Electricity & Magnetism Binder

pg 19History of Math pg 79 Review of Classical Electrodynamics book 3rd Ed pg 136 pg 176pg 160 ARidiculously Brief History ofElectricity and Magnetism .1foundthisatmaxwell.byu.edw~spencerr/phys442/noded.html ,brought itupinWord andhighlighted various things inbold andunderline. Ilaterfound aPDF atthesame site.Ithen didsome cutandpastes from thePDF intothis Word docforequations andvarious other symbols. Ifsomething isunreadable, seethePDF. Mostly from E.T.Whittaker’s AHistory oftheTheories ofAether andElectricity... Prehistory 900BC-Magnus, aGreek shepherd, walks across afield ofblack stones which pulltheironnails outofhissandals and theirontipfromhisshepherd's staff(authenticity notguaranteed). ThisregionbecomesknownasMagnesia. 600BC-Thales ofMiletos rubs amber (elektron inGreek) with catfurandpicks upbitsoffeathers.- 1269-PetrusPeregrinus ofPicardy, Italy, discovers thatnatural spherical magnets (lodestones) align needles with lines oflongitude pointing between twopole positions onthestone. The 17th Century 1600 -William Gilbert, court physician toQueen Elizabeth, discovers thattheearth isagiant magnet justlikeoneofthestonesofPeregrinus, explaining howcompasses work.Healsodiscusses staticelectricity andinventsanelectricfluid which isliberated byrubbing. @..1620-NiccoloCabeodiscoversthatelectricitycanberepulsiveaswellasattractive, 1630 -Vincenzo Cascariolo, aBolognese shoemaker, discovers fluorescence. 1638 -Rene Descartes theorizes thatlight isapressure wave through thesecond ofhisthree types ofmatter ofwhich the universeismade.Heinventsproperties ofthisfluidthatmakeitpossibletocalculate thereflection andrefraction oflight. The “modern” notion ofthe aether isborn, 1638 -Galileo attempts tomeasure thespeed oflight byalantern relay between distant hilltops. Hegetsavery large answer. 1644 -Rene Descartes theorizes that themagnetic poles areonthecentral axis ofaspinningvortexofoneofhisfluids. This vortex theory remains popular foralong time, enabling Leonhard Euler andtwooftheBernoullis toshare aprize of theFrench Academy aslateas1743. 1657-PierredeFermatshowsthattheprinciple ofleasttimeiscapableofexplaining refraction andreflection oflight. Fighting with theCartesians begins. (This principle forreflected light had been anticipated anciently byHero of Alexandria.) 1665-Francesco MariaGrimaldi, inaposthumous report,discovers andgivesthenameofdiffraction tothebendingof light around opaque bodies. 0.~RobertHookereportsinhisMicrographia thediscoveryofthe rings oflight formed byalayer ofairbetween two glass plates. These were actually first observed byRobert Boyle, which explains why they arenow called Newton's rings. 1 Inthesame work hegives thematching-wave-front derivation ofreflection andrefraction thatisstillfound inmost introductory physics texts. These waves travel through theaether. Healsodevelops atheory ofcolor inwhich white light isasimpledisturbance andcolorsarecomplexdistortionsofthebasicsimplewhiteform. ® 1671 -Isaac Newton destroys Hooke's theory ofcolor byexperimenting withprisms toshow thatwhite lightisamixtureofallthecolors andthatonceapurecolor isobtained itcannever bechanged intoanother color. Newton argues against lightbeingavibration ofthe ether, preferring thatitbesomething elsethatiscapable oftraveling through theaether. He doesn't insist thatthissomething elseconsist ofparticles, butallows thatitmay besome other kind ofemanation or impulse. InNewton's ownwords, “...let every manheretakehisfancy." 1675-OlafRoemer repeats Galileo's experiment using themoons ofJupiter asthedistant hilltop. Hemeasures ¢=2.3x 10°m/s. 1678~Christiaan Huygens introduces hisfamous construction andprinciple, thinksabouttranslating hismanuscript into Latin, thenpublishes itintheoriginal French in1690. Heuseshistheory todiscuss thedouble refraction ofIceland Spar. Hisisatheoryofpulses, however, notofperiodic waves. The 18th Century 1717 ~Newton shows thatthe“two-ness" ofdouble refraction clearly rules outlight being aether waves. (Allaether wavetheories weresound-like, soNewtonwasright;longitudinal wavescan'tbepolarized.) [Thepressure-sound wave theory through afluidcanonlysupport longitudinal compression anddoesnotsupport transverse polarization. Since light wasshowing twopolarization states inthisspar,thatruled outsimple pressure wave inaether idea.] 1728-JamesBradleyshowsthattheorbitalmotionoftheearthchangestheapparentmotionsofthe stars inaway th: isconsistent with light having afinite speed oftravel. (speed oflight) 1729~Stephen Gray shows thatelectricity doesn't havetobemade inplacebyrubbing butcanalsobetransferred from placetoplacewithconducting wires. Healsoshows thatthecharge onelectrified objects resides ontheirsurfaces. [Still inthestatic world only.] 1733 -Charles Francois duFaydiscovers thatelectricity comes intwokinds which hecalled resinous(-) and vitreous(+). 