precessing orbits, andsoon,andnowwearegoing toprove thatitisallwrong. Yesl~We aregoing toprove thatifyoufollow theclassical mechanics farenough, there arenosuch magnetic effects——they allcancel out. Ifyoustart aclassical argument inacertain place anddon’t gofarenough, youcangetanyanswer youwant. Buttheonly legitimate andcorrect proof shows thatthere isnomagnetic effect whatever. Itisaconsequence ofclassical mechanics thatifyouhave anykind ofsystem~ agaswith electrons, protons, andwhatever—kept inaboxsothatthewhole thing can’t turn, there willbenomagnetic effect. Itispossible tohave amagnetic effect ifyouhave anisolated system, likeastarheld together byitself, which canstart rotating when youputonthemagnetic ﬁeld. Butifyouhave apiece ofmaterial thatisheld inplace sothatitcan’t start spinning, then there willbenomagnetic effects. What wemean byholding down thespin issummarized thisway: Ata given temperature wesuppose thatthere isonly onestate ofthermal equilibrium Thetheorem then saysthatifyouturnonamagnetic ﬁeld andwait forthesystem togetintothermal equilibrium, there willbenoparamagnetism ordianiagnetism—— there willbenoinduced magnetic moment. Proof: According tostatistical nie- chanics, theprobability thatasystem willhave anygiven state ofmotion ispro- portional toe_U/I”, where Uistheenergy ofthatmotion. Now what istheenergy ofmotion? Foraparticle moving inaconstant magnetic ﬁeld, theenergy isthe ordinary potential energy plus mag/2, with nothing additional forthemagnetic ﬁeld. [You know thattheforces from electromagnetic ﬁelds areq(E+vXB), andthattherateofwork F-visjustqE~v,which isnotaffected bythemagnetic ﬁeld ]Sotheenergy ofasystem, whether itisinamagnetic ﬁeld ornot,isalways given bythekinetic energy plusthepotential energy. Since theprobability ofany motion depends onlyontheenergy—that is,onthevelocity andposition—it is thesame whether ornotthere isamagnetic ﬁeld. Forthermal equilibrium, there- fore, themagnetic ﬁeldhasnoeffect. lfwe have onesystem inabox,andthenhave another system inasecond box, thistime with amagnetic ﬁeld, theprobability ofanyparticular velocity atanypoint intheﬁrstboxisthesame asinthesecond. Iftheﬁrstboxhasnoaverage circulating current (which itwillnothave ifitisin equilibrium with thestationary walls), there isnoaverage magnetic moment. Since inthesecond boxallthemotions arethesame, there isnoaverage magnetic moment there either. Hence, ifthetemperature iskept constant andthermal equilibrium isre-established after theﬁeld isturned on,there canbenomagnetic moment induced bytheﬁe1d—according toclassical mechanics. Wecanonlygeta satisfactory understanding ofmagnetic phenomena from quantum mechanics. Unfortunately, wecannot assume thatyouhave athorough understanding of quantum mechanics, sothisishardly theplace todiscuss thematter. Ontheother hand, wedon’t always have tolearn something ﬁrstbylearning theexact rules and then bylearning how they areapplied indifferent cases. Almost every subject thatwehave taken upinthiscourse hasbeen treated inadifferent way. Inthe case ofelectricity, wewrote theMaxwell equations on“Page One” andthen de- duced alltheconsequences. That’s oneway. Butwewillnotnowtrytobegin anew "Page One,” writing theequations ofquantum mechanics anddeducing everything from them Wewilljusthave totellyousome oftheconsequences ofquantum mechanics, before youlearn where they come from. Sohere wego. 34»-7 Angular momentum inquantum mechanics Wehave already given youarelation between themagnetic moment andthe angular momentum. That’s pleasant. Butwhat dothemagnetic moment andthe angular momentum mean inquantum mechanics? Inquantum mechanics itturns outtobebesttodeﬁne things likemagnetic moments interms oftheother con- cepts such asenergy, inorder tomake surethatoneknows what itmeans. Now, 34-8 itiseasy todeﬁne amagnetic moment interms ofenergy, because theenergy of amoment inamagnetic ﬁeldis,intheclassical theory, it-B.Therefore, thefollow- ingdeﬁnition hasbeen taken inquantum mechanics: Ifwecalculate theenergy ofa system inamagnetic ﬁeld andweﬁndthatitisproportional totheﬁeld strength (forsmall ﬁeld), thecoefﬁcient iscalled thecomponent ofmagnetic moment in thedirection oftheﬁeld. (We don’t have togetsoelegant forourwork now. we canstillthink ofthemagnetic moment intheordinary, tosome extent classical, sense.) Now wewould liketodiscuss theidea ofangular momentum inquantum mechanics—or rather, thecharacteristics ofwhat, inquantum mechanics, iscalled angular momentum. You see,when yougotonewkinds oflaws, youcan’t just assume thateach word isgoing tomean exactly thesame thing. You may think, say,“Oh, Iknow what angular momentum is.lt’sthatthing thatischanged bya torque." Butwhat’s atorque? Inquantum mechanics wehave tohave new deﬁnitions ofoldquantities. Itwould, therefore, belegally besttocallitbysome other name such as“quantangular momentum,” orsomething likethat, because itistheangular momentum asdeﬁned inquantum mechanics Butifwecanﬁnda quantity inquantum mechanics which isidentical toouroldidea ofangular momentum when thesystem becomes large enough, there isnouseininventing anextra word. Wemight aswelljust callitangular momentum. With thatunder- standing, thisoddthing thatweareabout todescribe isangular momentum It isthething which inalarge system werecognize asangular momentum inclassical mechanics. First, wetake asystem inwhich angular momentum isconserved, such asan atom allbyitself inempty space. Now such athing (like theearth spinning onits axis) could, intheordinary sense, bespinning around anyaxisonewished tochoose. And foragiven spin, there could bemany different “states,” allofthesame energy, each “state” corresponding toaparticular direction oftheaxis ofthe angular momentum. Sointheclassical theory, with agiven angular momentum, there isaninﬁnite number ofpossible states, allofthesame energy. Itturns outinquantum mechanics, however, that several strange things happen. First, thenumber ofstates inwhich such asystem canexist islimited- there isonly aﬁnite number. lfthe system issmall, theﬁnite number isverysmall, andifthesystem islarge, theﬁnite number gets very, very large. Second, we cannot describe a“state” bygiving thedirection ofitsangular momentum, but onlybygiving thecomponent oftheangular momentum along some direction—say inthez-direction Classically, anobject with agiven total angular momentum Jcould have, foritsz-component, anyvalue from —l—Jto—J. Butquantum- mechanically, thez-component ofangular momentum canhave onlycertain discrete values. Any given system—a particular atom, oranucleus, oranything—with a given energy, hasacharacteristic number j,anditsz-component ofangular mo- mentum canonly beoneofthefollowing setofvalues: jri (1-1)ﬁ (i~2)h 1 (34.23) “(j—2)?» —(j—1)ﬁ _J-h Thelargest z-component isjtimes h;thenext smaller isoneunitofitless, andso ondown to—jh. Thenumber jiscalled “the spinofthesystem.” (Some people callitthe“total angular momentum quantum number”; butwe’ll callitthe“spin.”) You may beworried thatwhat wearesaying canonly betrueforsome “spe- cial” z-axis Butthatisnotso.Forasystem whose spin isj,thecomponent of angular momentum along anyaxiscanhave only oneofthevalues in(34.23) Although itisquite mysterious, weaskyoujusttoaccept itforthemoment We 34—9 willcome back anddiscuss thepoint later. Youmayatleast bepleased tohear that thez-component goes from some number tominus thesame number, sothatwe atleast don’t have todecide which 1Stheplusdirection ofthez-axis. (Certainly, if wesaidthatitwent from —l—/'tominus adifferent amount, thatwould beinﬁnitely mysterious, because wewouldn’t have been abletodeﬁne thez-axis, pointing the other way.) Now ifthez-component ofangular momentum must godown byintegers from +1to—j,then jmust beaninteger. No! Not quite; twice 1must be aninteger. Itisonlythedifference between —l—]and—jthatmust beaninteger. So, ingeneral, thespinjiseither aninteger orahalf-integer, depending onwhether 2]iseven orodd. Take, forinstance, anucleus likelithium, which hasaspinof three-halves,j =3/2. Then theangular momentum around thez-axis, inunits ofh,isoneofthefollowing: +3/2 +1/2-m -3/2. There arefour possible states, each ofthesame energy, ifthenucleus isinempty space with noexternal ﬁelds. Ifwehave asystem whose spin istwo, then the z-component ofangular momentum hasonly thevalues, inunits ofli, go--0»-to Ifyoucount howmany states there areforagiven j,there are(2j+1)possibilities. Inother words, ifyoutellmetheenergy andalso thespin1,itturns outthat there areexactly (2j+1)states with thatenergy, each state corresponding toone ofthedifferent possible values ofthez-component oftheangular momentum. Wewould liketoaddoneother fact. Ifyoupick outanyatom ofknown 1 atrandom andmeasure thez-component oftheangular momentum, then youmay getanyoneofthepossible values, andeach ofthevalues isequally likely. Allof thestates areinfactsingle states, andeach isjustasgood asanyother. Each one hasthesame “weight” intheworld. (Weareassuming thatnothing hasbeen done tosortoutaspecial sample.) This facthas,incidentally, asimple classical analog. Ifyouaskthesame question classically: What isthelikelihood ofaparticular z-component ofangular momentum ifyoutake arandom sample ofsystems, all with thesame total angular momentum?—the answer isthatallvalues from the maximum totheminimum areequally likely. (You caneasily work thatout.) Theclassical result corresponds totheequal probability ofthe(21+1)possi- bilities inquantum mechanics. From what wehave sofar,wecangetanother interesting andsomewhat surprising conclusion. Incertain classical calculations thequantity thatappears intheﬁnal result isthesquare ofthemagnitude oftheangular momentum J—-in other words, J~J.Itturns outthat itisoften possible toguess atthecorrect quantum-mechanical formula byusing theclassical calculation andthefollowing simple rule: Replace J2=J-Jby](j—l—l)h2. This rule iscommonly used, and usually gives thecorrect result, butnotalways. Wecangivethefollowing argument toshow whyyoumight expect thisruletowork. Thescalar product J-Jcanbewritten as J-J=J3+J,i+J3. Since itisascalar, itshould bethesame foranyorientation ofthespin. Suppose wepicksamples ofanygiven atomic system atrandom andmake measurements of Jf,orJ5,orJ3,theaverage value should bethesame foreach. (There isnospecial distinction foranyoneofthedirections.) Therefore, theaverage ofJ-Jisjust 34-10 equal tothree times theaverage ofanycomponent squared, sayofJ3; Butsince J-Jisthesame forallorientations, itsaverage is,ofcourse, justits constant value; wehave 1-J=3(.12>,,. (34.24) Ifwenow saythatwewillusethesame equation forquantum mechanics, we caneasily ﬁnd(J3),,v.Wejust have totakethesumofthe(2)+1)possible values ofJf,anddivide bythetotal number; <,g>,w2f+<1—if+ (1-1+oz+<—r>2,,2_(3425) Forasystem with aspinof3/2,itgoes likethis: <13)“Z<3/2)2+(1/2)?+4(—1/2)’ +(-3/2? ,2Z2he Weconclude that 1'1=3fJ§>.iv =3242=%(%+1)h2- Wewillleave itforyoutoshow thatEq.(34.25), together with Eq.(34.24), gives thegeneral result J-J=JU-l"l)h2. (34.26) Although wewould think classically thatthelargest possible value ofthez-com- ponent ofJisjustthemagnitude ofJ—name1y, \/J-J—quantum mechanically themaximum ofJ,isalways alittle lessthan that, because jhisalways lessthan h.Theangular momentum isnever “completely along thez-direction.” 34-8 Themagnetic energy ofatoms Now wewant totalkagain about themagnetic moment. Wehave saidthatin quantum mechanics themagnetic moment ofaparticular atomic system canbe written interms oftheangular momentum byEq.(34.6); fl.=—g J, (34.27) where —q,,andmarethecharge andmass oftheelectron. Anatomic magnet placed inanexternal magnetic ﬁeld willhave anextra magnetic energy which depends onthecomponent ofitsmagnetic moment along theﬁeld direction. Weknow that Um; =—;i-B. (34.28) Choosing ourz-axis along thedirection ofB, U,,,,,g =—,azB. (34.29) Using Eq.(34.27), wehave that _ .9: U,,,,,,_g(M)J,B. Quantum mechanics saysthatJ,canhave only certain values: jh,(j—l)h,..., —]h. Therefore, themagnetic energy ofanatomic system isnotarbitrary; itcan have only certain values. Itsmaximum value, forinstance, is g hjB. 34411 lUmag J2=+ JZ=+%ﬁ O > B JZ=--2+» “Z :_%’h Fig. 34-5. Thepossible magnetic en- ergies ofanatomic system with aspin of 3/2 inamagnetic ﬁeld B. Umoq J2=+L2f\ 5 I J2=-Th Fig. 34-6. The two possible energy states ofanelectron incimagnetic ﬁeld B.Thequantity qeh/2m isusually given thename “the Bohr magneton” andwritten ll-ll- =M.“B 2m Thepossible values ofthemagnetic energy are JzUmai; :g/-‘BB Z’ where J,/h takes onthepossible values j,(_]—1),(]——2),...,(—j+1),—j. Inother words, theenergy ofanatomic system ischanged when itisputina magnetic ﬁeld byanamount thatisproportional totheﬁeld, andproportional to J2.Wesaythattheenergy ofanatomic system is“split into2)-1-1levels” by amagnetic ﬁeld. Forinstance, anatom whose energy isU0outside amagnetic ﬁeld andwhose 1is3/2, willhave four possible energies when placed inaﬁeld. Wecanshow these energies byanenergy-level diagram likethat drawn inFig 34-5. Any particular atom canhave only oneofthefourpossible energies inany given ﬁeld B.That iswhat quantum mechanics says about thebehavior ofan atomic system inamagnetic ﬁeld. Thesimplest “atomic” system isasingle electron. Thespinofanelectron is 1/2,sothere aretwopossible states. J,=it/2andJ2=—h/2. Foranelectron atrest(noorbital motion), thespinmagnetic moment hasag-value of2,sothe magnetic energy canbeeither i/,L1gB. Thepossible energies inamagnetic ﬁeldare shown inFig.34-6. Speaking loosely wesaythattheelectron either hasitsspin “i1p” (along theﬁeld) or“down” (opposite theﬁeld). Forsystems with higher spins, there aremore states. Wecanthink thatthe spinis“up” or“down” orcocked atsome “angle” inbetween, depending onthe value ofJ2. Wewillusethese quantum mechanical results todiscuss themagnetic prop- erties ofmaterials inthenext chapter. 34-12 35 Paramagnetism and Magnetic Resonance 35-1 Quantized magnetic states Inthelastchapter wedescribed how inquantum mechanics theangular momentum ofathing does nothave anarbitrary direction, butitscomponent along agiven axiscantake ononly certain equally spaced, discrete values. Itis ashocking andpeculiar thing. You may think that perhaps weshould notgo intosuch things until your minds aremore advanced andready toaccept this kind ofanidea. Actually, your minds willnever become more advanced-in thesense ofbeing able toaccept such athing easily. There isn’t anydescriptive wayofmaking itintelligible that isn’t sosubtle andadvanced initsown form thatitismore complicated than thething youwere trying toexplain. Thebehavior ofmatter onasmall scale—as wehave remarked many times—is different from anything thatyouareused toandisvery strange indeed. Asweproceed with classical physics, itisagood idea totrytogetagrowing acquaintance with the behavior ofthings onasmall scale, atﬁrstasakind ofexperience without any deep understanding. Understanding ofthese matters comes very slowly, ifatall. Ofcourse, onedoes getbetter abletoknow what isgoing tohappen inaquantum- mechanical situation——-if thatiswhat understanding means—but onenever getsa comfortable feeling thatthese quantum-mechanical rules are“natural.” Ofcourse theyare,butthey arenotnatural toourownexperience atanordinary level. We should explain thattheattitude thatwearegoing totakewithregard tothisrule about angular momentum isquite different from many oftheother things wehave talked about. Wearenotgoing totryto“explain” it,butwemust atleast tellyou what happens; itwould bedishonest todescribe themagnetic properties ofmaterials without mentioning thefact that theclassical description ofmagnetism—of angular momentum andmagnetic moments—is incorrect. Oneofthemost shocking anddisturbing features about quantum mechanics isthatifyoutake theangular momentum along anyparticular axisyouﬁndthat itisalways aninteger orhalf-integer times h.This issonomatter which axisyou take. Thesubtleties involved inthatcurious fact—that youcantakeanyother axis andﬁndthatthecomponent foritisalsolocked tothesame setofvalues—we will leave toalater chapter, when youwillexperience thedelight ofseeing how this apparent paradox isultimately resolved. Wewillnowjustaccept thefactthatforevery atomic system there isanumber j,called thespinofthesystem—which must beaninteger orahalf-integer—and that thecomponent oftheangular momentum along anyparticular axis will always have oneofthefollowing values between +jl'iand-1h: j 1_1 1—2 J,=oneoft 5-ft. (35.1) \..\i.\-++'—‘l\) \_ Wehave also mentioned that every simple atomic system hasamagnetic moment which hasthesame direction astheangular momentum. This istruenot only foratoms andnuclei butalso forthefundamental particles. Each funda- mental particle hasitsown characteristic value ofjanditsmagnetic moment. 35-135-1 Quantized magnetic states 35-2 TheStern-Gerlach experiment 35-3 TheRabi molecular-beam method 35-4 Theparamagnetism ofbulk materials 35-5 Cooling byadiabatic demagnetization 35-6 Nuclear magnetic resonance Review. Chapter ll,IH.SIdG Dielectrics ul u ‘ 1=' j=I/2 X1? *\ --+\/21; =0iU0 >B U0 ' Z jz=‘I/2 ' lb) ’:=~, Fg35-l Anatomic system withspin (C) 1has(21-l—l)possible energy values ina magnetic ﬁeld BThe energy splitting is proportional toBforgmqll ﬁelds,-_%_ ul |‘L ./‘5j=3/2 \1 ‘='\'\/2 Tlwp ‘L-I‘ U9 Tlwp >'B jz=‘I/2 fiwp ~\1// <4Ii} (For some particles, both arezero.) What wemean by“the magnetic moment” inthisstatement isthat theenergy ofthesystem inamagnetic ﬁeld, sayin thez-direction, canbewritten as—)u,B forsmall magnetic ﬁelds. Wemust have the condition that theﬁeld should notbetoogreat, otherwise itcould disturb theinternal motions ofthesystem and theenergy would notbeameasure ofthemagnetic moment thatwasthere before theﬁeld wasturned on.Butifthe ﬁeld issufficiently weak, theﬁeld changes theenergy bytheamount AU=-—1.t,B, (35.2) with theunderstanding thatinthisequation wearetoreplace 1.12by I-12=14%)1.. (35.3) where J,hasoneofthevalues inEq.(35.1). Suppose Wetakeasystem with aspinj =3/2. Without amagnetic ﬁeld, the system hasfour different possible states corresponding tothedifferent values of J2,allofwhich have exactly thesame energy. Butthemoment weturnonthemag- netic ﬁeld, there isanadditional energy ofinteraction which separates these states intofourslightly different energy levels. Theenergies ofthese levels aregiven bya certain energy proportional toB,multiplied byfttimes 3/2,1/2,-l/2,and-3/2- thevalues ofJ,. Thesplitting oftheenergy levels foratomic systems with spins of 1/2,1,and3/2areshown inthediagrams ofFig.35-1. (Remember thatforany arrangement ofelectrons themagnetic moment isalways directed opposite tothe angular momentum.) You willnotice from thediagrams thatthe“center ofgravity” oftheenergy levels isthesame with andwithout amagnetic ﬁeld. Also notice thatthespacings from onelevel tothenextarealways equal foragiven particle inagiven magnetic ﬁeld. Wearegoing towrite theenergy spacing, foragiven magnetic ﬁeld B,as l”1a»,,—-which isjustadeﬁnition ofw,,.Using Eqs. (35.2) and(35.3), wehave hw,,=g%hB OI‘ a,=g5%B. (35.4) 35-2 Thequantity g(q/2rn) isjust theratio ofthemagnetic moment totheangular momentum-it isaproperty oftheparticle. Equation (35.4) isthesame formula thatwegotinChapter 34fortheangular velocity ofprecession inamagnetic ﬁeld, foragyroscope whose angular momentum isJandwhose magnetic moment ispt. YjI:i1 f1 oven . IMAGNET HOLE VACUUM \\-’\--"--’“\-4 Fig. 35-2. Theexperiment ofStern and Gerlach. 35-2 TheStern-Gerlach experiment Thefactthattheangular momentum isquantized issuch asurprising thing thatwewilltalkalittle bitabout ithistorically. Itwasashock from themoment itwasdiscovered (although itwasexpected theoretically). Itwasﬁrstobserved in anexperiment done in1922 byStern andGerlach. Ifyouwish, youcanconsider theexperiment ofStern-Gerlach asadirect justiﬁcation forabelief inthequantiza- tionofangular momentum. Stern andGerlach devised anexperiment formeasur- ingthemagnetic moment ofindividual silver atoms. They produced abeam of silver atoms byevaporating silver inahotoven andletting some ofthem come out through aseries ofsmall holes. This beam wasdirected between thepole tips ofaspecial magnet, asshown inFig. 35-2. Their idea wasthefollowing. If thesilver atom hasamagnetic moment ja,theninamagnetic ﬁeldBithasanenergy -u,B, where 2isthedirection ofthemagnetic ﬁeld. lntheclassical theory, 1.1, would beequal tothemagnetic moment times thecosine oftheangle between the moment andthemagnetic ﬁeld, sotheextra energy intheﬁeld would be AU=—1.iB cos6. (35.5) Ofcourse, astheatoms come outoftheoven, their magnetic moments would point inevery possible direction, sothere would beallvalues of9.Now ifthe magnetic ﬁeld varies very rapidly with z—if there isastrong ﬁeld gradient—then themagnetic energy willalsovary with position, andthere willbeaforce onthe magnetic moments whose direction willdepend onwhether cosine 6ispositive or negative. Theatoms willbepulled upordown byaforce proportional tothe derivative ofthemagnetic energy; from theprinciple ofvirtual work, 6U 6BF,=—-(E =/J.COS03Z~- (35.6) Stern andGerlach made their magnet with avery sharp edge ononeofthe poletipsinorder toproduce averyrapid variation ofthemagnetic ﬁeld. Thebeam ofsilver atoms wasdirected right along thissharp edge, sothattheatoms would feelavertical force intheinhomogeneous ﬁeld. Asilver atom with itsmagnetic moment directed horizontally would have noforce onitandwould gostraight pastthemagnet. Anatom whose magnetic moment wasexactly vertical would have aforce pulling ituptoward thesharp edge ofthemagnet. Anatom whose magnetic moment waspointed downward would feeladownward push. Thus, 35-3GLASS PLATE asthey leftthemagnet, theatoms would bespread outaccording totheir vertical components ofmagnetic moment. Intheclassical theory allangles arepossible, sothatwhen thesilver atoms arecollected bydeposition onaglass plate, oneshould expect asmear ofsilver along avertical line. Theheight ofthelinewould bepro- portional tothemagnitude ofthemagnetic moment. The3l)]6C[ failure ofclas*-aical ideas wascompletely revealed when Stern andGerlach sawwhat actually happened. They found ontheglass plate twodistinct spots. Thesilver atoms hadformed twobeams. That abeam ofatoms whose spins would apparently berandomly oriented gets split upinto twoseparate beams ismost miraculous. How does themagnetic moment know thatitisonlyallowed totakeoncertain components inthedirection ofthemagnetic ﬁeld” Well, thatwasreally thebeginning ofthediscovery ofthe quantization ofangular momentum, andinstead oftrying togiveyouatheoretical explanation, wewilljustsaythatyouarestuck with theresult ofthisexperiment _]USlasthephysicists ofthatdayhadtoaccept theresult when theexperiment was done. Itisanexperimentalfucr that theenergy ofanatom inamagnetic ﬁeld takes onaseries ofindividual values. Foreach ofthese values theenergy ispro- portional totheﬁeld strength. Soinaregion where theﬁeld varies, theprinciple ofvirtual work tellsusthatthepossible magnetic force ontheatoms willhave a setofseparate values. theforce isdifferent foreach state. sothebeam ofatoms is splitintoasmall number ofseparate beams. From ameasurement ofthedeﬂection ofthebeams, onecanﬁndthestrength ofthemagnetic moment. 35-3 TheRabi molecular-beam method Wewould now liketodescribe animproved apparatus forthemeasurement ofmagnetic moments which wasdeveloped byI.lRabi andhiscollaborators. intheStern-Gerlach experiment thedeﬂection ofatoms isvery small, andthe measurement ofthemagnetic moment isnotvery precise. Rabi’s technique per- mits afantastic precision inthemeasurement ofthemagnetic moments. The method isbased onthefactthat theoriginal energy oftheatoms inamagnetic ﬁeld issplit upintoaﬁnite number ofenergy levels. That theenergy ofanatom inthemagnetic ﬁeld canhave only certain discrete energies isreally notmore surprising than thefact that atoms ingeneral have only certain discrete energy levels—something wementioned often inVolume l.Why should thesame thing no!hold foratoms inamagnetic ﬁeld? Itdoes. Butitistheattempt tocorrelate this with theidea ofanorlenied magnetic moment that brings outsome ofthe strange implications ofquantum mechanics. When anatom hastwolevels which differ inenergy bytheamount AU,it canmake atransition from theupper level tothelower level byemitting alight quantum offrequency w,where hw=AU (35.7) Thesame thing canhappen with atoms inamagnetic ﬁeld. Only then, theenergy differences aresosmall that thefrequency does notcorrespond tolight. butto microwaves ortoradiofrequencies. Thetransitions from thelower energy level toanupper energy level ofanatom canalsotakeplace with theabsorption oflight or.inthecaseofatoms inamagnetic ﬁeld, bytheabsorption ofmicrowave energy. Thus ifwehave anatom inamagnetic ﬁeld, wecancause transitions from onestate toanother byapplying anadditional electromagnetic ﬁeldofthe proper frequency. Inother words, ifwehave anatom inastrong magnetic ﬁeld andwe“tickle” theatom with aweak varying electromagnetic ﬁeld. there willbeacertain prob- ability ofknocking ittoanother level ifthe frequency isnear tothetoinEq.(35.7). Foranatom inamagnetic ﬁeld, thisfrequency isJustwhat wehave earlier called w,,anditisgiven interms ofthemagnetic ﬁeld byEq(35.4). Ifthe atom istickled with thewrong frequency. thechance ofcausing atransition isvery small. Thus there isasharp resonance at(101)intheprobability ofcausing atransition. By measuring thefrequency ofthisresonance inaknown magnetic ﬁeld B,wecan measure thequantity g(q/2m)—and hence theg-factor—with great precision. 35—4 Itisinteresting thatonecomes tothesame conclusion from aclassical point ofview. According totheclassical picture, when weplace asmall gyroscope with amagnetic moment ]J.andanangular momentum Jinanexternal magnetic ﬁeld, thegyroscope willprecess about anaxisparallel tothemagnetic ﬁeld. (SeeFig 35—3.) Suppose weask: How canwechange theangle oftheclassical gyroscope withrespect totheﬁeld—namely, with respect tothez-axis? Themagnetic ﬁeld produces atorque around ahorizontal axis. Such atorque youwould think is trying tolineupthemagnet with theﬁeld, butitonly causes theprecession. lfwe want tochange theangle ofthegyroscope with respect tothez-axis, wemust exert atorque onitabout thez-axis. Ifweapply atorque which goes inthesame direction astheprecession, theangle ofthegyroscope willchange togiveasmaller component ofJinthez-direction lnFig. 35-3, theangle between Jandthe z-axis would increase. Ifwetrytohinder theprecession, Jmoves toward the vertical. Forourprecessing atom inauniform magnetic ﬁeld, how canweapply the kind oftorque wewant? Theanswer is:with aweak magnetic ﬁeld from theside You might atﬁrstthink that thedirection ofthismagnetic ﬁeld would have to rotate with theprecession ofthemagnetic moment, sothatitwasalways atright angles tothemoment, asindicated bytheﬁeld B’inFig.35—4(a). Such aﬁeld works very well, butanalternating horizontal ﬁeld isalmost asgood. Ifwehave asmall horizontal ﬁeld B’,which isalways inthex-direction (plus orminus) and which oscillates with thefrequency wp,then oneach one-half cycle thetorque on themagnetic moment reverses, sothat ithasacumulative eﬁfect which isalmost aseffective asarotating magnetic ﬁeld. Classically, then, wewould expect the component ofthemagnetic moment along thez-direction tochange ifwehave a veryweak oscillating magnetic ﬁeld atafrequency which isexactly 0),,Classically. ofcourse, #2would change continuously, butinquantum mechanics thez-com- ponent ofthemagnetic moment cannot adjust continuously. Itmustjump suddenly from onevalue toanother. Wehave made thecomparison between theeon- sequences ofclassical mechanics andquantum mechanics togiveyousome clue astowhat might happen classically andhowitisrelated towhat actually happens inquantum mechanics. You willnotice, incidentally, thattheexpected resonant frequency isthesame inboth cases. Oneadditional remark: From what wehave saidabout quantum mechanics, there 1Snoapparent reason whythere couldn’t alsobetransitions atthefrequency 2w,,. Ithappens thatthere isn’t anyanalog ofthisintheclassical case, andalso itdoesn’t happen inthequantum theory either—at least notfortheparticular method ofinducing thetransitions thatwehave described. With anoscillating horizontal magnetic ﬁeld, theprobability thatafrequency 205,,would cause ajump oftwosteps atonce iszero. Itisonly atthefrequency w,,thattransitions, either upward ordownward, arelikely tooccur. Now weareready todescribe Rabi’s method formeasuring magnetic mo- nients. Wewillconsider here onlytheoperation foratoms with aspinofl/2A diagram oftheapparatus isshown inFig.35-5. There isanoven which gives out astream ofneutral atoms which passes down alineofthree magnets. Magnet l U) b\B J mp F Fig. 35—3. The classical precession of anatom with themagnetic momentp. and theangular momentum J. B \\\ J F (Q) B//, \\ ) \ B \ \ J (bl " ___} B’=bcos(wt) Fig. 35—4. The angle ofprecession of anatomic magnet can bechanged by0 horizontal magnetic ﬁeld always atright angles top.,asin(a),orbyanoscillating ﬁeld, asin(b). Fig. 35~5. TheRabi molecular-beam apparatus 35-5 DETECTOR CURRENT -E-— /.1,_, E__‘U Fig. 35-6. The current ofatoms thebeam decreases when to=cop.is]UStliketheoneinFig.35-2, andhasaﬁeld with astrong ﬁeld gradient—say, with 8B2/62 positive. Iftheatoms have amagnetic moment. they willbedeﬂected downward ifJ2 =+5/2, orupward ifJ2 =-h/2 (since forelectrons itisdirected opposite toJ).Ifweconsider only those atoms which cangetthrough theslit S1,there aretwopossible tI‘2l_]€CiOI‘l€S, asshown. Atoms with J2=+fi/2 mtist goalong ctirve atogetthrough theslit,andthose withJ2=-h/2 must goalong curve b.Atoms which start outfrom theoven along other paths willnotget through theslit. Magnet 2hasauniform ﬁeld. There arenoforces ontheatoms inthis region, sothey gostraight through andenter magnet 3.Magnet 3is]LlSllike magnet lbutwith theﬁeld inverted, sothat6B2/62 hastheopposite sign. The atoms with J2=—l—h/2 (wesay“with spin up”), that feltadownward push in magnet l,getanupward push inmagnet 3;they continue onthepath aandgo through slitS2toadetector. Theatoms with J2:—h/2 (“with spin down") alsohave opposite forces inmagnets land3andgoalong thepath b,which also takes them through slitS2tothedetector. Thedetector may bemade invarious ways, depending ontheatom being measured. Forexample, foratoms ofanalkali metal likesodium, thedetector can beathin, hottungsten wireconnected toasensitive current meter. When sodium atoms land onthewire. they areevaporated offasNa+ ions, leaving anelectron behind. There isacurrent from thewire proportional tothenumber ofsodium atoms arriving persecond. Inthegapofmagnet 2there isasetofcoils thatproduces asmall horizontal magnetic ﬁeld B’.Thecoils aredriven with acurrent which oscillates atavariable freqtiency w.Sobetween thepoles ofmagnet 2there isastrong, constant, vertical ﬁeld B0andaweak, oscillating, horizontal ﬁeld B’. Suppose now that thefrequency woftheoscillating ﬁeld issetatw,,—the “precession” frequency oftheatoms intheﬁeld B.Thealternating ﬁeld willcause some oftheatoms passing bytomake transitions from oneJ2totheother An atom whose spin was initially “up” (J2=-l-ll/2) may beﬂipped “down” (J2=—h/2). Now thisatom hasthedirection ofitsmagnetic moment reversed, soitwillfeeladownward force inmagnet 3andwillmove along thepath a’, shown inFig. 35-5. Itwillnolonger getthrough theslitS2tothedetector. Similarly, some oftheatoms whose spins were initially down (J2=—h/2) will have their spins ﬂipped up(J2=+h/2) asthey pass through magnet 2.They willthen goalong thepath b’andwillnotgettothedetector. Iftheoscillating ﬁeld B’hasafrequency appreciably different from w,,,itwill notcause anyspin ﬂips. andtheatoms willfollow their undisttirbed paths to thedetector. Soyoucanseethat the“precession” frequency o.>,,oftheatoms intheﬁeld B,,canbefound byvarying thefrequency woftheﬁeld B’until ade- crease isobserved inthecurrent ofatoms arriving atthedetector. Adecrease in thecurrent willoccur when wis“inresonance” with w,,. Aplotofthedetector current asafunction oftomight look liketheoneshown inFig.35-6. Knowing wp,wecanobtain theg-value oftheatom. Such atoniic-beam or,astheyareusually called, “molecular” beam resonance experiments areabeautiful anddelicate wayofmeasuring themagnetic properties ofatomic objects. The resonance frequency 4,5,,canbedetermined with great precision—in fact, with agreater precision than wecanmeasure themagnetic ﬁeld B0,which wemust know toﬁndg. 35—4 Theparamagnetism ofbulk materials Wewould likenow todescribe thephenomenon oftheparamagnetism of bulk materials Suppose wehave asubstance whose atoms have permanent mag- netic nioments, forexample acrystal likecopper sulfate. Inthecrystal there are copper ionswhose inner electron shells have anetangular momentum andanet magnetic moment. Sothecopper ionisanobject which hasapermanent magnetic moment. Let’s sayJUSIaword about which atoms have magnetic moments and which ones don't. Any atom, likesodium forinstance, which hasanoddnumber 35-6 ofelectrons, willhave amagnetic moment. Sodium hasoneelectron initsun- ﬁlled shell. This electron gives theatom aspinandamagnetic moment. Ordinarily, however, when compounds areformed theextra electrons intheoutside shell are coupled together with other electrons whose spin directions areexactly opposite, sothatalltheangular momenta andmagnetic moments ofthevalence electrons usually cancel out. That’s why, ingeneral, molecules donothave amagnetic moment. Ofcourse ifyouhave agasofsodium atoms, there isnosuch cancella- tion.* Also, ifyouhave what iscalled inchemistry a“free radical”—an object with anoddnumber ofvalence electrons—then thebonds arenotcompletely satisﬁed, andthere isanetangular momentum. Inmost bulk materials there isanetmagnetic moment onlyifthere areatoms present whose inner electron shell isnotﬁlled. Then there canbeanetangular momentum andamagnetic moment. Such atoms arefound inthe“transition element” part oftheperiodic table—for instance, chromium, manganese, iron, nickel, cobalt, palladium, andplatinum areelements ofthiskind. Also, allofthe rareearth elements have unﬁlled inner shells andpermanent magnetic moments. There areacouple ofother strange things that also happen tohave magnetic moments, such asliquid oxygen, butwewillleave ittothechemistry department toexplain thereason. Now suppose thatwehave aboxfullofatoms ormolecules with permanent moments—-say agas,oraliquid, oracrystal. Wewould liketoknow what happens ifweapply anexternal magnetic ﬁeld. With nomagnetic ﬁeld, theatoms arekicked around bythethermal motions, andthemoments wind uppointing inalldirections. Butwhen there isamagnetic ﬁeld, itactstolineupthelittle magnets; then there aremore moments lying toward theﬁeld than away from it.The material is “magnetized.” Wedeﬁne themagnetization Mofamaterial asthenetmagnetic moment per unitvolume, bywhich wemean thevector sumofalltheatomic magnetic moments inaunitvolume. Ifthere areNatoms perunitvolume andtheir average moment is(M),,,,then Mcanbewritten asNtimes theaverage atomic moment: M=/V</U... (35-3) Thedeﬁnition ofMcorresponds tothedeﬁnition oftheelectric polarization P ofChapter l0. Theclassical theory ofparamagnetism isjustlikethetheory ofthedielectric constant weshowed youinChapter ll.Oneassumes thateach oftheatoms hasa magnetic moment pi,which always hasthesame magnitude butwhich canpoint inanydirection. Inaﬁeld B,themagnetic energy is—pt-B=—;.tB cos6,where 0istheangle between themoment andtheﬁeld. From statistical mechanics, the relative probability ofhaving anyangle ise_“"°”"'”°T, soangles near zero are more likely than angles near 77'.Proceeding exactly aswedidinSection ll-3, we ﬁndthatforsmall magnetic ﬁelds Misdirected parallel toBandhasthemagnitude lV]i2l?M=3kT (359) [SeeEq.(l1.20).] This approximate formula iscorrect only for/.tB/kT much less than one. Weﬁnd that theinduced magnetization—the magnetic moment perunit volume—is proportional tothemagnetic ﬁeld. This isthephenomenon ofpara- magnetism. Youwillseethattheeffect isstronger atlower temperatures andweaker athigher temperatures. When weputaﬁeld onasubstance, itdevelops, forsmall ﬁelds, amagnetic moment proportional totheﬁeld. Theratio ofM toB(forsmall ﬁelds) iscalled themagnetic susceptibility. Now wewant tolook atparamagnetism from thepoint ofview ofquantum mechanics. Wetakeﬁrstthecaseofanatom with aspinof1/2. Intheabsence of *Ordinary Navapor ismostly monatomic, although there arealsosome molecules of N32. 35-7 amagnetic ﬁeld theatoms have acertain energy, butinamagnetic ﬁeld there are twopossible energies, oneforeach value ofJ2.ForJ2=+h/2, theenergy is changed bythemagnetic ﬁeld bytheamount AU=—l-g (35.10)1 m (The energy shiftAUispositive foranatom because theelectron charge isnegative.) ForJ2=—h/2, theenergy ischanged bytheamount AU2=-4%)-%-B (35.11) Tosave writing, let’sset Ji 1v@=gQ§§-5; aim) AU==‘=}.l.0B. (35.13)then The meaning of/.t,,isclear: —;.t(, isthez-component ofthemagnetic moment in thetip-spin case, and-l-,LI.() isthez-component ofthemagnetic moment inthe down-spin case. Now statistical mechanics tellsusthattheprobability thatanatom isinone state oranother isproportional to e—(Energy ofstate)/kT With nomagnetic ﬁeldthetwostates have thesame energy; sowhen there isequilib- rium inamagnetic ﬁeld, theprobabilities areproportional to e_AU/"T. (35.14) Thenumber ofatoms perunitvolume withspinupis Nu], =aeT"°B/I”, (35.15) andthenumber with spindown is Ndown =ae*"“°B”“. (35.16) Theconstant aistobedetermined sothat N,,,,+Nd,,w,, =N, (35.17) thetotal number ofatoms perunitvolume. Sowegetthat N “-%mWiFmn- (“W What weareinterested inistheaverage magnetic moment along thez-axis. The atoms with spin upwill contribute amoment of—;.t.,, and those with spin down willhave amoment of+ttO. sotheaverage moment is (MW : w*(+“°). (35.19) Themagnetic moment perunitvolume Misthen N(;.t),,,..Using Eqs. (35.15), (35.16), and(35.17), wegetthat -1-#013/kT _ -no]!