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MAINLY MECHANICS, RADIATION, AND HEAT RICHARD P. FEYNMAN Richard Chace Tolman Professor ofTheoretical Physics California InstitueofTechnology ROBERT B. LEIGHTON Professor ofPhysics Calfornia InstituteofTechnology MATTHEW SANDS Professor Stanford University ADDISON-WESLEY PUBLISHING COMPANY A Reading, MassachusettsVV MenloPark,California -London-Amsterdam -DonMils,Ontario-Sydney Copyright ©1963 CALIFORNIA INSTITUTE OF TECHNOLOGY Printed intheUnited States ofAmerica MAY NOT BE REPRODUCED IN ANY FORM WITHOUT WRITTEN PERMISSION OF THE COPYRIGHT HOLDER. Library ofCongress Catalog Card No.63-20717 Sixth printing, February 1977 40 CRW 9695 de *ye yas By Feynman’s Preface These arethelectures inphysics that Igave lastyear and theyear before tothe freshman and sophomore classes atCaltech. The lectures are, ofcourse, not verbatim—they have been edited, sometimes extensively and sometimes less so. The lectures form only part ofthe complete course. The whole group of180 then they broke upinto small groups of15to20students inrecitation sections under theguidance ofateaching assistant. Inaddition, there was alaboratory The special problem wetried togetatwith these lectures was tomaintain the interest oftheveryenthusiastic andrathersmartstudents coming outofthe high schools and into Caltech, They have heard alotabout how interesting and exci ing physics isthe theory ofrelativity, quantum mechanics, and other modern and after two years itwas quite stultifying. The problem was whether ornot we could make acourse which would save themore advaneed and excited student by maintaining hisenthusiasm. serious. Ithought toaddress them tothemost intelligent intheclass and tomake sure, ifpossible, that even themost intelligent student was unable tocompletely encompass everything that wasinthelectures—by putting insuggestions ofappli- cations oftheideas and concepts invarious directions outside themain line of attack. For this reason, though, Itried very hard tomake allthestatements as accurate aspossible, topoint outinevery case where theequations andideas fitted into thebody ofphysics, and how—when they learned more—things would be modified. Ialso felt that forsuch students itisimportant toindicate what itis that they should——if they aresufficiently clever—be able tounderstand bydeduc- tion from what has been said before, and what isbeing put inassomething new. When new ideas came in,1would tryeithertodeducethemiftheywerededucible, ortoexplain that itwusanew idea which hadn't anybasis interms ofthings they had already learned and which was notsupposed tobeprovable—but wasjust added in, Atthestart ofthese lectures, |assumed that thestudents knew something when they came outofhigh school—such things asgeometrical optics, simple chemistryideas,andsoon,1alsodidn'tseethattherewasanyreasontomakethelectures 3 inadefinite order, inthesense that Iwould notbeallowed tomention something, until Iwasready todiscuss itindetail. There wasagreat deal ofmention 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 very qualitative way andarelater developed more completely. ‘Atthesame time that Iwas aiming atthemore active student, Ialso wanted totake care ofthefellow forwhom theextra fireworks and side applications are merely disquieting and who cannot beexpected tolearn most ofthematerial in the lecture atall. For such 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 him tounderstand everything, butonly thecentral and most direct features. Ittakes, ofcourse, @ certain intelligence onhispart toseewhich arethecentral theorems and central ideas, and which arethemore advanced side issues and applications which hemay understand only inlater years. Ingiving these lectures there wasoneserious difficulty: intheway thecourse was given, there wasn’t any feedback from thestudents tothelecturer toindicate how well thelectures were going over. This isindeed avery serious difficulty, and Idon't know how good thelectures really are. The whole thing wasessentially fanexperiment. 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 Iwas notsosatisfied. Inthefirst part ofthecourse, dealing with electricity and magnetism, Icouldn’t think ofany really unique ordifferent way ofdoing it—of any way that would beparticularly more exciting than the usual way ofpresenting it.SoIdon’t think Ididvery much inthelectures on electricity and magnetism. Attheend ofthesecond year Ihad originally intended togoon,after theelectricity and magnetism, 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. But since itwas notplanned that Iwould begiving these lec- ‘tures again, itwassuggested that itmight beagood idea totrytogive anintroduc- tion tothequantum mechanics—what you will find inVolume III. Itis perfectly clear that students who will major inphysics can wait until their third year forquantum mechanics. Ontheother hand, theargument was made that many ofthestudents inourcourse study physics asabackground fortheir primary interest inother fields. And theusual way ofdealing with quantum mechanics makes that subject almost unavailable forthegreat majority ofstudents because they have totakesolongtolearnit.Yet,initsrealapplications—espe- cially initsmore complex applications, such asinelectrical engineering and chem- istry—the fullmachinery ofthedifferential equation approach isnotactually used. SoItried todescribe theprinciples ofquantum mechanics inaway which wouldn't require that one first know themathematics ofpartial differential equa- tions. Even foraphysicist Ithink that isaninteresting thing totrytodo--to present quantum mechanics inthis reverse fashion—for several reasons which may beapparent inthelectures themselves. However, Ithink that theexperiment inthequantum mechanics part was notcompletely successful—in large part because Ireally didnothave enough time attheend (Ishould, forinstance, have had three orfour more lectures inorder todeal more completely with such matters asenergy bands and thespatial dependence ofamplitudes). Also, Ihad never presented thesubject this way before, sothelack offeedback was particularly serious. Inow believe thequantum mechanics should begiven atalater time. Maybe I'llhave achance todoitagain someday. Then I'lldoitright. ‘The reason there arenolectures onhow tosolve problems isbecause there were recitation sections. Although Ididputinthree lectures inthefirst year onhow to solve problems, they arenotincluded here. Also there was alecture oninertial 4 guidance which certainly belongs after thelecture onrotating systems, butwhich was, unfortunately, omitted, The fifth and sixth lectures areactually due to Matthew Sands, asIwas out oftown. The question, ofcourse, ishow well thisexperiment hassucceeded. Myownpointofview—which, however,doesnotseemtobesharedbymostofthepeople‘who worked with thestudents—is pessimistic. Idon’t thinkIdidverywellbythe students. When Ilook attheway themajority ofthestudents handled theproblems ‘ontheexaminations, Ithink that thesystem isafailure. Ofcourse, myfriends point outtomethatthere were oneortwodozen students who—very surprisingly —understood almost everything inallofthelectures, and who were quite active inworking with thematerial and worrying 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 itis almost superfluous.” (Gibbon) Still, Ididn’t want toleave any student completely behind, asperhaps Idid. think oneway wecould help thestudents more would bebyputting more hard work intodeveloping asetofproblems which would elucidate some oftheideas inthelectures. Problems give agood opportunity tofilloutthematerial ofthe lectures and make more realistic, more complete, and more settled inthemind theideas that have been exposed. Ithink, however, that there isn’t anysolution tothisproblem ofeducation other than torealize that thebest teaching canbedone only when there isadirect individual relationship between astudent andagood teacher—a situation inwhich thestudent discusses theideas, thinks about thethings, and talks about thethings. It’simpossible tolearn very much bysimply sitting inalecture, oreven bysimply doing problems that areassigned. Butinourmodern times wehave somany students toteach that wehave totrytofind some substitute fortheideal. Perhaps mylectures can make some contribution. Perhaps insome small place where there areindividual teachers and students, they may getsome inspiration orsome ideas from thelectures. Perhaps they willhave funthinking them through—or going ontodevelop some oftheideas further. RICHARD P,FEYNMAN June, 1963, 5 Foreword This book isbased upon acourse oflectures inintroductory physics given by Prof. R.P.Feynman attheCalifornia Institute ofTechnology during theacademic year 1961-62; itcovers thefirstyear ofthetwo-year introductory course taken by allCaltech freshmen andsophomores, andwasfollowed in1962-63 byasimilar series covering thesecond year. The lectures constitute amajor part ofafunda- mental revision oftheintroductory course, carried outover afour-year period. Theneed forabasic revision arose both from therapid development ofphysics, inrecent decades andfrom thefactthat entering freshmen have shown asteady increase inmathematical ability asaresult ofimprovements inhigh school mathe-maticscoursecontent.Wehopedtotakeadvantageofthisimprovedmathematical background, and also tointroduce enough modern subject matter tomake the ‘course challenging, interesting, andmore representative ofpresent-day physics. Inorder togenerate avariety ofideas onwhat material toinclude and how to present it,asubstantial number ofthephysics faculty were encouraged tooffer their ideas intheform oftopical outlines forarevisedcourse.Severalofthese were presented and were thoroughly and critically discussed. Itwas agreed almost atonce that abasic revision ofthecourse could notbeaccomplished either by merely adopting adifferent textbook, oreven bywriting one abinitio, butthat thenewcourse should becentered about asetoflectures, tobepresented atthe rate oftwoorthree perweek; theappropriate textmaterial would then beproduced asasecondary operation asthecourse developed, andsuitable laboratory experi- ments would also bearranged tofitthelecture material. Accordingly, arough outline ofthecourse wasestablished, butthiswas recognized asbeing incomplete, tentative, and subject toconsiderable modification bywhoever wastobear the responsibility foractually preparing thelectures. Concerning themechanism bywhich thecourse would finally bebrought to life,several plans were considered. These plans were mostly rather similar, involv- ingacooperative effort byNVstaff members who would share thetotal burden symmetrically and equally: each man would take responsibility for 1/N ofthe material, deliver thelectures, and write text material forhispart. However, the unavailability ofsufficient staff, and thedifficulty ofmaintaining auniform point ofviewbecauseofdifferences inpersonality andphilosophy ofindividual partici-pants, made such plans seem unworkable, The realization that weactually possessed themeans tocreate notjust anew anddifferent physics course, butpossibly aunique one, came asahappy inspira- tiontoProfessor Sands. Hesuggested that Professor R.P.Feynman prepare and deliver thelectures, and that these betape-recorded. When transcribed and edited, they would then become thetextbook forthenew course. This isessentially the plan that was adopted. Itwas expected that thenecessary editing would beminor, mainly consisting of supplying figures, andchecking punctuation andgrammar; itwastobedone by oneortwograduate students onapart-time basis. Unfortunately, thisexpectation wasshort-lived. Itwas, infact, amajor editorial operation totransform thever~ batim transcript into readable form, even without thereorganization orrevision ofthesubjectmatterthatwassometimes required. Furthermore, itwasnotajobforatechnical editor orforagraduate student, butonethat required theclose attention ofaprofessional physicist forfrom tentotwenty hours perlecture! 7 ‘The difficulty oftheeditorial task, together with theneed toplace thematerial inthehands ofthestudents assoon aspossible, setastrict limit upon theamount of“polishing” ofthematerial that could beaccomplished, and thus wewere forced toaim toward apreliminary buttechnically correct product that could be used immediately, rather than one that might beconsidered final orfinished. Because ofanurgent need formore copies forourstudents, andaheartening inter- estonthepart ofinstructors and students atseveral other institutions, wedecided topublish thematerial initspreliminary form rather than wait forafurther major revision which might never occur. Wehave noillusions astothecompleteness, smoothness, orlogical organization ofthematerial; infact, weplan several minor modifications inthecourse intheimmediate future, and wehope that itwill not become static inform orcontent. Inaddition tothelectures, which constitute acentrally important part ofthe course, itwas necessary also toprovide suitable exercises todevelop thestudents? experience and ability, and suitable experiments toprovide first-hand contact with thelecture material inthelaboratory. Neither ofthese aspects isinasad- vanced astate asthelecture material, butconsiderable progress hasbeen made ‘Some exercises were made upasthelectures progressed, and these were expanded and amplified foruseinthefollowing year. However, because wearenot yet satisfied that theexercises provide sufficient variety and depth ofapplication of thelecture material tomake thestudent fully aware ofthetremendous power being placed athisdisposal, theexercises arepublished separately inalessperma- nent form inorder toencourage frequent revision. ‘Anumber ofnew experiments forthenew course have been devised byProfessor H.V.Neher. Among these areseveral which utilize theextremely lowfriction exhibited byagasbearing: anovel linear airtrough, with which quantitative measurements ofone-dimensional motion, impacts, and harmonic motion can be made, andanair-supported, air-driven Maxwell top, with which accelerated rota-tionalmotionandgyroscopic precession andnutationcanbestudied.Thedevelop-ment ofnew laboratory experiments isexpected tocontinue foraconsiderable period oftime. The revision program was under thedirection ofProfessors R.B.Leighton, H.V. Neher, and M.Sands. Officially participating intheprogram were Professors R.P.Feynman, G.Neugebauer, R.M.Sutton, H.P.Stabler,* F,Strong, and R.Vogt, from thedivision ofPhysics, Mathematics andAstronomy, and Professors T.Caughey, M.Plesset, and C.H.Wilts from thedivision ofEngineering Science. The valuable assistance ofallthose contributing totherevision program isgrate- fully acknowledged. Weareparticularly indebted totheFord Foundation, without whose financial assistance this program could nothave been carried out. Roser B. LeiGHToN July, 1963 *1961-62, while onleave from Williams College, Williamstown, Mass. 8 Contents Curren 1, Etxcrronsowens Curren 6. Tu Furcrme Fi.o nx Vatoos 1-1Electrical forces1-1 (CIRCUMSTANCES1-2.Bcandmagusels1-3 61Equationsoftheelectrostaticpotential11-3)Chnraserisin ofwestonels1-4 2Theeeaicdpe2 {ckThelawsofclesromagnetim 1-3 3Remnsioontrequation4 1-3 Whataretetlt9 ©Theiplpecnl«paint6-4 15 Brctomagntim inscene andtechnology 1-10 &3.‘The ie apronmatio foranairy Vistnbuion6-6Curren2.Dirrensrax CaucousnovVacronFuze $$‘Tedsofbargedcondor6-62-1Understanding physis2-1 6-8Apintchargneaconductingplane6-922Senorandwerelds?andh2-2 6-5Aintchargnenconductingspe10 2-3 Derivatives offields—the gradient 2-4 6-10 Condensers; parallel plates 6-11 2-4 The operatory2-6 6-11High-voltage breakdown 6-13 23Opertionwih2-7 E12Theeldeminionmirscope614 2-6Thedifferential equationofheatflow2-8 27Seonderivates oftrBl29 Cumeven 7,Twxc FistosVanoos 2-8Pitfalls 2-11 CincumsTANces (Continued) Cuurren 3,Vecron Iertor CaLcutus 2-4.MethodsfoSingtecoatBk?) 3-1Vectorintegralstheineiteraofvy3-1 vasable2 32 Theta vecor eld 32 1-3. Plaa oilations 7-533Thefnfomcube;Gastheorem3-4 1-4Called!parteinanete7-83-4Henconductiontheitionexuaton3-6 4-3‘Thee feldofagd7-10 33Theciranton ofavectorBld8 5-6‘Tecleanaroundaaque Cumeren8,EuecrosrancEstacy 3-1 Curttree and divergencefees3-10 1Theelectroniceerofcharge.Aworm 3a Sommary 31 ieee8-2.Theenerayoacondenser,Foresoncharged Condon8-2 (Chapter4,Etectrostarics 8-3Theelectrostatic energyofanioniccrystal84 41 Stes1 SodHentaienegyincel8-6 £2Coulombaw;superposition#2 3Energyelect ld©-9 3 wt pout4 5‘Theergyofapointcarpe12 44B=46 SsCaeeedivergenceofE49 CHAPTER 9,ELECTRICITY INTHEATMOSPHERE 47 Field ofasphere ofcharge 4-10 9-1 Theelectric potential gradient ofthe {£8 Fallin cqiptental suc 4-1 umospere 1 9-2 Flere caret inthe atmosphere 9-2 : 9-3 Origin oftheatmospheric currents 9-4 Curren5.Arrticatow oFGas! Law 3-3.Orinoftheatmo 5-1Electrostatics isGauss'slawplus...S-1 9-5Themechanism ofchargeseparation 9-7 $2. Eaulirum nan ceevoxatetd$1 96Lighting910 53. Eyulvomwitconcer5-3, 5-4 Stability ofatoms 5-3 5-5Thefieldofalinecharge5-3 Cuseran 10,Drevecreycs5-6Asheetofcharge; twosheetsS—4 10-1Thedielectric constant 10-157Aspnereofcharge:sphereshell4 [0-2ThepolarizationvectorP10-2 58.Ibeeofpanechargecnaty 7275-5 10-3.Palaratoncares10-3 5-9Thefieldsofaconductor5-7 10-4Theelectrostatic equationswithdielectrics10-6 5-10 The field inacavity ofaconductor 5-8 10-5 Fields and forces with dielectrics 10-7 ° Cuarren 11, Ising Dietecrnics Cunpren 17. THE Laws oF INovcrion 11-1 Molecular dipoles 11-1 17-1 The physics ofinduction 17-1 11-2. Electronic polarization 11-1 17-2 Exceptions tothe“fux rule” 17-211-3Polarmolecules;orientation polarization 11-3 17-3Particleacceleration byaninducedelectricfield;11-4 Electri filds incavities ofa dielectric 11-5 the betatron 17-3,11-5.Thedielectricconstantofliquids;theClausius- 17-4Aparadox17-5 Mossotti equation 11-6 17-5. Alterating-current generator 17-611-6Soliddielectrics11-8 17-6Mutualinductance 17-9117 Ferroelectriety; BaTiO, 11-8 17-7 Seltinductance 17-11 17-8 Inductance and magnetic energy 17-12 Cuurren 12.Etectnostanic ANALoos Cuarren 18,TueMaxweLt Equarions 12-1 The same equations have thesame solutions 12-1 . 12-2. The flow ofheatapointsourcenearaninfinite 18-1Maxwell'sequations18-1 vinboundaay 12-2 18-2 Howthenewtermworks 18-312-3. Thosueiched meabeane 12-5 18-3. Allofclassical physics 18-5 12-4 Thediffusion ofneutrons; auniform spherical 18-4 Atraveling field18-5 source inahomogeneous medium 12-6 18-5 Thespeed oflight18-8 12-5. Irrotatonal fluid flow;theflowpastasphere12-8 18-6SolvingMaxwel’sequations;thepotentialsandthe 12-6 ilumination; theuniform lighting ofaplane 12-10 waveequation 18-9 12-7 The “underlying unity”ofnature12-12 Cuapten19,‘TuePrincieLeoFLeastAcTion Cnarten13,Maowerostarics Aspeciallecture—almost verbatim19-1 ‘Annote added after the lecture 19-14 13-1 The magnetic field 13-113-2Electriccurrent;theconservation ofcharge13-1 13-3‘Themagnetic forceonacurrent 13-2. (CHaPTER 20.SOLUTIONS OFMAXWELL’s EQUATIONS 13-4 The magnetic fieldofsteadycurrents; 16FansSrace Ampere'slaw13-3 20-1Wavesinfreespace;planewaves20-1 13-5 The magnetic field of straight wire and of8 20-2 Three-dimensional waves 20-8 solenoid; atomiccurrents 13-5 30-3Scientific imagination 20-9 13-6 Therelativity ofmagnetic andelectric fields 13-6 20-4 Spherical waves 20-1213-7Thetransformation ofcurrentsandcharges13-1113-8 Superposition; therighthandrule13-11 Cmpren21,Sotvniows orMaxowntt’s Equarions \witit CURRENTS AXD CHARGES Cwaeten 14,Tue: Maoneric Fito inVanious 21-1 Light andelectromagnetic waves 21-1Srrucrions 21-2Sphericalwavesfromapointsource21-2 14-1 Thevector potential 14-1 21-3 Thegeneral solution ofMaxwell's equations 21—4 14-2. Thevector potential ofknown currents 14-3 21-4 Theflelds ofanoscillating dipole 21-5 143 Astaightwire14-4 21-5‘Thepotentialsofamovingcharge;thegeneral 14-4 Along solenoid 14-5 solutionofLignardandWiechert21-9 14-5 Thefieldofasmall loop;themagnetic dipole 14-7 2~6 Thepotentials foracharge moving withconstant14-6Thevectorpotentialofacircuit14-8 velocity;theLorentzformula21-1214-7ThelawofBiotandSavart14-9 Cuarten22.ACCincurts Cuarren 15, Tue Vecror PorENTIAL 22-1 Impedances 22-1 22-2. Generators 22-515-1Thefreesonacurrentoopenerayof 22-3.Networksofidealelements;Kirchhof'srules22-78dipole15 22-4Equivalentcircuits22-10 15-2Mecha anelecene15-3 2kaun 15-3Theenergyofsteadycurrents15-6 shoeAadlernetwork22-12 IseByers A157 22-7Filters22-14 15-5.Thevector potentialandquantummechanics15-8 JaOthercielelements22-16 156 What istrue forstatics isfalse fordynamics 15-14 Cuarten 23. Cavity Resonators Cxarren16,INpucEDCURRENTS 23-1.Realcircuitelements23-116-1 Motors andgenerators 16-1 23-2 Acapacitor athigh frequencies 23-2 16-2 Transformers and inductances 16-8 23-3 Aresonant eavity 23-6 16-3 Forcesoninducedcurrents16-5 23-4Cavitymodes23-9 16-4 Electrical technology 16-8 23-5 Cavities andresonant circuits 23-10 10 Curren 24, Waveouroes CuaPren 30, ‘Tue INTERNAL GEOMETRY OF CavsTaLs 24-1 The transmission line 24-1 30-1 The internal geometry ofcrystals 30-124-2Therectangular waveguide24-4 30-2Chemicalbondsinerystals30-2 24-3Thecutofffrequency24-6 30-3.Thegrowthoferystals30-3 24-4Thespeedoftheguidedwaves24-7 30-4Crystallattes30-3 24-5 Observingguidedwaves24-7 30-5.Symmetriesintwodimensions30-4 24-6Waveguideplumbing24-8 30-6Symmetriesinthreedimensions30-7 24-7Waveguidemodes24-10 30-7Thestrengthofmetals30-8 24-8 Another way oflooking atthe guided waves 24-10 30-8 Dislocations anderystal growth 30-9 30-9 The Brage-Nye crystal model 30-10 Curren 25,Exucrnoovnases wwReLariisrc Curren 31,TensonsNotation 31-1 Thetensor ofpolarizability 31-1 25-1 Four-vectors25-1 3 25-2.Thesalarproduct25-3 Sioa.Tactfgrming thetensorcomponents31-3 25-3.Thefour-dimensional gradient 25-6 31-4Othertensors;thetensorofinertia31-6 25-4.Electrodynamics infour-dimensional notation 25-8 1754 tenor hetan 25-5.Thefour-potetial ofamovingcharge25-9 31rSThecrossproduct31-8 25-6Theinvarianceoftheequationsof Set eaelectrodynamics 25-10 31-8 Thefour-tensor ofelectromagnetic ‘omentum 31-12 Curren 26. Lonewrz, TRANSFORMATIONS OF THE FiELDsCuapren32.REFRACTIVE INDEXOF aren 26-1Thefour-potential ofamovingcharge26-1 vn 'BFRACTIVE INDEXOFDENSEMATERIALS 26-2.Thefieldsofapointchargewithaconstant 32-1Polarizationofmatter32-1 velocity26-2 32-2Maxwell'sequationsinadielectric32-3 26-3. Relativistic transformation ofthe lds 26-5 32-3 Waves inadiletric 32-526-4Theequationsofmotioninrelativistic 32-4Thecomplexindexofrefraction32-8 notation 2611 32-5Theindexofamixture32-8 32-6 Wavesinmetals32-10 32-7 Low-frequency and high-frequency approximations; (CuapTer 27. Fretp ENERGY AND FIELD MOMENTUM theskin depth and theplasma frequency 32-11 27-1Localconservation27-1 ‘27-2 Energy conservation andelectromagnetism 27-2 CuapTer 33. REFLECTION FROM SURFACES 27-3. Energy density andenergy flowinthe 33-1 Reflection andrefraction oflight33-1 electromagnetic field27-3 33-2Wavesindensematerials 33-2 27-4Theambiguityofthefieldenergy27-6 33-3.Theboundaryconditions33-4 21-5.Examples ofeneray flow 27-6 33-4 Thereflected andtransmitted waves 33-721-6Fieldmomentum27-9 33-5.Reflectionfrommetals33-1133-6Totalinternareflection33-12 CurTeR 28.ELectromacnenic Mass uneven 34,TheMacnenane orMa28-1Thefieldenergyof«pointcharge28-1 oa ™28-2Thefieldmomentum ofamovingcharge28-2 34-1Diamagnetism andparamagnetism 341 28-3.Electromagnetic mass28-3 gntemomentsandanamomen 28-4Theforceofanelectrononitself28-4 34-3Theprecessionofatomicmagnets 28-5AttemptstomodifytheMaxwelltheory28-6 esLamontther3-6 28-6Thenuclearforcefield28-12. ur “ores *34-6 Classical physics gives neither diamagnetism nor Paramagnetism 34-8 34-7 Angular momentum inquantum mechanics 34-8 Cuurrex 29. Tue Morion oF Cuancts 1s ELectnic Tae MorionoFCHA 34-8 Themagnetic energy ofatoms 34-11 29-1 Motion inauniform electric ormagnetic field29-1 oan nce 23- Movioninuniformde CHAPTER35.PARAMAGNETISM ANDMAGNETICRESON; 29-3. Anclectrostatic lens 29-2 35-1 Quantized magnetic states 35-129-4Amagneticlens29-3 35-2.TheStem-Gerlach experiment 35-329-5.Theelectronmicroscope29-3 35-3TheRabimolecular-beam method35-4 29-6 Accelerator guidelds29-4 35-4.Theparamagnetism ofbulkmaterials35-6 29-7 Aterating-gradient focusing29-6 35-5Coolingbyadiabaticdemagnetization 35-9 29-8Motionincrossedelecricandmagneticlds29-8 35-6Nuclearmagneticresonance35-10 n CHAPTER 36. FERROMAGNETISM Cuaprer 39, ELASTIC MATERIALS 36-1 Magnetization currents 36-1 39-1. The tensor ofstrain 39-136-2Thefield36-5 39-2Thetensorofelasticity39-4 36-3Themagnetization curve36-6 39-3.Themotionsinanelasticbody39-6 36-4Iron-coreinductances 36-8 39-4Nonelasticbehavior39-8 36-5 Electromagnets 36-9 39-5Calculating theelasticconstants39-10 36-6 Spontaneous magnetization 36-11 Cuapter 40, THE Flow oF Day WATER 40-1 Hydrostaties 40-1 CuarTeR37.MAGNETIC MATERIALS 40:2 Theequations ofmotion 40-237-1Understanding ferromagnetism 37-1 40-3Steadyflow—Bernoull’s theorem40-637-2Thermodynamic properties 37-4 40-4Circulation 40-937-3Thehysteresiscurve37-5 40-5Vortexlines40-10 37-4 Ferromagnetic materials 37-10 37-5. Extraordinary magnetic materials 37-11 Chapter 41. THE Flow orWet WATER 4I-1Viscosity41-141-2 Viscousflow41-4 (Cuarrer38.Exasriciry 41-3TheReynoldsnumber41-5 38-1 Hooke’s law38-1 41-4 Flow past acircular eylinder 41-738-2.Uniformstrains38-2 41-5Thelimitofzeroviscosity41-9 38-3Thetorsionbar;shearwaves38-5 41-6Couetteflow41-10 38-4Thebentbeam38-9 38-5Buckling38-11 INDEX R Ii Electromagnetism 1-1 Electrical forces Consider aforce likegravitation which varies predominantly inversely asthe 1-1. Electrical forees square ofthedistance, but which isabout abillin-billon-billon-billim timesstronger. Andwithanother diference. Therearetwokindsof“matte,” which 1-2Flectrle andmagnetic elds wecancallpositive and negative. Like kinds repel and unlike kinds attract— 1-3 Characteristicsofvectorfields unlike gravity where there isonly attraction. What would happen? ‘Abunchofpositiveswouldrepelwithanenormousforceandspreadoutin1-4Thelawsofelectromagnetism alldirections. Abunch ofnegatives would dothesame. Butanevenly mixed 1-5 What arethefields? bbunch ofpositives and negatives would dosomething completely different. The ,‘oppositepieceswouldbepulledtogetherbytheenormousattractions. Thenet '®iis leadscience resultwouldbethattheterrific forceswould balance themselves outalmost per- technol feetly,byformingtight,finemixturesofthepositiveandthenegative,andbetween twoseparatebunchesofsuchmixturestherewouldbepractically noattractionorrepulsion atall.Thereissuchaforce:theelectricalforce.Andallmatterisamixtureofposi-Review:Chapter12,Vol.1,Character- tiveprotonsandnegativeelectronswhichareattractingandrepellingwiththis isticsofForcearea force. Soperfect isthebalance, however, that when youstand near someonetseyoudon’tfeelanyforceatall.Itherewereevenalittlebitofunbalanceyouwould know it.Ifyou were standing atarm's length from someone and each of youhadonepercent more electrons thanprotons, therepelling force would bein- ‘edible. How great? Enough tolifttheEmpire State Building? No! Tolift Mount Everest? No! The repulsion would beenough tolifta“weight” equal to that ofthe entice earth! With such enormous forces soperfectly balanced inthisintimate mixture, it fgnothard tounderstand that matter, tying tokeep itspostive and negative charges inthefinest balance, canhave agreat stiffness andstrength. TheEmpireStateBuilding,forexample,swingsonlyeightfeetinthewindbecausetheelectricalforces hold every electron andproton more otlesintsproper place. Ontheother ‘band, ifwelook atmatter onascale small enough that weseeonly afewatoms, anysmall piece will not, usualy, have anequal number ofpositive and negative charges, andsothere willbe strong residual electrical forces. Even when there are ‘equal numbers ofboth charges intwo neighboring small pieces, there may stillbe large netelectrical forces because theforces between individual charges vary faverselyasthesquareofthe distance. Anetforce canarise ianegative charge of nepiece iscloser tothepositive than tothenegative charges oftheother piece.Theattractiveforcescanthenbelargerthantherepulsiveonesandtherecanbeanetattractionbetweentwosmallpieceswithnoexcesscharges.Theforcethatholdstheatoms together, and thechemical forces that hold molecules together, are really electrical forces acting inregions where thebalance ofcharge isnotperfect, orwhere thedistances arevery small You know, ofcourse, that atoms aremade with positive protons inthe sucleus and with electrons outside. You may ask: “Ifthis electrical force isso _tervifc, whydon’t theprotons andelectrons justgetontopofeach other? Ifthey want tobeinanintimate mixture, why isnt itstill more intimate?” The answer hastodowith thequantum effects. Ifwetrytoconfine ourelectrons inaregion fat isvery close totheprotons, then according totheuncertainty principle they must have some mean square momentum which islarger themore wetrytocon-finethem.Itisthismotion,requiredbythelawsofquantummechanics, thatkeepstheelectricalattractionfrombringingthechargesanyclosertogether. Mt ‘There isanother question: “What holds thenucleus together”? Inanucleus there areseveral protons, allofwhich arepositive. Why don't they push them- selves apart? Itturns outthatinnuclei there are,inaddition toelectrical forces, nonelectrical forces, called nuclear forces, which aregreater than theelectrical forces and which areable tohold theprotons together inspite oftheelectrical repulsion, The nuclear forces, however, have ashort range—their force falls off much more rapidly than 1/r2, And thishasanimportant consequence. Ifa nucleus hastoomany protons init,itgetstoobig,anditwillnotstay together. An example isuranium, with 92protons. The nuclear forces actmainly between each proton (orneutron) and itsnearest neighbor, while theelectrical forces actover larger distances, giving arepulsion between each proton andalloftheothers in thenucleus, The more protons inanucleus,thestrongeristheelectricalrepulsion, until, asinthecase ofuranium, thebalance issodelicate that thenucleus isalmost ready toflyapart from therepulsive electrical force. Ifsuch anucleus isjust “tapped”lightly(ascanbedonebysendinginaslowneutron),itbreaksintotwo aneaseetconte pieces,eachwithpositivecharge,andthesepiecesflyapartbyelectricalrepulsion.commonly“apna Theenergywhichisliberatedistheenergyoftheatomicbomb.Thisenergyis :bee usuallycalled“nuclear” energy,butitisreally“electrical” energyreleasedwhenr electrical forces haveovercome theattractive nuclear forces.a tied Wemayask,finally,whatholds negatively charged electron together (sinceiepsilon ithasnonuclearforces).Ifanelectronisallmadeofonekindofsubstance,eachtse Partshouldrepeltheotherparts,Why,then,doesn'titflyapart?Butdoestheoe electronhave“parts”?Perhapsweshouldsaythattheelectronisjustapointand 3@ theta thatelectrical forces onlyactbetween different pointcharges, sothattheelectron : foe doesnotactuponitself. Perhaps. Allwecansayisthatthequestion ofwhat‘kappa holdstheelectrontogetherhasproducedmanydifficultiesintheattemptstoformNAteks ‘acomplete theoryofelectromagnetism. ‘Thequestionhasneverbeenanswered.aa ‘Wewillentertainoverselvesbydiscussingthissubjectsomemoreinlaterchapters. , tu ‘Aswehaveseen,weshouldexpectthatitisacombination ofelectrical forces tz ni(ksiy andquantum-mechanical effects thatwilldetermine thedetailed structure of:Smoron materialsinbulk,and,therefore,theirproperties. Somematerialsarehard,somecn of aresoft.Someareelectrical “conductors”—because theirelectrons arefreeto Fo ‘moveabout;othersare“insulators”—because theirelectronsareheldtightlyto e=sigma individual atoms.Weshallconsiderlaterhowsomeoftheseproperties comeabout,‘ te butthatisaverycomplicated subject, sowewillbeginbylooking attheelectricalotupsiton forcesonlyinsimplesituations. Webeginbytreatingonlythelawsofelectricity—.. om including magnetism, whichisreallyapartofthesamesubject.amrcrr ny Wehavesaidthattheelectrical force,likeagravitational force,decreasesve psi inversely asthesquare ofthedistance between charges. Thisrelationship iscalledeo ones Coulomb's law.Butitisnotprecisely truewhencharges aremoving—the elec-tricalforcesdependalsoonthemotionsofthe charges inacomplicated way. One part oftheforce between moving charges wecallthemagnetic force. Itisreally one aspect ofanelectrical effect. That iswhy wecall thesubject “‘electromag- netism.” There isanimportant general principle that makes itpossible totreat elec- tromagnetic forces inarelatively simple way. Wefind, from experiment, that the force that actsonaparticular charge—no matter how many other charges there areorhow they aremoving—depends only ontheposition ofthat particular‘charge,onthevelocityofthecharge,andontheamountofcharge.Wecanwritetheforce Fonacharge gmoving with avelocity vas FagE+vXx B). ay Wecall Etheelectric field and Bthemagnetic field atthelocation ofthecharge. The important thing isthat theelectrical forces from alltheother charges inthe universe can besummarized bygiving just these two vectors. ‘Their values will depend onwhere thecharge is,and may change with time. Furthermore, ifwe replace that charge with another charge, theforce onthenew charge willbejustinproportion totheamountofchargesolongasalltherestofthechargesinthe 12 world donotchange their positions ormotions. (Inrealsituations, ofcourse, each charge produces forces onallother charges intheneighborhood andmay causetheseotherchargestomove,andsoinsomecasesthefieldscanchangeifwereplace‘our particular charge byanother.) ‘WeknowfromVol.Ihowtofindthemotionofaparticleifweknowtheforce onit.Equation (1.1) canbecombined with theequation ofmotion togive a mvalec] teevox, a2) SoifEandBaregiven, wecanfind themotions, Now weneed toknow how the E'sand B’sareproduced. One ofthemost important simplifying principles about thewaythefields are produced isthis: Suppose anumber ofcharges moving insome manner wouldproduceafield£1,andanothersetofchargeswouldproduceEz.Ifbothsetsof‘charges areinplace atthesame time (keeping thesame locations and motions they had when considered separately), then thefield produced isjust thesum En E+ Ex a3) Thisfactiscalledtheprincipleofsuperposition offields.Itholdsalsoformagneticfields. This principle means that ifweknow thelawfortheelectric and magnetic fields produced byasingle charge moving inanarbitrary way, then allthelaws of electrodynamics arecomplete. Ifwewant toknow theforce oncharge Aweneed onlycalculate theEandBproduced byeach ofthecharges B,C,D,etc.,andthen add theE'sand B’sfrom allthecharges tofind thefields, and from them the forces acting oncharge 4.Ifithad only turned outthat thefield produced bya single charge wassimple, thiswould betheneatest waytodescribe thelaws of electrodynamics. Wehave already given adescription ofthis law(Chapter 28, Vol. 1)anditis,unfortunately, rather complicated. Itturns outthat theform inwhich thelaws ofelectrodynamics aresimplest farenotwhat youmight expect. Itisnot simplest togiveaformula fortheforce that ‘onecharge produces onanother. Itistrue that when charges arestanding stillthe Coulomb force lawissimple, butwhen charges aremoving about therelations are complicated bydelays intime and bytheeffects ofacceleration, among others. Asaresult, wedonotwish topresent electrodynamics only through theforce laws between charges; wefind itmore convenient toconsider another point of view—a point ofview inwhich thelaws ofelectrodynamics appear tobethemost easily manageable, 1-2 Electric andmagnetic fields First, wemust extend, somewhat, our ideas oftheelectric and magnetic vectors, Eand B.We have defined them interms oftheforces that arefeltbya charge. Wewish now tospeak ofelectric andmagnetic fields atapoint even when there isnocharge present. Wearesaying, ineffect, that since there areforces “acting on” thecharge, there isstill “something” there when thecharge isremoved.Ifachargelocatedatthepoint(x,y,2)atthetime1feelstheforceFgivenbyEq,(1.1)weassociatethevectorsEandBwiththepointinspace(x,y,z).Wemay thinkofE(x,y,z,1)andB(x,y,2,1)asgivingtheforcesthatwouldbeexperienced atthetime1byachargelocatedat(x,y,2),withtheconditionthatplacingthecharge there didnotdisturb thepositions ormotions ofalltheother charges responsible forthefields. Following thisidea, weassociate with every point (x,»,2)inspace twovectors and B,which may bechanging with time. The electric and magnetic fields are, then,viewedasvectorfunctionsofx,y,z,andt.Sinceavectorisspecifiedbyits‘components, eachofthefieldsE(x,y,z,1)andB(x,yz,¢)representthreemathe-maticalfunctionsofx,y,z,and1. B Itisprecisely because E(orB)canbespecified atevery point inspace that itis called a“field.” A“field” isanyphysical quantity which takes ondifferent values atdifferent pointsinspace, Temperature, forexample, isafield—in thiscasea 7scalar field, which wewrite asT(x, y,z).The temperature could also vary intime, - andwewouldsaythetemperature fieldistime-dependent, andwriteT(x,y,2,1). —_ ‘Anotherexampleisthe“velocityfield”ofaflowingliquid.Wewriteu(x,y,z,1) Liedforthevelocity oftheliquid ateach point inspace atthetime 1.Itisavector field. —-— Le Returning totheelectromagnetic fields~although they areprodyced by + ‘charges according tocomplicated formulas, theyhave thefollowing important _ characteristic: therelationships between thevalues ofthefields atonepoint and thevaluesatanearbypointareverysimple.Withonlyafewsuchrelationships in — theform ofdifferential equations wecan describe thefields completely. Itisin terms ofsuch equations that thelaws ofelectrodynamics aremost simply written Fig.1-1,Avector fieldmaybe Therehavebeenvarious inventions tohelpthemindvisualize thebehavior ofrepresented bydrawing @vetoFarrows fields. Themostcorrect isalsothemostabstract; wesimply consider thefieldsas‘whosemogritvdes onddirectionindicore mathematical functionsofpositionandtime.Wecanalsoattempttogetamentalthevalues ofthevector fieldatthepoints picture ofthefieldbydrawing vectors atmany points inspace, each ofwhich gives j from which thearrows oredrawn. thefield strength and direction atthat point. Such arepresentation isshown in Fig. I-l. We can gofurther, however, and draw lines which are everywhere tangenttothevectors—which, sotospeak,followthearrowsandkeeptrackof WAthedirectionofthefield.Whenwedothiswelosetrackofthelengthsofthe ee‘vectors, butwecankeep track ofthestrength ofthefield bydrawing thelines far apart when thefield isweak and close together when itisstrong. Weadopt the a _sumeniton tatthemabeofnespraareatightangestothenesspro- = portional tothefield strength. This is,ofcourse, only anapproximation, and it ——-~___ sillrequire, ingenerat, thatnewtinessometimes startupinordertokeepthe numberuptothestrengthofthefield.ThefieldofFig.1-1isrepresentedby ———<a.fieldlinesinFig.1-2. TON1-3 Characteristics ofvector fields Fig:1-2. Avector field canbe There aretwomathematically important properties ofavector fieldwhichrepresented bydrawinglineswhichoreWeWilluseinourdescription ofthelawsofelectricity from thefield point ofview. angen!fothedirectionoftheflldvectorSupposeweimagineaclosedsurfaceofsomekindandaskwhetherwearelosing ‘ateachpoint,andbydrawingthedensity“something” fromtheinside;thatis,doesthefieldhaveaqualityof“outflow”? ‘oflines proportional tothemagnitude of Forinstance, foravelocity field wemight askwhether thevelocity isalways out- thefield vector. ward onthesurface or,more generally, whether more fluid flows out(per unittime)thancomesin.Wecallthenetamountoffluidgoingoutthroughthesurfaceperunit time the“flux ofvelocity” through thesurface. The flow through an clement ofasurface isjust equal tothecomponent ofthevelocity perpendicular tothesurface times thearea ofthesurface. For anarbitrary closed surface, the netoutward flow—or flux—is theaverage outward normal component ofthe vector Flux=(averagenormalcomponent) (surfacearea). ) 4J Inthecaseofanelectricfield,wecanmathematically definesomething4} analogoustoanoutflow,andweagaincallittheflux,butofcourseitisnotthe / cammuot perpndesier flowofanysubstance, because theelectric fieldisnotthevelocity ofanything. Ittothewurfoce turns out,however, thatthemathematical quantity which istheaverage normal\batace component ofthefieldstillhasausefulsignificance. Wespeak,then,oftheelectric flux—also defined byEq. (1.4). Finally, itisalso useful tospeak ofthe 4 fluxnotonlythroughacompletelyclosedsurface,butthroughanyboundedsur- /face. Asbefore, theflux through such asurface isdefined astheaverage normal component ofavector times thearea ofthesurface, These ideas areillustrated in nee Surtees nnoceme Thereisasecondproperty ofavectorfieldthathastodowithaline,rather feverage valve ofthenormal component than asurface. Suppose again thatwethink ofavelocityfieldthatdescribesthe ofthevectortimestheareaofthesurface. flowofaliquid.Wemightaskthisinterestingquestion:Istheliquidcirculating? 14 Bythatwemean: Isthere anetrotational motion around some loop? Suppose) that weinstantaneously freeze theliquid everywhere except inside ofatubewhich‘°) isofuniform bore, and which goes inaloop that closes back onitself asin Fig. 1-4, Outside ofthetube theliquid stops moving, butinside thetube itmaykeeponmovingbecauseofthemomentum inthetrappedliquid—that is,ifthereis‘moremomentumheadingonewayaroundthetubethantheother.Wedefineavequantity called thecirculation astheresulting speedoftheliquidinthetubetimesits circumference. Wecanagain extend ourideas anddefine the“circulation” forany vector field (even when there isn’t anything moving). For any vector field the circulation around anyimagined closed curve isdefined astheaverage tangential py component ofthevector (inaconsistent sense) multiplied bythecircumference ~=n.oftheloop(Fig.1-5). $$ Circulation =(averagetangential component)-(distance around). (1.5) = aan \SyYouwillseethatthisdefinitiondoesindeedgiveanumberwhichisproportional. =((----. -y\3J}?tothecirculation velocity inthequickly frozen tubedescribed above. array a Withjustthesetwoideas—fluxandcirculation—we candescribeallthelaws*7ST H ofelectricity andmagnetism atonce. Youmaynotunderstand thesignificance of a apt thelaws right away, butthey willgiveyousome ideaofthewaythephysics of oe electromagnetism willbeultimately described. te) 1-4Thelawsofelectromagnetism = The first law ofelectromagnetism describes theflux oftheelectric field:= TheHuxofEthroughanyclosedsurface=MEREcharEEinside,(yg) yy. where ¢9isaconvenient constant. (The constant ¢isusually read as“epsilon- bau zero” of“epsifon-naught"") Ifthere arenocharges inside thesurface, even though there arecharges nearby outside thesurface, theaverage normal component ofE ; iszero,sothereisnonetfluxthrough thesurface. Toshow thepower ofthis Fig:~4. (a)Thevelocity fleldino typeofstatement, wecanshowthatEq,(1.6)isthesameasCoulomb's law,pro- avid,Imagineatubeofwniformcross videdonlythatwealsoaddtheideathatthefieldfromasinglechargeisspherically coveosinte)"heliquidwere2sidenly symmetric.Forapointcharge,wedrawaspherearoundthecharge.Thenthefroven“everwhere ceeawtaeide,Mo averagenormalcomponent isjustthevalueofthemagnitude ofEatanypoint.tube,thegudintheibewouldcreulate sincethefieldmust bedirected radially andhave thesame strength forall points on gsshown in(c).thesphere.Ourrulenowsaysthatthefieldatthesurfaceofthesphere,timesthearea ofthesphere—that is,theoutgoing flux—is proportional tothecharge inside. Ifwewere tomake theradius ofthesphere bigger, thearea would increase as thesquare oftheradius. The average normal component oftheelectric field timesthatareamuststillbeequaltothesamechargeinside,andsothefieldmustdecreaseasthesquare ofthedistance—we getan“inverse square” field. Ifwehave anarbitrary curve inspace and measure thecirculation ofthe clectric field around thecurve, wewill find that itisnot, ingeneral, zero (although itisfortheCoulomb field). Rather, forelectricity there isasecond lawthat states: foranysurface S(not closed) whose edge isthecurve C, inten <7}Circulation ofEaroundC=4(fluxofBthrough5).a7 We Q, ‘Wecancomplete thelaws oftheelectromagnetic field bywriting two corre sponding equations forthemagnetic fieldB. \* FluxofBthrough anyclosedsurface =0. (1.8) gpioan saSinesCae For asurface Sbounded bythecurve C, ~ = Xi d 1.1-5, The circulation of@vectore%(irculation ofBaroundC)=(BuxofEthroughS) foldsteoveragetorncnbal compe,fluxofelectriccurrentthroughS rentofthevector(in@consistentsense) ++Hoteles currentBroughS.(1.9)nestecreoerance ofteloo. 1s ptorfan |)ta , toi ~reRMINAL ||LU exnnaoner Fig. 1-6. Abor magnet gives ofoldBotewire,Whenthreicent‘olong thewire, thewire moves becouse oftheforce F=qvXB. ‘Theconstantc?thatappearsinEq.(1.9)isthesquareofthevelocityoflight. Itappears because magnetism isinreality arelativistic effect ofelectricity. The constant éyhas been stuck intomake theunits ofelectric current come out in& ‘convenient way. Equations (1.6) through (1.9), together with Eq. (1.1), areallthelaws of electrodynamics*. Asyou remember, thelaws ofNewton were very simple to write down, butthey had alotofcomplicated consequences and ittook usalong time tolearn about them all. These laws arenot nearly assimple towrite down, whichmeansthattheconsequences aregoingtobemoreelaborate anditwilltake usquite alotoftime tofigure them allout. ‘Wecan illustrate some ofthelaws ofelectrodynamics byaseries ofsmall ‘experiments which show qualitatively the interrelationships ofelectric and magnetic fields. You have experienced thefirst term ofEq.(1.1) when combing yourhair,sowewon'tshowthatone.ThesecondpartofEq.(1.1)canbedemon- strated bypassing acurrent through awire which hangs above abar magnet, as shown inFig. 1-6. The wire willmove when acurrent isturned onbecause ofthe forceF=quXB.Whenacurrentexists,thechargesinsidethewirearemoving, ‘sothey have avelocity v,and themagnetic field from themagnet exerts aforce on ‘them, which results inpushing thewire sideways. ‘When thewire ispushed totheleft, wewould expect that themagnet must feelapush totheright. (Otherwise wecould putthewhole thing onawagon and have apropulsion system that didn't conserve momentum!) Although theforce is ‘too small tomake movement ofthe bar magnet visible, amore sensitively sup- Ported magnet, like acompass needle, willshow themovement. How does thewire push onthemagnet? The current inthewire produces a magnetic field ofitsown that exerts forces onthemagnet. According tothelast wae6||«Ba4 Hy ray Es Jaan uaanet Fig. 1-7. The magnetic field ofthe wire exer foes onthe magn *Weneedonlytoaddaremarkaboutsomeconventionsfrthesgnofthecirculation. M6 ! [o) + Fig. 1-8. Two wires, carrying cur- rent, exert forces oneach other. terminEq.(1.9),acurrentmusthaveacirculation ofB—inthiscase,thelinesofBare loops around thewire, asshown inFig. 1-7. This Befield isresponsible for theforce onthemagnet. Equation (1.9) tells usthat forafixed current through thewire thecirculation ofBisthesame foranycurve that surrounds thewire. Forcurves—say circles— that arefarther away from thewire, thecircumference islarger, sothetangential component ofBmust decrease. You canseethat wewould, infact, expect Bto decrease linearly with thedistance from along straight wire. ‘Now, wehave said that acurrent through awire produces amagnetic field, andthatwhen there isamagnetic field present there isaforce onawire carrying& current.Thenweshouldalsoexpectthatifwemakeamagneticfieldwithacurrent inone wire, itshould exert aforce onanother wire which also carries acurrent. This canbeshown byusing two hanging wires asshown inFig. 1-8. When thecurrentsareinthesamedirection, thetwowiresattract,butwhenthecurrentsare‘opposite, they repel. Inshort, electrical currents,aswellasmagnets,makemagneticfields.Butwait, what isamagnet, anyway? Ifmagnetic fields areproduced bymoving charges, is itnotpossible that themagnetic field from apiece ofiron isreally theresult of currents? Itappears tobeso. Wecanreplace thebarmagnet ofourexperimentwithacoilofwire,asshowninFig.1-9.Whenacurrentispassedthroughthe‘coil—as well asthrough thestraight wire above it—we observe amotion ofthewireexactly’as before,whenwehadamagnetinsteadofacoil.Inotherwords,thecurrent inthecoilimitates amagnet. Itappears, then, that apiece ofiron acts asthough itcontains aperpetual circulating current. Wecan, infact, understand‘magnetsintermsofpermanent currentsintheatomsoftheiron.Theforceonthe‘magnet inFig, 1-7isdue tothesecond term inEq,(1.1). .(tromcolt) emmaroelonwieed Bana, con.oFwine Nescret aa Fig. 1-9. Thebormagnet ofFig. 16 can bereplaced byacoil carrying on electrical current. Asimilar force acts fonthe wire. re] | ‘Wheredothecurrentscomefrom?Onepossibility wouldbefromthemotion oftheelectronsinatomicorbits.Actually,thatisnotthecaseforiron,although | itisforsome materials, Inaddition tomoving around inanatom, anelectron also spins about onitsown axis—something like thespin oftheearth—and itis thecurrent from this spin that gives themagnetic field iniron. (We say“some- thing like thespin oftheearth” because thequestion issodeep inquantum me- chanics that theclassical ideas donotreally describe things toowell.) Inmost substances, some electrons spin one way and some spin theother, sothemag- netism cancels out, but iniron—for amysterious reason which wewill discuss later—many oftheelectrons arespinning withtheiraxeslinedup,andthatisthe ‘source ofthemagnetism. ‘Since thefields ofmagnets arefrom currents, wedonothave toaddanyextra term toEqs. (1.8) or(1.9) totake care ofmagnets. Wejust take allcurrents, including thecirculating currents ofthespinning electrons, and then thelaw is right. You should also notice that Eq. (1.8) says that there arenomagnetic “charges” analogous totheelectrical charges appearing ontheright side of Eq.(1.6). None hasbeen found. °4 Currant caccant Fig. 1-10. The circulation of B ceround thecurveCisgiveneitherbythe ollcurrentpassingthroughthesurfaceSi, Wy -orbytherateofchangeofthefluxofE Curve through thesurface S2. Surtace 8;Surtooe Sp ‘Thefirstterm ontheright-hand sideofEq.(1.9) wasdiscovered theoretically ‘byMaxwell andisofgreat importance. Itsays that changing electric fields produce magnetic effects. Infact, without this term theequation would notmake sense, because without itthere could benocurrents incircuits that arenot complete loops.Butsuchcurrentsdoexist,aswecanseeinthefollowing example. Imagine acapacitor made oftwo flatplates. Itisbeing charged byacurrent that flows toward one plate and away from theother, asshown inFig. 1-10. Wedraw a ‘curveCaround oneofthewiresandfillitinwithasurface whichcrosses thewire,‘asshownbythesurfaceS,inthefigure.According toEq.(1.9),thecirculation ofBaround Cisgivenbythecurrentinthewire(timesc*).Butwhatifwefillinthe curve with adifferent surface Sz,which isshaped like abowl and passes between theplates ofthecapacitor, staying always away from thewire? There iscertainly nocurrent through this surface. But, surely, just changing thelocation ofan imaginary surface isnotgoing tochange areal magnetic field! The circulation of Bmust bewhat itwas before, The first term ontheright-hand side ofEq.(1.9) does, indeed, combine with thesecond term togive thesame result forthetwo surfaces S,and Sj. For S,thecirculation ofBisgivenintermsoftherateof ‘changeofthefluxofEbetweentheplatesofthecapacitor.Anditworksoutthat thechanging Eisrelated tothecurrent injustthewayrequired forEq.(1.9) tobecorrect. Maxwell sawthatitwasneeded,andhewasthefirsttowritethecomplete‘equation. . With thesetup shown inFig. 1-6wecandemonstrate another ofthelaws of electromagnetism. Wedisconnect theendsofthehanging wirefromthebatteryandconnect them toagalvanometer which tells uswhen there isacurrent through thewire. When wepush thewire sideways through themagnetic field ofthe magnet, weobserve acurrent. Such aneffect isagain just another consequence ofEq,(1.1)}the electronsinthewirefeeltheforceF=quXB.Theelectronshaveasidewise velocity because theymovewiththewire.Thisvwithavertical B from themagnet results inaforce ontheelectrons directed along thewire, which starts theelectrons moving toward thegalvanometer. Ls Suppose, however, that weleave thewire alone and move themagnet. We ‘guess from relativity that itshould make nodifference, and indeed, weobserve a similar current inthegalvanometer. How does themagnetic fieldproduce forces on charges atrest? According toEq,(1.1) there must beanelectric field. Amoving ‘magnet must make anelectric field. How that happens issaid quantitatively by Eq.(I.7). This equation describes many phenomena ofgreat practical interest, such asthose that occur inelectric generators and transformers. ‘Themost remarkable consequence ofourequations isthat thecombination ofEq.(1.7)andEq.(1.9)containstheexplanation oftheradiationofelectromagnetic effects over large distances. The reason isroughly something like this: suppose that somewhere wehave amagnetic field which isincreasing because, say, acurrent isturned onsuddenly inawire. Then byEq.(1.7) there must bea circulation ofanelectric field. Astheelectri field builds uptoproduce itscircula- tion, then according toEq.(1.9) amagnetic circulation willbegenerated. Butthe building upofhis magnetic field will produce anew circulation oftheelectric field, andsoon. Inthisway fields work their way through space without theneed ofcharges orcurrents except attheir source. That istheway weseeeach other!Itisallintheequationsoftheelectromagnetic fields. 1-5 What are the fields? Wenowmakeafewremarksonourwayoflookingatthissubject.Youmay bbesaying:“Allthisbusinessoffluxesandcirculations isprettyabstract. Thereareelectric fields atevery point inspace; then there are these ‘laws.’ But what is actually happening? Why can't you explain it,forinstance, bywhatever itisthat ‘goes between thecharges.” Well, itdepends onyour prejudices. Many physicists used tosay that direct action with nothing inbetween was inconceivable. (How could they find anidea inconceivable when ithadalready been conceived?) They would say: “Look, theonly forces weknow arethedirect action ofone piece of‘matteronanother.Itisimpossible thattherecanbeaforcewithnothingtotransmitit.”Butwhatreallyhappenswhenwestudythe“directaction”ofonepieceof‘matter right against another? Wediscover that itisnotone piece right against theother; they areslightly separated, and there areelectrical forces acting on@ tinyscale. Thus wefind that wearegoing toexplain so-called direct-contact action interms ofthepicture forelectrical forces. Itiscertainly notsensible totryto insist that anelectrical force has tolook like theold, familiar, muscular push or pill,whenitwillturnoutthatthemuscularpushesandpullsaregoingtobeinter-preted aselectrical forces! The only sensible question iswhat isthemost con- venient way tolook atelectrical effects. Some people prefer torepresent them as theinteraction atadistance ofcharges, andtouseacomplicated law. Others love thefield lines. They draw field lines allthetime, andfeelthat writing E'sandB's istooabstract. ‘The field lines, however, areonly acrude way ofdescribing afield,anditisverydifficulttogivethecorrect,quantitative lawsdirectlyintermsoffieldlines, Also, theideas ofthefield lines donotcontain thedeepest principle of electrodynamics, which isthesuperposition principle. Even though weknow how thefield lines look foronesetofcharges and what thefield lines look likeforan- other setofcharges, wedon’t getanyidea about what thefield line patterns will look like when both sets arepresent together. From themathematical stand- Point, ontheother hand, superposition iseasy—we simply add thetwo vectors. ‘The field lines have some advantage ingiving avivid picture, butthey also have some disadvantages. The direct interaction way ofthinking hasgreat advantages ‘when thinking ofelectrical charges atrest, buthasgreat disadvantages when dealing with charges inrapid motion. ‘Thebestwayistousetheabstractfieldidea.Thatitisabstractisunfortunate, butnecessary. The attempts totrytorepresent theelectric field asthemotion ofsomekindofgearwheels,orintermsoflines,orofstressesinsomekindofmate~tialhave used upmore effort ofphysicists than itwould have taken simply toget theright answers about electrodynamics. Itisinteresting thatthecorrect equationsforthebehavioroflightincrystalswereworkedoutbyMcCullough in1843.But 19 people said tohim: “Yes, butthere isnorealmaterial whose mechanical properties could possibly satisfy those equations, and since light isanoscillation that mustvibrateinsomething, wecannotbelievethisabstractequationbusiness.” Ifpeoplehhad been more open-minded, they might have believed intheright equations forthebehavioroflightalotearlierthantheydid.Tnthecase ofthemagnetic field wecan make thefollowing point: Suppose thatyoufinallysucceeded inmakingupapictureofthemagneticfeldintermsofsome kind oflines orofgear wheels running through space. ‘Then you tryto explain what happens totwocharges moving inspace, both atthesame speed andparalleltoeachother.Becausetheyaremoving,theywillbehaveliketwocurrentsand willhave amagnetic field associated with them (like thecurrents inthewires ofFig. 1-8). Anobserver who was riding along with thetwo charges, however, would seeboth charges asstationary, andwould saythat there isnomagnetic field. ‘The “gear wheels” or“lines” disappear when you ride along with theobject! All wwehave done istoinvent anew problem. How can thegear wheels disappear?! The people who draw field lines areinasimilar difficulty. Not only isitnotpos- sible tosaywhether thefield lines move ordonotmove with charges—they may disappear completely incertain coordinate frames ‘What wearesaying, then, isthat magnetism isreally arelativistic effect. In thecase ofthetwocharges wejustconsidered, travelling parallel toeach other, we ‘would expect tohave tomake relativistic corrections totheir motion, with terms of ‘order v2/c?. These corrections must correspond tothemagnetic force, Butwhat about theforce between thetwo wires inourexperiment (Fig. 1-8). There the magnetic force isthe whole force. Itdidn’t look like a“relativistic correction.” Also, ifweestimate thevelocities oftheelectrons inthewire (you can dothis yourself), wefind that their average speed along thewire isabout 0.01 centimeter persecond. Sov*/c? isabout 10*°. Surely anegligible “correction.” Butno! Although themagnetic force is,inthiscase, 10~** ofthe“normal” electrical force between themoving electrons, remember that the‘‘normal” electrical forces have disappeared because ofthealmost perfect balancing out—because thewires have thesame number ofprotons aselectrons. The balance ismuch more precise than‘onepartin102%,andthesmallrelativistic termwhichwecallthemagneticforceistheonly term left. Itbecomes thedominant term, Itisthenear-perfect cancellation ofelectricaleffectswhichallowedrelativityeffects (that is,magnetism) tobestudied and thecorrect equations—to order v/e?—to bediscovered, even though physicists didn’t know that’s what was happening. And that iswhy, when relativity wasdiscovered, theelectromagnetic laws didn’t need tobechanged. They—unlike mechanics—were already correct, toaprecision ofv?/c?, 1-6 Electromagnetism inscience andtechnology Letusend this chapter bypointing outthat among themany phenomena studied bytheGreeks there were twovery strange ones: that ifyourubbed apiece ofamber youcould liftuplittle pieces ofpapyrus, andthatthere wasastrange rock from theisland ofMagnesia which attracted iron. Itisamazing tothink that these were theonly phenomena known totheGreeks inwhich theeffects ofelec tricity ormagnetism were apparent. The reason that these were theonly phe- nomena that appeared isdue primarily tothefantastic precision ofthebalancingofchargesthatwementioned earlier.StudybyscientistswhocameaftertheGreeksuncovered onenewphenomena afteranotherthatwerereallysomeaspectoftheseamber and/or lodestone effects. Now werealize that thephenomena ofchemical interaction and, ultimately, oflifeitself aretobeunderstood interms ofelectro- magnetism. ‘Atthesame time that anunderstanding ofthesubject ofelectromagnetismwasbeingdeveloped, technicalpossibilities thatdefiedtheimagination ofthepeople thatcame before were appearing: itbecame possible tosignal bytelegraph over long distances, andtotalktoanother person miles away without anyconnections between, and torun huge power systems—a great water wheel, connected by 1410 filaments over hundreds ofmiles toanother engine that turns inresponse tothe master wheel—many thousands ofbranching filaments—ten thousand engines in tenthousand places running themachines ofindustries and homes—all turningbecauseoftheknowledge ofthelawsofelectromagnetism.Today weareapplying even more subtle effects. The electrical forces, enor- ‘mous asthey are,canalso bevery tiny, andwecancontrol them and usethem in very many ways. Sodelicate areour instruments that wecan tellwhat aman is doing bytheway heaffects theelectrons inathin metal rod hundreds ofmiles away. Allweneed todoistousetherodasanantenna foratelevision receiver! From along view ofthehistory ofmankind—seen from, say, tenthousand years from now—there canbelittle doubt that themost significant event ofthe 19th century willbejudged asMaxwel’s discovery ofthelaws ofelectrodynamics. TheAmerican Civil War willpale into provincial insignificance incomparison with thisimportant scientific event ofthesame decade. an 2 Differential Calculus of Vector Fields 21 Understanding physics Thephysicist needs afacility inlooking atproblems from several points of 2-1 Understanding physics view. ‘The exact analysis ofreal physical problems isusually quite complicated, andanyparticular physical situation maybetoocomplicated toanalyzedirectly 7?ScalarandvectorGelds—7-bysolvingthedifferential equation. Butonecanstillgetaverygoodideaofthebehaviorofasystemifonehassomefeelforthecharacterofthesolutionindifer- 2-3Derivatives offields—theentcircumstances. Ideas such asthefield lines, capacitance, resistance, and in- gradient ductance are, forsuch purposes, very useful. Sowewill spendmuch ofourtimeanalyzingthem,Tnthiswaywewilgetfelastohatshouldhappenindifferent 2-4Theoperator¥electromagnetic situations. Ontheotherhand,noneoftheheuristicmodels,such 2-5Operations withVasfield lines, isreally adequate andaccurate forallsituations. There isonly oneprecisewayofpresentingthelaws,andthatisbymeansofdifferential equations. 7-6heerential equationofThey have theadvantage ofbeing fundamental and, sofarasweknow, precise, Ifyouhave learned thedifferential equations youcanalways goback tothem. 2-7Second derivatives ofvector There isnothing tounlearn. fieldsItwilltakeyousometimetounderstand whatshouldhappenindiferent >pittacircumstances. You will have tosolve theequations. Each time you solve the equations, you willlearn something about thecharacter ofthesolutions. Tokeepthesesolutionsinmind,itwillbeusefulalsotostudytheirmeaningintermsoffield linesandofotherconcepts.Thisisthewayyouwillreally“understand” theequa-tions. Thatisthedifference between mathematics andphysics. Mathematicians, posi... Chapter 11,Vol.I,Veetorpeoplewhohaveverymathematical minds,areoftenledastraywhen“studying” R¢”ews Chapter11,Vol.I,Vectors physics because they lose sight ofthephysics. They say: “Look, these differential equations—the Maxwell equations—are allthere istoelectrodynamics; itis admitted bythephysicists that there isnothing which isnotcontained intheequa- tions. The equations arecomplicated, butafter allthey areonly mathematical equations and ifIunderstandthemmathematically insideout,Iwillunderstand thephysics inside out.” Only itdoesn't work that way. Mathematicians who study physics with that point ofview—and there have been many ofthem—usually ‘make litle contribution tophysics and, infact, little tomathematics. They fail because theactual physical situations inthereal world aresocomplicated that itis 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 ifIhaveawayoffiguring outthecharacteristics ofitssolutionWithoutactuallysolvingit.”Soifwehaveawayofknowingwhatshouldhappeningiven circumstances without actually solving theequations, then we“under- stand” theequations, asapplied tothese circumstances. Aphysical understanding, isacompletely unmathematical, imprecise, and inexact thing, butabsolutely neces- sary foraphysicist. Ordinarily, acourse likethisisgiven bydeveloping gradually thephysical ideas—by starting with simple situations and going ontomore andmore compli- cated situations. This requires that you continuously forget things you previouslylearned—things thataretrueincertainsituations, butwhicharenottrueingeneral.Forexample, the“law” that theelectrical force depends onthesquare ofthe distanceisnotalwaystrue,Weprefertheoppositeapproach.Weprefertotake first thecomplete laws, and then tostep back and apply them tosimple situa- tions, developing thephysical ideas aswegoalong. And that iswhat wearegoing todo. a Our approach iscompletely opposite tothehistorical approach inwhich one develops thesubject interms oftheexperiments bywhich theinformation was obtained. Butthesubject ofphysics hasbeen developed over thepast 200yearsbysomeveryingeniouspeople,andaswehaveonlyalimitedtimetoacquireourknowledge, wecannot possibly cover everything they did. Unfortunately oneof thethings that weshall have atendency tolose inthese lectures isthehistorical, experimental development. Itishoped thatinthelaboratory some ofthislackcan becorrected. You canalso fillinwhat wemust leave outbyreading theEncy- clopedia Brittanica, which hasexcellent historical articles onelectricity and on other parts ofphysics. You willalso find historical information inmany textbooks onelectricity and magnetism, 2-2 Scalar and vector fields—Tandh‘Webeginnowwiththeabstract,mathematical viewofthetheoryofelectricity andmagnetism. Theultimate idea istoexplain themeaning ofthelaws given in Chapter 1.Buttodothiswemust first explain anew and peculiar notation that wewant touse, Soletusforget electromagnetism forthemoment anddiscuss the mathematics ofvector fields. Itisofvery great importance, notonly forelectro- . ‘magnetism, butforallkinds ofphysical circumstances. Justasordinary differentialWarctaig vaeTone beyrows andintegral calculus issoimportant toallbranches ofphysics, s0alsoisthe differentialcalculusofvectors.Weturntothatsubject. Senssees Listedbelowareafewfactsfromthealgebraofvectors.Itisassumedthat moryou already know them. Ea E nye &.re A+Bescalar=ABs+AyBy+ABs @) Ortonprefer AXB=vector 2) £. (AX Bye=ABy ~AyBe . (AXBye=AyBe—ABy WeAidethepottery area= (AXBy=AB,~AB, ABCDE®R®F @& AXA=0 23) HIrEgJk uma A+(AXB)=0 4) orp aeRrsTty A+BXO)=(AXBC es) VWK YZ AX(BXC)=BA:C)—CAB) 2.6) Swot Lotere awfardar: ‘Alsowewillwant tousethetwofollowing equalities from thecalculus: < atLag Faye Hf aeaetg aft) =Fax+Lay+Lae, @n Rec Sk Amn 2 2Jd af.as. 8)axay~ayax op prea tw Thefirstequation (2.7)is,ofcourse, trueonlyinthelimitthatAx,4y,andAz wwe et gotowardzero. conThesimplestpossiblephysicalfieldisascalarfield.Byafield,youremember, {PtStreae -yeenounswemeanaquantitywhichdependsuponpositioninspace.Byascalarfieldwemerely mean afield which ischaracterized ateach point byasingle number—a scalar. Ofcourse thenumber may change intime, butweneed notworry about that forthemoment. Wewilltalkabout what thefield looks likeatagiven instant. Asanexample ofascalar field, consider asolid block ofmaterial which hasbeen heated atsome places and cooled atothers, sothat thetemperature ofthebody varies from point topoint inacomplicated way. Then thetemperature willbea funetion ofx,y,andz,theposition inspace measured inarectangular coordinate system. Temperature isascalar fel. 22 y >Hot me 7 te40° »4 ; -Tex Fig.2-1.Temperature Tisonexample ofo Toy) scolar field. With each point (x,y,2)inspace cod\ T=20% thereiansciatedonumberTix,ys2-_Allpointson ’7 thesurfacemarkedT=20°(shownasacurveat firete O)creettesametemperanre, Thearrows a] +‘oresamples oftheheat flow vector h. One way ofthinking about scalar fields istoimagine “contours” which are imaginary surfaces drawn through allpoints forwhich thefield basthesame value, justascontour lines onamap connect points with thesame height. Foratempera ture field thecontours arecalled “isothermal surfaces" orisotherms. Figure 2-1 illustrates @temperature field and shows thedependence ofTon xand ywhen z= 0.Several isotherms aredrawn, There arealso vector fields. The idea isvery simple. Avector isgiven foreach point inspace. The vector varies from point topoint. Asanexample, consider a fouating body. The velocity ofthematerial ofthebody atany point isavector which isafunction ofposition (Fig. 2-2). Asasecond example, consider theflow —eranionofheatinablockofmaterial.Ifthetemperature intheblockishighatoneplaceandlowatanother, there wilbeaflow ofheat from thehotter places tothecolder. ‘Theheat willbelowing indifferent directions indifferent parts oftheblock. ‘The heatflowisadirectional quantity which wecallf.Itsmagnitude isameasure of Fig,2-2, Thevelocity ofthetoms how much heat isflowing. Examples oftheheat flow vector arealso shown in@rotating object isenexomple of© inFig. 2-1. vector field. y a 6, to , ,1 Fig.2-3.Heatowisovectorld.Thevectorcation—BpelthalongheerectionofheRoweItmagntude isthe energy transported per ontfime across © surface element oriented perpendicuiar tothe Row, rr_dvided bythearea ofthe surface element. 4 Let'smaxeamareprecisedefinition ofh:Themagnitude ofthevectorheat , flow atapoint istheamount ofthermal energy that passes, perunit time and per unitarea through aninfinitesimal surface element, atright angles tothedirection 9 offlow. Thevector points inthedirection offlow(seeFig.2-3).Insymbols: If47 a isthethermal energythatpassesperunittimethrough thesurfaceelementda,then ZA fi he2% 9)RES he where eis aunitvector inthedirection offlow. i. ‘Thevectorhcanbedefinedinanother way—in termsofitscomponents. We aay askhow much heat flows through asmall surface atanyangle with respect tothe flow. InFig.2-4weshow asmall surface Aainclined with respect t0Aa, which Fig,2-4. Theheat flow through Aas isperpendicular totheflow. The unit vector misnormal tothesurface Aap. The isthe same asthrough day. 2 angle @between mand Aisthesame astheangle between thesurfaces (since hisnor- mal toAa,). Now what istheheat flow per unit area through Aa? The flow through ap isthesame asthrough Aaj; only theareas aredifferent. Infact, ay =Aa26c0s 8.The heat flow through dais A cos=hem 2.10)a3~Ba; Weinterpret thisequation: theheat flow (per unit time andperunit area) through any surface element whose unit normal.is #,isgiven byA’#.Equally, wecould Say: thecomponent oftheheat flow perpendicular tothesurface element da is hen,Wecan,ifwewish,considerthatthesestatements definehk.Wewillbeapply- ingthesame ideas toother vector fields. 2-3 Derivatives offields—the gradient ‘When fields vary intime, wecandescribe thevariation bygiving their deriva- ‘ives with respect to1.Wewant todescribe thevariations with position inasimilar way, because weareinterested intherelationship between, say, thetemperature in ‘oneplace andthetemperature atanearby place. How shall wetake thederivative ofthetemperature with respect toposition? Dowedifferentiate thetemperature with respect tox?Orwith respect toy,orz? ‘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 6T/ax? Isitascalar, oravector, orwhat? Itisneither ascalar nor avector, as you caneasily appreciate, because ifwetook adifferent x-axis, aT/ax would cer- tainly bedifferent. Butnotice: Wehave three possible derivatives: 47/dx, 07/ay, and 7/dz. Since there arethree kinds ofderivatives and weknow that ittakes three numbers toform avector, perhaps these three derivatives arethecomponents. ofavector: 2.2) avector. Qi) Ofcourse itisnotgenerally true that anythree numbers form avector. Itis true only if,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. The derivatives dotransform inthecorrect way when thecoordinate system isrotated. Wecan seths inseveral ways. One way istoask aquestion whose answer isindependent ofthecoordinate system,andtrytoexpresstheanswerinan“in- variant” form. For instance, ifS=A-B,andifAandBarevectors,weknow— ‘because weproved itinChapter 11ofVol. I—that Sis ascalar. We know that $ isascalar without investigating whether itchanges with changes incoordinatesystems. Itcan’s,becauseit’sadotproductoftwovectors,Similarly,ifweknow that Aisavector, andwehave three numbers B;,By,and Bg,andwefind outthat AB, +A,B2 +A.B =S, (2.12) where Sis thesame forany coordinate system, then itmust bethat thethree numbers B;,Bp,Byarethecomponents Bz,B,,B,ofsome vector B. ‘Now let’s think ofthetemperature field. Suppose wetake two points P;and a, separated bythesmall interval AR. The temperature atP,isTand atPais T+,and thedifference AT=Tz—T,. The temperatures atthese real, physical points certainly donotdepend onwhat axis wechoose formeasuring thecoordi- nates.Inparticular, ATisanumber independent ofthecoordinate system. Itisa‘scalar. 24 Ifwechoosesomeconvenientsetofaxes,wecouldwriteT;=T(x,y,2)and 7 Tz=T(x +Ax,y +Ay,z+Az),whereAx,ay,andAzarethecomponents of thevector AR(Fig.2-5). Remembering Eq,(2.7), wecanwrite . eonenaca Ty Tyg T Ninf Cogan ter ar Fax Tay+Fas, em SSHeae ‘TheleftsideofEq.(2.13) isascalar. Therightsideisthesumofthreeproducts , Bonney with4x,Ay,andAz,which arethecomponents ofavector. Itfollows thatthe .threenumbers wweae anat A eee ax"ay"82 arealsothex-,y»,andz-components ofavector. Wewritethisnewvector with _Fig.2-5. Thevector AR,whore com-thesymbolV7.Thesymbol¥(called‘“del”)isanupside-down 4,andissupposed _Ponen'soreAx,Ay,andAz.toremind usofdifferentiation, People read V7"invarious ways: “del-T,” or “gradient of7,”or“grad T;” ararat)* eur=or=(20,2). (@.14) Using thisnotation, wecanrewrite Eq.(2.13) inthemore compact form aT=VT-aR. 15) Inwords, thisequation says that thedifference intemperature between twonearby points isthedotproduct ofthegradient ofTandthevector displacement between thepoints. The form ofEq. (2.15) also illustrates clearly our proof above that ‘VT isindeed avector. Perhaps you arestill not convinced? Let’s prove itinadifferent way. (Al though ifyou look carefully, youmay beable toseethat it'sreally thesame proof inalonger-winded form!) Weshall show that thecomponents ofVTtransform in justthesame way that components ofRdo. Ifthey do,VT'is avector according to ‘ouroriginal definition ofavector inChapter 11ofVol. I.Wetake anew coordi- nate system x,y,7,and inthis new system wecalculate a7/ax’, aT/ay’, andy’ @T/az’. Tomake things alittle simpler, weletz=2’,sothat wecanforget about thez-coordinate. (You cancheck outthemore general caseforyourself.) 1 Wetakeanx’y’-system rotated anangle @with respect tothexp-system, as pect inFig.2-6(a). Forapoint (x,y)thecoordinates intheprime system are * vr x=xcos6+ysin8, 2.16) , y= =xsin@ +yc0s6. @7) * Or,solving forxandy, wh owx=x'cos#~sind, 2.18) oeywy=x'sin8+y/c0s8, 2.19) want, Ifany pair ofnumbers transforms with these equations inthesame way that x andydo,they arethecomponents ofavector. Now let'slook atthedifference intemperature between thetwonearby . points P;and Pa,chosen asinFig. 2-6(b). Ifwecalculate with thex-and »> coordinates, wewould write Fig.2-6, (a)Transformation to@ar rotatedcoordinate system.(b)Special aTjyox 2.20) coseofaninterval ARparallel tothe since Ayiszero. xeanis, *Inour notation, theexpression (a,b,)represents avector with components a,b, andc.Ifyoulket0usetheunitvectors,j,andk,youmaywrite ar, aT, yar wrettt hae 2s Ifwechoosesomeconvenientsetofaxes,wecouldwriteT;=T(x,y,2)and 7 Ty=T(x +Ox,y+Ay,z+A2),whereAx,ay,andAzarethecomponents of thevector AR(Fig.2-5). Remembering Eq,(2.7), wecanwrite . enna 7 or or Ni Torneyarn Fact Tay+Fas. 2.13)rl'as ‘TheleftsideofEq.(2.13) isascalar. Therightsideisthesumofthreeproducts i Wenn noe with 4x,Ay,and Az,which arethecomponents ofavector.Itfollowsthatthe : threenumbers wweae TT 4 me ax’ay"a2 arealsothex-,y-,andz-components ofavector. Wewritethisnewvector with __Fig.2-5. Thevector AR,whore com-thesymbolV7.Thesymbol¥(called‘“del”)isanupside-down 4,andissupposed Ponen'soreAx,Ay,andAz,toremind usofdifferentiation, People read VTinvarious ways: “del-T,” or “gradient of7,”or“grad T;” ‘aratat)* gadT=vr=@za). 2.14) Using thisnotation, wecanrewrite Eq.(2.13) inthemore compact form aT=VT- AR. 15) Inwords, thisequation says that thedifference intemperature between twonearby points isthedotproduct ofthegradient ofTandthevector displacement between thepoints, The form ofEq. (2.15) also illustrates clearly our proof above that, ‘VT isindeed avector. Perhaps you arestill not convinced? Let’s prove itinadifferent way. (Al- though ifyou look carefully, youmay beable toseethat it'sreally thesame proof inalonger-winded form!) Weshall show that thecomponents ofVTtransform injustthesamewaythatcomponents ofRdo.Iftheydo,VT'isavectoraccording to‘ouroriginal definition ofavector inChapter 11ofVol. I.Wetake anew coordi- natesystem x,y’,7,andinthisnewsystem wecalculate a7/ax’, aT/ay’, andy’ to @T/az’. Tomake things alittle simpler, weletz=2’,sothat wecanforget about thez-coordinate. (You cancheck outthemore general caseforyourself.) 1 Wetakeanx’y’-system rotated anangle @with respect tothexy-system, as pecan U inFig.2-6(a). Forapoint (x,y)thecoordinates intheprime system are * yr x=xcos6+ysin8, 2.16) , y= =xsin8+ycos6, 7 ° Or,solving forxandy, wh w x= x'cos# ~¥sing, 18) wthey=x'sin+y/cos8 2.19) Coes Ifany pair ofnumbers transforms with these equations inthesame way that x andydo,they arethecomponents ofavector. Now let'slook atthedifference intemperature between thetwonearby . points P;and Pa,chosen asinFig. 2-6(b). Ifwecalculate with thex-and »> coordinates, wewould write Fig.2-6. (a)Transformation to@ar rotatedcoordinate system.(b)Special ar=gyae 2.20) coseofaninterval ARporaliel tothe since Ayiszero. xeanis, *Inour notation, theexpression (a,b,)represents avector with components a,b, andc.Ifyoulket0usetheunitvectors,j,andk,youmaywrite ar, aT, yarwet tthae zs ‘What would acomputation intheprime system give? Wewould have written ar ar at=wea+ayay’, (2.21) Looking atFig. 2-6(b), weseethat Ax’ =Axcos 6 (2.22) and ay =-Axsing, 223) since Ayisnegative when Axispositive. Substituting these inEq.(2.21), wefind that ar T ar=Taxcose~2arsine 2.2 ar ar, =(Zeose ~2sind)ax (228) ‘Comparing Eq.(2.25) with (2.20), weseethat oT_aT aTFmTFcosa—Fsine 2.26) This equation says that 37/ax isobtained from 47/@x’ and T/ay’, just asxis ‘obtained from x’andy’inEq.(2.18). So47/ax isthex-component ofavector.Thesamekindofarguments wouldshowthat87/ayandaT/azarey-andz-com- ponents. SoPTis definitely avector. Itisa vector field derived from thescalar field 7. 244 The operator ¥ Now wecan dosomething that isextremely amusing and ingenious—and characteristic ofthethings that make mathematics beautiful. The argument that grad T,orV7, isa vector didnotdepend upon whut scalar feld wewere differ-entiating. Allthearguments wouldgothesameifT’werereplaced byanyscalar field. Since thetransformation equations arethesame nomatter what wediffer- tentiate, wecould just aswell omit the7and replace Eq.(2.26) bytheoperator equation Femgeen Bising, @27 Weleave theoperators, asJeans said, “hungry forsomething todifferentiate.” Since thedifferential operators themselves transform asthecomponents of& vector should, wecancallthem components ofavector operator. Wecanwrite _(8,8,8v=(¢-3-2): (2.28) which means, ofcourse, a a a weg wed wad 229) Wehave abstracted thegradient away from theT—that isthewonderful idea. You must always remember, ofcourse, that ¥isanoperator. 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 aTRe (2.30) which isnot «anumber, but isstill some kind ofoperator. However, according to thealgebra ofvectors wewould stillellTWavector. 26 ‘Nowlet'smultiplyVbyascalarontheotherside,sothatwehavetheproduct,(87). Inordinary algebra TA =AT, @31) butwehave toremember that operator algebra isalitle different from ordinary vector algebra. With operators wemust always keep thesequence right, sothattheoperations makepropersense.Youwillhavenodifficultyifyoujustrememberthat theoperator ¥obeys thesame convention asthederivative notation. What is tobedifferentiated must beplaced ontheright oftheV.The order isimportant. Keeping inmind thisproblem oforder, weunderstand that TVisanoperator, buttheproduct V7"isnolonger ahungry operator; theoperator iscompletely satisfied. Itisindeedaphysicalvectorhavingameaning.Itrepresentsthespatial rate ofchange ofT.The x-component ofWTishow fast Tchanges inthex-direc-tion.WhatisthedirectionofthevectorV7?Weknowthattherateofchangeof Tin any direction isthe component ofVT inthat direction (see Eq. 2.15). It follows that thedirection ofV7isthat inwhich ithas thelargest possible com- ponent—in other words, thedirection inwhich Tchanges thefastest. The gradientofThasthedirectionofthesteepestuphillslope(inT). 2-5 Operations with V Canwedoanyotheralgebrawiththevectoroperator€?Letustrycombining itwith avector. Wecancombine twovectors bymakingadotproduct.Wecould make theproducts (vector): ¥, or V+(avector) The first onedoesn’t mean anything yet, because itisstill anoperator. What it might ultimately mean would depend onwhat itismade tooperate on. The second product issome scalar field. (4Bis always ascalar.) Let's trythedotproduct of¥withavectorfieldweknow,sayA.Wewrite outthecomponents: Vek =Vahe +Vohy +Yoho (2.32) or vik Se ee (2.33) The sum isinvariant under acoordinate transformation. Ifwewere tochoose a different system (indicated byprimes), wewould have* the, dy, dhe Vehe+oe+Se (2.34) which isthesame number aswould begotten from Eq.(2.33), even though itlooksdifferent.Thatis,Voheveh 235) forevery point inspace. SoWh isascalar field, which must represent some physical quantity. You should realize that thecombination ofderivatives in V-hisrather special. There areallsorts ofother combinations likeh,/@x, which areneither scalars norcomponents ofvectors. Thescalar quantity V«(avector) isextremely useful inphysics. Ithasbeen given thename thedivergence. Forexample, Voh =divh =“divergence of.” 2.36) ‘AswedidforU7,wecanascribe aphysical significance toV«A.Weshall, how- ‘ever, postpone that until later. *We think ofAasaphysical quantity that depends onposition inspace, and not strictly asamathematical function ofthree variables. When his“differentiated” with respect tox,y, and =,orwith respect 10x,)’,and 2",themathematical expression for h ‘must frst beexpressed as«function oftheappropriate variables, 2 First, wewish toseewhat elsewecancook upwith thevector operator ¥. What about across product? Wemust expect that VX h=avector, 237) Itisvector whose components wecanwrite bytheusual ruleforcross products (GeeEq.2.2): ah, oh(8XBe=Vihy~Opty=GeSe (2.38) Similarly,ah oh (8 XWe=Vole—Vahey=Gy—Gy (2.39) and he_hy (8XBy=Vole—Vohy=She—Se (2.40) The combination ¥Xhis called “the curlof&."Thereasonforthename and thephysical meaning ofthecombination will bediscussed later. ‘Summarizing, wehave three kinds ofcombinations with V: WT =gradT=avector, Veh =divh =ascalar, VX =curlh =avector. Using these combinations, wecanwrite about thespatial variations offields ina ‘convenient way—in awaythat isgeneral, inthat itdoesn’t depend onanyparticularsetofaxes. ‘Asanexampleoftheuseofourvectordifferentialoperator¥,wewriteaset ZO‘ofvector equations which contain thesame laws ofelectromagnetism that wegave ¢ inwordsinChapter 1.TheyarecalledMaxwell’s equations, heWYY Maxwell'sEquationsY a)vE=2_oe 4 @) veB=0‘areaA j WA @evxea Fed bo° (0) wherep(cho).the“electricchargedensity,”istheamountofchargeperunitvolume, andj,the“electric current density,” istherate atwhich charge flows through aunit area persecond. These four equations contain thecomplete classical theory oftheelectromagnetic field. You seewhat anelegantly simple formwecangetwithournewnotation! ee h 2-6Thedifferentialequationofheatflow Aree8 Letusgiveanother example ofalawofphysicswntteninvectornotation. The law 1snotaprecise one, butformany metals and anumber ofother sub- stancesthatconductheatitisquiteaccurate.Youknowthatifyoutakeaslabof vwormeratmaterial and heat one face totemperature Tyand cool theother toadifferent temperature T,,theheat will flow through thematerial from TtoT;(Fig. 2~T(@)]- Theheatflowisproportional totheareaAofthefaces,andtothetemperaturetear Lr difference. Itisalsoinversely proportional tod,thedistance between theplates." " (For agiven temperature difference, thethinner theslabthegreater theheatflow.) ry LettingJbethethermalenergythatpassesperuntttimethroughtheslab,wewrite 4 Fig.2-7.(a)Heatflowthrough tale TG a2) slob. (b)Aninfritesimel slab parallel to fonisothermal surface in¢large block. Theconstant ofproportionality x(kappa) iscalled thethermal conductivity 28 ‘What willhappen inamore complicated case? Sayinanodd-shaped block of material inwhich thetemperature varies inpeculiar ways? Suppose welook ata tinypiece oftheblock andimagine aslab likethat ofFig.2-7(a) onaminiature scale, Weorient thefaces parallel totheisothermal surfaces, asinFig. 2-7(b), s0 that Eq.(2.42) iscorrect forthesmall slab. Ifthearea ofthesmall slab is4A, theheat flow perunit time is oa AJ=KatSB (2.43) where Asisthethickness oftheslab. Now 4//AA wehave defined earlier asthe magnitude of&,whose direction istheheat flow. The heat flow will befromT,+ATtowardT;,andsoitwillbeperpendicular totheisotherms, asdrawnin Fig. 2-7(b). Also, AT/As isjust therate ofchange ofTwith position. And since theposition change isperpendicular totheisotherms, ourAT/as isthemaximum rate ofchange. Itis,therefore, just themagnitude ofVT. Now since thedirection ofVTisopposite tothat ofA,wecan write (2.43) asavector equation: A= —KVT. (2.44) (The minus sign isnecessary because heat flows “downhill” intemperature.) Equation (2.44) isthedifferential equation ofheat conduction inbulk materials. ‘You seethat itisaproper vector equation. Each side isavector if«isjust anum- 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 notonly because itmakes theequations /ook simpler. Italso shows most clearly thephysical content oftheequations without reference toany arbitrarily chosen coordinate system. 2-7Second derivatives ofvector fields ‘Sofarwehave had only first derivatives. Why not second derivatives? We could have several combinations: (a) ¥-(¥T) @ +x) (©) ¥(v- A) (2.45) @ Vv xa) © Vx (vx hy You cancheck that these areallthepossible combinations. Let’s look first atthesecond one, (b). Ithas thesame form as AX(AT)=(AXAT=0, ‘sinceAXAisalways zero,Soweshould have curl (grad T)=¥X(¥T) =0. (2.46) ‘Wecanseehow thisequation comes about ifwegothrough once with thecom- oon: [¥X(WP. =VAT), —VAY: | “aaa-3@)-3@): ea Which iszero (byEq.2.8). Itgoes thesame fortheother components. So¥X (WT) =0,foranytemperature distribution—in fact, foranyscalar function. » Now letustake another example. Letusseewhether wecanfind another zero. The dot product ofavector with across product which contains that vector iszero: A(AX B)=0. (2.48) because AXBisperpendicular toA,andsohasnocomponents inthedirection 4. ‘The same combination appears in(4)of(2.45), sowehave V-(¥ X&)=div(curlA)=0. (2.49) Again, itiseasy toshow that itiszero bycarrying through theoperations with components. Now wearegoing tostate twomathematical theorems that wewillnotprove. ‘They arevery interestingandusefultheoremsforphysiciststoknow. Inaphysical problem wefrequently find that thecurl ofsome quantity—say ofthevector field 4—is zero. Now wehave seen (Eq. 2.46) that thecurl ofa Bradient iszero, which iseasy toremember because oftheway thevectors work Itcould certainly be.then. that Aisthegradient ofsome quantity. because then itscurl would necessarily bezero. The interesting theorem isthat ifthecurl 4is zero, then Aisalways thegradient ofsomething—there issome scalar field ¥(psi)suchthatAisequaltogrady.Inotherwords,wehavethe Tus0Rem: It vxa=0 there isa ¥ such that A=vy. (2.50) There isasimilar theorem ifthedivergence ofAiszero. Wehave seen in Eq,(2.49) that thedivergence ofacurl ofsomething isalways zero. Ifyoucome‘acrossavectorfieldDforwhichdivDiszero,thenyoucanconcludethatDisthe curl ofsome vector field C. ‘THEOREM: it vD=0 there isa c such that D= 0XC. asi Inlooking atthepossible combinations oftwo operators. wehave found that two ofthem always give zero. Now welook attheones that arenotzero. Take thecombination ¥«(7), which was frst onourlist. Itisnot, ingeneral, zero, Wewrite outthecomponents: WT =VT + +VT. Then VST) =VAGT) +VUGT) +VAT) er ar, er -etet ae 2.52) which would, ingeneral, come outtobesome number. Itisascalar field. You seethatwedonotneed tokeep theparentheses, butcanwrite, without any chance ofconfusion, WCW) =VT =(VOT =OF. 253) We lookatV?asanewoperator.Itisascalaroperator.Becauseitappearsoften imphysics, ithasbeen given aspecial name—the Laplacian. , 2 @ 8, aLaplacian=0?=+Se 2.54) 210 Since theLaplacian isascalar operator, wemay operate with itonavector— bywhich wemean thesame operation oneach component inrectangular coor- VA=(Vhz, V7hy, V7h,). Let'slookatonemorepossibility: ¥VX(¥XA),whichwas(e)inthelist(2.45).Nowthecurlofthecurlcanbewrittendifferently ifweusethevector equality (2.6): AX(BXC)=B(A-C) —C(A-B). (2.55) Inorder tousethis formula, weshould replace Aand Bbytheoperator Vand putC=A.Ifwedothat,weget YK (WX A= VOT)—MEY). 22 Wait aminute! Something iswrong. The first two terms arevectors allright (the operators aresatisfied), butthelastterm doesn’t come outtoanything. It’s stillanoperator. The trouble isthat wehaven't been careful enough about keeping. theorder ofourterms straight. Ifyou look again atEq.(2.55), however, you see that wecould equally well have written itas AX(BXC)=BAC) —(4°BIC. (2.56) ‘Theorderoftermslooksbetter. Nowlet'smakeoursubstitution in(2.56). Weget VX (VX A)=(VA) —(VO (2.57) This form looks allright. Itis,infact, correct, asyoucanverify bycomputing the ‘components. The last term istheLaplacian, sowecan equally well write Vv (VX A)=VIVA) —VR. (2.58) ‘Wehavehadsomething tosayaboutallofthecombinations inourlistof double V's,except for(c),¥(¥ «A).Itisapossible vector field, butthere isnothing special tosayabout it.It'sjust some vector feld which may occasionally come up. Itwill beconvenient tohave atable ofour conclusions: (@) (WT) =V°F=ascalar field ) Vx (wT) =0 (© V(V-#) =avector field @vwxh=0 O59) © VX WX A= vVeh) —v (©) (V+¥)k =V%h =avector field You may notice that wehaven't tried toinvent anew vector operator (VX¥). Doyouseewhy? 2-8 Pitfalls Wehave been applying ourknowledge ofordinary vector algebra tothealge-braoftheoperatorV.Wehavetobecareful,though,becauseitispossibletogoastray. There aretwopitfalls which wewillmention, although they willnotcome upinthiscourse. What would you sayabout thefollowing expression, that in- volves thetwo scalar functions ¥and¢(phi): (ev) x(6)? ‘You might want tosay: itmust bezero because it’sjust like (Aa) X(Ab), at which iszero because thecross product oftwoequal vectors AX. Aisalways zero. But inour example thetwo operators ¥arenotequal! The first one operates on onefunction, ¥;theother operates onadifferent function, g.Soalthough werep- resent them bythesame symbol ¥.they must beconsidered asdifferent operators. Clearly, thedirection ofTydepends onthefunction y,soitisnotlikely tobe parallel toV9. (WH) X(8) 0(generally). Fortunately, wewon't have tousesuch expressions. (What wehave said doesn’t change the fact that VX Vy =0forany scalar field, because here both ¥’s operate onthesame function.) Pitfall number two (which, again, weneed not getinto inour course) isthe following: The rules that wehave outlined here aresimple and nice when weuse rectangular coordinates. Forexample, ifwehave 2h andwewant thex-com- ponent, itis 2 ee a 2(Wh),=(&tat%)he=Why (2.60) The same expression would norwork ifwewere toask fortheradia? component ofUA, ‘Theradial component ofVis notequal toV'h,. The reason isthat when wearedealing with thealgebra ofvectors, thedirections ofthevectors are allquite definite, But when wearedealing with vector fields, their directions are different atdifferent places. Ifwetrytodescribe avector field in,say, polar coordi- nates, what wecallthe“radial” direction varies from point topoint. Sowecan getinto alotoftrouble when westart todifferentiate thecomponents. For ex- ample, even foraconstant vector field, theradial component changes from point topoint. Ttisusually safest and simplest just tostick torectangular coordinates and avoid trouble. butthere isone exception worth mentioning: Since theLaplacian V2.isascalar, wecanwrite itinanycoordinate system wewant to(forexample, inpolar coordinates). Butsince itis adifferential operator, weshould useitonly fonvectors whose components areinafixed direction—that means rectangular coordinates. Soweshallexpressallofourvectorfieldsintermsoftheirx-,»- andz-components when wewrite ourvector differential equations out incom- ponents. 22 3 Vector Integral Calculus 341Vectorintegrals; thelineintegralofVW" ‘Wefound inChapter 2that there were various ways oftaking derivatives of 3-1 Vector integrals; theline fields. Some gave vector fields; some gave scalar fields. Although wedeveloped integral ofV9" ‘many different formulas, everything inChapter 2could besummarized inonerule: theoperators 9/ax, 2/3), and /Az arethethree components ofavectoroperator >>Thefaxof«vectorfield ¥.Wewould now liketogetsome understanding ofthesignificance ofthederiva- 3-3 The flux from acube; Gauss? tives offields. Wewillthen have abetter feeling forwhat avector field equation theorem means.Wehavealreadydiscussedthemeaningofthegradientoperation(¥ona4Heatconduction; thediftusion scalar). Nowweturntothemeanings ofthedivergence andcurloperations. equat The interpretation ofthese quantities isbest done interms ofcertain vector 3- The circulationofavectorfield integrals and equations relating such integrals. These equations cannot, unfor- .tunately,beobtainedfromvectoralgebrabysomeeasysubstitution, oyouwill>6Theciscalation aroundasquaresjust have tolearn them assomething new. Ofthese integral formulas, one is practically trivial, buttheother twoarenot. Wewillderive them andexplain their 3-7 Curl-free anddivergence-free implications. The equations weshall study arereally mathematical theorems. fields Theywillbeuseful notonlyforinterpreting themeaning andthecontent ofthe gGan,divergence andthecurl,butalsoinworking outgeneral physical theories. These mary mathematical theorems are, forthetheory offields, what thetheorem ofthecon- servation ofenergy istothemechanics ofparticles. General theorems like these areimportant foradeeper understanding ofphysics. You willfind, though, that they arenotvery useful forsolving problems—except inthesimplest cases. Itis <elightful, however, that inthebeginning ofoursubject there willbemany simple problems which can besolved with the three integral formulas wearegoing to treat. Wewillsee,however, astheproblems getharder, thatwecannolonger use wy these simple methods. ce Wetakeupfirstanintegral formula involving thegradient. Therelation Comer contains avery simple idea: Since thegradient represents therate ofchange ofa field quantity, ifweintegrate that rate ofchange, weshould getthetotal change. Suppose wehave thescalar field ¥(x,y,z).Atanytwopoints(1)and(2),the as function¥willhavethevalues¥(1)and¥(2),respectively. [Weuseaconvenient 4, notation, inwhich (2)represents thepoint (x2, Y2,22)and ¥(2) means thesame thingasYc,ya,22).]I€T(gamma)isanycurvejoining(1)and(2),ainFig.3-1,fig,9-1,Thetermsusedina,(3.1) thefollowingrelationistrue: ThevectorVVisevaluatedoftheline ew1. clement ds.2)—vl)=f.(vy)«ds. Gl) ate ‘Theintegralisalineintegral,from(1)to(2)alongthecurveP,ofthedotproduct whsown w of¥y—a vector—with ds—another vector which isaninfinitesimal line element > ofthecurve I”(directed away from (1)andtoward (2)). FE cave P First, weshould review what wemean byalineintegral. Consider ascalarfunctionftx,y,2),andthecurveTjoiningtwopoints(1)and(2).Wemarkoff as,HOthecurveatanumberofpointsandjointhesepointsbystraight-line segments, asas,shown inFig.3-2. Each segment hasthelength As,,where iisanindex that runs ("4b1,2,3,.... Bythelineintegral As <2ffas Fig.3-2.Thelineintegralisthede limitofasum. a4 wemean thelimit ofthe sum Chasis wheref,isthevalueofthe function attheithsegment. The limiting value iswhatthesumapproaches asweaddmoreandmoresegments(inasensibleway,sothatthelargest as;—+0). The integral inour theorem, Eq. (3.1), means thesame thing, although it looks alittle different. Instead off,wehave another scalar—the component of ‘VyinthedirectionofAs,Ifwewrite(Vy).forthistangential component, itisclear that(PY).ds=(Y)as. 62) ‘The integral inEq.(3.1) means thesum ofsuch terms. Now let's seewhy Eq.(3.1) istrue. InChapter 1,weshowed that thecom- ponent ofVyalong 2small displacement ARwas therate ofchange ofyinthe direction ofAR. Consider theline segment Asfrom (1)topoint ainFig. 3-2. ‘According toourdefinition, a1 =¥(a) ~YD) =(WY “Ass. @3) Also, wehave (0) —¥(@) =(Ja As, G4) where, ofcourse, (Vy); means thegradient evaluated atthesegment As,, and (W¥)o, thegradient evaluated atAsa. IfweaddEqs.(3.3)and(3.4),weget Wb) —WL) =(WW)r Asa +(PH)2* Asa. G.5) You canseethat ifwekeep adding such terms, wegettheresult ¥2) —KD =LPH) As G66) ‘The left-hand side doesn’t depend onhow wechoose ourintervals—if(1)and(2) arekeptalwaysthesame—sowecantakethelimitofthe right-hand side. Wehave therefore proved Eq. (3.1). You canseefrom ourproof that just astheequality doesn’t depend onhow thepoints a,b,c, .....arechosen, similarly itdoesn’t depend onwhat wechooseforthecurveTtojoin(1)and(2).Ourtheoremiscorrectforanycurvefrom(1)102). ‘One remark onnotation: You will see that there isnoconfusion ifwewrite, forconvenience, (Wy) ds=vy ds, re) With this notation, our theorem is ‘THEOREM 1. @¥@)—vl)=f.Weds. G8) choses Surtace8 es 3-2Thefluxofavectorfield Vonune ¥ Before weconsider ournext integral theorem—a theorem about thedivergence —we would like tostudy acertain idea which hasaneasily understood physical significance inthecase ofheat flow. Wehave defined thevector h,which representstheheatthatflowsthroughaunitareainaunittime.Supposethatinsideablockofmaterial wehave some closed surface Swhich encloses thevolume V(Fig. 3-3). ‘Wewould liketofind outhow much heat islowing outofthis volume, Wecan, ofcourse,finditbycalculatingthetotalheatflowoutofthesurfaceS. defiSovolevontee ;Wewritedafortheareaofanelementofthesurface.Thesymbolstandsforshe cuwcrdfarcale terface *wostmensonal diferente, 1fforjnstane, theeenhappened tobeinthe‘loment do,andhistheheat-flow vector *”"P atthesurface element. da=dxdy. 32 Later weshall have integrals over volume and forthese itisconvenient tocon- sider adifferential volume that isalittle cube. Sowhen wewrite dV wemean dV=dxdydz. Some people liketowrite dainstead ofdatoremind themselves that itis kind ofasecond-order quantity. They would also write d*V instead ofdV. We willusethesimpler notation, andassume that youcanremember that anarea has two dimensions and avolume has three. ‘The heat flow outthrough thesurface element daisthearea times thecom- ponent ofhperpendicular toda, Wehave already definednasaunitvectorpointing ‘outward atright angles tothesurface (Fig. 3-3). The component ofhthat we want is fg=hom, 69) ‘Theheatflowoutthroughdaisthen honda, G.10) Togetthetotal heat flow through anysurface wesum thecontributions from all theelements ofthesurface. Inother words, weintegrate (3.10) over thewhole surface: Totalheatflowoutward through S=iAenda, Gu) Wearealsogoingtocallthissurfaceintegral“thefluxoffthroughthesur-face.”Originally thewordfluxmeantflow,sothatthesurfaceintegraljustmeansthe flow of&throughthesurface.Wemaythink:&isthe“currentdensity”of heat flow and thesurface integral ofitisthetotal heat current directed outofthe surface; that is,thethermal energy perunit time (joules persecond). ‘We would like togeneralize this idea tothecase where thevector does not represent theflow ofanything; forinstance, itmight betheelectric field. Wecan certainly stillintegrate thenormal component oftheelectric field over anarea ifwewish.Although itisnottheflowofanything, westilcallitthe“flux.”Wesay FluxofEthroughthesurfaceS=[Bomda. G.12) Wegeneralize theword “flux” tomean the“surface integral ofthenormal com- ponent” ofavector. Wewill also usethesame definition even when thesurface considered isnotaclosed one, asitis here. Returning tothe special case ofheat flow, letustake asituation inwhich heat isconserved. For example, imagine some material inwhich after aninitial heating nofurther heat energy isgenerated orabsorbed. Then, ifthere isanet heat flow out ofaclosed surface, the heat content ofthe volume inside must decrease. So,incircumstances inwhich heat would beconserved, wesaythat __4@ fn nda= ~2, G13) where Qistheheat inside thesurface. The heat fluxoutofSisequal tominus therateofchangewithrespecttotimeofthetotalheatQinsideofS.Thisinterpreta-tionispossiblebecausewearespeakingofheat flow and also because wesupposed that theheat was conserved. Wecould not, ofcourse, speak ofthetotal heat inside thevolume ifheat were being generated there. ‘Nowweshallpointoutaninterestingfactaboutthefluxofanyvector.You maythinkoftheheatflowvectorifyouwish,butwhatwesaywillbetrieforanyvector field C.Imagine that wehave aclosed surface Sthatencloses thevolume V. Wenow separate thevolume into two parts bysome kind ofa“cut,” asinFig. 3-4. Now wehave two closed surfaces and volumes. The volume Vis enclosedinthesurfaceS,,whichismadeupofpartoftheoriginalsurfaceS,andofthesurface ofthecut, Sy. The volume V2isenclosed byS2,which ismade upof therestoftheoriginal surface S,andclosed offbythecutS,», Now consider the 33 y 8,“ys S :)Yy\_Z a 1 Fig. 3-4. AvolumeVcontainedinsidethesurface "e Sisdivided intotwopieces bya“cut” atthesurface | asSat.WenowhavethevolumeVsenclosedinthe Ue!BY en tasesvesVoonanes UiFO inthesurfaceS:=Sb+Sob. KY; om following question: Suppose wecalculate theflux outthrough surface S;and add toittheflux through surface Sj. Does thesum equal theflux through the whole surface that westarted with? The answer isyes. The flux through thepartofthesurfacesSqcommontobothS,andSjustexactlycancelsout.Fortheflux ofthevector CoutofV;,wecan write thLe . Comda, 3.14) FluxthroughS;S.¢dafida, G14) and fortheflux outofV2, luxthrough Sy= onda Comyda. (3.15) Note that inthesecond integral wehave written m,fortheoutward normal for Saywhen itbelongs toS;,and mywhen itbelongs toSo,asshown inFig. 3-4. Clearly, m)=—my,sothat J, =Su,€1 G.16) IfwenowaddEqs.(3.14)and(3.15),weseethatthesumofthefluxesthrough'S,and Soisjust thesum oftwo integrals which, taken together, give theflux through theoriginal surface S=S,+Si. (swag, 8) “ Weseethat thefiux through thecomplete outer surface Scan beconsidered s ‘asthesum ofthefluxes from thetwo pieces into which thevolume was broken. « Wecansimilarly subdivide again—say bycutting V;intotwopieces. Yousee 5 thatthesame arguments apply. Soforanywayofdividing theoriginal volume, itA te. mustbegenerally truethatthefluxthroughtheoutersurface,whichistheoriginal aesaed integral,isequaltoasumofthefluxesoutofallthelittleinteriorpieces. Bttaeng(meena) a id 3.3Thefluxfrom acube; Gauss’ theorem ben 7 Wenowtakethespecial caseofasmallcubeandfindaninterestingformula fortheflux outofit.Consider acube whose edges arelined upwith theaxes asin Fig. 3-5. Computation ofthefluxof Fig. 3-5. Letussuppose that thecoordinates ofthecorner nearest theorigin Coutof«small cube. arex,y, 2.LetAxbethelength ofthecube inthex-direction, Aybethe length inthey-direction, and Azbethelength inthez-direction, Wewish tofind the flux ofavectorfieldCthroughthesurfaceofthecube.Weshalldothisbymaking asum ofthefluxes through each ofthesixfaces. First, consider theface marked in thefigure. The flux outward onthis face isthenegative ofthex-component ofC,integrated over thearea oftheface. This flux is -JCodydz. Since weareconsidering asmail cube, wecan approximate this integral bythe *The following development applies equally well toany rectangular parallelepiped. Fay value ofC,atthecenter oftheface—which wecallthepoint (1)—multiplied by thearea oftheface, AyAz: Flux outof|=—C,(1)ayaz. Similarly, fortheflux outofface 2,wewrite Flux out of2=C,(2)ayaz. Now C,(1) and C,(2) are, ingeneral, slightly different. IfAxissmall enough, we canwrite aC, C.Q) =Col) +Fax. ‘There are,ofcourse, more terms, butthey willinvolve (4,)* andhigher powers, and sowill benegligible ifweconsider only thelimit ofsmall Ax, Sotheflux through face 2is Flux out of2=[ea+%a]Ayaz. Summing thefoxes forfaces |and 2,weget Fluxoutof|and2=%AxayAz. The derivative should really beevaluated atthecenter offace 1;that is,at [xy+(Ay/2),z+(42/2)}-Butinthelimitofaninfinitesimal cube,wemake 4negligible error ifweevaluate itatthecorner (x,y, 2) Applying thesame reasoning toeach oftheother pairs offaces, wehave FluxoutofSand4=$+axayaz andac. FluxoutofSand6=92axayaz. Thetotal fluxthrough allthefaces isthesum ofthese terms. Wefindthat, =(a,Wy4aC [Conda=(G+ot962)axaya, andthesumofthederivatives isjustV-C.Also,AxAyAz=AV,thevolumeof thecube. Sowecansaythatforaninfinitesimal cube [Conda=(v-Cay. G17.) sasface We have shown that the outward flux from the surface ofaninfinitesimal cube is ‘equal tothedivergence ofthevector multiplied bythevolume ofthecube. We now seethe“meaning” ofthedivergence ofavector.Thedivergence ofavector atthepoint Pistheflux—the outgoing “flow” ofC—per unit volume, intheneigh-bothoodofP.‘Wehaveconnected thedivergence ofCothefluxofCoutofeachinfinitesimalvolume. For any finite volume wecanusethefact weproved above—that the total flux from avolume isthesum ofthefluxes outofeach part. Wecan, that is,integratethedivergence overtheentirevolume.Thisgivesusthetheoremthattheintegralofthe normal component ofanyvector over anyclosed surface canalso be written astheintegral ofthedivergence ofthevector over thevolume enclosed bythesurface. This theorem isnamed after Gauss. Gauss’TutoREM. [Onda=[w-cav, G.18) Is IY whereSisanyclosedsurfaceandVisthevolumeinsideit. 3s 3-4 Heat conduction; thediffusion equation Let's consider anexample oftheuseofthis theorem, just togetfamiliar with it.Suppose wetake again thecase ofheat flow in,say, ametal. Suppose we have asimple situation inwhich alltheheat hasbeen previously putinand the body isjust cooling off. There arenosources ofheat, sothat heat isconserved. ‘Then how much heat isthere inside some chosen volume atanytime? Itmust be decreasing byjust theamount that flows outofthesurface ofthevolume. Ifour volume isalittle cube, wewould write, following Eq.(3.17), Heatout=fa-mda=9av. G9)ee Butthismust equal therate oflossoftheheat inside thecube. Ifqistheheat per ‘unit volume, theheat inthecube isgAV, andtherate ofJossis ~dear) =-Hav. 6.20) Comparing (3.19) and (3.20), weseethat Hy.a7 A G.21) ‘Takecarefulnoteofthe form ofthisequation; theform appears often inphys- ics. Itexpresses aconservation law—here theconservation ofheat. We have expressed thesame physical factinanother wayinEq.(3.13). Here wehave thedifferential formofaconservation equation, whileEq.(3.13)istheintegral form.‘Wehave obtained Eq.(3.21) byapplying Eq.(3.13) toaninfinitesimal cube. Wecanalso gotheother way. For abigvolume Vbounded byS,Gauss’ law ‘saysthat f,bomda=[vokay. 3.22) Using (3.21), theintegral ontheright-hand side isfound tobejust —dQ/dt, and again wehave Eq. (3.13). Now let’s consider adifferent case. Imagine that wehave ablock ofmaterial and that inside itthere isavery tiny hole inwhich some chemical reaction is taking place and generating heat. Orwecould imagine that there aresome wiresrunningintoatinyresistorthatisbeingheatedbyanelectriccurrent. Weshall suppose that theheat isgenerated practically atapoint, and letWrepresent theenergyliberated persecondatthatpoint.Weshallsupposethatintherestofthe volume heat isconserved, and that theheat generation hasbeen going onfor long time—so that now thetemperature isnolonger changing anywhere. The .problemis:Whatdoestheheatvector&looklikeatvariousplacesinthemetal? wo276eosinneo©pontHowmuchheatNowisthereateachpoint? cewek ‘Weknowthatifweintegratethenormalcomponent offover aclosed surface that encloses thesource, wewillalways getW.Alltheheat that isbeing generated atthepoint source must flow outthrough thesurface, since wehave supposed that theflow issteady. We have thedifficult problem offinding avector field which, when integrated over anysurface, always gives W.Wecan, however, find thefield rather easily bytaking asomewhat special surface. Wetake asphere of radius R,centered atthesource, andassume that theheat flow isradial (Fig. 3-6). Our intuition tells usthat Ashould beradial iftheblock ofmaterial islarge and wedon’t gettooclose totheedges, and itshould also have thesame magnitude atallpoints onthesphere. You seethatweareadding acertain amount ofguess- work—usually called “physical intuition”—to ourmathematics inorder tofind the answer. When4isradialandspherically symmetric, theintegral ofthenormal com-ponent of&over thearea isvery simple, because thenormal component isjust 36 themagnitude of&andisconstant. The area over which weintegrate is4xR?. We have then that fybonda =heeaer? G23) (where histhemagnitude ofA).This integral should equal W,therate atwhich heat isproduced atthesource. Weget w hmoe or Ua a=ae G24) where, asusual, e,represents aunit vector intheradial direction. Our resultsaysthatAisproportional toWandvariesinverselyasthesquareofthedistancefrom the source, ‘The result wehave just obtained applies totheheat flow inthevicinity ofa point source ofheat. Let’s now trytofind theequations that hold inthemost general kind ofheat flow, keeping only thecondition that heat isconserved. ‘Wewill bedealing only with what happens atplaces outside ofany sources or absorbers ofheat. ‘The differential equation fortheconduction ofheat wasderived inChapter 2. According toEq.(2.44), hem ~«7. 3.25) (Remember that thisrelationship isanapproximate one, butfairly good forsome materials like metals.) Itisapplicable, ofcourse, only inregions ofthematerial where there isnogeneration orabsorption ofheat. Wederived above anotherrelation,Eq.(3.21),thatholdswhenheatisconserved. Ifwecombinethatequationwith (3.25), weget -4eoyh=—v-« WD, at2 or 4aeyr= x0, 6.26)atit ifxisaconstant. Youremember thatqistheamountofheatinaunitvolumeandV+ =?is theLaplacian operator 2 8 at eFWoantaftan Ifwenow make onemore assumption wecanobtain avery interesting equa- tion. We assume that thetemperature ofthematerial isproportional totheheat content perunit volume—that is,that thematerial hasadefinite specific heat. ‘When thisassumption isvalid (asitoften is),wecanwrite Aq=AT or uf.Gao, 7 8.27) The rate ofchange ofheat isproportional totherate ofchange oftemperature. ‘The constant orproportionality cyis,here, the specific heat per unit volume of thematerial, Using Eq.(3.27) with (3.26), weget a_kopFe kor, 6.28) Wefind that thetime rate ofchange ofT—at every point—is proportional tothe Laplacian ofT,which isthesecond derivative ofitsspatial dependence. We have 4differentialequation—in x,y,z,andforthetemperatureT. a7 The differential equation (3.28) iscalled theheat diffusion equation. Itis often written as qr 12;2~ovr, G29) where Discalled thediffusion constant, andishere equal tox/ex. ‘Thediffusion equation appears inmany physical problems—in thediffusion ofgases, inthediffusion ofneutrons, and inothers. Wehave already discussed thephysics ofsome ofthese phenomena inChapter 43ofVol. I.Now you have thecomplete equation thatdescribes diffusion inthemost general possible situa- tion, Atsome later time wewill take upways ofsolving thediffusion equation ‘tofind how thetemperature varies inparticular cases. Weturn back now to consider other theorems about vector fields. 3-5 The circulation ofavector field leo ¢ Wewishnowtolookatthecurlinsomewhat thesamewaywelookedattheyr; divergence. Weobtained Gauss’theorem byconsidering theintegral overa ge surface, although itwas notobvious atthebeginning that wewere going tobe a dealing with thedivergence. How didweknow that wewere supposed tointegrate over asurface inorder togetthedivergence? Itwasnotatallclear that thiswould betheresult.Andsowithanapparent equallackofjustification, weshall calculate something elseabout avector andshow that itisrelated tothecurl. This time we ?calculate what iscalled thecirculation ofavector field. IfCisany vector field, b wetake itscomponent along acurved lineandtake theintegral ofthiscomponent c alltheway around acomplete loop. Theintegral iscalled thecirculation ofthe vector field around theloop. Wehave already considered aline integral ofVy Fig.3-7. Thecirculation ofCaround earlier inthischapter. Now wedothesame kind ofthing foranyvector field C. thecurveI’isthelineintegral ofCi,the LetIbeanyclosedloopinspace—imaginary, ofcourse. Anexample isgiventangential component ofC. inFig. 3-7. Thelineintegral ofthetangential component ofCaround theloop iswritten as $Cds=§Crds. 8.30) You should note that theintegral istaken alltheway around, notfrom onepoint toanother aswedidbefore. The little circle ontheintegral sign istoremind us that theintegral istobetaken alltheway around. This integral iscalled thecirculation ofthevectorfieldaroundthecurveI.Thenamecameoriginally from considering thecirculation ofaliquid,Butthename—like flux—hasbeenextended toapplytoanyfieldevenwhenthereisnomaterial “circulating.” oyPlaying thesame kind ofgame wedidwith theflux, wecan show that the me i circulation around aloopisthesumofthecirculations around twopartial loops. ‘Suppose webreak upourcurve ofFig. 3-7into two loops, byjoining two points (1)and (2)ontheoriginal curve bysome line that cuts across asshown inFig. 3-8.Therearenowtwoloops,P';andI's,T';ismadeupofI,whichisthatpart oftheoriginal curve totheleftof(1)and(2),plus P49,the“short cut.” I’;ismade we upoftherestoftheoriginal curveplustheshortcut.The circulation around P;isthesum ofanintegral along ',and along Tas. -n. Similarly,thecirculation aroundT’3isthesumoftwoparts,onealongT,andthe wheleonisheceeatonroundteeotheralongFas.TheintegralalongTwillhave,forthecurveT's,theopposite round thetwoloops Ti=Ts+Ta, SignfromwhatithasforT',,because thedirection oftravel isopposite—we must ondT;=Te+Tob. take both ourlineintegrals with thesame “sense” ofrotation. Following thesame kind ofargument weused before, you canseethat the sum ofthetwocirculations willgivejust thelineintegral around theoriginal curve I.The parts duetoTscancel. The circulation around theonepart plus thecir- culation around the Second part equals thecirculation about theouter line, ‘Wecancontinue theprocessofcuttingtheoriginalloopintoanynumberofsmaller loops. When weadd thecirculations ofthesmaller loops, there isalways acan- cellation oftheparts ontheir adjacent portions, sothat thesum isequivalent tothe circulation around theoriginal single loop. a Nowletussupposethattheoriginalloopistheboundaryofsomesurface. KS Loop‘Thereare,ofcourse,aninfinitenumberofsurfaceswhichallhavetheoriginal Cott teloopsastheboundary. Ourresultswillnot,however, dependonwhichsurface bsewechoose. First,webreakouroriginalloopintoanumberofsmallloopsthatallAgLI(EHelsisisist fieonthesurfacewehavechosen,asinFig.3-9.NomatterwhattheshapeofCATELets srt thesurface,ifwechooseoursmallloopssmallenough, wecanassumethateach ITFPPfef8/3) ofthesmallloopswillencloseanareawhichisessentially lat.Also,wecanchoose eya ‘oursmall loops sothat each isvery nearly asquare. Now wecancalculate the circulation around thebigloop T’byfinding thecirculations around allofthe Fig,3-9. Some surface bounded by little squares andthen taking their sum. theloop Tis chosen. Thesurface is divided into anumber ofsmell oreos, cirenlation around Jach approximately @square.The How shall wefindthecirculation foreachlittlesquare? Onequestion is, _«itcvletions oround thelitleloops. how isthesquare oriented inspace? Wecould easily make thecalculation ifit hadaspecial orientation. Forexample, ifitwere inoneofthecoordinate planes. Since wehave notassumed anything asyetabout theorientation ofthecoordinate axes, wecanjust aswell choose theaxes sothat theone little square wearecon- centrating onatthemoment lisinthe2y-plane, asinFig. 3-10. Ifourresult is expressed invector notation, wecansaythat itwillbethesame nomatter What the particular orientation oftheplane. . Wewant now tofind thecirculation ofthefield Caround our little square. Gy bnay © Itwillbeeasytodothelineintegral ifwemake thesquare small enough thatthe a|s 7 vector Cdoesn’t change much along anyoneside ofthesquare. (The assumption isbetter thesmallerthesquare,s0wearereallytalkingaboutinfinitesimalsquares.) 2 Starting atthepoint (x,))—the lower leftcorner ofthefigure—we goaround in “, thedirection indicated bythearrows. Alongthefirstside—marked (I)—the iL Atangential component isC,(1)and thedistance isAx.Thefirstpartoftheintegral a %isC,(1)4x.Alongthesecondleg,wegetC,(2)4y. Alongthethird,weget #4} ~C,(@3) Ax, and along thefourth, —C,(4)Ay. ‘The minus signs arerequired ‘because wewant thetangential component inthedirection oftravel. The whole lineintegral isthen * fcds=-C,(1) Ax+C2)dy—C.3)4x—C4)ay.G31), Fig.3-10.Computing thecirculationofCaround«smallsquare. ‘Now let's look atthefirst and third pieces. Together they are [C.(1) —C,(3)]dx. (3.32) You might think that toourapproximation thedifference iszero. That istrue to thefirst approximation. Wecan bemore accurate, however, and take into account therateofchange ofC.. Ifwedo,wemay write . .aCe C28)=x0)+Fay. 833) Ifweincluded thenextapproximation, itwould involve terms in(ay)?, butsince ‘wewill ultimately think ofthelimit asAy—+0,such terms can beneglected. Putting (3.33) together with (3.32), wefind that [ce(1)—C.G))ay=—%Axay. 34) The derivative can, toourapproximation, beevaluated at(x,y). Similarly, fortheother two terms inthecirculation, wemay write GyQ)ay —GA)ay=26aay. 35) The circulation around our square isthen ay_aCe(@-acs)Axay, 8.36) 38 Which isinteresting, because thetwo terms intheparentheses arejust thez-com- ponent ofthecurl. Also, wenote that AxAyisthearea ofoursquare. Sowe can write our circulation (3.36) as (¥XCa. Butthez-component really means thecomponent normal tothesurface element. Wecan, therefore, write thecirculation around adifferential square inaninvariant vector form: $Cds=(VXCnda=(VXC)-mda. 37) © Ourresult is:thecirculation ofanyvector Caround aninfinitesimal squareisthecomponent ofthecurlofCnormaltothesurface,timestheareaofthesquare. =tall ‘Thecirculation aroundanyloopIcannowbeeasilyrelatedtothecurlof curtace g theVector field. Wefillintheloopwithanyconvenient surface S,asinFig.3-11, and addthecirculations around asetofinfinitesimal squares inthissurface. The sum canbewritten asanintegral. Our result isavery useful theorem called Stokes’ theorem (after Mr. Stokes). Stoxts’ THEOREM. vs, §,Cds=[09XOnda, @.38) y where Sisanysurface bounded byI. we Wemust now speak about aconvention ofsigns. InFig. 3-10 thez-axis,Fig.3-11.Thecirculation of —wouldpointrowardyouina“usual””—that is,“right-handed” —systemofaxes.‘around Tisthesurface integral ofthe When wetook ourlineintegral witha“positive” sense ofrotation, wefound that rormalcomponent ofVXC, thecirculation wasequaltothez-component ofVXC.Ifwehadgonearoundtheother way, wewould have gotten theopposite sign. Now how shall weknow, ingeneral, what direction tochoose forthepositive direction ofthe“normal” component of VXC? The “positive” normal must always berelated tothe sense ofrotation, asinFig. 3-10. Itisindicated forthegeneral case inFig. 3-11. ‘One wayofremembering therelationship isbythe“right-hand rule.” Ifyou make thefingers ofyour right hand goaround thecurve T,with thefingertips pointed inthedirection ofthepositive sense ofds,then your thumb points inthe direction ofthepositive normal tothesurface S. 3-7 Curkfree and divergence-free fields @ Wewould like, now, toconsider some consequences ofournew theorems.‘Takefirstthecaseofavectorwhosecutliseverywherezero.ThenStokes’theorem says that thecirculation around anyloop iszero. Now ifwechoose two points (i)and(2)onaclosed curve (Fig. 3-12), itfollows that thelineintegral ofthe tangential component from (1)to(2)isindependent ofwhich ofthetwopossible paths istaken, Wecanconclude that theintegral from (1)to(2)candepend only —©‘onthelocation ofthese points—that istosay, itissome function ofposition only. @ » ‘Thesame logicwasusedinChapter 14ofVol.I,whereweprovedthatiftheintegral ‘around aclosed loop ofsome quantity isalways zero, then that integral can beFig3-12,IfXCiszero,theTepresented asthedifference ofafunctionofthepositionofthe two ends. This ‘circulation around theclosed curve I’is fact allowedustoinventtheideaofapotential.Weproved,furthermore, thatthe ero.ThelineintegralofC+defrom(1)vectorfieldwasthegradientofthispotentialfunction(seeEq.14.13ofVol.1. to(2)along amust bethesome osthe Itfollows that anyvector field whose curliszero isequal tothegradient of fineintegralalongb. ‘somescalarfunction. Thatis,ifVXC=0,everywhere, thereissomey(psi)for which C=Vy—a useful idea. Wecan, ifwewish, describe thisspecial kind ofvectorfieldbymeansofascalarfield,Let's show something else. Suppose wehave anyscalar field ¢(phi). Ifwe take itsgradient, V@, theintegral ofthis vector around anyclosed loop must, be zero. Itslineintegral from point (1)topoint (2)is[6(2)~@(I)}.If(1)and(2) 310 arethesame points, ourTheorem 1,Eq.(3.8), tells usthat thelineintegral iszero: §ve-ds=0. Using Stokes’ theorem, wecan conclude that [9X (¥0)da =0 over any surface, But iftheintegral iszero over any surface, the integrand must bezero.So ¥X(¥6)=0,always. ‘Weproved thesame result inSection 2-7byvector algebra. Let’s look now ataspecial case inwhich wefillinasmall loop Twith alarge surface S,asindicated inFig. 3-13. Wewould like, infact, toseewhat happens when theloop shrinks down toapoint, sothat thesurface boundary disappears—thesurfacebecomesclosed.NowifthevectorCiseverywhere finite,theline @) Gif2integral around I’must gotozero asweshrink theloop—the integral isroughly proportional tothecircumference ofI,which goestozero.According toStokes’ L0PT theorem, thesurface integral of(¥XC),must alsovanish. Somehow, aswe ‘Surtoce S vac close the surface we add incontributions that cancel out what was there before.Sowehaveanewtheorem: Fig.3-13.Goingtothelimitof closedsurface,wefindthatthesurface f(FXOnda=0. B.39) integralof(VXChrmustvanish, 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 [(xOnda= ffvwxoar. 40) nea lume sirenvente Soweconclude that thesecond integral must also bezero: fviwx od=0, G41) voliioe and this istrue foranyvector field Cwhatever. Since Eq. (3.41) istrue foranyvolume,itmustbetruethatateverypointinspacetheintegrandiszero.Wehave ve(F XC) =0, always. But thisisthesame result wegotfrom vector algebra inSection 2-7. Now we begin toseehow everything fitstogether. 3-8 Summary Let ussummarize what wehave found about the vector calculus. These are really thesalient points ofChapters 2and 3: 1.The operators 8/ax, 4/89, and 8/az can beconsidered asthethree components ofavector operator V,and theformulas which result from vector algebra bytreating thisoperator as@vector arecorrect: aaa ve(322). 2.Thedifference ofthevaluesofascalarfieldattwopointsisequaltothe line integral ofthetangential component ofthegradient ofthat scalar along su any curve atallbetween thefirst and second points: ¥Q)— HD=f” Weeds. 42) 3.The surface integral ofthenormal component ofanarbitrary vector over aclosed surface isequal totheintegral ofthedivergence ofthevector over the volume interior tothe surface: [Cnda= fv-cav. 3.43)linea value 4.The line integral ofthetangential component ofanarbitrary vector around aclosed loop isequal tothesurface integral ofthenormal component ofthecurl ofthat vector over anysurface which isbounded bytheloop, {Cds=f(¥XC):nda. (G44) ary 4 Electrostatics 441 Staties Webegin now ourdetailed study ofthetheory ofelectromagnetism. Allof 4-1 Staties electromagnetism iscontained intheMaxwell equations. 42Coulomb's law;su Maxwell's equations: 4-3Electric potential ve=2, 4). 44E=-ve 45 The flux ofE vxe-- 8, 42)oF : 4-6 Gauss? law; thedivergence ofE evxe~ Hyd, (43)#7Fleldofsphereofcharge© 4-8Fieldlines; equipotential veB=0. (44) surfaces ‘Thesituationsthataredescribedbytheseequationscanbeverycomplicated Wewill consider first relatively simple situations, and learn how tohandle them before wetake upmore complicated ones. The easiest circumstance totreat isone inwhich nothing depends onthetime—called thestatic case. Allcharges are Review: Chapters 13and 14,Vol. I,permanently fixedinspace,oriftheydomove,theymoveasasteadyflowina WorkandPotentialEnergycircuit (60pandjareconstant intime). Inthese circumstances, alloftheterms in theMaxwell equations which aretime derivatives ofthefield arezero. Inthis case, theMaxwell equations become: Electrostatics: 7rT -En2,a viens 45) wae 1 VXE=0. 46|2=9x10ane Magnetostaties:[co]=coulomb?/newton-meter* ivxe-=4,, 47Pa an vB=0. (48) ‘You will notice aninteresting thing about thissetoffour equations. Itcan bbeseparated into two pairs. The electri field appears only inthefirst two, and themagnetic field Bappears only inthesecond two. The twofields arenotinter- connected. ‘This means that electricity and magnetism aredistinct phenomena so long ascharges and currents arestatic. The interdependence ofEand Bdoes not appear until there arechanges incharges orcurrents, aswhen acondensor is charged, oramagnet moved. Only when there aresufficiently rapid changes, so that thetime derivatives inMaxwell's equations become significant, willEand B depend oneach other. ‘Now ifyou look attheequations ofstatics you willseethat thestudy ofthe two subjects wecall electrostatics and magnetostatics isideal from thepoint of view oflearning about themathematical properties ofvector fields. Electrostatics isaneat example ofavector field with zero curl and agiven divergence. Magnet- ‘statics isaneat example ofafield with zero divergence andagiven curl. Themore conventional—and you may bethinking, more satisfactory—way ofpresenting ray thetheory ofelectromagnetism isto start firstwith electrostatics andthus tolearn about thedivergence, Magnetostatics and thecurl aretaken uplater. Finally, electricity and magnetism areput together. We have chosen tostart with the complete theory ofvector calculus. Now weshall apply ittothespecial case of electrostatics, thefield ofEgiven bythefirst pair ofequations. ‘Wewillbegin with thesimplest situations—ones inwhich thepositions ofall charges arespecified. Ifwehad only tostudy electrostatics atthis level (aswe shall dointhenext two chapters), lifewould bevery simple—in fact, almost trivial. Everything can beobtained from Coulomb's lawand some integration, asyou will see. Inmany real electrostatic problems, however, wedonot know, initially, where thecharges are. Weknow only that they have distributed them- selves inways that depend ontheproperties ofmatter. ‘The positions that the charges take updepend ontheEfield, which inturn depends onthepositions of thecharges, Then things cangetquite complicated. If,forinstance, acharged body isbrought near aconductor orinsulator, theelectrons and protons inthe conductor orinsulator will move around. The charge density pinEq. (4.5) may haveonepartthatweknowabout,fromthechargethatwebroughtup;buttherewill beother parts from charges that have moved around intheconductor. And allofthecharges must betaken into account. One can getinto some rather subtle andinteresting problems. Soalthough thischapter istobeonelectrostatics, itwill notcover themore beautiful andsubtle parts ofthesubject. Itwilltreat only the situation where wecan assume that the positions ofallthe charges are known. Naturally, you should beable todothat case before you trytohandle theother ones. 42 Coulomb's law; superposition Itwould belogical touseEqs. (4.5) and (4.6) asourstarting points. Itwill beeasier, however, ifwestart somewhere else and come back tothese equations. ‘The results willbeequivalent. Wewillstart with alawthat wehave talked about before, called Coulomb's law, which says that between two charges atrest there is1forcedirectlyproportional totheproductofthechargesandinverselypropor-tional tothesquare ofthedistance between. The force isalong thestraight line from onecharge totheother. Coulomb'slawspsagMen i) Fis theforce oncharge q1,¢12istheunit vector inthedirection tog1from qs,andr1isthedistancebetweeng,andqa.TheforceF»onq2isequalandopposite toFi. ‘The constant ofproportionality, forhistorical reasons, iswritten as1/4r¢9. Inthesystem ofunits which weuse—the mks system—it isdefined asexactly 10-7 times thespeed oflight squared. Now since thespeed oflight isapproxi- mately 3X10meters persecond, theconstant isapproximately 9x10°,and theunit turns outtobenewton-meter? percoulomb? orvoltmeter percoulomb. ae=107%?