1742 -Thomas LeSeur andFrancis Jacquier, inanotetotheedition ofNewton's Principia thattheypublish, show that theforce lawbetween twomagnets isinverse cube. 1749 -Abbe Jean-Antoine Nollet invents thetwo-fluid theory electricity. 1745 -Pieter vanMusschenbroek invents theLeyden jar,orcapacitor, andnearly killshisfriend Cunaeus. 1747~Benjamin Franklin inventsthetheoryofone-fluid electricity inwhichoneofNollet’s fluids exists and the other isjusttheabsenceofthefirst. Heproposes theprinciple ofconservation ofcharge andcallsthefluidthatexists andflows ~positive’. Thiseducated guessensures thatundergraduates willalways beconfused aboutthedirection ofcurrent flow. Healsodiscovers thatelectricity canactatadistance insituations where fluid flowmakes nosense. 1748-SirWilliamWatsonusesanelectrostaticmachineandavacuumpumptomakethefirstglowdischarge.His~~} vessel isthree feetlongandthree inches indiameter: thefirstfluorescent lightbulb. 2 TABI xe, =Ohare NN? 1750 -John Michell discovers thatthetwopoles ofamagnetareequalinstrengthandthattheforcelawforindividual poles isinverse square. . Bi aos Ou~JohannSulzerputsleadandsilvertogetherinhismouth,performing thefirstrecorded“tonguetest"ofabattery. 1759 -Francis Ulrich Theodore Aepinus shows thatelectrical effects areacombination offluid flow confined tomatter andaction atadistance. Healsodiscovers charging byinduction. 1762 -Canton reports thataredhotpoker placed close toasmall electrified body destroys itselectrification. 1764 -Joseph Louis Lagrange discovers thedivergence theorem inconnection with thestudy ofgravitation. Itlater becomes known asGauss's law. (See 1813). 1766 -Joseph Priestly, acting onasuggestion inaletter from Benjamin Franklin, shows thathollow charged vessels contain nocharge ontheinside andbased onhisknowledge thathollow shells ofmass have nogravity inside correctly deduces thattheelectric forcelawisinverse square. * ca1775 -Henry Cavendish invents theidea ofcapacitance andresistance (the latter without anyway ofmeasuring current other than thelevel ofpersonal discomfort). Butbeing indifferent tofame heiscontent towait forhiswork tobe published byLord Kelvin in1879. 1777-JosephLouisLagrange inventstheconceptofthe scalar potential forgravitational fields, 1780 -Luigi Galvani causes dead frog legstotwitch with static electricity, then alsodiscovers thatthesame twitching eo;becausedbycontactwithdissimilarmetals.Hisfollowersinventanotherinvisiblefluid,thatof“animalelectricity", Hodescribe thiseffect. [Hereally thinks animals arecreating theelectricity somehow, notjustresponding toit.] 1782 -Pierre Simon Laplace shows thatLagrange's potential satisfies V?V=0. 1785 -Charles Augustin Coulomb usesatorsion balance toverify thattheelectric force lawisinverse square. Healso proposes acombined fluid/action-at-a-distance theory likethat ofAepinus butwith twoconducting fluids instead ofone. Fighting breaks outbetween single anddouble fluid partisans. Healsodiscovers thattheelectric force near aconductor is proportional toitssurface charge density andmakes contributions tothetwo-fluid theory ofmagnetism. 1793 -Alessandro Volta makes thefirst batteries andargues thatanimal electricity isjust ordinary electricity flowing through thefrog legs under theimpetus oftheforce produced bythecontact ofdissimilar metals. Hediscovers the importance of““completing thecircuit." The period 1800-1825. --things heat upafter Volta invents thebattery socurrents may beproduced andstudied 1800 -In1800 Volta discovers theVoltaic pile(dissimilar metals separated bywetcardboard) which greatly increases themagnitude ofthe effect. 