/kT6‘ 6 M=”W;$Hj;mm' (“M This isthequantum-mechanical formula forMforatoms withj=l/2.Inciden- tally, thisformula canalso bewritten somewhat more concisely interms ofthe 35-8 hyperbolic tangent function: M=Npgtanh “B (35.21) AplotofMasafunction ofBisgiven inFig.35.7. When Bgetsvery large, thehyperbolic tangent approaches l,andMapproaches thelimiting value Np‘) Soathigh ﬁelds. themagnetization saturates. Wecanseewhy that is;athigh enotigh ﬁelds themoments arealllined upinthesame direction Inother words. theyareallinthespin-down state, andeach atom contributes themoment pi, Inmost normal cases—say, fortypical moments, room temperatures, and theﬁelds onecannormally get(like 10,000 gauss)—the ratio ,t.L0B//(T15 about 0.02. Onemust gotoverylowtemperatures toseethesaturation. Fornormal tempera- tures, wecanusually replace tanh xbyx,andwrite N2BM=~f7‘1—- (3522) Justaswesawintheclassical theory, Misproportional toB.Infact, the formula isalmost exactly thesame, except thatthere seems tobeafactor ofl/3 missing. Butwestillneed torelate the,u0inourquantum formula totheitthat appears intheclassical result, Eq(35.9). Intheclassical formula, what appears is/12=pi'pi.thesquare ofthevector magnetic moment, or (2-> itH=g—q,;’7> JJ. (3523) Wepointed outinthelastchapter thatyoucanvery likely gettheright answer from aclassical calculation byreplacing J-Jbyj(j+l)li2. Inourparticular example, wehavej =l/2,so 1'(j+nhz=at Substituting thisforJ~JinEq.(35.23), weget qp 2 “""=g2%T’ orinterms of[A0,deﬁned inEq.(35.12), weget M'I*=3I-Lg- Substituting thisfor,u.2intheclassical formula, Eq.(35.9), does indeed reproduce thecorrect quantum formula, Eq.(35.22). Thequantum theory ofparamagnetism iseasily extended toatoms ofany spinj.Thelow-ﬁeld magnetization is .. 1 2B M=Ng25L3‘f_) 9%. (35.24) where vi.- (35.25) isacombination ofconstants with thedimensions ofamagnetic moment. Most atoms have moments ofroughly thissize. Itiscalled theBohr magneton. The spinmagnetic moment oftheelectron isalmost exactly oneBohr magneton. 35-5 Cooling byadiabatic demagnetization There isavery interesting special application ofparamagnetism. Atvery lowtemperatures itispossible tolineuptheatomic magnets inastrong ﬁeld. Itisthen possible togetdown toextremely lowtemperatures byaprocess called adiabatic demagnetization. Wecantake aparamagnetic salt(for example, one 35-9M N/*0 — // — * / / / I I I J :o I z 3 4 ‘u.B/KT Fig. 35-7. Thevariation ofthepara- magnetic magnetization withthemagnetic ﬁeld strength B. containing anumber ofrare-earth atoms likepraseodynium-ammonium-nitrate), andstart bycooling itdown with liquid helium tooneortwodegrees absolute ina strong magnetic ﬁeld. Then thefactor ;.iB/kT islarger than l—say more like2or3. Most ofthespins arelined up,andthemagnetization isnearly saturated. Let’s say,tomake iteasy, thattheﬁeldisverypowerful andthetemperature isverylow, sothatnearly alltheatoms arelined up.Then youisolate thesaltthermally (say, byremoving theliquid helium andleaving agood vacuum) andturn offthemag- netic ﬁeld. Thetemperature ofthesaltgoes waydown. Now ifyouwere toturn oﬂtheﬁeld suddenly, thejiggling andshaking ofthe atoms inthecrystal lattice would gradually knock allthespins outofalignment. Some ofthem would beupandsome down. Butifthere isnoﬁeld (and disregard- ingtheinteractions between theatomic magnets, which willmake only aslight error), ittakes noenergy toturn over theatomic magnets. They could randomize their spins without anyenergy change and, therefore, without anytemperature change. Suppose, however, thatwhile theatomic magnets arebeing ﬂipped overbythe thermal motion there isstillsome magnetic ﬁeld present. Then itrequires some work toﬂipthem over opposite totheﬁeld—they must dowork against theﬁeld. This takes energy from thethermal motions andlowers thetemperature. Soifthe strong magnetic ﬁeld isnotremoved toorapidly, thetemperature ofthesaltwill decrease—it iscooled bythedemagnetization. From thequantum-mechanical view, when theﬁeldisstrong alltheatoms areinthelowest state, because theodds against anybeing intheupper state areimpossibly big. Butastheﬁeldislowered, itgetsmore andmore likely thatthermal ﬂuctuations willknock anatom intothe upper state. When thathappens, theatom absorbs theenergy AU=ii0B. Soif theﬁeld isturned offslowly, themagnetic transitions cantake energy outofthe thermal vibrations ofthecrystal, cooling itoil.Itispossible inthiswaytogofrom atemperature ofafewdegrees absolute down toatemperature ofafewthou- sandths ofadegree. Would youliketomake something even colder than that? Itturns outthat Nature hasprovided away. Wehave already mentioned thatthere arealsomag- netic moments fortheatomic nuclei. Ourformulas forparamagnetism work just aswellfornuclei, except thatthemoments ofnuclei areroughly athousand times smaller. [They areoftheorder ofmagnitude ofqh/2m,,, where mpistheproton mass, sothey aresmaller bytheratio ofthemasses oftheelectron andproton.] With such magnetic moments, even atatemperature of2°K, thefactor p.B/kT isonly afewparts inathousand. Butifweusetheparamagnetic demagnetiza- tion process togetdown toatemperature ofafewthousandths ofadegree, ;iB/kT becomes anumber near 1-—at these lowtemperatures wecanbegin to saturate thenuclear moments. That isgood luck, because wecanthen use theadiabatic demagnetization ofthenuclear magnetism toreach stilllower temperatures. Thus itispossible todotwostages ofmagnetic cooling. First we useadiabatic demagnetization ofparamagnetic ionstoreach afewthousandths of adegree. Then weusethecold paramagnetic salttocoolsome material which has astrong nuclear magnetism. Finally, when weremove themagnetic ﬁeldfrom this material, itstemperature willgodown towithin amillionth ofadegree ofabsolute zero—if wehave done everything very carefully. 35-6 Nuclear magnetic resonance Wehave saidthatatomic paramagnetism isvery small andthatnuclear mag- netism iseven athousand times smaller. Yetitisrelatively easy toobserve the nuclear magnetism bythephenomenon of“nuclear magnetic resonance.” Suppose wetake asubstance likewater, inwhich alloftheelectron spins areexactly bal- anced sothattheir netmagnetic moment iszero. Themolecules willstillhave a very, verytinymagnetic moment duetothenuclear magnetic moment ofthehydro- gennuclei. Suppose weputasmall sample ofwater inamagnetic ﬁeld B.Since theprotons (ofthehydrogen) have aspin of1/2, they willhave twopossible energy states. Ifthewater isinthermal equilibrium, there willbeslightly more 35-10 protons inthelower energy states with their moments directed parallel tothe ﬁeld. There isasmall netmagnetic moment perunitvolume. Since theproton moment isonly about one-thousandth ofanatomic moment, themagnetization which goes as,u2—using Eq.(35.22)—is only about one-millionth asstrong as typical atomic paramagnetism. (That’s why wehave topick amaterial with no atomic magnetism.) Ifyouwork itout, thedifference between thenumber of protons with spin upandwith spin down isonly onepart in108,sotheeffect isindeed very small! Itcanstillbeobserved, however, inthefollowing way. Suppose wesurround thewater sample with asmall coilthat produces a small horizontal oscillating magnetic ﬁeld. Ifthisﬁeld oscillates atthefrequency wp,itwillindtice transitions between thetwoenergy states—just aswedescribed fortheRabi experiment inSection 35-3. When aproton ﬂips from anupper energy state toalower one, itwillgiveuptheenergy /.i2Bwhich, aswehave seen, isequal toh.o.>,,. Ifitﬂips from thelower energy state totheupper one, itwill absorb theenergy h/wpfrom thecoil. Since there areslightly more protons inthe lower state than intheupper one,there willbeanetabsorption ofenergy from the coil. Although theeffect isvery small, theslight energy absorption canbeseen with asensitive electronic ampliﬁer. JustasintheRabi molecular-beam experiment, theenergy absorption willbe seenonly when theoscillating ﬁeld isinresonance, thatis,when ‘Iv “Z“"1gt?/1;)” Itisoften more convenient tosearch fortheresonance byvarying Bwhile keeping toﬁxed. Theenergy absorption willevidently appear when B:2;”Pw sq. Atypical nuclear magnetic resonance apparatus isshown inFig. 35-8. A high-frequency oscillator drives asmall coilplaced between thepoles ofalarge electromagnet. Two small auxiliary coils around thepole tipsaredriven with a 60-cycle current sothatthemagnetic ﬁeldis“wobb1ed” about itsaverage value by averysmall amount. Asanexample, saythatthemain current ofthemagnet isset togiveaﬁeld of5000 gauss, andtheauxiliary coils produce avariation of11gauss about thisvalue. Iftheoscillator issetat21.2megacycles persecond, itwillthen be attheproton resonance each time theﬁeld sweeps through 5000 gauss [using Eq. (34.13) with g=5.58fortheproton]. Thecircuit oftheoscillator isarranged togive anadditional output signal proportional toanychange inthepower being absorbed from theoscillator. This signal isfedtothevertical deﬂection ampliﬁer ofanoscilloscope. Thehorizontal sweep oftheoscilloscope istriggered once during each cycle oftheﬁeld-wobbling frequency. (Moreusually, thehorizontal deﬂection ismade tofollow inproportion tothewobbling ﬁeld.) Before thewater sample isplaced inside thehigh-frequency coil, thepower drawn from theoscillator issome value. (Itdoesn’t change withthemagnetic ﬁeld ) When asmall bottle ofwater isplaced inthecoil,however, asignal appears onthe oscilloscope, asshown intheﬁgure. Weseeapicture ofthepower being absorbed bytheﬂipping over oftheprotons! Inpractice, itisdifficult toknow how tosetthemain magnet toexactly 5000 gauss. What onedoes istoadjust themain magnet current until theresonance signal appears ontheoscilloscope. Itturns outthat thisisnow themost con- venient waytomake anaccurate measurement ofthestrength ofamagnetic ﬁeld. Ofcourse, atsome time someone hadtomeasure accurately themagnetic ﬁeld and frequency todetermine theg-value oftheproton. Butnowthatthishasbeen done, aproton resonance apparatus likethatoftheﬁgure canbeused asa“proton reso- nance magnetometer.” Weshould sayaword about theshape ofthesignal. Ifwewere towobble the magnetic ﬁeld very slowly, wewould expect toseeanormal resonance curve. Theenergy absorption would read amaximum when wparrived exactly atthe 35-11/// AUXILIARY MAGNET 0°15POLE i / WATER \>. /I‘ llll QQM SOURCEOSCILLATOR ui OUT LOSS SIGNAL OSCILiSCOPE V HSWEEP TRIGGER Fig 35-8. Anuclear magnetic reso- nance apparatus. oscillator frequency. There would besome absorption atnearby frequencies because alltheprotons arenotinexactly thesame ﬁeld—and different ﬁelds mean slightly different resonant frequencies. Onemight wonder, incidentally, whether attheresonance frequency weshould seeanysignal atall.Shouldn’t weexpect thehigh-frequency ﬁeld toequalize the populations ofthetwostates—so thatthere should benosignal except when the water isﬁrstputin?Notexactly, because although wearetrying toequalize the twopopulations, thethermal motions ontheir partaretrying tokeep theproper ratios forthetemperature T.Ifwesitattheresonance, thepower being absorbed bythenuclei isjustwhat isbeing losttothethermal motions. There is,however, relatively little “thermal contact” between theproton magnetic moments andthe atomic motions The protons arerelatively isolated down inthecenter ofthe electron distributions. Soinpure water, theresonance signal is,infact, usually toosmall tobeseen. Toincrease theabsorption, itisnecessary toincrease the “thermal contact.” This isusually done byadding alittle ironoxide tothewater. Theironatoms arelikesmall magnets; astheyjiggle around intheir thermal dance, theymake tinyjiggling magnetic ﬁelds attheprotons. These varying ﬁelds “couple” theproton magnets totheatomic vibrations andtend toestablish thermal equi- librium. Itisthrough this“coupling” thatprotons inthehigher energy states can lose their energy sothat they areagain capable ofabsorbing energy from the oscillator. Inpractice theoutput signal ofanuclear resonance apparatus does notlook likeanormal resonance curve. Itisusually amore complicated signal with oscilla- tions liketheonedrawn intheﬁgure. Such signal shapes appear because ofthe changing ﬁelds. Theexplanation should begiven interms ofquantum mechanics, butitcanbeshown thatinsuch experiments theclassical ideas ofprecessing mo- ments always givethecorrect answer. Classically. wewould saythatwhen wear- riveatresonance westart driving alotoftheprecessing nuclear magnets synchro- nously. Insodoing, wemake them precess together. These nuclear magnets, all rotating together, willsetupaninduced emfintheoscillator coilatthefrequency w,,.Butbecause themagnetic ﬁeldisincreasing withtime, theprecession frequency isincreasing also, andtheinduced voltage issoon atafrequency alittle higher than theoscillator frequency. Astheinduced emfgoes alternately inphase andoutof phase with theoscillator, the“absorbed” power goes alternately positive and negative. Soontheoscilloscope weseethebeatnote between theproton frequency andtheoscillator frequency. Because theproton frequencies arenotallidentical (different protons areinslightly different ﬁelds) andalsopossibly because ofthe disturbance from theironoxide inthewater, thefreely precessing moments soon getoutofphase, andthebeat signal disappears. These phenomena ofmagnetic resonance have been puttouseinmany ways astools forﬁnding outnew things about matter—especially inchemistry and nuclear physics. Itgoes without saying thatthenumerical values ofthemagnetic moments ofnuclei tellussomething about their structure. Inchemistry, much has been learned from thestructure (orshape) oftheresonances. Because ofmagnetic ﬁelds produced bynearby nuclei, theexact position ofanuclear resonance is shifted somewhat, depending ontheenvironment inwhich anyparticular nucleus ﬁnds itself. Measuring these shifts helps determine which atoms arenear which other ones andhelps toelucidate thedetails ofthestructure ofmolecules Equally important istheelectron spin resonance offreeradicals. Although notpresent toanyverylarge extent inequilibrium, such radicals areoften intermediate states ofchemical reactions. Ameasurement ofanelectron spinresonance isadelicate testforthepresence offreeradicals andisoften thekeytounderstanding the mechanism ofcertain chemical reactions. 35-12 36 Ferromagnetism 36-1 Magnetization currents Inthischapter wewilldiscuss some materials inwhich theneteffect ofthe magnetic moments inthematerial ismuch greater thaninthecaseofparamagnetism ordiamagnetism. Thephenomenon iscalledferromagnetism. Inparamagnetic and diamagnetic materials theinduced magnetic moments areusually soweak that wedon’t have toworry about theadditional ﬁelds produced bythemagnetic moments. Forferromagnetic materials, however, themagnetic moments induced byapplied magnetic ﬁelds arequite enormous andhave agreat eﬁ"ect ontheﬁelds themselves. Infact, theinduced moments aresostrong thatthey areoften the dominant effect inproducing theobserved ﬁelds. Sooneofthethings wewill havetoworry about isthemathematical theory oflarge induced magnetic moments. That is,ofcourse, JUSIatechnical question. The realproblem is,why arethe magnetic moments sostrong—how does itallwork? Wewillcome tothatquestion inalittle while. Finding themagnetic ﬁelds offerromagnetic materials issomething likethe problem ofﬁnding theelectrostatic ﬁeld inthepresence ofdielectrics. You will remember thatweﬁrstdescribed theinternal properties ofadielectric interms of avector ﬁeldP,thedipole moment perunitvolume. Then weﬁgured outthatthe effects ofthispolarization areequivalent toacharge density p,,.,1given bythedi- vergence ofP: ppol =—V-P. (36.1) Thetotal charge inanysituation canbewritten asthesum ofthispolarization charge plus allother charges, whose density wewrite* p,,ﬂ,..,. Then theMaxwell equation which relates thedivergence ofEtothecharge density becomes V.E=£=Q ,EQ 60 OI‘ V.E=_ ‘+PJlle_’.VP 60 60 Wecanthen pulloutthepolarization part ofthecharge andputitontheother sideoftheequation, togetthenewlaw VI(GOE + =potlicr- Thenewlawsaysthedivergence ofthequantity (eOE +P)isequal tothedensity oftheother charges. Pulling EandPtogether asinEq.(36.2). ofcourse. isuseful only ifweknow some relation between them. Wehave seen that thetheory which relates the induced electric dipole moment totheﬁeld wasarelatively complicated business andcanreally only beapplied tocertain simple situations, andeven then asan approximation. Wewould liketoremind youofoneoftheapproximate ideas weused. Toﬁndtheinduced dipole moment ofanatom inside adielectric, itis necessary toknow theelectric ﬁeld thatactsonanindividual atom. Wemade the approximation—which isnottoobadinmany cases—that theﬁeld ontheatom *Ifallofthe“other” charges were onconductors, pm“, would bethesame asour pfmofChapter 10. 36-136-1 Magnetization currents 36-2 TheﬁeldH 36-3 Themagnetization curve 36-4 Iron-core inductances 36-5 Electromagnets 36-6 Spontaneous magnetization Review: Chapter 10,Dielectrics Chapter 17,The Law ofIn- duction /Eh°|/,=/E (P/2,,’ Eé /a//.i/V .4 %/ / / / / / =E + //Fig. 36-1. The electric ﬁeld ina cavity inadielectric depends onthe shape ofthecavity.\\.\\\\'<~='\\isthesame asitwould beatthecenter ofthesmall holewhich would beleftifwe took outtheatom (keeping thedipole moments ofalltheneighboring atoms the same). You willalsoremember thattheelectric ﬁeld inahole inapolarized di- electric depends ontheshape ofthehole. Wesummarize ourearlier results in Fig. 36-1. Forathin, disc-shaped hole perpendicular tothepolarization, the electric ﬁeld inthehole isgiven by P Ehole =Edielectric +‘:0’ which weshowed byusing Gauss’ law. Ontheother hand, inaneedle-shaped slotparallel tothepolarization, weshowed—by using thefactthatthecurlofEis zero—~that theelectric ﬁelds inside andoutside oftheslotarethesame. Finally, wefound thatforaspherical holetheelectric ﬁeldwasone-third ofthe waybetween theﬁeld oftheslotandtheﬁeld ofthedisc: 51,01,=E.,,,,,,,.., +lg(spherical hole). (36.3) This wastheﬁeld weused inthinking about what happens toanatom insitf, a polarized dielectric. Now wehave todiscuss theanalog ofallthisforthecase ofmagnetism. Onesimple, short-cut wayofdoing thisistosaytheM,themagnetic moment per unitvolume, isjustlikeP,theelectric dipole moment perunitvolume, andthat, therefore, thenegative ofthedivergence ofMisequivalent toa“magnetic charge density” pm-——whatever thatmay mean. Thetrouble is,ofcourse, thatthere isn’t anysuch thing asa“magnetic charge” inthephysical world. Asweknow, the divergence ofBisalways zero. Butthatdoes notstopusfrom making anartiﬁcial analog andwriting V-M=—p..,, (36.4) where itistobeunderstood thatpmispurely mathematical. Then wecould make acomplete analogy with theelectrostatic caseanduseallouroldequations from electrostatics. People have often done something likethat. Infact, historically, people even believed thattheanalogy wasright. They believed thatthequantity pmrepresented thedensity of“magnetic poles.” These days, however, weknow that themagnetization ofmaterials comes from circulating currents within the atoms—either from thespinning electrons orfrom themotion oftheelectrons in theatom. Itistherefore nicer from aphysical point ofview todescribe things realistically interms oftheatomic currents, rather than interms ofadensity of some mythical “magnetic poles.” Incidentally, these currents aresometimes called “Amperian” currents, because Ampere ﬁrst suggested that themagnetism of matter came from circulating atomic currents. The actual microscopic current density inmagnetized matter is,ofcourse, very complicated. Itsvalue depends onwhere youlook intheatom——it’s large in some places andsmall inothers; itgoes onewayinonepartoftheatom andthe opposite way inanother part (just asthemicroscopic electric ﬁeld varies enor- mously inside adielectric). Inmany practical problems, however, weareinterested onlyintheﬁelds outside ofthematter orintheaverage magnetic ﬁeldinside ofthe matter—where wemean anaverage taken over many, many atoms. Itisonlyfor such macroscopic problems thatitisconvenient todescribe themagnetic state of thematter interms ofM,theaverage dipole moment perunitvolume. What we want toshow now isthattheatomic currents ofmagnetized matter cangiverise tocertain large-scale currents which arerelated toM. What wearegoing todo,then, istoseparate thecurrent density j—which is therealsource ofthemagnetic ﬁelds—into various parts: oneparttodescribe the circulating currents oftheatomic magnets, andtheother parts todescribe what other currents there may be.Itisusually most convenient toseparate thecurrents intothree parts. InChapter 32wemade adistinction between thecurrents which ﬂowfreely onconductors andtheones which areduetotheback andforth motions 36-2 ofthebound charges indielectrics. InSection 32-2 wewrote j:jpol Tl“jotlicrs where j,,,,1represented thecurrents from themotion ofthebound charges indi- electrics andj,,,),,., took care ofallother currents. Now wewant togofurther. Wewant toseparate j,,u,,., intoonepart, j,,,,,g, which describes theaverage currents inside ofmagnetized materials, andanadditional term which wecancallj,,,,,,,1 for whatever isleftover. Thelastterm willgenerally refer tocurrents inconductors, butitmay also include other currents—for example thecurrents from charges moving freely through empty space. Sowewillwrite forthetotal current density: .i:jpol +J-mag +jcond- Ofcourse itisthistotal current which belongs intheMaxwell equation forthe curlofB: 6 2 _L E. cV><B- 60+at (36.6) Now wehave torelate thecurrent j,,,,,g tothemagnetization vector M.So thatyoucanseewhere wearegoing, wewilltellyouthattheresult isgoing to bethat jmag=v><M. (36.7) Ifwearegiven themagnetization vector Meverywhere inamagnetic material, thecirculation current density isgiven bythecurlofM.Let's seeifwecanunder- stand whythisisso. First, let’stakethecaseofacylindrical rodwhich hasauniform magnetization parallel toitsaxis. Physically, weknow thatsuch auniform magnetization really means auniform density ofatomic circulating currents everywhere inside the material. Suppose wetrytoimagine what theactual currents would looklikein across section ofthematerial. Wewould expect toseecurrents something like those shown inFig.36-2. Each atomic current goesaround andaround inalittle circle, withallthecirculating currents going around inthesame direction. Now what istheeffective current ofsuch athing? Well, inmost ofthebarthere isno effect atall,because right next toeach current there isanother current going in theopposite direction. Ifweimagine asmall surface——but onestillquite abit larger than asingle atom——such asisindicated inFig. 36-2 bythelineATS’, thenetcurrent through such asurface iszero. There isnonetcurrent any- where inside thematerial. Note, however, thatatthesurface ofthematerial there areatomic currents which arenotcancelled byneighboring currents going the other way. Atthesurface there isanetcurrent always going inthesame direction around therod. Now youseewhy wesaidearlier thatauniformly magnetized rodisequivalent toalong solenoid carrying anelectric current. How does thisview ﬁtwith Eq.(36.7)" First, inside thematerial themagne- tization Misconstant, soallitsderivatives arezero. This agrees with ourgeometric picture. Atthesurface, however, Misnotreally constant~it isconstant upto theedge andthen suddenly collapses tozero. So,right atthesurface there are terriﬁc gradients which, according to(36.7), willgive ahigh current density. Suppose welook atwhat happens near thepoint CinFig.36-2. Taking thex- andy-directions asintheﬁgure, themagnetization Mis inthez-direction. Writing outthecomponents ofEq.(36.7), wehave HMZ . W :(]mag)1: (36.8) 6M; . _7,’; =(]mag)i/- Atthepoint C,thederivative 6M,/6y lSzero, but6M,/6x islarge andpositive. Equation (36.7) saysthatthere isalarge current density intheminus y-direction. Thisagrees with ourpicture ofasurface current going around thebar. 36-3on e Fig. 36-2. Schematic diagram ofthe circulating atomic currents asseen ina cross section ofanironrodmagnetized in thez-direction.COO y IQQ'-'5-3€ QQQQCQ QQQQCIQ L ea Z /// Q,XF- \\ \\ \\\I SURFACE AREA A Fig. 36-3. Thedipole moment ,uofa current loop isIA. M2 Fig. 36-4. Asmall magnetized block isequivalent toacirculating surface current.U\\\\ +-%-ANow wewant toﬁndthecurrent density foramore complicated caseinwhich themagnetization varies from point topoint inamaterial. Itiseasy toseequali- tatively thatifthemagnetization isdifferent intwoneighboring regions, there will notbeaperfect cancellation ofthecirculating currents sothatthere willbeanet current inthevolume ofthematerial. Itisthiseffect thatwewant towork out quantitatively. First, weneed torecall theresults ofSection 14-5 thatacirculating current Ihasamagnetic moment /.igiven by ,bL=IA, (36.9) where Aisthearea ofthecurrent loop (seeFig.36-3). Now let’sconsider asmall rectangular block inside ofamagnetized material, assketched inFig.36-4. We take theblock sosmall thatwecanconsider that themagnetization isuniform inside it.Ifthisblock hasamagnetization M,inthez-direction, theneteffect willbethesame asasurface current going around onthevertical faces, asshown. Wecanﬁndthemagnitude ofthese currents from Eq.(36.9). Thetotal magnetic moment oftheblock isequal tothemagnetization times thevolume: iu:Mz(abc)s from which weget(remembering thatthearea oftheloop isac) I=Mzb. Inother words, thecurrent perunit length (vertically) oneach ofthevertical surfaces isequal toM2. M, M14-AMZ C 0'ci Y14|"‘|Y“ ‘L-‘1_HI“-4-———I|‘ <- /’ ,/Cl I --—+-> ,4’ il-> ) / . .. /” i /’ 2Fig. 36-5. Ifthe magnetization of Z two neighboring blocks isnot thesame, l:y there isanetsurface current inbetween. X Now suppose thatweimagine twosuch little blocks next toeach other, as shown inFig.36-5. Because block 2isslightly displaced from block 1,itwillhave aslightly diﬂerent vertical component ofmagnetization, which wecallM,+AMZ. Now onthesurface between thetwoblocks there willbetwocontributions tothe total current. Block lwillproduce acurrent I1ﬂowing inthepositive y-direction, andblock 2willproduce asurface current I2ﬂowing inthenegative y-direction. Thetotal surface current inthepositive y-direction isthesum: I= 11- I2=Mzb -(M, —l—AM,)b =—AM,b. Wecanwrite AM, asthederivative ofM,inthex-direction times thedisplacement from block ltoblock 2,which isjustat AM,=§E&a_6x Thecurrent ﬂowing between thetwoblocks isthen 6M2I: —3c--(lb. 36-4 Torelate thecurrent Itoanaverage volume current density j,wemust realize thatthiscurrent 1isreally spread overacertain cross-sectional area. Ifweimagine thewhole volume ofthematerial tobeﬁlled with such little blocks, onesuch side face(perpendicular tothex-axis) canbeassociated with each block.* Then we seethatthearea tobeassociated with thecurrent 1isjustthearea abofoneof thefront faces. Wegettheresult ._l__6M, Jy_ab_ 6x Wehave atleast thebeginning ofthecurlofM. There should beanother term injgfrom thevariation ofthex-component of themagnetization withz.This contribution tojwillcome from thesurface between twolittle blocks stacked oneontopoftheother, asshown inFig.36-6. Using thesame arguments wehave justmade, youcanshow thatthissurface willcon- tribute toiytheamount 6M,/62. These aretheonlysurfaces which cancontribute tothey-component ofthecurrent sowehave thatthetotal current density inthe y-direction is ._GM,_6M, J1' dz dx Working outthecurrents ontheremaining faces ofacube—or using thefact that ourz-direction iscompletely arbitrary—we canconclude that thevector current density isindeed given bytheequation j=VXM. Soifwechoose todescribe themagnetic situation inmatter interms ofthe average magnetic moment perunitvolume M,weﬁndthatthecirculating atomic currents areequivalent toanaverage current density inmatter given byEq.(36.7). Ifthematerial isalsoadielectric, there may be,inaddition, apolarization current jllol=6P/6t. And ifthematerial isalsoaconductor, wemay have aconduction current j,,,,,.i aswell. Wecanwrite thetotal current as j:.i(‘0Il(l + VXM + at 36-2 The ﬁeld H Next, wewant toinsert thecurrent aswritten inEq.(36.10) intoMaxwell’s equations. Weget 2 __1_ g_i(. Q) aEcVXB—€0+6tTT€0jCO!](l+vXM+(9t _l'6t Wecanmove theterm inMtotheleft-hand side: 2 _a.-16»-=6 21:). cVX(B 6Oc2> -60+at(E+G0 (36.11) Asweremarked inChapter 32,many people liketowrite (E+P/co) asanew vector ﬁeld D/co. Similarly, itisoften convenient towrite (B—M/e0c2) asa single vector ﬁeld. Wechoose todeﬁne anewvector ﬁeld Hby H=B- (36.12)6002 Then Eq.(36.11) becomes 606% ><H=1......+335- (36.13) Itlooks simple, butallthecomplexity isjusthidden intheletters DandH. *Or,ifyouprefer, thecurrent Iineach faceshould besplit 50-50 with theblocks on thetwosides. 36-5if?O Z-'-'>Mx+AM ~t 1.:1--M.b i /1--—'-I z/’ ,’ / 1Y: x Fig. 36-6. Twoblocks, oneabove the other, may also contribute tojy.U' \ii?‘{M _’_5‘_.___\\,~to Table 36-1 Units ofmagnetic quantities [B]=weber/meterz =104gauss [H]=weber/meter2 =104gauss 0r104oersted [M] =ampere/meter [H'] =ampere/meter Convenient conversions B(gauss) =104B(weber/meterz) H(gauss) =H(oersted) =0.0126 H’(amp/meter)Now wehave togiveyouawarning. Most people who usethemksunits have chosen touseadifferent deﬁnition ofH.Calling their ﬁeld H’(ofcourse, they stillcallitHwithout theprime), itisdeﬁned by H’=e0c2B —M. (36.14) (Also, they usually write eocz asanewnumber 1/no; then they have onemore constant tokeep track of!) With thisdeﬁnition, Eq.(36.13) looks even simpler: VXHI :jcoiitl ‘l' ' Butthedifﬁculties with thisdeﬁnition ofH’are,ﬁrst, thatitdoesn’t agree withthe deﬁnition ofpeople who don’t usethemks units, andsecond, thatitmakes H’ andBhave different units. Wethink itismore convenient forHtohave thesame units asB—rather than theunits ofM,asH’does. Butifyouaregoing tobean engineer a_ndwork onthedesign oftransformers, magnets, andsuch, youwillhave towatch out. You willﬁndmany books which useforHthedeﬁnition ofEq. (36.14) rather than ourdeﬁnition ofEq.(36.12), andmany other books—especially handbooks about magnetic materials—that relate BandHthewaywehave done. You’ll have tobecareful toﬁgure outwhich convention they areusing. Onewaytotellisbytheunits they use. Remember thatinthemkssystem, B—and therefore ourH—are measured with theunit: oneweber persquare meter, equal to10,000 gauss. lnthemkssystem, amagnetic moment (acurrent times an area) hastheunit: oneampere-meterz. Themagnetization M,then, hastheunit: oneampere permeter. ForH’theunits arethesame asforM.You canseethat thisalsoagrees with Eq.(36.15), since Vhasthedimensions ofoneover alength. People who areworking with electromagnets alsogetinthehabit ofcalling the unitofH(with theH’deﬁnition) “one ampere turnpermeter”—thinking ofthe turns ofwireonawinding. Buta“turn” isreally adimensionless number, sothat doesn’t needtoconfuse you. Since ourHisequal toH’/eocz, ifyouareusing the mks system, H(inwebers/meterz) isequal to411'X10”’ times H’(inamperes permeter). Itisperhaps more convenient toremember that H(ingauss) = 0.0126 H’(inamp/meter). There isonemore horrible thing. Many people who useourdeﬁnition of Hhave decided tocalltheunits ofHandBbydiﬂerenz names! Even though they have thesame dimensions, theycalltheunitofBonegauss, andtheunitofHone oersted (after Gauss andOersted, ofcourse). So,inmany books youwillﬁnd graphs with Bplotted ingauss andHinoersteds. They arereally thesame unit-— l0_4 ofthemks unit. Wehave summarized theconfusion about magnetic units inTable 36-1. 36-3 Themagnetization curve Now wewilllook atsome simple situations inwhich themagnetic ﬁeld 1S constant, orinwhich theﬁelds change slowly enough thatwecanneglect 6D/61 in comparison withj,,,,,,1. Then theﬁelds obey theequations v~B=0, (36.16) VXH=j,,,,,,d/eocz, (36.17) H=B-M/GOC2. (36.18) Suppose wehave atorus (adonut) ofironwrapped with acoilofcopper wire, asshown inFig.36-7(a). Acurrent 1ﬂows inthewire. What isthemagnetic ﬁeld? Themagnetic ﬁeld willbemainly inside theiron; there, thelines ofBwill becircles, asdrawn inFig.36-7(b). Since theﬂuxofBiscontinuous, itsdivergence iszero. andEq(36.16) issatisﬁed Next. wewrite Eq(36.17) inanother form by 36-6 integrating around theclosed loop I‘drawn inFig. 36-7(b). From Stokes’s theorem, wehave that 1 .£H- ds=6062 /Sjcond -nda, where theintegral ofjistobecarried outover anysurface Sbounded byI‘.This surface iscutonce byeach turn ofthewinding. Each turncontributes thecurrent Itotheintegral, and, ifthere areNturns inall,theintegral 1SNI. From the symmetry ofourproblem, Bisthesame allaround thecurve I‘;ifweassume that themagnetization, andtherefore, theﬁeld Hisalsoconstant along F,Eq.(36.19) becomes(36.19) HI=GOC where Iisthelength ofthecurve I‘.So, 1NI Itisbecause Hisdirectly proportional tothemagnetizing current incases like thisonethatHissometimes called themagnetizing ﬁeld. Now allweneed isanequation which relates HtoB.Butthere isn’t anysuch equation! There is,ofcourse, Eq.(36.18), butitISnohelp because there isno direct relation between MandBforaferromagnetic material likeiron. Themag- netization Mdepends onthewhole pasthistory oftheiron, andnotonly onwhat Bisatthemoment. Allisnotlost, though. Wecangetsolutions incertain simple cases. Ifwe start outwith unmagnetized iron—let’s saywith iron that hasbeen annealed at hightemperatures-then inthesimple geometry ofthetorus, alltheironwillhave thesame magnetic history. Then wecansaysomething about M—and therefore about therelation between BandH—from experimental measurements. The ﬁeldBinthetorus is,from Eq.(36.20), given asaconstant times thecurrent I inthewinding. Theﬁeld Bcanbemeasured byintegrating over time theemfin thecoil(orinanextra coilwound over themagnetizing coilshown intheﬁgure). This emfisequal totherateofchange oftheﬂuxofB,sotheintegral oftheemf with time isequal toBtimes thecross-sectional area ofthetorus. Figure 36-8 shows therelation between BandH,observed with atorus of softiron. When thecurrent isﬁrstturned on,Bincreases with increasing Halong thecurve a.Note thedifferent scales onBandH;initially, ittakes onlyarelatively small Htomake alarge B.Why isBsomuch larger with theiron than itwould bewith air? Because there isalarge magnetization Mwhich isequivalent toa large surface current ontheiron—the ﬁeld Bcomes from thesumofthiscurrent andtheconduction current inthewinding. Why Mshould besolarge, wewill discuss later. Athigher values ofH,themagnetization curve levels off. Wesaythat the ironsaturates. With thescales ofourﬁgure, thecurve appears tobecome hori- zontal. Actually, itcontinues toriseslightly—for large ﬁelds, Bbecomes propor- tional toH,andwith aunitslope. There isnofurther increase ofM.Incidentally, weshould point outthatifthetorus were made ofsome nonmagnetic material, Mwould bezero andBwould equal Hforallﬁelds. Theﬁrstthing wenotice isthatcurve ainFig.36-8-which istheso-called magnetization curve—is highly nonlinear. Butit’sworse than that. If,after reaching saturation, wedecrease thecurrent inthecoiltobring Hback tozero, themagnetic ﬁeldBfallsalong curve b.When Hreaches zero, there isstillsome Bleft. Even with nomagnetizing current there isamagnetic ﬁeld intheiron—it hasbecome permanently magnetized. Ifwenow turn onanegative current inthecoil, the B-Hcurve continues along buntil theiron issaturated inthenegative direction. Ifwethen bring thecurrent back tozeroagain, Bgoesalong curve c.Ifwealternate thecurrent between large positive andnegative values, theB-H curve goes back andforth along very nearly thecurves bandc.Ifwevary Hinsome arbitrary 36-7\‘\\\\\§‘V§.\\\\*\to),/ I/I ._. nV ./1' I/1*’ 0 O ll n (bl .,/.~- g~ 0 . . /=3 cunvs 1" '1‘\ /=.- -:. 0 "" LINESOFB '~"\ \ ._-. ‘I’/o \ ='- .-, '3' / 0 K\ 22/ 0 6 0° Fig. 36-7. (a)Atorus ofironwound with acoilofinsulated wire. (b)Cross section oftorus showing ﬁeld lines.<_ 0,_._ W?-Z-°0 B1(gauss) Is,ooo- b iopoo G 5,000-LC 1 I I I I l L,-4 -3 -2 -i I 2 3 4 5 |-1(gauss) -io,ooo C --is,ooo Fig. 36-8. Typical magnetization andhysteresis curves forsoftiron. way, however, wecangetmore complicated curves which will, ingeneral. lie somewhere between thecurves bandc.The loop made byrepeated oscillation oftheﬁelds iscalled ahysteresis loop oftheiron. Weseethen thatwecannot write afunctional relationship likeB=f(H), because thevalue ofBatanyinstant depends notonly onwhat Hisatthattime, butonitswhole pasthistory. Naturally, themagnetization andhysteresis curves aredifferent fordifferent substances. Theshape ofthecurves depends critically on thechemical composition ofthematerial, andalsoonthedetails ofitspreparation andsubsequent physical treatment. Wewilldiscuss some ofthephysical explana- tions forthese complications inthenext chapter. 36-4 Iron-core inductances Oneofthemost important applications ofmagnetic materials isinelectrical circuits—for example, intransformers, electric motors, andsoon.Onereason is that with iron wecancontrol where themagnetic ﬁelds go,andalso getmuch larger ﬁelds foragiven electric current. Forexample, thetypical “toroidal” inductance ismade very much liketheobject shown inFig.36-7. Foragiven in- ductance, itcanbemuch smaller involume andusemuch lesscopper than an equivalent “air-core” inductance. Foragiven inductance, wegetamuch smaller resistance inthewinding, sotheinductance ismore nearly “ideal”—particularly forlowfrequencies. Itisvery easy tounderstand, qualitatively. how such an inductance works. IfIisthecurrent inthewinding, then theﬁeld Hwhich is produced intheinside isproportional toI—as given byEq.(36.20). Thevoltage *0across theterminals isrelated tothemagnetic ﬁeld B.Neglecting theresistance ofthewinding, thevoltage “Uisproportional to6B/61. Theinductance J3,which istheratio of'0todl/dt (seeSection 17-7), thus involves therelation between B andHintheiron. Since theBissomuch bigger thantheH,wegetalargefactor intheinductance. Physically, what happens isthat asmall current inthecoil, which would ordinarily produce asmall magnetic ﬁeld, causes thelittle “slave” magnets intheiron tolineupandproduce atremendously greater “magnetic” current than theexternal current inthewinding. Itisasifwehadalotmore current going through thecoilthan wereally have. When wereverse thecurrent, allthe little magnets ﬂipover—all those internal currents reverse—and wegetamuch higher induced emfthan wewould getwithout theiron. Ifwewant tocalculate theinductance, wecandosothrough theenergy-as described inSection 17-8. Therateatwhich energy isdelivered from thecurrent source is1'0.Thevoltage ’U isthecross-sectional area Aofthecore, times N,times dB/dt. From Eq.