(bydefinition)=9.0X10°(byexperiment). (4.10) Unit: newton-meter?/coulomb?, ‘or volt-meter/coulomb. ‘When there aremore than two charges present—the only really interesting times—we must supplement Coulomb's law with one other fact ofnature: the force onanycharge isthe vector sum oftheCoulomb forces from each oftheother charges. This factiscalled “the principle ofsuperposition.” That's allthere istoelectrostatics. IfwecombinetheCoulomb lawandtheprincipleofsuperposition,there isnothing else. Equations (4.5) and (4.6)—the electrostatic equations—say nomore and noless. a2 ‘When applying Coulomb's law, itisconvenient tointroduce theidea ofan electric field. We saythat thefield E(1) istheforce per unit charge ongy(due to allother charges). Dividing Eq.(4.9) byq1,wehave, foroneother charge besides Gs =, 2FU)=aghee aan Also, weconsider that E(1) describes something about thepoint (1)even ifgr ‘were notthere—assuming that allother charges keep their same positions. We say: E(1) istheelectric field atthepoint (1). Theelectric field Eisavector, sobyEq.(4.11) wereally mean three equations —one foreach component. Writing outexplicitly thex-component, Eq. (4.11) means a xa Ede Yu21)=77 (6.C002) =arei=aFO Feape I) and similarly fortheother components. Ifthere aremany charges present, thefield Eatanypoint (1)isasum ofthe contributions from each oftheother charges. Each term ofthesum willlook like(G.11)or(4.12).Lettingg;bethemagnitude ofthejthcharge,andr;,thedis- placement from q,tothepoint (1),wewrite =-ytu&0=Daze tor (4.13) Which means, ofcourse, 1 ax=x) Exe yu21)= De p~—— i=) at Cuyned=Lae aFOraPasa OM) and soon. ‘Often itisconvenient toignore thefact that charges come inpackages like electrons andprotons, andthink ofthem asbeing spread outinacontinuous smear —orina“distribution,” asitiscalled. ThisisO.K.solongaswearenotinterestedinwhat ishappening ontoosmall ascale, Wedescribeachargedistribution by the“chargedensity,” p(x,y,z).IftheamountofchargeinasmallvolumeAV located atthepoint (2)isAgo, then pisdefined by Ago =(2) AV. (4.15) TouseCoulomb's lawwithsuchadescription, wereplace thesumsofEqs. anogyep(4.13)or(4.14)byintegrals overallvolumes containing charges. Thenwehave .ee e BU)=gefeensds, 4.16)SNres |e \ Saneeonpeerwi SS somepeopleprefertowriteatt, ES) ne (2);0%%222) where rigisthevector displacement to(1)from (2),asshown inFig. 4-1. The integral forEisthenwritten as Fig.4-1. The electric field Eat point (1), from acharge distribution, isay=glfened. (4.17)Sbtained’fromvanintegraloverthem0oyTt distribution. Point(1)couldalsobeinsideothe thedistribution. When wewant tocalculate something with these integrals, weusually have to write them outinexplicit detail. For thex-component ofeither Eq. (4.16) or (4.17), wewould have nya|pO =Andon,yast2)deadyedea «(44g) ElenFu21)iid—uF+O taaeON) os Wearenotgoing tousethisformula much. Wewrite ithere only toempha-sizethefactthatwehavecompletely solvedalltheelectrostatic problems inwhich‘weknow thelocations ofallofthecharges. Given thecharges, what arethefields? Answer: Dothis integral. Sothere isnothing tothesubject; itis just acase of doing complicated integrals over three dimensions—strietly ajobforacomputing machine! Withourintegralswecanfindthefieldsproducedbyasheetofcharge,fromallineofcharge, from aspherical shell ofcharge, orfrom anyspecified distribution. tis important torealize, aswegoontodraw field lines, totalk about potentials, frtocalculate divergences, that wealready have theanswer here, Itismerely a matter ofitbeing sometimes easier todoanintegral bysome clever guesswork than byactually carrying itout. The guesswork requires learning allkinds of strange things. Inpractice, itmight beeasier toforget trying tobeclever and al- ‘ways todotheintegral directly instead ofbeing sosmart. Weare, however, going totrytobesmart about it.Weshall goontodiscuss some other features ofthe electric field. 4-3 Electric potential Firstwetakeuptheideaofelectricpotential,whichisrelatedtotheworkdone 5,incarrying acharge from one point toanother. There issome distribution of b charge, which produces anelectric field. Weaskabout how much work itwould 4 take tocarry asmall charge from oneplace toanother. ‘Thework done against ‘onepath, theelectrical forces incarrying acharge along some path isthenegative ofthecom- ‘another Ponent oftheelectrical force inthedirection ofthemotion, integrated along thepath path.Ifwecarryachargefrompointatopoint6, ° » ©chargefromatobisthenegativeoftheintegralofFdsalongthepathwhereFistheelectricalforceonthechargeateachpoint,anddsisthedifferentialtaken, vectordisplacement alongthepath.(SeeFig.4-2.) Itismore interesting forour purposes toconsider thework that would be done incarrying one unit ofcharge. Then theforce onthecharge isnumericallythesameastheelectricfield,Callingtheworkdoneagainstelectricalforcesinthiscase W(unit), wewrite . Wounit)=—fE-ds. 4.19) Now,ingeneral,whatwegetwiththiskindofanintegraldependsonthepathwetake. Butiftheintegral of(4.19) depended onthepath from atob,wecould get work outofthefield bycarrying thecharge tobalong onepath andthen back to.ontheother.WewouldgotobalongthepathforwhichWissmallerandbackalong theother, getting outmore work than weputin. There isnothing impossible, inprinciple, about getting energy outofafield.Weshall,infact,encounter fieldswhereitispossible.Itcouldbethatasyoumove ‘acharge you produce forces ontheother part ofthe“machinery.” Ifthe“ma- chinery” moved against theforce itwould lose energy, thereby keeping thetotal ‘energy intheworld constant. Forelectrostatics, however, there isnosuch “ma-chinery.” Weknowwhattheforcesbackonthesourcesofthefieldare.TheyaretheCoulomb forces onthecharges responsible forthefield. Iftheother charges arefixed inposition—as weassume inelectrostatics only—these back forces can donowork onthem. There isnoway togetenergy from them—provided, of course, that theprinciple ofenergy conservation works forelectrostatic situations. Webelieve that itwill work, butlet's ust show that itmust follow from Coulomb's law offorce. Weconsider first what happens inthefield due toasingle charge g.Let point abeatthedistance r;from q,and point batra.Now wecarry adifferent ‘charge, which wewillcallthe“test” charge, and whose magnitude wechoose to “4 bbeoneunit,fromatob.Letsstartwiththeeasiestpossiblepathtocalculate. We‘carryourtestchargefirstalongthearcofacircle, then alongaradius,asshownin part (a)ofFig. 4-3. Now onthat particular path itischild's play tofind thework Gone (otherwise wewouldn’t have picked it). First, there isnowork done atall onthepath from atoa’.The field isradial (from Coulomb's law), soit iatright angles tothedirection ofmotion. Next, onthepath from a’to6,thefeld isinthedirectionofmotionandvariesas1/r®.‘Thustheworkdoneonthetestchargeincarryingitfromatobwouldbe > . ae ian °sds = 4 =-,i-(1~1).~[ra~-e[ 8-8-4) am ‘Now let’s take another easy path. For instance, theone shown inpart (b)ofA Fig.4-3,Itgoesforawhilealonganarcofacircle,thenradiallyforawhile,thenalong anarcagain, thenradially, andsoon.Every timewegoalong thecircular 1 parts, wedonowork. Every time wegoalong theradial parts, wemust just integrate 1/r?, Along thefirst radial stretch, weintegrate from 7,tore’,then » along thenext radial stretch from rq:tora, and soon. The sum ofallthese in-tegralsisthesameas.asingleintegraldirectlyfromr,tor.Wegetthesameanswerforthispath that wedidforthefirst path wetried. Itisclear that wewould getthesameanswerforanypathwhichismadeupofanarbitrarynumberofthesame? 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. Fig.4-3.Incarryingotestcharge. fromatobthesameworkisdoneal me-few eitherpath. sd ab} ae Since thework done depends only ontheendpoints, itcanberepresented as thedifference between twonumbers. Wecanseethisinthefollowing way. Let’s choose areference point Poandagree toevaluate ourintegral byusing apath thatalwaysgoesbywayofpointPo,Let#(a)standfortheworkdoneagainstthefieldingoing from Potopoint a,and let4(6) bethework done ingoing from Potopointb(Fig.4-4).TheworkingoingfoPofroma(onthewaytob)isthenegative of¢(@), sowehave that .-fE-ds=(6)—4(2). G21)ameego)=0d > Since only thedifference inthefunction attwo points isever involved, we donot really have tospecify thelocation ofPo. Once wehave chosen some vr,=9)+H) reference point, however, anumber ¢isdetermined foranypoint inspace; is, then ascalar field. Itisafunction ofx,y,z.Wecallthisscalar function theelec- ‘trostatic potential atanypoint. 4,79 9) , Electrostatic potential: . Fig.4-4,Theworkdoneingoing‘alonganypathfromatobisthenegative «=~fBods. (422)oftheworkfromsomepointPotoaplusfe theworkfromPotob. For convenience, wewill often take thereference point atinfinity. Then, forasingle charge attheorigin, thepotential ¢isgiven foranypoint (x,y,2)— using Eq.(4.20): -41 0,92)=got 423) Theelectricfieldfromseveralchargescanbewrittenasthesumofthe electric field from thefirst, from thesecond, from thethird, etc. When weintegrate the sum tofind thepotential wegetasum ofintegrals. Each oftheintegrals isthe 4s potentialfromoneofthe charges. Weconclude that thepotential ¢from alotof charges isthesum ofthepotentials from alltheindividual charges. There isasuperposition principlealsoforpotentials. Usingthesamekindofarguments bywhichwefoundtheelectricfieldfromagroupofchargesandforadistribution ofcharges,wecangetthecompleteformulasforthepotential¢atapointwecall(1): 1g o)=Laeu. (4.24) 1f02)dV2 I= : (4.25) = ae) OE (4.25) Remember thatthepotential ¢hasaphysical significance: itisthepotential‘energywhichaunitchargewouldhaveifbroughttothespecifiedpointinspace from some reference point. 44 E=-v6 Whocaresabout4?ForcesonchargesaregivenbyE,theelectricfield.Thepoint isthat Ecan beobtained easily from ¢—it isaseasy, infact, astaking a derivative. Consider two points, oneatxand oneat(x+dx), butboth atthe sameyandz,andaskhowmuchworkisdoneincarryingaunitchargefromonepointtotheother.Thepathisalongthehorizontallinefromxtox-+dx.The ‘work done isthedifference inthepotential atthetwo points: AW=ox+Ax,¥,2)—665,952)=Bax. Butthework done against thefield forthesame path is AW=~[E-ds =—E,Ax. We see that 2%E=-%. (4.26) Similarly, E,=—89/ay, E,=—a9/42, or,summarizing with thenotation of vector analysis, E= vo. 427 This equation isthedifferential form ofEq.(4.22). Any problem with specified charges canbesolved bycomputing thepotential from (4.24) or(4.25) andusing (4.21) togetthefield. Equation (4.27) also agrees with what wefound from vector calculus: that forany scalar field ¢ .[Vode=(6)—$(a). (4.28) According toEq.(4.25) thescalar potential ¢isgiven byathree-dimensional integral similar totheone wehad forE.Isthere any advantage tocomputing ¢ rather than E?Yes. There isonly oneintegral for¢,while there arethree integrals forE—because itis avector. Furthermore, 1/risusually alittle easier tointegrate than x/r?. Itturns outinmany practical cases thatitiseasier tocalculate ¢and then take thegradient tofind theelectric field, than itistoevaluate thethree integrals forE.Itismerely apractical matter.‘Thereisalsoadeeperphysicalsignificance tothepotential¢.Wehaveshown that EofCoulomb's law isobtained from E=—grad¢,when¢isgivenby (4.22).ButifBisequaltothegradientofascalarfield,thenweknowfromthe vector calculus that the curl ofEmust vanish: VX E=0. (429) 46 Butthat isjust oursecond fundamental equation ofelectrostatics, Eq.(4.6). We have shown that Coulomb's law gives anEfield that staisfes that condition. So far,everything isallright. Wehad really proved that VXEwas zero before wedefined thepotential. Wehad shown that thework done around aclosed path iszero. That is,that fE-ds =0 foranypath. Wesaw inChapter 3that foranysuch field VX Emust bezero everywhere. Theelectric field inelectrostatics isanexample ofacurl-fre field ‘You canpractice your vector calculus byproving that VXEiszero inadif- ferent way—by computing thecomponents of¥XEforthefieldofapointcharge, asgiven byEq. (4.11). Ifyou getzero, thesuperposition principle says you would getzero forthefield ofanycharge distribution. Weshould point outanimportant fact. Foranyradial force thework done is independent ofthepath, and there exists apotential. Ifyou think about it,the entire argument wemade above toshow that thework integral was independentofthepathdependedonlyonthefactthattheforcefromasinglechargewasradial andspherically symmetric. Itdidnotdepend onthefactthat thedependence ondistance wasas1/r®—there could have been anyrdependence. Theexistence ofapotential, and thefact that thecurl ofEiszero, comes really only from the symmetry and direction oftheelectrostatic forces. Because ofthis, Eq. (4-28)— or(4.29)—can contain only part ofthelaws ofelectricity. 445 The flux ofE Wewill now derive afield equation that depends specifically anddirectly on thefact that theforce law isinverse square. That thefield varies inversely asthe squareofthedistanceseems,forsomepeople,tobe“onlynatural,”because“that's the way things spread out.” Take alight source with light streaming out: the amount oflight that passes through asurface cut out byacone with itsapex at the sourceithesamenomatteratwhatradiusthesurfaceisplaced.Itmustbeso ifthereistobeconservation oflightenergy.Theamountoflightperunitarea—theintensity—must vary inversely asthearea cutbythecone, i.e, inversely asthe square ofthe distance from thesource. Certainly theelectric field should vary inversely asthesquare ofthedistance forthesame reason! Butthere isnosuch thing asthe“same reason” here. Nobody can saythat theelectric field measures.theflowofsomething likelightwhichmustbeconserved. fwehadit“model”oftheelectri field inwhich theelectric field vector represented thedirection and speed—say thecurrent—of some kind oflittle “bullets” which were flying out, andifourmode! required that these bullets were conserved, that none could ever disappear once itwas shot outofacharge, then wemight saythat wecan“see” that theinverse square lawisnecessary. Ontheother hand, there would necessarilybbesomemathematical waytoexpressthisphysicalidea.Iftheelectricfieldwerelikeconserved bulletsgoingout,thenitwouldvaryinverselyasthesquareofthedistance and wewould beable todescribe that behavior byanequation—which ispurely mathematical. Now there isnoharm inthinking thisway, solong aswe donotsaythat theelectric field ismade outofbullets, butrealize that weare using amodel tohelp usfind theright mathematics. Suppose, indeed, that weimagine for amoment that theelectric field did represent theflow ofsomething that was conserved—everywhere, that is,exceptatcharges.(Ithastostartsomewhere!) Weimaginethatwhateveritislowsout‘ofacharge into thespace around. IfEwere thevector ofsuch aflow (ashifor heatflow), itwould have a1/r?dependence nearapoint source. Now wewish to usethis model tofind outhow tostate theinverse square law inadeeper ormore abstract way, rather than simply saying “inverse square.” (You may wonder ‘hy weshould want toavoid thedirect statement ofsuch asimple law, and want instead toimply thesame thing sneakily inadifferent way. Patience! Itwillturn outtobeuseful.) “7 on” ® + ete Ot- SurtoceSQOS Yor ae pa fooeo Fig.4-5.ThefluxofEoutofthe rae Fig.4-6.ThefuxofEoutofthePoint Charge surface $iszero. Point Chorge surface Siszero. Weask: What isthe“flow” ofEoutofanarbitrary closed surface inthe neighborhood ofapoint charge? First let’s take aneasy surface—the oneshown inFig. 4-5. Ifthe Efield islikeaflow, thenetflow outofthisboxshouldbezero. ‘Thatiswhatwegetifbythe“flow”fromthissurfacewemeanthesurfaceintegral Ofthenormalcomponent ofE—thatis,thefluxofE.Ontheradialfaces,thenor-mal component iszero. Onthespherical faces, thenormal component Eyisjust themagnitudeofE—minusforthesmallerfaceandplusforthelargerface.‘The LEmagnitude ofEdecreases as1/r?, butthesurface area isproportional tor?,so G rtoceS theproduct isindependent ofr.ThefluxofEintofaceaisjustcancelledbythe be fluxoutofface6,ThetotalflowoutofSiszero,whichistosaythatforthis yssurface eLEE VAES J,Fada=0. (430) [ABE ~~ ‘Nextweshowthatthetwoendsurfacesmaybetiltedwithrespecttothe radial linewithout changing theintegral (4.30). Although itistrue ingeneral, for¢ ‘ourpurposesitisonlynecessarytoshowthatthisistruewhentheendsurfacesare ‘small, sothat they subtend asmall angle from thesource—in fact, aninfinitesimal Fig. 4-7. Anyvolume canbethought angle. InFig.4-6weshowasurfaceSwhose“sides”areradial,butwhose“ends” ofascompletely mode upofinfritesimal aretilted. ‘Theendsurfaces arenotsmall inthefigure, butyouaretoimagine the truncated cones. ThefluxofEfromone situation forverysmall endsurfaces. Then thefield willbesufficiently uniform‘endofeachconicalsegmentisequolondoverthesurfacethatwecanusejustitsvalueatthecenter.Whenwetiltthesur-cpposteto,tefurfromtheotherwefacebyanangle8,theareaisincreased bythefactor1/cos@,ButE,,thecompo-therefore zero, ceSisnentofEnormaltothesurface, isdecreased [email protected] E,Saisunchanged. The flux outofthewhole surface Sisstillzero. ‘Now itiseasy toseethat theflux outofavolume enclosed byanysurface $ ‘must bezero, Any volume canbethought ofasmade upofpieces, like that inFig.4-6.Thesurfacewillbesubdivided completely intopairsofendsurfaces,and since thefluxes inand outofthese endsurfaces cancel bypairs, thetotal lux ‘outofthesurface willbezero. The idea isillustrated inFig. 4-7. Wehave the completely general result that thetotal lux ofEout ofany surface Sinthefield ofapointchargeiszero.Butnotice!OurproofworksonlyifthesurfaceSdoesnotsurroundthecharge. 5What would happen ifthepoint charge were inside thesurface? Wecould still divide oursurface into pairs ofareas that arematched byradial lines through the , charge, asshown inFig.4-8.Thefluxes through thetwosurfaces arestillequal—bythesame arguments asbefore—only now they have thesame sign. The flux outofasurfacethatsurroundsachargeisnotzero.Thenwhatisit?Wecanfind %outbyalittletrick.Supposewe“‘remove” thechargefromthe“inside”bysur-& rounding thecharge byalittlesurface S”totally inside theoriginal surface S,asx? ‘shown inFig.4-9.Nowthevolume enclosed between thetwosurfaces Sand S’ hasnocharge init.The total flux outofthisvolume (including that through 5’) Fig. 4-8. Ifacharge isinside a iszero, bythearguments wehave given above. The arguments tellus,infact, that surface, thefluxoutisnotzero. theflux into thevolume through Sis thesame astheflux outward through S. “ Wecanchoose anyshape wewish forS",solet's make itasphere centered on thecharge, asinFig. 4-10. Then wecaneasily calculate thefluxthrough it.Ifthe radius ofthelitle sphere isr,thevalue ofEeverywhere onitssurfaceis surtoce 14 reg 7?” andisdirected always normal tothesurface. Wefindthetotal fluxthrough S’if surface ‘wemultiply thisnormal component ofEbythesurfacearea: Go FluxthroughthesufaceS’=G&4)Grty=4,431) ‘anumber independent oftheradius ofthesphere! Weknow then that theflux‘outwardthroughSisalsoq/¢o—avalueindependent oftheshapeofSsolong25Fig.4-9,ThefxthroughSisthe thecharge gisinside. same osthefuxthrough S'. We can write our conclusions asfollows: 0;qoutside$ iFoda=V4.ainsideS G32) any wsitace « Let’s return toour“bullet” analogy andseeifitmakes sense. Our theorem says that thenetflow ofbullets through asurface iszero ifthesurface does not enclose thegunthat shoots thebullets. Ifthe gunisenclosed inasurface, whatever Gsizeandshapeitis,thenumberofbulletspassingthroughisthesame~itisgivenbytherate atwhich bullets aregenerated atthegun. Itallseems quite reasonable forconserved bullets. Butdoesthemodel tellusanything more thanweget eq simply bywriting Eq.(4.32)? Noonehassucceeded inmaking these “bullets” do anything else butproduce this one law. After that, they produce nothing but errors. Thatiswhytoday weprefer torepresent theelectromagnetic fieldpurely s abstractly. 4-6Gauss’Iaw;thedivergence ofE Figs4-10,Thefxtheoughospheri- (Our nice result, Eq,(4.32), was proved forasinglepointcharge.Nowsuppose0!surfacecontaining@pointcharge thattherearetwocharges, acharge g)atonepoint andacharge q2atanother. 9184/¢0- The problem looks more difficult. The electric field whose normal component we integrate fortheflux isthefield duetoboth charges. That is,ifErepresents the electric field that would have been produced byq,alone, and Eyrepresents theelectricfieldproducedbyg2alone,thetotalelectricfieldisE=Ey+Ez.Theflux through any closed surface Sis [Gm+Edda=fEinda+[Eanda. (433) sIs Is ‘The flux with both charges present istheflux due toasingle charge plus theflux duetotheother charge. Ifboth charges areoutside S,thefluxthrough Siszero.Ifq1isinsideSbutgoisoutside,thenthefirstintegralgivesq,/¢yandthesecondintegral gives zero. Ifthe surface encloses both charges, each willgive itscontribu tion andwehave that theflux is(g,+92)/éo. The general ruleisclearly that the total fluxoutofaclosed surface isequal tothetotal charge inside, divided by¢o Our result isanimportant general lawoftheelectrostatic field, called Gauss’ law. Ganstowsfg =Sfharesise, an any cloned° sug or foemda=2, 35) ie “NidaceSs where Qin=Dae (436) las 49 Ifwedescribe thelocation ofcharges interms ofacharge density p,wecancon- sider that each infinitesimal volume dVcontains a“point” charge pdV. The sum over allcharges isthen theintegral Om=fpay. 437) yhae From ourderivation you seethat Gauss’ lawfollows from thefact that the exponent inCoulomb's lawisexactly two. A1/r?field, oranyI/r*fieldwith 1n+2,would notgive Gauss’ law. SoGauss’ lawisjust anexpression, inadif- ferent form, oftheCoulomb lawofforces between twocharges. Infact, working back from Gauss’ law, you canderive Coulomb's law. The twoarequite equiva- lent50long aswekeep inmind therule that theforces between charges isradial. We would now like towrite Gauss’ law interms ofderivatives. Todothis, ‘weapply Gauss’ Jawtoaninfinitesimal cubical surface. Weshowed inChapter 3thatthefluxofEoutofsuchacubeisV»£timesthevolumed¥ofthecube.ThechargeinsideofdV,bythedefinition ofp,isequaltopd¥,soGauss’Jawgives vega ~2m, or vE=2. (4.38) ‘The differential form ofGauss’ lawisthefirst ofourfundamental field equations of electrostatics, Eq.(4.5). Wehave now shown that thetwo equations ofelectro- statics, Eqs. (4.5) and (4.6), areequivalent toCoulomb's law offorce. Wewill now consider oneexample oftheuseofGauss’ law. (We willcome later (omany more examples.) 4-7 Field ofasphereofcharge One ofthedifficult problems wehad when westudied thetheory ofgravita- -~\ tionalattractions wastoprovethattheforceproduced byasolidsphereofmatter 7o wasthesameatthesurfaceofthesphereasitwouldbeifallthematterwere / P concentrated atthecenter. For many years Newton didn't make public his \ theory ofgravitation, because hecouldn't besure thistheorem wastrue, We Re provedthetheoreminChapter13ofVol.Ibydoingtheintegralforthe Barution\, ‘Seyzsiog,potentialandthenfindingthegravitationalforcebyusingthegradient.Nowwe arcanprove thetheorem inamost simple fashion. Only thistime wewillprove the ~- corresponding theorem forauniform sphere ofelectrical charge. (Since thelaws ;.ofelectrostaticsarethesameasthoseofgravitation,thesameproofcouldbe Fi.ay1UsingGouss'lowtofinddoneforthegravitational field.) thefieldofceniformsphereofcharge, ‘Weask:WhatistheelectricfieldEatapointPanywhereoutsidethesurfaceofaspherefilledwithauniformdistribution ofcharge? Since there isno“special” direction, wecanassume that Eiseverywhere directed away from thecenter ofthe sphere. Weconsider animaginary surface that isspherical and concentric with thesphere ofcharge, and that passes through thepoint P(Fig. 4-11). For this surface, theflux outward is [iada =E-4rR. ‘Gauss’lawtellsusthatthisluxisequaltothetotalchargeQofthesphere(over€,): Earn? =2, or teEPie 39) 410 , 1 aoe .yN . / \ aqn c /nN, \ Lines of€ / “1 \ \ \ \\ !owe7\ o : N Y. ~_ / i \ Fig, 4-12. Feld nos and equpotetal surfaces fro postive pent chorge. which isthesame formula wewould have forapoint charge Q.Wehave proved Newton’s problem more easily than bydoing theintegral. Itis,ofcourse, afalse kindofeasiness—it hastakenyousometimetobeabletounderstand Gauss’ law, ‘soyoumay think thatnotime hasreally been saved. Butafter youhave used the ‘theorem more and more, itbegins topay. Itisaquestion ofefficiency. 4-8 Field lines; equipotential surfaces Wewouldlikenowtogiveageometrical description oftheelectrostatic field. Thetwolawsofelectrostatics, onethatthefluxisproportional tothechargeinside andtheotherthattheelectric fieldisthegradient ofapotential, canalsoberepre- sented geometrically. Weillustrate thiswith two examples.First,wetakethefieldofapointcharge. Wedrawlinesinthedirection ofthe field—lines which arealways tangent tothefield, asinFig. 4-12. These arecalled fieldlines.Thelinesshoweverywhere thedirection oftheelectric vector. Butwe ‘alsowish torepresent themagnitude ofthevector. Wecanmake therulethatthe strength oftheelectric field willberepresented bythe“density” ofthelines. By thedensity ofthelines wemean thenumber oflines perunitarea through asur- face perpendicular tothelines. With these tworules wecanhave apicture ofthe electric field. Forapoint charge, thedensity ofthelinesmust decrease as1/r*. Buttheareaofaspherical surface perpendicular tothelinesatanyradiusrincreases asr®,soifwealways keep thesame number oflines foralldistances from the charge, thedensity willremain inproportion tothemagnitude ofthefield. Wecan guarantee that there arethesame number oflines atevery distance ifweinsist ‘that thelines becontinuous—that once aline isstarted from thecharge, itnever stops. Interms ofthefield lines, Gauss’ lawsays that lines should start only at pluscharges andstop atminus charges. Thenumber which /eave acharge qmust beequal tog/€o. Now, wecanfindasimilar geometrical picture forthepotential ¢.Theeasiest waytorepresent thepotential istodraw surfaces onwhich ¢isaconstant. Wecallthemeguipotential surfaces—surfaces ofequalpotential. Nowwhatisthegeometri- ra | O(»~ 4 N A A 7 a N N <A t Sh Oo SIN INS \OOH; PAIK 7/ N\DN y, y od Fig. 4-13. Field lines and equipotentials fortwo equal and opposite point charges. calrelationship oftheequipotential surfaces tothefield lines? The electric field is. thegradient ofthepotential. The gradient isinthedirection ofthemost rapid change ofthepotential, andistherefore perpendicular toanequipotential surface. IfEwere not perpendicular tothe surface, itwould have acomponent inthe surface. The potential would bechanging inthesurface, butthen itwouldn't be aanequipotential. The equipotential surfaces must then beeverywhere atright angles totheelectric fieldlines. ‘ANoteaboutUnits Forapointchargeallbyitself,theequipotentialsurfacesarespherescentered Quantity Unie atthecharge.WehaveshowninFig.4-12theintersection ofthesesphereswithaF plane through thecharge.° coulomb ‘Asasecondexample,weconsiderthefieldneartwoequalcharges,apositiveL meter oneandanegativeone.Togetthefieldiseasy.Thefieldisthesuperposition ofw joule thefieldsfromeachofthetwocharges.So,wecantaketwopictureslikeFig.4-12p~Q/L?—_coulomb/meter? andsuperimpose them—impossible! Thenwewouldhavefieldlinescrossingeach1/eo~FL3/Q® newtonrmeter?/coulomb* other, andthat’s notpossible, because Ecan’t have twodirections atthe same point. E~ F/Q _newton/coulomb ‘Thedisadvantage ofthefield-line picture isnow evident. Bygeometrical argu- %~W/Q —_joule/coulomd =volt _ments itisimpossible toanalyze inaverysimple waywhere thenewlinesgo. E~/L |vol/meter Fromthetwoindependent pictures, wecan’tgetthecombined picture. The 1/eo~EL?/Q volt-meter/coulomb principle ofsuperposition, asimple anddeepprinciple about electric fields, does nothave, inthefield-line picture, aneasy representation. The field-line picture hasitsuses, however, sowemight still like todraw thepictureforapairofequal(andopposite)charges.IfwecalculatethefieldsfromEq. (4.13) and thepotentials from (4.23), wecan draw thefield lines and equi- potentials. Figure 4-13 shows theresult. But wefirst had tosolve theproblem mathematically! +n & Application ofGauss’ Law ‘5-1 Electrostatics isGauss?lawplus... ‘There are two laws ofelectrostatics: that the flux ofthe electric field from a -S-1_Electrostaties isGauss? law volume isproportional tothecharge inside—Gauss’ law, andthat thecirculation plus. ofthe eletrc field iszero—Eis agradient. From these twolaws, allthepredictions ofelectrostatics follow. Buttosaythesethingsmathematically isonething;to©"Fauilitrlum inanelectrostatic usethem easily, and withacertainamountofingenuity,isanother.Inthischapter wewillworkthroughanumberofcalculations whichcanbemadewithGauss’law5-3.Equilibrium withconductorsdirectly. Wewill prove theorems and describe some effects, particularly incon-ductors, thatcanbeunderstood veryeasilyfromGauss' law.Gauss' lawbyitself 4Stability ofatomscannotgivethesolutionofanyproblembecausetheotherlawmustbeobeyedto.$-SThefieldofalinechargeSowhen weuseGauss' lawforthe solution ofparticular problems, wewillhave to .‘addsomething toit.Wewillhavetopresuppose, forinstance,someideaofhow ©Asheetofcharge;twosheets thefield Iooks—based, forexample, onarguments ofsymmetry. Orwemay have 5-7. sphere ofcharge; aspherical tointroduce specifically theidea thatthe field isthegradient ofapotential. shell 5-8Isthefieldofapointcharge 5.2Equilibrium inanelectrostatic feld exactly1/P? Consider firstthefollowing question: When canapoint charge beinstable 5-9 Thefields ofaconductor mechanical equilibrium intheelectric field ofother charges? Asanexample,imaginethreenegativechargesatthecornersofanequilateraltriangleinahori-510Thefieldimacavityofazontalplane.Would2positivechargeplacedatthecenterofthetriangleremain conducthere? (Itwillbe simpler ifweignore gravity forthemoment, although including itwould notchange theresults.) The force onthepositive charge iszero, but istheequilibrium stable? Would thecharge return totheequilibrium position if displaced slightly? The answer isno. ‘There arenopoints ‘ofstable equilibrium inany electrostatic field—except right ontopofanother charge. Using Gauss’ law, itseasy toseewhy. First, fora charge tobeinequilibrium atany particular point Po,thefield must bezero. Second, iftheequilibrium istobeastable one, werequire that ifwemove the charge away from Poinany direction, there should bearestoring force directed ‘opposite tothedisplacement. ‘The electric field atallnearby points must be pointing inwaré—toward thepoint Pp. Butthat isinviolation ofGauss" lawif there isnocharge atPo,aswecaneasily see. Consider atinyimaginary surface thatencloses Po,asinFig.SI. Ifthe Pp,electricfieldeverywhere inthevicinityispointedtowardPo,thesurfaceintegral wdls ofthenormal component iscertainly notzero. Forthecaseshown inthefigure, feathefluxthroughthesurfacemustbeanegativenumber,ButGauss’lawsaysthat ieOfLetmaginorythefluxofelectric fieldthrough anysurface isproportional tothetotalcharge LY Beet 0inside. Ifthere isnocharge atPo,thefieldwehave imagined violates Gauss law. ote ‘ Itisimpossible tobalance apostive charge inempty space—at apoint where Fig.5-1, IfPywere opostion of there isnotsome negative charge. Apostive charge canbeinequilibrium ifitis stable equiibrivm for@postive chorge, inthemiddle ofadistributed negative charge. OFcourse, thenegative charge the electric. Reld everywhere inthe distribution would have tobeheld inplace byother than electrical forces! neighborhood would point toward Po. ‘Our result has been obtained forapoint charge. Does thesame conclusion hold foracomplicated arrangement ofcharges held together infixed relative positions—with rods, forexample? We consider thequestion fortwo equal charges fixed onarod. Isitpossible that thiscombination canbeinequilibrium insome electrostatic field? The answer isagain no. The total force onthe rod cannot berestoring fordisplacements inevery direction. Pa Call Fthetotal force ontherodinany position—F isthen avector field. Following theargument used above, weconclude that ataposition ofstable equi- librium, thedivergence ofFmust beanegative number. Butthetotal force onthe rodisthefirst charge times thefield atitsposition, plus thesecond charge times thefield atitsposition: Fm QE +q2Es 6.1) The divergence ofFisgiven by ViF=qi(VE)+92(¥Es). Ifeach ofthetwo chargesq,andqoisinfreespace,bothV“EyandVE,are zero, and V«Fiszero—not negative, aswould berequired forequilibrium. You canseethat anextension oftheargument shows that norigid combination ofany number ofcharges canhave aposition ofstable equilibrium inanelectrostatic fieldinfreespace. ' eeSe Fig.5-2. Acharge canbeinequili- 4bviumthereoremechanical contains, abe" Now wehave notshown that equilibrium isforbidden ifthere arepivots or other mechanical constraints. Asanexample, consider ahollow tube inwhich a charge canmove back and forth freely, butnotsideways. Now itisvery easy to devise anelectric field that points inward atboth ends ofthetube ifitis allowed that thefield may point laterally outward near thecenter ofthetube. Wesimply place positive charges ateach end ofthetube, asinFig. 5-2. There can now beanequilibrium pointeventhoughthedivergence ofEiszero.Thecharge,ofcourse,would notbeinstable equilibrium forsideways motion were itnot for“non- electrical” forces from the tube walls. 5-3 Equilibrium with conductors ‘There isnostable spot inthefield ofasystem offixed charges. What about 1system ofcharged conductors? Can asystem ofcharged conductors produce @ field that willhave astable equilibrium point forapoint charge? (We mean ata point other than onaconductor, ofcourse.) You know that conductors have the property that charges canmove freely around inthem. Perhaps when thepoint charge isdisplaced slightly, theother charges ontheconductors willmove inaway that will give arestoring force tothepoint charge? The answer isstill no—al- though theproof wehave just given doesn’t show it.The proof forthis case is more difficult, and wewillonly indicate how itgoes. First, wenote that when charges redistribute themselves ontheconductors, they can only dosoiftheir motion decreases their total potential energy. (Some‘energyislosttoheatastheymoveintheconductor.) Nowwehavealreadyshownthat ifthecharges producing afield arestationary, there is,near anyzero point Po inthefield, some direction forwhich moving apoint charge away from Powilldecreasetheenergyofthesystem(sincetheforceisawayfromPo).Anyreadjustment ofthecharges ontheconductors can only lower thepotential energy stillmore,80(bytheprincipleofvirtualwork)theirmotionwillonlyincreasetheforce inthat particular direction away from Po,and notreverse it ‘Our conclusions donotmean that itisnotpossible tobalance acharge by electrical forces. Itispossible ifone iswilling tocontrol thelocations orthesizesofthesupporting chargeswithsuitabledevices.Youknowthatarodstandingonitspoint inagravitational field isunstable, butthisdoes notprove that itcannotbebalancedontheendofafinger.Similarly, achargecanbeheldinonespotbyclectric fields ifthey arevariable. Butnotwith apassive—that is,astatic—system. 52 5-4 Stability ofatoms Ifcharges cannot beheldstably inposition, itissurely notproper toimagine freistetiteem mattertobemadeupofstaticpointcharges(electronsandprotons)governedonly_FTE uunpomnsuene bythelaws ofelectrostatics. Such astatic configuration isimpossible; itwould FEET HHH cuanoe collapse! BHIRARHTEItwasoncesuggested thatthepositive chargeofanatomcouldbedistributed HEOHTHTTTATTweearn RCE uniformlyinasphere,andthenegativecharges,theelectrons,couldbeatrestsestasseestaelfaeocerecrer insidethepositive charge, asshown inFig.5-3.Thiswasthefirstatomic model, HHH HHH Ht proposed byThompson. But Rutherford concluded from theexperiment ofGeiger Go and Marsden that thepositive charges were very much concentrated, inwhat he called thenucleus. Thompson's static model hadtobeabandoned. Rutherford _Fig.5-3. TheThompson model ofon andBohr then suggested thattheequilibrium might bedynamic, with theelectrons tom. revolving inorbits, asshown inFig. 5-4. The electrons would bekept from falling intoward the nucleus bytheir orbital motion. We already know atleast one difficulty with thispicture. With such motion, theelectrons would beaccelerating (because ofthecircular motion) and would, therefore, beradiating energy. They ‘would lose thekinetic energy required tostay inorbit, and would spiral intoward thenucleus. Again unstable! The stability oftheatoms isnow explained interms ofquantum mechanics.Theelectrostatic foreespulltheelectronasclosetothenucleusaspossible,butthe POSITIVEMUCLEUSelectron iscompelled tostay spread outinspace over adistance given bythe uncertainty principle. Ifitwere confined intoosmall aspace, itwould have a great uncertainty inmomentum. Butthatmeans thatitwould haveahighex- secourve pected energy—which itwould usetoescape from theelectrical attraction. The ELECTRONS. netresult isanelectrical equilibrium nottoodifferent from theideaofThompson PLANETARY ORBITS —only itisthenegative charge that isspread out(because themass oftheelectron issomuch smaller thanthemassoftheproton). Fig.4. The Bohrmodel ofanatom, ‘5-5Thefieldofalinecharge Gauss’ lawcanbeused tosolve anumber ofelectrostatic field problems in- volving aspecial symmetry—usually spherical, cylindrical, orplanar symmetry. Intheremainder ofthischapter wewillapply Gauss’ lawtoafewsuch problems.‘Theeasewithwhichtheseproblemscanbesolvedmaygivethemisleading impres-sion that themethod isvery powerful, and that one should beable togoonto many other problems. Itisunfortunately notso. One soon exhausts thelistof problems that can besolved easily with Gauss’ law. Inlater chapters wewill develop more powerful methods forinvestigating electrostatic fields, | ‘Asourfirst example, weconsider asystem with cylindrical symmetry. Suppose that wehave avery long, uniformly charged rod. Bythiswemean that electric charges aredistributed uniformly along anindefinitely long straight line, with the charge }perunitlength. Wewishtoknow theelectric field. Theproblem can,of | « course, besolved byintegrating thecontribution tothefield from every part of ~~ theline. Wearegoing todoitwithout integrating, byusing Gauss’ law and some guesswork. First, wesurmise that theelectric field willbedirected radially outward fromtheline.Anyaxialcomponent fromcharges ononesidewouldbeaccom- GS paniedbyanequalaxialcomponent fromchargesontheotherside.Theresult Z)couldonlybearadialfield.Italsoseemsreasonablethatthefieldshouldhavetheguysiay C>7samemagnitude atallpointsequidistant fromtheline.Thisisobvious. (Itmay ‘SURFACE, unenot beeasy toprove, butitis true ifspace issymmetric—as webelieve iti.) Eine Wecan useGauss’ law inthefollowing way. Weconsider animaginary . surface intheshapeofacylinder coaxial withtheline,asshowninFig.5-5.totomaNias wre ‘According toGauss’ law, thetotal fluxofEfrom thissurface isequal tothecharge 7 inside divided by€9. Since thefield isassumed tobenormal tothesurface, the normal component isthemagnitude ofthefield. Let’s callitE.Also, lettheradius ofthecylinder ber,and itslength betaken asone unit, forconvenience. The flux through thecylindrical surface isequal toEtimes thearea ofthesurface, which is 2mr. The flux through thetwo end faces iszero because theelectric field istan- 33 ‘gentialtothem.Thetotalchargeinsideoursurfaceisjust4,becausethelengthoftheline inside isone unit. Gauss’ law then gives E-2ar =Nem d §Eos 62) SSTheelectricfieldofalinechargedependsinverselyonthefirstpowerofthe \uronuy distance from theline. NANGEDsneer\\ 5-6Asheetofcharge;twosheets‘Asanother example, wewillcalculate thefield from auniform plane sheet of S charge. Suppose that thesheet isinfinite inextent and that thecharge perunit { areaiso. Wearegoing totakeanother guess. Considerations ofsymmetry leadBNO INS ustobelievethatthefielddirectioniseverywhere normaltotheplane,andifwe*WL SS ‘havenofieldfromanyotherchargesintheworld,thefieldsmustbethesame(inPRX magnitude) oneachside. Thistimewechoose forourGaussian surface arec- ARSS gaussian tangular boxthatcutsthrough thesheet,asshowninFig.5-6.ThetwofacesSQ” NACE parallel tothesheetwillhaveequalareas,sayA.Thefieldisnormal tothesetwo faces,andparallel totheotherfour.ThetotalfluxisEtimestheareaofthefirst L face,plusEtimestheareaoftheoppositeface—withnocontributionfromthe SSother four faces. The total charge enclosed intheboxis¢A. Equating thefluxto thechargeinside,wehave A S za+Ba=24, yfromwhich peg. 63) fia.5-6. |Theelectric fieldneor a4simple butimportant result. cnnivinyGacktlowtocninegieer bo "Youmayrememberthatthesamereultwasobtainediaanearlerchapterbyanintegration over theentire surface. Gauss’ lawgives ustheanswer, inthis instance, much more quickly (although itisnotasgenerally applicable asthe earlier method). Weemphasize that thisresult applies only tothefield due tothecharges on ,thesheet. Ifthere areother charges intheneighborhood, thetotal field near the { i sheet would bethesumof(5.3) andthefield oftheother charges. Gauss’ law +| - ‘wouldthentellusonlythat*4 A+h=%, (64) © (0) exo #0 where E,and Eyarethefields directed outward oneach side ofthesheet. The problem oftwo parallel sheets with equal and opposite charge densities, \ +oand ~g, isequally simple ifweassume again that theoutside world isquite a symmetric. Either bysuperposing twosolutions forasingle sheet orbyconstruct- ingagaussian boxthat includes both sheets, itiseasily seen that thefield iszero outsideofthetwosheets(Fig.5~Ta).Byconsidering aboxthatincludes onlyone co)_|surfaceortheother,asin(b)or(c)ofthefigure,itcanbeseenthatthefieldbetween thesheets must betwice what itisforasingle sheet. The result is | E(betweenthesheets)=/€o, 3)|E(outside) =0. 6H | teil; 5-7Asphereofcharge;asphericalshellWehave already (inChapter 4)used Gauss’ lawtofind thefield outside a uniformly charged spherical region. The same method canalso give usthefield H i atpoints inside thesphere. Forexample, thecomputation canbeused toobtain ‘ ‘ 12good approximation tothefield inside anatomic nucleus. Inspite ofthefact Fig.5-7. Thefield between two thattheprotons inanucleusrepeleachother,theyare,becauseofthestrongnu- charged sheets iso/¢>- clear forces, spread nearly uniformly throughout thebody ofthenucleus. ay Suppose that wehave asphere ofradius Rfilled uniformly with charge. Let1pbethechargeperunitvolume.Againusingarguments ofsymmetry, weassume Y thefieldtoberadialandequalinmagnitudeatallpointsatthesamedistanceWe from thecenter. Tofindthefieldatthedistance rfrom thecenter, wetakea mrone spherical gaussian surface ofradiusr(r<R),asshowninFig.5-8.Thefluxout So ofthis surface is A aan, ! Thecharge insideourgaussian surface isthevolume insidetimesp,or Hl 4rr°p. A a) «4 Using Gauss’ law,itfollows thatthemagnitude ofthefieldisgiven by iN 5 Ex <®. (7) H ‘Youcanseethatthisformula gives theproper result forr=R.Theelectric field . isproportional totheradius andisdirected radially outward. Fig.5-8. Govs’ lawcanbeusedt0 Thearguments wehavejustgiven forauniformly charged sphere canbe findthefieldinside ouniformly chorged applied alsotoathinspherical shell ofcharge. Assuming thatthefieldisevery- sphere, where radial and isspherically symmetric, one gets immediately from Gauss’ lawthat thefield outside theshell islike that ofapoint charge, while thefield everywhere inside theshell iszero. (Agaussian surface inside theshell willcon- tain nocharge.) 5-8Isthefieldofapoint charge exactly 1/7"? Ifwelook inalittle more detail athowthefield inside theshell gtstobezero, wwecanseemore clearly why itisthat Gauss’ lawistrue only because thecoulomb force depends exactly onthesquare ofthedistance. Consider any point Pinside1uniformsphericalshellofcharge.ImagineasmallconewhoseapexisatPandwhich extends tothesurface ofthesphere, where itcuts outasmall surface area ‘Aa, asinFig.5-9. Anexacily symmetric cone diverging from theopposite side ofPwouldcutoutthesurfaceareaAas.IfthedistancesfromPtothesetwoele-‘ments ofarea arer;and r,theareas areintheratio day_3 ‘oy4a,~7 ‘i (You canshow thisbygeometry foranypoint Pinside thesphere.) Ifthesurface ofthesphere isuniformly charged, thecharge Aqoneach oftheelementsofareaisproportional tothearea,so P ge_Aaa AAg: Bay Coulomb's lawthensaysthatthemagnitudes ofthefieldsproduced atPbythese ‘aoe two surface elements are inthe ratio Boalt1gir Fig.5-9. Thefieldiszeroatany : . pointPinside@spherical shellofcharge. ‘Thefieldscancelexactly.Sinceallpartsofthesurfacecanbepairedofinthesame way, thetotal field atPiszero. Butyou canseethat itwould notbesoifthe exponent of7inCoulomb's law were notexactly two. ‘The validity ofGauss’ lawdepends upon theinverse square lawofCoulomb. Iftheforce lawwere notexactly theinverse square, itwould notbetrue that the field inside auniformly charged sphere would beexactly zero. For instance, ifthe force varied more rapidly, like, say, theinverse cube ofr,that portion ofthesur- face which isnearer toaninterior point would produce afield which islarger than ‘that which isfarther away, resulting inaradial inward field forapositive surface ss ‘charge. These conclusions suggest anelegant way offinding outwhether thein- verse square lawisprecisely correct. Weneed only determine whether ornotthe field inside ofauniformly charged spherical shell isprecisely zero. Itislucky that such amethod exists. Itisusually difficulttomeasureaphysical quantity tohigh precision—a one percent result may notbetoodifficult, but howwouldonegoaboutmeasuring, say,Coulomb's lawtoanaccuracyofonepartin2billion? Itisalmost certainly notpossible with thebest available techniques to measure theforce between two charged objects with such anaccuracy. But by determining only that theelectric fields inside acharged sphere aresmaller than some value wecan make ahighly accurate measurement ofthecorrectness ofGauss’law,andhenceoftheinversesquaredependence ofCoulomb's law.Whatconedoes, ineffect, iscompare theforce lawtoanideal inverse square. Such com- parisons ofthings that areequal, ornearly so,areusually thebases ofthemost precise physical measurements. How shall weobserve thefield inside acharged sphere? One way istotry tocharge anobject bytouching ittotheinside ofaspherical conductor. You know that ifwetouch asmall metal balltoacharged object andthen touch itto anelectrometer themeter will become charged and thepointer will move from zero (Fig. 5-10a). The ball picks upcharge because there areelectric fields outsidethechargedspherethatcausechargestorunonto(oroff)thelitteball.Ifyoudothesame experiment bytouching thelitle balltotheinside ofthecharged sphere, a ~\ youfindthat nocharge iscarried totheelectrometer. With such anexperiment=es youcaneasilyshowthatthefieldinsideis,atmost,afewpercentofthefieldout-side, and that Gauss’ lawisatleast approximately correct. TtappearsthatBenjaminFranklinwasthefirsttonoticethatthefieldinsidea .,conductingshelliszero.Theresultseemedstrangetohim.Whenhereportedhis menaron eacrmoeren ‘observation toPriestley, thelatter suggested thatitmight beconnected withan inverse square law, since itwas known that aspherical shell ofmatter produced nogravitational field inside. But Coulomb didn’t measure theinverse square dependence until 18years later, and Gauss' lawcame even later still. » — ‘Gauss’ lawhasbeen checked carefully byputting anelectrometer inside a 7 large sphere and observing whether any deflections occur when thesphere is charged toahigh voltage. Anullresult isalways obtained. Knowing thegeometry oftheapparatus and thesensitivity ofthemeter, itispossible tocompute the ‘minimum field that would beobserved. From thisnumber itispossible toplace anupperlimitonthedeviationoftheexponentfromtwo.Ifwewritethattheelec-{trostaticforcedependsonr~?*,wecanplaceanupperboundon¢.Bythismethod cla k‘Maxwelldetermined that¢waslessthan1/10,000.‘Theexperiment wasrepeated snatideS-RO.,Thelactic Redis20°0 andimproved uponin1936byPimptonandLaughton.TheyfoundthatCoulomb's insideclosedconducting shell exponentdiffersfromtwobylessthanonepartinabillion.Now that brings upaninteresting question: How accurate doweknow this ] Coulomb lawtobeinvarious circumstances? Theexperiments wejustdescribedmeasure thedependence ofthefield ondistance fordistances ofsome tens of centimeters. But what about the distances inside anatom—in the hydrogen atom, forinstance, where webelieve theelectron isattracted tothe nucleus by thesame inverse square law? Itistrue that quantum mechanics must beused for themechanical part ofthebehavior oftheelectron, buttheforce istheusual electrostatic one. Intheformulation oftheproblem, thepotential energy ofanelectronmustbeknownasafunctionofdistancefromthenucleus,andCoulomb'slawgives apotential which varies inversely with thefirst power ofthedistance. How accurately istheexponent known forsuch small distances? Asaresult of very careful measurements in1947 byLamb and Retherford onthe relative positions oftheenergy levels ofhydrogen, weknow that theexponent iscorrect, ‘again toonepart inabillion ontheatomic scale—that is,atdistances oftheorder ofoneangstrom (10~® centimeter). The accuracy ofthe Lamb-Retherford measurement was possible again because ofaphysical “accident.” Two ofthestates ofahydrogen atom are expected tohave almost indentical energies only ifthepotential varies exactly as1/r.Ameasurement wasmadeoftheveryslightdifferenceinenergiesbyfinding $6 thefrequencywofthephotonsthatareemittedorabsorbedinthetransitionfrom conestate totheother, using fortheenergy difference AE=fw. Computations showed that AEwould have been noticeably different from what was observed iftheexponentintheforcelaw1/r?differedfrom2byasmuchasonepartinabillion. Isthesame exponent correct atstillshorter distances? From measurements in nuclear physics itisfound that there areelectrostatic forces attypical nuclear distances—at about 10~"? centimeter—and thattheystillvaryapproximately as theinverse square. Weshall look atsome oftheevidence inalater chapter. Coulomb's law is,weknow, still valid, atleast tosome extent, atdistances ofthe order of10-9 centimeter. How about 10" centimeter? This range canbeinvestigated bybombarding protons with very energetic electrons and observing how they arescattered. Re- sults todate seem toindicate that the law fails atthese distances. The electrical force seems tobeabout 10times tooweak atdistances lessthan 10~'* centimeter. Now there aretwo possible explanations. One isthat theCoulomb lawdoes not work atsuch small distances; the other isthat our objects, the electrons and protons, arenotpoint charges. Perhaps either theelectron orproton, orboth, is some kind ofasmear. Most physicists prefer tothink that thecharge oftheproton issmeared. We know that protons interact strongly with mesons. This implies thataproton will, from time totime, exist asaneutron with ax*meson around it,Such aconfiguration would act—on theaverage—like alittle sphere ofpositive charge. Weknow thatthefield from asphere ofcharge does notvary asI/r?all theway into thecenter. Itisquite likely that theproton charge issmeared, but thetheory ofpions isstillquite incomplete, soitmay also bethat Coulomb's law fails atvery small distances. The question isstill open. ‘One more point: The inverse square lawisvalid atdistances like onemeter andalso at10~"°m; butisthecoefficient 1/4r¢o thesame? The answer isyes; atleast toanaccuracy of15parts inamillion. Wegoback now toanimportant matter that weslighted when wespoke of theexperimental verification ofGauss’ law. You may have wondered how the experiment ofMaxwell orofPlimpton and Laughton could give such anaccuracy unless thespherical conductor they used was aperfect sphere. Anaccuracy of ‘onepart inabillion isreally somethingtoachieve,andyoumightwellaskwhether they could make asphere which was that precise. There arecertain tobeslight irregularities inanyrealsphere andifthere areirregularities, willthey notproduce fields inside? Wewish toshow now that itisnotnecessary tohave aperfect sphere. Itispossible, infact, toshow that there isnofield inside aclosed conducting shell ofanyshape. Inother words, theexperiments depended on1/r2, buthadnothing todowith thesurface being asphere (except that with asphere itiseasier tocal- culate what thefields would beifCoulomb had been wrong), sowetake upthat subject now. Toshow this, itisnecessary toknow some oftheproperties of electrical conductors. 5-9 The fields ofaconductor Anelectrical conductor isasolid that contains many “free” electrons. The electrons can move around freely inthematerial, butcannot leave thesurface. Inametal there are somany free electrons that any electric field will setlarge numbers ofthem into motion. Either thecurrent ofelectrons sosetupmust be continually kept moving byexternal sources ofenergy, orthemotion ofthe electrons will cease asthey discharge thesources producing the initial 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.) Iftherewereanyfieldleft,thisfieldwouldurgestillmoreelectrons tomove;theonly electrostatic solution isthat thefield iseverywhere zero inside. Now consider theinterior ofacharged conducting object. (By“interior” we ‘mean inthemetal itself.) Since themetal isaconductor, theinterior field must 7 bezero,andsothegradientofthepotential¢iszero.Thatmeansthat¢doesnotvary from point topoint. Every conductor isanequipotential region, and its surface isanequipotential surface. Since inaconducting material theelectricfieldiseverywhere zero,thedivergence ofEiszero,andbyGauss’lawthechargedensityintheinterioroftheconductor mustbezero.Iftherecanbenochargesinaconductor,howcaniteverbecharged?What dowemeanwhenwesayaconductoris“charged”? Wherearethecharges? hhh Theansweristhattheyresideatthesurfaceoftheconductor, wherethereare strongforcestokeepthemfromleaving—theyarenotcompletely“free.”When Oy7 westudysolid-state physics, weshallfindthattheexcesschargeofanyconductorisontheaverage within oneortwo atomic layers ofthesurface. Forourpresent .ASS#E purposes, itisaccurate enough tosaythatifanychargeisputon,orin,aconductor o, itallaccumulates onthesurface;thereisnochargeintheinteriorofaconductor. art‘Wenotealsothattheelectricfieldjustoutsidethesurfaceofaconductor must .bbenormal tothesurface. There canbenotangential component. Ifthere were a , oc tangential component, theelectrons would move along the surface; there areno Sins e™*°" forces preventing that. Saying itanother way: weknow thattheelectric fieldlines ‘must always goatright angles toanequipotential surface.‘eceldj ‘Wecanalso,usingGauss’law,relatethefieldstrengthjustoutsideaconductorsdneeraceotedetect ibet0thelocaldensityofthechargeathesurface.Foragaussiansurface,wetakeaionaltothelocelsurfoce density of smallcylindrical boxhalfinside andhalfoutside thesurface, liketheoneshownprong inFig.5-11.Thereisacontribution tothetotalfluxofEonlyfromthesideofthe boxoutsidetheconductor. Thefieldjustoutsidethesurfaceofaconductor isthen Outside aconduetor: a e-<. (6.8) where oisthelocal surface charge density. Why does asheet ofcharge onaconductor produce adifferent field than justasheetofcharge?Inotherwords,whyis(5.8)twiceaslargeas(5.3)?Thereason, ofcourse, isthat wehave nof said fortheconductor that there are no“other”chargesaround.Theremust,infact,besometomakeE=0intheconductor.‘Thecharges intheimmediate neighborhood ofapoint Ponthesurface do,infact, Bive afield Eiseai =Giocai/2¢o both inside and outside thesurface. But allthe restofthecharges ontheconductor “conspire” toproduce anadditional field at thepoint Pequal inmagnitude toEgat. The total field inside goes tozero and thefield outside to2Bient =6/€0- 5-10Thefieldinacavityofaconductor Wereturn now totheproblem ofthehollow container—a conductor with a cavity.Thereisnofieldinthemecal,butwhataboutinthecavity?Weshallshow PALLYthatifthe cavity isempty then there arenofields init, nomatter what theshape of A theconductororthecavity—sayfortheoneinFig.5-12.Consideragaussian Vi,“)surface,likeSinFig.5-12,thatenclosesthecavitybutstayseverywhereinthe Uyconducting material. Everywhere onSthefield iszero, sothere isnofluxthrough ‘SandtheroralchargeinsideSis2ero.Forasphericalshell,onecouldthenargue Ge. from symmetry that there could benocharge inside. But, ingeneral, wecanonly Ge saythatthereareequalamountsofpositiveandnegativechargeontheinnersurface ofthe conductor. There couldbeapositivesurfacechargeononepart *LB Pandanegativeonesomewhereelse,asindicatedinFig.512.Suchathingcannot /beruledoutbyGauss"law. ss, ‘Whatreallyhappens,ofcourse,isthatanyequalandoppositechargesonX theinner surface would slide around tomeet each other, cancelling outcompletely. Wecanshow that they must cancel completely byusingthelawthatthecirculationFig.5-12. What isthefieldinon ofEisalways zero(clectrostatics). Suppose therewerecharges onsomepartsofemptycavityofconductors foronytheinnersurface.Weknowthattherewouldhavetobeanequalnumberofop- thopetposite charges somewhere else. Now any lines ofEwould have tostart onthe 38 positive charges andendonthenegative charges (since weareconsidering only the case that there arenofree charges inthecavity). Now imagine aloop I'that crosses thecavity along alineofforce from some positive charge tosome negative charge, and returns toitsstarting point viatheconductor (asinFig. 5-12). The integralalongsuchaTineofforcefromthepositivetothenegativechargeswouldnotbezero. The integral through themetal iszero, since E=0.Sowewould have §Eds0277 But theline integral ofEaround any closed loop inanelectrostatic field isalways zero. Sothere canbenofields inside theempty cavity, noranycharges onthe inside surface, ‘You should notice carefully one important qualification wehave made. Wehave always said “inside anempty” cavity. Ifsome charges areplacedatsome fixed locations inthecavity—as onaninsulator oronasmall conductor insulated from themain one—then there canbefields inthecavity. Butthen that isnotan “empty” cavity. We have shown that ifacavity iscompletely enclosed byaconductor, no static distribution ofcharges outside can ever produce any fields inside. This explains theprinciple of“shielding” electrical equipment byplacing itinametal cat,Thesgggg canbewedtoshowthatsaehtibution ofsharps inside aclosed conductor can produce any fields outside. Shielding works both ways! Inelectrostatics—but notinvarying fields—the fields onthetwo sides ofa closed conducting shell arecompletely independent. Now you seewhy itwas possible tocheck Coulomb’s law tosuch agreat precision. The shape ofthehollow shell used doesn’t matter. Itdoesn’t need to bespherical; itcould besquare! IfGauss’ law isexact, thefield inside isalways zero. Now youalso understand why itissafe tositinside thehigh-voltage terminal ofamillion-volt van deGraaff generator, without worrying about getting a shock—because ofGauss” law. 39 6 The Electric Field in Various Circumstances 6-1 Equations oftheelectrostatic potential This chapter will describe the behavior oftheelectric field inanumber of 6-1 Equations oftheelectrostatic different circumstances. Itwillprovide some experience with theway theelectric potential field behaves, and will describe some ofthe mathematical methods which are i sede heathisehh 6-2 Theelectric dipole Webegin bypointing outthat thewhole mathematical problem isthesolution 6-3_Remarks onvector equations oftwoequations, theMaxwell equations forelectrostatics: 6-4Thedipelepotential 2 ve=2, (6.1) gradient 6-5Thedipoleapproximation for eXE=0. 62) anarbitrary distribution Infact,thetwocanbecombined intoasingleequation.Fromthesecondequation, 66‘Thefieldsofcharged weknow atonce that wecandescribe thefield asthegradient ofascalar (see conductors Section3-7): E=~vo. (63) 6-7Themethodofimages Wemay, ifwewish, completely describe anyparticular electric field interms 6-8 Apoint charge near a ofits potential ¢.Weobtain thedifferential equation that ¢must obey bysub- ‘conducting plane stituting Eq.(6.3)into(6.1),toget 6-9Apointcharge neara vive =P. 4) ‘conducting sphere Thedivergence ofthegradient of@isthesameasVoperating on¢: 6-10Condensers; parallel plates5%.o%.0% 6-11High-voltage breakdown Weve==Beet Gye gee” (65) 642‘Thefieldemission microscope wewrite Eq. (6.4) as “°#6) ve 2. (66) ‘0 Revew. Chapter 23,Vol.1,Resonance Theoperator V?1scalled theLaplacian, andEq_(66)1scalled thePoisson equa- tion. The entire subject ofelectrostatics, from amathematical point ofview, 1smerelyastudyofthesolutionsofthesingleequation(6.6).Once¢1sobtainedbysolving Eq.(6.6) wecanfind Eimmediately from Eq.(6.3). Wetakeupfirstthespectalclassofproblemsinwhichpisgwenasafunction ofx,y,z. Imthat case theproblem 1salmost trivial, forwealready know the solution ofEq.(6.6) forthegeneral case. Wehave shown that ifpisknown at every point, thepotential atpoint (1)is = [eae where p(2) isthecharge density, dV isthevolume element atpoint (2), and rzisthedistancebetweenpoints(I)and(2).Thesolutionofthedifferentialequation (6.6)isreducedtoaninregrationoverspace.Thesolution(6.7)shouldbeespecially noted, because there aremany situations inphysics that lead toequations like 1?(Gomething) =(something else), andEq.(6.7) 18aprototype ofthesolution forany ofthese problems. Thesolution ofelectrostatic field problems isthus completely straightforwardwhenthepositionsofall thecharges areknown. Let’s seehow itworks inafew examples 1 6-2 The electric dipole z First, take two point charges, +qand —g,separated bythedistance d.Let thez-axis gothrough thecharges, and pick theorigin halfway between, asshown Ptey.2)inFig. 6-1. Then, using (4.24), thepotential from thetwo charges isgiven by a 90%, 9,2) 1 q =4q — ~aliments erase! OO Wearenotgoing towrite outtheformula fortheelectric field, butwecan always calculate itonce wehave thepotential. Sowehave solved theproblem oftwo y charges. “a‘There isanimportant special case inwhich thetwo charges arevery close + together—which istosaythatweareinterested inthefields onlyatdistances from thecharges large incomparison with their separation. Wecallsuch aclose pair ofcharges adipole. Dipoles arevery common. , A“dipole”antennacanoftenbeapproximatedbytwochargesseparatedbya eeengthedinercedsport"8°*smalldistance—if wedon’taskaboutthefieldtooclosetotheantenna.(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 any material, theelectrons and protons feel opposite forces and aredisplaced relative toeach other. Inaconductor, you remember, some oftheelectrons move tothesurfaces, sothat thefield inside becomes zero. Inaninsulator the electronscannotmoveveryfar;theyarepulledbackbytheattraction ofthenu-cleus. They do, however, shift alittle bit. Soalthough anatom, ormolecule,remainsneutralinanexternalelectricfield,thereisaverytinyseparation ofitspositive and negative charges and itbecomes 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 and anetpositive = charge oneach ofthetwo hydrogen atoms, which arenotplaced symmetrically butasinFig.6-2.Although thechargeofthewholemolecule iszero,thereisacharge distribution with alittle more negative charge onone side and alittle more positive charge ontheother. ‘This arrangement iscertainly notassimple ‘astwopointcharges,butwhenseenfromfarawaythesystemactslikeadipole. (*) (+) Asweshallseealittlelater,thefieldatlargedistancesisnotsensitivetothe + + fine details. :Let’slook,then,atthefieldoftwooppositechargeswithasmallseparation ms62.ThewatermolecM0.d.Ifdbecomeszero,thetwochargesareontopofeachother,thetwopotentials thontee"oreSineelectronoouethecancel,andthereisnofield.Butiftheyarenotexactlyontopofeachother,we‘oxygen, slightly more cangetagoodapproximation tothepotential byexpanding theterms of(6.8)in ‘apower series inthesmall quantity d(using thebinomial expansion). Keeping terms only tofirst order ind,wecan write (---2-2 Itisconvenient towrite Pty tea? Then ?22 226 (:-9+xe4+y~eP-de=r (-#). and _—ees -vesteg 1-9)”: VE~ GDP + +? VFI dir 2 62 Using thebinomial expansion again for[1—(zd/r2)[-"!*—and throwing away terms with higher powers than thesquare ofd—we get 1 Lzr(:+3%) Similarly,similarly Lo(1-14) Ver@>Pprety + IP, Thedifference ofthese twoterms gives forthepotential 4(6,9,2) =gioSaad 69) 92)= ares 78 : The potential, and hence thefield, which isitsderivative, isproportional toad, theproduct ofthecharge and theseparation. This product isdefined asthe dipole moment ofthetwo charges, forwhich wewill usethesymbol p(donot confuse with momentum!): pa ad. (6.10) Equation (6.9) can also bewritten as _1peose (x,¥2)=tre (6.11) sincez/r=cos6,where0istheanglebetweentheaxisofthedipoleandtheradius vector tothepoint (x,y,z}—see Fig.6-1. Thepotential ofadipole decreases P asI/r? foragiven direction from theaxis (Whereas forapoint charge itgoes as 1/r). Theelectric field Eofthedipole willthen decrease as1/r’. 1 ‘Wecanputourformula intoavector form ifwedefine p2savector whose H magnitude ispandwhose direction isalong theaxisofthedipole, pointing from a7 q—toward q,.Then H cos0=pen, (6.12) Herwhere e,18theunitradial vector (Fig,6-3). Wecanalsorepresent thepoint ° (xy, 2)byr.Then Fig.6-3. Vector notation for0 dipole. Dipole potential: _ltpe lope on pole. WO~Se. Arey 13) This formula 1svalid foradipole with any orientation and position ifrrepresents thevector from thedipole tothepoint ofinterest. Ifwewant theelectric field ofthedipole wecan getitbytaking thegradient ofg.For example, thez-component ofthefield 1s—49/82. Foradipoleoriented along thez-axis wecan use(6.9): ~%._po(z)__ ip(L_ 2~~ meg G2 \r8)~~ Amen \r 78)” “ 3.cos?@—1 __P3cos?9— B= 6.14) The x-and y-components are =2 sx =2 32.Beire Ptarere These two can becombined togive one component directed perpendicular tothe z-axis, which wewill call thetransverse component E1: Ei.=VET B=Eghvere or =P Scossingd Eire 8 (615) os The transverse component E,isinthex-y plane and points directly away from theaxis ofthedipole. ‘The total field, ofcourse, is E=VE+E. The dipole field varies inversely asthecube ofthedistance from thedipole.Ontheaxis,at9=0,itistwiceasstrongasat@=90°,Atbothofthesespecialangles theelectric field hasonly az-component, butofopposite sign atthewo places (Fig. 6-4). 63 Remarks onvector equations This isagood place tomake ageneral remark about vector analysis. The fundamental proofs canbeexpressed byelegant equations inageneral form, but inmaking various calculations and analyses itisalways agood idea tochoose theaxes insome convenient way. Notice that when wewere finding thepotential 2¢,_ofadipote wechosethez-axisalongthedirectionofthedipole,ratherthanatsome Qso ‘arbitrary angle.Thismadetheworkmucheaster.Butthenwewrotetheequationsy= invectorformsothattheywouldnolongerdependonanyparticular coordinate(QO) system.Afterthat,weareallowedtochooseanycoordinatesystemwewish, knowing that therelation 1s,ingeneral, true. Itclearly doesn’t make any sense to bother with anarbitrary coordinate system atsome complicated angle when you ‘can choose aneat system fortheparticular problem—provided that theresult can finally beexpressed asavector equation, Sobyallmeans take advantage ofthe factthat vector equations areindependent ofanycoordinate system. Ontheotherhand,ifyouaretryingtocalculatethedivergence ofavector, fig.6-4.Theelectricfieldof©insteadofjustlookingat¥-Eandwonderingwhatitis,don’tforgetthatstcan dipole. always bespread outas OE,,Ey,Esxty te Ifyou can then work out thex-, and z-components oftheelectric field and differentiate them, youwillhave thedivergence. There often seems tobeafeeling, that there issomething inelegant—some kind ofdefeat involved—in writing out thecomponents; that somehow there ought always tobeaway todoeverything with thevector operators. There isoften noadvantage toit.The first time we ‘encounter aparticular kind ofproblem, itusually helps towrite outthecomponents tobesure weunderstand what isgoing on. There isnothing inelegant about put- ting numbers into equations, andnothing inelegant about substituting thederiva- tuves forthefancy symbols. Infact, there isoften acertain cleverness indoing justthat. Ofcourse when youpublish apaper inaprofessional journal itwilllook better—and bemore easily understood—ifyoucanwriteeverythinginvectorform. Besides, itsaves print. 6-4 The dipole potential asagradient Wewould like topoint outarather amusing thing about thedipole formula, Eq.(6.13). The potential canalso bewritten as 1 v Ifyoucalculate thegradient ofI/r,you get (=-4--%,r aR and Eq. (6.16) isthesame asEq.(6.13). How didwethink ofthat? Wejustremembered thate,/r® appeared inthe formula forthefield ofapoint charge, and that thefield was thegradient ofa potential which hasa1/rdependence. o4 There isaphysical reason forbeing able towrite thedipole potential inthe form ofEq.(6.16). Suppose wehave apoint charge qattheorigin. The potential atthepoint Pat(x,»,2)is =f.= 4 (Let’s leave offthe 1/47¢) while wemake these arguments; wecan stick itinat theend.) Now ifwemove thecharge +gupadistance Az,thepotential atPwill changealittle,by,say,Ag.Howmuchis4g?Well,itisjusttheamountthatthepotential would change ifwewere toleave thecharge attheorigin andmove Pdownward bythesamedistance Az(Fig.6-5).Thatis, = p 860 Tad bby=—BPaz, yZe fv where byAzwemean thesame asd/2. So,using @=g/r, wehave that thepo- “Y tential fromthepositive charge is he ag 4(9)4. re Applying thesame reasoning forthepotential from thenegative charge, wecan write % 744 8(-9)4. o=sted (eo)q (6.18) iali Fig.6-5.Thepotential atPfroma ‘Thetotalpotential isthesumof(6.17)and(6.18): pointcharge atAzobovetheorigins the a sameasthepotential atP’(AzbelowP) b=+e-= 5(Q)ad (6.19)fromthesamechargeattheorigin. a(t“-&()a. For other orientation ofthedipole, wecould represent thedisplacement of thepositive charge bythevector Ar,. Weshould then write Eq.(6.17) as G4=—Vd0°Ary, where Aristhen tobereplaced byd/2. Completing thederivation asbefore, Eq.(6.19) would then become r e=-0 (Sa This isthesame asEq.(6.16), ifwereplace qd=p,and putback the1/47€9. Looking atitanother way, weseethat thedipole potential, Eq. (6.13), can be interpreted as = —p Ve, (6.20) where ®=1/4zéor isthepotential ofaunitpointcharge. Although wecanalways find thepotential ofaknown charge distribution by anintegration, itissometimes possible tosave time bygetting theanswer with a clever trick. Forexample, onecanoften make useofthesuperposition principle. Ifwearegiven acharge distribution that canbemade upofthesum oftwo dis- tributions forwhich thepotentials arealready known, itiseasy tofind thede~ sired potential byjust adding thetwo known ones. One example ofthisisour derivation of(6.20), another isthefollowing. ‘Suppose wehave aspherical surface with adistribution ofsurface charge that varies asthecosine ofthepolar angle. The integration forthisdistribution is fairly messy. But, surprisingly, such adistribution can beanalyzed bysuper- position. For imagine asphere with auniform volume density ofpositive charge, and another sphere with anequal uniform volume density ofnegative charge, 6s Fig.6-6.TwouniformlychargedC7) spheres, superposed wih sight dspace: tren, oe equvelen! to noniform St NZ SEZ distribution ofsurface charge. (9) + (b) = (c) originally superposed tomake aneutral—that 1s,uncharged—sphere. Ifthe positive sphere isthen displaced slightly with respect tothenegative sphere, thebodyofthe uncharged sphere would remain nevtral, butalittle positive charge will appear ononeside, andsome negative charg. willappear ontheopposite side,asillustrated inFig.6-6.Iftherelativedisplacement ofthetwospheres1ssmall, thenetcharge isequivalent toasurface charge (onaspherical surface), and the surface charge density will beproportional tothecosine ofthepolar angle. Now ifwewant thepotential from this distribution, wedonotneed todoan integral. We know that thepotential from each ofthespheres ofcharge s—for points outside thesphere—the same asfrom apoint charge. The two displaced spheres arelike two point charges; thepotential 1sjust that ofadipole. Inthis way you can show that acharge distribution onasphere ofradius a with asurface charge density a=aces producesafieldoutsidethespherewhichisjustthatofadipolewhosemomentis Arooa®p=Staal. Itcanalso beshown that inside thesphere thefield isconstant, with thevalue B-$2. If6istheangle from thepositive z-axis, theelectric field inside thesphere 1sinthe negative z-direction. The example wehave just considered 1snotasartificial as itmay appear; wewillencounter itagain inthetheory ofdielectrics. 6-5 The dipole approximation foranarbitrary distribution ‘The dipole field appears inanother circumstance both interesting and im- portant. Suppose that wehave anobject that has acomplicated distribution of charge—like thewater molecule (Fig. 6-2)—and weareinterested only inthe fields faraway. Wewillshow that itispossible tofind arelatively simple expression forthefields which isappropriate fordistances large compared with thesize of theobject.‘Wecanthinkofourobyectasanassemblyofpointchargesq,nacertainlinuted region, asshown inFig. 6-7. (We can, later, replace q,bypdVifwewish.) Let each charge q,belocated atthedisplacement d,from anorigin chosen somewhere P 6 osBSf rtSs Fig.6-7.Computation ofthepor SE]oe/ tential atapoint Patalarge distancetromestofchargesos inthemiddleofthegroupofcharges. What isthepotential atthepoint P,located atR,where Rismuch larger than themaximum d,? The potential from the whole collection isgiven by _l qeoiez=oad (6.21) where r,isthedistance from Ptothecharge g,(the length ofthevector R—d,). ‘Now ifthedistance from thecharges toP,thepoint ofobservation, isenormous, each ofther,'scan beapproximated byR.Each term becomes g,/R, and we can take 1/R outasafactor infront ofthesummation. This gives usthesimple result -,! -2Fre REYTreR? 2) whereQisjustthetotalchargeofthewholeobject.Thuswefindthatforpoints farenough from any lump ofcharge, thelump looks like apoint charge. The result isnottoosurprising. But what ifthere areequal numbers ofpositive and negative charges? Then thetotal charge Qoftheobject iszero. This 1snotanunusual case; infact, aswe know, objects areusually neutral. ‘The water molecule isneutral, butthecharges arenotallatonepoint,soifwearecloseenough weshouldbeabletoseesome effects oftheseparate charges. Weneed abetter approximation than (6.22) for thepotential from anarbitrary distribution ofcharge inaneutral object. Equation (6.21) isstill precise, butwecan nolonger just setr,=R.Weneed amore accu- rate expression forr,.Ifthepoint Pisatalarge distance, r,will differ from Rto anexcellent approximation bytheprojection ofdonR,ascan beseen from Fig. 6-7. (You should imagine that Pisreally farther away than isshown inthe figure.) Inother words, ife,istheunit vector inthedirection ofR,then ournext approximation tor,is ne R~ dee, (6.23) ‘What wereally want is1/r,, which, since d,<R,canbewritten toourapproxima- tion as 14 4herROR(1+42). (6.24) Substituting this in(6.21), wegetthat thepotential is ~(2 4,fc). o>ire(g+zxRe+ (6.25) The three dots indicate the terms ofhigher order ind/R that wehave neglected. These, aswell astheones wehave already obtained, aresuccessive terms inaTaylor expansion ofI/r,about1/Rinpowers ofd,/R. The first term in(6.25) iswhat wegot before; itdrops out iftheobject 18 neutral. Thesecond term depends on1/R®, justasforadipole. Infact, ifwedefine p= lad (6.26) asaproperty ofthecharge distribution, thesecond term ofthepotential (6.25) is _1pe 6aeee 21) precisely adipole potential. The quantity piscalled thedipole moment ofthe distribution. Itisageneralization ofour earlier definition, and reduces toitfor thespecial case oftwo point charges. Our result isthat, farenough away from any mess ofcharges that isasa whole neutral, thepotential isadipole potential. Itdecreases as1/R? andvaries ‘ascos@—and itsstrength depends onthedipole moment ofthedistribution of charge. Itisforthese reasons that dipole fields areimportant, since the simple case ofapairofpointcharges isquiterare. o7 ‘The water molecule, forexample, hasarather strong dipole moment. The electric fields that result from this moment areresponsible forsome oftheim- portant properties ofwater. For many molecules, forexample CO,, thedipole moment vanishes because ofthesymmetry ofthemolecule. For them weshould expand still more accurately, obtaining another term inthe potential which de- creases as1/R*, andwhich iscalled aquadrupole potential. Wewilldiscuss such cases later. 6-6 The fields ofcharged conductors We have now finished with theexamples wewish tocover ofsituations in sn ee which thecharge distributions isknown fromthestart, IthasbeenaproblemKa oo without seriouscomplications, involving atmostsomeintegrations. Weturn » now toanentirely new kind ofproblem, thedetermmation ofthefields near charged conductors. <funy > ‘SupposethatwehaveasituationinwhichatotalchargeQisplacedonanSent arbitrary conductor. Nowwewillnotbeabletosayexactly wherethecharges\Lone) ,are,Theywillspreadoutinsomewayonthesurface.Howcanweknowhow|.>thechargeshavedistributedthemselvesonthesurface?Theymustdistribute (te)themselves sothat thepotential ofthesurface isconstant. Ifthesurface were not > anequipotential, there would beanelectric field inside theconductor, and the aa a charges would keepmoving untilitbecame zero. Thegeneral problem ofthis= i} ~ kindcanbesolvedinthefollowing way.Weguessatadistribution ofchargeand 1 calculate thepotential. Ifthepotential turns outtobeconstant everywhere on thesurface, theproblem isfinished. Ifthesurface 1snot anequipotential, we Fig.6-8. Thefieldlinesandequipo- have guessed thewrong distribution ofcharges, andshould guess again—hopefully tentials fortwopointcharges. withanimproved guess! Thiscangoonforever, unless wearejudicious about thesuccessive guesses. The question ofhow toguess atthedistribution 1smathematically difficult. Nature, ofcourse, hastime todoit;thecharges push andpulluntil they allbalance themselves. When wetrytosolve theproblem, however, ittakes ussolong to make each trial that that method isvery tedious With anarbitrary group of conductors and charges theproblem can bevery complicated, and ingeneral 1t cannot besolved without rather elaborate numerical methods. Such numerical computations, these days, aresetuponacomputing machine that will dothe work forus,once wehave told ithow toproceed, ‘Ontheother hand, there arealotoflittle practical cases where itwould 1 benicetobeabletofindtheanswer bysome more direct method—without having towrite aprogram foracomputer. Fortunately, there areanumber ofcases where theanswer canbeobtained bysqueezing itoutofNature bysome trick orother. Q The first trick wewill describe involves making useofsolutions wehave already ( obtained forsituations inwhich charges have specified locations. conoucToR 6-7Themethod ofimages Z Wehavesolved,forexample, thefieldoftwopointcharges. Figure6-8 shows some ofthefield lines and equipotential surfaces weobtained bythecom- Fig.6-9. Thefield outside ©con- putations inChapter 5.Now consider theequipotential surface marked 4.Sup- ductor shaped liketheequipotential A pose wewere toshapeathinsheetofmetalsothatitjustfitsthissurface. Ifwe ofFig.6-8. place itrightatthesurface andadjust itspotential totheproper value, noone would ever know itwasthere, because nothing would bechanged. But notice! We have really solved anew problem. We have asituation in which thesurface ofacurved conductor with agiven potential isplaced near a point charge. Ifthemetal sheet weplaced attheequipotentral surface eventuallyclosesonitself(or,inpractice,ifitgoesfarenough)wehavethekindofsituation considered inSection 5-10, inwhich ourspace isdivided into two regions, one inside and one outside aclosed conducting shell. Wefound there that thefields in thetwo regions arequite independent ofeach other. Sowewould have thesame fields outside ourcurved conductor nomatter what 1sinside. Wecaneven fillup os thewhole inside with conducting material. Wehave found, therefore, thefields forthearrangement ofFig. 6-9. Inthespace outside theconductor thefield is just like that oftwo point charges, asinFig. 6-8. Inside theconductor, itiszero ‘Also—as itmust be—the electric field just outside theconductor isnormal to the surface. Thus wecan compute thefields inFig. 6-9 bycomputing thefield due toq and toanimaginary point charge —gatasuitable point. The point charge we “imagine” existing behind theconducting surface iscalled animage charge. Inbooks you can find long lists ofsolutions forhyperbolic-shaped conductors and other complicated looking things, and you wonder how anyone ever solved these terrible shapes. They were solved backwards! Someone solved asimple problem with given charges. Hethen saw that some equipotential surface showed upina new shape, and hewrote apaper inwhich hepointed outthat thefield outside that particular shape can bedescribed inacertain way. 6-8 Apoint charge near aconducting plane Asthesimplest application oftheuseofthis method, let's make useofthe plane equipotential surface BofFig. 6-8. With it,wecan solve theproblem ofa charge infront ofaconducting sheet.Wejustcrossouttheleft-hand halfofthe picture. The field lines forour solution areshown inFig. 6-10. Notice that the plane, since stwas halfway between thetwo charges, haszero potential. Wehave solved theproblem ofapositivechargenexttoagroundedconducting sheet. Wehave now solved forthetotal field, butwhat about thereal charges that areresponsible forit? There are, inaddition toour positive point charge, some induced negative charges ontheconducting sheet that have been attracted bythe positive charge (from large distances away). Now suppose that forsome technical reason—or out ofcuriosity—you would like toknow how thenegative charges aredistributed onthesurface. You can find thesurface charge density byusing the result we worked out inSection 5-6 with Gauss’ theorem. The normal com- it S \ 1 foe \tlre ,\thas SNO cS ty?s SOV \hor, hynS XN \\I,aS heSN IRwooo RES ———macecnanctos—- ey 2 FOL TIIWAN ORYTPS SNS 4yori NN'Syor TEAS IN/yyy VyS / ) byy NSNpyELV SspTLV\S‘\ \“ES Fig. 6-10, Thefield ofacharge near aplane conducting surface, found bythe method ofimages, 69 ponentoftheelectricfieldjustoutsideaconductor isequaltothedensityofsurfacechargeodividedbyé»,Wecanobtainthedensityofchargeatanypointonthe surface byworking backwards from thenormal component oftheelectric field at thesurface. Weknow that, because weknow thefield everywhere. Consider apoint onthesurface atthedistance pfrom thepoint directly be- neath thepositive charge (Fig. 6-10). The electric field atthis point isnormal to thesurface andisdirected into st.‘The component normal tothesurface ofthe field from thepositive point charge is ot oqExt” ~egEOP (6.28) Tothiswemust addtheelectric field produced bythenegative image charge. That just doubles thenormal component (and cancels allothers), sothecharge density ©atany point onthesurface is - — ag ; O10)=e080)=~geargna +629) Aninteresting check onourwork Istointegrate over thewhole surface. We find that thetotal induced charge 1s—g, asitshould be. . ‘Onefurtherquestion:Isthereaforceonthepointcharge?Yes,becausethere 18anattraction from theinduced negative surface charge onthe plate. Now that weknow what thesurface charges are(from Eq.(6.29)), wecould compute the force onourpositive point charge byanintegral. Butwealso know that theforce acting onthepositive charge isexactly thesame asitwould hewith thenegativeimagechargeinsteadoftheplate,becausethefieldsintheneighborhood arethesame inboth cases. The point charge feels aforce toward theplate whose magni- tude is -,#476, Qa 6% Wehave found theforce much more easily than byintegrating over allthenega- tive charges. 6-9 Apoint charge near aconducting sphere ‘Whatothersurfacesbesidesaplanehaveasimplesolution? ‘Thenextmost ~< simple shape 1sasphere. Let’s find thefields around ametal sphere which hasa;IX “ pointchargeqnearit,asshowninFig.6-11.Nowwemustlookforasimplere physical situation which givesasphere foranequipotential surface. IfwelookSN ©aroundatproblemspeoplehavealreadysolved,wefindthatsomeonehasnoticedYaa thatthefieldoftwounequal point charges hasanequipotential thatisasphere Aha’ Ifwechoose the location ofanimage charge-—and pick the right amount ofcharge—maybe wecan make theequipotential surface fitoursphere. Indeed, * itcan bedone with thefollowing prescription, Fig.6-11. Thepoint charge qin ‘Assume thatyouwanttheequipotential surface tobeasphere ofradius @ duces chargeson2groundedconducting withitscenteratthedistancebfromthechargeg.Putanimagechargeofstrength Sphere whose fields are those ofan q’=—q(a/b) ontheline from thecharge tothecenter ofthesphere, and ata image charge q’placed atthepoint distance a/b from thecenter. Thesphere willbeatzero potential,shown. ‘Themathematical reasonstemsfromthefactthatasphereisthelocusofallpoints forwhich thedistances from two points areinaconstant ratio Referring toFig. 6-II, thepotential atPfrom qand q’isproportional to The potential will thus bezero atallpoints forwhich 49 gy Bl’ noon no | 6-10 Ifweplace q’atthedistance a*/b from thecenter, theratio r2/r, hastheconstant value a/b. Then if . “oag° 8 (6.31) thesphere 1sanequipotential. Itspotential 1s,infact, zero. ‘What happens ifweareinterested inasphere that 1snotatzero potential? ‘That would besoonly ifitstotal charge happens accidentally tobeg’ Ofcourse if'it isgrounded, thecharges induced onitwould have tobejust that. Butwhat ifit isinsulated, and wehave putnocharge on1t?Orifweknow that thetotal charge Qhasbeen putonit?Orjust that sthasagiven potential norequal tozero? All these questions areeasily answered. Wecan always add apoint charge g’atthe center ofthesphere The sphere still remains anequipotential bysuperposition: only themagnitude ofthepotent wall bechanged Ifwehave, forexample, aconducting sphere which 1sinitially uncharged and insulated from everything else, and webring near toitthe positive port charge q.thetotal charge ofthesphere will remain zero. The solution isfound byusing animage charge q’asbefore, but, inaddition, adding acharge q”atthe center ofthesphere, choosing =4=5a (6.32) The fields everywhere outside thesphere aregiven bythesuperposition ofthe fields ofg.q’,andq”. The problem issolved. We can seenow that there will beaforce ofattractton between thesphere and thepomt charge g.It1snot zero even though there 1snocharge ontheneutral sphere. Wheredoestheattraction comefrom?Whenyoubringapositive charge uptoaconducting sphere, the positive charge attracts negative charges tothe side closer toitself and leaves positive charges onthesurfitce ofthefarside. The attraction bythenegative charges exceeds therepulsion from thepositive charges. there 1sanetattraction. Wecanfind outhow large theattraction 1sbycomputing theforce onqinthefield produced byq’and q”. The total force 1sthesum ofthe attractive force between qand achargeq’=—(a/b)q.atthedistanceb—(a*/h), and therepulsive force between qand acharge q’=+(a/b)q atthedistance b. Those who were entertained inchildhood bythebaking powder box which hasonitslabel apicture ofabakingpowderboxwhichhasonitslabelapicture ‘ofabaking powder box which has. may beinterested inthefollowing problem. Two equal spheres, onewith atotal charge of+@ and theother with atotal charge of—Q. areplaced atsome distance from each other. What 1stheforce between them? The problem can besolved with aninfinite number ofimages. One first approximates each sphere byacharge atitscenter. These charges will have mage charges intheother sphere. The image charges will have images, etc,ete,etc ‘The solution 1slike thepicture onthebox ofbaking powder—and itconverges pretty fas anoea +4 6-10Condensers; parallel plates ae Fe,wkspronatsbieataphnvosbspsinticns Cuan Seggebaetharerdeent two large metal plates which areparallel toeach other and separated byadistance smallcompared withtheirwidth. Let'ssuppose thatequalandopposite charges Fig.6-12 Aparallel-plate —con- have been put ontheplates. The charges oneach plate will beattracted bythe denser. charges ontheother plate, and thecharges willspread outuniformly ontheinner surfaces oftheplates. The plates will have surface charge densities +oand —o, respectively. asinFig. 6-12. From Chapter 5weknow that thefield between the plates 15@/€), and that thefield outside theplates 1szero, The plates will have different potentials @)and $2. For convenience wewill call thedifference Vit isoften culled the“voltage”: orm ba (You will find that sometimes people use Vforthepotential, butwehave chosen touse¢.) on The potential difference Visthework per unit charge required tocarry @ small charge from one plate totheother, sothat a, 4 v=e=fa~- 40, (633) where + 1sthetotal charge oneach plate, A1sthearea oftheplates, and dis theseparation. Wefindthat thevoltage isproportional tothecharge, Such aproportionality between Vand Q1sfound forany two conductors inspace ifthere 1saplus charge fonone and anequal minus charge ontheother. The potentsl difference betweenthem—that is,thevoltage—will beproportional tothecharge.(Weareassumingthat there arenoother charges around.) Why this proportionality? Just thesuperposition principle. Suppose we know thesolution forone setofcharges. and then wesuperimpose two such solutions, The charges aredoubled, thefields aredoubled, and thework done in carrying aunit charge from one point totheother isalso doubled. ‘Therefore the potential difference between any two points isproportional tothecharges. In particular, the potential difference between thetwo conductors 1sproportional tothecharges onthem. Someone originally wrote theequation ofproportionality theother way. ‘That 1s,they wrote o-cy, where Cis aconstant This coefficient ofproportionality 1scalled thecapacity.andsuchasystemoftwo conductors 1scalled acondenser.* For ourparallel-plate condenser c=£4parallel plates). (6.34) This formula isnotexact, because thefield 1snotreally unsform everywhere between theplates, asweassumed. The field does notjust suddenly quit atthe edges, but really 1smore asshown inFig 6-13. The total charge isnot#4, aswe have assumed—there 1salitle correction fortheeffects attheedges. Tofind out what the correction 1s,wewill have tocalculate the field tore exactly and find out just what does happen attheedges. That 1sacomplicated mathematical problem which can, however, besolved bytechniques which wewall notdescribe now. The result ofsuch calculations 1sthat the charge density rises somewhat near theedges oftheplates This means that thecapacity oftheplates 1salttle hugher than wecomputed. [Avery good approximation forthecapacity 1sob- tained ifweuseEq, (6.34) buttake forAthearea one would getsftheplates were extended artificially byadistance 3/8oftheseparation between theplates.] We have talked about the capacity fortwo conductors only. Sometimes people talk about thecapacity ofasingle object. They say, forinstance, that the capactty ofasphere ofradius ais4eqa. What they magine 1sthat theother terminal isanother sphere ofinfinite radvus—that when there isacharge +Q on Fig.6-13, Theelecnie feldneorthe thEsphere. theopposite charge, —Q,isonan infimte sphere. Oneeanalsospeak edgeoftvsporal wines 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 and avery small separation Wecould put waxed paper between sheets ofaluminum foiland roll stup. (It weseal stinplastic, wehave atypical radio-type condenser.) What good 1s1? Its good forstoring charge. Ifwetrytostore charge onaball, forexample, sts potential rises rapidly aswecharge itup. {tmay even getsohigh that thecharge begins toescape into theairbyway ofsparks Butsfweputthesame charge on condenser whose capacity 1svery large. the voltage developed across. the con- denser will besmall. *Somepeoplethinkthewords“capacitance” and“capacitor”shouldbeused,instead of“capacity” and “condensor "We have decided tousetheolder terminology, because 11 still more commonly heard inthephysics laboratory—even ifnotin textbooks! 6-12 Inmany applications inelectronic circuits, itisuseful tohave something which canabsorb ordeliver large quantities ofcharge without changing itspo- tential much. Acondenser (or“capacitor”) does just that. ‘There arealso many applications inelectronic instruments and incomputers where acondenser 1s used togetaspecified change involtage inresponse toaparticular change in charge. We have seen asimilar application inChapter 23,Vol. I,where wede- scribed theproperties ofresonant circutts. From thedefinition ofC,weseethat itsunit 1sonecoul/volt. This unit is 1 of€9asfarad/meter, which istheunit most commonly used. Typical sizes of condensers runfromonemicro-microfarad (=1picofarad) tomillifarads. Smallcondensers ofafew picofarads are used inhigh-frequency tuned circuits, and capacities uptohundreds orthousands ofmicrofarads arefound inpower-supply filters. Apair ofplates onesquare centimeter inarea with aonemillimeter separa- tion have acapacity ofroughly onemicro-microfarad. 6-11 High-voltage breakdown Wewould like now todiscuss qualttatively some ofthecharacteristics ofthe fields around conductors. Ifwecharge aconductor that isnotasphere, butone that hasonitapoint oravery sharp end, as,forexample, theobject sketched 1nFig. 6-14, thefield around thepoint ismuch higher than thefield intheother —|_ regions. Thereason is,qualitatively, thatcharges trytospread outasmuch as he. possible onthesurface ofaconductor, andthetipofasharp point isasfaraway LLasitispossible tobefrommostofthesurface. Someofthecharges ontheplate. *~{~~|~ PI etpushed alltheway tothetip. Arelatively small amount ofcharge onthetip canstillprovide alarge surface density; ahigh charge density means ahigh field Yjustoutside. conpucTOR yOnewaytoseethatthefieldishighest atthose places onaconductor where of the radius ofcurvature issmallest istoconsider the combination ofabigsphere o\7 andalittlesphere connected byawire, asshown inFig.6-15. Itisasomewhat aa idealized version oftheconductor ofFig.6-14. Thewirewillhave little influence “ 4 onthefields outside; itisthere tokeep thespheres atthesame potential. Now, , which ballhasthebiggest fieldatitssurface? Iftheballonthelefthastheradius, < aand carries acharge Q,1tspotential isabout 1@ Fig.6-14.TheelectricfieldnearoO-Gea shorp pointon@conductor isveryhigh. (Ofcourse thepresence ofoneballchanges thecharge distribution ontheother, sothat thecharges arenot really spherically symmetric oneither. But ifweareinterested onlyinanestimateofthe fields, wecanusethepotential ofaspherical charge.) Ifthesmaller ball, whose radius is6,carries thecharge q,itspotential isabout 1g atre 5 ut dr= ba,80ane Q_g, ;a? wre —Q Ontheother hand, the field atthesurface (see Eq. 5.8) isproportional tothesurfacechargedensity,which1slikethetotalchargeovertheradiussquared. jp Wegetthat r ebzeoera (6.35)4 « Fig.6-15, Thefieldof@pointed ‘Therefore thefield ishigher atthesurface ofthesmall sphere. Thefields areinthe object canbeapproximated bythatof inverse proportion oftheradii. twospheres atthesame potential. This result istechnically very important, because airwill break down ifthe electric field istoo great. What happens 1sthat aloose charge (electron, orion) somewhere intheairisaccelerated bythefield, and ifthefield isvery great, the charge canpick upenough speed before ithitsanother atom tobeable toknock an os electron offthat atom. Asaresult, more and more ions are produced. Their motion constitutes adischarge, orspark. Ifyou want tocharge anobject toa high potential and not have itdischarge itself bysparks intheair, you must be sure that thesurface issmooth, sothat there isnoplace where thefield isab- normally large. 6-12 The field-emission microscope = ruugpescenr There1saninteresting application oftheextremely highelectricfieldwhich a= conn surrounds anysharpprotuberance onacharged conductor. Thefield-emission\4 ‘microscope depends foritsoperation onthehighfieldsproduced atasharpmetaley\|/ point.*Itisbuiltinthefollowingway.Averyfineneedle,withatipwhosediameter S\\ 77 isabout1000angstroms, isplacedatthecenterofanevacuated glasssphere(Fig.[-~-N\yvz= +H") 6-16.) Theinner surface ofthesphere iscoated with athinconducting layer of {> ——_]]} fluorescent material, andavery high potential difference 1sapplied between the Od ae fluorescent coating andtheneedle. \ Let's firstconsider what happens when theneedle isnegative with respect to rome thefluorescent coating. Thefieldlinesarehighly concentrated atthesharp point. . _ The electric field can beashigh as40million volts per centimeter. Insuch a assou intense fields,electrons arepulledoutofthesurface oftheneedleandaccelerated | across thepotential difference between theneedle and thefluorescent layer. WhenBaw theyarrivetheretheycauselight(obeemitted,justasinateleviston picturetube.pone ‘Theelectrons whicharriveatagivenpointonthefluorescent surfaceare,to anexcellent approximation, those which leave theother end oftheradial field line, Fewer vowrset because theelectrons will travel alorig thefield line passing from thepoint tothe surface. Thus weseeonthesurface some kind ofanimage ofthetipoftheneedle.Fig.6-16,Field-emission microscope. Moreprecisely, weseeapictureoftheemissivity ofthesurfaceofthe needle—that 1stheeasewithwhichelectrons canleavethesurfaceofthemetaltip.Iftheresolu-tion were high enough, one could hope toresolve thepositions oftheindividual atoms onthetpoftheneedle, With electrons, this resolution 1snotpossible for the following reasons. First, there isquantum-mechanical diffraction ofthe electron waves which blurs theimage. Second, due totheinternal motions ofthe electrons inthemetal they have asmall sideways initial velocity when they leave theneedle, and this random transverse component ofthe velocity causes some smearing ofthe image. Thecombination ofthese two effects limits theresolution .to25 Aorso. = aot If,however, wereverse thepolarity andintroduce asmall amount ofhelium ” * gasintothebulb, much higher resolutions arepossible. When ahelium atom col- : PAI lides with thetipoftheneedle, theintense field there strips anelectron offthe ed Sars eam=helium atom,leaving itpositively charged. Thehelium ion1sthenaccelerated ERI outward along.a field line tothefluorescent screen. Since thehelium ionissomuch :ae :heavierthananelectron, thequantum-mechanical wavelengths aremuchsmaller. NeSamm—Ifthetemperature 1snottoohigh,theeffectofthethermalvelocities isalsosmallerey Y AAthanintheelectroncase,Withlesssmearing ofthe image amuch sharper picture Sores Steps amote y Hof thepoint isobtained. Ithas been possible toobtain magnifications upto ify B FORE 2,000,000 times with thepositive ionfield-emission microscope—a magnification pr ae a! tentimes better than isobtained with thebest electron microscope. i tg TRE Figure 6-17isanexample oftheresults which wereobtained withafield- eet ionmicroscope, using atungsten needle. ‘Thecenter ofatungsten atom ionizes oe Qe Beta =2helium atom atashghtly different ratethanthespaces between thetungsten ue 5 on atoms. Thepattern ofspotsonthefluorescent screen shows thearrangement of ta et. theindividual atoms onthetungsten tip.Thereason thespotsappear inringscan ces aaa beunderstood byvisualizing alarge boxofballspacked inarectangular array, representing theatoms inthemetal. Ifyou cutanapproximately spherical section Fig.6-17.Imageproduced byaoutofthisbox,youwillseetheringpatterncharacteristic oftheatomicstructure.field-emission microscope. [Courtesy of The field-ion microscope provided human beings with themeans ofseeing atoms Erwin W. Mueller, Research Prof. of forthefirst time. This isaremarkable achievement, considering thesimplicity of Physics, Pennsylvania State University} theinstrument. *See E.W. Mueller: “The field-ion microscope,”” Advances mElectronicsandElectron Physics, 13, 83-179 (1960). Academic Press, New York os 7 The Electric Field in Various Circumstances (Continued) 7-1 Methods forfinding theelectrostatic field This chapter is@continuation ofour consideration ofthecharacteristics of 7-1 Methods forfinding the electric fields invarious particular situations. Weshall first describe some ofthe electrostatic field more elaborate methods forsolving problems with conductors. Itisnotexpected ,thatthesemoreadvancedmethodscanbemasteredatthistime.Yetitmaybeof7Treatimensional Belesinterest tohavesome ideaabout thekinds ofproblems thatcanbesolved, using c - techniques thatmaybelearned inmoreadvanced courses. Thenwetakeuptwo variable examples inwhich thecharge distribution isneither fixed noriscarried byacon- 7-3 Plasma oscillations ductor, butinstead isdetermined bysomeotherlawofphysics. +4Colloidal particles inan‘Aswefound inChapter 6,theproblem oftheelectrostatic field1sfundamen- Pat tallysimple whenthedistribution ofcharges isspecified ;itrequires onlytheevalua- electrolyte tion ofanintegral. When there areconductors present, however, complications 7-5 Theelectrostatic field ofagrid arise because thecharge distribution ontheconductors isnotinitially known;thechargemustdistributeitselfonthesurfaceofthe conductor insuch away that theconductor isanequipotential. The solution ofsuch problems isneither direct not simple.Wehavelookedatanindirectmethodofsolvingsuchproblems,inwhichwe find theequipotentials forsome specified charge distribution and replace oneof them byaconducting surface. Inthisway wecanbuild upacatalog ofspecial solutions forconductors intheshapes ofspheres, planes, ete. ‘The useofimages, described inChapter 6,isanexample ofanindirect method. Weshall describe another inthis chapter. Iftheproblem tobesolved does notbelong totheclass ofproblems forwhich wecanconstruct solutions bytheindirect method, weareforced tosolve theprob- lembyamore direct method. The mathematical problem ofthedirect method is thesolution ofLaplace's equation, vp =0, a subject tothecondition that ¢1sasuitable constant oncertain boundaries—the surfaces ofthe conductors, Problems which involve the solution ofadifferential field equation subject tocertain boundary conditions are called 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 ofachargedcylindricalmetalcan closed atboth ends—a beer can—presents formidable mathematical difficulties Itcanbesolved only approximately, using numerical methods. The only general methods ofsolution are numerical. There areafew problems forwhich Eq, (7.1) can besolved directly. For example, theproblem ofacharged conductor having theshape ofanellipsoid ofrevolution canbesolvedexactlyintermsofknown special functions. The solution forathin disc canbeobtained byletting theellipsoid become infinitely oblate. Inasimilar manner, thesolution foraneedie canbeobtained byletting theellipsoid become infinitely prolate. However, itmust bestressed that theonly direct methods ofgeneral applicability arethenumerical techniques. Boundary-value problems can also besolved bymeasurements ofaphysical analog. Laplace's equation arises inmany different physical situations: insteady- state heat flow, inirrotational fluid flow, incurrent flow inanextended medium, “4 become toodull.) Itisthis. For any “ordinary function” (mathematicians will define itbetter) thefunctions Uand Vautomatically satisfy therelations au_awnD A) av__aa ay (7.8) Itfollowsimmediately thateachofthefunctions UandVsatisfy Laplace's equation: eu, eu _ Gat Ge=o (79) av, avov av 7.get Geno (7.10) These equations areclearly true forthefunctions of(7.5) and(7.6). ‘Thus, starting with any ordinary function, wecan arrive attwo functions UC, »)and V(x, »),which areboth solutions ofLaplace’s equation intwo dimen- sions. Each function represents apossible electrostatic potential. Wecanpick any function F(a) and itshould represent some electric field problem—in fact, two problems, because Uand Veach represent solutions. Wecan write down asmany solutions aswewish—by just making upfunctions—then wejust have tofind the problem that goes with each solution. Itmay sound backwards, butit'sapossible approach. Key y vs Tt es z a S% 4< mine «04.0. ANY v %,Ae PX wa - ANAS TRE AAA Set ene ALL Is |e \s anil al 30 tr 7 ~fem fo Ben - JID a>Xho ard)ag lant” ane StS KOS) X.Lk Ve/ aKASALLE PNRSPAX eX Xx 31 2 “S hat ASD (S60 ame ee, Fig. 7-1. Two sets oforthogonal curves which can represent equipotentials inatwo-dimensional electrostatic field. Asanexample, let'sseewhat physics thefunction F(3) =3%gives us.From itwegetthetwo potential functions of(7.5) and (7.6). Toseewhat problem the function Ubelongs to,wesolve fortheequipotential surfaces bysetting U=4, aconstant: x?ayhe A, This istheequation ofarectangular hyperbola. For various values of4,weget thehyperbolas shown inFig. 7-1. WhenA=0,wegetthespecialcaseofdiagonal straight lines through theorigin. Such asetofequipotentials corresponds toseveral possible physical situations,First,itrepresents thefinedetailsofthefieldnearthepointhalfwaybetweentwo 73 CONDUCTOR + Siva ws awaww aa awa wa awa waa Ta aww a ee sree LLEZIZIEIZZ= RyMK. Fig.7-2.ThefieldnearthepointC RQ isthesomeosthotinFig.7-1. XY“ eonoucTor — equal point charges. Second, itrepresents thefield ataninside right-angle corner ofaconductor. Ifwehave two electrodes shaped like those inFig. 7-2, which are held atdifferent potentials, thefield near thecorner marked Cwilllook just like thefield above theorigin inFig. 7-1. The solid lines aretheequipotentials, and thebroken lines atright angles correspond tolines ofE.Whereas atpoints or protuberances theelectric field tends tobehigh, ittends tobeow indents or hollows, ‘The solution wehave found also corresponds tothat forahyperbola-shaped electrode near aright-angle corner, orfortwo hyperbolas atsuitable potentials. You will notice that thefield ofFig. 7-1 hasaninteresting property. The x-com- ponent oftheelectric field, E,,isgiven by 86 E,=~S= 2x. Theelectric field isproportional tothedistance from theaxis. This factisused to make devices (called quadrupole lenses) that areuseful forfocusing particle beams gat (seeSection 29-9). Thedesired field isusually obtained byusing four hyperbola- shaped electrodes, asshown inFig. 7-3. Fortheelectric field lines inFig. 7-3, wehave simply copied from Fig. 7-1 thesetofbroken-line curves that represent V=constant. We have abonus! The curves forV=constant areorthogonal totheones forU=constant because oftheequations (7.7) and (7.8), Whenever ge-v g=-v_ wechoose afunction F(a),wegetfrom UandVboththeequipotentials andfield lines. And you will remember that wehave solved either oftwoproblems, depend- ‘ingonwhich setofcurves wecalltheequipotentials. ‘Asa second example, consider thefunction ‘Conoucton FO)=V3. (AD)gat Ifwewrite b= x+ iy=pe", Fig. 7-3. Thefield inaquadrupole wherelens, p-VeEy and tan6=y/x, then, FQ) =plte"? =pl?(cos3+isin’). from which 2g yee ype 2g yaa qeFQ)=[erpe ses+{ety =4]-@2) 14 = y =x4 te f023, !\ae8 - /‘ L7 /2|axe !/ 2 . 4 i Le tS !Tey|& ! 1 < aso! “NN_|eno} __ _|_x i \ 1» \\ ~ \s) ~. \S > \ \s \C SS Fig.7-4.Curvesofconstant Ulx,y) \N NS condV\x,y)from Eq.(7.12). \ ~ \ me \ N ThecurvesforU(x,»)=AandV(x,y)=B,usingUand¥from Eq.(7.12), areplotted inFig. 7-4. Again, there aremany possible situations that could bedescribed bythesefields.Oneofthemostinterestingisthefieldneartheedgeofa thinplate.IfthelineB=0—totherightofthe y-axis—representsathincharged plate, thefield lines near itaregiven bythecurves forvarious values ofA.The physical situation isshown inFig. 7-5. Further examples are F(a)=25/7, (7.13) which yields thefield outside arectangular corner FQ) =log3, 7.14) which yields thefield foralinecharge, and FQ) =1/8, (7.15) which gives thefield forthetwo-dimensional analog ofanelectric dipole, ie., ‘twoparallel linecharges withopposite polarities, veryclosetogether. ae ‘Wewillnotpursue thissubject further inthiscourse, butshould emphasize that although thecomplex variable technique isoften powerful, itislimited to two-dimensional problems; andalso, itisanindirect method. . 7-3Plasma oscillations Fig.7-5. Theelectric field near the Weconsider nowsomephysical situations inwhich thefieldisdetermined °49®of¢thingroundedplate, neither byfixed charges norbycharges onconducting surfaces, butbyacom- bination oftwo physical phenomena. Inother words, thefield willbegoverned simultaneously bytwo sets ofequations: (I)theequations from electrostatics relating electric fields tocharge distribution, and (2)anequation from another part ofphysics that determines thepositions ormotions ofthecharges inthe presence ofthefield. The first example that wewill discuss isadynamic one inwhich themotion ofthecharges isgoverned byNewton's laws. Asimple example ofsuch asituation occurs inaplasma, which isanionized gasconsisting ofions and free electrons distributed overaregioninspace.Theionosphere—an upperlayeroftheatmos- phere—is anexample ofsuch aplasma. The ultraviolet rays from thesun knock 1s electrons offthemolecules oftheair,creating free electrons and ions. Insuch a plasma the positive 1ons are very much heavier than theelectrons, sowemay neglect theionic motion, incomparison tothat oftheelectrons. Letmobethedensity ofelectrons intheundisturbed, equilibrium state. This must also bethedensity ofpositive ions, since the plasma iselectrically neutral (when undisturbed). Now wesuppose that theelectrons aresomehow moved from equilibrium andaskwhat happens. Ifthedensity oftheelectrons in ‘one region isincreased, they will repel each other and tend toreturn totheir equilibrium positions. Astheelectrons move toward their original positions they pick upkinetic energy, and instead ofcoming torestintheir equilibrium configura- tion, they overshoot themark. They will oscillate back and forth. The situation 1ssimilar towhat occurs insound waves, inwhich therestoring force isthegas pressure.Inaplasma,therestoringforceistheelectricalforceontheelectrons. qlTosimplify thediscussion, wewillworry only about asituation inwhich the motions areallinone dimension, sayx.Letussuppose that theelectrons origi- nally atxare,attheinstant 1,displaced from their equilibrium positions byasmallamounts(x,1).Sincetheelectronshavebeendisplaced, theirdensitywill,ingeneral,bechanged. Thechange indensity iseasily calculated. Referring toFig.7-6. o—*a on— theelectronsmitiallycontained betweenthetwoplanesaandbhavemovedand 1ay yy #2nowcontained between theplanesa’and6’,Thenumber ofelectrons thatHWYVS77/7)werebetweenaandbsproportionaltonox;thesamenumberarenowcontainedims yestds imthespace whose width isAx+As.Thedensity haschanged to' UY n=—ToAx_ to (716)1 +8ofax+4s——-| ax+as”1¥(as/Ox) Fig.7-6.Motioninaplasmawave.—Ifthechangeindensityissmall,wecanwrite[usingthebinomial expansion forTheelectrons attheplane amove toa’, (1+¢)~1]andthoseatbmovetob’. asn=NM(:-2). (7.17) Weassume that thepositive ions donotmove appreciably (because ofthemuch larger inertia), sotheir density remains mo. Each electron carries thecharge —qe, sotheaverage charge density atanypoint isgiven by p= ~(0=MMe or =ng,p=noeFe (7.18) (where wehave written thedifferential form for4s/Ax). ‘The charge density isrelated totheelectric field byMaxwell's equations, in particular, veeE=2. (7.19) & Iftheproblem isindeed one-dimensional (and ifthere arenoother fields butthe ‘one due tothedisplacements oftheelectrons), theelectric field Ehas asingle component E,. Equation (7.19), together with (7.18), gives BE,_Mode asRte (7.20) Integrating Eq.(7.20) gives E,=toles+. 7.21) Since E,=0when s=0,theintegration constant Kiszero. ‘The force onanelectron inthedisplaced position is FoMBs, (7.22) 1-6 7-4 Colloidal particles inanelectrolyte Weturn toanother phenomenon inwhich thelocations ofcharges isgoverned byapotential that arises inpart from thesame charges. The resulting effects influence inanimportant way thebehavior ofcolloids. Acolloid consists ofa ‘suspension inwater ofsmall charged particles which, though microscopic, from anatomic point ofview arestill very large. Ifthecolloidal particles were not charged, they would tend tocoagulate into large lumps: but because oftheir charge, they repel each other and remain insuspension. Now ifthere isalso some salt dissolved inthe water, itwill bedissociated into positive and negative ions, (Such asolution ofions iscalled anelectrolyte.) The negative ionsareattracted tothecolloidparticles (assuming theirchargeispositive) and thepositive ions arerepelled. Wewill determine how theions which surround such acolloidal particle aredistributed inspace. Tokeep theideas simple, wewill again solve only aone-dimensional case. Ifwe think ofacolloidalparticleasaspherehavingaverylargeradius—on an atomic scale!—we canthen treat asmall part ofitssurface asaplane. (Whenever oneistrying tounderstand anew phenomenon itisagood idea totake asomewhat oversimplified model; then, having understood theproblem with that model, one isbetter able toproceed totackle themore exact calculation.) ‘Wesuppose that thedistribution ofions generates acharge density p(x), and anelectrical potential ¢,related bytheelectrostatic lawV26 =—p/ey or,for fields that vary inonly onedimension, by 7__pfea (7.28) Now supposing there were such apotential (x), how would theions dis- tribute themselves init? This wecan determine bythe principles ofstatistical mechanics. Our problem then istodetermine ¢sothat theresulting charge density from statistical mechanics also satisfies (7.28). According tostatistical mechanics (seeChapter 40,Vol. 1),particles inthermal equilibrium inaforce field aredistributed insuch away that the density nof particles attheposition xisgiven by n(x) =nyeVOT, (7.29) whereU(x)isthepotential energy, kisBoltzmann’s constant, andTistheabsolute temperature, We assume that theions carry one electronic charge, positive ornegative. Atthedistance xfrom thesurface ofacolloidal particle, apositive ionwill have potential energy g(x), sothat U(x) =geo(x). Thedensity ofpositive ions, m,,isthen glx)=getrent? Similarly, thedensity ofnegative tons is n(x) =gee, The total charge density is P= at —Ginn or p= qanletlht ete, 720) ‘Combining this with Eq. (7.28), wefind that thepotential @must satisfy #6 GeO(gueskT__gtacdihT fe-me —e’). (731) 1. ‘This equation isreadily solved ingeneral [multiply both sides by2(dg/dx), and integrate with respect tox],buttokeep theproblem assimple aspossible, wewill consider here only thelimiting case inwhich thepotentials aresmall orthetem-peratureTishigh.Thecasewhereoissmallcorresponds toadilutesolution. Forthese cases theexponent issmall, and wecan approximate eter 1aAB. (7.32) Equation (7.31) then gives 2 fe=+meoe). 733) Notice that this time thesign ontheright ispositive. The solutions for¢arenot oscillatory, butexponential. ‘The general solution ofEq. (7.33) is $= Aem!? +Bet*!?, (734) with 2€okT| Dt=on 7.35) The constants Aand Bmust bedetermined from theconditions oftheproblem. Inourcase, Bmust bezero; otherwise thepotential would gotoinfinity forlarge x. Sowehave that $= Ae, (7.36) inwhich Aisthepotential atx=0,thesurface ofthecolloidal particle. % Fig. 7-7. The variation ofthe po- tential near the surface ofacolloidal particle. DistheDebye length. ol ° ° z The potential decreases byafactor 1/eeach time thedistance increases byD, asshowninthegraphofFig.7-7.Thenumber DiscalledtheDebyelength,and isameasure ofthethickness oftheionsheath that surrounds alarge charged particle inanelectrolyte. Equation (7.36) says that thesheath gets thinner with increasing concentration oftheions (70) orwith decreasing temperature. Theconstant AinEq.(7.36) iseasily obtained ifweknow thesurface chargeo onthecolloid particle, Weknow that E,=Ex)=2. 737) But Eis also thegradient of¢: ; =—%) 244 £0)=~3,7 +5 (7.38) from which weget 4-2. 7.39) ay Using this result in(7.36), wefind (bytaking x=0)that thepotential ofthe colloidal particle is 60-2. (740) ‘You willnotice that this potential isthesame asthepotential difference across a condenser with aplate spacing Dand asurface charge density Wehave said that thecolloidal particles arekept apart bytheir electricalrepulsion. Butnowweseethatthefieldalittlewayfromthesurfaceofaparticle 1sreduced bytheionsheath thatcollects around it.Ifthe sheaths getthin enough, theparticles have agood chance ofknocking against each other. They will then stick, and thecolloid willcoagulate and precipitate outoftheliquid. From our analysis, weunderstand why adding enough salt toacolloid should cause itto precipitate out. The process iscalled “salting outacolloid.” Another interesting example istheeffect that asaltsolution hasonprotein molecules. Aprotein molecule isalong, complicated, and flexible chain ofamino acids. The molecule has various charges onit,and itsometimes 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 effects. Wecan, therefore, have asuspensionofchainmoleculesinaliquid.Butifweaddsaltotheliquidwechangetheproper-ties ofthesuspension. Assalt isadded tothesolution, decreasing theDebye distance, thechain molecules canapproach one another, and can also coil up. Ifenough salt isadded tothesolution, thechain molecules will precipitate outof thesolution. There aremany chemical effects ofthiskind that canbeunderstoodintermsofelectrical forces. 7-5 The electrostatic field ofagrid AAsourlastexample, wewould liketodescribe another interesting property ofelectric fields. Itisone which ismade useofinthedesign ofelectrical instru- ments, intheconstruction ofvacuum tubes, and forother purposes. This isthe character oftheelectric field near agrid ofcharged 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 @ plane. Asweapproach thegrid ofwires, thefield begins todeviate from the uniform field wefound atlarge distances from thegrid. Wewould like toestimate how close tothegrid wehave tobeinorder toseeappreciable variations inthe potential. Figure 7-8 shows arough sketch oftheequipotentials atvarious distances from thegrid. The closer wegettothegrid, thelarger thevariations. ‘Aswetravel parallel tothegrid, weobserve that thefield fluctuates inaperiodic manner, tz DDB TAN TR PPP Bop gs _io —j Fig.7-8. Equipotential surfaces x ebove @uniform grid ofcharged wires. 710 Now wehave seen (Chapter 50,Vol. I)that anyperiodic quantity canbé‘expressed asasumofsinewaves(Fourier’s theorem). Let’sseeifwecanfinda suitable harmonic function thatsatisfies ourfield equations. Ifthewires lieinthexy-plane and runparallel tothey-axis, then wemight tryterms like $64,2)=Fy(z)cos22, ED) where aisthespacing ofthewires andnistheharmonic number. (We have as- sumed long wires, sothere should benovariation with y.) Acomplete solution wouldbemadeupofasumofsuchtermsform=1,2,3,.... Ifthisistobeavalid potential, itmust satisfy Laplace's equation inthe region above thewires (where there arenocharges). That is, Fo, aeoa+a Trying thisequation onthe¢in(7.41), wefind that 4x? 2anx5dFog2anx —iTFa(z)cos TS+Ttcos X=0, (7.42) orthat F,(z) must satisfy PF, Atono Fe (7.43) So we must have Fy=Age", 44) where 20=Tan" (7.45) Wehave found that ifthere isaFourier component ofthefield ofharmonic n, that component will decrease exponentially with acharacteristic distance 2)= a/2nn. Forthefirst harmonic (n=1),theamplitude falls bythefactor e~?* (alarge decrease) each time weincrease zbyonegrid spacing a.The other har- monics falloffeven more rapidly aswemove dway from thegrid. Weseethat if weareonly afewtimes thedistance aaway from thegrid, thefield isvery nearly uniform, ie., theoscillating terms aresmall. There would, ofcourse, always remain the “zero harmonic” field 0=—Eqz togive theuniform field atlarge z.Foracomplete solution, wewould combine this term with asum ofterms like (7.41) with F,from (7.44). The coefficients A, would beadjusted sothat thetotal sum would, when differentiated, give anelectric field that would fitthecharge density \ofthegrid wires. The method wehave just developed canbeused toexplain why electrostatic shielding bymeans ofascreen isoften just asgood aswith asolid metal sheet. Except within adistance from thescreen afewtimes thespacing ofthescreen wires, thefields inside aclosed screen arezero. Weseewhy copper screen— lighter and cheaper than copper sheet—is often used toshield sensitive electrical equipment from external disturbing fields. oan 8 Electrostatic Energy 8-1Theelectrostatic energyofcharges.Auniformsphere Inthestudy ofmechanics, one ofthemost interesting anduseful discoveries 8-1 The electrostatic energy of was thelawoftheconservation ofenergy. ‘The expressions forthekinetic and charges. Auniform sphere potential energies ofamechanical system helped ustodiscover connections betweenthestatesofasystemattwodifferent timeswithouthavingtolookintothedetails &2ow,shes=a‘ ofwhatwasoccurring inbetween. Wewishnowtoconsider theenergyofelectro- targedconductors static systems. Inelectricity alsotheprinciple oftheconservation ofenergy will 8-3Theelectrostatic energy ofan beuseful fordiscovering anumber ofinteresting things. ionic crystal ‘The lawoftheenergy ofinteraction inelectrostatics isvery simple; wehave, isinfact,already discussed it.Suppose wehavetwocharges g1andqsseparated by 4Electrostatic energyinnuclei thedistance r. There issome energy inthesystem, because acertain amount of 8-5 Energy intheelectrostatic field work wasrequired tobring thecharges together. Wehave already calculated theworkdoneinbringing twocharges together fromalargedistance. Itis 8-6Theenergy ofapointcharge a2Areoria ey Wealso know, from theprinciple ofsuperposition, that ifwehave many charges Review: Chapter 4,Vol. I,Conservationpresent,thetotalforceonanychargeisthesumoftheforcesfromtheothers.It ofEnergy follows, therefore, that thetotal energy ofasystem ofanumber ofcharges isthe Chapters 13and 14,Vol. I,sumoftermsduetothemutualinteraction ofeachpairofcharges.Ifq,and9; WorkandPotentialEnergyareanytwoofthechargesandr,isthedistancebetweenthem(Fig.8-1),theenergyofthatparticular pairis 995Trev 62) The total electrostatic energy Uisthesum oftheenergies ofallpossible pairs of ° ° charges: ° °U= ts. 63) ° “ so ° Ifwehaveadistribution ofchargespecified byachargedensityp,thesumofEq. 00oN ° (8.3) is,ofcourse, tobereplaced byanintegral. YS ‘Weshallconcern ourselves withtwoaspects ofthisenergy. Oneistheapplica- ° ° NY tionoftheconcept ofenergytoelectrostatic problems; theotheristheevaluation oOUoftheenergy indifferent ways. Sometimes itiseasier tocompute thework done ° ° forsome special case than toevaluate thesum inEq.(8.3), orthecorresponding integral. Asanexample, letuscalculate theenergy required toassemble asphere Fig.6-1. Theelectrostatic energy of ofcharge with auniform charge density. Theenergy isjustthework done in ©System ofparticles isthesumofthe gathering thecharges together from infinity. electrostatic energy ofeachpair. Imagine that weassemble thesphere bybuilding upasuccession ofthin spherical layers ofinfinitesimal thickness. Ateach stage oftheprocess, wegather asmall amount ofcharge and putitinathin layer from rtor +dr.Wecontinue theprocess until wearrive atthefinal radius a(Fig. 8-2). IfQ,isthecharge ofthe sphere when ithasbeen built uptotheradius r,thework done inbringing acharge dQtoitis =QedQ aU= (8.4) a Ifthedensity ofcharge inthesphere isp,thecharge Q,is 4 Q.=p-$xr, VeyandthechargedQis Y dQ=p-4nr*dr.RO Equation(8.4)becomes VARa 245 = ay=Setar, 5)€o Cj)‘ThetotalenergyrequiredtoassemblethesphereistheintegralofdUfromr= LDPOtor =aor 2a v=See. @6) Fig. 8-2. The energy ofauniform sphere ofcharge canbecomputed by Orifwewish toexpress theresult interms ofthetotal charge Qofthesphere, imagining that itixassembled from successive spherical shells. _3 2U=5Trea” en The energy isproportional tothesquare ofthetotal charge and inversely pro- portional totheradius. Wecanalso interpret Eq.(8.7) assaying that theaverage of(1/r,) forallpairs ofpoints inthesphere is3/5a. 8-2 The energy ofacondenser. Forces oncharged conductors ‘Weconsider now theenergy required tocharge acondenser. Ifthecharge Q hasbeentakenfromoneofthe conductors ofacondenser andplacedontheother, thepotential difference between them is =2v=§. (8.8) where Cisthecapacity ofthecondenser. How much work isdone incharging thecondenser? Proceeding asforthesphere, weimagine that thecondenser has been charged bytransferring charge from oneplate totheother insmall increments dQ. The work required totransfer thecharge dQis dU=vg. Taking Vfrom Eq.(8.8), wewrite QdQ au~240. Orintegrating from zero charge tothefinal charge Q,wehave -12.U=356 (8.9) This energy can also bewritten as U=4cV?. (8.10) Recalling that thecapacity ofaconducting sphere (relative toinfinity) is Cupore =41°€0a, ‘wecanimmediately getfrom Eq.(8.9) theenergy ofacharged sphere, =1 2.U=sae @.11) 22 This, ofcourse, isalso theenergy ofathin spherical shell oftotal charge Qandis just5/6oftheenergy ofauniformly charged sphere, Eq.(8.7). Wenow consider applications oftheidea ofelectrostatic 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, togetherwiththeprinciple ofvirtual work(Chapters 4,13,and14ofVol.I). Let’s usethis method fordetermining theforce between theplates ofa parallel-plate condenser. Ifweimagine that thespacing oftheplates isincreased bythesmall amount Az,then themechanical work done from theoutside in moving theplates would be AW =Faz, (8.12) where Fistheforce between theplates. This work must beequal tothechange intheelectrostatic energy ofthecondenser. ByEq.(8.9), theenergy ofthecondenser wasoriginally 1g?U-7e Thechange inenergy (ifwedonotletthecharge change) is Lo (l au=50a(3)- (8.13) Equating (8.12) and(8.13), wehave o 4(1Faz=5-a(z)- (8.14) This can also bewritten as --2Paz=~7GqAC. (8.15) The force, ofcourse, results from theattraction ofthecharges ontheplates, but weseethat wedonothave toworry indetail about how they aredistributed; everything weneed istaken care ofinthecapacity C. Itiseasy toseehow theidea isextended toconductors ofanyshape, andfor other components oftheforce. InEq.(8.14), wereplace Fbythecomponent we arelooking for,and wereplace Azbyasmall displacement inthecorresponding direction. Orifwehave anelectrode with apivot andwewant toknow thetorque T,wewrite thevirtual work as AW=7A6, ry where48isasmallangulardisplacement. Ofcourse,A(1/C)mustbethechangein¢* 1/Cwhich corresponds toA.Wecould, inthisway,findthetorque onthemov- ‘ ableplatesinavariable condenser ofthetypeshowninFig.8-3. Returning tothespecial case ofaparallel-plate condenser, wecan usethe formula wederived inChapter 6forthecapacity: 1 d Gowed? (8.16) where Aistheareaofeachplate. Ifweincrease theseparation byAz, Fig.8-9. What isthetorque on@ 1\ az variable capacitor?a)*ed From Eq.(8.14) wegetthat theforce between theplates is or-2: @.17) ry Let’slookatEq.(8.17)alittlemorecloselyandseeifwecantellhowtheforcearises.Ifforthechargeononeplatewewrite Q=o, Eq. (8.17) can berewritten as -19%F=502. Or,since theelectric field between theplates is « Ey=eo then F=$0Eo. (8.18) ‘One would immediately guess that theforce acting ononeplate isthechargeontheplatetimesthefieldactingonthecharge.Butwehaveasurprising factorofone-half. The reason isthat Episnotthefieldatthecharges.Ifweimaginethat \Ythecharge atthesurface ofthe plate occupies athin layer, asindicated inFig. 8-4,thefieldwillvaryfromzeroattheinnerboundaryofthelayertoEyinthespace compare NS LAYEROFoutsideoftheplate.TheaveragefeldactingonthesurfacechargesisEo/2.Thatsueree eis whythefactor one-half isinEq.(8.18). You should notice that incomputing thevirtual work wehave assumed that . thecharge onthecondenser wasconstant—that itwasnotelectrically connectedfe tootherobjects,andsothetotalchargecouldnotchange.‘Suppose wehadimagined that thecondenser washeld ataconstant potential difference aswemade thevirtual displacement. Then weshould have taken tel Eo u=4cv? and inplace ofEq. (8.15) wewould have had Faz =4V?.AC, Fig.8-4.ThefieldofthesurfaceofwhichgivesaforceequalinmagnitudetotheoneinEq.