1800 -William Nicholson andAnthony Carlisle discover thatwater may beseparated intohydrogen andoxygen bythe action ofVolta's pile. O60:-ThomasYounggivesatheoryofNewton'sringsbasedonconstructiveanddestructiveinterferenceofwaves.He explains thedark spot inthemiddle byproposing thatthere isaphase shift onreflection between alessdense andmore 3 densemedium, thenusesessenceofsassafras (whoseindexofrefraction isintermediate betweenthoseofcrownandflintglass)togetalightspotatthecenter.[diffraction ] |1803-ThomasYoungexplainsthefringesattheedgesofshadowsbymeansofthewavetheoryoflight.The ) theory begins itsascendance, buthasoneimportant difficulty: light isthought ofasalongitudinal wave, which makes it difficult toexplain double refraction effects incertain crystals. [scalar theory bypasses polarization question] 1807 -Humphrey Davy shows thattheessential element ofVolta’s pileischemical action since pure water gives no effect. Heargues thatchemical effects areelectrical innature. 1808 -Laplace gives anexplanation ofdouble refraction using theparticle theory, which Young attacks asimprobable. 1808 -Etienne Louis Malus, amilitary engineer, enters aprize competition sponsored bytheFrench Academy “To furnish amathematical theory ofdouble refraction, andtoconfirm itbyexperiment." Hediscovers that light reflected at certain angles from transparent substances aswell astheseparate raysfrom adouble-refracting crystal have thesame property ofpolarization. In1810 hereceives theprize andemboldens theproponents oftheparticle theory oflightbecausenooneseeshowawavetheorycanmakewavesofdifferent polarizations. 1811 -Arago shows thatsome crystals alter thepolarization oflight passing through them. [rotation! ] |1812-BiotshowsthatArago'scrystalsrotatetheplaneofpolarization about thepropagation direction. 1812 -Simeon Denis Poisson further develops thetwo-fluid theory ofelectricity, showing thatthecharge onconductors must reside ontheir surfaces and besodistributed that theelectric force within theconductor vanishes. This surface chargedensitycalculation iscarriedoutindetailforellipsoids. Healsoshowsthatthepotentialwithinadistribution °electricity satisfies theequation V?@=-p/eo. 1812 -Michael Faraday, abookbinders apprentice, writes toSirHumphrey Davy asking forajobasascientific assistant. Davy interviews Faraday andfinds thathehaseducated himself byreading thebooks hewassupposed tobe binding. Hegets thejob. ca.1813 -Laplace shows thatatthesurface ofaconductor theelectric force isperpendicular tothesurface andthatE= leo, 1813 -Karl Friedrich Gauss rediscovers thedivergence theorem ofLagrange. Itwilllater become known asGauss's law 1815 -David Brewster establishes hislawofcomplete polarization upon reflection ataspecial angle now known as Brewster's angle. Healsodiscovers thatinaddition touniaxial cystals there arealsobiaxial ones. Foruniaxial crystals |there isthefaintpossibility ofawave theory oflongitudinal-type, butthisappears tobeimpossible forbiaxial ones. | 1816-DavidBrewster inventsthekaleidoscope. 1816 -Francois Arago, anassociate ofAugustin Fresnel, visits Thomas Young anddescribes tohimaseries of experiments performed byFresnel andhimself which shows thatlight ofdiffering polarizations cannot interfere. Reflecting lateronthiscurious effect Young seesthatitcanbeexplained iflightistransverse instead oflongitudinal. Thisideaiscommunicated toFresnelin1818andheimmediately seeshowitclearsupmanyoftheremaining difficulties ofthewavetheory.Sixyearslatertheparticletheoryisdead. r) 4 Sattarahea atiiaien ee 1817 -Augustin Fresnel annoys theFrench Academy: TheAcademy, hoping todestroy thewave theory once andfor all,proposes diffraction astheprize subject for1818. Tothechagrin oftheparticle-theory partisans intheAcademy the on: memoirin1818isthatofAugustin Fresttelwhéexplainsdiffraétion asthemutualinterferénce ofthesecondary wavesemittedbytheunblocked portionsoftheincident wave, inthestyle ofHuygens. Oneofthejudges from theparticlecampofthe Academy isPoisson, who points outthatifFresnel's theory were tobeindeed correct, then there should bea brightspotatthecenteroftheshadow ofacircular disc.This,hesuggests toFresnel, mustbetestedexperimentally. The experiment doesn't goasPoisson hopes,however, andthespotbecomes knownas‘Poisson's spot." 1820 -Hans Christian Oersted discovers thatelectric current in_a wire causes acompass needle toorient itself perpendicular tothewire. 