(36.20), I=(coczl/N)H. Sowehave dU_ _ 2 dBI —{OI —(€()C Integrating over time, wehave U=(6062121) IHdB. (36.21) Notice thatIAisthevolume ofthetorus, sowehave shown thattheenergy density u=U/vol inamagnetic material isgiven by U=@0621 HdB. (36.22) Aninteresting feature isinvolved here. When weusealternating currents, theironisdriven around ahysteresis loop. Since Bisnotasingle-valued function ofH,theintegral offHdB around onecomplete cycle isnotequal tozero. It isthearea enclosed inside thehysteresis curve. Thus, thedriving source delivers acertain netenergy each cycle—an energy proportional tothearea inside the hysteresis loop. And that energy is“lost.” Itislostfrom theelectromagnetic goings on,butturns upasheatintheiron. Itiscalled thehysteresis loss. Tokeep such energy losses small, wewould likethehysteresis loop tobeasnarrow as 36-8 possible. Onewaytodecrease thearea oftheloop istoreduce themaximum ﬁeld thatisreached during each cycle. Forsmaller maximum ﬁelds, wegetahysteresis curve liketheoneshown inFig.36-9. Also, special materials aredesigned tohave averynarrow loop. Theso-called transformer ir0ns—which areiron alloys with asmall amount ofsilicon-—have been developed tohave thisproperty. When aninductance isrunover asmall hysteresis loop, therelationship between BandHcanbeapproximated byalinear equation. People usually write B=/.tH. (36.23) Theconstant itisnotthemagnetic moment wehave used before. Itiscalled the permeability oftheiron. (Itisalsosometimes called the“relative permeability”) Thepermeability ofordinary irons istypically several thousand. There arespecial alloys alike “supermalloy” which canhave permeabilities ashigh asamillion. Ifweusetheapproximation that B=].LHinEq.(36.21), wecanwrite the energy inatoroidal inductance as 2 U=(@0621/1),. fHdH=(@.,¢2iA) #- (36.24) Sotheenergy density isapproximately 2 u~% ;iH2. Wecannowsettheenergy ofEq.(36.24) equal totheenergy £12/2 ofaninductance, andsolve for.13.Weget 2 H2is=(EQC T ' Using H/Ifrom Eq.(36.20), wehave ,u.N2A£=-E-OFT ' Theinductance isproportional to[.I..Ifyouwant inductances forsuch things as audio ampliﬁers, youwilltrytooperate them onahysteresis loop where the B-Hrelationship isaslinear aspossible. (You willremember thatwespoke in Chapter 50,Vol. I,about thegeneration ofharmonics innonlinear systems.) Forsuch purposes, Eq.(36.23) isauseful approximation. Ontheother hand, ifyouwant togenerate harmonics, youmay useaninductance which isintention- allyoperated inahighly nonlinear way. Then youwillhave tousethecomplete B-Hcurves. andanalyze what happens bygraphical ornumerical methods. A“transformer” isoften made byputting twocoils onthesame torus—or core—of amagnetic material. (For thelarger transformers, thecoreismade with rectangular proportions forconvenience.) Then avarying current inthe“primary” winding causes themagnetic ﬁeld inthecore tochange, which induces anemfin the“secondary” winding. Since theﬂuxthrough each turnofboth windings is thesame, theemf’s inthetwowindings areinthesame ratio asthenumber of turns oneach. Avoltage applied totheprimary istransformed toadifferent voltage atthesecondary. Since acertain netcurrent around thecore isneeded to produce therequired change inthemagnetic ﬁeld, thealgebraic sumofthecurrents inthetwowindings willbeﬁxed andequal totherequired “magnetizing” current. Ifthecurrent drawn from thesecondary increases, theprimary current must in- crease inproportion—there isa“transformation” ofcurrents aswellasvoltage. 36-5 Electromagnets Now let’s discuss apractical situation which isalittle more complicated. Suppose wehave anelectromagnet oftherather standard form shown inFig. 36—10—there isa“C-shaped” yoke ofiron, with acoilofmany turns ofwire wrapped around theyoke. What isthemagnetic ﬁeld Binthegap? 36-9Bl(Wl-'53) /g-” '- '- / / // / / / / / I0,000-/ / / / /_ / / / I l -4 -3-2 -i 1234 I / H(qauss) / ’/’/ I I ///// // // Z4? Fig. 36-9. Ahysteresis loop that doesn't reach saturation. \\\’ Fig. 36-10. Anelectromagnet.YL—> (cl I=O d2 Curve I" B..H. Baa“ *6*7'7“*/ / ~:.\\ee\\\.;§'\::“‘§\i56asé\\~:-:l3<§:“?:- .\--_I’ —‘II,’ //’//////// \__———___—_\-111‘-—1/Surface S COPPER CURRENT6Fig. 36-1 l.Cross section ofanelectromagnet. Ifthegapthickness issmall compared with alltheother dimensions, wecan, asaﬁrstapproximation, assume that thelines ofBwillgoaround through the loop, justasthey didinthetorus They willlook more orlessasshown inFig. 36-ll(a). They tend tospread outsomewhat inthegap, butifthegapisnarrow, thiswillbeasmall effect. Itisafairapproximation toassume thattheﬂuxof Bthrough anycross section oftheyoke isaconstant Iftheyoke hasauniform cross-sectional area—and ifweneglect anyedge effects atthegaps oratTti‘E: corners —we cansaythatBisuniform around theyoke. Also, Bwillhave thesame value inthegap. This follows from Eq.(36.16). Imagine theclosed surface S,shown inFig.36-l1(b), which hasonefaceinthe gapandtheother intheiron. Thetotal ﬂuxofBoutofthissurface must bezero. Calling B1theﬁeld inthegapandB2theﬁeld intheiron, wehave that B1A1 —B2A2 =0. O E,13627, thegap. Wehave that NI PO H1l1+ H212 =mm.‘> eocl Fig. 36-12. Solving for the field in anelectromagnet.NI H namely, theonewhich relates BtoHintheiron.15 Since A1=A2(toourapproximation), itfollows thatB1=B2. \ Now let’slook atH.Wecanagain useEq.(36.19), taking thelineintegral around thecurve I‘inFig.36-1l(b). Asbefore, theright-hand sideisNI,the number ofturns times thecurrent. Now, however, Hwillbedifferent intheiron andintheair.Calling H2theﬁeld intheiron andl2thepath length around the yoke, thispartofthe curve willcontribute theamount H212 totheintegral. Calling C H1theﬁeldinthegapandl1thegapthickness, wegetthecontribution H111 from (36.26) eon“; Now weknow something else: thatintheairgap,themagnetization isnegligi- ble,sothatB1=H1. Since B1=B2,Eq.(36.26) becomes B211+H212= (36.27) Westillhave twounknowns. ToﬁndB2andH2,weneed another relationship- Ifwecanmake theapproximation thatB2=/.iH2, wecansolve theequation algebraically. However, let’sdothegeneral case, inwhich themagnetization curve oftheironisonelikethatshown inFig.36-8. What wewant isthesimultaneous solution ofthisfunctional relationship together with Eq.(36.27). Wecanﬁndit byplotting agraph ofEq.(36.27) onthesame graph with themagnetization curve, asisdone inFig.36-12. Where thetwocurves intersect, wehave oursolution. Foragiven current I,thefunction (36.27) isthestraight linemarked I>0 inFig.36-12. Thelineintersects theH-axis (B2=0)atH2=NI/e11c2l2, and theslope is—l2/l1. Different currents justshift thelinehorizontally. From Fig. 36-10 36-12, weseethatforagiven current there areseveral different solutions, depending onhow yougotthere. Ifyouhave justbuilt themagnet andturned thecurrent uptoI,theﬁeldB2(which isalsoB1)willhave thevalue given bypoint a.If youhave runthecurrent tosome very high value andcome down toI,theﬁeld willbegiven bypoint b.Or,ifyouhave justhadahigh negative current inthe magnet andthen come uptoI,theﬁeld istheoneatpoint c.Theﬁeld inthegap willdepend onwhat youhave done inthepast. When thecurrent inthemagnet iszero, therelation between B2andH2in Eq.(36.27) isshown bythelinemarked I=Ointheﬁgure. There arestillvarious possible solutions. Ifyouhave ﬁrstsaturated theiron, there maybeaconsiderable residual ﬁeldinthemagnet asgiven bypoint d.You cantakethecoiloff,andyou have apermanent magnet. You canseethatforagood permanent magnet, you would want amaterial with awide hysteresis loop. Special alloys, such asAlnico V,have very wide loops. 36-6 Spontaneous magnetization Wenow turn tothequestion ofwhy itisthatinferromagnetic materials a small magnetic ﬁeld produces such alarge magnetization. Themagnetization of ferromagnetic materials likeiron andnickel comes from themagnetic moment oftheelectrons intheinner shell oftheatom. Each electron hasamagnetic moment itequal toq/2m times itsg-factor, times itsangular momentum J.Forasingle electron with nonetorbital motion, g=2,andthecomponent ofJinanydirec- tion—say thez-direction—is ih/2, sothecomponent of,ualong thez-axis is M2= =0.92s><10-2“amp-m2. (36.28) Inaniron atom, there areactually twoelectrons that contribute totheferro- magnetism, sotokeep thediscussion simpler wewilltalkabout nickel, which is ferromagnetic likeironbutwhich hasonlyoneelectron intheinner shell. (Itis easytoextend thearguments toiron.) Now thepoint isthatinthepresence ofanexternal ﬁeldB,theatomic magnets tend tolineupwith theﬁeld, butareknocked about bythermal motions justas wedescribed forparamagnetic materials. Inthelastchapter wefound outthatthe balance between amagnetic ﬁeld trying tolineuptheatomic magnets andthe thermal motions trying toderange them produced theresult thatthemean mag- netic moment perunitvolume willendupas M=Nittanh (36.29) ByB,wemean theﬁeld acting attheatom, andkTistheBoltzmann energy. Inthetheory ofparamagnetism weused forB1,justBitself, neglecting thepartof theﬁeld atanygiven atom contributed bytheatoms nearby. Intheferromagnetic case, there isacomplication. Weshouldn’t usetheaverage ﬁeld intheiron for theBaacting onanindividual atom. Instead, wemust doaswedidinthecaseof dielectrics—we have toﬁndthelocal ﬁeld acting atasingle atom. Foranexact calculation weshould adduptheﬁelds attheatom inquestion contributed byall oftheother atoms inthecrystal lattice. Btitaswedidfordielectrics, wewillmake theapproximation thattheﬁeld atanatom isthesame aswewould ﬁndinasmall spherical hole inthematerial-—assuming that themoments oftheatoms inthe neighborhood arenotchanged bythepresence ofthehole. Following thearguments wemade inChapter 11,wemight think thatwe could write 1MB11611. =B+§26;, (wroiigl). Butthatisnottight. Wecan,however, make useoftheresults ofChapter 11if wemake acareful comparison oftheequations ofChapter 11with theequations 36-ll forferromagnetism inthischapter. Let’s puttogether thecorresponding equations. Forregions where there arenoconduction currents orcharges wehave: Electrostatics Static ferromagnetism V-(E+€£>=0 V-B=O 0 (36.30) VXE=0 VX<B—l2)=0EQC ._\\ These twosetsofequations canbethought ofasanalogous ifwemake thefallow- ingpurely mathematical correspondences: GQC 60 This isthesame asmaking theanalogy E->H, P_>M/c2. (36.31) Inother words, ifwewrite theequations offerromagnetism as M 0 (36.32) VXH=0, they look liketheequations ofelectrostatics. This purely algebraic correspondence hasledtosome confusion inthepast. People tended tothink thatHwas“the magnetic ﬁe1d.” But, aswehave seen, BandEarephysically thefundamental ﬁelds. andHisaderived idea. Soalthough theequations areanalogous, thephysics isnotanalogous. However, thatdoesn’t need tostopusfrom using theprinciple thatthesame equations have thesame solutions. Wecanuseourearlier results fortheelectric ﬁeld inside ofholes ofvarious shapes indielectrics—summarized inFig. 36—1—to ﬁnd theﬁeld Hinside of corresponding holes. Knowing H,wecandetermine B.Forinstance (using the results wesummarized inSection 1),theﬁeld Hinaneedle-shaped hole parallel toMisthesame astheHinthematerial, Hhole :IIrnaterial- Butsince Minthehole iszero, wehave MBhole :Bmaterial '—Z0? ' Ontheother hand, foradisc-shaped hole, perpendicular toM,wehave PEhole :Edioloctric +6_’ O which translates into M Hhole :Hinittorial 'l' * 2 EQC Or,interms ofB, Bliolc =Bm:iterial- Finally, foraspherical hole, bymaking ouranalogy with Eq.(36.3) wewould have Hliolc =Hiiiiitcriiil +36062 OI‘ 2M Bholc =Biiiiiti-rial —52062' This result isquite different from what wegotforE. 36-12 Itis,ofcourse, possible togetthese results inamore physical way, byusing theMaxwell equations directly. Forexample, Eq.(36.34) follows directly from V-B=0.(You useagaussian surface thatishalfinthematerial andhalfout.) Similarly, youcangetEq.(36.33) byusing alineintegral along acurve thatgoes upinside thehole andreturns through thematerial. Physically, theﬁeld inthe hole isreduced because ofthesurface currents—which aregiven byV><M. Wewillleave itforyoutoshow thatEq.(36.35) canalsobeobtained byconsidering theeffects ofthesurface currents ontheboundary ofthespherical cavity. lnﬁnding theequilibrium magnetization from Eq.(36.29), itturns outtobe most convenient todealwith H;sowrite MBa_:H+>\*€‘£)c2' (36.36) Inthespherical hole approximation, wewould have A=2,but, asyouwillsee, wewillwant later tousesome other value, soweleave itasanadjustable parameter. Also, wewilltakealltheﬁelds inthesame direction sothatwewon’t need toworry about thevector directions. Ifwewere now tosubstitute Eq.(36.36) intoEq. (36.29), wewould have oneequation thatrelates themagnetization Mtothemag- netizing ﬁeld H: 2 M=N,utanh (L+2:4/66¢ Itis,however, anequation thatcannot besolved explicitly, sowewilldoitgraph- ically. Let’s puttheproblem inageneralized form bywriting Eq.(36.29) as MM-2; -tanh x, (36.37) where M,1,isthesaturation value ofthemagnetization, namely, Nu,andxrepresents ;.iB,,/kT. Thedependence ofM/Mm onxisshown bycurve ainFig. 36-13. Wecanalsowrite xasafunction ofM—using Eq.(36.36) forB,,—as ,,:ea2em,kT kT e11c2kT M321 Foranygiven value ofH,thisisastraight-line relationship between M/Mm and x.Thexintercept isatx=;.iH/kT, andtheslope ise0c2kT/u t\M,,,1. Forany particular H,wewould have alineliketheonemarked binFig. 36-13. The intersection ofcurves aandbgives usthesolution forM/M,,,,. Wehave solved theproblem. Let’s look athow thesolutions willgoforvarious circumstances. Westart with H=0.There aretwopossible situations, shown bythelines b1andb2 inFig.36-14. You willnotice from Eq.(36.38) thattheslope ofthelineispro- portional totheabsolute temperature T.So,athigh temperatures wewould have alinelikeb1.Thesolution isM/Mm =O.When themagnetizing ﬁeld Hiszero, themagnetization isalsozero. Butatlowtemperatures, wewould have alinelikeb2, andthere aretwosolutions forM/M,,,,—one with M/M82, =0andonewith M/Mm near one. Itturns outthatonly theupper solution isstable-as youcan seebyconsidering small variations about these solutions. According tothese ideas, then, amagnetic material should magnetize itself spontaneously atsuﬂiciently lowtemperatures. Inshort, when thethermal motions aresmall enough, thecoupling between theatomic magnets causes them allto lineupparallel toeach other—we have apermanently magnetized material anal- ogous totheferroelectrics wediscussed inChapter 11. Ifwestart athigh temperatures andcome down, there isacritical temperature. called theCurie temperature T,,where theferromagnetic behavior suddenly setsin. This temperature corresponds tothelineb3ofFig.36-14, which istangent tothe curve a,andhas,therefore, aslope of1.TheCurie temperature isgiven by €QC2kTc —-- =l. 36.39 ”>\Msat ( )(36.38) 36-13M 11 Msat SOLUTION 1.0——--—--—-——- —-- Eq(3637) D 0.5- Eq136381 O i \ 1 ; 0 O5 H I0 I5 x I Fig. 36-13. Agraphical solution of Eqs.(36.37) and (36.38). _"_ Msat 1,HIGH T c Low T 1.0 ’' Q *__ — T T“ _ :_ D‘ b3 a be O. 1 1 1 > O O5 I0 I5 I Fig. 36-14. Finding themagnetiza- tionwhen H=O. Wecan,ifwewish, write Eq.(36.38) more simply interms ofT,as \ _£1BM. Now wewant toseewhat happens forsmall magnetizing ﬁelds H.Wecan seefrom Fig.36-14 how things willgoifweshift ourstraight lines alittle tothe right. Forthelow-temperature case, theintersection point willmove outalittle bitalong thelow-slope partofcurve a,andMwillchange relatively little. Forthe high-temperature case, however, theintersection point runs upthesteep part of curve a,andMwillchange relatively rapidly. Infact, wecanapproximate this partofcurve abyastraight lineofunitslope, andwrite: M H_(M) t. uﬁZ =="‘_ __2_ . Msut X + Msu Now wecansolve forM/M,,,,1: M _ ;.iH iv;"1' (“"4" Wehave alawthatissomething liketheonewehadforparamagnetism. For paramagnetism, wehad M = Onedifference nowisthatwehave themagnetization interms ofH,which includes some oftheeffects oftheinteraction oftheatomic magnets, butthemain difference isthat themagnetization isinversely proportional tothediﬂerence between T andTC,instead oftotheabsolute temperate T,alone. Neglecting theinteractions between neighboring atoms corresponds totaking >1=0,which from Eq.(36.39) means taking T,=0.Then theresults arejustwhat wehadinChapter 35. Wecancheck ourtheoretical picture withtheexperimental datafornickel. Itisobserved experimentally thattheferromagnetic behavior ofnickel disappears when itstemperature israised above 63l°K. Wecancompare thiswith T,,cal- culated from Eq.(36.39). Remembering thatMW =;.iN,wehave N2 Tc = 3i . Xk€0C2 From thedensity andatomic weight ofnickel, weget N=9.1X1028 m‘3. Taking ufrom Eq.(36.28), andsetting A=5,weget T,=0.24°K. There isadiscrepancy ofafactor ofabout 2600! Ourtheory offerromagnetism failscompletely. Wecantryto“patch up”thetheory asWeiss didbysaying thatforsome unknown reason )1isnotone-third, but(2600) X;?,——or about 900. Itturns out that onegetssimilar values forother ferromagnetic materials likeiron. Tosee what thismeans, let’sgoback toEq.(36.36). Weseethatalarge Xmeans that Ba,thelocal ﬁeld ontheatom, appears tobemuch, much larger than wewould think. Infact, writing H=B—M/e1,c2, wehave __ ()1-l)M Bi»"B+W' According toouroriginal idea-with )1=-§—the local magnetization Mreduces theeffective ﬁeld Babytheamount —%M/co. Even ifourmodel ofaspherical holewere notvery good, wewould stillexpect some reduction. Instead, toexplain 36-14 thephenomenon offerromagnetism, wehave toimagine that themagnetization oftheﬁeld enhances thelocal ﬁeld bysome large factor—like onethousand or more. There doesn’t seem tobeanyreasonable waytomanufacture such tremen- dous ﬁelds atanatom noreven ﬁelds oftheproper sign! Clearly, our“magnetic” theory offerromagnetism isadismal failure. Wemust conclude, then, thatferro- magnetism hastodowith some nonmagnetic interaction between thespinning electrons inneighboring atoms. This interaction must generate astrong tendency forallofthenearby spins tolineupinonedirection. WeWlllseelater thatithas todowith quantum mechanics andthePauli exclusion principle. Finally, welook atwhat happens atlowtemperatures—for T<T,.. We have seenthatthere willthen beaspontaneous magnetization—even withH=0— given bytheintersection ofthecurves aandb2ofFig.36-14. Ifwesolve forM forvarious temperatures—-by varying theslope ofthelineb2—~we getthetheoretical curve shown inFig.36-15. This curve should bethesame forallferromagnetic materials forwhich theatomic moment comes from asingle electron. Thecurves forother materials areonly slightly ditferent. Inthelimit, asTgoes toabsolute zero, Mgoes toMW. Asthetemperature isincreased, themagnetization decreases, falling tozero attheCurie temperature. Thepoints inFig.36-15 aretheexperimental observations fornickel. They ﬁtthe theoretical curve fairly well. Even though wedon’t understand thebasic mecha- nism, thegeneral features ofthetheory seem tobecorrect. Finally, there isonemore disturbing discrepancy inourattempt tounder- stand ferromagnetism. Wehave found thatabove some temperature thematerial should behave likeaparamagnetic substance with amagnetization Mpropor- tional toH(orB).andthatbelow thattemperature itshould become spontane- ously magnetized. Butthat’s notwhat wefound when wemeasured themag- netization curve foriron. Itonly became permanently magnetized after wehad “magnetized” it.According totheideas justdiscussed, itwould magnetize itself! What iswrong? Well, itturns outthatifyou look atasmall enough crystal ofiron ornickel, itisindeed completely magnetized! Butinlarge pieces ofiron, there are many small regions or“domains” thataremagnetized indiﬁerent directions, so thatonalarge scale theaverage magnetization appears tobezero. Ineach small domain, however, theiron hasalocked-in magnetization with Mnearly equal to Mm. The consequences ofthisdomain structure arethat gross properties of large pieces ofmaterial arequite different from themicroscopic properties that wehave really been treating. Wewilltake upinthenext lecture thestory ofthe practical behavior ofbulk magnetic materials. 36—l5L‘ Msnt l. O EXPERIMENT O O O5_ THEORY O l 0 05 i.o’ T/Tc Fig. 36-15. Spontaneous magnetiza- tion cisufunction oftemperature for nickel. 37 Magnetic Materials 37-1 Understanding ferromagnetism Inthischapter wewilldiscuss thebehavior andpeculiarities offerromagnetic materials andofother strange magnetic materials. Before proceeding tostudy magnetic materials, however, wewillreview veryquickly some ofthethings about thegeneral theory ofmagnets thatwelearned inthelastchapter. First, weimagine theatomic currents inside thematerial thatareresponsible forthemagnetism, andthen describe them interms ofavolume current density J-“lag =VXM. Weemphasize thatthisisnotsupposed torepresent theactual currents. When themagnetization isuniform thecurrents donotreally cancel outprecisely; thatis,thewhirling currents ofoneelectron inoneatom andthe whirling currents ofanelectron inanother atom donotoverlap insuch away that thesum isexactly zero. Even within asingle atom thedistribution of magnetism isnotsmooth. For instance, inaniron atom themagnetization isdistributed inamore orlessspherical shell, nottooclose tothenucleus and nottoofaraway. Thus, magnetism inmatter isquite acomplicated thing inits details; itisvery irregular. However, weareobliged now toignore thisdetailed complexity anddiscuss phenomena from agross, average point ofview. Then itistruethat theaverage current intheinterior region, over anyﬁnite area that isbigcompared with anatom, iszero when M=0.So,what wemean by magnetization perunit volume andj,,,,,g and soon,atthelevel wearenow considering, isanaverage over regions that arelarge compared with thespace occupied byasingle atom. Inthelastchapter, wealsodiscovered thataferromagnetic material hasthe following interesting property: above acertain temperature itisnotstrongly magnetic, whereas below this temperature itbecomes magnetic. This fact is easily demonstrated. Apiece ofnickel wire atroom temperature isattracted bya magnet. However, ifweheat itabove itsCurie temperature with agasﬂame, it becomes nonmagnetic andisnotattracted toward themagnet—even when brought quite close tothemagnet. Ifweletitlienear themagnet while itcools off,atthe instant itstemperature fallsbelow thecritical temperature itissuddenly attracted again bythemagnet! Thegeneral theory offerromagnetism thatwewillusesupposes thatthespin oftheelectron isresponsible forthemagnetization. Theelectron hasspinone-half andcarries oneBohr magneton ofmagnetic moment /.t=[LB=qeh/2m. The electron spincanbepointed either “up” or“down.” Because theelectron hasa negative charge, when itsspinis“up” ithasanegative moment, andwhen itsspin is“down” ithasapositive moment. With ourusual conventions, themoment pt oftheelectron isopposite itsspin. Wehave found thattheenergy oforientation ofamagnetic dipole inagiven applied ﬁeld Bis—;4-B,buttheenergy ofthe spinning electrons depends ontheneighboring spin alignments aswell. Iniron, ifthemoment ofanearby atom is“up,” there isavery strong tendency thatthe moment oftheonenext toitwillalsobe“up.” That iswhat makes iron, cobalt, andnickel sostrongly magnetic-the moments allwant tobeparallel. Theﬁrst question wehave todiscuss iswhy. Soon after thedevelopment ofquantum mechanics, itwasnoticed thatthere isavery strong apparent force-—not amagnetic force oranyother kind ofactual force, butonly anapparent force—trying tolinethespins ofnearby electrons opposite tooneanother. These forces areclosely related tochemical valence forces. There isaprinciple inquantum mechanics-——called theexclusion principle—that 37-137-1 Understanding ferromagnetism 37-2 Thermodynamic properties 37-3 Thehysteresis curve 37-4 Ferromagnetic materials 37-5 Extraordinary magnetic materials References." Bozorth, R.M,“Magne- tism,” Encyclopaedia Bri- tatmtca, Vol. l4, l957, pp.636-667. Kittel, C.,Introduction to Solid State Physics, John Wiley andSons, Inc., New York, 2nded.,1956. twoelectrons cannot occupy exactly thesame state, thattheycannot beinexactly thesame condition astolocation andspin orientation.* Forexample. iftheyare atthesame point, theonly alternative istohave their spins opposite. So,ifthere isaregion ofspace between atoms where electrons liketocongregate (asinachem- icalbond) andwewant toputanother electron ontopofonealready there, the only waytodoitistohave thespinofthesecond onepointed opposite tothespin oftheﬁrstone. Tohave thespins parallel isagainst thelaw,unless theelectrons stayaway from each other. This hastheelfect thatapairofparallel-spin electrons near toeach other have much more energy than apairofopposite-spin electrons; theneteffect isasthough there were aforce trying toturn thespin over. Some- times thisspin-turning force iscalled theexchange force, butthat only makes it more mysterious—it isnotavery good term. ItisJustbecause oftheexclusion principle that electrons have atendency tomake their spins opposite. Infact, that istheexplanation ofthelaclc ofmagnetism inalmost allsubstances! The spins ofthefreeelectrons ontheoutside oftheatoms have tremendous tendency tobalance inopposite directions. Theproblem istoexplain why formaterials likeironitisjustthereverse ofwhat weshould expect. Wehave summarized thesupposed alignment effect byadding asuitable term intheenergy equation, bysaying thatiftheelectron magnets intheneighborhood have amean magnetization M,then themoment ofanelectron hasastrong tendency tobeinthesame direction astheaverage magnetization oftheatoms in theneighborhood. Thus, wemay write forthetwopossible spin orientations,’[ it *9 Spin upenergy =+;i(H+-6)?) , (37.1) ’ $6 ss Spin down energy =-it(H+E362) When itwasclear thatquantum mechanics could supply atremendous spin- orientating force-even if,apparently, ofthewrong sign—it wassuggested that ferromagnetism might have itsorigin inthissame force, thatduetothecomplexi- tiesofironandthelarge number ofelectrons involved, thesignoftheinteraction energy would come outtheother wayaround. Since thetime thiswasthought of-- inabout 1927 when quantum mechanics wasﬁrstbeing understood—many people have been making various estimates andsemicalculations, trying togetatheoretical prediction forA.Themost recent calculations oftheenergy between thetwoelec- tron spins iniron—assuniing thattheinteraction isadirect onebetween thetwo electrons inneighboring atoms—still givethewrong sign. Thepresent understand- ingofthisisagain toassume that thecomplexity ofthesituation issomehow responsible andtohope thatthenext man who makes thecalculation with amore complicated situation willgettheright answer! Itisbelieved thattheup-spin ofoneoftheelectrons intheinside shell, which ismaking themagnetism, tends tomake theconduction electrons which ﬂyaround theoutside have theopposite spin. Onemight expect thistohappen because the conduction electrons come intothesame region asthe“magnetic” electrons. Since they move around, they cancarry their pI‘6_]LldlC€ forbeing upside down over to thenext atom; thatis,one“magnetic” electron tries toforce theconduction elec- trons tobeopposite, andtheconduction electron then makes thenext “magnetic” electron opposite toit.Thedouble interaction isequivalent toaninteraction which triestolineupthetwo“magnetic” electrons. Inother words, thetendency tomake parallel spins istheresult ofanintermediary thattends tosome extent tobeop- posite toboth. This mechanism does notrequire thattheconduction electrons be completely “upside down.” They could _|US[have aslight pI‘€]LI(llC€ tobedown, justenough toloadthe“niagnetic” odds theother way. This isthemechanism that *SeeChapter 43. TWewrite these equations with H=B—M/e¢ic'~' instead ofBtoagree with thework ofthe lastchapter. You might prefer towrite UI=t=uB,, :1l'l}l.(B +i\’M/ent-2), where N=A—1.It’sthesame thing. 37-2 thepeople who have calculated such things now believe isresponsible forferro- magnetism. Butwemust emphasize that tothisdaynobody cancalculate the magnitude of)\simply byknowing thatthematerial isnumber 26intheperiodic table. Inshort, wedon’t thoroughly understand it. Now letuscontinue with thetheory, andthen come back later todiscuss a certain error involved inthewaywehave setitup.Ifthemagnetic moment ofa certain electron is“up,” energy comes both from theexternal ﬁeld andalsofrom thetendency ofthespins tobeparallel. Since theenergy ISlower when thespins areparallel, theeﬁect issometimes thought ofasduetoan“etlective internal ﬁeld.” Butremember, itisnotduetoatruemagnetic force; itisaninteraction thatismore complicated. Inanycase, wetake Eqs. (37.1) astheformulas forthe energies ofthetwospinstates ofa“magnetic” electron. Atatemperature T,the relative probability ofthese twostates isproportional toe"°“°‘gY”"T, which we canwrite ase”, with x=;i(H +>\M/12002)/kT. Then, ifwecalculate the mean value ofthemagnetic moment, weﬁnd(asinthelastchapter) thatitis M=Natanh x. (37.2) Now wewould liketocalculate theinternal energy ofthematerial. Wenote thattheenergy ofanelectron isexactly proportional tothemagnetic moment, sothatthecalculation ofthemean moment andthecalculation ofthemean energy arethesame—except thatinplace of/.tinEq.(37.2) wewould write —,u.B, which is—;t(H +>\M/e002). Themean energy isthen av =—N/.4 (H+ tanh x.EQC2 Now thisisnotquite correct. Theterm AM/e002 represents interactions of allpossible [M11118 ofatoms, andwemust remember tocount each paironly once. (When weconsider theenergy ofoneelectron intheﬁeld oftherestandthen the energy ofasecond electron intheﬁeldoftherest,wehave counted partofthe ﬁrstenergy once more.) Thus, wemust divide themutual interaction term bytwo, andourformula fortheenergy then turns outtobe XM(U)_,V -—N/.¢ (H—l—2-6-£5) tanh x. (37.3) Inthelastchapter wediscovered aninteresting thing—that below acertain temperature thematerial ﬁnds asolution totheequations inwhich themagnetic moment isnotzero, even with noexternal magnetizing ﬁeld. When wesetH=0 inEq.(37.2), wefound that M TcM M5 1 where M.-tt =Nu, and Tc=it>\M.,,,t/ke0c2. When wesolve this equation (graphically orotherwise), weﬁndthattheratio M/Mm asafunction ofT/T, is acurve likethatlabeled “quantum theory” inFig.37-1. Thedashed curve marked “cobalt. nickel” shows theexperimental results forcrystals ofthese elements. Thetheory andexperiment areinreasonably good agreement. Theﬁgure also shows theresult oftheclassical theory inwhich thecalculation iscarried out assuming that theatomic magnets canhave allpossible orientations inspace. You canseethatthisassumption gives aprediction thatisnoteven close tothe experimental facts. Even thequantum theory deviates from theobserved behavior atboth high andlowtemperatures. Thereason forthedeviations isthatwehave made arather sloppy approximation inthetheory: Wehave assumed that theenergy ofan atom depends upon themean magnetization ofitsneighboring atoms. Inother words, foreach onethatis“up” intheneighborhood ofagiven atom, there will beacontribution ofenergy duetothat quantum mechanical alignment effect. Buthow many arethere pointed “up”? Ontheaverage, thatismeasured bythe 37-3 Fig. 37-1. Thespontaneous magne- tization (H=O)offerromagnetic crystals asafunction oftemperature. [Permission from Encyclopaedia Britannica] U1 rc _ T (a) cvl Tc T> lb) 0,,‘ -1 '30""K"-" -4> Fig. 37-2. The energy perunitvol- ume andspeciﬁc heat ofaferromagnetic crystal._._ ==~mi‘: 09 - \ \\ mo" \\ M \ ‘T 01COBA LY\NICKEL CLASSICAL ‘\ Tusonv \\ 06 ‘ M/Msat‘\ \\ \ \\ Quantum Tncoav3:- z’ -—'03 OZ o 0 0| oz 03 04 05 os 01 oa 09 1.0 T/Tc magnetization M—but only ontheaverage. Aparticular atom somewhere might ﬁndallitsneighbors “up.” Then itsenergy willbelarger than theaverage. Another onemight ﬁndsome upandsome down, perhaps averaging tozero, anditwould have noenergy from thatterm, andsoon.What weought todoistousesome more complicated kind ofaverage, because theatoms indiﬁerent places have (lllTCI‘CI‘ll1 environments, andthenumbers upanddown arediﬁerent fordifferent ones. Instead ofjust taking oneatom subjected totheaverage inﬂuence, weshould take each oneinitsactual situation, compute itsenergy, andﬁnd theaverage energy. Buthowdoweﬁndouthowmany are“up” andhowmany are“down” intheneighborhood? That is,ofcourse, justwhat wearetrying tocalculate- thenumber “up” and“down”—so wehave avery complicated interconnected problem ofcorrelations, aproblem which hasnever been solved. Itisanintriguing andexciting onewhich hasexisted foryears andonwhich some ofthegreatest names inphysics have written papers, buteven theyhave notcompletely solved it. Itturns outthatatlowtemperatures, when almost alltheatomic magnets are “up” andonly afeware“down,” itiseasy tosolve; andathigh temperatures, far above theCurie temperature T,when they arealmost allrandom, itisagain easy. Itisoften easy tocalculate small departures from some simple, idealized situation, soitisfairly wellunderstood why there aredeviations from thesimple theory at lowtemperature. Itisalso understood physically thatforstatistical reasons the magnetization should deviate athigh temperatures. Buttheexact behavior near theCurie point hasnever been thoroughly ﬁgured out. That’s aninteresting problem towork outsome dayifyouwant aproblem thathasnever been solved. 9/; 37-2 Thermodynamic properties Inthelastchapter welaid thegroundwork necessary forcalculating the thermodynamic properties offerromagnetic materials. These are,naturally, related totheinternal energy ofthecrystal, which includes interactions ofthevarious spins, given byEq.(37.3). Fortheenergy ofthespontaneous magnetization below theCurie point, wecansetH=0inEq.(37.3), and—noticing thattanhx = M/Ms,,t—we ﬁndamean energy proportional toM2: _ iv,.xM2av -' Ag ' Ifwenow plottheenergy duetothemagnetism asafunction oftemperature, we getacurve which isthenegative ofthesquare ofthecurve ofFig.37-1, asdrawn inFig.37-2(a). Ifwewere tomeasure then thespeciﬁc heat ofsuch amaterial wewould obtain acurve which isthederivative of37-2(a). Itisshown inFig. 37-4 37-2(b). Itrises slowly with increasing temperature, butfallssuddenly tozero at T=Tc.Thesharp drop isduetothechange inslope ofthemagnetic energy and isreached right attheCurie point. Sowithout anymagnetic measurements at allwecould have discovered thatsomething wasgoing oninside ofironornickel bymeasuring this thermodynamic property. However, both experiment and improved theory (with ﬂuctuations included) suggest that thissimple curve is wrong andthat thetrue situation isreally more complicated. The curve goes higher atthepeak andfalls tozero somewhat slowly. Even ifthetemperature is high enough torandomize thespins ontheaverage, there arestilllocal regions where there isacertain amount ofpolarization, andinthese regions thespins still have alittle extra energy ofinteraction—which only diesoutslowly asthings get more andmore random with further increases intemperature Sotheactual curve looks likeFig. 37-2(0). One ofthechallenges oftheoretical physics today isto ﬁndanexact theoretical description ofthecharacter ofthespeciﬁc heat near the Curie transition-an intriguing problem which hasnotyetbeen solved. Naturally, thisproblem isvery closely related totheshape ofthemagnetization curve inthe same region. Now wewant todescribe some experiments, other than thermodynamic ones, which show thatthere issomething right about ourinterpretation ofmagnetism When thematerial ismagnetized tosaturation atlowenough temperatures, Mis very nearly equal toM,.,t—nearly allthespins areparallel, aswellastheir mag- netic moments. Wecancheck thisbyanexperiment. Suppose wesuspend abar magnet byathinﬁber andthen surround itbyacoilsothatwecanreverse the Xmagnetic ﬁeld without touching themagnet orputting anytorque onit.This isa ‘every difficult experiment because themagnetic forces aresoenormous that any irregularities, anylopsidedness, oranylack ofperfection intheiron willproduce accidental torques. However, theexperiment hasbeen done under careful con- ditions inwhich such accidental torques areminimized. Bymeans ofthemagnetic ﬁeld from acoilthatsurrounds thebar,weturn alltheatomic magnets over at once. When wedothiswealsochange theangular momenta ofallthespins from “up” to“down” (seeFig.37-3). Ifangular momentum istobeconserved when the spins allturn over, therestofthebarmust have anopposite change inangular momentum. Thewhole magnet willstart tospin. And sureenough, when wedo theexperiment, weﬁnd aslight turning ofthemagnet. Wecanmeasure the total angular momentum given tothewhole magnet, andthisissimply Ntimes h, thechange intheangular momentum ofeach spin. Theratio ofangular momentum tomagnetic moment measured thiswaycomes outtowithin about 10percent of what wecalculate. Actually, ourcalculations assume thattheatomic magnets are duepurely totheelectron spin, butthere is,inaddition, some orbital motion alsoin most materials. Theorbital motion isnotcompletely freeofthelattice anddoes notcontribute much more than afewpercent tothemagnetism. Asamatter of fact, thesaturation magnetic ﬁeld thatonegetstaking Mm =Nuandusing the density ofiron of7.9andthemoment i.iofthespinning electron isabout 20,000 gauss. Butaccording toexperiment, itisactually intheneighborhood of21,500 gauss. This isatypical magnitude oferror—5 or10percent-due toneglecting thecontributions oftheorbital moments thathave notbeen included inmaking theanalysis. Thus, aslight discrepancy with thegyromagnetic measurements is quite understandable. 37-3 Thehysteresis curve Wehave concluded from ourtheoretical analysis thataferromagnetic material should spontaneously become magnetized below acertain temperature sothat allthemagnetism would beinthesame direction. Butweknow thatthisisnottrue foranordinary piece ofunmagnetized iron. Why isn’t alliron magnetized? We canexplain itwith thehelp ofFig.37-4. Suppose theiron were allabigsingle crystal oftheshape shown inFig.37-4(a) andspontaneously magnetized allinone direction. Then there would beaconsiderable external magnetic ﬁeld, which would have alotofenergy. Wecanreduce thatﬁeld energy ifwearrange thatonesideof 37-5//// /// ////// ELECTRON SPINS L/r Fig. 37-3. When themagnetization ofabarofiron isreversed, thebaris given some angular velocity. _- <_ <—-1.4-1./Q /1/\/‘\{gnu} ivisi ~' ssss .s@JN/N (=1) (b) (C) -1»I»,. _,_ZU,I 1 Q2)- Q‘)-:- /<,_.