(8.15)(because=9/C), 2conductor varies from zero toE= butwith theopposite sign! Surely theforce betweenthecondenserplatesdoesn’t 2/¢o, asonepasses through thelayer of reverse insign aswedisconnect itfrom itscharging source. Also, weknow that surface charge. twoplates with opposite electrical charges must attract. Theprinciple ofvirtual work has been incorrectly applied inthesecond case—we have not taken into account thevirtual work done onthecharging source. That is,tokeep thepo- tential constant atVasthecapacity changes, acharge VACmust besupplied by ‘asource ofcharge. But this charge issupplied atapotential V,sothework done bytheelectrical system which keeps thepotential constant isV?AC.Themechan- icalwork FAzplusthiselectrical work V?ACtogether make upthechange inthe total energy #¥?ACofthecondenser. Therefore FAzis—4¥? AC,asbefore. ‘8-3Theelectrostatic energyofanioniccrystal Wenowconsider anapplication oftheconcept ofelectrostatic energyinatomic physics. Wecannot easily measure theforces between atoms, butweareoften interested intheenergy differences between one atomic arrangement and another, as,forexample, theenergy ofachemicalchange.Sinceatomicforcesarebasically electrical, chemical energies areinlarge part justelectrostatic energies. Let’s consider, forexample, theelectrostatic energy ofanionic lattice. An ionic crystal likeNaCl consists ofpositive andnegative ions which canbethoughtofasrigidspheres. Theyattractelectrically untiltheybegintotouch;thenthereisarepulsiveforcewhichgoesupveryrapidlyifwetrytopushthemclosertogether.For our first approximation, therefore, weimagine asetofrigid spheresthatrepresenttheatomsinasaltcrystal.Thestructureofthelatticehasbeendetermined byx-ray diffraction. Itisacubic lattice—like athree-dimensional 4 checkerboard. Figure 8-5shows across-sectional view. The spacing oftheions is, 281A(=2.81X107%cm).Ifourpicture ofthis system iscorrect, weshould beable tocheck itbyasking, thefollowing question: How much energy will ittake topull allthese ions apart— that is,toseparate thecrystal completely into ions? This energy should beequal totheheatofvaporization ofNaCIplustheenergyrequiredtodissociatethe|-~=2@-~S2-S 2SSLOKmoleculesintoions.ThistotalenergytoseparateNaCltoionsisdeterminedexperi-aaan a\)mentallytobe7.92electronvoltspermolecule. Usingtheconversion RAs <<< andAvogadro'snumberforthenumberofmoleculesinamole, WaennoS|=600x0 tooooos theenergyofvaporizationcanalsobegivenas neEES xZSZN ZS zs W=7.64X10°joules/mole. : tai Physical chemists prefer foranenergy unit thekilocalorie, which is4190 joules; sothat1evpermoleculeis23kilocalories permole.Achemistwouldthensay_Fig.8-5.Cross.section,of9ol cy Pera . crystal onanatomicscale, recker- thatthedissociation energy ofNaClis Sreecrceeront ofNeondGhiowe W=183kcal/mole. thesame inthetwocross sections per- pendicular totheone shown. (See Vol. |, Canweobtainthischemical energytheoretically bycomputing howmuch_Fig.1-7.) work itwould take topull apart thecrystal? According toourtheory, this work is thesum ofthepotential energies ofallthepairs ofions. Theeasiest way tofigure outthissum istopick outaparticular ionandcompute itspotential energy witheachofthe other ions. That willgive us‘wice theenergy perion, because theenergybelongstothepairsofcharges. Ifwewanttheenergytobeassociated withoneparticular ion, weshould take half thesum. But wereally want theenergy per molecule, which contains two ions, sothat thesum wecompute will give directly theenergy permolecule. Theenergy ofanionwith oneofitsnearest neighbors is?/a, where e?= g2/4reo andaisthecenter-to-center spacing between ions. (We areconsidering monovalent ions.) This energy is5.12 ev,which wealready seeisgoing togive usaresultofthecorrectorderofmagnitude. Butitisstillalongwayfromtheinfinitesum ofterms we need. Let’s begin bysumming alltheterms from theions along astraight line. Considering that theionmarked NainFig. 8-isourspecial ion, weshall consider first those ions onahorizontal line with it. There are two nearest Clions with negative charges, each atthedistance a.Then there aretwo positive ions atthe distance 2a,etc. Calling theenergy ofthis sum Uj,wewrite @(_2,2 2,2 Y=S(-Ft+5-5tGt 2? r,t--2(-h44-d4-)- 8.19) Theseries converges stowly, soitisdifficult toevaluate numerically, butitisknown tobeequal toIn2.So uy,=22na=-13862- (8.20) a a ‘Now consider thenext adjacent line ofions above. ‘The nearest isnegative andatthedistance a.Then there aretwopositives atthedistance +/2.a. Thenext pair areatthedistance1/5a,thenextat1/10a,andsoon.Soforthewholeline wegettheseries efi, 2 2 2€(-1,2_2,2...). (8.21)@(ITVav3"Vi0) &s ‘There arefour such lines: above, below, infront, and inback. Then there arethe four lines which arethenearest lines ondiagonals, andonand on. Ifyou work patiently through forallthelines, and then take thesum, you find that thegrand total is 2 uv=1m7e, which isjust somewhat more than what weobtained in(8.20) forthefirst line. Using e?/a =5.12ev,weget U=894ev. Our answer isabout 10% above theexperimentally observed energy. Itshows that ouridea that thewhole lattice ishéld together byelectrical Coulomb forces is fundamentally correct. This isthe first time that wehave obtained aspecific property ofamacroscopic substance from aknowledge ofatomic physics. We will domuch more later. The subject that tries tounderstand thebehavior of bulk matter interms ofthelaws ofatomic behavior iscalled solid-state physics. ‘Now what about theerror inourcalculation? Why isitnot exactly right? Itisbecause oftherepulsion between theions atclose distances. They arenot perfectly rigid spheres, sowhen they areclose together they arepartly squashed. ‘Theyarenotverysoft,sotheysquashonlyalittlebit.Someenergy,however, isusedindeforming them,andwhentheionsarepulledapartthisenergyisreleased.‘The actual energy needed topulltheions apart isalittle lessthan theenergy that wecalculated; therepulsion helps inovercoming theelectrostatic attraction. Isthere anyway wecanmake anallowance forthiscontribution? Wecould ifweknew thelawoftherepulsive force. Wearenotready toanalyze thedetails, ofthisrepulsive mechanism, butwecangetsome idea ofitscharacteristics from some large-scale measurements. From ameasurement ofthecompressibility ofthewholecrystal,itispossibletoobtainaquantitative ideaofthelawofrepulsionbetween theions and therefore ofitscontribution totheenergy. Inthis way it, hasbeenfoundthatthiscontribution mustbe1/9.4ofthecontribution fromtheelectrostatic attraction and,ofcourse,ofoppositesign.Ifwesubtractthiscontribu-tionfromthepureelectrostatic energy,weobtain7.99evforthedissociation energypermolecule. Itismuch closer totheobserved result of7.92 ev,butstill not in perfect agreement. There isonemore thing wehaven’t taken into account: we have made noallowance forthekinetic energy ofthecrystal vibrations. Ifacor- rection ismade forthiseffect, very good agreement with theexperimental number isobtained. The ideas arethen correct; themajor contribution totheenergy ofa crystal likeNaCl iselectrostatic. 84 Electrostatic energy innuclei We will now take upanother example ofelectrostatic energy inatomic physics, theelectrical energy ofatomic nuclei. Before wedothiswewillhave todiscusssomeproperties ofthemainforces(callednuclearforces)thatholdtheprotons and neutrons together inanucleus. Intheearly days ofthediscovery of nuclei—and oftheneutrons and protons that make them up—it was hoped that thelaw ofthestrong, nonelectrical part oftheforce between, say, aproton and another proton would have some simple law, liketheinverse square lawofelec tricity.Foronceonehaddetermined thislawofforce,andthecorresponding onesbetween aproton andaneutron, andaneutron andaneutron, itwould bepossible todescribe theoretically thecomplete behavior ofthese particles innuclei. There- fore abigprogram wasstarted forthestudy ofthescattering ofprotons, inthe hope offinding thelawofforce between them; butafter thirty years ofeffort,nothingsimplehasemerged. Aconsiderable knowledge oftheforcebetweenprotonandproton hasbeen accumulated, butwefind that theforce isascomplicated as itcan possibly be. What wemean by“ascomplicated asitcan be” isthat theforce depends on asmany things asitpossibly can. 86 First,theforceisnotasimplefunctionofthedistancebetweenthetwoprotons. Atlarge distances there isanattraction, butatcloser distances there isarepulsion. Thedistance dependence isacomplicated function, stillimperfectly known.Second,theforcedependsontheorientation ofthe protons’ spin. The protons have aspin, andanytwointeracting protons may bespinning with their angular g > ‘momenta inthesamedirection orinopposite directions. Andtheforceisdifferentwhenthespinsareparallelfromwhatitiswhentheyareantiparallel, asin(a)fo)ro) i)C)and (b)ofFig. 8-6. The difference isquite large; itisnotasmall effect. Third, theforce isconsiderably different when theseparation ofthetwo protons isinthedirection parallel! totheir spins, asin(c)and(4)ofFig. 8-6, than ¢ 4itiswhentheseparationisinadirectionperpendicular tothespins,asin(a)and(b).fo) fo)Fourth, theforce depends, asitdoes inmagnetism, onthevelocity ofthe protons,onlymuchmorestronglythaninmagnetism. Andthisvelocity-<dependent fe) fo) force isnotarelativistic effect; itisstrong even atspeeds much lessthan thespeed oflight. Furthermore, thispart oftheforce depends onother things besides the‘magnitude ofthevelocity. Forinstance,whenaprotonismovingnearanotherProton,theforceisdifferentwhentheorbitalmotionhasthesamedirection of|©»—— u-_-~rotationasthespin,asin(¢)ofFig.8-6,thanwhenithastheoppositedirection } fo) ofrotation,asin(f).Thisiscalledthe“spinorbit”partoftheforce. Coe The force between aproton and aneutron and between aneutron and a neutron arealsoequally complicated. Tothisdaywedonotknow themachinery figg-6,Theforce between two behind these forces—that istosay,anysimple wayofunderstanding them. protons depends onevery possible Thereis,however, oneimportant wayinwhichthenucleon forcesaresimpler parameter. than they could be.That isthat thenuclear force between twoneutrons isthesame astheforce between aproton andaneutron, which isthesame astheforce between two protons! If,inany nuclear situation, wereplaceaprotonbyaneutron(orvice versa), thenuclear interactions arenotchanged. The “fundamental reason” for thisequality isnotknown, butitisanexample ofanimportant principle thatcan ostbeextendedalsototheinteraction lawsofotherstronglyinteracting particles— =| relsuch asthe-mesons and the“strange” particles. 2Thisfactisnicelyillustrated bythelocation oftheenergylevelsinsimilar 2-328—|bieae-4 nuclei. Consider anucleus likeB'? (boron-eleven), which iscomposed offive a 6 protonsandsixneutrons. Inthenucleustheelevenparticlesinteractwithone [799 f2——|another inamostcomplicated dance. Now,thereisoneconfiguration ofallthe p2——possible interactions which hasthelowest possible energy; this isthenormal state £20ofthenucleus,andiscalledthegroundstate.Ifthenucleusisdisturbed(forexam- festa—ple,bybeingstruckbyahigh-energyprotonorotherparticle)iteanbeputinto | anynumber ofother configurations, called excited states, each ofwhich willhave P 1acharacteristic energy thatishigher than thatoftheground state. Innuclear . physics research, such asiscarried onwith Van deGraaff generator (for example, inCaltech’s Kellogg and Sloan Laboratories), theenergies and other properties ofthese excited states aredetermined byexperiment. The energies ofthefifteen Jowest known excited states ofB'!areshown inaone-dimensional graph onthe 2a a lefthalf ofFig. 8-7. The lowest horizontal line represents theground state. The first excited state has anenergy 2.14 Mev higher than the ground state, thenextanenergy4.46Mevhigherthanthegroundstate,andsoon.Thestudy ofnuclear physics attempts tofind anexplanation forthis rather complicated pattern ofenergies; there isasyet,however, nocomplete general theory of 8 1.982 c" such nuclear energy levels. IfwereplaceoneoftheneutronsinB'!withaproton,wehavethenucleus Fig.8-7.TheenergylevelsofB'' ofanisotope ofcarbon, C1, Theenergies ofthelowest sixteen excited states of andC''(energies inMev). ThegroundCC?"havealsobeenmeasured; theyareshownintherighthalfofFig.8-7.#ateofC'!is1.982Mevhigherthan(The broken lines indicate levels forwhich theexperimental information is "etof8’. questionable.) Looking atFig. 8-7, weseeastriking similarity between thepattern ofthe energy levels inthetwonuclei. The first excited states areabout 2Mev above the ground states. There isalarge gapofabout 2.3Mev tothesecond excited state, then asmall jump ofonly 0.5Mev tothethird level. Again, between thefourthandfifthlevels,abigjump;butbetweenthefifthandsixthatinyseparation ofthe “7 tobeaccounted forbyelectrostatic energy, isthus more than 1.982 Mev; itis 1.982+0.784=2.786Mev. UsingthisenergyinEq.(8.23),fortheradiusofeitherB**orC1wefind r= 312 X10-cm. (8.24) Does this number have any meaning? Toseewhether itdoes, weshould ‘compare itwith some other determination ofthe radius ofthese nuclei. For example, wecanmake another measurement oftheradius ofanucleus byseeing how itscatters fastparticles. From such measurements ithasbeen found, infact, that thedensity ofmatter inallnuclei isnearly thesame, i.e.,their volumes are proportional tothenumberofparticlestheycontain.IfweletAbethenumberofprotons and neutrons inanucleus (anumber very nearly proportional toitsmass), itis found that itsradius isgiven by r= Aro, (8.25) where ro=1.2X1078 em. 8.26) Fromthesemeasurements wefindthattheradiusofaB*?(oraC")nucleus isexpected tobe r= (12 X10-1)" =2.7X10-1 om. Comparing this result with (8.24), weseethat our assumptions that the energy difference between B'? and C'" iselectrostatic isfairly good; thedis- crepancy isonly about 15% (not bad forourfirst nuclear computation!). The reason forthediscrepancy isprobably thefollowing. According tothe currentunderstanding ofnuclei,anevennumberofnuclearparticles—in thecaseofB1?, fiveneutrons together with fiveprotons—makes akind ofcore; when one more particle isadded tothiscore, itrevolves around ontheoutsidetomakeanew sphericalnucleus,ratherthanbeingabsorbed.Ifthisisso,weshouldhavetaken adifferent electrostatic energy fortheadditional proton. Weshould have takentheexcessenergyofC!*overB*!tobejust Zea 4reqa” which istheenergy needed toadd one more proton tothe outside ofthecore. This number isjust 5/6ofwhat Eq.(8.23) predicts, sothenew prediction forthe radius is5/6 of(8.24), which isinmuch closer agreement with what isdirectly measured. Wecandraw twoconclusions from thisagreement. One isthat theelectrical lawsappear tobeworking atdimensions assmall as10" cm.Theother isthatwwehaveverifiedtheremarkable coincidence thatthenonelectrical partofthe forces between proton and proton, neutron and neutron, and proton and neutron are allequal. 8-5 Energy intheelectrostatic field ‘Wenowconsiderothermethodsofcalculatingelectrostatic energy.Theycan allbederived from thebasic relation Eq.(8.3), thesum, over allpairs ofcharges, ofthemutual energies ofeach charge-pair. First wewish towrite anexpression fortheenergy ofacharge distribution. Asusual, weconsider that each volumeelementdVcontainstheelementofchargepdV.ThenEq.(8.3)shouldbewritten _1fe@p@)UR5IFreoradV,dV. (8.27) rey ofgravitational attraction. Wealsoknow, byE=mc, thatmass andenergy are equivalent. Allenergyis,therefore, asourceofgravitational force.Ifwecouldnot locate theenergy, wecould notlocate allthemass. Wewould notbeable tosaywherethesourcesofthegravitational fieldarelocated. Thetheoryofgravitation would beincomplete. Ifwerestrict ourselves toelectrostatics there isreally noway totellwhere the energy islocated. The complete Maxwell equations ofelectrodynamics give us much more information (although even then theanswer is,strictly speaking, not unique.) Wewilltherefore discuss thisquestion indetail again inalater chapter. Wewill give you now only theresult fortheparticular case ofelectrostatics. ‘The energy islocated inspace, where theelectric field is.This seems reasonable because weknow that when charges areaccelerated they radiate electric fields.Wewouldliketosaythatwhenlightorradiowaves travelfromonepointtoanother,they carry their energy with them. But there arenocharges inthewaves. Sowe would like tolocate theenergy where theelectromagnetic field isand notatthe charges from which itcame. We thus describe theenergy, not interms ofthe charges, butinterms ofthefields they produce. Wecan, infact, show that Eq. (8.28) isnumerically equal to U=8fe-ea. (8.30) ‘Wecanthen interpret thisformula assaying that when anelectric field ispresent, there islocated inspace anenergy whose density (energy perunit volume) is =p. 2F.u=PE-E=% (831) Thisideaisillustrated inFig.8-8. AToshow that Eq.(8.30) isconsistent with ourlaws ofelectrostatics, webegin byintroducing intoEq.(8.28)therelation between pand¢thatweobtained in Chapter 6: p=60%. E Weget --2 [ovrU=zfor‘odV. (8.32) aw Writingoutthecomponents oftheintegrand, weseethat,2,(ao,ae#8) OW ove=¢ &+50 oa eA ~2 (6%)-(+26%)-(+269)-(8) 8-8.EachvolumealementdV= *ox(#)@)+away,ay)*32\?ae32)deddeinonelectricReldcontainsthe =V+ ¥6)—(¥8): (V9). (8.33) eneray (o/2)E av. ur energy integral isthen U=$[evey-cwoyav —2[o-wvow. WecanuseGauss’ theorem tochange thesecond integral into asurface integral: fVv@Vd)dV=f(V6)+nda. (8.34) va. atte ‘Weevaluate thesurface integral inthecase that thesurface goes toinfinity (Gothevolume integrals become integrals over allspace), supposing that allthe charges arelocated within some finite distance. The simple way toproceed isto take aspherical surface ofenormous radius Rwhose center isattheorigin of coordinates. Weknow that when wearevery faraway from allcharges, ¢varies as1/Rand¥¢as1/R®. (Both willdecrease even faster with Rifthere thenet eu charge inthedistribution iszero.) Since thesurface area ofthelarge sphere in- creases asR?,weseethatthesurface integral fallsoffas(1/R)(1/R*)R? =(1/R)astheradiusofthesphereincreases. Soifweinclude allspaceinourintegration (R— 2),thesurface integral goes tozero andwehave that v-2f wowoa=% few. (8.35) au an, Weseethat itispossible forustorepresent theenergy ofanycharge distribution asbeing theintegral over anenergy density located inthefield. 8-6 The energy ofapoint charge Our new relation, Eq.(8.35), says that even asingle point charge qwillhave some electrostatic energy. Inthiscase, theelectric field isgiven by -—_4_.E-4reor? Sotheenergy density atthedistance rfrom thecharge is, oe 2~3iwteort Wecantake foranelement ofvolume aspherical shell ofthickness drand area 4xr?, Thetotal energy is -{fia-- 01[-. uJSregr©~Fre7hoo (836) Now thelimit atr=cogives nodifficulty. Butforapoint charge weare supposed tointegrate down tor=0,which gives aninfinite integral. Equation (8.35)saysthatthereisaninfiniteamountofenergyinthefieldofapointcharge, although webegan with theidea that there wasenergy only between point charges. Inouroriginal energy formula foracollection ofpoint charges (Eq. 8.3), wedid notinclude anyinteraction energy ofacharge with itself. What hashappened is that when wewent over toacontinuous distribution ofcharge inEq.(8.27), we counted theenergy ofinteraction ofevery infinitesimal charge with allother infinitesimal charges. The same account isincluded inEq. (8.35), sowhen we apply ittoafinite point charge, weareincluding theenergy itwould take to assemble that charge from infinitesimal parts. You willnotice, infact, that we would also gettheresult inEq.(8.36) ifweused ourexpression (8.11) fortheenergy ofacharged sphereandlettheradiustendtoward zero. ‘Wemust conclude that theidea oflocating theenergy inthefield isincon- sistent with theassumption oftheexistence ofpoint charges. One way outofthe difficulty would betosaythat elementary charges, such asanelectron, arenot points butarereally small distributions ofcharge. Alternatively, wecould say that there issomething wrong inourtheory ofelectricity atvery small distances, orwith theidea ofthelocal conservation ofenergy. There aredifficulties with either point ofview. These difficulties have never been overcome; they exist tothis day. Sometime later, when wehave discussed some additional ideas, such asthe momentum inanelectromagnetic field, wewill give amore complete account of these fundamental difficulties inourunderstanding ofnature. an 9 Electricity inthe Atmosphere 9-1Theelectric potential gradient oftheamosphere Onanordinary dayover flatdesert country, orover thesea,asonegoes up- 9-1Theelectric potential gradient ward from thesurface oftheground theelectric potential increases byabout 100 oftheatmosphere volts permeter. Thus there isavertical electric field Eof100volts/m intheair. The signofthefieldcorresponds toanegative chargeontheearth’ssurface. This °-?Meefriecurrentsinthe means thatoutdoors thepotential attheheight ofyournoseis200voltshigher rosphere than thepotential atyour feet! You might ask: “Why don't wejuststick apair of 9-3 Origin oftheatmospheric electrodes outintheaironemeter apart and usethe100volts topower ourelectric currents, lights?” Oryou might wonder: “Ifthere isreally apotential difference of200voltsbetween mynoseandmyfeet,whyisitIdon’tgetashockwhenIgooutinto 9-4‘Thunderstorms thestreet?” 9-5Themechanism ofcharge ‘Wewill answer thesecond question first. Your body isarelatively good separation conductor. Ifyouaeincontact withtheground, youandthe ground willend106 smimake oneequipotential surface. Ordinarily, theequipotentials areparallel tothe . surface, asshown inFig, 9-1(@), butwhen you arethere, theequipotentials are distorted, andthefield looks somewhat asshown inFig. 9-1(b). Soyoustillhave very nearly zero potential difference between your head and your feet. There arechargesthatcomefromtheearthtoyourhead,changingthefield.SomeofthemReference: Chalmers, J.Alan,Atmos-may bedischarged byions collected from theair,butthecurrent ofthese isvery pheric Electricity, Pergamon small because airisapoor conductor. Press, London (1957). te — Tots e +300V ao “>> ~tye__-____-~ ~~~ 2 S200 eas ~~2 L ~ ~ tevUNootpS. ~ le=100¥/m ne [- sS toy fF (~__i___ - - oo ---- ~----- 77777 erowna 777777 ROUND ) ) " Fig. 9-1. (a)Thepotential distribution above theearth. (b)Thepotential distribution near amaninanopen flatplace. How canwemeasure such afield ifthefield ischanged byputting something there? There areseveral ways. One way istoplaceaninsulatedconductor atsome distance above theground andleave itthere until itisatthesame potential asthe air. Ifweleave itlong enough, thevery small conductivity intheairwillletthe charges leak off(oronto) theconductor until itcomes tothepotential atitslevel.Thenwecanbringitbacktotheground,andmeasuretheshiftofits potential as wedos0. Afaster way isto lettheconductor beabucket ofwater with asmall leak. Asthewater drops out, itcarries away anyexcess charges and thebucket willapproach thesame potential astheair. (The charges, asyouknow, reside on thesurface, and asthedrops come off“pieces ofsurface” break off.) Wecan meas- urethepotential ofthebucket with anelectrometer. m4 ‘There isanother way todirectly measure thepotential gradient. Since there isanelectric field, there isasurface charge ontheearth (¢=€9£). Ifweplace aflatmetal plate attheearth’s surface andground it,negative charges appear on ||€| it(Fig.9-2a).IfthisplateisnowcoveredbyanothergroundedconductingcoverB,thecharges will appear onthecover, and there will benocharges ontheoriginal plateA.IfwemeasurethechargethatflowsfromplateAtotheground(by,say, sommerron 8 u SOSmuseagalvanometer inthegroundingwire)aswecoverit,wecanfindthesurface STI oechargedensitythatwasthere,andthereforealsofindtheelectricfield. ay Having suggested how wecan measure theelectric field intheatmosphere, ‘wenow continue ourdescription ofit.Measurements show, first ofall,that the fieldcontinuestoexist,butgetsweaker,asonegoesuptohighaltitudes.Byabout ||e| 50kilometers,thefieldisverysmall,somostofthepotentialchange(theintegralof£)isatlower altitudes. The total potential difference from thesurface ofthe _- _ ___OVENAATE® earth tothetopoftheatmosphere isabout 400,000 volts. 777AT FTFFFF io) 9-2 Electric currents intheatmosphere Fig.9-2. (a)Agrounded metalplate Another thingthatcanbemeasured, inaddition tothepotential gradient, is willhave thesame surface charge osthe thecurrent intheatmosphere. Thecurrent density issmall—about 10micromicro- earth. (b)Iftheplate iscovered with a amperes crosses each square meter parallel totheearth. The airisevidently nota grounded conductor itwill have no perfect insulator, andbecause ofthisconductivity, asmall current—caused bythe surface charge. electric field wehave just been describing—passes from theskydown totheearth. ‘Why does theatmosphere have conductivity? Here andthere among theair molecules there isanion—a molecule ofoxygen, say, which has acquired an extra electron, orperhaps lost one. These ions donotstay assingle molecules; ‘because oftheir electric field they usually accumulate afewother molecules around them. Each ionthen becomes alittle lump which, along with other lumps, drifts inthefield—movingslowlyupwardordownward—making theobservedcurrent. \. bw‘Wheredotheionscomefrom?Itwasfirstguessedthattheionswereproducedby 1+tons7>~airtheradioactivity oftheearth.(Itwasknownthattheradiationfromradioactive =v Ser =, materials would make airconducting byionizing the airmolecules.) Particles = = likeB-rayscomingoutoftheatomicnucleiaremovingsofastthattheytearelec- trons from theatoms, leaving ions behind. This would imply, ofcourse, that if ELECTROMETER wwewere togotohigher altitudes, weshould find lessionization, because theradio- activity isallinthedirtontheground—in thetracesofradium, uranium, po- Fig.9-3.Measuring theconductivity tassium, etc. ofairduetothemotion ofions. Totestthistheory, some physicists carried anexperiment upinballoons to measure theionization oftheair(Hess, in1912) anddiscovered that theopposite ‘was true—the ionization perunit volume increased with altitude! (The apparatuswaslikethatofFig.9-3.Thetwoplateswerechargedperiodically tothepotentialV.Due totheconductivity oftheair,theplates slowly discharged; therate of discharge was measured with the electrometer.) This was amost mysterious result—the most dramatic finding intheentire history ofatmospheric electricity. Itwas sodramatic, infact, that itrequired abranching offofanentirely new subject—cosmic rays. Atmospheric electricity itself remained less dramatic. Tonization was evidently being produced bysomething from outside theearth; theinvestigation ofthissourceledtothediscovery ofthecosmicrays.Wewillnotdiscuss thesubject ofcosmic rays now, except tosaythat they maintain the supply ofions. Although theions arebeing swept away allthetime, new ones are being created bythecosmic-ray particles coming from theoutside. Tobeprecise, wemust saythat besides theions made ofmolecules, there are alsootherkindsofions.Tinypiecesofdirt,likeextremely finebitsofdust,floatintheairand become charged. They aresometimes called “nuclei.” Forexample, when awave breaks inthesea,little bitsofspray arethrown into theair. When ‘oneofthese drops evaporates, itleaves aninfinitesimal crystal ofNaC! floating in the air, These tiny crystals can then pick upcharges and become ions; they arecalled “large ions.” ‘The small ions—those formed bycosmic rays—are themost mobile. Because ‘they aresosmall, they move rapidly through theair—with aspeed ofabout 1 92 cm/sec inafield of100volts/meter, or1volt/em. Themuch bigger andheavier ionsmovemuchmoreslowly. Itturnsoutthatiftherearemany“nuclei,” theywillpick upthecharges from thesmall ions. Then, since the“large ions” move so slowly inafield, thetotal conductivity isreduced. Theconductivity ofair,there- fore, isquite variable, since itisvery sensitive totheamount of“dirt” there isinit. There ismuch more ofsuch dirt over land—where thewinds canblow updust ‘orwhere man throws allkinds ofpollution intotheair—than there isover water. Itisnotsurprising thatfrom daytoday, from moment tomoment, from place toplace, theconductivity near theearth’s surface varies enormously. Thevoltage gradient observed atanyparticular place ontheearth’s surface also varies greatly ‘because roughly thesame current flows down from highaltitudes indifferent places, and thevarying conductivity near theearth results inavarying voltage gradient. ‘Theconductivity oftheairduetothedrifting ofions alsoincreases rapidly with altitude—for tworeasons. First ofall,theionization from cosmic rays in- creases with altitude. Secondly, asthedensity ofairgoes down, themean freepathoftheionsincreases, sothattheycantravelfartherintheelectricfieldbeforetheyhave acollision—resulting inarapid increase ofconductivity asonegoes up. Although theelectric current-density intheairisonly afewmicromicro- amperes persquare meter, there arevery many square meters ontheearth’s surface. The total electric current reaching theearth’s surface atany time isvery nearly constant at1800 amperes. This current, ofcourse, is“positive”—it carries plus charges totheearth. Sowehave avoltage supply of400,000 volts with acurrent of1800 amperes—a power of700megawatts! With such alarge current coming down, thenegative charge ontheearth should soon bedischarged. Infact, itshould take only about halfanhour todis- condStuviry charge theentire earth. Buttheatmospheric electric fieldhasalready lasted more sopoom— P= = thanahalf-hoursinceitsdiscovery.Howisitmaintained? Whatmaintainsthe 4 Jourvoltage? Andbetweenwhatandtheearth?Therearemanyquestions. 000 SeaTheearthisnegative,andthepotentialintheairispositive. Ifyougohigh vours snrenough, theconductivity issogreat that horizontally there isnomore chance for voltage variations. Theair,forthescale oftimes that wearetalking about, be- 4V8,1 comes effectively aconductor. This occurs ataheight intheneighborhood of50 tairn'ssinrace kilometers. This isnotashigh aswhat iscalled the“ionosphere,” inwhich there ; Aareverylargenumbersofionsproducedbyphotoelectrcity fromthesun.Never-,,Fi0.9-4.Typicaltence! condetheless, forourdiscussions ofatmospheric electricity, theairbecomes sufficiently 7 conductive atabout 50kilometers that wecanimagine that there ispractically a perfect conducting surface atthis height, from which thecurrents come down. Our picture ofthesituation isshown inFig. 9-4. The problem is:How isthe positive charge maintained there? How isitpumped back? Because ifitcomes down totheearth, ithastobepumped back somehow. That was one ofthe greatest puzzles ofatmospheric electricity forquite awhile. quem Each piece ofinformation wecangetshould give aclue or,atleast, tellyou something about it.Here isaninteresting phenomenon: Ifwemeasure thecurrent (which ismore stable than thepotential gradient) over thesea,forinstance, orin we carefulconditions, andaverageverycarefully sothatwegetridofthe irregularities, wediscover that there isstilladaily variation, Theaverage ofmany measurements over the oceans has avariation with time roughly asshown inFig. 9-5. The currentvariesbyabout+15percent, anditislargestat7:00p.m.inLondon. The strange part ofthething isthatnomatter where youmeasure thecurrent—in the t Atlantic Ocean, thePacific Ocean, ortheArctic Ocean—it isatitspeak value ° yours our when theclocks inLondon say7:00 P..! Allover theworld thecurrent isatits maximum at7:00 P.s.London time anditisataminimum at4:00amt.London Fig,9-5. Theaverage daily vario- time. Inother words, itdepends upon theabsolute time ontheearth, notupon _tionoftheatmospheric potential gradient thelocal time attheplace ofobservation. Inonerespect thisisnotmysterious; onaclear dayover theoceans; referred itchecks with ouridea that there isavery high conductivity laterally atthetop, 10Greenwich time, because that makes itimpossible forthevoltage difference from theground to thetoptovary locally. Any potential variations should beworldwide, asindeed they are. What wenow know, therefore, isthat thevoltage atthe“top” surface isdropping and rising by15percent with theabsolute time ontheearth. 33 9-3 Origin oftheatmospheric currents We must next talk about the source ofthe large negative currents: which ust beflowing from the“top” tothesurface oftheearth tokeep charging itup negatively. Where arethebatteries thatdothis? The“battery” isshown inFig.9-6.Itisthethunderstorm anditslightning. Itturnsoutthattheboltsoflightningdonot“discharge” thepotential wehave been talking about (asyoumight at firstguess).Lightning stormscarrynegativechargestotheearth.Whenalightningboltstrikes,ten-to-oneitbringsdownnegativechargestotheearthinlargeamountsItisthethunderstorms throughout theworld thatarecharging theearth with an average of1800 amperes, which isthen being discharged through regions of fair weather. There areabout 300thunderstorms perday allover theearth, and wecan think ofthem asbatteries pumping theelectricity totheupperlayerandmaintainingthevoltage difference. Then takeintoaccount thegeography oftheearth— there are thunderstorms inthe afternoon inBrazil, tropical thunderstorms in Africa, andsoforth. People have made estimates ofhowmuch lightning isstriking world-wide atanytime, andperhaps needless tosay,their estimates more oFless agree withthevoltage difference measurements: thetotal amount ofthunderstorm activity ishighest onthewhole earth atabout 7:00 P.M.inLondon. However, thethunderstorm estimates arevery difficult tomake andwere 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 ispeak intheactivity at7:00 p.at. Greenwich Mean Time. a nani ‘ % 24 «ft a Fig.9-6. Themechanism thatgenerates theatmospheric electric field. [Photo byWilliamL.Widmayer.) 34 Inorder tounderstand how these batteries work, wewill look atathunder- ‘storm indetail. What isgoing oninside athunderstorm? We will describe this insofarasitisknown. Aswegetintothismarvelous phenomenon ofrealnature— instead oftheidealized spheres ofperfect conductors inside ofother spheres that ‘wecansolve soneatly—we discover that wedon’t know very much. Yetitisreally quite exciting. Anyone who hasbeen inathunderstorm hasenjoyed it,orhasbeen frightened, oratleasthashadsome emotion, Andinthose places innature where j= wegetanemotion, wefind that there isgenerally acorresponding complexity and > mysteryaboutit.Itisnotgoingtobepossibletodescribeexactlyhowathunder-_|e22.— ttttt si storm works, because wedonotyetknow verymuch. Butwewilltrytodescribe fort VAAN AYalittlebitaboutwhathappens. gies|cgactennonnten | Parr FTA 9-4Thunderstorms aprATTBNAeeeInthefirstplace, anordinary thunderstorm ismade upofanumberof“cells”|...{_s-2--~=-s ont fairlyclosetogether,butalmostindependentofeachother.Soitisbesttoanalyze rr onecellatatime.Bya“cell”wemeanaregionwithalimitareainthehorizontal 5<aceeeeeeeenn sdirectioninwhichallofthebasicprocesses occur.Usuallythereareseveralcellssam—{__a2-22-=="===n==.side byside, andineach oneabout thesame thing ishappening, although perhaps with adifferent timing. Figure 9-7indicates inanidealized fashion what such a celllooks likeintheearly stage ofthethunderstorm. Itturns outthatinacertain, —<=S=mSReK=RENEE ENR place intheair,under certain conditions which weshall describe, there isageneral he oom risingoftheair,withhigherandhighervelocitiesnearthetop.Asthewarm,[2sweersea!bn moist airatthebottom rises, itcools and condenses. Inthefigure thelittle crosses indicate snow andthedotsindicate rain, butbecause theupdraft currents aregreat Fig.9-7. Athunderstorm cellinthe enough andthedrops aresmall enough, thesnow andraindonotcome down ateatly stages ofdevelopment. [From U.S. thisstage. Thisisthebeginning stage, andnottherealthunderstorm yet—in the Department ofCommerce WeatherBureausensethatwedon’thaveanything happening attheground. Atthesametimethat RePort, June1949.) thewarm airrises, there isanentrainment ofairfrom thesides—an important point which wasneglected formany years. Thus itisnotjust theairfrom below which isrising, butalso acertain amount ofother airfrom thesides. ‘Why does theairriselikethis? Asyou know, when yougoupinaltitude the airiscolder. ‘The ground isheated bythesun, and there-radiation ofheat tothe ‘skycomes from water vapor high intheatmosphere; soathigh altitudes theair iscold—very cold—whereas lower down itiswarm. You may say, “Then it's very simple. Warm airislighter than cold; therefore thecombination ismechan- ‘ icallyunstable andthewarmairrises.”Ofcourse,ifthetemperature isdifferent > Natdifferent heights, theairisunstable thermodynamically. Lefttoitselfinfinitely - long, theairwould allcome tothesame temperature. Butitisnotlefttoitself; N € thesunisalways shining (during theday). Sotheproblem isindeed notoneof \ thermodynamic equilibrium, butofmechanical equilibrium. Suppose weplot—as_ x ¢ inFig.9-8—the temperature oftheairagainst height above theground. In WN > ordinary circumstances wewould getadecrease along acurve liketheonelabeled \ (a);astheheight goes up,thetemperature goes down. How cantheatmosphere bestable? Whydoesn’t thehotairbelow simply riseupintothecoldair?The ALTITUDE. answer isthis: iftheairwere togoup,itspressure would godown, and ifwe .consideraparticularparce!ofairgoingup,itwouldbeexpanding adiabatically. (al"eene8.omennean ee (Therewouldbenoheatcominginoroutbecauseinthelargedimensions con-coolingofdrycir;(c)adicbaticcooling sideredhere,thereisn'ttimeformuchheatflow.)Thustheparcelofairwould ofwetair;(d)wetairwithsomemixing cool asiit rises. Such anadiabatic process would giveatemperature-height relation-ofambientair. ship likecurve (b)inFig. 9-8. Any airwhich rose from below would becolder than theenvironment itgoes into. Thus there isnoreason forthehotairbelow torise; ifitwere torise, itwould cool toalower temperature than theairalready there,wouldbeheavier thantheairthere,andwouldjustwanttocomedownagain. Onagood, bright daywith very little humidity there isacertain rate atwhich the temperature intheatmosphere falls, and this rate is,ingeneral, lower than the “maximum stable gradient,” which isrepresented bycurve (b). The airisin stable mechanical equilibrium. os fax Ontheotherhand,ifwethinkofaparcelofairthatcontainsalotofwateree aoneas vapor beingcarriedupintotheair,itsadiabatic coolingcurvewillbedifferent. Asa itexpands andcools, thewater vapor initwillcondense, andthecondensing water fowillliberateheat.Moistair,therefore,doesnotcoolnearlyasmuchasdryair TonTee, idoes.Soifairthatiswetterthantheaveragestartstorise,itstemperature will afollowacurvelike(c)inFig.9-8.Itwillcooloffsomewhat, butwillstillbewarmerxsaan reer thanthesurrounding airatthesamelevel.IfwehavearegionofwarmmoistSee asad {Jairandsomethingstartsitrising,itwillalwaysfinditselflighterandwarmerthanaeRRNNCENT yaa—“ theairarounditandwillcontinuetoriseuntilitgetstoenormousheights.ThisSONA ete+isthemachinery thatmakestheairinthethunderstorm cellrise. 0sath ae2a peFormanyyearsthethunderstorm cellwasexplainedsimplyinthismanner.fet eae Butthenmeasurements showedthatthetemperature ofthecloudatdifferent go} ategeAVA Yodheights wasnotnearlyashighasindicated bycurve(c).Thereason isthatasthetantaa et|moistair“bubble” goesup,itentrainsairfromtheenvironment andiscooledle ae ELTTLE Tdoftbyit.Thetemperature-versus-height curvelooksmorelikecurve(4),which LST parte ere ismuch closer totheoriginal curve (a)thantocurve (¢).7za.UML r|Aftertheconvectionjustdescribedgetsunderway,thecrosssectionofa Ey aE thunderstorm celllookslikeFig.9-9.Wehavewhatiscalleda“mature”thunder- jaaaBeeadieaeestorm,Thereisaveryrapidupdraftwhich,inthisstage,goesuptoabout10,000 aaeaereimerpeniene _ "|to15,000 meters—sometimes even much higher. The thunderheads, with their —Hee condensation, climbwayupoutofthegeneralcloudbank,carriedbyanupdraftPeeteteFreememe thatisusuallyabout60milesanhour.Asthewatervaporiscarriedupand; condenses, itforms tiny drops which arerapidly cooled totemperatures below Fig.9-9. Amature thunderstorm cell. zerodegrees, Theyshould freeze, butdonotfreeze immediately—they are“super-fromUS.Department, ofwee cooled.”Waterandotherliquidswillusuallycoolwellbelowtheirfreezingpointsport ! before crystallizing ifthere areno“nuclei” present tostartthecrystallization process. Only ifthere issome small piece ofmaterial present, like atinycrystal of ‘NaCl, willthewater drop freeze into alittle piece ofice. Then theequilibrium is such that thewater drops evaporate and theicecrystals grow. Thus atacertainpointthereisarapiddisappearance ofthewaterandarapidbuildupofice.Also,there may bedirect collisions between thewater drops and theice—collisions 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, they begin tofallthrough therising air—they gettooheavy tobesupported anylonger intheupdraft. Asthey come down, they draw alittle airwith them and start adowndraft, And surprisingly enough, itiseasy toseethat once thedowndraft isstarted, itwill maintain itself. The air now drives itself down! Noticethatthecurve(4)inFig.9-8fortheactualdistribution oftemperature inthecloudisnotassteepascurve(c),whichappliestowetair.Soifwehavewetairfalling, itstemperature willdrop with theslope ofcurve (c)and willgobelow thetemperature oftheenvironment ifitgets down farenough, asindicated by curve (¢)inthefigure. The moment itdoes that, itis denser than theenvironment and continues tofallrapidly. You say, “That isperpetual motion. First, you argue that theairshould rise, andwhen youhave itupthere, youargue equally well that theairshould fall.” But itisn’t perpetual motion. When thesituation isunstable andthewarm airshould rise, then clearly something hastoreplace thewarm air. Itisequally true that cold aircoming down would energetically replace thewarm air,butyou realize that what iscoming down isnottheoriginal air. The early arguments, that had aparticular cloud without entrainment going upand then coming down, had some kind ofapuzzle. They needed therain tomaintain thedowndraft—an argument whichishardtobelieve.Assoonasyourealizethatthereisalotoforiginal airmixed inwith therising air,thethermodynamic argument shows that there canbeadescent ofthecold airwhich was originally atsome great height. This explains thepicture oftheactive thunderstorm sketched inFig. 9-9. ‘Astheaircomes down, rain begins tocome outofthebottom ofthethunder- storm. Inaddition, therelatively cold airspreads outwhen itarrives attheearth's surface. Sojust before therain comes there isacertain little cold wind that gives, 96 usaforewarning ofthecoming storm. Inthestorm itself there arerapid andir- regular gusts ofair,there isanenormous turbulence inthecloud, and soon, But basically wehave anupdraft, then adowndraft—in general, avery complicated 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 cellafter about one-half anhour toanhour. Thecelllooks asshown inFig.9-10. Theupdraft stops because there isnolonger enough warm airtomaintain it.Thedownward precipitation continues forawhile, thelastlittle bitsofwater come out,andthings getquieter andquieter—although there aresmall icecrystals leftwayupintheair. Because thewinds atvery great altitude areindifferent directions, thetopofthecloud usually spreads into an anvil shape. The cellcomes totheendofitslife. far —SSSSS oe ez 7 wePte!zh =etereD| PORN ABELL1.1 ANNO eee EETat Risers Fig.9-10. Thelatephase ofathunderstorm Fig.9-11. Thedistribution ofelectrical charges ina cell. [From U.S,DepartmentofCommerceWeather maturethunderstorm cell.[FromU.S.Department ofCom- Bureau Report, June 1949.) merce Weather Bureau Report, June 1949.) 9-5 The mechanism ofcharge separation Wewant now todiscuss themost important aspect forourpurposes—the development oftheelectrical charges. Experiments ofvarious kinds—including flying airplanes through thunderstorms (the pilots who dothisarebrave men!)— tellusthat thecharge distribution inathunderstorm cellissomething like that 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 why itisthere, 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 sign forthebattery needed todrive theearth negative. The positive charges are 6or7kilometers upintheair,where thetemperature isabout —20°C, whereas thenegative charges are3or4kilometers high, where thetemperature isbetween zero and —10°C. Thechargeatthebottomofthecloudislargeenoughtoproducepotentialdifferences of20, of30,oreven 100 million volts between thecloud and theearth— much bigger than the0.4million volts from the“sky” totheground inaclear 7 atmosphere. These large voltages break down theairand create giant arcdis- charges. When thebreakdown occurs thenegative charges atthebottom ofthe thunderstorm arecarried down totheearth inthelightning strokes. Now wewill describe insome detail thecharacter ofthelightning. First of all,there arelarge voltage differences around, sothat theairbreaks down. There arelightning strokes between one piece ofacloud and another piece ofacloud, orbetween one cloud and another cloud, orbetween acloud and theearth. In eachoftheindependent discharge flashes—the kindoflightningstrokesyousee—thereareapproximately 20or30coulombs ofcharge brought down. One question is:How long does ittake forthecloud toregenerate the20or30coulombs which aretaken away bythelightning bolt? This can beseen bymeasuring, farfrom a cloud, theelectric field produced bythecloud’s dipole moment. Insuch measure~ ‘ments you seeasudden decrease inthefield when thelightning strikes, and then ‘anexponential return totheprevious value with atime constant which isslightly differentfordifferentcasesbutwhichisintheneighborhood of5seconds.Ittakes athunderstorm only 5seconds after each lightning stroke tobuild itscharge up again. That doesn’t necessarily mean that another stroke isgoing tooccur in exactly 5seconds every time, because, ofcourse, thegeometry ischanged, and soon. ferns Thestrokes occur more orlessirregularly, buttheimportant point isthatittakesVa Sa about5secondstorecreatetheoriginalcondition. Thusthereareapproximately~\ 4amperes ofcurrent inthegenerating machine ofthethunderstorm. This means thatanymodelmadetoexplainhowthisstormgenerates itselectricity mustbeonewith plenty ofjuice—it must beabig,rapidly operating device. Before wegofurther weshall consider something which isalmost certainly completely irrelevant, butnevertheless interesting, because itdoes show theeffect vy ofanelectric field onwater drops. Wesaythat itmay beirrelevant because it YA relates toanexperiment onecandointhelaboratory with astream ofwater to show therather strong effects oftheelectric field ondrops ofwater. Inathunder- storm there isnostream ofwater; there isacloud ofcondensing iceanddrops of water. Sothequestion ofthemechanisms atwork inathunderstorm isprobably Tomater notatallrelated towhat youcanseeinthesimple experiment wewill describe. surrey Ifyoutakeasmallnozzleconnected toawaterfaucetanddirectitupward ataFig.9-12.AjetofwaterwithanSteDangle,asinFig.9-12,thewaterwillcomeoutinafinestreamthateventuallyelectric feld near thenozzle. breaks upinto aspray offinedrops. Ifyounow putanelectric field across the stream atthenozzle (bybringing upacharged rod, forexample), theform ofthe stream will change. With aweak electric field you will find that thestream breaks upinto asmaller number oflarge-sized drops. Butifyou apply astronger field, . thestream breaks upinto many, many finedrops—smaller than before.* With a weak electric field there isatendency toinhibit thebreakup ofthestream into drops. With astronger field, however, there isanincrease inthetendency tosepa~ rate into drops. The explanation ofthese effects isprobably thefollowing. Ifwehave the stream ofwater coming outofthenozzle and weputasmall electric field across itonesideofthe water gets slightly positive andtheother sidegets slightly negative. Then, when thestream breaks, thedrops ononeside may bepositive, andthose on theother side may benegative. They will attract each other and will have atend- ency tostick together more than they would have before—the stream doesn’t break upasmuch. Ontheother hand, ifthefield isstronger, thecharge ineach oneofthedrops gets much larger, and there isatendency forthecharge itself to help break upthedrops through their own repulsion. Each drop willbreak into many smaller ones, each carrying acharge, sothat they areallrepelled, and spread outsorapidly. Soasweincrease thefield, thestream becomes more finely separated. The only point wewish tomake isthatincertain circumstances electric fields can have considerable influence onthedrops. The exact machinery by which something happens inathunderstorm isnotatallknown, and isnotatall necessarily related towhat wehave justdescribed. Wehave included itjustsothat, *Ahandywaytoobservethesizesofthedropsistoletthestreamfallonalargethin ‘metal plate. The larger drops make alouder noise. os you will appreciate thecomplexities that could come into play. Infact, nobody hhasatheory applicable toclouds based onthat idea. We would like todescribe two theories which have been invented toaccount fortheseparation ofthecharges inathunderstorm. Allthetheories involve the ideathatthereshouldbesomechargeontheprecipitation particlesandadifferentcharge intheair.Then bythemovement oftheprecipitation particles—the water ortheice—through theairthere isaseparation ofelectric charge. The only ques- tion is:How does thecharging ofthedrops begin? One oftheolder theories is called the“breaking-drop” theory. Somebody discovered that ifyou have adrop ofwater that breaks into twopieces inawindstream, there ispositive charge onthe water and negative charge intheair. This breaking-drop theory has several disadvantages, among which themost serious isthat thesign iswrong. Second, inthelarge number oftemperate-zone thunderstorms which doexhibit lightning, theprecipitation effects athigh altitudes areinice,notinwater. ‘From whatwehavejustsaid,wenotethatifwecould imagine somewayfor Bans thecharge tobedifferent atthetopandbottom ofadropandifwecouldalsosee some reason why drops inahigh-speed airstream would break upinto unequal le pieces—a largeoneinthefrontandasmaller oneinthebackbecause ofthe motion through theairorsomething—we would have atheory. (Different from anyknown theory!) Then thesmall drops would notfallthrough theairasfastasthebig ‘ones,because oftheairresistance, andwewouldgetachargeseparation. You “ oyse¢,itispossible toconcoct allkindsofpossibilities. @Oneofthemoreingenious theories, whichismoresatisfactory inmanyre- lv spects than thebreaking-drop theory, isduetoC.T.R.Wilson. Wewilldescribe LARGE 10s, it,asWilson did, with reference towater drops, although thesame phenomenon would also work with ice. Suppose wehave awater drop that isfalling intheelectricfieldofabout100voltspermetertowardthenegatively chargedearth.The_Fig.9-13._C.T.R.Wilton'stheoryof drop willhave aninduced dipole moment—with thebottom ofthedrop positive charge separation inathundercloud.andthetopofthedropnegative,asdrawninFig.9-13.Nowthereareintheairthe “nuclei” that wementioned earlier—the large slow-moving ions. (The fast ions donothave animportant effect here.) Suppose that asadrop comes down, itapproaches alargeion.Iftheionispositive, itisrepelled bythepositive bottom ofthedrop and ispushed away. Soitdoes notbecome attached tothedrop. Iftheionwere toapproach from thetop, however, itmight attach tothenegative, topside. Butsince thedrop isfalling through theair,there isanairdrift relative toit,going upwards, which carries theions away iftheir motion through theair isslow enough. Thus thepositive ions cannot attach atthetopeither. This would apply, you see,only tothelarge, slow-moving ions. The positive ions ofthistypewillnotattachthemselves eithertothefrontorthebackofafallingdrop.Ontheother hand, asthelarge, slow, negative ions areapproached byadrop,theywillbeattractedandwillbecaught.Thedropwillacquirenegativecharge—thesign ofthecharge having been determined bytheoriginal potential difference ontheentire earth—and wegettheright sign. Negative charge will bebroughtdowntothebottompartofthecloudbythedrops,andthepositively chargedionswhich areleftbehind willbeblown tothetopofthecloud bythevarious updraft currents. The theory looks pretty good, and itatleast gives theright sign. Also it doesn’t depend onhaving liquid drops. Wewillsee,when welearn about polariza- tion inadielectric, that pieces oficewilldothesame thing. They also willdevelop positive andnegative chargesontheirextremities whentheyareinanelectricfield. ‘There are, however, some problems even with this theory. First ofall,the total charge involved inathunderstorm isvery high. Afterashorttime,thesupply oflarge ions would getused up. SoWilson andothers have hadtopropose thatthereareadditional sourcesofthe large ions. Once thecharge separation starts, very large electric fields aredeveloped, andinthese large fields there may beplaces where theairwillbecome ionized. Ifthere isahighly charged point, oranysmall object likeadrop, itmay concentrate thefield enough tomake a“brush discharge.” When there isastrong enough electric field—Iet ussay itispositive—electrons will fallinto thefield and will pick upalotofspeed between collisions. Their speed willbesuch that inhitting another atom they willtear electrons offatthat 2 atom, leaving positive charges behind. These new electrons also pick upspeed and collide with more electrons. Soakind ofchain reaction oravalanche occurs, and there isarapid accumulation ofions. ‘The positive charges areleftnear their original positions, sotheneteffect istodistribute thepositive charge onthepoint intoaregionaroundthepoint.Then,ofcourse,thereisnolongerastrongfield, andtheprocess stops. This isthecharacter ofabrushdischarge. Itispossiblethat thefieldsmaybecomestrongenoughinthecloudtoproducealittlebitofbrushdischarge; there may also beother mechanisms, once thething isstarted, topro- duce alarge amount ofionization. Butnobody knows exactly how itworks. So thefundamental origin oflightning isreally notthoroughly understood. Weknow ©cy a jitcomesfromthethunderstorms. (Andweknow, ofcourse, thatthunder comes P* fromthelightning—from thethermal energy released bythebolt.) ‘ Atleast wecanunderstand, inpart, theorigin ofatmospheric electricity. Due \ totheaircurrents, ions, andwater drops oniceparticles inathunderstorm, positiveFis|andnegativechargesareseparated. Thepositivechargesarecarriedupwardto ’y thetopofthecloud(seeFig.9-11),andthenegativechargesaredumpedintothegroundinlightning strokes. Thepositivechargesleavethetopofthecloud,enter / thehigh-altitude layersofmorehighlyconductingair,andspreadthroughoutthe cx earth.Inregionsofclearweather,thepositivechargesinthislayerareslowly <i conducted totheearth bytheionsintheair—ions formed bycosmic rays,bythe i sea,andbyman’s activities. The atmosphere isabusy electrical machine! \ 9-6Lightning . 4 The first evidence ofwhat happens inalightning stroke wasobtained in photographs taken with acamera held byhand andmoved back and forth with theshutter open—while pointed toward aplace where lightning was expected. a ‘The first photographs obtained this way showed clearly that lightning strokes are usually multiple discharges along the same path. Later, the “Boys” camera, Fig.9-14,Photograph ofalightning Whichhaszwolensesmounted180°apartonarapidlyrotatingdisc,wasdeveloped.fashtaken witha"Boys" camera, [From Theimage made byeachlensmoves across thefilm—the picture isspread outin ‘Schonland, Malan, andCollens, Proc. Roy. time. If,forinstance, thestroke repeats, there willbetwoimages sidebyside. Soe. London, Vol. 152(1935).] Bycomparing theimages ofthetwolenses, itispossible towork outthedetails ofthetimesequence oftheflashes. Figure 9-14showsaphotograph takenwitha “Boys” camera. ‘Wewillnowdescribethelightning.Again,wedon’tunderstandexactlyhow J72/LLitworks.WewillgiveaqualitativedescriptionofwhatitJookslike,butwewon't NcLoup 80intoanydetailsofwhyitdoeswhatitappearstodo.Wewilldescribeonlythe4 ordinary caseofthecloud with anegative bottom overflatcountry. Itspotential ra ismuch more negative thantheearth underneath, sonegative electrons willbeaccelerated toward theearth. What happens isthefollowing. Itallstarts with a thing called a“step leader,” which isnotasbright asthestroke oflightning. On thephotographsonecanseealittlebrightspotatthebeginningthatstartsfromthe i. cloudandmovesdownward veryrapidly—at asixthofthespeedoflight!Itgoes y only about 50meters and stops. Itpauses forabout 50microseconds, and then SS takes another step. Itpauses again andthen goes another step, andsoon. It moves inaseries ofsteps toward theground, along apath likethat shown inFig. 9-15. Intheleader there arenegative charges from thecloud; thewhole column . isfullofnegative charge. Also, theairisbecoming ionized bytherapidly moving chargesthatproducetheleader,sotheairbecomesaconductoralongthepath TRAIL TTT tracedout.Themomenttheleadertouchestheground,wehaveaconducting EARTH “wire”thatrunsallthewayuptothecloudandisfullofnegative charge. Now, atlast, thenegative charge ofthecloud can simply escape and run out. TheFig.9-15.Theformationofthe“stepelectronsatthebottomoftheleaderarethefirstonestorealizethis;theydump leader.” out, leaving positive charge behind that attracts more negative charge from higher upintheleader, which initsturn pours out,etc. Sofinally allthenegative charge inapart ofthecloud runs outalong thecolumn inarapid and energetic way. Sothelightning stroke youseeruns upwards from theground, asindicated inFig. 9-16. Infact, thismain stroke—by farthebrightest part—is called thereturn 910 stroke. Ytiswhat produces thevery bright light, andtheheat, which bycausing arapid expansion oftheairmakes thethunder clap. \ ‘Thecurrent inalightning strokeisabout10,000amperes atitspeak,andit |ncarriesdownabout20coulombs. / } LButwearestillnotfinished.Afteratimeof,perhaps,afewhundredthsofa/JLL second,whenthereturnstrokehasdisappeared, anotherleadercomesdown. <=Zz Butthis time there arenopauses. Itiscalled a“dark leader” thistime, and it 0¢s allthewaydown—from toptobottom inoneswoop. Itgoes fullsteam on S exactly theoldtrack, because there isenough debris there tomake ittheeasiest route. The new leader isagain fullofnegative charge. The moment ittouches the ground—zing!—there isareturn stroke going straight upalong thepath. Soyou y seethelightning strike again, andagain, andagain. Sometimes itstrikes only an ‘once ortwice, sometimes fiveortentimes—once asmany as42times onthesame track wasseen—but always inrapid succession. ‘Sometimes things geteven more complicated. Forinstance, after oneofitspausestheleadermaydevelopabranchbysendingouttwosteps—both towardtheground butinsomewhat different directions, asshown inFig. 9-15. What happens thendepends onwhether onebranch reaches theground definitely before theother. TTR NTATATVW Ifthatdoes happen, thebright return stroke (ofnegative charge dumping into theground) worksitswayupalongthebranchthattouchestheground, andwhenit_Fig:9-16.Theretumlightning strokereachesandpassesthebranching pointonitswayuptothecloud,abrightstroke"Wsbackupthepathmadebytheleader.appears togodown theother branch. Why? Because negative charge isdumping outand that iswhat lights upthebolt. ‘This charge begins tomove atthetopof thesecondary branch, emptying successive, longer pieces ofthebranch, sothe bright lightning bolt appears towork itsway down that branch, atthesame time asitworks uptoward thecloud. If,however, oneofthese extra leader branches happens tohave reached theground almost simultaneously with theoriginal leader, itcansometimes happen that thedark leader ofthesecond stroke will take the second branch. Then you will seethefirst main flash inone place and thesecond flash inanother place. Itisavariant oftheoriginal idea. Also, ourdescription isoversimplified fortheregion very near theground. When thestep leader getstowithin ahundred meters orsofrom theground, there isevidence that adischarge rises from theground tomeet it.Presumably, the field gets bigenough forabrush-type discharge tooccur. If,forinstance, there is asharp object, like abuilding with apoint atthetop, then astheleader comes down nearby thefields aresolarge that adischarge starts from the sharp point andreaches uptotheleader. The lightning tends tostrike such apoint. Ithasapparently been known foralong time that high 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 how God with hislightning always smites thebigger animals and will notsufferthemtowaxinsolent,whiletheseofalesserbulkchafehimnot.Howlike-wisehisboltsfalleveronthehighesthousesandtallesttrees.””Andthenheexplainsthereason: “So, plainly, doth helove tobring down everything that exalts itself.” Doyouthink—now thatyouknowatrueaccount oflightning striking tall trees—that youhave agreater wisdom inadvising kings onmilitary matters than didArtabanis 2300 years ago? Donotexalt yourself. You could only doitless poetically. oat 10 Dielectrics 10-1 The dielectric constant Here webegin todiscuss another ofthepeculiar properties ofmatter under 10-1 The dielectric constant theinfluence oftheelectric field.Inanearlierchapter weconsidered thebehaviorofconductors, inwhichthecharges movefreelyinresponse toanelectric fieldto 1-2Thepolarization vectorP such points that there isnofield left inside aconductor. Now wewill discuss 10-3 Polarization charges insulators, materials which donotconduct electricity. One might atfirst believe .thatthereshouldbenoeffectwhatsoever. However,usingasimpleelectroscope 10-4Theelectrostatic equationsandaparallel-plate capacitor, Faraday discovered thatthiswasnotso.Hisexperi- lectricsmentsshowedthatthecapacitance ofsuchacapacitorisincreasedwhenanin-10-5Fieldsandforceswith sulator isputbetween theplates. Iftheinsulator completely illsthespace between dielectrics theplates, thecapacitance isincreased byafactor xwhich depends only onthe nature oftheinsulating material. Insulating materials arealso called dielectrics;thefactorxisthenaproperty ofthedielectric, andiscalledthedielectric constant.Thedielectric constant ofavacuumis,ofcourse,unity. Ourproblem now istoexplain why there isanyelectrical effect iftheinsulators areindeed insulators and donotconduct electricity. Webegin with theexperi- ‘mental fact that thecapacitance isincreased and trytoreason out what might begoing on. Consider aparallel-plate capacitor with some charges omthesurfacesoftheconductors, leussaynegativechargeonthetopplateandpositivechargeonthebottom plate. Suppose that thespacing between theplates isdand thearea of each plate isA.Aswehave proved earlier, thecapacitance is c=sf, (10.1) and thecharge andvoltage onthecapacitor arerelated by o=cy. (10.2) Now theexperimental fact isthat ifweput apiece ofinsulating material like lucite orglass between theplates, wefindthat thecapacitance islarger. That means, ofcourse, that thevoltage islower forthesame charge. But thevoltage difference istheintegral oftheelectric field across thecapacitor; sowemust conclude that inside thecapacitor, theelectric field isreduced even though thecharges onthe plates remain unchanged. ree conoucTo WLLALLEALROMAEE pTpeorTitertei: NNSSS‘SN pitivyititia ira Fig.10-1.Aporallel-plate capaci- (ALEFAAZTY REALTY EFEATYLALIT) —torwithadielectric. ThelinesofEareOrRee CONBUCTOR shown, ‘Now how can that be? We have alaw due toGauss that tells usthat the flux oftheelectric field isdirectly related totheenclosed charge. Consider thegaussian surface Sshown bybroken lines inFig. 10-1. Since theelectric field isreduced withthedielectric present, weconclude thatthenetcharge inside thesurface must 104 ‘belower than itwould bewithout thematerial. There isonly onepossible conclu- sion, andthat isthat there must bepositive charges onthesurface ofthedielectric, Since thefield isreduced butisnotzero, wewould expect thispositive charge to besmaller than thenegative charge ontheconductor. Sothephenomena can be explained ifwecould understand insome way that when adielectric material is placed inanelectric field there ispositive charge induced ononesurface andnega- tive charge induced ontheother. conouctor CLALOTELOIELLTIAL - freceesecoseesaa Fig.10-2.Ifweputaconducting 4aplateinthegapofaparallel-plate con- fTtF111] 11417144 denser,theinducedchargesreducethe=[YAHAARAAAAAAAAT) field intheconductor tozero. ‘CONDUCTOR Wewould expect that tohappen foraconductor. Forexample, suppose that wehad acapacitor with aplate spacing d,and weputbetween theplates aneutral conductor whose thickness is6,asinFig. 10-2. Theelectric field induces apositive charge ontheupper surface and anegative charge onthelower surface, sothere is nofield inside theconductor. The field intherestofthespace isthesame asit waswithout theconductor, because itisthesurface density ofcharge divided by €0;butthedistance over which wehave tointegrate togetthevoltage (the potential difference) isreduced, The voltage is a v=Z-d). The resulting equation forthecapacitance islike Eq. (10.1), with (d—6)sub- stituted for d: =_fsC-aI wah (103) ‘The capacitance isincreased byafactor which depends upon (b/d), theproportion ofthevolume which isoccupied bytheconductor. This gives usanobvious model forwhat happens with diclectrics—that inside ‘thematerial therearemanylittlesheetsofconducting material. Thetrouble with such amodel isthat ithasaspecific axis, thenormal tothesheets, whereas most dielectrics have nosuch axis. However, this difficulty can beeliminated ifwe SSPE EPP assume that allinsulating materials contain small conducting spheres separatedBRREORNOOD fromeachotherbyinsulation,asshowninFig.10-3.Thephenomenon oftheSAL ERACL REA dielectric constant isexplained bytheeffectofthecharges whichwouldbeinducedZROCRRECOLES oneachsphere. Thisisoneoftheearliest physical models ofdielectrics usedto explain thephenomenon that Faraday observed. More specifically, itwasassumed Fig.10-3. Amodelofadielectric: thateachoftheatomsofamaterial wasaperfect conductor, butinsulated fromsmall conducting spheres embedded in theothers. The dielectric constant xwould depend ontheproportion ofspace «anidealized insulator. which wasoccupied bytheconducting spheres. This isnot,however,themodel that isused today. 10-2 The polarization vector P Ifwefollow theabove analysis further, wediscover that theidea ofregionsofperfectconductivity andinsulation isnotessential. Eachofthesmallspheresacts like adipole, themoment ofwhich isinduced bytheexternal field. The only thing that isessential totheunderstanding ofdielectrics isthat there aremany little dipoles induced inthematerial. Whether thedipoles areinduced because there aretinyconducting spheres orforanyother reason isirrelevant. 102 Why should afield induce adipole moment inanatom iftheatom isnota conducting sphere? This subject will bediscussed inmuch greater detail inthe next chapter, which will beabout the inner workings ofdielectric materials. However, wegive here oneexample toillustrate apossible mechanism. Anatom hasapositive charge onthenucleus, which issurrounded bynegative electrons. Inanelectric field, thenucleus willbeattractedinonedirectionandtheelectronsin theother. The orbits orwave patterns oftheelectrons (orwhatever picture isusedinquantummechanics) willbedistortedtosomeextent,asshowninFig.10-4; thecenter ofgravity ofthenegative charge will bedisplaced and will nolonger coincide with thepositive charge ofthenucleus. Wehave already discussed such distributions ofcharge.Ifwelookfromadistance,suchaneutralconfiguration :isequivalent, toafirstapproximation, toalittledipole.Itseemsreasonable thatifthefieldisnottooenormous, theamountofinduced ELECTRON DrsTRIBUTIONdipole moment willbeproportional tothefield. That is,asmall field willdisplace thecharges alittle bitand alarger field will displace them further—and inpropor- tion tothefield—unless thedisplacement getstoolarge. Fortheremainder ofthis chapter, itwillbesupposed that thedipole moment isexactly proportional tothe field. Wewillnowassume thatineach atom there arecharges qseparated bya E distance 8,sothat 96isthedipole moment peratom. (We use6because weare already using dfortheplate separation.) Ifthere are Natoms perunit volume, there will beadipole moment per unit volume equal toNg’. This dipole moment perunit volume willberepresented byavector, P.Needless tosay, itisinthe direction oftheindividual dipole moments, i.e., inthedirection ofthecharge separation 8: Fig. 10-4, Anatom inanelectric P= Ng. (10.4) field hasitsdistribution ofelectrons dis- placed with respect tothenucleus. Ingeneral, Pwillvary from place toplace inthedielectric. However, atany point inthematerial, Pisproportional totheelectric field E.The constant of proportionality, which depends ontheease with which theelectron aredisplaced, willdepend onthekinds ofatoms inthematerial. ‘What actually determines how this constant ofproportionality behaves, how accurately itisconstant forvery large fields, and what isgoing oninside different materials, wewilldiscuss atalater time. Forthepresent, wewillsimply suppose that there exists amechanism bywhich adipole moment isinduced which is proportional totheelectric field. 10-3 Polarization charges ‘Now letusseewhat thismodel gives forthetheory ofacondenser with adi- electric. First consider asheet ofmaterial inwhich there isacertain dipole moment perunitvolume. Will there beontheaverage anycharge density produced bythis? Not ifPisuniform. Ifthepositive andnegative charges being displaced relative toeach other have thesame average density, thefactthat they aredisplaced does notproduceanynetchargeinsidethevolume.Ontheotherhand,ifPwerelarger atoneplace andsmaller atanother, thatwould mean thatmore charge would be moved into some region than away from it;wewould then expect togetavolume densityofcharge.Fortheparallel-plate condenser, wesupposethatPisuniform,soweneed tolook only atwhat happens atthesurfates. Atonesurface thenega- tivecharges, theelectrons, have effectively moved outadistance 6;attheother surface they have moved in,leaving some positive charge effectively outadistance 4.Asshown inFig. 10-5, wewillhave asurface density ofcharge, which willbe called thesurface polarization charge. a ee aa tt 2¢t 8it+tf Fig.10-5.Adielectricslabina 5 uniform field. Thepositive charges dis- .LF lacedthedistance§withrespecttoWee SaaS aeSa=SeS2— Sa the negatives. 10-3 Thischargecanbecalculated asfollows.IfAistheareaoftheplate,thenumber ofelectrons that appear atthesurface istheproduct ofAand N,the number perunit volume, and thedisplacement 5,which weassume here isper- pendicular tothesurface. The total charge isobtained bymultiplying bythe electronic charge q,.Togetthesurface density ofthepolarization charge induced onthesurface, wedivide byA.The magnitude ofthesurface charge density is p01 =Ngo 8. Butthisisjustequal tothemagnitude Pofthepolarization vector P,Eq.(10.4): Opa =P. (10.5) ‘The surface density ofcharge isequal tothepolarization inside thematerial. The surface charge is,ofcourse, positive ononesurface andnegative ontheother. ‘Now letusassume that ourslab isthedielectric ofaparallel-plate capacitor. The plates ofthecapacitor also have asurface charge, which wewill call cfrecy because they canmove “freely” anywhere ontheconductor. This is,ofcourse,thechargethatweputonwhenwechargedthecapacitor. Itshouldbeemphasizedthat7,0)exists only because OfGiree- IfGiree isremoved bydischarging thecapacitor, then ¢0i willdisappear, notbygoing outonthedischarging wire, butbymoving back into thematerial—by therelaxation ofthepolarization inside thematerial, ‘We can now apply Gauss’ law tothegaussian surface SinFig. 10-1. The electric field Einthedielectric isequal tothesotal surface charge density divided by¢.Itisclearthatopo1andcireehaveoppositesigns,so E=Zire—Spot, (10.6)€ ‘Note that thefield Eybetween themetal plate andthesurface ofthedielectric ishigherthanthefieldE;itcorresponds toreealone.Buthereweareconcerned with thefield inside thedielectric which, ifthedielectric nearly fillsthegap, isthe field over nearly thewhole volume. Using Eq.(10.5), wecanwrite B=Siro=P, (10.7)€ This equation doesn’t telluswhat theelectric field isunless weknow what Pis. Here, however, weareassuming thatPdepends onE—in fact, thatitisproportional toE.This proportionality isusually written as P=X€oE. (10.8) ‘Theconstant x(Greek “khi”) iscalled theelectric susceptibility ofthedielectric. Then Eq. (10.7) becomes =Mee 1 EoTED (10.9) which gives usthefactor 1/(1 +x)bywhich thefield isreduced. The voltage between theplates istheintegral oftheelectric field. Since the field isuniform, theintegral isjusttheproduct ofEandtheplate separation d. We have that a oeVe = tx ‘The total charge onthecapacitor isord, $0that thecapacitance defined by(10.2) becomes =eA+%)_Keo, C= ri a (10.10) Wehave explained theobserved facts. When aparallel-plate capacitor is filled withadielectric, thecapacitance isincreased bythefactor k=1tx ao.11) 104 whichisaproperty ofthematerial. Ourexplanation, ofcourse, isnotcomplete until wehave explained—as wewilldolater—how theatomic polarization comes about. Let’s now consider something alittle bitmore complicated—the situation in which thepolarization Pisnoteverywhere thesame. Asmentioned earlier, ifthe polarization isnotconstant, wewould expect ingeneral tofind acharge densityinthevolume,becausemorechargemightcomeintoonesideofasmallvolume‘elementthanleavesitontheother.Howcanwefindouthowmuchchargeisgained orlostfromasmall volume? \Firstlet’scomputehowmuchchargemovesacrossanyimaginarysurface BN NNwhenthematerial ispolarized. Theamount ofchargethatgoesacrossasurface 4 SASSisjustPtimes thesurface areaifthepolarization isnormal tothesurface. DOX“e x OFcourse, ifthepolarization issangential tothesurface, nocharge moves écos8acrossit NA SAASFollowing thesame arguments wehave already used, itiseasy toseethat thechargemovedacrossanysurfaceelementisproportional tothecomponent ofP rs «perpendicular tothesurface.CompareFig.10-6withFig.10-5,Wescethat4,71910-6Alateheha rovedacrossEq.(10.5) should, inthegeneral case,bewritten dielectric iproportional to.thecom Ponent ofPnormal tothesurface, Opt =Pom (10.12) Ifwearethinkingofanimaginedsurfaceelementinsidethedielectric,Eq. \Xd \ (10.12) gives thecharge moved across thesurface butdoesn’t result ina net peLecttic P surface charge, because there areequal and opposite contributions from thedi- yselectriconthetwosidesofthesurface. IN ‘The displacements ofthe charges can, however, result inavolume charge density.Thetotalchargedisplaced outofanyvolumeVbythepolarization isthe ~integraloftheoutwardnormalcomponent ofPoverthesurfaceSthatboundsthe Girtooe volume (sceFig.10-7). Anequalexcesscharge oftheopposite signisleftbehind. ExeDenoting thenetchargeinsideVbyAQpo1wewrite Nes Y8Qpat==f,Ponda, 40.13)N Wecanattribute AQyoi toavolume distribution ofcharge with thedensity pjoi, Fig. 10-7. Anonuniform polariza- and so. tion Pcan result inanetchorge inthe AQ=iPootdV. (10.14) bodyofadielectric. Combining thetwoequations yields JpProid¥=—|P-mda. (10.15) ‘Wehave akind ofGauss’ theorem that relates thecharge density from polarized materials tothepolarization vector P.We can secthat itagrees with theresult wegotforthesurface polarization charge orthedielectric inaparallel-plate capaci- tor. Using Eq.(10.15) with thegaussian surface ofFig. 10-1, thesurface integral gives PAA, andthecharge inside isp01AA, sowegetagain that o=P. Just aswedidforGauss’ lawofelectrostatics, wecanconvert Eq.(10.15) to adifferential form—using Gauss’ mathematical theorem: I,Ponda=//v-Pav.Is Iv We get Poot =—V°P. (10.16) Ifthereisanonuniform polarization, itsdivergence givesthenetdensityofcharge‘appearing inthematerial. Weemphasize thatthisisaperfectly realchargedensity; wecallit“polarization charge” only toremind ourselves how itgotthere. 10-5 Today welook upon these matters from another point ofview, namely, that wehave simpler equations inavacuum, and ifweexhibit inevery case allthe charges, whatever their origin, theequations arealways correct. Ifweseparate some ofthecharges away forconvenience, orbecause wedonot want todiscuss whatisgoingonindetail,thenwecan,ifwewish,writeourequationsinanyother form that may beconvenient. Onemorepointshouldbeemphasized. AnequationlikeD=¢Eisanattempt todescribe aproperty ofmatter. Butmatter isextremely complicated, andsuch anequation isinfact notcorrect. For instance, ifEgets toolarge, then Disno longer proportional toE.Forsome substances, theproportionality breaks down even with relatively small fields. Also, the“constant” ofproportionality may de- pend onhow fast £changes with time. Therefore thiskind ofequation isakindofapproximation, likeHooke’slaw.Itcannotbeadeepandfundamental equation.Ontheother hand, ourfundamental equations forE,(10.17) and(10.19), representourdeepestandmostcomplete understanding ofelectrostatics. 10-5 Fields and forces with dielectrics ‘Wewillnow prove some rather general theorems forelectrostatics insituations wheredielectrics arepresent.Wehaveseenthatthecapacitance ofaparallel-platecapacitor isincreased byadefinite factor ifitis filled with adielectric. Wecan show that this istrue foracapacitor ofanyshape, provided theentire region in theneighborhood ofthetwoconductors isfilled with auniform linear dielectric. Without thedielectric, theequations tobesolved are VE=P and VXBo=0. € Withthedielectric present,thefirstoftheseequations ismodified;wehaveinstead theequations VB)=PaseandVXE=0. (10.26) Now since wearetaking xtobeeverywhere thesame, thelasttwoequations can bewritten as V(x)=PissandWX(¢B)=0. (10.27) WethereforehavethesameequationsforxEasforEo,sotheyhavethesolu- tion xE=Ep. Inother words, thefield iseverywhere smaller, bythefactor 1/x, than inthecase without thedielectric. Since thevoltage difference isaline integral ofthefield, thevoltage isreduced bythis same factor. Since thecharge onthe electrodes ofthecapacitor hasbeentakenthesameinbothcases,Eq.(10.2)tellsusthat thecapacitance, inthecase ofaneverywhere uniform dielectric, isin- creased bythefactor x. Letusnow askwhat theforce would bebetween two charged conductors ina dielectric. Weconsider aliquid dielectric that ishomogeneous everywhere. We have seen earlier that one way toobtain theforce istodifferentiatetheenergywith respect totheappropriate distance. Iftheconductors have equal and opposite charges, theenergy U=Q?/2C, where Cistheir capacitance. Using theprinciple ofvirtual work, anycomponent isgiven byadifferentiation; forexample, _2_@ai) raUe-F2(1). (10.28) Since thedielectric increases thecapacity byafactor x,allforces will bereduced bythis same factor. One point should beemphasized. What wehave said istrue only ifthedi- electricisaliquid.Anymotionofconductors thatareembedded insoliddielectricchanges themechanical stress conditions ofthedielectric and alters itselectrical 10-7 properties, aswell ascausing some mechanical energy change inthedielectric. Moving theconductors inaliquid does notchange theliquid. The liquid moves toanew place butitselectrical characteristics arenotchanged. Many older books onelectricity start with the “fundamental” law that the force between twocharges is = 9192Fofe,, 10.29) point ofview which isthoroughly unsatisfactory. For one thing, itisnot true ingeneral; itistrue only foraworld filled with aliquid. Secondly, itdepends on thefactthatxisaconstant,whichisonlyapproximately trueformostrealmaterials.Itismuch better tostart with Coulomb’s law forcharges inavacuum, which is always right (for stationary charges). What does happen inasolid? This isavery difficult problem which hasnot been solved, because itis,inasense, indeterminate. Ifyou putcharges inside a dielectric solid, there aremany kinds ofpressures andstrains. You cannot deal with virtual work without including also themechanical energy required tocom- press thesolid, and itisadifficult matter, generally speaking, tomake aunique distinction between the electrical forces and the mechanical forces due tothe solid ‘material itself. Fortunately, nooneever really needs toknow theanswer tothe question proposed. Hemay sometimes want toknow how much strain there is goingtobeinasolid,andthatcanbeworkedout.Butitismuchmorecomplicatedthan thesimple result wegotforliquids. Asurprisingly complicated problem inthetheory ofdielectrics isthefollow- ing:Whydoesachargedobjectpickuplitlepiecesofdielectric? Ifyoucombyourhair onadryday, thecomb readily picks upsmall scraps ofpaper. Ifyouthought casuallyaboutit,youprobablyassumedthecombhadonechargeonitandthe eepaper hadtheopposite charge onit.Butthepaper isinitially electrically neutral. Ithasn’t anynetcharge, butitisattracted anyway. Itistrue that sometimes the paper willcome uptothecomb and then flyaway, repelled immediately after it E touchesthecomb.Thereasonis,ofcourse,thatwhenthepapertouchesthecomb,itpicks upsome negative charges andthen thelikecharges repel. Butthat doesn’t > answer theoriginal question. Why didthepaper come toward thecomb inthe firstplace? pievectaic‘Theanswerhastodowiththepolarizationofadielectricwhenitisplacedin oevect anelectric field. There arepolarization charges ofboth signs, which areattracted andrepelledbythecomb.Thereisanetattraction, however,becausethefield \ nearer thecomb isstronger than thefield farther away—the comb isnotaninfinite sheet. Itscharge islocalized. Aneutral piece ofpaper will notbeattracted to itor. is Fig.10-8.Adielectricobjectinocheraeaetheartoechoes ‘ThevariationoftheFeldnoneniform eldfoolsswergnfoward AsillustratedinFig.10-8,adielectricisalwaysdrawnfromaregionofweak i7 7 fieldtowardaregionofstronger field.Infact,onecanprovethatforsmallobjects theforce isproportional tothegradient ofthesquare oftheelectric field. Whydoesitdependonthesquareofthefield?Becausetheinducedpolarization chargesareproportional tothefields, andforgiven charges theforces areproportional tothefield.However, aswehavejustindicated, therewillbeanetforceonlyifthesquare ofthefield ischanging from point topoint. Sotheforce isproportional to thegradient ofthesquare ofthefield. ‘The constant ofproportionality involves, among other things, thedielectric constant oftheobject, and italso depends uponthesizeandshapeoftheobject.There isarelated problem inwhich theforce onadielectric canbeworked out quite accurately. Ifwehave aparallel-plate capacitor with adielectric slab only partially inserted, asshown inFig. 10-9, there willbeaforce driving thesheet in. Adetailed examination oftheforce isquite complicated; itisrelated tononuni- formities inthefield near theedges ofthedielectric andtheplates. However, if wedonotlookatthedetails,butmerelyusetheprincipleofconservation ofenergy, wecan easily calculate theforce. Wecan find theforce from theformula wede- 108 epupuston NJTASSRE Fig. 10-9, The force onadielectric‘ 1 t) sheetinaparallel-plate capacitor canbex= computed byapplyingtheprincipleof Lenergy conservation, rived earlier. Equation (10.28) isequivalent to au vac R=-E-4+se (10.30) Weneed only findouthow thecapacitance varies with theposition ofthedielectric slab. Let'ssupposethatthetotallengthoftheplatesisL,thatthewidthoftheplates isW,that theplate separation anddielectric thickness ared,andthat thedistance towhich thedielectric hasbeen inserted isx.The capacitance istheratio ofthe total free charge ontheplates tothevoltage between theplates. Wehave seen above thatforagiven voltage Vthesurface charge density offreecharge isxegV/d. Sothetotal charge ontheplates is =Kook Wg o=“oFww+Foxm, from which wegetthecapacitance: c=Hex tL—». 00.31) Using (10.30), wehave V?EW FaeBOG, (10.32) Now thisequation isnotparticularly useful foranything unless you happen to need toknow theforce insuch circumstances. Weonly wished toshow that the theory ofenergy canoften beused toavoid enormous complications indetermining theforces ondielectric materials—as there would beinthepresent case. ur discussion ofthetheory ofdielectrics hasdealt only with electrical phe- nomena, accepting thefactthat thematerial hasapolarization which isproportional totheelectric field. Why there issuch aproportionality isperhaps ofgreater interesttophysics. Onceweunderstand theoriginofthe dielectric constants from anatomic point ofview, wecan useelectrical measurements ofthedielectric constants in varying circumstances toobtain detailed information about atomic ormolecular structure, This aspect willbetreated inpart inthenext chapter. 10-9 ad Inside Dielectrics 11-1 Molecular dipoles Inthischapter wearegoing todiscuss why itisthat materials aredielectric. 11-1 Molecular dipoles We said inthelast chapter that wecould understand theproperties ofelectrical i ‘isystems withdielectrics onceweappreciated thatwhenanelectric fieldisapplied 14-2Electronic potarization toadielectric itinduces adipole moment intheatoms. Specifically, iftheelectric 11-3 Polar molecules; orientation field Einduces anaverage dipole moment perunit volume P,then x,thedielectric polarization constant, isgivenby 11-4Electric fieldsincavities ofa ron any dielectric“ 11-5 Thedielectric constant of Wehavealready discussed howthisequation isapplied; nowwehavetodis- liquids; theClausius-Mossotti cussthemechanism bywhich polarization arises when there isanelectric field equation inside amaterial. Webegin with thesimplest possible example—the polarization 446 Solid dielectrics ofgases. Buteven gases already have complications: there aretwo types. The molecules ofsomegases, likeoxygen, which hasasymmetric pairofatoms ineach 11-7 Ferroelectricity; BaTiO, molecule, have noinherent dipole moment. Butthemolecules ofothers, likewater vapor (which hasanonsymmetric arrangement ofhydrogen and oxygen atoms) carry apermanent electric dipole moment. Aswepointed outinChapters 6and 7, there isinthewater vapor molecule anaverage plus charge onthehydrogen atoms andanegative charge ontheoxygen. Since thecenter ofgravity ofthenega- Review: Chapter 31,Vol.1,TheOrigin tivecharge andthecenter ofgravity ofthepositive charge donotcoincide, the oftheRefractive Index totalcharge distribution ofthemolecule hasadipole moment. Such amolecule is Chapter 40,Vol. I,ThePrin-calledapolarmolecule. Inoxygen, because ofthesymmetry ofthemolecule, the ciplesofStatistical Mechanics centers ofgravity ofthepositive and negative charges arethesame, soitisa nonpolar molecule. Itdoes, however, become adipole when placed inanelectric field. The forms ofthetwo types ofmolecules aresketched inFig. 11-1. 11-2Electronicpolarization C) Wewillfirstdiscuss thepolarization ofnonpolar molecules. Wecanstart with thesimplest case ofamonatomic gas(forinstance,helium).Whenanatomof CENTEROF suchagasisinanelectric field, theelectrons arepulled onewaybythefieldwhile = Guam thenucleus ispulled theother way, asshown inFig. 10-4. Although theatoms are @ very stiffwith respect totheelectrical forces wecanapply experimentally, there isa slight netdisplacement ofthecenters ofcharge, and adipole moment isinduced. For small fields, theamount ofdisplacement, and soalso thedipole moment, is proportional totheelectric field. The displacement oftheelectron distribution which produces thiskindofinduced dipole moment iscalled electronic polarization. 4Wehavealreadydiscussed theinfluence ofanelectricfieldonanatomin 17 ° Chapter 31ofVol. I,when wewere dealing with thetheory oftheindex ofrefrac- _ceNTER oF tion. Ifyouthink about itforamoment, youwillseethatwhat wemust donowis ~“ARSE exactly thesameaswedidthen. Butnowweneedworry onlyabout fieldsthatdo GENTER.oF notvarywithtime, while theindex ofrefraction depended ontime-varying fields. rn InChapter 31ofVol. Iwesupposed that when anatom isplaced inanoscilla~ tingelectric field thecenter ofcharge oftheelectrons obeys theequation Fig.11-1. (a)Anoxygen molecule with zero dipole moment. (b)Thewater 28amabe=a8 (11a) maeapment ptmen nm Foranestimate ofthenatural frequency «wo,wecansetthisenergy equal tohioy— theenergy ofanatomic oscillator whose natural frequency iswo. Weget wy=32a Ifwenowusethisvalue ofwinEq,(11.7), wefindfortheelectronic polarizability 27anver[25]: (ai.a2y ‘Thequantity (h2/me*) istheradius oftheground-state orbit ofaBohr atom (see Chapter 38,Vol. 1)andequals 0.528 angstroms. Inagasatstandard pressure and temperature (1atmosphere, 0°C) there are2,69 X10'®atoms/cm®, soEq.(11.9) gives us kK=1+(2.69X10!%)16m (0.528X107%)?=1.00020. (11.13) The dielectric constant forhydrogen gasismeasured tobe Kexp =1.00026. Weseethat our theory isabout right. Weshould notexpect any better, because themeasurements were, ofcourse, made with normal hydrogen gas, which has diatomic molecules, notsingle atoms. Weshould notbesurprised ifthepolariza- tion oftheatoms inamolecule isnotquite thesame asthat oftheseparate atoms. The molecular effect, however, isnot really that large. Anexact quantum- mechanical calculation ofaforhydrogen atoms gives aresult about 12% higher than (11.12) (the 167ischangedto187),andthereforepredictsadielectricconstant somewhat closer totheobserved one. Inanycase, itisclear that ourmodel ofa dielectric isfairly good. ‘Another check onourtheory istotryEq.(11.12) onatoms which havea .higherfrequency ofexcitation. Forinstance,ittakesabout24.5voltstopullthe ~ $2 electronoffhelium,compared withthe13.5voltsrequired toionizehydrogen. \ 7 ~ Wewould, therefore, expect thattheabsorption frequency woforhelium would be & xabouttwiceasbigasforhydrogen andthatawouldbeone-quarter aslarge. We 4yur expect that ~- Khelum*1.000050. sgx» Experimentally, X 3, Kyelium =1.000068, (@) soyou seethat ourrough estimates arecoming outontheright track. Sowehave understood thedielectric constant ofnonpolar gas,butonlyqualitatively, because swwehavenotyetusedacorrectatomictheoryofthemotionsoftheatomicelectrons. sre ‘ teys 11-3Polarmolecules;orientationpolarization afA ‘Nextwewillconsideramoleculewhichcarriesapermanent dipolemoment od4 ‘ Po—such asawatermolecule. Withnoelectricfield,theindividual dipolespoint firwtuwinrandom directions, sothenetmoment perunit volume iszero. But when an electric field isapplied, twothings happen: First, there isanextra dipole moment induced because oftheforces ontheelectrons; this part gives just thesame kind of (v) electronic polarizability wefound for anonpolar molecule. For very accurate work, this effect should, ofcourse, beincluded, but wewill neglect itforthe Fig. 11-2. (a) In@gas ofpolar moment. (Itcanalways beadded inattheend.) Second, theelectric field tends to molecules, the individual moments cre lineuptheindividual dipoles toproduce anetmoment perunitvolume. Ifallthe oriented atrandom; theaverage moment dipoles inagasweretolineup,there would beaverylargepolarization, butthat inasmallvolume iszero. (b)When there doesnothappen, Atordinary temperatures andelectric fields thecollisions ofthe _i#anelectric field,thereissomeaverage ‘molecules intheirthermal motion keepthemfromliningupverymuch. Butthere __#i@"ment ofthemolecules. issome netalignment, andsosome polarization (seeFig. 11-2). The polarization that does occur can becomputed bythemethods ofstatistical mechanics we described inChapter 40ofVol. I. 13 Tousethismethodweneedtoknowtheenergyofadipoleinanelectricfield.Consider adipole ofmoment poinanelectric field, asshown inFig. 11-3. The energy ofthepositive charge is96(1), and theenergy ofthenegative charge is —40(2). Thus theenergy ofthedipole is U=ao)—96(2)=adVe, or wE U=poE=—poEcos0, (14) +a where0istheanglebetweenpyandE.Aswewouldexpect,theenergyislower when thedipoles arelined upwith thefield, -O%e Wenowfindouthowmuch lining upoccurs byusing themethods ofstatisticalmechanics. WefoundinChapter40ofVol.Ithatinastateofthermalequili-brium,therelativenumberofmoleculeswiththepotentialenergyUisproportional Fig. 11-3. The energy ofodipole to ointhefleld Eis—po-E. eure, (is) whereU(x,y,2)isthepotentialenergyasafunctionofposition. Thesameargu-ments would saythat using Eq. (11.14) forthepotential energy asafunction of angle, thenumber ofmolecules at@perunitsolidangle isproportional toe~¥/*", Letting n(@) bethenumber ofmolecules perunit solid angle at8,wehave (6) =nge*oBsoeier, (116 For normal temperatures and fields, theexponent issmall, sowecan approximate byexpanding theexponential: oFcos6 0)=no(1+?Ecos). quip Wecan find moifweintegrate (11.17) over allangles; theresult should bejust N,thetotal number ofmolecules perunit volume. The average value ofcos@over allangles iszero, sotheintegral isjust motimes thetotal solid angle 4x. Weget N mah. (11.18) ‘Weseefrom (11.17) that there will bemore molecules oriented along thefield (cos @=1)than against thefield (cos @=—1). Soinanysmall volume contain- ingmany molecules there will beanetdipole moment perunit volume—that is, apolarization P.Tocalculate P,wewant thevector sum ofallthemolecular moments inaunit volume, Since weknow that theresult isgoing tobeinthe direction of£,wewilljust sum thecomponents inthat direction (the components atright angles toEwillsum tozero): P= ¥pocoss, volume Wecanevaluate thesum byintegrating over theangular distribution. The solid angle at@is2msin@d6, so P=fn(@)pocos82xsin0db. (aia) Substituting forn(@) from (11.17), wehave xf" JE P=2(:+cos0)pocs#eo which iseasily integrated togive —NpsEp=Mise (11.20) 14 The polarization isproportional tothefield E,sothere will benormal dielectric behavior. Also, asweexpect, thepolarization depends inversely onthetempera- ture,because athighertemperatures thereismoredisalignment bycollisions. This 1/TdependenceiscalledCurie’slaw.Thepermanent momentpoappearssquared forthefollowing reason: Inagiven electric field, thealigning force depends upon ‘Po,andthemean moment that isproduced bythelining upisagain proportional topo.Theaverage induced moment isproportional top3. m /Weshould now trytoseehow well Eq.(11.20) agrees with experiment. Le’s 0004 # lookatthecaseofsteam. Since wedon’t know what pois,wecannot compute P ra directly, butEq.(11.20) does predict thatx—1should varyinversely asthetem- ¥ perature, and thisweshould check. ¥ From(11.20)weget 0008 / -2 Ne / KO1Oe3k?” (1121) / sox—1should varyindirectproportion tothedensity N,andinversely asthe0.008! / absolute temperature. Thedielectric constant hasbeen measured atseveral / different pressures andtemperatures, chosensuchthatthenumber ofmolecules in / unit volume remained fixed.* [Notice that ifthemeasurements had allbeen / taken atconstant pressure, thenumber ofmolecules perunitvolume would po / decrease linearly withincreasing temperature andx—1would varyasT-? /instead ofasT~1.]InFig.11-4weplottheexperimental observations forx—1 asafunction of1/T. The dependence predicted by(11.21) isfollowed quite well. There isanother characteristic ofthe dielectric constant of polarmolecules— otitsvariationwiththefrequency oftheappliedfield.Duetothefomentofinertia °‘001 0.008 9008ofthemolecules, ittakes acertain amount oftimefortheheavy molecules toturn VTeK) towardthedirection ofthe field. Soifweapply frequencies inthehigh microwave . .regionorabove,thepolarcontribution tothedielectricconstantbeginstofall/i0+11-4Experimenta! measure-away because themolecules cannot follow. Incontrast tothis,theelectronic Ypor etyerious tomperatorer, polarizability still remains the same uptooptical frequencies, because ofthe smaller inertia inthe electrons. 11-4Electricfieldsincavitiesofadielectric Ys,Yi ‘Wenowturntoaninteresting butcomplicated question—the problem ofthe CR ZLdielectric constantindensematerials. SupposethatwetakeliquidheliumorWy e|pf OF,omIiquid argon orsome other nonpolar material. Westillexpect electronic polari- wy a zation, Butinadense material, Pcanbelarge, sothefield onanindividual atom y, will beinfluenced bythepolarization oftheatoms initsclose neighborhood. The question is,what electric field actsontheindividual atom? Zc Imagine thattheliquid isputbetween theplates ofacondenser. Iftheplates oy te) arecharged they will produce anelectric field intheliquid, But there arealso chargesintheindividualatoms,andthetotalfieldisthesumofbothoftheseCA 7S/S, effects.Thistrueelectricfieldvariesvery,veryrapidlyfrompointtopointinthe te4liquid. Itisveryhighinsidetheatoms—particularly rightnexttothenucleus~and rk, Otad relatively small between theatoms. Thepotential difference between theplates is e thelineintegral ofthistotal field. Ifweignore allthefine-grained variations, we PycanthinkofanaverageelectricfieldE,whichisjustV/d.(Thisisthefieldwewere VA Wf usinginthelastchapter.) Weshouldthinkofthisfieldastheaverage overaspace LL. YL containing many atoms. (b) (a)‘Nowyoumightthinkthatan“average”atominan“average”locationwould feelthisaverage field. Butitisnotthatsimple,aswecanshowbyconsideringwhatFig.11-5,Thefeldinaslotcutina happens ifweimagine different-shaped holes inadielectric. Forinstance, suppose dielectric depends onthe shape ond that wecutaslotinapolarized dielectric, with theslot oriented parallel tothe orientation oftheslot. field, asshown inpart (a)ofFig. 11-5. Since weknow that VXE=0,theline integral ofEaround thecurve, I,which goes asshown in(b)ofthefigure, should *Singer, Steiger, and Gachter, Helvetica Physica Acta $,200 (1932). ws bezero. The field inside theslot must give acontribution which just cancels the part from thefield outside, Therefore thefield Epactually found inthecenter of along thin slotisequal toE,theaverage electric field found inthedielectric. Now consider another slotwhose large sides areperpendicular toE,asshown inpart (C)ofFig. 11-5. Inthis case, thefield Eointheslot isnotthesame asE because polarization charges appear onthesurfaces. Ifweapply Gauss’ lawto 1surface Sdrawn asin(d)ofthefigure, wefind that thefield Eqintheslot is given by PEo=E+=—> (11.22) rs where Eisagain theelectri field inthedielectric. (The gaussian surface contains thesurface polarization charge gpq1 =P.) Wementioned inChapter 10that€oE+PisoftencalledD,80€o£=DoisequaltoDinthedielectric.Earlier inthehistory ofphysics, when itwassupposed tobevery important todefine every quantity bydirect experiment, people were delighted todiscover that they could define what they meant by£and Dinadielectric without having tocrawl around between theatoms. The average field Eisnumerically equal tothefieldEothatwouldbemeasuredinaslotcutparalleltothefield.AndthefieldDcould bemeasured byfinding Eyinaslotcutnormal tothefield. Butnobody ever measures them that way anyway, soitwas just one ofthose philosophical things. VAT 4 Fig.11-6.ThefleldatanypointAGy - + inadielectriccanbeconsideredostheCXy) VIDsumofthefieldin@sphericalholeplus VA thefieldduetoaspherical plug. For most liquids which arenottoocomplicated instructure, wecould expect that anatom finds itself, ontheaverage, surrounded bytheother atoms inwhat ‘would beagood approximation toaspherical hole. And soweshould ask: “What would bethefield inaspherical hole?” Wecanfind outbynoticing that ifwe imagine carving outaspherical hole inauniformly polarized material, wearejust removing asphere ofpolarized material. (We must imagine that thepolarization is“frozen in”before wecutoutthehole.) Bysuperposition, however, thefields inside thedielectric, before thesphere was removed, isthesum ofthefields from allcharges outside thespherical volume plus thefields from thecharges within the polarized sphere. That is,ifwecall Ethefield intheuniform dielectric, wecan write E= Eros +Epes (123) POLE FEL Where nore isthefield inthehole andEy1ug isthefield inside asphere which isOuTSibe uniformly polarized(seeFig.11-6).Thefieldsduetoauniformly polarizedsphere areshown inFig. 11-7. The electric field inside thesphere isuniform, and its ,Be valueis { ) --2(ep Baus=3° te)Ws Using (11.23), weget Fro=E+= (a.2sy ‘The field inaspherical cavity isgreater than theaverage field bythe amount P/3eo. (The spherical hole gives afield 1/3oftheway between aslot parallel to fcle thefieldandaslotperpendicular tothefield.)' 11-5 Thedielectric constant ofliquids; theClausius-Mossotti equation Fig.11-7.Theelectricfieldof©Inaliquidweexpectthatthefieldwhichwillpolarizeanindividual atomisuniformly polarized sphere. more likeEyoie than justE,IfweusetheEo of(11.25) forthepolarizing fieldin 11-6 Eq.(11.6), then Eq.(11.8) becomes P PeNawo(E+£) (11.26) or Na P=Hay oF (11.27) Remembering that x—1isjust P/éoE, wehave No «-1=7am’ (11.28) which gives usthedielectric constant ofaliquid interms ofa,theatomic polar- izability. This iscalled theClausius-Mossotti equation. Whenever Naisvery small, 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) with some experimental results. Itisfirst necessary tolook atgases forwhich, using themeasurement ofx,wecanfind afrom Eq. (11.29). Forinstance, forcarbon disulfide atzero degrees centigrade thedielectric constant is1.0029, soNais0.0029. Now thedensity ofthegasiseasily worked out andthedensity oftheliquid canbefound inhandbooks. At20°C, thedensity of liquid CS is381times higher than thedensity ofthegasat°C. This means that Nis381timeshigher intheliquidthanitisinthegasso,that—if wemakethe approximation that thebasic atomic polarizability ofthecarbon disulfide doesn’t change when itiscondensed into aliquid—Na intheliquid isequal to381times 0.0029, or1.11. Notice that theNa/3 term amounts toalmost 0.4,soitisquite significant. With these numbers wepredict adielectric constant of2.76, which agrees reasonably well with theobserved value of2.64. InTable 11-1 wegive some experimental data onvarious materials (taken fromtheHandbook ofChemistry andPhysics), together withthedielectric constantscalculated from Eq. (11.28) intheway just described. The agreement between observation and theory iseven better forargon and oxygen than forCSy—and not sogood forcarbon tetrachloride. Onthewhole, theresults show that Eq. (11.28) works very well. Table 11-1 ‘Computation ofthedielectricconstantsofliquidsfrom thedielectric constant ofthegas. es Big Substance |«(exp) Na|Density|Density|Ratiot|Na|x(predict)|x(exp)|cs:|1.0029|0.0029 |0.00339|1.293|98|iit 2.16 26|Os 1.000823 |0.000823 |0.00143|1.19 332|043s|1.509 1507|ch|1.0030 |0.0030 |0.00889|1.59 325|os77|245 224|ALoess|oooos4s|o.o17e|vas|aio|oa|1817 16| *Ratio =density ofliquid/density ofgas. Ourderivation ofEq.(11.28) isvalid only forelectronic polarization inliquids. Itisnotrightforapolarmolecule likeH2O.Ifwegothrough thesamecalcu-lations forwater, weget13.2 forNa,which means that thedielectric constant for theliquid isnegative, while theobserved value ofxis80.Theproblemhastodo with thecorrect treatment ofthepermanent dipoles, andOnsager haspointed out theright way togo.Wedonothave thetime totreat thecase now, butifyouare interested itisdiscussed inKittel’s book, Iniroduction toSolid State Physics. na 11-6 Solid dielectrics Now weturn tothesolids. The first interesting factabout solids isthat there canbeapermanent polarization built in—which exists even without applying an electric field. Anexample occurs withamateriallikewax,whichcontainslong molecules having apermanent dipole moment. Ifyoumelt some wax and puta strong electric field onitwhen itisaliquid, sothat thedipole moments getpartly lined up,they willstay that way when theliquid freezes. The solid material will have apermanent polarization which remains when thefield isremoved. Such a solid iscalled anelectret. iohooh pg ‘Anelectrethaspermanent polarization chargesonitssurface.Itstheelectrical - ttt i analog ofamagnet. Itisnotasuseful, though, because freecharges fromtheair areattracted toitssurfaces, eventually cancelling thepolarization charges. The electret is“discharged” and there arenovisible external fields. ---Pesoles setegs ‘Apermanent internal polarizationPisalsofoundoccurringnaturallyinsome SISIISI(6)crystallinesubstances.Insuchcrystals,eachunitcellofthelatticehasanidentical _..[9@|66|60| 00|eo permanent dipolemoment,asdrawninFig.11-8.Allthedipolespointinthesame gigigiglel “"direction,evenwithnoappliedelectricfield.Manycomplicatedcrystalshave,infact, such apolarization; wedonotnormally notice itbecayse theexternal fields ---[QOLC0}C0100) G2!aredischarged, justasfortheelectrets. Ifthese internal dipole moments ofacrystalarechanged, however, external56|36|54|5e|36 fieldsappearbecausethereisnottimeforstraychargestogatherandcancelthe+ ——--polarization charges.Ifthedielectric isinacondenser, freechargeswillbeinducedi ontheelectrodes. For example, themoments canchange when adielectric isheated,becauseofthermal expansion. Theeffectiscalledpyroelectricity. Similarly, Fig.11-8. Acomplex crystal lattice ifwechange thestresses inacrystal—for instance, ifwebend it—again themo-canhave@permanent intrinsicpolarize- mentmaychangealittlebit,andasmallelectrical effect,calledpiezoelectricity,tionP. canbedetected. Forcrystals that donothave apermanent moment, onecanwork outatheory ofthedielectric constant that involves theelectronic polarizability oftheatoms. Itgoes much thesame asforliquids. Some crystals also have rotatable dipoles inside, andtherotation ofthese dipoles willalso contribute tox.Inionic crystals such asNaCl there isalsoionic polarizability. Thecrystal consists ofacheckerboard ofpositive and negative ions, and inanelectric field thepositive ions arepulled ‘oneway and thenegatives theother; there isanetrelative motion oftheplus and minus charges, and soavolume polarization. We could estimate themagnitude oftheionic polarizability from ourknowledge ofthestiffness ofsaltcrystals, but wewillnotgointo that subject here. a 311-7 Ferroelectricity; BaTiO, @ Wewanttodescribe nowonespecial classofcrystals which have,justby a ' 0 accident almost, abuilt-in permanent moment. The situation issomarginal that Qi @ ifweincrease thetemperature alittlebittheylosethepermanent moment com-+h-- pletely. Ontheotherhand, iftheyarenearly cubiccrystals, sothattheirmoments POSS canbeturnedindifferentdirections,wecandetectalargechangeinthemoment oe TN» ‘whenanapplied electric fieldischanged. Allthemoments flipoverandwegeta& t a large effect. Substances which have thiskind ofpermanent moment arecalled Q ferroelectric, afterthecorresponding ferromagnetic effects which werefirstdis- q covered iniron. = ‘Wewouldliketoexplainhowferroelectricity worksbydescribing aparticular ahexample ofaferroelectric material. There areseveral ways inwhich theferro- ® electric property canoriginate; butwewilltake uponly onemysterious case—that en ont Dot ofbarium titanate, BaTiOs. Thismaterial hasacrystal lattice whose basiccelissketched inFig. 11-9. Itturns outthat above acertain temperature, specifically Fig.11-9. TheunitcellofBoTiO;, 118°C, barium titanate isanordinary dielectric withanenormous dielectric con- Theatoms really fillupmostofthespace, _Stant. Below thistemperature, however, itsuddenly takes onapermanent moment.forclarity,onlythepositionsoftheir Inworkingoutthepolarization ofsolid material, wemust first find what are centers areshown. thelocal fields ineach unitcell. Wemust include thefields from thepolarization 18 topick outchains ofions along vertical lines. One ofthem consists ofalternating, oxygen and titanium ions. There areother lines made upofeither barium or ‘oxygen ions, butthespacing along these lines isgreater. Wemake asimple modelf—2—| toimitatethissituationbyimagining, asshowninFig.11-10(a), aseriesofchains4 : ofions. Along what wecallthemain chain, theseparation oftheions isa,which 4“4 ishalfthelatticeconstant; thelateraldistancebetweenidenticalchainsis2a, T ‘There areless-dense chains inbetween which wewill ignore forthemoment. Toe maketheanalysisalittleeasier,wewillalsosupposethatalltheionsonthemainLe $ chainareidentical. (Itisnotaserioussimplification becausealltheimportanteffects will still appear. This isone ofthetricks oftheoretical physics. One does adifferentproblembecauseitiseasiertofigureoutthefirsttime—thenwhenone é.¢ understands howthethingworks,itistimetoputinallthecomplications.) Now let’s trytofindoutwhat would happen with ourmodel. Wesuppose that thedipolemomentofeachatomispandwewishtocalculatethefieldatoneof $ $ theatomsofthechain,Wemustfindthesumofthefieldsfromalltheotheratoms,‘Wewill first calculate thefield from thedipoles inonly one vertical chain; wewill talk about theother chains later. The field atthedistance rfrom adipole ina $b -4 direction alongitsaxisisgivenby =» to) E=aah (11.32) Atany given atom, thedipoles atequal distances above and below itgive fields in thesamedirection,soforthewholechainweget $+' Bane=23-0424 24340) -298.quan ¢ $ Itisnottoohardtoshowthatifourmodelwerelikeacompletely cubicerystal— that is,ifthenext identical lines were only thedistance aaway—the number 0.383 wouldbechangedto1/3.Inotherwords,ifthenextlineswereatthedistancea $ ¢$ theywouldcontribute only—0.050unittooursum.However, thenextmain chain weareconsidering isatthedistance 2aand, asyou remember from Chapter 7, thefieldfromaperiodicstructurediesoffexponentially withdistance.Therefore $ $ theselinescontribute muchlessthan—0.050andwecanjustignorealltheother chains. Itisnecessarynowtofindoutwhatpolarizability aisneededtomakethe $ ¢$ runaway processwork,Suppose thattheinducedmomentpofeachatomofthe chain isproportional tothefield onit,asinEq,(11.6). Wegetthepolarizing field 1) ontheatom from Eehain, using Eq.(11.32). Sowehave thetwoequations Fig.11-10. Models ofaferroelec- P=aEcysin tric: (a)corresponds toanantiferro- and : ‘electric, and(b)to«normal ferroelectric. 0.383 pFoun = Therearetwosolutions: andpbothzero,or a = O58" with Eandpboth finite. Thus ifaisaslarge asa°/0.383, apermanent polarization sustained byitsown field willsetin.This critical equality must bereached for barium titanate atjustthetemperature T...(Notice thatif«were larger than the critical value forsmall fields, itwould decrease atlarger fields andatequilibrium thesame equality wehave found would hold.) ForBaTiOg, thespacing ais2X10-* cm,sowemust expect that @= 21.8 X10-*cm®. Wecancompare thiswith theknown polarizabilities ofthe individual atoms. Foroxygen, a=30.2 X10~** cm®; we're ontheright track! Butfortitanium, a=2.4X10-*cm’;rathersmall.Touseourmodelweshould probably take theaverage. (We could work outthechain again foralternating 11410 42 Electrostatic Analogs 12-1 The same equations have thesame solutions The total amount ofinformation which hasbeen acquired about thephysical 12-1 The same equations have the world since thebeginning ofscientific progress isenormous, and itseems almost same solutions impossible that anyoneperson could know areasonable fraction ofit.Butitis .actuallyquitepossibleforaphysicisttoretainabroadknowledge ofthephysical 12-2Theflowofheat;apointn ‘source nearaninfinite plane world rather than tobecome aspecialist insome narrow area. The reasons for thisarethreefold: First,therearegreatprinciples which applytoallthedifferent boundary kinds ofphenomena—such astheprinciples oftheconservation ofenergy and of 12-3 The stretched membraneangularmomentum. Athorough understanding ofsuchprinciples givesanunder- .standing ofagreatdealallatonce.Second, thereisthefactthatmanycompli- 12-4mediffusioneons catedphenomena, suchasthebehavior ofsolidsundercompression, really waiformspherical sourcebasically depend onelectrical andquantum-mechanical forces, sothatifone jomogeneousmedi understands thefundamental laws ofelectricity andquantum mechanics, there is 12-5 Irrotational fluid flow; the atleast some possibility ofunderstanding many ofthephenomena that occur flow past asphere incomplex situations. Finally, there isamost remarkable coincidence: The .equationsformanydifferentphysicalsituationshaveexactlythesameappearance. 12-6Tuaminations theuniformOfcourse, thesymbols maybedifferent—one letterissubstituted foranother— ighting ofaplane butthemathematical form oftheequations isthesame. This means that having 12-7 The“underlying unity” of studied one subject, weimmediately have agreat deal ofdirect and precise nature knowledge about thesolutions oftheequations ofanother. Wearenow finished with thesubject ofelectrostatics, and will soon goonto study magnetism and electrodynamics. But before doing so,wewould like to show that while learning electrostatics wehave simultaneously learned about a large number ofother subjects. Wewill find that theequations ofelectrostatics appear inseveral other places inphysics. Byadirect translation ofthesolutions (ofcourse thesame mathematical equations must have thesame solutions) itis possible tosolve problems inother fields with thesame ease—or with thesame difficulty—as inelectrostatics. The equations ofelectrostatics, weknow, are V+(KE)=Pe, (24) ‘0 VX E=0. (12.2) (We take theequations ofelectrostatics with dielectrics soastohave themost general situation.) The same physics can beexpressed inanother mathematical form: E=-—Vv¢, (12.3) v-(9)=—Pee (12.4) Now the point isthat there are many physics problems whose mathematical equations have thesame form. There isapotential (¢)whose gradient multiplied byascalar function (x)hasadivergence equal toanother scalar function (—p/€o). Whatever weknow abdut electrostatics canimmediately becarried over into that other subject, and viceversa. (Itworks both ways, ofcourse—if theother subject hassome particular characteristics that areknown, then wecan apply that knowledge tothecorresponding electrostatic problem.) Wewant toconsider faseries ofexamples from different subjects thatproduce equations ofthisform. ma Thecylinder iscovered with aconcentric sheath ofinsulating material which hasa conductivity K.“Say theoutside radius oftheinsulation isband the outside is kept attemperature T>(Fig. 12-1a). Wewant tofind outatwhat rate heat will belostbythewire,orsteampipe, orwhatever itisinthecenter. Letthetotal FoR,amountofheatlostfromalengthLofthepipebecalledG—whichiswhatweare Vn Q>) tryingtofind. YsPD <o‘Howcanwesolvethisproblem? Wehavethedifferential equations, butsince Z PATA,thesearethesameasthoseofelectrostatics, wehavereallyalreadysolvedthe 4Ch mathematical problem. The analogous problem isthat ofaconductor ofradiusa SSyY| atthepotential¢,separatedfromanotherconductorofradiusbatthepotential VWWA 2,with aconcentric layer ofdielectric material inbetween, asdrawn inFig. Ra 12-1(b). Now since theheat flow hcorresponds totheelectric field £,thequantity Gthatwewant tofindcorresponds tothefluxoftheelectric figldfrom aunit 1) ength (inother words, totheelectric charge perunit length over €9). Wehave solved theelectrostatic problem byusing Gauss’ law.Wefollow thesamepro- XPcedure forourheat-flow problem. >Fromthesymmetryofthesituation,weknowthatdependsonlyonthe SSXQ,distancefromthecenter.Soweenclosethepipeinagaussiancylinderoflength iMPSYLandradiusr.FromGauss’law,weknowthattheheatflow4multipliedby ANes) y thearea2xrLofthesurface mustbeequal tothetotalamount ofheatgenerated Q QPL KYinside,whichiswhatwearecallingG: NSSyQelh=G oth=5S-- (129) aaywih = >Fark ° Theheatflowisproportional tothetemperature gradient: (oy k=—KVT, Fig.12-1.(a)Heatflowinacylin-drical geometry. (b)The corresponding or,inthiscase, themagnitude of is electrical problem. ar h=-KE. This, together with (12.9), gives ar G~~ Tak (12.10) Integrating fromr=ator=b,weget GF ne Ta—Tr=—sag in2 a2.) Solving forG,wefind 2xKL(T, —T2) 6 (12.12) This result corresponds exactly totheresult forthecharge onacylindrical conden- ser: g=2rtobln =#2).In(b/a) Theproblems arethesame,andtheyhavethesamesolutions. Fromourknowledgeofelectrostatics, wealso know how much heat islost byaninsulated pipe. Let's consider another example ofheat flow. Suppose wewish toknow the heat flow intheneighborhood ofapointsourceofheatlocatedalittlewaybeneath thesurface oftheearth, ornear thesurface ofalarge metal block. The localized heat source might beanatomic bomb that was setoffunderground, leaving an intense source ofheat, oritmight correspond toasmall radioactive source inside ablock ofiron—there arenumerous possibilities. Wewilltreat theidealized problem ofapoint heat source ofstrength Gatthe distance abeneath the surface ofaninfinite block ofuniform material whose thermal conductivity isK.And wewill neglect thethermal conductivity ofthe 123 airoutsidethematerial. Wewanttodetermine thedistribution ofthetemperaturecnthesurface oftheblock. How hotisitright above thesource and atvarious places onthesurface oftheblock? How shall wesolve it?_It islike anelectrostatic problem with two materials withdifferentdielectriccoefficients xonoppositesidesofaplaneboundary. Aha!Perhaps itistheanalog ofapoint charge near theboundary between adielectric and aconductor, orsomething similar. Let’s seewhat thesituation isnear the surface. The physical condition isthat thenormal component of&onthesurface iszero, since wehave assumed there isnoheat flow outoftheblock. Weshould ask: Inwhat electrostatic problem dowehave thecondition that the normal component oftheelectric field E(which istheanalog ofA)iszero atasurface? There isnone! ‘That isone ofthethings that wehave towatch outfor. For physical reasons, there may becertain restrictions inthekinds ofmathematical conditions which arise inanyonesubject. Soifwehave analyzed thedifferential equation only for certain limited cases, wemay have missed some kinds ofsolutions that can occur inother physical situations. For example, there isnomaterial with adielectric constant ofzero, whereas avacuum does have zero thermal conductivity. Sothere Che isnoelectrostatic analogy foraperfect heatinsulator. Wecan,however, silluse ny 'fs 7 thesamemethods. Wecantrytoimagine whatwould happen ifthedielectric SON NS constant were zero. (Ofcourse, thedielectric constant isnever zero inanyreal we situation. Butwemight haveacaseinwhich thereisamaterial withaveryhighSSSA _--F0 dielectric constant, sothatwecouldneglect thedielectric constant oftheairout- qmsINT side.) PORTE How shall wefind anelectric field that hasnocomponent perpendicular todFERRY thesurface?Thatis,onewhichisalwaystangentatthesurface?YouwillnoticeLaer yrthatourproblemisoppositetotheoneofapointchargenearaplaneconductor. SOAS‘There wewanted thefield tobeperpendicular tothesurface, because theconductor LEO wasallatthesamepotential. Intheelectrical problem, weinvented asolution T=Copa byimagining apoint charge behind theconducting plate. Wecan usethesame idea again. Wetrytopick an“image source” that will automatically make the normal component ofthefield zero atthesurface. The solution isshown in 1 Fig. 12-2. Animage source ofthesame signandthesame strength placed atthe ‘TEMPERATURE distanceaabovethesurfacewillcausethefieldtobealwayshorizontal atthesur-face. The normal components ofthetwosources cancel out. +— Thus our heat flow problem issolved. ‘The temperature everywhere isthe |_same,bydirectanalogy,asthepotentialduetotwoequalpointcharges!The thersalsrece@netheatsureatagg"emperature Tatthedistancerfromasingle point source Ginaninfinite medium is distonce abelow thesurface ofagood G thermal conductor. Animage souree is T=ae (12.13) shown outside the material. (This, ofcourse, isjust theanalog of¢=q/47€or.) The temperature forapoint source, together with itsimage source, is G G T=Fak tGaRr oy ‘This formuta gives usthetemperature everywhere intheblock. Several isothermal surfaces areshown inFig. 12-2. Also shown arelines of&,which canbeobtained from h=—KVT. Weoriginally asked forthetemperature distribution onthesurface. For a point onthesurface atthedistance pfrom theaxis, ry=r2=V/p? +a2, so 1 26 Testes)=aRSS (12.15) This function isalso shown inthefigure. ‘The temperature is,naturally, higher right above thesource than itisfarther away. This isthekind ofproblem that geophysicists often need tosolve. Wenow seethat itisthesame kind ofthing we have already been solving forelectricity. 24 12-3 The stretched membrane Now letusconsider acompletely different physical situation which, never-theless,givesthesameequationsagain.Considerathinrubbersheet—amembrane “een \aS —whichhasbeenstretched overalargehorizontal frame(likeadrumhead). aS Ae‘Supposenowthatthemembrane ispushedupinoneplaceanddowninanother; 280i an BS a NNpe asshowninFig,12-3.Canwedescribetheshapeofthesurface? Wewilshow “FY howtheproblemcanbesolvedwhenthedefections ofthemembrane arenotwooAEA V-\-\ AVA ‘There are forces inthe sheet because itisstretched. Ifwe were tomake a a small cutanywhere, thetwosides ofthecutwould pull apart (see Fig. 12-4). So there isasurface tension inthesheet, analogous totheone-dimensional tensioninastretchedstring.Wedefinethemagnitude ofthesurfacetension7astheforcecerunitlength which willjust hold together thetwo sides ofacutsuch asoneof those shown inFig, 12-4 Suppose now that welook atavertical cross section ofthemembrane. It will appear asacurve, like theone inFig. 12-5. Letubetheverticaldisplacement Fig.12-3.Athinrubbersheet ofthemembrane from itsnormal position, and xand ythecoordinates inthe stretched over @cylindrical frame (like horizontal plane. (Thecrosssection shown isparallel tothex-axis.) adrumhead). Ifthesheetispushed upConsideralittepieceofthe surface oflength Axandwidth 4y.There willbe 9!Aanddown otB,what istheshape forces onthepiece from thesurface tension along eachedge. Theforce along °Fthesurface?edge|ofthefigurewillbe7sAy,directedtangenttothesurface—that is,attheangle 6;from thehorizontal. Along edge 2,theforce will berAyattheangle 6, (There will besimilar forces ontheother two edges ofthepiece, butwewill forget themforthemoment.) Thenetupward forceonthepiecefromedges|and2is ANMEASA, Wewilllimitourconsiderationstosmalldistortionsofthemembrane,i.,to SS small slopes: wecanthen replace sin@bytan6,which canbewritten asdu/dx. The \\ forceisthen SNe Seau) ‘au MA x ar=[ro(24),—71(4)Jay. S ‘Thequantityinbracketscanbeequallywellwritten(forsmallAx)as XAX< \Y cs¢2)ax: Fig.12-4,Thesurfacetensionrofax\"ax, «@stretchedrubbersheetistheforceper thenunit length across aline.a(_aw % There will beanother contribution toAFfrom theforces ontheother two 2 ‘edges; thetotalisevidently i a (a), a(_aones =[2(-2)42(, lo aF=[2(x)+%(2)AxAy. (1216) Thedistortionsofthediaphragm arecausedbyexternalforces.Let'slet-= Frepresent theupward force per unit area onthesheet (akind of“pressure") {from theexternal forces. When themembrane isinequilibrium (the static case), _Fig. 12-5. Cross section ofthede- thisforce must bebalanced bytheinternal force wehave just computed, Eq fected sheet. (12.16), That is ar I>"ey Equation (12.16) can then bewritten fe-vewrvw, (2179) Where by¥wenow mean, ofcourse, the two-dimensional gradient operator (@/ax, a/ay). Wehave thedifferential equation that relates u(x,))totheapplied 2s forces f(x, y)and thesurface tension 7(x, y),which may, ingeneral, vary from place toplace inthesheet. (The distortions ofathree-dimensional elastic body are also governed bysimilar equations, but wewill stick totwo-dimensions.) We will worry only about thecase inwhich thetension 7isconstant throughout the sheet, Wecan then write forEq. (12.17), vue —f. (12.18) 7 Wehave another equation that isthesame asforelectrostatics!—only this time,limited totwo-dimensions. Thedisplacement ucorresponds to¢,andf/rcorresponds top/€o.Soalltheworkwehavedoneforinfiniteplanechargedsheets, orlong parallel wires, orcharged cylinders isdirectly applicable tothestretched ‘membrane. Suppose wepush themembrane atsome points uptoadefinite height—that is,‘wefixthevalueofwatsomeplaces.Thatistheanalogofhaving adefinite potential atthecorresponding places inanelectrical situation. So,forinstance, wemay make apositive “potential” bypushing uponthemembrane with anobject having thecross-sectional shape ofthecorresponding cylindrical conductor. For example, ifwepush thesheet upwith around rod, thesurface willtake ontheshape shown inFig. 12-6. The height uisthesame astheelectrostatic potential ¢ofacharged cylindrical rod. Itfalls offasIn(I/r). (The slope, which corresponds tothe electric field E,drops offas1/r.) Fig. 12-6, Cross section of a an >> stretched rubber sheet pushed upbya round rod. Thefunction u(x,y)isthesame astheelectric potential dix,y)near a very long charged rod. ‘Thestretched rubber sheethasoftenbeenusedasawayofsolving complicated electrical problems experimentally. The analogy isused backwards! Various rods and bars arepushed against thesheet toheights that correspond tothepo- tentials ofasetofelectrodes. Measurements oftheheight then give theelectrical potential fortheelectrical situation. The analogy hasbeen carried even further. Iflittle balls areplaced onthemembrane, their motion corresponds approximately tothemotion ofelectrons inthecorresponding electric field. One canactually watch the“electrons” move ontheir trajectories. This method wasused todesign thecomplicated geometry ofmany photomultiplier tubes (such astheones used forscintillation counters, andtheoneused forcontrolling theheadlight beams on Cadillacs). The method isstillused, buttheaccuracy islimited. Forthemost accurate work, itisbetter todetermine thefields bynumerical methods, using the large electronic computing machines. 12-4 The diffusion ofneutrons; auniform spherical source inahomogeneous medium Wetake another example that gives thesame kind ofequation, this time having todowith diffusion. InChapter 43ofVol. Iweconsidered thediffusion ofions inasingle gas, and ofone gasthrough 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 arefree towander around, They travel inastraight line forseveral centimeters, ontheaverage, before being scattered byanucleusanddeflected intoanewdirection. Soifwehavealargeblock—many meterson aside—the neutrons initially atoneplace willdiffuse toother places. Wewant to findadescription oftheiraverage behavior—that is,theiraverage flow. 126 Let N(x,y,2)AVbethenumberofneutronsintheelementofvolumeAV.°NN ADQeaprire atthepoint(x,y,z).Because oftheirmotion,someneutrons willbeleavingAV, \ GRAPHITEandotherswillbecomingin.Iftherearemoreneutrons inoneregionthanina. LOZnearbyregion,moreneutronswillgofromthefirstregiontothesecondthancome PESYROMrox\ SSX_INXA back;therewillbeanetflow.Following thearguments ofChapter 43inVol.I, AON hewwedescribetheflowbyaflowvectorJ.Itsx-component Jyisthenetnumberof SNYXRESneutrons thatpassinunittimeaunitareaperpendicular tothex-direction. We XN \\foundthat WIAan INAS =-D (12.19) dso?! - 1 ALwherethediffusionconstantDisgivenintermsofthemeanvelocityv,andthe SQx mean-free-path /between scatterings isgiven by YX ’ ‘ 1 1 D=3b. N H ‘Thevector equation forJis , J=—-DWN. (12.20) i o $ - The rate atwhich neutrons flow across any surface element daisJ-ada (where, asusual,mistheunitnormal). Thenetflowoutofavolumeelementisthen (a) (followingtheusualgaussianargument) V+Jd¥.Thisflowwouldresultin 1adecrease with time ofthenumber inAVunless neutrons arebeing created in AV(bysome nuclear process). Ifthere aresources inthevolume thatgenerate S ‘ , ‘neutrons perunit time inaunit volume, then thenetflow outofAVwill beequal - to(S—AN/at)AV.Wehavethenthat BEDS7 N aN LIARSves=s-F. (12.21) \ —H ; 7d) ‘Combining (12.21) with(12.20), wegettheneutron diffusion equation \wau7 | H 1 ve(-pym =5-2. (2.22) Z |rN H \ Inthestaticcase—where N/at=O—wehaveEq.(12.4)alloveragain! ‘| ‘Wecanuseourknowledge ofelectrostatics tosolve problems about thediffusion ¢ 1 ofneutrons. Solet’s solve aproblem. (You may wonder: Why doaproblem if ' wehave already done alltheproblems inelectrostatics? Wecandoitfaster this ‘ time because wehave done theelectrostatic problems!) i Suppose wehave ablock ofmaterial inwhich neutrons arebeing generated— i)saybyuranium fission—uniformly throughout aspherical regionofradius« of + (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 Fig.12-7. (a)Neutrons areproduced atthesurface ofthesource region? Finding theanswers iseasy. The source uniformly throughout asphereofradivs@ density Soreplaces thecharge density p,soourproblem isthesame astheproblem inalarge graphite block and diffuse ofasphereofuniform chargedensity. FindingNisjustlikefindingthepotentisl__ outward. TheneutrondensityNisfound$.Wehave already worked outthefields inside andoutside ofauniformlycharged0@functionofr,thedistancefromthe sphere;wecanintegratethemtogetthepotential. Outside,thepotentialiscenterofhesource,{Peclogs ; ; . el F«uniforms Q/4meor,withthetotalcharge Qgivenby4xa%p/3. So charge, whoreNcorresponds '0'@ond corresponds toE.~oa " outside=ep (12.23) Forpoints inside, thefield isdueonly tothecharge Q(r) inside thesphere ofradius 7,Q(r) =4xr¥p/3, so =f.E30 (12.24) na imagine acase offluid flow that isanalogous toelectrostatics. Sowetake vev=0 (12.28) and vxv=0. (12.29) Wewant toemphasize that thenumber ofcircumstances inwhich liquid flow follows these equations isfarfrom thegreat majority, butthere areafew. They must becases inwhich wecanneglect surface tension, compressibility, and viscosity, and inwhich wecanassume that theflow isirrotational. These assump- tions arevalid sorarely forreal water that themathematician John von Neumann said that people who analyze Eqs. (12.28) and (12.29) arestudying “dry water"! (We take uptheproblem offluid flow inmore detail inChapters 40and41.) Because VXv=0,thevelocity of“dry water” can bewritten asthe gradient ofsome potential: v= —vy. (12.30) What isthephysical meaning ofy?There isn’t any very useful meaning. The velocity canbewritten asthegradient ofapotential simply because theflow is, irrotational. And byanalogy with electrostatics, yiscalled thevelocity potential, butitisnotrelated toapotential energy inthewaythat¢is.Since thedivergence ofviszero, wehave Vi(wy) =Vy =0. (12.31) The velocity potential yobeys thesame differential equation astheelectrostatic potential infree space (p=0). Let's pick aproblem inirrotational flow and seewhether wecan solve itby themethods wehave learned. Consider theproblem ofaspherical ball falling through aliquid. Ifitisgoing too slowly, theviscous forces, which wearedis- regarding, will beimportant. Ifitisgoing toofast, little whirlpools (turbulence) willappear initswakeandtherewillbesomecirculation ofthewater. Butifthe Bty ball isgoing neither toofast nortooslow, itismore orlesstrue that thewater flow will fitourassumptions, and wecandescribe themotion ofthewater byour simple equations. Itisconvenient todescribe what happens inaframe ofreference fixed inthesphere.Inthisframeweareaskingthequestion: Howdoeswaterflowpastasphere atrest when theflow atlarge distances isuniform? That is,when, farfrom the sphere, theflow iseverywhere thesame. Theflow near thesphere willbeasshownbythestreamlines drawninFig.12-8.Theselines,alwaysparalleltov,correspondtolines ofelectric field. We want togetaquantative description forthevelocity field, ie.,anexpression forthevelocity atanypoint P. | ‘Wecan find thevelocity from thegradient of¥,sowefirst work outthepo- tential. Wewant apotential that satisfies Eq. (12.31) everywhere, and which Fig. 12-8. The velocity field ofir- alsosatisfies tworestrictions: (1)there isnoflow inthespherical region inside rotational fividflowpastasphere. thesurface oftheball, and (2)theflow isconstant atlarge distances. Tosatisfy (1), thecomponent ofvnormal tothesurface ofthesphere must bezero. That means that 8//ar iszero atr=a.Tosatisfy (2),wemust have ay/8z =voat allpoints where >>a.Strictly speaking, there isnoelectrostatic case which corresponds exactly toour problem. Itreally corresponds toputting asphere of dielectric constant zero inauniform electric field. Ifwe had worked out the solution totheproblem ofasphere ofadielectric constant xinauniform field, then byputting x=0wewould immediately have thesolution tothis problem. Wehave notactually worked outthis particular electrostatic problem inde- tail, butlet's doitnow. (We could work directly onthefluid problem with vand ¥,butwewill useEand ¢because wearesoused tothem.) The problem is:Find asolution of¥? =0such that E=—vo isacon- stant,sayEp,forlarger,andsuchthattheradialcomponent ofEisequaltozeroatr =a.Thatis, 8%|=0. 