1820 -Andre Marie Ampere, oneweek after hearing ofOcrsted’s discovery, shows thatparallel currents attract each otherandthatofDpositecurrents attract. 1820 -Jean-Baptiste Biot andFelix Savart show thatthemagnetic force exerted onamagnetic pole by awirefallso_like=randisoriented perpendicular tothe’wire. Whittaker thensaysthat\Thisresultwas soon further analyzed," toobtain Idsxr eB.St 1820 -John Herschel shows thatquartz samples thatrotate theplane ofpolarization oflight inopposite directions have different crystalline forms. This difference ishelical innature. eo~FaradaybeginselectricalworkbyrepeatingOersted'sexperiments. 1821 -Humphrey Davy shows thatdirect current iscarried throughout thevolume of@conductor andestablishes that : : Ras a forlong wires. Healso discovers thatresistance isincreased asthetemperature rises. 1822 -Thomas Johann Seebeck discovers thethermoelectric effect byshowing thatacurrent willflow inacircuit made ofdissimilar metals ifthere isatemperature difference between themetals. 1824 -Poisson invents theconcept ofthemagnetic scalar potential andofsurface andvolume pole densities described bytheformulas r=) vem Men’ va-/[™.dv’=[Raw -[Se { je-rP fer] Jfr-r] .Healso finds themagnetic field inside aspherical cavity within magnetized material. 1825 -Ampere publishes hiscollected results onmagnetism. Hisexpression forthemagnetic field produced byasmall segment ofcurrent isdifferent from thatwhich follows naturally from theBiot-Savart lawbyanadditive term which or tozeroaroundclosedcircuit[wasthisthetimeterminthecurlBequation? ]Itisunfortunate thatelectrodynamics andrelativity decide infavor ofBiot andSavart rather than forthemuch more sophisticated Ampere, 5 whose memoir contains both mathematical analysis andexperimentation, artfully blended together. Inthismemoir are given some special instances oftheresult wenow callStokes theorem. LfBeds=pol. r) Maxwell describes thisworkas“oneofthe most brilliant achievements inscience. Thewhole, theory andexperiment, seemsasifithadleaped, full-grown andfull-armed, fromthebrainofthe"Newton ofelectricity’. Itisperfect informand unassailable inaccuracy; anditissummed upinaformula from which allthephenomena may bededuced, andwhich must always remain thecardinal formula ofelectrodynamics." ‘The period 1825-1850 1825 -Fresnel shows thatcombinations ofwaves ofopposite circular polarization traveling atdifferent speeds canaccountfortherotationofthe plane ofpolarization. 1826-GeorgSimon Ohmestablishes theresultnowknown asOhm's law.V=IRseemsaprettysimple lawtonameafter someone, buttheimportance ofOhm's work does notlieinthissimple proportionality. What Ohm didwas develop the ideaofvoltage asthedriverofelectric current. Hereasoned bymaking ananalogy between Fourier’s theoryofheatflow andelectricity. Inhisscheme temperature andvoltage correspond asdoheat flow andelectrical current. [twasnotuntil some years later that Ohm's electroscopic force (Vinhislaw) andPoisson's electrostatic potential were shown tobe identical. [again,thenewworldofDCcurrents frombatteries isnotquitelinkedtothestaticelectricity world] 1827 -Augustin Fresnel publishes adecade ofresearch inthewave theory oflight. Included inthese collected papers areexplanations ofdiffraction effects, polarization effects, double refraction, andFresnel's sine andtangent laws for,reflectionattheinterfacebetweentwotransparent media. [) 1827 -Claude Louis Marie Henri Navier publishes thecorrect equations forvibratory motions inonetype ofelastic solid. This begins thequest foradetailed mathematical theory oftheacther based ontheequations ofcontinuum mechanics. 1827 -F.Savery, after noticing thatthecurrent from aLeyden jarmagnetizes needles inalternating layers, conjectures thattheelectric motion duringthedischarge consists ofaseriesofoscillations, 1828 -George Green generalizes andextends thework ofLagrange, Laplace, andPoisson andattaches thename potential totheir scalar function. Green's theorems aregiven, aswell asthedivergence theorem (Gauss's law), butGreendoesn'tknowoftheworkofLagrange andGaussandonlyreferences Priestly's deduction ofthe inverse square law from Franklin's experimental work onthecharging ofhollow vessels, 1828-Augustine LouisCauchy presents atheorysimilartoNavier's, butbasedonadirectstudyofelasticproperties rather than using amolecular hypothesis. These equations aremore general thanNavier’s. InCauchy's theory, andin muchofwhatfollows, theaetherissupposed tohavethesameinertiaineachmedium, butdifferent elastic properties. 