\ (d) (Q) Fig. 37-4. Theformation ofdomains inasingle crystal ofiron. [From Charles Kittel, Introduction toSolid State Physics, John Wiley andSons, Inc.,New York, 2nd ed.,1956.1 theblock ismagnetized “up” andtheother sidemagnetized “down,” asinFig. 37—4(b). Then, ofcourse, theﬁelds outside theironwould extend overlessvolume, sothere would belessenergy there. Ah,butwait! Inthelayer between thetworegions wehave up-spinning electrons adjacent todown-spinning electrons. Butferromagnetism appears only inthose materials forwhich theenergy isreduced iftheelectrons areparallel rather than opposite. So,wehave added some extra energy along thedotted lineinFig. 37—4(b); thisenergy issometimes called wallenergy. Aregion having only one direction ofmagnetization iscalled adomain. Attheinterface-the “wall”- between twodomains, where wehave atoms onopposite sides which arespinning indifferent directions, there isanenergy perunitarea ofthewall. Wehave de- scribed itasthough twoadjacent atoms were spinning exactly opposite, butit turns outthatnature adjusts things sothatthetransition ismore gradual. But wedon’t need toworry about such ﬁnedetails atthispoint. Now thequestion is:When isitbetter orworse tomake awall? Theanswer isthatitdepends onthesizeofthedomains. Suppose thatwewere toscale upa block sothatthewhole thing wastwice asbig. Thevolume inthespace outside ﬁlled with agiven magnetic ﬁeld strength would beeight times bigger, andthe energy inthemagnetic ﬁeld, which isproportional tothevolume, would alsobe eight times greater. Butthesurface areabetween twodomains, which willgivethe wallenergy, would beonlyfour times asbig. Therefore, ifthepiece ofironisbig enough, itwillpaytosplit itintomore domains. This iswhy only thevery tiny crystals canhave butasingle domain. Anylarge object—one more than about a hundredth ofamillimeter insize—will have atleast onedomain wall; andany ordinary, “centimeter-size” object willbesplitintomany domains, asshown inthe ﬁgure. Splitting intodomains goes onuntil theenergy needed toputinoneextra wallisaslarge astheenergy decrease inthemagnetic ﬁeld outside thecrystal. Actually nature hasdiscovered stillanother waytolower theenergy: Itisnot necessary tohave theﬁeldgooutside atall,ifalittle triangular region ismagnetized sideways, asinFig.37—4(d).* Then withthearrangement ofFigY;_37—4(d) wesee thatthere isnoexternal ﬁeld, butinstead only alittle more domain’-wall. Butthatintroduces anewkind ofproblem. Itturns outthatwhen asingle crystal ofironismagnetized, itchanges itslength inthedirection ofmagnetization, soan“ideal” cube with itsmagnetization, say,“up,” isnolonger aperfect cube. The“vertical” dimension willbedifferent from the“horizontal” dimension. This eﬂect iscalled magnetostriction. Because ofsuch geometric changes, thelittle triangular pieces ofFig.37-4(d) donot,sotospeak, “ﬁt” intotheavailable space anymore—the crystal hasgottoolong onewayandtooshort theother way. Of course, itdoes ﬁt,really, butonly bybeing squashed in;andthisinvolves some mechanical stresses. So,thisarrangement also introduces anextra energy. It isthebalance ofallthese various energies which determines how thedomains ﬁnally arrange themselves intheir complicated fashion inapiece ofunmagnetized iron. Now, what happens when weputonanexternal magnetic ﬁeld? Totakea simple case, consider acrystal whose domains areasshown inFig.37-4(d). If weapply anexternal magnetic ﬁeld intheupward direction, inwhat manner does thecrystal become magnetized? First, themiddle domain wall canmove over sideways (totheright) andreduce theenergy. Itmoves oversothattheregion which is“up” becomes bigger than theregion which is“down”. There aremore elemen- tarymagnets lined upwith theﬁeld, andthisgives alower energy. So,forapiece ofironinweak ﬁelds—at thevery beginning ofmagnetization—the domain walls begin tomove andeatintotheregions which aremagnetized opposite totheﬁeld. Astheﬁeld continues toincrease, awhole crystal shifts gradually into asingle *You may bewondering how spins that have tobeeither “up” or“down” canalso be“sideways”! That’s agood question. butwewon’t worry about itright now. We’ll simply adopt theclassical point ofview, thinking oftheatomic magnets asclassical dipoles which canbepolarized sideways. Quantum mechanics requires considerable expertness tounderstand how things canbequantized both “up-and-down,” and“right- and-left,” allatthesame time. 37-6 large domain which theexternal ﬁeld helps tokeep lined up.Inastrong ﬁeld the crystal “likes” tobeallonewayjustbecause itsenergy intheapplied ﬁeldisreduced -itisnolonger merely thecrystal’s ownexternal ﬁeld which matters. What ifthegeometry isnotsosimple? What iftheaxes ofthecrystal andits spontaneous magnetization areinonedirection, butweapply themagnetic ﬁeld insome other directi'on—say at45°? Wemight think thatdomains would reform themselves with their magnetization parallel totheﬁeld, andthen asbefore, they could allgrow into onedomain. Butthisisnoteasy fortheiron todo,forthe energy needed tomagnetize acrystal depends onthedirection ofmagnetization relative tothecrystal axis. Itisrelatively easy tomagnetize iron inadirection parallel tothecrystal axes, butittakes more energy tomagnetize itinsome other direction—like 45°with respect tooneoftheaxes. Therefore, ifweapply amag- netic ﬁeld insuch adirection, what happens ﬁrstisthatthedomains which point along oneofthepreferred directions which isnear totheapplied ﬁeld grow until themagnetization isallalong oneofthese directions. Then withmuch stronger ﬁelds. themagnetization isgradually pulled around parallel totheﬁeld, assketched inFig.37-5. InFig. 37-6 areshown some observations ofthemagnetization curves of single crystals ofiron. Tounderstand them, wemust ﬁrstexplain something about thenotation that isused indescribing directions inacrystal. There aremany ways inwhich acrystal canbesliced soastoproduce afacewhich isaplane of atoms. Everyone who hasdriven past anorchard orvineyard knows this—it is fascinating towatch. Ifyoulook oneway, youseelines oftrees——if youlook an- other way, youseediﬂerent lines oftrees, andsoon.Inasimilar way, acrystal hasdeﬁnite families ofplanes that hold many atoms, andtheplanes have this important characteristic (weconsider acubic crystal tomake iteasier): Ifwe observe where theplanes intersect thethree coordinate axes—we ﬁnd that the reciprocals ofthethree distances from theorigin areintheratio ofsimple whole numbers. These three whole numbers aretaken asthedeﬁnition oftheplanes. Forexample, inFig.37-7(a), aplane parallel totheyz-plane isshown. This is called a[100] plane; thereciprocals ofitsintersection ofthey-andz-axes areboth zero. Thedirection perpendicular tosuch aplane (inacubic crystal) isgiven the same setofnumbers. Itiseasy tounderstand theideainacubic crystal, forthen theindices [100] mean avector which hasaunitcomponent inthex-direction and none inthey-orz—directions. The[110] direction isinadirection 45°from the x-andy-axes, asinFig.37-7(b); andthe[111]direction isinthedirection ofthe cube diagonal, asinFig.37-7(c). 1800 W -'' -;g"v'.4 M0: i1 ll j /0 _-// M00,’ r 1200‘: ' M1000/ 1r ll 4 800‘ ooo‘~]‘ M H M H MH Fig. 37-5. Amagnetizing ﬁeld Hat anangle with respect tothecrystal axis willgradually change thedirection ofthe magnetization without changing itsmagni- tude. ;i Fig. 37-6. Thecomponent ofMpar- wofl allel toH,fordifferent directions ofH mall — (with respect tothecrystal axes). [From . F.Bitter, introduction toFerromagnetism, °<>inwewv~00w==,,w<> MnoanrowMcGraw-Hill BookCo.,Inc.,1937.] Returning now toFig. 37-6, weseethemagnetization curves ofasingle crystal ofironforvarious directions. First, note thatforvery tinyﬁelds—so weak thatitishard toseethem onthescale atall-the magnetization increases extremely rapidly toquite large values. Iftheﬁeld isinthe[100] direction—namely along oneofthose nice, easy directions ofmagnetizati0n—the curve goes uptoahigh value, curves around alittle, andthen issaturated. What happened isthat the 37-7 ii 1 1‘V [I001 to) (D) (C) 1 -—7 t IOOPLANE / 7 1 Z , Fig 37-7 Theway thecrystal planes arelabeled Fig. 37-8. Magnetization curves for single crystals ofiron, nickel, and cobalt. [From Charles Kittel, Introduction toSolid State Physics, John Wiley and Sons, Inc., New York, 2nded.,1956.]0 ~ domains which were already there arevery easily removed. Only asmall ﬁeldis required tomake thedomain walls move andeatupallofthe“wrong-way” domains. Single crystals ofiron areenormously permeable (magnetic sense), much more sothan ordinary polycrystalline iron. Aperfect crystal magnetizes extremely easily. Why isitcurved atall?Why doesn’t itjust goright uptosatura- tion? Wearenotsure. Youmight study thatsome day. Wedounderstand whyit isﬂatforhighﬁelds. When thewhole block isasingle domain, theextra magnetic ﬁeldcannot make anymore magnetization—it isalready atM,,,,,, with alltheelec- trons lines up. Now, ifwetrytodothesame thing inthe[110] direction-which isat45° tothecrystal axes—what willhappen? Weturn onalittle bitofﬁeld andthe magnetization leaps upasthedomains grow. Then asweincrease theﬁeldsome more, weﬁndthatittakes quite alotofﬁeldtogetuptosaturation, because nowthemagnetization isturning away from an“easy” direction. Ifthisexplanation iscorrect, thepoint atwhich the[l10]curve extrapolates back tothevertical axis should beat1/\/2 ofthesaturation value. Itturns out,infact, tobevery, very close to1/\/2. Similarly, inthe[111] direction—which isalong thecube diagonal —we ﬁnd, aswewould expect, that thecurve extrapolates back tonearly 1/\/3 ofsaturation. Figure 37-8 shows thecorresponding situation fortwoother materials, nickel andcobalt. Nickel isdifferent from iron. Innickel, itturns outthatthe[111] direction istheeasy direction ofmagnetization. Cobalt hasahexagonal crystal form, andpeople have botched upthesystem ofnomenclature forthiscase. They want tohave three axes onthebottom ofthehexagon andoneperpendicular to these, sothey have used four indices. The[0001] direction isthedirection ofthe axisofthehexagon, and[1010] isperpendicular tothataxis. Weseethatcrystals ofdifferent metals behave indifferent ways. Now wemust discuss apolycrystalline material, such asanordinary piece of iron. Inside such materials there aremany, many little crystals with their crystal- lineaxespointing every which way. These arenotthesame asdomains. Remember thatthedomains were allpart ofasingle crystal, butinapiece ofironthere are A ﬁ 0 200 400 600 0 100 200 H(qaussl _>M/‘urea ..slsIl..ae_§_eeIIl\§_!!!_an§§“,t1a||\_§IBlll§,IIss_|,II|uc2QCIUSS§E RE20,, 37-8 many dzﬂerent crystals with axes atdifferent orientations, asshown inFig.37—9. Within each ofthese crystals, there willalsogenerally besome domains. When weapply asmall magnetic ﬁeldtoapiece ofpolycrystalline material, what happens isthatthedomain walls begin tomove, andthedomains which have afavorable direction ofeasy magnetization grow larger. This growth isreversible solong as theﬁeld stays very small—if weturn theﬁeld off,themagnetization willreturn to zero. This partofthemagnetization curve ismarked ainFig.37-10. Forlarger ﬁelds—-in theregion bofthemagnetization curve shown—things getmuch more complicated. Inevery small crystal ofthematerial, there arestrains anddislocations; there areimpurities, dirt, andimperfections. And atallbutthe smallest ﬁelds, thedomain wall, inmoving, getsstuck onthese. There isaninter- action energy between thedomain wall andadislocation, oragrain boundary. oranimpurity. Sowhen thewallgetstooneofthem, itgetsstuck; itsticks there atacertain ﬁeld. Butthen iftheﬁeld israised some more, thewallsuddenly snaps past. Sothemotion ofthedomain wall isnotsmooth thewayitisinaperfect crystal—it getshung upevery once inawhile andmoves injerks. Ifwewere to look atthemagnetization onamicroscopic scale, wewould seesomething likethe insert ofFig.37—lO. Now theimportant thing isthatthese jerks inthemagnetization cancause an energy loss. Intheﬁrstplace, when aboundary ﬁnally slips pastanimpediment, itmoves very quickly tothenext one, since theﬁeld isalready above what would berequired fortheunimpeded motion. Therapid motion means that there are rapidly changing magnetic ﬁelds which produce eddy currents inthecrystal. These currents loose energy inheating themetal. Asecond effect isthatwhen adomain suddenly changes, part ofthecrystal changes itsdimensions from themagneto- striction. Each sudden shift ofadomain wallsetsupalittle sound wave thatcarries away energy. Because ofsuch effects, thesecond part ofmagnetization curve isirreversible, andthere isenergy being lost. This istheorigin ofthehysteresis effect, because tomove aboundary wallforward—snap—and then tomove itback- ward———snap—produces adifferent result. lt’slike“jerky” friction, andittakes energy. Eventually, forhighenough ﬁelds, when wehave moved allthedomain walls andmagnetized each crystal initsbestdirection, there arestillsome crystallites which happen tohave their easy directions ofmagnetization notinthedirection ofourexternal magnetic ﬁeld. Then ittakes alotofextra ﬁeld toturn those magnetic moments around. Sothemagnetization increases slowly, butsmoothly, forhigh ﬁelds—namely intheregion marked cintheﬁgure. Themagnetization does notcome sharply toitssaturation value, because inthelastpartofthecurve theatomic magnets areturning inthestrong ﬁeld. Soweseewhythemagnetization curve ofanordinary polycrystalline materials, such astheoneshown inFig.37-10, rises alittle bitandreversibly atﬁrst, then rises irreversibly, andthen curves over slowly. Ofcourse, there isnosharp break-point between thethree regions~they blend smoothly, oneintotheother. Itisnothard toshow thatthemagnetization process inthemiddle partofthe magnetization curve isjerky—that thedomain walls jerk andsnap asthey shift Allyouneed isacoilofwire—with many thousands ofturns—connected toan ampliﬁer andaloudspeaker, asshown inFig.37-11. Ifyouputafewsilicon steel sheets (ofthetype used intransformers) atthecenter ofthecoilandbring abar magnet slowly near thestack, thesudden changes inmagnetization willproduce impulses ofemfinthecoil, which areheard asdistinct clicks intheloudspeaker. Asyoumove themagnet nearer totheironyouwillhear awhole rush ofclicks that sound something likethenoise ofsand grains falling over each other asa canofsand istilted. Thedomain walls arejumping, snapping, andjiggling asthe ﬁeld isincreased. This phenomenon iscalled theBarkhausen eﬂecz. Asyoumove themagnet even closer totheironsheets, thenoise grows louder andlouder forawhile butthen there isrelatively little noise when themagnet gets veryclose. Why? Because nearly allthedomain walls have moved asfarasthey cango. Any greater ﬁeld ismerely turning themagnetization ineach domain, which isasmooth process. 37-9Q’70$ it40"§'_e-“ft\ ‘_ \ ‘ 1“, € \\\ ,li1-»\ Fig. 37—9. The microscopic structure ofanunmagnetized ferromagnetic ma- terial. Each crystal grain hasaneasy direction ofmagnetization and isbroken upintodomains which arespontaneously magnetized (usually) parallel tothis direction. 8 c b __a_____. H Fig. 37—lQ. The magnetization curve forpolycrystalline iron. stucou°°"- steer STRIP ilfﬂ €AfﬁfT - MOTION AMPLIFIER I SPEAKER Fig. 37—ll. The sudden changes in the magnetization ofthe steel strip are heard asclicks intheloudspeaker.$4((‘D)‘\\ B tgauss)l ~|5,ooo Brio,ooo _He *s,ooo II ll I IIt> -eoo -400 o 400 aoo H 1 (gauss) Fig 37—l2. The hysteresis curve of Alnico V.Ifyounowwithdraw themagnet, soastocome back onthedownward branch ofthehysteresis loop, thedomains alltrytogetback tolowenergy again, andyou hear another rush ofbackward-going jerks. You canalsonote thatifyoubring themagnet toagiven place andmove itback andforth alittle bit,there isrelatively little noise. Itisagain liketilting acanofsand—once thegrains shift intoplace, small movements ofthecandon’t disturb them. intheironthesmall variations inthemagnetic ﬁeld aren’t enough tomove anyboundaries over anyofthe “humps.” 37-4 Ferromagnetic materials Now wewould liketotalkabout thevarious kinds ofmagnetic materials that there areinthetechnical World andtoconsider some oftheproblems involved in designing magnetic materials fordifferent purposes. First, theterm “the magnetic properties ofiron," which oneoften hears, isamisnomer—there isnosuch thing. “Iron” isnotawell-deﬁned niaterial———the properties ofirondepend critically on theamount ofimpurities andalsoonhowtheironisformed. You canappreciate thatthemagnetic properties willdepend onhoweasily thedomain walls move and thatthislSagross property, notaproperty oftheindividual atoms. Sopractical ferromagnetism isnotreally aproperty ofanironat0m—it isaproperty ofsolid iron inacertain form. For example, iron can take ontwo different crystalline forms. Thecommon form hasabody-centered cubic lattice, butitcanalsohave aface-centered cubic lattice, which is,however, stable only attemperatures above ll00°C. Ofcourse, atthat temperature thebody-centered cubic structure is already past theCurie point. However, byalloying chromium andnickel with theiron (one possible mixture is18percent chromium and8percent nickel) we cangetwhat iscalled stainless steel, which, although itismainly iron, retains the face-centered lattice even atlowtemperatures. Because itscrystal structure is different, ithascompletely different magnetic properties. Most kindsTIT‘tainless steel arenotmagnetic toanyappreciable degree, although there aresome kinds which aresomewhat magnetic—it depends onthecomposition ofthealloy. Even when such analloy ismagnetic, itisnotferromagnetic likeordinary iron—even though itismostly justiron. Wewould likenowtodescribe afewofthespecial materials which have been developed fortheir particular magnetic properties. First, ifwewant tomake a permanent magnet, wewould likematerial with anenormously wide hysteresis loop sothat, when weturn thecurrent offandcome down tozero magnetizing ﬁeld, themagnetization willremain large. Forsuch materials thedomain bounda- riesshould be“frozen” inplace asmuch aspossible Onesuch material isthere- markable alloy “Alnico V”(51% Fe,8%Al,l4‘/f, Ni,24% Co,3‘/OCu). (The rather complex composition ofthisalloy isindicative ofthekind ofdetailed effort thathasgone intomaking good magnets. What patience ittakes tomixﬁvethings together andtestthem until youﬁndthemost ideal substance!) When Alnico solidiﬁes, there isa“second phase" which precipitates out,making many tinygrains andvery high internal strains. Inthismaterial, thedomain boundaries have a hard time moving atall. Inaddition tohaving aprecise composition, Alnico is mechanically “worked” inaway that makes thecrystals appear intheform of long grains along thedirection inwhich themagnetization isgoing tobe. Then themagnetization willhave anatural tendency tobelined upinthese directions andwillbeheld there from theanisotropic effects. Furthermore, thematerial is even cooled inanexternal magnetic ﬁeldwhen itismanufactured, sothatthegrains willgrow with theright crystal orientation. Thehysteresis loop ofAlnico Vis shown inFig37-12. You seethatitisabout 500times wider than thehysteresis curve forsoftironthatweshowed inthelastchapter inFig.36-8. Let’s turnnow toadifferent kind ofmaterial. Forbuilding transformers and motors, wewant amaterial which ismagnetically “soft"—one inwhich theiiiag— netism iseasily changed sothat anenormous amount ofmagnetization results from avery small applied ﬁeld. Toarrange this, weneed pure, well-annealed material which willhave very fewdislocations andimpurities sothatthedomain 37-10 walls canmove easily. Itwould also benice ifwecould make theanisotropy small. Then, even ifagrain ofthematerial sitsatthewrong angle with respect to theﬁeld, itwillstillmagnetize easily. Now wehave saidthatironprefers tomag- netize along the[lO0] direction, whereas nickel prefers the[lll]direction; soif wemixiron andnickel invarious proportions, wemight hope toﬁndthat with justtheright proportions thealloy wouldn't prefer anydirection—the [100] and [ll1]directions would beequivalent. Itturns outthatthishappens with amixture of70percent nickel and30percent iron. lnaddition—possibly byluck ormaybe because ofsome physical relationship between theanisotropy andthemagneto- striction effects—~it turns outthatthemagiietostriction ofiron andnickel hasthe opposite sign. And inanalloy ofthetwometals, thisproperty goes through zero atabout 80percent nickel. Sosomewhere between 70and80percent nickel weget very“soft“ magnetic materials—alloys thatarevery easy tomagnetize. They are called theperiiialloys. Perinalloys areuseful forhigh-quality transformers (atlow signal levels), butthey would benogood atallforpermanent magnets. Perm- alloys must bevery carefully made andhandled. Themagnetic properties ofa piece ofpernialloy aredrastically changed ifitisstressed beyond itselastic limit——it niustn’t bebent. Then. itspermeability isreduced because ofthedislocations, slip bands, andsoon,which areproduced bythemechanical deformations. The domain boundaries arenolonger easy tomove. Thehigh permeability can, how- ever, berestored byannealing athigh temperatures. Itisoften convenient tohave some numbers tocharacterize thevarious magnetic materials. Two useful numbers aretheintercepts ofthehysteresis loop withtheB-andH-axes, asindicated inFig.37—l2. These intercepts arecalled the remanent magnetic field B,andthecoercive force H,.InTable 37—l welistthese numbers forafewmagnetic materials. 11)l1rjlTable 37-1 Properties ofsome ferromagnetic materials Material Supermallov Silicon steel (transformer) Armco iron Alnico VB7 Residual magnetic ﬁeld (gauss) (~5000) 12,000 4()t)U i3p00 Fig. 37—l3. Relative orientation of ferrite, (d)yttrium-iron alloy. (Broken arrows show direction oftotal angular+ lelectron spins invarious materials: (a) 1 1 | |ferromagnetic, (b)antiferromagnetic, (cl I l (cl (dl momentum, including orbital motion.) 37-5 Extraordinary magnetic materials Wewould now liketodiscuss some ofthemore exotic magnetic materials. There aremany elements intheperiodic table which have incomplete inner electron shells andhence have atomic magnetic moments Forinstance, right next tothe ferromagnetic elements iron, nickel, andcobalt youwillﬁndchromium andmanga- nese. Why aren’t i/ieyferromagnetic‘? Theanswer isthatthe)\term inEq.(37.1) hastheopposite .SIgI1forthese elements. Inthechromium lattice, forexample, the spins ofthechromium atoms alternate atom byatom, asshown inFig.37—l3(b). Sochromium is“magnetic” from itsown point ofview, butitisnottechnically interesting because there arenoexternal magnetic effects. Chromium, then, isan example ofamaterial inwhich quantum mechanical effects make thespins alter- nate. Such amaterial iscalled anziferromagneiic. Thealignment inantiferromag- netic materials isalso temperature dependent. Below acritical temperature, all thespins arelined upinthealternating array, butwhen thematerial isheated above acertain temperature—which isagain called theCurie temperature——the spins suddenly become random. There is,internally, asudden transition. This transition canbeseeninthespeciﬁc heatcurve. Also itshows upinsome special “magnetic" effects. Forinstance, theexistence ofthealternating spins canbeveriﬁed byscatter- ingneutrons from acrystal ofchromium. Because aneutron itself hasaspin 37-11Hr Coercive force (gauss) O004 O05 06 550. "W -<2Mgu 1 O Mu Fig. 37-l4. Crystal structure ofthe mineral spinel (MgA|;O,,); theMg” ions occupy tetrahedral sites, each surrounded byfouroxygen ions; theA|+3 ionsoccupy octahedral sites, each surrounded bysix oxygen ions. [From Charles Kittel, Intro- duction toSo/id State Physics, John Wiley and Sons, lnc,New York, 2nd ed., l956]- V)(and amagnetic moment), ithasadifferent amplitude tobescattered, depending on whether itsspinisparallel oropposite tothespinofthescatterer. Thus, wegeta different interference pattern when thespins inacrystal arealternating thanwe dowhen they have arandom distribution There isanother kind ofsubstance inwhich quantum mechanical effects make theelectron spins alternate, butwhich isnevertheless ferr0magnetic—that is,the crystal hasapermanent netmagnetization. The idea behind such materials is shown inFig.37-l4. Theﬁgure shows thecrystal structure ofspinel, aiiiagnesium- aluminum oxide, Which—as itlSshown—is notmagnetic. Theoxide hastwokinds ofmetal atoms: magnesium andaluminum. Now ifwereplace themagnesium andthealuminum bytwomagnetic elements likeiron andzinc, orbyzincand manganese—in other words, ifweputinmagnetic atoms instead ofthe nonmagnetic ones—an interesting thing happens. Let’s callonekind ofmetal atom aandthe other kind ofmetal atom b;then thefollowing combination offorces iii%"§qbe considered. There isana-binteraction which tries tomake theaatoms andthe hatoms have opposite spins~because quantum mechanics always gives theoppo- sitesign (except forthemysterious crystals ofiron, nickel, andcobalt). Then, there isadirect a-ainteraction which tries tomake thea’sopposite, andalsoa b-binteraction which tries tomake theb’sopposite. Now, ofcourse wecannot have everything opposite everything else—a opposite b,aopposite a,andhop- posite bPresumably because ofthedistances between thea’sandthepresence of theoxygen (although wereally don’t know why), itturns outthatthea-binter- action isstronger than thea-aortheb-b. Sothesolution thatnature usesinthis caseistomake allthea'sparallel toeach other. andalltheb’sparallel toear/iother, butthetwosystems opposite That gives thelowest energy because ofthestronger a-binteraction. Theresult: allthea’sarespinning upandalltheb’sarespinning down—or viceversa, ofcourse. Butifthemagnetic moments oftheii-type atom andtheb-type atom arenotequal, wecangetthesituation shown inFig.37-l3(c), andthere canbeanetmagnetization inthematerial. Thematerial willthen be ferroinagnetic—although somewhat weak Such materials arecalled ferrites. They donothave ashigh asaturation magnetization asiron—for obvious reasons —so they areonly useful forsmaller ﬁelds. Buttheyhave avery important differ- ence—they areinsulators; theferrites areferromagnetic insulators. Inhigh- frequency ﬁelds, they willhave very small eddy currents andsocanbeused, for example, inmicrowave systems. Themicrowave ﬁelds willbeable togetinside such aninsulating material, whereas they would bekept outbytheeddy currents inaconductor likeiron. There isanother class ofmagnetic materials which hasonly recently been discovered—inembers ofthefamily oftheorthosilicates called garnets. They are again crystals inwhich thelattice contains twokinds ofmetallic atoms, andwe have again asituation inwhich twokinds ofatoms canbesubstituted almost at Wlll. Among themany compounds ofinterest there isonewhich iscompletely ferromagnetic. Ithasyttrium andironinthegarnet structure, andthereason itis ferromagnetic isvery curious. Here again quantum mechanics ismaking the neighboring spins opposite, sothat there isalocked-in system ofspins with the electron spins oftheirononewayandtheelectron spins oftheyttrium theopposite way Buttheyttrium atom iscomplicated. ltisarare-earth element andgetsa large contribution toitsmagnetic moment from orbital motion oftheelectrons. Foryttrium, theorbital motion contribution is0])[)()S'll€ thatofthespinandalso isbigger. Thus, although quantum mechanics, working through theexclusion principle, makes thespins‘ oftheyttrium opposite those oftheiron, itmakes the total magnetic moment oftheyttiiuin atom parallel totheiron because ofthe orbital effect—as sketched inFig.37-l3(d) Thecompound istherefore aregular ferromagnet. Another interesting example offerromagnetism occurs insome oftherare- earth elements. Ithastodowith astillmore peculiar arrangement ofthespins. Thematerial isnotferromagnetic inthesense thatthespins areallparallel. noris itantiferroiiiagnetic inthesense thatevery atom isopposite. lnthese crystals all ofthespins inonelayer areparallel andlieintheplane ofthelayer. lnthenext 37-12 layer allspins areagain parallel toeach other, butpoint inasomewhat different direction. lnthefollowing layer they areinstillanother direction, andsoon. Theresult isthatthelocal magnetization vector varies intheform ofaspiral—the magnetic moments ofthesuccessive layers rotate asweproceed along aline perpendicular tothelayers. Itisinteresting totrytoanalyze what happens when a ﬁeld isapplied tosuch aspiral—all thetwistings andturnings thatmust goonin allthose atomic magnets. (Some people liketoamuse themselves with thetheory ofthese things!) Notonly arethere cases of“ﬁat” spirals, butthere arealsocases inwhich thedirections themagnetic moments ofsuccessive layers map outacone, sothat ithasaspiral component andalso auniform ferromagnetic component inonedirection! Themagnetic properties ofmaterials, worked outonamore advanced level than wehave been abletodohere, have fascinated physicists ofallkinds. lnthe ﬁrstplace, there arethose practical people who lovetowork outways ofmaking things inabetter way—they love todesign better andmore interesting magnetic materials. Thediscovery ofthings likeferrites, ortheir application, immediately delights people who liketoseeclever new ways ofdoing things. Besides this, there arethose who ﬁndafascination intheterrible complexity thatnature can produce using afewbasic laws. Starting with oneandthesame general idea, nature goes from theferromagnetism ofiron anditsdomains, totheantiferro- magnetism ofchromium, tothemagnetism offerrites andgarnets, tothespiral structure oftherareearth elements, andon,andon.Itisfascinating todiscover experimentally allthestrange things thatgooninthese special substances. Then, tothetheoretical physicists, ferromagnetism presents anumber ofveryinteresting, unsolved, andbeautiful challenges. Onechallenge istounderstand whyitexists atall.Another istopredict thestatistics oftheinteracting spins inanideal lattice. Even neglecting anypossible extraneous complications, thisproblem has,sofar, deﬁed fullunderstanding. Thereason thatitissointeresting isthatitissuch an easily stated problem: Given alotofelectron spins inaregular lattice, interacting withsuch-and-such alaw,what dotheydo? Itissimply stated, butithasdeﬁed complete analysis foryears. Although ithasbeen analyzed rather carefully for temperatures nottooclose totheCurie point, thetheory ofthesudden transition theCurie point stillneeds tobecompleted. Finally, thewhole subject ofthesystem ofspinning atomic magnets—in ferromagnetic, orinparamagnetic materials andinnuclear magnetism, hasalso been afascinating thing toadvanced students inphysics. Thesystem ofspins can bepushed onandpulled onwith external magnetic ﬁelds, soonecandomany tricks with resonances, with relaxation effects, with spin-echoes, andwith other effects. Itserves asaprototype ofmany complicated thermodynamic systems. Butinparamagnetic materials thesituation isoften fairly simple, and people have been delighted both todoexperiments andtoexplain thephenomena theo- retically. Wenow close ourstudy ofelectricity andmagnetism. Intheﬁrstchapter, wespoke ofthegreat strides thathave been made since theearly Greek observation ofthestrange behaviors ofamber andoflodestone. Yetinallourlong andin- volved discussion wehave never explained whyiiisthatwhen werubapiece of amber wegetacharge onit,norhave weexplained whyalodestone ismagnetized’ You may say, “Oh, wejustdidn’t gettheright sign.” No,itisworse than that Even ifwedidgettheright sign, wewould stillhave thequestion: Why isthepiece oflodestone intheground magnetized? There istheearth’s magnetic ﬁeld, of course, butwhere does theearth’s ﬁeld come from? Nobody really knows—there have only been some good guesses. Soyousee,thisphysics ofours isalotof fakery-we start outwith thephenomena oflodestone andamber, andweendup notunderstanding either ofthem very well. Butwehave learned atremendous amount ofvery exciting andvery practical information intheprocess! 37-13 38 Elasticity 38-1 Hooke’s law Thesubject ofelasticity deals with thebehavior ofthose substances which have theproperty ofrecovering their sizeandshape when theforces producing deformations areremoved. Weﬁnd thiselastic property tosome extent inall solid bodies. Ifwehadthetime todealwith thesubject atlength, wewould want tolook intomany things: thebehavior ofmaterials, thegeneral laws ofelasticity, thegeneral theory ofelasticity, theatomic machinery thatdetermine theelastic properties, andﬁnally thelimitations ofelastic laws when theforces become so great thatplastic ﬂow andfracture occur. Itwould take more time than wehave tocover allthese subjects indetail, sowewillhave toleave outsome things Forexample, wewillnotdiscuss plasticity orthelimitations oftheelastic laws. (Wetouched onthese subjects brieﬂy when wewere talking about dislocations in metals.) Also, wewillnotbeabletodiscuss theinternal mechanisms ofelasticity- soourtreatment willnothave thecompleteness wehave tried toachieve inthe earlier chapters. Ouraimismainly togiveyouanacquaintance with some ofthe ways ofdealing with such practical problems asthebending ofbeams. When youpush onapiece ofmaterial, it“gives”—the material isdeformed. Iftheforce issmall enough, therelative displacements ofthevarious points inthe material areproportional totheforce—we saythebehavior iselastic. Wewill discuss only theelastic behavior. First, wewillwrite down thefundamental laws ofelasticity, andthen wewillapply them toanumber ofdifferent situations Suppose wetake arectangular block ofmaterial oflength l,width w,and height h,asshown inFig.38-l. Ifwepullontheends with aforce F,then the length increases byanamount Al.Wewillsuppose inallcases thatthechange in length isasmall fraction oftheoriginal length. Asamatter offact, formaterials likewood andsteel, thematerial willbreak ifthechange inlength ismore than a fewpercent oftheoriginal length. Foralarge number ofmaterials, experiments show thatforsufficiently small extensions theforce isproportional totheextension FocAl. (38.1) Thisrelation isknown asHooke’s law. Thelengthening Alofthebarwillalsodepend onitslength. Wecanﬁgure out howbythefollowing argument. Ifwecement twoidentical blocks together, end toend.thesame forces actoneach block, each willstretch byAl.Thus, thestretch ofablock oflength 21would betwice asbigasablock ofthesame cross section. butoflength l.Inorder togetanumber more characteristic ofthematerial, and lessofanyparticular shape, wechoose todealwith theratio Al/loftheextension totheoriginal length. This ratio isproportional totheforce butindependent ofl: F0(#. (38.2) Theforce Fwillalsodepend onthearea oftheblock. Suppose thatweput twoblocks sidebyside. Then foragiven stretch Alwewould have theforce F oneach block, ortwice asmuch onthecombination ofthetwoblocks. Theforce, foragiven amount ofstretch, must beproportional tothecross-sectional area A oftheblock. Toobtain alawinwhich thecoefficient ofproportionality isinde- pendent ofthedimensions ofthebody, wewrite Hooke’s lawforarectangular 38-138-1 Hooke’s law 38-2 Uniform strains 38-3 Thetorsion bar;shear waves 38-4 The bent beam 38-5 Buckling Review: Chapter 47,Vol. I,Sound; theWave Equation. M 5F F ___--~_--~-- -AREAA p (+b!::l*l Fig. 38-1. The stretching ofabar under uniform tension. P I’ Pblock intheform F=YA (38.3) Theconstant Yisaproperty only ofthenature ofthematerial; itisknown as Young’s modulus. (Usually youwillseeYoung’s modulus called E.ButWe’ve used Eforelectric ﬁelds, energy, andemf’s,soweprefer touseadifferent letter.) Theforce perunitarea iscalled thestress, andthestretch perunitlength—the fractional stretch—is called thestrain. Equation (38.3) cantherefore berewritten inthefollowing way: F AlA_YX-I-, (38.4) Stress =(Young’s modulus) X(Strain). There isanother part toHooke’s law: When youstretch ablock ofmaterial inonedirection itcontracts atright angles tothestretch. The contraction in width isproportional tothewidth wandalsotoAl/l. Thesideways contraction is inthesame proportion forboth width andheight, andisusually written /T /T /F /( g :Q;=_,7Bil, (335) P Fig. 38-2. Abar under uniform hydrostatic pressure. I F| F2 F2 Fig 38-3. Hydrostatic pressure is the superposition ofthree longitudinal compressions.where theconstant 0isanother property ofthematerial called Poisson’s ratio. Itis always positive insignandisanumber lessthan l/2. (Itis“reasonable” thato' should begenerally positive, butitisnotquite clear thatitmust beso.) Thetwoconstants Yandaspecify completely theelastic properties ofaho- mogeneous’ isotropic (that is,noncrystalline) material. Incrystalline materials the stretches andcontractions canbedifferent indifferent directions. sothere canbe many more elastic constants. Wewillrestrict ourdiscussion temporarily tohomo- geneous’ isotropic materials whose properties canbedescribed byYanda.Asusual there aredifferent ways ofdescribing things—some people liketodescribe the elastic properties ofmaterials bydifferent constants. Italways takes two, and theycanberelated to0andY. Thelastgeneral lawweneed istheprinciple ofsuperposition. Since thetwo laws (384)and(38.5) arelinear intheforces andinthedisplacements, superposition willwork. Ifyouhave onesetofforces andgetsome displacements, andthen youaddanewsetofforces andgetsome additional displacements, theresulting displacements willbethesumoftheones youwould getwith thetwosetsofforces acting independently. Now wehave allthegeneral principles-the superposition principle andEqs. (38.4) and(38.5)-and that’s allthere istoelasticity. Butthatislikesaying that once youhave Newton’s laws that’s allthere istomechanics. Or,given Maxwell’s equations, that’s allthere istoelectricity. Itis,ofcourse, true thatwith these principles youhave agreat deal, because with your present mathematical ability youcould goalong way. Wewill, however, work outafewspecial applications. 38-2 Uniform strains Asourﬁrstexample let'sﬁndoutwhat happens toarectangular block under uniform hydrostatic pressure Let's putablock under water inapressure tank. Then there willbeaforce acting inward onevery faceoftheblock proportional tothearea (seeFig.38-2). Since thehydrostatic pressure isuniform, thestress (force perunitarea) oneach faceoftheblock isthesame. Wewillwork outﬁrst thechange inthelength. Thechange inlength oftheblock canbethought ofas thesumofchanges inlength thatwould occur inthethree independent problems which aresketched inFig.38-3. 