02.32) 29 s re ANS ge g Fig. 12-9, The illumination J,ofa H surface isthe radiant energy per unit 1 time arriving ataunitarea ofthesurface. symmetric, sothat light isradiated equally inalldirections. Then theamount of radiant energy whichpassesthrough aunitareaatrightanglestoalightflowvariesinversely asthesquareofthe distance. Itisevident that theintensity ofthelight in thedirection normal totheflow isgiven bythesame kind offormula asforthe electric field from apoint source. Ifthelight rays meet thesurface atanangle @to thenormal, then J,theenergy arriving perunitarea ofthesurface, isonly cos@as great, because thesame energy goes onto anarea larger by1/cos 0.Ifwecallthe strength ofourlight source S,then J,,theillumination ofasurface, is n=Seen, (1239) wheree,istheunitvectorfromthesource andaistheunitnormal tothesurface. The illumination /,,corresponds tothenormal component oftheelectric field from apoint charge ofstrength 4m€oS. Knowing that, weseethat forany distribution of light sources, wecan find theanswer bysolving thecorresponding electrostatic problem. Wecalculate thevertical component ofelectric field ontheplane due to adistribution ofcharge inthesame way asforthat ofthelight sources.* Consider thefollowing example. Wewish forsome special experimental situation toarrange thatthetopsurface ofatablewillhaveaveryuniformillumina- tion. Wehave available long tubular fluorescent lights which radiate uniformly along their lengths. Wecan illuminate thetable byplacing thefluorescent tubes jinaregular array ontheceiling, which isattheheight zabove thetable. What is thewidest spacing bfrom tube totube that weshould useifwewant thesurface illumination tobeuniform to,say, within one part inathousand? Answer; (1) Find theelectric field from agrid ofwires with thespacing 6,each charged uni~ formly; (2)compute thevertical component oftheelectric field;(3)findoutwhat‘bmustbesothattheripplesofthefieldarenotmorethanonepartinathousand. ‘InChapter 7wesawthat theelectric field ofagrid ofcharged wires could be represented asasum ofterms, each oneofwhich gave asinusoidal variation of thefield with aperiod ofb/n, where nisaninteger. The amplitude ofany one of these terms isgiven byEq.(7.44): Fy=Ane? ‘Weneed consider only n=1,solong asweonly want thefield atpoints nottoo close tothegrid. For acomplete solution, wewould still need todetermine the coefficients Aq, which wehave not yetdone (although itisastraightforward calculation). Since weneed only A,wecanestimate that itsmagnitude isroughly thesame asthat oftheaverage field. The exponential factor would then give us directly therelative amplitude ofthevariations. Ifwewant thisfactor tobe10~°, wefindthat bmust be0.91z. Ifwemake thespacing ofthefluorescent tubes 3/4 sheannlogous lors cares wlaay hve thesere sa Alsou analy apps only tothelight energy arriving atthetopofanopaque surface, sowemust include in. our integral only the sources which shine onthe surface (and, naturally, not sources located below thesurface!). 13 Magnetostatics 13-1 The magnetic field The force onanelectric charge depends notonly onwhere itis, butalso on ‘13-1 The magnetic leld howfastitismoving. Every point inspace ischaracterized bytwovector quantities iwhichdeterminetheforceonanycharge.First,thereistheelectricforce,which 1"Electriccarenthegivesaforcecomponent independent ofthemotionofthecharge.Wedescribeit conservation ofchargebytheelectric field, E.Second, there isanadditional force component, called the 13-3 The magnetic force ona ‘magnetic force, which depends onthevelocity ofthecharge. This magnetic force current hhas astrange directional character: Atany particular point inspace, both thedirectionoftheforceanditsmagnitudedependonthedirectionofmotionofthe4Themagneticsecaparticle: atevery instant theforce isalways atrightangles tothevelocity vector; currents; Ampere’s law also, atanyparticular point, theforce isalway’ atright angles toafixed direction 13-8 Themagnetic field ofa inspace (see Fig. 13-1); andfinally, themagnitude oftheforce isproportional to straight wire andofasolenoids thecomponent ofthevelocity atright angles tothis unique direction. Itispossible atomic currents todescribe allofthis behavior bydefining themagnetic field vector B,which speci- fiesboththeuniquedirection inspaceandtheconstant ofproportionality withthe 13-6Therelativity ofmagnetic andvelocity, andtowritethemagnetic forceasquXB.Thetotalelectromagnetic lectrie force onacharge can, then, bewritten as 13-7 The transformation ofcurrents andcharges F=qE+0xB). a3.) 13-8Superposition; theright-hand This iscalled theLorentz force. rale “The magnetic force iseasily demonstrated bybringing abarmagnet close toa cathode-ray tube. The deflection oftheelectron beam shows that thepresence of themagnet results inforces ontheelectrons transverse totheir direction ofmotion, aswedescribed inChapter 12ofVol. I. . .TheunitofmagneticfieldBisevidentlyonenewton'second perReviewsTheosofaanTheSpecial coulomb:meter. Thesame unitisalsoonevolt-second permeter®. Itisalso a o called one weber persquare meter. 13-2 Electric current; theconservation ofcharge Weconsider firsthow wecanunderstand themagnetic forces onwires carrying clectric currents, Inordertodothis,wedefine whatismeant bythecurrent density. 8 Electric currents areelectrons orother charges inmotion with anetdrift orflow. Wecanrepresent thecharge flow byavector which gives theamount ofcharge passing perunit area andperunittime through asurface element atright angles to theflow (just aswedidforthecase ofheat flow). Wecallthisthecurrent density ¥ andrepresent itbythevector j.Itisdirected along themotion ofthecharges. laos Ifwetake asmall area ASatagiven place inthematerial, theamount ofcharge flowing across that area inaunit time is jonas, (13.2) wherenistheunitvectornormaltoAS. cononent lotTheforeonamown Thecurrentdensityisrelatedtotheaverageflowvelocityofthecharges. Sharee's ahvightanglesfovondtothe Suppose that wehave adistribution ofcharges whose average motion isadrift Girection ofB.Itsalsoproportional to with thevelocity v.Asthisdistribution passes overasurfaceelementAS,thechargethecomponentofvatrightanglestoB, 44gpassing through thesurface element inatime Arisequal tothecharge contained thati,10vsin0. inaparallelepiped whose base isASandwhose height isvAt,asshown inFig. 13-2. The volume oftheparallelepiped istheprojection ofASatright angles tovtimes Bt Af, which when multiplied bythecharge density pwillgive Ag. ‘Thus — Aq=pu:wASaAt. Ui.Ys_ ‘ThechargeperunittimeisthenpumAS,fromwhichwegetGf.‘\ i=pv. (13.3)CX : Ifthechargedistribution consists ofindividual charges, sayelectrons, each ga) with thecharge qandmoving with themean velocity v,thenthecurrent density isAt Wheyy,“ j=Nav, (3.4) 7“ Cd on where Nisthenumber ofcharges perunitvolume. ~ The total charge passing perunit time through any surface Siscalled the Fig.13-2. Ifacharge distribution of electric current, I.Itisequal totheintegral ofthenormal component oftheflowdensitypmoveswiththevelocityv,thethroughalloftheelementsofthe surface: charge per unit time through AS is ;pvenS. T=[jonds (13.5) (Gee Fig. 13-3). The current Jout ofaclosed surface Srepresents therate atwhich charge a leaves thevolume Venclosed byS,One ofthebasic laws ofphysics isthat i o electric charge isindestructible; itisnever lostorcreated. Electric charges canx<S, <I ‘movefromplacetoplacebutneverappearfromnowhere.Wesaythatchargeisconserved. Ifthere isanet current out ofaclosedsurface,theamountofcharge inside must decrease bythecorresponding amount (Fig. 13-4). Wecan, therefore, ACES writethelawoftheconservation ofcharge as ; aFig.13-3.Thecurrent|throughthe fJsmdS=—F,(Qiasie). (13.6)surfaceSis{j-dS. sayHonea ‘The charge inside canbewritten asavolume integral ofthecharge density: i \x inside=fpa. (3.7)a SO K ina8 AN wal Ifweapply(13.6)toasmallvolume AV,weknowthattheleft-hand integral isV-JAV.ThechargeinsideispAV,sotheconservation ofchargecanalsobe ~—written as p= 2ay D vi- -2 (13.8)7} \‘SURFACE .‘ ap MH (Gauss’mathematics onceagain!). Fig. 13-4, The integral ofj-mover@cloted surface istherateofchange of 13-3Themagnetic forceonacurrent thetotal charge @inside. Now weareready tofind theforce onacurrent-carrying wire inamagnetic field, ‘The current consists ofcharged particles moving with thevelocity valong thewire. Each charge feels atransverse force F-qxB (Fig. 13-Sa). Ifthere areNsuch charges perunit volume, thenumber inasmall volume AVofthewire isNAV. The total magnetic force AFonthevolume AV isthesum oftheforces ontheindividual charges, that is, AF=(NAV)qu XB). Butgu isjustj,so OF=jx Bav (13.9) (Fig. 13-Sb). The force perunit volume isjXB. 32 Ifthecurrent isuniform across awire whose cross-sectional area isA,we may take asthevolume element acylinder with thebase area 4and thelength AL. Then AF=jXBAAL. (13.10) 5 rye Now wecancalljAthevector current Zinthewire. (Itsmagnitude istheelectric | currentinthewire,anditsdirectionisalongthewire.)Then TL : OF=1XBAL. 3.11) hee fee! TheforceperunitlengthonawireisIXB. F This equation gives theimportant result that themagnetic force onawire, duetothemovement ofcharges init,depends only onthetotal current, andnoton (0) theamount ofcharge carried byeach particle—or even itssign! The magnetic force onawire near amagnet iseasily shown byobserving itsdeflection when a current isturned on,aswasdescribed inChapter 1(seeFig.1-6). 8 ry“ 13-4Themagnetic fieldofsteadycurrents; Ampere’s law : \oy We have seen that there isaforce onawireinthepresenceofamagneticfield,Cy2/0 _ produced, say, byamagnet. From theprinciple that action equals reaction we might expect that there should beaforce onthesource ofthemagnetic field, ie., BF onthemagnet, when there isacurrent through thewire.* There areindeed such forces, asisseen bythedeflection ofacompass needle near acurrent-carrying () wire. Now weknow that magnets feel forces from other magnets, sothat means that when there isacurrent inawire, thewire itself generates amagnetic field. Fig13-5. Themagnetic force ona Moving charges, then, produce amagnetic field. We would like now totryto current-carrying wire isthesum ofthe discover thelaws thatdetermine how such magnetic fields arecreated. Thequestion forces ontheindividual moving charges. is:Given acurrent, what magnetic field does itmake? Theanswer tothisquestion was determined experimentally bythree critical experiments and abrilliant theoretical argument given byAmpere. Wewillpass over thisinteresting historical development andsimply saythatalargenumberofexperiments havedemonstrated thevalidity ofMaxwell’s equations. Wetake them asourstarting point. Ifwe drop theterms involving time derivatives inthese equations wegettheequations of ‘magnetostatics: v-B=0 (03.12) and evxB=s. (13.13) © These equations arevalid only ifallelectric charge densities areconstant andall currents aresteady, sothat theelectric and magnetic fields arenotchanging with time—all ofthe fields are “static.” Wemay remark that itisrather dangerous tothink that there issuch athing asastatic magnetic situation, because there must becurrents inorder togeta magnetic field atall—and currents cancome only from moving charges. “Mag- netostatics” is,therefore, anapproximation. Itrefers toaspecial kind ofdynamic situation with large numbers ofcharges inmotion, which wecanapproximate by asteady flow ofcharge. Only then canwespeak ofacurrent density jwhich does notchange with time, Thesubject should more accurately becalled thestudy of steady currents. Assuming that allfields aresteady, wedrop allterms in4£/01 and@B/at from thecomplete Maxwell equations, Eqs. (2.41), and obtain the twoequations (13.12) and(13.13) above. Also notice that since thedivergence of thecurlofanyvector isnecessarily zero, Eq.(13.13) requires that -j=0.Thisistrue,byEq.(13.8),onlyif4p/ariszero.ButthatmustbesoifEisnotchangingwith time, soourassumptions areconsistent. *Wewillseelater,however,thatsuchassumptions arenorgenerallycorrectforelectro-magnetic forces! 133 The requirement that V-j=0means that wemay only have charges which flow inpaths that close back onthemselves. They may, forinstance, flow inwires that form complete loops—called circuits. The circuits may, ofcourse, contain generators orbatteries that keep thecharges flowing. But they may notinclude condensers which arecharging ordischarging. (We will, ofcourse, extend the theory later toinclude dynamic fields, butwewant firsttotake thesimpler case of steady currents.) Now letuslook atEqs. (13.12) and (13.13) toseewhat they mean. The first one says that thedivergence ofBis zero. Comparing ittotheanalogous equation inelectrostatics, which says that V-E=p/eo, wecan conclude that there isno magnetic analog ofanelectric charge. There arenomagnetic charges from which lines ofBcanemerge. Ifwethink interms of“lines” ofthevector field B,they can never start and they never stop. Then where dothey come from? Magnetic fields “appear” inthepresence ofcurrents; they have acurl proportional tothecurrent density. Wherever there arecurrents, there arelines ofmagnetic field making loops around thecurrents. Since lines ofBdonotbegin orend, they will often close back onthemselves, making closed loops. But there canalso becomplicated 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 istrue notonly formagnetostatics, itisalways true— even fordynamic fields. 8 Theconnection between theBfield andcurrents iscontained inEq.(13.13). Herewehaveanewkindofsituation whichisquitedifferent fromelectrostatics, dsLoo?rwherewehadVXE=0.ThatequationmeantthatthelineintegralofEaround \ any closed path iszero: SURFACE S\ fE-ds=0. op aLYS WegotthatresultfromStokes’theorem,whichsaysthattheintegralaroundany vxB closed pathofanyvector fieldisequal tothesurface integral ofthenormal com- ponent ofthecurl ofthevector (taken over anysurface Which hastheclosed loop Fig.13-6. Thelineintegral ofthe 2itsperiphery). Applying thesametheorem tothemagnetic fieldvector and tangential component ofBisequal tothe vsing thesymbols shown inFig.13-6, weget surface integral ofthenormal component ofVXBL fw-de=[owxB)-nas. (13.14) Taking thecurl ofBfrom Eq.(13.13), wehave I fedsrdfuads. (13,15) Theintegral overj,according to(13.5), isthetotal current /through thesurface S. Since forsteady currents thecurrent through Sisindependent oftheshape ofS, solong asitisbounded bythecurve I’,oneusually speaks of“the current through theloop I." Wehave, then, ageneral law: thecirculation ofBaroundanyclosed curve isequal tothecurrent /through theloop, divided by€9c?: fds=Homage, (13.16) fi €oc? . This law—called Ampere’s Jaw—plays thesame roleinmagnetostatic thatGauss’ lawplayed inelectrostatics. Ampere’s lawalone does notdetermine Bfrom cur- rents; wemust, ingeneral, also use ¥-B =0. But, aswewill see inthe next section, itcanbeused tofindthefield inspecial circumstances which have certain simple symmetries. 134 13-5 The magnetic field ofastraight wire and ofasolenoid; atomic currents Wecanillustrate theuseofAmpere’s lawbyfinding themagnetic field near awire. Weask: What isthefield outside along straight wire with acylindrical cross section? Wewillassume something which may notbeatallevident, butwhich isnevertheless true: that thefield lines ofBgoaround thewire inclosed circles. Ifwemake thisassumption, then Ampere’s law, Eq.(13.16), tellsushow strong the field is.From thesymmetry oftheproblem, Bhasthesame magnitude atall points onacircleconcentric withthewire(seeFig.13-7). Wecanthendotheline wDintegralofB-dsquiteeasily;itisjustthemagnitude ofBtimesthecircumference. s Ifristheradius ofthecircle, then ¢B-ds=B-2nr. Thetotal current through theloop ismerely thecurrent /inthewire, so Q Bim-,,con ae or _tow Fig,13-7. Themagnetic fleldoutside 8Set To (13.17) ofalongwirecarrying thecurrent 1. ‘The strength ofthemagnetic field drops offinversely asr,thedistance from the axisofthewire.Wecan,ifwewish,writeEq.(13.17)invectorform.Remembering that Bisatright angles both toJand tor,wehave _ 1uxe B=Frei (13.18) Wehave separated outthefactor 1/4meoc?, because itappears often. Itis worth remembering that itisexactly 10-7 (inthemks system), since anequation like(13.17) isused todefine theunit ofcurrent, theampere. Atonemeter from a current ofoneampere themagnetic field is2X10-7 webers persquare meter. Since acurrent produces amagnetic field, itwill exert aforce onanearby wire which isalsocarrying acurrent. InChapter 1wedescribed asimple demonstration oftheforces between two current-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. poe .i, aPOPPED RPPERPPREP PEERS7 CUTEST 4 \WATT TET g yVT ed TTTSsJB Lokbledgeltsaktabslalatstalals's 272 Fig.13-8. Themagnetic fieldofa LINES: Jong solenoid. ors Let’s take another example that canbeanalyzed byAmpere's lawifweadd some knowledge about thefield. Suppose wehave along coil ofwire wound ina tight spiral, asshown bythecross sections inFig. 13-8. Such acoil iscalled a solenoid. Weobserve experimentally that when asolenoid isvery long compared with itsdiameter, thefield outside isvery small compared with thefield inside. Using just that fact, together with Ampere’s law, wecanfind thesizeofthefield inside. Since thefield stays inside (and haszero divergence), itslines must goalong parallel totheaxis, asshown inFig. 13-8. That being thecase, wecanuseAmpere’sJawwiththerectangular “curve” T'showninthefigure.Thisloopgoesthedistance Bs Linside thesolenoid, where thefield is,say, Bo,then goes atright angles tothe field, and returns along theoutside, where thefield isnegligible. The line integral ofBforthiscurve isjustBoL, anditmustbe1/egc?timesthetotalcurrentthrough T,which isNVfthere areNturns ofthesolenoid inthelength L.Wehave NI Bol=a Or, letting nbethenumber ofturns perunit length ofthesolenoid (that is,n= N/L), weget al Bo=aay (13.19) What happens tothelines ofBwhen they gettotheend ofthesolenoid? 8Presumably, theyspreadoutinsomewayandreturntoenterthesolenoidatthe —S= otherend,assketchedinFig.13-9.Suchafieldisjustwhatisobservedoutsideofabarmagnet. Butwhat isamagnetanyway? OurequationssaythatBcomesfrom thepresence ofcurrents. Yetweknow that ordinary bars ofiron (nobatteries or generators) also produce magnetic fields. You might expect that there should be some other terms ontheright-hand side of(13.12) or(13.13) torepresent “the density ofmagnetic iron” orsome such quantity. Butthere isnosuch term. Our theory says that themagnetic effects ofiron come from some internal currents which arealready taken care ofbythejterm. Fig.13-9. Themagnetic fieldoutside Matter isverycomplex when looked atfrom afundamental point ofview—as ofasolenoid. wesaw when wetried tounderstand dielectrics. Inorder nottointerrupt ourpres- entdiscussion, wewillwait until later todeal indetail with theinterior mechanisms ofmagnetic materials likeiron. You willhave toaccept, forthemoment, that all magnetism isproduced from currents, and that inapermanent magnet there arepermanent internalcurrents. Inthecaseofiron, these currents come from electrons spinning around their own axes. Every electron hassuch aspin, which corresponds toatinycirculating current. Ofcourse, oneelectron doesn’t produce much mag- netic field, butinanordinary piece ofmatter there arebillions andbillions ofelec- trons. Normally these spin and point every which way, sothat there isnonet effect. The miracle isthat inavery fewsubstances, like iron, alarge fraction of theelectrons spin with their axes inthesame direction—for iron, twoelectrons from ‘each atom takes part inthiscooperative motion. Inabarmagnet there arelarge ‘numbers ofelectrons allspinning inthesame direction and, aswewill see,their total effect 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 that abarmagnet isequivalent toasolenoid. 13-6 The relativity ofmagnetic andelectric fields When wesaid that themagnetic force onacharge was proportional toits velocity, you may have wondered: “What velocity? With respect towhich refer- ence frame?” Itis,infact, clear from thedefinition ofBgiven atthebeginning of thischapter that what thisvector iswilldepend onwhat wechoose asareference frame forourspecification ofthevelocity ofcharges. Butwehave said nothing about which istheproper frame forspecifying themagnetic field. Itturns outthat any inertial frame will do. Wewill also seethat magnetism andelectricity arenotindependent things—that they should always betaken to- getherasonecomplete electromagnetic field.Although inthestaticcaseMaxwell'sequations separate into two distinct pairs, one pair forelectricity and one pair for magnetism, with noapparent connection between thetwo fields, nevertheless, in nature itself there isavery intimate relationship between them that arises from the principle ofrelativity. Historically, theprinciple ofrelativity was discovered after Maxwell's equations. Itwas, infact, thestudy ofelectricity andmagnetism which ledultimately toEinstein’s discovery ofhisprinciple ofrelativity. But let's see 136 arty@ q ‘| s s _ w vy,=0 wav G, )vymeov “=0GY a 0) > ral Fig. 13-10. The interaction ofacurrent-carrying wire and aparticlewiththe charge qasseen intwo frames. Inframe S(part al,thewire isatrest; infrome S'(part b),thecharge isatrest. what ourknowledge ofrelativity would tellusabout magnetic forces ifweassume that therelativity principle isapplicable—as itis—to electromagnetism. ‘Suppose wethink about what happens when anegative charge moves with velocity vyparallel toacurrent-carrying wire, asinFig. 13-10. Wewilltrytounder- stand what goes onintwo reference frames: one fixed with respect tothewire,asinpart(a)ofthefigure,andonefixedwithrespecttotheparticle,asinpart(b).We will call the first frame Sand the second S’. IntheS-frame, there isclearly amagnetic force ontheparticle. The force is directed toward thewire, soifthecharge ismoving freely wewould seeitcurve in toward thewire. ButintheS’-frame there canbenomagnetic force ontheparticle, because itsvelocity iszero. Does it,therefore, stay where itis? Would wesee different things happening inthetwosystems? The principle ofrelativity would saythat inS’weshould also seetheparticle move closer tothewire. Wemust trytounderstand why that would happen. Wereturntoouratomicdescription ofawirecarryingacurrent.Inanormal conductor, like copper, theelectric currents come from themotion ofsome ofthe negative electrons—called theconduction electrons—while the positive nuclear charges andtheremainder oftheelectrons stay fixed inthebody ofthematerial. Weletthedensity oftheconduction electrons bep_and their velocity inSbev. Thedensity ofthecharges atrestinSisp, which must beequal tothenegative ofp_,since weareconsidering anuncharged wire. There isthus noelectric field outside thewire, and theforce onthemoving particle isjust F=quoX B. UsingtheresultwefoundinEq.(13.18)forthemagneticfieldatthedistance 1from theaxis ofawire, weconclude that theforce ontheparticle isdirected toward thewire and hasthemagnitude = 1Uaro PoFred UsingEqs.(13.4)and(13.5),thecurrentJcanbewrittenasp_vd,whereAisthe area ofacross section ofthe wire. Then = 12ap_Arvo |rene (03.20) Wecould continue totreat thegeneral case ofarbitrary velocities forvand 9, butitwillbejust asgood tolook atthespecial case inwhich thevelocity vof theparticle isthesame asthevelocity voftheconduction electrons. Sowewrite t=v,and Eq.(13.20) becomes =9PAFa,f eae (13.21) Now weturn ourattention towhat happens in’,inwhich theparticle isat rest and thewire isrunning past (toward theleftinthefigure) with thespeed v. The positive charges moving with thewire will make some magnetic field B”at theparticle. But theparticle isnow atrest, sothere isnomagnetic force onit! Ifthere isany force ontheparticle, itmust come from anelectric field. Itmust BT bethat themoving wire hasproduced anelectric field. Butitcandothat only ifit appears charged—it mustbethataneutralwirewithacurrentappearstobecharged when set inmotion. ‘Wemust look into this. Wemust trytocompute thecharge density inthe wire inS’from what weknow about itinS.One might, atfirst, think they arethe same; butweknow that lengths arechanged between Sand S’(see Chapter 15, Vol. 1),sovolumes will change also. Since thecharge densities depend onthe volume occupied bycharges, thedensities willchange, too. Before wecandecide about thecharge densities inS’,wemust know what happens totheelectric charge ofabunchofelectrons whenthecharges aremoving.Weknowthattheapparent massofaparticlechanges by1/./1—02/c2.Does itscharge dosomething similar? No! Charges arealways thesame, moving or not. Otherwise wewould notalways observe that thetotal charge isconserved. ‘Suppose that wetake ablock ofmaterial, sayaconductor, which isinitially uncharged. Nowweheatitup.Because theelectrons haveadifferent massthan theprotons, thevelocities oftheelectrons andoftheprotons willchange bydiffer- entamounts. Ifthecharge ofaparticledependedonthespeedoftheparticlecarry- ingit,intheheated block thecharge oftheelectrons andprotons would nolonger balance. Ablock would become charged when heated. Aswehave seen earlier, a very small fractional change inthecharge ofalltheelectrons inablock would give rise toenormous electric fields. No such effect has ever been observed. Also, wecanpoint outthat themean speed oftheelectrons inmatter depends ‘onitschemical composition. Ifthecharge onanelectron changed withspeed,the netcharge inapiece ofmaterial would bechanged inachemical reaction. Again, astraightforward calculation shows that even avery small dependence ofcharge ‘onspeed would give enormous fields from thesimplest chemical reactions. No sucheffectisobserved, andweconclude thattheelectricchargeofasingleparticle isindependent ofitsstate ofmotion. Sothechargegonaparticleisaninvariantscalarquantity,independent of theframe ofreference. That means that inany frame thecharge density ofa distribution ofelectrons isjust proportional tothenumber ofelectrons perunit volume. Weneed only worry about thefactthat thevolume canchange because ofthe relativistic contraction ofdistances. ‘Wenow apply these ideas toourmoving wire. Ifwetake alength Loofthe wire, inwhich there isacharge density poofstationary charges, itwill contain thetotal charge @=poLoAo. Ifthesame charges areobserved inadifferent frame tobemoving with velocity v,they willallbefound inapiece ofthematerial with theshorter length L=LyV1— ee, (13.22) butwith thesame area Ao(since dimensions transverse tothemotion areun- changed), SeeFig. 13-11. Ifwecallpthedensity ofcharges intheframe inwhich they aremoving, thetotalchargeQwillbepLAp.ThismustalsobeequaltopoLoA,becausechargeis thesame inanysystem, sothat pL=poLo or,from (13.22), p=. (13.23)VI =02/e? ~ hh! os al deena el ° ” Area A ‘Area A@ v0 Q. vw Fig. 13-11. Ifadistribution ofcharged particles atresthasthecharge density Po,thesame charges willhave thedensity p=po/\/T —vi/e? when seen from a frame with therelative velocity v. 8 One way ofseeing thisistoaskaquestion like: What transverse momentumwilltheparticlehaveaftertheforcehasactedforalitlewhile?WeknowfromChapter16ofVol.Ithatthetransverse momentum ofaparticleshouldbethesameinboth the S-and S’-frames. Calling thetransverse coordinate y,wewant to compare Ap, and Ap;. Using therelativistically correct equation ofmotion, F=dp/dt, weexpect that after thetime Atour particle will have atransverse ‘momentum Ap,intheS:system given by Ap, =Fat. (1331) s IntheS’-system, thetransverse momentum willbe 4p, =Fear. (13.32) > Wemust,ofcourse,compare Ap,andAp,forcorresponding timeintervals Arand 8YAr,WehaveseeninChapter15ofVol.Ithatthetimeintervalsreferredtoa OSSZY‘movingparticleappeartobelongerthanthoseintherestsystemoftheparticle. (o) G Sinceourparticle isinitially atrestinS’,weexpect,forsmallAs,that -—_*_. 13.33 ataa (13.33) and everything comes outO.K. From (13.31) and (13.32), s 4p)_Flat’ap” Far’ > whichisjust=1ifwecombine(13,30)and(13.33). Pa GY Wehavefoundthatwegetthesamephysicalresultwhetherweanalyzethe G ‘motionofaparticlemovingalongawireinacoordinatesystematrestwithrespectZ tothewire, orinasystem atrestwith respect totheparticle. Inthefirstinstance, theforce waspurely “magnetic,” inthesecond, itwaspurely “electric.” The two (b) points ofview areillustrated inFig. 13-12 (although there isstill amagnetic field BYinthesecond frame, itproduces noforces onthestationary particle). Fig.13-12. Infrome $thecharge Ifwehadchosen stillanother coordinate system, wewould have found a density iszero and thecurrent density is different mixture ofEandBfields,Electricandmagneticforcesarepartofone i.There isonly @magnetic field. InS', physical phenomenon—the electromagnetic interactions ofparticles. Thesepara- there is@charge density p',andodiffer- tionofthisinteraction intoelectric andmagnetic parts depends verymuch ontheentcurrentdensity'.Themagneticfleldreferenceframechosenforthedescription. Butacompleteelectromagnetic de- Haiiterent ondthereisanelectric scriptionisinvariant; electricity andmagnetism takentogetherareconsistent eldwith Einstein's relativity. Since electric andmagnetic fields appear indifferent mixtures ifwechange our frame ofreference, wemust becareful about how welook atthefields Eand B.Forinstance,ifwethinkof“lines”ofEorB,wemustnotattachtoomuchreality tothem. The lines may disappear ifwetrytoobserve them from adifferent co- ordinate system. Forexample, insystem S’there areelectric feld lines, which we donotfind “moving past uswith velocity vinsystem S.” Insystem Sthere areno electric field lines atall!Therefore itmakes nosense tosaysomething like: WhenImoveamagnet,ittakesitsfieldwithit,sothelinesofBarealsomoved.Thereisnoway tomake sense, ing*reral, outoftheidea of“the speed ofamovingfield line.” The fields areourway ofdescribing what goes onatapoint inspace. In particular, Eand Btellusabout theforces that will actonamoving particle. The {question “What istheforce onacharge from amoving magnetic field?” doesn't mean anything precise, The force isgiven bythevalues ofEandBatthecharge, andtheformula (13.1) isnottobealtered ifthesource ofEorBismoving (itis thevalues ofEand Bthat will bealtered bythemotion). Our mathematical de- scription deals only with thefields asafunction ofx,y,z,and £with respect tosomeinertialframe.Wewilllaterbespeakingof“awaveofelectricandmagneticfieldstravellingthroughspace,"as,forinstance,alightwave.Butthatislikespeakingofawavetravellingonastring.Wedon'tthenmeanthatsomepartofthesiringismoving 1340 4 The Magnetic Field in Various Situations 14-1 The vector potential Inthischapter wecontinue ourdiscussion ofmagnetic fields associated with 14-1 The vector potential steady curtents—the subject ofmagnetostatics. Themagnetic fieldisrelated 0 44.»-Theyectorpotentialofknown electric currents byour basic equations vB=0, (4.1) 14.3Astraightwire evxs-i. (14.2) 14-4Alongsolenoid° 14-5Thefieldofasmallloop;the Wewant nowtosolve these equations mathematically inageneral way, thatis, magnetic dipole without requiring anyspecial symmetry orintuitive guessing. Inelectrostatics, 44. Thevector potential ofawefoundthattherewasastraightforward procedure forfindingthefieldwhenthe per Pom positions ofallelectric charges are known: One simply works out the scalar potential @bytaking anintegral over thecharges—as inEq.(4.25). Then ifone 14-7 ThelawofBiotandSavart wants theelectric field, itis obtained from thederivatives of¢.Wewill now show thatthere isacorresponding procedure forfinding themagnetic field Bifweknow thecurrent density jofallmoving charges. Inelectrostatics wesaw that (because thecurl ofEwas always zero) itwas possible torepresent Easthegradient ofascalar field g.Now thecurl ofBisnot always zero, soitisnotpossible, ingeneral, torepresent itasagradient. However, thedivergence ofBisalwayszero,andthismeansthatwecanalwaysrepresent Basthecur!ofanother vector field. For, aswesawinSection 2-8, thedivergence ofa ccurl isalways zero. ‘Thus wecan always relate Btoafield wewill call4by B=VXA. (14.3) Or,bywriting outthecomponents, Ody_dy B=(VXA=GtSe OAs_OAs B=(VXAy=GFHE (14.4) ody_ade Bea(VXAye=FEFe Writing B=VXAguarantees that Eq.(14.1) issatisfied, since, necessarily, VB=V-(VX4)=0. Thefield Aiscalled thevector potential. You willremember that thescalar potential ¢was notcompletely specifiedbyitsdefinition. Ifwehavefound¢forsomeproblem,wecanalwaysfindanotherpotential ¢'that isequally good byadding aconstant: emote ‘The new potential ¢’gives thesame electric fields, since thegradient VC iszero;@!and¢represent thesamephysics.Similarly, wecan have different vector potentials Awhich give thesame magnetic fields. Again, because Bisobtained from Abydifferentiation, adding a “4 Itisclear that foranyparticular field B,thevector potential isnotunique; there aremany possibilities. The third solution, Eq. (14.8), hassome interesting properties. Since the x-component isproportional to—yand they-component isproportional to+x, Amust beatright angles tothevector from thez-axis, which wewillcallr’(the y “prime” istoremind usthatitisnorthevector displacement from theorigin) _— Also, themagnitude ofAisproportional to/x?+y?and,hence,tor’.SoA ' ‘canbesimply written (forouruniform field) as i\a by A=4BXr. (49) Thevector potential 4hasthemagnitude By’/2 and rotates about thez-axis asshowninFig.14-1.If,forexample,theBfieldistheaxialfieldinsideasolenoid, (Ppthenthevectorpotential circulates inthesamesenseasdothecurrentsofthe —4 xsolenoid. |CNThe vector potential forauniform field can beobtained inanother way. Thecirculation of4onanyclosed loop I’canberelated tothesurface integral of VX AbyStokes’ theorem, Eq. (3.38): a §.4rd= f(VvXA)-nda. (14.10)iniiae But theintegral ontheright isequal totheflux ofBthroughtheloop,so Fig.14-1.Auniformmagnetic field Binthez-direction corresponds toa f.Aw= [Benda (14.11)VectorpotentialAthatrotatescbouttheinside z-axis, with themagnitude A=Br'/2 Sothecirculation ofAaround anyloopisequaltothefluxofBthrough theloop. _!"#thedisplacement fromthez-axis). Ifwetakeacircular loop,ofradius’inaplaneperpendicular toauniform field B,thefluxisjust r?B, Ifwechooseouroriginonanaxisofsymmetry, sothatwecantakeAascircum-ferential andafunction only ofr’,thecirculation willbe GfAds=2nd=7B. Weget,asbefore, Br ak, Intheexample wehave justgiven, wehave calculated thevector potential from themagnetic field, which isopposite towhat onenormally does. Incomplicated problems itis usually easier tosolve forthevector potential, andthen determine themagnetic field from it.Wewillnow show how thiscanbedone. 14-2 Thevector potential ofknown currents Since Bisdetermined bycurrents, soalso isA.Wewant now tofind4in terms ofthecurrents. Westart with ourbasic equation (14.2): evxs=d, rn which means, ofcourse, that evx(wxa=Zz. (14.12) This equation isformagnetostatics what theequation veven 2% (14.13) © was for electrostatics. 43 Ourequation(14.12)forthevectorpotentiallooksevenmorelikethatfor ¢ifwerewrite ¥X(¥XA)using thevector identity Eq.(2.58): vx (VX A)=(VA) —VPA (14.14) Since wehave chosen tomake V+4=0(and now you seewhy), Eq. (14.12) becomes 7 ae va-- 4, (14.15) This vector equation means, ofcourse, three equations: v4=-2, v4,--2, v4.--25. 14.16)oe oc! eoc! ‘a: . ‘Andeachoftheseequations ismathematically identical to pia!Be vg= 2. (14.17) fo Allwehave learned about solving forpotentials when p1sknown canbeused for solving foreach component ofAwhenjisknown! Wehave seen inChapter 4that ageneral solution fortheelectrostatic equation Fig.14-2, Thevector potential Aat (14-17) is point1isgivenbyanintegraloverthe «= f(2)dV2 current elements [dV atallpoints 2. Fre) ra ‘Soweknow immediately that ageneral solution forA;is = |4@ah Ad)=ma! nm! asi) and similarly for4,and A,. (Figure 14-2 will remind you ofourconventions for rigand dV.) Wecan combine thethree solutions inthevector form ol ‘JQ)dV2 AD)=Fee/ne oe (You canverify ifyouwish, bydirect differentiation ofcomponents, that thisinte- gral forAsatisfies VA=0solong asV-j =0,which, aswesaw, must happen forsteady currents.) Wehave, then, ageneral method forfinding themagnetic field ofsteady cur- rents. The principle is:thex-component ofvector potential arising from acurrent density jisthesame astheelectric potential ¢that would beproduced byacharge density pequal toj,/c?—and similarly forthey-andz-components. (This principle works only with components infixed directions. The “radial” component ofA does notcome inthesame way from the“radial” component ofj,forexample.) Sofrom thevector current density j,wecanfind Ausing Eq.(14.19)}—that is,we find each component ofAbysolvingthreeimaginaryelectrostatic problemsfor thecharge distributions p,=j./e2, pz=jy/c®, andps=j./c. Then weget Bbytaking various derivatives ofAtoobtainVXA.It’salittlemorecompli- cated than electrostatics, butthesame idea. Wewill now illustrate thetheory by solving forthevector potential inafewspecial cases. 14-3Astraight wire Forourfirstexample, wewillagain findthefield ofastraight wire—which we solved inthelast chapter byusing Eq. (14.2) and some arguments ofsymmetry. ‘Wetake along straight wire ofradius a,carrying thesteady current J.Unlike the charge onaconductor intheelectrostatic case, asteady current inawire 1suni- formly distributed throughout thecross section ofthewire. Ifwechoose our 44 coordinates asshown inFig. 14~3, thecurrent density vector jhasonly az-com- ponent. Itsmagnitude is I a (14.20) inside thewire, and zero outside. Since j,andj,areboth zero, wehave immediately 4,=0, Ay=0. 2 Toget4,wecanuseoursolution fortheelectrostatic potential ¢ofawirewitha 14 uniformchargedensityp=j,/c2.Forpointsoutsideaninfinitechargedcylinder, qfbr theelectrostatic potentialis Wd"% nN (jo fa o>re Bg , . , » Ae wherer’=4/x?+y3and)isthechargeperunitlength,wa%p.SoA,mustbe Y==ra ag A=—-225ne aa2areoc? ij4forpointsoutside alongwirecarrying auniform current. Sincema%j.=J,we 4 can also write --,2, Fig.14-3. Alongcylindrical wire As—Fraga (14.21) longthez-axis with«uniform current density j.NowwecanfindBfrom(14.4).Thereareonlytwoofthesixderivatives that vk arenotzero. Weget --any-- 4 Bemey =ee (14.22) __1 a ~_/ x, By=spectHel=eeeoa (14.23) B,=0. Wegetthesame result asbefore: Bcircles around thewire, andhasthemagnitude 1om yoree (14.24) vf5, -Te14-4Alongsolenoid ces/in Next, weconsider again theinfinitely long solenoid with acircumferential>EY \ currentonthesurfaceofn/perunitlength. (Weimagine therearenturnsofwire = 7 perunitlength, carrying thecurrent/,andweneglecttheslightpitchofthewinding.) \NEB,/ Justaswehavedefineda“surfacechargedensity”0,wedefineherea“sur- \Le facecurrent density” Jequal tothecurrent perunitlength onthesurface ofthe ~-|- solenoid (which is,ofcourse, justtheaveragejtimesthethicknessofthethin winding).Themagnitude ofJis,here,nf.Thissurfacecurrent(seeFig.14~4)has | thecomponents. J=Ising, J,=Joosd, Jp=0. Fig.14-4. Along solenoid witha face currentdensity J. NowwemustfindAforsuchacurrent distribution. surface current density J First, wewish tofindA,forpoints outside thesolenoid. The result isthesame astheelectrostatic potential outside acylinder with asurface charge o=aosing, withoo=J/c?, Wehave notsolved such acharge distribution, butwehave done something similar. This charge distribution isequivalent totwo solid cylinders of charge, onepositive and onenegative, with aslight relative displacement oftheir 4s axes inthey-direction. The potential ofsuch apair ofcylinders isproportional tothederivative with respect toyofthepotential ofasingle uniformly charged cylinder. Wecould work outtheconstant ofproportionality, butlet's notworry about itfor the moment. The potential ofacylinder ofcharge isproportional toIn»’;thepotential ofthepair isthen ainy_yOyo So we know that Ae=Ko (14.25) where Kis some constant. Following thesame argument, wewould find x Ay=KS (14.26) Although wesaid before that there was nomagnetic field outside asolenoid, we findnow that there isanA-field which circulates around thez-axis, asinFig. 14-4. The question is:Isitscurl zero? Clearly, B,andByarezero, and =2 (K%)—-2 (Kx 12x?) 1 ay") _aK(B- +A =0. Sothemagnetic field outside avery long solenoid isindeed zero, even though the vector potential isnot. ‘Wecancheck ourresult against something elseweknow: The circulation ofthevectorpotentialaroundthesolenoidshouldbeequaltothefluxofBinsidethecoil(Eq.14.11).Thecirculation isA-2xr’or,sinceA=K/r’,thecirculation is2nK,Noticethatitisindependent ofr’.Thatisjustasitshouldbeifthere isno Boutside, because theflux isjust themagnitude ofBinsidethesolenoidtimes \ ' a,Itisthesame forallcircles ofradius r’>a,Wehavefound inthelastchapter 1 1 that thefield inside isnJ/eoc?, sowecandetermine theconstant K: y . 2K=ra?1, Jeov foe EIS or nla? | J KoJeet as Sothevectorpotentialoutsidehasthemagnitudenla?1 Anyaay (4.27) andisalways perpendicular tothevector r’, We have been thinking ofasolenoidal coil ofwire, but wewould produce : } thesame fields ifwerotated alongcylinder withanelectrostatic charge onthe i i surface. Ifwehave athin cylindrical shell ofradius awith asurface charge 0, rotating thecylinder makesasurfacecurrentJ=or,where»=awisthevelocity ofthesurface charge. There willthen beamagnetic field B=gaw/egc? inside ; |thecylinder. Fig.14-5.Arotating charged cylin Nowwecanraiseaninteresting question. Suppose weputashortpieceof derProdvcesomagneticFeldinside.AwireWperpendiculartotheaxisofthecylinder,extendingfromtheaxisoutto phonfediedreratingwi‘necylinder thesurface,andfastenedtothecylindersothatitrotateswithit,asinFig.14-5. taschargesindloced onHisnde, Thiswireismoving inamagnetic field,sothevXBforces willcause theendsof thewire tobecharged (they willcharge upuntil theE-field from thecharges just balances the v Bforce). Ifthecylinder hasapositive charge, theendofthewire attheaxiswillhave anegative charge. Bymeasuring thecharge ontheendofthe 146 wire, wecould measure thespeed ofrotation ofthesystem. Wewould have an “angular-velocity meter"! Butareyouwondering: “What ifIputmyselfintheframeofreferenceofthe rotating cylinder? Then there isjustachargedcylinderatrest,andIknowthatthe electrostatic equations saythere will benoelectric fields inside, sothere will beno force pushing charges tothecenter. Sosomething must bewrong.” Butthere is nothing wrong. There isno“relativity ofrotation.” Arotating system isnotan inertial frame, and thelaws ofphysics aredifferent. Wemust besure touseequa- tions ofelectromagnetism only with respect toinertial coordinate systems. Itwould benice ifwecould measure the absolute rotation ofthe earth with such acharged cylinder, butunfortunately theeffect ismuch toosmall toobserve even with the most delicate instruments now available. 14-5 The field ofasmall loop; themagnetic dipole Let’s use thevector-potential method tofind themagnetic field ofasmall loop ofcurrent. Asusual, by“small” wemean simply that weareinterested in thefields only atdistances large compared with thesizeoftheloop. Itwillturn outthat any small loop isa“magnetic dipole.” That is,itproduces amagnetic field liketheelectric field from anelectric dipole. Pp z a yy fe y en eed 7 a Teese ee of ro Fig. 14-6. Arectangular loop ofwire with the Fig. 14-7. The distribution ofjxin currentJ.Whatisthemagnetic fieldatP#(R>a,orb.)thecurrentloopofFig.14-6. We take first arectangular loop, and choose our coordinates asshown in Fig.14-6. There arenocurrents inthez-direction, soAziszero. There arecurrents inthex-direction onthetwo sides oflength a.Ineach leg, thecurrent density (and current) isuniform. Sothesolution forAzisjust like theelectrostatic po- tential from twocharged rods (see Fig. 14-7). Since therods have opposite charges, their electric potential atlarge distances would bejust thedipole potential (Section 6-5). Atthepoint PinFig, 14-6, thepotential would be =1peer, =aa (14.28) wherepisthedipolemomentofthechargedistribution. Thedipolemoment,in this case, isthetotal charge onone rodtimes theseparation between them: p=rab, (14.29) The dipole moment points inthenegative y-direction, sothecosine oftheangle between Randpis—y/R(whereyisthecoordinate ofP).Sowehave go- my.4reg R? RK WegetAzsimply byreplacing )by1/c?: Jaby A=-Zhe (14,30) “7 Bythesame reasoning, labx Ay=Sree RS (14.31) Again, A,isproportional toxandA,isproportional to—y,sothevector potential (atlarge distances) goes incircles around thez-axis, circulating inthesame sense asTin theloop, asshown inFig. 14-8, The strength of isproportional tofab, which isthecurrent times thearea oftheloop. This product iscalled themagnetic dipole moment (or, often, just : “magnetic moment”) oftheloop. Werepresent itbyu: b= Tab. (14,32) The vector potential ofasmallplaneloopofanyshape(circle,triangle,etc.)is also given byEqs. (14.30) and(14.31) provided wereplace Jabby ¥ b=T°(areaofloop). (1433) Weleavetheproofofthistoyou. aWecanputourequation invector form ifwedefine thedirection ofthevector 4110 bethenormal totheplane oftheloop, with apositive sense given bytheright- . hand rule(Fig. 14-8). ‘Then wecanwrite Tt * ~1wXR__ 1XenA=Freoct R?~Frege? RP (14.34) Fig. 14-8. The vector potential ofa small current loop attheorigin (inthe Wehave stiltofindB,Using(14.33)and(14.34),togetherwith(14.4),weget xy-plane); «magneticdipolefield. cy-plane);«magneticdipolefield. et kde as*=~52FregeRS BS : (where by... wemean 4/4zreo¢*), 9(_...¥) 2... DF 8=2( &)- BS 2 (.8) 2 (oeB=aa)5( #) (436) --ae-#). ‘The components 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 wecall theloop amagnetic dipole. The word “dipole” isslightly misleading when applied toamagnetic field because there arenomagnetic “poles” that correspond toelectric charges. The magnetic “dipole field” isnotproduced bytwo “charges,” butbyanelementary current loop. Itiscurious, though, that starting with completely different laws, V-E=p/¢. and¥XB= j/eoc?, wecanendupwith thesame kind ofafield.Whyshould that be? 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 one case, it’sacirculating cur- rent; intheother, apair ofcharges, oneabove andonebelow theplane oftheloop forthecorresponding field. 14-6 The vector potential ofacircuit Weareoften interested inthemagnetic fields produced bycircuits ofwire in which thediameter ofthewire isvery small compared with thedimensions ofthe whole system. Insuch cases, wecan simplify theequations forthemagnetic field 48 For athin wire wecan write our volume element as av=Sds, where Sisthecross-sectional areaofthewireanddsistheelement ofdistance alongthewire.Infact,sincethevectordsisinthesamedirection as,asshownin ¥,Fig. 14-9 (and wecan assume thatjisconstant across anygiven cross section), Se, wecanwrite avector equation: 5 jdV =jSds. (14.37) S. (ds But/Sisjustwhatwecallthecurrent/inawire,soourintegralforthevectorpotential (14.19) becomes 1 Ids; AQ)=ie| (14.38) Fig,14-9.Forfine wire {dV isthe same asIds. (eeFig. 14-10). (We assume thatFisthesame throughout thecircuit. Ifthere are several branches with different currents, weshould, ofcourse, usetheappropriate 1for each branch.) Again, wecanfind thefields from (14,38) either byintegrating directly orby solving thecorresponding electrostatic problems. 14-7ThelawofBiotandSavart fie ‘ Instudying electrostatics wefound that theelectric field ofaknown charge distribution could beobtained directly with anintegral (Eq. 4-16): BU)=are 7 Fig.14-10.Themagneticfieldofa Aswehaveseen,itisusuallymoreworktoevaluatethisintegral—there arereallywireconbeobtainedfromonintegralthree integrals, one foreach component—than todotheintegral forthepotential around thecircuit. and take itsgradient. There isasimilar integral which relates themagnetic field tothecurrents. Wealready have anintegral forA,Eq.(14.19); wecangetanintegral forBby taking thecurl ofboth sides: - =ume) BI)=VXA=9«lal ra (1439) Now wemust becareful: The curl operator means taking thederivatives of A(1),thatis,itoperates onlyonthecoordinates (x1,y1,21).Wecanmovethe VX operator inside theintegral sign ifweremember that itoperates only on variables with thesubscript 1,which ofcourse, appear only in ne=(G1~m+ O1- mt —2)? (14.40) Wehave, forthex-component ofB, aA, dAyBo He a! ,a (1 ,8 (1=ea! [oa(E)- kG) asen -- Y=dea4] ‘The quantity inbrackets isjust thex-component of iXre _ixen ia te “9 1s The Vector Potential 15-1 The forces onacurrent loop; energy ofadipole Inthelastchapter westudied themagnetic field produced byasmall rec- 1-1 The forces onacurrent loop; tangular current loop. Wefound thatitisadipole field, with thedipole moment energy ofadipole Biven by =A. (15.1) 15-2Mechanical andelectricalwealA ; ee where /isthecurrent andAisthearea oftheloop. The direction ofthemoment 18-3 The energy ofsteady currents isnormal totheplaneoftheloop,sowecanalsowrite 13-48 ‘4 n= Ida, 15-5 Thevector potential and quantum mechanies where mis the unit normal tothe area A. ‘Acurrent loop—or magnetic dipole—not onlyproduces magnetic fields, but 15-6What istrueforstatics is willalsoexperience forces when placed inthemagnetic fieldofother currents. falsefordynamics ‘Wewill look first attheforces onarectangular loop inauniform magnetic field. Letthez-axis bealong thedirection ofthefield, and theplane oftheloop be placed through they-axis, making theangle 6with thexy-plane asinFig. 15-1 Then themagnetic moment oftheloop—which isnormal toitsplane—will make theangle 6with themagnetic field. Since thecurrents areopposite onopposite sides oftheloop, theforces are also opposite, sothere isnonetforce ontheloop (when thefield isuniform). Because offorces onthetwosides marked Iand2inthefigure, however, there isa torque which tends torotate theloop about they-axis. The magnitude ofthese forces Fand Fyis Fy=Fa=IBb. Theit moment arm is asin 4, sothetorque is 1=[abBsin8, z or,since Jabisthemagnetic moment oftheloop, y 8 r=uBsind. F,Wsmn ‘Thetorquecanbewritteninvectornotation: MsWY pd xrex (15.2)iWyeeEW Although'we haveonlyshownthatthetorqueisgivenbyEq.(15.2)inonerather >specialcase,theresultisrightforasmallJoopofanyshape,aswewillsee.Youwill °\e remember that wefound thesame kind ofrelation forthetorque onanelectric dipole: TH=PXE Fig.15-1. Arectangular loopcarry- ingthecurrent Isitsin@uniform field B Wenow askabout themechanical energy ofourcurrent loop. Since there is {inthez-direction). Thetorque onthe atorque, theenergy evidently depends ontheorientation. Theprinciple ofvirtual loop is+= XB,where themagnetic work saysthatthetorque istherateofchange ofenergy withangle, sowecanwrite moment »=lab. dU = -7d0. 154 Setting +=—nB sin6,and integrating, wecanwrite fortheenergy U=—uBcos6+aconstant. (15.3) (CThe sign isnegative because thetorque tries tolineupthemoment with thefield; theenergy islowest when uand Bareparallel.) Forreasons which wewilldiscuss later, thisenergy isnotthetotal energy ofa current loop. (We have, foronething, nottaken into account theenergy required tomaintain thecurrent intheloop.) Wewill, therefore, callthisenergy Umecns toremind usthat itis only part oftheenergy. Also, since weareleaving outsome oftheenergy anyway, wecansettheconstant ofintegration equal tozero inEq. (15.3). Sowerewrite theequation: Unnech =—H°B. (15.4) Again, thiscorresponds toourresult foranelectric dipole: U= rE (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 that thecurrent intheloop—or atleast u—is kept constant. Wecan show forourrectangular loop that Use also corresponds tothe mechanical work done inbringing theloop into thefield. The total force onthe oop iszero only inauniform field; inanonuniform field there arenetforces ona current loop. Inputting theloop into aregion with afield, wemust have gone through places where thefield was notuniform, and sowork was done. Tomake thecalculation simple, weshall imagine that theloop isbrought into thefield with itsmoment pointing along thefield. (Itcanberotated toitsfinal position after it isin place.) Imagine that wewant tomove theloop inthex-direction—toward aregion of stronger field—and that theloop isoriented asshown inFig. 15-2. Westart somewhere where thefield iszero and integrate theforce times thedistance aswe bring theloop into thefield. B Fig. 15-2. Aloop iscarried along“ s x thex-direction through thefield B,at % ie Fight angles tox. First, let's compute thework done oneach side separately and then take the sum (rather than adding theforces before integrating). The forces onsides 3and 4 areatright angles tothedirection ofmotion, sonowork isdone onthem. The force onside 2is16B(x) inthex-direction, and togetthework done against the magnetic forces wemust integrate thisfrom some xwhere thefield iszero, sayat Xx=—0, t0x2,itspresent position: W.=~[Cia —Ib[”BG)ax. (15.6) Similarly, thework done against theforces onside 1is Ww,=-ftFidx=bfB(x)dx. (15.7) 152 Jy x 8 8, ~ NTS 482 1, LoopT, q q = a Te (a) (b) Fig. 15-3. Finding theenergy ofasmall loop inamagnetic feld. Wecould wait until thenext chapter tofind outabout thisnew energy term, but wecan alsoseewhatitwillbeifweusetheprincipleofrelativityinthefollowing way. When wearemoving theloop toward thestationary coilweknow that its electrical energy isjust equal and opposite tothemechanical work done. So Unnech +Usteor(loop) =0. Supposenowwelookatwhatishappeningfromadifferentpointofview, inwhich theloop isatrest, andthecoilismoved toward it.Thecoilisthen moving intothefield produced bytheloop. The same arguments would give that Urea. +Untee(Coil) =0. Themechanical energy isthesame inthetwocases because itcomes from theforce between the two circuits. Thesum ofthetwo equations gives Waech +Uctec(loop) +Usreer(coil) =0. The total energy ofthewhole system is,ofcourse, thesum ofthetwo electrical energies plus themechanical energy taken only once. Sowehave Vrotat =Uciect(l0op) +Ustoct(COil) ++Uec =—Urnechs (15.13) The total energy oftheworld isreally thenegative ofUmecn- Ifwewant the true energy ofamagnetic dipole, forexample, weshould write Vion =tH B. Itisonly ifwemake thecondition that allcurrents areconstant that wecanuse only apart oftheenergy, Umer (Which isalways thenegative ofthetrue energy), tofind themechanical forces. Inamore general problem, wemust becareful to include allenergies. Wehave seen ananalogous situation inelectrostatics. Weshowed that the energy ofacapacitor isequal toQ?/2C. When weusetheprinciple ofvirtual work tofindtheforce between theplates ofthecapacitor, thechange inenergy isequal to07/2 times thechange in1/C. That is, ~2@ (1)__ gac av=¥a(t)=-$46. (15.14) Now suppose that wewere tocalculate thework done inmoving two con- ductors subject tothedifferent condition that thevoltage between them isheld constant. Then wecangettheright answers forforce from theprinciple ofvirtual work ifwedosomethingartificial.SinceQ=CV,therealenergyis}CV?.But ifwedefineanartificialenergyequalto—}CV®,thentheprincipleofvirtualworkcanbeused togetforces bysetting thechange intheartificial energy equal tothe iss mechanical work, provided that weinsist that thevoltage Vbeheld constant. Then a a AUec=a(-&)=-Fac (15.15) which isthesame asEq.(15.14). Wegetthecorrect result even though weare neglecting thework done bytheelectrical system tokeep thevoltage constant. Again, thiselectrical energy isjust twice asbigasthemechanical energy and of theopposite sign. Thus ifwecalculate artificially, disregarding thefact that thesource ofthe potential hastodowork tomaintain thevoltages constant, wegettheright answer. Itis exactly analogous tothesituation inmagnetostatics. 15-3 The energy ofsteady currents Wecannow useourknowledge that Uioust =—Unech tofind thetrue energy ofsteady currents inmagnetic fields. Wecanbegin with thetrue energy ofasmall current loop. Calling Uyorat just U,wewrite U=eB (15.16) Although wecalculated thisenergy foraplane rectangular loop, thesame result 2 holds forasmall plane loop ofany shape. —— Wecanfindtheenergy ofacircuitofanyshapebyimaginingthatitismade ASE N00 F upofsmall current loops. Saywehave awire intheshape oftheloop T’ofFig. Aa ~~ 15-4,WefillinthiscurvewiththesurfaceS,andonthesurfacemarkoutalargeLESrimann numberofsmallloops,eachofwhichcanbeconsideredplane.Ifweletthecurrent a! a rH circulate aroundeachofthelittleloops,thenetresultwillbethesameasacurrent Nacane ry aroundT,,sincethecurrents willcancelonalllinesinternaltoT.Physically, theAE system oflittlecurrents isindistinguishable fromtheoriginal circuit, ‘TheT Sutoces energymustalsobethesame,andsoisjustthesumoftheenergiesofthe little loops. Ifthearea ofeach little loop isAa,itsenergy is/AaB, where B,isthecom- Fig.15-4. The energy of@large ponent normal toAa.Thetotal energy is loop in@magnetic field can beconsidered ‘asthesumofenergies ofsmaller loops. U=DIB, ba. Going tothelimit ofinfinitesimal loops, thesum becomes anintegral, and U=1B,da=IfB-mda, as.17) where misthe unit normal toda, Ifweset B=VX A,wecanconnect thesurface integral toaline integral, using Stokes’ theorem, If (wx A)-nda=1gAds, 15.18 if,0XAden $, (15.18) where dsistheline element along I’.Sowehave theenergy foracircuit ofany shape: Un §And. (15.19) irouit Inthis expression Arefers, ofcourse, tothevector potential due tothose currents (other than the/inthewire) which produce thefield Batthewire. Now anydistribution ofsteady currents canbeimagined tobemade upof filaments that runparallel tothelines ofcurrent flow. Foreach pairofsuch circuits, theenergy isgiven by(15.19), where theintegral istaken around onecircuit, using thevector potential Afrom theother circuit. For thetotal energy wewant the sumofallsuch pairs. If,instead ofkeeping track ofthepairs, wetake thecomplete sum over allthefilaments, wewould becounting theenergy twice (we saw a similar effect inelectrostatics), sothetotal energy canbewritten U=afi-aav. (15.20) 156 x Qq NN N N bert NEON some2S -N-----. osar- tS @ 3 wit] L Fig. 15-5. Aninterference experiment with electrons (see also Chapter 37ofVol. I). trons arediffracted bytwo slits. The arrangement isshown again inFig. 15-5. Electrons, allofnearly thesame energy, leave thesource and travel toward awall with twonarrow slits. Beyond thewall isa“backstop” with amovable detector. The detector measures therate, which wecallJ,atwhich electrons arrive atasmall region ofthebackstop atthedistance xfrom theaxis ofsymmetry. The rate is proportional totheprobability that anindividual electron that leaves thesourcewillreachthatregionofthe backstop. This probability hasthecomplicated-looking distribution shown inthefigure, which weunderstand asduetotheinterference of two amplitudes, one from each slit. The interference ofthetwo amplitudes depends ontheir phase difference. That is,iftheamplitudes areCes andCze"*2, thephase difference §=©;—#2determines their interference pattern [see Eq. (29.12) inVol. 1].Ifthedistance between thescreen and theslits isL,and ifthe difference inthepath lengths forelectrons going through thetwo slits isa,as shown inthefigure, then thephase difference ofthetwo waves isgiven by a o-f. (527) Asusual,weletX=d/27,where)isthewavelength ofthespacevariationoftheprobability amplitude. For simplicity, wewill consider only values ofxmuch less than L;then wecan set a=td L and xd e=55- (15.28) ‘Whenxiszero,6iszero;thewavesareinphase,andtheprobability hasamaxi-mum. When 6is=,thewaves areoutofphase, they interfere destructively, and the probability isaminimum. Sowegetthewavy function fortheelectron intensity. ‘Now wewould like tostate thelawthat forquantum mechanics replaces the force lawF=qvXB.Itwillbethelawthatdeterminesthebehaviorofquantum- mechanical particles inanelectromagnetic field. Since what happens isdetermined byamplitudes, thelaw must tellushow themagnetic influences affect theampli-tudes;wearenolongerdealingwiththeacceleration ofaparticle.Thelawisthefollowing: thephase oftheamplitude toarrive viaany trajectory ischanged by thepresence ofamagneticfieldbyanamountequaltotheintegralofthevector potential along thewhole trajectory times thecharge oftheparticle over Planck's constant. That is, ‘Magneticchangeinphase=i|Avds. (15.29) trajectory 159 The same conclusion isevident ifweuse the results ofSection 14-1. There wefound that thelineintegral ofAaround aclosed path isthefluxofBthrough thepath, which here istheflux between paths (1)and (2). Equation (15.33) can, ifwewish, bewritten as b=(B=0)+F(fuxofBbetween(1)and(2)},(15.34) 8wherebythefluxofBwemean,asusual,thesurfaceintegralofthenormalcom-ponent ofB.The result depends only onB,andtherefore only onthecurl ofA. ch Now because wecan write theresult interms ofBaswell asinterms ofA,youmightbeinclinedtothinkthattheBholdsitsownasa“real”fieldandthat ae theAcanstill bethought ofasanartificial construction. Butthedefinition of“eal”fieldthatweoriginallyproposedwasbasedontheideathata“real”field LT would notactonaparticle from adistance. Wecan, however, give anexampleinwhichBiszero—oratleastarbitrarilysmall—atanyplacewherethereissome ET chance tofind theparticles, sothat itisnotpossible tothink ofitacting direcily onthem. A ‘You remember that foralong solenoid carrying anelectric current there is bs aB-fieldinsidebutnoneoutside,whilethereislotsofAcirculating aroundoutside, ttasshown inFig.15-6. Ifwearrange asituation inwhich electrons aretobefound ee onlyoutside ofthesolenoid—only where thereisA—there willstillbeaninfluence in onthemotion, according toEq.(15.33). Classically, thatisimpossible. Classically, theforce depends only onB;inorder toknow that thesolenoid iscarrying current, Fig. 15-6, The magnetic fleld and theparticle must gothrough it.Butquantum-mechanically youcanfindoutthat __vector potential ofalongsolenoid. there isamagnetic field inside thesolenoid bygoing around it—without ever going close toit! ‘Suppose that weputavery long solenoid ofsmall diameter just behind the wall andbetween thetwoslits, asshown inFig. 15-7. Thediameter ofthesolenoid istobemuch smaller than the distance dbetween the two slits. Inthese circum- stances, thediffraction oftheelectrons attheslitgives noappreciable probability that theelectrons willgetnear thesolenoid. What willbetheeffect onourinter- ference experiment? __-- 777 A SO LENOID INES OFB L Fig. 15-7. Amagnetic field can influence themotion ofelectrons even thoughitexistsonlyinregionswherethereisonarbitrarily smallprobability offindingthe ‘electrons. Wecompare thesituation with and without acurrent through thesolenoid. If'we have nocurrent, wehave noBorAand wegettheoriginal pattern ofelec~ tron intensity atthebackstop. Ifweturn thecurrent oninthesolenoid and build upamagnetic field Binside, then there isanAoutside. There isashift inthephasedifference proportional tothecirculation of4outsidethesolenoid,whichwillmean thatthepattern ofmaxima andminima isshifted toanew position. Infact, since the flux ofBinsideisaconstantforanypairofpaths,soalsoisthecircula~ tionofA.Foreveryarrivalpointthereisthesamephasechange;thiscorresponds 1541 5 N « y 2sore iOg aeeen TAH Pat’ peT--- gr| - =--- SSro--- Bea ers LINESOF XSLbLv Fig,15-8, Theshiftoftheinterference pattern duetoastripofmagnetic field. over alarger region behind theslits, asshown inFig. 15-8. Wewilltake theideal- ized case where wehave amagnetic field which isuniform inanarrow strip of width w,considered small ascompared with L.(That caneasily bearranged; the backstop canbeputasfaroutaswewant.) Inorder tocalculate theshift inphase,wemusttakethetwointegrals ofAalongthetwotrajectories (1)and(2).They differ, aswehave seen, merely bytheflux ofBbetween thepaths. Toourapproxi- mation, thefluxisBwd. ‘The phase difference forthetwo paths isthen 3=B=0)+$FBod. (1537) Wenote that, toourapproximation, thephase shift isindependent oftheangle. Soagain theeffect will betoshift thewhole pattern upward byanamount Ax. Using Eq.(15.28), Ix 4,18 ax=Bas=Bos-a=oy. Using(15.37) for8—8(B=0), en ax=1x2Bw. (15.38) weet. ‘Suchashiftisequivalent todeflecting allthetrajectories bythesmall angle a -= . (seeFig.15-8), where en « a=B=Kooy. (15.39) Rance 1 Ep LINES OFB Nowclassically wewouldalsoexpectathinstripofmagnetic fieldtodeflect Ee alltrajectories through some small angle, saya’,asshown inFig.15-9(a). Asthe j jelectronsgothroughthemagneticfield,theyfeelatransverse forcegoXBwhich w_lastsfor’atimew/v.Thechange intheirtransverse momentum isjustequalto (a)thisimpulse, so Ape =quB. (15.40) Theangular deflection [Fig. 15-9(b)] isequal totheratio ofthistransverse mo- "mentumtothetotalmomentum p.Wegetthat et P Ape_qwB d= =f 15.41- aay) (o) Wecancompare thisresult with Eq.(15.39), which gives thesame quantity Fig.15-9, Deflection of particle computed quantum-mechanically. Buttheconnection between classical mechanics due topassage through @strip of andquantum mechanics isthis: Aparticle ofmomentum pcorresponds toaquan- magnetic feld. 113 Table 15-1 FALSE INGENERAL (trueonlyforstatics) ‘TRUEALWAYS Fe«see (Coulomb's law) F=qE+0x B) (Lorentzforce) wvene (Gauss’ law) vxXE=0 avxe--2 (Faraday's law) oA E=-veB=-ve-F 1_pQer2 BU)=aeYs For conductors,E=0,=constant. Q=CV Inaconductor, Emakescurrents. 7VvB=0 (No magnetic charges) | BavxAevxp=t (Ampere’s law) mevxp-L4% _f2Xe2 Bil)ret a,a2 2, 2 2,10% 2 ve--£ (Poisson’s equation) Ve-aa--% and : 2g od ay-1ea_ _iWAoa V4ae~~cat withwith ‘A= 2y.4 4b- | vea=0 ovat Sno | 1f2@ 1fae00)=gig[Bar 0Fre)ona |and Or ta! ava Aa.gs[22a with nfer-@ a |umsfooay+afs-aar u-|(geese a) ‘The equations marked byanarrow (=) areMaxwell's equations. .154s rents.Soinvaryingfieldsaconductor isnotanequipotential. Italsofollowsthattheideaofacapacitance isnolongerprecise.Since there arenomagnetic charges, thedivergence ofBisalways zero. So Ban always beequated toVXA.(Everything doesn’t change!) Butthegenera- tion ofBisnotonlyfromcurrents:VXBisproportional tothecurrentdensity plus anewterm dE/at. This means that4isrelated tocurrents byanewequation. Itisalso related tog.IfwemakeuseofourfreedomtochooseV«Aforourown convenience, theequations forAor@canbearranged totake onasimple and ele- gant form. Wetherefore make thecondition that ¢°¥-4 =—a¢/at, andthe differential equations for4or¢appear asshown inthetable. The potentials 4and ¢can still befound byintegrals over thecurrents and charges, butnotthesame integrals asforstatics. Most wonderfully, though, the true integrals arelike thestatic ones, with only asmall and physically appealing modification. When wedotheintegrals tofind thepotentials atsome point, say point (1)inFig. 15-10, wemust usethevalues ofjandpatthepoint(2)atan earlier time t'=1—ry2/e. Asyouwould expect, theinfluences propagate from point (2)topoint (1)atthespeed e.With this small change, one cansolve forthe fields ofvarying currents and charges, because once wehave 4and 4,wegetB from ¥XA,asbefore, and Efrom —Vé —a4/at. Qe he Fig. 15-10. The potentials atpoint (1)andatthetimetaregiven bysum- # ming thecontributions from each element ofthesource attheroving point (2), using thecurrents and charges which were present attheearlier time #—ri2/c. Finally, you willnotice that some results—for example, that theenergy density inanelectric field is¢9£?/2—are true forelectrodynamics aswell asforstatics. You should notbemisled into thinking that this isatall“natural.” The validity ofanyformula derived inthestatic case must bedemonstrated over again forthe dynamic case. Acontrary example istheexpression fortheelectrostatic energy intermsofavolumeintegralofpg.Thisresultistrueonlyforstatics.Wewillconsider allthese matters inmore detail induetime, butitwillperhaps beuseful tokeep inmind thissummary, soyou will know what you canforget, and what you should remember asalways true, 1516 16 Induced Currents 16-1 Motors andgenerators ‘Thediscovery in1820thatthere wasaclose connection between electricity 16-1 Motors andgeneratorsandmagnetism wasveryexciting—until then,thetwosubjectshadbeenconsidered 46-2Transformers andinductancesasquite independent. Thefirst discovery wasthat currents inwires make magnetic fields; then,inthesameyear,itwasfound thatwirescarrying current inamagnetic 16-3 Forces oninduced currents fieldhaveforces onthem. 16-4 Electrical technology One oftheexcitements 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 inbare outline inFig. 16-1. Apermanent magnet—usually with some pieces ofsoft iron—is used toproduce amagnetic field intwo slots. Across each slotthere isanorth andsouth pole, asshown. Arectangular coilofcopper igplaced with onesideineach slot. When acurrent passes through thecoil, itflows inopposite directions inthetwo slots, 80theforces arealso opposite, producing atorque onthecoil about theaxis shown. Ifthecoil ismounted onashaft sothat itcan turn, itcan becoupled to pulleys orgears andcandowork. The same idea can beused formaking asensitive instrument forelectrical measurements. Thus themoment theforce lawwas discovered theprecision of electrical measurements was greatly increased. First, thetorque ofsuch amotor can bemade much greater foragiven current bymaking thecurrent goaround ‘many turns instead ofjustone. Then thecoilcanbemounted sothat itturns with ‘very little torque—either bysupporting itsshaft onvery delicate jewel bearings or byhanging thecoil onavery finewire oraquartz fiber. Then anexceedingly small current willmake thecoilturn, andforsmall angles theamount ofrotation will sou beproportional tothecurrent. The rotation canbemeasured bygluing @pointer ‘tothecoilor,forthemostdelicateinstruments, byattachingasmallmirrortotheé| cer\YoORREPcoilandlooking attheshiftoftheimage ofascale, Suchinstruments arecalled Ql NGgalvanometers. Voltmeters andammeters workonthesameprinciple. te|ARS‘Thesameideascanbeappliedonalargescaletomakelargemotorsforpro- RK tgviding mechanical power. Thecoilcanbemade togoaround andaround byar- - ranging that theconnections tothecoil arereversed each half-turn bycontacts mounted ontheshaft. Then thetorque isalways inthesame direction. Small wanedemotors aremadejustthisway.Larger motors, deorac,areoftenmadeby rome replacing thepermanent magnet byanelectromagnet, energized from theelectrical power source, With therealization thatelectric currents make magnetic fields, people im- _Fig.16-1. Schematic outline of mediately suggested that, somehow orother, magnets might alsomake electric _simple electromagnetic motor. fields. Various experiments were tried. Forexample, twowires were placed parallel toeach other and acurrent was passed through one ofthem inthehope offinding acurrent intheother. Thethought wasthat themagnetic field might insome way drag theelectrons along inthesecond wire, giving some such lawas“likes prefer tomove alike.” With thelargest available current and themost sensitive gal- vanometer todetect anycurrent, theresult was negative. Large magnets next to wires also produced noobserved effects. Finally, Faraday discovered in1840 the essential feature that had been missed—that electric effects exist only when there issomething changing. Ifoneofapair ofwires hasachanging current, acurrent isinduced intheother, orifamagnet ismoved near anelectric circuit, there isa current, Wesaythat currents areinduced. This wastheinduction effect discovered 16-41 |ay,aeed a) [=JcawanonereR GRIVANOMETER Fig. 16-2. Moving awire through amagnetic field Fig. 16-3. Acoil with current produces produces acurrent, osshown bythegalvanometer. current inasecond collifthefstcollismoved orifitscurrent ischanged. Thecoilofthe generator hasaninduced emffrom itsmotion. The amount of theemf isgiven byasimple rule discovered byFaraday. (We will just state the rulenow andwait until later toexamine itindetail.) Theruleisthatwhen themag- netic ux that passes through theloop (this flux isthenormal component ofB integrated over thearea oftheloop) ischanging with time, theemf isequal totherateofchangeoftheflux.Wewillrefertothisas“thefluxrule.”Youseethatwhen thecoil ofFig. 16-1 isrotated, theflux through itchanges. Atthestart some flux goes through oneway; then when thecoil hasrotated 180° thesame fluxgoes through theother way. Ifwecontinuously rotate thecoil theflux is firstpositive, then negative, then positive, andsoon.The rateofchange ofthe fluxmust alternate also. Sothere isanalternating emf inthecoil. Ifweconnect thetwo ends ofthecoil tooutside wires through some sliding contacts—called slip-rings—(just sothewires won't gettwisted) wehave analternating-current generator. (Orwecanalso arrange, bymeans ofsome sliding contacts, that after every one-half rotation, theconnection between thecoil ends and theoutside wires isreversed,$0thatwhentheemfreverses,sodotheconnections. Thenthepulsesofemfwillalways push currents inthesame direction through theexternal circuit. Wehave what iscalled adirect-current generator. ‘The machine ofFig, 16-1 iseither amotor or@generator. The reciprocity ‘between motors andgenerators isnicely shown byusing twoidentical de“motors”‘ofthepermanent magnetkind,withtheircoilsconnected bytwocopperwires. When theshaft ofone isturned mechanically, itbecomes agenerator and drives ‘theother asamotor. Iftheshaft ofthesecond isturned, itbecomes thegenerator and drives thefrst asamotor. Sohere isaninteresting example ofanew kind of equivalence ofnature: motor andgenerator areequivalent. Thequantitative agesow |sunpressune equivalence is,infact, notcompletely accidental. Itisrelated tothelawofcontarvationofenergy. S S S_N"Anotherexampleofdevicethatcanoperateeithertogenerateemsorto=NSSGorenoonrespondtoemf’sisthereceiverofastandard telephone—that is,an“earphone.” S SS ‘Theoriginal telephone ofBellconsisted oftwosuch “earphones” connected by b>‘twolongwires.ThebasicprincipleisshowninFig.164.Apermanentmagnet WHproducesamagneticfieldintwo“yokes”ofsoftironandinathindiaphragm that ena takismoved bysound pressure. When thediaphragm moves, itchanges theamount‘ofmagnetic fieldintheyokes.Therefore acoilofwirewoundaroundoneofthe Fig.16-4.Atelephone tronsmitter yokes willhave thefluxthrough itchanged when asound wave hitsthediaphragm. orreceiver. 163 Sothere isanemf inthecoil. Iftheends ofthecoil areconnected toacircuit, a ‘current which isanelectrical representation ofthesound issetup. Iftheends ofthecoil ofFig. 16-4 areconnected bytwo wires toanother identical gadget, varying currents willflow inthesecond coil. These currents will produce avarying magnetic field and will make avarying attraction ontheiron Giaphragm. The diaphragm will wiggle and make sound waves approximately similar totheones thatmoved theoriginal diaphragm. With afewbitsofiron and copper thehuman voice istransmitted over wires! (The modern home telephone uses areceiver liketheonedescribed butuses ‘animproved invention togetamore powerful transmitter. Itisthe“carbon- button microphone,” that uses sound pressure tovary theelectric current from abattery.) 16-2 Transformers and inductances One ofthemost interesting features ofFaraday’s discoveries isnotthat an emf exists inamoving coil—which wecan understand interms ofthemagnetic force qvXB—but that achanging current inonecoilmakes anemf inasecond coil. And quite surprisingly theamount ofemfinduced inthesecond coilisgiven bythesame “flux rule”: that theemf isequal totherate ofchange ofthemagnetic fluxthroughthecoil.Supposethatwetaketwocoils,eachwoundaroundseparate (YETTA Ligier bundles ofironsheets (these helptomake stronger magnetic fields), asshown in=o Fig.16-5.Nowweconnectoneofthecoils—coil (a)—toanalternating-current8\t \ generator. Thecontinually changing current produces acontinuously varying magnetic field. This varying field generates analternating emfinthesecondcoil— coil(b).Thisemfcan,forexample,produceenoughpowertolightanelectricbulb. xv“Theemfalternatesincoil(b)atafrequencywhichis,ofcourse,thesameasthe /]frequency oftheoriginal generator. But thecurrent incoil (b)can belarger or i smaller than thecurrent incoil(a).Thecurrent incoil(b)depends ontheemfse induced initandontheresistance andinductance oftherestofitscircuit. Ther— (~)eniSiron emfcambelessthanthatofthegeneratorif,say,thereislittlefluxchange.Orthe eZ emfincoil(b)canbemademuchlargerthanthatinthegeneratorbywindingcoil a(b)withmanyturns,sinceinagivenmagneticfieldthefluxthroughthecoilis =7 thengreater.(Orifyouprefertolookatitanotherway,theemfisthesameineachturn, and since the total emfisthesumoftheemf'softheseparateturns,many turns inseries produce alarge emf.) Such acombination oftwocoils—usually with anarrangement ofiron sheets toguidethemagnetic fields—is calledatransformer. Itcan“transform” oneem Fig.16-5.Twocoils, wrapped (alsocalleda“voltage”) toanother. roundbundles ofironsheets, allow a There arealsoinduction effects inasingle coil.Forinstance, inthesetupin Generator tolightabulbwithnodirect Fig.16-5thereisachanging fluxnotonlythrough coil(b),which lightsthebulb,connection, butalsothroughcoil(a).Thevaryingcurrentincoil(a)produces avaryingmagneticfieldinsideitselfandthefluxofthisfieldiscontinually changing, sothereisaselfsinduced emfincoil(a).Thereisanemfactingonanycurrentwhenitisbuilding upamagnetic field—or, ingeneral, when itsfield ischanging inanyway. The effect iscalled self-inductance When wegave “the fluxrule” that theemfisequal totherateofchange ofthe flux linkage, wedidn’t specify thedirection oftheemf. There isasimple rule,calledLen2’srule,forfiguringoutwhichwaytheemfgoes:theemftries10opposeany flux change. ‘That is,thedirection ofaninduced emfisalwayssuchthatifa current were toflow inthedirection oftheemf, itwould produce aflux ofBthat ‘opposes thechange inBthat produces theemf. Len2’s rule can beused tofind the direction ofthe emfinthegeneratorofFig.16-1,orinthetransformer winding ofFig. 16-3. Inparticular, ifthere isachanging current inasingle coil (orinany wire) there isa“back” emf inthecircuit. This emf acts onthecharges flowing incoil (a)ofFig. 16-5 tooppose thechange inmagnetic field, andsointhedirection to ‘oppose thechange incurrent. Ittries tokeep thecurrent constant; itisopposite to thecurrent when thecurrent isincreasing, anditisinthedirection ofthecurrent 164 <7 sure Ss—Fh yauP [|Fig.16-6.Circuitconnectionsforan 1—s) electromagnet. Thelampallowsthe A> BATTERY Possage ofcurrentwhentheswitchisopened, preventing the appearance of excessive emf's. when itisdecreasing. Acurrent inaself-inductance has“inertia,” because the inductive effects trytokeep theflow constant, just asmechanical inertia tries to keepthevelocity ofanobjectconstant. Any large electromagnet will have alarge self-inductance. Suppose that a battery isconnected tothecoil ofalarge electromagnet, asinFig. 16-6, and that a strong magnetic field hasbeen built up. (The current reaches asteady value deter- mined bythebattery voltage andtheresistance ofthewire inthecoil.) Butnow ‘suppose that wetrytodisconnect thebattery byopening theswitch. Ifwereally ‘opened thecircuit, thecurrent would gotozero rapidly, and indoing soitwould generate anenormous emf. Inmost cases thisemf would belarge enough tode- velop anarcacross theopening contacts oftheswitch. The high voltage that ap- pears might also damage theinsulation ofthecoil—or you, ifyou aretheperson ‘who opens theswitch! Forthese reasons, electromagnets areusually connected in acircuit liketheoneshown inFig. 16-6. When theswitch isopened, thecurrent does not change rapidly but remains steady, flowing instead through thelamp, being driven bytheemf from theself-inductance ofthecoil. 16-3 Forces oninduced currents You have probably seen thedramatic demonstration ofLenz’s rulemade with thegadget shown inFig. 16-7. Itisanelectromagnet, just likecoil (a)ofFig. 16-5. Analuminum ring isplaced ontheend ofthemagnet. When thecoilis connected toanalternating-current generator byclosing theswitch, thering flies into theair. The force comes, ofcourse, from theinduced currents inthering. ‘The factthat thering flies away shows that thecurrents initoppose thechange of thefield through it.When themagnet ismakinganorthpoleatitstop,theinduced current inthering ismaking adownward-point north pole. The ring and thecoil arerepelled just like twomagnets with likepoles opposite. Ifathin radial cutis made inthering theforce disappears, showing that itdoes indeed come from the currents inthering. S fZ( CONDUCTING RING Ved L a We SSA ZA = Samer Yy cou—\ |GeneRAtton ALSS owe joeSSS ZZZLLIILLLL LILILLLL UmPERFECTLY CONDUCTING PLATE Fig. 16-7. Aconducting ringisstrongly repelled Fig. 16-8. Anelectromagnet near aperfectly byanelectromagnet with ovarying ewrent. conducting plate. 165 If,instead ofthering, weplace adisc ofaluminum orcopper across theend oftheelectromagnetofFig.16-7,itisalsorepelled;inducedcurrentscirculatein ewthematerial ofthedisc, and again produce arepulsion..LY) ‘Aninterestingeffect,similarinorigin,occurswithasheetofaperfectcon- GS ductor.Ina“perfectconductor”thereisnoresistancewhatevertothecurrent.SocK ifcurrentsaregeneratedinit,theycankeepgoingforever.Infact,theslightest emf would generate anarbitrarily large current—which really means that there ; ; canbenoems atall, Any attempt tomake amagnetic flux gothrough such a Fig.16-9. Abarmagnet issus- sheetgenerates currents thatcreate opposite Bfields—all withinfinitesimal emf's,pended above osuperconducting bowl, ate catebytherepulsion ofeddy currents. sowithnofluxentering. ;Ifwehave asheet ofaperfect conductor and putanelectromagnet next toit, ‘when weturn onthecurrent inthemagnet, currents called eddy currents appear in SS pivor thesheet, sothatnomagnetic fluxenters. Thefield lines would look asshown inro Fig.16-8. Thesamething happens, ofcourse, ifwebring abarmagnet neara \ perfect conductor. Sincetheeddycurrents arecreating opposing fields, the \ magnets arerepelled from theconductor. This makes itpossible tosuspend abar ‘magnet inairabove asheet ofperfect conductor shaped like adish, asshown in | Fig.16-9.Themagnetissuspendedbytherepulsionoftheinducededdycurrents {intheperfectconductor.Therearenoperfectconductorsatordinarytempera~ <s tures,butsomematerialsbecomeperfectconductorsatlowenoughtemperatures. A-| Forinstance,below3.8°Ktinconductsperfectly.Itiscalledasuperconductor.eo IftheconductorinFig.16-8isnotquiteperfecttherewillbesomeresistance NStoflowoftheeddycurrents.‘Thecurrentswilltendtodieoutandthemagnetwill SSslowly settle down. The eddy currents inanimperfect conductor need anemf to A keep them going, and tohave anemf theflux must keep changing. The flux of—a== themagneticfieldgraduallypenetratestheconductor.== Inanormalconductor, therearenotonlyrepulsiveforcesfromeddycurrents, butthere canalso besidewise forces, Forinstance, ifwemove amagnet sideways along aconducting surface theeddy currents produce aforce ofdrag, because the induced currents areopposing thechanging ofthelocation offlux. Such forces are =¥ proportional tothevelocity and arelike akind ofviscous force.EF-Siaren TheseeffectsshowupnicelyintheapparatusshowninFig.16-10.AsquareBATTERY sheetofcopperissuspended ontheendofarodtomakeapendulum, Thecopper ; swingsbackandforthbetween thepolesofanelectromagnet. Whenthemagnet Fig.16-10. ThebrokingofthePen- isturnedon,thependulum motionissuddenly arrested, Asthemetalplateenters dulumshowstheforcesduetoeddycur-thepanofthemagnet,thereisacurrentinducedintheplatewhichactstooppose rents. thechange influx through theplate. Ifthesheet were aperfect conductor, the currents would besogreat that they would push theplate outagain—it would ounce back. With acopper plate there issome resistance intheplate, so thecurrents atfirst bring theplate almost toadead stop asitstarts toenter thefield.Then,asthecurrentsdiedown,theplateslowlysettlestorestinthemagneticfield . ‘The nature oftheeddy currents inthecopper pendulum isshown inFig. Si 16-11.Thestrengthandgeometryofthecurrentsarequitesensitivetotheshape rBiews oftheplate.If,forinstance, thecopperplateisreplacedbyonewhichhasseveral \narrowslotscutinit,asshowninFig.16-12,theeddy-currenteffectsaredrastically \(Q) reduced.Thependulumswingsthroughthemagneticfieldwithonlyasmall © retardingforce.ThereasonisthatthecurrentsineachsectionofthecopperhaveSs lessfluxtodrivethem,sotheeffectsoftheresistance ofeachlooparegreater.— Thecurrents aresmaller andthedragisless.Theviscous character oftheforce isseen even more clearly ifasheet ofcopper isplaced between thepoles ofthemagnetofFig.16-10andthenreleased.Itdoesn’tfal;itjustsinksslowlydown-ward. The eddy currents exert astrong resistance tothemotion—just like the viscous drag inhoney. If,instead ofdragging aconductor past amagnet, wetrytorotate itina magnetic field, there willbearesistive torque from thesame effects. Alternatively, ifwerotate amagnet—end over end—near aconducting plate ofring, thering is Fig.16-11. Theeddy currents inthe dragged around; currents inthering willcreate atorque that tends torotate copper pendulum, thering with themagnet. 166 y 2 3 2 2 3 st ‘ B 6 5 & 50) (by (G) 2 3 2 3 2 3 la orig AN Yl Ais NGS |y—* ZX 6563é5 Co)(e) " Fig. 16-12. Eddy-current effects ore drasti- Fig. 16-13. Making arotating magnetic field. cally reduced bycutting slots intheplate. Afieldjustlikethatofarotatingmagnetcanbemadewithanarrangement ofcoils such asisshown inFig. 16-13. Wetake atorus ofiron (that is,aring of ironlikeadoughnut) andwindsixcoilsonit.Ifweputacurrent,asshowninpart (a),through windings (1)and (4),there will beamagnetic field inthedirection shown inthefigure. Ifwenow switch thecurrent towindings (2)and (5),the magnetic field willbeinanew direction, asshown inpart (b)ofthefigure. Con- tinuing theprocess, wegetthesequence offields shown intherest ofthefigure. Iftheprocess isdone smoothly, wehave a“rotating” magnetic field. Wecaneasily gettherequired sequence ofcurrents byconnecting thecoils toathree-phase power line, which provides justsuch asequence ofcurrents. ““Three-phase power” ismade inagenerator using theprinciple ofFig. 16-1, except that there arethree loops fastened together onthesame shaft inasymmetrical way—that is,with an angle of120° from one loop tothenext. When thecoils arerotated asaunit, the emf isamaximum inone, then inthenext, and sooninaregular sequence. Therearemanypracticaladvantagesofthree-phasepower.Oneofthemisthepossibility CED ofmaking arotating magnetic field. The torque produced onaconductor bysuch ‘arotating field iseasily shown bystanding ametal ring onaninsulating table just above thetorus, asshown inFig. 16-14. The rotating field causes thering tospin _Fig. 16-14, The rotating field of about avertical axis. Thebasic elements seen here arequite thesame asthose at Fig.16-13 canbeused toprovide torque playinalarge commercial three-phase induction motor. ‘onaconducting ring. Another form ofinduction motor isshown inFig. 16-15. The arrangement shown isnotsuitable forapractical high-efficiency motor butwillillustrate the principle. The electromagnet M,consisting ofabundle oflaminated iron sheets wound with asolenoidal coil, ispowered with alternating current from agenerator. ‘The magnet produces avarying fluxofBthrough thealuminum disc. Ifwehave just these two components, asshown inpart (a)ofthefigure, wedonotyethave motor. There areeddy currents inthedisc, butthey aresymmetric andthere is notorque. (There will besome heating ofthedisc due totheinduced currents.) If wenow cover only one-half ofthemagnet pole with analuminum plate, asshown inpart (b)ofthefigure, thedisc begins torotate, and wehave amotor. The operation depends ontwoeddy-current effects. First, theeddy currents inthe aluminum plate oppose thechange offlux through it,sothemagnetic field above theplate always lags thefield above that halfofthepolewhichisnotcovered.This so~alled “shaded-pole” effect produces afield which inthe“shaded” region varies 167 | sate ' bal ily Basec Base GH{NTI (9) {MTD (b)NIT scmane NU Fig. 16-15. Asimple example ofashaded-pole induction motor. much likethat inthe“unshaded” region except that itisdelayed aconstant amount. intime. The whole effect isasifthere were amagnet only half aswide which is continually being moved from theunshaded region toward theshaded one. Then thevarying fields interact with theeddy currents inthedisc toproduce thetorque onit. 16-4 Electrical technology ‘When Faraday first made public hisremarkable discovery that achanging magnetic flux produces anemf, hewas asked (asanyone isasked when hedis- covers anew fact ofnature), “What istheuseofit?” Allhehad found wasthe ‘oddity that atiny current was produced when hemoved awire near amagnet. . Ofwhat possible “use” could that be? Hisanswer was: “What istheuseofanew- born baby?” Yetthink ofthetremendous practical applications hisdiscovery hasledto. Whatwehavebeendescribing arenotjusttoysbutexamples chosen inmostcases torepresent theprinciple ofsome practical machine. Forinstance, therotating ring intheturning field isaninduction motor. There are, ofcourse, some differences between itand apractical induction motor. The ring hasavery small torque; it can bestopped with your hand. For agood motor, things have tobeputtogether ‘more intimately: there shouldn't besomuch “wasted” magnetic field out inthe air. First, thefield isconcentrated byusing iron. Wehave notdiscussed how iron does that, butiron canmake themagnetic field tens ofthousands oftimes stronger than copper coils alone could do. Second, thegaps between thepieces ofiron are made small; todothat, some iron iseven built into therotating ring. Everything isarranged soastogetthegreatest forces and thegreatest efficiency—that is, conversion ofelectrical power tomechanical power—until the“ring” can no longer beheld still byyour hand. This problem ofclosing thegaps and making thething work inthemost practical way isengineering. Itrequires serious study ofdesign problems, although there arenonew basic principles from which theforces areobtained. But there isalong way togofrom thebasic principles toapractical andeconomic design. Yetitisjust such careful engineering design that hasmade possible such atre- ‘mendous thing asBoulder Dam and allthat goes with it. ‘What isBoulder Dam? Ahuge river isstopped byaconcrete wall. Butwhat awall itis!Shaped with aperfect curve that isvery carefully worked outsothat theleast possible amount ofconcrete will hold back awhole river. Itthickens at thebottom inthat wonderful shape that theartists like butthat theengineers can appreciate because they know that such thickening isrelated totheincrease of pressure with thedepth ofthewater. But wearegetting away from electricity. Thenthewaterofthe river isdiverted into ahuge pipe. That’s anice engineer- ingaccomplishment initself. The pipe feeds thewater into a“waterwheel”—a huge turbine—and makes wheels turn, (Another engineering feat.) But why turn wheels? They arecoupled toanexquisitely intricate mess ofcopper and iron, all 168 M7 The Laws of Induction 17-1 The physics ofinduction Inthelastchapter wedescribed many phenomena which show that theeffects 17-1 The physics ofinduction ofinduction arequitecomplicated andinteresting. Nowwewanttodiscuss the 19ceptions tothefluxrule””fundamental principles whichgoverntheseeffects.Wehavealreadydefinedtheemf mt inaconducting circuit asthetotal accumulated force onthecharges throughout _-17-3 Particle acceleration byan thelength oftheloop. More specifically itsthetangential component oftheforce induced electric field; the perunit charge, integrated along thewire once around thecircuit. This quantity betatron isequal,therefore, tothetotalworkdoneonasinglecharge thattravels once 4944paradoxaround the circutt. Wehave also given the“flux rule,” which says that theemfis equal totherate —-17-§ Alternating-current generator atwhich themagnetic flux through such aconducting circuit ischanging. Let’s iseeifwecanunderstand whythatmight be.First,we'llconsider acaseinwhich ‘17-6Mutual inductance theflux changes because acirourt ismoved inasteady field 17-7 Self-induetance InFig. [7-1 weshowasimpleloopofwirewhosedimensions canbechanged. i ‘Theloophastwoparts,afixedU-shapedpart(a)andamovablecrossbar(b)17-8Inductanceandmagnetic thatcanslidealongthe1wolegsoftheU.Thereisalwaysacomplete circuit,but energy itsarea 1svariable. Suppose wenow place theloop inauniform magnetic field with theplane oftheUperpendicular tothefield. According totherule, when thecross- barismoved there should beintheloop anemfthat 1sproportional totherateof change oftheflux through theloop. This emf willcause acurrent intheloop. Wewill assume that there isenough resistance inthewire that thecurrents are small, Then wecanneglect anymagnetic fieldfrom thiscurrent. ww. Thefluxthrough theloopiswl.B, sothe“flux rule” would givefortheemf— f(a Whichwewriteas6— tf. ar at 4 ——— , ,meee Ne where nisthespeed oftranslation ofthecrossbar, Lines oF& Now weshould beable tounderstand this result from themagnetic vXB forces onthecharges inthemoving crossbar. These charges will feel aforce, _Fig. 17-1. Anemf isinduced ina tangential tothewire, equal tovBperunitcharge. Itisconstant along thelength loopifthefluxischanged byvarying the wofthecrossbar andzeroelsewhere, sotheintegral is ‘rea ofthecircuit. &=ws, which isthesame result wegotfrom therate ofchange oftheflux. The argument just given can beextended toanycase where there isafixed magnetic field and thewires aremoved. One canprove, ingeneral, that forany careuit whose parts move inafixed magnetic field theemf 1thetime derivative oftheflux, regardless oftheshape ofthecircuit. Ontheother hand, what happens iftheloop 1sstationary and themagnetic field ischanged? We cannot deduce theanswer tothis question from thesame argument. Itwas Faraday's discovery—from experiment—that the“flux rule” issull correct nomatter why theflux changes. The force onelectric charges 1s given incomplete generality by F=q(E +vXB); there are nonew special “forces due tochanging magnetic fields.” Any forces oncharges atrest ina stationary wire come from theEterm. Faraday’s observations ledtothediscovery thatelecirie and magnetic fields arerelated byanew law: inaregion where the magnetic field ischanging with tume, electric fields aregenerated. Iisthis electric a exyer 2 Ee \ oyiCOPPERDISC |Fig.17-2.Whenthediserotatesthere isanemf from vXB,but with GALVANOMETER nochange inthelinked flux. Now wewill describe asituation inwhich theflux through acircuit does not change, butthere 15nevertheless anemf. Figure 17-2 shows aconducting dise Which can berotated onafixed axis inthepresence ofamagnetic field. One contact ismade totheshaft and another rubs ontheouter periphery ofthedisc. Acircuit 1scompleted through agalvanometer. Asthedise rotates, the“circuit,” inthesense oftheplace inspace Where thecurrents are, 18always thesame. But thepart ofthe“circuit” inthedisc 1sinmaterial which ismoving. Although the fluxthrough the“circuit” 1sconstant, there 1sstilanemf, ascan beobserved bythedeflection ofthegalvanometer. Clearly,here1sacasewherethevXBforcein COPPERPLATESthemoving dise gives risetoanemf which cannot beequated toachange offlux. { Now weconsider. asanopposite example, asomewhat unusual situation in - ag whichthefluxthrough a“circuit” (againinthesenseoftheplacewherethecurrent tf~\ 1s)changesbutwherethere1snoemf.Imaginetwometalplateswithslightlycurved WPoes edges,asshowninFig,17-3,placedinauniformmagnetic fieldperpendicular to os hk theirsurfaces.Eachplate1sconnected tooneofthe terminals ofagalvanometer, \ asshown. Theplates make contact atonepoint P.sothere isacomplete circurt x Iftheplates arenow rocked through asmall angle, thepoint ofcontact willmove . a toP’,Ifweimagine the“circuit” tobecompleted through theplates onthedotted lineshown inthefigure, themagnetic flux through this circuit changes byalarge amount astheplates arerocked back and forth. Yet therocking can bedone with small mottons, $othat vXB1svery small andthere ispractically noemf. The — “fluxrule”doesnotworkinthiscase. Itmustbeapplied focircuits inwhich the = ‘material ofthecircuit remains thesame. When thematerial ofthecircuit 1schang- ing,wemust return tothebasic laws. Thecorrect physics isalways given bythe ‘GALVANOWETER two basic laws ; Fig.17-3. When the plates are Foqe+x By rockedinuniform magnetic Fel,here OB canbealargechange intheflux vxe~ —98 linkage without the generation ofanar emt. 17-3 Particle acceleration byaninduced electric field; thebetatron Wehave said that theelectromotive force generated byachanging magnetic field can exist even without conductors; that 1s,there can bemagnetic induction without wires. Wemay still imagine anelectromotive force around anarbitrary mathematical curve inspace. Itisdefined asthetangential component ofE inlegrated around thecurve. Faraday’s law says that this line integral 1sequal to therate ofchange ofthemagnetic flux through theclosed curve, Eq. (17.3). Asanexampleoftheeffectofsuchaninducedelectricfield,wewantnowtoconsider themotion ofanelectron inachanging magnetic field. We imagine a magnetic field which, everywhere onaplane, points inavertical direction, asshown, 1nFig. 17-4. The magnetic field isproduced byanelectromagnet, butwewill not worry about thedetails Forourexample wewillmagine that themagnetic field issymmetric about some axis, 1€., that thestrength ofthe magnetic field will depend only onthedistance from theaxis. The magnetic field 1salso varying with ume Wenow imagine anelectron that 1smoving inthisfield onapath that 1sa circle ofconstant radius with itscenter attheaxis ofthefield. (We will see later m3 + %—~R. q 8 : : ge\ ++ Nines oF8 sive view oP view Fig. 17-4. An electron accelerating inan axially symmetric, time-varying magnetic field. howthismotioncanbearranged.) Becauseofthe changing magnetic field, there will beanelectric field £tangential totheelectron’s orbit which willdrive itaround thecircle. Because ofthesymmetry, thiselectric field will have thesame value everywhere onthecitcle. Iftheelectron’s orbit hastheradius r,theline integralofEaroundtheorbit1sequaltotherateofchangeofthemagneticfluxthroughthecircle. The lineintegral ofEisjust stsmagnitude tumes thecircumference of thecircle, 2x7. The magnetic flux must, ingeneral, beobtained from anintegral. For themoment, weletByyrepresent theaverage magnetic field intheinterior of thecircle; then thefluxisthisaverage magnetic field times thearea ofthecircle. We wall have a 2 Derk=9Bye wr). Since weareassuming r1sconstant, Esproportional totheume derivative of theaverage field: 1dBay enjee. (74) ‘Theelectron willfeltheelectric force gEandwillbeaccelerated byit.Remember-ungthattherelauvistially correctequationofmotionisthattherateofchange of themomentum isproportional totheforce, wehave =aE=BP. (7s) For thecircular orbit wehave assumed, the electric force onthe electron isalwaysinthedirectionofits motion, soststotal momentum will beincreasing at therategiven byEq.(17.5). Combining Eqs. (17.5) and (17.4), wemay relate the rateofchange ofmomentum tothechange oftheaverage magnetic field dp_a7dB.a7 2di 07g Integrating with respect to1,wefind fortheelectron's momentum = met 1P=pot FAB, (77) Wherepoisthemomentum withwhichtheelectronsstartout,and&B,.isthesub-sequent change inBy. The operation ofabetatron—a machine foraccelerating electrons tohigh energies—is based onthis idea. Toseehow thebetatron operates indetail, wemust now examine how the electron can beconstrained tomove onacircle. Wehave discussed inChapter 11 ofVol. Itheprinciple involved. Ifwearrange that there 1samagnetic field Bat theorbit oftheelectron, there will beatransverse force quXBwhich, forasuit- m4 actinonly one way (and that 1stheright way, naturally). Soinphysics aparadox 1sonly aconfusion inour own understanding. Here isour paradox. Imagine that weconstruct adevice like that shown inFig, 17-5, There 1sa thin, circular plastic dise supported onaconcentric shaft with excellent bearings, sothat itisquite free torotate. Onthedisc 1sacoil ofwire intheform ofashort solenoid concentric with theaxis ofrotation. This solenoid carries asteady current 1providedbyasmallbattery,alsomountedonthedisc.Neartheedgeofthedisc Heeoeres cowofwine aNdspaceduniformly arounditscircumference areanumberofsmallmetalspheres:nsulated from each other andfrom thesolenord bytheplastic material ofthedise. rm} ee Eachofthesesmallconducting spheres ischarged withthesameelectrostatic e e charge Q.Everything isquite stationary, andthediscisatrest. Suppose nowthat e @) bysomeaccident—or byprearrangement—the current inthesolenoid 1sinter-fearreNy rupted,without,however,anyintervention fromtheoutside.Solongasthecurrent ee.r ee continued, therewasamagnetic fluxthrough thesolenoid moreorlessparalleleee totheaxisofthedisc. When thecurrent isinterrupted, thisfluxmust gotozero. There will, therefore, beanelectric field induced which will circulate around in puastic ose circles centered attheaxis. Thecharged spheres ontheperimeter ofthediscwill allexperience anelectric field tangential totheperimeter ofthedisc. This electric force 1sinthesame sense forallthecharges andsowillresult inanettorque onthe ise, From these arguments wewould expect that asthecurrent inthesolenoid Fig.17-5. Willthediscrotate ifthe disappears, thediscwould begin torotate. Ifweknew themoment ofinertia of current lisstopped? thedisc,thecurrent inthesolenord, andthecharges onthesmallspheres, wecould compute theresulting angular velocity. Butwecould also make adifferent argument. Using theprinciple ofthecon- servation ofangular momentum, wecould saythat theangular momentum ofthe dise with all1tsequipment isintially zero, and sotheangular momentum ofthe assembly should remain zero. There should benorotation when the current 1s stopped. Which argument 1scorrect? Will thedise rotate orwill itnot? Wewill leave this question foryou tothink about. Weshould warn you that thecorrect answer does notdepend onany non- essential feature, such astheasymmetric position ofabattery, forexample. In fact, you can imagine anideal situation such asthefollowing: The solenoid 1s made ofsuperconducting wire through which there isacurrent. After thedise hasbeencarefullyplacedatrest,thetemperature ofthe solenoid 1sallowed toriseslowly When thetemperature ofthewire reaches thetransition temperature between superconductivity and normal conductivity, the current inthe solenoid will be brought tozero bytheresistance ofthewire. The flux will, asbefore, falltozero, aandthere willbeanelectric field around theaxis. Weshould also warn you thatthe solution 1noteasy, nor 1sitatrick, When you figure itout, you will have dis- covered animportant principle ofelectromagnetism. db 17-5Alternating-current generator Intheremainder ofthis chapter weapply theprinciples ofSection 17-1 to i analyzeanumberofthe phenomena discussed inChapter 16.Wefirstlook inmore = detail atthealternating-current generator. Such agenerator consists basically ofa coil ofwire rotating inauniform magnetic field. The same result can also be = toao|achievedbyafixedcoilinamagnetic fieldwhosedirection rotatesinthemanner ems\\ey/2 described inthelastchapter. Wewillconsider onlytheformer case. Suppose we sy haveacircular coilofwirewhichcanbeturned onanaxisalongoneofitsdiam-eters. Letthiscoilbelocated inauniform magnetic field perpendicular totheaxis ofrotation, asinFig. 17-6 We also imagine that thetwo ends ofthecoil are brought toexternal connections through some kind ofsliding contacts. Fig.17-6. Acoilofwirerotating ina Duetotherotation ofthecoil, themagnetic fluxthrough itwillbechanging. uniform magnetic field—the basic idea The circutt ofthecoil will therefore haveanemfinit.Let$betheareaofthecotl oftheacgenerater. and@theangle between themagnetic field andthenormal totheplane ofthecotl.* *Now that weareusing theleter forthevector potential, weprefer toletSstand for aSurface area. 126 The flux through thecoi! isthen BScos6. (07.13) Ifthecoil isrotating atthe uniform angular velocity «, varies with time as 6=at. The emf&inthecollisthen __d d 6=~5,(lux)=—5;(BSc08wt), or &=BSwsinwt (7.14) Ifwebring thewires from thegenerator toapoint some distance from the rotating coil, where themagnetic field iszero, oratleast 1snotvarying with me, thecurl of£inthis region will bezero and wecan define anelectric potential. Infact, ithere isnocurrent being drawn from thegenerator, thepotential differ- ence Vbetween thetwowires willbeequal totheemfintherotating coil. That is, V=BSwsin wt=Vosinat. The potential difference between thewires varies assinwt.Such avarying potential difference 1scalled analternating voltage. Since there isanelectric field between thewires, they must beelectrically charged. It1sclear that theemf ofthegenerator haspushed some excess charges outtothewire until theelectric field from them isstrong enough toexactly counter~ balance theinduction force. Seen from outside thegenerator, thetwo wires appear asthough they had been electrostatically charged tothepotential difference V, andasthough thecharge wasbeing changed withtumetogiveanalternating po- t tential difference. There isalso another difference from anelectrostatic situation. + Ifweconnect thegenerator toanexternal circuit that permits passage ofacurrent,wefindthattheemfdoesnotpermitthewirestobedischarged butcontinues to acprovide charge tothewires ascurrent isdrawn from them, attempting tokeepthegenerator! Rwires always atthesame potential difference. Ifinfact, thegenerator isconnected inacircuit whose total resistance isR,thecurrent through thecircuit willbepro- portional totheemfofthegenerator andinversely proportional toR.Since the re£2 sinwtemfhasasinusoidal timevariation, soalsodoesthecurrent.Thereisanalternating, RoRcurrent y Fig.17-7.A.circuit.withanocTez=FPsinot, generator andaresistance. Theschematic diagram ofsuch acircuit isshown inFig. 17-7. Wecan also seethat theemf determines how much energy issupplied bythe generator. Each charge inthewire isreceiving energy attherate Fv. where Fis theforce onthecharge andv1sitsvelocity. Now letthenumber ofmoving charges perunit length ofthewire bem;then thepower being delivered into any element dsofthe wire is Feunds. Forawire, visalways along ds,sowecanrewrite thepower as mF +ds, The total power being delivered tothecomplete circuit isthe integral ofthis expression around thecomplete loop: Power=fmFds. (7.15) Now remember that gnv1sthecurrent /,and that theemf isdefined astheintegral ofF/q around thecircuit. Wegettheresult Power from agenerator =&1. (17.16) m9 Wemay also point outthat Eq.(17.22) shows that theforce from induced currents—that is,any eddy-current force—is inversely proportional tothe re- sistance. The force willbelarger, thebetter theconductivity ofthematerial. The reason, ofcourse, isthat anemf produces more current iftheresistance islow, and thestronger currents represent greater mechanical forces. Wecanalso seefrom ourformulas how mechanical energy isconverted into electrical energy. Asbefore, theelectrical energy supplied totheresistance ofthe circuit istheproduct &/.The rateatwhich work isdone inmoving theconducting ‘crossbar istheforce onthebartimes itsvelocity. Using Eq. (17.21) fortheforce, therate ofdoing work is dW_Bw? dR Weseethat thisisindeed equal totheproduct &/wewould getfrom Eqs. (17.19) and (17.20). Again themechanical work appears aselectrical energy. 17-6Mutualinductance \|ga\_\j-__EWenowwanttoconsiderasituationinwhchtherearefixedcolsofwirebut CTTDeoes changing magnetic fields. When wedescribed theproduction ofmagnetic fields by Ke,currents,weconsidered onlythecaseofsteadycurrents. Butsolongasthecurrents. §—§“=.arechangedslowly,themagneticfieldwillateachinstantbenearlythesameasthe Szmagneticfieldofasteadycurrent.Wewillassumeinthediscussion ofthissection con2 that thecurrents arealways varying sufficiently slowly that thisistrue. SZInFig.17-8isshownanarrangementoftwocoilswhichdemonstratesthe Sy basic effects responsible fortheoperation ofatransformer. Coil |consists ofa conducting wire wound intheform ofalong solenoid, Around this coil—and insulated fromit—iswoundcoil2,consisting ofafewturnsofwire.Ifnowaca =current ispassed through coilI,weknow thatamagnetic fieldwillappear inside it. S Thismagnetic fieldalsopasses through coil2.Asthecurrent incoil11svaried, |themagnettefluxwillalsovary,andtherewillbeaninducedemfincoil2.Wewill inow calculate this induced emf. Wehave seen inSection 13-5 that themagnetic field inside along solenoid is, uniform and hasthemagnitude Fig. 17-8. Acurrent incoil 1pro- duces amagnetic fieldthrough coil2. 1Mh =1Mh, (17.23) where Nyisthenumber ofturns incoil 1,1,isthecurrent through it,andJsits length. Let’s saythat thecross-sectional area ofcoil 1is3then theflux ofBis itsmagnitude times S.Ifcoil 2has Noturns, this flux links thecoil V3times. Therefore theemf incoil2isgiven by fy=—Nps28. fy=MsF (17.24) Theonly quantity inEq.(17.23) which varies with time is/,.The emfistherefore given by MNS dh,fe el dt (17.25) Weseethattheemfincoil2isproportional totherateofchangeofthe current incoil 1.The constant ofproportionality, which isbasically ageometric factor of thetwo coils, iscalled themutual inductance, and isusually designated 91. Equa- tion (17.25) 1sthen written diy 82=MarGP (17.26) Suppose now that wewere topass acurrent through coil 2and askabouttheemfincoil1.Wewouldcompute themagnetic field,whichiseverywhere 119 proportional tothecurrent J.The flux linkage through coil 1would depend on thegeometry, butwould beproportional tothecurrent Jy. The emf incoil | would, therefore, again beproportional tod//dt: Wecan write =m222 f= MasGP (17.27) The computation of9% would bemore difficult than thecomputation wehave just done for91.1. Wewill notcarry through that computation now, because we ‘will show later inthis chapter that 31; isnecessarily equal toM21. Since forany coil itsfield isproportional toitscurrent, thesame kind of result would beobtained forany two coils ofwire. The equations (17.26) and (17.27) would have thesame form; only theconstants 321 and 94 would be different. Their values would depend ontheshapes ofthecoils and their relative positions. ds, ds, ny Fig. 17-9. Any two coils have @ ' mutual inductance 3.proportional tothe integral ofds; -ds2/n2. Suppose that wewish tofind themutual inductance between any two arbitrary coils—for example, those shown inFig. 17-9, Weknow that thegeneral expression for the emf incoil 1can bewritten as d &=4],B-nda, where B1sthemagnetic field and themtegral istobetaken over asurface bounded bycircuit 1.Wehave seen inSection 14-1 that such asurface integral ofBcan be related toalineintegral ofthevectorpotential. Inparticular, fB-nda= }Ads, w wy where Arepresents thevector potential and ds,isanelement ofcircuit 1.The ine integral istobetaken around circuit 1.The emfincoil 1cantherefore bewritten as d --4 sds. 7. a=-%fads (17.28) Now let’s assume that thevector potential ate1reuit 1comes from currents incireuit 2.Then itcan bewritten asaline integral around circuit 2: L_§ beds, = = , 7.29) where /»isthecurrent incircuit 2,andr,,isthedistance from theelement ofthe circuit ds»tothepoint oncircuit |atwhich weareevaluating thevector potential. (See Fig. 17-9.) Combining Eqs. (17.28) and (17.29), wecanexpress theemf in circuit 1asadouble line mtegral: Lodf¢.hds,&=-pha4 BBs ds,* Aree? dtJey)Jayria Inthisequation theintegrals arealltaken with respect tostationary circuits, The only variable quantity isthecurrent /2,which does notdepend onthevariables of 17-10 coil wemust overcome thisinertia byconnecting thecoil tosome external voltage source such asabattery oragenerator, asshown intheschematic diagram ofFig. 1 17-10(a). Insuch acircuit, thecurrent /depends onthevoltage Vaccording to = the relation sa ell 2 v=2a: (17.35) ae Thisequation hasthesameformasNewton's lawofmotion foraparticle in ‘onedimension.Wecanthereforestudyitbytheprinciplethat“thesameequations (o) have thesame solutions.” Thus, sfwemake theexternally applied voltageVcorre- spond toanexternally applied force F,and thecurrent /inacoilcorrespond tothevelocity»ofaparticle,theinductance &ofthecoilcorresponds tothemassmoftheparticle." SeeFig. 17-10(b). Wecanmake thefollowing table ofcorresponding 7 quantities. F Particle Coit F(force) (potential difference) »(velocity) 1(current) (byx(displacement) (charge) dv dt Fig.17-10 (a)A.circuit with@ Fame vedvoltage source andaninductance. (b)An d J Sparco mechonical systems ‘me(momentum) stAmv? (kinetic energy) 441 (magnetic energy) 17-8 Inductance and magnetic 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. Wehave noright, ofcourse, osaythat£/1s therealmomentum ofthe circurt; infact, itisn't, The whole circuit may bestanding still and have nomo- mentum, Itisonly that £7analogous tothemomentum minthesense ofsatisfy- ingcorresponding equations. Inthesame way, tothekinetic energy hme, there corresponds ananalogous quantity 47°. Butthere wehave asurprise. This 421° isreally theenergy mntheelectrical case also. This 1sbecause therateofdoing work ontheinductance 18‘O/, and inthemechanical system it1sFr,thecorre- sponding quantity. Therefore, inthecase oftheenergy, thequantities notonly correspond mathematically, butalso have thesame physical meaning aswell. Wemay seethis inmore detail asfollows. Aswefound inEq. (17.16), the rate ofelectrical work byinduced forces istheproduct oftheelectromotive force and the current: aw Woo, Replacing &byitsexpression intermsofthecurrentfromEq.(17.34),wehave aw_dlan a (17.36) Integrating thisequation, wefind that theenergy required from anexternal sourcetoovercometheemfintheselfnductance whilebuildingupthecurrent}(whichmust equal theenergy stored, U)is -Weu=ser (1737) ‘Therefore theenergy stored inaninductance is3/2. *This i,inewdentally, northeonly way acorrespondence can besetupbetween me- chanical and electrical quantities. +Weareneglecting any energy loss toheat from thecurrent inthe resistance ofthecoil Such losses require addtional energy from thesource butdonotchange theenergy which goes anto theinductance. 2 18 The Maxwell Equations 18-1 Maxwell’s equations Inthischapter wecome backtothecomplete setofthefourMaxwell equations «18-1 Maxwell’s equations thatwetookasourstarting point inChapter 1.Until now, wehavebeenstudying 4.» ‘Howthenewterm works Maxwell’s equations inbits and pieces; itistime toadd one final piece, and toput them alltogether. Wewillthen have thecomplete andcorrect story forelectro- _-«‘18-3. Allofclassical physics magnetic fields that may bechanging with time inanyway. Anything said inthischapter thatcontradicts something saidearlieristrueandwhatwassaidearlierig 1-4 traveling eld false—because what wassaidearlier applied tosuch special situations as,for (18-5 Thespeed oflight instance, steady currents orfixedcharges. Although wehavebeenverycareful to 4Solving Maxwell’s equations; pointouttherestrictions whenever wewroteanequation, itiseasytoforget allof thepotentials andthewave thequalifications andtolearntoowellthewrong equations. Nowweareready equationtogive thewhole truth, with noqualifications (oralmost none). The complete Maxwell equations arewritten inTable 18-1, inwords aswell asinmathematical symbols. The fact that thewords areequivalent totheequations should bythis time befamiliar—you should beable totranslate back and forth from one form tothe other. The first equation—that thedivergence ofEisthecharge density over €o—is true ingeneral. Indynamic aswell asinstatic fields, Gauss’ lawisalways valid, The flux ofEthrough any closed surface isproportional tothecharge inside. The third equation isthecorresponding general aw formagnetic fields. Since there arenomagnetic charges, theflux ofBthrough any closed surface isalways zero. The second equation, that thecurl ofEis—2B/at,isFaraday’slawandwas discussed inthelasttwochapters. Italso isgenerally true. The lastequation has something new. Wehave seen before only thepart ofitwhich holds forsteady currents. Inthatcase wesaidthatthecurlofBisj/eoc?, butthecorrect general equation hasanew part that was discovered byMaxwell. Until Maxwell’s work, the known laws ofelectricity and magnetism were those wehave studied inChapters 3through 17.Inparticular, theequation for themagnetic field ofsteady currents was known only as -AvxB=s. (18.1) Maxwell began byconsidering these known laws and expressing them asdiffer- ential equations, aswehave done here. (Although the Vnotation was not yet invented, itismainly duetoMaxwell that theimportance ofthecombinations of derivatives, which wetoday callthecurlandthedivergence, firstbecame apparent.) Hethen noticed that there wassomething strange about Eq.(18.1). Ifonetakes the divergence ofthisequation, theleft-hand sidewill bezero, because thedivergence ofacurl isalways zero. Sothisequation requires that thedivergence ofjalso bezero.Butifthedivergence ofjiszero,thenthetotalfluxofcurrentoutofanyclosed surface isalso zero. The flux ofcurrent from aclosed surface isthedecrease ofthecharge inside thesurface. This certainly cannot ingeneral bezero because weknow that the charges canbemoved from oneplace toanother. The equation __vi--F (18.2) has,infact, been almost ourdefinition ofj.This equation expresses thevery funda- 184 Table 18-1 Classical Physics ‘Maxwell's equations LVE=Pa (FluxofEthroughaclosedsurface)=(Chargeinside)/¢o eB i a Wvxe--F (Lineintegral ofEaroundaloop)=~4(FluxofBthroughtheloop) ML v-B=0 (Flux ofBthroughaclosedsurface)=0 2 i,9E W.cyxB=24 Adntegralof Baroundaloop)=(Currentthroughtheloop)/éo +2tuxof£throughtheloop) ‘Conservation ofcharge vie -% (Fluxofcurrentthroughaclosedsurface)=—2(Chargeinside) Force law F=qE+vXB) Law ofmotion fey =Fr where p= me (Newton's law,withEinstein's modification)a . Vi=ye . Gravitation P=-GMe, mental law that electric charge isconserved—any flow ofcharge must come from some supply. Maxwell appreciated this difficulty and proposed that itcould be avoided byadding theterm dE/at totheright-hand side ofEq,(18.1); hethen got, thefourth equation inTable 18-1: 2 aL, 2, Ww.evxBe t4+o Itwas notyetcustomary inMaxwell’ time tothink interms ofabstract fields. Maxwell discussed his ideas interms ofamodel inwhich the vacuum was like an elastic solid. Healso tried toexplain themeaning ofhisnew equation interms of themechanical model. There was much reluctance toaccept histheory, first be- cause ofthemodel, and second because there was atfirst noexperimental justi- fication. Today, weunderstand better that what counts aretheequations themselves, andnotthemodel used togetthem. Wemay only question whether theequations aretrue orfalse. This isanswered bydoing experiments, and untold numbers of experiments have confirmed Maxwell's equations. Ifwetake away thescaffolding heused tobuild it,wefind that Maxwell’s beautiful edifice stands onitsown. He ‘broughttogetherallofthelawsofelectricity andmagnetism and made onecomplete and beautiful theory. Letusshow that theextra term isjust what isrequired tostraighten outthe difficulty Maxwell discovered. Taking thedivergence ofhisequation (IVinTable 18-1), wemust have that thedivergence oftheright-hand side iszero: gy Fevidtv a0. (18.3) 182 Inthesecond term, theorder ofthederivatives with respect tocoordinates and time canbereversed, sotheequation canberewritten as ViteodvE=0. (18.4) ButthefirstofMaxwell’s equations says that thedivergence ofEisp/eo.Inserting this equality inEq. (18.4), wegetback Eq, (18.2), which weknow istrue. Con- versely, ifweaccept Maxwell's equations—and wedobecause noone hasever found anexperiment that disagrees with them—we must conclude that charge is always conserved. ‘The laws ofphysics have noanswer tothequestion: “What happens ifa charge issuddenly created atthis point—what electromagnetic effects arepro- duced?” Noanswer canbegiven because ourequations sayitdoesn’t happen. Ifitwere tohappen, wewould need new laws, butwecannot saywhat they would be. Wehave nothad thechance toobserve how aworld without charge con- servation behaves. According toourequations, ifyousuddenly place acharge at some point, you had tocarry itthere from somewhere else. Inthat case, wecan saywhat would happen. ‘When weadded anew term totheequation forthecurl ofE,wefound that a whole new class ofphenomena was described. Weshall seethat Maxwell's little addition totheequation forVXBalso hasfar-reaching consequences. Wecan touch ononly afew ofthem inthis chapter. 18-2 How the new term works Asourfirstexample weconsider whathappenswithaspherically symmetric \ epradial distribution ofcurrent. Suppose weimagine alittlesphere withradioactive /material onit.Thisradioactive material issquirting outsome charged particles. \ ’ — (Or wecould imagine alarge block ofjello with asmall hole inthecenter into \ i which some charge had been injected with ahypodermic needle and from which r thechargeisslowlyleaking out.)Ineithercasewewouldhaveacurrent thatis>< DXeverywhere radially outward. Wewillassume thatsthasthesame magnitude in ~ ‘a alldirections. Letthetotalcharge insideanyradiusrbeQ(r).Iftheradialcurrent density 10:atthesameradius1sj(e),thenEq,(18.2)requiresthatQdecreasesattherate Ne!3g=-497?i). (ss)7ZTS Wenowaskaboutthemagneticfieldproducedbythecurrentsinthissituation. / ‘ E‘Suppose wedrawsomeloopI’onasphere ofradius r,asshown inFig.18-1. ‘ \There issomecurrent through thisloop,sowemight expect tofindamagnetic , \ field circulating inthedirection shown. Butwearealready indifficulty. How cantheBhave anyparticular direction Fig, 18-1. What isthe magneticconthesphere?Adifferent choice ofI’would allow ustoconclude that itsdirection field ofaspherically symmetric current? isexactly opposite tothat shown. Sohow canthere beany circulation ofBaround the currents? Wearesaved byMaxwell's equation. The circulation ofBdepends notonly fonthetotal current through T°but also ontherate ofchange with time ofthe electric flux through it.Itmustbethatthesetwopartsjustcancel.Let'sseeifthat works out, Theelectric field attheradius rmust beQ(+)/4:req'?—so long asthecharge issymmetrically distributed, asweassume. Itisradial, and itsrate ofchange isthen oE 1 3Treg cs) Comparing thiswith Eq.(18.5), weseethat atanyradius edead. (08.7) 183 Loor F, :Loop 1, & q © \ {r y | a weN een >SEE*e ieeeeee & ba) (9) 4 (by Fig. 18-2. Themagnetic field nearachargingcapacitor. InEq.IVthetwosource terms cancel andthecurlofBisalways zero. There is nomagnetic field inourexample. ‘Asour second example, weconsider themagnetic field ofawire used to charge aparallel-plate condenser (see Fig. 18-2). Ifthecharge Qontheplates is changing with time (but nottoofast), thecurrent inthewires isequal todQ/dt. ‘Wewould expect that thiscurrent willproduce amagnetic field that encircles the wire. Surely, thecurrent close tothewire must produce thenormal magnetic field—it cannot depend onwhere thecurrent isgoing. Suppose wetake aloop Iwhich isacircle with radius r,asshown inpart (a) ofthefigure. The lineintegral ofthemagnetic field should beequal tothecurrent Idivided byege?.Wehave 1 2mrB=a (18.8) This iswhat wewould getforasteadycurrent,butitisalsocorrectwithMaxwell's addition, because ifweconsider theplane surface Sinside thecircle, there areno electric fields onit(assuming thewire tobe@very good conductor). The surface integral of3E/ar iszero. Suppose, however, that wenow slowly move thecurve I’downward. Weget always thesame result until wedraw even with theplates ofthecondenser. Then thecurrent Jgoes tozero. Does themagnetic field disappear? That would be quite strange. Let’s seewhat Maxwell’s equation says forthecurve I'g,which isa circle ofradius rwhose plane passes between thecondenser plates (Fig. 18-2(b)]. ‘The line integral ofBaround Iis2rB. This must equal thetime derivative of the flux ofEthroughtheplanecircularsurface5».ThisfluxofE,weknowfrom Gauss’aw,mustbeequalto1/¢9timesthechargeQononeofthe condenser plates. Wehave 4(02oOrB=5(2)- (18.9) That isvery convenient. Itisthesame result wefound inEq.(18.8). Inte- grating over thechanging electric field gives thesame magnetic field asdoes inte- gratingoverthecurrentinthewire.Ofcourse,thatisjustwhatMaxwell's equationsays. Itiseasy toseethat thismust always besobyapplying oursame arguments tothetwo surfaces S;and S{that arebounded bythesame circle TyinFig. 18-2(b). Through S,there isthecurrent J,butnoelectrie flux, Through S{there isnocurrent, butanelectric fluxchanging attherateJ/ép. Thesame Bisobtained ifweuseEq.IVwith either surface. From ourdiscussion sofarofMaxwell’s new term, you may have theim- pression that itdoesn’t addmuch—that itjustfixes uptheequations toagree withwhatwealreadyexpect.ItistruethatifwejustconsiderEq.IVbyitself,nothingparticularly new comes out. The words “by itself” are, however, all-important. Maxwell’s small change inEq. IV,when combined with theother equations, does indeed produce much that isnew and important. Before wetake upthese matters, however, wewant tospeak more about Table 18-1. 184 18-3Allofclassicalphysics InTable 18-1 wehave allthat was known offundamental classical physics, thatis,thephysics that wasknown by1905. Here italls, inonetable. With these equations wecanunderstand thecomplete realm ofclassical physics. First wehave theMaxwell equations—written inboth theexpanded form and theshort mathematical form. Then there istheconservation ofcharge, which is even written inparentheses, because themoment wehave thecomplete Maxwell equations, wecandeduce from them theconservation ofcharge. Sothetable isevenalittleredundant. Next,wehavewrittentheforcelaw,becausehavingalltheelectric andmagnetic fields doesn’t tellusanything until weknow what theydotocharges.Knowing EandB,however,wecanfindtheforceonanobjectwiththecharge qmoving with velocity v.Finally, having theforce doesn’t telusany- thing until weknow what happens when aforce pushes onsomething; weneed the lawofmotion, which isthat theforce isequal totherate ofchange ofthemo- mentum. (Remember? Wehad that inVolume I.)Weeven include relativity effects bywriting themomentum asp=mgo//T—02/3. Ifwereally want tobecomplete, weshould add one more law—Newton’s lawofgravitation—so weputthatattheend. Therefore inone small table wehave allthe fundamental laws ofclassical physics—even with room towrite them outinwords and with some redundancy. This isagreat moment. Wehave climbed agreat peak. Weareonthetopof K-2—we arenearly ready forMount Everest, which isquantum mechanics. We have climbed thepeak ofa“Great Divide,” and now wecangodown theother side, Wehave mainly been trying tolearn how tounderstand theequations. Now thatwehave thewhole thing puttogether, wearegoing tostudy what theequations‘mean—what newthingstheysaythatwehaven'talreadyseen.We'vebeenworkinghard togetuptothispoint. Ithasbeen agreat effort, butnow wearegoing tohave nice coasting downhill asweseealltheconsequences ofouraccomplishment. 18-4 Atravelling field Now forthenew consequences. They come from putting together allof Maxwell's equations. First, let's seewhat would happen inacircumstance which wepick tobeparticularly simple. Byassuming that allthequantities vary only in ‘onecoordinate, wewillhave aone-dimensional problem. The situation isshown inFig. 18-3. Wehave asheet ofcharge located ontheyz-plane. The sheet isfirst atrest, then instantaneously given avelocity winthey-direction, andkept moving with this constant velocity. You might worry about having such an“infinite” acceleration, butitdoesn't really matter; justimagine thatthevelocity isbrought to very quickly. Sowehave suddenly asurface current J(Jisthecurrent petunit y MOWING BOUNDARY :OFFIELDS, . eae wee npanaH p\WS Len zepocsrtce- | B NSoSploois! an<7 E a ' .Sx.ee SpT Fig.18-3.Aninfinitesheetofcharge N1-7EEaPES™ issoidenlysotintomotionparcillto > a= ------- te itself. There aremagnetic andelectric 7 z fields that propagate outfrom thesheetwo t=% ctaconstantspeed. 18s width inthez-direction). Tokeep theproblem simple, wesuppose that there is, also astationary sheet ofcharge ofopposite sign superposed ontheyz-plane, so that there arenoelectrostatic effects. Also, although inthefigure weshow only what ishappening inafinite region, weimagine that thesheet extends toinfinity in*yand =z. Inother words, wehave asituation where there isnocurrent, and then suddenly there isauniform sheet ofcurrent. What will happen? Well, when there isasheet ofcurrent intheplus y-direction, there is,aswe know,amagneticfieldgeneratedwhichwillbeintheminusz-direction forx>0 Bore]and intheopposite direction for x<0.Wecould find themagnitude ofBbyusingthefactthatthelineintegralofthemagneticfieldwillbeequaltothecurrent .over éqc?. Wewould getthat B=J/2«9¢? (since thecurrent /inastrip ofwidth wisJwandthelineintegral ofBis2Bw). 