1828 -Poisson shows that theequations ofNavier andCauchy have wave solutions oftwo types: transverse and longitudinal. Mathematical physicists spend thenext50years trying toinvent anelasticaether forwhich thelongitudinal waves areabsent. [Bythistime, itwasknown thatsolids cansupport transverse polarizations, soitprobably seemed only amatter oftime until thenature oftheaether assome kind ofsolid was elucidated. Butwhy then didtheether nsupportlongitudinal polarization, asallsolidsdo?] e 6 SET Renee 1831 -Faraday shows that[thetransformer }changing currents inonecircuit induce currents inaneighboring circuit. Over thenext several years heperforms hundreds ofexperiments andshows thatthey canallbeexplained bytheideaof oz magneticflux.Nomathematics isinvolved,justpictutethinking.tusing hisficld-lines.[+ 1831 -Ostrogradsky rediscovers thedivergence theorem ofLagrange, Gauss, andGreen. 1832 -Joseph Henry independently discovers induced currents. 1833 -Faraday begins work ontherelation ofelectricity tochemistry. Inoneofhisnotebooks heconcludes after a series ofexperiments, “...there isacertain absolute quantity oftheelectric power associated with each atom ofmatter.” 1834-Faraday discovers selfinductance 1834 -Jean Charles Peltier discovers theflipside ofSeebeck’s thermoelectric effect. Hefinds thatcurrent driven ina circuit made ofdissimilar metals causes thedifferent metals tobeatdifferent temperatures. > 1834 -Emil Lenz. formulates hisrulefordetermining thedirection ofFaraday’s induced currents. Initsoriginal form it wasaforce lawrather than aninduced emflaw: “Induced currents flow insuch adirection astoproduce magnetic forces thattrytokeep themagnetic flux thesame." SoLenz would predict that ifyoutrytopush aconductor intoastrong magnetic field, itwillberepelled. Hewould alsopredict thatifyoutrytopullaconductor outofastrong magnetic field thatthemagnetic forces ontheinduced currents willoppose thepull. 1835 -James MacCullagh andFranz Neumann extend Cauchy's theory tocrystalline media oe”-Faradaydiscoverstheideaofthedielectricconstant, 1837 -George Green attacks theclastic aether problem from anew angle. Instead ofderiving boundary conditions between different media byfinding which onesgiveagréefnent withtheexperimental lawsofoptics, hederives the correct boundary conditions from general dynamical principles. This advance makes theelastic theories notquite fitwith light. 1838 -Faraday shows thattheeffects ofinduced electricity ininsulators areanalogous toinduced magnetism in magnetic materials. Those more mathematically inclined immediately appropriate Poisson's theory ofinduced magnetism, inventing P,D,ande. 1838 -Faraday discovers Faraday's dark space, adark region inaglow discharge nearthenegative electrode. 1839 -James MacCullagh invents anelastic aether inwhich there arenolongitudinal waves. Inthisaether thepotential energy ofdeformation depends only ontherotation ofthevolume elements andnotontheir compression orgeneral distortion. This theory gives thesame wave equation asthatsatisfied byEand BinMaxwell's theory. [interesting...] 1839-WilliamThomson (LordKelvin)removessomeofthe objections toMacCullagh’s rotation theory byinventing a mechanical model which satisfies MacCullagh's energy ofrotation hypothesis. Ithasspheres, rigid bars, sliding contacts, and flywheels. [Itderives itsname from thetitle, Baron Kelvin ofLargs, that Thomson received from theBritish government in1892,andnamed afterThomson because ofhisproposal inthis1848paper. SirWilliam waslaternamed . Lord Kelvin when hewas made aBaron andamember ofthieHouse ofLords. Headopted thename Kelvin from the eo: KelvininScotland withdueregardtotheassociation hehadwithGlasgow University andthearea.] 7 1839 -Cauchy andGreen present more refined elastic aether theories, Cauchy's removing thelongitudinal waves by postulating anegativecompressibility, andGreen'susinganinvolveddescription ofcrystalline solids. 1841-MichaelFaraday iscompletely exhausted byhiseffortsoftheprevious2decades,soherestsfor4years. e 1841 -James Prescott Joule shows thatenergy isconserved inelectrical circuits involving current flow, thermal heating, and chemical transformations. 