38-2 Problem I.Ifwepush ontheends oftheblock with apressure p,thecom- pressional strain isp/Y,anditisnegative, %=_el Y Problem 2.Ifwepush onthetwosides oftheblock with pressure p,thecom- pressional strain isagain p/Y,butnow wewant thelengthwise strain. Wecanget thatfrom thesideways strain multiplied by—o. Thesideways strain is A_W__£-w— Y’ SO Al2_ 2 T‘+"Y' Problem 3.Ifwepush onthetopoftheblock, thecompressional strain is once more p/Y,andthecorresponding strain inthesideways direction isagain -op/ Y.Weget A13 __ P 7'-+”r' Combining theresults ofthethree problems-—that is,taking Al=All—l- A12—l—Al3——we get . Al_ p _T_—Y(l 20). (38.6) Theproblem is,ofcourse, symmetrical inallthree directions; itfollows that Thechange inthevolume under hydrostatic pressure isalsoofsome interest. Since V=lwh,wecanwrite, forsmall displacements, AV Al Aw All 7_7+7+7' Using(38.6)and(38.7), wehave VA7 :-3 {/1(1*20') (388) Fig 38-4 Acube inuniform shear People liketocallAV/Vthevolume strain andwrite iv=—K~AV Thevolume stress pisproportional tothevolume strain—Hooke’s lawonce more. F7_''_”__‘I Thecoefﬁcient Kiscalled thebulk modulus," itisrelated totheother constants by YK-§(Ti 2U)~ (38.9) Since Kisofsome practical interest, many handbooks give YandKinstead ofY and0.Ifyouwanto' youcanalways getitfrom Eq.(38.9). Wecanalsoseefrom Eq.(38.9) that Poisson’s ratio, o,must belessthan one-half. Ifitwere not,the bulk modulus Kwould benegative, andthematerial would expand under increas- L___—TT__T ingpressure. That would allow ustogetmechanical energy outofanyoldblock— itwould mean thattheblock wasinunstable equilibrium. Ifitstarted toexpand itwould continue byitself with arelease ofenergy. i=,g_3g_5_ Awbe withwmpressmg Now wewant toconsider what happens when youputa“shear” strain on fQ['Ce5 ontQpand boftgm and equal___c____z;,93-=¥= -!1(i-20). (38.7) ﬂ% ew li Y --> F F F F something. Byshear strain wemean thekind ofdistortion shown inFig.38-4. Asa stretching forces Ontwosides. preliminary tothis, letuslook atthestrains inacube ofmaterial subjected tothe forces shown inFig.38-5. Again wecanbreak itupintotwoproblems: thevertical 38-3 FI_- G _ :l!Il|IIIIIIIm, ll'1> an\-II-1pushes, andthehorizontal pulls. Calling Athearea ofthecube face, wehavefor thechange inhorizontal length -<.--3'11Al_lF _1+aF Thechange inthevertical height isjustthenegative ofthis. \/2-G »/2-G AREA= ,/2'1: ---|||l ,‘I (01 /T6 /5'6 ~"!!!!:|“iW REAAllil!l: Q_1i_‘I’\_iiIIll'V015, Fig. 38-6. Thetwo pairs ofshecir forces in(o)produce thesome stress cs thecompressing and stretching forces oflb). AD G H. '19 | U -4?-ioﬁ:<i— { l ~/////////// /// Fig. 38-7. The shear strain 6is 2AD/D.Now suppose wehave thesame cube andsubject ittotheshearing forces shown inFig.38—6(a). Note thatalltheforces have tobeequal ifthere aretobe nonettorques andthecube istobeinequilibrium. (Similar forces must also exist inFig.38-4, since theblock isinequilibrium. They areprovided through the“glue” thatholds theblock tothetable.) Thecube isthen saidtobeinastate ofpure shear. Butnote thatifwecutthecube byaplane at45°—say along the diagonal Aintheﬁgure——the total force acting across theplane isnormal toplane andisequal to\/56. Theareaoverwhich thisforce actsis\/2A; therefore, the tensile stress normal tothisplane issimply G/A. Similarly, ifweexamine aplane atanangle of45°theother way—the diagonal Bintheﬁgure—we seethatthere isacompressional stress normal tothisplane of—G/A. From this, weseethat thestress ina“pure shear” isequivalent toacombination oftension andcom- pression stresses ofequal strength andatright angles toeach other, andat45°to theoriginal faces ofthecube Theinternal stresses andstrains arethesame as wewould ﬁndinthelarger block ofmaterial with theforces shown inFig.38—6(b). Butthisistheproblem wehave already solved. Thechange inlength ofthediagonal isgiven byEq.(38.10), AD_ l—l—aG WT? <38“) (One diagonal isshortened; theother iselongated.) Itisoften convenient toexpress ashear strain interms oftheangle bywhich thecube istwisted—the angle 6inFig.38-7. From thegeometry oftheﬁgure you canseethatthehorizontal shift 6ofthetopedge isequal to\/2’AD. So 5\/2AD AD0-7-A-l_~_ 2eD~- (38.12) Theshear stress gisdeﬁned asthetangential force ononeface divided bythe area, g=G/A. Using Eq.(38.11) in(38.12), weget 1+00=2—~—- . Y8 Or,writing thisintheform “stress =constant times strain,” g=,u0. (38.13) 38~4 Theproportionality coefﬁcient itiscalled theshear modulus (or,sometimes, the coefﬁcient ofrigidity). Itisgiven interms ofYand0by it=K%?- (38.14) Incidentally, theshear modulus must bepositive——otherwise youcould getwork outofaself-shearing block. From Eq.(38.14), 0must begreater than —1.We know, then, thata must bebetween —-land+%; inpractice, however, itisalways greater than zero. Asalastexample ofthetypeofsituation where thestresses areuniform through thematerial, let’sconsider theproblem ofablock which isstretched, while itis atthesame time constrained sothatnolateral contraction cantake place. (Tech- nically, it’salittle easier tocompress itwhile keeping thesides from bulging out— butit’sthesame problem.) What happens? Well, there must besideways forces which keep itfrom changing itsthickness-—forces wedon’t know off-hand but willhave tocalculate. It’sthesame kind ofproblem wehave already done, only with alittle different algebra. Weimagine forces onallthree sides, asshown in Fig.38-8; wecalculate thechanges indimensions, andwechoose thetransverse forces tomake thewidth andheight remain constant. Following theusual argu- ments, wegetforthethree strains: 1F, F Fz "*-_""*"_"-Yh7“kﬁ+Zﬂ’Gm” _Alt=_-__ _A]I1! -A-‘=--~--+ (38.17)..~\vii ~<-~<---<~'s.~qa.~1§"..?1M"@§Qq“<q/.\/3:s’“11>5:$5‘H"-i—l-§__q~<qan(38.16) Now since A1,,andAI,aresupposed tobezero, Eqs. (38.16) and(38.17) give twoequations relating F,,andF,toF,.Solving them together, wegetthat F11_F=aInfa. 7”_I_1 Ax (38.18)—cr Substituting in(38.15), wehave Al, 1 202 F, 1l-0--262)F, *3'7<1_iii?) 32_Y<'_i—-“¢_ A1‘ <38”) Often, youwillseethisturned around, andwith thequadratic in0factored out,it isthen written S=(T+J&7(l‘Q:~§;) 1/9,5 (38.20) When weconstrain thesides, Young’s modulus getsmultiplied byacomplicated function ofa. Asyoucanmost easily seefrom Eq.(38.19), thefactor infront of Yisalways greater than 1.Itisharder tostretch theblock when thesides are held—which also means that ablock isstronger when thesides areheld than when they arenot. 38-3 Thetorsion bar;shear waves Let’s nowturnourattention toanexample which ismore complicated because different parts ofthematerial arestressed bydifferent amounts. Weconsider a twisted rodsuch asyouwould ﬁndinadrive shaft ofsome machinery, orina quartz ﬁber suspension used inadelicate instrument. Asyouprobably know from experiments with thetorsion pendulum, thetorque onatwisted rodisproportional totheangle—the constant ofproportionality obviously depending upon the length oftherod, ontheradius oftherod,andontheproperties ofthematerial. Thequestion is:Inwhat way? Wearenow inaposition toanswer thisquestion; it’sjustamatter ofworking outsome geometry. 38-5F! ,;nQ GN1Q Fy Fig. 38-8. Stretching without lateral contraction.F: "/g_;<(((<<i,a<l“Ar I’-i'Iq"‘.. __,/i"->13i D'11Al’ (C) T Fig 38-9 aAcylindrical barintorsion. (blAcylindrical shell intorsion. (c)Each small piece oftheshell isinshear. Fig. 38-9(a) shows acylindrical rodoflength L,andradius a,with oneend twisted bytheangle ¢with respect totheother. Ifwewant torelate thestrains to what wealready know, wecanthink oftherodasbeing made upofmany cylindrical shells andwork outseparately what happens toeach shell. Westart bylooking at athin, short cylinder ofradius r(less than a)andthickness Ar—as drawn inFig. 38-9(b). Now ifwelook atapiece ofthiscylinder thatwasoriginally asmall square, weseethatithasbeen distorted intoaparallelogram. Each such element ofthecylinder isinshear, andtheshear angle 6is _£‘2.9“L Theshear stress ginthematerial is,therefore [from Eq.(38.l3)], ¢> 8'=M9=urf- (38-21) Theshear stress isthetangential force AFontheendofthesquare divided bytheareaAlAroftheend[seeFig.38-9(0)] _A5.g‘AlAr Theforce AFontheendofsuch asquare contributes atorque A-raround theaxis oftherodequal to Ar=rAF=rgAlAr. (38.22) Thetotal torque 1-isthesumofsuch torques around acomplete circumference of thecylinder. Soputting together enough pieces sothattheAl’saddupto21rr, weﬁndthatthetotal torque, forahollow tube, is rg(21i'r) Ar. (38.23) Or,using (38.21), 3 T=271'/J._~’1"’? (38.24) Wegetthattherotational stiffness, 1'/¢>, ofahollow tube isproportional tothe cube oftheradius randtothethickness Ar,andinversely proportional tothe length L. Wecannowimagine asolid rodtobemade upofaseries ofconcentric tubes, each twisted bythesame angle ¢(although theinternal stresses aredifferent for each tube). The total torque isthesum ofthetorques required torotate each shell; forthesolid rod T=Zrrujgfrddr, 38-6 where theintegral goes from r=0tor=a,theradius oftherod. Integrating, wehave 4 T=[L%¢. (38.25) Forarodintorsion, thetorque isproportional totheangle andisproportional to thefourth power ofthediameter—a rodtwice asthick issixteen times asstiff fortorsion. Before leaving thesubject oftorsion, letusapply what wehave justlearned toaninteresting problem: torsional waves. Ifyoutake along rodandsuddenly twist oneend, awave oftwist works itway along therod, assketched inFig. 38—10(a). That’s alittle more exciting than asteady twist—let’s seewhether we canwork outwhat happens. (bl —"- X~** Tl I-('r+A'r) (0) \ ,-’ ENDI ' IEND2 l Z >l l zl-t-Az Fig. 38-IO. (a)Atorsional wave onarod. (blAvolume Letzbethedistance tosome point down therod. Forastatic torsion the torque isthesame everywhere along therod,andisproportional to¢/L, thetotal torsion angle over thetotal length. What matters tothematerial isthelocal torsional strain, which is,youwillappreciate, 6¢/62. When thetorsion along the rodisnotuniform, weshould replace Eq.(38.25) by 4a1(2)=,31’;_F‘? (38.26) Now let’s look atwhat happens toanelement oflength Azshown magniﬁed in Fig.38—l0(b). There isatorque 1-(z)atend1ofthelittle hunk ofrod,andadiffer- enttorque 'r(z+Az)atend2.IfAzissmall enough, wecanuseaTaylor ex- pansion andwrite T(Z+AZ)=1'(z)+ AZ. (38.27) Thenettorque Aracting onthelittle piece ofrodbetween zandz+Azis clearly thedifference between T(Z) and1'(z+Az), orAr=(61/62) Az. Differ- entiating Eq.(38.26), weget 1ra462¢AT = [1,T 523* AZ. Theeffect ofthisnettorque istogive anangular acceleration tothelittle slice oftherod. Themass oftheslice is AM =(1ra2 Az)p, where pisthedensity ofthematerial. Weworked outinChapter 19,Vol. I,that themoment ofinertia ofacircular cylinder ismr2/2; calling themoment ofinertia ofourpiece AI,wehave AI=—gpa4AZ. (38.29) Newton’s lawsays thetorque isequal tothemoment ofinertia times theangular acceleration, or 2 AT=81%? (38.30) 38-7element oftherod. Pulling everything together, weget 4 2 21ra6¢ tr461¢) or "T32-Y‘“=r"““Z5iT 62¢ p02¢_53-5—I:-3-1-Z -0. (38.31) You willrecognize thisastheone-dimensional wave equation. Wehave found thatwaves oftorsion willpropagate down therodwith thespeed at...= <38-32> Thedenser therod—for thesame stiffness—the slower thewaves; andthestzﬂer therod,thequicker thewaves work their waydown. Thespeed does notdepend upon thediameter oftherod. Torsional waves areaspecial example ofshear waves. Ingeneral, shear waves arethose inwhich thestrains donotchange thevolume ofanypartofthematerial. Intorsional waves, wehave aparticular distribution ofsuch shear stresses—namely, distributed onacircle. Butforanyarrangement ofshear stresses, waves will propagate with thesame speed—the onegiven inEq.(38.32). Forexample, the seismologists ﬁndsuch shear waves travelling intheinterior oftheearth. Wecanhave another kind ofawave intheelastic world inside asolid material. Ifyoupush something, youcanstart “longitudinal” waves—also called“ compres- sional” waves. They arelikethesound waves inairorinwater—the displace- ments areinthesame direction asthewave propagation. (Atthesurfaces ofan elastic body there canalsobeother types ofwaves—called “Rayleigh waves” or “Love waves.” Inthem, thestrains areneither purely longitudinal norpurely transverse. Wewillnothave time tostudy them.) While we’re onthesubject ofwaves, what isthevelocity ofthepure com- pressional waves inalarge solid body liketheearth” Wesay“large” because the speed ofsound inathick body isdifferent from what itis,forinstance, along a thinrod. Bya“thick” body wemean oneinwhich thetransverse dimensions are much larger than thewavelength ofthesound. Then, when wepush ontheobject, itcannot expand sideways—-it canonly compress inonedimension. Fortunately, wehave already worked outthespecial case ofthecompression ofaconstrained elastic material. Wehave also worked outinChapter 47,Vol. I,thespeed of sound waves inagas. Following thesame arguments youcanseethatthespeed ofsound inasolid isequal to\/Y’/p, where Y’isthe“longitudinal modulus”- orpressure divided bytherelative change inlength—for theconstrained case. This isjusttheratio ofAl/ltoF/A wegotinEq.(38.20). Sothespeed ofthe longitudinal waves isgiven by 2_2:1 1—01.C‘°"“'p‘<1+oo—20>p 68'”) Solong as0isbetween zeroandl/2,theshear modulus uislessthan Young’s modulus Y,andalso Y’isgreater than Y,so ,u<Y<Y’. This means thatlongitudinal waves travel faster than shear waves. Oneofthemost precise ways ofmeasuring theelastic constants ofasubstance isbymeasuring the density ofthematerial andthespeeds ofthetwokinds ofwaves. From this information onecangetboth Yandcr.Itis,incidentally, bymeasuring thediffer- ence inthearrival times ofthetwokinds ofwaves from anearthquake thata seismologist canestimate—even from thesignals atonly onestation—the distance tothequake. 38-8 38-4 Thebentbeam Wewant now tolook atanother practical matter-—the bending ofarodora beam. What aretheforces when webend abarofsome arbitrary cross section? Wewillwork itoutthinking ofabarwith acircular cross section, butouranswer willbegood foranyshape. Tosave time, however, wewillcutsome corners, so ourtheory wewillwork outisonly approximate. Ourresults willbecorrect only when theradius ofthebend ismuch larger than thethickness ofthebeam. Suppose yougrab thetwoends ofastraight barandbend itintosome curve liketheoneshown inFig.38-l1.What goes oninside thebar? Well, ifitiscurved, thatmeans thatthematerial ontheinside ofthecurve iscompressed andthema- terial ontheoutside isstretched. There issome surface which goes along more or lessparallel totheaxisofthebarthatisneither stretched norcompressed. This is called theneutral surface. You would expect thissurface tobenear the“middle” ofthecross section. Itcanbeshown (but wewon’t doithere) that, forsmall bending ofsimple beams, theneutral surface goes through the“center ofgravity” ofthecross section. This istrueonlyfor“pure” bending—if youarenotstretching orcompressing thebeam atthesame time. Forpure bending, then, athintransverse slice ofthebarisdistorted asshown inFig. 38-l2(a). The material below theneutral surface hasacompressional strain which isproportional tothedistance from theneutral surface; andthematerial above isstretched, alsoinproportion toitsdistance from theneutral surface. So thelongitudinal stretch Alisproportional totheheight y.Theconstant ofpro- portionality isjustlover theradius ofcurvature ofthebar—see Fig.38-12: Al_1lFR Sotheforce perunitarea—the stress—in asmall strip atyisalsoproportional to thedistance from theneutral surface AF_ XK2-YR- (38.34) Now let’s look attheforces thatwould produce such astrain. Theforces acting onthelittle segment drawn inFig.38-12 areshown intheﬁgure. Ifwe think ofanytransverse cut,theforces acting across itareoneway above the neutral surface andtheother waybelow. They come inpairs tomake a“bending moment” &m—by which wemean thetorque about theneutral line. Wecancom- pute thetotal moment byintegrating theforce times thedistance from theneutral surface foroneofthefaces ofthesegment ofFig.38-12: :m= fydf‘. (38.35) cross sect From Eq.(38.34), dF=Yy/R dA,so _Y 2 811 -jitulf JVGL4. 2Theintegral ofydAiswhat wecancallthe“moment ofinertia” ofthegeometric cross section about ahorizontal axisthrough its“center ofmass”;* wewillcall itI: on=-lg (38.36) 1=/y2dA. (38.37) *Itis,ofcourse, really themoment ofinertia ofaslice with unitmass perunitarea. 38-9¢@§R Fig. 38-11. Abent beam. e\ e --3-1 ‘<- i>l n__ »lINEUTRAL SURFACE (<1)—ll——:o Ay y NEUTRAL SURFACE lb) Fig. 38-12. (clSmall segment ofa bent beam. (blCross section ofthebeam. Fig. 38-13. An"I" beam. -—- i_/ X Z / A W Fig. 38-14. Acantilevered beam with aweight atoneend.Equation (38.36), then, gives ustherelation between thebending moment Em andthecurvature l/Rofthebeam. The“stiffness” ofthebeam isproportional toYandtothemoment ofinertia I.Inother words, ifyouwant thestiffest possible beam with agiven amount of,say,aluminum, youwant toputasmuch ofitaspossible asfarasyoucanfrom theneutral surface, tomake alarge moment ofinertia. You can’t carry thistoanextreme, however, because then thething willnotcurve aswehave supposed-it willbuckle ortwist andbecome weaker again. Butnow youseewhystructural beams aremade intheform ofanIoran H—as shown inFig.38-13. Asanexample oftheuseofourbeam equation (38.36), let’s work outthe deﬂection ofacantilevered beam with aconcentrated force Wacting atthefree end, assketched inFig.38-14. (By“cantilevered” wesimply mean thatthebeam issupported insuch awaythatboth theposition andtheslope areﬁxed atone end-—it isstuck intoacement wall.) What istheshape ofthebeam? Let’s call thedeﬂection atthedistance xfrom theﬁxed endz;wewant toknow z(x). We’ll work itoutonly forsmall deﬂections. Wewillalsoassume thatthebeam islong incomparison with itscross section. Now, asyouknow from your mathematics courses, thecurvature 1/Rofanycurve z(x)isgiven by 1:_ d2z/dx2‘_~_ (3838) R[1+(dz/dx)1]3/2 Since weareinterested only insmall slopes—this isusually thecaseinengineering structures—we neglect (dz/dx)2 incomparison with 1,andtake 1 dzzi-3;? (38.39) Wealsoneed toknow thebending moment EJTL.Itisafunction ofxbecause itis equal tothetorque about theneutral axisofanycross section. Let’s neglect the weight ofthebeam andtakeonlythedownward force Wattheendofthebeam. (You canputinthebeam weight yourself ifyouwant.) Then thebending moment atxis iTIZ(x) =W(L —x), because that isthetorque about thepoint atx,exerted bytheweight W—the torque which thebeam must support ofx.Weget YI dzz Of dzz W This onewecanintegrate without anytricks; weget 2WL 3Z=Y7 - (38.41) using ourassumptions thatz(0) =0andthatdz/dx isalsozero atx=0.That istheshape ofthebeam. Thedisplacement oftheendis WL3z(L)=-Y7?; (38.42) thedisplacement oftheendofabeam increases asthecube ofthelength. Inderiving ourapproximate beam theory, wehave assumed thatthecross section ofthebeam didnotchange when thebeam wasbent. When thethickness ofthebeam issmall compared totheradius ofcurvature, thecross section changes verylittle andourresult isO.K. Ingeneral, however, thiseffect cannot beneglected, asyoucaneasily demonstrate foryourselves bybending asoft-rubber eraser in your ﬁngers. Ifthecross section wasoriginally rectangular, youwillﬁndthatwhen 38-10 itisbent itbulges atthebottom (seeFig.38-15). This happens because when we compress thebottom, thematerial expands sideways—as described byPoisson’s ratio. Rubber iseasy tobend orstretch, butitissomewhat likealiquid inthat it’shard tochange thevolume asshows upnicely when youbend theeraser. For anincompressible material, Poisson’s ratio would beexactly l/2—for rubber itis nearly that. 38-5 Buckling Wewant now touseourbeam theory tounderstand thetheory ofthe“buck- ling” ofbeams, orcolumns, orrods. Consider thesituation sketched inFig. 38-16 inwhich arodthatwould normally bestraight isheld initsbent shape by twoopposite forces thatpush ontheends oftherod. Wewould liketocalculate theshape oftherodandthemagnitude oftheforces ontheends. Letthedeﬂection oftherodfrom thestraight linebetween theends bey(x), where xisthedistance from oneend. Thebending moment rmatthepoint P intheﬁgure isequal totheforce Fmultiplied bythemoment arm, which isthe perpendicular distance y, EtlZ(x) =Fy. (38.43) Using thebeam equation (38.36), wehave Y1-F=Fy. (38.44) Forsmall deﬂections, wecantake 1/R=—d2y/dx2 (theminus signbecause the curvature isdownward). Weget d2y F which isthedifferential equation ofasinewave. Soforsmall deﬂections, thecurve ofsuch abent beam isasinecurve. The“wavelength” Xofthesinewave istwice thedistance Lbetween theends. Ifthebending issmall, thisisjusttwice the unbent length oftherod. Sothecurve is y=Ksin rrx/L. Taking thesecond derivative, weget d2y 1r2 dx‘~’ L2y' Comparing thistoEq.(38.45), weseethattheforce is V2‘Y1_F-7TL2 (38.46) Forsmall bendings theforce isindependent ofthebending displacement yl Wehave, then, thefollowing thing physically Iftheforce islessthan theF given inEq.(38.46). there willbenobending atall. Butifitisslightly greater than thisforce, thematerial willsuddenly bend alarge amount—that is,for forces above thecritical force Tr2Y1/L2 (often called the“Euler force”) thebeam will“buckle.” Iftheloading onthesecond ﬂoor ofabuilding exceeds theEuler force forthesupporting columns. thebuilding willcollapse. Another place where thebuckling force ismost important isinspace rockets. Ononehand, therocket must beabletohold itsown weight onthelaunching padandendure thestresses during acceleration, ontheother hand, itisimportant tokeep theweight ofthe structure toaminimum, sothat thepayload andfuelcapacity may bemade as large aspossible. Actually abeam willnotnecessarily collapse completely when theforce exceeds theEuler force. When thedisplacements getlarge, theforce islarger than 38-11s (<1) ¢_ﬂI II..~Q ’@ Q~‘ ’O 0., S (bl Fig. 38-15 lal Abent eraser; (bl cross section. P Y 5 En --lL Fig. 38-16. Abuckled beam. “x x\>“6€ P6 0/ R Fig. 38-17. Thecoordinates Sand 0 forthecurve ofabent beam. QFl Fl .-lNT? <i~ ~——-> F3 F3 Fig. 38-18. Curves ofabent rodwhat wehave found because oftheterms in1/RinEq.(38.38) thatwehave ne- glected. Toﬁndtheforces foralarge bending ofthebeam, wehave togoback to theexact equation, Eq.(38.44), which wehadbefore weused theapproximate relation between Randy.Equation (38.44) hasarather simple geometrical prop- erty.* It’salittle complicated towork out, butrather interesting. Instead of describing thecurve interms ofxandy,wecanusetwonew variables: S,the distance along thecurve, and0theslope ofthetangent tothecurve. SeeFig.38-17. Thecurvature istherateofchange ofangle with distance: l_Z2.R_dS Wecan, therefore write theexact equation (38.44) as ea__Las_Y11* Ifwetakethederivative ofthisequation withrespect toSandreplace dy/dS by sin0,weget 8120 F.F§-2 = —YT S111 0. [If6issmall, wegetback Eq.(38.45). Everything isO.K.] Now itmay ormay notdelight youtoknow thatEq.(38.47) isexactly the same oneyougetforthelarge amplitude oscillations ofapendulum—with F/Y1 replaced byanother constant, ofcourse. Welearned wayback inChapter 9,Vol.I, how toﬁndthesolution ofsuch anequation byanumerical calculationft The answers yougetaresome fascinating curves-known asthecurves ofthe“Elastica.” Figure 38-18 shows three curves fordifferent values ofF/Y1. *Thesame equation appears, incidentally, inother physical situations—for example, themeniscus atthesurface ofaliquid contained between parallel p1anes—and thesame geometrical solution canbeused. tThesolutions canalsobeexpressed interms ofsome functions, called the"Jacobian elliptic functions,” thatsomeone elsehasalready computed. 38-12 39 Elastic Materials 39-1 Thetensor ofstrain Inthelastchapter wetalked about thedistortions ofparticular elastic objects. Inthischapter wewant tolook atwhat canhappen ingeneral inside anelastic material. Wewould liketobeabletodescribe theconditions ofstress andstrain inside some bigglob ofjello which istwisted andsquashed insome complicated way. Todothis,weneed tobeabletodescribe thelocal strain atevery point inan elastic body; wecandoitbygiving asetofsixnumbers—which arethecomponents ofasymmetric tensor—for each point. Earlier, wespoke ofthestress tensor (Chapter 31);now weneed thetensor ofstrain. Imagine that westart with thematerial initially unstrained andwatch the motion ofasmall speck of“dirt” embedded inthematerial when thestrain is applied. Aspeck thatwasatthepoint Plocated atr=(x,y,z)moves toanew position P’atr’=(x’,y?,z’)asshown inFig.39-1. Wewillcalluthevector displacements from Pto Then u=r’—r. (39.1) Thedisplacement udepends, ofcourse, onwhich point Pwestart with, souisa vector function ofr-—or, ifyouprefer, of(x,y,z). Let’s look ﬁrstatasimple situation inwhich thestrain isconstant over the material—so wehavewhat iscalled ahomogeneous strain. Suppose, forinstance, thatwehave ablock ofmaterial andwestretch ituniformly. Wejustchange its dimensions uniformly inonedirection—say, inthex-direction, asshown inFig. 39-2. Themotion u,ofaspeck atxisproportional tox.Infact, ux AlY=T. Wewillwrite u,thisway: H; =€IxX. AFTER BEFORE \.\\\(\\\ \P\\ \\\ .\ .\ \ \ SPECK \ \ \ \\ \ \ \SPECK // //_‘////// //_\ // //6//l_____._._./§//////,.//// //// V./A./_\//\‘u/// T /// / /// //////////” ,\‘ \\\\\\ \\\ \ \\ \\\ \\\~ ________\ \ —— K .\ '1 ‘\\39-1 Thetensor ofstrain 39-2 Thetensor ofelasticity 39-3 Themotions inanelastic body 39-4 Nonelastic behavior 39-5 Calculating theelastic constants Reference: C.Kittel, Introduction to Solid State Physics, John Wiley andSons, Inc., New York, 2nded.,1956. BEFORE \\ \ \\ \ \ \\\ .\\ \. /// ////.// ///'0/////E:¢""'7‘€*1/.// //*1// ///)Tl"~ -PAFTER /--7/./.i-—-ll--—-1 ‘_"l l"_ux Fig. 39-1. Aspeck ofthematerial atthepoint Pinanunstrained block Fig. 39-2. Ahomogeneous stretch-type strain. moves toP'where theblock isstrained. 39-1 |, I up’ P’. l BEFORE AFTER _Q/'1 2l‘\ I ITheproportionality constant enis,ofcourse, thesame thing asAl/l. (You will seeshortly whyweuseadouble subscript.) Ifthestrain isnotuniform, therelation between u,,andxwillvaryfrom place toplace inthematerial. Forthegeneral situation, wedeﬁne theex,byakindof local Al/l, namely by en=6u,,/8x. (39.2) This number—which isnow afunction ofx,y,andz—describes theamount of stretching inthex-direction throughout thehunk ofjello. There may, ofcourse, alsobestretching inthey-andz—directions. Wedescribe them bythenumbers 8 6Ze,,,,=ﬁi e.,= (39.3) Weneed tobeabletodescribe alsotheshear-type strains. Suppose weimagine alittle cube marked outintheinitially undisturbed jello. When thejello ispushed outofshape, thiscube may getchanged intoaparallelogram, assketched inFig. 39-3.* Inthiskind ofastrain, thex-motion ofeach particle isproportional to itsy-coordiiiate, ii,=gy. (39.4) And there isalsoay-motion proportional tox, u,=git. (39.5) Sowecandescribe such ashear-type strain bywriting u,=ewy, up=ewx with 6GI!/= By; =E Now youmight think thatwhen thestrains arenothomogeneous wecould describe thegeneralized shear strains bydeﬁning thequantities en,andev,by 614 Ou6,,” : "6-5 > 6;,” : ' Q_____ _____J£7 2 Fig. 39-3. Ahomogeneous shear strain. Butthere isonedifficulty. Suppose thatthedisplacements umanduywere given by 0 9 utziys uy:_i’ *Wechoose forthemoment tosplit thetotal shear angle 6into twoequal parts and make thestrain symmetric with respect toxandy. 39-2 A BEFORE AFTER | l 8"".K I Fig. 39-4. Ahomogeneous rotation—there isnostrain. They arelikeEqs. (39.4) and(39.5) except thatthesignofu,,isreversed. With these displacements alittle cube inthejello simply getsshifted bytheangle 0/2, asshown inFig.39-4. There isnostrain atall—just arotation inspace. There is nodistortion ofthematerial; therelative positions ofalltheatoms arenotchanged atall. Wemust somehow make ourdeﬁnitions sothat pure rotations arenot included inourdeﬁnitions ofashear strain. Thekeypoint isthatif6a,,/6x and 6u,/6yareequal andopposite, there isnostrain; sowecanﬁxthings upbydeﬁning / if era=911$=%(3ui//ax '1'3"x/3y)- Forapure rotation they areboth zero, butforapure shear wegetthat em,is equal toey,,aswewould like. Inthemost general distortion—which may include stretching orcompression aswellasshear-—we deﬁne thestateofstrain bygiving theninenumbers err =%6x’ @,,,,= (39.7) ea,=%(<9uy/ax +ﬂut/6y), These aretheterms ofatensor ofstrain. Because itisasymmetric tensor—our deﬁnitions make em,=eyx,always—there arereally only sixdifferent numbers. You remember (seeChapter 31)thatthegeneral characteristic ofatensor isthat theterms transform liketheproducts ofthecomponents oftwovectors. (If AandBarevectors, C”=A,B, isatensor.) Each term ofe,,isaproduct (orthesumofsuch products) ofthecomponents ofthevector u=(ux,uy,uz),and oftheoperator V=(6/6x, 6/6y, 6/62), which weknow transforms likeavector. Let’s letx1,x2,andx3stand forx,y,andzandul,a2,andu3stand foru,,uh, anduz;then wecanwrite thegeneral term e,-,ofthestrain tensor as e,,=%(6u,/6x, +6u,/6x,), (39.8) where iandj canbe1,2,or3. When wehave ahomogeneous strain—which may include both stretching andshear-—all ofthee,,areconstants, andwecanwrite use:exzx "l"ext/y +earzZ- (Wechoose ourorigin ofx,y,2atthepoint where uiszero.) Inthiscase, thestrain tensor e,,gives therelationship between twovectors: thecoordinate vector r= (x,y,z)andthedisplacement vector u=(u,,,uy,u,). 39-3i\ U When thestrains arenothomogeneous, anypiece ofthejello mayalsoget somewhat twisted-there willbealocal rotation. Ifthedistortions areallsmall, wewould have Au,=Z(3,,-w,,)Ax,, (39.10)J where (73,,isanantisymmetric tensor, 03,,=%(6u,/6x, —6u,/6x,), (39.11) which describes therotation. Wewill, however, notworry anymore about rota- tions, butonly about thestrains described bythesymmetric tensor e,,. 39-2 Thetensor ofelasticity Now thatwehave described thestrains, wewant torelate them totheinternal forces—the stresses inthematerial. Foreach small piece ofthematerial, we assume Hooke’s lawholds andwrite that thestresses areproportional tothe strains. InChapter 31wedeﬁned thestress tensor S”astheithcomponent ofthe force across aunitarea perpendicular tothej-axis. Hooke’s lawsaysthateach component ofS”islinearly related toeach ofthecomponents ofstrain. Since Sandeeach have nine components, there are9X9=81possible coefficients which describe theelastic properties ofthematerial. They areconstants ifthe material itself ishomogeneous. Wewrite these coefficients asC,,;,; anddeﬁne them bytheequation St;=2Culclekli (39-12) 18,1 where i,j,k,lalltake onthevalues 1,2,or3.Since thecoeﬂfcients CH1.) relate onetensor toanother, they alsoform atensor—a tensor ofthefourth rank. We cancallitthetensor ofelasticity. Suppose thatalltheC’sareknown andthatyouputacomplicated force on anobject ofsome peculiar shape. There willbeallkinds ofdistortion, andthe thing willsettle down with some twisted shape. What arethedisplacements? You canseethatitisacomplicated problem. Ifyouknew thestrains, youcould ﬁndthestresses from Eq.(39.12)—or viceversa. Butthestresses andstrains you endupwith atanypoint depend onwhat happens inalltherestofthematerial. Theeasiest waytogetattheproblem isbythinking oftheenergy. When there isaforce Fproportional toadisplacement x,sayF=kx,thework required for anydisplacement xiskx2/2. Inasimilar way, thework wthat goes intoeach unitvolume ofadistorted material turns outtobe W=gZc,,,,@,,@,,. (39.13)zjlcl Thetotal work Wdone indistorting thebody istheintegral ofwover itsvolume: W=/gZc,,,,,e,,@,,, dVo1. (39.14)ijlcl This isthen thepotential energy stored intheinternal stresses ofthematerial. Now when abody isinequilibrium, thisinternal energy must beataminimum. Sotheproblem ofﬁnding thestrains inabody canbesolved byﬁnding thesetof displacements uthroughout thebody which willmake Waminimum. InChapter 19wegave some ofthegeneral ideas ofthecalculus ofvariations thatareusedin tackling minimization problems likethis. Wecannot gointotheproblem inany more detail here. What wearemainly interested innow iswhat wecansayabout thegeneral properties ofthetensor ofelasticity. First, itisclear thatthere arenotreally 81 different terms inCUM. Since both S,,ande,,aresymmetric tensors, eachwith only sixdifferent terms, there canbeatmost 36different terms inCHM. There are, however, usually many fewer than this. 39-4 Let’s look atthespecial case ofacubic crystal. Init,theenergy density w starts outlikethis: +++999W:%{CrxIxe2r xzyexxezy + Crxxzexxexz Zwemezy —l—cxxyyexxeyy ...etc... 2yyyeyy —l—...etc...etc...}, (39.15) with81terms inall!Now acubic crystal hascertain symmetries. Inparticular, if thecrystal isrotated 90°,ithasthesame physical properties. Ithasthesame stiffness forstretching inthey-direction asforstretching inthex-direction. There- fore, ifwechange ourdeﬁnition ofthecoordinate directions xandyinEq.(39.15), theenergy wouldn’t change. Itmust bethatforacubic crystal Caxcaca: :C1/712/11 =Czzzz- Next wecanshow thattheterms likeCum, must bezero. Acubic crystal has theproperty thatitissymmetric under areﬂection about anyplane perpendicular tooneoftheaxes. Ifwereplace yby—y,nothing isdifferent. Butchanging yto —ychanges em,to—e,,,,—a displacement which wastoward +yisnowtoward —y. Iftheenergy isnottochange, cajxzy must gointo -Cum, when wemake areﬂec- tion. Butareﬂected crystal isthesame asbefore, soCmy must bethesame as —CxzIy- This canhappen only ifboth arezero. You say,“But thesame argument willmake C,,,,,,,, =0!”No,because there arefour y’s. Thesignchanges once foreach y,andfour minuses make aplus. If there aretwoorfour y’s,theterm does nothave tobezero. Itiszero only when there isone,orthree. So,foracubic crystal, anynonzero term ofCwillhave only aneven number ofidentical subscripts. (The arguments wehave made foryob- viously hold alsoforxandz.)Wemight then have terms likeC,,,,,,,,, C,,,,,,,,, Cxyyx, andsoon.Wehave already shown, however, thatifwechange allx’stoy’sand viceversa (orallz’sandx’s,andsoon)wemust get—for acubic crystal—the same number. Thismeans thatthere areonlythree diﬂerent nonzero possibilities: Cxzxz (= Cyyyy = C2222): Cum, (=Cyym =Cu”, etc.), (39.17) T, Cm/:cy (:C1/xya: =Cxzxza eta)- Foracubic crystal, then, theenergy density willlook likethis: W:%{C:ca:xx(eg:c +9?/y +93:) +2Cm,i,(@m@i,i, +ewe” +meta) (39-18) +4Czyxy(e:%y +839+631)}- For anisotropic—that is,noncrystalline—materia1, thesymmetry isstill higher. TheC’smust bethesame foranychoice ofthecoordinate system. Then itturns outthatthere isanother relation among theC’s,namely, that Cxxxx = Cxxyy '1‘ Czyxy- Wecanseethatthisissobythefollowing general argument. Thestress tensor S,,-hastoberelated toe,,inawaythatdoesn’t depend atallonthecoordinate directions—it must berelated only byscalar quantities. “That’s easy,” yousay. “The only way toobtain S”from e,,isbymultiplication byascalar constant. It’sjustHooke’s law. Itmust bethat S”=(const)e,,.” Butthat’s notquite right; there could also betheunittensor 5,,multiplied bysome scalar, linearly related toe,,.Theonly invariant youcanmake thatislinear inthee’sisZen (Ittransforms likex2—l—y2+22,which isascalar.) Sothemost general form fortheequation relating S,,toe,,—for isotropic materials——is s,,=2,.i@,,+x(Zem.)8,,. (39.20)lo (The ﬁrstconstant isusually written astwotimes u;then thecoefﬁcient uisequal 39-5 \\ MOLUME V ll {Q\SURFACE A .8Fig. 39-5. Asmall volume element V bounded bythesurface A.totheshear modulus wedeﬁned inthelastchapter.) Theconstants uand)\are called theLamé elastic constants. Comparing Eq.(39.20) with Eq.(39.12), you seethat Céxyy ==A, /7 C,,,,,,,, =2/i, Zr’ (39.21) Czxxx :2|“ + Sowehave proved thatEq.(39.19) isindeed true. You alsoseethattheelastic properties ofanisotropic material arecompletely given bytwoconstants, aswe saidinthelastchapter. TheC’scanbeputinterms ofanytwooftheelastic constants wehave used earlier—for instance, interms ofYoung’s modulus YandPoisson’s ratio a.We willleave itforyoutoshow that Y 0 CW-W;(1+ Y aCm”! =Ti‘: (4) , (39.22) Y 7(Ta 39-3 Themotions inanelastic body Wehave pointed outthat foranelastic body inequilibrium theinternal stresses adjust themselves tomake theenergy aminimum. Now wetakealookat what happens when theinternal forces arenotinequilibrium. Let’s saywehave asmall piece ofthematerial inside some surface A.SeeFig.39-5. lfthepiece isin equilibrium, thetotal force Facting onitmust bezero. Wecanthink ofthisforce asbeing made upoftwoparts. There could beonepartdueto“external” forces likegravity, which actfrom adistance onthematter inthepiece toproduce a force perunitvolume y’s,“. Thetotal external force Fe,“istheintegral offmover thevolume ofthepiece: Foxt : 7/./cxt Inequilibrium, thisforce would bebalanced bythetotal force Fm,from theneigh- boring material which actsacross thesurface A.When thepiece isnotinequili- brium—if itismoving—the sumoftheinternal andexternal forces isequal tothe mass times theacceleration. Wewould have Foxt: 'l"Fint : fPl’dVs where pisthedensity ofthematerial, andrisitsacceleration. Wecannowcom- bine Eqs. (39.23) and(39.24), writing Fm, =/_(—feXt —l—pr)(IV. (39.25) Wewillsimplify ourwriting bydeﬁning f=—f@xt +p'r- (39-26)Then Eq.(39.25) iswritten Fm, =/fdV. (39.27) What wehave called Fm,isrelated tothestresses inthematerial. Thestress tensor SHwasdeﬁned (Chapter 31)sothatthex-component oftheforce dFacross asurface element da,whose unitnormal isn,isgiven by dF,,=(Sun,+s,,n,,+S,,,,nz)da. (39.28) 39-6 Thex-component ofFm,onourlittle piece isthen theintegral ofdF,,over the surface. Substituting thisintothex-component ofEq.(39.27), weget £1(Sun, +Swny —l—Sxznz) da=/pf, dV. (39.29) Wehave asurface integral related toavolume integral—and that reminds usofsomething welearned inelectricity. Note thatifyouignore theﬁrstsubscript xoneach oftheS’sintheleft-hand sideofEq.(39.29), itlooks justliketheintegral ofaquantity “S”-n—that is,thenormal component ofavector—over the surface. Itwould betheﬂuxof“S”outofthevolume. And thiscould bewritten, using Gauss law, asthevolume integral ofthedivergence of“S”. Itis,infact, truewhether thex-subscript isthere ornot-it isjustamathematical theorem yougetbyintegrating byparts. Inother words, wecanchange Eq.