7 t This gives usthefield next tothesheet—for small x—but since weareim- agining aninfinite sheet, wewould expect thesame argument togive themagneticfieldfartheroutforlargervaluesofx.However, thatwouldmeanthatthemoment os weturn onthecurrent, themagnetic field issuddenly changed from zero toa finitevalueeverywhere. Butwait!Ifthemagneticfieldissuddenlychanged,it Bore!will produce tremendous electrical effects. (Ifitchanges inany way, there are electrical effects.) Sobecause wemoved thesheet ofcharge, wemake achanging wet? magnetic field, andtherefore electric fields must begenerated. Ifthere areelectric = fields generated, they hadtostart from zero andchange tosomething else. There » will besome 4/01 that willmake acontribution, together with thecurrentJ, tothe production ofthemagneticfield.Sothroughthevariousequations thereisabigintermixing, and wehave totrytosolve forallthefields atonce. BylookingattheMaxwellequationsalone,itisnoteasytoseedirectlyhow foretogetthesolution. Sowewill first show you what theanswer isand then verify thatitdoesindeedsatisfytheequations. Theansweristhefollowing: ThefieldB *that wecomputed is,infact, generated right next tothecurrent sheet (Forsmall x). Itmust beso,because ifwemake atiny loop around thesheet, there isnoroom a * foranyelectric fluxtogothrough it.ButthefieldBoutfarther—for larger x—is, atfirst, zero. Itstays zero forawhile, andthen suddenly turns on. Inshort, we ’ turn onthecurrent andthemagnetic field immediately next toitturns ontoa constant value B;then theturning onofBspreads out from thesource region. Fig.18-4. (a)Themagnitude ofB After acertain time, there isauniform magnetic field everywhere outtosome (or£)2safunction ofxatthetimefaftervaluex,andthenzerobeyond.Becauseofthesymmetry,itspreadsinboththe thecharge sheet issetinmotion. (b)The plusandminus x-directions.fieldsforchargesheetsetinmotion, ‘TheE-fielddoesthesamething.Before¢=0(whenweturnonthecurrent), towardnegative yatt=T.(c)Thesumhefieldiszeroeverywhere. Thenafterthetime1,bothEandBareuniform out of(0)ond(b}. tothedistance x=vt,andzerobeyond. Thefieldsmake theirwayforward like atidal wave, with afront moving atauniform velocity which turns outtobec, butforawhile wewilljustcallit».Agraph ofthemagnitude ofEorBversus x, asthey appear atthetime 1,isshown inFig. 18-4(a). Looking again atFig. 18-3, atthetime 1,theregion between x=oris“filled” with thefields, butthey have notyetreached beyond. Weemphasize again that weareassuming thatthecurrent sheet and, therefore thefields EandB,extend infinitely farinboth they-andz-di- rections. (We cannot draw aninfinite sheet, sowehave shown only what happens inafinite area.) Wewant now toanalyze quantitatively what ishappening. Todothat, we want tolook attwocross-sectional views, atopview looking down along they-axis, asshown inFig. 18-5, and aside view looking back along thez-axis, asshown in Fig. 18-6. Suppose westart with theside view. Weseethecharged sheet moving up;themagnetic field points into thepage for+x, and outofthepage for—x, and theelectric field isdownward everywhere—out tox=vt. Let's seeifthese fields areconsistent with Maxwell's equations. Let's first draw oneofthose loops that weusetocalculate alineintegral, saytherectangle Tzshown inFig. 18-6. You notice that oneside oftherectangle isintheregion where there arefields, butoneside isintheregion thefields have stillnotreached. There issome magnetic fluxthrough thisloop. Ifitis changing, there should be anemf around it.Ifthewavefront ismoving, wewillhave achanging magnetic, 186 oPview ty__SIDEMEW Proeorer Tel rypeyayepeg ns rye: ; 3 eft i 1ye E iy r y .Arl Tetpete |Zea]y RENT Z gamer]peal Y gamer] [Te[x[x GY * MEETS] | Ys tt vt wat wt vatbey [Teteaaly JP roa X=0XX ° % Y z Fig. 18-5. Top view ofFig. 18-3, Fig. 18-6. Side view ofFig. 18-3. flux, because theareainwhich Bexists isprogressively increasing atthevelocity v. The flux inside I'zisBtimes thepart ofthearea inside I’,which has amagnetic field. Therateofchange oftheflux, since themagnitude ofBisconstant, isthemagnitude timestherateofchangeofthearea.Therateofchangeoftheareaiseasy. Ifthewidth oftherectangle T'zisL,thearea inwhich Bexists changes by LoAtinthetime At. (See Fig. 18-6.) The rate ofchange offlux isthen BLv. According toFaraday’s law, this should equal theline integral ofEaround T, which isjust EL. Wehave theequation E= vB. (18.10) Soiftheratio ofEtoBisv,thefields wehave assumed will satisfy Faraday’s equation. Butthatisnottheonly equation; wehave theother equation relating EandB: 2 aL HE.eux Bel 4 8.11) Toapply thisequation, welook atthetopview inFig. 18-5. Wehave seen that thisequation willgive usthevalue ofBnext tothecurrent sheet. Also, forany Joop drawn outside thesheet butbehind thewavefront, there isnocurl ofBnor anyjorchanging £,50theequation iscorrect there. Now let’s look atwhat hap- pens forthecurve I’;that intersects thewavefront, asshown inFig. 18-5. Here there arenocurrents, soEq.(18.11) canbewritten—in integral form—as sfoa-4 [fp efaas=5,Enda, (18.12)inside ry Thelineintegral ofBisjust Btimes L.The rate ofchange ofthefluxofEisdue only totheadvancing wavefront. ‘The area inside Ty,where Eisnotzero, isin-creasingattheratevL.Theright-hand sideofEq.(18.12)isthenvLE.Thatequa- tionbecomes 2B=Ev, (18.13) We have asolution inwhich we have aconstant Band aconstant Ebehind thefront, both atright angles tothedirection inwhich thefront ismoving andat right angles toeach other. Maxwell's equations specify theratio ofEtoB.From Eqs. (18.10) and (18.13), 2 E=vB,andE=oB. But one moment! Wehave found twodifferent conditions ontheratio E/B. Can such afield aswedescribe really exist? There is,ofcourse, only one velocity vfor : which both ofthese equations can hold, namely v=c.The wavefront must travel with thevelocity c.Wehave anexample inwhich theelectrical influence from acurrent propagates atacertain finite velocity c. 187 “Inother words, thelaws ofNewton could bestated notintheform F=ma butintheform: theaverage kinetic energy lesstheaverage potential energy isas little aspossible forthepath ofanobject going from onepoint toanother. “Let meillustrate alitle bitbetter what itmeans. Ifyou take thecase ofthe ‘gravitational field, then ifthe particle hasthepath x(¢) (let's just take onedimension foramoment; wetake atrajectory that goes upand down and notsideways), where xistheheight above theground, thekinetic energy is4m(dx/di)*, andthe potential energy atany time ismex. Now Itake thekinetic energy minus the potential energy atevery moment along thepath and integrate that with respect totime from theinitial time tothefinal time. Let’s suppose that attheoriginal time 1,westarted atsome height and attheend ofthetime 1wearedefinitely ending atsome other place. —— “Then theintegral is Theactual motion issome kind ofacurve—it’s aparabola ifweplotagainst the MM) time—and gives acertain value fortheintegral. Butwecould imagine some other [iy motion that went very high andcame upanddown insome peculiar way. . — FF Wecan calculate thekinetic energy minus thepotential energy and integrate for Ji such apath ...orforanyother path wewant. The miracle isthat thetrue path is theone forwhich that integral isleast. “Let's tryitout. First, suppose wetake thecase ofafreeparticle forwhich d thereisnopotential energy atall.Thentherulesaysthatingoing fromonepoint } toanother inagiven amount oftime, thekinetic energy integral isleast, soitmust . 2goatauniformspeed.(Weknowthat’stherightanswer—to goatauniformspeed.) . Why isthat? Because iftheparticle were togoanyother way, thevelocities would ‘besometimes higher and sometimes lower than theaverage. The average velocity isthesame forevery case because ithastogetfrom ‘here’ to‘there’ inagiven amountoftime.“Asan example, sayyour jobistostart from home and gettoschool inagiven length oftime with thecar. You can doitseveral ways: You can accelerate like madatthebeginning andslowdown withthebrakes neartheend,oryoucango (IN atauniform speed,oryoucangobackwards forawhileandthengoforward, Hallandsoon. The thing isthat theaverage speed hasgottobe,ofcourse, thetotal tne distance thatyouhavegone overthetime, Butifyoudoanything butgoatauni- ne A form speed, then sometimes youaregoing toofastandsometimes youaregoing fone tooslow.Nowthemeansquare ofsomething thatdeviates around anaverage, as I ‘youknow, isalways greater than thesquare ofthemean; sothekinetic energy [i integral would always behigher ifyou wobbled your velocity than ifyou went ata uniform velocity. Soweseethat theintegral isaminimum ifthevelocity isa Lawconstant(whentherearenoforces).Thecorrectpathislikethi.—_> “Now, anobject thrown upinagravitational fielddoesrisefaster firstand La - — * thenslowdown. Thatisbecause thereisalsothepotential energy, andwemust * * have theleast difference ofkinetic and potential energy ontheaverage. Because thepotential energy rises aswegoupinspace, wewillgetalower difference ifwe ‘cangetassoonaspossible uptowhere thereisahighpotential energy. Thenwe [EAN Ply cantake thatpotential away from thekinetic energy andgetalower average. So \ itisbetter totake apath which goes upand gets alotofnegative stuff from the \ potential energy. ——> Be “Ontheotherhand, youcan’tgouptoofast,ortoofar,because youwillthen. joore [ have toomuch kinetic energy involved—you have togovery fast togetway | >upandcomedownagaininthefixedamountoftimeavailable. Soyoudon’twant [il |0gotoofarup,butyou want togoupsome. Soitturns outthat thesolution is some kindofbalance between trying togetmore potential energy withtheleast 4amountofextrakineticenergy—trying togetthedifference, kineticminusthe|‘_ ~ potential, assmall aspossible. — x * 192 “That isallmyteacher told me,because hewasavery good teacher andknew whentostoptalking.ButIdon’tknowwhentostoptalking.Soinsteadofleaving itasaninteresting remark, Iamgoing tohorrify anddisgust youwith thecomplexi- tiesoflifebyproving that itisso.The kind ofmathematical problem wewill have isvery difficult and anew kind. We have acertain quantity which iscalled theaction, S.Itisthekinetic energy, minus thepotential energy, integrated over time. Action=S=[Ke—PE)dt. Remember that the PE and KE are both functions oftime. For each different possible path you getadifferent number forthisaction. Our mathematical problem istofind out for what curve that number isthe least. “You say—Oh, that’s just the ordinary calculus ofmaxima and minima. You calculate theaction and just differentiate tofind theminimum. “But watch out. Ordinarily wejust have afunction ofsome variable, andwe have tofind the value ofthat variable where the function isleast ormost. For instance, wehave arodwhich hasbeen heated inthemiddle andtheheat isspread around. Foreach point ontherodwehave atemperature, andwemust find the point atwhich that temperature islargest. Butnow foreach path inspace wehave anumber—quite adifferent thing—and wehave tofindthepath inspace forwhich thenumber istheminimum. That isacompletely different branch ofmathematics.Itisnottheordinarycalculus. Infact,itiscalledthecalculusofvariations.“There aremany problems inthis kind ofmathematics. For example, the circle isusually defined asthelocus ofallpoints ataconstant distance from a fixed point, butanother wayofdefining acircle isthis: acircle isthatcurve of [IN siven length which encloses thebiggest area. Any other curve encloses lessarea for posta agiven perimeter than thecircle does. Soifwegive theproblem: find that curve which encloses thegreatest area foragiven perimeter, wewould have aproblem y x ofthecalculus ofvariations—a different kind ofcalculus than you're used to. “So wemake thecalculation forthepath ofanobject. Here istheway we ‘ aregoing todoit.The idea isthat weimagine that there isatrue path and that anyother curve wedraw isafalse path, sothatifwecalculate theaction forthe false path wewillgetavalue thatisbigger than ifwecalculate theaction forthe é true path. —e “Problem: Find thetruepath. Where isit?Oneway, ofcourse, istocalculate ‘ theaction formillions and millions ofpaths and look atwhich one islowest. ‘When you find thelowest one, that’s thetrue path, “That's apossible way. Butwecandoitbetter than that. When wehave a quantity which hasaminimum—for instance, inanordinary function like thetemperature—one oftheproperties ofthe minimum isthat ifwegoaway from theminimum inthefirstorder,thedeviation ofthefunctionfromitsminimum valueisonly second order. Atanyplace else onthecurve, ifwemove asmall distance thevalue ofthefunction changes also inthefirst order. Butataminimum, atiny motion away makes, inthefirst approximation, nodifferenct. —> “That iswhat wearegoing tousetocalculate thetrue path. Ifwehave the true path, acurve which differs only alittle bitfrom itwill, inthefirst approxima- tion, make nodifference intheaction. Any difference will beinthesecond approximation, ifwereally have aminimum. “That iseasy toprove. Ifthere isachange inthefirst order when Ideviate thecurve acertain way, there isachange intheaction that isproportional tothe deviation. Thechange presumably makes theaction greater; otherwise wehaven't gotaminimum. Butthen ifthechange isproportional tothedeviation, reversing thesign ofthedeviation will make theaction less. Wewould gettheaction to increase oneway andtodecrease theother way. The only way that itcould really beaminimum isthat inthefirst approximation itdoesn’t make anychange, thatthechangesareproportional tothesquareofthedeviations fromthetruepath. 193 “Soweworkitthisway:Wecallx(¢)(withanunderline) thetruepath—the‘onewearetrying tofind. Wetakesometrialpathx(#)thatdiffers fromthetrue x! pathbyasmallamount whichwewillcalln(#)(etaof1). ————p> | “Now theidea isthat ifwecalculate theaction Sforthepath x(7), then the| /~ difference between thatSandtheaction thatwecalculated forthepathx()—to lowe A simplify thewriting wecancallitS—the difference ofSandSmust bezeroin “ef thefirst-order approximation ofsmall ».Itcandiffer inthesecond order, but | Lag inthe first order the difference must bezero. “And thatmust betrue forany»atall.Well, notquite. The method doesn’tmeananything unlessyouconsider pathswhichallbeginandendatthesametwo [iam a points—each path begins atacertain point atf,andends atacertain other point atf2,andthose points andtimes arekept fixed. Sothedeviations inour7have to ‘bezeroateachend,n(t,)=Oandn(t2)=0.Withthatcondition, wehavespeci- fied ourmathematical problem. “Ifyou didn’t know anycalculus, you might dothesame kind ofthing to find theminimum ofanordinary function f(x). You could discuss what happens ifyoutakef(x)andaddasmallamount /toxandarguethatthecorrection tof(x) inthefirst order inAmust bezero attheminimum. You would substitute x+h forxandexpand outtothefirst order inA....just aswearegoing todowith n. “The idea isthen that wesubstitute x(?) =x(1) +n(#) intheformula for ‘the action:~ ‘m(dx\? s-f[s(¢)-veo]at, where Icall thepotential energy V(x). The derivative dx/dt is,ofcourse, the derivative ofx(¢)plusthederivative of»(#),sofortheactionIgetthisexpression: =[fm(as4an)? s-f[(@+9 —V+nae “Now Imust write thisoutinmore detail. Forthesquared term Iget dx?,4dxda(ay @)+aatla): Butwait. I'mnotworrying about higher than thefirst order, soIwilltake allthe termswhichinvolve »?andhigherpowers andputtheminalittleboxcalled ‘second andhigher order.’ From thisterm Igetonly second order, butthere will ‘bemore from something else. Sothekinetic energy part is m(dx\? dxdq2(4)+m48.condanhigheroe “Now weneed thepotential Vatx+9.Iconsider »small, soIcanwrite V(x) asaTaylor series. Itisapproximately V(x); inthenextapproximation(fromtheordinary natureofderivatives) thecorrection is1timestherateofchangeofVwithrespecttox,andsoon: Vet n=VO+VOtPVOt Ihave written V’forthederivative ofVwith respect toxinorder tosave writing. ‘Theterm in7?andtheones beyond fallintothe‘second andhigher order’ category andwedon’t have toworry about them. Putting italltogether, =["[m(asy dxdaSeI,[3(@)-reasgs—AV")+(condandhigherorder)dr 194 n(t1) =0,and n(t2) =0.Sotheintegrated term iszero. Wecollect theother terms together and obtain this: as=J.[-mfaveoa0de. The variation inSisnow theway wewanted it—there isthestuff inbrackets, say F,allmultiplied by»(¢) and integrated from 1;tot9. “We have that anintegral ofsomething orother times (1)isalways zero: froa(t)dt=0. ‘ Ihave some function of1;Imultiply itby9(); and Iintegrate itfrom oneendto ‘theother. Andnomatter whatthe»is,Igetzero.Thatmeansthatthefunction F(i) iszer0, That’s obvious, butanyway I'llshow you onekind ofproof. “Suppose that forn(#) 1took something which was zero forall¢except right i}nearoneparticular value. Itstayszerountilit getstohis, —- ———___}|_-___, then itblips upforamoment andblips right back down. When wedotheintegralofthisytimesanyfunctionF,theonlyplacethatyougetanything otherthanzero ‘waswhere »(1)wasblipping, andthen yougetthevalue ofFatthatplace times the integral over theblip. The integral over theblip alone isn't zero, butwhen multi- plied byFithastobe;sothefunction Fhastobezero where theblip was. But theblip was anywhere Iwanted toputit,soFmust bezero everywhere. “We seethat ifour integral iszero forany ,then thecoefficient of»must be zero. The action integral willbeaminimum forthepath that satisfies thiscompli- cated differential equation: fz yp)[ms -va}=0. It'snotreally socomplicated; youhave seen itbefore. Itisjust F=ma, Thefirst term isthemass times acceleration, and thesecond isthederivative ofthepotential energy, which istheforce. “So, foraconservative system atleast, wehave demonstrated that theprinciple ofleast action gives theright answer; itsays that thepath that hastheminimum action istheonesatisfying Newton’s law. “Oneremark:|didnotproveitwasaminimum—maybe it’samaximum. In. fact, itdoesn’t really have tobeaminimum. Itisquite analogous towhat wefound forthe‘principle ofleast time’ which wediscussed inoptics. There also, wesaid atfirstitwas‘least’ time. Itturned out,however, thatthere were situations inwhich itwasn’t theJeast time. The fundamental principle was that foranyfirst-order variation away from theoptical path, thechange intime was zero; itisthesame story. What wereally mean by‘least’ isthat thefirst-order change inthevalue ofS,when youchange thepath, iszero. Itisnotnecessarily a‘minimum.’ “Next, Iremark onsome generalizations. Inthefirst place, thething canbe done inthree dimensions. Instead ofjust x,1would have x,y,and zasfunctionsof1;theactionismorecomplicated. Forthree-dimensional motion,youhaveto usethecomplete kinetic energy—(m/2) times thewhole velocity squared. That is, _m{(ax\? |(dy\?,(dz\? . xe3((G)+@)+GY] Also, thepotential energy isafunction ofx,y,and z.And what about thepath? ‘The path issome general curve inspace, which isnotsoeasily drawn, buttheidea isthesame. And what about the»?Well, 1can have three components. You could shift thepaths inx,oriny,orinz—or you could shift inallthree directionssimultaneously. So»wouldbeavector.Thisdoesn’treallycomplicate thingstoo much, though. Since only thefirst-order variation has tobezero, wecan dothe calculation bythree successive shifts. Wecan shift »only inthex-direction and 6 minus thepotential energy. That's only true inthenonrelativistic approximation. Forexample, theterm mgc*\/F—02/e?isnotwhatwehavecalledthekinetic energy. The question ofwhat theaction should beforanyparticular case must bedetermined bysome kind oftrial and error. Itisjust thesame problem asdeter- mining what arethelaws ofmotion inthefirstplace. You justhave tofiddle around with theequations that youknow andseeifyoucangetthem into theform ofthe principle ofleast action. “One other point onterminology. The function that isintegrated over time togettheaction Siscalled theLagrangian, £,which isafunction only ofthevelocities andpositions ofparticles. Sotheprincipleofleastactionisalsowritten S=[oC 00d, where byx;and »,aremeant allthecomponents ofthepositions and velocities. Soifyouhear someone talking about the‘Lagrangian,’ you know they aretalking about the function that isused tofind S. For relativistic motion inanelectro- magnetic field £=—moc?VT= v8/eF—ge+vA). “Also, Ishould saythat Sisnotreally called the‘action’ bythemost precise andpedantic people. Itiscalled ‘Hamilton’s first principal function.’ Now Ihate togive alecture on‘the-principle-of-least-Hamilton’s-frst-principal-function.”SoIcall it‘theaction.’ Also, more andmore people arecalling ittheaction. You see,historically something elsewhich isnotquite asuseful wascalled theaction, butIthink it'smore sensible tochange toanewer definition. Sonow you too willcallthenew function theaction, andpretty soon everybody willcallitbythat simple name. “Now Iwant tosaysome things onthissubject which aresimilar tothedis- cussions Igave about theprinciple ofleast time. There isquite adifference inthe characteristic ofalawwhich saysacertain integral from oneplace toanother isa minimum—which tells something about thewhole path—and ofalawwhich says that asyougoalong, there isaforce that makes itaccelerate. The second way tells, howyouinchyour wayalong thepath, andtheother isagrand statement about the whole path. Inthecase oflight, wetalked about theconnection ofthese two. Now, Iwould like toexplain why itistrue that there aredifferential laws when there isaleast action principle ofthiskind. Thereason isthefollowing: Consider theactual path inspace and time. Asbefore, let’s take only onedimension, so wecanplot thegraph ofxasafunctionoft.Alongthetruepath,Sisaminimum. Let’s suppose that wehave thetrue path and that itgoes through some point a inspace andtime, andalsothrough another nearby point. —> Nowiftheentireintegralfrom1;tofzisaminimum, itisalsonecessary thattheintegralalongthelittlesectionfromato6isalsoaminimum. Itcan’tbethatthepartfromatobisalittlebitmore.Otherwise youcouldjustfiddlewithjustthatpieceofthepathandmakethewholeintegralalittlelower.“So every subsection ofthepath must also beaminimum. And thisistrue ‘nomatter how short thesubsection. Therefore, theprinciple that thewhole path sives aminimum canbestated also bysaying that aninfinitesimal section ofpath also has acurve such that ithas aminimum action. Now ifwetakeashortenough sectionofpath—between twopointsaandbveryclosetogether—how thepotential varies from oneplace toanother faraway isnottheimportant thing, because you arestaying almost inthesame place over thewhole little piece ofthepath. The only thing that you have todiscuss isthefirst-order change inthepotential. The answer canonly depend onthederivative ofthepotential andnotonthepotential everywhere, Sothestatement about thegross property ofthewhole path becomes astatement ofwhat happensforashortsectionofthe path—a differential statement. ‘And thisdifferential statement only involves thederivatives ofthepotential, that is,theforce atapoint. That's thequalitative explanation oftherelation between thegross lawandthedifferential law. 98 Inorder forthisvariation tobezero foranyf,nomatter what, thecoefficient of ‘Ffmustbezeroand,therefore,v6=—p/eo. ‘Wegetback ouroldequation. Soour‘minimum’ proposition iscorrect. “Wecangeneralizeourpropositionifwedoouralgebrainalittledifferent way. Let’s goback and doourintegration byparts without taking components. Westart bylooking atthefollowing equality: V:(f¥8) =Wf¥b+S0%. Ifdifferentiate outtheleft-hand side, Icanshow that itisjustequal totheright- hand side. Nowwecanusethisequationtointegratebyparts.InourintegralAU*, wereplace—vg-Vfby[76—V-(f¥$), whichgetsintegrated overvolume. ‘The divergence term integrated over volume canbereplaced byasurface integral: fo-usear =[rve-ndo Since weareintegrating over allspace, thesurface over which weareintegrating is atinfinity. There, fis zero and wegetthesame answer asbefore. “Only now weseehow tosolve aproblem when wedon’t know where allthe charges are. Suppose that wehave conductors with charges spread outonthem in some way. Wecanstill useourminimum principle ifthepotentials ofallthe conductors arefixed. Wecarry outtheintegral forU*only inthespace outsideofallconductors. Then,sincewecan'tvary¢ontheconductor, fiszeroonallthose surfaces, andthesurface integral =~ ff¥e-nda isstill zero. ‘The remaining volume integral aut=foeve—pg)faV isonlytobecarriedoutinthespacesbetweenconductors. Ofcourse,wegetPoisson's equation again, v6=—p/eo Sowehave shown that ouroriginal integral U*isalso 2minimum ifweevaluate itover thespace outside ofconductors allatfixed potentials (that is,such thatany trial 4(x, y,2)must equal thegiven potential oftheconductors when x,y,zisa point onthesurface ofaconductor). “There isaninteresting case when theonly charges areonconductors. Then ut=2[ewotav. Ourminimum principle says thatinthecase where there areconductors setat certain given potentials, thepotential between them adjusts itself so’that integral U*isleast. What isthisintegral? The term V¢istheelectric field, sotheintegral istheelectrostatic energy. Thetruefield istheone, ofallthose coming from the aa gradient ofapotential,withtheminimumtotalenergy. / “Iwould liketousethisresult tocalculate something particular toshow you that these things arereally quite practical. Suppose Itake twoconductors inthe . form ofacylindrical condenser. ——=> Uy V7 ‘Theinside conductor hasthepotential V,andtheoutside isatthepotential zero. , Let theradius oftheinside conductor beaand that oftheoutside, b.Now wecan suppose any distribution ofpotential between thetwo. Ifweusethecorrect g, andcalculate €o/2f (vg)? dV,itshould betheenergy ofthesystem, CV?. ist 20 Solutions ofMaxwell’s Equations in Free Space 20-1 Waves infree space; plane waves InChapter 18wehadreached thepoint where wehadtheMaxwell equations «20-1. Waves infree spaces plane incomplete form. Allthere 15toknow about theclassical theory oftheelectric waves andmagnetic fieldscanbefound inthefourequations: 20-2Three-dimensional waves Loven? IoovVXE=— 20-3Scientific imagination 5ag 20.1) 20-4Spherical waves Ulv-B=0 W.yxp=b+© oF When weputalltheseequations together, aremarkable newphenomenon occurs: References: Chapter 47,Vol.I:Sound: fields generated bymoving charges canleave thesources andtravel alone through TheWave Equation space. Weconsidered aspecial example inwhich aninfinite current sheet is Chapter 28,Vol.1:Electro- suddenly turned on.After thecurrent hasbeenonforthetime¢,there areuniform magnetic Radiation electric and magnetic fields extending outthedistance crfrom thesource. Suppose that thecurrent sheet liesintheyz-plane with asurface current density Jgoing toward positive y.‘The electric field will have only ay-component, and themag- eis81 netic field, only az-component. The magnitude ofthefield components isgiven by : J By=Be=~50 (20.2) forpositive values ofxlessthanct,Forlargerxthefieldsarezero.Thereare, —_. ——|—, ofcourse, similar fields extending thesame distance from thecurrent sheet inthe “ negative x-direction. InFig.20-1weshow agraph ofthemagnitude ofthefields Fig.20-1, Theelectric ond mag- a8.a function ofxattheinstantr.Astimegoeson,the“wavefront” atcfMOVESneticfieldasafunctionofxotthetimet ‘outward inxattheconstant velocity ¢. after thecurrent sheet isturned on. Now consider thefollowing sequence ofevents. Weturn onacurrent ofunit strength forawhile, thensuddenly increase thecurrent strength tothreeunits, ©|andhold1constantatthisvalue.Whatdothefieldsooklikethen?Wecanseewhat thefields willfook likeinthefollowing way. First, weimagine acurrent of | unitstrength thatisturned onat¢=Oand leftconstant forever. Thefields for ( positive xarethen given bythegraph inpart (a)ofFig. 20-2. Next, weaskwhat © ws would happen ifweturnonasteady current oftwounits atthetime1). et Thefieldsinthiscasewillbetwiceashighasbefore, butwillextend outinh xonly thedistance e(¢—f,),asshown inpart (b)ofthefigure. When weadd these twosolutions, using theprinciple ofsuperposition, wefindthatthesumof afthetwosourcesisacurrentofoneunitfortheumefromzerotof,andacurrent, |© atheofthreeunitsfortimesgreaterthan1,.Atthetime¢thefieldswillvarywithx€‘asshown inpart(c)ofFig.20-2. iNowlet’stakeamorecomplicated problem.Consideracurrentwhichis1 turned ontooneunit forawhile, then turned uptothree untts, andlater turned offtozero. What arethefields forsuchacurrent? Wecanfindthesolution in| ayy thesame way—by adding thesolutions ofthree separate problems. Furst, wefind co) thefieldsforastepcurrentofunitstrength,(Wehavesolvedthatproblemalready.) ig.29-2,TheelectricfieldofoNext, wefindthefields produced byastepcurrent oftwounits. Finally, WeSO€ current sheet, (a)One unitofcurrentforthefieldsofastepcurrentofminusthreeunits.Whenweaddthethreesolutions, turnedonatt=Q;(b)Twonitsofwewill have acurrent which 1one unit strong from ¢=0tosome later time, current turned onatt=th;(e)Super- sayf,then three units strong until astilllater time fg,and then turned off—that position of(a)and(b). 204 J ~ey 2 2 i f T ° = =i 2 t eit e=ty oe (o) Oo) Fig. 20-3. Ifthecurrent source strength varies osshown in(a),then atthetime#shownbythearrowtheelectricfieldasafunctionofxisasshownin(b). 1s,tozero. Agraph ofthecurrent asafunction oftime isshown inFig.20-3(a). When weadd thethree solutions fortheelectric field, wefind that itsvariation with x,atagiven instant ,isasshown inFig. 20-3(b). The field isanexact, representation ofthecurrent. The field distribution inspace isanice graph of thecurrent variation with time—only drawn backwards. Astime goes onthewholepicturemovesoutwardatthespeedc,sothereisalittlebloboffield,travellingtoward 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 had varied atthe source. You will also notice that long 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 clectric field atagiven place andagiven time isproportional tothecurrent atthe source, only notatthesame time, butattheearlier time ¢—x/e, Wecanwrite =—t=x/0) By=—X. 203) 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 byathin layer of oscillating dipoles inasheet ofdielectric material with thedipoles setinmotion bytheelectric field ofanincoming electromagnetic wave. Our problem wasto calculate thecombined fields oftheoriginal wave and thewaves 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. Ifyou willlook inChapter 31ofVol. I,youwillseethatEq.(31.10) there isjustthesame astheEq.(20.3) that wehave just written down. Although ourearlier derivation wascorrect only atlarge distances from thesource, weseenow that thesame result continues to becorrect even right uptothesource. Wewant now tolook inageneral way atthebehavior ofelectric andmagnetic fields inempty space faraway from thesources, i..,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. 2 twoequations will then bethesame (except forthefactor ¢”). Sowefind that E,satisfies theequation @E, 1aE, 7ae 2oe7 20.19) Wehave seen thesame differential equation before, when westudied thepropaga- tion ofsound. Itisthewave equation forone-dimensional waves. ‘You should note that intheprocess ofourderivation wehave found something ‘more than iscontained inEq. (2011). Maxwell's equations have given usthe further information that electromagnetic waves have field components only at rightanglestothedirection ofthewavepropagation.Let’s review what we know about the solutions ofthe one-dimensional wave equation. Ifanyquantity ¥satisfies theone-dimensional wave equation ay Layfe oe TO (20.20) then one possible solution isafunction 9(x, 1)oftheform V(x.) =flx et), (20.21) that is,some function ofthesingle variable (x~ct). The function f(x —ct) represents a“rigid” pattern inxwhich travels toward positive xatthespeed ¢ (ee Fig. 20-4). For example, ifthefunction fhasamaximum when itsargument iszero, then for¢=0themaximumofywilloccuratx=0.Atsomelatertime, say f=10, will have itsmaximum atx=10c.Astimegoeson,themaximum thomoves toward positive xatthespeedi ‘Sometimes itismoreconvenient tosaythatasolutionofthe one-dimensional Sadi ant wave equation isafunction of(1—x/c). However, thisissaying thesame thing, i)Yo, becauseanyfunctionof(t—x/c)isalsoafunctionof(x—ct):~Ft—x/e)=#{-aaa]=fle=ct). Fig.20-4, The function flx—ef) Let’s show thatf(x—ct)isindeed asolution ofthewave equation. Since represents aconstant “shope” thattravels itisafunction ofonlyonevariable—the variable (x~ci)—we willletf”representtowardpositivexwiththespeedc. thederivative offwithrespecttoitsvartableand/”representthesecondderivativeoff.Differentiating Eq. (20.21) with respect tox,wehave Wy%=se-ob, since thederivative of(x—ct)with respect toxis1.The second derivative of ¥with respect toxisclearly OY_pm 20.22 oe=px-ot. (20.22) Taking derivatives ofywith respect to1,wefind oy x= et(—%-Ux-a0, 2, ad=efx —et) (20.23) Weseethat ¥-does indeed satisfy theone-dimensional wave equation. ‘You may bewondering: “IfIhavethewaveequation,howdoIknowthat Ishould take f(x —ct)asasolution? Idon’t like this backward method. Isn't there some forward way tofind thesolution?” Well, one good forward way is, toknow thesolution. Itispossible to“cook up” anapparently forward mathe- matical argument, expecially because weknow what thesolution issupposed to be,but with anequation assimple asthis wedon’t have toplay games. Soon you will getsothat when you see Eq. (20.20), you nearly simultaneously see 20-6 something which isnew, butwhich isconsistent with everything which hasbeen seen before, 1sone ofextreme difficulty. While I'monthis subject |want totalk about whether itwill ever bepossible toimagine beauty that wecan’t see Itisaninteresting question. When welook atarainbow, itlooks beautiful tous.Everybody says, “Ooh, arainbow.” (You seehow scientific |am. 1amafraid tosaysomething 1sbeautiful unless Ihave an experimental way ofdefining it.)But how would wedescribe arainbow ifwewere blind? We are blind when we measure the infrared reflection coefficient ofsodium chloride, orwhen wetalkabout thefrequency ofthewaves that arecoming from some galaxy that wecan’t see—we make adiagram swe make aplot. For instance, fortherainbow, such aplot would betheintensity ofradiation vs.wavelength measured with aspectrophotometer foreach direction inthesky. Generally, such measurements would give acurve that was rather flat. Then some day, someone would discover that forcertain conditions oftheweather, and atcertain angles in thesky, thespectrum ofintensity asafunction ofwavelength would behave strangely; itwouldhaveabump.Astheangleofthe instrument was varied only a little bit,themaximum ofthebump would move from onewavelength toanother.Thenonedaythephysicalreviewofthe blind men might publish atechnical article with thetitle “The Intensity ofRadiation asaFunction ofAngle under Certain Conditions oftheWeather.” Inthis article there might appear agraph such as theone inFig. 20-5 The author would perhaps remark that atthelarger angles there was more radiation atlong wavelengths, whereas forthesmaller angles themaximum intheradiation cameatshorterwavelengths. (Fromourpointofview, ‘wewould saythat thelight at40° 1spredominantly green and thelight at42° is predominantly red.) > Ss a= Sscy x Fig.20-5,Theintensityofelectro- ¢ ! magnetic waves asafunction ofwave- = y length forthree angles (measured from ~ thedirection opposite thesun), observed only with certain meteorological con- Wavelength ditions. Now dowefind thegraph ofFig. 20-5 beautiful? Itcontains much more de- tailthan weapprehend when welook atarainbow, because oureyes cannot see theexact details intheshape ofaspectrum. The eye, however, finds therainbow beautiful. Dowehave enough imagination toseeinthespectral curves thesame beauty weseewhen welook directly attherainbow? 1don’t know. Butsuppose Ihave agraph ofthereflection coefficient ofasodiumchloride crystal asafunction ofwavelength intheinfrared, and also asafunction 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 Ican ever look atagraph ofthereflection coefficient ofNaCl measured with some instrument andsaythat ithasthesame 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 abouttheregularity oftheappearance ofthex,they,thez,andthe¢.Andthis nice symmetry nappearance ofthex.y,z,and fsuggests tothemind stillagreater beauty which hastodowith thefour dimensions, thepossibility that space has four-dimensional symmetry, thepossibility ofanalyzing thatandthedevelopments ofthespecral theory ofrelativity. Sothere 1splenty ofintellectual beauty asso- ciated with theequations. 20-1 weask, then, what functions Y(r, 1)aresolutions ofthethree-dimensional wave equation 2, La VD ~HO =0. (2033) Since ¥(r, )depends only onthespatial coordinates through r,wecanusetheequa- tion fortheLaplacian wefound above, Eq. (20.32). Tobeprecise, however, since ¥isalso afunction of1,weshould write thederivatives with respect toraspartial derivatives. Then thewave equation becomes le 1a 7M ~aipy=O ‘Wemust now solve thisequation, which appears tobemuch more complicated than theplane wave case. But notice that ifwemultiply this equation byr,weget a 1 *mw-1LXm=o. 20.34a)~25a =0 (2034) This equation tells usthat thefunction rysatisfies theone-dimensional wave equa- tion inthevariable r.Using thegeneral principle which wehave emphasized so often, that thesame equations always have thesame solutions, weknow that if Vis afunction only of(r—ct)then itwill beasolution ofEq.(20.34). Sowe know that spherical waves must have theform Mr) =Mr~ed. Or,aswehave seen before, wecanequally well saythat rycan have theform = ft=1/0). Dividing byr,wefind that thefield quantity (Whatever itmay be)hasthefollow- ingform: ye (20.35) Such afunction represents ageneral spherical wave travelling outward from the ongin atthespeed c.Ifweforget about therinthedenominator foramoment, theamplitude ofthewave asafunction ofthedistance from theorigin atagiven lume hasacertain shape that travels outward atthespeed c.The factor rinthe denominator, however, says that theamplitude ofthewave decreases inproportion toI/ras thewave propagates. Inother words, unlike aplane wave inwhich the amplitude remains constant asthewave runs along, inaspherical wave theampli- tude steadily decreases, asshown inFig. 20-6. This effect iseasy tounderstand from asimple physical argument. \ \ a\ q\ ba SS ue i1 ~.0 2 h eg AR ‘ty es " 2 r 1 te 1’ b}—eltg—4h)ed(0) (by Fig. 20-6. Aspherical wave y=flt—r/el/r. (a)¥as@functionofrfort=fyondthe same wave for the later time ta. (b) ¥as@function of#for r=r;and the some wave seen ofr2. 2013 21 Solutions ofMaxwell's Equations with Currents and Charges 21-1 Light and electromagnetic waves Wesaw inthelast chapter that among their solutions, Maxwell’s equations ‘21-1 Light and electromagnetic have waves ofelectricity and magnetism. ‘These waves correspond tothephe- waves nomena ofradio, light, x-rays, and soon,depending onthewavelength. WehaveateadystudiedightimgreatdealinVal.Inthischapterwewanttotictogether 21-2Sphericalwavesfromapointthetwosubyects—we wanttoshowthatMaxwell’s equations canindeed formthe source base forourearlier treatment ofthephenomena oflight. 21-3 Thegeneral solution of When westudied light, webegan bywriting down anequation fortheelectric Maxwell’s equations fieldproduced byacharge which moves inanyarbitrary way.Thatequation WS 94.4Thefieldsofanoscillating _gferard(er1d| dipole Emtre[s+edt(«)+age| CLD) 44-5Thepotentialsofamoving Beer XE charge; thegeneral solution [see Eq. (28.3), Vol. L]* ofLignard andWiechert Ifacharge moves inanarbitrary way, theelectric field wewould findnowat 24-6 Thepotentials foracharge some point depends only ontheposition and motion ofthecharge notnow, but moving with constant velocity; atanearlier time—at aninstant which isearher bythetume itwould take light, theLorentz formula going atthespeed c,totravel thedistance r’from thecharge tothefield point. Inother words, ifwewant theelectric field atpoint (1)atthetime1,wemustcal- culatethelocation (2’)ofthecharge and1tsmotion atthetime (1—r’/c), where . _ 7isthedistance tothepoint (1)from theposition ofthecharge (2’)attheumeRewewsChapter28,Vol.h,Eleciro- (1=r'/c).Theprimeistoremindyouthatr’istheso-called “retarded distance” Cele Pee Theofromthepoint(2')tothepoint(1),andnottheactualdistancebetweenpoint(2),the ofthe“RefractiveIndegin position ofthecharge atthetime ¢,and thefield point(1)(seeFig.21-1).Note eee eTRetarwistic thatweareusingadifferent convention nowforthedirection oftheunitvector 5piers"Rodeo ela¢,.InChapters 28and36ofVol.Iitwasconvenient totaker(andhencee,) fectsinRadiation pointing oward thesource. Now wearefollowing thedefinition wetook forCou- Tomb’ law, inwhich risdirected from thecharge,at(2),cowardthefieldpointat(1) The only difference, ofcourse, 1sthat our new r(and e,)arethenegatives ofthe old ones. Wehave also seen that ifthevelocity »ofachargeisalwaysmuchlessthan ¢,and ifweconsider only points atlarge distances from thecharge, sothat only the fastterm ofEq.(21.1) 1simportant, thefields canalso bewritten as 4g._[acceleration ofthechargeat(t—r'/e) , » PeFreer|projectedatrightanglestor’|:en oo andPyaoe oO cB eyXE. rassronl aS oa Tete 4 Let’s lookatwhat thecomplete equation, Eq.(21.1), saysinahitlemore a detail. Thevector e,/1stheunit vector topomnt(1)fromtheretardedposition(2'). ii Thefirstterm, then, 1swhat wewould expect fortheCoulomb fieldofthecharge Postionot atitsretarded position—we may call this “the retarded Coulomb field.” The electricfielddependsinverselyonthesquareofthedistanceandisdirectedaway a.21 heldsot(1)otthefromtheretarded position ofthecharge(thatis,inthedirection ofe,). tinydeendovineeeitiont2nccepied Butthat1sonlythefirstterm.Theothertermstellusthatthelawsofelectricity bythechargeqatthetime(t—r’/e). donotsaythat allthefields arethesame asthestatic ones, butjust retarded (which iswhat people sometimes liketosay). Tothe“retarded Coulomb field” wemust at —using, ofcourse, theinstantaneous dipole moment p(#). But ifwegovery far ‘out, weought tofind aterm inthefield that goes as1/rand depends ontheac-celeration ofthe charge perpendicular totheline ofsight. Let’s seeifwegetsuch aresult Webegin bycalculating thevector potential 4,using Eq. (21.16). Supposethatourmovingchargeisinasmallblobwhosechargedensityisgivenbyp(x,92)sand thewhole thing ismoving atany anstant with thevelocity v.Then thecurrent z density J(4,9,2)willbeequaltoup(x,»,2).Itwillbeconvenenttotakeour coordinate system sothat thez-axis 181nthedirection ofv;then thegeometry of our problem isasshown inFig. 21-2. Wewant theintegral i) pee5fuels)Ws, ui? av, ot a Nowifthesizeofthecharge-blob isreallyverysmallcomparedwithra,We ficansetthe7)»term inthedenominator equal tor,thedistance tothecenter ofthe (s,y,2) ¥blob,andtakeroutside theintegral, Neat,wearealsogoingtosetrz=rimm thenumerator, although thatisnotreally quite right. It1snotrightbecause we should take jat,say, thetopoftheblob ataslightly different time than weused 4 forjatthebottomoftheblob.Whenwesetry2=rimjt—r12/c),wearetaking thecurrent density forthewhole blob atthesame time (1—r/c). That 1s Fig.21-2. Thepotentials ot(1)ere approximation thatwillbegood onlyifthevelocity»ofthechargeismuch given byintegrals over thecharge lessthan c,Sowearemaking anonrelativistic calculation. Replacing jbypv, density p. theintegral (21.17) becomes fever —r/c)dV. Since allthecharge hasthesame velocity, thisintegral isjust v/rumes thetotal charge g.Butqv1sjustdp/at,therateofchangeofthedipolemoment—which 1, ofcourse, tobeevaluated attheretarded tume (¢— r/c). Wewill write itas PU ~r/c). Sowegetforthevector potential ~~! wt=re) MLD=ee SO 1.18) ‘Our result says that thecurrent inavarying dipole produces avector potential 1ntheform ofspherical waves whose source strength 1sp/4€,¢* Wecannow getthemagnetic field from B=VX A.Since pstotally inthe z-direction, Ahasonly az-component; there areonly two nonzero derivatives in thecurl SoB,=a4,/ay andB,=~AA,/Ax. Let's firstlook atBe: ad, 10ptr/o) Be=“ay=Sweet ay 21.19) Tocarry outthedifferentiation, wemust remember thatr=\x?+y?+23,so 1 a(t Lia bo Be=gragMEFO)5()+agpaptr/o.1.20) Remembering that dr/ay =y/r, thefirst term gives 1 ype =r/e)a 2 21.21Aree rt etal which drops offasI/r?likethefields ofastaticdipole(becausey/r1sconstantfor agiven direction). The second term inEq. (21.20) gives usthenew effects. Carrying out the differentiation, weget ly -5 —r 21.areae?giht—1/0 (21.22) where pmeans, ofcourse, thesecond derivative ofpwith respect 101.This term, 21-6 which comes from differentiating thenumerator, 1responsible forradiation, First, tdescribes afield which decreases with distance only asI/r. Second, it depends ontheacceleration ofthecharge. You canbegin toseehow wearegoingtogetaresultlikeEq.(211’),whichdescribestheradiationoflight Let’s examine inalittle more detail how this radiation term comes about—1t tssuch aninteresting and rmportant result. Westart with theexpression (21.18), which hasaI/rdependence and istherefore like aCoulomb potential, except for thedelay term inthenumerator. Why isitthen that when wedifferentiate with respect tospace coordinates togetthefields, wedon’t justgetaI/r? field—with,ofcourse,thecorresponding timedelays’?Wecanseewhy inthefollowing way: Suppose that weletourdipole oscillate upand down inasinusoidal motion, Then wewould have P= Pe=posinat and A,=<b,#Paeos w(t=r/c) Free? r Itweplotagraphof4,asafunctionofrat agiven instant, wegetthecurve shown, inFig 21-3. The peak amplitude decreases asI/r, but there is,inaddition, «net oscillation inspace, bounded bytheI/renvelope. When wetake thespatial de- | yy rivatives, theywillbeproportional totheslope ofthecurve. From thefigure we |). seethat there areslopes much steeper than theslope oftheI/rcurve itself. Its. | .infact,evidentthatforagivenfrequencythepeakslopesareproportional tothe [\Ln~-amplitudeofthewave,whichvariesasI/r.Sothatexplainsthedrop-offrateof||\mnead the radiation term.|fVs\ZM4 Uallcomesaboutbecausethevariations withtimeatthesourcearetranslated ||Le intovariations ispaceasthewavesarepropagated outward, andthemagnet Va fields depend onthespatial derivatives ofthepotential , L'sgobackandfinish ourcalculation ofthemagnetic field, Wehavefor |’ B,thetwo terms (21.21) and (21.22), s0 Fig. 21-3. Themagnitdve ofAos© b= [-yet=r/c)_wm=2/0), functionofrottheinstanttforthe eS Frege? re oF spherical wove from onoscillating dipole. With thesame kind ofmathematics, weget 1[ape—r/c),x(t~r/o) Byareal?aea Orwecanputitalltogether inanicevector formula: can pet _WtC/O Xr ers lerece? rs B Now let’s look atthis formula. First ofall,ifwegovery faroutinr,only the term counts. The direction ofBis given bypXr,which isatright angles tothe 5 radiusrandalsoatrightanglestotheacceleration, asinFig.21-4.Everything 1s °re) coming outright; that 1salso theresult wegetfrom Eq,21.1"). Now let’s look atwhat wearenotused to—at what happens closer in. In fig. 21-4. Theradiation fieldsBond Section 14-9 weworked outthelawofBiot andSavart forthemagnetic field ofanofanoscillatingdipole. element ofcurrent, Wefound that acurrent element jdVcontributes tothemag- netic field the amount 1jxr aB=gts wv. (21.24) You seethat thisformula looks very much likethefirst term ofEq(21.23) ifwe remember that pisthecurrent, Butthere isonedifference, InEq,(21.23), the current isto beevaluated atthetime (¢—r/e), which doesn’t appear inEq.(21.24). Actually, however, Eq. (21.24) 1sstill very good forsmall r,because thesecond 217 o—4 Ay; qTGOA 4o Z| 50) (0) chew (py Fig.21-5.(0)A“point”charge—consdered as«smallcubicaldistributionof charge—moving with thespeed vtoward point (1) (b)The volume element AV, used forcalculating thepotentials. follow, wewill make thecalculation first fora“point” charge which isintheformt ofalittlecubeofchargemovingtowardthepoint(1)withthespeed»,asshown inFig. 21-S(a). Letthelength ofaside ofthecube bea,which wetake tobe much, much less than ry, the distance from thecenter ofthe charge tothe pomnt (1). Now toevaluate theintegral ofEq. (21.28), wewill return tobasic principles; wwe will write itasthe sum reat, 21.30) where r;1thedistance from point (1)totheithvolume element AV,andp,isthe sen chargedensityatAV,attheime1,=¢—r,/e.Sincer,>a,always,itwillbe qlaTTTfq ()convenienttotakeourAV,intheformofthin,rectangularshesperpendicular toath rr asshowninFig.21-5(b). ltiti} SupposewestartbytakingthevolumeelementsAV,withsomethickness yl . much less thana.Theindividual elementswillappearasshowninFig.21-6(a), ul. wherewehaveputinmorethanenoughtocoverthecharge.Butwehavenot WZ7«wy__Shownthecharge,andforagoodreason.Whereshouldwedraw11?Foreach os"volume elementAY,wearetotakepatthetime1,=(¢—r,/¢),butsincethe YZj charge1smoving,wtisinadifferentplaceforeachvolumeelementAV,! uf‘1 Let’ssaythatwebeginwiththevolumeelementlabeled“1”inFig.21-6(a), th i chosensothatatthetimef,=(1—r1/c)the“back”edgeofthechargeoccupies W777' AV,asshowninFig,21-6(b). ThenwhenweevalutepyAVg,Wemustusethe ey_-—,—2-£) positionofthechargeattheslightlylatermety=(t—rz/c),whenthecharge W,F ' willbeintheposition shown inFig.21-6(c). Andsoon,forAV,AV4,ete.Nowt ' wecan evaluate the sum. " f Sincethethickness ofeachAV,151,tsvolume iswa®,TheneachvolumeWY, i elementthatoverlipsthechargedistributioncontainstheamountofcharge 5.1!) sup,wherepisthedensityofchargewithinthecube—which wetaketobe oo: R44 a °density BsVA, uniform. Whenthedistancefromthechargetopomt(1)1slarge,wewillmakeanas | negligible errorbysettingallther,"sinthedenominators equaltosomeaverage‘neatapa value,saytheretardedpositionr’ofthecenterofthecharge.Thenthesum(21.30) yy .» Xp? weGZpone pee, ae where AV,isthelastAV,thatoverlaps thecharge distributions, asshown inFig, Fig.21-6, Integrating plt—r’c)dv 2!-6(e). Thesumis,clearly, for omoving charge. a)for@movingcharg yema_patae).rp Xa Now pais justthetotal charge qandNwisthelength bshown inpart (e)ofthe figure. Sowehave =,4,(° 2 21-10 21-6 The potentials foracharge moving with constant velocity; theLorentz formula Wewant next tousetheLiénard-Wiechert potentials foraspecial case—to find thefields ofacharge moving with uniform velocity inastraight line. Wewill dostaga later, using theprinciple ofrelativity. Wealready know what thepo- tentials arewhen wearestanding intherest frame ofacharge. When thecharge 1smoving, wecan figure everything out byarelativistic transformation from one system totheother. But relativity had utsorigin inthetheory ofelectricity and magnetism, The formulas oftheLorentz transformation (Chapter 15,Vol. 1)werediscoveries madebyLorentz whenhewasstudying theequations ofelectricity and magnetism. Sothat you can appreciate where things have come from, we would liketoshow that theMaxwell equations dolead totheLorentz transforma- tion. We begin bycalculating thepotentials ofacharge moving with uniform velocity, directly from the electrodynamics ofMaxwell's equations. We have shown that Maxwell’s equations lead tothepotentals foramoving charge that we gotinthelast section. Sowhen weusethese potentials, weareusing Maxwell's theory. 'y Pp f(6922) "RETARDEDPOSITION 1 GEER) cy 1)64 “fe lsvi} 17 Fig.21-7. Finding thepotential ot hear aaa Pofacharge moving withuniform eaten velocity along thex-axis. 2 ‘Supposewehaveachargemovingalongthex-axiswiththespeed7.Wewant thepotentials atthepoint P(x,»,2),asshowninFig.21-7.If=O1sthemoment whenthecharge1sattheorigin,atthetimefthecharge1satx="y=z=0 What weneed toknow, however, 1sitsposition attheretarded time vert, 21.35) where risthedistance tothepoint Pfrom thecharge attheretarded time. Atthe earlier time #’,thecharge was atx=vt’,so r=Va-WP Fyre (21.36) Tofind 7’or£wehave tocombine this equation with Eq. (21.35). First, we ‘ehmunate /’bysolving Eq. (21.35) forr’and substituting inEq. (21.36). Then, squaring both sides, weget et —2=xe— $y? 4+2, which isaquadratic equation in’, Expanding thesquared binonualsandcollecting like terms in1’,weget (0?=e2y'? —2 —cr 4x?+y?422=(CN? =O. Solving for1’, fa mt fea Gea oyee a—ety? =P42 x (:ar1Boen +(1Not+)137) a2 22 AC Circuits 22-1 Impedances Most ofour work inthis course has been aimed atreaching thecomplete 22-1 Impedances equations ofMaxwell. Inthelasttwochapters wehavebeendiscussing the€On- 99» Generators sequences ofthese equations. Wehave found that theequations contarn allthe static phenomena wehad worked outearlier, aswell asthephenomena ofelectro- 22-3 Networks ofideal elements; ‘magnetic waves and light that wehad gone over insome detail inVolume I.The Kirchhoff's rules Maxwell equations give both phenomena, depending upon whether one computes valent eireuitheeldsclosetothecurrents andcharges, orveryfurfromthemThereienot 22-4Eauivalent circuits much interesting tosay about the intermediate region; nospecial phenomena 22-5 Energy appear there. There stillremain, however, several subjects inelectromagnetism thatwe 22-6Aladder network want totake up. We want todiscuss thequestion ofrelativity and theMaxwell 22-7 Filters equations—what happens when onelooks attheMaxwell equations withrespect 99@(hercircuitelements tomoving coordinate systems. There isalso thequestion oftheconservation of energy inelectromagnetic systems. Then there isthebroad subject oftheelectro- magnetic properties ofmaterials; sofar,except forthestudy ofthepropertiesofdetects, wehaveconsideredonlytheelectromagneticfieldsm freespaceAndpeveyChapter22,Vol1,Algebraalthough wecovered thesubject oflightinsomedetailinVolume I,thereare Chapter 23:Vol1,Resonance stilafewthings wewould liketodoagainfromthepointofviewofthefield Ghani 25,vol’1Enoequations. . SystemsandReview Inparticular, we want totake upagain the subject ofthe index ofre- fraction, particularly fordense materials. Finally, there are thephenomena associated with waves confined inalimited region ofspace. We touched onthis kind ofproblem briefly when wewere studying sound waves. Maxwell's equationsleadalsotosolutionswhichrepresentconfinedwavesofthe electric and magneue fields. We will take upthis subject, which hasimportant technical applications, insome ofthe following chapters. Inorder tolead uptothat subject, wewill begin byconsidering theproperties ofelectrical circuits atlowfrequencies. We will then beable tomake acomparison between those situations inwhich the almost static approximations ofMaxwell's equations are applicable and those situations inwhich high-frequency effects aredominant. Sowedescend from thegreat and esoteric heights ofthelast few chaptersandturntotherelativelylow-levelsubjectofelectrical circuits. We will see, how- ever, that even such amundane subject, when looked atinsufficient detail, can contain great complications We have already discussed some ofthe properties ofelectrical circuits inChapters23and25ofVol.1.Nowwewillcoversomeofthesamematerialagain,butin greater detail. Again wearegoing todeal only with linear systems and with voltages and currents which allvary sinusosdally; wecanthen represent allvoltages and currents bycomplex numbers, using theexponential notation described in Chapter 22ofVol. 1.Thus atime-varying voltage ¥() will bewritten VW) =Pe", ea where represents aeomplex number that isindependent of1.Itis,ofcourse, understood that theactual time-varying voltage V(1) isgiven bythereal part of thecomplex function ontheright-hand side oftheequation. 2 Similarly, allofour other time-varying quantities will betaken tovary sinusoidally atthesame frequency w.Sowewrite T= Te (current), 6=&e* (emf), (22.2) E=Ee (electric field), and soon. Most ofthetime wewillwrite ourequations interms ofV,I, ...(instead of imterms ofV,7,&...), remembering, though, that thetime variations areas given in(22.2).Inourearlierdiscussion ofcircuits weassumed that such things asinductances, 1 ‘capacitances, andresistances werefamiliartoyou.Wewantnowtolookinalittle=. more detail atwhat ismeant bythese idealized circuit elements. Webegin with the inductance. ‘Aninductance ismade bywinding many turns ofwire intheform ofacoil and bringing thetwo ends outtoterminals atsome distance from thecoil, asshown inFig. 22-1. Wewant toassume that themagnetic field produced bycurrents in V___ thecoildoes notspread outstrongly allover space andinteract with other parts of thecircuit. This isusually arranged bywinding thecoil inadoughnut-shaped form, orbyconfining themagnetic field bywinding thecoil onasuitable sron core, corbyplacing thecoil insome suitable metal box, asindicated schematically in Fig.22-1.Inanycase,weassume thatthere1sanegligible magnetic fieldinthe bexternal region near theterminals aand 6.Wearealso going toassume that we can neglect anyelectrical resistance inthewire ofthecoil. Finally. wewilassume that wecanneglect theamount ofelectrical charge that appears onthesurface of Fig. 22-1. Aninductance. awire inbuilding uptheelectric fields. With allthese approximations wehave what wecall an“ideal” inductance. (We will come back later and discuss what happens inareal inductance.) For an ideal inductance wesaythat thevoltage across theterminals isequal toL(dl/d). Let's seewhy that isso.When there isacurrent through theinductance, amagnetic field proportional tothecurrent isbuilt upinside thecoil. Ifthecurrent changes with time, themagnetic field also changes. Ingeneral, thecurl ofEisequal to —dB/dr; or,putdifferently, theline integral ofEalltheway around any closedpath1sequaltothenegativeoftherateofchangeofthefluxofBthroughtheloop Now suppose weconsider thefollowing path: Begin atterminal aand goalong thecoil (staying always inside thewire) toterminal 6;then return from terminal 6 toterminal athrough theairinthespace outside theinductance. The hine integral ofEaround this closed path can bewritten asthesum oftwo parts: . [Easfeaefeds. (223) ‘Aswehave seen before, there can benoelectric fields inside aperfect conductor. (The smallest fields would produce infinite currents.) Therefore theintegral fromatobviathecoil18zero.‘Thewholecontribution tothelineintegralofEcomesfrom thepath outside theinductance from terminal btoterminal a.Since wehave assumed that there arenomagnetic fields inthespace outside ofthe“box,” this part oftheintegral isindependent ofthepath chosen and wecandefine thepo- tentials ofthetwo terminals. The difference ofthese two potentials iswhat we callthevoltage difference, orsimply thevoltage V,sowehave Vm—[Beds=—§Bods. The complete line integral iswhat wehave before called theelectromotive force &and 1s,ofcourse, equal totherate ofchange ofthemagnetic flux inthe coil. We have seen earlier that this emfisequaltothenegativerateofchangeof n2 the current, sowehave at Va-5-249 where Listheinductance ofthe coil, Since d//di =ial, wehave V=iwLl. 22.4) ‘Thewaywehave described theideal inductance illustrates thegeneral approach toother ideal circuit elements—usually called “lumped” elements. The properties oftheelement aredescribed completely interms ofcurrents and voltages that appear atthe terminals. Bymaking suitable approximations, 1tispossible to ignorethegreatcomplextties ofthe fields that appear inside theobject. Aseparation ismade between what happens inside and what happens outside Forallthecircuit elements wewillfind arelation liketheoneinEq.(22.4), in whichthevoltage isproportional tothecurrent withaproportionality constant a° that is,ingeneral, acomplex number. This complex coefficient ofproportionality iscalledtheimpedance andisusually written asz(nottobeconfused withthe \z-coordinate). Its,ingeneral,afunctionofthe frequency w.Soforanylumped clementwewrite \ vie \ toate 2.5) vy Foraninductance,wehave /2(inductance) =2;,=iwl. 22.6) / Now let’s took atacapacitor from thesame point ofview.* Acapacitor con- sistsofapairofconducting platesfromwhichtwowiresarebrought outtosuitable r6terminals. Theplatesmaybeofanyshapewhatsoever, andareoftenseparated —_—_I bysomedielectric material. Weillustrate suchasituation schematically inFig, Fig:22-2. Acapacitor (orcom22-2. Again wemake several simphfying assumptions. Weassume thatthe gency), plates and thewires areperfect conductors. Wealso assume that theinsulation between theplates 1sperfect, sothat nocharges can flow across theinsulation from one plate totheother. Next, weassume that thetwo conductors areclose toeach other butfarfrom allothers, sothat allfield lines which leave one plate end upontheother. Then there arealways equal and opposite charges onthetwo plates and thecharges ontheplates aremuch larger than thecharges onthesur- faces ofthelead-in wires. Finally, weassume that there arenomagnetic fields close tothecapacitor. ~Suppose now weconsider theline integral ofEaroundaclosedloopwhich starts atterminal a,goes along inside thewire tothetop plate ofthecapacitor, jumps across thespace between theplates, passes from thelower plate toterminal»throughthewire.andreturnstoterminalainthespaceoutsidethecapacitor.Since there isnomagnetic field, theline integral ofEaround this closed path 1s zero. The integral can bebroken down into three parts: GEds= [Eeds+ fBeds+ f°Bods. 2.»oie “oles ourente The integral along thewires iszero, because there arenoelectric fields inside per-fectconductors. Theintegralfrom6toaoutsidethecapacitor 1sequaltothenega-tive ofthepotential difference between theterminals. Since weimagined that the two plates areinsome way isolated from therest oftheworld, thetotal charge on *There arepeople who say weshould call theobyeets bythenames “inductor” and “capacitor” and call thear propernes “inductance” and “capacitance” (byanalogy with “resistor” and “resistance”), We would rather use the words you wll hear inthe labora tory. Most people sull say“inductance” forboth thephysical coil and itsinductance L. ‘The word “capacitor” seems tohave caught on—although you will sull hear “condenser” fairly often—and most people stillprefer thesound of“capacity” to“capacitance.” - 23 thetwo plates must bezero; ifthere isacharge Qontheupper plate, there isan equal, opposite charge —Qonthelower plate. We have seen earlier that iftwo conductors have equal and opposite charges, plus and minus Q,the potential difference between theplates isequal toQ/C, where C1s called thecapacity ofthe two conductors. From Eq. (22.7) thepotential difference between theterminals aandbisequaltothepotential difference between theplates.Wehave,therefore, that -2ves: a, Theelectric current Jentering thecapacitor through terminal a(andleaving through terminal 6)isequal todQ/dr, therate ofchange oftheelectric charge on theplates. Writing dV/dt as1wV, wecan putthevoltage current relationship for acapacitor inthefollowing way: ‘ I \ 4 i=Z or L /VeBe (22.8) a Theimpedance zofacapacitor, isthenFd ; 1 2(capacitor) =20=Ze 22.9) Fig,22-3. Aresistor. ‘Thethird clement wewant toconsider isaresistor. However, since wehave notyetdiscussed theelectrical properties ofreal materials, wearenotyetready totalk about what happens inside areal conductor. Wewill just have toaccept asfact that electric fields can exist inside real materials, that these clectric fields ave rise toaflow ofelectric charge—that is,toacurrent—and that this current 1sproportional totheintegral oftheelectric field from one end oftheconductor totheother. Wethen imagine anideal resistor constructed asinthediagram of Fig.22-3. Two wires which wetake tobeperfect conductors gofrom theterminals aandbtothetwoendsofabarofresistive material. Following ourusuallineof argument, thepotential difference between theterminals aand 6isequal tothe luneintegral oftheexternal electric field, which isalso equal tothelineintegral of theelectric field through thebarofresistive material. Itthen follows that thecur- rent /through theresistor isproportional totheterminal voltage V: (9) (b) () @ I=ke 8 where R1scalled theresistance. We will seelater that therelation between the \ if currentandthevoltageforrealconductingmaterials1sonlyapproximately linear.qz{v L C Weill alsoscethatthisapproximate proportionality 1sexpected tobeindependenta] i ofthefrequencyofvariationofthecurrentandvoltageonlyifthefrequency18 nottoohigh,Foralternating currents then,thevoltageacrossaresistor1sinphase with thecurrent, which means that theimpedance isareal number. - uz= wt Tt R 2(resistance) =zy=R. (22.10) Fig. 22-4. The ideal lumped circuit Our results forthethree lumped circutt elements—the inductor, thecapacitor, elements (passive). and theresistor—are summarized inFig. 22-4. Inthis figure, aswell asinthe preceding ones, wehave indicated thevoltage byanarrow that 1sdirected from one terminal toanother. Ifthevoltage 1s“positive” —that1s,1ftheterminal a1sata Augher potential than theterminal b—the arrow indicates thedirection ofapositive “voltage drop.” Although wearetalking about alternating currents, wecanofcourse include thespecial case ofcircuits with steady currents bytaking thelimit asthefrequency. ©goes tozero. Forzero frequency—that is,forbc—the impedance ofaninduc- tancegoestozero;1tbecomes ashortcircuit.Forpc,theimpedance ofacondenser 24 g0es toinfinity; itbecomes anopen circuit. Since theimpedance ofaresistor is independent offrequency, it1stheonly element left when weanalyze acircuit forpe. Inthe eircut elements wehave deseribed sofar, thecurrent and voltage are proportional toeach other. Ifoneis ero, soalso istheother. Weusually think in terms likethese: Anapplied voltage is“responsible” forthecurrent, oracurrent “ives riseto”avoltage across theterminals; soinasense theelements “respond” tothe“applied” external conditions. For this reason these elements arecalled passive elements, They canthus becontrasted with theactive elements, such as thegenerators wewill consider inthenext section, which arethesources ofthe oscillating currents orvoltages inacircuit L — 22-2Generators = Now wewant totalk about anactive circuit element—one that 1sasource ofthecurrentsandvoltagesinacircuit—namely, agenerator. (ss y Suppose that wehave acoil like aninductance except that sthas very fewturns,sothatwemayneglectthemagneticfieldofitsowncurrent,Thiscoil, (Ss however, sitsinachanging magnetic field such asmight beproduced byarotating‘magnet,assketchedinFig.22-5.(Wehaveseenearlierthatsucharotatingmag- ( netic feld canalso beproduced byasurtable setofeoils with alternating currents.) o ‘Again wemust make several simplifying assumptions. The assumptions wewill makearealltheonesthatwedescribedfortheaseoftheinductance. Inparticular, weassume thatthevarying magnetic fieldisrestricted toadefinite region inthe 5. A-generator consivicinityofthecouanddoesnotappearoutsidethegeneratormthespacebetween ggylyecilend-»Saratngmasneteele the terminals. Following closely theanalysis wemade fortheinductance, weconsider the lineintegral ofEaround acomplete oop that starts atterminal a,goes through the coil toterminal band returns toslsstarting point inthespace between thetwo terminals. Again weconclude that thepotential difference between theterminals, isequal tothetotal line integral ofEaround theloop: Ve~fed P Thislineintegralisequaltotheemfinthecircuit,sothepotentialdifference \\across theterminals ofthegenerator isalso equal totherate ofchange ofthemag- v eticfluxlinkingthecoil /)ve-8=4aun, 2.)q > Foranidealgenerator weassume thathemagnetic fluxlinking thecolisdete- gx.90.4 Symbol foronidealgemmined byexternal conditions—such astheangular velocity ofarotating magneue 4,4 22% field—and isnot influenced anany way bythecurrents through thegenerator. Thus agenerator—at least the ideal generator weare considering—1s not an impedance, The potential difference across itsterminals isdetermined bythe arbitcarly assigned electromotive force &(). Such anideal generator srepresentedbythesymbolshowninFig,22-6,Thelittearrowrepresents thedireetionofthe emf when 1is positive, Aposttive emf inthe generator ofFig. 22-6 will produce avvoltage V=&,with theterminal aatahigher potential than the terminal b. ‘There isanother way tomake agenerator which isquite different onthe inside bytwhich isindistinguishable fromtheonewehavejustdescribed insofar aswhat’happens beyond itsterminals. Suppose wehave acoil ofwire 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 coil are made 10theoutside world bymeans ofsliding contacts or“slip rings.” Again, weareinterested inthepotential difference that appears across thetwo terminals ns |aca Fig.22-7. Agenerator consisting of » @coil rotating inafixed magnetic field. aand 6,which isofcourse theintegral oftheelectric field from terminal atoter- minal 6along apath outside thegenerator. Now inthesystem ofFig. 22-7 there arenochanging magnetic fields, sowe might atfirst wonder how any voltage could appear atthegenerator terminals Infact, there arenoelectric fields anywhere inside thegenerator. Weare, asusual, assuming forourideal elements that thewires inside aremade ofaperfectly con ducting material, and aswehave said many times, theelectric field inside aperfect conductor isequal tozero. Butthat 1snottrue. Itisnottrue when aconductor ismoving inamagnetic field.Thetruestatement isthatthetotalforceonany. charge inside aperfect conductor must bezero. Otherwise there would beaninfiniteflowofthe freecharges. Sowhat isalways true isthat thesum ofthe electric field Eand thecross product ofthevelocity oftheconductor and themagnetic field. B—which 1sthe total force onaunit charge—must have the value zero inside the conductor: F=E+vX B=0 (inaperfect conductor), 22.12) where vrepresents thevelocity oftheconductor. Our earher statement that there isnoelectric field inside aperfect conductor 1sallright ifthevelocity vofthe conductor iszero; otherwise thecorrect statement isgiven byEq. (22.12). Returning toour generator ofFig. 22-7, wenow seethat theline integral of theelectric field Efrom terminal gtoterminal 6through theconducting path of thegenerator mustbeequaltothelineintegral ofvXBonthesamepath, a ‘ foBaa=[0xByes (22.13) insidesonido Itisstill true, however, that theline integral ofEaround acomplete loop, including thereturn from 6toaoutside thegenerator, must bezero, because there areno changing magnetic fields. Sothefirst integral inEq. (22.13) 1salso equal toV, thevoltage between thetwo terminals. Itturns out that theright-hand integral ofEq.(2213)1sjusttherateofchangeofthefluxlinkage through thecoilandis therefore—by theflux rule—equal totheemf inthecoil. Sowehave again that thepotential difference across theterminals 1sequal totheelectromotive force in thecircuit, inagreement with Eq.(22.11). Sowhetherwehaveageneratorinwhich ‘amagnetic field changes near afixed coil, orone inwhich acoil moves inafixed magnetic field, theexternal properties ofthegenerators arethesame. There isa voltage difference Vacross theterminals, which isindependent ofthecurrent in the circuit but depends only onthe arbitrarily assigned conditions inside the generator. Solong aswearetrying tounderstand theoperation ofgenerators from the point ofview ofMaxwell’s equations, wemight also askabout theordinary chemi- calcell, like aflashlight battery It1salso agenerator, 1.€., avoltage source, al- though itwill ofcourse only appear inpccircuits. The simplest kind ofcellto understand 1sshown inFig. 22-8. Weimagine two metal plates immersed insome 2-6 chemical solution. Wesuppose that thesolution contains positive and negative ions. Wesuppose also that onekind ofion,saythenegative, ismuch heavier than theoneofopposite polarity, sothat itsmotion through thesolution bytheprocessofdiffusion ismuchslower.Wesupposenextthatbysomemeansorotheritisarranged that theconcentration ofthesolution ismade tovary from onepart of theliquid totheother, sothat thenumber ofions ofboth polarities near, say, the lower plate ismuch larger than theconcentration ofions near theupper plate. a Becauseoftheirrapidmobilitythepositiveionswilldriftmorereadilyintothei lsregionoflowerconcentration, sothattherewillbeaslightexcessofpositive charge ——arriving attheupperplate. Theupperplatewillbecome positively charged and | |\ thelowerplatewillhaveanetnegative charge. | |\Asmoreandmorecharges diffusetotheupperplate.thepotential ofthis plate | | \ willriseuntil theresulting electric fieldbetween theplates produces forces onthe | te yt v ionswhich justcompensate fortheir excess mobility, sothetwoplates ofthecell Steiett J quicklyreachapotentialdifferencewhichischaracteristicoftheinternalcon-|a ||struction. | |7Arguing justaswedidfortheidealcapacitor, weseethatthepotential differ- | —————;, encebetween theterminals @and6isjustequaltothelineintegral oftheelectric | ’ _]fieldbetweenthetwoplateswhenthereisnolongeranynetdiffusionoftheions.§—§(|_______—There 1s,ofcourse, anessential difference between acapacitor andsuch achemical , ,it Fig.22-8.Achemicalcell. cell. Ifweshort-circuit theterminals ofacondenser foramoment, thecapacitor isdischarged and there isnolonger anypotential difference across theterminals. Inthe case ofthe chemical cell acurrent can bedrawn from the terminals con- tinuously without anychange intheemf—until, ofcourse, thechemicals inside thecell have been used up. Inareal cell itisfound that thepotential difference across the terminals decreases asthe current drawn from the cell increases. In keeping with theabstractions wehave been making, however, wemay imagine an ideal cell inwhich thevoltage across theterminals isindependent ofthecurrent. ‘Areal cell can then belooked atasan ideal cell inseries with aresistor. 22-3 Networks ofideal elements; Kirchhoff’s rules Aswehave seen inthelastsection, thedescription ofanideal circuit element interms ofwhat happens outside theelement isquite simple. The current and thevoltagearelinearlyrelated.Butwhatisactuallyhappeninginsidetheelement a[a|bisquite complicated, and itisquite difficult togiveaprecisedescriptionintermsof ~ ‘Maxwell’sequations. Imaginetryingtogiveaprecisedescription oftheelectric Nees A andmagnetic fields oftheinside ofaradio which contains hundreds ofresistors, » | capacitors, andinductors. Itwould beanimpossible tasktoanalyze suchathing v vfbyusingMaxwell's equations. Butbymakingthemanyapproximations wehave z/ 3! described inSection 22-2andsummarizing theessential features ofthereal ) \circuit elements intermsofidealizations, itbecomes possible toanalyze anelec~ \ tricalcircuit inarelatively straightforward way. Wewillnowshowhowthat an \ isdone. tv, a Suppose wehaveacircuit consisting ofagenerator andseveral impedances 1 aconnected together, asshown inFig.22-9. Accordingtoourapproximations there y27 » isnomagneticfieldintheregionoutsidetheindividual circuitelements. Therefore . thelineintegral ofEaround anycurve which doesnotpassthrough anyofthe \ elements 1szero.Consider thenthecurveIshown bythebroken linewhichgoes z1¥, *allthewayaroundthecircuitinFig.22-9,Thelineintegral of£aroundthiscurve «lie \ ismadeupofseveral pieces. Each piece isthelineintegral from oneterminal ofa Yscircuitelementtotheother.Thislineintegralwehavecalledthevoltagedrop wa==>across thecircuit element. Thecomplete lineintegral isthenjustthesumofthe 4voltage drops across alloftheelements inthecircuit: ¢E-ds=Vp. Fig.22-9. Thesumofthevoltagedrops around any closed path iszero. Since theline integral iszero, wehave that thesum ofthepotential differences 21 around acomplete loop ofacircuit isequal tozero: Dhao 22.14) shy ioe This result follows from one ofMaxwell's equations—that inaregion where there ‘arenomagnetic fields theline integral ofEaround any complete loop iszero. ° A ¢ 4 Suppose weconsider now acircuit like that shown inFig. 22-10. The hori- zontallinejoining theterminals a,b,¢,anddisintended toshowthattheseter- tr,Meits414jinalsareallconnected,orthattheyarejoinedbywiresofneghgibleresistance. [Inany case, thedrawing means that terminals a,b,¢,and dareallatthesame v(ef Zz, zy potential and,similarly, thattheterminals e,f,g,andAarealsoatonecommon potential. ThenthevoltagedropVacrosseachofthefourelements isthesame. \Nowoneofoursdealizations hasbeenthatnegligible electrical chargesac- \[tn He [Hs|4cumulate ontheterminals oftheimpedances. Wenowassumefurtherthatany rr % electrical charges onthewires joining terminals canalso beneglected. Then the conservation ofchargerequiresthatanychargewhichleavesonecircuitelement gqfide22-10. Thesumofthecurrents immediately enterssomeothercircuttelement. Or,whatisthesamething,we intoonynodeiszero, require thatthealgebraic sumofthecurrents which enteranygivenjunction must bezero, Byajunction, ofcourse, wemean any setofterminals such asa,b,¢, and dwhich areconnected. Such asetofconnected terminals 1susually called a “node.” The conservation ofcharge then requires that forthecircuit ofFig. 22-10, ho h-b-h=0. (2.15) ‘The sum ofthecurrents entering thenode which consists ofthefour terminals e.f,8, and hmust also bezero: “ht htht+h=0. (22.16) _Thisis,ofcourse,thesameasEq.(22.15).Thetwoequationsarenotindependent. sta} —S{=}§Thegeneralrule1sthattheswmofthecurrentsintoanynodemustbezero. P122 7 - DLn=0. (22.17) ats ot]ay atte: t : Oureartier conclusion thatthesumofthevoltage drops around aclosed loop a & 1szero must apply toanyloop inacomplicated circuit. Also, ourresult thattheq 4,\__1}sumofthecurrentsintoanodeiszeromustbetrueforanynode.Thesetwoequa-= tionsareknownasKirchhoff’srules.Withthesetworulesitispossibletosolvefor |thecurrents and voltages inany network whatever. | Suppose weconsider themore complicated circuit ofFig. 22-11. How shall | Tf|2s 44]2)wefindthecurrents andvoltages inthiscircuit? Wecanfindtheminthefollowing | L straightforward way. Weconsider separately each ofthefoursubsidiary closed y loops which appear inthecircuit. (For instance, oneloop goes from terminal ato dorm. terminalbtoterminaletoterminaldandbacktotermunala.)Foreachoftheloops wa wewrite theequation forthefirstofKirchhoff’s rules—that thesum ofthevoltages — around each loop 1sequal tozero, Wemust remember tocount thevoltage drop Fig. 22-11. Analyzing acircuit with aspositive ifwearegoing inthedirection ofthecurrent and negative ifweare Kirchhof's rules. going across anelement inthedirection opposite tothecurrent; andwemust remember that thevoltage drop across agenerator isthenegarive oftheemf in that direction. Thus ifweconsider thesmall loop that starts and ends atterminal awe have theequation zal +zals +24h —£1=0. Applying thesame ruletotheremaining loops, wewould getthree more equations ofthe same kind. Next,wemustwritethecurrentequationforeachofthenodesintheeircut. Forexample,summing thecurrentsintothenodeatterminalbgivestheequation ha-h-h=0. 2-8 Similarly, forthenode labeled ¢wewould have thecurrent equation Ia— Ik+Ig—Tp=0. Forthecircuit shown there arefivesuch current equations. Itturns out, however, that any one ofthese equations can bederived from theother four; there are, therefore, only four independent current equations. Wethus have atotal ofeight independent, linear equations: the four voltage equations and the four current equations. With these eight equations wecansolve fortheeight unknown currents. Once thecurrents areknown thecircuit issolved. ‘The voltage drop across any clement isgiven bythecurrent through that element times itsimpedance (or,in thecase ofthevoltage sources, itisalready known). Wehave seen that when wewrite thecurrent equations, wegetone equation which isnotindependent oftheothers. Generally itisalso possible towrite down toomany voltage equations. For example, inthecircuit ofFig. 22-11, although wwehave considered only thefour small loops, there arealarge number ofother loops forwhich wecould write thevoltage equation. There 1s,forexample, the loop along the path abefeda. There isanother loop which follows the pathser Youcarseat nea maynop. anevseg compres Sr —_——cuitsitisveryeasytogettoomanyequations.Thereareruleswhichtellushowtoiz |proceed sothat only theminimum number ofequations iswritten down, but ]usuallywithalittlethoughtitispossibletoseehowtogettherightnumberof|22 ul}|z%equations inthesimplest form. Besides, writing anextra equation ortwo doesn’t doany harm. They will notlead toany wrong answers, only perhaps alittle unnecessary algebra. | InChapter 25ofVol. Iweshowed that ifthetwo impedances 2,and 2»are —inseries,theyareequivalent toasingleimpedance z,givenby | neatz2 (22.18)‘G)E*}a| Wealso showed that ifthetwo impedances areconnected inparallel, they are equivalent tothesingle impedance z,given by 1 2122 i , *»=(yay+We)i+7 ee ameey conte Ifyou lookbackyouwillseethatinderiving theseresults wewereineffectmaking Combinations. useofKirchhoff's rules. Itisoften possible toanalyze acomplicated circuit by repeated application oftheformulas forseries and parallel impedances. For in- stance, thecircuit ofFig. 22-12 can beanalyzed that way. First, theimpedances 2,and 25can bereplaced bytheir parallel equivalent, and soalso can zoand 7. Then theimpedance z»can becombined with theparallel equivalent ofz»and 27 bytheseries rule. Proceeding inthis way, thewhole circuit can bereduced toa generator inseries with asingle impedance Z.‘The current through thegenerator isthen just /Z. ‘Then byworking backward one can solve forthecurrents in each oftheimpedances. There are, however, quite simple circuits which cannot beanalyzed bythis method, asforexample thecircuit ofFig. 22-13. Toanalyze this circuit wemust 1©4+©+ aa|mm Fig. 22-13. Acircuit that cannot be conalyzed interms ofseries and parollel J € ' combinations. 29 write down thecurrent andvoltage equations from Kirchhoff’s rules. Let’s doit. There 1sjust onecurrent equation: htht+h=0 soweknow immediately that Ip=—h +1). Wecan save ourselves some algebra ifweimmediately make useofthis result in writing thevoltage equations. For this circuit there aretwo independent voltage equations; they are —8) +hte —hz, =0 and 82—Uh+Indes —Inte =0. There aretwo equations and two unknown currents. Solving these equations for aha TyandIs,weget a 2282 —(22+za)€1f=238 Get a 22.20)\=2iGe¥2a)+Z2%3 2.20) and | =rat 2081ANVYGS 2Seth)+ah e220) 5) 25 ‘Thethirdcurrent isobtained fromthesumofthesetwo. 1 Another example ofacircuit that cannot beanalyzed byusing therules forz seriesandparallelimpedance isshowninFig.22-14.Suchacircuttiscalleda“bridge.” Itappears inmany instruments used formeasuring impedances. With such acircuit one isusually interested inthequestion: How must thevarious . impedances berelated ifthecurrent through theimpedance zistobezero? We—— leave itforyou tofind theconditions forwhich this isso. Fig. 22-14. Abridge circuit, 22-4 Equivalent circuits Supposeweconnectagenerator &toacircuitcontaining somecomplicated1 interconnection ofimpedances, asindicated schematically inFig.22-15(a). All +e oftheequations wegetfrom Kirchhoff’s rules arelinear, sowhen wesolve them forthecurrent Jthrough thegenerator, wewill getthat /isproportional to&. Any Wecan write () v|circuit 8of Toae Z's wherenow2,issomecomplexnumber,analgebraicfunctionofalltheelements cy inthecircuit.(Ifthecircuitcontainsnogenerators otherthantheoneshown,thereisnoadditional term independent of&.)Butthisequation isjustwhat wewould t writeforthecircuit ofFig.22-15(b). Solongasweareinterested onlyinwhatog happens sotheleftofthetwo terminals aand b,thetwo circuits ofFig. 22-15 are equivalent, Wecan, therefore, make thegeneral statement that anytwo-terminal network ofpassive elements can bereplaced byasingle impedance z.. withoutrn(te) changingthecurrentsandvoltagesintherestofthecircuit.Thisstatement1s,of course, just aremark about what comes outofKirchhoff’s rules—and ultimately from thelinearity ofMaxwell’s equations. Theideacanbegeneralized toacircuitthatcontainsgeneratorsaswellas 6 impedances. Suppose welook atsuch acircust “from thepoint ofview” ofoneof fig. 22. ;theimpedances, which wewillcallz,,asinFig. 22-16(a). Ifwewere tosolve the wane15.aytwoterminalnetequationforthewholecircuit,wewouldfindthatthevoltageV,,betweenthetwohehe ecaenents #eavivelent 10terminals aandbisalinearfunction ofJ,whichwecanwriteconeffectiveimpedance. Vi=A- Bh, (22.22) where 4and Bdepend onthegenerators and impedances inthecircutt totheleft 2-10 oftheterminals. For instance, forthecircuit ofFig. 22-13,wefindVy=iz) 1, Thiscanbewritten(byrearranging Eq.(22.20)] as any at 22 2379 reutt"-[GFe)&-%]- neal 223) wae|/ond €'s ‘Thecomplete solution isthenobtained bycombining thisequation withtheone (0) Yn fortheimpedance 2,namely,V;=1,21,orinthegeneralcase,bycombining \ Eq,(22.22) with Va =Intn- $ Ifnow weconsider that 2,isattached toasimple series circuit ofagenerator andacurrent, asinFig.22-15(b), theequation corresponding toEq.(22.22) is oh Vu=boi ~Inzetts Which isidentical toEq.(22.22) provided weset8.=Aand244=B.Soifwe areinterested only inwhat happens 10theright oftheterminals aand 6,thearbi- trarycircuit ofFig.22-16 canalways bereplaced byanequivalent combination of (gy 4generator inseries with animpedance. 22-5Energy (&) Wehave seen that tobuild upthecurrent /inaninductance, theenergy U=4L2? must beprovided bytheexternal circuit, When thecurrent falls back > tozero, thisenergy isdelivered backtotheexternal circuit. There isoenergy-I0% Fig.29-16, Anytworterminol net mechanism inanideal inductance. When there isanalternating current through workconbereplaced by@generator in aninductance, energy flows back andforth between itandtherestofthecircutt, series withonimpedance butthearerage rateatwhich energy isdelivered tothecircuit iszero. Wesaythat aninductance isanondissipative element; noelectrical energy isdissipated—that 1s, “ost"—in it, Similarly, theenergy ofacondenser, U=}CV?,isreturnedtotheexternal circuit when acondenser 1sdischarged. When acondenser isinanAccircuit energy flows inand outof1,butthenetenergy flow ineach cycle iszero. Anideal condenser isalso anondissipative element, Weknow that anemf isasource ofenergy. When acurrent Jflows inthe direction oftheemf, energy isdelivered totheexternal circuit attherate dU/dt = 81. ICcurrent isdriven agaist theemf—by other generators inthecircuit—theemfwillabsorbenergyattherate&f;since1snegative,dU/drwillalsobenegativeIfagenerator 1connected toaresistor R,thecurrent through theresistor is=6/R. The energy being supplied bythegenerator attherate &/1sbeing absorbed bytheresistor. This energy goes into heat intheresistor and islost from theelectrical energy ofthecircuit. Wesaythat electrical energy isdissipated inaresistor. Therateatwhich energy isdissipated inaresistor isdU/dt =RIP. InanACcircuit theaverage rate ofenergy lost toaresistor 1stheaverage of RP?over onecycle. Since |=Je™'—by which wereally mean that [varies as ‘coswot—the average of?over onecycle 1s|/|?/2, since thepeak current is|/|and theaverage ofcos?wtis1/2. R What about theenergy loss when agenerator isconnected toanarbitrary impedance 2?(By“loss” wemean, ofcourse, conversion ofelectrical energy into = thermal energy.) Any impedance 2can bewritten asthesum ofitsreal and im-= ginary parts, ‘That is, Z= RIX, (22.24) where Rand Xarerealnumbers. From thepoint ofview ofequivalent circuits we can say that any impedance isequivalent toaresistance inseries with apure imaginary impedance—called areactance—uas shown inFig. 22-17, ‘Wehave seen earlier that anycircuit that contains only L’sand C’shasan Fig. 22-17. Anyimpedonce isequiv-impedance that1sapureimaginary number.Sincethereisnoenergylossintoanyalentto@seriescombination of«pureoftheL'sandC'sontheaverage, apure reactance containing only L'sandC's Fesistance andapure reactance.willhavenoenergyloss.Wecanseethatthismustbetrueingeneralforareactance. zu Ifa generator with theemf&isconnectedtotheimpedance=ofFig.22-17, theemf must berelated tothecurrent Jfrom thegenerator by 8=UR +iX). 22.25) Tofind theaverage rate atwhich energy isdelivered, wewant theaverage ofthe product &/. Now wemust becareful. When dealing with such products, wemustdealwiththerealquantities &(1)and1().(Therealpartsofthe complex functions will represent theactual physical quantities only when wehave /inear equations: now weareconcerned with products, which arecertainly notlinear.) Suppose wechoose our origin of sothat theamplitude /isareal number, let’s sayZp;then theactual time variation Isgiven by T= [pcos at. The emf ofEq. (22.25) isthereal part of Toe“"(R +1X) or 8=[pcos wt—IoX sinat. (22.26) The two terms inEq. (22.26) represent thevoltage drops across Rand X inFig. 22-17. Weseethat thevoltage drop across theresistance isnphase with ten, Q thecurrent, whilethevoltage dropacrossthepurely reactive part1soutofphase“7 withthecurrent, flaBlyearn Theaveragerateofenergy1085,(P)..,fromthegenerator1stheintegralof vg theproduct&/overonecycledividedbytheperiod7;inotherwords, ny2 if? UL” pcos? If" p .oi. ¢ Pov=pf,lar=7)Geos"ardr—7JWXcosetsinerdh ~ Thefirstintegral is4/$R,andthesecondintegral iszero.Sotheaverage o=o energy lossinanimpedance z=R+iXdepends onlyontherealpartof2, andis3R/2, which 1sinagreement with ourearlier result fortheenergy lossina resistor. There 18noenergy loss inthereactive part. 22-6 Aladder network =m yao By Wewould likenowtoconsider aninteresting circuit which canbeanalyzed a e interms ofseries andparallel combinations. Suppose westart withthecircuit of Fig. 22-18(a). Wecanseeright away that theimpedance from terminal atoter- ebek menen minalbissimply21+zz.Nowlet’stakealittlehardercircunt,theoneshownin anFig. 22-18(b). We could analyze this circuit using Kirchhoff’s rules, but it1s Fig. 22-18. Theeffective impedance also easy tohandle with series and parallel combinations. Wecan replace the ofaladder. two impedances ontheright-hand end byasingle impedance z3=21+22,as inpart (¢)ofthefigure. Then thetwo impedances 2and 24can bereplaced by their equivalent parallel impedance z,,asshown inpart (d)ofthefigure. Finally, 2,and 2,areequivalent toasingle impedance zs,asshown inpart (¢). Now wemay askanamusing question: What would happen ifinthenetwork ofFig. 22-18(b) wekept onadding more sections forever—as weindicate bythe dashed linesinFig.22-19(a)? Canwesolvesuchaninfinitenetwork? Well,that’s a 8 ¢ 8 (0) | al ete, o Fabl= (b) ih b, a b b Fig. 22-19. The effective impedance ofaninfinite ladder. 22 notsohard. First, wenotice that such aninfinite network 1sunchanged ifweadd one more section atthe“front” end. Surely, ifweadd one more section toan infinite network itisstillthesame infinite network. Suppose wecalltheimpedance between thetwo terminals aandbofthe infinite network zo;then theimpedance of allthestuff totheright ofthetwo terminals cand disalso zy.Therefore, sofaras thefront endisconcerned, wecanrepresent thenetwork asshown inFig.22-19(b). Combining theparallel combinations 2229 and adding theresult inseries with 24, wecan immediately write down theimpedance ofthis combination: 1wee 2220 sa+oEyedE % FTE EE Butthis impedance isalso equal tozo,sowehave theequation - 220pratee Wecan solve forzytoget 20= B+VGA) a. 2.27) ‘Sowehave found thesolution fortheimpedance ofaninfinite ladder ofrepeated series and parallel impedances. The impedance zoiscalled the characteristic ‘impedance ofsuch aninfinite network. enh L u Let’snowconsideraspecificexampleinwhichtheserieselementisanin-hosel hadeshidden MOductance Landtheshunt element isacapacitance C,asshown inFig.22-20(a). (o c c CceteInthiseasewefindtheimpedanceoftheinfinitenetworkbysetting2=ww=2-1 aandz2=1/uwC.Noticethatthefirstterm,2/2,inEq.(22.27)isjustone-half La Leuntheimpedance ofthefirstelement. Itwouldtherefore seemmorenatural, orat gsngfe ROK.leastsomewhat simpler, ifweweretodrawourinfinitenetwork asshowninFig. Tr22-20(b). Looking attheinfinite network fromtheterminal a’wewould seethe (® ad cotecharacteristic impedance lL... zo=VEO) —@L4). (22.28) Fig.22-20. AnL-Cladder drawn ‘Nowtherearetwointeresting cases, depending onthefrequency w.Ifw"isless '"twoequivalent ways, than 4/LC, the second term inthe radical will besmaller than the first, and the impedance z»willbearealnumber. Ontheother hand, ifw?1sgreater than 4/LC theimpedance zowill beapure imaginary number which wecanwrite as 2=IVE) —C/O) Wehave said earlier that acircut which contains only imaginary impedances, such asinductances and capacitances, will have animpedance which ispurely imaginary. How can1tbethen thatforthecircuit wearenow studying—which has onlyL’sandC’s—the impedance isapureresistance forfrequencies below V/4/LC? For higher frequencies theimpedance ispurely imaginary, inagreement with our earlier statement. For lower frequencies theimpedance 1sapure resistance and willtherefore absorb energy. Buthow canthecircuit continuously absorb energy, asaresistance does, ifitismade only ofinductances andcapacitances? Answer: Because there isaninfinite number ofinductances and capacitances, sothat when ‘asource isconnected tothecircuit, itsupplies energy tothefirst inductance and capacitance, then tothesecond, tothethird, and soon. Ina circuit ofthis kind, energy iscontinually absorbed from the generator ataconstant rate and flows constantly out into thenetwork, supplying energy which isstored intheinduc- tances and capacitances down theline. This idea suggests aninteresting point about what ishappening inthecircut.Wewouldexpectthatifweconnectasourcetothefrontend,theeffectsofthissource will bepropagated through thenetwork toward the infinite end. The propagation ofthewaves down theline ismuch like theradiation 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 \/47LC, wewould notexpect toseeany such propagation. 243 22-7 Filters Wesaw inthelast section that theinfinite ladder network ofFig. 22-20 absorbs. energy continuously ifitisdriven atafrequency below acertain critical frequency V4/LC, whichwewillcallthecutofffrequency wo.Wesuggested thatthiseffect could beunderstood interms ofacontinuous transportofenergydowntheline. ‘Ontheother hand, athigh frequencies, forw>«wo,there isnocontinuous ab- sorption ofenergy; weshould then expect that perhaps thecurrents don’t “pene- trate” very fardown theline. Let's seewhether these ideas areright. ‘Suppose wehave thefront endoftheladder connected tosome Acgenerator and weask what thevoltage looks like at,say, the754th section oftheladder, Since thenetwork isinfinite, whatever happens tothevoltage from one section to thenext 1salways thesame; solet’sjust look atwhat happens when wegofrom some section, saythenthtothenext. Wewill define thecurrents /,,and voltages V,,asshown inFig. 22~21(a). hook Ow ty Tye vy V, — —_ ~eo ee “2“OHBH——iBhd Fig. 22-21. Finding hepropagotion fecor ofaladder. Wecan getthevoltage V,.,;from V,,byremembering that wecanalways replace therestoftheladderafterthenthsection byitscharacteristic impedance zy; then weneed only analyze thecircuit ofFig. 22-21(b). First, wenotice that any Vu,Since itisacross zo,must equal J,z9. Also, thedifference between V,,and Vj.4 1 isjust Jn242 . Vn—Vag=Int=VanZo Sowegettheratio Vous oy2 font,Va Ze” Eo Wecancallthis ratio thepropagation factor foronesection oftheladder; we'll call it. Itis, ofcourse, thesame forallsections: a= EH. (22.29) The voltage after thenthsection isthen V,=a6. (22.30) You can now find thevoltage after 754 sections: itisjust «(othe754th power times 6. Supposeweseewhata1slikefortheL-CladderofFig.22-20(a) Usingzy from Eq. (22.27), and 2,=iwL, weget a=YUO) =(oil?4)—Hol/2) 22.31) VLIC) —(L7/4) +(wl /2) Ifthedriving frequency isbelow thecutoff frequency wy=\/4/LC, theradical isareal number, and themagnitudes ofthecomplex numbers inthenumerator and denominator areequal. Therefore, themagnitude ofaisone; wecan write whichmeans thatthemagnitude ofthevoltage isthesameateverysection, only a itsphasechanges.Thephasechange81s,infact,anegativenumberandrepresentsthe“delay” ofthevoltage asitpasses along thenetwork. For frequencies above thecutofl frequency wyit1sbetter toFactor out an1 from thenumerator and denominator ofEq. (2231)and rewrite itas a-VELD=WO~(wl2) ayVW) —(LC)+(ol/2) ' | Thepropagation factoraisnowarealnumber, andanumber lessthurtone.That | mieans thatthevoltage atanysection isalways lessthan thevoltage atthepre- ! ceding section bythefactor aForanyfrequency above wy,thevoltage dies de o away rapidly aswegoalong thenetwork. Aplot oftheabsolute value of ay« function offrequency looks likethegraph inFig. 22-22. Fig 22-22. Thepropagation foctor Weseethat thebehavior ofa,both above and below wy,agrees with our of@section ofanL-Cladder interpretation that thenetwork propagates energy for@<wyand blocks itfor ©>wy. Wesaythat thenetwork “passes” low frequencies and “rejects” oF “filters out” thehigh frequencies. Any network designed tohave itscharacteristics varymaprescribed waywith frequency 1scalled a“filter.” Wehave been analyzing coe c c a“low-pass filter.”~ You may bewondering why allthis discussion ofaninfinite network which ‘obviously cannotactually occur. Thepomtisthatthesamecharacteristics are iL iL L Lfoundinafinitenetworkifwefinishstoffattheendwithanimpedence equalivthecharacteristic impedence zy. Now inpractice it1snotpossible toexucily ~ reproduce thecharacteristic impedance with afewsimple elements—Itke R's (@) L's,and C’s. Butitisoften possible todosowithaFarrapproximation foracertain rangeoffrequencies. Inthiswayonecanmakeafinitefilternetworkwhose Japroperties arevery nearly thesame asthose fortheinfinite case For instance, the L-C ladder behaves much aswehave described 1tifit1sterminated inthepure resistance R=y'L/C.IfinourL-Cladderweinterchange thepositionsofthe L’s and C's, tomake theladdershowninFig.22-23(a), wecanhaveafilterthatpropagates ughfre-quencies andrejects fowfrequencies. Itiseasy toseewhat happens with thisnet- work byusing theresultswealreadyhave.Youwillnoticethatwhenever wechange Ol Tay Te anLtoaCand viceversa, wealsochange every 1to1/ie. Sowhatever happenei| ib) atwbefore willnow happen at1/«. Inparticular, wecanseehow awillvary with frequency byusing Fig. 22-22 and changing thelabel ontheaxis toI/aa, aswe ‘Fig. 22-23. (a)Ahigh-pass filter; havedone inFig.22-23(b). (b)itspropagation factor as@function Thelow-pass andhigh-pass filters wehave described have various technical fT applications. AnL-C low-pass filter isoften used asa“smoothing” filterinaDc power supply. Ifwewant tomanufacture Dcpower from anAcsource, webegin with arectifier which permits current toflow only inone direction, From the reclifier wegetaseries ofpulses that look like the function ¥(1) shown in Fig22-24, which 1slousy DC,because itwobbles upand down. Suppose wewould like anice pure pc,such asabattery provides. Wecan come close tothat by putting alow-pass filter between therectifier and theload. Weknow from Chapter$0ofVol.IthatthetimefunctioninFig.22-24canbe represented asasuperposition ofaconstantvoltageplusasinewave,plusahigher~ frequency sine wave, plus astill higher-frequency sine wave, etc.—by aFourier series. Ifourfilter islinear (if,aswehave been assuming, theL’sandC’sdon't 4yi vary with thecurrents orvoltages) then what comes outofthefilter 18thesuper- ; _ - positionoftheoutputsforeachcomponentattheinput.Ifwearrangethatthely>4\fo\v cutoff frequency wyofourfilter 1swell below thelowest frequency inthefunction Yo V(),theDe(forwhichw=0)goesthrough fine,buttheamplitude ofthefirst iu ‘harmonte willbecutdownalot.Andamplitudes ofthehigherharmonies willbecutdown even more. Sowecangettheoutput assmooth aswewish, depending _Fig.22-24. Theoutput voltage ofa onlyonhow many filter sections wearewilling tobuy. full-wave rectifier. Ahigh-pass filter isused ifone wants (oreject certain low frequencies. For instance, inaphonograph amplifier ahigh-pass filter may beused toletthemuste 221s through, while keeping out thelow-pitched rumbling from themotor ofthe turntable, Itisalso possible tomake “band-pass” filters that reject frequencies below some frequency w;andabove another frequency ws(greater than w), butpass the frequencies between «;and w2. This canbedone simply byputting together a high-passandalow-passfilter,butitismoreusuallydonebymakingaladderin { which theimpedances z,andzyaremore complicated—-being each acombination| HRA ofL'sandC’s.Suchaband-pass filtermighthaveapropagation constant like | ity thatshown inFig.22-25(a). Itmight beused, forexample, inseparating signalsL Mia thatoccupyonlyanintervaloffrequencies, suchaseachofthemanyvoicechannels —}++———5+ inahigh-frequency telephone cable, orthemodulated carrier ofaradio trans- rit mission. {it WehaveseeninChapter 25ofVol.Ithatsuchfilteringcanalsobedoneusing oJi\ theselectivity ofanordinary resonance curve,whichwehavedrawn forcomparisonao 1mFig.22-25(b). Buttheresonant filter 1snotasgood forsome purposes asthe La =____ band-pass filter. Youwillremember (Chapter 48,Vol.1)thatwhenacarrier of ae = frequency «ismodulated with a“signal” frequency «,,thetotal signal contains notonly thecarrier frequency butalso thetwo side-band frequencies w,+0 Fig. 22-25. (a)Abond-poss filter. andw,—«,.Witharesonantfilter,theseside-bands arealwaysattentuated some (b)Asimple resonant filter. what, andtheattenuation ismore, thehigher thesignal frequency, asyoucansee from thefigure. Sothere isapoor “frequency response.” The higher musical tones don’t getthrough. Butifthefiltering isdone with aband-pass filter designed sothat thewidth wz—ayisatleast twice thehighest signal frequency, thefre- quency response will be“flat” forthesignals wanted. Wewant tomake one more point about theladder filter: theL-C ladder of Fig. 22-20 isalso anapproximate representation ofatransmission line. Ifwe hhave along conductor that runs parallel toanother conductor—such asawire ina coaxtal cable, orawire suspended above theearth—there willbesome capacitance between thetwo conductors and also some inductance due tothemagnetic field between them. Ifweimagine thelineasbroken upinto small lengths A/,each length will look like one section oftheL-C ladder with aseries inductance ALand ashunt capacitance AC. Wecan then useour results fortheladder filter. Ifwe 1,__takethelimitas4¢goestozero,wehaveagooddescription ofthetransmissionL line. Notice thatasA¢ismade smaller andsmaller, both ALandACdecrease, butoct oece5 °inthesameproportion, sothattheratioAL/ACremainsconstant. Soifwetake=) Cpa thelimitofEq.(22.28)asALandACgotozero,wefindthatthecharacteristic= CAPE impedance 2»isapureresistance whosemagnitude is/AL/AC. Wecanalso°- ‘<a> writetheratioAL/ACasLo/Co, whereLoandCyaretheinductance andcapaci-ai ©tanceofaunitlengthofthe line; then wehave n=VE 2233) ( Youwillalsonotice thatasALandACgotozero,thecutoff frequency wy=VA/LE goes toinfinity. There isnocutofT frequency foranideal transmission line. q, Tp = = 22-8 Other circuit elements uy le Wehave sofardefined only theideal circuit impedances—the inductance, thecapacitance, andtheresistance—as well astheideal voltage generator. Wewant now toshow that other elements, such asmutual inductances ortransistors or vacuum tubes, canbedescribed byusing only thesame basic elements. Suppose that wehave twocoils and that onpurpose, orotherwise, some flux from one of thecoilslinkstheother,asshowninFig.2226(a).Thenthetwocoilswillhavea (bymutual inductance Msuch that when thecurrent varies inone ofthe coils, there will beavoltage generated intheother. Can wetake into account such aneffect Fig. 22-26. Equivalent circuit ofo inour equivalent circuits? Wecan inthefollowing way. Wehave seen that the mutual inductance. 2216 induced emt’s ineach oftwointeracting coils can bewritten asthesum oftwoparts: a-1.4aMe, (22.34) dl dh, b=LaGe MG The first term comes from theself-inductance ofthecoil, and thesecond term comes from itsmutual inductance with theother cou. The sign ofthesecond term canbeplusorminus, depending ontheway thefluxfrom onecoillinks theother. Making thesame approximations weused indescribing anideal inductance, we would saythat thepotential difference across theterminals ofeach coil isequal to theelectromotive force inthecoil. Then thetwoequations of(22.34) arethesame astheones wewould getfrom thecircuit ofFig. 22-26(b), provided theelectro- motive force ineach ofthetwo circuits shown depends onthecurrent imthe ‘opposite circuit according totherelations n e £1=+iwMIz, 62=*ioMh. (22.35) == Sowhatwecandoisrepresenttheeffectoftheselinductance inanormalwaybut {job replace theeffect ofthemutual inductance byanauxiliary ideal voltage generator. (© | |( mi Wemustinaddition, ofcourse, havetheequation thatrelates thisemftothe \|es currentinsomeotherpartofthecircuit;butsolongasthisequation islinear,we|stbit ts havejustaddedmorelinearequations toourcircuitequations, andallofour |earlierconclusions aboutequivalent circuitsandsofortharestillcorrect. le °Inaddition tomutual inductances there may also bemutual capacitances. Sofar,when wehave talked about condensers wehave always imagined that there wereonlytwoelectrodes, butinmany situations, forexample inavacuum tube, ay ° there may bemany electrodes close toeach other. Ifweputanelectric charge on anyoneoftheelectrodes. itselectric field willinduce charges oneach oftheother — electrodes andaffect itspotential. Asanexample, consider thearrangement of eo oF four plates shown inFig. 22-27(a). Suppose these four plates areconnected toexternalcircuitsbymeansofthewires4,B,C,andD.Solongasweareonly mosworried about electrostatic effects, theequivalent circuit ofsuch anarrangement oh ° ofelectrodes 1sasshown inpart (b)ofthefigure. The electrostatic interaction of anyelectrode with each oftheothers isequivalent toacapacity between the Fig. 22-27. Equivalent circuit of two electrodes. mutual capacitance. Finally, let’s consider how weshould represent such complicated devices as transistorsandradiotubesinanAccircuit,Weshouldpointoutatthestartthat such devices areoften operated insuch away that therelationship between the currents and voltages isnot atalllinear. Tnsuch cases, those statements wehave made which depend onthelinearity ofequations are,ofcourse, nolonger correct.Ontheotherhand,inmanyapplications theoperating characteristics aresufficiently linear thatwemayconsider thetransistors andtubes tobelinear devices. Bythis PLATE P ‘wemean thatthealternating currents in,say,theplate ofavacuumtubearelinearly proportional tothe voltages that appear onthe other electrodes, say thegrid GRI 6, voltage andtheplate voltage. When wehave such linear relationships, wecan fs incorporate thedeviceintoourequivalent circuitrepresentation. *Asin thecase ofthemutual inductance, ourrepresentation will have toinclude auuliaryvoltagegenerators whichdescribetheinfluenceofthevoltagesorcurrents IrHODE ,8inonepartofthe device onthecurrents orvoltages inanother part. Forexample, em-pqtheplatecircuitofatriodecanusuallyberepresented byaresistance inserieswithanideal voltage generator whose source strength isproportional tothegrid voltage. Fig. 22-28, Alow-frequency equiv- Wegettheequivalent circuit shown inFig.22-28.* Similarly, thecollector circuit alent cirevit ofavacuum triode, *The equivalent eircunt shown 1scorrect only forlow frequencies. For high frequencies theequivalent circuit gets much more complicated and will include various so-called “parasitic” eapacitances and inductances. 217 EMITTER coECTOR =E cOo"Base8 Fig.22-29. Alow-frequency equiv- elent circuit ofatransistor. mKIy ofatransistor 18conveniently represented asaresistorinserieswithanidealvoltage generator whose source strength isproportional tothecurrent from the emitter tothebase ofthetransistor. The equivalent circuit isthen like that inFig. 22-29. Solong astheequations which describe theoperation arelinear, wecan usesuch representations fortubes ortransistors. Then, when they areincorporated inacomplicated network, ourgeneral conclusions about theequivalent representa~ tion ofanyarbitrary connection ofelements isstillvalid. There isone remarkable thing about transistor and radio tube circuits which isdifferent from circuits containing only impedances: thereal part oftheeffective impedance z,i,canbecome negative. Wehave seen that therealpart ofzrepresents theloss ofenergy. But itistheimportant characteristic oftransistors and tubes that they supply energy tothecitcuit. (Ofcourse they don’t just “make” energy; they take energy from thenccircurts ofthepower supplies and convert itinto Acenergy.) Soit18possible tohave acircuit with anegative resistance. Such a circuit hastheproperty that ifyou connect ittoanimpedance with apositive real part, i.e., apositive resistance, and arrange matters sothat thesum ofthetwo real parts isexactly zero, then there isnodissipation mnthecombined circuit. If there isnolossofenergy, anyalternating voltage once started willremain forever. This isthebasic idea behind theoperation ofanoscillator orsignal generator which can beused asasource ofalternating voltage atany desired frequency. 218 24 Cavity Resonators 23-1 Real circuit elements When looked atfrom any one pair ofterminals, any arbitrary evcuit made 23-1. Real circuit elements upofideal impedances and generators 1s,atany given frequency, equivalent toa i ‘ ,generator &inseries withanimpedance =.Thatcomesuboutbocauseifwe puta22-2Aeapwctor athighfrequencies voltage Vacross theterminals and solve alltheequations tofind thecurrent f, 23-3 Aresonant cavity wemust getalinear relation between thecurrent and thevoltage. Since allthe naviequations arelineur, theresultfor#mustalsodepend onlylinearly onVThe 23-4Cavity modes most general linear form can beexpressed as 23-5 Cavities and resonant circuits r=lv-e, 3.) .Review: Chapter 23,Vol.1.Resonance Ingeneral,bothzand&maydependinsomecomplicated wayonthefrequencyw. Chapter49,Vol1,Modes Equation (23 1),however, 1stherelation wewould getifbehind thetwo terminals there wasjust thegenerator &(a) anseries with theimpedance 2(4). There 1salso theopposite kind ofquestion” Ifwehave anyelectromagnetic device atallwith two terminals and wemeasure the relation between Jand V10 determine fand >asfunctions offrequency, canwefind acombination ofourWeal elements that 1sequivalent totheinternal impedance 2?‘The answer 18that for any reasonable—that 1s,physically meaningfiel—function 2(a), 1t«8possible to uppronamute thesituation toashigh anaccuracyasyouwishwithaeireuitcontaining L atfinite setofideal elements, Wedon’t want toconsiderthegeneralproblemnow butonlylook atwhat might beexpected from phystcal arguments forafewcases c Itwe think ofarealresistor,weknowthatthecurrentthrough1twillproduce R amagnetic field, Soany real resistor should also have some inductance. Also. when aresistor hasapotential difference across it,there must becharges onthe ends oftheresistor toproduce thenecessary electric fields Asthevoltage changes. thecharges will change inproportion, sotheresistor will also have some capact- tance. Weexpect that areal resistor might have theequivalent eireust shown 1n Fig 23-1 Ina well-designed resistor,theso-called“parasitic”elementsLandCFig.23-1.Equivalentcircuitof aresmall, sothatatthefrequencies forwhich itisintended, wL1smuch lessthan F@2!resistor. Rand 1/CismuchgreaterthanR.Itmaythereforebepossibletoneglectthem AAsthefrequency isrased, however, they willeventually become important, anda resistor begins tolook like aresonant circutt Areal inductance isalso notequal totheidealized inductance, whose impe- dance 1s1wL. Arealcoilofwire willhave some resistance, soatlowfrequencies the coilisreally equivalent toaninductance inseries with some resistance, asshown 1n Fig. 23-2(a) But, you arethinking, theresistance and inductance arerogerher ina real coil—the resistance 1sspread allalong thewire, soit1smixed inwith the inductance. Weshould probably useacircuit more like theone inFig. 23-2(b). Which hasseveral little R'sand L'sinseries. Butthetotal impedance ofsuch a circuit isjust ZR +SSiwL., which 1sequivalent tothesimpler diagram ofpart (a)‘Aswegoupinfrequency witharealcoil,theapproximation ofan inductance plus aresistance 1snolonger very good. The charges that must build uponthe Wirestomakethevoltageswillbecomeimportant. Itisasiftherewereittlecondensers across theturns ofthecoil, assketched inFig 23-3(a). Wemight tryto approximate therealcoilbythecircuit inFig.23-3(b). Atlowfrequencies, this (0) v) circuit canbeimmtated fairly well bythesimpler oneinpart (¢)ofthefigure (which isagain thesame resonant exrcuit wefound forthehigh-frequency model of Fig.23-2, Theequivalent circuit ofresistor) Forhigherfrequencies. however, themorecomplicated circuitof—@realinductance atlowfrequencies. Pray Fig,23-3(b)isbetter.Infac,themoreaccuratelyyouwishtorepresenttheactual Kaumpedance ofareal, physical inductance, themore sdeal elements you will have to useintheartificialmodelofit =Let'slookalittlemorecloselyatwhatgoesoninarealcoil.‘Theimpedance xeofananductance goes asaL,soitbecomes zero atlowfrequencies—it isa“shortcurcurt”:allwesee1stheresistanceofthewire.Aswegoupinfrequeney. «Lsoono) becomes much larger thanR,andthecoillooks pretty much likeanideal indue- tance. Aswegostill higher, however, thecapacities become important. Their impedance isproportional to1/wC, which islarge forsmall w.For small enough frequencies acondenser 1san“open circuit,” andwhen itisinparallel with some-thingelse,itdrawsnocurrent,Butathighfrequencies, thecurrentpreferstoflowinto thecapacitance between theturns, rather than through theinductance. So thecurrent inthecoiljumps from one turn totheother and doesn't bother tog0 around and around where athas tobuck theemf Soalthough wemay have‘tendedthatthecurrentshouldgoaroundtheloop,itwilltaketheeasterpath—thepath ofleast impedance. Ifthesubject had been one ofpopular interest, this effect would have been called “the high-frequency barrier,” orsome such name. The same kind ofthing happens snallsubjects. Inaerodynamics, sfyou trytomake things gofaster than thespeed ofsound when they were designed forlower speeds, they don't work. w © Itdoesn’t mean thatthereisagreat“barrier” there, 1justmeans thattheobject should beredesigned. Sothis coil which wedesigned usan“inductance” 1snotFig.23-3.Theequivalent circuitofgoingtoworkasagoodinductance, butassomeotherkindofthingatveryhigh«real inductance athigher frequencies. frequencies. For high frequencies, wehave tofind #new design 23-2 Acapacitor athigh frequencies Nowwewanttodiscussindetailthebehaviorofacapacitor—a geomettically ideal eapacitor—as thefrequency gets larger and larger, sowecanseethetransitionofitsproperties. (Weprefertouseacapacitorinsteadofaninductance, becausethegeometry ofapairofplatesismuchlesscomplicatedthanthegeometryof coil.) Weconsider thecapacitor shown inFig. 23-4(a), which consists oftwo par- allel circular plates connected toanexternal generator byapair ofwires. Ifwe charge thecapacitor with Dc,there will beapositive charge onone plate anda negative charge ontheother; and there will beauniform electric field between the plates. Now suppose that instead ofpc,weputanAcoflow frequency ontheplates. (We will find outlater what is“low” and what is“high",) Say weconnect theea- pacitor toalower-frequency generator. Asthevoltage alternates, thepositive charge onthetopplate istaken offand negative charge 18puton. While that 1s happening, theelectric field disappears and then builds upintheopposite direction. Ry > s =e SS aaaPetes Pele|[feet Pos cunve jeA 4XCD elo}9]! o/s as SS L. BLY) URVETy | LINESOF8 r—| ines OF € (@) rc) Fig, 23-4, Theelectric and magnetic felds between theplates of@capacitor. 22 ‘TheintegralsaresimpleifwetakethemforthecurveI's,showninFig23-4(b).which goes upalong theaxis, out radhally thedistancerslongthetopplate,down vertically tothebottom plate, andback totheaxls Thelineintegral ofEaround this curve 18,ofcourse, 7e70; soonly Eycontributes, and itsintegral 1sjust E40)", where fis thespacing between theplates. (We call Epositive if points upward.) This 1sequal totherateofchange ofthefluxofB,which wehave togetbyanintegral over theshaded area Sinside tyinFig. 23-4(b). Theflux through avertical step ofwidth drs B(ryh dr.sothetotal flux 1s hffBor) dr, Setting —9/01 ofthefluxequal tothelineintegral ofEx,wehave a £0=2|aan. 0236) Notice thatthecancels out, thefields don’t depend ontheseparation oftheplates Using Eq(23.5) forB(r), wehave bay =2 ppt200)=5)gen Foe. te “Theumederivative justbrings down another factor uo:weget ae ese . uN Ex(r)=—8Ee", ay) ya| Se ez 7“Ss : Vv | mY Asweexpect. theinduced fieldtendstoreduce theelectric fieldfarther out,The| H corrected fieldE=Ey+Ey1sthen : 4 —e ExBtB=(i-yer)Ew". (38) Fig. 29-5. Theelectric field between ‘The electric field inthecapacitor isnolonger uniform: ithastheparabohe thecopecior plates ofhigh frequency. shape shown bythebroken line inFig. 23-5 You seethat oursimple capacitor ss (Edge effects areneglected.) getting slightly complicated, We could now use our results tocalculate theimpedance ofthecapacitor ‘athigh frequencies. Knowing theelectric field, wecould compute thecharges on theplates and find out how thecurrent through thecapacitor depends onthe frequency «,butwearenotinterested inthat problem 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 that themagneve field wehave calculated isnolonger right. The magnetic field ofEq.23.5) is approximately right, butit1sonly afirst approximation Solet's calltBy We should then rewrite Eq. (23.5) as By=5Exe" 23.9) You willremember that thisfield was produced bythevariation ofE,. Now the correct magnetic field will bethat produced bythetotal electric field Ey+Ex Ifwewrite themagnetic field asB=By+By,thesecond term isjust theaudt- tional field produced byEx Tofind B,wecangothrough thesame arguments wehave used tofind By,theneintegral ofByaround thecurve I;18equal to therate ofchange oftheflux ofE,through €). Wewillust have Eq(234)again with Breplaced byByand Ereplaced byEs: a a :2B, -2nr=&(uxofE;throughP). Since E»varies with radius, toobtain itsfluxwemust integrate over thecwrcular 24 Then wecanwrite oursolution asEye" times thisfunction, with x=ar/e: saty, (20 E=Exe''Io(“)- (23.17) The reason wehave called ourspecial function Jyisthat, naturally, thsisnot thefirst time anyone hasever worked outaproblem with oscillations inacylinder. The function has come upbefore and isusually called Jp. Italways comes up whenever you solve aproblem about waves with cylindrical symmetry. The fune- ton Jyistocylindrical waves what thecosine function istowaves onastraight lune, Soitisanportant function, invented along time ago. Then aman named Bessel gothisname attached toit.The subscript zero means that Bessel invented awhole lotofdifferent functions and this 1sjust thefirst ofthem The other functions ofBessel—J;, J2,and soon—have todowith cylindrical waves which have avariation oftheir strength with theangle around theaxis of thecylinder. The completely corrected electric field between the plates ofour circular capacitor, given byEq.(23.17), 18plotted asthesolid line inFig. 23-5. For frequencies that are not too high, our second approximation was already quite good. The third approximation was even better—so good, infuct. that sfwehad plotted it,you would nothave been able toseethedifference between itand the solid curve. You will seeinthenext section, however, that thecomplete series is needed togetanaccurate description forlarge radii, orforhigh frequencies. 23-3 Aresonant cavity Wewant tolook now atwhat oursolution gives fortheelectric field between theplates ofthecapacitor aswecontinue togotohigher and higher frequencies. Forlarge w,theparameter x=ar/e also gets large, and thefirstfewterms nthe series forJoofxwill increase rapidly. That means that theparabola wehave drawn inFig. 23-5 curves downward more steeply athigher frequencies. Infact, itlooks asthough thefield would fallalltheway tozero atsome high frequency. perhaps when c/w 1sapproximately one-half ofa.Let’s seewhether J,does indeed {0through zero and become negative. Webegin bytrying x=2 IQ)=11+b=he=022 wan ‘The function issull notzero, solet'stryahighervalueofx,say,x=25Putting -innumbers, we write »\ Jo(25)=1=1.56+0.61—009=—0.04, \2008ko> ‘ThefunctionJyhasalreadygonethroughzerobytheumewegettox=2.5. NK Nat =—Comparingtheresultsforx=2andx=25,itlooksasthoughJ,goesthrough + xapproximately equal to2.4. Let's seewhat that value ofxguves: Jo24) =1=144 +0.52 —008 =0.00 Fig.23-6. TheBesselfunction Jolx). Wegetzerototheaccuracy ofourtwodecimal places. Ifwemakethecalculation more aecurate (oFsince J1sawell-known function, iFwelook itupit book), we find that stgoes through zero atx=2405 Wehave worked itout byhand to show you that you toocould have discovered these things rather than having to borrow them from abook Aslong aswearelooking upJy1nabook, itisinteresting tonotice how it goes forlarger values ofx.itlooks like thegraph inFig 23-6. Asxincreases, J.(3) oscillates between positive and negative values with adecreasing amplitude ofoscillation Wehavegottenthefollowing interesting result:If'wegohighenoughinfre-quency, theelectric field atthecenter ofourcondenser willbeone way and the electric field near theedge will point intheopposite direction, For example. 26 suppose that wetake anwhigh enough sothat x=ur/e attheouter edge ofthe capacitor isequal to4;then theedge ofthecapacitor corresponds totheabscissa x=4/in Fig, 23-6, This means that ourcapacitor isbeing operated atthefre- quency @=4e/a Attheedge oftheplates, theelectric field wall have arather high magnitude opposite thedirection wewould expect. That istheterrible thing that canhappen toacapacitorathighfrequencies. Ifwegotoveryhighfrequencies, thedirection oftheelectric field oscillates back and forth many times asweg0 outfrom thecenter ofthecapacitor. Also there arethemagnetic fields associated with these electric fields. Itisnotsurprising that ourcapacitor doesn’t look like theideal capacitance forhigh frequencies. Wemay even start towonder whether itlooks more likeacapacitor oraninductance Weshould emphasize that thereareevenmorecomplicated effectsthatwehaveneglected whichhappenattheedgesofthecapacitor. For instance, there will bearadiation ofwaves outpast theedges, sothefields areeven more complicated than theones wehave computed, butwe will notworry about those effects now. Wecould trytofigure outanequivalent circuit forthecapacitor, butperhaps Atisbetter ifwejust admut that thecapacitor wehave designed forlow-frequency fields 1syust nolonger satisfactory when thefrequency istoohigh. Ifwewant to treat theoperation ofsuch anobject athigh frequencies, weshould abandon the approximations toMaxwell’s equations that wehave made fortreating circuits and return tothecomplete setofequations which describe completely thefields inspace Instead ofdealing with idealized circuit elements, wehave todeal with thereal conductors asthey are, taking into account allthefields inthespaces in between. Forinstance, sfwewantaresonant circuit athighfrequencies wewill LINES OF8 nottrytodesign oneusing acoilandaparallel-plate capacitor. a a aa a Wehave already mentioned that theparallel-plate capacitor wehave been [0 0 analyzing hassomeoftheaspects ofbothacapacitor andaninductance. Withthe fy) 4electricfieldtherearechargesonthesurfacesoftheplates,andwiththemagneue el»ef fieldstherearebackemis.Isitpossible thatwealreadyhavearesonant circuit? LeleJoDS i Wedoindeed. Suppose wepickafrequency forwhichtheelectricfieldpattern === tietteee falls tozero atsome radius inside theedge ofthedisc; thatis,wechoose wa/e (o) greater than 2.405 Everywhere onacircle coaxial with theplates theelectric field will bezero. Now suppose wetake athin metal sheet and cutastrip just wide enough tofitbetween theplates ofthecapacitor. Then webenditintoacylinder 5thatwillgoaround attheradius where theelectric field iszero Since there are * noelectric fields there, when weputthis conducting cylinder inplace, nocurrents wo ' willflowinit:andtherewillbenochanges intheelectric andmagnetic fields. We | havebeenabletoputadirect short circutt across thecapacitor without changing (b) ‘ 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 eaoscnut and magnetic fields inside, asshown inFig. 23-7(a). ‘The electric fields areos- Be! cillating back and forth atthefrequency «w—which, don't forget, determined the 1 diameter ofthecan Theamplitude oftheoscillating Efieldvaries withthedistance 'fromtheaxisofthecan,asshowninthegraphofFig.23-7(b).Thiscurvessjust (¢) :thefirstarch oftheBessel function ofzero order. There isalsoamagnetic field . which goes incircles around theaxis and oscillates intime 90°outofphase with : the electric field Wecanalsowrite outasertes forthemagnetic fieldandplotit.asshown in r thegraph ofFig. 23-1(c). Fig. 23-7. Theelectric andmagnetic How isitthat wecan have anelectric and magnetic field inside acan with no fields inanenclosed cylindrical con.externalconnections? It1sbecausetheelectricandmagneticfieldsmaintainthem-selves: thechanging Emakes aBand thechanging Bmakes anE—all according totheequations ofMaxwell, The magnetic field hasaninductive aspect, and the electric field acapacitive aspect; together they make something like aresonant circuit. Notice that theconditions wehave described would only happen 1ftheradiusofthecanisexactly2.405¢/o.Foracanofagivenradius,theoscillatingelectric andmagnetic fields wallmaintain themselves—in theway wehave described 27 —only atthat particular frequency. Soacylindrical can ofradhus risresonant at thefrequency ¢ wo=2.405£ 3.18) We have said that thefields continue tooscillate inthesame way after thecan 1scompletely closed. That 1snotexactly right. Itwould bepossible ifthewalls ofthecan were perfect conductors. For areal can, however, theoscillating cur- rents which exist ontheinside walls ofthecan lose energy because oftheresistance ofthematerial, The oscillations ofthefields wall gradually dieaway. Wecan See from Fig. 23-7 that there must bestrong currents associated with electric and magnetic fields inside thecavity. Because thevertical electrical field stops suddenly aittheLopand bottom plates ofthecan, ithasakarge divergence there: sothere must beposttive and negative electric charges ontheinner surfaces ofthecan, a shown inFig. 23-7(a) When theelectric field reverses, thecharges must reverse | , also,sotheremustbeanalternating current between thetopandbottom phates Pot| ofthecanThesechargeswillflowinthesidesofthecan,asshowninthefigure meuymeet [E-¢-ourgyr WecanalsoseethattheremustbecurrentsinthesidesofthecanbyconsideringSETA TALAF696%"whathappenstothemagneticfieldThegraphofFig.23-7(e)tellsusthatthepoe magneticfieldsuddenlydropstozeroattheedgeofthecanSuchasuddenchange lprTrcb | inthemagnetic fieldcanhappen onlyifthere 1s.acurrentinthewallThiscurrent 7ul iswhat gives thealternating electric charges onthetopandbottom plates ofthe Fig.23-8.Couplingintoandoutof Youmaybewondering aboutourdiscoveryofcurrentsintheverticalsidesof ©resonant cavity. thecan What about ourearlier statement that nothing woul] bechanged when we introduced these vertical sides inaregion where theelectric field was zero” Re- member, however, that when wefirst put inthesides ofthecan, thetop and bottom plates extended out beyond them, sothat there were alo magnetic fields, ‘ontheoutside ofour can was only when wethrew away the parts ofthe capacitor plates beyond theedges ofthecan that netcurrents hadtoappear onthe insides ofthe vertical wally nF Although theelectric and!magnet fields inthecompletely enclosed eanwillsina gradually dieawaybecauseoftheenergylosses,wecanstopthisfromhappening ‘GENERATORifwemake alittle hole inthecan and putina little bitofelectrical energy tomake locreeror¢|upthelossesWetakeasmiallwire,pokettthroughtheholeinthesideoftheean, © |AMPUFIER|andfastensttotheinsidewallsothatitmakesasmallloop,asshowninFig,23-8,s == Ifwenowconnectthiswiretoasourceofhigh-frequency alternatingcurrent.this cate current willcouple energy into theelectric andmagnetic fields ofthecavity and keep theoscillations going. This will happen, ofcourse, only ifthefrequeney ofthe Fig. 23-9. Asetup forobserving the riving source iyattheresonant frequency ofthecan IFthe source rst thewrong cavity resonance. frequency, theelectric and magneue fields will not resonate, and thefieldy anthe can will bevery weak Theresonant behavior caneasilybeseenbymaking another smallholein 4thecanandhooking inanother coupling loop,aswehavealsodrawninFig,23-8. 