1842 -F.Neumann andMatthew O’Brien suggest thatoptical properties inmaterials arise from differences intheamount offorcethattheparticles ofmatterexertontheaetherasitflowsaroundandbetweenthem. 1842 -Julius Robert Mayer asserts thatheatandwork areequivalent. Hispaper isrejected byAnnalen derPhysik. 1842-JosephHenryrediscovers theresultofF.Saveryabouttheoscillation oftheelectriccurrentinacapacitivedischarge andstates,“Thephenomena requireustoadmittheexistence ofaprincipaldischargeinonedirection,andthen several reflex actions backward and forward, each more feeble than thepreceding, until equilibrium isrestored." 1842-ChristianDopplergivesthetheoryofthe Doppler effect. 1845 -Faraday quits resting anddiscovers thattheplane ofpolarization oflightisrotated when ittravels inglass alongthedirectionofthemagnetic linesofforceproducedbyanelectromagnet (Faradayrotation). 1845 -Franz Neumann uses (i)Lenz’s law, (ii)theassumption thattheinducedemfisproportional tothemagnetic force onacurrent element, and(iii)Ampere's analysis todeduce Faraday's law.Intheprocess hefinds apotential function from whichtheinducedelectricfieldcanbeobtained,namelythevectorpotentialA(intheCoulombgauge),thussiscoveringgy the result which Maxwell wrote as E=-Vo~-dA/at. 1846 -George Airy modifies MacCullagh's elastic acther theorytoaccountforFaradayrotation. 1846 -Faraday, inspired byhisdiscovery ofthemagnetic rotation oflight, writes ashort paper speculating thatlight might beelectro-magnetic innature. Hethinks itmight betransverse vibrations ofhisbeloved field lines. [asopposed to awave traveling inanether? } 1846 -Faraday discovers diamagnetism. Heseestheeffect inheavy glass, bismuth, andother materials. 1846 -Wilhelm Weber combines Ampere's analysis, Faraday’s experiments, andtheassumption ofFechner thatcurrents consist ofequal amounts ofpositive andnegative electricity moving opposite toeach other atthesame speed toderive an electromagnetic theory based onforces between moving charged particles. This theory hasavelocity-dependent potential energy andiswrong, butitstimulates much work onelectromagnetic theory which eventually leads tothework of ‘Maxwell_and Lorenz. Italso inspires anew look atgravitation byWilliam Thomson toseoifavelocity-dependent correction tothegravitational energy could account fortheprecession ofMercury's perihelion. 1846 -William Thomson shows thatNeumann's electromagnetic potential isinfactthevector potential from which B may beobtained via B=VxA. 1847-Weberproposesthatdiamagnetism isjustFaraday'slawactingonmolecularcircuits.InansweringtheoreoAD that this would mean that everything should bediamagnetic hecorrectly guesses that diamagnetism ismasked in 8 Bee ee NG paramagnetic andferromagnetic materials because they have relatively strong permanent molecular currents. This work ridstheworld ofmagnetic fluids, @.., -HermannvonHelmboltz writesamemoir“OntheConservation ofForce"whichemphatically statesthe principle ofconservation ofenergy: “Conservation ofenergy isauniversal principle ofnature. Kinetic andpotential energy ofdynamical systems maybeconverted intoheataccording todefinite quantitative lawsastaught byRumford, Mayer, andJoule. Any ofthese forms ofenergy may beconverted intochemical, electrostatic, voltaic, andmagnetic forins." Hereads itbefore thePhysical Society ofBerlin whose older members regard itastoospeculative andreject itfor publication inAnnalen derPhysik. 1848-9-GustavKirchoff [misspelled] extendsOhm'sworktoconduction inthreedimensions, giveshislawsforcircuit networks, andfinally shows that Ohm's “electroscopic force" which drives current through resistors andtheold electrostatic potential ofLagrange, Laplace, andPoisson arethesame. Healso shows thatinsteady state electrical currents distribute themselves soastominimize theamount ofJoule heating. 1849 -A,Fizeau repeats Galileo's hilltop experiment (9kmseparation distance) witharapidly rotating toothed wheel. [ speedoflight]andmeasures c=3.15x10*m/s. 1849 -George Gabriel Stokes studies diffraction around opaque bodies both theoretically andexperimentally andshows thatthevibration ofaether particles areexecuted atright angles totheplane ofpolarization. Three years later hecomes to thesame conclusion byapplying aether theory tolight scattered from thesky. This result is,however, inconsistent with optics incrystals. eo period1850-1875 ca,1850-Stokesovercomes someofthedifficulties withcrystalsbyturningCauchy'shypothesis aroundandlettingthe elastic properties oftheaether bethesame inallmaterials, butallowing theinertia todiffer. This gives risetothe conceptual difficulty ofhaving theinertia bedifferent indifferent directions (inanisotropic crystals). ca.1850 -JeanFoucault improves onFizeau's measurement anduseshisapparatus toshow thatthespeed oflightisloss inwater than inair. 