(39.29) into 'as... as,,, as... _IA(TX +337-+Y) dV_Of,dV (39.30) Now wecanleave offthevolume integrals andwrite thedifferential equation for thegeneral component offas f,=Z (39.31)x] This tellsushow theforce perunitvolume isrelated tothestress tensor S,,. Thetheory ofthemotions inside asolid works thisway. Ifwestart outknow- ingtheinitial displaceinents—given by,say,u—we canwork outthestrains e,,. From thestrains wecangetthestresses from Eq.(39.12). From thestresses we cangettheforce densityfin Eq.(39.31). Knowingﬁ Wecanget,from Eq.(39.26), theacceleration rofthematerial, which tells ushow thedisplacements willbe changing. Putting everything together, wegetthehorrible equation ofmotion foranelastic solid. Wewilljustwrite down theresults thatcome outforan isotropic material. Ifyouuse(39.20) forS”,andwrite thee,,as;§_~6u._/6x, -l- 6u,/6x,, youendupwith thevector equation f=(x+74)v(v-ll)+itVzu. (39.32) You can,infact, seethattheequation relating fandumust have thisform. Theforce must depend onthesecond derivatives ofthedisplacements uWhat second derivatives ofuarethere thatarevectors? OneisV(V -u);that’s atrue vector. Theonly other oneisVzu. Sothemost general form is '_ f=aV(V-u)—l—bV2u, which isjust (39.32) with adifferent deﬁnition oftheconstants. You may be wondering why wedon’t have athird term using VXVXu,which isalso a vector. Butremember that VXVXuisthesame thing asV2u—V(V su), soitisalinear combination ofthetwoterms wehave. Adding itwould addnothing new. Wehave proved once more that isotropic material hasonly twoelastic constants. Fortheequation ofmotion ofthematerial, wecanset(39.32) equal to p6211/6t2—neglecting fornow anybody forces likegravity-and get 2 p%=(>.+0)v(v~ll)+[.LV2ll. (39.33) Itlooks something likethewave equation wehadinelectromagnetism, except thatthere isanadditional complicating term. Formaterials whose elastic proper- tiesareeverywhere thesame wecanseewhat thegeneral solutions look likeinthe following way. You willremember thatanyvector ﬁeld canbewritten asthesum oftwovectors: onewhose divergence iszero, andtheother whose curliszero. ln 39-7 / POL AROIDS 9/. )4-'/5'Z?’ 'i_ /» / '\._\\\\=.!'\\BRIGHT SCREEN LUGITE MODEL UNDER STRESS Fig. 39-6. Measuring internal stresses with polarized light. Fig. 39-7. Astressed plastic model asseen between crossed polaroids. [From F.W. Sears, Optics, Addison- Wesley Publishing Co., Reading, Mass., 1949.1.//A /// ><z:/ ../,¢¢</1-. /I///. ’/ 4 \ . ;- \ Z3 .other words, wecanput u="1+U2, (39.34) where V-ul =0, VXu2=0. (39.35) Substituting U1+u2foruin(39.33), weget P'32/3t2l"i '1'142]=(A*1‘ft)V(V ''12)+I1V2("i '1‘"2)- (39-35) Wecaneliminate ulbytaking thedivergence ofthisequation, pat/arttv -42)=0+l1)V2(V'"2) +IJV-V2112- Since theoperators (V2) and(V-)canbeinterchanged, wecanfactor outthedi- vergence toget v-{p02142/at’ -(x+2u)V2u2} =0. (39.37) Since VXu-2iszero bydeﬁnition, thecurlofthebracket {}isalsozero; sothe bracket itself isidentically zero, and p62u2/612 =(x+2}.t)V2ll2. Z (39.38) This isthe vector wave equation forwaves which move atthespeed C2=\/(xi; 2a)/p. Since thecurlofU2iszero, there isnoshearing associated withthiswave; thiswave isjustthecompressional—sound-type-wave wediscussed inthelastchapter, andthevelocity isjustwhat wefound forclung: Inasimilar way—by taking thecurlofEq.(39.36)—we canshow thatI41 satisﬁes theequation p621:1/6t2 =;.tV2u1. (39.39) This isagain avector wave equation forwaves with thespeed C2=\/I/E. Since V-u1iszero, ulproduces nochanges indensity; thevector ulcorresponds tothetransverse, orshear-type, wave wesawinthelastchapter, andC2=C,he,,,. Ifwewished toknow thestatic stresses inanisotropic material, wecould, inprinciple, ﬁndthem bysolving Eq.(39.32) withfequal tozero—or equal tothe static body forces from gravity such aspg-—under certain conditions which are related totheforces acting onthesurfaces ofourlarge block ofmaterial. Thisis somewhat more difﬁcult todothan thecorresponding problems inelectromagne- tism. Itismore difficult, ﬁrst, because theequations arealittle more difﬁcult to handle, andsecond, because theshape oftheelastic bodies wearelikely tobe interested inareusually much more complicated. Inelectromagnetism, weare often interested insolving Maxwell’s equations around relatively simple geometric shapes such ascylinders, spheres, andsoon,since these areconvenient shapes forelectrical devices. Inelasticity, theobjects wewould liketoanalyze mayhave quite complicated shapes-like acrane hook, oranautomobile crankshaft, orthe rotor ofagasturbine. Such problems cansometimes beworked outapproxi- mately bynumerical methods, using theminimum energy principle wementioned earlier. Another wayistouseamodel oftheobject andmeasure theinternal strains experimentally, using polarized light. Itworks thisway: When atransparent isotropic material-for example, a clear plastic likelucite—is putunder stress, itbecomes birefringent. Ifyouput polarized light through it,theplane ofpolarization willberotated byanamount related tothestress: bymeasuring therotation, youcanmeasure thestress. Figure 39-6 shows how such asetup might look. Figure 39-7 isaphotograph ofa photoelastic model ofacomplicated shape under stress. 39-4 Nonelastic behavior Inallthathasbeen saidsofar,wehave assumed thatstress isproportional tostrain; ingeneral, that isnottrue. Figure 39-8 shows atypical stress-strain curve foraductile material. Forsmall strains, thestress isproportional tothe 39-8 strain. Eventually, however, after acertain point, therelationship between stress andstrain begins todeviate from astraight line. Formany materials—the ones wewould call“brittle” —the object breaks forstrains only alittle above thepoint where thecurve starts tobend over. Ingeneral, there areother complications in thestress-strain relationship. Forexample, ifyoustrain anobject, thestresses maybehigh atﬁrst, butdecrease slowly with time. Also ifyougotohigh stresses, butstillnottothe“breaking” point, when youlower thestrain thestress will return along adifferent curve. There isasmall hysteresis effect (like theonewe sawbetween BandHinmagnetic materials). Thestress atwhich amaterial willbreak varies widely from onematerial to another. Some materials willbreak when themaximum tensile stress reaches a certain value. Other materials willfailwhen themaximum shear stress reaches a certain value. Chalk isanexample ofamaterial which ismuch weaker intension than inshear. Ifyoupullontheends ofapiece ofblackboard chalk, thechalk will break perpendicular tothedirection oftheapplied stress, asshown inFig.39—9(a). Itbreaks perpendicular totheapplied force because itisonly abunch ofparticles packed together which areeasily pulled apart. Thematerial is,however, much harder toshear, because theparticles getineach other’s way. Now youwillre- member thatwhen wehadarodintorsion there wasashear allaround it.Also, we showed thatashear wasequivalent toacombination ofatension andcompression at45°. Forthese reasons, ifyoutwist apiece ofblackboard chalk, itwillbreak along acomplicated surface which starts outat45°totheaxis. Aphotograph ofa piece ofchalk broken inthiswayisshown inFig.39—9(b). Thechalk breaks where thematerial isinmaximum tension.FRACTURE OCURRED STRESS HERE LINEAR REGION STRAIN Fig. 39-8. Atypical stress-strain re- lation forlarge strains. ("I \Vl Fig. 39-9. (alApiece ofchalk broken bypulling ontheends; (blapiece broken bytwisting. Other materials behave instrange andcomplicated ways. Themore compli- cated thematerials are,themore interesting their behavior. Ifwetake asheet of “Saran-Wrap” andcrumple itupintoaballandthrow itonthetable, itslowly unfolds itself andreturns toward itsoriginal ﬂatform. Atﬁrstsight, wemight betempted tothink thatitisinertia which prevents itfrom returning toitsoriginal form. However, asimple calculation shows thattheinertia isseveral orders of .magnitude toosmall toaccount for,theeffect. There appear tobetwoimportant competing eﬂ'ects: “something” inside thematerial “remembers” theshape ithad initially and“tries” togetback there, butsomething else“prefers” thenewshape and“resists” thereturn totheoldshape. Wewillnotattempt todescribe themechanism atplay intheSaran plastic, butyoucangetanideaofhowsuch aneffect might come about from thefollowing model. Suppose youimagine amaterial made oflong, ﬂexible, butstrong, ﬁbers mixed together with some hollow cells ﬁlled with aviscous liquid. Imagine also thatthere arenarrow pathways from onecelltothenext sotheliquid canleak slowly from acelltoitsneighbor. When wecrumple asheet ofthisstuff, we distort thelong ﬁbers, squeezing theliquid outofthecellsinoneplace andforcing itintoother cells which arebeing stretched. When weletgo,thelong ﬁbers tryto 39-9 return totheir original shape. Buttodothis,theyhave toforce theliquid back to itsoriginal location—which willhappen relatively slowly because oftheviscosity. Theforces weapply incrumpling thesheet aremuch larger than theforces exerted bytheﬁbers. Wecancrumple thesheet quickly, butitwillreturn more slowly. Itisundoubtedly acombination oflarge stiffmolecules andsmaller, movable ones intheSaran-Wrap thatisresponsible foritsbehavior. This ideaalsoﬁtswiththe factthatthematerial returns more quickly toitsoriginal shape when it’swarmed upthan when it’scold—-the heat increases themobility (decreases theviscosity) ofthesmaller molecules. Although wehave been discussing how Hooke’s lawbreaks down, there- markable thing isperhaps notthatHooke’s lawbreaks down fq;large strains but thatitshould besogenerally true. Wecangetsome ideaofwlgythismight beby looking atthestrain energy inamaterial. Tosaythatthestress isproportional to thestrain isthesame thing assaying thatthestrain energy varies asthesquare of thestrain. Suppose wehave arodandwetwist itthrough asmall angle 0.If Hooke’s lawholds, thestrain energy should beproportional tothesquare of0. Suppose wewere toassume thattheenergy were some arbitrary function ofthe angle; wecould write itasaTaylor expansion about zero angle U(0) =U(O) +U’(0)0 +%U”(O)62 +%U”’(0)03 ... (39.40) Thetorque risthederivative ofUwith respect toangle; wewould have t(0)=U’(0) +U”(0)0 —l—%U"’(0)02 —l—--- (39.41) Now ifwemeasure ourangles from theequilibrium position, theﬁrstterm iszero. Sotheﬁrstremaining term isproportional to0;andforsmall enough angles, it willdominate theterm in02. [Actually, materials aresufficiently symmetric internally sothat7-(0) =-r(—6); theterm in62willbezero, andthedepartures from linearity would come only from the03term. There is,however, noreason whythisshould betrueforcompressions andtensions.] Thething wehavenot explained iswhymaterials usually break soonafterthehigher-order terms become signiﬁcant. 39-5 Calculating theelastic constants Asourlasttopic onelasticity wewould liketoshow how onecould tryto calculate theelastic constants ofamaterial, starting with some knowledge ofthe properties oftheatoms which make upthematerial. Wewilltake onlythesimple case ofanionic cubic crystal likesodium chloride. When acrystal isstrained, its volume oritsshape ischanged. Such changes result inanincrease inthepotential energy ofthecrystal. Tocalculate thechange instrain energy, wehave toknow where each atom goes. Incomplicated crystals, theatoms willrearrange themselves inthelattice inverycomplicated ways tomake thetotal energy assmall aspossible. This makes thecomputation ofthestrain energy rather difficult. Inthecaseofa simple cubic crystal, however, itiseasy toseewhat willhappen. Thedistortions inside thecrystal willbegeometrically similar tothedistortions oftheoutside boundaries ofthecrystal. Wecancalculate theelastic constants foracubic crystal inthefollowing way. First, weassume some force lawbetween each pairofatoms inthecrystal. Then, we calculate thechange intheinternal energy ofthecrystal when itisdistorted from itsequilibrium shape. This gives usarelation between theenergy andthestrains which isquadratic inallthestrains. Comparing theenergy obtained thiswaywith Eq.(39.13), wecanidentify thecoefficient ofeach term with theelastic constants Ci]lcl- Forourexample weW111assume asimple force law: thattheforce between neighboring atoms isacentral force, bywhich wemean thatitactsalong theline between thetwoatoms. Wewould expect theforces inionic crystals tobelike this, since they arejustprimarily Coulomb forces. (The forces ofcovalent bonds areusually more complicated, since they canexert asideways push onanearby 39-10 atom; wewillleave outthiscomplication.) Wearealsogoing toinclude only the forces between each atom anditsnearest andnext-nearest neighbors. Inother words, wewillmake anapproximation which neglects allforces beyond thenext- nearest neighbor. Theforces wewillinclude areshown forthexy-plane inFig. 39-10(a). The corresponding forces intheyz-andzx-planes also have tobe included. Since weareonly interested intheelastic coefﬁcients which apply tosmall strains, andtherefore only want theterms intheenergy which vary quadratically with thestrains, wecanimagine that theforce between each atom pair varies linearly with thedisplacements. Wecanthen imagine thateach pairofatoms is joined byalinear spring, asdrawn inFig.39—10(b). Allofthesprings between a sodium atom andachlorine atom should have thesame spring constant, saykl. Thesprings between twosodiums andbetween twochlorines could have different constants, butwewillmake ourdiscussion simpler bytaking them equal; wecall them k2.(Wecould come back later andmake them different after wehave seen how thecalculations go.) Now weassume that thecrystal isdistorted byahomogeneous strain de- scribed bythestrain tensor e,-,. Ingeneral, itwillhave components involving x,y,andz;butwewillconsider now only astrain with thethree components en,cw,andem,sothatitwillbeeasy tovisualize. Ifwepick oneatom asour origin, thedisplacement ofevery other atom isgiven byequations likeEq.(39.9): uz=ezrx 'l'ezuys ull=e-‘tux '1'ewJ'- (i) Suppose wecalltheatom atx=y=0“atom 1”andnumber itsneighbors in thexy-plane asshown inFig.39-11. Calling thelattice constant a,wegetthex andydisplacements u,andu,,listed inTable 39-1. Now wecancalculate theenergy stored inthesprings, which isk2/2 times thesquare oftheextension foreach spring. Forexample, theenergy inthehori- zontal spring between atom landatom 2is 2 m . (39.43) Note thattoﬁrstorder, they-displacement ofatom 2does notchange thelength of thespring between atom 1andatom 2.Togetthestrain energy inadiagonal spring, such asthattoatom 3,however, weneed tocalculate thechange inlength dueto both thehorizontal andvertical displacements. Forsmall displacements from the 34 -___ Q ~Q‘ °"° i/ ' \ _, 3\_,/exyo U 2 4 1 , / \/ xx 6 /\ (T\ FAQ)-4) / \/9ii-ci--i(,,,\<v)/ \<")/ \/ % NO <—-> C1 <—-> <*- /~ ii\\ 1-\ \~ ~\1/-@- -0)-T\ / 795%>@§9%/\ >\ (8) Na ~- Cl i-- Na/\ 1 \K2 _ \- K1. , |( T \2 k’, ‘\ at g 2 , oC») kl - kl - 2klV _ *9 -7k2 ~ 'I 4 \ 1 1 Na.qp0.N9 Fig. 39-10. (al The interatomic forces wearetaking intoaccount; lbla model inwhich theatoms areconnected bysprings. , Fig. 39-l I.Thedisplacements ofthe } 8 nearest and next-nearest neighbors of 7 atom 1(exaggerated). 39-11 Table 39-1 Location Atom x,y u, uy k \OOO\IO‘\U1-l>UJI\J'—*O,a 0 0 —- a,0 ena ewa k1 a,a (en+e,u)a (e,,,—l—e,,,,)a kg O,a e,,,,a ewa k1 —a,a (—e,,, +e,,,)a (—e,,,, +e,,,,)a k2 —a,0 —e,,,a —e,,,a k1 —a,—a —(e,, +e,,,,)a —(e,,,, +e,,,)a /C2 0,—a —e,,,a —em,a k1 a,-11 (em—emu (ew—Ema k2 original cube, wecanwrite thechange inthedistance toatom 3asthesumofthe components ofu,,andu,,inthediagonal direction, namely as % (um+"21)- Using thevalues ofu,anduyfrom thetable, wegettheenergy 2 2 =5?‘?(en+e,,,+em,+@,,,,)2. (39.44) Forthetotal energy forallthesprings inthexy-plane, weneed thesumof eight terms like(39.43) and(39.44). Calling thisenergy U0,weget 412 2 k2 2U0 :7 klexr +7(ex: +e]/I +exy +er/11) k +klejy +“Z2(err "@111 *@111+em/)2 k +klefz: +“Z2(exx +eya: +exy +em/)2 k+1<1e,3,,+’2l(e1;1; -em,-em,+@Z,,,)2}- (39.45) Togetthetotal energy ofallthesprings connected toatom l,wemust make one addition totheenergy inEq.(39.45). Even though wehave only x-andy-com- ponents ofthestrain, there arestillsome energies associated with thenext-nearest neighbors offthexy-plane. This additional energy is k2(e§,a2 +eﬁua2). (39.46) Theelastic constants arerelated totheenergy density wbyEq.(39.13). The energy wehave calculated istheenergy associated with oneatom, orrather, itis twice theenergy peratom, since one-half oftheenergy ofeach spring should be assigned toeach ofthetwoatoms itjoins. Since there are1/a3 atoms perunit volume, wandU0arerelated by U “’=ﬁ' Toﬁndtheelastic constants Cum. Weneed onlytoexpand outthesquares in Eq.(39.45)—adding theterms of(39.46)——and compare thecoefﬁcients ofe,-,-eh; with thecorresponding coefﬁcient inEq.(39.13). Forexample, collecting theterms 39-12 ine2,andineﬁy,wegetthefactor (k1+2k2)a2, so Cram =gm” = a Fortheremaining terms, there isaslight complication. Since wecannot distin- guish theproduct oftwoterms likeewe,” from ewe", thecoefﬁcient ofsuch terms inourenergy isequal tothesum oftwoterms inEq.(39.13). Thecoefﬁcient of emew inEq.(39.45) is2k2, sowehave that 2k(C111/y +Ci/i/Ir) :713' Butbecause ofthesymmetry inourcrystal, Cm”, =C,,,,,,,, sowehave that Crruy =Cw/Ir Ia Byasimilar process, wecanalsoget k2Cxyacy =Cyxyx : a Finally, youwillnotice thatanyterm which involves either xoryonly once is zero—as weconcluded earlier from symmetry arguments. Summarizing ourresults: k1+2k2Czxxz = Cyyyy ='_€a'i ’ _ _k2Cm”"CW‘“7’ (39.47) /<2Cwwuu =Ci/1m= =Cum: =C1/rev =7’ C,,,,,,, =C,,,,,,,, =etc.=0. Wehave been abletorelate thebulk elastic constants totheatomic properties which appear intheconstants klandk2.Inourparticular case, C,,,,,,, =CIIW. Itturns out—as youcanperhaps seefrom theway thecalculations went—that these terms arealways equal foracubic crystal, nomatter how many force terms aretaken into account, provided only that theforces actalong thelinejoining each pairofatoms—that is,solong astheforces between atoms arelikesprings anddon’t have asideways part such asyoumight getfrom acantilevered beam (and youdogetincovalent bonds). Wecancheck thisconclusion with theexperimental measurements ofthe elastic constants. InTable 39—2 wegivetheobserved values ofthethree elastic COCl11CiCI1lS forseveral cubic crystals.* You willnotice that Cm”, andC,,,,,,, are, ingeneral, notequal. Thereason isthatinmetals likesodium andpotassium the interatomic forces arenotalong thelinejoining theatoms, asweassumed inour model. Diamond does notobey thelaweither, because theforces indiamond are covalent forces andhave some directional properties—the bonds would prefer to beatthetetrahedral angle. Theionic crystals likelithium ﬂuoride, sodium chloride, andsoon,dohave nearly allthephysical properties assumed inourmodel, and thetable shows thattheconstants Cm”, andC1,,” arealmost equal. Itisnotclear whysilver chloride should notsatisfy thecondition thatCum, =C,,,,,,,,. *Intheliterature youwilloften ﬁndthat adifferent notation 1Sused. Forinstance, people usually write C11,, =C11, Cm”, =C12, andC,,,,,,, =C44. 39-13Table 39—2* Elastic Moduli ofCubic Crystals in10” dynes-cmz Crzzz Na 0.055 K 0.046 Fe 2.37 Diamond 10.76 Al 108 LiF 1.19 NaCl 0.486 KCI 0.40 NaBr 0.33 KI 0.27 AgCl 0.60CIZII/I] 0.042 0.037 1.41 1.25 062 0.54 0.127 0.062 0.13 0.043 036_Cﬂ’"'. 0.049 0026 1.16 5.76 0.28 0.53 0.128 0.062 0.13 0.042 0.062 *From C.Kittel, Introduction toSolid State Physics, John Wiley andSons, Inc, New York, 2nd. ed,1956, p.93. 40 The Flow ofDry Water 40-1 Hydrostatics Thesubject oftheﬂow ofﬂuids, andparticularly ofwater, fascinates every- body. Wecanallremember, aschildren, playing inthebathtub orinmud puddles with thestrange stuff. Aswegetolder, wewatch streams, waterfalls, andwhirl- pools, andwearefascinated bythissubstance which seems almost alive relative tosolids. Thebehavior ofﬂuids isinmany ways veryunexpected andinteresting- itisthesubject ofthischapter andthenext. Theefforts ofachild trying todam a small stream ﬂowing inthestreet andhissurprise atthestrange way thewater works itswayouthasitsanalog inourattempts over theyears tounderstand the ﬂow ofﬂuids. Wehave tried todam thewater up—in ourunderstanding—by getting thelaws andtheequations thatdescribe theﬂow. Wewilldescribe these attempts inthischapter. Inthenext chapter, wewilldescribe theunique wayin which water hasbroken through thedam and escaped ourattempts tounder- stand it. Wesuppose that theelementary properties ofwater arealready known to you. Themain property thatdistinguishes aﬂuid from asolid isthataﬂuid cannot maintain ashear stress foranylength oftime. Ifashear isapplied toaﬂuid, it willmove under theshear. Thicker liquids likehoney move lesseasily than ﬂuids likeairorwater. Themeasure oftheeasewithwhich aﬂuidyields isitsviscosity. Inthischapter wewillconsider onlysituations inwhich theviscous effects canbe ignored. Theeffects ofviscosity willbetaken upinthenextchapter. Webegin byconsidering hydrostatics, thetheory ofliquids atrest. When liquids areatrest, there arenoshear forces (even forviscous liquids). Thelaw ofhydrostatics, therefore, isthat thestresses arealways normal toanysurface lI1SlClCi theﬂuid. Thenormal force perunitarea iscalled thepressure. From the factthatthere isnoshear inastatic ﬂuid itfollows thatthepressure stress isthe same inalldirections (Fig. 40-1). Wewillletyouentertain yourself byproving thatifthere isnoshear onanyplane inaﬂuid, thepressure must bethesame in anydirection. Thepressure inaﬂuid may vary from place toplace. Forexample, inastatic ﬂuid attheearth ’ssurface thepressure willvary with height because oftheweight oftheﬂuid. Ifthedensity poftheﬂuid isconsidered constant, andifthepressure atsome arbitrary zero level iscalled po(Fig. 40-2), then thepressure ataheight habove thispoint isp=p0—pgh, where gisthegravitational force perunit mass. Thecombination iv+pg/1 is,therefore, aconstant inthestatic ﬂuid. This relation isfamiliar toyou, butwe willnow derive amore general result ofwhich itisaspecial case. Ifwetakeasmall cube ofwater, what isthenetforce onitfrom thepressure? Since thepressure atanyplace isthesame inalldirections, there canbeanet force perunitvolume only because thepressure varies from onepoint toanother. Suppose thatthepressure isvarying inthex-direction—and wetakethecoordinate directions parallel tothecube edges. Thepressure onthefaceatxgives theforce pAyAz(Fig. 40-3), and thepressure ontheface atx+Axgives theforce -[p +(dp/6x) Ax]AyAz,sothat theresultant force is——(8p/6x) AxAyAz. If wetake theremaining pairs offaces ofthecube, weeasily seethatthepressure force perunitvolume is—Vp.Ifthere areother forces inaddition—such asgravity —then thepressure must balance them togiveequilibrium. 40-140-1 Hydrostatics 40-2 Theequations ofmotion 40-3 Steady ﬂow—Bernoulli’s theorem 40-4 Circulation 40-5 Vortex lines \\I§i\§\w/PIe<'We\\\ iQ\\\ /F \\\\ i \\\ \\\\\\\\\\ \ \\\ \\\ Fig. 40-1. Inastatic ﬂuid theforce per unit area across any surface is normal tothesurface and isthesame for allorientations ofthesurface. SURFACE T/_ 3. /LIQUID // // //Fig. 40-2. The pressure inastatic liquid. 59p p‘i':x'AX A)’ x Ax x+Ax Fig. 40-3. Thenetpressure force on acube is—Vpperunitvolume.Let’s take acircumstance inwhich such anadditional force canbedescribed byapotential energy, aswould betrueinthecase ofgravitation; wewilllet¢ stand forthepotential energy perunitmass. (Forgravity, forinstance, ¢isjustgz.) Theforce perunitmass isgiven interms ofthepotential by—V¢, andifpisthe density oftheﬂuid, theforce perunitvolume is-pV¢. Forequilibrium this force perunitvolume added tothepressure force perunitvolume must givezero: —Vp —pV¢=0. (40.1) Equation (40.1) istheequation ofhydrostatics. Ingeneral, ithasnosolution. Ifthedensity varies inspace inanarbitrary way, there isnowayfortheforces to beinbalance, andtheﬂuid cannot beinstatic equilibrium. Convection currents willstart up.Wecanseethisfrom theequation since thepressure term isapure gradient, whereas forvariable ptheother term isnot. Only when pisaconstant isthepotential term apure gradient. Then theequation hasasolution p+pqs=const. Another possibility which allows hydrostatic equilibrium isforptobeafunction only ofp.However, wewillleave thesubject ofhydrostatics because itisnot nearly sointeresting asthesituation when ﬂuids areinmotion. 40-2 Theequations ofmotion First, wewilldiscuss ﬂuid motions inapurely abstract, theoretical wayand then consider special examples. Todescribe themotion ofaﬂuid, wemust givt itsproperties atevery point. Forexample, atdiﬂerent places, thewater (letus calltheﬂuid “water”) ismoving with different velocities. Tospecify thecharacter oftheﬂow, therefore, wemust givethethree components ofvelocity atevery point andforanytime. Ifwecanﬁndtheequations thatdetermine thevelocity, thenwe would know how theliquid moves atalltimes. Thevelocity, however, isnotthe only property thattheﬂuid haswhich varies from point topoint. Wehave just discussed thevariation ofthepressure from point topoint. And there arestill other variables. There may also beavariation ofdensity from point topoint. Inaddition, theﬂuid may beaconductor andcarry anelectric current whose densityj varies from point topoint inmagnitude anddirection. There may bea temperature which varies from point topoint, oramagnetic ﬁeld, andsoon.So thenumber ofﬁelds needed todescribe thecomplete situation willdepend onhow complicated theproblem is.There areinteresting phenomena when currents and magnetism play adominant part indetermining thebehavior oftheﬂuid; the subject iscalled magnetohydrodynamics, andgreat attention isbeing paid toitat thepresent time. However, wearenotgoing toconsider these more complicated situations because there arealready interesting phenomena atalower level of complexity, andeven themore elementary level willbecomplicated enough. Wewilltakethesituation where there isnomagnetic ﬁeldandnoconductivity, andwewillnotworry about thetemperature because wewillsuppose that the density andpressure determine inaunique manner thetemperature atanypoint. Asamatter offact, Wewillreduce thecomplexity ofourwork bymaking theas- sumption that thedensity isaconstant—we imagine that theﬂuid isessentially incompressible. Putting itanother way, wearesupposing thatthevariations of pressure aresosmall thatthechanges indensity produced thereby arenegligible. Ifthatisnotthecase, wewould encounter phenomena additional totheones we willbediscussing here—for example, thepropagation ofsound orofshock waves. Wehave already discussed thepropagation ofsound andshocks tosome extent, sowewillnow isolate ourconsideration ofhydrodynamics from these other phenomena bymaking theapproximation thatthedensity pisaconstant. Itis easy todetermine when theapproximation ofconstant pisagood one Wecan saythatifthevelocities ofﬂowaremuch lessthan thespeed ofasound wave inthe ﬂuid, wedonothave toworry about variations indensity. Theescape thatwater makes inourattempts tounderstand itisnotrelated totheapproximation of 40-2 constant density. Thecomplications thatdopermit theescape willbediscussed inthenext chapter. Inthegeneral theory ofﬂuids onemust begin with anequation ofstate for theﬂuid which connects thepressure tothedensity. Inourapproximation this equation ofstate issimply p=const. This then istheﬁrstrelation forourvariables. Thenext relation expresses the conservation ofmatter—if matter ﬂows away from apoint, there must beadecrease intheamount leftbehind. Iftheﬂuid velocity isv,then themass which ﬂows ina unittime across aunitarea ofsurface isthecomponent ofpvnormal tothesur- face. Wehave hadasimilar relation inelectricity. Wealsoknow from electricity thatthedivergence ofsuch aquantity gives therateofdecrease ofthedensity per unittime. Inthesame way, theequation v-(pv)=-git’ (40.2) expresses theconservation ofmass foraﬂuid; itisthehydrodynamic equation of continuity. Inourapproximation, which istheincompressible ﬂuid approximation, pisaconstant, andtheequation ofcontinuity issimply v-v=0. (40.3) Theﬂuid velocity v-like themagnetic ﬁeld B—-has zero divergence. (The hydro- dynamic equations areoften closely analogous totheelectrodynamic equations; that’s why westudied electrodynamics ﬁrst. Some people argue theother way; theythink thatoneshould study hydrodynamics ﬁrstsothatitwillbeeasier to understand electricity afterwards. Butelectrodynamics isreally much easier than hydrodynamics.) Wewillgetournextequation from Newton’s lawwhich tellsushowthe velocity changes because oftheforces. Themass ofanelement ofvolume ofthe ﬂuidtimes itsacceleration must beequal totheforce ontheelement. Taking an element ofunitvolume, andwriting theforce perunitvolume asf,wehave pX(acceleration) =/. Wewillwrite theforce density asthesum ofthree terms. Wehave already con- sidered thepressure force perunitvolume, ——Vp. Then there arethe“external” forces which actatadistance—like gravity orelectricity. When they arecon- servative forces with apotential perunitmass, ¢>,theygiveaforce density -pV¢. (Iftheexternal forces arenotconservative, wewould have towrite fex,forthe external force perunitvolume.) Then there ISanother “internal” force perunit volume, which isduetothefactthatinaflowing ﬂuid there canalsobeashearing stress. This iscalled theviscous force, which wewillwrite fmc. Ourequation of motion is pX(acceleration) =—Vp -pV¢—l—fem. (40.4) Forthischapter wearegoing tosuppose thattheliquid lS“thin” inthesense thattheviscosity isunimportant, sowewillomitf,.,,L.. When wedrop theviscosity term, wewillbemaking anapproximation which describes some ideal stuff rather than realwater. John vonNeumann waswellaware ofthetremendous difference between what happens when youdon’t have theviscous terms andwhen youdo, andhewasalso aware that, during most ofthedevelopment ofhydrodynamics until about 1900, almost themain interest wasinsolving beautiful mathematical problems with thisapproximation which hadalmost nothing todowith realﬂuids. Hecharacterized thetheorist who made such analyses asaman who studied “dry water ”Such analyses leave outan€SY(’l’1Zl(1l property oftheﬂuid Itis because weareleaving thisproperty outofourcalculations inthischapter that wehave given itthetitle“The Flow ofDryWater.” Wearepostponing adis- cussion ofrealwater tothenext chapter. 40-3 v+Av v(x,y,z,t) / “. P D\/(V PARTICLE VAY PATH\ Fig. 40-4. The acceleration ofa ﬂuid particle. Ifweleave outf,.,,.,,, wehave inEq.(40.4) everything weneed except anex- pression fortheacceleration. You might think thattheformula fortheaccelera- tionofaﬂuid particle would bevery simple, foritseems obvious thatifvisthe velocity ofaﬂuid particle atsome place intheﬂuid, theacceleration would just be6v/61. Itisn0t—and forarather subtle reason. Thederivative 6v/61, isthe rateatwhich thevelocity v(x,y,z,t)changes ataﬁxed point inspace. What we need ishow fastthevelocity changes foraparticular piece ofﬂuid. Imagine that wemark oneofthedrops ofwater with acolored speck sowecanwatch it.In asmall interval oftime At,thisdrop willmove toadifferent location. Ifthedrop ismoving along some path assketched inFig. 40-4, itmight inAtmove from P1toP2.Infact, itwillmove inthex-direction byanamount 0,At,inthey-direc- tion bytheamount 11,,At,andinthez-direction bytheamount v,At.Wesee that, ifv(x,y,z,t)isthevelocity oftheﬂuid particle which isat(x,y,z)atthe time t,then thevelocity ofthesame particle atthetime t+Atisgiven byv(x+ Ax,y+Ay,z—l—Az,t+At)—with Ax=v,At, Ay=221,At, and Az=0,At. From thedeﬁnition ofthepartial derivatives—recall Eq.(2.7)—we have, toﬁrst order, that v(x+v,At,y —l—vyAt,z +0,At,t+At) O 6 6 =v(x,y,z, t)+%UIA1+ ivyAt+£1)ZAl +gm. Theacceleration Av/At is 8 6 6 8 4e+aa+»e-5Wecanwrite thissymbolically——treating Vasavector—as (v-V)v+ (40.5) Note that there canbeanacceleration even though 6v/at =0sothat velocity atagiven point isnotchanging. Asanexample, water ﬂowing inacircle ata constant speed isaccelerating even though thevelocity atagiven point isnot changing. Thereason is,ofcourse, thatthevelocity ofaparticular piece ofwater which isinitially atonepoint onthecircle hasadifferent direction amoment later; there isacentripetal acceleration. Therestofourtheory isjustmathematical—ﬁnding solutions oftheequation ofmotion wegetbyputting theacceleration (40.5) intoEq.(40.4). Weget %+(v-v)v=-¥-va, (40.6) where viscosity hasbeen omitted. Wecanrearrange thisequation byusing the following identity from vector analysis: (v-V)v= (VXv)>< v+ %V(v-v). 40-4 Ifwenow deﬁne anewvector ﬁeld Q,asthecurlofv, Q=VXv, (40.7) thevector identity canbewritten as (v-V)v =QXv+ %Vii2, » fndourequation ofmotion (40.6) becomes @+o><v+lvi»*=-E-v¢ (408)61 2 p ' ' You canverify that Eqs. (40.6) and(40.8) areequivalent bychecking that the components ofthetwosides oftheequation areequal—and making useof(40.7). Thevector ﬁeldQiscalled thevorticity. Ifthevorticity iszeroeverywhere, we saythattheﬂow isirrotational. Wehave already deﬁned inSection 3-5athing called thecirculation ofavector ﬁeld. Thecirculation around anyclosed loop ina ﬂuid isthelineintegral oftheﬂuid velocity, atagiven instant oftime, around that loop: (Circulation) =§v-ds. The circulation perunit area foraninﬁnitesimal loop isthen—using Stokes’ theorem—-equal toVXv.Sothevorticity Qisthecirculation around aunit area (perpendicular tothedirection of£2).Italsofollows thatifyouputalittle piece ofdirt—n0t aninﬁnitesimal point—-at anyplace intheliquid itwillrotate with theangular velocity Q/2. Trytoseeifyoucanprove that. You canalso check itoutthatforabucket ofwater onaturntable, Qisequal totwice thelocal angular velocity ofthewater. Ifweareinterested only inthevelocity ﬁeld, wecaneliminate thepressure from ourequations. Taking thecurlofboth sides ofEq.(40.8), remembering that pisaconstant andthatthecurlofanygradient iszero, andusing Eq.(40.3), weget ‘:,—£:+v><(n><v)=0. (40.9) This equation, together with theequations £2=VXv (40.10) and V-v =0, (40.11) describes completely thevelocity ﬁeld v.Mathematically speaking, ifweknow £2 atsome time, then weknow thecurl ofthevelocity vector, andwealso know thatitsdivergence iszero, sogiven thephysical situation wehave allweneed to determine veverywhere. (Itisjustlikethesituation inmagnetism where wehad V-B=0andVXB=j/eocz.) Thus, agiven £2determines vjustasagiven jdetermines B.Then, knowing v,Eq.(40.9) tellsustherateofchange ofQfrom which wecangetthenewQforthenext instant. Using Eq.(40.10), again weﬁnd thenewv,andsoon.You seehow these equations contain allthemachinery for calculating theﬂow. Note, however, thatthisprocedure gives thevelocity ﬁeld only; wehave lostallinformation about thepressure. Wepoint outonespecial consequence ofourequation. IfQ=0everywhere atanytime t,69/6t alsovanishes, sothatQisstillzero everywhere att+At. Wehave asolution totheequation; theﬂow ispermanently irrotational. Ifa ﬂow wasstarted with zero rotation, itwould always have zero rotation. The equations tobesolved then are V-v=O, VXv=0. They arejust liketheequations fortheelectrostatic ormagnetostatic ﬁelds in freespace. Wewillcome back tothem andlook atsome special problems later. 40-5 i;-7""*‘" ‘Wl ‘s I m __ __ A__ Fig.40-5. Streamlines insteady ﬂuid ﬂow. (0) VI./r __.Al40-3 Steady flow—-Bernoulli’s theorem Now wewant toreturn totheequation ofmotion, Eq.(40.8), butlimit our- selves tosituations inwhich theﬂowis“steady.” Bysteady ﬂowwemean that atanyoneplace intheﬂuid thevelocity never changes. Theﬂuid atanypoint is always replaced bynewﬂuid moving inexactly thesame way. Thevelocity picture always looks thesame—v isastatic vector ﬁeld. Inthesame waythatwedrew “ﬁeld lines” inmagnetostatics, wecannow draw lines which arealways tangent totheﬂuid velocity asshown inFig. 40-5. These lines arecalled streamlines. Forsteady ﬂow, theyareevidently theactual paths ofﬂuid particles. (Inunsteady ﬂow thestreamline pattern changes intime, andthestreamline pattern atany instant does notrepresent thepath ofaﬂuid particle.) Asteady ﬂow does notmean thatnothing ishappening—atoms intheﬂuid aremoving andchanging their velocities. Itonly means that 6v/6t =0.Then ifwetake thedotproduct ofvintotheequation ofmotion, theterm v~(QXv) drops out,andweareleftwith v-v{§4-¢+-%ﬁ]=0. (m1n This equation saysthatforasmall displacement inthedirection oftheﬂuid velocity thequantity inside thebrackets doesn’t change. Now insteady ﬂow alldisplace- ments arealong streamlines, soEq(40.12) tellsusthatforallthepoints along a streamline, wecanwrite l%+EU2—l—¢=const (streamline). (40.13) This isBernoulli’s theorem. Theconstant may ingeneral bedifferent fordifferent streamlines; allweknow isthattheleft-hand sideofEq.(40.13) isthesame all along agiven streamline. Incidentally, wemay notice thatforsteady irrotational motion forwhich Q=0,theequation ofmotion (40.8) gives ustherelation vE+%ﬁ+d=Q sothat ~e+|\))1l— —112—l—¢==const (everywhere). (40.14)‘O It’sjustlikeEq.(40.13) except thatnowtheconstant hasthesame value throughout theﬂuid. vAt1/i , "2‘,4 2 _/ (b) u W -'* A2 _,, Fig. 40-6. Fluid motion inClﬂowtube. Thetheorem ofBernoulli isinfactnothing more than astatement ofthecon- servation ofenergy. Aconservation theorem such asthisgives usalotofinforma- tion about aﬂow without ouractually having tosolve thedetailed equations. Bernoulli’s theorem issoimportant andsosimple thatwewould liketoshow you how itcanbederived inawaythatisdifferent from theformal calculations we have justused. Imagine abundle ofadjacent streamlines which form astream tube assketched inFig.40-6. Since thewalls ofthetube consist ofstreamlines, noﬂuid ﬂows outthrough thewall. Let’s callthearea atoneendofthestream 40-6 tube A1,thefluid velocity there 2'1,thedensity oftheﬂuid p1,andthepotential energy ¢1. Attheother endofthetube, wehave thecorresponding quantities A2,v2.p2,and¢2.Now after ashort interval oftimeAt,theﬂuid atA1hasmoved adistance 7'1At,andtheﬂuid atA2hasmoved adistance 02At[Fig. 40-6(b)]. Theconservation ofmass requires thatthemass which enters through A1must be equal tothemass which leaves through A2.These masses atthese twoends must bethesame: AM =p1A1l)1Af =/J2/12122 Al. Sowehave theequality pl/11111 =p2/1202. This equation telljausthatthevelocity varies inversely with thearea ofthestream tube ifpisconstfint. Now wecalculate thework done bytheﬂuid pressure. Thework done onthe ﬂuid entering atA1isp1A101At,andthework given upatA2isp2A2i12 AtThe network ontheﬂuid between A1andA2is,therefore, P1/11741 A1— P2/4292 Al, which must equal theincrease intheenergy ofamass AMofﬂuid ingoing from A1toA2. Inother words, ])1A1l)]At * [)2/42U2 I * E1), where E1istheenergy perunitmass ofﬂuid atA1,andE2istheenergy perunit mass atA2.Theenergy perunitmass oftheﬂuid canbewritten as E=%v2+¢+U. where 229isthekinetic energy perunitmass, ¢isthepotential energy perunit mass, andUisanadditional term which represents theinternal energy perunitmass ofﬂuid. Theinternal energy might correspond, forexample, tothethermal energy inacompressible ﬂuid, ortochemical energy. Allthese quantities can vary from point topoint. Using thisform fortheenergies in(40.16), wehave [)é1/(ill/‘Y/[iit—!)