5 The changing magnetic field through this loop willgenerate aninduced electro~Fa j motiveforceintheloop.Ifthisloop1snowconnected tosomeexternalmeasuring3 | circuit, thecurrents willbeproportional tothestrength ofthefields inthecavity 5 ‘Suppose wenow connect themput loop ofourcavity toanRFsignal generator, £ asshowninFig.23-9.Thesignalgenerator containsasourceofalternating current8 Awea,/0 whose frequency canbevaried byvarying theknobonthefrontofthegenerator Then weconnect theoutput loop ofthecavity toa“detector,” which 1saninsteu- Ge Frequengy ‘Mentthatmeasures thecurrent fromtheoutput loop. Itgivesameter reading pro- portional tothis current. Ifwenow measure theoutput current asafunction of Fig. 23-10. Thefrequency response thefrequencyofthesignalgenerator, wefindacurvelikethatshowninFig23-10, curveofaresonant cavity. Theoutput current issmall forallfrequencies except those very near thefrequency 4,which 1stheresonant frequency ofthe cavity. Theresonance curve 1svery much luke those wedescribed inChapter 23ofVol. I.The width oftheresonance 1s, however, much narrower than weusually find forresonant circuits made ofinduc~ tances and capacitors: that is,the@ofthecavity 1svery high. It1snotunusualtofindQ'sashighas100,000ormoreiftheinsidewallsofthe cavity aremade of some material with avery good conductivity, such assilver 23.8 23-4 Cavity modes Suppose wenow trytocheck our theory bymaking measurements with anactualcan,Wetakeacanwhichisacylinderwithadiameterof3.0inchesandheightofabout2.5inches.Thecanisfittedwithaninputandoutputloop,as| shown inFig.23-8. Ifwecalculate theresonant frequency expected forthiscan | >.according toEq(23.18),wegetthatfy=wo/2=3010megacycles When=|—| sawesetthefrequency ofoursignalgenerator near3000megacycles andvaryit&i|slightlyuntlwefindtheresonance, weobserve thatthe maximum outputcurrent. °/|||‘occursforafrequency of3050megacycles, whichisquiteclosetothepredictes =L_y itresonant frequency, butnotexactly thesame, There areseveral possible reasons ortwomenen forthediscrepancy. Perhaps theresonant frequency ischanged alitle bitbecause oftheholes wehave cuttoputinthecoupling loops. Alittle thought, however. Fig.23-11. Observed resonant fre- shows thattheholes should lower theresonant frequency alittle bit,sothatcannot quencies ofacylindrical cavity bethereason, Perhaps there 1ssome slight error nthefrequency calibration ofthe signal generator, oFperhaps our measurement ofthediameter ofthecavity #8not accurate enough. Anyway, theagreement isfairly close. Much more important issomething that happens ifwevary thefrequency of ‘oursignal generator somewhat further from 3000 megacycles. When wedothat wegettheresults shown inFig.23-11. Wefindthat, inaddition totheresonance E weexpected near 3000 megacycles, there isalso aresonance near 3300 megacyCles and one near 3820 megacycles. What dothese extra resonances mean? WemightgetacluefromFig.23-6.Althoughwehavebeenassumingthatthefirstzeroof [rere ||theBessel function occurs attheedge ofthecan, itcould also bethat thesecond zero oftheBessel funetion corresponds totheedge ofthecan, sothatthere isone complete oscillation oftheelectric field aswemove from thecenter ofthecan out totheedge, asshown inFig,23-12. Thisisanother possible mode fortheoscillating to fields. We should certainly expect thecan toresonate insuch amode. But notice, the second zero ofthe Bessel function occurs atx=5.52, which isover twice aslarge asthevalue atthefirst zero. The resonant frequency ofthis mode ic should therefore behigher than 6000 megacycles. Wewould, nodoubt, find st 4 there, butitdoesn’t explain theresonance weobserve at3300, ' '“Thetroubleisthatinouranalysisofthebehaviorofaresonantcavitywehave ' rasseca! considered only onepossible geometric arrangement oftheelectric andmagnetic ' : fields. Wehave assumed that theelectric fields arevertical and that themagnetic \ i fields lieinhorizontal circles, Butother fields arepossible. The only requirements H arethat thefields should satisfy Maxwell's equations inside thecanand that the ‘ t—+ electric field should meet thewallatright angles. Wehave considered theease in : 'whichthetopandthebottomofthecanareflat,butthings would notbecompletely ' : different sfthetopand bottom were curved. Infact, how isthecan supposed to H know which isitstopandbottom, andwhich areitssides? Itis,infact, possible () toshow thut there isamode ofoscillation ofthe fields inside the ean inwhich the electric fields gomore orlesacross thediameter ofthe can,asshown inFig.23-13. Fig.23-12 Ahigher-frequency mode. Itisnot too hard tounderstand why thenatural frequency ofthis mode should benotvery different from thenatural frequency ofthefirst mode wehave considered. Suppose that instead ofour cylindrical cavity wehad taken acavity — which was acube 3inches onaside. Itisclear that thiscavity would have three dierent modes, butallwith thesame frequency. Amode with theelectric field going more orless upand down would certainly have thesume frequency asthe mode inwhich theelectric field wasdirected right and left Ifwenow distort the cube into acylinder, wewill change these frequencies somewhat. Wewould still expect them nottobechanged toomuch, provided wekeep thedimensions ofthecavitymoreorlessthesame.Sothefrequencyofthe mode ofFig, 23-13 should not betoodifferent from themode ofFig. 23-8. Wecould make adetailed cal-culationofthe natural frequency ofthemode shown inFig. 23-13, butwewall not dothat now. When thecalculations arecarried through, itis found that, forthe dimensions wehave assumed, theresonant frequency comes outveryclose tothe Fig,23-13. Atransverse mode of observed resonance at3300 megacycles thecylindrical cavity Bysimilar calculations 1ispossible toshow that there should bestillanother mode attheother resonant frequency wefound near 3800 megacycles For this 29 mode, theelectric and magnet fields areasshown inFig. 23-14, The electric a field does notbother togoalltheway across thecavity Itgoes Irom thesides to the ends, asshown, S = ‘Asyouwillprobably nowbelieve, ifwegohigher andhigher infrequency we should expect tofind more and more resonances. There aremany diferent modes. cach ofwhich will have adiferent resonant frequency corresponding tosome par=ticularcomplicated arrangement oftheelectricandmagneticfields.Eachofthese fieldarrangements scalledaresonantmode.Theresonancefrequencyofeachmode ey a canbecalculated bysolving Maxwell's equations fortheelectric and magneticna, fieldsinthecavity.When wehave aresonance atsome particular frequency, how can weknowwhichmodeisbeingexcited”Oneway1topokeahtewireintothecavity Fig. 23-14, Another mode of@cy- through asmall hole Iftheelectric field 1salong thewive, asinFig 23-15(a), lindrical cavity there willberelauively large currents inthewire, supping energy from thefields. aand the resonance will besuppressed. Iftheelectric field 1sasshown anFig 23-15(b), the wire will have amuch smaller effeet. We could find which way the field points inthismode bybending theendofthewire. asshown 1nFig 23-15(c) Then, aswerotate thewire, there will beabigeffect when theend ofthewire 1s parallel toEand asmall effect when 11s votated soastobeat90° 10E. Sh Se ( a) ir) i) Fig. 23-15. Ashort metal wire inserted into acavity wlldisturb the resonance much more when tisparallel toEthan when iisotright angles. 23-5 Cavities and resonant circuits Although theresonant cavity wehave been describing seems 10bequite diferent from theordinary resonant circuit consisung ofanmductance and acapacitor, thetworesonantsystemsare,ofcourse, closely related They areboth members ofthesume famuly: they arejust two extreme eases ofelectromagnetic resonators—and there are many intermediate cases between these two extremes.Supposewestartbyconsidering theresonantcircuttofacapacitorinparallelwithaninductance, asshowninFig.23-16(a).Thiscircuitwallresonateatherequencyoy= VEC. Iwe want toraise theresonant frequency ofthis exrcuit, wecan dosobylowering theinductance L.One way istodectease thenumber ofturns 1nthecoil.Wecan,however,goonlysofarnthisdirection. Eventually wewillgetdown tothelastturn, and wewill have just apiece ofwire joining thetopand bottom plates ofthecondenser. Wecould ratse theresonant frequency still further bymaking thecapacitance smaller: however, wecan also continue todecrease theinductance byputtingseveralinductances inparallelTwoone-turninductances inparallel will have only half theinductance ofeach turn. Sowhen our mductance has been reduced toasingleturn,wecancontinuetoraisetheresonantfrequency byaddingothersingleloopsfromthetopplatetothebottomplateofthecondenser.For instance, Fig 23-16(b) shows the condenser plutes connected bysixsuch ~single-turn inductances.” Ifwecontinue toadd many such piecesofwire.wecan make thetransition tothecompletely enclosed resonant system shown inpart (€) ofthefigure, which isadrawing ofthecross section ofacylindrically symmetrical 210 TT 7 f |10g f | le @|eoetbe| las> om f k\nto offTf)e eo] u € 4 \\ \ {|KO |Jooftorch oo|WN aye” st jot JWT og Ta —— (9) 7 (b) (c) Fig. 23-16. Resonators ofprogressively higher resonant frequencies. ‘obyect. Our inductance 1snow acylindrical hollow can attached totheedges of thecondenser plates. The electric and magnetic fields will beasshown inthe figure. Such anobject is,ofcourse, aresonant cavity. Itiscalled a“loaded” cavity. But we can sull think of1tasanZ-Ccircuitinwhichthecapacity section1sthe region where wefind most ofthe electric fiekd and the inductance section 1s that region where wefind most ofthemagnetic field. Ifwewant tomake thefrequency oftheresonator inFig, 23-16(c) still higher, wecan dosobycontinuing todecrease theinductance L,Todothat, wemust decrease the geometric dimensions ofthe inductance section, for example by decreasing thedimension /inthedrawing. Asfhisdecreased, theresonant fre- quency will beincreased Eventually, ofcourse, wewill gettothesituation in Which theheight /isjust equal totheseparation between thecondenser plates We then have just acylindrical can, our resonant circut has become thecavity resonator ofFig, 23-7. You will notice that intheoriginal L-C resonant circuit ofFig. 23-16 the electric and magnetic fields arequite separate, Aswehave gradually modified the resonant system tomake higher and higher frequencies, themagnetic field hasbeen brought closer andcloser totheclectric field until inthecavity resonator thetwo arequite intermixed ‘Although thecavity resonators wehave talked about inthischapter have been Oy cylindrical cans, there isnothing magic about thecylindrical shape Acanofany / shape will have resonant frequencies corresponding tovarious possible modes of . oscillations ofthe electric and magnetic fields Forexample, the“cavity” shown a inFig 23-17 will have ttsown particular setofresonant frequencies—although they would berather dificult tocalculate, Fig.23-17. Another resonant covity. zu 24 Waveguides 24-1 The transmission line Inthelastchapter westudied what happened tothelumped elements ofcircuits 24-1 The transmission line when they were operated atvery high frequencies, and wewere ledtoseethat a iresonant exrcuit couldbereplaced byacavitywiththefieldsresonating inside, 4-2Therectangular waveguide Another interesting technical problem istheconnection ofone object toanother, 24-3 The eutoff frequency sothat electromagnetic energy can betransmitted between them. Inlow-frequency ;circuits theconnection ismacewithwires,butthismethod doesn’t workveryweil 4-4Thespeedoftheguidedwaves athigh frequencies because thecircuits Would radiate energy into allthespace 24-8 Observing guided waves around them, and 11shard tocontrol where theenergy willgo.The fields spread ,coutaround thewires; thecurrents andvoltages arenot“guided” verywellby 4-®Waveguide plumbing thewires. Inthis chapter wewant tolook into theways that objects can be 24-7 Waveguide modes imrconnected athighfrequenciesAtleast,that'sonewayofpresentingOU24.Anotherwayoflookingatthe‘Another wayistosaythatwehavebeendiscussing thebehavior ofwaves in guided waves freespace. Now 118time toseewhat happens when oscillating fields areconfined inone ormore dimensions. We will discover theinteresting new phenomenon ‘when thefields areconfined inonly twodimensions andallowed togofree inthe third dimension, they propagate inwaves. These are“guided waves"—the subject ofthis chapter. Webegin byworking outthegeneral theory ofthesransmission line. The ‘ordinary power transmission line that runs from tower totower over thecountry- side radiates away some ofitspower. butthepower frequencies (50-60 cycles/sec) aresolowthat thisloss1snotserious. The radiation could bestopped bysurround ingthelinewith ametal pipe, butthismethod would notbepractical forpower lunes because thevoltages and currents used would require avery large, expensive, and heavy pipe, Sosimple “open lines” areused. For somewhat higher frequencies—say afew kilocycles—radiation can al- ready beserious However, itcanbereduced byusing “twisted-pair” transmission lunes. as1sdone forshort-run telephone connections. Athigher frequencies, how- ever, theradiation soon becomes intolerable, either because ofpower losses or because theenergy appears inother circuits where itisn’t wanted Forfrequenctes from afew kilocycles tosome hundreds ofmegacycles, electromagnetic signals andpower areusually transmitted viacoaxial lines consisting ofawire inside a ee cylindrical “outer conductor” or“shield “Although thefollowing treatment will to apply toatransmission lineoftwoparallelconductors ofanyshape,wewillcarry ee itout referring toacoaxial line. Wetake thesimplest coaxial line that hasacentral conductor, which wesup- a pose1sathinhollow cylinder, andanouter conductor which isanother thin 5.244, coaxial transmisuontine. cylinder onthesame axis astheinner conductor, asinFig. 24-1 We begin by figuring out approximately how the line behaves atrelatively low frequencies We have already described some ofthe low-frequency behavior when wesaid earlier that two such conductors had acertain amount ofinductance per unit length oracertain capacity per unit length. Wecan, infact, describe thelow- frequency behavior ofany transmission fine bygiving itsimductance per unit length, Lyand itscapacity perunit length, Co, Then wecan analyze theline as thelimiting caseoftheL-Cfilter asdiscussed inSection 22-6. Wecanmake a filter which imitates theInne bytaking small series elements LoAxand small shunt capacities CyAx,where Ax1sanelement oflength oftheline. Using our resultsfortheinfinitefilter.weseethattherewouldbeapropagation ofelectric m4 signals along theline. Rather than following that approach, however, wewould now rather look atthelinefrom thepoint ofview ofadifferentialequation. Suppose that wesee what happens attwo neighboring points along the transmission line, say atthedistances xand x+Axfrom thebeginning ofthe line. Let’s call thevoltage difference between thetwoconductors V(x). and the current along the“hot” conductor (x) (see Fig. 24-2). Ifthecurrent intheline isvarying, theinductance will give usavoltage drop across thesmall section of line from xtox+Ax inthe amount AV=Vox+Ax)—VOX)=~Lodx F Or,taking thelimit asAx—0,weget ov a wv ye. 24.1wine1MayEsan) ax~hoay eawR)4:\ ‘Thechangingcurrentgivesagradientofthevoltage. wy, |Vix+dx) Referring againtothefigure,ifthevoltage atxischanging, theremustbe wine2\‘ VA somechargesupplied tothecapacity inthatregion. Ifwetakethesmallpieceofass C lunebetween xandx+Ax,thecharge onitisq=CoAxV. Thetimerate-of- change ofthischarge isCy4xdV/dr,butthecharge changes only ifthecurrent Fig.24-2. Thecurrents andvoltages (x)intotheelementisdifferentfromthecurrentI(x++x)out.Callingthediffer- of@transmission line. cence Al,wehave wv Al=—Cyax Taking thelimit asAx—0,weget al wv Rr OO (242) Sotheconservation ofcharge implies that thegradient ofthecurrent ispropor-tionaltothetimerate-of-change ofthe voltage. 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 wewilljust stay with thesimple example. The two transmission line equations can becombined bydifferentiating one with respect torand theother with respect toxand ehminating either Vor1. Then wehave either av av Se=Coloae (24.3) or ar aroToy,ot. 4axt~Coloaps may ‘Once more werecognize thewave equation inx.Forauniform transmission line, thevoltage (and current) propagates along theline asawave, The voltagealongthelinemustbeoftheformV(x,1)=f(x~11)orVix,1)=gC+rsorasum ofboth. Now what isthevelocity »?We know that thecoefficient of the4?/at? term isjust 1/v?, so 1 _ 45) VEC We will leave itforyou toshow that thevoltage for each wave inaline is proportional tothecurrent ofthat wave and that theconstant ofproportionality isjust thecharacteristic impedance zo, Calling Vand /..thevoltage and current forawave going intheplus x-direction, youshould get Va=Zoly: (24.6) m2 ~ The factor 1/eg¢ has thedimensions ofaresistanceandisequalto120%ohms. The geometric factor In(b/a) depends only logarithmically onthedimensions. so H forthecoaxial line—and most lines—the characteristic impedance hastypical ty values offrom50ohms orsotoafewhundred ohms. NY weno 24-2Therectangular waveguide~ Thenextthing wewanttotalkabout seems, atfirstsight, tobeastriking aN \. phenomenon: ifthecentral conductor isremoved fromthecoaxial line,itcanstill . carry electromagnetic power. Inother words, athigh enough frequencies ahollow a tubewillwork justaswellasonewithwires. Itisrelated tothemysterious wayin \., which aresonant circuit ofacondenser andinductance getsreplacedbynothing. butacanathigh frequencies. . Although itmay seem tobearemarkable thing when onehasbeen thinking 5 interms ofatransmission line asadistributed inductance and capacity, weall “ know thatelectromagnetic waves cantravel along inside ahollow metal pipe. Fig.24-3. Coordinates chosen for Ithepipeisstraight, wecanseethrough it!Socertainly electromagnetic waves therectangular waveguide. gothrough apipe. Butwealso know thatitisnotpossible totransmit low-fre-quencywaves(powerortelephone) throughtheinsideofasinglemetalpipe.So itmustbethatelectromagnetic waveswillgothrough iftheirwavelength isshort yenough. Therefore wewant todiscuss thelimiting case ofthelongest wavelength o—| (orthelowestfrequency) thatcangetthroughapipeofagivensize.Sincethepipeisthenbeingusedtocarrywaves,itiscalledawaveguide. a}Wewill begin with arectangular pipe, because itisthesimplest case to b analyze. Wewillfirstgiveamathematical treatment andcome back latertolook attheprobleminamuchmoreelementaryway.‘Themoreelementaryapproach. 4d.however, canbeapplied easily only toarectangular guide. The basic phenomena (a) % arethesame forageneral guide ofarbitrary shape, sothemathematical argument 5 isfundamentally more sound.y Ourproblem, then, istofindwhat kindofwaves canexistinside arectangular pipe. Let’s first choose some convenient coordinates: wetake thez-axis along the length ofthepipe, and thex-and y-axes parallel tothetwo sides, asshown in Fig. 24-3. We know that when light waves godown thepipe, they have atransverse ) © —-X electric field; sosuppose welook firstforsolutions inwhich Eisperpendicular to ; z,saywithonlyay-component, E,.Thiselectricfieldwillhavesomevartation ray ete,Mectricfeldintheacrosstheguide;infut,itmustgotozeroatthesidesparalleltothey-axis,because waveguideofsomevalve ofz thecurrents andcharges inaconductor always adjust themselves sothatthere is notangential component oftheelectric field atthesurface ofaconductor. So E,will vary with xinsome arch, asshown inFig. 24-4. Perhaps it1stheBessel function wefound foracavity? No, because theBessel function hastodowith » cylindrical geometries. Forarectangular geometry, waves areusually simple harmome functions, soweshouldtrysomething likesinKex. I‘Sincewewantwavesthatpropagate downtheguide,weexpect thefieldto T Ts![8 alternatebetweenpositiveandnegativevaluesaswegoalongin2,asmFig.24-5, 'otfs andtheseoscillations willtravelalongtheguidewithsomevelocity r.Ifwehave Lioscillations atsome definite frequency «,wewould guess that thewave might vary — (o with zlike cos(wf —k,2), ortouse the more convenient mathematical form, ey likee"“‘-*) Thisz-dependence represents awave travelling withthespeed v= ok, (see Chapter 29,Vol. 1). Sowemight guess that thewave inthe guide would have the following ha mathematical form: = Ey=Eosinkzxe'*'—*, (24.12) a Let's seewhether thisguess satisfies thecorrect field equations. First, the electric field should have notangential components attheconductors. Our field Fig.24-5. Thez-dependence ofthe _SAtsfies thisrequirement; it1perpendicular tothetopandbottom faces andis field inthewaveguide. zero atthetwosidefaces. Well, itisifwechoose k,sothatone-half acycle of m4 sinkxjust fitsinthewidth oftheguide—that ss,if koa =7. 24.13) There areother possibilities, ike Kea =2,3... oF, ingeneral, ka =nm, (2414) whereaisanyinteger.Theserepresentvariouscomplicated arrangements ofthefield, but fornow let’s take only thesimplest one, where k=/a, where ais thewidth oftheinside oftheguide. Next, thedivergence ofEmust bezero inthefree space inside theguide, since there arenocharges there. Our Ehasonly ay-component, and itdoesn’t change with y,sowedohave that V+E=0. Finally, ourelectric field must agree with therestofMaxwell's equations in thefree space inside theguide. That 1sthesame thing assaying that atmust satisfy thewave equation #E,,OE,,aE, 1aE,aetoy+oe7@oe~ 24.15) Wehavetoseewhetherourguess,Eq.(24.12),willwork.Thesecondderivative ofE,with respect tox1sjust —A2E, The second derivative with respect toy1s zero, since nothing depends ony.Thesecond derivative withrespect to218—K7Ey. andthesecond derivative with respect tois—w®E,. Equation (24.15) then says that 2 apKE, +KGE, ~%E,=0. Unless £,iszero everywhere (which isnotvery interesting), thisequation iscorrect it K+ =0. C416)Wehavealreadyfixedk,.sothisequationtellsusthattherecanbewavesofthei.EN, typewehaveassumedif&,isrelatedtothefrequency«sothatEq.(2416)8|Mp\satisfied—in other words, if uxSSN ee a kz=V(w/e?) —(w?/a"). 24.17) NS\ \ Thewaveswehavedescribed arepropagated inthez-direction withthisvalueofks BND) . ‘The wave number&,wegetfromEq.(24.17)tellsus,foragivenfrequencyw,*aN \ thespeed withwhich thenodes ofthewave propagate down theguide. The “te NEIN NaSphasevelocityis yXNNeon ®, (24.18) %kK aN Sai Youwillremember thatthewavelength dofatravelling wave1sgivenby Ne! A=2mu/, soke18alsoequal to277/y, where dyisthewavelength oftheoscilla- X . tons along thez-direction—the “guide wavelength.” The wavelength intheguide1sdifferent,ofcourse,fromthefree-space wavelength ofelectromagnetic waves \ ofthesamefrequency. Ifwecallthefree-space wavelength Xo,which1sequalto \ 2me/w, wecan write Eq. (24.17) as y Fig.24-6.Themagneticfieldinthe ),=—— (24.19) waveguide. VI=Qo/2aP? Besides theelectric fields there aremagnetic fields that will travel with the wave. butwewil notbother towork outanexpression forthem right now. Since fv XB= dE/ar, thelines ofBwillcirculate around theregions inwhich E/ar islargest, that 1s,halfway between themaximum and minimum ofE.The loops ofBwill lieparallel tothexz-plane and between thecrests and troughs of E,asshown inFig. 24-6. m5 24-3 The cutoff frequency Insolving Eq.(24.16) fork,,there should really betwo roots—one plus and fone minus. We should write ke=*V(w?Je?) =7a"). (24.20) The twosigns simply mean that there canbewaves which propagate with anega- five phase velocity (toward —2), aswell aswaves which propagate inthepositive direction intheguide. Naturally, stshould bepossible forwaves togoineither direction. Since both types ofwaves can bepresent atthesame time, there will be thepossibility ofstanding-wave solutions. Our equation fork;also tells usthat higher frequencies give larger values of k,,and therefore smaller wavelengths, until inthehmit oflarge w,kbecomes equal tow/c, which isthevalue wewould expect forwaves infree space. The light we“see” through apipe stilltravels atthespeed c.Butnow notice that ifwe gotoward lowfrequencies, something strange happens. Atfirstthewavelength gets longer and longer, butifwgets toosmall thequantity inside thesquare root ofEq.(24.20) suddenly becomes negative. This willhappen assoon as«gets to belessthan xe/a—or when Aobecomes greater than 2a. Inother words, when thefrequency gets smaller than acertain critical frequency w,=c/a, thewave number k:(and also A,)becomes imaginary and wehaven't gotasolution any more. Ordowe? Who said that k,has tobereal? What ifitdoes come out imaginary? Our field equations arestill satisfied. Perhaps animaginary k.also represents awave. ‘Suppose w1sless than we;then wecan write kp=ik, (2421) where A’isapositive real number K=V@a) —OP). (24.22) Ifwenow goback toourexpression, Eq. (24.12), forEy,wehave E,=Epsin k.xe*™", (24.23) whichwecanwriteas Fy E,=Epsinkexe***e™', (24.24) This expression gives anE-field that oscillates with time ase"**butwhich varies with zase**. Itdecreases orincreases with 2smoothly asarealexponent ial Inourderivation wedidn’t worry about thesources that started thewaves, butthere must, ofcourse, beasource someplace intheguide. ‘The sign thatgoes with k’must betheone that makes thefield decrease with increasing distance from the source ofthe waves. Soforfrequencies below w,=e/a, waves donorpropagate down theguide; theoscillating fields penetrate into theguide only adistance oftheorder of1/k’. For this reason, thefrequency «.1scalled the“cutoff frequency” oftheguide Looking atEq.(2422),weseethat forfrequencies just alittle below we.thenum- ¢4dszs serk’1ssmallandthefieldscanpenetratealongdistanceintotheguide.Butif2 # 18much lessthanw,,theexponential coefficient k’1sequal to7/aandthefield dies offextremely rapidly, asshown inFig.24-7. The field decreases byI/eintheFig.24-7.Thevariationof&,with—distancea/,orinonlyaboutone-third ofthe guide width. ‘The fields penetrate 2foreK we very little distance from thesource Wewant toemphasize aninteresting feature ofouranalysis oftheguidedwaves—the appearance oftheimaginary wavenumberk,.Normally, ifwesolveanequation inphysics and getanimaginary number, 1tdoesn't mean anything physical. Forwaves, however, animaginary wave number does mean something. The wave equation jsstill satisfied; itonly means that thesolution gives expo- nentially decreasing fields instead ofpropagating waves. Soinanywave problem where kbecomes imaginary forsome frequency, itmeans that theform ofthewave cchanges—the sine wave changes into anexponential. 46 FROM TODETECTOR Fig.24-8. Awaveguide withadriv- toe ingstuband@pickup probe. en oe InFig, 24-8, weshow aguide with some cutaways toshow adriving stub and a pickup “probe”. The driving stub can beconnected toasignal generator viaa coaxial cable, and the pickup probe can beconnected byasimilar cable toa intheguide, asshown inFig. 24-8. Then theprobe canbemoved back and forth along theguide tosample thefields atvarious positions. Ifthesignal generator issetatsome frequency wgreater than thecutoff frequency w.,there will bewaves propagated down theguide from thedriving stub, These will betheonly waves present iftheguide 1sinfinitely long, which can effectively bearranged byterminating the guide with acarefully designed absorber insuch away that there arenoreflections from thefarend. Then, since thedetector measures thetime average ofthefields near theprobe, itwill pick upasignal which 1smdependent oftheposition along theguide: itsoutput will Ifnow thefarend oftheguide isfinshed offimsome way that produces a reflected wave—as anextreme example. ifweclosed itoffwith ametal plite—there will interfere and produce astanding wave intheguide similar tothestanding waves onastring which wediscussed inChapter 49ofVol. I.Then, asthepickup probe 1smoved along theline, thedetector reading willriseand fallperiodically, showing amaximum inthefields ateach loop ofthe standing wave andatminimum, ateach node The distance between two successive nodes (orloops) tsjust \v/2. ‘This gives aconvenient way ofmeasuring theguide wavelength. Ifthefrequency 1snow moved closer tow,,thedistances between nodes increase, showing that the guide wavelength increases aspredicted byEq.(24.19). Suppose now thesignal generator 1ssetatafrequency just atlittle below a, Then thedetector output willdecrease gradually asthepickup probe 1smoved down theguide Ifthefrequency issetsomewhat lower, thefield strength will fallrapidly, following thecurve ofFig. 24-7. and showing that waves arenot propagated, 24-6 Waveguide plumbing ‘Animportant practical useofwaveguides isforthetransmission ofhigh frequency power, as,forexample, incoupling the high-frequency oscillator or output amplifier ofaradar settoanantenna. Infact, theantenna itself usually consists ofaparabolicreflectorfedatstsfocusbyawaveguideflaredoutatthe endtomake a“horn” that radiates thewaves coming along theguide, Although high frequencies can betransmitted along acoaxtal cable, awaveguide isbetter fortransmitting large amounts ofpower. First. themaximum power that can betransmitted alongaline1slimitedbythebreakdown ofthe msulation (solid orgas) between theconductors. For agiven amount ofpower, thefield strengths ina guide are usually less than they areinacoaxtal cable, sohigher powers can be transmitted before breakdown occurs. Second, thepower losses inthecoaxial cable areusually greater than inawaveguide. Inacoaxtal cable there must beinsulating material tosupport thecentral conductor, and there isanenergy loss inthis material—particularly athigh frequencies. Also, the current densities onthe central conductor arequite high, andsince thelosses goasthesquare ofthecurrent density, thelower currents that appear onthewalls oftheguide result inlower a Figure 24-13 isadrawing ofaunidirectional coupler: aprece ofwaveguide ABhasanother piece ofwaveguide CDsoldered toitalong oneface. The guide CDiscurved away sothat there isroom fortheconnecting flanges. Before the guides aresoldered together, two(ormore) holes have been drilled ineach guide (matching each other) sothat some ofthefields inthemain guide 4Bcan becoupledintothesecondary guideCD.Eachoftheholesactslikealittleantennathat produces awave inthesecondary guide. Ifthere were only onehole, waves would besent inboth directions and would bethesame nomatter which way the . wave wasgoing intheprimary guide. Butwhen there are‘woholes with asepara~ —~~ _-©[> tuonspaceequaltoone-quarter oftheguidewavelength,theywillmaketwosources |iSfe~~op7 90°outofphaseDoyourememberthatweconsideredinChapter29ofVol1the5 ST, interference ofthewavesfromtwoantennas spaced\/4apartandexcited 90°a 7 outofphase intime? Wefound thatthewaves subtract inonedirection andadd Sr 1mtheopposite direction Thesamethingwillhappen here.Thewaveproduced~{Les intheguideCDwillbegoinginthesamedirectionasthewaveinAB. 3 Ifthewave intheprimary guide istravelling from 4toward B,there willbe 4wave attheoutput Dofthesecondary guide. Ifthewave intheprimary guide Fig 24-13. Aunidirectionol coupler. goes from Btoward A.there willbeawave going toward theend Cofthe secondary guide. This endisequipped with atermination, sothat thiswave 1sabyorbed and there 1snowave attheoutput ofthecoupler 24-7 Waveguide modes y Thewave wehave chosen toanalyze 1saspecial solution ofthefieldequations. There aremany more. Each solution iscalled awaveguide “mode ”Forexample, ourx-dependence ofthefield wasjustone-halfacycleofasinewave.There1san equally good solution with afullcycle, then thevariation ofE,with x15asshown mmFig24-14 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 fick! has (a) = components onlyinxandy—so thatthetotalelectric fieldisalways atright angles tothe-lirection—the mode 1scalled a“transverse electric™ (orTE) mode. ey Themagnetic field ofsuch modes willalways have az-component. Itturns out that rfEhasacomponent inthez-direction (along thedirection ofpropagation). then themagnetic field willalways have only transverse components. Sosuch ficlds arecalled transverse magnetic (TM) modes. For arectangular gurde, all theother modes have ahigher cutoff frequency than thesimple TEmode wehave % described. It1s,therefore, possible—and usual—to useaguide with frequency Just above thecutoff forthis lowest mode butbelow thecutoff frequency forall theothers, sothatjust theonemode ispropagated. Otherwise. thebehavior gets «) complicated anddificult tocontrol. Fig. 24-14, Another possible vari: 24-8 Another way oflooking attheguided waves tionofEywithx Wewantnowtoshowyouanother wayofunderstanding whyawaveguide attenuates thefields rapidly forfrequencies below thecutoff frequeney w,. Then you willhave amore “physical” idea ofwhy thebehavior changes sodrastically between lowand high frequencies Wecandothisfortherectangular guide by analyzing thefields interms ofreflections—or mages—in thewalls oftheguide The approach only works forrectangular guides, however, that’s why westarted with themore mathematical analysis which works, inprinciple. forguides ofany shape. For themode wehave described. thevertical dimension (in»)had noeffect, sowecanignore thetopand bottom oftheguide and imagine that theguide 1s extended indefinitely inthevertical direction. We imagine then that theguide Just consists oftwo vertical plates with theseparation a. Let's saythat thesource ofthefields isavertical wire placed inthemiddle of the guide, with the wire carrying acurrent that oscillates atthe frequency w. Intheabsence ofthegurle walls such awire would radiate cylindrical waves 2410 Now weconsider that theguide walls areperfect conductors. Then, justasin electrostatics, the conditions atthe surface will becorrect ifweadd tothe field of thewire thefield ofone ormore suitable image wires. The image idea works just aswell forelectrodynamics asitdoes forelectrostatics, provided, ofcourse, that we also include the retardations. We know that istrue because we have often seenamirror producing animage ofalightsource. Andamirror isjusta“perfect” °#*~ conductor forelectromagnetic waves with optical frequencies. Nowlet’stakeahorizontalcrosssection,asshowninFig.24-15,whereW,Sy. and W,arethetwo guide walls and Spisthesource wire. Wecallthedirection of SSmace thecurrent inthewirepositive. Now ifthere wereonlyonewall,sayW,,wecould Se Somes remove itifweplaced animage source (with opposite polarity) attheposition wi marked S;.ButwithbothwallsinplacetherewillalsobeanimageofSointhe gone ce OXwallW’,which weshowastheimage So.Thissource, too,willhaveanimage in Pama W,,which wecallSs.Now both S;andSswillhave images in1»atthepositions We marked S,andSq,andsoon,Forourtwoplaneconductors withthesource *"\ suage halfway between, thefieldsarethesameasthoseproduced byaninfinite hneof 2omesources, allseparated bythedistance a.(Itis,mfactjustwhat youwould seeif See you looked atawire placed halfway between (wo parallel mirrors.) Forthefields tobezero atthewalls, thepolarity ofthecurrents intheimages must alternate sje- from one image tothenext. Inother words, they oscillate 180° outofphase. The waveguide field is,then, just thesuperposition ofthefields ofsuch aninfinite ‘Fig. 24-15. The line source Sobe- setofline sources tween the conducting plane walls Ws Weknow thatifweareclose tothesources, thefield 1svery much likethe andWa. Thewalls canbereplaced by static fields. Weconsidered inSection 7-5thestatic fieldotagridoflinesources _*¢infinite sequence ofimage sources. andfound that it1slikethefield ofacharged plate except forterms that decrease exponentially with thedistance from thegrid. Here theaverage source strength iszero, because thesign alternates 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. Sonow weseewhy thewaveguide below cutoff fre~ quency gives anexponentially decreasing field. Atlow frequencies, inparticular. thestatic approximation isgood, and itpredicts arapid attenuation ofthefields with distance. Now wearefaced with theopposite question: Why arewaves propagated , \ atall? That isthemysterious part! The reason isthat athigh frequencies the . retardation ofthefields canintroduce additional changes inphase which cancauses... \thefields oftheout-of-phase sources toaddinstead ofcancelling. Infact, in ey Chapter 29ofVol.Iwehavealready studied, justforthisproblem, thefields \Agenerated byanarrayofantennas orbyanoptical grating. There wefound that °** \ fox x. \ when several radio antennas aresuitably arranged, they cangiveaninterference \, wan pattern thathasastrong signal insome direction butnosignal inanother. Seo » @ \ ‘Suppose wegobacktoFig.24-15 andlookatthefieldswhich arrive ata VX ee \largedistance fromthearrayofimagesources. Thefieldswillbestrong onlyin SS Nee N. certain directions which depend onthefrequency—only inthose directions for KN . which thefieldsfromallthesources addinphase. Atareasonable distance from Ke Oe \ thesources thefieldpropagates inthese special directions asplane waves. Wehave . S\N sketched suchawaveinFig.24-16, where thesolidlinesrepresent thewavecrests Koy \SN andthedashed linesrepresent thetroughs. ‘Thewave direction willbetheone S* yi\a\ forwhich thedifference intheretardation fortwoneighboring sources tothecrest KAA ofawave corresponds toone-halfaperiodofoscillation. Inotherwords,the54 Vs\o\y differencebetween r2androinthefigureisone-half ofthefree-space wavelength: NN\A none Fig.24-16,Onesetofcoherent The angle #sthen gwen by . wove:fromonarrayofinesources, sing=32. (24.33) There is,ofcourse, another setofwaves travelling downward atthesymmetric angle with respect tothearray ofsources. Thecomplete waveguide field (not too Pray close tothesource) 1sthesuperposition ofthese two sets ofwaves, asshown 1n Fig. 24-17. The qctual fields arereally ikethis, ofcourse, only between thetwo walls ofthewaveguide. Atpomnts like Aand C,thecrests ofthetwo wave patterns comeide, and the field will have amaximum, atpoints hike B,both waves have their peak negative value,andthefieldhasitsminimum(largestnegative)value.Astimegoeson See, . thefield intheguide appears tobetravelling wlong theguide with awavelength :JNY, %wwhichisthedistancefromAtoC,Thatdistanceisrelatedto#by —" ef 2 cos9=2. (2434) Sort 8 SFaK UsingEq,(24.33) for8,wegetthat we CR s 43s\/a cos0VT=(hg2a? ee “ which1sjustwhatwefoundinEq,(2419) Fig.24-17. Thewaveguide field can Now weseewhy there isonly wave propagation above thecutoff frequencybeviewedasthesuperposition oftwo» I'thefree-space wavelength 1slongerthan2u,there1snoanglewherethewaves,trains ofplone waves. shown inFig 24-16 canappear. The necessary constructive interference appearssuddenly When\ydropsbelow2a,orWhenwgoesabovewy~¥C/d.Ifthefrequency 1shigh enough, there can betwo ormore possible directions inwhich thewaves willappear. For ourcase, thiswill happen ifXy<3a In general, however, itcould also happen when Xy<a. ‘These additional waves correspond tothehigher guide modes wehave mentioned. Ithas also been made evident byour analysis why thephase velocity ofthe guided waves 1sgreater than cand why this velocity depends on@Asis changed, theangle ofthefree waves ofFig. 24-16 changes, and therefore sodoes thevelocity along theguide. Although wehave described theguided wave asthesuperposition ofthefields ofaninfinite array oflinesources, you can seethat wewould arrive atthesame result afweimagined two sets offree-space waves being continually reflected back and forth between two perfect mirrors—remembering that areflection means a reversal ofphase. These sets ofreflecting waves would allcancel each other unlesstheyweregoingatjust theangle @given inEq. (24.33) There aremany ways of Joking atthesame thing. 12 25 Electrodynamies inRelativistic Notation 25-1 Four-vectors Wenow discuss theapplication ofthespecial theory ofrelativity toelectro- ——-25-1. Four-veetors dynamics. Since wehavealready studied thespectal theory ofrelativity inChapters lar product15through17ofVol.1,wewilljustreviewquicklythebasicideas. 25-2ThescalarProductItisfound experimentally that thelaws ofphysics areunchanged ifwemove 25-3. The four-dimensional gradient Withuniform velocity. Youcan’ttellifyouareinside aspaceship moving With 95gEtectrodynamies inuniform velocity inxstraight ne,unless youlookoutside thespaceship, orat fou-dimeaziceal notation Teast make anobservation having todowith theworld outside. Any true lawof physics wewrite down must bearranged sothat this fact ofnature 1Sbuilt an 28-5 The four-potential ofa The relationship between thespace and time oftwo systems ofcoordinates. moving charge one,$nuniformmationthesehretionwithspedrrelative(otheother.S.356hevarianceoftheequationsgivenby reneetransformatee ofelectrodynamics 1 tom , t= yanvie i Lea(5.1) Inthischapter: c=1 oxo , ne oevi-w® The laws ofphysics must besuch that after aLorentz transformation, thenew form ofthelaws looks justliketheoldform. This isjustliketheprinciple that ; thelawsofphysicsdon’tdependontheorientation ofourcoordinate system. In Fer nev uneChapter11ofVol1,wesawthatthewaytodescribemathematically theinvariance Seca ThornofRelay ofphysieswithrespecttorotations wastowriteourequations intermsofvectors. neenevelaeForexample, ifwehavetwovectors fannie Energy and Mo- mentum A= (AeAg A) and B= (Be.ByBE, Chapter 17,Vol.1,Space- Tune wefound that thecombination Chapter 13.Vol. I,Mag- netostaties AB =AB, +AyBy +ABs was notchanged ifwetransformed toarotated coordinate system. Soweknow that ifwehave ascalar product likeA«Bonbothsidesofanequation,theequation will have exactly thesame form inallrotated coordinate systems. Wealso dis- covered anoperator (see Chapter 2), aaa ve(22-2): which, when applied toascalar function, gave three quantities which transform gust lke avector With this operator wedefined thegradient, and incombination with other vectors, thedivergence and theLaplacian, Finally wediscovered that bytaking sums ofcertam products ofpairs ofthecomponents oftwo vectors we could getthree new quantities which behaved like anew vector. Wecalled itthecrossproductoftwovectorsUsingthecrossproductwithouroperatorVwethendefined the curl ofavector Since wewill bereferring back towhat wehave done invector analysis. we have put inTable 25-1 asummary ofalltheimportant vector operations in three dimensions that wehave used inthepast. The point isthat stmust bepossible towrite theequations ofphysics sothat both sides transform thesame way under 25-1 rotations. Ifone side isavector, the other side must also beaveetor, and both sides willchange together 1nexactly thesume way ifwerotate ourcoordinate sys tem Similarly, ifone side 1sascalar, theother side must also beascalar, sothat Table 25-1 neither side changes when werotate coordinates, andsoon. “Theimportant quantities andoperations Nowinthecaseofspecial relativity, timeandspace areinextricably nved., oftector analysis inthreedimensions.» #nktwemustdotheanalogous things forfourdimensions Wewantourequtations sore ase nes" toremain thesame notonly forrotations, but also forany inertial frame. That Definition ofa means that our equations should besnvariant under theLorentz transformation vector A=AndyA)|ofequations(25.1).Thepurposeofthischapter1stoshowyouhowthatcanbe |Scalar product 4-8 done, Before wegetstarted, however. wewant todosomething thatmakes our 1fesene work aloteasier (and saves some confusion) And that 15tochoose our units ofjParent eeor Tengthandtimesothatthespeedoflight¢1sequalto1Youcanthinkof1tas| Greene te takingourunitoftumetobethetimethat1takeslight10goonemeter(which1s | * about3X10~"sec)Wecanevencallthistimeunit“onemeter.”Usingthis Divergenceva unit,allofourequationswillshowmoreclearlythespace-timesymmetryAlso, |Laplacian vevare allthec’swilldisappear fromourrelativistic equations. (Ifthisbothersyou. Cross product AX B you can always putthec'sback into any equation byrephieing every 1byef.oF.in Curl wx general, bysticking ina¢wherever it15needed tomake thedimensions ofthel- ~ equations comeoutright.) Withthisgroundwork weareready10begin Ourprogramistodointhefourdimensions ofspace-time allofthe things wedidwith vectors forthree dimensions. Itisreally quite asimple game. weyust work by analogy The only real complications 1sthenotation (we've already used upthe vector symbol forthree dimenstons) and one slight twist ofsigns Fist, byanalogy with vectors inthree dimensions, wedefine afowr-neeror ay set ofthefour quantities [email protected] ds,which transform likefx, ¥.and =when wechange toamoving coordinate system. There areseveral different notations people useforafour-vector: wewill write g,,bywhich wemean thegroup offour numbers (@%, dr.dy.d.)—in other words, the subscript #can take onthefour “values” 1.x,y.2Itwallalsobeconvenient,attimestoindicatethethreespace components byathree-vector, lkethis: a,=(ana) We have already encountered one four-vector, which consists oftheenergy and momentum ofaparticle(Chapter 17,Vol.1).Inournewnotationwewrite Pr.=(Ep), (25.2) which means that thefour-vector p,18made upoftheenergy Eand thethree components ofthethree-veetor pofaparticle, Ttlooks asthough thegame 1sreally very simple—for each three-vector in physics allwehave todoisfind what theremaining component should be,andwe have afour-vector Toseethat this 1snot thecase, consider thevelocity vector with components ar ar re2 eG a Ue The question 1s:What 18thetime component? Instinct should give theright answer. Since four-vectors arelike 1x, y.z,wewould guess that theume com- ponent is a waGel Tins 15wrong The reason isthat the #ineach denominator isnot aninvariant when we makeaLorentztransformation Thenumerators havetherightbehavior tomake afour-vector, butthedrinthedenominator spoils things: 11sunsymmetrie and1snotthesame intwodifferent systems. Itturns outthat thefour “velocity” components which wehave written down will become thecomponents ofafour-vectorifwejustdividebyVT=r,We can see that that istrue because ifwe start with the momentum four-veetor a=(Ep)=(o@-a): 53)Vie VP 252 Sowewillfollow theconvention and write thedotproduct simply as«jb. So, bydefinition, bg=ib—abe=yy~ashe 25.7) Whenever you seetwo identical subscripts together (we will occasionally havetouse¥oFsomeotherletterinsteadofu)itmeansthatyouaretotakethefourproducts and sum, remembering theminus sign for the products ofthe space components. With thisconvention theinvariance ofthescalar product under a Lorentz transformation can bewritten as a,b, =ayby. Since thelast three terms in(25.7) arejust thescalar dot product 1nthree dimensions, itisoften more convenient towrite Abe =aby ~ab, Itisalso obvious that the four-dimensional length wedescribed above can be written asaay: aya, =a—a3=ay~as=a—aa. (25.8) Itwillalsobeconventent tosometimes write thisquantity asaf: G5=aay. Wewillnow giveyouanillustration oftheusefulness offour-vector dot products, Antiprotons (P) are produced inlarge accelerators bythe reaction P+PoP+P+P+P. That is,anenergetic proton collides with aproton atrest (for example, inahy- drogen target placed inthebeam), and ifthemeident proton hasenough energy, aproton-antiproton parmay beproduced, inaddition tothetwooriginal protons.” The question is:How much energy must begiven totheincident proton tomake thsreaction energetically possible? The easiest way togettheanswer istoconsider what thereaction looks hike 1mthecenter-of-mass (CM) system (see Fig. 25-1). We'll call theinculent proton and usfour-momentum py Similarly, we'll al thetarget proton band itsfour BEFORE AFTER a o © q|% e a S2)e- os) Fig. 25-1. The reaction P+P — — — 3P+P viewed inthelaboratory and o eCMsystems. Theincidentprotonissup-23, Pa 4 PaposedtohovejustborelyenoughenergySe|@—t—= OB 3fomake the reaction go. Protons ore denoted bysolid circles; antiprotons, by F open circles. ee *You may well ask: Why not consider the reactions PHPoP+ PHP, or even PEPOPHEP which clearly require less energy? The answer 1sthat aprinciple called conservation of darvons tes usthequantity “number ofprotons minus number ofuntiprotons” cannot change. Thus quantity 1s2ontheleft side ofour reaction. Therefore, sfwewant an antiproton ontheright side, wemust have also three protons (orother baryons). 25-4 theD'Alembertian and hasaspecial notation: at-vy,- 2-92 (25.20) We=3— From itsdefinition itisaninvariant scalar operator; ifitoperates onafour-vector field. itproduces anew four-vector field. (Some people define theD’Alembertian with theopposite sign toEq.(25.20), soyou will have tobecareful when reading thehiterature.) We have now found four-dimensional equivalents ofmost ofthethree- dimensional quantities wehad listed inTable 25-1. (We donot yethave the ‘equivalents ofthecross product andthecurl operation; wewon't gettothem until thenext chapter )Itmay help you remember how they goifweputalltheimpor- tantdefinitions andresults together inoneplace, sowehave made such asummary inTable 25-2. Table 25-2 “The important quantities ofvector analysis inthree and four dimensions. i- — | j ‘Threedimensions Fourdimensions | |Vector A= (An AyAD = (Wisdendayds)=(aes) |Scalarproduct A+B=A,B,+AyBy+A.B:Uyby=aby—abe—ayy—ab,=ay~ab| Vector operator V=@/Ax,0/Ay,0/02) Vu@/a1,—O/Ax, —A/dy, —0,82) =O/H, —T) OA,,Ay,AA, ar,Ste,day,da.ae | |Divergence vid oetay tae Nae an aeFay Hae aFM Laplacianand eee ee ee eeD'Alembertian Vv=gatgatan =a8—ae~aeanaeY79 25-4 Electrodynamics infour-dimensional notation ‘Wehave already encountered theD'Alembertian operator, without giving it that name, inSection 18-6. thedifferential equations wefound there forthepo- tentials can bewritten inthe new notations as: m8, gyed.Oe=PaoAPa (25.21) The four quantities ontheright-hand side ofthetwo equations in(25.21) are .JesJoJodivided by€o,which 1sauniversal constant which wall bethesume inallcoordinate systems ifthesame unit ofcharge isused inallframes. Sothefour quantities p/€9, j+/€0. Jy/€0s js/€o also transform asafour-vector Wecan write them asy,/ey TheD’Alembertian doesn’t change when thecoordinate system 1schanged, sothequantities @A,,A,4:must alsotransform likeafour-vector— which means that they arethecomponents ofafour-vector. Inshort. Ay=(¢.A) 1safour-vector. What wecallthescalar andvector potentials arereally different aspects ofthesame physical thing. They belong together And ifthey arekept together therelativistic invariance oftheworld 1sobvious Wecall4,thefour- 258 Inthefour-vector notation Eqs. (25.21) become simply a4, =2, (25.22) © “The physics ofthisequation isjustthesame asMaxwell's equations, Butthere issomepleastireinbeingabletorewritetheminanelegantform.Theprettyform isalso meaningful, itshows directly theinvariance ofelectrodynamics under the Lorentz transformation. Remember that Eqs (25.21) could bededuced from Maxwell’s equations only ifweimposed thegauge condition Bevo, (25.23)ar which just says 0.4, =0;thegauge condition says that thedivergence ofthe four-vector 4,18zero. This condition iscalled theLorentz condition. It1svery convenient because it1saninvariant condition and therefore Maxwell's equations stay intheform ofEq.(25.22) forallframes. 25-5 The four-potential ofamoving charge Although 1tisimphcit inwhat wehave already said, letuswrite down the transformation laws which give ¢and Ainamoving system interms of and A ina stationary system. Since 4,=(6,A)isafour-vector, theequations must Took just like Eqs. (25.1), except that ¢18replaced by@,and xisreplaced byA Thus, ” !@~vA, v of=SE A=Ay Je tsvize Pots .Ay—1% 5.24) ~ PAn Ba, am A, ae view ;poe ae' This assumes that theprimed coordinate system ismoving with speed vin the + Tal f positive x-direction, asmeasured intheunprimed coordinate system. “% mSWewillconsider oneexample oftheusefulness oftheideaofthefour-potential . x What arethevector andscalar potentials ofachargegmovingwithspeed1along Fig.25-2.ThefromeS!moveswith thex-axis? Theproblem 1seasy inacoordinate system moving with thecharge. _yeloeity vlinthex-direction) withrespect since inthissystem thecharge isstanding still. Let's saythat thecharge isatthe toS.Acharge atrestattheorigin ofS’originofthe S’-frame, asshown inFig. 25-2. The scalar potential inthemoving isatx =vtinS.Thepotentials atPcan system isthen given by ‘bbecomputed ineither frame. v=Trev? (25.25) being thedistance from qtothefield point, asmeasured inthemoving system The vector potential A’1s,ofcourse, zero Now it1sstraightforward tofind ¢and A,thepotentials asmeasured inthe stationary coordinates. The inverse relations toEqs. (2524)are oO+ode oH Ste, A=,vir " a (25.26) 4p=BAM, A,=A vi= Using the9"given byEq. (25.25), and 4’=0,weget -4.1 _ ** Gre pio re ire DWE 259 26 Lorentz Transformations of the Fields 26-1 The four-potential ofamoving charge Wesaw inthelast chapter that thepotential a,=(¢,A)is@four-vector. 26-1 The four-potential ofa Thetime component 1sthescalar potential 4,andthethree space components are moving charge thevector potential 4.Wealso worked outthepotentials ofaparticle moving with iruniformspeedonastraightlinebyusingtheLorentztransformation (Wehad22ieetdofpoichargealready found thembyanother method inChapter 21.)Forapointcharge whose y position atthetime ris(0,0, 0),thepotentials atthepoint (x.y.2)are 26-3 Relativistic transformation aofthe fields o> ew ae 26-4 Theequations ofmotion indreovi=[a tere=| relativisticnotation A=— eamF i @6.) Inthis chapter:¢=1inet a(SOFyy2° Inthischapter:c= Ay=As=0 Equations (261)givethepotentials atx.y,andzatthetime1,foracharge Review: Chapter 20,Vol.I,Solution whose “present” position (bywhich wemean theposition atthetimet)isatx =vr ofMaxwell's Equations in Notice that theequations areintermsof(x—rv),y,andz,whicharethecoordi FreeSpace nates measured from thecurrent position Pofthemoving charge (see Fig. 26-1) ‘Theactual influence weknow really travels atthespeed c,soitisthebehavior of thecharge back attheretarded position Pthat really counts. The pointP”isat x=ri!(where, =1—~1’/cisthe retardedtime)Butwesardthatthechargewasmoving with uniform velocity inastraight line, sonaturally thebehavior atP”andthecurrentpositionaredirectlyrelated,Infact,ifwemaketheaddedassumptionthatthepotentials depend only upon theposition and thevelocity attheretarded moment, wehave inequations (26.1) acomplere formula forthepotentials for charge moving anyway. Itworks thisway. Suppose thatyouhaveacharge "| moving insome arbitrary fashion, saywith thetrajectory inFig. 26-2, and you aretrying tofind thepotentials atthepomnt (x,»,2).First, you find theretarded position P”andthevelocity 1atthat point. Then youimgaine that thecharge pen) would keep onmoving with thisvelocity during thedelay time (1°—1),sothat RETARDED ‘ itwould thenappear atanimaginary position P,,,,,, Which wecancallthe“pro- nemion a » jected position,” and would arrive there with thevelocity+’.(Ofcourse,itdoesn’t|) Amse| dothat:itsrealpositionat1isatP.)Thenthepotentials at(x,y,2)arejustwhat ey asiequations(261)wouldgiveforthesmaginary chargeattheprojectedposition fa! nerPry.Whatwearesayingisthatsincethepotentials dependonlyonwhatthe|[7vt) :charge 1sdoing attheretarded ume, thepotentials will bethesame whether the | charge continued moving ataconstant velocity orwhether 1tchanged itsvelocity! after /—that 15,after thepotentials that were going toappear at(x,y,2)atthetime(werealeadydetermined. 10Boe|FinchnatheFeesotPdue Youknow,ofcourse,thatthemomentthatwehavetheformulaforthepo-WahtheconstontspeedvThefield tentialsfrom acharge moving inanymanner whatsoever, wehave thecomplete “now” ‘etthepoint (x,y,2)canbeex: electrodynamics; wecangetthepotentials ofanycharge distribution bysuper- pressed interms ofthe“present” position —— P,aswell asin terms ofP’the“retarded”+Theprimesusedheretoindicatetheretardedpositionsand timesshouldnotbeconfused position(attf=t—e'/e). with theprimes referring toaLorentz-transformed frameintheprecedingchapter.26-41 position. Therefore wecan summarize allthephenomena ofelectrodynamics either bywriting Maxwell's equations orbythefollowing series ofremarks. (Remember them incase youareever onadesert island. From them, allcan be reconstructed. You will, ofcourse, know the Lorentz transformation; you will never forget shar onadesert sskand oranywhere else ) First, A,18afour-vector. Second, theCoulomb potential forastationary charge isq/4m¢yr. Third, thepotentials produced byacharge moving inanyway (sox) depend only upon thevelocity and position atthe retarded time With those A threefactswehaveeverything Fromthefactthat4,1safour-vector, wetransform LY. theCoulomb potential, whichweknow,andgetthepotentials foraconstantx velocity. Then,bythelaststatement thatpotentials dependonlyuponthepast 4 velocity attheretarded time, wecanusetheprojected position game tofind them. 4 Itsnotaparticularlyusefulwayofdoingthings.butitisinterestingtoshowthat ssnempwore. foo) thelaws ofphysics canbeputinsomany different wayseen sonergo Itissometimes said,bypeoplewhoarecareless,thatallofelectrodynamics 3 sanecanbededuced solely from theLorentz transformation and Coulomb's law. OF tansecromYe PB course, that 1scompletely false. First, wehave tosuppose that there 1sascalar “ potential andavector potential that together make afour-vector That tells us how thepotentials transform Then why isitthat theeffects attheretarded Fig. 26-2. Acharge moves onon timearetheonlythingsthatcount?Betteryet,why1sitthatthepotentialsdepend corbitrary trajectory. The potentials at only ontheposition andthevelocity and not, forinstance, ontheacceleration? (x,y,2}atthetimetaredetermined byTheficldsEandBdodependontheacceleration. Ifyoutrytomakethesame thepositionP’andvelocityv’attheandofanargument withrespecttothem,youwouldsaythattheydependonlyretarded time1’—«They orecom upontheposition andvelocity attheretarded timeButthenthefieldsfromanvemently expressedintermsoftheco”accelerating chargewouldbethesameasthefieldsfromachargeattheproyected ordinates from the “projected” positionPoa{theactual postion attsP) position—which isfalse.Thefieldsdepend notonlyonthepositionandthevelocity “along thepath butalso ontheacceleration. Sothere areseveral additional tacit assumptions inthis great statement that everything can bededuced from theLorentztransformation (Whenever youseeasweeping statement thatatremen-dous amount cancome from avery small number ofassumptions, you always find that itisfalse. There areusually alarge number ofimplied assumptions that arefarfrom obvious ifyou think about them sufficiently carefully.) 26-2 The fields ofapoint charge with aconstant velocity Now that wehave thepotentials from apoint charge moving atconstant velocity, weought tofind thefields—for practical reasons There aremany cases where wehave uniformly moving particles—for instance, cosmic rays going through acloud chamber, oreven slow-moving electrons inawire. Solet's atleast see what thefields actually dolook like forany speed—even forspeeds nearly thatoflight—assuming onlythatthereisnoacceleration. Itisaninteresting question‘Wegetthefields from thepotentials bytheusual rules E=-ve- 4,BoyxXa Furst, forE; 6_ade Ee~a ar But 4,15zero: sodifferentiating @inequations (261),weget E,=—_W_ a (26.2) 4reVT==|a [* Similarly, forEy, E,=— oF J... (26.3) ‘The x-component isaInttle more work. The derivanve of¢ismore complicated 26-2 and A,isnot zero. First, x—nid= ~#-—— afaa.aoe 26.4)dreVT— [SS ey+2i Then, differentiating 4,with respect to1,wefind 22 =wed =o-te=— &oeWA=Re (265)dregVT=%(&SF+ee=| And finally. taking thesum, &=—4_ ,.4 (26.6)rev=8[Se=et=f r mA We'll look atthephysics ofEinaminute.let'sfirstfindB_Forthez-compo-| wale, nent, | J) pwtyOde j i=BS | J : SinceA,iszero.wehavejustonederivative toget.Notice, however, that4, | Bo' L isjustr4,and3/ay ofr@isjust —rE,. So v2 ~ \pneseurB.=Ey (26.7)| verre Simitarly, 2B,=ths ade, 88 Fig.26-3. For@charge moving with= a oe tar? constant speed, theelectric field points and radially from the “present” position of B,=~0B. (26.8) thecharge. Finally, Bis zero. since Ayand4,areboth zero. Wecanwrite themagnetic field simply as B=uXE (26.9) ie Nowlet'sseewhatthefieldslookItke.Wewilltrytodrawapictureofthe rs fieldatvarious positions around thepresent position ofthecharge. Itsteuethat N\ 1/4 theinfluence ofthecharge comes, inacertain sense. fromtheretarded position, ~ \I/ =butbecause themotion isexactly specified, theretarded position 1suniquely given NA 1mtermsofthepresentpositionForuniformvelocities,it’snicertorelatethebveoIK afieldstothecurrentposition,becausethefieldcomponents at(x.y.2)depend '9)VO ae)onlyon(x—v1),y.andz—which arethecomponents ofthedisplacements LIN rpfromthepresent position to(x,»,2)(seeFig.26-3). ‘4 yyConsider firstapoint with 2=0.Then Ehasonly x-andy-components. yy \ FromEqs.(26.3) and(26.6), theratioofthesecomponents 1sjustequal tothe cratio ofthex-andy-components ofthedisplacement. That means that E1sin \ thesumedirection asFr,asshown1nFig.26-3.SinceEsalsoproportional to2. Vditisclearthatthisresult holdsinthreedimensions. Inshort, theelectric field1s yfradial fromthecharge, andthefieldlinesradiate directly outofthecharge, just Why astheydoforastationary charge. Ofcourse, thefieldisn’texactly thesameas. BN2 v forthestationary charge. because ofalltheextra factors of(I—2) Butwe oo Dh OT canshowsomething rather interesting. Thedifference 1sjustwhatyouwould get. (P)¥=O.9C ¥//\\ ifyouweretodraw theCoulomb fieldwithapecuhar setofcoordinates inwhich N\y the scale ofxwassquashed upbythefactor\/T—12.Ifyoudothat,thefield lyy Iineswillbespread outahead andbehind thecharge andwillbesqueezed together ‘an around thesides, asshown inFig. 26-4.IfwerelatethestrengthofEtothedensityofthefieldlinesintheconventional 4,Theelectticfieldof@ way,weseeastrongerfieldatthesidesandaweakerfieldaheadandbehind. 4,40,eatingeamineconsentspeed whichisjustwhattheequations say.Furst,ifwelookatthestrength ofthefield """“Q.gc, part(b),compared with,the atright angles tothelineofmotion, thatis,for(x—vf)=0,thedistance from fieldofacharge ofrest,par(a). 26-3 thecharge is(y?+2%). Here thetotal field strength is/E}+E%,whichis a a 26.10) dren1—2YF e510) The field isproportional totheinverse square ofthedistance—yust liketheCou- Iomb field except increased bytheconstant, extra factor 1//T =02, which 1s always greater than one. Soatthesides ofamoving charge, theelectric field 1s stronger than you getfrom theCoulomb law. Infact, thefield inthesidewise direction isbigger than theCoulomb potential bytheratio oftheenergy ofthe particle toitsrest mass. ‘Ahead ofthecharge (and behind), yand zarezero and ~a=0) EnB=op (26.11) The field again varies astheinverse square ofthedistance from thecharge but1s nowreduced bythefactor (I—12),inagreement with thepicture ofthefieldlines. Ifn/cissmall, r?/c? isstillsmaller, andtheeffect ofthe(I—?)terms isvery small; wegetback toCoulomb's law. But ifaparticle ismoving very close to thespeed oflight, thefield intheforward direction isenormously reduced, and thefield inthesidewise direction isenormously increased. Our results fortheelectric field ofacharge can beput this way: Suppose youwere todraw onapiece ofpaper thefield lines foracharge atrest, andthen setthepicture totravelling with thespeed v.Then, ofcourse, thewhole picture would becompressed bythe Lorentz contraction; that is,thecarbon granules onthepaper would appear indifferent places The miracle ofit1sthat thepicture you would seeasthepage flies bywould stillrepresent thefield Ines ofthepoint charge. The contraction moves them closer together atthesides and spreads them outahead and behind, just intheright way togive thecorrect line densities. We have emphasized before that field lines arenotrealbutareonly oneway ofrepre- senting thefield. However, here they almost seem tobereal. Inthis particular case, ifyou make themistake ofthinking that thefield lines aresomehow really there inspace, andtransform them, yougetthecorrect field. That doesn’t, however,makethefieldlinesanymorerealAllyouneeddotoremindyourselfthatthey 7 aren'trealistothinkabouttheelectricfieldsproducedbyachargetogetherwith [4magnet; when themagnet moves, new electric fields areproduced, anddestroy { thebeautiful picture Sotheneat idea ofthecontracting picture doesn’t work 1n —|—\y> general. It1s,however, ahandy waytoremember what thefields from afast- 4 moving charge areIke.we) ‘Themagneticfield1s»XE[fromEq,(26.9)}.Ifyoutakethevelocitycrossedinto aradial E-field, you get aBwhich circles around theline ofmotion, asshown Fig. 26-5. The magnetic field near inFig. 26-5. Ifweputback thec’s,youwillseethat 1'sthesame result wehad ©moving chorge isvXE. (Compare forlow-velocity charges. Agood waytoseewhere thec’smust go1storefer back withFig.26-4.) totheforce law, F= QE +x B). You seethat avelocity times themagnetic field hasthesame dimenstons asan electric field. ‘Sotheright-hand sideofEq.(26.9) must have afactor 1/c*: Ba2X. (2612) For aslow-moving charge (<c),wecan take forEtheCoulomb field: then 4 xr 8sree or! 26.13) This formula corresponds exactly toequations forthemagnetic field ofacurrent that we found inSection 14-7. 26-4 ‘We would liketopointout,inpassing,somethinginterestingforyoutothink ett 4 about.(Wewillcomebacktodiscussitagainlater.)Imaginetwoelectronswith(gy i ‘.velocities atright angles, sothat onewillcross over thepath oftheother, butin 2 front ofit,sothey don’t collide. Atsome instant, their relative positions wallbe asinFig 26-6(a), Welook attheforce onq;due toqaand vice versa. Onga there isonly theelectric force from qx,since1makesnomagneticfieldalongitsF4y,x8, lineofmotion. Onq;,however, there1sagain theelectric force but,inaddition, et . amagnetic force,sinceitismoving inaB-field madeby2.Theforcesareasdrawn (®) |NY EQ h inFig. 26-6(b). The electric forces onq;andq»areequalandopposite.However, Ge,8,ig thereisasidewise (magnetic) force onq,andnosidewise force onqa.Does action notequal reaction? Weleave itforyou toworry about. Fig. 26-6. The forces between two 26-3 Relativistic transformation ofthefields moving charges arenetalways equal andoppotite. Itappears that “action” isnot Inthelast section wecalculated theelectric and magnetic fields from the equal fo“reaction.” transformed potentials, ‘The fields areimportant, ofcourse, inspite oftheargu ments given earlier that there isphysical meaning and reality tothepotentials Thefields, to0, arereal. Itwould beconvenient formany purposes tohaveaway tocompute thefields inamoving system ifyou already know thefields insome “est” system. We have thetransformation laws for and A,because A,isa four-vector. Now we would like toknow the transformation laws ofEand B. Given Eand Binoneframe, how dothey look inanother frame moving past? Iisa convenient transformation tohave. Wecould always work back through the potentials, but it1suseful sometimes tobeable totransform thefields directly Wewillnow seehow that goes. How can wefind the transformation laws ofthe fields? We know the trans- formation laws ofthe@and A,and weknow how thefields aregiven interms of 6and A—it should beeasy tofind thetransformation fortheBand E,(You might think that with every vector there should besomething tomake itafour- vector, sowith Ethere's gottobesomething else wecan useforthefourth com- ponent. And also forB.Butit'snotso. It’squite different from what you would expect.) Tobegin with, let's take just amagnetic field B,which is,ofcourse ¥Xd. Now weknow that thevector potential with itsx-,-,and z-components isonly apiece ofsomething; there isalso acomponent. Also weknow that for derivatives likeV,besides thex,»,zparts, there 1salso aderivative with respect to 1.Solet's trytofigure outwhat happens ifwereplace a‘ bya“1”, ora“2” bya“ ofsomething like that First, notice the form ofthe terms in¥XAwhen wewrite out the com- ponents aA._Ady ade_Ay Ody_As 3,=Me eg =eg (26.14) The x-component isequal toacouple ofterms that involve only yand 2-com= ponents, Suppose wecall this combination ofderivatives and components “zy-thing,” and guve itashorthand name, Fy. Wesimply mean that 0A,_aA, FoySe—2h, (26.15) Similarly, B,1equal tothesame kind of“thing,” butthistime itisan“xz-thing.”” [And B.is,ofcourse, thecorresponding ‘px-thing.” We have Be=Foy By=Fey Be=Fae (26.16) Now what happens ifwesimply trytoconcoct also some “r"-type things hikeF,,and Fi,(since nature should benice and symmetric mx,y,z,and 1)? For instance, what isF,.? Its, ofcourse, A_ade“eH 26-5 Butremember that 4,=¢,soit1salso a6 OAL, am You've seen that before. It1sthez-component ofE. Well, almost—there 1sa sign wrong Butweforgot that snthefour-dimensional gradient thederivative comeswiththeopposite signfromx,y,and=Soweshould reallyhavetakenthe more consistent extension ofF,,as aA, OAs Fi=“az.+on (26.17) Then it1sexactly equal to—E. Trying also F;,and F;,, wefind that thethree possibilities give Fy =~Ey Fy =—E, Fry = Es. (26.18) What happens ifboth subscripts are/?Or, forthat matter, ifboth arex? Wegetthings ike ad, adefem an and a4, Ade FeaOn ~Gx” which give nothing butzero. Wehave then sixofthese F-things. There aresixmore which you getby reversing thesubscripts, butthey give nothing really new, since Fey =—Fyp- ind soon. So,outofsixteen possible combinations ofthefour subscripts taken inpairs, wegetonly sixdifferent physical objets: and thev arethecomponents ofBand EB, Torepresent thegeneral term ofF,wewill usethegeneral subscripts uand», where each canstand for0,1,2,or 3—meaning inourusual four-veetor notation t.x,y.and 2Also,everything willbeconsistent withourfour-vector notation if wedefine F,,by Table 26-1 Fav=Wade —Vode 2619) Thecomponents ofFy remembering that¥,=(8/at,~0/dx,~d/ay, —4/A2)andthatAy=(4.AeAye A) Fu=Fg : What wehave found 1sthat there aresixquantities that belong together in boo| nature—that aredifferent aspects ofthesame thing. Theelectric andmagnetic imm= fields which wehave considered asseparate vectors inour slow-moving world |Fy=-Bo Fu=E.| (where wedon’tworryaboutthespeedoflight)arenotvectors infour-space,re knot Theyarepartsofanew“thing.”Ourphysical“field”tsrellythesix-component w=BePum Ey object F,,.That1sthewaywemustlookatitforrelativity Wesummatize our Fu=-By Fu Ee results on£,,1nTable 26-1 You scethat what wehave done here 1stogeneralize thecross product We began with thecurl operation, and thefact that thetransformation properties of thecurl arethesame asthetransformation properties ofvoveetors—the ordinary three-dimenstonal vector4andthegradient operator whichweknowalsobehaves like avector Let’s look foramoment atanordinary cross product ithree di- mensions, forexample, theangular momentum ofaparticle When anobject 1s moving inaplane. thequantity (xr, —pv) 8important. For motion inthree dimensions, there arethree such important quantities, which wecall theangular momentum: Lay =mxry =veo, Ly, =mor ara) Lb =mcr are) Then (although you may have forgotten bynow) wediscovered inChapter 20 ofVol Ithemiracle that these three quantities could beidentified with thecom- 26-6 S-frame with thespeed »=—u. Wewillletyoushow that thetransformations ofTables 26-3 and 26-4 give thesame electric and magnetic fields wegotinSection 26-2. The transformation ofTable 26-2 gives usaninteresting and simple answerforwhatweseeifwemovepastanysystemoffixed charges. Forexample. suppose wewant toknow thefields inourframe $’ifwearemoving along between the plates ofacondenser, asshown inFig. 26-7. (Itis,ofcourse, thesame thing ifwesaythatachargedcondenser ismovingpastus.)Whatdowesee?Thetrans _ formation iseasy 1nthis case because theB-field intheoriginal system 1szero. TST STSTs TSTeTSTSS Suppose, first,thatourmotton isperpendicular to£,thenwewillseean”=eo y E/\/1 —12/2whichisstillcompletely transverse. Wewillsee,inaddition, a yo 1fe 17magnetic fieldBY=—vXE’/e’, (TheV/T=¢? doesn't appear inourformula t forB’because wewrote itinterms ofE”rather than E;butit’sthesame thing.) yet et. 1 || 1 Sowhen wemove along perpendicular toastatic electric field, weseeareduced Eandanadded transverse B.Ifourmotion isnotperpendicular toE,webreakFig.26-7.Thecoordinate frameS’—£intoEy,andEz.Theparallelpartisunchanged, £”,=Ej),andtheperpendicular moving through @static electric field. ‘component doesasjustdescribed. Let’s take theopposite case, and imagine wearemoving through apure static magnetic field. This time wewould seeanelectric field E’equal tovXB’, andthemagnetic fieldchanged bythefactor 1/\/T—°27e?(assumingit1strans- verse). Solong as«ismuch lessthan c,wecanneglect thechange inthemagnetic field, and themain effect isthat anelectric field appears. Asone example ofthis effect, consider thisonce famous problem ofdetermining thespeed ofanairplane. It’snolonger famous, since radar can now beused todetermine theairspeed from ground reflections, butformany years itwas very hard tofind thespeed of anairplane inbad weather. You could notseetheground and you didn’t know which way was up,and soon, Yet stwas important toknow how fast you were moving relative totheearth. How can this bedone without seeing theearth?Manywhoknewthetransformation formulasthoughtofthe idea ofusing thefact that theairplane moves inthemagnetic field oftheearth Suppose that anairplane 1sflying where there isamagnetic field more orlessknown, Let's just take the simple case where themagnetic field 1svertical. Ifwewere flying through 1twith ahorizontal velocity r,then, according toour formula, weshould seeanelectric field which isvXB, ie. perpendicular totheline ofmotion Ifwehang an insulated wire across thearrplane, this electric field will andluce charges ontheends ofthewire, That 1snothing new. From thepoint ofview ofsomeone ontheground, wearemoving awire through afield, and the»XBforce causes charges tomove totheends ofthewire The transformation equations yust saythesame thing in adifferent way. (The fact that wecansaythething more than oneway doesn’t mean that one way 1sbetter than another We are getting somany different methods and tools that wecan usually getthesame result in65different ways!) Sotomeasure 1,allwehave todoismeasure thevoltage between theends of the wire. We can’t doitwith avoltmeter because the same fields will act onthe wires inthevoltmeter, but there areways ofmeasuring such fields. Wetalkedaboutsomeofthemwhenwediscussed atmospheric electricity inChapter 9.Soitshould bepossible tomeasure thespeed oftheairplane. ‘This important problem was, however, never solved this way. The reason is that theelectric field that isdeveloped 1softheorder ofmillivolts permeter. It ispossible tomeasure such fields, butthetrouble 1sthat these fields are, unfortun- ately, notany different from any other electric fields. The field that 1sproduced bymotion through the magnetic field can’t bedistinguished from some electric field that was already intheairfrom another cause, sayfrom electrostatic charges intheair,orontheclouds Wedescribed inChapter 9that there are, typically, electric fields above thesurface oftheearth with strengths ofabout 100volts per meter Butthey arequite irregular. Soastheairplane Mies through theaur, it sees fluctuations ofatmospheric electric fields which areenormous incomparison tothetinyfields produced bythe»XBerm, anditturns outforpractical reasons. tobeimpossible tomeasure speeds ofanairplane byitsmotion through theearth's magnetic field. 26-10 27 Field Energy and Field Momentum 27-1 Local conservation Ikisclear that theenergy ofmatter isnotconserved. When anobject radiates 27-1 Local conservation lightitlosesenergy. However, theenergy lostispossibly describablein someother 47>Fnergy conservation andform,sayinthelight.Thereforethetheoryoftheconservation ofenergy1s detenamnetionincomplete without aconsideration oftheenergy which isassociated with thelight or,ingeneral, with theelectromagnetic field, Wetake upnow thelawofconserva- ‘27-3 Energy density andenergy tion ofenergy and, also, ofmomentum forthefields, Certainly, wecannot treat flow intheelectromagnetic conewithout theother, because intherelativity theory they aredifferent aspects of field the same four-vector. oui VeryearlyinVolume 1,wediscussed theconservation ofenergy: wesaid 27-4Theambiguity ofthefeld thenmerely thatthetotalenergy intheworld isconstant. Nowwewanttoextend energy theidea oftheenergy conservation lawinanimportant way—in away that says 27-5 Examples ofenergy flow something inderail about how energy isconserved. ‘The new law will saythat if ‘ energy goesawayfromaregion, itisbecause itflowsawaythrough theboundaries 27-6Fieldmomentum ofthat region, Itisasomewhat stronger law than theconservation ofenergy without such arestriction, Toseewhat thestatement means, let's look athow thelaw oftheconservation ofcharge works. Wedescribed theconservation ofcharge bysaying that there is acurrent density jand acharge density p,and that when thecharge decreases at some place there must beaflow ofcharge away from that place. Wecalthat the conservation ofcharge. The mathematical form oftheconservation law is ;.v= -%. @11) Tinslawhastheconsequence thatthetotal charge intheworld isalways constant— @ there1sneveranynetgainorlossofcharge. However, thetotalcharge inthe —_\'? LD)worldcouldbeconstant inanother way.Suppose thatthereissomechargeQ,//// Minearsomepoint(1)whilethereisnochargenearsomepoint(2)somedistance Ya UWaway(Fig.27-1).Nowsuppose that,astimegoeson,thechargeQ;wereto Q, (a) Qgradually fade away and that simultaneously with thedecrease ofQsome charge Q2would appear near point (2),and insuch away that atevery instant thesum of Q,and Qzwasaconstant. Inother words, atanyintermediate state theamount a ofchargelostbyQ,wouldbeaddedtoQa.ThenthetotalamountofchargeinLZa—— the world would beconserved. That'sa“world-wide” conservation, butnotwhat —, LY wewilclla“local”conservation, becauseinorderforthechargetogetfrom—7/5» —~///, (1)to(2.itdidn'thavetoappearanywhere inthespacebetween pont(I)and 2—_ apoint(2).Locally,thechargewasjust“lost.” Q,Se 2 Thereisadifficultywithsucha“world-wide” conservation lawinthetheory ofrelativity. The concept of“simultaneous moments”atdistantpointsisonewhich ) isnot equivalent indifferent systems. Two events that aresimultaneous inone system arenotsimultaneous foranother system moving past. For“world-wide” Fig.27-1. Two ways toconserve conservation ofthekinddescribed, itisnecessary thatthechargeTostfromQ,charge: (0),Q1+Q2consort,) should appear simultaneously inQ2.Otherwise there would besome moments nidt=Ji-ado = iat when thecharge was notconserved. ‘There seems tobenoway tomake the Jawofcharge conservation relativstically 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 wecall a“nonlocal” interaction—where something here m4 27-4 ‘The ambiguity ofthefield energy Before wetake upsome applications ofthePoynting formulas (Eqs. (27.14) and (27.15)], wewould like tosaythat wehave notreally “proved” them. All Wwedid was tofind apossible “u” and apossible “S." How doweknow that by juggling theterms around some more wecouldn't find another formula for*u" and another formula for“$"? The new Sand the new uwould 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 like asecond derivative orthesquare ofafirst derivative) There are, infact, aninfinite number ofdifferent possibilities forwand S,andsofarnoonehasthoughtofan experimental way totellwhich one isright! People have guessed that thesimplest oneisprobably thecorrect one, butwemust say that wedonotknow forcertain what istheactual location inspace oftheelectro~ ‘magnetic field energy. Sowetoowilltake theeasy way outandsaythat thefield energy isgiven byEq.(27.14). Then theflow vector $must begiven byEq.(27.15). Itisinteresting that there seems tobenounique way toresolve theindefinite-nessinthelocationofthefieldenergy.Itissometimes claimedthatthisproblemcan beresolved byusing the theory ofgravitation inthe following argument, Inthetheory ofgravity, allenergy isthesource ofgravitational attraction. There fore theenergy density ofelectricity must belocated properly ifwearetoknow in which direction thegravity force acts. Asyet, however, noone hasdone such a delicate experiment that theprecise location ofthegravitational influence of clectromagnetic fields could bedetermined, ‘That electromagnetic fields alone can bethesource ofgravitational force isanidea itis hard todowithout. Tthas, in fact, been observed that light isdeflected asitpasses near thesun—we could saythat thesunpulls thelight down toward it.Doyou notwant toallow that the light pulls equally onthesun? Anyway, everyone always accepts thesimple expressions wehave found forthelocation ofelectromagnetic energy and itsflow. And although sometimes the results obtained from using them seem strange, noboby hasever found anything wrong with them—that is,nodisagreement with experiment, Sowewill follow therest oftheworld—besides, webelieve that itis probably perfectly right Weshould make onefurther remark about theenergy formula, Inthefirst place, theenergy perunit volume inthefield isvery simple: It1stheelectrostaticenergyplusthemagneticenergy,ifwewritetheelectrostatic energyintermsofE?and themagnetic energy asB*, Wefound two such expressions aspossible expressions fortheenergy when wewere doing static problems. Wealso found a number ofother formulas fortheenergy intheelectrostatic field, such aspé, which isequal totheintegral ofEE intheelectrostatic case However, 1nan. electrodynamic field theequality failed, and there was noobvious chorce astowhichwastherightone,Nowweknowwhichistherightone,Similarly, wehavefound theformula forthemagnetic energy that 1scorrect ingeneral The right formula fortheenergy density ofdynam fields isEq.(27.14) 27-5 Examples ofenergy flow E Ourformula fortheenergy flowvector $issomething quitenew.Wewant now toseehow itworks insome special cases and also toseewhether stchecks ] s outwithanything thatweknewbefore. Thefirstexample wewilltakeislightInalight wave wehave anEvector and aBvector atright angles toeach other (fe andtothedirection ofthewave propagation. (See Fig27-2.) Inanelectromag-£—wrcenonoFwave neticwave,themagnitude ofBisequalto1/ctimesthemagnitudeofE,andsince PROPAgatONthey areatright angles, Fig.27-2.ThevectorsE,B,and$ ExBj-© foralight wave. Therefore, forlight, theflow ofenergy perunit area persecond is S=ec? 27.16) 26 Foralight wave inwhich E=Epcos w(t—x/¢), theaverage rate ofenergyflowperunitarea,(5)—which iscalledthe“intensity” ofthe light—is themean. value ofthesquare oftheelectric field times €9¢: Intensity =(S)ay =€o¢(E*)aye 7.17) Believe itornot, wehave already derived thisresult inSection 31-3 ofVol. I, when wewere studying light. Wecanbelieve that itisright because italso checks against something else When wehave alight beam, there isanenergy density in spacegivenbyEq.(27.14). UsingcB=Efor alight wave, wegetthat 2 =£0 peg f° (EY _ gs ButEvaries inspace, sotheaverage energy density is Qi)av =€0o(E*Daw (27.18) Now thewave travels atthespeed c,soweshould think that theenergy that goes through asquare meter inasecond is¢times theamount ofenergy inonecubic meter. Sowewould saythat (Sav =€0e(E*)o . ‘And it’sright; itisthesame asEq.(27.17). _ ‘Now wetake another example. Here isarather curious one. Welook atthe energy flow inacapacitor that wearecharging slowly. (We don’t want frequencies sohigh that thecapacitor isbeginning tolook like aresonant cavity, butwedon’t wantDceither.)Supposeweuseacircularparallelplatecapacitorofourusual SS ‘ kind, asshown inFig.27-3. There isanearly uniform electric field inside which 1s ‘ qichangingwithume.Atanyinstantthetotalelectromagnetic energyinside1su kebsa,times thevolume. Iftheplates have aradius aandaseparation f,thetotal energy ~H e between theplates is we u=(eBean, 27.19) ;+ This energy changes when Echanges. When thecapacitor isbeing charged, the volume between theplates isreceiving energy attherate Fig.27-3. Near @charging capaci- w tor,thePoynting vector §points inward Gf=corathbb. 27.20) toward theoxis. Sothere must beaflow ofenergy into that volume from somewhere. Ofcourse youknow that itmust come inonthecharging wires—not atalll Itcan’t enter thespace between theplates from that direction, because E1sperpendicular to theplates; EXBmust beparallel! totheplates. You remember, ofcourse, that there 1samagnetic field that circles around theaxis when thecapacitor ischarging. Wediscussed that inChapter 23. Using thelastofMaxwell's equations, wefound that themagnetic field attheedge ofthe capacitor 1sgiven by 2rac*B =E+xa, or BaGk Itsdirection 1sshown inFig. 27-3. Sothere isanenergy flow proportional toEXBthat comes inallaround the edges, asshown inthe figure. The energy isn’t actually coming down thewires, butfrom thespace surrounding the capacitor. Let’scheckwhether ornotthetotalamount offlowthrough thewholesurfacebetween theedgesoftheplatescheckswiththerateofchange oftheenergy inside— ithad better; wewent through allthat work proving Eq.(27.15) tomake sure, 24 bbutlet’ssee.Theareaofthesurface1s2rah,and$=€yc*EXBasinmagnitude ee _— paces2). _~~ sothetotalfluxofenergy1s — raheEE. Itdoes check with Eq. (27.20). But ittells usapeculiar thing: that when weare charging acapacitor. theenergy isnotcoming down thewires; itiscoming in through theedges ofthegap. That's what this theory says! \ How canthatbe?That's noraneasyquestion, buthereisonewayofthinking aboutit.Supposethatwehadsomechargesaboveandbelowthecapacitorand ~~faraway. When thecharges arefuraway, there 1saweak butenormously spread- out field that surrounds thecapacitor. (See Fig. 27-4.) Then, asthecharges _ ee cometogether, thefieldgetsstronger nearer tothecapacitor. Sothefieldenergy —_=~ which 1swayoutmoves toward thecapacitor andeventually ends upbetween the plates. ‘Asanother example, weaskwhat happens inapiece ofresistance wire when it sscarrying acurrent. Since thewire hasresistance, there isanelectric field along it,Fig.27-4.Thefieldsoutsideacapacitor drivingthecurrent.Becausethere1sapotentialdropalongthewire,thereisalsowhen itisbeing cherged bybringing two anelectric fieldjustoutside thewire, parallel tothesurface. (See Fig.27-5.) charges fromalarge distance. There is,inaddition, amagnetic fieldwhich goesaround thewirebecause ofthe current. The Eand Bareatright angles; therefore there 1saPoynting vector —— directed radially inward, asshown inthefigure. There isaflow ofenergy into the wire allaround, Itis,ofcourse, equal totheenergy being lostinthewire inthe a form ofheat. Soour “crazy” theory says that theelectrons aregetting their energy togenerate heat because oftheenergy flowing into thewire from thefield ian outside. Intuition would seemtotellusthattheelectrons gettheirenergy from Mite |p beingpushed alongthewire,sotheenergy should beflowing down(orup)alongi thewire. Butthetheory saysthatthee1ectrons arereally being pushed byanelectric $ 8 field, which hascome from some charges very faraway, andthattheelectrons getD theirenergy forgenerating heatfromthesefields. Theenergy somehow flows | fromthedistantchargesintoawideareaofspaceandtheninwardtothewire. Finally,inordertoreallyconvinceyouthatthistheoryisobviouslynuts, ?‘wewill take one more example—an example inwhich anelectric charge and a _ magnet areafrestnear each other—both sitting quite still. Suppose wetake the example ofapoint charge sitting near thecenter ofabar magnet, asshown in Fig. 27-5 ThePoynting vector Snear Fig. 27-6 Everything isatrest, sotheenergy 1snotchanging with time. Also, @wirecarrying«current. EandBarequitestatic.ButthePoyntingvectorsaysthatthereisaflowofenergy, because there isan EXBthat isnotzero. Ifyoulook attheenergy flow, youfind that itjust circulates around and around, There isn’t anychange intheenergy €__anywhere—everything which flows into one volume flows outagain It1slike <7 incompressible water flowing around. Sothere isacirculation ofenergy inthis > w so-called static condition. How absurd itgets!‘ i ty Perhapsitisn’tsoterriblypuzzling, though,whenyouremember thatwhat \ wecalleda“static”magnetisreallyacirculatingpermanentcurrent.Inaperma- =_rentmagnettheelectronsarespinningpermanently inside.Somaybeacirculation 3softheenergy outside isn’t soqueer after all. Fig.27-6. Acharge ondamagnet Younodoubt begintogettheimpression thatthePoynting theoryatleastproduce Poynting setter atcieocrey Partially violates yourintuition astowhere energy 1located inanelectromagneticInclosed loops. field. Youmight believe thatyoumustrevamp allyourintuitions, and,thereforehavealotofthings tostudy here. But itseems really notnecessary You don't need tofeelthat youwillbeingreat trouble ifyou forget once inawhile that the energy inawire isflowing into thewire from theoutside, rather than along thewire.Itseemstobeonlyrarelyofvalue,whenusingtheideaofenergy conserva- tion, tonotice indetail what path theenergy 1staking. The circulation ofenergy around amagnet and acharge seems, inmost circumstances, tobequite unimpor- tant. It1snotavital detail, but itisclear that our ordinary intuitions arequite wrong. 278 which 1sjust 1/c? times theenergy flow—as thetheorem says. Sothetheorem 1s true forabunch ofparticles. Itisalso true forlight. When westudied light inVolume I,wesaw that when theenergy 1sabsorbed from alight beam, acertain amount ofmomentum 1sde- livered totheabsorber. Wehave, infact, shown inChapter 36ofVol. Ithat the momentum isI/ctimes theenergy absorbed (Eq. (36.24) ofVol. I].IfweletUo betheenergy arriving ataunit area persecond, then themomentum arriving ata unit area persecond isUp/c. Butthemomentum istravelling atthespeed c,s0its,— densityinfrontoftheabsorber mustbeUo/c?. Soagainthetheorem isright : Finally wewill give anargument due toEinstein which demonstrates the samethingoncemore.Supposethatwehavearailroadcaronwheels(assumed v.frictionless) with acertain bigmass M. Atone end there isadevice which will shoot outsome particles orlight (oranything, itdoesn’t make anydifference what 1is),which arethen stopped attheopposite end ofthecar. ‘There was some energy originally atone end—say theenergy Uindicated inFig. 27-7(a)—and then o |lateritisattheopposite end,asshown inFig.27-7(c). Theenergy Uhasbeen |displaced thedistance L,thelength ofthecar. Now theenergy Uhasthemass (a) U/e2, soifthecarstayed still, thecenter ofgravity ofthecarwould bemoved.|Einstein didn’tliketheideathatthecenterofgravity ofanobjectcouldbemoved |byfooling around onlyontheinside, soheassumed thatitisimpossible tomove a |thecenter ofgravity bydoing anything inside. Butifthatisthecase, when we = |movedtheenergyUfromoneendtotheother,thewholecarmusthaverecoiled u | some distance x,asshown inpart (c)ofthefigure. You cansee,infact, that the | _ |totalmassofthecar,times x,mustequal themassoftheenergy moved, U/c? (c)=(c)|timesZ(assumingthatU/e?ismuchlessthanM):wo) ' Mx=Zo. 27.22) Let's now look atthespecial case oftheenergy being carried byalight flash (The argument would work aswell forparticles, butwewillfollow Einstein, who y| was interested intheproblem oflight )What causes thecartobemoved” Einstein |argued asfollows: When thelightisemitted theremustbearecoil, someunknownrecoil with momentum p.Itisthis recoil which makes thecar roll backward. = |The recoil velocity »ofthecarwillbethismomentum divided bythemass ofthe 5 bx car: (©) Me Fig. 27-7. Theenergy Uinmotion ot The carmoves with this velocity until thelight energy Ugets totheopposite end. thespeed ¢carries themomentum Ue. Then, when ithits, itgives back itsmomentum and stops thecar. Ifx1ssmall, then thetime thecarmoves isnearly equal toL/c; sowehave that ema ka Btxauavta he Putting this xinEq.(27.22), wegetthat au pare Again wehave therelation ofenergy and momentum forlight. Dividing bycto getthemomentum density g=p/c, wegetonce more that u e-J¥. 27.23) You may well wonder: What issoimportant about thecenter-of-gravity theorem? Maybe itiswrong. Perhaps. butthen wewould also losetheconserva- tuon ofangular momentum. Suppose that our boxcar ismoving along atrack at some speed vand that weshoot some light energy from thetoptothebottom of thecar—say, from AtoBin Fig. 27-8. Now welook attheangular momentum of thesystem about thepoint PBefore theenergy Uleaves A,ithasthemass 27-10 m=U?/c andthevelocity 1,soithastheangular momentum mr, When it arrives atB,sthas thesame mass and, ifthelinear momentum ofthewhole boxcar isnottochange, itmust stillhave thevelocity v.It'sangular momentum about P| isthen morg. The angular momentum wall bechanged unless theright recotl momentum was given tothecarwhen thelight was emitted—that is,unless the light carries themomentum U/c, Itturns outthat theangular momentum con- SSservationandthetheoremofcenter-of-gravity arecloselyrelatedintherelativityIr ms |theory.Sotheconservation ofangularmomentum wouldalsobedestroyedifour| ,i LetheoremwerenottrueAtanyrate,itdoesturnouttobeatruegenerallaw,and| Nu| inthecase ofelectrodynamics wecan useittogetthemomentum inthefield ij Bo Wewillmention two further examples ofmomentum intheelectromagnetic t field.Wepointed outinSection 26-2thefailureofthelawofaction andreaction O° j oO} when twocharged particles were moving onorthogonal trajectories. The forces syonthetwoparticles don’tbalance out,sotheactionandreaction arenotequal. | . thereforethenetmomentumofthemattermustbechanging.Itisnotconserved || Butthemomentum imthefieldisalsochanging insuchasituation. Ifyouwork LBL ‘outtheamount ofmomentum givenbythePoynting vector, st1snotconstant Fig.27-8. the energyUmustHowever, thechangeoftheparticlemomenta isjustmadeupbythefieldmomen- 4,18 ven4raulnetum,sothetotalmomentum ofparticlesplusfieldisconserved. mentumaboutPistobeworsened.Finally, another example isthesituation with themagnet and thecharge. " shown inFig.27-6. Wewere unhappy tofind that energy was flowing around in circles, butnow, since weknow that energy flow andmomentum areproportional, wweknow also that there ismomentum circulating inthespace. But acirculating ‘momentum means that there isangular momentum, Sothere isangular momentum, inthefield. Doyouremember theparadox wedescribed inSection 17-4 about a solenoid and some charges mounted onadisc? Itseemed that when thecurrent turned off,thewhole disc should start toturn The puzzle was: Where did the angular momentum come from? Theanswer isthat ifyouhave amagnetic field and some charges, there will besome angular momentum inthefield. Itmust have been putthere when thefield wasbuilt up.When thefield isturned off,theangular momentum isgiven back. Sothedisc intheparadox would start rotating, This mystic circulating flow ofenergy, which atfirst seemed soridiculous, isab- solutely necessary. There isreally amomentum flow. Itisneeded tomaintain the conservation ofangular momentum inthewhole world. 21 28 Electromagnetic Mass 28-1 The field energy ofapoint charge Inbringing together relativity andMaxwell's equations, wehave finished our --28-1 The field energy ofapoint main work onthetheory ofelectromagnetism, There are, ofcourse, some details chargewehaveskippedoverandonelargearcathatwewillbeconcerned withinthefuture 9>-Thefieldmomentum of—theinteraction ofelectromagnetic fieldswithmatter.Butwewanttostopfora mmoringchargemoment toshow you that this tremendous edifice, which issuch abeautiful suceess inexplaining somany phenomena, ultimately falls onitsface. When 28-3 Electromagnetic mass youfollow anyofourphysics toofar,youfindthatitalways getsntosome kind 98. -theforce ofanelectron on Oftrouble. Nowwewanttodiscuss aserious trouble—the failure oftheclassical hetelectromagnetic theory.Youcanappreciate thatthereisafailureofallclassicalphysics because ofthequantum-mechanical effects. Classical mechanics isamathe- 28-5 Attempts tomodify the ‘matically consistent theory; itjust doesn't agree with experience, Itisinteresting, Maxwell theory though, that theclassical theory ofelectromagnetism isanunsatisfactory theoryallbyitself,Therearedifficulties assocrated withtheideasofMaxwell's theory 28-®Thenuclear forcefield Which arenotsolved byand not directly associated with quantum mechanics. You may say, “Perhaps there's nouseworrying about these difficulties. Since the quantum mechanics 1sgoing tochange thelaws ofelectrodynamies, weshould wait toseewhat difficulties there are after the modification.” However, when electromagnetism isjoined toquantum mechanics, thedifficulties remain. Soit Will not beawaste ofour time now tolook atwhat these difficulties are. Also, they areofgreat historical importance, Furthermore, you may getsome feeling ofaccomplishment from being able togofarenough with thetheory toseeevery thing—including allofststroubles. Thedifficulty wespeak ofisassociated with theconcepts ofelectromagnetic momentum and energy, when applied totheelectron orany charged particle The concepts ofsimple charged particles and theelectromagnetic field areinsome way inconsistent. Todescribe thedifficulty, webegin bydoing some exercises with ourenergy and momentum concepts. First, wecompute theeneray ofacharged particle. Suppose wetake asimplemodelofan electron inwhich allofstscharge qisuniformly distributed onthe surface ofasphereofradiusa,whichwemaytaketobezeroforthespecialcaseof apoint charge, Now let’s calculate theenergy intheelectromagnetic field. If thecharge isstanding still, there isnomagnetic field, and theenergy per unit volume isproportional tothesquare oftheelectric field. The magnitude ofthe electric fieldisq/4m€or*, andtheenergy density is -% pigua EB=pte Togetthetotal energy, wemust integrate this density over allspace. Using the volume element 4? dr,thetotal energy, which wewillcallUy. is ¢ Vetec=feat This isreadily integrated. The lower limit isa,and theupper limit is22,so -l@lUse=3ihea" (28.1) 4 Ifweusetheelectronic charge q.forgandthesymbol e?forq?/47¢o, then le Uses=3G" (28.2) Itisallfine until 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, but must staythereforever,isthereanyrealdifficultywithaninfiniteenergy? Ofcourse,aquantity that comes outinfinite may beannoying, butwhat really matters isonly whether there areany observable physical effects. Toanswer that question, we must turn tosomething else besides theenergy. Suppose weaskhow theenergy changes when wemove thecharge. Then, ifthechanges areinfinite, wewillbe 1mtrouble, 28-2 The field momentum ofamoving charge Suppose anelectron ismoving atauniform velocity through space, assuming foramoment that thevelocity islowcompared with thespeed oflight. Associated with thismoving electron there isamomentum—even iftheelectron hadnomass t. beforeitwascharged—because ofthemomentum intheelectromagnetic field. meee ¢__Wecanshowthatthefieldmomentum 1inthedirection ofthevelocityvoftheAle 7 chargeandis,forsmallvelocities, proportional tov.ForapointPatthedistance&[aTBA, rfromthecenterofthechargeandattheangle@withrespecttothelineofmotionCem fi (seeFig.28-1) theelectric fieldisradial and,aswehaveseen, themagnetic fieldeeHe isvXE/c?,Themomentumdensity,Eq.(27.21),is |Rese g=GEX B. It1sdirected obliquely toward theineofmotion, asshown inthefigure, andhas Fig. 28-1. Thefields EandBand the themagnitude momentum density gforapositive elec- g=2EBsing,tron. For@negative electron, EandB e crereversed butgisnot ‘Thefieldsaresymmetric aboutthelineofmotion, sowhenweintegrate over space, thetransverse components will sum tozero, giving aresultant momentum parallel tov.The component ofginthisdirection isgsin6.which wemust inte~ grate over allspace. Wetake asourvolume element aring with itsplane per- pendicular tov,asshown inFig.28-2. Itsvolume is2zr? sin@dédr.‘Thetotal or ‘momentum isthen ° 0B gin?92m?ee eeoNiad p=[itesin?92mr?sin0dodr. || ENasine SinceEisindependent of@(forv<¢),wecanimmediately integrateover6;the{“sh integral is NZ, [sinodo=~fa~cos?8)dtcos8)=cos6+S258. Fig. 28-2. The volume element The limits of@are0and 7,sotheé-integral gives merely afactor of4/3, and 2nr? sin6d8drused forcalculating the fieldmomentum. p-ee|Edn. The integral (for »<c)is theone wehave just evaluated tofind theenergy; itis @?/N6x¢a, and 24 v P* 3Gre, act’ or _2¢p=55. (28.3) 22 Clearly, assoon aswehave toputforces ontheinside oftheelectron, the beauty ofthewhole idea begins todisappear. Things getvery complicated. You would want toask: How strong arethestresses? How does theelectron shake? Doesitoscillate? Whatarealltsinternalproperties? Andsoon.Itmightbepossible that anelectron does have some complicated internal properties. Ifwe ‘made atheory oftheelectron along these lines, itwould predict odd properties, likemodes ofoscillation, which haven't apparently been observed. Wesay“ap-parently” becauseweobservealotofthingsinnaturethatstilldonotmakesense. ‘Wemay someday find outthat oneofthethings wedon't understand today (for example, themuon) can, infact, beexplained asanoscillation ofthe Poincaré ‘stresses. Itdoesn’t seem likely, butnoone can sayforsure. There aresomany things about fundamental particles that westildon’t understand. Anyway, the complex structure implied bythis theory isundesirable, and theattempt toexplain allmass interms ofelectromagnetism—at least intheway wehave described—has ledtoablind alley. ‘Wewould liketothink alittle more about why wesaywehave amass when themomentum inthefield 1sproportional tothevelocity. Easy! The mass 1sthe Coefficient between momentum and velocity. Butwecanlook atthemass inanother way: aparticle hasmass sfyou have toexert aforce inorder toaccelerate it.So itmayhelpourunderstanding ifwelookalittlemorecloselyatwheretheforces come from. How do we know that there has tobeaforce? Because we have proved thelawoftheconservation ofmomentum forthefields. Ifwehave a charged particle and push onitforawhile, there will besome momentum inthe electromagnetic field. Momentum must have been poured into thefield somehow. Therefore there must have been aforce pushing ontheelectron inorder togetit going—a force inaddition tothat required byitsmechanical inertia, aforce due toitselectromagnetic interaction, And there must beacorresponding force back onthe“pusher."" But where does that force come from? tiF 2i 7 XeF oF 1 =| a z \ P xX va i (0) >) te) Fig. 28-3. The self-force onanaccelerating electron inotzero because ofthe retardation, (BydFwemean theforce onasurface element da;byd?Fwemean the force onthesurface element da,from thecharge onthesurface element day.) ‘The picture issomething like this. Wecan think oftheelectron asacharged sphere. When itisatrest, each piece ofcharge repels electrically each other piece, buttheforces allbalance inpairs, sothat there isnonetforce. (See Fig. 28-3(a).] However, when theelectron isbeing accelerated, theforces will nolonger bein balance because ofthe fact that theelectromagnetic influences take time tog0 from one piece toanother. For instance, theforce onthepiece ainFig. 28-3(b) from apiece8ontheopposite sidedepends ontheposition of8atanearliertime, asshown. Both themagnitude and direction oftheforce depend onthemotion ofthecharge. Ifthecharge 1saccelerating, theforces onvarious parts ofthe electron might beasshown inFig. 28-3(c). When allthese forces areadded up, they don’t cancel out. They would cancel forauniform velocity, even though itlooks atfirst glance asthough theretardation would give anunbalanced force even forauniform velocity. But itturns outthat there isnonetforce unless the electron isbeing accelerated. With acceleration, ifwelook attheforces between 28-5 willthen begiven bytheintegral ofj,times thisfunction over allspace: AW)=[jsQ)fri2) dV. That's all. Nodifferential equation, nothing else. Well, one more thing. Wealso askthattheresult should berelativistically invariant. Soby“distance” weshould take theinvariant “distance” between two points inspace-time. This distance squared (within asign which doesn’t matter) is she= =12)?~ri =0%(ty =t2)?=(x1=42)? —On=¥2)? —G1~22), (28.14) So,forarelativistically invariant theory, weshould take some function ofthe magnitude of5,2,orwhat isthesame thing, some function ofsf. SoBopp's (84) theory isthat ACht)=fjsQ,te)F(si2) aedts. 28.15) (The integral must, ofcourse, beover thefour-dimensional volume df2dxdy2dz2) Allthatremains 1stochoose asuitable function forF,Weassume only one thing about F—that itisvery small except when itsargument isnear zero—so that a graph ofFwould beacurve liketheoneinFig. 28-4, Itisanarrow spike with a finite areacentered ats?=0,andwith awidth which wecansayisroughly a”. ‘Wecansay,crudely, that when wecalculate thepotential atpoint (1),only those points (2)produce anyappreciable effect ifsz=c%(tg —11)? ~riziswithin a?ofzero.Wecanindicate thisbysaying thatFisimportant onlyfor — ~~° Fr sie=Ot—t2)?—ry=0% (28.16) (0) You canmake itmore mathematical ifyou want to,butthat’s theidea, ‘Now suppose that aisvery small incomparison with thesize ofordinary objects likemotors, generators,andthelikesothatfornormalproblemsr12>>a. r ' Then Eq.(28.16) says that charges contribute totheintegral ofEq. (28.15) only _v whenf;—12isinthesmallrange ce[—a Vee ci~4)~Vaz=@=nai«2. 2 12 (b) Sincea/rj,<«1,thesquarerootcanbeapproximated by|+a?/2r}z, 30 Fig.28-4, Thefunction F(s”)usedin > > thenonlocal theory ofBopp. hole 2) ew. c22, c Ire ‘What isthesignificance? This result says that theonly times f2that areim- portant intheintegral ofA,arethose which differ from thetime ,,atwhich we want thepotential, bythedelay r12/c—with anegligible correction solong as riz>a.Inother words, thistheory ofBopp approaches theMaxwell theory—so Jong aswearefaraway from any particular charge—in thesense that itgives the retarded wave effects. ‘Wecan, infact, seeapproximately what theintegral ofEq. (28.15) isgoing togive. Ifwemntegrate first over 12from —20 to+2—keeping r12 fixed—then sizisalsogoingtogofrom~2to+20.Theintegral willallcomefromf’sin‘asmall interval ofwidth At, =2Xa?/2ry2c, centered atty—riz/e. Say thatthefunction F(s?) hasthevalue Kats?=0;then theintegral over¢2gives approximately KjjAt2, or Ka’he © ne Weshould, ofcourse, take thevalue ofj,att2=1)—r12/¢, sothat Eq.(28.15) becomes Ad,4)=Ke|AQh=rile)gy, ema 2-9 electromagnetic forces aredifferent; electrically theproton and neutron areas different asnight and day. This 1sjust what wewanted There aretwo particles,identicalfromthepointofviewofthe strong interactions, butdifferent electrically ‘And they have asmall difference inmass. The mass difference between theproton andtheneutron—expressed asthedifference intherest-energy mc? inunits of Mev—is about 1.3Mev, which 1sabout 2.6times theelectron mass. The classical theory would then predict aradius ofabout 4to}theclassical electron radius, orabout 10" cm. Ofcourse, oneshould really usethequantum theory, butby some strange accident, alltheconstants—2z’s and fs,ete.—come outsothat the quantum theory gives roughly thesame radius astheclassical theory. The only trouble isthat thesign1swrong! The neutron isheavier than theproton. Table 28-1 Particle Masses |Charge Mass|Am* Particle|electronic)|(Mev)|(Mev)| 1m(neutron) 0 939.5P(proton) +1 9382|-13x(meson)0|135.0| a“ 1396|+46| K(K-meson) 0 497.8“1 439|39| (sigma) ) 1191.5| + 11894) -24 *4m =(mass ofcharged) —(mass ofneutral). Nature hasalso given usseveral other pairs—or triplets—of particles which appear tobeexactly thesame except fortheir electrical charge. They interact with protons and neutrons, through theso-called “strong” interactions ofthenuclear forces. Insuch interactions, theparticles ofagiven kind—say the7-mesons— behave inevery way like one object except fortheir electrical charge. InTable28-1wegivealistofsuchparticles, togetherwiththeirmeasured masses.Thecharged x-mesons—positive ornegative—have amass of139.6 Mev, but the neutral x-meson 1s4.6Mev lighter. Webelieve that this mass difference iselectro- magnetic; itwould correspond toaparticle radius of3104 X10"*em, You will seefrom thetable that themass differences oftheother particles areusually ofthe same general size. Now thesizeofthese particles canbedetermined byother methods, forin- . stancebythediameters theyappear tohaveinhigh-energy collisions. Sothe J Negative electromagnetic mass seems tobeingeneral agreement with electromagnetic . : theory,ifwestopourintegralsofthefieldenergyatthesameradiusobtainedby oeQ: . these other methods. That's why webelieve that thedifferences dorepresent a electromagnetic mass. oN You arenodoubt worried about thedifferent signs ofthemass differences in “PROTON thetable. Itiseasy toseewhy thecharged ones should beheavier than theneutral ‘ones. Butwhat about those pars like theproton and theneutron, where themea- sured mass comes outtheother way? Well, itturns outthatthese particles are Fig.28-5. Aneutron may exist, at complicated, andthecomputation oftheclectromagnetic mass must bemore Times @s@proton surrounded bya elaborate forthem. Forinstance, although theneutron hasnonercharge, itdoes Pegative T-meson. have acharge distribution inside t—it isonly thenercharge that iszero Infact, webelieve that theneutron looks—at least sometimes—like aproton with anega- tiver-meson ina “cloud” around it,asshown inFig. 28-5. Although theneutron is“neutral,” because itstotal charge 1szero, there arestillelectromagnetic energies 241 where6couldbeadifferentfour-vectororperhapsascalar,Itturnsoutthatthe pion hasnopolarization, so@should beascalar. With thesimple equation 224 ~0,themeson field would vary with distance from asource as1/r?, just astheelectricfielddoes.Butweknowthatnuclearforceshavemuchshorterdis-tances ofaction, sothesimple equation won’t work. There isoneway wecan change things without disrupting therelativistic invariance: wecanadd orsubtract from theD'Alembertian aconstant, times 4.SoYukawa suggested that thefree quanta ofthenuclear force field might obey theequation O% ~u*e =0, 2817) where4?isaconstant—that is,aninvariantscalar.(Since1)?isascalardiffer- ential operator infour dimensions, itsinvariance isunchanged ifweaddanother scalar toit.) Let’s seewhat Eq. (28.17) gives forthenuclear force when things are not changingwithtime.Wewantaspherically symmetric solutionof Ve —wp =0 around some point source at,say, theorigin. If¢dependsonlyonr,weknowthat 1 2%1 V6=553(18). Sowehave theequation 13? 2Fan (re)—wd=0 or ae 5p70)=(78). Thinking of(r@)asourdependent variable, thisisanequation wehave seen many times. 11's solution is ro=Ke, \ Clearly, ¢cannot become infinite forlarger,sothe+signintheexponent is \ruledout.Thesolution 1s i o=Ko. (28.18)|\ This function1scalledtheYukawapotential.Foranattractiveforce,Kisanegative ‘ number whose magnitude must beadjusted tofittheexperimentally observed Se strength oftheforces. so TheYukawapotential ofthenuclearforcesdiesoffmoreraprdlythanU/r—_) ye bytheexponential factor. The potential—and therefore theforce—falls to’zero wae muchmorerapidly than1/rfordistances beyond 1/y,asshown inFig.28-6 ———p 3 jooThe“range”ofnuclearforces1smuchlessthanthe“range”ofelectrostatic forces Itisfound experimentally that thenuclear forces donotextend beyond about Fig. 28-6, The Yukowa potential 10" em, sou =10" m~? e""/r, compared with the Coulomb Finally, tet’s look atthefree-wave solution ofEq(2817) Ifwesubstitute potential 1/r. 6=bee into Eq.(2817), wegetthat ganoR =0. Relating frequency toenergy and wave number tomomentum, aswedidatthe endofChapter 36ofVol. 1,wegetthat Bo »a7 =er, Which says that theYukawa “photon” hasamass equal todi/c. Ifweuseforw 26.13 29 The Motion ofCharges inElectric and Magnetic Fields 29-1 Motion inauniform electric ormagnetic field We want now todeseribe—mainly inaqualitative way—the motions of 29-1 Motion inauniform electric charges invarious circumstances. Most oftheinteresting phenomena inwhich formagnetic field charges aremoving infields occur 1nverycomplicated situations. withmany. 99>Momentumanalysis ‘many charges allinteracting with each other For instance, when anelectromagne- ticwave goes through ablock ofmaterral oraplasma, billions and billions of 29-3 Anelectrostatic lens charges areinteracting withthewaveandwitheachother. Wewillcometosuch 9g.4-4mapnetie lensproblems later, butnowwejustwanttodiscuss themuch simpler problem ofthe ba motions ofasingle charge imagiven field. Wecanthen disregard allother charges _-29-5 The electron microscope srsacen: ofsun,thonchargesandcorens whichevstsomewhere toProd¥ee 99.ceeeratar guidefields Weshould probably askfirst about themotion ofaparticlem4uniformelec-_-29-7_Alternating-gradient focusing tricfield Atlowvelocities, themotion isnotparticularly interesting—it isjust @ .uniform acceleration inthedirection ofthefield. However, itheparticle picks 29-8Motion incrossed electricandmagnetic fields upenough energy tobecome relativistic, then themotion gets more complicated. Butwewill leave thesolution forthat case foryou toplay with Next, weconsider themotion inauniform magnetic field with zero electric field. Wehavealready solved thisproblem—one solution isthattheparticle goes new Chapter30,Vo Fractiimaccrele Themagnetic forcequXBisalways atrightangles tothemotion, Reve Chapter 30.Vol1,Diffraction sodp/dt 1sperpendicular topand hasthemagnitude rp/R, where Ristheradius ofthe circle »p r= qb.R te ‘Theradius ofthecircular orbit18then | r-5 29.1 ~\2 29.1) y —) That1sonlyonepossibility. Iftheparticlehasacomponentof1tsmotion Ci t R \ alongthefielddirection, thatmotionisconstant, sincetherecanbenocomponent — ofthemagnetic forceinthedirection ofthe field. The general motion ofaparticleYL 7cl~ inauntform magnetic field1saconstant velocity parallel toBand acircular motion 1 atrightanglestoB—the trayectory 1sacylindrical helix(Fig.29-1). Theradius _ aSofthehelix 1sgiven byEq(291)ifwereplace pbyps,thecomponent ofmo- \ mentum atright angles tothefield 1 —! 29-2 Momentum analysis \ i ni is (9) (b) Auniform magnetic field 1soften used inmaking a“momentum analyzer.” or“momentum spectrometer,” forhigh-energy charged particles. Suppose that Fig. 29-1. Mohon of particle ino charged particles areshot into &uniform magnetic field atthepoint 4inFig. uniform magnetic field. 29-2(a), themagnetic field being perpendicular totheplane ofthedrawing. Each particle will gointo anorbit which isacircle whose radius isproportional toits momentum. Ifalltheparticles enter perpendicular totheedge ofthefield, they willleave thefield atadistance x(rom A)which isproportional totheir momentum p.Acounter placed atsome point such asCwill detect only those particles whose ‘momentum 1simaninterval Apnear themomentum p=gBx/2 Its, ofcourse, notnecessary that theparticles gothrough 180° before theyarecounted.buttheso-called “180°spectrometer” hasaspecialproperty It1snot ww (777 77777)necessary thatalltheparticlesenteratrightanglestothefieldedge.Figure29-2(b)[UNIFORMMAGNETFIELD showsthetrajectories ofthreeparticles,allwiththesamemomentumbutentering, YAy Y,thefieldatdifferentangles.Youseethattheytakedifferenttrajectories,butall> // leave thefieldveryclose tothepoint C.Wesaythatthere isa““focus.”” Sucha / focusing property hastheadvantage thatlarger angles canbeaccepted atA— although some limit isusually imposed, asshown inthefigure. Alarger angular 4 Z Aacceptance usuallymeansthatmoreparticles arecounted inagiventime,decreasing._Ej! theumerequired foragivenmeasurement.te, SS Byvarying themagnetic field, ormoving thecounter along inx,orbyusing @ many counters tocover arange ofx,the“spectrum” ofmomenta intheincoming beam can bemeasured. [By the“momentum spectrum” f(p), wemean that the as - F number ofparticles with momenta between pand (p+dp) isf(p)dp.] Such Y energies intheé-decay ofvarious nuclei i ) There aremany other forms ofmomentum spectrometers, butwewilldescribejustonemore,whichhasanespeciallylargesolidangleofacceptance.It1sbased Ly onthehelical orbits inauniform field, lke theone shown inFig. 29-1. Let’s%OL) thinkofacylindricalcoordinatesystem—p,8,2—setupwiththez-axisalongthe rf C4 directionofthefield.Ifaparticleisemittedfromtheoriginatsomeanglea somewith respect tothez-axis, itwill move alongaspiralwhoseequation1s Fig. 29-2. Auniform-field, momen- p=asinkz, 0=bz, tum spectrometer with 180° focusing: (0)different momenta; (b)different wherea,b,andkareparameters youcaneasilyworkoutintermsofp,a,andtheangles.(Themagnetic fieldisdirected —magnetic fieldB.Ifweplotthedistancepfromtheaxisasafunction of=foraPerpendicular totheplaneoftheFigure.)—sivenmomentum, butforseveralstartingangles,wewillgetcurveslikethesolid ones drawn inFig. 29-3. (Remember that this isjust akind ofprojection ofa helical trajectory.) When theangle between theaxis and thestarting direction1slarger,thepeakvalueofpislargebutthelongitudinal velocityisless,sothetrajectories fordifferent angles tend tocome toakind of“focus” near thepoint 4m thefigure. Ifweputanarrow aperture ofA,particles with arange ofinitial‘ anglescanstillgetthroughandpassontotheaxis,wheretheycanbecountedby ’thelong detector D. een Particles which leave thesource attheorigin with ahigher momentum but es ‘atthesameangles, follow thepaths shown bythebroken linesanddonotget (te SS throughtheapertureat4.Sotheapparatus selectsasmallintervalofmomenta 3 - The advantage over thefirst spectrometer described 1sthat theaperture A—and oF theaperture 4’—can beanannulus, sothatparticles which leave thesource ina Fig.29-3. Anaxiolfeld spectrom ther largesolidangleareaccepted. Alargefraction oftheparticles fromtheeter, sourceareused—an important advantage forweaksourcesorforveryprecise‘measurements, One pays aprice forthis advantage, however, because alarge volume of uniform magnetic field isrequired, and this isusually amly practical forlow-energy particles One way ofmakingauniformfield,youremember,istowindacot!on asphere, with asurface current density proportional tothe sine ofthe angle You canalso show that thesame thing 1strue foranellipsoid ofrotation. Sosuch 990002097, spectrometers areoften made bywinding anelliptical coilonawooden (oralumi-eee af num)frame. Allthat1srequired isthatthecurrent ineachinterval ofaxualdistance A Axbethesame, asshown inFig29-4 q a re SU 29-3Anelectrostatic lens Popa Particle focusing hasmany applications. Forinstance, theelectrons thatleave| Ae thecathodeinaTVpicturetubearebroughttoafocusatthescreen—to makeafinespot. Inthiscase, onewants totake electrons allofthesame energy butwith Fig.29-4. Anellipsoidal coil with _cifferent initial angles andbring them together inasmall spot. ‘The problem is equal currents ineach oxial interval Ax like focusing light with alens, and devices which dothecorresponding jobfor produces ouniform magnetic field inside. particles arealso called lenses. 292 SSSSSSSSSSSS SSSxy atecw - ca ‘ a ead Fig. 29-5. Anelectrostatic lens. Thefleld lines shown are “lines of force,” thal is,ofgE. One example ofanelectron lens issketched inFig 29-5. Itisan“electro- static” lens whose operation depends ontheelectric field between two adjacent electrodes. Itsoperation can beunderstood byconsidering what happens toa parallel beam that enters from theleft. When theelectrons arrive attheregion a, theyfeelaforce with asidewise component andgetacertainimpulsethatbendsthem toward theaxis You might think that they would getanequal and opposite im- pulseintheregionb,butthatisnotso.Bythetimetheelectronsreachtheyhave gained energy andsospend lessueintheregion 6.Theforces arethesame, but p> >thetimeisshorter,sotheimpulseisless.Ingomngthroughtheregionsaand6,(ZAZA LL,thereisanetaxialpulse,andtheelectronsarebenttowardacommonpointZInleavingthehigh-voltage region,theparticlesgetanotherkicktowardtheaxisZ 4 AThe force isoutward inregioneandinwardinregiond,buttheparticlesstaylongery\ Y 1mthelatter region, sothere tsagainanetimpulseFordistancesnottoofarfromg{D.0 theaxis,thetotalimpulsethroughthelensisproportional tothedistancefromthe TY ants(Canyouseewhy”),andthisisjusttheconditionnecessaryforlens-type LLL von]focusing ig,29-6.Amagneticlens Youcanusethesamearguments toshowthatthere1sfocusing ifthe Fig,29-6. Amagnetic| potential ofthemiddle electrode iseither posttive ornegative with respect tothe other two. Electrostatic lenses ofthis type arecommonly used incathode-ray tubesandinsomeelectronmicroscopes. |29-4Amagneticlens 4“ps7 Anotherkindoflens—often foundinelectronmicroscopes—is themagnetic foo.~ lenssketched schematically inFig.29-6.Acylindrically symmetric electromagnet | hasvery sharp circular pole ups which produce astrong, nonuniform field imasmallregion,Electrons whichtravelvertically throughthisregionarefocused |3.‘Youcanunderstand themechanism bylooking atthemagnified view ofthepole-tip / V\ regiondrawninFig.29-7,Consider twoelectrons aand6thatleavethesource {{ \/)\eB |‘Satsomeanglewithrespecttotheaxis.Aselectronareachesthebeginningofthe\\oyJ\)|} field,it1sdeflectedawayfromyoubythehorizontal component ofthefieldBut nN { thenitwillhavealateralvelocity,sothatwhenitpassesthroughthestrongvertical YY field, twill getanimpulse toward theaxis. Itslateral motion 1taken outbythe bk do magnetic force asitleaves thefield, sotheneteffect isanimpulse toward the A-[--> axis,plusa“rotation” about theaxis, Alltheforces onparticle bareopposite, Y Y soitalso 1sdeflected toward theaxis. Inthefigure, thedivergent electrons are Lowe brought into parallel paths. The action 1slikealens with anobject atthefocal point. Another similar lens upstream can beused tofocus theelectrons back toa single point, making animage ofthesource S. Fig. 29-7. Electron motion inthe 29-5Theelectron microscope magnetic len. You know that electron microscopes can“see objects toosmal tobeseen byoptical microscopes. Wediscussed inChapter 30ofVol. Ithebasic limitations ofanyoptical system duetodiffraction ofthelensopening Ifalensopening sub- ws tends theangle 26from asource (seeFig.29-8), twoneighboring spots atthesource Lens cannot beseen asseparate ifthey arecloser than about, OPENING d sna> 8 where )isthewavelength ofthehight. With thebest optical microscope, @ap- proaches thetheoretical limit of90°, so41sabout equal to,orapproximately 5000angstroms. sobnce‘The same Timitation would also apply toanelectron microscope, butthere thewavelength 1s—for 50-kilovolt electrons—about 0.05 angstrom. Ifone could Fig. 29-8 Theresolution ofamicro- use&lens opening ofnear 30°, itwould bepossible toseeobjects only }ofanscopeislimitedbytheanglesubtended angstrom apart.Sincetheatomsinmoleculesaretypically|or2angstromsapart. fromthesource. wecould getphotographs ofmolecules. Biology would beeasy; wewould have 4photograph oftheDNA structure. What atremendous thing that would be! Mostofpresent-day researchinmolecularbiologyisanattempttofigureoutthe BLURREDshapesofcomplexorganicmolecules. Ifwecouldonlyseethem! fa"MAGE Unfortunately, thebestresolvingpowerthathasbeenachievedinanelectron HK\ microscope ismorelike20angstroms. Thereasonisthatnoonehasyetdesigned alens with alarge opening. Alllenses have “spherical aberration,” which means that rays atlarge angles from theaxis have adifferent point offocus than therays nearer theaxis. asshown inFig. 29-9 Byspecial techniques, optical microscope \\\\ ga4EtSive lenses eanbemade withanegligible spherical aberration, butnoonehasyet ° beenabletomake anelectron lenswhich avoids spherical aberration, \\\I Infact,onecanshow thatanyelectrostatic ormagneuc lensofthetypes we \\| havedescribedmusthaveanirreducibleamountofsphericalaberration. This\\ aberration—together with diffraction—limits the resolving power ofelectron \\ microscopes totheirpresent value. TheInmutation wehavementioned doesnotapply toelectric andmagnetic STPOINTSOURCE fieldswhicharenotaxially symmetric orwhicharenotconstant intime.Perhaps Fig. 29-9. Spherical aberration of some day someone will think ofanewkindofelectronlensthatwillovercomethe 6lens. inherent aberration ofthesimple electron lens. Then wewillbeabletophotograph atoms directly. Perhaps one daychemical compounds will beanalyzed bylooking atthepositions oftheatoms rather than bylooking atthecolor ofsome pre- cipitatel 29-6 Accelerator guide fields Magnetic fields arealso used toproduce special particle trayectories inhigh- energy particle accelerators. Machines like thecyclotron and synchrotron bring particles tohigh energies bypassing the particles repeatedly through astrong electric field. The particles areheld intheir cyclic orbits byamagnetic field. Wehave seen that aparticle inauniform magnetic field willgoinacircular orbit. This, however, 1strue only foraperfectly uniform field, Imagine afield Bwhich isnearly uniform over alarge area butwhich 1sshghtly stronger inone region than inanother. Ifweputaparticle ofmomentum pinthis field, itwillgo FIELD STRONGER imanearly circular orbit withtheradius R=p/qB. Theradius ofcurvature will, HERE however, beslightly smaller intheregion where thefield 1sstronger. The orbit 1s notaclosed circle butwill “walk” through thefield, asshown inFig. 29-10.Fig.29-10.Particlemotionin@—Wecan,ifwewish,considerthattheslight“error”inthefieldproduces anextraslightly nonuniform field, angular kick which sends theparticle offonanewtrack. Iftheparticles aretomake millions ofrevolutions inanaccelerator, some kind of“radial focusing” isneeded which willtend tokeep thetrajectories close tosome design orbit. Another difficulty with auniform field 1sthat theparticles donotremain ina plane Ifthey start outwith theslightest angle—or aregiven aslight angle by any small error inthefield—they willgoinahelical path that willeventually take them into themagnet pole ortheceiling orfloor ofthevacuum tank Some arrangement must bemade toinhibit such vertical drifts; thefield must provide “vertical focusing” aswell asradial focusing. 4 a magne mone on a —_—_— oo cy S\ SN Z . J,\\) My \\ /la(S| iy} \\ J} \ Jy | _ Pan SD A : mya Lot ona ef LA ot — a NS! Ld a oe Loi. Fig. 29-11. Radial motion of@par- Fig. 29-12. Radial motion of@par- Fig. 29-13. Radial motion of@por- ticle in@magnetic field with alarge ticle inamagnetic field wit @small ticle in@magnetic field with alarge positive slope. negative slope. negative slope. One would, atfirst, guess that radial focusing could beprovided bymaking amagneticfieldwhichincreaseswithincreasing distancefromthecenterofthe design path Then ifparticle goes outto#large radius, 1willbe mastronger field which will bend itback toward thecorrect radius, Ifitgoes totoosmall aradius, the bending will beless, and itwill bereturned toward thedesign radius. If'aparticle isonce started atsome angle with respect totheideal circle, 1will oscillate abouttheidealcircularorbit,asshowninFig.29-11.‘Theradialfocusingwouldkeeptheparticles near theerreular path. Actually there 1sstillsome radial focusing even with theopposite field slope This can happen sftheradius ofcurvature ofthetrajectory does notincrease more rapidly than theincrease 1nthedistance oftheparticle from thecenter ofthefieldTheparticleorbatswillbeasdrawninFig29-12.Ifthegradientofthefieldistoolarge, however. theorbits will notreturn tothedesign radius butwillspiral inward oroutward, asshown inFig. 29-13. Weusually describe theslope ofthefield interms ofthe“relative gradient” aB/B SS v\ oy n=HB, (29.2) ae Y\bgare WA | | Aguideficldgivesradial focusing ifthisrelative gradient isgreater than—1. Sy papAradial fiedgradient willalsoproduce vertical forces ontheparticles )74sem Supposewehavesfieldthatisstrongernearertothecenteroftheorbitandweaker oh yy attheoutside,Averticalcrossseetionofthemagnetatrightanglestotheorbit nf /mightbeasshowninFig.29-14,(Forprotonstheorbitswouldbecomingoutof "PAL/ thepage)Ifthefieldistobestrongertotheleftandweakertotheright,thelines yy ofthe magnetic field must becurved asshown. Wecanseethat thismust besobyusingthelawthatthecirculation ofB1szeroinfreespace.Ifwetakecoordinates Fig.29-14.Averticalguidefieldasasshown inthefigure, then seen inacross section perpendicular 10 ab.ab. theorbits. (¥xB),=Fe~=0, or OB,_aB,2B OB (293) Since weassume that 4B./0x 1snegative, there must beanequal negative 0B,/02. ws Ifthe“nominal” planeofthe orbit isplane ofsymmetry where B,~0,then the radial component B,will benegative above theplane and positive below The lines must becurved asshown.Suchafieldwillhaveverticalfocusingproperties. Imagine«protonthatwstravelling more orless parallel tothecentral orbit but above it.The horizontal component ofBwillexertadownwardforceonit.Iftheprotonisbelowthecentral orbit, theforce 1sreversed. Sothere 1saneffective “restoring force” toward the central orbit. From our arguments there will bevertical focusing, provided thattheverticalfielddecreaseswithincreasingradius;butifthefieldgradientispositive,there will be“vertical defocusing.” Soforvertical focusing, thefield index nmust beless than zero. We found above that forradial focusing nhad tobegreater than —1, The twoconditions together give thecondition that -l<n<0 iftheparticles aretobekept instable orbits. Incyclotrons, values very near zero areused; inbetatrons and synchrotrons, thevalue n=—0.6 istypically used. 29-7 Alternating-gradient focusing Such small values ofmgive rather “weak” focusing. Itisclear that much more effective radial focusing would begiven byalarge positive gradient (n>1),but then thevertical forces would bestrongly defocusing Similarly, large negative slopes (<<—1) would give stronger vertical forces but would cause radial de- focusing. Itwasrealized about 10years ago, however, that aforce that alternates between strong focusing and strong defocusing can still have anetfocusing force Toexplain how alternating-gradient focusing works, wewillfirstdescribe the operation ofaquadrupole lens,which1sbasedonthesameprinciple. Imaginethat auniform negative magnetic field 1sadded tothefield ofFig 29-14, with the strength adjusted tomake zero field attheorbit. The resulting field—for small displacements from theneutral pomt—would belikethefield shown inFig 29-15 Such afour-pole magnet 1scalled a“quadrupole lens.” Apositive particle that enters (from thereader) totheright oFleftofthecenter 1spushed back toward thecenter, Ifthe particle enters above orbelow, itwspushed aay from thecenter This 18ahorizontal focusing lens Ifthehorizontal gradient 1sreversed—as can bedone byreversing allthepolarities—the signs ofalltheforces arereversed and wehave avertical focusing lens, asinFig. 