1850 -Stokes lawisstated without proof byLord Kelvin (William Thomson), Later Stokes assigns theproof ofthis theorem aspartoftheexamination fortheSmith's Prize. Presumably, heknows howtodotheproblem. Maxwell, who wasacandidate forthisprize, later remembers thisproblem, traces itback toStokes andcalls itStokes theorem. 1850 -William Thomsori (Lord Kelvin) invents theideaofmagnetic permeability andsusceptibility along withthe separate concepts ofB,M,andH. 1851 -Thomson gives ageneral theory ofthermoelectric phenomena, describing theeffects seen bySeebeck andPeltier. 1853 -Thomson usesPoisson's magnetic theory toderive thecorrect formula formagnetic energy: U=fwHPav/2. r)HealsogivestheformulaU=LI7/2andgivestheworldthepowerful,butconfusing,analysiswheretheforcesoncircuitsareobtained bytaking either thepositive ornegative gradient ofthemagnetic energy. Knowing which sign.to useis,of course, theconfusing part. 9 1853 -Thomson gives thetheory oftheRLC circuit providing amathematical description fortheobservations ofHenry andSavery. e 1854-Faraday clearsuptheproblem ofdisagreements inthemeasured speedsofsignalsalongtransmission linesbyshowingthatitiscrucialtoincludetheeffectofcapacitance. 1854 -Thomson, inaletter toStokes, gives theequation oftelegraphy ignoring theinductance: ,where Risthecable resistance andwhere Cisthecapacitance perunitlength. Since thisisthediffusion equation, thesignal does nottravel at definite speed. {transmission linetheory] 1855 -Faraday retires, living quietly inahouse provided bytheQueen until hisdeath in1867. {good work! ] 1855 ~James Clerk Maxwell writes amemoir inwhich heattempts tomarry Faraday's intuitive field line ideas with Thomson's mathematical analogies. Inthis memoir thephysical importance ofthedivergence andcurl operators for electromagnetism firstbecome evident. ‘Theequations V-(eB)=4xp, VxA=B,andVxH=4rJappear inthismemoir. 1857 -Gustav Kirehoff derives theequation oftelegraphy foranaerial coaxial cable where theinductance isimportant andderives thefulltelegraphy equation. 8°V/Ox* =LCHV/At +ROAV/dt. Herecognizes thatwhen theresistance issmall, thisisthewave equation withpropagation speed v=sqrt(LC), which for acoaxial cable turns outtobevery close tothespeed oflight. Kirchoff notices thecoincidence, andisthusthefirstdiscoverthatelectromagnetic signalscantravelatthespeedoflight. eo |1861-Bernhard Riemann develops avariantofWeber'selectromagnetic theorywhichisalsowrong. 1861, -Maxwell publishes amechanical model oftheelectromagnetic field. Magnetic fields correspond torotating vortices with idle wheels between them and electric fields correspond toelastic displacements, hence displacement currents. Theequation forHnowbecomes VxH=4nJq, where Jiaisthetotalcurrent, conduction plusdisplacement, andisconserved VeJ=0.Thisadditionofadisplacement currentcompletes Maxwell's equations anditisnoweasyforhimto derive thewave equation exactly asdone inourtextbooks onelectromagnetism andtonote thatthespeed ofwave propagation wasclose tothemeasured speed oflight. Maxwell writes, “"We canscarcely avoid theinference thatlightinthetransverse undulations ofthe same medium which isthecause ofelectric andmagnetic phenomena."Thomson, onthe other hand, saysofthedisplacement current, ‘(itisa)curious andingenious, butnotwholly tenable hypothesis.” 1864 -Maxwell reads amemoir before theRoyal Society inwhich themechanical model isstripped away andjustthe equations remain. Healsodiscusses thevector andscalar potentials, using theCoulomb gauge. Heattributes physical significance tobothofthesepotentials. Hewants topresent thepredictions ofhistheory onthesubjects ofreflection and refraction, buttherequirements ofhismechanical mode! keep himfrom finding thecorrect boundary conditions, sohe never does this calculation. 1867 -Stokes performs experiments thatkillhisownanisotropic inertia theory. 