‘2*/‘gill,’/I2'g:'%_"gJr<l>2"l' U2—;Y‘:i—¢i— U1 Btitwehave seen thatAM =pAi~At,soweget 1 1>Bl+—vi+<t>1+U1=H3+—1'§+¢2-l-U2, (40.17)Pi 2 P2 2 which istheBernoulli result with anadditional term fortheinternal energy. If theﬂuid isincompressible, theinternal energy term isthesame onboth sides, and wegetagain thatEq.(40.14) holds along anystreamline. Weconsider now some simple examples inwhich theBernoulli integral gives usadescription oftheﬂow. Suppose wehave water ﬂowing outofahole near thebottom ofatank. asdrawn inFig.40-7. Wetake asituation inwhich the ﬂow speed ii,,,,,atthehole ismuch larger than theﬂow speed near thetopofthe tank; inother words, weimagine thatthediameter ofthetank issolarge that wecanneglect thedrop intheliquid level. (We could make amore accurate calculation ifwewished.) Atthetopofthetank thepressure ispu, theatmospheric pressure, andthepressure atthesides ofthejetisalsop11. Now wewrite our Bernoulli equation forastreamline, such astheoneshown intheﬁgure. Atthe topofthetank, wetake 0equal tozero andwealsotake thegravity potential ¢tobezero. Atthespeed ii,,,,1,and¢>=-gh, sothat P0=P0+311113111 _pgh, or ii,,1,1 =\/Zgli. (40.18) 40-7 it[1, 1 A _ ‘L _ _ WATER l l_._ __ + __ ___ \ \ -- _*\__ \ srrﬁw was-:'\_ 9\_----~\_. /\ Po 1_ Vout \ Fig. 40-7. Flow fromatank. Fig. 40-8. With are-entrant dis- L: -.1 _ 1 T i, A i_ 5 ___":-- 5 Fig.40-9. The pressure islowest where thevelocity ishighest.charge tube, thestream contracts toone- half thearea oftheopening. This velocity isjustwhat wewould getforsomething which falls thedistance h. Itisnottoosurprising, since thewater attheexitgains kinetic energy attheex- pense ofthepotential energy ofthewater atthetop. Donotgettheidea, however, thatyoucanﬁgure outtheratethattheﬂuid ﬂows outofthetank bymultiplying thisvelocity bythearea ofthehole. Theﬂuid velocities asthejetleaves thehole arenotallparallel toeach other buthave components inward toward thecenter ofthestream—the jetisconverging. After thejethasgone alittle way, thecon- traction stops andthevelocities dobecome parallel. Sothetotal ﬂowisthevelocity times theareaatthatpoint. Infact,ifwehaveadischarge opening which isjusta round holewithasharp edge, thejetcontracts to62percent oftheareaofthehole. Thereduced effective area ofthedischarge varies fordiﬂerent shapes ofdischarge tubes, andexperimental contractions areavailable astables ofeﬁlux coeﬂicients. Ifthedischarge tubeisre-entrant, asshown inFig.40-8, itispossible toprove inamost beautiful waythattheeﬂlux coefﬁcient isexactly 50percent. Wewill givejustahintofhow theproof goes. Wehave used theconservation ofenergy togetthevelocity, Eq.(40.18), butthere isalso momentum conservation to consider. Since there isanoutﬂow ofmomentum inthedischarge jet,there must beaforce applied over thecross section ofthedischarge tube Where does the force come from? Theforce must come from thepressure onthewalls. Aslong astheeﬁ’lux holeissmall andaway from thewalls, theﬂuid velocity near thewalls ofthetank willbevery small. Therefore, thepressure onevery faceisalmost exactly thesame asthestatic pressure inaﬂuid atrest——from Eq.(30.14). Then thestatic pressure atanypoint onthesideofthetank must bematched byan equal pressure atthepoint ontheopposite wall, except atthepoints onthewall opposite thecharge tube. Ifwecalculate themomentum poured outthrough the jetbythispressure, wecanshow thattheeﬂlux coefﬁcient isl/2. Wecannot use thismethod foradischarge hole likethatshown inFig.40-7, however, because thevelocity increase along thewall right near thediscglarge area gives apressure fallwhich wearenotabletocalculate. Let’s look atanother example—a horizontal pipe with changing cross section, asshown inFig. 40-9, with water ﬂowing inoneend and outthe other. Theconservation ofenergy, namely Bernoulli’s formula, saysthatthepres- sure islower intheconstricted area where thevelocity ishigher. Wecaneasily demonstrate thiseffect bymeasuring thepressure atdifferent cross sections with small vertical columns ofwater attached totheﬂow tube through holes small enough sothatthey donotdisturb theﬂow. Thepressure isthen measured by theheight ofwater inthese vertical columns. Thepressure 1Sfound tobelessat theconstriction than itisoneither side. Iftheareabeyond theconstriction comes back tothesame value ithadbefore theconstriction, thepressure rises again. 40-8 Bernoulli’s formula would predict that thepressure downstream ofthecon- striction should bethesame asitwasupstream, btitactually itisnoticeably less. Thereason thatourprediction iswrong isthatwehave neglected thefrictional, viscous forces which cause apressure drop along thetube. Despite thispressure drop thepressure isdeﬁnitely lower attheconstriction (because oftheincreased speed) than itisoneither side ofit—as predicted byBernoulli. Thespeed 1:2 must certainly exceed 1/1togetthesame amount ofwater through thenarrower tube. Sothewater accelerates ingoing from thewide tothenarrow part. The force thatgives thisacceleration comes from thedrop inpressure. Wecancheck ourresults with another simple demonstration Suppose we have onatank adischarge tube which throws ajetofwater upward asshown in Fig40-10. Iftheelllux velocity were exactly \/257/1', thedischarge water should risetoalevel even with thesurface ofthewater inthetank. Experimentally, it falls somewhat short. Ourprediction isroughly right, butagain viscous friction which hasnotbeen includejj inourenergy conservation formula hasresulted in alossofenergy Have youever held twopieces ofpaper close together andtried toblow them apart? Tryit!They come together. Thereason, ofcourse. isthattheairhas ahigher speed going through theconstricted space between thesheets than it does when itgetsoutside. Thepressure between thesheets islower than atmos- pheric pressure, sothey come together rather than separating. 40-4 Circulation Wesawatthebeginning ofthelastsection thatifwehave anincompressible ﬂuid with nocirculation, theﬂow satisﬁes thefollowing twoequations: V-v=O, VXv=0. (40.19) They arethesame astheequations ofelectrostatics ormagnetostatics inempty space. Thedivergence oftheelectric ﬁeld iszero when there arenocharges, and thecurloftheelectrostatic ﬁeld isalways zero. Thecurlofthemagnetic ﬁeld is zero ifthere arenocurrents. andthedivergence ofthemagnetic ﬁeld isalways zero. Therefore, Eqs. (40.19) have thesame solutions astheequations forEin electrostatics orforBinmagnetostatics. Asamatter offact, wehave already solved theproblem oftheﬂow ofaﬂuid pastasphere, asanelectrostatic analogy, inSection 12-5. Theelectrostatic analog isauniform electric ﬁeld plus adipole ﬁeld. Thedipole ﬁeld issoadjusted thattheﬂow velocity normal tothesurface ofthesphere iszero. Thesame problem fortheﬂow pastacylinder canbeworked outinasimilar waybyusing asuitable linedipole with auniform ﬂowﬁeld. This solution holds forasituation inwhich theﬂuid velocity atlarge distances iscon- stant—both inmagnitude anddirection. Thesolution issketched inFig.40-11(a). There isanother solution fortheﬂow around acylinder when theconditions aresuch thattheﬂuid atlarge distances moves incircles around thecylinder. The ﬂowis,then, circular everywhere, asinFig.40-1l(b). Such aﬂow hasacirculation around thecylinder, although VXvisstillzero intheﬂuid. How canthere be circulation without acurl? Wehave acirculation around thecylinder because the lineintegral ofvaround anyloop enclosing thecylinder isnotzero. Atthesame time, thelineintegral ofvaround anyclosed path which does notinclude thecyl- inder iszero. Wesawthesame thing when wefound themagnetic ﬁeld around a wire. ThecurlofBwaszero outside ofthewire, although alineintegral ofB around apath which encloses thewire didnotvanish. Thevelocity ﬁeld inanir- rotational circulation around acylinder isprecisely thesame asthemagnetic ﬁeldaround awire. Foracircular path with itscenter atthecenter ofthecylinder, thelineintegral ofthevelocity is fv-ds=27171‘. Forirrotational ﬂow theintegral must beindependent ofr.Let’s calltheconstant 40-9;____i_, -- \/ ,-/"T ,1/'/ZZZ ;T’1' ,// miY- _ _ l1 1 Fig. 40-10. Proof that visnotequal to\/2gh. -—§>-—--——i _\- »—\ J \__.>_._ _ /W -3‘it _'>in‘\ Li} xiigig _\ K//\_ iZ, gm, ______>_ .7 g xii,’ \ (1%;lF _ -é (7?-. {AZ -+-\ R_A- _ A > Li; .__mm--//I IL. > >. ._._All /“T *—-(.7 _>_ 7 Fig. 40-l 1. la)Ideal ﬂuid flow past acylinder. (bl Circulation around a cylinder. (c)The superposition ofla) and (bl. /\ / Z 3 \ //*\\ \\ // \\‘f, if/’>\&____\ _( /\\ ll ///\\ \\®’l/ l -\LL’|| Fig. 40-12. Water with circulation draining from atank.value C,then wehave that C1)=-2-E9 (40.20) where visthetangential velocity, andristhedistance from theaxis. There isanicedemonstration ofaﬂuid circulating around ahole. Youtakea transparent cylindrical tank with adrain holeinthecenter ofthebottom. Youﬁll itwith Water, stirupsome circulation with astick, andpullthedrain plug. You getthepretty effect shown inFig. 40-12. (You’ve seen asimilar thing many times inthebathtub!) Although youputinsome watbeginning, itsoon diesdown because ofviscosity andtheﬂow becomes irrotational-although stillwith some circulation around thehole. From thetheory, wecancalculate theshape oftheinner surface ofthewater. Asaparticle ofthewater moves inward itpicks upspeed. From Eq.(4020)the tangential velocity goes as1/r-—it’s just from theconservation ofangular mo- mentum, liketheskater pulling inherarms. Also theradial velocity goes as l/r. Ignoring thetangential motion, wehave water going radially inward toward ahole; from V~v=0,itfollows thattheradial velocity isproportional tol/r. Sothetotal velocity alsoincreases as1/r,andthewater goesinalong Archimedean spirals. Theair-water surface isallatatmospheric pressure, soitmust have—from Eq.(40.l4)—the property that gz—l—%mv2 =const. Butvisproportional to1/r,sotheshape ofthesurface is k (Z—Z0)=;2' Aninteresting point—-which isnottrueingeneral butistrueforincompressible, irrotational ﬂow—is thatifwehave onesolution andasecond solution, then the sum isalso asolution. This istrue because theequations in(40.19) arelinear. Thecomplete equations ofhydrodynamics, Eqs. (40.8), (40.9), and(40.10), are notlinear, which makes avast difference. Fortheirrotational ﬂow about the cylinder, however, wecansuperpose theﬂow ofFig. 40-ll(a) ontheﬂow of Fig.40-1l(b)andgetthenewﬂow pattern shown inFig.40-1 l(c). This ﬂow is ofspecial interest. Theﬂow velocity ishigher ontheupper sideofthecylinder than onthelower side. Thepressures aretherefore lower ontheupper sidethan onthelower side. Sowhen wehave acombination ofacirculation around a cylinder andanethorizontal ﬂow, there isanetverticalforce onthecylinder-—it iscalled aliftforce. Ofcourse, ifthere isnocirculation, there isnonetforce on anybody according toourtheory of“dry” water. 40-5 Vortex lines Wehave already written down thegeneiial equations fortheﬂow ofanin- compressible ﬂuid when there may bevorticity. They are I.V-v=O, II.Q=V><v, III.%‘t-’+v><(o><v)=0. Thephysical content ofthese equations hasbeen described inwords byHelmholtz interms ofthree theorems. First, imagine thatintheﬂuid wewere todraw vortex lines rather than streamlines. Byvortex lines wemean ﬁeld lines that have the direction ofQandhave adensity inanyregion proportional tothemagnitude of £2.From IIthedivergence ofQisalways zero (remember—Section 3-7-—that the divergence ofacurlisalways zero). Sovortex lines arelikelines ofB—they never start orstop, andwilltend togoinclosed loops. Now Helmholtz described III 40-10 inwords bythefollowing statement: thevortex lines move with theﬂuid. This means that ifyouwere tomark theﬂuid particles along some vortex lines—by coloring them with ink,forexample—then astheﬂuid moves andcarries those particles along, they willalways mark thenew positions ofthevortex lines. In whatever way theatoms oftheliquid move, thevortex lines move with them That isonewaytodescribe thelaws. Italso suggests amethod forsolving anyproblems. Given theinitial ﬂow pattern-say veverywhere-then youcancalculate Q.From thevyoucanalso tellwhere thevortex lines aregoing tobealittle later—they move with thespeed v.With thenewQyoucanuselandlltoﬁndthenewv.(That’s justlikethe problem ofﬁnding B,given thecurrents.) Ifwearegiven theﬂow pattern atone instant wecaninprinciple calculate itforallsubsequent times. Wehave thegeneral solution fornonviscous ﬂow. Wewould liketoshow howHelmholtz’s statement—and, therefore, III—can beatleast partly understood. Itisreally justthelawofconservation ofangular momentum applied totheﬂuid. Suppose weimagine asmall cylinder oftheliquid whose axisisparallfil tothevortex lines, asinFig.40-l3(a). Atsome time later, thissame piece ofﬂilid willbesomewhere else. Generally itwilloccupy acylinder with adifferent diameter andbeinadifferent place. Itmay alsohave adifferent orientation, sayasinFig.40—l3(b). Ifthediameter haschanged, however, the length willhave increased tokeep thevolume constant (since weareassuming an incompressible ﬂuid). Also, since thevortex lines arestuck with thematerial, their density willgoupasthecross-sectional area goes down. Theproduct ofthe vorticity Qand area Aofthecylinder willremain constant, soaccording to Helmholtz, weshould have 02A2 =t21A1. (40.21) Now notice thatwith zero viscosity alltheforces onthesurface ofthecy- lindrical volume (oranyvolume, forthatmatter) areperpendicular tothesurface Thepressure forces cancause thevolume tobemoved from place toplace, or cancause ittochange shape; butwith notangential forces themagnitude ofthe angular momentum ofthematerial inside cannot change. Theangular momentum oftheliquid inthelittle cylinder isitsmoment ofinertia Itimes theangular velocity oftheliquid, which isproportional tothevorticity S2.Foracylinder, the moment ofinertia isproportional tomr2. Sofrom theconservation ofangular momentum, wewould conclude that (MiRi)9i =(M2Rg)92- Butthemass isthesame, M1=M2, andtheareas areproportional toR2,so wegetagain just Eq.(40.21). Helmholtz’s statement—which isequivalent to III-—is justaconsequence ofthefactthatintheabsence ofviscosity theangular momentum ofanelement oftheﬂuid cannot change. ti=2‘ F_/i O 1‘ _\,\ -—>/ /////// //// ,4/// (0) //// AREAA / /// /// / ///// / / / AREAA’,/, .. (bl / /// ///// /’/ Fig. 40-13. la) Agroup ofvortex lines att;(blthe same lines atalater time ll. Fig. 40-14. Making atravelling vor- ////////////////// texring. There isanice demonstration ofamoving vortex which ismade with the simple apparatus ofFig40-14. Itisa“drum” twofeetindiameter andtwofeet long made bystretching athick rubber sheet over theopen endofacylindrical “box.” The“bottom”—the drum istipped onitsside—is solid except fora3-inch diameter hole. Ifyougiveasharp blow ontherubber diaphragm with your hand, avortex ringisprojected outofthehole. Although thevortex isinvisible, youcan tellit’sthere because itwillblow outacandle 10to20feetaway. Bythedelay in 40-1 l TX VORTEX '/ \ V K‘)1') iv) 0)(I) V (bl vontcx __€____ LINES\ v DIRECTION\ OFMOTION Q0 /as®e V Fig. 40-15. Amoving vortex ring fasmoke ring). (a)Thevortex lines. (blA cross section ofthering.theeffect, youcantellthat“something” istravelling ataﬁnite speed. You can seebetter what isgoing onifyou ﬁrst blow some smoke into thebox. Then you seethevortex asabeautiful round “smoke ring.” Thesmoke ring ISatorus-shaped bundle ofvortex lines, asshown inFig 40-l5(a). Since £2=VXv,these vortex lines represent alsoaciiculation ofv asshown inpart (b)oftheﬁgure. Wecanunderstand theforward motion ofthe ring inthefollowing way: The circulating velocity around the/iottom ofthering extends uptothetopofthering, having there aforward motion. Since thelines of$2move with theﬂuid, they alsomove ahead with thevelocity v.(Ofcourse, thecirculation ofvaround thetoppart oftheringisresponsible fortheforward motion ofthevortex lines atthebottom. Wemust now mention aserious diﬂiculty Wehave already noted thatEq. (409)says that, ifs:isinitially zero, itwillalways bezero. This result isagreat failure ofthetheory of“dry” water, because itmeans that once Qiszero itis always zero—it ISimpossible toproduce anyvorticity under anycircumstance. Yet, inoursimple demonstration with thedrum, wecangenerate avortex ring starting with airwhich wasinitially atrest. (Certainly, v10,$2:Oeverywhere intheboxbefore wehitit.)Also, weallknow thatwecanstart some vorticity ina lakewithapaddle. Clearly, wemust gotoatheory of“wet” water togetiicomplete understanding ofthebehavior ofaﬂuid. Another feature ofthedrywater theory which isincorrect 1Sthesupposition wemake regarding theﬂow attheboundary between itandthesurface ofasolid. When wediscussed theﬂow past acylinder—-as inFig.40-ll, forexample—we permitted theﬂuid toslide along thesurface ofthesolid. lnourtheory, the velocity atasolid surface could have anyvalue depending onhowitgotstarted. andwedidnotconsider any“friction” between theﬂuid andthesolid. ltisan experimental fact, however, thatthevelocity ofarealﬂuid always goes tozero at thesurface ofasolid object. Therefore, oursolution forthecylinder, with or without circulation, iswrong—as isourresult regarding thegeneration ofvorticity. Wewilltellyouabout themore correct theories inthenext chapter. 40-l 2 41 The Flow ofWet Water 41—1 Viscosity Inthelastchapter wediscussed thebehavior ofwater, disregarding the phenomenon ofviscosity. Now wewould liketodiscuss thephenomena ofthe ﬂowofﬂuids, including theeffects ofviscosity. Wewant tolook attherealbehavior ofﬂuids. Wewilldescribe qualitatively theactual behavior oftheﬂuids under various different circumstances sothatyouwillgetsome feelfortheSUb)€Ct. Al- though youwillseesome complicated equations andhear about some complicated things, itisnotourpurpose thatyoushould learn allthese things. This is,ina sense, a“cultural” chapter which Wlllgiveyousome ideaofthewaytheworld is. There isonly oneitem which isworth learning, andthatisthesimple deﬁnition of viscosity which wewillcome toinamoment. Therestisonly foryour entertain- ment. Inthelastchapter wefound thatthelaws ofmotion ofaﬂuid arecontained intheequation %l;+(v-V)v=——Vpl—V¢+%- (41.1) Inour“dry” water approximation weleftoutthelastterm, sowewere neglecting allviscous effects. Also, wesometimes made anadditional approximation by considering theﬂuid asincompressible; then wehadtheadditional equation V-v=O. This lastapproximation isoften quite good~particularly when ﬂow speeds are much slower than thespeed ofsound. Butinrealﬂuids itisalmost never truethat wecanneglect theinternal friction thatwecallviscosity; most oftheinteresting things thathappen come from itinonewayoranother. Forexample, wesawthat in“dry” water thecirculation never changes——if there isnone tostart outwith, there willnever beany. Yet, circulation inﬂuids isaneveryday occurrence. We must ﬁxupourtheory. Webegin with animportant experimental fact. When weworked outthe ﬂow of“dry” water around orpastacylinder——the so-called “potential ﬂow”——We hadnoreason nottopermit thewater tohave avelocity tangent tothesurface; only thenormal component hadtobezero. Wetook noaccount ofthepossibility thatthere might beashear force between theliquid andthesolid. Itturns out— although itisnotatallself-evident—that inallcircumstances where ithasbeen experimentally checked, thevelocity ofaﬂuid isexactly zero atthesurface ofa solid. You have noticed, nodoubt, thattheblade ofafanwillcollect athinlayer of dust—and thatitisstillthere after thefanhasbeen churning uptheair. You canseethesame eﬂect even onthegreat fanofawind tunnel. Why isn't thedust blown oﬂ"bytheair? Inspite ofthefactthatthefanblade ismoving athigh speed through theair,thespeed oftheairrelative tothefanblade goes tozero right at thesurface. Sothevery smallest dust particles arenotdisturbed.* Wemust modify thetheory toagree with theexperimental factthatinallordinary ﬂuids, themolecules next toasolid surface have zero velocity (relative tothesurface).T *You canblow large dust particles from atable top,butnotthevery ﬁnest ones. The large ones stick upintothebreeze. TYou canimagine circumstances when itisnottrue: glass istheoretically a“liquid,” butitcancertainly bemade toslide along asteel surface. Soourassertion must break down somewhere. 41-141-1 Viscosity 41-2 Viscous ﬂow 41-3 TheReynolds number 41-4 Flow pastacircular cylinder 41-5 Thelimit ofzeroviscosity 41-6 Couette ﬂow l>21F71Dl> v_9_, \///////////////////////////I4]-i—> .I_/l.Tl‘, T Vl'-—-W’ " ‘ I~ d .' ‘ FLUID Fig. 41-1. Viscousdragbetweentwo ‘L _ I '__' /////////////parallel plates. r .~ . MAF ., ___\t___geav, _._AyI'_F_--ii_' _-.X_>_ V___.-_____> —|.— : —r — 4; Fig. 4l—2. The shear stress ina viscous ﬂuid. ti [ZZZ T/]~]_)\ Q ,f/ \. \\<\ \ I\v»7 I I \>\ KY1. / \< \< \Q4»-{ FLUID 1142* my Vb;>ixmx XXX /”"_i“~.\U //U\\ 5 \\\~ k/‘-__,,/'/ \J¢"\§§\‘§9">1 Fig. 4l—3. The ﬂow inafluid be- tween two concentric cylinders rotating atdif¥erent angular velocities.'l/// ///////////_’j/A v=O Weoriginally characterized aliquid bythefactthat ifyouputashearing stress onit—no matter howsmall—it would giveway. ltﬂows. lnstatic situations, there arenoshear stresses. Butbefore equilibrium isreached—as long asyoustill push onit~there canbeshear forces. Viscosity describes these shear forces which exist inamoving ﬂuid. Togetameasure oftheshear forces during themotion ofaﬂuid, weconsider thefollowing kind ofexperiment. Suppose thatwehave two solid plane surfaces with water between them, asinFig.4l—l, andwekeep one stationary while moving theother parallel toitattheslow speed 110.lfyou measure theforce required tokeep theupper plate moving, youﬁndthatitISproportional totheareaoftheplates andtozit./d. where disthedistance between theplates. So theshear stress F/A isproportional to00/dz E- 91».A_”d Theconstant ofproportionality 1;iscalled thecoefficient ofviscosity. Ifwehave amore complicated situation, wecanalways consider alittle, ﬂat, rectangular cellinthewater with itsfaces parallel totheﬂow, asinFig.41-2. The shear force across thiscellisgiven by AF at ai» _= __’= -4. 4|. AA ”Ay ”6y l2) Now, 61',/6y istherateofchange oftheshear strain wedeﬁned inChapter 38,so foraliquid, theshear stress isproportional totherateofchange oftheshear strain. Inthegeneral casewewrite s,,,=17 + - (41.3) Ifthere isauniform rotation oftheﬂuid, 61‘,/6y isthenegative of60,,/6x andS,,,, iszero—as itshould besince there arenostresses inauniformly rotating ﬂuid. (We didasimilar thing indeﬁning em,inChapter 39.) There are,ofcourse, the corresponding expressions forSy,andS_.,,. Asanexample oftheapplication ofthese ideas, weconsider themotion ofa ﬂuid between twocoaxial cylinders. Lettheinner onehave theradius aandthe peripheral velocity va,andlettheouter onehave radius bandvelocity ii/,.See Fig.4l—3. Wemight ask,what isthevelocity distribution between thecylinders ‘? Toanswer thisquestion, webegin byﬁnding aformula fortheviscous shear in theﬂuid atadistance rfrom theaxis From thesymmetry oftheproblem, wecan assume thattheﬂow isalways tangential andthatitsmagnitude depends only on r;Z)=v(r). Ifwewatch aspeck inthewater attheradius r,itscoordinates asa function oftime are x=rcos wt, y=rsin wt, where (.0=ii/r, Then thex-andy-components ofvelocity are ii,=—rw sinwt=-—wy and try=rwcoswt=wx. (41.4) From Eq.(41.3), wehave ((A7 6 6 3 6 Sm]IVila; (Kw) —53)(y¢°)jl =Tllx (ix”J’ (41-5) 41-2 Forapoint aty=0,aw/6y =0,andx6w/6x isthesame asrdw/dr. Soatthat point dw (S11/)1/=0 :ll"‘ (41-6)c/r (Itisreasonable that Sshould depend on6w/61-: when there isnochange in0: with r,theliquid isinuniform rotation andthere arenostresses.) The stress wehave calculated isthetangential shear which isthesame all around thecylinder Wecangetthetorque acting across acylindrical surface at theradius rbymultiplying theshear stress bythemoment arm randthearea 21rrl. Weget T=21ir2l(s,,,),:(, =27T'I1lI‘3 (41.7) Since themotion ofthewater issteady——there isnoangular acceleration—the nettorque onthecylindrical shell ofwater between randr—l—drmust bezero; thatis,thetorque atrmust bebalanced byanequal andopposite torque atr—l—dr, sothatTmust beindependent ofr.Inother words, r3dw/dr isequal tosome con- stant, sayA,and dw A Integrating, weﬁndthattovaries with ras Aw=—Trz+B. (41.9) The constants AandBaretobedetermined toﬁttheconditions that to=w,, atr=a,andw =w,,atr=b.Wegetthat 2a2b2 A=BY, (wt_ma): 2 2 (41.10) bwi,—aw,,, B-Wigs’ SoWeknow wasafunction ofr,andfrom itv=wr. Ifwewant thetorque, wecangetitfrom Eqs. (41.7) and(41.8): T=27I"r;lA Or 41rlaih“T=-5?"a,(w1, -41,). (41.11) Itisproportional totherelative angular velocities ofthetwocylinders. Onestand- ardapparatus formeasuring thecoefﬁcients ofviscosity isbuilt thisway. One cylinder——say theouter one-—is onpivots butisheld stationary byaspring balance which measures thetorque onit,while theinner oneisrotated ataconstant angular velocity. Thecoefﬁcient ofviscosity isthen determined from Eq.(41.11). From itsdeﬁnition, youseethattheunits of17arenewton~sec/m2. Forwater at20°C, 11=103newton-sec/m2. Itisusually more convenient tousethespeciﬁc viscosity, which is1;divided by thedensity p.Thevalues forwater andairarethen comparable: water at20°C, 1;/p=lO“’m2/sec, (41.12) airat20°C, 1;/p=15X10”“ m2/sec. Viscosities usually depend strongly ontemperature. Forinstance, forwater just above thefreezing point, 17/pis1.8times larger than itisat20°C. 4l»3 41-2 Viscous ﬂow Wenow gotoageneral theory ofviscous ﬂow—at least inthemost general form known toman Wealready understand thattheshear stress components are proportional tothespatial derivatives ofthevarious velocity components such as611,,/6y or811,,/0x. However, inthegeneral case ofacompressible ﬂuid there is another term inthestress which depends onother derivatives ofthevelocity. Thegeneral expression is s.,= + +7,’is,,(v U), (41.13) where x,isanyoneoftherectangular coordinates x.y,orz,andv,isanyoneof therectangular coordinates ofthevelocity. (The symbol 6,,istheKronecker delta which islwhen i=/‘and Ofor1;é/.)Theadditional term adds i7’V-v toallthediagonal elements S,,ofthestress tensor. Iftheliquid isincompressible V-v=O,andthisextra term doesn’t appear. Soithastodowith internal forces during compression. Sotwoconstants arerequired todescribe theliquid, Just aswehadtwoconstants todescribe ahomogeneous elastic solid. Thecoellicient 17isthe“ordinary” coeﬂicient ofviscosity which wehave already encountered. Itisalsocalled theﬁrst coefﬁcient ofviscosity orthe“shear viscosity coefficient,” andthenewcoeﬂicient 77’iscalled thesecond coe/ﬁcient ofviscosity. Now wewant todetermine theviscous force perunitvolume,f.,..., sowecan putitintoEq(411)togettheequation ofmotion forarealﬂuid. Theforce on<1 small cubical volume element ofaﬂuid istheresultant oftheforces onallthesix faces. Taking them twoatatime, wewillgetdiﬁerences that depend onthe derivatives ofthestresses, and, therefore, onthesecond derivatives ofthevelocity. This isnicebecause itwillgetusback toavector equation. Thecomponent of theviscous force perunit volume inthedirection oftherectangular coordinate x,is 38S(fVlSO)l =E ‘E/‘L: /=1 _ QatQt Q7. - 7xptM>l+Ma7vv) one :1 KM“.-Q:></1’? /5Q: Usually, thevariation oftheviscosity coeﬂicients with position isnotsigniﬁcant andcanbeneglected. Then, theviscous force perunitvolume contains onlysecond derivatives ofthevelocity. WesawinChapter 39thatthemost general form of second derivatives thatcanoccur inavector equation isthesum ofaterm inthe Laplacian (V-Vv:Vzv), andaterm inthegradient ofthe divergence (V(V -v)). Equation (41.14) 1S_]USlsuch asumwith thecoeﬂicients 77and(77-1-77’).Weget .ma=7W»+o+7awvv) Mum Intheincompressible case, V*v=O,andtheviscous force perunitvolume is just77Vzv. That isallthatmany people use;however, ifyoushould want tocal- culate theabsorption ofsound inaﬂuid, youwould need thesecond term. Wecannowcomplete ourgeneral equation ofmotion forarealﬂuid. Sub- stituting Eq.(4115)intoEq.(41.1), weget pl§~f+ (v-V)v} I~Vp—pVd>+ 17V2v—l— (i7+ i7’)V(V-v) It’scomplicated. Butthat’s thewaynature is. Ifweintroduce thevorticity £2IVXv,aswedidbefore, wecanwrite our equation as pl¥+£2><v—l—;Vi/2}: —V])—pV¢—l—i7V2l) +(ii+ti’)V(V'v)- (41-16) 41-4 Wearesupposing again thattheonly body forces acting areconservative forces likegravity. Toseewhat thenew term means, let’s look attheincompressible ﬂuid case. Then, ifwetake thecurlofEq.(41.16), weget ‘ll;+v><(o><v)=gV2Q. (41.17) This islikeEq.(40.9) except forthenewterm ontheright-hand side. When the right-hand sidewaszero, wehadtheHelmholtz theorem that thevorticity stays with theﬂuid. Now, wehave therather complicated nonzero term ontheright- hand side which, however, hasstraightforward physical consequences. Ifwe disregard forthemoment theterm VX(QXv),wehave adiﬂusion equation. Thenew term means thatthevorticity Qdiﬂuses through theﬂuid. Ifthere isa large gradient inthevorticity, itwillspread outintotheneighboring ﬂuid. This istheterm thatcauses thesmoke ring togetthicker asitgoes along. Also, itshows upnicely ifyousend a“clean” vortex (a“smokeless” ringmade by theapparatus described inthelastchapter) through acloud ofsmoke. When it comes outofthecloud, itwillhave picked upsome smoke, andyouwillseea hollow shell ofasmoke ring. Some oftheS2diffuses outward into thesmoke, while stillmaintaining itsforward motion with thevortex. 41-3 TheReynolds number Wewillnow describe thechanges which aremade inthecharacter ofﬂuid ﬂow asaconsequence ofthenewviscosity term. Wewilllook attwoproblems insome detail. Theﬁrstofthese istheﬂow ofaﬂuid pastacy1inder—a ﬂowwhich wetried tocalculate intheprevious chapter using thetheory fornonviscous ﬂow. Itturns outthattheviscous equations canbesolved byman today only forafew special cases. Sosome ofwhat wewilltellyouisbased onexperimental measure- ments——assuming thattheexperimental model satisﬁes Eq.(41.17). Themathematical problem isthis:Wewould likethesolution fortheﬂowof anincompressible, viscous ﬂuid pastalongcylinder ofdiameter D.Theﬂowshould begiven byEq.(41.17) andby Q=VXv (41.18) with theconditions thatthevelocity atlarge distances issome constant velocity, sayV(parallel tothex-axis), andatthesurface ofthecylinder iszero. That is, 1),;=vy=vz=O (41.19) for 2x2_|__y2=_€__ That speciﬁes completely themathematical problem. Ifyoulook attheequations, youseethatthere arefour diﬂerent parameters totheproblem: 77,p,D,andV.You might think thatwewould have togivea whole series ofcases fordifferent V’s,diﬂerent D’s,andsoon.However, thatis notthecase. Allthedifferent possible solutions correspond todifferent values of oneparameter. This isthemost important general thing wecansayabout viscous ﬂow. Toseewhy thisisso,notice ﬁrstthattheviscosity anddensity appear only intheratio 77/p—the specific viscosity. That reduces thenumber ofindependent parameters tothree. Now suppose wemeasure alldistances intheonly length thatappears intheproblem, thediameter Dofthecylinder; thatis,wesubstitute forx,y,z,thenewvariables x’,y’,2’with x=x’D, y=y’D, z=z’D. Then Ddisappears from (41.19). Inthesame way, ifwemeasure allvelocities in terms ofV—that is,weset1'=2"V"—W€ getridoftheV,and11'1Sjustequal to1 atlarge distances. Since wehave ﬁxed ourunits oflength andvelocity, ourunit 41—5 oftime isnow D/V; soweshould set t=z (41.20) With ournewvariables, thederivatives inEq.(41.18) getchanged from 6/6x to(1/D) 8/ax’, andsoon:soEq(41.18) becomes Q=VXv=l;V’Xv’=-ESE’. (41.21) Ourmain equation (41.17) then reads 8Q, i 1 / ll 2> =---~VQ’. 6,,-l—V X(Q Xv) PVD Alltheconstants condense intoonefactor which wewrite, following tradition, as 1/(R: (ii=5;VD. (41.22) Ifwejustremember thatallofourequations aretobewritten with allquantities inthenewunits, wecanomit alltheprimes. Ourequations fortheﬂow arethen 85‘;+v><(o><v)Z3;vzo (41.23) and £2=VXv with theconditions v=O for x2—l—y2=1/4 (41.24) and 0,,=1, 21,,=112=O for x2+y2+z2>>l. What thisallmeans physically isveryinteresting Itmeans, forexample. that ifwesolve theproblem oftheﬂow foronevelocity V1andacertain cylinder diameter D7,andthen askabout theﬂowforadifferent diameter D2andadillerent ﬂuid, theﬂow willbethesame forthevelocity V2which gives thesame Reynolds number——that is,when (it,=%V7D7=($12=%%V202. (41.25)l 2 Foranytwosituations which have thesame Reynolds ntimber, theﬂows will “look” thesame~in terms oftheappropriate scaled x’,y’,2’,andt’.This isan important proposition because itmeans thatwecandetermine what thebehavior oftheﬂowofairpastanairplane wing willbewithout having tobuild anairplane andtryit.Wecan, instead, make amodel andmake measurements using avelocity that gives thesame Reynolds number. This istheprinciple which allows usto apply theresults of“wind-tunnel“ measurenients onsmall-scale airplanes, or “model-basin“ results onscale model boats. tothefull-scale objects. Remember, however, thatwecanonly dothisprovided thecompressibility oftheﬂuid canbe neglected. Otherwise, anewquantity enters~the speed ofsound. And ditlerent situations willreally correspond toeach other only iftheratio ofVtothesotiiid speed isalsothesame This latter ratio iscalled theMac/1 number So,forveloci- tiesnear thespeed ofsound orabove, theﬂows arethesame intwosituations ifbot/i the Mach number and the Reynolds number are the same for both situations. 41-6 ,1 2.. Co STEADY |_ lPERl0DlC (LAMINAR) I PERIODIC I (TURBULENT) E_ F._ OTUR BULENT BOUNDARY LAYER 1 10 IO (-3N, Ul %on 6uils I7> IO IO Fig. 4l-4. Thedrag coefficient C1)ofcicircular cylinder asafunction oftheReynolds number. 41-4 Flow pastacircular cylinder Let’s goback totheproblem oflow-speed (nearly incompressible) ﬂow over thecylinder. Wewillgive aqualitative description oftheﬂow ofarealﬂuid. There aremany things wemight want toknow about such aﬂow—for instance, what isthedrag force onthecylinder? Thedrag force onacylinder isplotted in Fig.41-4 asafunction of(R—which isproportional totheairspeed Vifeverything elseisheldﬁxed. What isactually plotted istheso-called drag coeﬁicient C7,, which isadimensionless number equal totheforce divided by%pV2Dl, where Disthediameter, listhelength ofthecylinder, andpisthedensity oftheliquid: F CD~ Thecoeﬂicient ofdrag varies inarather complicated way, giving usapre-hint thatsomething rather interesting andcomplicated ishappening intheﬂow. Wewill now describe thenature ofﬂow forthediﬂerent ranges oftheReynolds number. First, when theReynolds number isvery small, theﬂow isquite steady; thatis, thevelocity isconstant atanyplace, andtheﬂow goes around thecylinder. The actual distribution oftheﬂow lines is,however, notlikeitisinpotential ﬂow. They aresolutions ofasomewhat different equation. When thevelocity isvery lowor,what isequivalent, when theviscosity isveryhigh sothestuff islikehoney, then theinertial terms arenegligible andtheﬂow isdescribed bytheequation V29=0. This equation wasﬁrstsolved byStokes. Healsosolved thesame problem fora sphere. Ifyouhave asmall sphere moving under such conditions oflowReynolds number, theforce needed todrag itisequal to611-i7aV, where aistheradius ofthe sphere andVisitsvelocity. This isaveryuseful formula because ittellsthespeed atwhich tiny grains ofdirt(orother particles which canbeapproximated as spheres) move through aﬂuid under agiven force-as, forinstance, inacentrifuge, orinsedimentation. ordiffusion InthelowReynolds number region—for titless than 1-the lines ofvaround acylinder areasdrawn inFig41-5. Ifwenowincrease theﬂuid speed togetaReynolds number somewhat greater than 1,weﬁndthattheﬂow isdifferent. There isacirculation behind thesphere, asshown inFig.41—6(b). Itisstillanopen question astowhether there isalways 41-7 - _i>-_i X/2 -Fig. 41-5. Viscous ﬂow (low veloci- ties) oround cicircular cylinder. > - *\/——: Tax./f” ’ _¢-\Q:-.lOO l‘_!§;_?_ t. %<55€>‘;=—? Q?’-f - 4'2,V4 m'5';$,'5~,. :55 I‘ reef}: _,. iR(-=10“ W 1.-=~-~~*~- " (Rz|O6 Fig.41-6.L7* Flow past acylinder forvarious Reynolds numbers. acirculation there even atthesmallest Reynolds number orwhether things sud- denly change atacertain Reynolds number. Itused tobethought thatthecir- culation grew continuously. Butitisnow thought thatitappears suddenly, and itiscertain thatthecirculation increases with (R.Inanycase, there isadiﬂerent character totheﬂow for(Rintheregion from about 10to30.There isapairof vortices behind thecylinder. Theﬂow changes again bythetime wegettoanumber of4Uorso.There is suddenly acomplete change inthecharacter ofthemotion. What happens isthat oneofthevortices behind thecylinder getssolong thatitbreaks offandtravels downstream with theﬂuid. Then theﬂuid curls around behind thecylinder and makes anewvortex. Thevortices peeloffalternately oneach side, soaninstan- taneous view oftheﬂow looks roughly assketched inFig.41-6(c). Thestream of 41-8 pp /" 1‘ .\\ / ,_ ,;» t’ ~V(1I//. f é/0 2 _ix.4.