29-16 For such lenses, thefield strength—and therefore thefocusing forces—increase linearly with thedistance ofthe lens from the axis. Fig.29-15. Ahorizontal focusing Fig.29-16. Avertical focusingquad quadrupole lens. rupote lens. 26 IwonizonTa nowARS Gray on . ° oerance|° DistancezoRTAL HORIZONTAL VERTICAL Verticarreo” hired FewNS rete"® (@) tb) Fig. 29-17. Horizontal andverticalfocusingwith@pairofquadrupole lenses. Nowimagine thattwosuchlensesareplacedinseries. Ifaparticle enterswith =.some horizontal displacement from theaxis, asshown inFig.29-17(a), itwillbe \/ 'deflected towardtheaxisinthefirstlens.Whenitarrivesatthesecondlensitis|\ Nucloser totheaxis, sotheforce outward 1slessandtheoutward deflection isless WoThere1sanetbendingtowardtheaxis;theaverageeffect1shorizontally focusing \WorsOntheotherhand,ifwelookataparticlewhichentersofftheaxisinthevertical \Woredirection, thepathwillbeasshowninFig.29-17(b). ‘Theparticle1sfirstdeflected yy‘awayfromtheaxis,butthenitarrives atthesecond lenswithalargerdisplacement, WWfeels.astrongerforce,andsosbenttowardtheaxis.Againtheneteffectisfocusing, \Wiy Thus apairofquadrupole lenses actsindependently forhorizontal andvertical Vit,motion—very muchlikeanopticallens,Quadrupole lensesareusedtoformand \\= Licontrolbeamsofparticlesinmuchthesamewaythatopticallensesareusedfor fal— -TTAlightbeams. apa ie|¥ Weshouldpointoutthatanalternating-gradient systemdoesnotaways IKA_i|produce focusing. Ifthegradientsaretoolarge(inrelationtotheparticlemomen- iNOanl } tumortothespacingbetweenthelenses),theneteffectcanbeadefocusingone.fo eDy/ 2 Youcanseehowthatcouldhappen ifyouimagine thatthespacing between the| Wy twolenses ofFig.29-17 were increased, say,byafactor ofthree orfour. — - Let’s return now tothesynchrotron guide magnet. Wecan consider that it consists ofanalternating sequence of“positive” and “negative” lenses with a Fig.29-18. Apendulum with ansuperimposed uniformfield.Theuniformfieldservestobendtheparticles, onthe_stillating pivotcanhaveastableposi-average, inahorizontal circle(withnoeffectonthevertical motion), andthe__0”withthebobabovethepivot. alternating lenses actonanyparticles that might tend togoastray—pushing them always toward thecentral orbit (on theaverage). There isanice mechanical analog which demonstrates that aforee which alternates between a“focusing” force and a“defocusing” force can have anet “focusing” effect. Imagine amechanical “pendulum” which consists ofasolidrodwithaweightontheend,suspended fromapivotwhichisarranged tobemovedrapidly upand down byamotor driven crank. Such apendulum hasrwo equih- brium positions. Besides thenormal, downward-hanging position, thependulum1salsoinequilibrium “hanging upward”—with its“bob”abovethepivot!Sucha oependulum isdrawninFig.29-18. a \Bythefollowing argument youcanseethat thevertical pivot motion is \ ' ‘equivalent toanalternating focusing force. When thepivot isaccelerated down- AG ward, the“bob” tends tomove inward, asindicated inFig. 29-19. When the \\\\ pivot isaccelerated upward, theeffect isreversed. The force restoring the‘*bob” \ toward theaxisalternates, buttheaverage effect isaforce toward theaxis. Sothe \ pendulum willswing back andforth about aneutral position which isjustopposite \\ thenormal one. \ There is,ofcourse, amuch easier wayofkeeping apendulum upside down, \\ andthatisbybalancing itonyourfinger! Buttrytobalance twoindependent \»sticksonthesamefinger! Oronestickwithyoureyesclosed! Balancing involves \|making acorrection forwhat 1sgoing wrong. And thisisnotpossible, ingeneral. ifthereareseveral things going wrong atonce. Inasynchrotron there arebillions Fig.29-19. Adownward accelera- ofparticles going around together, each oneofwhich maystart outwith adifferent ionofthepivot causes thependulum to “error.” The kind offocusing wehave been describing works onthem all move toword thevertical. 27 29-8 Motion incrossed electric and magnetic 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 Band anelectric field Eat right angles. Particles that start outperpendicular toBwill move ina % curve lketheoneinFig29-20 (The figure isaplane curve, notahex!) Wecan ¥ i understand thismotion qualitatively. When theparticle (assumed positive) moves 1nthedirection ofE,1tpicks upspeed, and soit1sbent less bythe magnetic field. f WhenitisgoingagainsttheE-field, losesspeedand1scontinually bentmoreby \t themagnetic field. ‘The neteffect isthat ithasanaverage “drift” inthedirection19 \ ofEXB. 8 vy, Wecan.infact,showthatthemotion1sauniformcircularmotionsuper- imposed onauniform sidewise motion atthespeed ry=E/B—the trajectory in Fig. 29-20. Path of@particle in Fig, 29-20 18acycloid, Imagine anobserver who ismoving totheright atacon- crossed electric andmagnetic fields. stant speed. Inhisframe ourmagnetic fieldgetstransformed toanewmagnetic field plus anelectric field inthedownward direction. Ifhehasyusttheright speed,histotalelectricfieldwillbezero,andhewillseetheelectron going inacircle. So the motion ivesee 1sacircular motion, plus atranslation atthe drift speed fy=E/B_Themotionofelectrons incrossed electric and magnetic fields 1sthe basts ofthemagnetron tubes,1€.,oscillators usedforgeneratingmicrowave energy. Therearemany other interesting examples ofparticle mottons inelectric and magnetic fields—such astheorbits oftheelectrons and protons trapped intheVanAllenbelts—but wedonot,unfortunately, havethetimetodealwiththemhere 2-8 30 The Internal Geometry ofCrystals 30-1 The internal geometry ofcrystals Wehave finished thestudy ofthebasic laws ofelectricity andmagnetism, and 30-1 Theinternal geometry of wearenow going tostudy theelectromagnetic properties ofmatter. Webegin crystals bydescribing solids—that is,crystals. When theatoms ofmatter arenotmoving 35.>Chemicalbondsincrystals around very much, they getstuck together and arrange themselves inaconfigura- tion with aslowanenergy aspossible. Iftheatoms inacertain place have founda 30-3 The growth ofcrystals pattern which seems tobeoflow energy, then theatoms somewhere else will jiprobably makethesamearrangement. Forthesereasons, wehaveinasolidma- 20-4Crystal lattices terial arepetitive pattern ofatoms 30-5 Symmetries intwo dimensions Inother words, theconditions ina erystal arethis way: The environment ofa34.6symmetriesinthreedimensions particular atom inacrystal hasacertain arrangement, and ifyou look atthesame kind ofanatom atanother place farther along, you will find one whose surround- 30-7 The strength ofmetals ings areexactly thesame. Ifyou pick anatom farther along bythesame distance, islocaticYyouwilfindtheconditions exactlythesameoncemore.Thepatternisrepeated 20-8Dislocations andcrystalgrowth ‘over and over again—and, ofcourse, inthree dimensions. 30-9 The Bragg-Nye crystal model Imagine theproblem ofdesigning awallpaper—or acloth, orsome geometric design foraplane area—in which youaresupposed tohave adesign element which repeats and repeats and repeats, sothat you canmaketheareaaslargeasyouwant.Reference:C.Kittel,Introduction 10 This 1sthetwo-dimensional analog ofaproblem which aerystal solves inthree Solid State Physics, John dimensions. Forexample, Fig. 30-1(a) shows acommon kind ofwallpaper design. Wiley and Sons, Inc., New There isasingle element repeated inapattern thatcangoonforever. Thegeometric York, 2nded.,1956 characteristics ofthis wallpaper design, considering only itsrepetition properties and notworrying about thegeometry oftheflower itself oritsartistic merit, are contained inFig. 30-1(b). Ifyou start atany point, you canfind thecorresponding point bymoving thedistance aalong thedirection ofarrow |You canalso getto.acorrespordingpointifyoumovethedistance61nthedirectionoftheothervee ®8® arrow. There are, ofcourse, many other directions. You can go,forexample. from point «topoint 8and reach acorresponding position, butsuch astep can beconsidered asacombination ofastepalongdirection1,followedbyastep along direction 2.One ofthebasic properties ofthepattern can bedescribed bythetwoshorteststepstonearbyequalpositions.By“equal”positionswemeanthat28 aaifyouwere tostand inany oneofthem and look around you, you would seeexactly thesame thing asifyou were tostand inanother one. That's thefundamental property ofacrystal. Theonlydifference isthatacrystal 1sathree-dimenstonal () arrangement instead ofatwo-dimensional arrangement; andnaturally,insteadof flowers, each element ofthelattice issome kind ofanarrangement ofatoms. P , perhaps sixhydrogen atoms andtwocarbon atoms—in some kindofpattern = f= f= fe Thepattern ofatoms inacrystal canbefound outexperimentally byx-ray diffrac /tion.Wehavementioned thismethodbrieflybefore,andwon'tsayanymorenow fe e-£ leexceptthattheprecise arrangement oftheatomsinspacehasbeenworked outfor is 7most simple crystals andalsoforsome fairly complex ones. / > / Theinternalpatternofacrystalshowsupinseveralways.First,thebinding >anfot festrength oftheatoms incertain directions 1susually stronger than inother direc- Sotions.Thismeansthattherearecertainplanesthroughthecrystalwhereitsmore. LL Leasilybrokenthanothers.Theyarecalledthecleavageplanes.Ifyoucracka Tf— crystal withaknifeblade itwilloftensplitapartalong suchaplane. Second, the wointernal structure often appears atthesurface because oftheway thecrystal was formed. Imagine acrystal being deposited outofasolution. There aretheatoms Fig.30-1. Arepeating pattern in floating around inthesolution and finally settling down when they find aposition twodimensions. 30-1 oflowest energy. (It’sasifthewallpaper gotmadebyflowers drifting around auntil onedrifted accidentally into place andgotstuck, andthen thenext, and the next sothat thepattern gradually grows.) You can appreciate that there will be certaindirectionsinwhichstwillgrowatadifferentspeedthaninotherdirections, | therebygrowingintosomekindofgeometrical shape.Becauseofsuch effects, the outside surfaces ofmany crystals show some ofthecharacter ofthe internal arrangement oftheatoms For example, Fig. 30-2(a) shows theshape ofatypical quartz erystal whose / internal pattern ishexagonal. Ifyoulook closely atsuch aerystal, youwill notice that theoutside does notmake avery good hexagon because thesides arenotall ofequal length—they are, nfact, often very unequal. But inone respect it18 very good hexagon: theangles between thefaces areexactly 120°. Clearly, thesize ofany particular face 1sanaccident ofthegrowth, buttheangles arearepresenta- tion oftheinternal geometry Soevery crystal ofquartz hasadifferent shape, even though theangles between corresponding faces arealways thesame (c) ‘The internal geometry ofacrystal ofsodium chloride isalso evident from ts external shape Figure 30-2(b) shows theshape ofatypical grain ofsalt. Again thecrystal 1snotaperigct cube, but thefaces areexactly atright angles toone another ‘Amore complicated crystal 1smica, which hastheshape shown inFig 30-2(c) Iisa highly anisotropic crystal, asiseasily seen from thefact that sts very tough Afyou teytopull itapart inone direction (hortzontally inthefigure), butvery easy tosplit bypulling apact intheother direction (vertically) Ithascommonly been usedtoobtainverytough,thinsheetsMicaandquartzare{woexamplesof rr)natural minerals containing silica, Athird example ofamineral with sihca 1s asbestos, which has theinteresting property that itiseasily pulled apart intwo directions butnotinthethird. Itappears tobemade ofvery strong, linear fibers. fe Sle 30-2.Chemical bonds incrystals \’\Na Themechanicalpropertiesofcrystals clearly depend onthekind ofchemical ay. bindings betweentheatoms.Thestrikingly different strengthofmicaalongdiffer- e entdirections depends onthekindsofinteratomic binding inthedifferent directions. You have already learned inchemustry, nodoubt, about the different kinds of chemical bonds Furst,thereare1oniebonds,aswehavealready discussed for (c)sodium chloride Roughly speaking, thesodium atoms have lost anelectron and Fig.30-2. Notural crystals (a) DESOME posite tons,thechlorine atoms havegained anelectron andbecomequan, 1b)sadiors chloride: (3sic. negative ions. Thepositive andnegative ionsarearranged inathree-dimensional checkerboard and areheld together byelectrical forces. The covalent bond—in which electrons are shared between two atoms—s more common and 1susually very strong. Inadiamond, forexample, thecarbon atoms have covalent bonds 1nallfour directions tothe nearest neighbors. sothe crystal 1svery hard indeed. There 1salso covalent bonding between silicon and oxygen inaquartz crystal, but there thebond 1sreally only partially covalentBecausethere1snotcompletesharingoftheelectrons, theatomsarepartlycharged,and thecrystal 1ssomewhat tonic Nature 1snotassimple aswetrytomake 1t; there are really allpossible gradations between covalent and ome bondingx ‘Asugarcrystalhassullanotherkindofbinding.Inttherearelargemolecules1mwhich theatoms areheld strongly together bycovalent bonds, sothat themole- culeisatoughstructure. Butsincethestrongbondsarecompletely satisfied, there xareonly relatively weak attractions between theseparate, individual molecules Insuch molecular erystals themolecules keep their individual identity, sotospeak, ; and theinternal arrangement might beasshowninFig.30-3.Sincethemolecules ee7% Trelatticeof@moleculor arenotheldstrongly toeachother,theerystals areeasytobreakTheyarequite Uiflerent from something like diamond, which 1sreally one guant molecule that cannot bebroken anywhere without disrupting strong covalent bonds. Parifin 1another example ofamolecular crystal. Anextreme example ofamolecularcrystalaccursinasubstancelikesolid argon. There isvery littl attraction between theatoms—each atom 1sacompletely 302 saturated monatomic molecule. Butatvery lowtemperatures, thethermal motionisverysmall,sotheshghtinteratomic forcescancausetheatomstosettledownintoaregulararraylikeapileofcloselypackedspheres.The metals form acompletely different class ofsubstances The bonding is ofanentirely different kind. In.a metal thebonding 1snotbetween adjacent atoms butisaproperty ofthewhole crystal. The valence electrons arenotattached to ‘oneatom ortoapair ofatoms butareshared throughout thecrystal. Each atom contributes anelection toauniversal pool ofelectrons, and theatomic positive ionsreside intheseaofnegative electrons. The electron seaholds theions together likesome kind ofglue Inthemetals, since there arenospecial bonds inanyparticular direction, there isnostrong directionality inthebinding. They arestill crystalline, however, be- cause thetotal energy islowest when theatomic ions arearranged insome definitearray—although theenergyofthe preferred arrangement isnotusually much lower than other possible ones. Toafirst approximation, theatoms ofmany metals are likesmall spheres packed inastightly aspossible. 30-3 The growth ofcrystals Trytosmagine thenatural formation ofcrystals intheearth, Intheearth’s surface there 1sabigmixture ofallkinds ofatoms. They arebeing continually churned about byvolcanic action, bywind, and bywater—continually being moved about and mixed. Yet, bysome trick, silicon atoms gradually begin tofind each other, andtofind oxygen atoms, tomake silica, One atom atatime 1sadded to theothers tobuild upacrystal—the mixture gets unmixed. And somewhere nearby, sodium and chlorine atoms arefinding each other and building upacrystal ofsat. How does ithappen that once acrystal 1sstarted, itpermits only aparticular kind ofatom tojoin on? Ithappens because thewhole system 1sworking towardthelowestpossibleenergy.Agrowingcrystalwillacceptanewatomifit1sgoingtomake theenergy aslow aspossible. But how does itknow that astlicon—or fanoxygen—atom atsome particular spot isgomng toresult inthelowest possible energy? Itdoesitbytrialanderror.Intheliquid,alloftheatomsareinperpetual (2) motion. Each atom bounces against itsneighbors about 10" times every second. Fithitsagainst theright spor ofgrowing crystal, ithas asomewhat smaller chance ofjumping offagain iftheenergy 1slow. Bycontinually testing over periods of mulhons ofyears atarateof10" tests persecond. theatoms gradually build up atheplaces where they find their lowest energy. Eventually they grow into big crystals. 30-4Crystallattices AS Thearrangement oftheatomsinacrystal—the crystallartice—can takeon J1manygeometric forms, Wewould liketodescribe firstthesimplest lattices, which lofarecharacteristic ofmostofthemetalsandofthesolidformoftheinertgases. \‘Theyarethecubiclattices whichcanoccurintwoforms: thebody-centered cubic, Q QshowninFig.30-4(a),andtheface-centered cubicshowninFig.30-4(b).The NW Ndrawings show, ofcourse, onlyonecubeofthelattice; youaretoimagine thatthe (>)pattern1srepeatedindefinitelyinthreedimensions.Also,tomakethedrawingYe clearer,onlythe“centers”oftheatomsareshown.Inanactualcrystal,theatoms INaremorelikespheresincontactwitheachother.Thedarkandlightspheresin J— ONthedrawings may, ingeneral, stand fordifferent kinds ofatoms ormay bethe samekind, Forinstance, ironhasabody-centered cubiclatice atlowtemperatures. Fig.30-4. Theunitcellofcubic butaface-centered cubic lattice athigher temperatures. Thephysical properties crystels. (a)body-contered, (b)face, arequite different inthetwocrystalline forms. centered How dosuch forms come about? Imagine that you have theproblem of packing spherical atoms together astightly aspossible. One way would betostartbymakingalayerina“hexagonal close-packed array,”asshowninFig.30-S(a)Then you could build upasecond layer like thefirst, butdisplaced horizontally, 303 —_ ass cTN ShoN ~{} RASSRR \INN\Qy XNSRNYVY~ SsSSK(NR AVgJ } (0 e(xX woSY _/ Janke —— == /WKWQQg \ftWE ;\\ySANS < RR NST,XY_/ \/\/- _ —_ Fig.30-5.Buildingupohexogonolclose-packed lace. asshown inFig. 30-(b) Next, you can put onthe third layer. But notice! There are‘wodistinct ways ofplacing theurd layer Ifyou start thethird layer byplacing anatom at inFig 30-5) each atom inthe third layer 1sdirectly above anatom ofthe bottom layer Ontheother hand, s'you start thethird layer byputting anatom attheposition B,theatoms ofthethird layer will becentered atpoints exactly inthemiddle ofatriangleformedbythreeatomsofthebottom layer. Any other starting place 1sequivalent toAorB,sothere areonly two ways ofplacing the third layer. Ifthethird layer hasanatom atpoint B,thecrystal lattice 1saface-centered cubro—but seen atanangle Itseems funny that starting with hexagons you can outline Forinstance, Fig.30-6couldrepresent aplanehexagon ofacubeseenin If'athird layer isadded toFig. 30-5(b) bystarting with anatom atA,there 1s nocubical structure, and thelattice hasinstead only ahexagonal symmetry. It1s clear that both possibilities wehave described areequally close-packed Fig, 30-6. Isthis ahexagon oracube Somemetaly—for example, copperandsilver—choose thefirstalternative, seenfrom onecorner? theface-centered cubic. Others—for example, beryllum andmagnesiumi—choose theother alternatives; they form hexagonal crystals. Clearly, which crystal latticeappearscannotdependonlyonthepackingoflittle spheres, butmust also bedeter mined inpart byother factors Inparticular, stdepends ontheslight remaining angulardependence ofthe interatomic forces (or, 1nthe case ofthe metals, onthe energy oftheelectron pool) You will, nodoubt, learn allabout such things your chemistry courses. 30-5 Symmetries intwo dimensions Wewouldnowliketodiscusssomeoftheproperties ofcrystals from thepoint, ‘ofview oftheir internal symmetries. The main feature ofacrystal 1sthat ifyou tareagain inthesame kind ofanenvironment. That'sthefundamental proposition. But ifyou were anatom, there would beanother kind ofchange that could take you again tothe same environment—that is,another possible “symmetry.” Figure 30-7(a) shows another possible “wallpaper-type” design (though one you have probably never seen). Suppose wecompare theenvironments forpointsAandBYoumight,atfirst,thinkthattheyarethesame—but notquitePoints Cand Dareequivalent toA,buttheenvironment ofB1slikethatof4onlyifthe surroundings are reversed, asinamurror reflection. 04 yR OR dle lg dors0/61slools8 le2 Ie tg Ree ~el asd{e| asye ae ~ =k 9 b r+--b-\ ~-|---}-R dio ols ato Nis Cop =] i a '' ~| ab!dle|oledle | I> oR aT ele le 9 YR R °) () Fig. 30-7. Apottern ofhigh symmetry. There areother kinds of“equivalent” points inthepattern. For instance, thepoints £andFhave the“same” environments except that oneisrotated 90° with respect totheother. ‘The pattern 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 spectal examples, let’s trytofigure outall thepossible symmetries acrystal can have. First, weconsider what happens ina plane. Aplane lattice can bedefined bythetwo so-called primitive vectors that go from onepoint ofthelattice tothetwo nearest equivalent points. The two vectors Land2aretheprimitivevectorsofthelatticeofFig.30-1.Thetwovectorsaand ‘bofFig.30-7(a)aretheprimitivevectorsofthepatternthereWecould,ofcourse, , equally well replace aby—a, orbby—b. Since aand 6areequal inmagnitude , andatrightangles,arotation of90°turnsainto6,andbinto—a,givingthesame D /6latticeonceagain.% ett etd taiatad Wescethattherearelatticeswhichhavea“four-sided” symmetry. Andwe ad havedescribed earheraclose-packed arraybasedonahexagon whichcouldhave \, Soeasixsided symmetry. ArotationofthearrayofcirclesinFig.30-S(a)byanangle og| —of60°aboutthecenterofanycirclebringsthepatternbacktoitself. aoe ‘Whatotherkindsofrotationalsymmetryarethere?Canwehave,forexample, 1) 4fivefold oraneightfold rotational symmetry” Itiseasy toseethat they are impossible. The only symmetry with more sides than four 1sasix-sided symmetry. ¢ First,le’sshowthatmorethansixfoldsymmetryisimpossible.Supposewetryto imaginealatticewithtwoequalprimtivevectorswithanenclosedanglelessthan ry > 60°,asinFig,30-8(a). Wearetosuppose thatpointsBandCareequivalent re toA,andthataandbarethetwoshortest vectorsfromAtotsequivalent neighbors bi Butthats clearly wrong, because thedistance between BandCrsshorter than from 2° eitheronetoA.There mustbeaneighbor atDequivalent toAwhich iscloser _g’ 62 — than BorC.Weshould have chosen 4’asoneofourprimitive vectors. Sothe E A o 8 angle between thetwoprimitive vectors must be60”orlarger. Octagonal symmetry rt) isnotpossible. What about fivefold symmetry? Ifweassume that theprimitive vectors @ Fig, 30-8. (a)Rotational symmetries and6have equal lengths and make anangle of2/5 =72°, asinFig. 30-8(b). greater than sixfold are notpossible.thenthereshouldalsobeanequivalent latticepomtatD,at72°fromC.Butthe(b)Fivefoldrotational symmetry isnotvector 6’from EtoDisthen lessthanb,so6isnotaprimitive vector. There can Possible. benofivefold symmetry. The only possibilities that donotgetusinto this kind ofdifficulty are @~60°, 90°, or120°. Zero or180° arealso clearly possible. Oneway ofstating ourresult isthat thepattern canbeleftunchanged byarotation ‘ofone full turn (no change atall), one-half ofaturn, one-third, one-fourth, or one-sixth ofaturn And those areallthepossible rotational symmetries ina plane—a total offive. If@=2x/n, wespeak ofan“nfold” symmetry. Wesay 30-8 5 (0) (b) hfe hkhe 997y - an/[a a _A 6 yo ee a ane/ / / 7 7 / (c) (a) Fig. 30-9. Symmetry under inversion. Pattern (b)isunchanged ifR—»—R, but pattern (a)ischanged. Inthree dimensions pattern (d)1ssymmetric under aninversion but(c)isnot. that apattern with nequal to4orto6hasa“higher symmetry” than one with Returning toFig. 30-7(a), weseethat thepattern hasafourfold rotational symmetry. We have drawn inFig. 30-7(b) another design which hasthesame symmetry properties aspart (a). The little commaclike figures are asymmetric objects which serve todefine thesymmetry ofthedesign inside ofeach square Notice that thecommas arereversed inalternate squares, sothat theunit cell 1s suillhavefourfoldsymmetry, buttheunitcellwouldbesmaller. ‘Thepatternsof Fig. 30-7 also have other symmetry properties. Forinstance, areflection about anyofthebrokenlinesR-Rreproduces thesamepattern The patterns ofFig. 30-7 have still another kind ofsymmetry. Ifthepattern, tyreflected about theline Y-Yand shifted one square totheright (orleft), weget back theoriginal pattern The line ¥-Y 1scalled a““ghde™ line. ‘spatial symmetry operation which 1sequivalent imfwodimensions toa180° rotation, butwhich isaquite distinct operation inthree dimenstons, It1sinversion. Byan [forinstance, thepoint AinFig. 30-9(b)} 1smoved tothepoint at—R Aninversion ofpattern (a)ofFig. 30-9 produces anew pattern, butanin- version ofpattern (b)reproduces thesame pattern. Foratwo-dimensional pattern (asyou can seefrom thefigure), anmversion ofthepattern (b)through thepoint A15equivalent toarotation of180° about thesame point Suppose, however, ‘wemake thepattern inFig. 30-9(b) three dimenstonal byimagining that thelittle 6sand9'seach have an“arrow” pointing outofthepage. After aninversion in three dimensions allthearrows will bereversed, sothepattern 18nofreproduced. Ifweindicate theheads and tails ofthe arrows bydots and crosses, respectively, wecanmake athree-dimensional pattern, asinFig. 30-9(c), which isnorsymmetric under aninversion, orwecan make apattern ike theone shown in(d),which does have such asymmetry. Notice that it15norpossible toimitate athree dimensional inversion byany combination ofrotations. ws symmetry inFig.30-1, andoneofhigh symmetry inFig. 30-7. Wewillleave you with thegame oftrying tofigure outallofthe17possible patterns. Itispeculiar how fewofthe17possible patterns areused inmaking wall- paper andfabrics. One always sees thesame three orfour baste patterns. Isthis because ofalack ofimagination ofdesigners, orbecause many oftheposstble patterns arenotpleasing totheeye? 30-6 Symmetries inthree dimensions a Sofarwehavetalked onlyabout patterns intwodimensions, Whatweare Fonereallyinterested in,however, arepatterns ofatomsinthreedimensions. Furst, sg ///itisclearthatathree-dimensional crystalwillhavethreeprimitivevectors.If/7--anewethenaskabout thepossible symmetry operations inthree dimensions, wefind . - that there are230different possible symmetries! For some purposes, these 230 TRICLINIC types canbegrouped intoseven classes, which aredrawn inFig.30-10. Thelattice ercte7 withtheleastsymmetry iscalled thesriclinic. Itsunitcell1saparallelepiped. The aaaprimitive vectorsareofdifferent lengths,andnotwooftheanglesbetween themare fffequal. There 1snopossibility ofanyrotational orreflection symmetry. There are, Food however, stilltwo possible symmetries—the unit cell1s,orisnot, changed byan o inversion through thevertex. (Byaninversion inthree dimensions, weagain mean TRIGONAL thatspatial displacements Rarereplaced by—R—in other words, that(x,y,2) art goesinto(—x, —y,—z) Sothetriclinic lattice hasonlytwopossible symmetries, lee) ' unlessthere1ssomespecialrelation amongtheprimitive vectors, Forexample, if c| '|H allthevectors areequal andareseparated byequal angles, onehasthetrigonal ~4d lattice shown inthefigure. This figure can have anadditional symmetry, itmay oe beunchanged byarotation about thelong, body diagonal. 3 Ifoneoftheprimitive vectors, saye,18atright angles totheother two, we MONOCLINIC getamonoctinic unitcell. Anewsymmetry 1spossible—a rotation by180°about ¢ wea Tre Thehexagonal cellisaspecial case inwhich thevectors aand6areequal and the oan angle between them 1s60°, sothat arotationof60°,or120°,or180°aboutthevector qiyI ¢repeats thesamelattice(forcertain internal symmetries). oi --JIfallthree primitive vectors areatright angles, butofdifferent lengths, we pe gettheorthorhombic cell. The figure 1ssymmetric forrotations of180° about the a three axes. Higher-order symmetries arepossible with thererragonal cell, which HEXAGONAL hasallright angles and two equal primitive vectors Finally, there 1sthecubic ers cell,which 1sthemost symmetric ofall. f-t---y 1Thepointofallthis discussion about symmetries isthattheinternal symmetries it ofthecrystals showup—sometimes insubtleways—in themacroscopic physical ty 15properties ofthecrystal Forinstance, acrystal will, imgeneral, have atensor / ws electric polarizability. Ifwedescribe thetensor interms oftheellipsoid ofpolari- “ zation, weshould expect that some ofthecrystal symmetries should show up ORTHORHOMBIC also intheellpsoid. Forexample, acubic crystal 1ssymmetric with respect to are arotation of90°aboutanyoneofthreeorthogonal directions. Clearly, the aaa onlyellipsoid withthisproperty isasphere. Acubiccrystal mustbeanisotropic q tydielectric. aaOntheother hand, atetragonal crystal hasafourfold rotational symmetry Ye Itsellipsoid must have twoofitsprincipal axesequal, andthethird must be TETRADONAL parallel totheaxis ofthecrystal. Similarly, since theorthorhombic crystal has om twofold rotational symmetry aboutthreeorthogonal axes,itsaxesmustcoincide oO 71 with theaxes ofthepolarization ellipsoid. Inalikemanner, oneoftheaxes ofa fo | monocliniccrystalmustbeparalleltooneoftheprincipalaxesoftheellipsoid,|| thoughwecan’tsayanything abouttheotheraxes.Sinceatriclinic crystalhasno 9ao-L-J rotational symmetry, theellipsoid canhave anyorientation atall, 17 Asyoucansee, wecanmake abiggame offiguring outthepossible sym- metries andrelating them tothepossible physical tensors. Wehaveconsidered cusic only thepolarization tensor, butthings getmore complicated forothers—for instance, forthetensor ofelasticity. There isabranch ofmathematics called Fig.30-10. The seven classes of “group theory” thatdeals with such subyects, butusually youcanfigure outwhat —_¢rystal lattices. you want with common sense. 30-7 12 3 4 Fagetie (a) 0) Fig. 30-11. Slippage ofcrystal planes. 30-7 The strength ofmetals Wehave said that metals usually have asimple cubse crystal structure; we want now todiscuss their mechanical properties—which depend onthis structure Metals are, generally speaking, very “soft,” because itiscusy toslide one layeroftheerystaloverthenext.Youmaythink:“That'sridiculous;metalsaretrong.” Not so,asingle crystal ofametal can bedistorted very easily Suppose welook attwo layers ofacrystalsubjectedtoashearforce,asshown snthediagram ofFig. 30-11(a). You might atfirst think thewhole layer would resist motion until theforce was bigenough topush thewhole layer “over the hump,” sothat ishifted one notch totheleft. Although slipping does occur alongaplane,doesn’thappenthatway(Iftdid,youwouldeaiculatethatthemetal 1smuch stronger than itreally 1s.) What happens ismore likeone atom going ata ‘ume; first theatom ontheleftmakes itsjump, then thenext, and soon,asindicated anFig. 30-11(b). Ineffect it1sthevacant spuce between twoatonis that quickly travels totheright, with thenet result that thewhole second layer has moved over one atomic spacing. The slipping goes this way because ittakes much less energy tohitoneatom atatime over thehump than toliftawhole row. Once theforce 1senough tostart theprocess, itgoes therest oftheway very fast Itturns outthat inareal crystal, slipping wil occur repeatedly atone plane,Fig.30-12.Aphotograph ofasmatithenwillstopthereundstartatsomeotherplane.‘Thedetailsofwhyitstartsandcrystel ofcopper afterstretching. [Cour StOPS ATEquite mysterious. Itis,infact,quite strange thatsuccessive regions of fesy ofSS. Brenner, Senior ‘Scientut, HPareoften fairly evenly spaced. Figure 30-12 shows aphotograph ofauny, United States Steel Research Center, thin copper crystal that hasbeen stretched. You canscethevarious planes where Monroeville, Pa} Slipping hasoccurred, The sudden slipping ofindividual erystal planes 1squite apparent sfyou take 41prece oftunwire that haslarge erystals initand stretch atwhile holding stnext twyour ear. You can hear arush of“tucks” astheplanes snap totheir new post- tions, one after the other. Theproblem ofhavinga“missing” atominonerowissomewhat moredificult a than stmight appear from Fig. 30-11, When there aremore layers, thesituation aiyeynyyt ‘mustbesomething likethatshown inFig.30-13. Suchanimperfection unacrystal AA iscalledadislocation. It18presumed thatsuchdislocations areeitherpresentY ) whenthecrystalwasformed oraregenerated atsomenotchorcrackatthesurface cyyAYAyy\ grossdistortions resultfromthemotionsofmanyofsuchdislocations.OrOF») ( Dislocations canmovefreely—that is,theyrequirelittleextiaenergy—soC1GAt ~longastherestofthecrystal hasaperfectlattice.Buttheymayget“stuck”ifthey OTSAANZAencountersomeotherkindofimperfectioninthecrystal,Ifittakesalotofenergy C RC, OC))forthemtopasstheimperfection, theywillbestopped.This18preciselytheJOS ZIGICY mechamism thatgives strength tounperfect metal crystals. Purewoncrystals are } ‘NiD)quitesoft,butasmallconcentration ofimpurityatomsmaycauseenoughimper-_)—fectionstoeffectivelyimmobilizethedislocations. Asyouknow,steel,which1s DOOYOKDIC primaryiron,1sveryhard.Tomakesteel,«smallamountofcarbonwsdissolved vs oe intheironmelt; ifthemelt 1scooled rapidly, thecarbon precipitates outinlittle ons grains,makingmanymicroscopic distortions inthelattice.Thedislocations can stent 2O1S:& dislocation in@17° nolonger moveabout, andthemetalishard. ure copper isvery soft, butcanbe“work-hardened."" ‘This isdone byham- mering onitorbending itback andforth. Inthiscase, many new dislocations of various kinds aremade which interfere with oneanother, cutting down their 308 mobility. Perhaps you've seenthetrickoftaking abarof“dead soft”copper OSS. andgently bending itaround someone's wrist asabracelet. Intheprocess, it LSbecomes work-hardened andcannoteaslybeunbentagain!Awork-hardened << OOD metalikecopper canbemadesoftagabyannealing atahightemperature, (GIA 59Thethermal motion oftheatoms “irons out”thedislocations andmakes large |KR LTsingle crystals again, Wehave,sofar,described onlytheso-called slipdislocation am et ‘Therearemanyotherkinds, oneofwhichisthescrewdislocation shown inFig. Kert 30-14. Such dislocations often playanimportant partincrystal growth. Y | 30-8 Dislocations and crystal growth Fig. 30-14. A screw dislocation [From Charles Kittel, Introduction to OneofthegreatpuzzlesforalongtimewashowcrystalscanpossiblyBroW.SondStatePhysics,JohnWileyandSons, Wehave described howit1sthateach atom might, byrepeated testing, determine Ine,,New York, 2nded.,19564 whether itwas better tobeinthecrystal ornot. But that means that each atom must find aplace oflow energy. However, anatom putonanew surface 1sonly bound byoneortwo bonds from below, anddoesn't have thesame energy it \ would haveifwtwereplacedinacorner,whereitwouldhaveatomsonthreesides a ‘Supposeweimagineagrowingcrystalasastackofblocks,asshowninFig.30-15, f<wa. Ifwetry anewblock at,say,position A,itwillhave only oneofthesixneighbors, Ce. .itshouldulumatelyget.Withsomanybondslacking,itsenergy1snotverylow. POX! KeItwouldbebetteroffatpositionB,whereitalreadyhasone-halfof1tsquotaof AN. wa)bonds. Crystals doindeed grow byattaching new atoms atplaces likeB. OK \Whathappens,though,whenthatlineisfinished?Tostartanewline,an SotSLSY]ftommustcometorestwithonlytwosidesattached,andthat1sagamnotveryNIWNTlikely. Evenifwtdid,whatwouldhappen whenthelayerwasfinished? How NIWA!couldanewlayergetstarted?OneansweristhatthecrystalpreferstogrowataINSK Dr4dislocation, forinstance around ascrew dislocation liketheoneshown inFig. NOY30-14.Asblocksareaddedtothiscrystal,thereisalwayssomeplacewherethere NUarethree available bonds. Thecrystal prefers, therefore, togrow with adislocation buat in,Such aspiral pattern ofgrowth isshown inFig, 30-16, which 1saphoto- graphofasinglecrystalofparaffin Fig.30-15.Crystalgrowth z Bg ge Bis we : E ee ze Fig.30-16.AporatfincrystalwhichBi Re *~ 1 hasgrown around @screw dislocation. ‘f - WJ. [From Charles Kittel, Introduction 10Solid ;<.|StePrec,ohnWeyondSom,ne q New York, 2nded,1956.) 30-9 The Bragg-Nye crystal model Wecannot, ofcourse, seewhat goes onwith theindividual atoms inacrystal. ‘Also, asyou realize bynow, there aremany complicated phenomena that arenot easy totreat quantitatively. SirLawrence Bragg and J.F.Nye have devised a scheme formaking amodel ofametallic crystal which shows inastriking way many ofthephenomena that arebelieved tooccur inareal metal, 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 Soctety ofLondon, Vol. 190, September 1947, pp.474-481 —with thepermission oftheauthors and oftheRoyal Society.) 309 LL—— Adynamical modelofacrystalstructure N‘ol | eS aaaaso,PRS.anJ.7.Xe <a . 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Diameter 141 mm Bocce SSMaeubeeeneyeyes ie) hhh aA eeyenakeee: eres: fectcrystalline raftofbubbles. Diameter 0-3=e TREN ITYIIISEET ISAARERR ON LUIRIAN RG227LOISPLATESOT ERE EEEURTNCa ATIpaBeEAERIEGREIDFRI I NR NS FA RR KRIS SAMAR REID IDIIID III RID I RR NAMM eS ini nbn PRR oDDMPRRA SEH hw mERYMoshighvierNEDy RAN ARmAMa ERRTti PHAMKniineDieiiiDERPiayaPy satscatatetatoo arecaEetateteetCECE SENATE COREAEEREeaEtattte Moe RONaon eR NaS Ly RXR KK ERR ei) PEERED NBDAP) DPEEDPEDODD EEEDDD 30:16 *SSSA ae NaSety Naha hySyAy hyRyRe wor Hercecedeoereec ccs cee ASSAM ANNA BABAR ARBRE RARERRARER EEE SOC Seleedeatanciteitdatt iat _ Ley sapneaess Heleeieeleeiorl EEEeiswmceaiereiey SEES vewieleieleietel OO * f ao Tensors 31-1 The tensor ofpolarizability Physicists always have ahabit oftaking thesimplest example ofany phenome- 31-1 The tensor ofpolarizability non and calling 1t“physics,” leaving themore complicated examples tobecome ;theconcernofotherfields—say ofappliedmathematics, lectrialengineering, 34-2Transforming thetensorchemistry, orcrystallography. Evensolid-state physics isalmost onlyhalfphysics ‘components because itworries toomuch about special substances. Sointhese lectures wewill 31-3 The energy ellipsoid beleaving outmany interesting things. For mstance, one oftheimportant proper- tes ofcrystalsorofmostsubstancesisthattheirelectricpolarzabiity 1s3I#Othertensors;thetensorof different indifferent directions. Ifyouapplyafieldinanydirection, theatomic inertia charges shift alittle and produce adipole moment, butthemagnitude ofthe 31-5 Thecross product moment depends very much onthedirection ofthefield. That is,ofcourse, quiteacomphcation. Butinphysics weusually startoutbytalking aboutthe 3!76Thetensorofstress special case inwhich thepolarizability isthesame inalldirections, tomake life 31-7 Tensors ofhigher rank easier. We leave theother cases tosome other field. Therefore, forour later work, wewillnotneedatallwhatwearegomgtotalkaboutinthischapter. 31-8Thefour-tensor of ‘Themathematics oftensors isparticularly useful fordescribing properties lectromagnetic momentumofsubstances whichvaryindirection—although that'sonlyoneexampleoftheiruse, Since most ofyou arenotgoing tobecome physicists, butaregoing togo intothereal world, where things depend severely upon direction, sooner orlater ,youwillneedtousetensors. Inordernottoleaveanything out,wearegoingtoReve” Chapter 11,Voll, Vectorsdescribe tensors, although notingreatdetail. Wewantthefeeling thatourtreat- Chapter 20,Vol.1.Rotation in mentofphysics iscomplete. Forexample, ourelectrodynamics iscomplete—as ‘Space complete asanyelectricity and magnetism course, even agraduate course. Our mechanics isnotcomplete, because westudied mechanics when youdidn't have a high level ofmathematical sophistication, andwewere notable todiscuss subjects liketheprinciple ofleast action, orLagrangians, orHamiltonians, and soon, which aremore elegant ways ofdescribing mechanics. Except forgeneral relativity, however, wedohavethecompletelawsofmechanics. Ourelectricity and magnetism iscomplete, andalotofother things arequite complete. Thequantum mechanics, naturally, willnotbe—we have toleave something forthefuture. Butyoushould atleast know what atensor is. Weemphasized inChapter30thattheproperties ofcrystalline substances are different indifferent directions—we say they are anisotropic. 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 perunit volume P isproportional tothestrength oftheapplied field E.(This isagood approximation formany substances ifEisnottoo large.) Wewill call theproportionality constant a.* We want now toconsider substances inwhich adepends onthe direction oftheapplied fild, as,forexample, incrystals likecalcite, which make double images when you look through them. Suppose, inaparticular crystal, wefindthatanelectric field Eyinthex-diree- tionproduces thepolarization P,inthex-direction. Then wefind that anelectric fieldEyinthey-direction, with thesame strength, asEproduces adifferent polar *InChapter 10wefollowed theusual convention and wrote P=¢oxE and called x(*khi) the“susceptibility.” Here, itwill bemore convenient touseasingle letter soWewriteaforeoxForisotropicdiclectics,a~(x-1)eo,wherex1thedieletricconstant {Gee Section 10-4), a ization P»inthey-direction. What would happen ifweputanelectric field at 45°? Well, that’s asuperposition oftwo fields alongxandy,sothepolarization Pwill bethevector sum ofP,and Ps,asshown inFig. 31-I(a). The polarization isnolonger inthesame direction astheelectric field. You canseehow that might come about. There may becharges which can move easily upand down, but which arerather stiff forsidewise motions. When aforce isapplied at45°, the charges move farther upthan they dotoward theside. The displacements are notinthedirectionoftheexternalforce,becausethereareasymmetric internal e,|elastic forces. There is,ofcourse, nothing special about 45°. Itisgenerally true that the induced polarization ofacrystal isnorinthedirection oftheelectric field. Inour R example above, wehappened tomake a“lucky” choice ofourx-andy-axes,forwhich Pwas along Eforboth thex-and y-directions. Ifthecrystal were rotated with respect tothecoordinate axes, theelectric field Ezinthey-direction would have produced apolarization Pwith both anx-and ay-component. P E, Similarly, thepolarization duetoanelectric field inthex-direction would have (0) produced apolarization withanx-component andaj-component. Thenthepolarizations would beasshown inFig. 31-1(b), instead ofasinpart(a).Things getmore complicated—but foranyfield E,themagnitude ofPisstillproportional tothemagnitude ofE. € ‘Wewantnowtotreatthegeneral caseofanarbitrary orientation ofaerystal—_ withrespect tothecoordinate axes. Anelectric fieldinthex-direction willproduce y 4polarization Pwith x-,»-,and z-components; wecan write Pe=desk, Py=GyeEs, Pe=OeEx. GL) Allwearesayinghereisthatiftheelectric fieldisinthex-direction, the &polarization does nothave tobeinthat same direction, butrather hasanx-,ay- ©) and a2-component—each proportional toE,. Wearecalling theconstants of proportionality azz,ays, andazz,respectively (the first letter totelluswhich com- Fig. 31-1. The vector addition of ponent ofPisinvolved,thelasttorefertothedirectionoftheelectricfield). olarizations inonanisotropic crystal. Similarly, forafield inthey-direction, wecan write Pym aryEy Py=ayEy Pe=ayy 1.2) and forafield inthe z-direction, Pes aE Py=ayBy Pe=eke G13) Now wehave said that polarization depends linearly onthefields, soifthere isan electric field Ethat hasboth anx-and ay-component, theresulting x-component ofPwill bethesum ofthetwo P,’s ofEqs. (31.1) and (31.2). IfEhascomponents along x,y,and z,theresulting components ofPwill bethesum ofthethree contributions inEqs. (31.1), (31.2), and (31.3). Inother words, Pwill begiven by Pe=casks +aayEy +aves Py=ayeEs +ayEy +ayes G14) Pz=ausEs +aay, +isk: The dielectric behavior ofthecrystal isthen completely described bythenine quantities (azz, ayy azz, ayes --.),Which wecan represent bythesymbol @,,. (The subscripts iandjeach stand forany oneofthethree possible letters x,y, andz.)Any arbitrary electric field Ecanberesolved with thecomponents E;,Ey, and E,;from these wecan usethea,,tofind P,,P,,and P,,which together give thetotal polarization P.The setofnine coefficients ay,iscalled afensor—in thisinstance,thetensorofpolarizability. Justaswesaythatthethreenumbers(E>,E,,E,)“form thevector E,”wesaythat thenine numbers (aig ay )“form thetensor a,” 32 areallthepoints onanellipse (Fig. 31-2). (Itmust beanellipse, rather than a parabola orahyperbola, because theenergy forany field isalways positive and finite.) The vector Ewith components E,and E,can bedrawn from theorigin totheellipse. Sosuch an“energy ellipse” isanice way of“visualizing” thepolar- ization tensor. 2 Ifwenow generalize toinclude allthree components, theelectric vector Ein anydirection required togive aunit energy density gives apoint which willbeon thesurface ofanellipsoid, asshown inFig. 31-3. ‘The shape ofthis ellipsoid of constant energy uniquely characterizes thetensor polarizability.Nowanellipsoidhasthenicepropertythatitcanalwaysbedescribedsimply wy bygiving thedirections ofthree “principal axes” and thediameters oftheellipse along these axes. The “principal axes” are the directions ofthe longest and shortest diameters and thedirection atright angles toboth. They areindicated bytheaxes a,b,and cinFig. 31-3. With respect tothese axes, theellipsoid has theparticularly simpleequation Fig.31-2. Locusofthevector E= eaaEs +anEd +axecE? =uo. (E.,£,)thotgives aconstant energy of polarization. Sowith respect tothese axes, thedielectric tensor hasonly three components, that arenotzer0: age, as,anda, That istosay, nomatter how complicated crystal is,itis always possible tochoose asetofaxes (not necessarily thecrystal axes) forwhich thepolarization tensor hasonly three components. With such a ° setofaxes, Eq. (31.4) becomes simplyPr=toaBnPy=aEPe=acc G19).COSAnelectricfieldalonganyoneoftheprincipalaxesproducesapolarizationalong YX thesameaxis,butthecoefficients forthethreeaxesmay,ofcourse, bedifferent. NSOften,atensorisdescribedbylistingtheninecoefficientsinatableinsideof 4pair ofbrackets: [=ow=| GIO)Fig.31-3,Theenergyellipsoidofte Gay ee thepolarization tensor. Fortheprincipal axes a,6,and c,only thediagonal terms arenotzero; wesay then that “the tensor isdiagonal.” The complete tensor is ae «00 0 as Of GLI 0 0 ae Theimportant point isthat anypolarization tensor (infact, anysymmetric tensor ofrank two inany number ofdimensions) can beputinthis form bychoosing a suitable set ofcoordinate axes. Ifthethree elements ofthepolarization tensor indiagonal form areallequal, that is,if Aaa =084=Ace=0 @1.12) theenergy ellipsoid becomes asphere, and thepolarizability isthesame inall directions. The material isisotropic. Inthetensor notation, ay=aby G13) where 4,,istheunit tensor 1 0 0 a=|0 1 o}- @LI4) oo 1 That means, ofcourse, wel if Tas a)<0, iffey GUS ais three principal axes, then wand Lare, ingeneral, not inthesame direction (see Fig. 31-4). They arerelated inaway analogous totherelation between and P.Ingeneral, wemust write 0 Lz=Inside +Ieyity +Testes = Ly=Leyte +Iydry +Iyer @1.16) L Ls=Insite +Teylty +Tes The nine coefficients /,,arecalled thetensor ofinertia. Following theanalogy with thepolarization, thekinetic energy forany angular momentum must. be some quadratic form inthecomponents Ww,yyandwe: KE=$0how, QGI.I7) Ne ‘Wecanusetheenergy todefine theellipsoid ofinertia, Also, energy arguments Fig. 31-4. The angular momentum canbeused toshow thatthetensor issymmetric—that /,,=Jy Lof@solid object isnot,ingeneral, Thetensor ofinertia forarigid body canbeworked outiftheshape ofthe Parallel toitsangular velocity w. object isknown. Weneed only towrite down thetotal kinetic energy ofallthe particles inthebody. Aparticle ofmassmandvelocityvhasthekineticenergy 4mv?, and thetotal kinetic energy isjust thesum Lym? over alloftheparticles ofthebody. The velocity vofeach particle isrelated to theangular velocity wofthesolid body. Let's assume that thebody 1srotating about itscenter ofmass, which wetake tobeatrest. Then ifristhedisplacement ofaparticlefromthecenterofmass,itsvelocity visgivenby#Xr.Sothetotal kinetic energy is KE=0jm Xn) (31.18) Now allwehave todo1swrite #Xroutinterms ofthecomponents «;,«).Ws. and x,y,z,and compare theresult with Eq. (31.17); wefind /,,byidentifying terms. Carrying outthealgebra, wewrite Xn =X NE+ @X Ni+X NE =(oz —wy? +(ox —22) +(Wey =wx)? =+uke? —2wywzy +oly? +ix? —2wswerz +wiz? +why? —2w,wyyx +whx?, Multiplying this equation bym/2, summing over allparticles, and comparing with Eq.(31.17), weseethat /,,,forinstance, isgiven by lez=DDmO* +2). This istheformula wehave had before (Chapter 19,Vol. I)forthemoment of inertia ofabodyaboutthex-axis.Sincer?=x?+y?+2°,wecanalsowrite this term as Tez=Ym? —x*). Working outalloftheother terms, thetensor ofinertia can bewritten as Lime?—x2)~Em ~Smxz h,=| -Xomyx Xme? —y?) -Lm: | G19) — mx —Xomzy mr? —2%): Ifyou wish, this may bewritten in“tensor notation” as Ly=Dome? 6,—nr). 1.20) 37 31-6 The tensor ofstress The symmetric tensors wehave described sofararose ascoefficients inre- lating one vector toanother. Wewould like tolook now atatensor which hasa different physical significance—the tensor ofsiress. Suppose wehave asold object with various forces onit.Wesaythat there arevarious “stresses” inside, bywhich wemean thatthere areinternal forces between neighboring parts ofthe f/ Hmaterial.Wehavetalkedalittleaboutsuchstressesinatwo-dimensional case Ti! ifwhen weconsidered thesurface tension inastretched diaphragm inSection As,12-3.Wewillnowseethattheinternalforcesinthematerialofathree-dimensional /7/| /bodycanbedescribed intermsofatensor. V(7's) (Consider abody ofsome elastic material—say ablock ofjello. Ifwemake en 5 acutthroughtheblock,thematerialoneachsideofthecutwill,ingeneral.get|ea displaced bytheinternal forces. Before thecutwasmade, there must have been Me : forcesbetweenthetwopartsoftheblockthatkeptthematerralinplace:wecan t Ndefinethestressesintermsoftheseforces.Supposewelookatanimaginaryplane|WaaN)perpendicular tothex-axis—hike theplaneoinFig31-S—andaskabouttheforce Y acrossasmallareaAyAzinthisplaneThematerialontheleftoftheareaexerts 7 theforce AF, onthematerial totheright, asshown inpart (b)ofthefigure co) o There1s,ofcourse, theopposite reuction force—AFexerted ontheMari! © pg315 Thematenol totheleftoftheletofthesurface,Iftheareu1ssmallenough,weexpectthatAF,1spropor—heJaneexertsacrosstheareatonal totheares AyAz.| AyAztheforceAF,onthematerial to Youarealreadyfamuliarwithonekindofstress—the pressureinastaticjherightoftheplone Iiguid. There theforce 1sequal tothepressure times thearea and 1satright angles tothesurface element. For solids—also forviscous hquids inmotion—the force need notbenormal tothesurface; there areshear forces inaddition topressures (positive ornegative) (By a“shear” force wemean therangential components oftheforce across asurface.) Allthree components oftheforce must betaken into account. Notice also that ifwemake our cutonaplane with some other orientation, theforces will bedifferent. Acomplete description oftheinternal stress requires atensor. oFyiS gra)Wy os‘GoW yee elementofareaAyAzperpendicular 10 dF,v the x-axis isresolved. info the three a components SFxi,AFyi,andAFei. Wedefine thestress tensor inthefollowing way: First, weimagine acut perpendicular tothex-axis and resolve theforce AF, across thecutinto itscom- ponents AF,), SF,, AF:1, asinFig. 31-6. The ratio ofthese forces tothearca AyAz,wecall Sze, Syx, and Szz. For example, Fn See=apas” The first index yrefers tothedirection force component; thesecond index xis normal tothearea. Ifyou wish, you can write thearea AyAzasAa,, meaning an element ofarea perpendicular tox.Then AF Sw=Ta Next, wethink ofanimaginary cutperpendicular tothey-axis. Across asmall 319 areaAxAztherewillbeaforceAF.Againweresolvethisforceintothreecom- OF2) ponents, asshowninFig31-7,anddefinethethreecomponents ofthestress, \ Senn Sys Sey astheforce perunit area inthethree directions. Finally, wemake an . imaginary cutperpendicular tozand define thethree components Sy2y Sys, ind Sx Sowehave the nine numbers OF Siz Sey See Si=|Sie Sy Sul @1.23) Sir Sey Se. LIM 4Fx2 pletelytheinternal stateofstress,andthatS,,18indeedatensor Suppose wewantLL, toknowtheforceacrossasurface oriented atsomearbitrary angle Canwefind/atfrom5,,2-Yes,inthefollowing way:Weimaginealittlesolidfigurewhichhas Afv Jf onefaceNVinthenewsurface, andtheotherfacesparallel tothecoorcmate axes. Iftheface Nhappened tobeparallel tothez-axis, wewould have thetriangular \ pieceshowninFig.31-8.(This1sasomewhatspectalcase,butwillillustratewell OF, enough thegeneral method.) Nowthestressforcesonthehitilesolidtrrangle an 7Fig 31-8 are inequilibrium (atleast inthe limit ofinfinitesimal chmensions), Fig.31-7. Theforce across anele- >?thetotal force onitmust bezero. Weknow theforces onthefaces parallel to ment ofarea perpendiculer toyisre- thecoordinate axes directly from S,, ‘Their vector sum must equal theforce on solved intothree rectangular components theIaice N,sowecan express this force interms ofS,, Our assumption that thesurfuce forces onthesmall triangular volume arein equilibrium neglectsanyotherbodyforcesthatmightbepresent,suchasgravity ar,‘01pseudo forces ifourcoordinate system isnotaninertial frame Notice, however,yn thatsuchbodyforceswillbeproportional tothevolumeofthe little triangle and, 'Ine, therefore. toAx,Ay,Az,whereasallthesurfaceforcesareproportional tothe” areas such asAxAy,AyAz,ete. Soifwetake thescale ofthettle wedge small a> enough, thebody forces canalways beneglected incomparison withthesurface LY, ~ Let's now adduptheforces onthehttle wedge. Wetake firstthex-component, by KEK OFxnwhichisthesumoffiveparts—one fromeachfaceHowever,ifAzissmallenough, OF in Y theforces onthetriangular faces (perpendicular tothez-axis) willbeequal and m opposite, sowecanforget them. Thex-component oftheforce onthebottom rectangle is Fig.31-8. The force Fyacross the MFea =SeyAxAz foce N(whose unit normal isa) isresolved intocomponents The x-component oftheforce onthevertical rectangle is Fer=SeeAyAz. These two must beequal tothex-component oftheforce ourward across theface NV. Let’s call mthe unit vector normal tothe face NV,and the force onitF,,, then we have Fan =SezAYAZ+SeyAxAz. The x-component S;»ofthestress across thisplane isequal to4F,,, divided by thearea, which 1sAy/zAx? +Ay?, of Son=Sux = BhatSpyBe Vax +ay? Vax +Ay? Now Ax/\/Ax?+Ay?isthecosineoftheangle@betweenmandthey-axis,as showninFig.31-8,soitcanalsobewrittenasn,,they-component of#.Similarly, Av/V/Ax? +Ay? issin@=m,.Wecanwrite Sen =Seatte +Sesty Ifwenow generalize toanarbitrary surface element, wewould getthat Son =Seatte +SeyMy +Seats a0 or,ingeneral, Sim=D Suny 31.28= oz on Wecanfind theforce across any surface element interms oftheS,,, s0itdoes describe completely thestate ofmternal stress ofthematerial. Equation (3124)saysthatthetensor S,,relates theforceS,,totheunitvector Syx 1,justasa,,relates PtoE.Since mandS,arevectors, thecomponents ofS,,must stransform asatensor with changes incoordinate axes. SoS,,isindeed atensor. ay WecanalsoshowthatS,,isasymmetrictensorbylookingattheforcesona s,EZ hittlecubeofmaterial.Supposewetakealittlecube,orientedwithitsfacesparallel = toourcoordinateaxes,andlookatitincrosssection,asshowninFig31-9.If @W Sux welettheedgeofthecubebeoneumt,thex-andy-components oftheforces on xy thefuces normal tothex-and y-axes might beasshowninthefigure.Ifthecube issmall, thestresses donotchange appreciably from onesideofthecube tothe Sys opposite side, sotheforce components areequal and opposite asshown Now there must benotorque onthecube, oritwould start spinning. ‘The total torque about thecenter is(Siz—Sz3)(times theunitedgeofthecube), andsince the Sy total iszero, Sj.isequal toS,,, and thestress tensor 1ssymmetric. Since S,,isasymmetric tensor, tcan bedescribed byanellipsoid which will Fig. 31-9. The x-and y-forces on have three principal axes. For surfaces normal tothese axes, thestresses are four faces of@small unitcube. particularly simple—they correspond topushes orpulls perpendicular tothesur- faces There arenoshear forces along these faces. For any stress, wecan always choose ouraxes sothat theshear components arezero. Iftheellipsord 1sasphere, there areonly normal forces inanydirection. This corresponds toahydrostatic pressure (positive ornegative). Soforahydrostatic pressure, thetensor isdiagonal and allthree components areequal; they are, infact, just equal tothepressure p. We can write Sy=Poiy (1.25) “The stress tensor—and also itsellpsoid—will, ingeneral, vary from point to point inablock ofmaterial; todescribe thewhole block weneed togive thevalue ‘ofeach component ofS,,asafunction ofposition. Sothestress tensor 18afield. Wehave had scalar fields, like thetemperature T(x, »,2),Which give one number foreach point inspace, and vector fields like E(x, ,2),which give three numbers foreach point. Now wehave arensor field which gives nine numbers foreach point inspace—or really sixforthesymmetric tensor S,,. Acomplete description oftheinternal forces inanarbitrarily distorted solid requires sixfunctions of x,Jyand 2 31-7Tensorsofhigher rank The stress tensor S,,describes theinternal forces ofmatter. Ifthematertal 18elastic,tisconventent todescribetheinternaldistortion intermsofanother tensor T,called thestrain tensor. For asimple object lke abarofmetal, you know thatthechange inlength, AL, 1sapproximately proportional totheForce, sowe sayitobeys Hooke's law: AL =F. Forasolid elastic body with arbitrary distortions, thestrain 7,,1srelated tothe stress S,,byasetoflinear equations: Ty=DLYomSer (31.26) Also,youknowthatthepotential energyofaspring(orbar)is 3PAL =YF. ‘Thegeneralization fortheelastic energy density inasolid body is Ueiasue =DYF%41SSee 1.27)om 3ktl 32 Refractive Index of Dense Materials 32-1 Polarization ofmatter Wewant now todiscuss thephenomenon oftherefraction oflight—and also, 32-1 Polarization ofmatter therefore, theabsorption oflight—by dense materials. InChapter 31ofVolume|55>\4axwel’s equationsina wediscussed thetheoryoftheindexofrefraction, butbecause ofourlimited dielectric eat mathematical abihties atthat time, wehad torestrict ourselves tofinding theindex nly formaterials oflowdensity, likegases. The physical principles that produced 32-3 Waves inadielectric theindex were, however, made clear Theelectric fieldofthelightwave polarizes complex index ,themolecules ofthegas,producing oscillating dipolemoments. Theacceleration 92-4The complexindexofrefractionoftheoscillating chargesradiatesnewwavesofthe field. This new field, interfering 32-5 The index ofamixture with theoldfield, produces achanged field which isequivalent toaphase shift of saves itheoriginal wave. Because thisphaseshiftisproportional tothethickness ofthe 92-6Wavesinmetals material, theeffect isequivalent tohaving adifferent phase velocity mthematerial. 32-7 Low-frequency and When welooked atthesubject before, weneglected thecomplications that arise high-frequency approximations; from such effects asthenew wave changing thefields attheoscillating dipoles theskin depth andtheplasma Weassumed that theforces onthecharges intheatoms came justfrom theincoming frequency wave, whereas, infact, their oscillations aredriven notonly bytheincoming wave butalso bytheradiated waves ofall theother atoms Itwould have been difficult forusatthat time toinclude this effect, sowestudied only the rarefied gas, where such effects arenotimportant. Review: SeeTable 32-1 Now, however, wewillfind that itisvery easy totreat theproblem bytheuse ofdiferential equations. ‘This method obscures thephysical origin oftheindex (@scoming from there-radiated waves interfering with vieoriginal waves), but itmakes the theory fordense materials much simpler. This chapter will bring togetheralargenumberofpiecesfromourearlierwork.We'vetakenuppracticallyeverything wewillneed, othere arerelatively fewreally new ideas tobetroduced. Since you may need t0refresh your memory about what wearegoing toneed,wegiveinTable32-1alistoftheequationswearegoingtouse,togetherwith@teference totheplace where each canbefound. Inmost instances, wewillnottake thetime togive thephysical arguments again, butwilljust usetheequations. Table 32-1 ‘Our work inthis chapter will bebased onthefollowing materia, already covered inearlier chapters Subject Reference Equation |amrotto Vol1,Chap.23.|mis+7x+ais)=F|neta vatnoms|atehla|naw me |Mobitty Vol. 1,Chap. 41) mx+yx=F |tectricalconductivity |Vol.Chap.43|=250=Ne Insidedielectnes Voth,Chap1|eyB+LP| mt 32-4 The complex index ofrefraction We want tolook now attheconsequences ofour result, Eq (32.33). Furst. wwenotice thatascomplex, sotheindex 11sgoing tobeacomplex number. What does that mean? Let’s saythat wewrite 1asthesum ofareal and animaginary part n= ne~inn, 235) where myand nyarereal functions of Wewrite iwith aminus sign, sothat ny will beapositive quantity inallordinary optical materials. (Inordinary inactive \ materials—that arenot,likelasers,lightsourcesthemselves~7 isapositivenumber, \ and that makes theimaginary part of negative.) Our plane wave ofEq.(32.21) VS aeumeve iswritten interms ofmas \Se Eom yeti i \OAL Writing nasinEq.(32.35), wewould have Aen Em Eger igtatnale (236) ue The term e'"!"e!” represents awave travelling with thespeed c/n, 80me OO. atten) represents whatwenormallythinkofastheindexofrefraction. Buttheamplitude “ ofthiswaveis Lv Een, if whichdecreases exponentially withzAgraphofthestrengthoftheelectricfield/ atsome instant asafunction ofzisshowninFig.32-1,forny~my/2m.The /imaginary part oftheindex represents theattenuation ofthewave due tothe Fig.32-1. Agraph ofExforsome energy losses intheatomic oscillators. Theintensity ofthewave 18proportionalinstanttifny~ne/2. tothesquareoftheamplitude, so Intensity xe~24"!* This isoften written as Intensity <e™*, where6=2wny/ciscalledtheabsorptioncoefficient. ThuswehaveinEq.(32.33) notonly thetheory oftheindex ofrefraction ofmatermals, butthetheory oftheir absorption oflight aswel. Inwhat weusually consider tobetransparent material, thequantity ¢/on— which has thedimensions ofalength—is quite large incomparison with the thickness ofthe material 32-5 The index ofamixture ‘There isanother prediction ofour theory oftheindex ofrefraction that we can check against experiment. Suppose weconsider amixture oftwo materials The index ofthemixture 15nottheaverage ofthetwo indexes, butshould begivenintermsofthesumofthetwopolarizabilties, asinEq.(32.34).Ifweaskabout theindex of,say, asugar solution, thetotal polarizability 1sthesum ofthe polarizability ofthewater and that ofthesugar. Each must, ofcourse, becal-culatedusingforWVthenumberperunitvolumeofthemolecules ofthe particular kind. Inother words, ifgiven solution hasN',molecules ofwater, whose polariz- ability 18a,and Nzmolecules ofsucrose (CyzH201),whosepolarizabrlity 1saa,weshould have that ni=13("5-5) =May+Neos. 32.37 (4) Nyay+Noe 8237) Wecan usethis formula totest our theory against experiment bymeasuring theindex forvarious concentrations ofsucrose inwater. Wearemaking several assumptions here, however. Our formula assumes that there 1snochemical action when the sucrose 1sdissolved and that the disturbances tothe individual atomic 32.8 Table 32-2 Refractive indexofsucrosesolutions,andcomparison withpredictions ofEq.(32.37). Data from Handbook A Boj ec|op|ie Fol oG 4|J| Molesof|Molesof(:_‘j w Fractionofsucrose|density nsucrose!|waters|3(5—5)| vaca byweight tgm/em') |at20°C|porter,|perter, wee2D)Mey|Neeamyier)|N2/No Mi/No | | | o 0.9982 1.333 0 55.5 0617 0617 of030 1.1270 13811 0970 43.8 0698 |0487|O21 02130.30 12296|14200|798|ats 0759|0379|0380|021108s Naase |1.5033) 3.99 1202 0.88 |0.1335) 0.752 02101.00 |Less iss|464 0|0960 |00.960 0.207 *pure water ®sugar erystals average (see (ext) “molecular weight ofsucrose =342 molecular weight ofwater =18 oscillators arenottoodifferent forvarious concentrations. Soourresult iscertainly only approximate. Anyway, let’s seehow good itis. Wehave picked theexample ofasugar solution because there isagood table ofmeasurements ofthe index ofrefraction inthe Handbook ofChemistry and Physics and also because sugar 1samolecular crystal that goes into solution with- ‘out1onizing orotherwise changing itschemucal state. Wegive inthefirst three columns ofTable 32~2 thedata from thehandbookColumnAisthepercentofsucrosebyweight,columnBisthemeasured density(gm/cm®), and column Cisthemeasured index ofrefraction forhight whose wavelength is589.3 millimicrons. For pure sugar wehave taken themeasured index ofsugar crystals. The crystals arenot isotropic, sothemeasured index is different along different directions. The handbook gives three values: my=1.5376, ng=1.5651, mg=1.5705. Wehave taken theaverage. Now wecould trytocompute nforeach concentration, butwedon’t know what value totake fora:oraz. Let’s testthetheory thisway: Wewillassume that thepolarizability ofwater (a,) isthesame atallconcentrations andcompute thepolarizability ofsucrose byusing theexperiment ofvalues formand solving Eq. (38.27) fora2. Ifthetheory iscorrect, weshould getthesame agforall concentrations. First, weneed toknow NV,andN2:let's express them interms ofAvogadro's number, No. Let's takeoneliter(1000 cm*) forourunttofvolume. Then N,/No 18 theweight per liter divided bythegram-molecular weight. And theweight per liter isthedensity (multiplied by1000 togetgrams perliter) times thefractional weight ofeither thesucrose orthewater. Inthis way, wegetV2/Ny and N;/NoasincolumnsDandEofthetable.Incolumn Fwehave computed 3(n? —1)/(n? +2)from theexperimental values of mincolumn C.Forpure water, 3(n*? —1)/(n? +2)is0.617, which is equal tojust Nya. Wecan then fillintherest ofColumn G,since foreach row row G/E may beinthesame ratio—namely, 0.617:55.5. Subtracting column G from column F,wegetthecontribution Nga. ofthesucrose, shown incolumn H Dividing these entries bythevalues ofN2/No incolumn D,wegetthevalue of Noa2 shown incolumn J From ourtheory wewould expect allthevalues ofNaz tobethesame They arenotexactly equal, butpretty close, Wecan conclude that our ideas arefairly correct. Even more, wefind that thepolarizability ofthesugar molecule doesn't seem todepend much onitssurroundings—its polarizability isnearly thesame ina dilute solution asitisinthecrystal. 329 32-6 Waves inmetals ‘The theory wehave worked outinthischapter forsolid materials canalso beapplied togood conductors, likemetals, with very little modification. Inmetals some oftheelectrons have nobinding force holding them toanyparticular atom; itis these “free” electrons which areresponsible fortheconductivity. ‘There are other electrons which arebound, and thetheory above isdirectly applicable to them. Their influence, however, isusually swamped bytheeffects ofthecon- duction electrons. Wewillconsider now only theeffects ofthefreeelectrons Ifthere isnorestoring force onanelectron—but still some resistance tots motion—its equation ofmotton differs from Eq. (32.1) only because theterm in aieislacking. Soallwehave todo1sseta=0intherestofourderivations— except that there isone more difference. The reason that wehad todistinguish between theaverage field and thelocal field inadielectric isthat inaninsulator each ofthedipoles isfixed inposition, sothat ithasadefinite relationship tothe position oftheothers. But because theconduction electrons inametal move around allover theplace, thefield onthem onrheaverage isjust theaverage field E.Sothecorrection wemade toEq.(325)byusing Eq.(32.28) should notbe made for conduction electrons Therefore the formula for the index ofrefraction formetals should look like Eq.(32.27), except with wosetequal tozero, namely, 2 Ng? 12yyNae 23 WTeySw?+He G238) This isonly thecontribution from theconduction electrons, which wewill assume 1sthemajor term formetals io Nowweevenknow howtofindwhatvaluetousefor7,because itisrelatedfarts totheconductivity ofthemetal.InChapter43ofVolumeIwediscussed howtheconductivity ofametal comes from thediffusion ofthefree electrons through the crystal. The electrons goonajagged path from one scattering tothenext, and between scatterings they move freely except foranacceleration duetoanyaverage electric field (asshown inFig 32-2), We found imChapter 43ofVolume Ithat ‘AVERAGE TIME BETWEEN theaverage drift velocity isjust theacceleration times theaverage time 7between COLLISIONS IS© collisions, Theacceleration isg.E/m, so Fig. 32-2. The motion of @freeelectron. tae=4x, (2.39) This formula assumed that Ewas constant, sothat rin Was asteady velocity. Since there 1snoaverage acceleration, thedrag force 1sequall totheapplied force Wehave defined ¥bysaying that Ym» isthedrag force (see Eq.(32.1)], which is qeE; therefore wehave that 1 y=. (32.40) Although wecannot eastly measure 7directly, wecandetermine itbymeasur-ingtheconductivity ofthe metal. Itisfound experimentally thatanelectric field E inametal produces acurrent with thedensity jproportional to (For ssotropic materials): jE. The proportionality constant¢1scalledtheconductivity. This1sjustwhatweexpect from Eq. (32.39) ifweset J=Naor Then o=Me, G241) m Sor—and therefore 1—can berelated totheobserved electrical conductivity. Using Eqs (32.40) and (32.41), wecan rewrite ourformula fortheindex, Eq. s210 (82.38), inthefollowing form: 2 o/€=14—% : 32.42Welt open Gai) where 1 mo ra5c ly 82.43) 7 NG This isaconvenient formula fortheindex ofrefraction ofmetals 32-7 Low-frequency andhigh-frequency approximations; theskin depth and the plasma frequency Our result, Eq.(32.42), fortheindex ofrefraction formetals predicts quite different characteristics forwave propagation atdifferent frequencies. Let's first seewhathappensatveryJonfrequencies. Ifwissmallenough,wecanapproximateEq.(32.42) by Wa. (2.44) eo Now, asyoucancheck bytaking thesquare,* 1-i / : vai= ts, v2 soforlow frequencies, n=Va) (1~i). (24s) ener The real and imaginary parts ofmhave thesame magnitude. With such alarge imaginary partton, thewave israpidly attenuated inthemetal. Referring to aa Eq.(32.36), theamplitude ofawave going inthez-direction decreases as x exp[=Vow 2eoc? 2]. (32.46) Let’s write this as en, 247) where51sthenthedistance inwhichthewaveamplitude decreases bythefactor siemace ° ae~!=1/2.72—or roughlyone-third. Theamplitude ofsuch awave asafunction ofzisshown inFig. 32-3. Since electromagnetic waves will penetrate into a Fig. 32-3. Theamplitude of@trans- metal only thisdistance, 6iscalled theskin depth. Itisgiven by verse electromagnetic wave asofunction of distance into «metal. 6=V2e?/ou. (32.48) Now what dowemean by“low” frequencies? Looking atEq. (32.42), we seethat itcan beapproximated byEq.(32.44) only ifwr1smuch lessthan one andifweo/@ 18also much lessthan one—that is,our low-frequency approximation applies when 1 eK, and o«t. (32.49) Let’s seewhat frequencies these correspond toforatypical metal like copper. Wecompute 7byusing Eq.(32.43), anda/¢o, byusing themeasured conductivity Wetake thefollowing data from ahandbook: @=5.76 X10"(ohm-meter)~', atomic weight =63.5 grams, density =8.9grams —em™', Avogadro’s number =6.02 X10** (gram atomic weight). ©Orwating <1=2; VST=e-1E cose/4 —rnx/4,which givesthe same result, sz several metals theexperimental observed wavelength atwhich they begin tobecome transparent. Inthe second column wegive thecalculated critical wavelength dy=2nc/wy. Considering that the experimental wavelength isnot too well defined, thefitofthetheory isfairly good. You may wonder why theplasma frequency «,should have anything todo with thepropagation ofelectromagnetic waves inmetals. The plasma frequency came upinChapter 7asthenatural frequency ofdensity oscillations ofthefree electrons. (Aclump ofelectrons isrepelled byelectric forces, andtheinertia ofthe Table 32-3* electrons leads toanoscillation ofdensity.) Solongitudinal plasma waves are resonant at«2p.Butwearenowtalking about ‘ransverse electromagnetic waves, Wavelengths below which themetal andwehave found thattransverse waves areabsorbed forfrequencies below w». becomes transparent (It'saninterestingandnoraccidentalcoincidence.) [Metal]XTexpermenial) [Xp=Prey Althoughwehavebeentalkingaboutwavepropagation inmetals,youap-|————|——“2 at preciatebythistimetheuniversalityofthephenomenaofphysies—thatitdoesn't|Tt|1550A|130.4| makeanydifferencewhetherthefreeelectronsareinametalorwhethertheyarePal 3150 3870 intheplasmaoftheionosphere oftheearth,orintheatmosphere ofastar.TO|Ry|3400 |jounderstand radiopropagation intheionosphere, wecanusethesameexpressions—_ |_SO1300 |_3220using, ofcourse, theproper values forNand 7.Wecanseenowwhylongradio» grom: C,Kittel, Inreduction toSolud waves areabsorbed orreflected bytheionosphere, whereas short waves goright Syate Physics, John Wiley andSons, Inc, through. (Short waves must beused forcommunication with satelhtes.) New York, 2nded.,1956, p.266, Wehave talked about thehigh- andlow-frequency extremes forwave propaga- tion inmetals. For thein-between frequencies the full-blown formula ofEq. (62.42) must beused. Ingeneral, theindex willhave real and imaginary parts; thewave isattenuated asitpropagates intothemetal. Forvery thin layers, metals aresomewhat transparent even atoptical frequencies. Asanexample, special goggles forpeople who work around high-temperature furnaces aremade by evaporating athin layer ofgold onglass. The visible light 1transmitted faurly 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 xdiscussed inChapter 10. Thedielectric constant xmeasures theresponse ofthematerial toaconstant field, that is,for»=0.Ifyou look carefully atthedefinition ofmandxyouseethat xissimply thelimit ofn?as«—+0.Indeed,placing»~Oandn?=«inequa- tions ofthischapter willreproduce theequations ofthetheory ofthedielectric constant ofChapter 11. a3 33 Keflection from Surfaces 3341Reflection andrefraction oflight ‘Thesubjectofthis chapter isthereflection andrefraction oflight—or electro- 33-1. Reflection andrefraction of magnetic waves ingeneral—at surfaces. We have already discussed thelaws of Tight teflection andrefraction inChapter 38ofVolume 1.Here'swhatwefoundOU 35>Wayesindensematerials 1,The angle ofreflection isequal totheangle ofincidence. With theangles 33-3 The boundary conditionsfinedasshown1nFig.33-1definedasshowninFig.33-1, 33-4Thereflectedandtransmitted%=0, G3.) waves 2.The product 1sin@isthesame forthe incident and transmitted beams 33-5 Reflection from metals (Snell’s aw). inter inysind,=ngsin0, 33.2) 33-6Totalinternal reflection 3.The intensity ofthereflected light depends ontheangle ofincidence and alsoonthedirection ofpolarization. ForEperpendicular totheplane of a 7neidence, thereflection coefficient Ryis Review. Chapter 35,Vol.1,Polarization J,_sin?(0,~0) =i _ nO.= 33 Re 1sine@+0,) OS) ForEparalleltotheplaneofincidence, thereflection coefficient Ri1s fp _tan?(0,—6)R= 7tane@, +0) GH) 4.Fornormal incidence (anypolarization, ofcourse!), a < I,_(mz—ms)? : ai>Geen) 09) tei: S (Earlier,weused1forthemeidentangleandrfortherefractedangleSincewe fave a’,can’tuserforboth“refracted” and“reflected” angles, wearenowusing6= aa ‘ incident angle, 6,=reflected angle, and 9=transmitted angle.) 7 Oureatherdiscussion isreallyaboutasfarasanyonewouldnormally need">@\ xe| togowiththesubject, butwearegoingtodoialloveragainadifferent way “> SARKan SURFACE,Why” Onereason isthatweassumed before thattheindexes werereal(noab- Cu sorption inthematerials) Butanother reason isthatyoushould know howto - n ny deal with what happens towaves atsurfaces from thepoint ofview ofMaxwell's p . equations. We'll getthesame answers asbefore, butnow from astraightforward, : solution ofthewave problem, rather than bysome clever arguments. - Wewant toemphasize thattheamplitude ofasurface reflection 1Snot& Fig.33-1, Reflection andrefractionproperty ofthemarertal, asistheindex ofrefraction It1sa“surface property.” ofightwaves atosurface. (Thewave ‘onethatdepends precisely onhow thesurface ismade. Athinlayer ofextraneous gireetions erenormal 10thewave crests) junk onthesurface between two materrals ofindices m,and mywill usually change thereflection. (There areallkinds ofpossibilities ofinterference here—hke the colors ofoilfilms Suitable thickness can even reduce thereflected amplitude to zero foragiven frequency: that's how coated lenses are made.) The formulas wewillderive arecorrect only ifthechange ofidex 1ssudden—within adistance very small compared with one wavelength. For light, the wavelength isabout 5000 A,sobya"'smooth” surface wemean one inwhich theconditions change in 3 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 you about theconvenient way ofdescribing asinusoidal \ planewaveweusedinChapter36ofVolumeI.Anyfieldcomponentinthewave \\\ \\GweuseBasanexample)canbewrittenintheForm \\\\B \ E=Eyeth", (3.6)\\\\A\whereErepresentstheamplitudeatthepointr(fromtheorigin)atthetime4. \‘\ \-4 \Thevector&pointsinthedirection thewaveistravelling, anditsmagnitude\\K SXx \|k|=k=2x/Xisthewavenumber.Thephasevelocityofthe wave 18fy=@/ky \\} \ foralightwaveinamaterial ofindex1,vy,=¢/n,80 Cranes kaon (3.7) \ ‘Supposekisinthez-direction, thenk«rsjustkz,aswehaveoftenuseditFor \\ WAVECRESTS &inanyotherdirection, weshouldreplace zbyr;,thedistance fromtheorigin intheA-divection; that 18,weshould replace kzbykri, which 15yust kr. (SeeFig,33-2.)SoEq.(33.6)isaconvenient representation ofawaveinanydirectionWe must remember, ofcourse, that Fig. 33-2. Forawave moving inthe direction kthephoseotanypointPi Keon=hex+ky+kes (at — ker. where k,,ky,and k,arethecomponents of&along thethree axes. Infact, we pointed outonce that (w,ks,ky.k:)isafour-vector, and that itsscalar product with (f,x,»,2)isaninvariant. Sothephase ofawaveisaninvariant,andEq. (33.6) could bewritten B= Eye, Butwedon’t need tobethat fancy now. Forasinusoidal E,asinEq.(33.6), 0£/ar isthesame asiwE, and a£/ax is ~tk,E, and soonfortheother components. You can seewhy itis very convenient tousetheform inEq.(336)when working with differential equations—differentia- tions arereplaced bymultiplications. One further useful point: The operation V=(a/ax, /ay, 8/22) gets replaced bythethree multiplications (—tk,, —iky, —ik,). But these three factors transform asthecomponents ofthevector &,so theoperator ¥gets replaced bymultiplication with —7k 2.yom Vo ik. 38) This remains true forany Voperation—whether it1sthegradient, orthediver- gence, orthecurl. For instance, thez-component of¥XEis aE,_aE, ‘ax ay Ifboth E,andE,vary ase~"*", then weget ~tkeEy +tkyEx which 1s,you see, thez-component of—k XE. Sowehave thevery useful general fact that whenever you have totake thegradientofavectorthatvariesasawaveinthreedimenstons(theyareanimportant part ofphysics), you can always take thederivations quickly and almost without thinking byremembering that theoperation V1sequivalent tomultiplication by ~ik. 332 Forinstance, theFaraday equation _ OB VKE= —a becomes for awave ~ik XE=~iwB. This tells usthat poEXE, G39) which corresponds totheresult wefound earlier forwaves infreespace—that B, imawave, 1satright angles toEandtothewave direction. (Infreespace, w/k = ¢.)Youcanremember thesigninEq.(339)fromthefactthat&isinthedirectionofPoynting’s vector S$=€,c*7E XB. Ifyou usethesame rule with theother Maxwell equations, you getagain the results ofthelastchapter and, inparticular, that ant kok= =SE (G310) Butsince weknow that, wewon't doitagain. Ifyouwant toentertain yourself, youcantrythefollowing terrifying problem that was theultimate test forgraduate students back in1890: solve Maxwell's equations forplane waves inananisotropre crystal, that is,when thepolarization P's related totheelectric field Ebyatensor ofpolarizability. You should, of course, choose your axes along theprincipal axes ofthetensor, sothat therelations aresimplest (then P,=agE,, Py=ay£,, and P,=acE,), but letthe waves have anarbitrary direction and polarization. You should beable tofind therela- tionsbetween EandB,andhow&varieswithdirection andwavepolarization. Then you will understand theoptics ofananisotropic crystal. Itwould bebest tostart with thesimpler case ofabirefringent crystal—like caleite—for which twoofthepolarrzabilities areequal (say, a,=a),and seeifyou can understand why you seedouble when you look through such acrystal Ifyoucandothat, then trythehardest case, inwhich allthree a’saredifferent. Then you will know whether you areuptothelevel ofagraduate student of1890. Inthis chapter, however, wewillconsider only isotropic substances .afy E, voy & Koh oy LENS 7Ke . Ke Loe LF, % ky] -7 wa tos Fig. 33-3. The propagation vectors Bo ala k,K’,andk”fortheincident, reflected,.'bd andtransmitted waves. Weknow from experience that when aplane wave arrives attheboundary between two different materials—say, arrand glass, orwater and o1l—there isa wave reflected and awave transmitted Suppose weassume nomore than that and seewhat wecan work out. Wechoose our axes with theyz-plane inthesurfaceandthexy-plane perpendicular totheincident wavesurfaces, asshowninFig.33-3. ms The electric vector ofthe incident wave can then bewritten as E,=Eqe's'"#?, 3.1) Since kisperpendicular tothez-axis, kor =kx+ky. 312) We write the reflected wave as E,=Eye’ wn, 33.13) sothat itsfrequency isw’,itswave number isk’, anditsamphtude 1sEj. (We know, ofcourse, that thefrequency isthesame andthemagnitude ofk1sthesame asfortheincident wave, butwearenotgoing toassume even that. Wewill letit come outofthemathematical machinery.) Finally, wewrite forthetransmitted wave, Fy=Efess"k, aaa) ‘Weknow that oneofMaxwell’s equations gives Eq_(33.9), soforeach ofthe waves we have BaRXR, geEXE, geEE ons) Also, ifwecallthemdexes ofthetwo media m,andnz,wehave from Eq.(33.10) ake a=OP (3316) Since thereflected wave 1sinthesame material, then wnt2m, 33.17 Kew @3.17) whereas forthetransmitted wave, : we. (33.18) y oa 33-3 The boundary conditions oP Allwehavedonesofaristodescribe thethreewaves; ourproblem now1s oe towork outtheparameters ofthereflected and transmitted waves interms of -Efe |Pye those oftheincident wave. How canwedothat? Thethree waves wehave de- ” scribed satisfy Maxwell's equations intheuniform material, but Maxwell’s equa- : Lf: tions must also besatisfied aftheboundary between thetwodifferent materials. eres Sowemust nowlookatwhat happens right aftheboundary. Wewillfindthat ny Maxwell's equations demand thatthethree waves fittogether inacertain way. oe ‘Asanexample ofwhat wemean, they-component oftheelectric field Emust —t. bethesameonbothsidesoftheboundary. Thisisrequired byFaraday’s law,1 [Pe x B- 8, 33.19)Fig.33-4. Aboundary condition VXER ay C19) Ey,=Fyisobtained fromf,Eds=0. . i aswecan seeinthefollowing way. Consider alittle rectangular loop 1°which straddles theboundary, asshown inFig 33-4, Equation (33.19) says that theline integral ofEaround I’isequal totherate ofchange ofthefluxofBthrough the loop: a few--2/e nda. Now imagine that therectangle isvery narrow, sothat theloop encloses an1n- finitesimal area. IfBremains finite (and there's noreason itshould beinfinite attheboundary!) theflux through thearea iszero Sotheline integral ofEmust 34 Now these equations must allhold inregion 1(totheleftoftheboundary) andinregion2(totherightofthe boundary). Wehave already written thesolu- tuons inregions 1and2.Finally, they must also besatisfied 1the boundary, which wecan call region 3.Although weusually think oftheboundary asbeing sharply discontinuous, inreality st1snot. The physical properties change very rapidly but notinfinitely fast. Inany case, wecan imagine that there isavery rapid, but continuous, transition oftheindex between region Iand 2,inashort distance we cancallregion 3.Also, anyfield quantity likeP,,orE,,ete.. willmake asimilar kind oftransition inregion 3.Inthis region, thedifferential equations must still besatisfied, and it1sbyfollowing thedifferential equations inthis region that we can arrive attheneeded “boundary conditions.” ' i Forinstance, suppose thatwehave aboundary between vacuum (region 1) HlPi e andglass(region2).There1snothing topolarize imthevacuum, soP;=0.1 - 2 Let's saythere 1ssome polarization P»intheglass. Between thevacuum andthe (0)| ' glassthereisasmooth, butrapid,transition Ifwelookatanycomponent of' P,sayP,,1might vary asdrawn inFig.33-5(a). Suppose now wetake thefirst : ofourequations, Eq(33.21). Itinvolves derivatives ofthecomponents ofPwith e<o! H respect tox,y,and2.They-andz-derivatives arenotinteresting; nothing spec-ad tacularishappening inthosedirections. Butthex-derivative ofP,willhavesomenecion!resign3!recone —_*YETYHargevaluesinregion3,becauseofthetremendous slopeofP,..Thederivative! H aP,/ax willhave asharp spike attheboundary, asshown inFig.33-5(b). Ifwe 'oP imagine squashing theboundary toaneventhinner layer,thespikewouldget 1 | mach higher Iftheboundary isreally sharp forthewaves weareinterested 1m, t f themagnitude ofaP,/dx inregion 3will bemuch, much greater than anycontribu- 1 ' tuons wemight have from theVariation ofP1nthewaveawayfromtheboundary— 1' soweignoreanyvarrations otherthanthoseduetotheboundary. 1' NowhowcanEq.(3321)besatisfiedifthere1sawhoppingbigspikeonthe \ \ right-hand side? Onlyifthere1sanequally whopping bigspikeontheotherside. I i Something ontheleft-hand side must also bebig. The only candidate is4£,/@x,r 1 Tbecausethevariations withy’and2areonlythosesmalleffectsinthewavewejust' H mentioned. So~e(dE/ax) mustbeasdrawn inFig.33-5(c)—just acopyof ' * aP,/ax. Wehavethat1[oe eqEe.ORs 3' i °°Ox ax i ©! H Ifweintegrate thisequation withrespecttoxacrossregion3,weconclude that : 1 j €0(Ex2 —Ext)=~(Psa ~Pes) 3.25) ' \ Inother words, thejump 1n€yE, ingoing from region Itoregion 2must beequal F tothejump in—P,. Wecanrewrite Eq.(33.25) as terialsinregions(1)and(2). whichsaysthatthequantity(€)E,-+Pz)hasequalvaluesinregion2.andregion1People say: thequantity (€)E, +P.)iscontinuous across theboundary. Wehave, inthis way, one ofour boundary conditions. Although wetook asanillustration thecase inwhich P,was zero because region Iwas avacuum, it1sclear that thesame argument apphes forany two materials inthetwo regions, soEq. (33.26) 1strue ingeneral. Let’s now gothrough therestofMaxwell's equations and seewhat each of them tells us.Wetake next Eq.(33.22a), There arenox-derwvatives, sottdoesn’t tellusanything. (Remember that thefields rhemselves donotgetespecialy large attheboundary; only thederivatives with respect toxcan become sohuge that they dominate theequation.) Next, welook atEq. (3322b). Ah There 1san arderivative! We have a£./ax onthe lefichand side. Suppose ithas ahuge de- rivative But wait amoment! There 1snothing ontheright-hand side tomatch it with; therefore E.cannot have anyjump ingoing from region Itoregion 2. [Ifatdid,therewouldbeaspikeontheleftofEq.(33.228)butnoneontheright, 336 andtheequation would befalse ]Sowehave @new condition Fa =Ey 3.27) Bythesame argument, Eq_(33.22c) gives Eyx ~En (33.28) This lastresult isjust what wegotinEq.(3320)byalineintegral argument. WegoontoEq.(3323) The only term that could have aspike 180B,/0x Butthere's nothing ontheright tomatch it,soWeconclude that Bes =Ber (33.29) Ontothelast ofMaxwetl’s equations! Equation (3324a) gives nothing,becausetherearenov-derivatives Equation (3323b)hasone,—c*48./ax, but again, there 1snothing tomatch itwith. Weget Be =Ber. (33.30) The lastequation 1squite simular, and gives Byz =Byx (3.31) Thelast three equations gives usthat By=By. Wewant toemphasize, ‘Table 33-1 however, that wegetthis result only when thematerials onboth sides ofthe ” ;boundary arenonmagnetie—or rather,whenwecanneglectanymagnetic effects Boundary conditions atthesurfaceof+ofthematerials, Thiscanusually bedone formost materials, except ferromagnetic - opes (We willtreat themagnetic properties ofmaterials insome later chapters.) (coEs +Pods =(euk's +Pade fOurprogram hasnetted usthesixrelations between thefields inregion 1and (Ey), =(Ea), those inregion 2,Wehave putthem alltogether inTable 33-1. Wecannow use (Ey): =(Es). them tomatch the wavesinthetworegions.Wewanttoemphasize, however,that BiBs theidea wehave just used will work inany physical situation inwhich you have curt —differential equations andyouwantasolution thatcrosses asharp boundary (Thesurface 15inthey2-plane) between two regions where some property changes. For our present purposes, wecould have easily derived thesame equations byusing arguments about the fluxes and circulations attheboundary. (You might seewhether you can getthe same result that way.) But now you have seen amethod that willwork incase you ever getstuck and don’t seeanyeasy argument about thephysics ofwhat 1shappen- ingattheboundary—you canjust work with theequations. 33-4 The reflected and transmitted waves Now weareready toapply our boundary conditions tothewaves wewrote down inSection 33-2. We had: E,=Eyeis!teenbon, (332) E,=Bie" atin, 3.33) E,=Efe" Metin, (33.34) B=boxe, (33.35) a= EXE, (33.36) BeEXE (3337) We have one further bitofknowledge: E1sperpendicular toitspropagation vector kfor each wave, 337 ‘The results willdepend onthedirection oftheE-vector (the “polarization”) oftheincoming wave. The analysis ismuch simplified ifwetreat separately thecase ofanincident wave with itsE-vector parallel tothe“plane ofincidence” (that 1s, thexy-plane) and thecase ofanineident wave with theE-vector perpendicular toy theplaneofincidence. Awaveofany other polarization 1sjust alinear combina- tion oftwo such waves. Inother words, the reflected and transmitted intensities yO ‘ aredifferent fordifferent polarizations, and 1tiseasiest topick thetwo simplest K.W casesandtreatthemseparately. er Ey Wewillcarrythrough theanalysis foranincoming wavepolarized per- pendicular totheplane ofmeidence and then just give you theresult fortheother. . 8 Wearecheating attlebytaking thesimplest case, buttheprinciple 1sthesame %forboth, Sowetake that E,hasonly az-component, and since alltheE-vectors k areinthesamedirectionwecanleaveoffthevectorsigns. g,a Solong asboth materials areisotropic, theinduced oscillations ofchargesin ~~SURFAGE thematertal willalsobeinthez-direction, andtheE-fieldofthetransmutted and ABradiated waves will have only 2-components. Soforallthewaves, E,and E, .° and P,and P,arezero, The waves will have their E-and B-vectors asdrawn in. ave one Fig.33-6(Wearecuttingacornerhereonouroriginalplanofgettingeverythingfrom theequations. This result would also come outoftheboundary conditions, Fig.33-6, Polarization ofthere- bulwecansave alotofalgebra byusing thephysical argument When youhave fiected ond transmitted waves when the Some spare time, seeifyou can getthesame result from theequations. Itisclear E-field oftheincident wave isperpendicu- that what wehave said agrees with theequations; it1sustthat wehave notshown lartotheplane ofincidence. that there arenoother possibilities.) Now our boundary conditions, Eqs. (3326) through (33.31), give relations between thecomponents ofEandBinregions |and2.Forregion 2wehave only thetransmitted wave, butinregion 1wehave sowaves. Which one doweuse?Thefieldsinregion1are,ofcourse,thesuperposition ofthefieldsoftheincidentand reflected waves. (Since cach satisfies Maxwell’s equations, sodoes thesum.) Sowhen weusetheboundary conditions, wemust usethat E,=E +k, Ey=Es and similarly fortheB's.Forthepolarization weareconsidering, Eqs.(33.26)and(33.28)giveusnoS new information; only Eq(33.27) isuseful. Itsays that E+h=B attheboundary, that is,forx=0.Sowehavethat Eyet@e 4Bho he Bye eK, (33.38) which mustbetrueforall¢andforally.Supposewelookfirstaty=0.Thenwe have Eye’! +Ege"! =Eye’ Thisequation saysthattwooscillating termsareequaltoathirdoscillation.‘That canhappen only afalltheoscillations have thesame frequency. (Itissm- possible forthree—or any number—of such terms with different frequencies to ‘add tozero foralltimes.) So w= wee, 33.39) Asweknew allalong, thefrequencies ofthereflected and transmutied waves are the same asthat ofthe incident wave. Weshould really have saved ourselves some trouble byputting that inatthe beginning, butwewanted toshow youthat itcanalso begotoutoftheequations. When youaredoing arealproblem, it1susually thebestthing toputeverything you know into theworks right atthestert and save yourselfalotoftrouble. Bydefinition, themagnitude ofk1sgiven byk?=n*w?/c?, sowehave also that ee weee (33.40) none ot a8 From Eqs. (33.35) through (33.37). © ra o Recalling that w”=w’=wand ky)=Ki,=k,,wegetthat Ey+Bh=Eb. Butthis isjust Eq.(3348)allover again! We've just wasted time getting something wealready knew. Wecould tryEq. (33.30), Bee =B.1, but there arenoz-components ofBt Sothere's only one equation left: Eq. (33.31), By2 =By1. For thethree waves. By=kes, By=~KE, By=-ME 3.49) Putting forE.,E,,and E;thewave expression for x=0(tobeattheboundary), theboundary condition 1s, ar?) Again all«’sandky’sareequal, sothisreduces to _ ‘ key +KLE) =KEY. (33.50) je 7%Kos hs Thisgivesusanequationforthe£’sthat1sdifferentfromEq.(3348).Withthe .Br u two,wecansolve for£;and£,'.Remembering thatk,=—k,,weget E, ke=Ke = Ee, Eo, 33 -Bomae 351) gE,: 2%,ye Oke 33 “ Z BS=TgBo (33.52) 8 SURFACE : , These, together with Eq.(33.45) orEq.(3346)fork””,give uswhat wewanted to know. Wewill discuss theconsequences ofthis result inthenext section. mo Ne Ifwebegin with awave polarized with itsE-vector parallel totheplane of incidence, Ewill have both x-and y-components, asshown inFig. 33-7. The Fig.33-7. Polonzation ofthewoves algebra isstraightforward butmore complicated (The work canbesomewhat when theE-field oftheincident wave is reduced byexpressing things inthis case interms ofthemagnetic fields, which are porallel totheplane ofincidence. ailinthez-direction.) One finds that nk,—nik B5|=ee IE 353) nik, +mike and 2nyns sl=—tetas (3354) ike +mk? Let's seewhether ourresults agree with those wegotearlier Equation (333) 1stheresult weworked outinChapter 35ofVolume Ifortheratio oftheintensity ofthereflectedwavetotheintensityoftheincidentwaveThen,however,wewereconsidering only real indexes For real indexes (and k’s), wecan write kz=kos8,=“THcos8, Ky=K"cos8,="2cos0 Substituting inEq. (33.51), wehave Eb_cos8,—2.608% Eb _ 16080, —nzc0s Or, 3Ey~mycos9,+nz60s0, (3358) 33:10 which does notlook thesame asEq.(33.3). Itwill, however, ifweuseSnell's law togetridofthen's.Setting ny=m,sin6,/sin @,,andmuluplying thenumerator anddenominator bysin6,weget Eh_0089,sin0;—sin.cos0 Ey~ cos 8,sin 6,+sin 0,cos 8, The numerator and denominator arejust the sines of(6,—6)and (8,+4); weget E%_sin(@,~64) Ey sin(i,+0) (5350) Since Ej,andEoareinthesame material, theintensities areproportional tothe squares oftheelectric fields, andwegetthesame result asbefore. Similarly, Eq. 83.53) 1thesame asEq. (33.4). Forwaves which arrive atnormal incidence, 8,=0and ,=0.Equation (33.56) gives 0/0, which isnotvery useful. Wecan, however, goback toEq (33.55), which gives te_(EY?_(m—12)? a CES) eas ‘This result, naturally, applies for“either” polarization, since fornormal incidencethereisnospecial“planeofincidence.” 33-5 Reflection from metals Wecan now useourresults tounderstand theinteresting phenomenon of reflection from metals. Why isitthat metals areshiny? Wesawinthelastchapterthatmetalshaveanindexofrefraction which,forsomefrequencies, hasalargeimaginary part. Let's seewhat wewould getforthereflected intensity when light shines from air(with n=1)onto amaterial with n=—in). Then Eq. (33.55) gives (fornormal incidence) By _i+in \ Y E,1in, SANS WZ Fortheintensity ofthereflected wave, wewantthesquare oftheabsolute values GREEN S|ReD ofEyandEy: Y\y -fy(EI? =|b+om? CNA 1B [in — or; Jo|cuapoate fateoy, (3358) oa 5 Pat DRIED REDINK Foramaterial with anindex which isapure imaginary number, there is100per- cent reflection! Fig. 33-8. Amaterial which absorbs Metals donotreflect 100percent, butmany doreflect visible light very well. light strongly atthe frequency wcloInotherwords,theimaginary partoftheir indexes isvery large Butwehave seen reflects light ofthatfrequency. that alarge imaginary part oftheindex means astrong absorption. Sothere 1sa general rule that ifany material gets tobeavery good absorber atany frequency, thewaves arestrongly reflected atthesurface and very little gets inside tobeab- sorbed You can seethis effect with strong dyes Pure crystals ofthestrongest dyes have a“metallic” shine. Probably youhave noticed thatattheedge ofabottle ofpurple inkthedried dyewillgive agolden metallic reflection, orthat dried red inkwill sometumes give agreenish metallic reflection. Red ink absorbs out the greens oftransmitted light, soifthe ink1svery concentrated, itwillexhibit astrong surface reflection forthefrequencies ofgreen light. You can easily show this effect bycoating aglass plate with red ink and letting1dry.Ifyoudirectabeamofwhitelightatthebackoftheplate,asshown inFig. 33-8, there willbeatransmitted beam ofrédlight andareflected beam of green light, sar oy \eyl PoE 7 Geh :. ra hy oo Em|ome Fig. 33-9. Total internal reflection. 33-6 Total internal reflection Iflight goes from amaterial likeglass, with arealsmdex greater than 1, toward, say,air,with anindex myequal to1,Snell's lawsays that sin6=sin 8 The angle 4,ofthetransmitted wave becomes 90°when theincident angle 8,is equal tothe“critical angle” 8,given by sin, =1 (33.59) What happens for9,greater than thecritical angle? You know that there 1stotal internal reflection. But how does that come about? Let’s goback toEq.(33.45) which gives thewave number &’’forthetrans mitted wave, We would have wee Bk ” Nowk,=ksin0,andk=wn/e,so oo fo. Ifnsin 6,18 greater than one, k’’?1snegasive andKi?isapure imaginary, say . ik. You know bynow what that means! The“transmitted” wave (Eq. 33.34) . acy will have the form os : Ey=Bettie, : vs Thewaveamplitude either grows ordrops offexponentially withincreasing x an ps Clearly, what wewant here1sthenegative sign. Then theamplitude ofthewave My totheright oftheboundary willgoasshown inFig.33-9. Notice thatAy18of lf Ln theorder «/e—which 18Xo,thefree-space wavelength ofthehght. When light1s . noe totally reflected from theinside of#glass-airsurface,therearefieldsinthearr, “nem. singzol’ agen buttheyextend beyond thesurface onlyadistance oftheorder ofthewavelength ofthelight Fig. 23-10. Ifthere is@smell gop, Wecannow seehow toanswer thefollowing question: Ifalight wave inglass internal reflection isnot“Yotol”; @trans- arrives atthesurface atalarge enough angle, it1sreflected, sfanother piece of mitted wave appears beyond thegap. glass isbrought uptothesurface (sothatthe“surface” ineffect disappears) the light istransmitted, Exactly when does this happen? Surely there must becon- tinuous change from total reflection tonoreflection’ The answer, ofcourse, 1s that iftheairgap15sosmall that theexponential talofthe wave intheairhasan appreciable strength atthesecond piece ofglass, itwall shake theelectrons there andgenerate &new wave, asshown inFig. 33-10, Some light willbetransmitted. (Clearly, oursolution isincomplete, weshould solve alltheequations again fora thin layer ofairbetween two regions ofglass.) saa I] . | , (6)TRANSMITTER veTEcTOR oETECTOR || 8 li]| 8 II} hry] | " (] " fay](ay TRANSMITTER DETECTOR DETECTOR TRANSWITTER OETECTOR ——-oETECTOR Fig. 39-11. Ademonstration ofthepenetration ofinternally reflected waves. ‘This transmission effect can beobserved with ordinary light only iftheair gapisvery smalll (oftheorder ofthewavelength oflight, like10~* cm), butitis easily demonstrated with three-centimeter waves. Then the exponentially de- creasing field extends several centimeters. Amicrowave apparatus that shows the effect 1sdrawn inFig. 33-11 Waves from asmall three-centimeter transmitter are directed ata45°prism ofparaffin, The index ofrefraction ofparaffin forthese frequencies is1.50, and therefore thecritical angle is41.5°. Sothewave 1stotally reflected from the 45° face and ispicked upbydetector A,asindicated in Fig. 33-11(a). Ifa second paraffin prism 1splaced incontact with thefirst, as shown inpart (b)ofthefigure, thewave passes straight through and 1spicked up atdetector B.If gap ofafewcentimetersisleftbetweenthetwoprisms,asin part (c),there areboth transmitted and reflected waves. The electric field outside the45°face oftheprism inFig, 33-11(a) can also beshown bybringing detector Btowithin afew centimeters ofthe surface. aa 34 The Magnetism ofMatter 341 Diamagnetism and paramagnetism Inthischapter wearegoing totalkabout themagnetic properties ofmaterials. 34-1 Diamagnetism and ‘The material which hasthemost striking magnetic properties 1s,ofcourse, iron. paramagnetism Similar magnetic propertiesaresharedalsobytheelementsnickel,cobalt,and-—at meticmomentsandangal suficiently lowtemperatures (below16°C)—bygadobinium, aswellasbyanumber 4"?Magnetic momentsandangular ofpeculiar alloys. ‘That kind ofmagnetism, called ferromagnetism, issufficiently stciking and complicated that wewill discuss itinaspecial chapter. However, 343 Theprecession ofatomic allordinary substances doshow some magnetic effects, although very small magnets ‘ones—a thousand toamullion times lessthan theeffects inferromagnetic materials : ,34-4Diamagnetism Herewearegoingtodescribeordinarymagnetism, thatistosay,themagnetism ofsubstances other than theferromagnetic ones. 34-5 Larmor’s theorem ‘This small magnetism isoftwo kinds. Some materials areattracted toward ical physies gives neilassical physicsgivesneither magnetic fields; others arerepelled. Unlike theelectrical effectinmatter, which 4®Classical physics givesne7 , ‘ diamagnetism nor alwayscausesdielectrics tobeattracted, therearetwosignstothemagnetic amas effect,Thesetwosignscanbecasilyshownwiththehelpofastrongelectromagnet paramagr which has one sharply pointed pole piece and one flat pole piece, asdrawn in 347 Angular momentum inquantumFig.34-1.Themagnetcfieldismuchstrongernearthepointedpolethannearthe mechanicsflatpole. If@small piece ofmaterial isfastened toalongstring andsuspended sagnetic energybetween thepoles,therewill,ingeneral, beasmallforceonit.‘Thissmallforce 34-8Themagnetic energy ofatoms canbeseenbytheslight displacement ofthehanging material when theMAgMet pose. Section 15-1, “The forces on isturned on,‘Thefewferromagnetic materials areattracted verystrongly toward Sen eae foressonthepointedpole:allothermaterials feelonlyaveryweakforce.Someareweakly Sigle ey attracted tothepointed pole;andsomeareweakly repelled. pole. STRING fie /SMALL PIECE OFMATERIAL WYBELtSWiz LINES OF 8 Fig.34-1.Asmallcylinder ofbis: Za~rovesof4strane ZA ruthisweaklyrepelledbythesharppole;ELECTAOMAGNET «@pieceofaluminumisottracted. The effect ismost easily seen with asmall cylinder ofbismuth, which is repelled from thehigh-field region. Substances which arerepelled inthisway are called diamagnetic. Bismuth isone ofthestrongest diamagnetic materials, but even with it,theeffect isstill quite weak. Diamagnetism isalways very weak. Ifasmall piece ofaluminum issuspended between thepoles, there isalso aweak force, but roward thepointed pole. Substances like aluminum arecalled para- ‘magnetic. (Insuch anexperiment, eddy-current forces arise when themagnet is turned onand off, and these can give offstrong impulses. You must becareful tolook forthenetdisplacement after thehanging object settles down.) ra astowhat might happen—even though thereally honest way tostudy thissubject would betolearn quantum mechanics first andthen tounderstand themagnetism 1mterms ofquantum mechanics. Ontheother hand, wedon’t want towait until welearn quantum mechanics inside out tounderstand asimple thing like diamagnetism We will have to Jean ontheclassical mechanics askind ofhalf showing what happens, realizing, however, that thearguments arereally notcorrect. Wetherefore make aseries of theorems about classical magnetism that will confuse you because they will prove different things. Except forthelast theorem, every one ofthem will bewrong, Furthermore, they willallbewrong asadescription ofthephysical world, because quantum mechanics isleftout. 34-2 Magnetic moments and angular momentum Thefirsttheorem wewant toprove from classical mechanics 1sthefollowing: JItanelectronismovinginacircularorbit(forexample,revolving aroundanucleusunder the influence ofacentralforce),thereisadefiniteratiobetweenthemagnetic a ‘moment and theangular momentum. Let’s call Jtheangular momentum and xthemagnetic moment oftheelectron intheorbit. The magnitude oftheangular momentum 1sthemass oftheelectron times thevelocity times the radius (See Fig.34-2.) Itisdirected perpendicular totheplane oftheorbit. J=mor. G4. "a (This is,ofcourse, anonrelativistic formula, butitisagood approximation for Fig. 34-2. Foranycircular orbit the atoms, because fortheelectrons involved v/c1sgenerally oftheorder ofe?/he = magnetic moment wisq/2m times the 1/137, orabout 1percent) fengularmomentumJ. The magnetic moment ofthesame orbit 1sthecurrent times thearea. (See Section 14-5) The current isthecharge perunit time which passes any point on theorbit, namely, thechargegtimesthefrequencyofrotation.Thefrequencyisthe velocity divided bythecircumference oftheorbit; so T=a5- ‘Thearea 18772,sothemagnetic moment 1s, ant (342) Itisalso directed perpendicular totheplane oftheorbit. SoJandyeareinthe same direction: w=3hJ(orbit) 43) Their ratio depends neither onthevelocity norontheradius. Forany particle moving inacircular orbit themagnetic moment isequal tog/2m times theangular momentum. For anelectron, thecharge isnegative—we can call it~g,: sofor anelectron Bee &J(electron orbit). (G44) That's what wewould expect classically and, miraculously enough, itisalsotruequantum-mechanically It’soneofthosethings.However, ifyoukeepgoingwith theclassical physics, you find other places where itgives thewrong answers, and itisagreat game totrytoremember which things areright and which things arewrong. We might aswell give you immediately what 1strue mgeneral m quantummechanics. First,Eq.(344)1struefororbizalmotion,butthat’snottheonly magnetism that exists. The electron also hasaspin rotation about itsown axis (something like theearth rotating onitsaxis), and asaresult ofthat spin it hasboth anangular momentum and amagnetic moment Butforreasons that are purely quantum-mechanical—there isnoclassical explanation—the ratio ofye cre figure, weseethat thechange ofangular momentum inthetime Atis AJ=Usin (wy A). Sotherateofchange oftheangular momentum is a G7 rdSi, (348) whichmustbeequaltothetorque: Lucas)7=uBsin’. G49) NCO. 1) ‘Theangular velocity ofprecession isthen Ms aHoy=FB, 4.10) =PSubstitutingu/JfromEq.(34.6),weseethatforanatomicsystem ‘|}=fh = 8, 4p theprecession frequency isproportional toB. Itishandy toremember that Fig. 34-3. Anobject with angular foranatom (orelectron) momentum Jand aparallel magnetic , moment1placedin@magnetic fieldB Jy=32=(1Amegacycles/gauss)gB, G4.12) recesses withtheangular velocity wp. and that for anucleus Sy=xt=(0.76kilocycles /gauss)gB. (34.13) (The formulas foratoms and nuclei aredifferent only because ofthedifferent conventions forgforthetwo cases.) According totheclassical theory, then, theelectron orbits—and_spins—in anatom should precess inamagnetic field. Isitalso true quantum-mechanically? Itsessentially true, butthemeaning ofthe“precession” isdifferent. Inquantum mechanics one cannot talk about thedirection oftheangular momentum inthe same sense asone does classically, nevertheless, there 1savery close analogy—so close that wecontinue 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, butoneoftheniceways isthefollowing. 8 ‘Suppose that weslowly turn onamagnetic field inthevicinity ofanatom.As 77,b> themagneticfieldchangesanelectricfieldisgeneratedbymagneticinduction. YOZWy>>PothFromFaraday’slaw,thelineintegralofEaroundanyclosedpathistherateofGQOchangeofthemagneticfluxthroughthepath,SupposewepickapathI’which1s Wo4YAacircle ofradius rconcentric withthecenter oftheatom, asshown inFig.344 << Theaveragetangential electricfield£aroundthispathisgivenby F| E2ar=-4(Br), Fig.34-4.Theinducedelectric and there isacirculating electric field whose strength is forces ontheelectrons inanatom, _rab PoTa The induced electric field acting onanelectron intheatom produces atorque equal to—g.Er, which must equal therate ofchange oftheangular momentum dI/dr: wu arabfiw. 4.14)a~ 2dt Guy aes Thequantityg-h/2m1susuallygiventhename“theBohrmagneton” andwritten Umogart The possible values ofthemagnetic energy are atth a Unag=guuB2s —— a Imag=SHBG: — ~J,=-$h —whereJ./htakesonthepossible valuesj,(J—1), —2),-.-.(-7 +D,=i. ~~ Inother words, theenergy ofanatomic system ischanged when itisputina — 4 Magnetic fieldbyanamount thatisproportional tothefield,andproportional to“25 -$h J.Wesaythattheenergy ofanatomic system is“split into2)+Ilevels” by 4magnetic field. For instance, anatom whose energy isUyoutside amagnetic Fig.34-5. Thepossible mognetic en- fieldandwhose yis3/2,willhave fourpossible energies when placed inafield.ergiesofanatomicsystemwithaspinof_Wecanshowtheseenergiesbyanenergy-level diagramlikethatdrawninFig3/2 inamagnetic field B. 34-5. Any particular atom can have only one ofthefour possible energies inany gwenfieldB.That1whatquantummechanicssaysaboutthebehaviorofan Unesatomic system inamagnetic field. eat th Thesimplest “atomic” systemisasingleelectron. Thespinofanelectron 1s — 1/2,sotherearetwopossible states.J,=/2.and J,=—1/2.Foranelectron —atrest(noorbital motion), thespinmagnetic moment hasag-value of2,sothe —magnetic energy canbeeither +B. The possible energies inamagnetic field are a 8 shown inFig. 34-6. Speaking loosely wesaythat theelectron either hasstsspin Se “up” (along thefield)or“down” (opposite thefield). — For systems with higher spins, there aremore states. Wecan think that the Su, =-¢h spinis“up” or“down” orcocked atsome “angle” inbetween, depending onthe value ofJ. Fig.34-6, Thetwopossible energy Wewillusethese quantum mechanical results todiscuss themagnetic prop- states ofonelectron inamagnetic field B. erties ofmaterials inthenext chapter. se Bs Paramagnetiam and Magnetic Resonance 35-1 Quantized magnetic states Inthelastchapter wedescribed how inquantum mechanics theangular 35-1 Quantized magnetic states momentum ofathing does not have anarbitrary direction, but stscomponent ;alonggivenaviscantakeononlycertainequallyspaced,discrete values, Its 32THESterm-Gerlach experiment ashocking and peculiar thing. You may think that perhaps weshould notgo 35-3 The Rabi molecular-beam into such things until your minds aremore advanced and ready toaccept this, method kind ofanidea, Actually, your minds will never become more advanced—in , thesenseofbeingabletoaccept suchathingeasily. Thereisn'tanydescriptive 38-4TheParamagnetism ofbulk.wayofmaking itintelligible thatisn’tsosubtle andadvanced initsownform ‘materials thatitismore complicated than thething youwere trying toexplain, The behavior 35-8 Cooling byadiabatic ofmatter onasmall scale—as wehave remarked many times—is different from demagnetization anything that you areused toand isvery strange indeed. Asweproceed with classical physics, it18agoodideatotrytogetagrowing acquaintance withthe 35-6Nuclear magnetic resonance behavior ofthings onasmall scale, atfirstasakind ofexperience without any deep understanding. Understanding ofthese matters comes very slowly, ifatalOfcourse,onedoesgetbetterabletoknowwhatisgoingtohappeninaquantum- hante -Des mechanical stuation-—ifthatiswhatunderstanding means—butonenevereetsaRev#e™:Chapter11,InsideDielectrics comfortable feeling that these quantum-mechancal rules are“natural.” Ofcourse they are, butthey arenotnatural toourown experience atanordinary level. We should explain that theattitude that wearegoing totake with regard tothisrule about angular momentum isquite different from many oftheother things wehave talked about. Wearenotgoing totryto“explain” it,butwemust atleast ellyouwhathappens;itwouldbedishonesttodescribethemagnetieproperties ofmaterials without mentioning the fact that theclassical description ofmagnetism—of angular momentum and magnetic moments—is incorrect. One ofthemost shocking and disturbing features about quantum mechanics isthat ifyou take theangular momentum along anyparticular axis you find that itisalways aninteger orhalf-integer times f.This issonomatter which axisyou take, Thesubtleties involved inthatcurious fact—that youcantake anyother axis and find that thecomponent foritisalso locked tothesame setofvalues—we will leave toalater chapter, when you will experience thedelight ofseeing how this apparent paradox isultimately resolved. Wewill now just accept thefact that forevery atomic system there isanumber j,called thespin ofthesystem—which must beaninteger orahalf-integer—and that thecomponent oftheangular momentum along any particular axis willalwayshaveoneofthe following values between +h and jh: {a jn I= oneof} ff oh 5.1) ai+2)itl jo) We have also mentioned that every simple atomic system has amagnetic moment which hasthesame direction astheangular momentum. This istrue not only foratoms and nuclei butalso forthefundamental particles. Each funda- mental particle has itsown characteristic value ofjand itsmagnetic moment. 3541 u v ist jee \eD eavei’ j,=0 Uo ——————+8 Uo| z 8 LT >(b) >> () ’ iacal ui ait \e ina Ss jer?Hyte Ue +8Jr*~v2 twp Fig.35-1. Anatomic system withspin (e) Le thas(2)+1)possible energy values ina Che magnetic field B.The energy splitting is proportional to8for small fields. (For some particles, both arezero.) What wemean by“the magnetic moment” inthis statement isthat the energy ofthesystem inamagnetic field, say in thez-direction, can bewritten a8—y,B forsmall magnetic fields. Wemust have the condition that the field should not betoo great, otherwise stcould disturb the internal motions ofthe system and the energy would not beameasure ofthemagnetic moment that was there before thefield was turned on. Butifthe field 1ssufficiently weak, thefield changes theenergy bytheamount AU =—u,B, 35.2) with theunderstanding that inthis equation wearetoreplace u,by we=(sh)Jo G53) where J;hasone ofthevalues inEq, (35.1). Suppose wetakeasystemwithaspinj=3/2.Without amagnet field,the system hasfour different possible states corresponding tothedifferent values of Jayallofwhich have exactly thesame energy. Butthemoment weturn onthemag- netic field, there 1sanadditional energy ofinteraction which separates these states into four slightly different energy levels. The energies ofthese levels aregiven by@ certain energy proportional toB,multiplied by4times 3/2, 1/2,~1/2, and —3/2— thevalues ofJ.The splitting oftheenergy levels foratomic systems with spins of 1/2, 1,and 3/2areshown inthediagrams ofFig. 35-1. (Remember that forany arrangement ofelectrons themagnetic moment isalways directed opposite tothe angular momentum.) You will notice from thediagrams that the“center ofgravity” ofthe energy. levels isthesame with and without amagnetic field. Also notice that thespacings from onelevel tothenext arealways equal foragiven particle inagiven magnetic field. Wearegoing towrite theenergy spacing, foragiven magnetic field B,as ‘has,—which 1syust adefinition ofw,. Using Eqs. (35.2) and (35.3), wehave ag4tnfay=gs AB al op=sk (35.4) 352 The quantity g(g/2m) 1sjust theratio ofthemagnetic moment totheangular momentum—it isaproperty oftheparticle. Equation (35.4)1sthesameformula thatwegotinChapter34fortheangularvelocityofprecessioninamagnetic field,foragyroscope whose angular momentum 1sJand whose magnetic moment isn Qo F—7 a SS —;po|MAGNET HOLE GLASS” VACUUM. PLaTe Fig. 35-2. Theexperiment ofStern and Gerlach. 35-2 The Stern-Gerlach experiment ‘The fact that theangular momentum isquantized issuch asurprising thing that wewill talk alittle bitabout ithistorically. Itwas ashock from themoment itwasdiscovered (although itwasexpected theoretically). Itwasfirstobserved in anexperiment done in1922 byStern and Gerlach. Ifyou wish, you can consider theexperiment ofStern-Gerlach asadirect justification forabeliefinthequantiza- tionofangular momentum. Stern andGerlach devised anexperiment formeasur- ingthemagnetic moment ofindividual silver atoms. They produced abeam of silver atoms byevaporating silver inahotoven and letting some ofthem come out through aseries ofsmall holes. This beam was directed between thepole tips ofaspecial magnet, asshown inFig. 35-2. Their idea was thefollowing. If thesilveratomhasamagnetic moment w,theninamagnetic fieldBithasanenergy —n,B, where =1sthedirection ofthemagnetic field. Intheclassical theory, 4. would beequal tothemagnetic moment times thecosine oftheangle between the moment and themagnetic field, sotheextra energy inthefield would be AU =~uB cos 6. (35.5) Ofcourse, asthe atoms come out ofthe oven, their magnetic moments would point inevery possible direction, sothere would beallvalues of6.Now ifthe magnetic field varies very rapidly with z—if there 1sastrong field gradient—then themagnetic energy will also vary with position, and there will beaforce onthe magnetic moments whosedirection willdependonwhether cosine@ispositive or negative. The atoms will bepulled upordown byaforce proportional tothe derivative ofthemagnetic energy; from theprinciple ofvirtual work, Fe=~2h=pcos0SB. (35.6) Stern and Gerlach made their magnet with avery sharp edge ononeofthe pole tipsinorder toproduce avery rapid variation ofthemagnetic field. The beam ofsilver atoms was directed right along this sharp edge, sothat theatoms would feelavertical force intheinhomogeneous field. Asilver atom with itsmagnetic moment directed horizontally would have noforce onitand would gostraight past themagnet. Anatom whose magnetic moment was exactly vertical would have aforce pulling ituptoward thesharp edge ofthemagnet. Anatom whose magnetic moment was pointed downward would feel adownward push. Thus, 33 Itisinteresting that one comes tothesame conclusion from aclassical point 8 ofview. According totheclassical picture, when weplace asmall gyroscope with amagnetic moment uand anangular momentum Jinanexternal magnetic field, thegyroscope will precess about anaxis parallel tothemagnetic field. (See Fig 35-3.) Suppose weask: How can wechange theangle oftheclassical gyroscope with respect tothefield—namely, with respect tothez-axis? The magnetic field produces atorque around ahorizontal axis. Such atorque youwould think 1 4 trying tolineupthemagnet withthefield, butitonlycauses theprecession. Ifwe wywanttochange theangle ofthegyroscope withrespect tothez-axis, wemust - exert atorque onitabout thez-axis. Ifweapply atorque which goes inthesame direction astheprecession, theangle ofthegyroscope will change togive asmaller component ofJinthez-direction InFig.35-3,theanglebetweenJandthe_Fig.35-3._The classicalprecession ofZaxis Would increase. Ifwetrytohinder theprecession, Jmoves toward the onatom withthemagnetic moment4and vertical,theangular momentum J. For our precessing atom mauniform magnetic field, how can weapply the kindoftorque wewant? Theanswer is:withaweak magnetic fieldfrom theside 8 You might atfirst think that thedirection ofthis magnetic field would have to rotate with theprecession ofthemagnetic moment, sothat itwas always atright angles tothemoment, asindicated bythefield B”inFig. 35-4(a). Such afield . works verywell, butanalternating horizontal fieldisalmost asgood. Ifwehave és asmall horizontal field B’,which tsalways inthex-direction (plus orminus) and a whichoscillates withthefrequency «,,thenoneachone-halfcyclethetorqueonthemagnette moment reverses, sothat ithasacumulative effect which isalmost a aseffective asarotating magnetic field. Classically, then, wewould expect the ‘component ofthemagnette moment along thez-direction tochange ifwehave@ Cevery weak oscillating magnetic field atafrequency which 1sexactly «,Classically, ad ofcourse, jz;Would change continuously, butimquantum mechanics thez-com- 8 ponent ofthemagneticmomentcannotadjustcontinuously. Itmustjumpsuddenly from one value toanother. We have made the comparison between the con- sequences ofclassical mechanics and quantum mechanics togive you some clue astowhat might happen classically andhow it1srelated towhat actually happens ~. inquantum mechanics. You willnotice, incidentally, that theexpected resonant A frequency isthesame inboth cases ‘Oneadditional remark: Fromwhatwehavesaidaboutquantum mechanics, we) {there 1snoapparent reason why there couldn't also betransitions atthefrequency 2up. Ithappens thatthere isn’tanyanalog ofthisintheclassical case, andalso coe itdoesn’t happen inthequantum theory either—at least notfortheparticular B=bcoset) method ofinducing thetransitions that wehave described. With anoscillatinghorizontal magneticfield,theprobability thatafrequency 2w,wouldcauseajump Fig.35-4.Theoraleofprecessionofaftwostepatonce1zero.TeisonlyattheFeguency w,thattransivons, ether o”famie magnetcanbechangedByc upward ordownward, arelikely tooccur. tangles to1,8in(a),oFbyanoscillating Now weareready todescribe Rabi’s method formeasuring magnetic mo- field, asintb) ments. Wewill consider here only theoperation foratoms with aspin of1/2. A diagram oftheapparatus 1sshown inFig. 35-5. There isanoven which gives out astream ofneutral atoms which passes down aline ofthree magnets, Magnet | MLS ELA MEZA KSwathy2, { cy Ww p28OSeAte = B DETECTOR OVEN pSbo? fl ~ = Je a \one SOMA sateCASS, MBGNET| MBE \Swacner'sL/S YW) 2 NNN sus, Fig. 35-5. The Rabi molecular-beom apparatus. 355 1sjust liketheoneinFig. 35-2, and hasafield with astrong field gradient—say, with @B,/az positive. Iftheatoms have amagnetic moment. they will bedeflected downward J. =+h/2, orupward ifJ, =—A/2 (since forelectronsy1sdirected opposite toJ). Ifweconsider only those atoms which cangetthrough theslit 5;,there aretwo possible trajectories, asshown. Atoms with J,=-+//2 must goalong curveatogetthroughtheslit,andthosewithJ,=—A/2mustgoalong curve b.Atoms which start outfrom theoven along other paths will notget through theshit. Magnet 2has «uniform field. There are noforces ontheatoms inthis region, sothey gostraight through and enter magnet 3.Magnet 31just lke magnet 1butwith thefield inverted. sothat @B./az hastheopposite sign. The atoms with J,=+/2 (we say “with spin up”), that felt adownward push in magnet I,getanupward push inmagnet 3;they continue onthepath «and go through shtSztoadetector. The atoms with J,=—A/2 (“with spin down”) also have opposite forces inmagnets 1and 3and goalong thepath 6,which also takes them through slitS.tothedetector. Thedetector may bemade invarious ways. depending ontheatom being measured. For example, foratoms ofanalkalimetallikesodium,thedetectorcan beathin, hottungsten wire connected toasensitive current meter. When sodium atoms land onthewire, they areevaporated offasNa* ions, leaving anelectron behind. There 1sacurrent from thewire proportional tothenumber ofsodium atoms arriving persecond. Inthegap ofmagnet 2there 1sasetofcouls that produces asmall horizontal magnetic field B’.The cous aredriven with acurrent which oscillates atavariable frequency w. Sobetween the poles ofmagnet 2there isastrong, constant, verticalfieldByandaweak,oscillating,horizontalfieldBY. perectorSupposenowthatthefrequencyaoftheoscillatingfield1ssetat«,—the ‘CORRENT “precession” frequency oftheatomsmnthefieldB,Thealternating fieldwallcausesome oftheatoms passing bytomake transitions from oneJ.totheother An j atom whose spin was initially “up” (z= +A/2) may befipped “down” i Ue=h/2). Now this atom hasthedirection ofitsmagnetic moment reversed, vy soitwillfeeladownwardforceinmagnet3andwillmovealongthepatha’,shown inFig. 35-5. Itwill nolonger getthrough theslitS$)tothe detector. H Similarly, some oftheatoms whose spins were imially down J.=—/2) will have their spins flipped up(J,=+A/2) asthey pass through magnet 2.They ' willthen goalong thepath b”andwillnotgettothedetector. H Ithe oscillating fieldB’hasafrequency appreciably different from wit willt+, o> notcauseanyspinfips,andtheatomswillfollowtheirundisturbed pathst0 thedetector. Soyoucanseethatthe“precession” frequency «,oftheatoms Fig.35-6.ThecurrentofatomsininthefieldBycanbefoundbyvaryingthefrequencywofthefieldB’unuilade- thebeam decreases when w=tp. crease 1sobserved inthecurrent ofatoms arriving atthedetector. Adecrease in thecurrent walloccur when w1s“inresonance” with «,, Aplot ofthedetector current asafunction ofwmight look liketheoneshown inFig. 35-6. Knowing p,wecan obtain theg-value oftheatom, Such atomic-beam or,asthey areusually called, “molecular” beam resonance experimentsareabeautifulanddelicatewayofmeasuringthemagneticproperties ofatomic objects. The resonance frequency w,can bedetermined with great precision—in fact, with agreater precision than wecan measure themagnetic field Bo,which wemust know tofindg. 35-4 The paramagnetism ofbulk materials Wewould like now todescribe thephenomenon oftheparamagnetsm of bulk materials Suppose wehave asubstance whose atoms have permanent mag- netic moments, forexample acrystal likecopper sulfate. Inthecrystal there are copper 1ons whose inner electron shells have anetangular momentum andanet ‘magnetic moment. Sothecopper 1on1sanobject which hasapermanent magnetic moment, Let's sayjust aword about which atoms have magnetic moments and which ones don’t. Any atom, likesodium forinstance, which hasanoddnumber 35-6 hyperbolic tangent funetion: - 0B M=Nugtanh 422 5.21) Aplot ofMasafunctionofBisgiveninFig.35.7.WhenBgetsverylarge, thehyperbolic tangent approaches 1,and Mapproaches thehmiting value Nip Soathigh fields. themagnetization saturates. We can seewhy that 18;athigh enough fields themoments arealllined upinthesame direction Inother words, they areallinthespin-down state, and each atom contributes the moment jy Inmost normal cases—say, fortypical moments, room temperatures, andy thefields onecannormally get(like 10,000 gauss)—the ratio oB/KT1sabout0.02. ‘One must gotovery low temperaturestoseethesaturation, Fornormaltempera- "He|=—$=== tures, wecanusually replace tanh xbyx,andwrite ) _Mabe a, om=Mu (3522) Just aswesaw intheclassical theory, Misproportional toB.Infact, the formula isalmost exactly thesame, except thatthereseems tobeafactor of1/3 — _ missing. But westill need torelate thewoinour quantum formula tothewthat o. *appears intheclassical result, Eq(35.9). wee Intheclassical formula, what appears 1su?=jej,thesquare ofthevector _Fig.35-7. Thevariation ofthepara- magnetic moment, or magnetic magnetization withthemagnetic a? fieldstrength 8.”w~(¢)ys. (3523) Wepointed out inthelast chapter that you can very likely gettheright answer from aclassical calculation byreplacing J:Jbyj(j+1)A?. Inourparticular example, wehave j=1/2, so JG+Wn? =gn. Substituting this forJ-JinEq.(35.23), weget yp)? 3h? orinterms ofuo,defined inEq.(35.12), weget won =Bub. Substituting this foruintheclassical formula, Eq. (35.9), does indeed reproduce thecorrect quantum formula, Eq. (35.22). The quantum theory ofparamagnetism iseasily extended toatoms ofany spinj.The low-field magnetization 1s 2iG+1)wa M=ng TDD ene. 35.24) where ihn=3 85.25) 18acombination ofconstants with thedimensions ofamagnetic moment. Most aloms have moments ofroughly this size. Itxscalled the Bohr magneton. ‘The spinmagnetic momentofthe electron 1salmost exactly one Bohr magneton. 35-5 Cooling byadiabatic demagnetization There isavery interesting special application ofparamagnetism. Atvery lowtemperatures 1tispossible toline uptheatomic magnets inastrong field Itisthen possible togetdown toextremely lowtemperatures byaprocess called adiabatic demagnetizanon. Wecan take aparamagnetic salt (for example, one 35-9 protons inthelower energy states—with their moments directed parallel tothe field. ‘There isasmall netmagnetic moment perunit volume. Since theproton moment isonly about one-thousandth ofanatomic moment, themagnetization which goes asu?—using Eq.(35.22)—is only about one-millionth asstrong as, typical atomic paramagnetism. (That's why wehave topick amaterial with no atomic magnetism.) Ifyou work itout, thedifference between thenumber of protons with spin upandwith spin down isonly onepart in10%,sotheeffect isindeed very small! Itcan still beobserved, however, inthefollowing way. Suppose wesurround the water sample with asmall coil that produces a small horizontal oscillating magnetic field. Ifthisfield oscillates atthefrequency ‘Wpitwillinduce transitions between thetwoenergy states—just aswedescribed forthe Rabi experiment inSection 35-3. When aproton flips from anupper energy state toalower one, itwillgive uptheenergy u,Bwhich, aswehave seen, 38equal tofio, Ifitfips from thelower energy state totheupper one, itwill absorb theenergy hw,from thecoil. Since there areshghtly more protons inthe lower state than intheupper one, there will beanetabsorption ofenergy from the coil. Although theeffect isvery small, theslight energy absorption can beseen with asensitive electronic amplifier. Just asintheRabi molecular-beam experiment, theenergy absorption will be seen only when theoscillating field 1sinresonance, that 1s,when 4 w=oy=8(3m,) 8 Itisoften more convenient tosearch fortheresonance byvarying Bwhile keeping wfixed. The energy absorption willevidently appear when 4, Aypical nucle: c 7 LAA mppuany yypical nuclear magnetic resonance apparatus 1sshown inFig. 