}/ll‘,/,;ij/ ‘J -rI ‘l5 . (~. titty \.,‘, a-a, ,;;,/ vortices iscalled a“Karman vortex street.” They always appear for(R>40. Weshow aphotograph ofsuch aﬂow inFig.41-7. Thedifference between thetwoﬂows inFig.4l—6(c) and41—6(b) or4l—6(a) isalmost acomplete difference inregime InFig.4l—6(a) or(b),thevelocity is constant. whereas inFig4l—6(c), thevelocity atanypoint varies with time There isnosteady solution above G1=40-which wehave marked onFig.41-4 bya dashed line. Forthese higher Reynolds numbers, theﬂowvaries with time butina regular, cyclic fashion. Wecangetaphysical idea ofhow these vortices areproduced Weknow thattheﬂuid velocity nitist bezero atthesurface ofthecylinder andthatitalso increases rapidly away from thatsurface. Vorticity iscreated bythislarge local variation inﬂuid velocity. Now when themain stream velocity islowenough, there issulhcient time forthisvorticity todiffuse outofthethinregion near thesolid surface where itisproduced andtogrow into alarge region ofvorticity. This physical picture should help toprepare usforthenext change inthenature ofthe ﬂow asthemain stream velocity, or(Pi,isincreased stillmore. Asthevelocity getshigher andhigher, there islessandlesstime forthe vorticity todiﬂuse intoalarger region ofﬂuid. Bythetime wereach aReynolds number ofseveral hundred, thevorticity begins toﬁllinathinband, asshown in Fig.4l—6(d). Inthislayer theﬂow ischaotic andirregular. Theregion iscalled theboundary layer andthisirregular ﬂow region works itswayfarther andfarther upstream as(tiisincreased. Intheturbulent region, thevelocities areveryirregular and“noisy”; alsotheﬂow isnolonger two-dimensional buttwists andturns in allthree dimensions. There isstillaregular alternating motion superimposed on theturbulent one. AstheReynolds number isincreased further, theturbulent region works its wayforward until itreaches thepoint where theﬂow lines leave thecy1inder—for ﬂows somewhat above (Pi=105. Theﬂow isasshown inFig. 41—6(e), andwe have what iscalled a“turbulent boundary layer.” Also, there isadrastic change inthedrag force; itdrops byalarge factor, asshown inFig.41-4. Inthisspeed region, thedrag force actually decreases with increasing speed. There seems to belittle evidence ofperiodicity. What happens forstilllarger Reynolds numbers? Asweincrease thespeed further, thewake increases insizeagain andthedrag increases. Thelatest experi- ments—-which goupto(R=107orso-indicate thatanew periodicity appears inthewake. either because thewhole wake isoscillating back andforth inagross motion orbecause some newkind ofvortex isoccurring together with anirregular noisy motion. Thedetails areasyetnotentirely clear, andarestillbeing studied experimentally. 41-5 Thelimit ofzeroviscosity Wewould liketopoint outthat none oftheﬂows wehave described are anything likethepotential ﬂow solution wefound inthepreceding chapter. This is,atﬁrstsight, quite surprising. After all,61isproportional to1/77. So77going to zero isequivalent to(llgoing toinﬁnity. And ifwetake thelimit oflarge (Piin 41-9Fig. 4l—7. Photograph by Ludwig Prqndtl Qfthe vgrtex Street" inthe ﬂgw behind acylinder s Q -i ‘Z *7 si / ‘K -/ \‘__- L 2 \ ___, \\ / ‘ \_ _,—- \ __// __ C\L/‘ \___ ,‘___ //_._ \_”__ /\ __4, _ § T r , :"/ /1*“ . (c) (d)(<11 C _) Fig. 4l-8. Liquid ﬂow patterns be- tween two fransparent rotating cylinders.’‘/‘ \ O///////////M '5.'1\\‘x\\\\\\\\\\/Eq.(41.23), wegetridoftheright-hand sideandgetjusttheequations ofthelast chapter. Yet, youwould ﬁndithard tobelieve thatthehighly turbulent ﬂowat (R=107wasapproaching thesmooth ﬂowcomputed from theequations of“dry” water. How canitbethat asweapproach (Pt=vs,theﬂow described byEq. (41.23) gives acompletely different solution from theoneweobtained taking 77=0tostart outwith? Theanswer isveryinteresting. Note thattheright-hand term ofEq.(41.23) hasl/(Rtimes asecond derivative. Itisahigher derivative than anyother derivative intheequation. What happens isthatalthough thecoefﬁcient 1/(Piissmall, there arevery rapid variations ofQinthespace near thesurface. These rapid variations compensate forthesmall coefficient, and theproduct does notgotozerowith increasing (R.Thesolutions donotapproach thelimiting case asthecoefﬁcient ofV29 goes tozero. You may bewondering, “What istheﬁne-grain turbulence andhowdoes it maintain itself? How canthevorticity which ismade somewhere attheedge of thecylinder generate somuch noise inthebackground?” Theanswer isagain interesting. Vorticity hasatendency toamplify itself. Ifweforget foramoment about thediffusion ofvorticity which causes aloss, thelaws ofﬂowsay(aswehave seen) thatthevortex lines arecarried along with theﬂuid, atthevelocity v.We canimagine acertain number oflines ofQ which arebeing distorted andtwisted bythecomplicated ﬂow pattern ofv.This pulls thelines closer together andmixes them allup. Lines that were simple before willgetknotted andpulled close together. They willbelonger andtighter together. Thestrength ofthevorticity will increase and itsirregularities-—the pluses and minuses—will, ingeneral, increase. Sothemagnitude ofvorticity inthree dimensions increases aswetwist theﬂuid about. You might wellask,“When isthepotential ﬂow asatisfactory theory atall?” Intheﬁrstplace, itissatisfactory outside theturbulent region where thevorticity hasnotentered appreciably bydiﬂusion. Bymaking special streamlined bodies, wecankeep theturbulent region assmall aspossible; theﬂow around airplane wings—which arecarefully designed—is almost entirely truepotential ﬂow. 41-6 Couette ﬂow Itispossible todemonstrate thatthecomplex andshifting character ofthe ﬂow past acylinder isnotspecial butthatthegreat variety ofﬂow possibilities occurs generally. Wehave worked outinSection 1asolution fortheviscous ﬂow between twocylinders, andwecancompare theresults with what actually happens. Ifwetaketwoconcentric cylinders with anoilinthespace between them andputaﬁnealuminum powder asasuspension intheoil,theﬂowiseasytosee. Now ifweturn theouter cylinder slowly, nothing unexpected happens; seeFig. 41-8(a). Alternatively, ifweturn theinner cylinder slowly, nothing very striking occurs. However, ifweturn theinner cylinder atahigher rate, wegetasurprise. The ﬂuid breaks into horizontal bands, asindicated inFig.41-8(b). When the outer cylinder rotates atasimilar ratewith theinner oneatrest, nosuch eﬂect occurs. How canitbethatthere isadifference between rotating theinner orthe outcylinder? After all,theﬂow pattern wederived inSection 1depended only onwi,—o.t,,. Wecangettheanswer bylooking atthecross sections shown in Fig.41-9. When theinner layers oftheﬂuid aremoving more rapidly than the outer ones, they tend tomove outwara'—the centrifugal force islarger than the pressure holding them inplace. Awhole layer cannot move outuniformly because theouter layers areintheway. Itmust break intocells andcirculate, asshown in Fig.4l—9(b). Itisliketheconvection currents inaroom which hashotairatthe bottom. When theinner cylinder isatrestandtheouter cylinder hasahighvelocity, thecentrifugal forces build upapressure gradient which keeps everything in equilibrium—see Fig.4l—9(c) (asinaroom with hotairatthetop). Now let’sspeed uptheinner cylinder. Atﬁrst, thenumber ofbands increases. Then suddenly youseethebands become wavy, asinFig.4l—8(c), andthewaves travel around thecylinder. Thespeed ofthese waves iseasily measured. Forhigh rotation speeds they approach 1/3thespeed oftheinner cylinder. And noone 41-10 CENTRIFUGAL FORCES 1\* {i ‘ 4 IQUOI (BIO!(0) (bl4 t _ / Fig. 4l-9. Why theﬂow breaks upinto ba knows why! There’s achallenge. Asimple number like1/3,andnoexplanation Infact, thewhole mechanism ofthewave formation isnotvery wellunderstood, yetitissteady laminar ﬂow. lfwe nowstart rotating theouter cylinder a1so—but intheopposite direction— theﬂowpattern starts tobreak up.Wegetwavy regions alternating withapparently quiet regions, assketched inFig.41-8(d), making aspiral pattern. lnthese “quiet” regions, however, wecanseethat theﬂow isreally quite irregular; itis,infact completely turbulent. Thewavy regions also begin toshow irregular turbulent ﬂow Ifthecylinders arerotated stillmore rapidly, thewhole ﬂow becomes chaotically turbulent. Inthissimple experiment weseemany interesting regimes ofﬂow which are quite different, andyetwhich areallcontained inoursimple equation forvarious values oftheoneparameter (R.With ourrotating cylinders, wecanseemany of theeffects which occur intheﬂowpastacylinder: ﬁrst, there isasteady ﬂow,second, aﬂow setsinwhich varies intime butinaregular, smooth way; ﬁnally, theﬂow becomes completely irregular. You have allseen thesame effects inthecolumn ofsmoke rising from acigarette inquiet air. There isasmooth steady column followed byaseries oftwistings asthestream ofsmoke begins tobreak up,ending ﬁnally inanirregular churning cloud ofsmoke Themain lesson tobelearned from allofthisisthatatremendous varietv ofbehavior 1Shidden inthesimple setofequations in(41.23). Allthesolutions areforthesame equations, only with different values of(RWehave noreason tothink thatthere areanyterms missing from these equations. Theonly difﬁculty isthatwedonothave themathematical power today toanalyze them except for very small Reynolds nunibers—that is,inthecompletely viscous case. That we have written anequation does notremove from theﬂow ofﬂuids itscharm or mystery oritssurprise. Ifsuch variety ispossible inasimple equation with only oneparameter, how much more ispossible with more complex equations! Perhaps thefundamental equation that describes theswirling nebulae andthecondensing, revolving, and exploding stars and galaxies isjust asimple equation forthehydrodynamic behavior ofnearly pure hydrogen gas. Often, people insome unjustiﬁed fearof physics sayyoucan’t write anequation forlife. Well, perhaps wecan. Asamatter offact, wevery possibly already have theequation toasuﬂicient approximation when wewrite theequation ofquantum mechanics: 6 _h¢H¢——i6t Wehave justseen thatthecomplexities ofthings cansoeasily anddramatically escape thesimplicity oftheequations which describe them. Unaware ofthescope ofsimple equations, man hasoften concluded thatnothing short ofGod, notmere equations, isrequired toexplain thecomplexities oftheworld. 4l—llCENTFHFUGAL - l nds.(C) T; I FORCES s Wehave written theequations ofwater ﬂow. From experiment, weﬁndaset ofconcepts andapproximations tousetodiscuss thesolution—vortex streets, turbulent wakes, boundary layers. When wehave similar equations inaless familiar situation, andoneforwhich wecannot yetexperiment, wetrytosolve theequations inaprimitive, halting, andconfused waytotrytodetermine what newqualitative features may come out,orwhat newqualitative forms areacon- sequence oftheequations. Ourequations forthesun, forexample, asaballof hydrogen gas,describe asunwithout sunspots, without therice-grain structure of thesurface, without prominences, without coronas. Yet, allofthese arereally intheequations; wejusthaven’t found thewaytogetthem out. There arethose who aregoing tobedisappointed when nolifeisfound on other planets. NotI—I want tobereminded anddelighted andsurprised once again, through interplanetary exploration, with theinﬁnite variety andnovelty of phenomena thatcanbegenerated from such simple principles. Thetestofscience isitsability topredict. Had younever visited theearth, could youpredict the thunderstorms, thevolcanos, theocean waves, theauroras, andthecolorful sunset? Asalutary lesson itwillbewhen welearn ofallthat goes ononeach ofthose dead planets——those eight ortenballs, each agglomerated from thesame dustcloud andeach obeying exactly thesame laws ofphysics. Thenextgreat eraofawakening ofhuman intellect maywellproduce amethod ofunderstanding thequalitative content ofequations. Today wecannot. Today wecannot seethatthewater ﬂowequations contain such things asthebarber pole structure ofturbulence thatoneseesbetween rotating cylinders. Today wecannot seewhether Schrodinger’s equation contains frogs, musical composers, ormorality —or whether itdoes not. Wecannot saywhether something beyond itlikeGod isneeded, ornot. And sowecanallhold strong opinions either way. 41-l2 Inclvx __ __l _ I i 1 ' Aberration. l-27—7, I-34-I0 Absolute zero, l—l-5 Absorption, l—3l—8 ff Absorption coefﬁcient. ll~32-8 Acceleration, l—8v8 ff components of,I-9-3 ofgravity, l~9-4 Accelerator guide ﬁeld, ll—29~4 ff Activation energy, l-42-7 Active circuit element, ll—Z2—5 Adams, JC,l—7—5 Adiabatic compression, l—39—5 Adiabatic demagnetization, ll~35—9 f Adiabatic expansion, l—44—5 Affective future, l—l7—4 Aharanov, ll—l5—l2 Air trough, l—l0—5 Algebra, l—22—l ff Alternating-current circuits, ll—22—l ff Alternating-current generator, ll—l7—(i ff Alnico V,ll-37—l0 Amber, ll—l—lO Ammeter, ll—l6-1 Ampere, A,ll—l3—3 Ampere's law, ll-l3—4 Amperian current, ll-36-2 Amplitudes ofoscillation, l—2l—3 Amplitude modulation, l-48-3 Analog computer, l—25—8 Anderson, CD,l-52-10 Angle, ofincidence, l—26—3 otprecession, ll—34—4 otreﬂection, l-26—3 Angstrom (unit), l—l—3 Angular frequency, l—2l—3. Angular momentum, I-7-7, -- I—20—l conservation of,l—4—7, I—18—6 ff. I-20—5 ofrigid body, l-20-8 Anomalous refraction, I—33—9 f Antiferromagnetic material, ll—37—ll Antimatter, l-52—l0 f Antiparticle, l—2—8 Aristotle, l—5—l Atom, I—l—2 metastable, l—42—l0 Rutherford-Bohr model, ll-5—3 stability of,ll—5—3 Thompson model, ll—5—3 Atomic clock, l—5—5 Atomic currents, ll—l3—5 f Atomic hypothesis, I-1-2 Atomic orbits, ll-l—8 Atomic particles, I—2—9 f Atomic polarizability, ll-32-2 Atomic processes, l—l—5 f Attenuation. l—3l—8 Avogadro, A,l—39—2%4i—l -—-I\>oo\OU1 “P-HNAvogadro's number, I—4l—l0 Axial vector, l—52—6 fCenter ofmass, l—l8—l f,l—l9—l ff Centrifugal force, I-7-5, l—l2-1 l Cerenkov, P.A.,l—5l—2 Cerenkov radiation, l—5l—2 Charge, conservation of.l—4—7, ll—l3—l f onelectron, I~l2—7 lineof,ll—5—3 f motion of,ll—29—l ff sheet of,ll—5—4 sphere of,ll—5—4 f Charge density, ll-5—4 Charge separation, ll-9-7 if Charged conductor, ll—8-2 ff Chemical energy, l—4-2 Chemical kinetics, l—42—7 f Chemical reaction, l-l—6 ff Chromaticity, l~35—6 f Circuits, alternating-current, ll—22—l ff equivalent, ll—22—l0 f Circuit elements, ll-23—l f active, Il—22—5Barkhausen effect, ll—37—9 Battery, lI—22—6 Becquerel, AH.,l—28—3 Bell, AG,ll—l6—3 Benzene molecule, lll—49—l0 IT Bernoulli’s theorem, lI—40—6 ff Bessel function, lI—23-6 Betatron, ll-17-5 Biot-Savart law, Il—l4—l0 Birefringence, l—33—3 ff Blackbody radiation, l—4l—5 f Boehm, I—52—l0 Bohm, ll~7—7, ll-15-12 Bohr. N.,I—42—9, ll-5—3 Bohr magneton, lI—34—l2 Bohr radius, l-38-6 Boltzmann, L,I-41-2 Boltzmann’s law, l—40—2 f Bopp’ H_28_8 assive, II-22-5Born, M,I-37—1, I-38-9, II-28-7 Ofcum motion, I_2,_4 Boundary layer’ H_41—9 Circulation. lI—l—5, ll—3-8 ffBoundary-value problems, lI—7—l Cl 1elt ,d H283QSSICH eC TOII ra IUS, — — P§Z1'=S,§ L’;”lYr;e:4(I’;f9_10 ClaUSlUS, R,I-44-2, l—44—3Bray LH__3’O 9 Clausius-Clapeyron equation, I-45-6 ff ggii - _ ll-11-6r,Bragg-Nye crystal model, Il—30—9 fr C'a“S',“S M°Ss°"‘ °q“““°"’ I—32—7Breaking-drop theory, II-9-9 Cleavagﬁ plane, H_3O_1 Bremsstrahlung, I-34-6 f C I1 II24I Brewsteris angle’ I—33~6 Cgzgiiiierirtieabsorptjon, ll—32-8Briggs, H,I—22—6 ’l ll—l7—14Brown RI—4l—l ofcoup mg’ii ff ,I—l2—4BFOWIIIIZT I}"l(;¥lOf1, I—1—8, I—6—5, grafllgignnal, I__7_9 Brush discharge, ll—9—9 ofviscosity’? 412 Bulkmodulus II~38-3 C°“‘S‘°"’I 166’ elastic, I—l0—7 Colloidal particles, ll—7—8 ff Calculus, differential, I—8—4, ll—Z-l if Color vision, I—35-1 ff integral, II—3—l if physiochemistry of,I—35—9 f ofvariations, II—l9—3 Complex impedance, l—23—7 Cantilever beam, II—38—l0 Complex numbers, l—22—7 ff,l—23—l if Capacitance, I—23—5 Complex variable, II—7—2 ff mutual, II-22-l7 Compound eye, I-36-6 ff Capacitor, I—l4—9, I—23-5, lI—22-3 ff, Compression, adiabatic, l—39—5 II—23—2 ﬁ‘ isothermal, l—44—5 parallel-plate, I—l4—9, ll—6—l1 ff, Condenser, parallel-plate, I—l4—9, II—8—3 ll-6-ll ff,ll-8-3 Capacity, Il—6—l2 Conductivity, ll—32—l0 ofacondenser, II—8—2 thermal, ll—2—8, ll—l2-2 Capillary action, I—5l—8 Conductor, lI—l—2 Carnot, S,I—4—2, I—44—2 ff Cones, I—35—l Carnot cycle, I—44—5 f,I-45—2 Conservation, ofangular momentum, Carrier signal, I—48—3 1-4-7, I*13—6 Ff,I-10-5 Catalyst, I—42—8 ofcharge, l—4—7, ll—l3—l f Cavendish, H.,I-7-9 ofenergy, l—3—2, l—4—l ff,lI—Z7—1 f Cavendish’s experiment, I—7—9 oflinear momentum, l~4-7, Cavity resonator, II-23—l ff I—lO—l ff INDEX l Contraction hypothesis, l—l5-3 Dynamics, I-7-2 -— Copernicus, l-7-l relativistic, I-1-- Coriolis force, l-19-8 f _s- 5252;?’ EH5; IlI_9_2 Eddy current, ll-l6-6 Couette ﬂow’, ll_4l_l0 ff Eﬁiciencylff l;j€2Bi6€I1%lfl€, I-44-7 f\’\\:-ﬁ\Q)—4""032 Emissivity, ll-6-l4 Energy, ll—22—ll t chemical, l-4-2 ofacondenser, ll-8-2 ff conservation of,I-3-2, l-4-1 ff, -27-1 f ii cciilcmue law,l-28-2, ll-4-2 ff, ‘“$“"“’ "i“"i'74" H242’ elastic, l-4-2, l-4-6“_5_6 l-l5-l, l—l6—l, I-41-8, l-42-8, Coupling. coefﬁcient oi,ll-l7—l4 E l_42_9 rovalent bciici ll-30-2 lam“°°“'s'°“’ “10-7Cross product ’“_2_8 “_3l_8 Elastic constants, II—39—6, II—39-l0 f Cross section forscattering, I-32-7 E2152‘: enetrgy’lI'1il—23’9I“l4?f6, ( _ _l as Cma $113 S, — — (rfzgigelti 33,,£30_l f Elastica, ll-38-12 CF2;/still iiirireciieii I-38-4f E'a“‘°"Y’ “"384 ‘TCrystal lamce H_'30_3 t Elasticity tensor, Il—39—4 ffelectrical, I-4-2, ll-15-3 ff electromagnetic, I-29-2 electrostatic, ll-8-l ff inelectrostatic ﬁeld, ll-8-9 ff gravitational, I-4-2 ff heat, I-4-2 -- --7,l—l0—8 kinetic, l-l- —— —4—5 f, l-39-4 magnetic ll-l7-l2 ff>1»-i-44>es»wlw,_,._.O cubic cell,ll-30-7 E"’°"°" ""''“8 . mass,I-Li-2, I-4-7(Line law HAl_5 Electric charge density, II-2-8, ll-4-3 (, ’It 7H_36_13 Electric current, ll-l3-l f in i Cur] Oarmor l’I_¢_8 H_3_l Electric current density, ll-2-8 Curran‘: Am’ena“’g36_2 Electric d|p0le, ll-6-2 n ’ p P’H Electric ﬁeld I2-4 l—l2 7f II-l 2atomic, ll-13-3 I ’_ ’ _’ _’ eddy. II-16-6 ll-l—3, ll-6-l ff,II-7—l ffelectric “_l3_] f relativity of,II—l3—6 ff ’ Electric ﬂux, II-I-4induced, ll-16-l ff ,. Electric potential, II-4-4 guinfafné (ienzgg “T_32_2l_l4 Electric susceptibility, ll-10-4 UOrqu y’ Electrical energy, I-4-2, lI—l5—3 if Electrical forces, ll-l—l ff,II-13-1 D"Alembertian, ll-25-8 Electrodynamics, II-1-3 Debye length, ll-7-9 relativistic notation, ll-25-I ff Dedekind, R,I-22-4 Electromagnet, II-36-9 ff Degrees offreedom, l—25-2, I-39-l2 Electfgmagnetic energy, I-29-2 Demagnetization, adiabatic, ll-35-9 f Electromagnetic ﬁeld, I_2_2, 1-2-5, Density, l—l-4 l—lO—9 Derivative, I~8~5 fl Electromagnetic mass, ll-28-3 f partial, l—l4—9 Electromagnetic radiation, I-26-1, Diamagnetism, ll-34-l ff ]_28_1 ff Dickc, RH,P7’! l Electromagnetic waves, ll—2l-l f DlClCCIf|C, ll——l()—l ll,l|—l l—l if cQS[‘n|Q rays, I-Z-5 Dielectric constant, ll-I0-l f gamma rays, 1-2-5 Differential calculus, I—8—4, ll-2-1 if mfrared, I-2-5, I-23-8, I_26_l Diffraction, I-30-l fl lrghr, I-2-5 byScreen, 1-31-10 f ultraviolet, I—2—5, l—26-l Ditlraction grating, l—29—5, l—30—3 ff x-rays, I_2_5, I-Z6-1 DlfTLl§IOn, l—43—l if Electromagnetism, ll-l—l ff otneutrons, ll-12-6 ff laws Of,[I-1-5 ff DIp0lc, ll—Zl—5 ff Electromotive force, II-16-2 electric, II—6—2 fl Electron, I-2-4, I-37-1, I-37-4 if magnetic, ll-14-7 f charge on,I-I2-7 Dipole m0mem» l—l2'6i II-6—7 radius of,classical, I-32-4 Dipole pO[ential, ll-6-4 if Elgctron cloud, I-6-ll D1p0l<3 f=1dl8IOF, l—38—5 f,l—Z9—3 if Electron microscope, ll—29—3 f Dirac, P,l—52—l(), ll—Z—l. ll‘-28—7 Electron-ray tubg, I-12-9 Dirac equation, I-20-6 Electron volt(unit), I-34-4 Disl0C21ti0n, 1l—3U*8, ll—30—9 Electronic polarization, ll-1l-Iff Dispersion, l-3l-6ff Electrostatic energy, ll-8-l ff Distance, l-5-5 ff ofcharges, ll-8—l f Distance measurement, color brightness, ofionic crystal, ll-8-4 ff I—5—6 innuclei, ll—8—6 ff Ifl1iﬂgLll<ltlOﬂ, l—5~(i ofapoint charge, ll-8-12 Divergence, ll-25-7 Electrostatic equations, II-10-6 fmechanical, ll-15-3 ff nuclear, l-4-2 potential, I-4-4, l—l3-l ff,I-14-l ff radiant, l—4—2 relativistic. l-I6-l ff Energy density, ll—27—2 Energy ﬂux, ll—27—2 Energy levels, l-38-7 f Energy theorem, I-50-7 f Enthalpy, l-45-5 Entropy, I-44-10 ff,l—46-7 ff Eotvos, L.,I-7-l l Equilibrium, I—l—6 Equipotentlal surfaces, ll-4-ll f Equivalent circuits, ll—22-I0 f Euclid, I-5-6 Euclidean geometry, l—l2—3 Euler force, ll—38—l l Evaporation, l—l—5 f ofaliquid, l-40-3 f,l—42-l ff Exchange force, ll—37-2 Excited state, ll—8—7 Expansion, adiabatic, I-44-5 isothermal, l-44-5 Exponential atmosphere, l-40-l f Eye, compound, I-36-6 ff human, I-35-l f,l—36—3 ff Farad (unit), I-25-7, ll-6-I3 Faraday, M,ll-I0-l Faraday's lawofinduction, ll—l7-2 Fermat, P,I-26-3 Fermi (unit), I-5-10 Fermi, E,l—5—l0 Ferrite, ll—37-I2 Ferroelectricity, ll-11-8ff Ferromagnetic insulators, ll-37-12 Ferromagnetism, ll-34-l f,ll-36—l if. ll—37-1 ff Feynman, R.,ll—28—8 Fields, I-2-2, I-2-4, I-2-5, l-l0-9, I-12-7 ff,l—l3—8 f,I-14-7 ff inacavity, l[—5—8 f ofacharged conductor, ll-6-8 ofaconductor, ll-5-7 f Divergence operal0Fi 11-2-7, ll—3—l Electrostatic ﬁeld, ll-5-1 ff,ll-7-l f electric, l-2-4, I-12-7 f,ll-1-2, Domain, ll-37-6 energy in,ll-8-9 ff Doppler effect, I-l7-8, I-23-9, ofagrid, Il—7—lO f l—34—7 f,I-38-6 Electrostatic lens, II-29-2 f Dot product, ll—2—4, ll-25-3 Electrostatic potential, equations of, Double stars, l—7—6 ll-6-1 Drag coeﬁlcient, ll—4l—7 Electrostatics, ll-4-1 ff,lI—S-l "Dry" water, ll—4(1—I ff Ellipse, l-7-l INDEX 2ll-1-3, ll-6-l ff,Il-7-I if electrostatic, ll-5-l if,ll-7-l f magnetic, ll-l-2, ll-1-3, II-13-1, II—l4-l ff magnetizing, lI—36-7 scalar, ll—2-2 ff superposition of,l—l2-9 ‘_l\)2::two-dimensional, ll—7— vector, ll-l—4 f,ll—2- Field energy, lI—27-l ff ofapoint charge, ll-28-1 f Field index, II—29—S Field-ion microscope, ll—6—l4 Field lines, II—4—l l Field momentum, II-27-9 ff ofamoving charge, II-28-2 f Field strength, II-l-4 Filter, ll—22-l4 ff Flow, ﬂuid, ll-l2—8 ff irrotatiorul. ll-40-5 viscous, ll-4l—4 f Fluid ﬂow, ll-l2—8 ff Flux, ll-4-7 ff electric, ll-1-4 ofavector ﬁeld, II-3-2 ff Flux rule, ll—l7—l ff Focal length, I-27-l ff Focus, I-26-5 Force, centritiigal. I-7-5, I-12-ll components of,l-9-3 conservative, l~l4-3 ff Coriolis, I-19-8 f electrical, I-2-3 ff,ll-l—l ff,I-I3-l electromotive, ll-16-2 gravitational, I-2-3 Lorentz, ll—l3—l, II-l5—l4 magnetic, ll—l-2, ll-13-1 molecular, l—l—3, l-12-6 f moment of,I-I8-5 nonconservative, l-l4—6 f nuclear, I-l2—l2 pseudo, l-l2-l0 ff Fourier, J,I-50-2 f Fourier analysis, l—50—2 ff Fourier theorem, ll-7—l l Fourier transform, I-25-4 Four-vectors, I—l5—8 f,I—l7—5 fl, ll—25—l ff Fovca, I-35-l Frank, I,I-51-2 Franklin, B.,ll—5—6 Frequency, angular, l—2l—3, l—29—2 otoscillation, I-2-5 plasma, ll—7—6, II-32-I2 Fresnel’s reﬂection formulas, l—33—8 Friction, I-l0—5, I-l2-3 ff coefﬁcient of,I-l2-4 Galileo, I-5-l, I-7-2, l-9-I, I-52-3 Galilean relativity, l-l0-3 Galilean transformation, l—l2—ll Galvanometcr, ll—l-8, II-l6—l Garnet, ll—37—l2 Gauss (unit), I-34-4 Gauss, K,II-l6—2 Gauss‘ law, II—4—9 f,ll-5-I ff (iaiiss' theorem, ll—3—5 (iaussian surface. ll-l0-l Geiger, ll-5-3 Gell-Mann, M,I—2—9 Generator, alternating-current, Il—l7—6 ff electric, ll-l6—l ff,II-22-5 ff van deGraaff, II-5-9, lI—8—7 Geometrical optics, I-26-l, I-27-l f Gerlach, Il—35-3 Gradient operator. II-2-4, II—3—l\Q\Or-1)r\ LBJIQ Gravitation, I—2—3, I-7-l ff,-— Ionization energy, I-42-5 Gravitational acceleration, I—— Ionosphere, ll-7-5, lI—9—3 Gravitational coeﬂﬁcient, I-7- Irrotational ﬂow, II-40-5 Gravitational energy, I-4-2 ff Isotherm, II-2-3 Gravitational ﬁeld, I-I2-8 ff,I—l3—8 f Isothermal atmosphere, I-40-2 Gravity, I-l3-3 ff Isothermal compression, I-44-5 acceleration of,I—— Isothermal expansion, I-44-5 Green"s function, —— Isothermal surfaces, II—2—3 Ground state, II—- Isotopes, I-3-4 ff Gyroscope, I-20- UlOO—i:R\1t\>U1\OLP Jeans, J,I-40-9, I—4l—6 f,Il—2—6 Hamilton’s ﬁrst principal function, Johnson H0156’ I_4l_2’ I_4l_8H_19_8 Joule (unit), I—l3—3 Harmonic motion, I-21-4, I-23-l ff Joule heating’ l_24_2 Harmonic oscillator I—lO—l I—2l—l ff , S 9 ' i forced’ I_2]_5 f’I_23_3 if Karman vortex street, Il9 Kepler, J,I—7—lHarmonics, I-50-1 ff ,Keplers laws, I-7-l f,I-9-l, I-18-6 Heat’ ‘A-3’ I*l3—3 Kerr cell I335Heat conduction, ll-3-6 If ’. Kilocalorie (unit), II—8—5Htdff t,lI—3-8 H22, 1I—l()—8 Heat engines, I-44-l ff K1:3?$Kt)1:§g)’rI_EZ;_IT ff Heatﬂow,ll-2-8 r,II-12-2 ii ofgases,_Y59_, ff H°'Se“be‘g’ W"HMO’ HM’ Kirchhoff’s laws,I-25-9, ll-22-7 ff I—37_9’ 1-37-11’ I_37—12’ I_38—9 Kronecker delta ll-3|-6Helmholtz, H., I—35—7, II—40—lO ’ Henry (unit), I-—25—7 Lamb ][_5__6 Hess’ H_9_2 Laméielastic constants, Il—39-6Hexagonal Cﬁll, II—30—7 Landé g_fact0r, “_34_4 High-voltage breakdown, II-6-l3 f Laplace, P’|_47_7 H°°k°'S1aWi I-12-6’ ‘F384 f Laplace €qU8llOn, ll-6-l, II-7-l Huygens’ C",I_l5‘2’ I“26"2 Laplacian operator, ll—2—lO HYdr°dY"aml°Si H'4O*2 if Larmor frequency, ll—34—7 Hydrostatlcsi 11-404 ff Larmor's theorem, ll—34-6 f HYP°°Y°1°'di I-34-3 Laser, I-32-6, l-42-10Hysteresis curve. II—37-5 ffLaughton, ll—5—6 Hysteresls loop’ “_36"8 Laws, ofelectromagnetism, ll—l-5 ff ofinduction, ll—l7—l ﬂ Least action, principle of,ll—l9—l ff Least time, principle of,I-26-3 ff,Ideal gaslaw, I-39-10 ff Illumination, II-l2—lO ff Image charge, II-6-9 I-26-8 Impedance, I-25-8 f,II—22-1 ff Leibnitz, G.W,I—8—4 complex, I-23-7 Lens formula, I-27-6 Incidence, angle of,I-26-3 Lenz"s rule, ll-lo—4, II-34-2 Inclined plane, I—4—4 Leverrier, U,l—7—5 Index ofrefraction, I-31-l if Liénard-Wiechert potentials, II-2l—l1 Induced currents, II—l6—l ff Light, II-21-l f Inductance, I-23-6, II-l6-4 f, momentum of.I—34—l0 f ll-l7-I2 ff,II—22-2 f polarized, I-32-9 mutual, II-l7-9 ff,Il—22—16 scattering of,I-32-5 ﬂ‘ self-, II-l6—4, II—|7-ll f speed Of.I-l5—l. lI—l8—8 f Induction, laws of,II—l7-l ff Light waves, I-48-I Inductor, I-23-6 Lightning, Il—9-l0f Inertia, I-2-3, I-7-ll Line ofcharge, II-5-3 f moment of,I-l8—7, I—l9—5 ff Line integral, ll—3—l principle of,I-9-l Linear momentum, conservation of, Infeld. II—28—7 I—4—7, I-lO—l fl Infrared radiation, I—23—8, I-26-l Linear systems, I-25-l ﬂ Integral, I-8-7 f Lodestone, ll—l—lO Integral calculus, II—3—l ff Logarithms, I-22-4 Insulator, II—l-2, II—lO—l Lorentz, HA,l-l5—3 Interference, I-28-6, I-29—l ff Lorentz condition, II-25-9 Interfering waves, I-37-4 Lorentz contraction, l-l5—7 Interferometer, I-15-5 Lorentz force, ll—l3—l, II-l5—l4 Internal reﬂection, II—33-l2 Lorentz formula, Il—2l—l2 f Ion, I-I-6 Lorentz gauge, ll-l8—ll Ionic bond, II—30—2 Lorentz transformation, l-l5—3, Ionic conductivity, I-43-6 f l—l7—l, I-34-8, -52-2, lI—25-l Ionic polarizability, ll-1I-8 ofﬁelds, II—26-l ff INDEX 3 McCullough, ll—l—‘) Mach number, ll-41-6 Magnetic dipole, ll-l4-7 t Magnetic dipole moment, ll—l4—8 Magnetic energy, ll-l7-l2 ﬂ‘ Magnetic ﬁeld l-l2-9 f,ll—l-2, ll-l-3, ll-l3—l, ll-l4-l ﬂ rcliitiviiy of.ll-I3-6 fl ofsteady currents, ll-13-3 t Magnetic lorcc, ll—l-2, ll—l3-l on.icurrent, ll-I3-2 l Magnetic induction. l-I2-ll) Magnetic lens, ll-29-3 Magnetic lIIctt€I'l2ll\, ll—37-l fl Magnetic moments, ll—34—3 t Magnetic resonance, ll—35-l ff Magnetic susceptibility, ll-35-7 Magnetism, I-2-4, ll-34-l ff Magnetization currents, ll—36-l fl Magnetizing ﬁelds, ll-36-7 Magnetostatics, ll-4-I, ll-I3-l ff Magnetostriction, ll-37-6 Magniﬁcation. I-27-5 Marsden, ll—5-3 Maser, I-42-10 Mass, l-9-l, I-l5—l center of,l-18-l f,I-I9-l ll electromagnetic, ll-28-3 f relativistic, l-l6-6ffMomentum spectrometer, ll-29—l Momentum spectrum, ll-29-2 Monatomic gas, I-39-5 Monoclinic cell, ll-30-7 Motion, I-5-I, I-8—l ff ofcharge, ll-29-l ff circular, l—2l—4 constrained, l—l4-3 harmonic, l—2l-4, I-23-l ff parabolic. l—8—lO planetary, l-7-l fl,I-9-6 f,I-l3-5 Motors, electric, ll-l6—l Ff Moving charge, ﬁeld momentum of, II-28-2 f Music, I-50—l Mutual inductance, ll—l7-9 ff, ll-22—l6 Nernst heat theorem, I-44-l l Neuman, Jvon, ll-l2—9 Neutrons, I-2-4 diffusion of,II—l2-6 ff Neutron diffusion equation, ll-l2-7 Newton, I,I-8-4, I-I5-I, I-37-l, ll-4-l0 Newton-meter (unit), —-.. "G _‘bsJ--oi=12 \,.:'~’“Newton's laws, l—2—6, I-- -7-l l, I-9—l ll‘,I-lO—l ff,I- — ___’! '9 Mass energy,l 4._,I-4-7 I-12-l, I—39—i., I-4l—l, I-46-l, Mass-energy equivalence, I-l5-IO f ll-7-5 Maxwell,J C,l-6-l, l-6-9, l-28-l, Nishiiima, l—2—9 I-40-8, l--4|-7, l-46-5, ll-l-8, Nodes, l-49-2 ll-1-ll, ll-5-6, ll-I8-l ff Noise, l—50—| Maxwell's equations, l—l5-2, l-25-3, Nonpolar molecule, ll-ll-l l-47-7, ll-2-l, ll-2-8, ll-4-l, Nuclear cross section, I-5-9 ll—6—l, ll-18-l ll,ll-32-3 ff Nuclear energy, I-4-2 currents andcharges, ll-2|-l Ff Nuclear forces, I-l2—l2 treespace, Il—2()-l ff Nuclear g-factor, ll-34-4 Mayer, JR,I-3-2 Nuclear interactions, II-8-7 Mean free path. l-43-3 f Nuclear magnetic resonance, Mean square distance, I-6-5, I—4l—9 ll-35-l0 ff Mechanical energy, ll—l5-3 ff Nucleus, I-2-4, I-2-8 ﬂ° Mendcléev, I—2—9 Numerical analysis, I-9-6 Metastable atom, l-42—l() Nutation, I-20-7 Meter (unit), I—5—l() Nye, JF,II-30-9 Mev (unit), l-2-9 Michelson-Morley experiment, l-l5—3 ff Oersted (unit), II-36-6 Miller, W(‘,I-35-2 Ohm (unit), I-25-7 Minkowski, l—l7-8 Ohm’s law, I-25-7, I-43-7 Minkowski space, ll-31-12 Operator, curl, II-2-8, ll-3-l Modes, I-49-l ﬂf divergence, II-2-7, II-3-1 Mossbauer, R,I-23-9 gradient, ll-2-4, II-3-l Mole (unit), I-39-l0 Laplacian, II-2-I0 Molecular attraction, l-l-3, I-l2-6 f vector, Il—2—6 Molecular crystal, ll-30-2 Optic axis, I-33-3 Molecular diﬂusion, I-43-7 ff Optic nerve, I-35-2 Molecular dipole, II—ll-l Optics, I-26-l ff Molecular motion, l—4l—l geometrical, I-26-l, I-27-l ff Molecule, I-l-3 Orbital motion, II-34-3 Moment, dipole, I-l2-6 Orientation polarization, I[—ll-3ﬂ° offorce, I-I8-5 Oriented magnetic moment, ll-35—4 ofinertia, I-I8-7, I-l9-5 ﬂf Orthorhombic cell, II-30—7 Momentum, I-9-l t,l—38—2 ff Oscillation, amplitude, of,I-21-3 angular, I-7-7, l—l8-5 ff,I-NO damped, I-24-3 f I-20-5 frequency of,I-2-5 oflight, I-34-l() f period of,I—2l-3 linear, I-4-7, I-l0-l ff periodic, I-9-4 I'Cli1flVlsllC, I-lt)-8 f,I-I6-l ff phase of,I-2l—3 INDLX 4Oscillator. l-5-2 harmonic, I-10-l, l-2l—l, I-21-5 f, I-23-3 fl Pappus, theorem ofl-I9-4 Parabolic antenna, l-30-6 t Parabolic motion, I-8-l() Parallel-axis theorem, l—l9-(i Parallel-plate capacitor, I-l4-9, ll-6—ll ﬂ,ll-8-3 Paramagnetism, ll-34-l fl,ll-35-1 ff Paraxial rays, I-27-2 Partial derivative, I-l4-9 Particles, “strange", ll-8-7 Permalloy, Il—37—l l Permeability, ll-36-9 Pascal's triangle, I-6-4 Passive circuit element, ll-22-5 Pendulum, I-49-6 f Pendulum clock, l-5-2 Period ofoscillation, I-2l-3 Periodic time, I-5-l f Perpetual motion, l—46—2 Phase ofoscillation, I-2l-3 Phase shift, l—2l—3 Phase velocity, I-48-6 Photon, I-2-7, l—26—l, I-37-8 Physiochemistry ofcolor vision, I-35-9 t Piezoelectricity, ll-ll-8 Pines, ll-7-7 Planck, M,l—4l-6, I—42—8, I-42-9 Planck's constant, I-5-10, l—6—l0, I-l7-8, I-37-l l Plane lattice, ll—30-5 Plane waves, II-2|-l ff Planetary motion, I-7-l ff,I-9-6 f, I—l3-5 Plasma frequency, ll—7—6, ll-32-l2 Plasma oscillations, ll-7-5 ff Plimpton, ll-5-6 Poincaré, H,I-15-3, I-l5-S, I-l6-l Polncaré stress. ll-28-4 Point charge, electrostatic energy of, II—8-I2 ﬁeld energy of.ll-28-l f Poisson's ratio, ll-38-2 Polar molecule, ll-ll-l, ll-1l-3ff Polarization, l-33-l ff.Il—32-l ff Polarization charges, l[—l0—3 ff Polarization vector, ll-l0—2 f Polarized light, I-32-9 Potential energy, I-4-4, l—l3-l fl, I-14-l ff Potential gradient oftheatmosphere, II-9-2 f Power, I—l3—2 Poynting, J,ll—27—3 Precession, angle of,ll—34-4 ofatomic magnets, ll-34-4 f Pressure, I-l-3 Priestly, J,ll-5-6 Principle ofleast action, II-l9-l ff Principle ofsuperposition, II-l-3, II-4-2 Probability, I-6-l ff Probability density, I—6—8 f Probability distribution, I-6-7 ff Propagation factor, ll—22-l4 Proton, I-2-4 Proton spin, II—8—7 Pseudo force, l-l2-l0 ff Ptolemy, I-26-2 Purkinje effect, I-35—2 Pyroelectricity, ll-1I-8 Quadrupole lens, ll—7—4, lI—29—6 Quadrupole potential, II-6-8 Quantized magnetic states, II—35—l ff Quantum electrodynamics, I-2-7, I~28—3 Quantum mechanics, I-2-2, I—2—6 ff, l—6—l0, I-10-9, I—37—1 ff, I—38—1 ff Rabi, II.,ll-35—4 Rabi molecular-beam method, ll—35—4 ff Radiant energy, l—4—2 Radiation, infrared, I—— —- relativistic effects, l-—— synchrotron, l—34—3 -— ultraviolet, I-26-1 Radiation damping, I~32—3 f Radiation resistance, l—32—l ff Radioactive clock l—5—3 ff Radius ofelectron, I—32—4 Ramsey, N.,I—5—5 Random walk, l—6—5 ﬂ’,I—4l—8 ff Ratchet andpawl machine, I—46—l ff Rayleigh‘s criterion, l—30-6 Rayleigh‘s law, l—4l—6 Rayleigh waves, ll—38—8 Reactance, ll—22—l l Reciprocity principle, I—30—7 Rectiﬁcation, l—50—9 Rectiﬁer, ll—22—1S Reﬂected waves, lI—33-7 ff Reﬂection, I-26-2 f angle of,l—26-3 internal, ll—33—l2 oflight, ll—33-l ff Refraction, l—26—2 f anomalous, l—33—9 f index of,l—3l—l ff oflight, II—33—l ﬂ° Refractive index, ll—32—l if Relative permeability, ll—36—9 Relativistic dynamics, I-15-9 f Relativistic energy, I—l6—l if Relativistic mass, l-l6—l ff Relativistic momentum, I-IO-8 f, l—l6—l ff Relativity, ofelectric ﬁeld, II—l3-6 ff Galilean, l—l0—3 ofmagnetic ﬁeld, ll—l3—6 ff special theory of,I—l5—1 ﬂ theory of,l-7-l 1,I—l7—l Resistance, I-23-5 Resistor, I-23-5, ll—22—4 Resonant cavity, II—23—6 ff Resonant circuits, ll-23—10 f Resonant mode, ll-23-l0 Resonator, cavity, ll-23-l ff Resolving power, l—27—7 f,l—30—5 f Resonance, l—23—l ff electrical, l—23—5 ff innature, I—23—7 ff Resonance interaction, 1-2-9 Retarded time, l—28—2l\-) .=@§7’i<~v—1._.PQus-mm“l\) C'\U\Retherford, ll—5—6 Retina, I—35—] Reynolds’ number, II—41—5 f Rigid body, l—l8—l angular momentum of,I—20—8 rotation of,I-18-2 if Ritz combination principle, I—38—8 Rods, I—35—l, l—36—6 Roemer, O.,l—7—5 Root-mean-square distance, I-6-6 Rotation, ofaxes, I-I1-3f plane, l—l8—l ofarigid body, I—l8—2 ff inspace, I-20-1 ff intwodimensions, I—18—1 if Rushton, I—35—9 Rutherford, Il—5—3 Rutherford-Bohr atomic model, Il—5—3 Rydberg (unit), I—38—6 Scalar, I-11-5 Scalar ﬁeld, Il~2—2 ff Scalar product, ll—25— Scattering oflight, I—— Schrodinger, E.,I—35— —- I—38—9 Schrodinger equation, ll-15-12 Scientiﬁc method, I-2-1 f Screw dislocation, ll—30—9 Screw ]aCk, I—4~5 Second (unit), l-5-5 Seismograph, I—5l—5 Self-inductance, ll—l6—4, ll~l7—1l f Shannon, C,I—44-Z Shear modulus, Il—38-5 Shear wave, I-51-4, II-38-8 Sheet ofcharge, lI—5—4 Side bands, I—48—4 f Simultaneity, I-15-7 f Sinusoidal waves. l—29—2 f Skin depth, II—32-ll Slipdislocation, ]I—30—9 Smoluchowski, I-41-8 Smooth muscle, I—14—2 Snell, W.,l—26—3 Snell’s law, I-26-3, I—3l—2, II-33—l Solenoid, lI—13—5 Solid-state physics, II—8—6 Sound, I—2—3, I—47—l ﬂ‘,I—50-I speed of,l-47—7 f Space, 1-8-2 Space-time, 1-2-6, I—l7—l ff,II-26-12 Special theory ofrelativity, I—l5—l ff Speciﬁc heat, I—40—7 f,l—45—2, Il—37—4 Speed, I—8—2 ff,I-9-2 oflight, l—l5—l, Il—l8—8 f ofsound, I—47—7 f Sphere ofcharge, II—5—4 f Spherical waves, ll—ZO-l 2ﬂ°,Il—2l—2 ff Spinel, lI—37—l2 Spin orbit, ll—8—7 Spontaneous emission, I—42—9 Standard deviation, 1-6-9 Statics, II-4—l f Statistical ﬂuctuations, l—6—3 ff Statistical mechanics, 1-3-1, l—40—l ff Steady ﬂow, lI—40—6 ff Step leader, II-9—10 Stern, lI—35—3 Stern-Gerlach experiment, ll—35-3 ffU) ‘O'\Nw>—<\/1;};t,o:§\lStevinus, S.,I-4-5 Stokes’ theorem, Il—3—10 Strain, ll—38—2 Strain tensor, ll—3l—l 1,ll—39—l ff “Strange” particles, lI-8-7 “Strangeness” number, I—2—9 Streamlines, ll—40—6 Stress, ll—38—2 Stress tensor, lI—3l—9 ff Striated muscle, I~l4—2 Supermalloy, ll—36—9 Superposition, II—13-11 f ofﬁelds, I-12-9 principle of,I—25—2 ff,II—1—3, ll-4-2 Surface, equipotential, II-4-11 f gaussian, ll-l0-l isothermal, Il—2-3 Surface tension, ll—l2—5 Symmetry, I—l—4, l—l1-1ff ofphysical laws, I-l6—3, I-52—1 ff Synchrotron, I-Z—5, I—l5—9, I-34-3 ff I-34-6, ll-1 7-5 Tamm, I.,I-5l-2 Taylor expansion, ll—6—7 Temperature, l—39—6 ff Tensor, lI—26—7, Il—31-1 if Tensor ﬁeld, ll—3l—ll Tetragonal cell, ll—30—7 Thermal conductivity. ll—2—8. ll—l2-2 ofagas, l—43—9 f Thermal equilibrium, I—4l—3 ff Thermal ionization, I—42—5 ff Thermodynamics, I—39—2, I-45-1 ff, II—37—4 f laws of,l-44—l if Thompson, Il—5—3 Thompson atomic model, Il—5—3 Thompson scattering cross section, I—32—8 Three-body problem, I-10-1 Three-dimensional waves, lI—20—8 f Thunderstorms, ll—9-5 ff Tides, I-7-4 f Time, I-2-3, l—5—l ff,I-8-l, I-8—2 retarded, l—Z8—Z standard of,I—5—5 transformation of,l—l5—5 ﬂ Torque, I—l8—4, l—20—l ff Torsion bar, ll—38-5 if Total internal reﬂection, ll—33—l2 f Transformation, Fourier, l—25—4 Galilean, l—l2—ll linear, l—l1-6 Lorentz, [—l5—3, I-17-l, I—34—8, I—52-2, ll-25—l, II-26-1 ff oftime, I-l5—5 ff ofvelocity, l—l6—4 ff Transformer, ll—l6-4 f Transient, I-24-1 ft" electrical, I-24~5 f Transient response, l—2l—6 Translation ofaxes, l-ll—lff Transmission line, ll—24—l if Transmitted waves, lI—33—7 ff Travelling ﬁeld, ll—l8-5ff Triclinic lattice, ll—30—7 Trigonal lattice, ll-30-7 Twin paradox, I—l6-3 f INDEX 5 Two-dimensional ﬁeld, ll—7~2 If Viscous ﬂow, ll—4l—4 f Waveguides, ll—24—l ft Tycho Brahe, l—7—l Vision, I—36—l ff Wavelength, I—l9—3, I-26-l binocular, I—36-4 Wave number, I-29-2 Ultraviolet radiation. l—26—l color, I—35—l ff Weber, ll—l6—2 Uncertainty principle, l—2~6, l—6—lO t, Visual cortex, I—36—4 Weber (unit), ll—l3—l l—37—9, l—37—l l,l~38—8 f,ll—5—3 Visual purple, I—35—9 “Wet“ water, ll—4l—l ff Unit cell, l—38—5 Voltmeter, ll-l6—l Weyl, H.,l—ll—l Unit vector, l—ll-l(), ll—2—3 Volume strain, II—38—3 Wheeler, lI—28—8 Unworldliness, ll—25—l0 Volume stress, lI—38—3 Wilson, CT.R,ll—9-9 vonNeumann, J,II-40-3 Work, I—l3—l ff,I—l4»l ff vandeCiraafl’ generator, ll-5—9, ll-8-7 Vortex lines, ll-40-10 ff Vector, l—ll—5ff Vorticity, Il—40—5 Vector algebra, l—ll~6f X'rayS» I‘2_5’ I"26_l Vector analysis, —— ——_ Wall energy, I[—37—6 X'ray dlﬁracuon’ “-304 Vector ﬁeld, ll—— -— Wapstra, I~52—lO ﬂUX Oi, ll—3—2 ll Watt (Ufllll, I—l3—3 Young I_35_7 Vector integrals, ll-3—l i‘ Wave, I-51-1 ff,ll—20—l if Youngls modlllw H_38_Z Vector operator, Il—2—6 electromagnetic, ll—2l—l f Yukawa H,[_2_8 n_28_l3 Vector potential, ll—4—l ll,ll-l5—l ff light, I—48—l Yukawa’pO[cmla|’ i|_28_l3 Vector product, l—20~4 plane, ll—20—l ff Yustova‘ 1_35_8 Velocity, l—8—3, l—9~Z t reﬂected, l[—33-7 ff components of,l~9—3 shear, l—5l—4, lI—38-8 transformation of,l—l6—4 fl sinusoidal, l—29—2 f Zeno, I-8-3 Velocity potential, ll—l2-9 spherical, ll—20—l2 ff,]l—2l—2 ff Zero, absolute, l—l—5 Vinci, Leonardo da,I-36-2 three-dimensional, lI—20-8 f Zero cur], ll—3~l0 f,ll—4—l Virtual work, principle oi,l-4—5 transmitted, Il—33—7 ff Zero divergence, ll—3-10 t,ll-4-l Viscosity, ll—4l—l ll‘ Wave equation, I—47—l ff,ll—l8—9 ff Zero mass, I-2-10 coeﬂicient ol,ll—4l—2 Wavefront, l—47—34>_ :U1 t\)'—< _uiN I32Q INDEX 6 l§(‘l)l t1')8Tt3'>t