35-8. A waGner cas high-frequency oscillator drives asmall coilplaced between thepoles ofalarge rove electromagnet. Two small auxihary coils around thepoleupsaredriven witha joscuaton 60-cyclecurrentsothatthemagnetic fieldis“wobbled” aboutitsaveragevaluebywater=p 7averysmallamount.Asanexample,saythatthemaincurrentofthemagnet1sset on|togivea field of5000 gauss, and theauxiliary coils produce avariation of+1gauss oss about thisvalue, Iftheoscillator issetat21.2megacycles persecond, itwillthenbe S16NA attheproton resonance eachtimethefieldsweeps through 5000 gauss (using Eq. SASL) (B4.13) withg=5.58fortheproton}. pe ‘Thecircuitoftheoscillatorisarrangedtogiveanadditionaloutputsignal |tan)proportional toanychange inthepower beingabsorbed fromtheoscillator. This Wysignal 1sfedtothevertical deflection amplifier ofanoscilloscope. Thehorizontal uy sweep oftheoscilloscope istriggered once during each cycle ofthefield-wobbling, frequency. (Moreusually, thehorizontal deflection ismadetofollowinproportion once tater tothewobbling field.) Before thewater sample isplaced inside thehigh-frequency coil, thepower Fig 35-8. Anuclear magnetic reso- drawn from theoscillator issome value, (Itdoesn’t change with themagnetic field ) nance apparatus. When asmall bottle ofwater isplaced inthecoil, however, asignal appears onthe oscilloscope, asshown inthefigure. Weseeapicture ofthepower being absorbed bytheflipping over oftheprotons! Inpractice, itis difficult toknow how tosetthemain magnet toexactly $000 gauss. What one does 1stoadjust themain magnet current until theresonance signal appears ontheoscilloscope. Itturns out that this isnow themost con venient way tomake anaccurate measurement ofthestrength ofamagnetic field. Ofcourse, atsome time someone hadtomeasure accurately themagnetic field and frequency todetermine theg-value oftheproton. Butnow that this hasbeen done, proton resonance apparatus like that ofthefigure can beused asa“proton reso- nance magnetometer.” Weshould sayaword about theshape ofthesignal. Ifwewere towobble the magnetic field very slowly, wewould expect tosee anormal resonance curve. ‘The energy absorption would read amaximum when w,arrived exactly atthe 35-11 36 Ferromagnetism 36-1 Magnetization currents Inthischapter wewilldiscuss some materials inwhich theneteffect ofthe 36-1 Magnetization currents ‘magnetic momentsinthematerial1smuchgreaterthaninthecaseofparamagnetism ordiamagnetism. Thephenomenon iscalledferromagnetism. Inparamagnetic and 36"?Thefield diamagnetic materials theinduced magnetic moments areusually soweak that 36-3 The magnetization curve wedon’t havetoworry about theadditional fields produced bythemagnetic 1 inmoments. Forferromagnetic materials, however, themagnetic moments induced 3&4 Hron-core inductances byapplied magnetic fields arequite enormous and have agreat effect onthefields 36-8 Electromagnets themselves. Infact,theinduced moments aresostrong thattheyareoftenthe s sizatdominant effectinproducing theobserved fields. Sooneofthethings wewill 36-6Spontaneous magnetization have toworry about isthemathematical theory oflarge induced magnetic moments. That is,ofcourse, just atechnical question. The real problem is,why aretheticmomentssostrong—how work?W. on magnetic moments 0strong—how doesitallwork’WewilcometothatquestRenew:Chapter10,Dielectrcs Finding themagnetic fieldsofferromagnetic materials issomething likethe Chapter 17,TheLawofIneproblem offinding theelectrostatic fieldinthepresence ofdielectrics. Youwill lwctionremember thatwefirstdescribed theinternalproperties ofadielectric intermsofavector field P,thedipole moment perunit volume, Then wefigured outthat the effects ofthis polarization areequivalent toacharge density p,.1 given bythedi- vergence ofP: Poot =VP. (36.1) The total charge inany situation can bewritten asthesum ofthis polarization charge plus allother charges, whose density WeWrite* oxic. ‘Then theMaxwell equation which relates thedivergence ofEtothecharge density becomes eB =P xProttRosier, © © or vib=—SIP4Potner,«6 € Wecanthen pull outthepolarization part ofthecharge and putitontheother side oftheequation, togetthenew law V+€oE +P)=Rother 36.2) Thenew Jawsays thedivergence ofthequantity (¢£+P)isequal tothedensity oftheother charges. Pulling £andPtogether asinEq.(36.2), ofcourse, 1suseful only ifweknow some relation between them. We have seen that the theory which relates the induced electric dipole moment tothefield was arelatively complicated business andcan really only beapplied tocertain simple situations, and even then asan approximation. We would like toremind you ofone oftheapproximate ideas weused. Tofind theinduced dipole moment ofanatominsideadielectric,118 necessary toknow the electric field that acts onanindividual atom. We made the approximation—which 1snottoobadinmany cases—that thefield ontheatom *Ifall ofthe“other" charges were onconductors, porn Would bethesame asour fiomofChapter 10. 364 1sthesame asitwould beatthecenter ofthesmall hole which would beleftifwe took outtheatom (Keeping thedipole moments ofalltheneighboring atoms the same). You will also remember that theelectric field inahole inapolarized di- Aardgale keene electricdepends ontheshapeofthehole.Wesummarize ourearlierresultsinEnoie=E+P/Eo “iyFig.36-1.Forathin,disc-shapedholeperpendiculartothepolarization,theyy (7eaefieldintheholeisgivenbyVA ig Bate=Enotoorne+2‘A Y whichweshowedbyusingGauss’law.Ontheotherhand,inaneedle-shapedYLslotparalleltothepolarization, weshowed—by usingthefactthatthecurlofBiszero—that theelectricfieldsmsideandoutsideoftheslotarethesame.Finally, ZL wefoundthatforasphericalholetheelectricfieldwasone-third ofthewaybetween YL,Yio Yjthefieldofthe slot and the field ofthe disc: yy> “te Bsoie=Exwiecne+5©(opericlhole). 63)1 This was thefield weused inthinking about what happens toanatom insidf. a polarizeddielectric. G4Now wehave todiscuss theanalog ofallthis forthecase ofmagnetism. ‘One simple, short-cut way ofdoing this istosaytheM,themagnetic moment per unit volume, isjust hkeP,theelectric dipole moment perunit volume, andthat,berry therefore, thenegativeofthedivergence ofMf1sequivalent toa“magnetic chargeEngi=E+P/3E0 density”p,,—whatever thatmaymean.Thetrouble1s,ofcourse,thatthereisn't / any suchthingasa“magnetic charge” inthephysical world. Asweknow,the ED| divergence ofBisalways zero.Butthatdoesnotstopusfrommaking anartificial1P, analog andwriting ow VM=~pay (364) where itistobeunderstood that pqispurely mathematical. Then wecould make acompleteanalogywiththeelectrostatic caseanduseallouroldequationsfrom Y (/, electrostatics. People haveoftendonesomething likethat.Infact,historically,people even believed that theanalogy was right. They believed that thequantity Fig.36-1. Theelectric field inaPmfepresented thedensity of“magnetic poles.” ‘These days, however, weknow cavity inadielectric depends onthe thatthemagnetization ofmaterials comes from circulating currents within the shape ofthecavity. 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 1s,ofcourse, very complicated. Itsvalue depends onwhere you look intheatom—it's large in some places and small inothers; itgoes one way inone part oftheatom and the ‘opposite way inanother part (just asthemicroscopic electric field varies enor- mously inside adielectric). Inmany practical problems, however, weareinterested only 1nthefields outside ofthematter orintheaverage magnetic field inside ofthe matter—where wemean anaverage taken over many, many atoms. Itisonly for such macroscopic problems that rtisconvenient todescribe themagnetic state of thematter interms ofM,theaverage dipole moment perunit volume. What we want toshow now isthat theatomic currents ofmagnetized matter can give rise tocertain large-scale currents which arerelated toM. What wearegoing todo,then, istoseparate thecurrent density j—which is thereal source ofthemagnetic fields—into various parts: one part todescribe the circulating currents oftheatomic magnets, and theother parts todescribe what other currents there may be. Itisusually most convenient toseparate thecurrents into three parts. InChapter 32wemade adistinction between thecurrents which flow freely onconductors and theones which aredue totheback and forth motions 362 oftheboundchargesindielectrics. InSection32-2wewrote 4=Soot +Gotwers where jj: represented thecurrents from themotion ofthebound charges indi- electrics andj,rnee took care ofallother currents. Now wewant togofurther. Wewant toseparate juaier into Onepart, juysgy Which describes theaverage currentsinsideofmagnetized materials, andanadditional termwhichwecancalljeowaforwhatever isleftover. The last term will generally refer tocurrents inconductors, butitmay also include other currents—for example thecurrents from charges moving freely through empty space. Sowewill write forthetotal current density F=Joo +fang +Joona- 36.5) OfcourseitisthistotalcurrentwhichbelongsimtheMaxwellequation forthecurl ofB: 2, ~i4 2%.evxBe ds (36.6) Now wehave torelate thecurrent jag tothemagnetization vector M. So that you can seewhere wearegoing, wewill tellyou that theresult isgoing to bethat Jinag =VX M. (36.7) Ifwearegiventhemagnetization vectorMeverywhere inamagnetic material,thecirculation currentdensity1sgivenbythecurlofM.Let'sseeifwecanunder- stand whythisisso ede.First,let'stakethecaseofacylindrical rodwhichhasauniformmagnetization parallel toitsaxis. Physically, weknow that such auniform magnetization really eae53> meansauniformdensityofatomiccirculating currentseverywhere insidethe “material. Suppose wetrytoimagine whattheactual currents would looklikein YY OBB 73701 across section ofthematerial. Wewould expect toseecurrents something like WIA S6//0 those shown inFig.36-2. Each atomic current goes around andaround inalittle 4 citcle,withallthecirculating currents goingaround inthesamedirection. Now egcgavenenene whatistheeffectivecurrentofsuchathing?Well,inmostofthebarthereisno 56Z effectatall,because rightnexttoeachcurrent thereisanother current goinginy loge; theoppositedirection. Ifweimagineasmallsurface—but onestillquiteabitL feyos larger than asingle atom—such asisindicated inFig. 36-2 bytheline 4B, z thenetcurrent through such asurface 1szero. There 18nonetcurrent any- * where inside thematerial. Note, however, thatatthesurface ofthematerial there Fig,36-2, Schematic diagram ofthe areatomie currents which arenotcancelled byneighboring currents going the circulating atomic currents osseen in@ other way. Atthesurface there 1sanetcurrent always going inthesame direction cross section ofanironrodmagnetized in around therod. Now you seewhy wesaid earlier that auniformly magnetized thez-direction. rod18equivalent toalong solenoid carrying anelectric current. How does this view fitwith Eq.(36.7) First, inside thematertal themagne-tizationMisconstant, soall itsderivatives arezero. This agrees with ourgeometric picture. Atthesurface, however, Misnot really constant—it isconstant upto theedge and then suddenly collapses tozero, So,right atthesurface there are terrific gradients which, according to(36.7), will give ahigh current density Suppose welook atwhat happens near thepoint CinFig. 36-2. Taking the x- andy-directions asinthefigure, themagnetization Mis mnthez-direction. Writing ‘outthecomponents ofEq. (36.7), wehave oe=mele (36.8) =OMeGnwel Atthepoint C,thederivative aM,/ay 1szero, butaM,/ax islarge and positive. Equation (36.7) says that there isalarge current density intheminus y-direction. This agrees with our picture ofasurface current going around thebar. 36-3 Now wewant tofindthecurrent density foramore complicated case inwhich themagnetization varies from point topoint inamaterial. Itiseasy toseequali- » tatively that ifthemagnetization isdifferent intwoneighboring regions, there will notbeaperfect cancellation ofthecirculating currents sothat there willbeanet current inthe Volume ofthe material. Itisthis effect that we want towork out quantitatively. QHAY First, weneed torecall theresults ofSection 14-5 that acirculating current Wee Thasamagnetic moment ugiven by SURFACE AREA A nal, G65) Fig.36-3. Thedipole moment of@WhereAistheareaofthecurrentloop(seeFig.36-3).Nowlet'sconsiderasmallcurrent loopis1A. rectangular block inside ofamagnetized material,assketchedinFig.36-4.We take theblock sosmall that wecan consider that themagnetization 1suniform My inside it.Ifthisblock hasamagnetization M,inthez-direction, theneteffect will bethesame asasurface current going around onthevertical faces, asshown. Wecan find themagnitude ofthese currents from Eq.(36.9). The total magnetic A moment oftheblock isequal tothemagnetization umes thevolume: 1 ae el5T from which weget(remembering that thearea oftheloop isac) a T=Mab. id Inotherwords,thecurrentperunitlength(vertically) oneachofthevertical5 surfaces isequal toM,. Fig.36-4.Asmallmagnetized block M, Mz+OM, isequivalent to@circulating surface a 3 \ :eet Thqe [ee DTH lh Wh eo Fig.36-5. Ifthemagnetization of 2, <1 2 twoneighboring blocksisnotthesame, t2 there isanet surface current inbetween. . Now suppose that weimagine two such little blocks next toeach other, asshowninFig.36-5.Becauseblock2isslightlydisplaced fromblock1,1twillhaveaslightly different vertical component ofmagnetization, which wecall M,++AM,. Now on the surface between the two blocks there will betwo contributions tothe total current, Block |willproduce acurrent /;flowing inthepositive y-direction, and block 2will produce asurface current J»flowing inthenegative y-direction. The total surface current inthepositive y-direction isthesum T= 1,=I=Mab ~(M, +AM, =-aM.b. Wecanwrite AM, asthederivative ofM,inthex-direction tumes thedisplacement from block 110block 2,which isjust a: OM, am, = a. The current flowing between thetwo blocks isthen aM, r=—2ab. 36-4 Torelate thecurrent Jtoanaverage volume current density j,wemust realizethatthiscurrent/isreallyspreadoveracertaincross-sectional area.Ifweimaginethewhole volume ofthe material tobefilled with such little blocks, one such side face (perpendicular tothex-axis) can beassociated with each block.* ‘Then we seethat thearea tobeassociated with thecurrent /isjust thearea abofone of thefront faces. Wegettheresult =-t__aM,I~~ ax Wehave atleast thebeginning ofthecurl ofM. There should beanother term inj,from thevariation ofthex-component of themagnetization with z.This contribution towillcome from thesurface between twolittle blocks stacked oneontopoftheother, asshown inFig.36-6. Using LH thesame arguments wehave justmade, youcanshow that thissurface willcon- f My+BMy tribute toj,theamount aM,/a2. ‘These aretheonlysurfaces which cancontribute 72! tothey-component ofthecurrent sowehave thatthetotal current density inthe 1 y-direction is aTenj,=OMe_OM: <2“ae ox H Workingoutthecurrentsontheremaining facesofacube—or usingthefact a1 Mythatourz-direction iscompletely arbitrary—we canconclude thatthevector L currentdensity1indeedgivenbytheequation Ly iygoUxXM. 5 Soifwechoose todescribe themagnetic situatron inmatter interms ofthe Fig,36-6. Twoblocks, oneabove the average magnetic moment perunit volume M,wefind that thecirculating atomic other, may also contribute tojy. currents areequivalent toanaverage current density inmatter given byEq.(36.7). Ifthe material 1salso adielectric, there may be,inaddition, apolarization current jini =OP/at. And ifthematerial isalso aconductor, wemay have aconduction current j,,. a8well. Wecan write thetotal current as Fajot vxMae. (36.10) 36-2 The field H Next, wewant toinsert thecurrent aswritten inEq. (36.10) into Maxwell's equations. Weget 2 i,21, a)aE| VXBRLHS=elemVXMHS) +S We can move the term inMtothe left-hand side: a MY_fous2(’) ‘Asweremarked inChapter 32,many people like towrite (E+P/eo) asanew vector field D/éo. Similarly, itisoften convenient towrite (B—M/eoc*) asa single vector field, Wechoose todefine anew vector field Hby u-p-™. (36.12)coe! Then Eq. (36.11) becomes e00"V XH joa + 6.13) Itlooks simple, butallthecomplexity isjust hidden intheletters Dand H. *Or,sfyou prefer, thecurrent Fineach face should besplit 80-50 with theblocks on the two sides 365 Now wehave togive you awarning. Most people who usethemks units have chosen touseadifferent definition ofH. Calling their field H’(ofcourse, they stillcall1Hwithout theprime), itisdefined by H’=c?B —M. (36.14) (Also, they usually write égc? asanew number I/uo: then they have onemore constant tokeep track of!) With thisdefinition, Eq.(36.13) looks even simpler: VXH=jon+O- 6.15) Butthedifficulties withthisdefimtion ofH”are,first,that1doesn’tagreewiththedefinition ofpeople who don’t usethemks units, and second, that 1makes H! and Bhave different units. We think itismore convemient for Htohave the same units asB—rather than theunits ofM,asH’does. But ifyou aregoing tobean engineer andwork onthedesign oftransformers, magnets, andsuch, youwillhave towatch out. You will find many books which useforHfthedefinition ofEq. (36.14) rather than ourdefinition ofEq.(36.12), and many other books—especially handbooks about magnetic matertals—that relate Band Htheway wehave done. You'll have tobecareful tofigure outwhich convention they areusing. Table 36-1 ‘One way totellisbytheunits they use. Remember that imthemks system, sai ‘B—and therefore ourH—are measured with theunit: one weber persquare meter,nits quant Unitsofmagneticquantities equalto10,000gauss.Inthemkssystem,amagneticmoment(acurrenttimesan {B]=weber/meter® =104gauss area)hastheunit:oneampere-meter®, Themagnetization M,then,hastheunit: [#1]=weber/meter® =104gauss ‘oneamperepermeter.ForH’theunitsarethesameasforM.Youcanseethator10%oersted thisalsoagreeswithEq.(36.15),since¥hasthedimensionsofoneoveralength. in=ampere/meter Peoplewhoareworking withelectromagnets alsogetinthehabitofcallingthe [EP]~ampere/meter unitofH(with theH’definition) “oneampere turnpermeter”—thinking ofthe turnsofwireonawinding. Buta“turn”’1sreallyadimensionless number,sothat Convenient " conversionsdoesn’t need toconfuse you. Since ourH'1s equal toH’/egc2, ifyouareusing the B(gauss) =104B(weber/meter?) mkssystem, #(inwebers/meter?) 1sequal to4xX10? times H’(inamperes H(gauss) =H(ocrsted) permeter). Itisperhaps more convenient toremember that H(ingauss) = =0.0126 H'(amp/meter) 0.0126 H’(inamp/meter). There isone more horrible thing. Many people who useourdefinition of Hhave decided tocalltheunits ofHand Bbydifferent names! Even though they have thesame dimensions, they call theunit ofBonegauss, and theunit ofHone oersted (after Gauss and Oersted, ofcourse). So,inmany books you willfind graphs with Bplotted ingauss and Hinoersteds. They arereally thesame unit— 10~* ofthemks unit, Wehave summarized theconfusion about magnetic units inTable 36-1. 36-3 The magnetization curve Now wewill ook atsome simple situations inwhich themagnetic field 18 constant, orinwhich thefields change slowly enough that wecanneglect D/A in comparison withj.....Thenthefieldsobeytheequations vB=0, (36.16) VX =joona/€oc*, (36.17) H= B~ M/eg?. (36.18) Suppose wehave atorus (adonut) ofiron wrapped with acoil ofcopper wire, asshown inFig, 36-7(a). Acurrent Jflows inthewire. What 1sthemagnetic field? The magnetic field will bemainly inside theiron; there, thelines ofBwillbecircles,asdrawninFig.36-7(b).SincethefluxofB1scontinuous, itsdivergence1szero, andEq(36.16) 18satisfied Next. wewrite Eq.(36.17) 1nanother form by 36-6 integrating around theclosed loop Tdrawn inFig. 36-7(b). From Stokes’s <theorem,wehavethat eBOQfds |janamat 6.19) LZ ?» f oeJy Lk Gq wheretheintegralofjistobecarriedoutoveranysurfaceSboundedbyI.This Y Cysurfaceiscutoncebyeachturnofthewinding, Eachturncontributes thecurrent. =—V/) iJtotheintegral,and,ifthereareNturnsinall,theintegral1sNJ.Fromthe %symmetryofourproblem,Bisthesameallaroundthecurve[’;ifweassumethat > LYthemagnetization, andtherefore,thefieldHisalsoconstantalongI’,Eq.(36.19) oy>ed becomesry ww AO a=. where /isthelength ofthecurve I.So, Lr - SN He (52) 9” ~ Itisbecause Hisdirectly proportional tothemagnetizing current incases like i . thisonethatHissometimes calledthemagnetizing field. Ii \\ Nowallweneedisanequation which relates HtoB.Butthereisn’tanysuch My \i\pequation!There1s,ofcourse,Eq.(36.18),butit1snohelpbecausethereisnoq\'TA idirectrelationbetweenMandBforaferromagneticmateriallikeiron.Themag- i/|Pnetization Mdependsonthewholepasthistoryoftheiron,andnotonlyonwhat Xs yyBisatthemoment. WK wyAllisnotlost,though. Wecangetsolutions incertainsimplecases.Ifwe SS LA startoutwith unmagnetized iron—Iet’s saywithironthathasbeen annealed at ad hightemperatures—then inthesimplegeometry ofthetorus, allthetronwillhave ° thesame magnetic history. Then wecansaysomething about M-and therefore Fig.36-7. (a)Atorus ofironwound about therelation between Band H—from experimental measurements. The with acoilofinsulated wire. (b)Cross field Binthetorus is,from Eq. (36.20), given asaconstant times thecurrent J section oftorus showing field lines inthewinding. The field Bcanbemeasured byintegrating over time theemf in thecoil(orinanextra coilwound over themagnetizing coilshown inthefigure). 8 Thisemfisequal totherateofchange ofthefluxofB,sotheintegral oftheemf (amuse) with time isequal toBtimes thecross-sectional area ofthetorus. " » Figure 36-8 shows therelation between Band H,observed with atorus of softiron. When thecurrent isfirst turned on,Bincreases with increasing Halong 1 thecurve a.Note thedifferent scales onBand H;initially, ittakes only arelatively small Htomake alarge B,Why isBsomuch larger with theiron than itwould bewith air? Because there isalarge magnetization Mwhich isequivalent toa soot [fe large surface current ontheiron—the field Bcomes from thestn ofthis current and theconduction current inthewinding. Why Mshould besolarge, wewill discuss later. at a ‘Athigher values ofH,themagnetization curvelevels off.Wesaythatthe 1(gov) iron sasurates. With thescales ofourfigure, thecurve appears tobecome hori- zontal. Actually, itcontinues toriseshightly—for large fields, Bbecomes propor- tional toH,and with aunit slope. There isnofurther increase ofM.Incidentally, wweshould point outthatifthetorus were made ofsome nonmagnetic material, 10,000 ‘Mwould bezero andBwould equal Hforalfields. % Thefirstthing wenotice isthatcurve ainFig.36-8—which istheso-called ”‘magnetization curve—is highly nonlinear. Butt'sworse thanthat.If,afterreaching “ene saturation, wedecrease thecurrent inthecoiltobring Hback tozero, themagnetic fieldBfallsalong curve 6,When Hreaches zero,there isstillsome Bleft.Even ig,36-8, Typical magnetization withnomagnetizing current there isamagnetic fieldintheiron—it hasbecome andhysteresis curves forsoftiron, permanently magnetized. Ifwenow turn onanegative current inthecoi, the B-H curve continues along 6until theiron issaturated inthenegative direction.Itwethenbringthecurrentbacktozeroagain,Bgoesalongcurvec.Ifwealternate thecurrent between large positive and negative values, theB-H curve goes back andforth along very nearly thecurves 6and c.Ifwevary Hinsome arbitrary 36-7 possible. One way todecrease thearea oftheloop istoreduce themaximum field thatisreached during each cycle. Forsmaller maximum fields, wegetahysteresis curve like theone shown inFig. 36-9. Also, special materials aredesigned tohave 2 averynarrow loop. Theso-called transformer irons—which areironalloys with -_ _. asmall amount ofsilicon—have beendeveloped tohavethisproperty. a When aninductance isrunover asmall hysteresis loop, therelationship vor between BandHcanbeapproximated byalinearequation. Peopleusuallywrite rope,’ B= ul. (36.23) / !Theconstant uisnotthemagnetic moment wehaveusedbefore.Itiscalledthe in permeability oftheiron. (Itisalsosometimes called the“relative permeability.") fli The permeability ofordinary irons istypically several thousand. There arespecial alloysalike“supermalloy” whichcanhavepermeabilities ashighasamillion. =“**“*“"f)'3aayIfweusetheapproximation that B=uA!inEq.(36.21), wecanwrite the hi energy inatoroidal inductance as iA! uP wy U=(ecufHadH=(697A) (36.24) J : Sotheenergy density isapproximately woe wa86ua Fig.36-9.Ahysteresisloopthot WecannowsettheenergyofEq.(36.24)equaltotheenergy£/*/2ofaninductance, doesn'treachsaturation. and solve for£.Weget“u(y£=(eA (>7 Using H/I from Eq. (36.20), wehave NAoe (36.25) The inductance isproportional towu.Ifyou want inductances forsuch things as audio amplifiers, you will trytooperate them onahysteresis loop where the B-H relationship isaslinear aspossible. (You will remember that wespoke in Chapter 50,Vol. 1,about thegeneration ofharmonics innonlinear systems.) For such purposes, Eq. (36.23) isauseful approximation. Ontheother hand, ifyouwant togenerate harmonics, youmay useaninductance which isintention- allyoperated inahighly nonlinear way. Then you willhave tousethecomplete B-H curves, and analyze what happens bygraphical ornumerical methods.‘A“transformer” isoftenmadebyputtingtwocoilsonthesametorus—or —————core—ofamagneticmaterial.(Forthelargertransformers, thecoreismadewithGY arectangularproportionsforconvenience.)Thenavaryingcurrentinthe“primary” LL)V7 windingcausesthemagneticfieldinthecoretochange,whichinducesanemfin Atly the“secondary” winding. Since theflux through each turn ofboth windings is IEEthesame,theemf’sinthetwowindingsareinthesameratioasthenumberof HPAturnsoneach.Avoltageappliedtotheprimaryistransformed toadifferent {(GGa,voltageatthesecondary.SinceacertainnefcurrentaroundthecoreisneededtoVA,i 7producetherequiredchangeinthemagneticfield,thealgebraicsumofthecurrents Hlainthetwowindingswillbefixedandequaltotherequired“magnetizing”current. Mil\IZL) Ifthecurrent drawnfromthesecondary increases, theprimary current mustin- Hel) crease inproportion—there isa“transformation” ofcurrents aswellasvoltage. T Fig.36-10. Anelectromagnet. 36-5 Electromagnets Now Jet'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? 369 Buti. Baste cuven —_LT_-surtoce s (EA FZIZ HVLDL iZZZIIZ ZN ) + Vf~i(o) ut «WYSeyWSL Y)AAT corPer ‘CURRENT Fig. 36-11. Cross section ofanelectromagnet. Ifthegapthickness issmall compared with alltheother dimensions, wecan, asafirst approximation, assume that thelines ofBwill goaround through the loop, just asthey didinthetorus They will look more orless asshown inFig, 36-I1(a). ‘They tend tospread outsomewhat inthegap, butifthegap isnarrow, this will beasmall effect. Itisafair approximation toassume that theflux of Bthrough anycross section oftheyoke isaconstant Iftheyoke hasauniform cross-sectional area—and ifweneglectanyedgeeffectsatthegapsoratticorners —we can saythat Bisuniform around theyoke. ‘Also, Bwillhave thesame value inthegap. This follows from Eq.(36.16). Imagine theclosed surface S,shown inFig. 36-11(b), which has one face inthe gap and theother intheiron, The total flux ofBoutofthis surface must bezero. Calling B;thefield inthegapand Bythefield intheiron, wehave that BiAy —BaAz =0. a Since 4,=Ag(toourapproximation), itfollows that By=Ba. Now Iet’s look atH. Wecan again useEq. (36.19), taking theline integral around thecurve I’inFig. 36-11(b). Asbefore, theright-hand side isNJ,the number ofturns times the current. Now, however, Hwill bedifferent inthe iron and intheair. Calling M»thefield intheiron and /,thepath length around thered \ yoke,thispartofthe curve willcontribute theamount 1» totheintegral. Calling ( ‘H,thefield inthegapand/,thegapthickness, wegetthecontribution fromeg(3627) thegap.Wehavethat 4 N \\r-0 Wh+Hale=BE (36.26) ir *ae Nowweknowsomething else:thatintheairgap,themagnetization 1sneghgi-ble, sothat By=Hy. Since By=Bs,Eq.(36.26) becomes / NL /Bali+Hale=Ces (36.27) ZL ‘Westillhavetwounknowns. TofindByandH2,weneedanotherrelationship—a | namely,theonewhichrelatesBtoHinthetron,Ifwecanmake theapproximation that B,=M2, wecan solve theequation algebraically. However, les dothegeneral case, inwhich themagnetization curve Fig.36-12. Solving forthefieldinoftheironisonelikethatshowninFig.36-8.Whatwewant1sthesimultaneous ‘onelectromagnet. solution ofthisfunctional relationship together withEq.(36.27). Wecanfindit byplotting agraph ofEq.(36.27) onthesame graph with themagnetization curve, asisdone inFig. 36-12. Where thetwo curves intersect, wehave oursolution. For agiven current /,thefunction (36.27) 1sthestraight line marked />0 inFig,36-12. Thelineintersects theH-axis (By=0)atHy=Mi/eyc*ls, and theslope is—/2/l,. Different currents just shift thelinehorizontally. From Fig. 36-10 Itis,ofcourse, possible togetthese results inamore physical way, byusing theMaxwell equations directly. For example, Eq.(36.34) follows directly from¥-B=0.(Youuseagaussiansurfacethatishalfinthematerialandhalfout.)Similarly, youcangetEq.(36.33) byusing alineintegral along acurve that goes upinside thehole and returns through thematerial, Physically, thefield inthe hole 1sreduced because ofthe surface currents—which are given by VXM.WewillleaveitforyoutoshowthatEq.(36.35)canalsobeobtainedbyconsideringtheeffects ofthesurface currents onthe boundary ofthespherical cavity. Infinding theequilibrium magnetization from Eq.(36.29), 1turns outtobe most convenient todeal with H;sowrite M B=H+neE (36.36) Inthespherical hole approximation, wewould have =4,but, asyou will see, ‘wewill want later tousesome other value, soweleave itasanadjustable parameter. ‘Also,wewilltakeallthefieldsinthesamedirectionsothatwewon'tneedtoworry=| about thevector directions. Ifwewere now tosubstitute Eq.(36.36) into Eq (36.29), wewould have one equation that relates themagnetization Mtothemag- sousrionnetizingfieldH: |a 2 2M=Nutanh(4Mle) exes { Itis,however, anequation thatcannot besolved explicitly, sowewilldoitgraph- caveically. e168) Let’s puttheproblem inageneralized form bywriting Eq. (36.29) as ‘ 63sisws it=tanhx, (36.37) og Mow WhereM,..isthesaturation valueofthemagnetization, namely,Nu,and.xrepresents Fig.36-13.Agraphical solutionof4B,/KT. ‘Thedependence ofM/My. onxisshown bycurve an Fig,36-13. Eqs.(36.37) and(36.38). Wecan also write xasafunction ofM—using Eq,(36.36) forB,—as ByuH,(uXMue)M- xaip t(Ga)Moa’ (6.38) For any given value ofH,this 1sastraight-line relationship between M/M,.. and x.Thexintercept isatx=wH/KT, andtheslope is€o¢*KT/u XM,ux. Forany particular H,wewould have aline like theone marked inFig. 36-13. The intersection ofcurves aand bgives usthesolution forM/M... Wehave solved theproblem. Let's look athow thesolutions willgoforvarious circumstances. Westart |withH=0.Therearetwopossiblesituations,shownbythelines6,andbycaecwx inFig.36-14. YouwillnoticefromEq.(36.38)thattheslopeofthelineispro- ys we portional totheabsolute temperature7.So,athightemperatures wewouldhave“}~~~/-~~5~= aline likeby.Thesolution 1sM/M,.. =0.When themagnetizing fieldHiszero, % by themagnetization 1salso zero. Butatowfemperatures, wewould havealinelikeb, % and there are rwo solutions for M/M,..—one with M/Myqx = and one with 9 M/M,.. near one. Itturns outthat only theupper solution isstable—as you can seebyconsidering small variations about these solutions. According tothese ideas, then, amagnetic material should magnetize itselt | spontancously atsufficiently lowtemperatures. Inshort, when thethermal motions Cn aresmall enough, thecoupling between theatomic magnets causes them allto line upparallel toexch other—we have apermanently magnetized material anal- Fig. 36-14. Finding the magnetiza- ‘ogous totheferroelectrics wediscussed inChapter 1. tionwhen H=0, Ifwestart athigh temperatures and come down, there isacritical temperature, called theCurie temperature T,,where theferromagnetic behavior suddenly sets in. This temperature corresponds tothelinebyofFig. 36-14, which istangent tothe curve a,and has, therefore, aslope of1.The Curie temperature isgiven by eohT et) 36.39) 36-13 thephenomenon offerromagnetism, wehave toimagine that themagnetization ofthefield enhances thelocal field bysome large factor—like one thousand or more. There doesn’t seem tobeany reasonable way tomanufacture such tremen- dous fields atanatom—nor even fields oftheproper sign! Clearly, our“magnetic” theory offerromagnetism isadismal failure. Wemust conclude, then, that ferro- magnetism hastodowith some nonmagnetic interaction between thespinning electrons inneighboring atoms. This interaction must generate astrong tendency forallofthenearby spins tolineupinonedirection. Wewillseelater that ithas Wea todowith quantum mechanics and thePauli exclusion principle. 19a Finally, welook atwhat happens atlow temperatures—for T<T.. We > have seen that there will then beaspontaneous magnetization—even with H=0— given bytheintersection ofthecurvesaand63ofFig.36-14.IfwesolveforM forvarious temperatures—by varying theslope ofthelineb2—we getthetheoretical o curve shown inFig. 36-15. This curve should bethesame forallferromagnetic 9 materials forwhich theatomic moment comes from asingle electron. ‘The curves ° forother materials areonly slightly different. Inthelimit, asTgoes toabsolute zero, Mgoes toM,.... Asthetemperature isincreased, themagnetization decreases, falling tozero attheCurie temperature. Thepoints inFig.36-15 aretheexperimental observations fornickel. They fitthe t 3 5 theoretical curve fairly well. Even though wedon’t understand thebasic mecha- uw nism, thegeneral features ofthetheory seem tobecorrect. Finally, there isone more disturbing discrepancy inourattempt tounder- Fig. 36-15. Spontaneous magnetiza-standferromagnetism. Wehavefoundthatabovesometemperature thematerial tionasafunctionoftemperature forshould behave likeaparamagnetic substance withamagnetization Mpropor- _nickel. tional toH(orB),and that below that temperature itshould become spontane- ously magnetized. But that’s not what wefound when wemeasured themag- netization curve foriron. Itonly became permanently magnetized after wehad “magnetized” it.According totheideas just discussed, itwould magnetze 1s! What iswrong? Well, 1tturns outthatifyoulook atasmalll enough crystal ofiron ornickel, 11sindeed completely magnetized! Butinlarge pieces ofiron, there are many small regions or“domains” that aremagnetized indifferent directions, so that onalarge scale theaverage magnetization appears tobezero. Ineach small domain, however, theiron hasalocked-in magnetization with Mnearly equal to May. The consequences ofthis domain structure arethat gross properties of large pieces ofmaterial arequite different from themicroscopic properties that wwehave really been treating. Wewilltake upinthenext lecture thestory ofthe practical behavior ofbulk magnetic materials 36-15 37 Magnetic Materials 37-1 Understanding ferromagnetism Inthischapter wewilldiscuss thebehavior andpeculiarities offerromagnetic 37-1 Understanding ferromagnetism materials andofother strange magnetic materials. Before proceeding tostudy 37.9Thermodynamic properties magnetic materials, however, wewill review very quickly some ofthethings about thegeneral theory ofmagnets thatwelearned inthelastchapter. 37-3 Thehysteresis curve First, weimagine theatomic currents inside thematerial that areresponsible Jc materforthemagnetism, andthendescribethemintermsofavolumecurrentdensity 37-4Ferromagnetic materials Jing =VX M. Weemphasize thatthisisnotsupposed torepresent theactual 37-5 Extraordinary magnetic currents. When themagnetization isuniform thecurrents donotreally cancel materials outprecisely; that is,thewhirlng currents ofone electron inone atom and the Whirling currents ofanelectron imanother atom donotoverlap insuch away that thesum isexactly zero. Even within asingle atom thedistribution of magnetism isnotsmooth. For instance, inaniron atom themagnetization References: Bozorth, R.M.,“Magne- isdistributed inamore orlessspherical shell, nottooclose tothenucleus and usm," Encyclopaedia Bri- nottoofaraway. Thus, magnetism inmatter isquite acomplicated thing in1s tannica, Vol. 14 1957, details; itisvery irregular. However, weareobliged now toignore thisdetailed PP.636-667. complexity anddiscuss phenomena from agross, average point ofview. Then Kittel, C.,Introduction 10 111struethattheaverage current intheinterior region, overanyfiniteareathat Solid State Physea, John 'sbigcompared withanatom, 18zerowhen M=0.So,whatwemean by Wiley andSons, Inc,Newmagnetization perunitvolumeandjing,andsOon,atthelevelwearenow Yor’,Indof.1956considering, 1sanaverage over regions that arelarge compared with thespace occupied byasingle atom. Inthelast chapter, wealso discovered that aferromagnetic material has the following interesting property: above acertain temperature it18not strongly 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, 1t becomes nonmagnetic and isnotattracted toward themagnet—even when brought quite close tothemagnet. Ifweletitlienear themagnet while itcools off,atthe instant itstemperature falls below thecritical temperature itissuddenly attracted again bythemagnet! ‘The general theory offerromagnetism that wewill usesupposes that thespin oftheelectron isresponsible forthemagnetization. Theelectron hasspin one-half and carries one Bohr magneton ofmagnetic moment 4=sin=qch/2m. ‘The electron spin can bepointed either “up” or“down.” Because theelectron hasa negative charge, when itsspin is“up” ithasanegative moment, and when itsspin is“down” ithasapositive moment. With our usual conventions, themoment 4. oftheelectron 1sopposite stsspin. Wehave found that theenergy oforientation ofamagnetic dipole inagiven applied field Bis—a-B,buttheenergy ofthe spinning electrons depends ontheneighboring spin alignments aswell, Iniron, ifthemoment ofanearby atom is“up,” there isavery strong tendency that the moment oftheonenext toitwillalso be“up.” That iswhat makes iron, cobalt, and nickel sostrongly magnetic—the moments allwant tobeparallel. The first question wehave todiscuss 18why. Soon after thedevelopment ofquantum mechanics, itwas noticed that there isavery strong apparent force—not amagnetic force orany other kind ofactual force, but only anapparent force—trying toline the spins ofnearby electrons apposite tooneanother. These forces areclosely related tochemical valence forces. There isaprinciple inquantum mechanies—called theexclusion principle—that 4 PasLAT RSet7I LET INNe]ETTTTseNTT 3EE2S17 |TTTTey] Fig.37-1.Thespontaneousmagne.‘EeeEEEfiatin(h~Oettenemoynarecnses =ELEEEE| ‘05afunction oftemperature. [Permission a fromEncyclopaedia Britannica.) ar) magnetization M—but only ontheaverage. Aparticular atom somewhere mightfindaifitsneighbors “up.”Thenitsenergywillbelargerthantheaverage.Another‘onemight find some upandsome down, pethaps averaging tozero, anditwould have noenergy from that term, andsoon, What weought todoistousesome more complicated kind ofaverage, because theatoms indifferent places have different environments, and the numbers upand down are different fordifferent ones. Instead ofjust taking one atom subjected totheaverage influence, weshould takeeachoneinstsactualsituation, computeitsenergy,andfindtheaverage vf energy. Buthow dowefind outhow many are“up” and how many are“down” intheneighborhood? Thatis,ofcourse, justwhatwearetrying tocalculate— xthenumber “up” and “down”—so wehave avery complicated interconnected T problem ofcorrelations, aproblem which hasnever been solved. Itisanintriguing and exciting one which has existed foryears and onwhich some ofthegreatest names inphysics have written papers, buteven they have notcompletely solved it. Ttturnsoutthatatlowtemperatures, whenalmostalltheatomic magnets are (0“up” andonly afeware“down,” itiseasy tosolve; andathigh temperatures, far ey above theCurie temperature T,when theyarealmost allrandom, 1tisagain easy. Itis often easy tocalculate small departures from some simple, idealized situation, 50itisfairly well understood why there aredeviations from thesimple theory at low temperature, It1salso understood physically that forstatistical reasons the magnetization should deviate athigh temperatures, But theexact behavior near the Curie point has never been thoroughly figured out. That's aninteresting + 77 problem toworkoutsomedayifyouwantaproblem thathasnever beensolved. o = 37-2 Thermodynamic properties wy Inthelastchapter welatdthegroundwork necessary forcalculating the I thermodynamic properties offerromagnetic materials. These are, naturally, related fi totheinternal energyofthecrystal, whichincludes interactions ofthevarious| spins, given byEq.(37.3). Fortheenergy ofthespontaneous magnetization below theCurie point, wecansetH=0inEq.(37.3), and—noticing that tanhx= } ‘M/Mya.—we findamean energy proportional toM?: + 2 ©* " (Wy=—ee 75) Fig.37-2. Theenergy perunitvol- Ifwenow plottheenergy duetothemagnetism asafunction oftemperature, we lumeandspecific heatofoferromagnetic getacurve which isthenegative ofthesquare ofthecurve ofFig.37-1, asdrawncrystal. inFig.37-2(a). Ifweweretomeasure thenthespecificheatofsuchamaterial wewould obtain acurve which isthederivative of37-2(a). Itisshown inFig. a4 37-2(b). Itrises stowly with increasing temperature, butfalls suddenly tozero at T=T..The sharp drop isdue tothechange inslope ofthemagnetic energy and isreached right attheCurie point. Sowithout any magnetic measurements at allwecould have discovered that something was going oninside ofiron ornickel bymeasuring this thermodynamic property. However, both experiment and improved theory (with fluctuations included) suggest that this simple curve 1s wrong and that the true situation isreally more complicated. The curve goes higher atthepeak and falls tozero somewhat slowly. Even ifthetemperature is high enough torandomize thespins ontheaverage, there arestilllocal regions where there isacertain amount ofpolartzation, and inthese regions thespins stillhavealittleextraenergyofinteraction—which only dies outslowly asthings get moreandmorerandom withfurtherincreases intemperature Sotheactualcurve aelooks like Fig. 37-2(c). One ofthechallenges oftheoretical physics today isto ] find anexact theoretical description ofthecharacter ofthespecific heat near the Curie transition—an intriguing problem which hasnotyetbeen solved. Naturally, ~ thisproblem isveryclosely related totheshapeofthemagnetization curveinthe vig : same region.; i — Nowwewanttodescribesomeexperiments, otherthanthermodynamic ones, {|eugge”|| which show that there issomething right about ourinterpretation ofmagnetism 4Whenthematerialismagnetized tosaturation atlowenoughtemperatures, MisI{ |IverynearlyequaltoM..c—nearlyallthespinsareparallel,aswellastheirmag-abee4.-f--4netic moments. Wecancheck thisbyanexperiment. Suppose wesuspend abar “ ~ magnet byathinfiber andthen surround itbyacoilsothatwecanreverse the _ ee Magnetic field without touching themagnet orputting anytorque onst.This 1sa — ‘very difficult experiment because themagnetic forces aresoenormous thatany Fig,37-3. When themagnetization irregularities, any lopsidedness, orany lack ofperfection intheiron will produce ofabar ofiron isreversed, thebar is accidental torques. However, theexperiment has been done under careful con- given some angular velocity. ditions inwhich such accidental torques areminimized. Bymeans ofthemagnetic field from acoil that surrounds thebar, weturn alltheatomic magnets over at ‘once. When wedothis wealso change theangular momenta ofallthespins from “up”to“down”(seeFig.37-3).Ifangularmomentum istobeconserved whenthespins allturn over, therestofthebarmust have anopposite change inangular momentum, The whole magnet will start tospin. And sure enough, when wedo theexperiment, wefind aslight turning ofthe magnet. We can measure the total angular momentum given tothewhole magnet, and this issimply Ntimes A, thechange intheangular momentum ofeach spin. The ratio ofangular momentum tomagnetic moment measured this way comes outtowithin about 10percent of what wecalculate. Actually, ourcalculations assume thattheatomic magnets are Br 205 duepurely totheelectron spin, butthere is,inaddition, some orbital motion alsoin ~ 7 most materials. The orbital motion isnotcompletely free ofthelattice and does notcontribute much more thanafewpercent tothemagnetism. Asamatter of a) fact, thesaturation magnetic field that onegets taking My. =Nuandusing the : density of1ronof7.9andthemoment xofthespinning electron 1sabout 20,000 Fi wy ae gauss. Butaccording toexperiment, itisactually intheneighborhood of21,500 we gauss. This isatypical magnitude oferror—S or10percent—due toneglecting, « « ° thecontributions oftheorbital moments that have not been included inmaking = theanalysis. Thus, ashght discrepancy with thegyromagnetic measurements 1s yquiteunderstandable, !i' 37-3 Thehysteresis curve ° ° Wehave concluded from ourtheoretical analysis that aferromagnetic material _Fig. 37-4. Theformation ofdomains should spontaneously become magnetized belowacertain temperature sothat inasinglecrystalofiron.[FromCharlesallthemagnetism would beinthesame direction. Butweknow thatthisisnottrue Kittel, introduction foSolid State Physics, foranordinary pieceofuxmagnetized iron.Whyisn’tallironmagnetized? We—40hnWileyandSons,Inc.,NewYork,2ndcanexplain itwiththehelpofFig.37-4. Suppose theironwereallabigsingle °4»1956.) crystal oftheshape shown inFig.37-4(a) andspontaneously magnetized allinone direction. Then there would beaconsiderable external magnetic field, which would havealotofenergy.Wecanreducethatfieldenergyifwearrangethatonesideof ars large domain which theexternal field helps tokeep lined up. Inastrong field the crystal “likes” tobeallonewayjust because itsenergy intheapplied field isreduced ~it isnolonger merely thecrystal’s own external field which matters.Whatifthegeometry isnotsosimple? Whatiftheaxesofthecrystalandits insomeotherdirection—say at45°?Wemightthinkthatdomainswouldreform | themselves with their magnetization parallel tothefield, andthen asbefore, they HM could allgrow into onedomain. Butthisisnoteasy fortheiron todo,forthe energy needed tomagnetize acrystal depends onthedirection ofmagnetization relative (0thecrystal axis. Itisrelatively easy tomagnetize iron inadirection parallel tothecrystal axes, butittakes more energy tomagnetize itinsome other direction—hike 45°with respect toone oftheaxes. Therefore, ifweapply amag- Mneticfieldinsuchadirection,whathappensfirstisthatthedomainswhichpoint izalong oneofthepreferred directions which isnear totheapplied field grow until themagnetization 1sallalong one ofthese directions. Then with much stronger fields, themagnetization isgradually pulled around parallel tothefield, assketched inFig, 37-5. InFig. 37-6 areshown some observations ofthemagnetization curves of singlecrystals ofiron.Tounderstand them,wemustfirstexplain something about M.4 thenotation that isused indescribing directions inacrystal. There are many ways inwhich acrystal can besliced soastoproduce aface which isaplane of atoms. Everyone who hasdriven past anorchard orvineyard knows this—it is fascinating towatch. Ifyou look one way, you seelines oftrees—if you look an- otherway,youseedifferent linesoftrees,andsoon.Inasimilar way,acrystal fig.37-5, Amagnetizing feldHof hasdefinite families ofplanes thathold many atoms, andtheplanes have this gngngie withrespect tothecrystol Oxis important characteristic (weconsider acubic crystal tomake iteasier): IfWe willgraduelly chonge thedirection ofthe observe where theplanes intersect thethree coordinate axes—we find that the magnetization without changing itsmagni- reciprocals ofthethree distances from theorigin areintheratio ofsimple whole tude. numbers. ‘These three whole numbers aretaken asthedefinition oftheplanes. For example, inFig. 37-7(a), aplane parallel totheyz-plane isshown. This iscalleda[100]plane;thereciprocals ofits intersection ofthey-and z-axes areboth zeto. Thedirection perpendicular tosuch aplane (inacubic crystal) isgiven the same setofnumbers. Itiseasy tounderstand theidea inacubic crystal, forthen theindices [100] mean avector which hasaunit component inthex-direction and noneinthey-orz-directions. The[110]direction isinadirection 45°fromthe x-andy-axes, asinFig. 37-7(b); and the[111]direction isinthedirection ofthe ‘cube diagonal, asinFig. 37-7(c). 1409 SefA sea aaeueanuce*paSnESEEEEE “7 allel toH,fordifferent directions ofH 200} (withrespect tothecrystal axes). [From Eerc] rie. teentaleentron aase = McGraw-Hill Book Co,Inc,,1937.) Returning now toFig. 37-6, weseethemagnetization curves ofasingle crystal ofiron forvarious directions. First, note that forvery tinyfields—so weak that itishard toseethem onthescale atall—the magnetization increases extremely rapidly toquite large values. Ifthefield isinthe[100] direction—namely along ‘oneofthose nice, easy directions ofmagnetization—the curve goes uptoahigh value, curves around alittle, and then issaturated. What happened isthat the , f —/} Ss =YA(100) Ww Fig. 37-7, The way thecrystal planes ore labeled. domains which were already there arevery easily removed. Only asmall eld it required tomake thedomain walls move and eatupallofthe“wrong-way” domains. Single crystals of1ron are enormously 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. You might study thatsome day. Wedounderstand whyit isflatforhigh fields. When thewhole block isasingle domain, theextra magnetic field cannot make any more magnetization—it isalready atMeas, With alltheelec- trons fines up. Now, ifwetrytodothesame thing inthe[110] direction—which isat45° tothecrystal axes—what willhappen? Weturnonalittlebitoffieldandthe magnetization leaps upasthedomains grow. Then asweincrease thefield some more, wefind that ittakes quite alotoffield togetuptosaturation, becausenowthemagnetization isturningawayfroman“easy”direction. Ifthis explanation iscorrect, thepoint atwhich the[110] 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 fortwo other materials, nickel and cobalt. Nickel isdifferent from iron. Innickel, itturns outthat the[111] direction istheeasy direction ofmagnetization. Cobalt hasahexagonal crystal form, andpeople have botched upthesystem ofnomenclature forthiscase. They want tohave three axes onthebottom ofthehexagon and one perpendicular to these, sothey have used four indices. The [0001] direction isthedirection ofthe axis ofthehexagon, and [1010] isperpendicular tothat axis. 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- line axes pointing every which way. These arenotthesame asdomains, Remember that thedomains were allpart ofasingle crystal, butinapiece ofiron there are “Ge oo feoFe ||ST alaSA] she "AEaafa.27-8.mogntiatn cmstr||_|al elt1 [FromCharles Kittel,Introduction toSolid fel ||%errd KLLE] StatePhysics, JohnWiley andSons,Inc., a ee New York, 2nded.,1956.) H(gauss) —= a8 manydifferent crystals withaxesatdifferent orientations, asshown inFig.37-9. (HAN byeWithin eachofthesecrystals, therewillalsogenerally besomedomains. When 7 t —veapplyasmallmagneticfieldtoapieceofpolyerystalinematerial,whathappens4ct=1: =isthat thedomain walls begin tomove, andthedomains which have afavorable ~sdirectionofeasymagnetization growlarger.Thisgrowthisreversibles0longas<== (ir),— thefieldstays verysmall—if weturnthefieldoff,themagnetization willreturn 0 —2a= — zero.Thispartofthemagnetization [email protected]. =~ =For larger fields—in theregion bofthemagnetization curve shown—things ‘getmuchmorecomplicated. Ineverysmallcrystalofthematerial,therearestrains "4 Sanddislocations; thereareimpurities, dirt,andimperfections. Andatallbutthe Y, x\smallest fields,thedomain wall,inmoving, getsstuckonthese,Thereisaninter- OS 4{14]t{/=yssaction energy between thedomain wall and adislocation, oragrain boundary, oranimpurity. Sowhenthewallgtstooneofthem, 1getsstuck; itstickshere gs39ahicroneopie struct atacertainfield.Butthenifthefieldisraisedsomemore,thewallsuddenlysnaps4=oomagynatizes"ferromagnetic ma. past.Sothemotion ofthedomain wallisnotsmooth thewayit1sinaperfect ferigt, Eacherystal greinneoncoy crystal—it getshung upevery onceinawhile andmoves injerks. Ifwewere{0 Girection ofmagnetization ondisbroken Jookatthemagnetization onamicroscopic scale, wewould seesomething likethe ypintodomains which orespontaneously insert ofFig.37-10. magnetized (usvally) parallel tothis Now theimportant thing isthat these jerks inthemagnetization can cause an direction. energy loss. Inthefirst place, when aboundary finally slips past animpediment, itmoves very quickly tothenext one, since thefield isalready above what would berequired fortheunimpeded motion. The rapid motion means that there are rapidly changing magnetic fields which produce eddy currents inthecrystal. These currents loose energy inheating themetal, Asecond effect isthat when adomain suddenly changes, part ofthecrystal changes itsdimensions from themagneto-striction. Eachsuddenshiftofadomainwallsetsupalitlesoundwavethatcarriesaway energy. Because ofsuch effects, thesecond part ofmagnetization curve isirreversible, andthere isenergy being lost. This 18theorigin ofthehysteresis. ® effect, because tomoveaboundary wallforward—snap—and thentomoveitback- ay ward—snap—produces adifferent result, It’slike “Jerky” friction, and ittakes 7 \ energy. — \ Eventually, forhigh enough fields, when wehave moved allthedomain walls — } andmagnetized eachcrystal initsbestdirection, therearestillsomecrystallites ~%--- ~ 7 which happen tohave their easy directions ofmagnetization notinthedirection x ofour external magnetic field. Then 1ttakes alotofextra field toturn those magnetic moments around, Sothemagnetrzation increases slowly, butsmoothly, Fig. 37-10. Themagnetization curve forhigh fields—namely intheregion marked cinthefigure. Themagnetization forpolycrystalline iron does notcome sharply toitssaturation value, because inthelastpart ofthecurvetheatomicmagnetsareturninginthestrongfield.Soweseewhythemagnetizationcurve ofanordinary polycrystalline materials, such astheoneshown inFig.37-10, rises alittle bitand reversibly atfirst, then rises irreversibly, and then curves over slowly. OFcourse, there isnosharp break-point between thethree regions— they blend smoothly, one into theother. Itsnothardtoshowthatthemagnetization processinthemiddlepartofthe coy BEESae magnetization curve 1sjerky—that thedomain walls jerk andsnap asthey shift .Allyouneed1sacoilofwire—with manythousandsofturns—connected toan (C0)amplifier and aloudspeaker, asshown inFig. 37-11. Ifyou putafewsilicon steel“ —sheets(ofthetypeusedintransformers) atthecenterofthecoilandbringabar siren‘magnet slowly near thestack, thesudden changes inmagnetization will produce a _ impulses ofemfinthecoil,whichareheardasdistinctclicksintheloudspeaker. _ Z Asyou move themagnet nearer totheiron you will hear awhole rush ofclicks awruries that sound something likethenoise ofsand grains falling over each other asa - lvestencanofsandistilted.‘Thedomainwallsarejumping, snapping, andjigglingasthe .field isincreased, This phenomenon iscalled theBarkhausen effect. Fig.37-11. Thesudden changes in‘Asyoumovethemagnetevenclosertothetronsheets,thenoisegrowslouderthemagnetization ofthesteelstripare and louder forawhilebutthenthereisrelativelylittlenoisewhenthemagnetgetsheardosclicksintheloudspeaker. very close. Why? Because nearly allthedomain walls have moved asfarasthey cango. Any greater field 1smerely turning themagnetization ineach domain, which isasmooth process. 319 Ifyou now withdraw themagnet, soastocome back onthedownward branch ofthehysteresis loop, thedomains alltrytogetback tolow energy again, and you hear another rush ofbackward-going jerks. You canalso note that ifyou bring themagnet toagiven place and move itback and forth alittle bit,there isrelatively little noise, Its again lke tilting acan ofsand—once thegrams shift into place, small movements ofthe can don’t disturb them. Inthe iron the small variations imthe magnetic field aren’t enough tomove any boundaries over any ofthe “humps.” 37-4 Ferromagnetic materials Now wewould like totalk about thevarious kinds ofmagnetic materials that there areinthetechnical world and toconsider some oftheproblems involved in designing magnetic materials fordifferent purposes. Furst, theterm “the magnetic properues ofiron,” which one often hears, isamsnomer—there 1snosuch thing “Iron” isnotawell-defined matertal—the properties ofiron depend critically on theamount ofmpurities and also onhow theiron 1sformed. You can appreciate that themagnetic properties willdepend onhow easily thedomain walls move and that this1sagross property, notaproperty oftheindividual atoms. Sopracticalferromagnetism isnotreallyaproperty ofanironatom—it isaproperty ofsold sron inacertain form. For example, iron can take ontwo diferent crystalline forms. The common form hasabody-centered cubic lattice, butitcan also have aface-centered cubic lattice, which is,however, stable only attemperatures above 1100°C. OFcourse, atthat temperature the body-centered cubte structure is, already past theCurie point. However, byalloying chromium and nickel with theiron (one possible mixture 1s18percent chromium and 8percent nickel) we ‘can getwhat 1scalled stainless steel, which, although it1smainly tron, retains the face-centered lattice even atlow temperatures. Because itscrystal structure 1sdifferent, 1hascompletely differentmagnetic properties. MostkindsU=~tainlesssteel arenot magnetic toany appreciable degree, although there aresome kinds which aresomewhat magnetic—it depends onthecomposition ofthealloy. Even when such analloy ismagnetic, it18notferromagnetic like ordinary 1ron—even though 11s mostly just iron. Wewould likenow todescribe afewofthespecial materials which have been developed fortheir particular magnetic properties. First, 1fwewant tomake a permanent magnet, wewould like material with anenormously wide hysteresis a loopsothat,when weturnthecurrent offandcome down tozeromagnetizing(gauss) field,themagnetization willremainlarge.Forsuchmaterials thedomainbounda- niesshouldbe“frozen”inplaceasmuchaspossibleOnesuchmaterialisthere- 18,000markable alloy “Alnico V”(51% Fe,8% Al, 14% Ni,24% Co, 3% Cu). (The 8 rather complex composition ofthisalloy18indicative ofthekindofdetailed effort 10,000 that hasgone into making good magnets. What patience ittakes tomixfivethingstogetherandtestthemuntilyoufindthemostidealsubstance!)WhenAlnico (|5,000 solidifies,there1sa“secondphase”whichprecipitatesout,makingmanytinygrains “Heand very high internal strams, Inthis matertal, thedomain boundaries have a hard time moving atall, Inaddition tohaving aprecise composition, Alnico 1s#04009=40000H mechanically “worked” mawaythatmakesthecrystalsappearintheformof(90uss) Jonggrains along thedirection inwhich themagnetization 1sgoing tobe.Then themagnetization will have anatural tendency tobelined upinthese directions and will beheld there from theanisotropic effects Furthermore, thematerial is even cooled inanexternal magnetic field when 11ismanufactured, sothat thegrains will grow with theright crystal onentation, The hysteresis loop ofAlmeco Vis shown inFig 37-12. You seethat 1¢1sabout 500times wider than thehysteresis curve forsoft iron that weshowed inthelast chapter inFig. 36-8. Fig 37-12, The hysteresis curve of Let’s turn now toadifferentkindofmateral.Forbuildingtransformers and AlnicoV. motors, WeWant amaterial Which 1smagnetically “Soit"—one inwhich themag- nnetiym 18easily changed sothat anenormous amount ofmagnetization results from avery small apphed field, Toarrange this, weneed pure, well-annealed material which will have very fewdislocations and impurities sothat thedomain 37-10 walls can move easily. Itwould also benice sfwecould make the anisotropy small. Then, even ifagrain ofthemateral sitsatthewrong angle with respect to thefield, itwill sull magnetize easily. Now wehave said that iron prefers tomag- netize along the[100] direction, whereas nickel prefers the[111] direction; soaf wemixiron and nickel invarious proportions, wemight hope tofind that with justtheright proportions thealloy wouldn't prefer anydirection—the [100] and [111]directions wouldbeequivalent. Itturnsoutthatthishappenswithamixtureof70percent nickel and30percent iron. Inaddition—possibly byluck ormaybe Table 37-1 because ofsome physical relationship between theanisotropy andtheMagNetO- Properties ofsomeferromagnetic materialsstriction effects—it turns out that themagnetosiriction ofiron and nickel hasthe opposite sign. Andinanalloy ofthetwometals, thisproperty goesthrough zero B Ae atabout 80percent nickel. Sosomewhere between 70and80percent nickel weget Residual Coercive very“soft”magnetic materials—alloys thatareveryeasytomagnetize. ‘Theyare magnet forcecalledthepermalloys. Permalloys areusefulforhigh-quality transformers (atlow Material as (gauss)signal levels), butthey would benogood atallforpermanent magnets. Perm- alloys must bevery carefully made and handled. The magnetic properties ofa Superalloy (~s000) 0004 pieceofpermalloy aredrastically changed if11sstressed beyond itselastic limit—st__ Silicon steel musta’t bebent, ‘Then, 1spermeability 1sreduced because ofthedislocations, shp (transformer) ‘12,000,005 bands, and soon,which areproduced bythemechanical deformations, The Armeo iron 400006 domain boundartes arenolonger easytomove. Thehighpermeability can,how- _Almico V 13,000 550. ever, berestored! byannealing athigh temperatures, It1soften convenient tohave some numbers tocharacterize the various magnetic materials. Two useful numbers aretheintercepts ofthehysteresis loop with theB-and H-axes, asindicated inFig, 37-12. ‘These interceptsarecalledthe remanent magnetic field B,and the coerewe force H,. InTable 37-1 welistthese numbers forafew magnetic materials. (@) (b) Fig.37-13. Relative orientation of '}electronspinsinvariousmaterials:(a) | | | |ferromagnetic, (b)antiferromagnetic, (c) | |fertite,(d)yitrium-iron alloy.—(Broken arrows showdirection oftotalangular )cy momentum, including orbital motion.) 37-5 Extraordinary magnetic materials Wewould now like(odiscuss some ofthemore exotic magnetic materials,Therearemanyelementsintheperiodictablewhichhaveincomplete innerelectronshells and hence have atomic magnetic moments For instance, right next tothe ferromagnetic elements iron, nickel, andcobalt youwillfindchromium andmanga- nese. Why aren't hey ferromagnetic? The answer isthat the\terminEq.(37.1) hastheopposite signforthese elements, Inthechromium lattice, forexample, the spins ofthechromium atoms alternate arom byatom, asshown inFig. 37-13(b) Sochromium 1s“magnetic” from itsown point ofview, butit1snottechnically interesting because there arenoexternal magnetic effects. Chromium, then, 1san example ofamaterialinwhichquantummechanical effectsmakethespinsalter- nate. Such amaterial iscalled annferromagnetic. The alignment inanuferromag-neticmaterials isalsotemperature dependent. Belowacriticaltemperature, allthespins arelined upinthealternating array, butwhen thematerial isheated above acertain temperature—which isagain called theCurie temperature—the spinssuddenlybecomerandom.Thereis,internally, asuddentransition,Thistransition ‘canbeseen inthespecific heat curve. Also 1tshows upinsome special “magnetic” effects. Forinstance, theexistence ofthealternating spins canbeverified byscatter- ngneutrons from acrystal ofchromium. Because @neutron itself hasaspin st (andamagnetic moment), ithasadifferentamplitude tobescattered, depending onwhether itsspin 1sparallel oropposite tothespin ofthescatterer. Thus, wegeta different interference pattern when thespins inacrystal arealternating than we dowhen they have arandom distribution ‘There 1sanother kind ofsubstance inwhich quantum mechanical effects make theelectron spins alternate, but which isnevertheless ferromagnetic—that is,the crystal has apermanent netmagnetization, The idea behind such materials is shown inFig.37-14. Thefigure shows thecrystal structure ofspine/, amagnestum- aluminum oxide, which—as it1sshown—1s normagnetic. The oxide hastwokinds ofmetal atoms: magnesium and aluminum, Now ifwereplace themagnesium e+ andthealuminum bytwomagnetic elements likeironandzine, orbyzineand‘manganese—in otherwords.ifweputinmagneticatomsinsteadofthenonmagnetic ‘ones—an interesting thinghappens. Let’scallonekindofmetalatomaandthe @ «other kind ofmetal atom 5;then thefollowing combination offorces nmetbe eax considered. ‘There 1sanabinteraction which triestomake theaatoms andthe‘atoms have opposite spins—because quantum mechanics always gives theoppo- site sign (except forthemysterious crystals ofiron, nickel, and cobalt). Then, there isadirect a-a interaction which tries tomake the a'sopposite, and also a ‘hobinteraction which tries tomake the6'sopposite. Now, ofcourse wecannot Fig.37-14. Crystal structure ofthe have everything opposite everything else—a opposite b,aopposite a,andhop- mineral spinel (MgAl;0,); theMg"? ions powite bPresumably because ofthedistances between theu'sandthepresence of‘occupytetrohedral sites,eachsurrounded theoxygen(although wereallydon’tknowwhy),1turnsoutthatthea-binter-byfouroxygen ions;theAl*? ionsoccupy action isstronger than thea-aortheh-b, Sothesolution thatnature usesinthis octahedral sites, each surrounded bysix case1stomake allthea'sparallel 10eachother. andalltheb'sparallel 10euchother, oxygen tons.[From Charles Kittel, Intro- butthetwosystems opposire Thatgivesthelowest energy because ofthestronger duction toSolidStatePhysics, JohnWiley 4.interaction, Theresult: allthea'sarespinning upandallthe6’sarespinning ‘ondSons,nc,NewYork,2nded.19561 down—or viceversa, ofcourse. Butifthemagnetic moments ofthea-type atom andtheb-type atom arenorequal, wecangetthesituation shown inFig. 37-13), and there can beanetmagnetization inthematerial. The material will then be ferromagnetic—although somewhat weak Such materials are called ferries. They donothave ashigh asaturation magnetization asiron—for obvious reasons —so they areonly useful forsmaller fields. But they haveaveryimportantdiffer- ence—they are insulators: the ferrites are ferromagnetic insulators. Inhigh- frequency fields, they will have very small eddy currents and socan beused, for example, inmicrowave systems. The microwave fields will beable togetinside such aninsulating material, whereas they would bekept out bytheeddy currents inaconductor like iron. There 18another class ofmagnetic matertals which has only recently been discovered—memibers ofthefamily oftheorthosilicates called garnets. They are again crystals inwhich thelaitice contains two kinds ofmetallic atoms, and we have again asituation inwhich two kinds ofatoms can besubstituted almost at will. Among themany compounds ofinterest there 1sone which 1scompletely ferromagnetic, Ithasyttrium and iron inthegarnet structure, and thereason 1 ferromagnetic 1$very curious. Here again quantum mechanics 18making the neighboring spins opposite, sothat there 1salocked-in system ofspins with the electronspinsoftheirononewayandtheelectronspinsoftheyttriumtheoppositeway But theyttrium atom 1scomplicated. It1sarare-earth element and gets a large contribution toitsmagnetic moment from orbital motion oftheelectrons. For yttrium, theorbital motion contribution isopposite that ofthespin and also 1sbigger. Thus, although quantum mechamies, working through theexclusion principle, makes thespiny oftheyttrium opposite those oftheiron, 1tmakes the total magnetic moment oftheyttium atom parallel totheiron because ofthe orbital effect—as sketched inFig. 37-13(d) ‘The compound 1stherefore «regular ferromagnet Another interesting example offerromagnetism occurs insome oftherare- earth elements. Ithas todowith astill more peculiar arrangement ofthespins. The matertal isnotferromagnetic inthesense that thespins areallparallel. noris, Atantiferromagnet inthesense that every atom 1sopposite. Inthese crystals all ofthespins 1onelayer areparallel and lieintheplane ofthelayer. Inthenext ara ae Elasticity 38-1 Hooke’s law The subject ofelasticity deals with thebehavior ofthose substances which 38-1. Hooke’s law have theproperty ofrecovering their size and shape when theforces producing deformations areremoved. Wefindthiselastic property tosomeextent inall 38-2Uniform strains solid bodies. Ifwehad thetime todeal with thesubject atlength, wewould want 38-3 The torsion bar; shear waves tolook into many things: thebehavior ofmaterials, the general laws ofelasticity,thegeneraltheoryofelasticity, theatomicmachinery thatdetermine theela 38-4Thebentbeam properties, and finally thelimitations ofelastic laws when theforces become so 38-5 Buckling great that plastic low and fracture occur. Itwould take more time than wehave tocover allthese subjects 1ndetail, sowewill have toleave out some things For example, wewill notdiscuss plasticity orthelimutations oftheelastic laws. (We touched onthese subjects briefly when wewere talking about dislocations im Reviews Chapter 47,Vol. 1,Sound: metals.) Also, wewillnotbeable todiscuss theinternal mechanisms ofelastcity— theWave Equation. 80our treatment will not have thecompleteness wehave tried toachieve inthe earlier chapters, Our aimismainly togwve you anacquaintance with some ofthe ways ofdealing with such practical problems asthebending ofbeams. When you push onapiece ofmaterial, it“grves"—the material isdeformed. Iftheforce issmall enough, therelative displacements ofthevarious points inthe material areproportional totheforce—we say thebehavior 1selastic. We wall discuss only theelastic behavior. First, wewillwrite down thefundamental laws ofelasticity, and then wewillapply them toanumber ofdifferent situations Suppose wetake arectangular block ofmaterial oflength /,width w,and height/,asshown inFig.38-1. IfwepullontheendswithaforceF,thenthe eolength increases byanamount A/. Wewill suppose inallcases that thechange 19 4 length1sasmallfraction oftheoriginal length. Asamatteroffact,formaterials |“|likewoodandsteel,thematerial willbreakifthechangeinlength1smorethan3 noefewpercentoftheoriginallength.Foralargenumberofmaterials,experiments Caan opt showthatforsufficrently smallextensions theforceisproportional totheextension : hbnean| ECF.4H |Heol Feal (8.1) ialaiaalait iaisiaah (teaaaat This relation 1sknown asHooke's law. ‘The lengthening4/ofthebarwillalsodependonitslength.Wecanfigureout‘Fig.38-1.Thestretchingof«bar howbythefollowing argument. Ifwecement twoidentical blocks together, end under uniform tension.toend,thesameforcesactoneachblock,eachwillstretchbyA/.Thus,thestretchofablock oflength 2/would betwice asbigasablock ofthesame cross section,butoflengthJ.Inordertogetanumbermorecharacteristic ofthematerial,and lessofanyparticular shape, wechoose todeal with theratio A//I oftheextension totheoriginal length. This ratio isproportional totheforce butindependent of/: pall (382) The force Fwill also depend onthearea oftheblock. Suppose that weput two blocks side byside. Then foragiven stretch a/wewould have theforce F ‘oneach block, ortwice asmuch onthecombination ofthetwo blocks. The force, foragiven amount ofstretch, must beproportional tothecross-sectional area A oftheblock. Toobtain alaw inwhich thecoefficient ofproportionality 1sinde- pendent ofthedimensions ofthebody, wewrite Hooke's lawforarectangular 3841 block inthe form revat (83) The constant ¥isaproperty only ofthenature ofthematerial; 1isknown as Young’s modulus. (Usually you will seeYoung’s modulus called E. Butwe've used Eforelectric fields, energy, and emt’s, soweprefer touseadifferent letter) Theforce perunitarea iscalled thestress, andthestretch perunit length—the ‘fractional stretch—is called thestrain. Equation (38.3) can therefore berewritten inthefollowing way: F al fox d. (84) Stress =(Young's modulus) x(Strain). There isanother part toHooke’s law: When you siretch ablock ofmaterial ° inonedirection itcontracts atright angles tothestretch. Thecontraction in widthisproportional tothewidthwandalsotoA//l.Thesidewayscontraction is a*Tr TrS- 1nthesameproportion forbothwidthandheight,and1susuallywrittenaaa E, wehegM, G35) P where theconstant ¢isanotherproperty ofthematerial calledPoisson's ratio.Itis Fig. 38-2. Abar under uniform always positive insignand1sanumber lessthan 1/2. (It1s“reasonable” that¢ hydrostatic pressure. should begenerally positive, butitisnotquite clear that 1tmust beso.) Thetwoconstants ¥andospecifycompletely theelasticproperties ofaho-‘mogeneous’ isotropic (that 1S,noncrystalline) material. Incrystalline materials the stretches and contractions can bedifferent indifferent directions, sothere can be many more elastic constants, Wewill restrict ourdiscussion temporarily tohomo- geneous’ isotropic materials whose properties canbedescribed byYanda. Asusual there aredifferent ways ofdescribing things—some people like todescribe the elastic properties ofmaterials bydifferent constants. Italways takes two, and they can berelated too and ¥. The last general law weneed istheprinciple ofsuperposition, ‘Since thetwo Jaws (384)and(38.5) arelinear intheforces andinthedisplacements, superposition F,willwork. Ifyouhave onesetofforces andgetsome displacements, andthen you add anew setofforces and getsome additional displacements, theresulting displacements willbethesum oftheones youwould getwith thetwosetsofforces actingindependently. FeNow wehave allthegeneral principles—the superposition principle andEqs. (38.4) and (38.5)—and that’s allthere istoelasticity. Butthat islikesaying that once you have Newton’s laws that’s allthere1stomechanics.Or,givenMaxwell's equations, that’s allthere istoelectricity. Itis,ofcourse, true that with these principles you have agreat deal, because with your present mathematical ability‘youcouldgoalongway.Wewill,however,workoutafewspecialapplications. Fe 38-2Uniform strains a Asourfirstexample let'sfindoutwhathappens toarectangular blockunderuniform hydrostatic pressure Let's putablock under water inapressure tank, Then there willbeaforce acting inward onevery face oftheblock proportional (force perunit area) oneach face oftheblock isthesame. Wewillwork outfirst Fig38-3.Hydrostatic pressureigtheChangeinthelength.‘Thechange1lengthoftheblockcanbethoughtofasthesuperposition ofthree longitudinal thesumofchanges inlength thatwould occur inthethree independent problemscompressions whicharesketchedinFig.38-3.382 Problem 1.Ifwepush ontheends oftheblock with apressure p,thecom- pressional strain isp/Y, and itisnegative, oho e, 77 ¥ Problem 2.Ifwepush onthetwosides oftheblock with pressure p,thecom- presstonal strain isagain p/Y, butnow wewant thelengthwise strain, Wecan get that from thesideways strain multiplied by—c. The sideways strain is Awe wo C¥? so Mb _ GPT>4ty Problem 3.Ifwepush onthetop oftheblock, thecompressional strain is ‘once more p/Y, and thecorresponding strain inthesideways direction isagain —op/Y. Weget 4s_4gB Ba tod. Combining the results ofthe three problems—that is,taking Al=Aly + Ala+Als—we get . a 7Vo2a—2), (38.6) The problem is,ofcourse, symmetrical inallthree directions; itfollows that Aw_ah Pp an a 3 EE S2S #(I~20). (38.7) Thechange inthevolume under hydrostatic pressure isalsoofsome interest. fs] Since V=Iwh, wecanwrite, forsmall displacements, ' AV_Al,Aw,dh :VO Tt wt ; A Using(38.6)and(38.7),wehi - ae sing(38.6)and(38.7),wehave = _—— WP ap=34 -20, (388) Fig38-4Acubeinuniformshear People like tocall AV/V thevolume strain and write av F p=Ky, ‘The volume stress pisproportional tothevolume strain—Hooke’s law once more. [oie alate Rate Ba The coefticient K1scalled thebulk modulus; 1isrelated totheother constants by =? F F K= am 8.9) Since Kisofsome practical interest, many handbooks give Yand Kinstead of¥ anda. Ifyou want¢youcanalwaysgetitfromEq.(38.9).Wecanalsoseefrom Eq.(38.9)thatPoisson’s ratio,o,mustbelessthanone-half. Ifitwerenot,the L — bulk modulus Kwould benegative, andthematerial would expand under increas- sn 1ngpressure. Thatwould allowustogetmechanical energy outofanyoldblock— Fitwouldmeanthattheblockwasinunstableequilibrium. Ifitstartedtoexpanditwould continue byitself witharelease ofenergy. Fig.38-5. Acube withcompressing Now wewant toconsider what happens when you puta“shear” strain on forces ontop and bottom and equalsomething. Byshearstrainwemeanthekindofdistortion showninFig.38-4.Asastretchingforcesontwosides. preliminary tothis, letuslook atthestrains inacube ofmaterial subjected tothe forces shown inFig.38-5. Again wecanbreak itupito twoproblems: thevertical 383 pushes, and thehorizontal pulls. Calling 4thearea ofthecube face, wehave for thechange inhorizontal length ALUF UF itortayateypan ee: (38.10) ‘The change inthevertical height isjust thenegative ofthis. FG 6 actor 6 A! 8AA OER ) areasEA = m / ai =)SH vEG VEGCOTTON areaa 6 Fig. 38-6. The two pairs ofshear forces in(a)produce the same stress as thecompressing and stretching forces of(b). Now suppose wehave thesame cube and subject ittotheshearing forces shown inFig. 38-6(a). Note that alltheforces have tobeequal ifthere aretobe nonettorques and thecube istobeinequilibrium, (Similar forces must also exist inFig. 38-4, since theblock isinequilibrium. They areprovided throughthe“glue”thatholdstheblocktothetable.)Thecubeisthensaidtobeinastateofpureshear.Butnotethatifwecutthecubebyaplaneat45°—sayalongthediagonal inthefigure—the total force acting across theplane isnormal toplane andisequal to2G. Thearea over which thisforce actsis\/24; therefore,the tensile stress normal tothis plane 1ssimply G/A. Similarly, ifweexamine aplane atanangle of45°theother way—the diagonal Binthefigure—we seethat there isacompressional stress normal tothis plane of—G/A. From this, weseethat thestress ina“pure shear” isequivalent toacombination oftension and com- pression stresses ofequal strength and atright angles toeach other, and at45°to theoriginal faces ofthecube The internal stresses and strains arethesame as ‘wewould find inthelarger block ofmaterial with theforces shown inFig. 38-6(b).Butthistheproblemwehavealreadysolved.Thechangeinlengthofthe diagonal 1sgiven byEq.(38.10), AD_l+0G apitee 8.11) aD ‘ 6 (One diagonal isshortened; theother iselongated.) Itisoften convenient toexpress ashear strain interms oftheangle bywhich pk thecubeistwisted—the [email protected] ofthefigureyoufl i canseethatthehorizontal shift6ofthetopedgeisequaltoy/2AD.Soi '| 6_vid_4Di =$2v2ad_ “0.t| o-5 p= 245 (38:12) 1 '|1Theshearstressgisdefinedasthetangentialforceononefacedividedbythe =—t 1 area,g=G/A.UsingEq.(38.11)in(38.12),weget _jltea=21ty, Fig.38-7. The shear strain @is Or,writing thisintheform “stress =constant times strain,” 2.0/0. B= wo 38:13) 384 The proportionality coefficient iscalled theshear modulus (or, sometimes, the coefficient ofrigidity). Itisgiven interms of¥and oby y “=ats 8.14) Incidentally, theshear modulus must bepositive—otherwise you could getwork‘outofaself-shearing block.FromEq.(38.14),omustbegreaterthan—1.We know, then, that¢mustbebetween—Iand+4;inpractice,however,itisalways sreater than zero. Asalastexampleofthetypeofsituation where thestresses areuniform through thematerial, let’sconsidertheproblemofablockwhichisstretched, whileitisatthesame time constrained sothat nolateral contraction cantake place. (Tech- nically, it’salittle easier tocompress itwhile keeping thesides from bulging out— 5 butit’sthesameproblem.) What happens? Well,theremustbesideways forces tf which keep ttfrom changing itsthickness—forces wedon’t know off-hand but ———— st willhavetocalculate. It’sthesamekindofproblem wehavealready done,only to og |withalittledifferent algebra, Weimagine forcesonallthreesides,asshown inf ~~]! ° ip Fig.38-8; wecalculate thechanges indimensions, andwechoose thetransverse ! j forces tomake thewidth and height remain constant. Following theusual argu- ~ ments, wegetforthethree strains: 5 Me_1FeofoF,[Fe Fy4Fe i iEVR -YROVE re-o(&+))-8.15)Fig,38-8.Shetchingwitoutlateral aly[Fy_(FeyFe 1Y[fe“(i+ap 810) al, _U[Fe Fey Fyieri-o(&+%)]- (38.17) Now since Al,andAl;aresupposed tobezero, Eqs. (38.16) and (38.17) gwe twoequations relating F,andF,toF,.Solving them together, wegetthat Fy Feo beA, a,10 Ay” CRIS) Substituting in(38.15), wehave Ale 1 2? \F,_1(L-020°) oY(:ot3)A.Y&)e C819) Often, you will seethis turned around, and with thequadratic ino factored out, t 1sthenwritten FE l-o ala>(Fo =2)” (3820) ‘When weconstrain thesides, Young's modulus gets multiplied byacomplicated function of¢.AsyoucanmosteasilyseefromEq.(38.19),thefactorinfrontof Yisalways greater than 1.It1sharder tostretch theblock when thesides are held—which also means that ablock isstronger when thesides areheld than when they arenot. 38-3 The torsion bar; shear waves Let’s now turn ourattention toanexample which ismore complicated because different parts ofthematerial arestressed bydifferent amounts. Weconsider a twisted rod such asyou would find inadrive shaft ofsome machinery, orina quartz fiber suspension used inadelicate instrument. Asyou probably know from experiments with thetorsion pendulum, theforque onatwisted rodisproportional totheangle—the constant ofproportionality obviously depending upon the length oftherod, ontheradius oftherod, and ontheproperties ofthematerial. Thequestion 1s:Inwhat way? Wearenow inaposition toanswer thisquestion; it’sjust amatter ofworking outsome geometry. 38-5 (a; iZa7 }_____, —___* JJ Po ()Cast 0 t ne“Y oF OF Bw tt4 | Fig. 38-9. (a)Acylindrical bar intorsion. (b)Acylindrical shell intorsion. (c)Each small piece oftheshell isinshear. Fig, 38-9(a) shows acylindrical rodoflength Z,andradius a,with oneend twisted bytheangle ¢with respect totheother. Ifwewant torelate thestrains to what wealready know,wecanthinkoftherodasbeingmadeupofmanycylindrical shells and work outseparately what happens toeach shell. Westart bylooking at athin, short cylinder ofradius r(less than a)and thickness Ar—as drawn inFig. 38-9(b). Now ifwelook atapiece ofthis cylinder that was originally asmall square, weseethat ithasbeen distorted into aparallelogram. Each such element ofthecylinder 1sinshear, and theshear angle @is -%a= L The shear stress ginthematerial is,therefore [from Eq.(38.13), 18 gown? (38.21) ‘The shear stress isthetangential force AFontheend ofthesquare divided bythearea AlAroftheend [see Fig. 38-9(c)] _AF.8~War The force AFontheendofsuch asquare contributes atorque Araround theaxis oftherodequal toAr=rAF=rgAlar, (38.22) ‘Thetotaltorque+isthesumofsuchtorquesaroundacomplete circumference of thecylinder. Soputting together enough pieces sothat theAl's add upto2zr, wefind that thetotal torque, forahollow tube, 1s rg(2mr) ar. (38.23) Or,using (38.21), : 7=Dep2. 8.24) Wegetthat therotational stiffness, 7/6, ofahollow tube isproportional tothe cube oftheradius rand tothe thickness Ar,and inversely proportional tothe length Z. ‘Wecannowimagineasolidrodtobemadeupofaseriesofconcentrictubes, eachtwisted bythesameangle¢(although theinternal stresses are different for each tube). The total torque isthesum ofthetorques required torotate each shell; forthe solid rod . ws wheretheintegralgoesfromr~0tor=a,theradiusofthe rod. Integrating, we have rane (38.25) Forarodintorsion, thetorque isproportional totheangle andisproportional to thefourth power ofthediameter—a rod twice asthick issixteen times asstiff for torsion. Before leaving thesubject oftorsion, letusapply what wehave just learned toaninteresting problem: torsional waves. Ifyou take along rod and suddenly twist one end, awave oftwist works itway along therod, assketched inFig.38-10(a). That'salittlemoreexcitingthanasteadytwist—Iet’s seewhetherwecan work outwhat happens. we) —--3 st + nr+a) ° oS— eno |Senve jj___ ,—_. z reaz Fig. 38-10. {a}Atorsional wave onarod. (b)Avolume element oftherod. Letzbethedistance tosome point down therod. For astatic torsion the torque isthesame everywhere along therod, and isproportional to¢/L, thetotal torsion angle over thetotal length. What matters tothematerial isthe local torsional strain, which is,you will appreciate, 84/22. When thetorsion along the rod isnotuniform, weshould replace Eq. (38.25) by =pT, 1)=4 SB (38.26) Now let’s look atwhat happens toanelement oflength Azshown magnified in Fig,38-10(b). Thereisatorque7(z)atend|ofthe little hunk ofrod, and adiffer- enttorque r(z+Az) atend 2.IfAzissmall enough, wecan useaTaylor ex- pansion and write an az+a2)=102)+(2)az. (38.27) The nettorque Aracting onthelittle piece ofrodbetween zand z+Azisclearlythedifference between7(2)and7(z+42),orAr=(61/82)Az.Differ-entiating Eq. (38.26), weget _ratae ar=AYFEas. (38.28) The effect ofthis nettorque istogive anangular acceleration tothelittle slice ofthe rod. The mass ofthe slice is AM =(na? 2)p, where pisthedensity ofthematerial, Weworked outinChapter 19,Vol. I,that,themomentofinertiaofacircularcylinderismr?/2;callingthemomentofinertia ofour piece A/,wehave Al=Fpataz. (38.29) Newton’s law says thetorque isequal tothemoment ofinertia umes theangular acceleration, or =a 28.ar= alFe (38.30) 387 38-4 The bent beam Wewant now tolook atanother practical matter—the bending ofarodora beam. What aretheforces when webend abarofsome arbitrary cross section? Wewill work itoutthinking ofabarwith acircular cross section, butouranswer will begood forany shape. Tosave time, however, wewill cutsome corners, so Leourtheorywewillworkoutisonlyapproximate. Ourresultswillbecorrectonly———— —whentheradiusofthebendismuchlargerthanthethickness ofthebeam. <Suppose you grab thetwo ends ofastraight barand bend itinto some curve / liketheoneshowninFig.38-11.Whatgoesoninsidethebar?Well,ifitiscurved, y,that means that thematerial ontheinside ofthecurve iscompressed andthema- / terial ontheoutside isstretched. ‘There issome surface which goes along more or /lessparalleltotheaxisofthebarthatisneither stretched norcompressed. Thisis /calledtheneutralsurface.Youwouldexpectthissurfacetobenearthe“middle” /ofthe cross section. Itcan beshown (but wewon't doithere) that, for smallbending ofsimplebeams,theneutralsurfacegoesthrough the“centerofgravity” / ofthecross section. This istrue only for“pure” bending—if you arenotstretching / orcompressing thebeam atthesame time. Forpure bending, then, athin transverse slice ofthebarisdistorted asshown Fig. 38-11. Abent beam inFig. 38-12(a). The material below the neutral surface has acompressional strain which isproportional tothedistance from theneutral surface; and thematerial above isstretched, also inproportion toitsdistance from theneutral surface. So thelongitudinal stretch AJisproportional totheheight y.The constant ofpro- portionality isjust /over theradius ofcurvature ofthebar—see Fig. 38-12: ay. TR iP Sotheforceperunitarea—thestress—inasmallstripatyisalsoproportional to \Caml thedistancefromtheneutralsurface _Fa _—aF_ yy ——oR (38.34) = SJE Nowlet’slookattheforces thatwould produce suchastrain, Theforces 1 I acting onthelittlesegment drawn inFig.38-12 areshown inthefigure. Ifwe Rf\weurrac.thinkofanytransverse cut,theforcesactingacrossitareonewayabovethe ||surraceneutral surface and theother way below. They come inpairs tomake a“bending moment” st—by whichwemeanthetorque abouttheneutral line.Wecancom- T pute thetotal moment byintegrating theforce times thedistance from theneutral ) surface foroneofthefacesofthesegment ofFig.38-12: ay m=fydF. (88.35)eae FromEq.(38.34), dF=Yy/RdA, so yeuTeale mazfyaa. () Fig. 38-12. (a)Small segment of Theintegral ofy?daiswhat wecancallthe“moment ofinertia” ofthegeometric bentbeam. (b)Cross section ofthebeam. cross section about ahorizontal axis through its“center ofmass”;*wewillcall ith: YI a=2 (38.36) I=[>dA. (38.37) *Ts, ofcourse, really themoment ofinertia ofaslice with unit mass perunt area. 389 Equation (38.36), then, gives ustherelation between thebending moment 3 and thecurvature 1/R ofthebeam, The “stiffness” ofthebeam isproportional to¥and tothemoment ofmertia Inother words, sfyou want thestifest possible beam with @given amount of,say, aluminum, you want toputasmuchofitaspossibleasfarasyoucanfromtheneutralsurface,tomakealargemomentofinertia, You can't carry this toanextreme, however, because then thething will notcurve aswehave supposed—t will buckle ortwist and become weaker again. But now you seewhy structural beams aremade intheform ofanIoran H—as shown inFig. 38-13. Fig. 38-13. An“I”beam. Asanexample oftheuseofour beam equation (38.36), let’s work outthe deflection ofacantilevered beam with aconcentrated force Wacting atthefree end, assketched inFig. 38-14. (By “cantilevered” wesimply mean that thebeam 1ssupported insuch away that both theposition and theslope arefixed atone end—it 1sstuck into acement wall.) What istheshape ofthebeam? Let’s call thedeflection atthedistance xfrom thefixed end 2;wewant toknow 2(x).. We'll work itoutonly forsmall deflections. Wewillalso assume that thebeam islong incomparison with itscross section. Now, asyou know from your mathematics courses, thecurvature 1/R ofany curve 2(x) isgiven by 1 @z/dx*Ba 3 7 7 RO +(dz/dxyy? 6838) —ae|sinceweareinterestedonlyinsmallslopes—thisisusuallythecaseinengineering Y—structures—weneglect(dz/dx)?incomparisonwith1,andtake Z,w Lid, 3R-ge (38.39) Fig. 38-14. A.cantilevered beam Wealso need toknow thebending moment si Itisafunction ofxbecause itis with @weight atoneend. equal tothetorque about theneutral axis ofanycross section. Let's neglect theweightofthe beam and take only thedownward force Wattheend ofthebeam. (You canputinthebeam weight yourselfifyouwant.)Thenthebendingmoment atxis mx) =WL =2), because that isthetorque about thepoint atx,exerted bytheweight W—the torque which thebeam must support ofx.Weget YI a WL-= B= or a:_Wwfatyt-9. (38.40) This one wecanintegrate without any tricks; weget W (Le xrt -2), (sai) using ourassumptions that 2(0) =0andthat dz/dx isalso zero atx=0.That istheshape ofthebeam, The displacement oftheend is wh aL)=ppt (38.42) thedisplacement oftheend ofabeamincreasesasthecubeofthelength. Inderiving ourapproximate beam theory, wehave assumed that thecross section ofthebeam didnotchange when thebeam was bent. When thethickness ofthebeam issmall compared totheradius ofcurvature, thecross section changes very little andourresult is O.K.Ingeneral, however, thiseffect cannot beneglected, asyou can easily demonstrate foryourselves bybending asoft-rubber eraser in your fingers. Ifthecross section wasoriginally rectangular, youwillfindthatwhen 38-10 itisbentitbulges atthebottom (seeFig.38-15). Thishappens because when we 8 compress thebottom, thematerial expands sideways—as described byPoisson's (@) ratio. Rubber iseasy tobend orstretch, butitissomewhat likealiquid inthat it’shard tochange thevolume—as shows upnicely when you bend theeraser, For anincompressible material, Poisson's ratio would beexactly 1/2—for rubber ttis nearly that. | ~ s 38-5 Buckling Wewant nowtouseourbeam theory tounderstand thetheory ofthe‘“buck- ling” ofbeams, orcolumns, orrods. Consider thesituation sketched inFig. (b) 38-16 inwhich arod that would normally bestraight isheld initsbent shape by ‘wo opposite forces that push ontheends oftherod. Wewould like tocalculate theshape oftherodandthemagnitude oftheforces ontheends. Fig.38-15. (0)Abenteroser; (b) Letthedeflection oftherodfrom thestraight linebetween theends be(3). cross section. where x18thedistance from oneend. The bending moment :itatthepoint P inthefigure 1sequal totheforce Fmultiplied bythemoment arm, which 1sthe perpendicular distance », ama) =Fy. (38.43) Using thebeam equation (38.36), wehave vr ThomFy. (38.44) . Forsmalldeflections, wecantake1/R=—d®y/dx* (themmussignbecausethe ai wecurvature isdownward), Weget oo ¥ & @y__F f — 7de ~yr (38.45) |,__.| |newhichisthedifferential equationofasinewave.Soforsmalldeflections, thecurveofsuch abent beam 1sasine curve. The “wavelength” \ofthesine wave 1stwice Fig. 38-16. Abuckled beam. thedistance 1between theends. Ifthebending issmall, this 1sjust twice the unbent length ofthe rod. Sothecurve 1s y=Ksin wx/L. ‘Taking thesecond derivative, weget ay de> BY ‘Comparing thistoEq.(38.45), weseethat theforce is ret, (38.46)E For small bendings theforce 1sindependent ofthebending displacement y! Wehave, then, thefollowing thing physically Iftheforce 1slessthan theF given inEq, (38.46), there will benobending atall, But sfit1sshghtly greater than this force, thematerial will suddenly bend alarge amount—that is,for forces above theeritical force x®Y//L? (often called the“Euler force”) thebeam will “buckle.” Iftheloading onthesecond floor ofabuildingexceedstheEuler force forthesupporting columns, thebuilding willcollapse. Another place where thebuckling force 18most important isinspace rockets. Onone hand, therocket must beable tohold 11sown weight onthelaunching pad and endure thestresses during acceleration, ontheother hand, it1simportant tokeep theweight ofthe structure toaminimum, sothat thepayload and fuel capacity may bemade as large aspossible. Actually abeam will not necessarily collapse completely when theforce exceeds theEuler force. When thedisplacements getlarge, theforce islarger than se ‘ oea P, Ss Fig. 38-17. The coordinates $and 6 for the curve ofabent beam. what wehave found because oftheterms in1/RinEq.(38.38) that wehave ne- lected. Tofind theforces foralarge bending ofthebeam, wehave togoback to theexact equation, Eq. (38.44), which wehad before weused theapproximate relation between Rand y.Equation (38.44) hasarather simple geometrical prop- erty." It'salittle complicated towork out, but rather interesting. Instead of describing thecurve interms ofxand y,wecan usetwo new variables: S,the distance alongthecurve,and#theslopeofthetangent(othecurve,SeeFig.38-17. ‘The curvature istherate ofchange ofangle with distance: tw R- dS Wecan, therefore write theexact equation (38.44) as Olt aeds~~yr a —_ Ifwetakethederivative ofthisequationwithrespecttoSandreplacedy/dsbyF F sin8,weget ao FF. ase~~yrs" a (38.47) {IfGissmall, wegetback Eq.(38.45). Everything isO.K.] NowitmayormaynotdelightyoutoknowthatEq.(38.47)isexactlythe F, Fy same oneyougetforthelargeamplitude oscillations ofapendulum—with F/YI replaced byanother constant, ofcourse. Welearned way back inChapter 9,Vol. I, hhow tofind thesolution ofsuch anequation byanumerical calculation.t The answers yougetaresomefascinating curves—known asthecurvesofthe“Elastica.” Figure 38-18 shows three curves fordifferent values ofF/YJ. *The same equation appears, incidentally, mother physical situations—for example, = = themeniscus atthesurface of@quid contained between parallel planes—and thesame3 3 geometrical solution can beused. +The solutions can also beexpressed interms ofsome funcuons, called the“Jacobian Fig. 38-18. Curves ofabent rod elliptic functions,” that someone elsehasalready computed. 3812 39 Elastic Materials 39-1 The tensor ofstrain Inthelastchapter wetalked about thedistortions ofparticular elastic objects. 39-1 The tensor ofstrain Inthis chapter wewant tolook atwhat can happen ingeneral inside anelasticmaterial. Wewould liketobeabletodescribe theconditions ofstress andstrain 392Thetensor ofelasticity inside some bigglob ofjello which istwisted and squashed insome complicated 39-3 The motions inanelastic body way. Todothis,weneed tobeabletodescribe the/ocal strain atevery pomt inan °l clastic body;wecandoitbygivingasetofsixnumbers—which arethecomponents 39-4Nomelastic behavior ofasymmetric tensor—for each point. Earlier, wespoke ofthestress tensor 39-5 Calculating theelastic constants, (Chapter 31); now weneed thetensor ofstrain, Imagine that westart with thematerial initially unstrained and watch the motion ofasmall speck of“dirt” embedded inthematerial when thestrain is applied. Aspeck that was atthepoint Plocated atr=(x,y, 2)moves toa new Reference: C.Kittel, Introduction 10 position P’atr’=(x’,y’,2’)asshown inFig. 39-1, Wewillcallwthevector Solid State Physics, John displacements from PtoA. Then Wiley and Sons, Inc., New waPror 9.1) York,2nded.,1956. ‘The displacement #depends, ofcourse, onwhich point Pwestart with, so isa vector function ofr—or, ifyou prefer, of(x,9,2) Let’s look first atasimple situation inwhich thestrain 1sconstant over the material—so wehave what iscalled ahomogeneous strain. Suppose, forinstance, that wehave ablock ofmaterial and westretch ituniformly. Wejust change its dimensions uniformly inone direction—say, inthex-direction, asshown inFig. 39-2. The motion u,ofaspeck atxisproportional tox.Infact, Me Al, xT Wewill write u,this way: te = eax serore 4P AFTER H BEFORE.. ' — xSS oN SSS \ ‘\ \ i i N\ aN f L 'P i Y 1 +t peeSS ha ' aerer| iN SPECK Woy PECK | CNN TN \\ Joayy to Nor \4,. a ip am ia ' Noetle }\ iy> aN \}! ov aN . Sl et bu Fig.29-1.AspeckofthematerialatthepointPinanunstrainedblock Fig.39-2.Ahomogeneous stretch.type strain,moves toP?where the block isstrained. 4 The proportionality constant e-,is,ofcourse, thesame thing asAl/I, (You will seeshortly why weuseadouble subscript.) Ifthestrain isnotuniform, therelation between tu,andxwillvary from place toplace inthematerial. For thegeneral situation, wedefine the¢,.byakind of local A//I, namely by ese=au,/ax. G92) This number—which isnow afunction ofx,y,and z—describes theamount ofstretching inthex-direction throughout thehunkofjello.Theremay,ofcourse, also bestretching inthey-and z-directions. Wedescribe them bythenumbers uy _uy c=Gite eae=SE G93) ‘Weneed tobeable todescribe also theshear-type strains. Suppose weimagine alittle cube marked outintheinitially undisturbed jello. When thejello ispushed outofshape, this cube may getchanged into aparallelogram, assketched inFig. 39-3.* Inthis kind ofastrain, thex-motion ofeach particle isproportional to itsy-coordinate, w=$y. 694) And there isalso ay-motion proportional tox, ty=$x G9.) Sowecandescribe such ashear-type strain bywriting Me =ayy ty =ak with = =. fay=Ce=5 Now you might think that when thestrains arenothomogeneous wecould describe thegeneralized shear strains bydefining thequantities eyand ey,by ate any,c=ey ey= 696) J r Ve X\' NX\' 1“PX ‘ BeFoRe\)seter |\\\) ge.\NX\ aKAA A qKS [Ss soF z Fig. 39-3. Ahomogeneous shear strain, Butthere isonedifficulty. Suppose thatthedisplacements uzandw,were given by -2 =!Mt =F *Wechoose forthemoment tosplitthetotalshearangle@intotwoequalpartsandmake thestain symmetric with respect toxand y. 92 X] ree1 SPRN XN ' y \\! 1Pe SN crore arrer! ‘ 7i “I $n SSIX IN NY Fig. 39-4. Ahomogeneous rotation—there isnostrain They arelikeEqs. (39.4) and (39.5) except that thesign ofw,isreversed. With these displacements alittle cube inthejello simply gets shifted bytheangle @/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. The keypoint isthatifduy/ax and ‘au,/dy areequal and opposite, there isnostrain; sowecanfixthings upbydefining “ Gey=lye=H(Oty/Ax +Ate/Ay). Forapure rotation they areboth zero, butforapure shear wegetthat ez,is equal toyz, aswewould like. Inthemost general distortion—which may include stretching orcompression aswell asshear—we define thestate ofstrain bygiving thenine numbers =ie er=HE, ay en=Fe (39.7) Cry=HOuy/Ox +duz/dy), These aretheterms ofatensor ofstrain. Because itisasymmetric tensor—our definitions make ezy=eye,always—there arereally only sixdifferent numbers. You remember (see Chapter 31)that thegeneral characteristic ofatensor 1sthat theterms transform like theproducts ofthecomponents oftwo vectors. (If Aand Bare vectors, C,,=A,B, isatensor.) Each term of¢,,isaproduct (orthesum ofsuch products) ofthecomponents ofthevector #=(lz,ty,Us),and oftheoperator V=(3/dx, 4/dy, 4/82), which weknow transforms like avector. Let’s letx1,x2,and xgstand forx,y,and zand w,,ua,and uystand forue,ty, andu,;then wecanwrite thegeneral term e;,ofthestrain tensor as ey=h(Gu,/dx, +du,/dx,), 9.8) where jandjcanbe1,2,or3. When wehave ahomogeneous strain—which may include both stretching andshear—all ofthee,,areconstants, andwecanwrite Uy=xxx +Cay +erate 39.9) (Wechooseouroriginofx,y,zatthepointwherewiszero.)Inthiscase,thestrain tensor e,,gives therelationship between two vectors: thecoordinate vector r= (x,»,z)and thedisplacement vector w=(uz, Uy,Us). 33 totheshear modulus wedefined inthelastchapter.) The constants yand)are called theLamé elastic constants. Comparing Eq.(39.20) with Eq.(39.12), you see that Cory =+ A Cayzy =2, f 921) Corse = td. Sowehave proved that Eq.(39.19) isindeed true, You also seethat theelastic properties ofanisotropic material arecompletely given bytwo constants, aswe said inthelastchapter. ‘TheC’scanbeputintermsofanytwooftheelasticconstants wehaveused earlier—for instance, interms ofYoung's modulus Yand Poisson’s ratioa. We will leave itforyou toshow that Y © Cou=49(14+7S): Y o Con=EG(x) , (39.22) Y SS Cony=Ee)’ oyVOLUME Van\ 39-3Themotionsinanelasticbody ‘SURFACE A ‘We have pointed outthat foranelastic body inequilibrium theinternal \ stresses adjust themselves tomake theenergy aminimum. Now wetakealookat \ what happens when theinternal forces arenotinequilibrium. Let's saywehave _ ‘asmallpieceofthematerial insidesomesurfaceA.SeeFig.39-S.Ifthepieceisin \ equilibrium, thetotalforceFactingonitmustbezero.Wecanthinkofthisforce asbeingmadeupoftwoparts.Therecouldbeonepartdueto“external”forces cLTike gravity, which actfrom adistance onthematter inthepiece toproduce@ forceperunitvolumefax.ThetotalexternalforceFoxistheintegraloffix,over thevolume ofthepiece: Fig. 39-5. Asmoll volume element V boundedbythesurfaceA. Fou=|foad¥. (39.23) Inequilibrium, thisforce would bebalanced bythetotal force F,,,from theneigh boring material which acts across thesurface A.When thepiece isnotinequili- brium—ifitismoving—the sumoftheinternal andexternal forcesisequaltothe mass times the acceleration. We would have Fox+Fine=fprav, 9.24) where pisthedensity ofthematerial, andr1sitsacceleration. Wecannowcom bine Eqs. (39.23) and (39.24), writing Fn=f(“fous+pi)dV. (39.25) Wewill simplify our writing bydefining S= ~foxs +Br. 69.26) Then Eq. (39.25) iswritten Fin=fsav. @9.27) What wehave called Fy. isrelated tothestresses inthematerial. The stress tensor S,,was defined (Chapter 31)sothat thex-component oftheforce dFacross asurface element da,whose unit normal isn,isgiven by AF, =(Sette +Spy +Seats) da. 9.28) 96 other words, wecan put way tm, 934 where Vou =0, VX m= 0. 6935) Substituting 1,+1»forwin (39.33), weget pd?/at7[uy+wo)=(+w)(Va) +uV?(uy+wa).39.36) Wecan eliminate u,bytaking thedivergence ofthis equation, p.99/A02(W =wa)=(X+uw)VT+wa)+weV2u2. Since theoperators (V*) and(¥*) canbeinterchanged, wecanfactor outthedi- vergence toget POLAROIDS V+{p.d?u2/at?—(A+2u)V¥uz}=0. 39.37) Aw\Since¥Xwyiszerobydefinition,thecurlofthebracket{}isalsozero;sothe 2g\ bracket itself 1sidentically zero, and iat al a p.9%us/at?=(0+2)Vay :9.38) Li|Vad Thisisthevectorwaveequationforwaveswhichmoveatthespeed ||fia]( Cy=V0+2w)/p.Sincethecurlofus1szero,thereisnoshearingassociatedey ( withthiswave;thiswave1sjustthecompressional—sound-type—wavewediscussed | aH AA inthelastchapter, andthevelocity isjustwhat wefound forCiong: | Age Inasimilarway—by takingthecurlofEq.(39.36)—we canshowthatayxEee satisfiestheequation omen sco crs pd%u/at? =Va, 939) Fig.39-6. Measuring internal This isagain avector wave equation forwaves with thespeed C,=Vu/o.stresseswithpolarizedlight. Since¥+u;18zero,wxproducesnochangesindensity;thevectoraycorrespondstothetransverse, orshear-type, wave wesaw inthelastchapter, and Cz=Cykeae Ifwewished toknow thestatic stresses inanisotropic material, wecould, 1principle, find them bysolving Eq.(39.32) with fequal tozero—or equal tothet V7 staticbodyforcesfromgravitysuchaspg—under certainconditions whichare\ Ws related totheforces acting onthesurfaces ofourlarge block ofmaterial. This is , e %somewhatmoredifficulttodothanthecorrespondingproblems1nelectromagne- ((tism. Itismore difficult, first, because theequations arealittle more difficult to ‘ } handle, and second, because theshape oftheelastic bodies wearelikely tobe ' \ interested inareusually much more complicated. Inelectromagnetism, weare ' often interested insolving Maxwell’s equations around relatively simple geometric i\\//A| shapes such ascylinders, spheres, and soon,since these areconvenient shapes 'le 4|forelectricaldevices.Inelasticity,theobjectswewouldliketoanalyzemayhave) | {quite complicated shapes~like acrane hook, oranautomobile crankshaft, orthe | |rotorofagasturbine.Suchproblemscansometimes beworkedoutapproxi- mately bynumerical methods, using theminimum energy principle wementioned\ } earlier.Anotheray1stouseamodelofthe object andmeasure theinternal strains experimentally, using polarized light. ys Itworks thisway: When @transparent isotropic material—for example, a \\ iy clear plastic like lucite—is putunder stress, itbecomes birefringent. Ifyouput polarized light through tt,theplane ofpolarization will berotated byanamountq | relatedtothestress:bymeasuring therotation, youcanmeasure thestress.Figure 39-6 shows how such asetup might look. Figure 39-7 isaphotograph ofa Fig.39-7. Astressed plasc model photoelastic model ofacomplicated shapeunderstress, as seen between crossed polarcids. [From F,W. Sears, Optics, Addison: Wesley Publishing Co,Reading, Moss. 39-4 Nonelastic behavior 1949.) Inallthathasbeensaidsofar,wehaveassumed thatstress isproportional tostrain; ingeneral, that isnor true, Figure 39-8 shows atypical stress-strain curve foraductile material. For small strains, thestress isproportional tothe Bs atom; wewillIeave outthiscomplication.) Wearealsogoing toinclude only the forces between each atom and itsnearest and next-nearest neighbors. Inother words, wewillmake anapproximation which neglects allforces beyond thenext- nearest neighbor. Theforces wewillinclude areshown forthexy-plane inFig. 39-10(a). Thecorresponding forcesinthey2-andzx-planes alsohavetobe j | |included, ot4*A N14 stb Sinceweareonlyinterested intheelastic coefficients which applytosmall ~(No) -—~(cr)++ (we)~ strains, andtherefore onlywantthetermsintheenergy whichvaryquadratically 7s areswiththestrains,wecanimaginethattheforcebetweeneachatompairvaries | Slinearly withthedisplacements. Wecanthenimagine thateachpairofatomsis | LASJoined byalinearspring, asdrawninFig.39-10(b). Allofthesprings between aSf NNsodiumatomandachlorineatomshouldhavethesamespringconstant,sayk1. iD) aamaThespringsbetweentwosodiumsandbetweentwochlorinescouldhavedifferent 74 as |constants, butwewillmakeourdiscussion simplerbytakingthemequal;wecall | i. themka,(Wecouldcomebacklaterandmakethemdifferent afterwehaveseen . SION howthecalculations go.) ~(We)——~ (er)==(Ne) = Nowweassume thatthecrystalisdistorted byahomogeneous strainde- ae\SYS LIXscribed bythestrain tensor e;,.Ingeneral, itwillhavecomponents involving \ i x,y,and 2;butwewillconsider now only astrain with thethree components Cex,Cry,ANAeyy$0thatitwillbeeasytovisualize. Ifwepickoneatomasour Ne origin,thedisplacement ofeveryotheratomisgivenbyequationslikeEq.39.9): we Seiten Avete are LMe=Cask+Cay) 9.42) 3an+S,9My=Cay+Cy ies AL CM| Suppose wecalltheatomatx=y=0“atom1”andnumber itsneighbors in SE eS)thexy-plane asshowninFig.39-11. Calling thelatticeconstant a,wegetthex Dm Se |andydisplacements u,andu,listedinTable39-1. 3>WX 3‘Nowwecancalculatetheenergystoredinthesprings,whichisk?/2times LSwn,1%Pp thesquare oftheextension foreachspring. Forexample, theenergy inthehori- (e$-crer ter) Line)zontalspringbetween atom1andatom2is - Fig.39-10.(a)Theinteratomic 2eed). (29.43) forces wearetaking intoaccount; (b)amodel inwhich the atoms are connected Notethattofirstorder,they-displacement ofatom2doesnotchangethelengthof|°Y*PIing®thespring between atom |and atom 2.Togetthestrain energy inadiagonal spring, such asthat toatom 3,however, weneed tocalculate thechange inlength due to both thehorizontal and vertical displacements. For small displacements from the x 4 — ~eve3: | ) a7 ~ ey ° ~ 1 > 3;t — b o—4 ot \ n&)) 7s ’ \79 : Fig. 39-11. The displacements ofthe i 8 nearest ondnext-nearest neighbors of 7 atom 1(exaggerated). sur inef,andine3,,wegetthefactor (ky+2k2)a?, so Corse=Cry=Mth,a@ Fortheremaining terms, there isaslight complication. Since wecannot distin- Buish theproduct oftwoterms likeecz¢yy from eyyezz, thecoefficient ofsuch terms 1nour energy isequal tothesum oftwo terms inEq. (39.13). The coefficient of €zzeyy inEq.(39.45) is2k2, $0wehave that 2ky (Cons +Caves) =FP Butbecause ofthesymmetry inourcrystal, Czayy =Cyyze» 80wehave that = =k Con =Crys = Byasimilar process, wecan also get Coury=Cyzve=ksa Finally, you will notice that any term which involves either xoryonly once iszero—asweconcluded earlierfromsymmetry arguments. Summarizing ourresults: Coxe=Cony=+22,a ke Coy=Cue=2 oan Com=Cones=Coe=Cray="2s Cassy=Coy=ete.=0. Table39-2 Wehavebeenabletorelatethebulkelastic constants totheatomic properties . jwhichappearintheconstantskyandk».Inourparticularcase,Czyzy=Cesyw- ElasticMoatofCubicCrystalsItturnsout—asyoucanperhapsseefromthewaythecalculations went—that in101°dynesem these terms arealways equal foracubic crystal, nomatter howmany forceterms c c c aretaken into account, provided only that theforces actalong thelinejoining Gee Gy Gum, each pair ofatoms—that is,solong astheforces between atoms areIikesprings Na 0.055 0.042 0.049 and don’t have asideways part such asyou might getfrom acantilevered beam K 0.086 0.037 0026 (and youdogetincovalent bonds). Fe 237 Ll L1G ‘Wecancheck thisconclusion with theexperimental measurements ofthe Diamond 10.76 1.25 5.76 elastic constants. InTable 39-2wegivetheobserved values ofthethree elastic AL 108 0G (0.28 coefficients forseveral cubiccrystals.* Youwillnotice thatCeryyandCeyeyare, LF. 119 0,540.53. Nae ce ‘ NaCl 04860.1270.128ingeneral, notequal. Thereason isthatinmetals likesodium andpotassium the & ae Obes oe interatomic forces arenotalong thelinejoining theatoms, asweassumed inour Nowe O33 OL OS model. Diamond does notobey thelaweither, because theforces indiamond are xy 027 onda aoe covalent forces andhave some directional properties—the bonds would prefer to agi 060-036 0,062 beatthetetrahedral angle. The ionic erystals likelithium fluoride, sodium chloride, _ andsoon,dohave nearly allthephysical properties assumed inourmodel, and «From. Kittel, Inrodvction toSoldthetableshowsthattheconstants CeryyandCyyzyarealmostequal.[tisnotclear gratePhysics,JohnWileyandSons,Inc,why silver chloride should notsatisfy thecondition that Cesyy =Cayey- New York, 2nd. ed,1956, p.93 *Intheliterature you will often find that adifferent notation isused. For instance, people usually write Crser =City Cony =C12 tind Cayay =Code 33 4 The Flow of Wet Water 41-1 Viscosity Inthe last chapter wediscussed the behavior ofwater, disregarding the 41-1 Viscosity phenomenon ofviscosity. Nowwewould liketodiscuss thephenomena ofthe 441-2Viscous fomflowoffluids,meludingtheeffectsofviscosity. Wewant tolook atthereal behavior offluids. Wewill describe qualitatively theactual behavior ofthefluids under 41-3 The Reynolds number various different circumstances sothat you will getsome feel for the subject. Al-thoughyouwllseesomecomplicated equationsandhearaboutsomecomphieated 41-4Flowpastacircularcylinder things,iisnotourpurposethatyoushouldlearnallthesethings.This1s,ina41-5Thelimitofzeroviscosity sense,a“cultural” chapterwhichwillgiveyousomeideaofthe way theworld is.‘Thereisonlyoneitemwhichiswortheanning, andthatisthesimpledehmtion of 41-6Couette flow viscosity which wewill come toinamoment. ‘The rest 1sonly foryour entertain- ment, Inthelastchapterwefoundthatthelawsofmotionofafluidarecontained 1mtheequation aw_ Lass, 4wevw=—SP—v9+Le (aul) Inour“dry” water approximation weleftoutthelast term, sowewere neglecting allviscous effects. Also, wesometimes made anadditional approximation by considering thefluid asincompressible; then wehad theadditional equation vevn0. This lastapproximation isoften quite good—particularly when flow speeds are much slower than thespeed ofsound. Butinrealfluids iisalmost never true that wecan neglect theinternal friction that wecall viscosity; most oftheinteresting things that happen come from itinone way oranother. Forexample, wesaw that in“dry” water thecirculation never changes—if there isnone tostart out with, there will never beany. Yet, circulation influids isaneveryday occurrence. We must fixupourtheory. Webegin with animportant experimental fact. When weworked out the flow of“dry” water around orpasta eylinder—the so-called “potential flow"—we hhadnoreason nottopermit thewater tohave avelocity tangent tothesurface; only thenormal component had tobezero. Wetook noaccount ofthepossibility that there might beashear force between theliquid and thesolid, Itturns out— although it1smot atallself-evident—that inallcircumstances where ithas been experimentally checked, thevelocity ofafluidisexacilyzeroatthesurfaceofa solid, You have noticed, nodoubt, that theblade ofafanwillcollect athin layer of dlust-—and that itisstill there after thefan has been churning uptheair. You canseethesame effect even onthegreat fanofawind tunnel. Why isn’t thedust blown offbytheair? Inspite ofthefact that thefanblade ismoving athigh speed through theair, thespeed oftheairrelative tothefanblade goes tozero right at thesurface. Sothevery smallest dust particles arenot disturbed.* Wemust modify thetheory toagree with theexperimental fact that inallordinary 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 upinto thebreeze.#Youcanimaginecircumstances whenitisnottrue:glastheoretically a“liquid,”butatcan certainly bemade toslide along asteel surface. Soourassertion must break down somewhere. au AREA A Mo, F, Y — ' Ca Coat ne Fig.41-1,Viscousdragbetween two4 tor .porallel plates. veneer v=0 Weoriginally characterrzed aliquid bythefact that fyou putashearing stressonit—nomatterhowsmall—itwouldgiveway.Itflows.Instaticsituations, there arenoshear stresses. But before equilibrium 1sreached—as long asyou still push ontt—there canbeshear forces. Viscosity describes these shear forces which exist ina moving fluid. ‘Togetameasure oftheshear forces during themouion ofafluid,weconsiderthefollowing kindofexperiment. Supposethatwehavetwo ee solid plane surfaces with water between them, asinFig. 41-1, and wekeep one _stationary while moving theother parallel (o1attheslow speedry.If'youmeasure —_~~ theforce required tokeep theupper plate moving, youfindthat11sproportional gp theareaoftheplates andtor/d,where disthedistance between theplates. So 77SS, etoe theshearstressF/A1sproportional tov9/d: — heat} Sa room Ana The constant ofproportionality iscalled thecoefficient ofviscosity. Ifwehave amore complicated situation, wecanalways consider ahittle, fat, a —— rectangularcellinthewaterwithitsfacesparalleltotheflow,asinFig.41-2.The F€actcelenby Fig.41-2. Theshear stress ing steForceacross thiscellsgiven by viscous fuid, AF ane aAF _ Ate _ tte aa4” "ay 7ay (41.2) Now, ar,/ay istherate ofchange oftheshear strain wedefined inChapter 38,so Foraquid, theshear stress isproportional totherate ofchange oftheshear strain. Inthegeneral case wewrite ‘ary yar = (ey 3 Seyo(%+) (13a) bv Ifthere 1sauniform rotation ofthefluid, 4”,/ay18thenegativeofav,/Ax andSey eee, 1zero—as 1should besince there arenostresses inauniformly rotating fluid.foiion (Wedidasimilarthingindefininge,,inChapter39.)Thereare,ofcourse,the fap Ghee corresponding expressions forS,.andS..INS xO& Asanexampleoftheapplication oftheseideas,weconsiderthemotionofaAYaN|Sn fluidbetweentwocoaxialcylinders. Lettheinneronehavetheradius«andthePOF AEY 4peripheral velocityvq,andlettheouteronehaveradiusbandvelocity1»,SeeA‘|NL_44=_Fig.41-3.Wemightask,what1sthevelocitydistributionbetweenthecylinders?an Pi *Toanswerthisquestion, webeginbyfindingaformulafortheviscousshearin ~\N f, thefluidatadistance rfromtheaxis From thesymmetry oftheproblem, wecan. Xe AY assume thattheflowisalways tangential andthatitsmagnitude depends onlyonSS OSE JL rv=u(r).Ifwewatchaspeckinthewaterattheradiusr,itscoordinates asa eefunctionoftimeare “esate X=reosar,y=rsinwt, * where @=0/r.Then thex-andy-components ofvelocity are Fig. 41-3. The flow inofluid be- ; tween twoconcentric cylinders rotating re=sresinet =-ey and ry=rwcoset =wx. (41.4) «otdifferent angular velocities.cies. From Eq.(41.3), wehave a a ao do Sa=[Zo aHow|=fsax7?bal (415) 412 5) 2 \ ©' ' sor 1} . 1b 0 nenoore | '*Canina f\ ' Croneitem | ' 1\ ' ' Tummutexr' ' ' ! sounpany tavenoa a: 10 0 eg oF 10 io o Fig.41-4,Thedragcoefficient Cpof@circulareylinderas©functionoftheReynoldsnumber. 41-4 Flow past acircular cylinder Let's goback totheproblem oflow-speed (nearly incompressible) flow over thecylinder. We will give aqualitative description oftheflow ofareal fluid. ‘There aremany things wemight want toknow about such aflow—for instance, what isthedrag force onthecylinder? The drag force onacylinder isplotted in Fig. 41-4 asafunction of6—which1sproportional totheairspeedVsfeverything, else isheld fixed. What isactually plotted 1stheso-called drag coefficient Cr, which isadimensionless number equal totheforce divided by4pV2DI, where Dis thediameter, 1sthelength ofthecylinder, and p1sthedensny oftheliquid: F Co=VEDI The coefficient ofdrag varies inarather complicated way, giving usapre-hint that something rather interesting andcomplicated 1shappening intheflow, Wewill, nowdescribe thenature offlowforthedifferent ranges oftheReynolds number. —_First,whentheReynoldsnumberisverysmall,theflowisquitesteadythat1s,|[==thevelocity isconstant atanyplace, and theflow goes around thecylinder. The —— actualdistribution oftheflowlinesis,however,notlikeit1sinpotentialflow.=p =They aresolutions ofasomewhat different equation. When thevelocity isvery —=~ Ge = low or,what 1sequivalent,whentheviscosityisveryInghsothestuff1sikehoney,—>—=—Z ———— thentheinertialtermsarenegligibleandtheflowisdescribedbytheequation SSSva =0, od This equation wasfirstsolved byStokes. Healsosolved thesame problem fora fag,41-5, Viscous flow (lowveloci- sphere, Ifyouhave asmall sphere moving under such conditions oflowReynolds ties) ground @cirevlor cylinder. number, theforce needed todrag itisequal to6xnaV, where a1stheradius ofthe sphere and V1sitsvelocity. This 1savery useful formula because ittells thespeed alwhich tiny grains ofdirt (or other particles which can beapproximated asspheres)movethroughafluidunderagivenforce—as, forinstance,inacentrifuge,orinsedimentation, ordiffusion Inthelow Reynolds number region—for itless than I—the lines ofvaround aeylinder areasdrawn inFig 41-5 Ifwenow increase thefluid speed togetaReynolds number somewhat greater than 1,wefind that theflow 1sdifferent, There isacirculation behind thesphere, asshown inFig, 41-6(b). Itistill anopen question astowhether there isalways 47 eT WWRE Se Rz20 —GY) DpoS R=100 At BOO =C.. , = REO eaeey Se — R=108 acirculation there even atthe smallest Reynolds number orwhether things sud- denly change atacertain Reynolds number. Itused tobethought that thecir- culation grew continuously. But it1snow thought that itappears suddenly, and character tothe flow for&intheregionfromabout10to30.Thereisapairof vortices behind thecylinder. ‘The flow changes again bythetime wegettoanumber of4Uorso.There is suddenly acomplete change inthecharacter ofthemotion. What happens isthat one ofthevortices behind thecylinder gets solong that itbreaks offand travels downstream with thefluid. Then thefluid curls around behind thecylinder and makes anew vortex. The vortices peel offalternately oneach side, soaninstan- taneous view oftheflow looks roughly assketched inFig. 41-6(c). The stream of Bes Ge Rae age TAG IENCRAA Bako ASSRE 7 ESS AS TED eee aaoe SM f0,417.Ptooroph bytwaBhigtte RES. SEE Prandtlofthe“vortexstreet”intheflow vortices iscalled a“Karmén vortex street.” They always appear for®@>40. Weshow aphotograph ofsuch aflow inFig. 41-7. The difference between thetwoflows inFig. 41-6(c) and 41-6(b) or41-6(a) isalmost acomplete difference inregime InFig. 41-6(a) or(b),thevelocity isconstant, whereasinFig41-6(c), thevelocityatanypointvarieswithumeThere isnosteady solution above ®=40—-which wehave marked onFig. 41-4 bya dashed line. For these higher Reynolds numbers, theflow varies with time butina regular, cyclic fashion. We can getaphysical dea ofhow these vortices areproduced We know that thefluid velocity must bezero atthesurface ofthecylinder and that italso increases rapidly away from that surface. Vorticity iscreated bythis large local variation influid velocity, Now when themain stream velocity islowenough, there 1ssufficient time forthis vorticity todiffuse out ofthethin region near thesolid surface where itisproduced and togrow into alarge region ofvorticity. This physical picture should help toprepare usforthenext change inthenature ofthe flow asthemain stream velocity, orst,isincreased stillmore. Asthevelocity gets higher and higher, there isless and less time forthe vorticity todiffuse into alarger region offluid. Bythetime wereach aReynolds number ofseveral hundred, thevorticity begins tofillinathin band, asshown in Fig. 41-6(d). Inthis layer theflow ischaotic and irregular. The region iscalled theboundary layer andthisirregular flow region works itsway farther and farther upstream astis inereased. Intheturbulent region, thevelocities arevery irregular and “noisy”; also theflow 1snolonger two-dimensional buttwists and turns in allthree dimensions. There isstill aregular alternating motion superimposed on the turbutent one, ‘AstheReynolds number isincreased further, theturbulent region works its way forward until itreaches thepoint where theflow lines leave thecylinder—for flows somewhat above ®=10°, The flow isasshown inFig. 41-6(e), andwe have what 1scalled a“turbulent boundary layer.” Also, there isadrastic change inthedrag force; stdrops byalarge factor, asshown inFig. 41-4. Inthis speed region, thedrag force actually decreases with increasing speed. There seems to belittle evidence ofperiodicity. ‘What happens forstill larger Reynolds numbers? Asweincrease thespeed further, thewake increases insize again and thedrag increases. Thelatest experi- ments—which goupto®=107orso—indicate thatanew periodicity appears inthewake. either because thewhole wake isoscillating back and forth inagross motion orbecause some new kind ofvortex isoccurring together with anirregular noisy motion. The details areasyetnotentirely clear, and arestill being studied experimentally. 41-5 The limit ofzero viscosity We would like topoint out that none ofthe flows wehave described are anything likethepotential flow solution wefound inthepreceding chapter. Thisis,atfirstsight,quitesurprising. Afterall,«tisproportional to1/n.Sogoingto zero isequivalent toétgoing toinfinuy. And ifwetake thelimit oflarge otin 49 Eq, (41.23), wegetridoftheright-hand side and getjust theequations ofthelast chapter. Yet, you would find ithard tobelieve that thehighly turbulent flow at G=107wasapproaching thesmooth flowcomputed from theequations of“dry” water. How can itbethat asweapproach @=2»,theflow described byEq. (41.23) gives acompletely different solution from the one weobtained taking 7=Otostartoutwith?Theanswerisveryinteresting. Notethattheright-handterm ofEq.(41.23) has 1/6 times asecond derivarve. Itis ahigher derwvative than anyother derivative intheequation. What happens isthat although thecoefficient 1/0 18small, there arevery rapid variations of@inthespacenearthesurface. ‘These rapid variations compensate for the small coefficient, and the product does notgotozero with increasing &.The solutions donotapproach thelimiting case asthecoefficient of20 goes tozero. You may bewondering, “What isthefine-grain turbulence and how does it maintain itself? How can thevorticity which 1smade somewhere attheedge of thecylinder generate somuch noise inthebackground?” ‘The answer 1sagain interesting. Vorticity hasatendency toamplify itself. Ifweforget for amoment about thediffusion ofvorticity which causes aloss, thelaws offlow say(aswehave seen) that thevortex lines arecarried along with thefluid, atthevelocity v.We canimagine acertain number oflines of&whicharebeingdistortedandtwisted bythecomplicated flow pattern ofv.This pulls thelines closer together andmixes - themallup.Linesthatweresimplebeforewillgetknotted andpulledclose a > topetiner. Theywillbelongerandtightertogether. Thestrength ofthevorticity~ ‘~===7| —willincrease anditsirregularities—the plusesandminuses—will, ingeneral, Pp= SS] merease. Sothemagnitude ofvorticity inthree dimensions increases aswetwist ~ Te SS =| thefluid about. ~ -LAS Se ‘You might well ask, “When isthepotential flow asatisfactory theory atall?” a SS~== ==] Inthefirstplace, itissatisfactory outside theturbulent region where thevorticity —_ ISS -= =] hasnotentered appreciably bydiffusion. Bymaking special streamlined bodies,= ===)—wecankeeptheturbulent regionassmallaspossible; theflowaroundairplanete eo wings—which arecarefully designed—is almost entirely truepotential flow. 41-6 Couette flow <>) <>) Itispossibletodemonstrate thatthecomplex andshiftingcharacter ofthe==)—fowpastacylinderisnotspecialbutthatthegreatvarietyofflowpossibilities —s=|_occursgenerally.WehaveworkedoutinSectionIasolutionfortheviscous | <= flowbetweentwocylinders,andwecancomparetheresultswithwhatactually ——A =— happens.Ifwetaketwoconcentriccylinderswithanoilinthespacebetweenthem |andputafinealuminum powder asasuspension intheoil,theflow iseasy tosee. Now ifweturn theouter cylinder slowly, nothing unexpected happens; seeFig. = 41-8(a). Alternatively, ifweturntheinner cylinder slowly, nothing verystriking occurs. However, ifweturn theinner cylinder atahigher rate, wegetasurprise. © ro The fluid breaks into horizontal bands, asindicated inFig. 41-8(b). When the outercylinder rotatesatasimilarratewiththeinneroneatrest,nosucheffect Fig.41-8. Liquidflowpatterns be- ‘occurs. Howcanitbethatthereisadifference between rotating theinnerorthe weentwotransparent rotating cylinders. utcylinder? After all,theflow pattern wederived inSection 1depended only onw—ay. Wecan gettheanswer bylooking atthecross sections shown in Fig. 41-9, When theinner layers ofthefluid aremoving more rapidly than the outer ones, they tend tomove outward—the centrifugal force 18larger than the pressure holding them inplace. Awhole layer cannot move outuniformly because theouter layers areintheway. Itmust break into cells and circulate, asshown in Fig. 41-9(b). Itislike theconvection currents inaroom which hashotairatthe bottom. When theinner cylinder isatrestandtheouter cylinder hasahigh velocity, thecentrifugal forces build upapressure gradient which keeps everything in equilibrium—see Fig. 41-9(c) (asinaroom with hotairatthetop). Now tet’s speed uptheinner cylinder. Atfirst, thenumber ofbands increases. Then suddenly you seethebands become wavy, asinFig. 41-8(c), and thewaves travelaroundthecylinder. Thespeedofthese waves iseasily measured. For high rotation speeds they approach 1/3thespeed oftheinner cylinder. And noone 41-40 CENTRIFUGAL FORCES [. : iOl « to)| (b) :C) ) “J ) | - A cenrrirusar FORCES Fig. 41-9. Why theflow breaksupintobands. knows why! There'sachallenge. Asimplenumberlike1/3,andnoexplanation Infact, thewhole mechanism ofthewave formation isnotvery well understood, yetitis steady laminar flow. Ifwenow start rotatingtheoutereylinderalso—butintheopposttedirection— theflowpatternstartstobreakup.Wegetwavyregionsalternatingwithapparently quietregions,assketchedinFig.41-8(d).makingaspiralpattern.Inthese“quiet” regions, however, wecan seethat theflow 1really quite irregular: 1t1s,infact completely turbulent. The wavy regions also begin toshow regular turbulent flow Ifthecylinders arerotated still more rapidly, thewhole flow becomes chaotically turbulent. Inthissimple experiment weseemany interesting regimes offlow which are quite different, andyetwhich areallcontained inoursimple equation forvarious values oftheoneparameter, With ourrotating cylinders, wecanseemany oftheeffectswhichoccurintheflowpastacylinder:first,thereisasteadyflowsecond,aflow sets inwhich 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 imanirregular churning cloud ofsmoke The main lesson tobelearned from allofthis isthat atremendous varieww ofbehavior 1shidden inthesimple setofequations in(41.23). Allthesolutions areforthesame equations, only with different values of«We have noreason tothink that there areany terms missing from these equations, The only difficulty 1that wedonothave themathematical power today toanalyze them except for very small Reynolds numbers—that 1s,inthecompletely viscous case. That we have written anequation does notremove from theflow offluids 1tscharm 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 and thecondensing, revolving, and exploding stars and galaxies 1sjust asimple equation for the hydrodynamic behavior ofnearly pure hydrogen gas. Often, people insome unjustified fear of physics sayyoucan’t write anequation forlife. Well, perhaps wecan. Asamatter offact, wevery possibly already have theequation toasufficient approximation when wewrite theequation ofquantum mechanics: hoy We We have just seen that thecomplexities ofthings can soeasily and dramatically escape thesimplicity oftheequations which describe them, Unaware ofthescope ofsimple equations, man hasofien concluded that nothing short ofGod, notmere equations, 1srequired toexplain thecomplexities oftheworld ait