1867-JosephBoussinesq suggests thatinsteadofaether beingdifferent indifferent media,perhaps theaetherisi)same everywhere, butitinteracts differently withdifferent materials, similartothemodemelectromagnetic wavetheory. 10 6Aie eh ere P 1867-Riemann proposesasimpleelectrictheoryoflight inwhich Poisson's equation isreplaced by e WV=(1/c?)0°V £00?2dep. oF 1867 -Ludwig Lorenz develops anelectromagnetic theory oflightinwhich thescalar andvector potentials, inretarded form, arethestarting point. Heshows that these retarded potentials each satisfy thewave equation andMaxwell's ‘equations forthefields and canbederived from hispotentials. Hisvector potential does notobey theCoulomb gauge, however, butanother relation nowknown astheLorenz gauge. Although heisabletoderive Maxwell's equations from hisretarded potentials, hedoes notsubscribe toMaxwell's view thatlight involves electromagnetic waves intheaether. Hefeels,rather, thatthefundamental basisofallluminous vibrations iselectric currents, arguing thatspacehasenough Inatter inittosupport thenecessary currents. 1868 -Maxwell decides thatgiving physical significance tothescalar andvector potentials isabadidea andbases his further workonlightonEandH.[Au] es 1869 -Maxwell presents thefirst calculation inwhich adispersive medium ismade upofatoms with natural frequencies. This makes possible detailed modeling ofdispersion with refractive indices having resonant denominators. 1869-Hittorffindsthatcathoderayscancastashadow. 1870 -Helmholtz derives thecorrect laws ofreflection andrefraction from Maxwell's equations byusing thefollowing boundary conditions: Dn,Et,andBarecontinuous.. Once these boundary conditions aretaken Maxwell's theory isjusta Tepeat ofMacCullagh's theory. The details were notgiven byHelmholtz himself, butappear rather intheinaugural dissertation ofH.A.Lorentz. 1870-1900 -Thehunt isonforphysical models oftheaether which arenatural andfrom which Maxwell's equations canbederived.Thephysicists whoworkonthisproblemincludeMaxwell, Thomson, Kirchoff,Bjerknes, Leahy,FitzGerald, Helmholtz, and Hicks. 1872 -E.Mascart looks forthemotion oftheearth through theaether bymeasuring therotation oftheplane of polarization oflight propagated along theaxisofaquartz crystal. Nomotion isfound withasensitivity ofv/c=10°. 1873 -Maxwell publishes hisTreatise onElectricity andMagnetism, which discusses everything known atthetime about electromagnetism from theviewpoint ofFaraday. Hisown theory isnotvery thoroughly discussed, buthedoes introduce hiselectromagnetic stress tensor inthiswork, including theaccompanying ideaofelectromagnetic momentum. The period 1875-1900 1875 -John Kerr shows that ordinary dielectrics subjected tostrong electric fields become double refracting, showing. directly thatelectric fields andlight areclosely related. [Kerr effect, anelectric Faraday rotation effect} 1876-HenryRowland performs anexperiment inspiredbyHelmholtz, whichshowsforthefirsttimethatmoving electric charge isthesame thing asanelectric current, 1876-A.Bartoliinfersthenecessity oflightpressurefromthermalarguments,thusbeginningtheexplorationofthe eoasectionbetween electromagnetism andthermodynamics. IL 1879 -J.Stefan’discovers theStefan-Boltzmann law,i.e.,thatradiant emission isproportional toT‘. 1879 -Edwin Halll performs anexperiment thathadbeen suggested byHenry Rowland anddiscovers theHall effec including itstheoretical description bymeansoftheHallterminOhm'slaw. @ 1879 -SirWilliam Crookes invents theradiometer andstudies theinteraction ofbeams ofcathode rayparticles in vacuum tubes. 1879 -Ludwig Boltzmann uses Hall's result toestimate thespeed ofcharge carriers (assuming thatcharge carriers are onlyofonesign.) 1880 -Rowland shows thatFaraday rotation canbeobtained bycombining Maxwell's equations andtheHall term in ‘Ohm's law, assuming thatdisplacement currents areaffected inthesame way asconduction currents. 1881-J.J.Thomson attempts toverifytheexistence ofthedisplacement currentbylookingformagnetic effects produced bythechanging electric field made byamoving charged sphere. 1881 -George Fitz Gerald points outthatJ.J.Thomson's analysis isincorrect because heleftouttheeffects ofthe conduction current ofthemoving sphere. Including both currents makes theseparate effect ofthedisplacement current disappear. |1881-Helmholtz, inalectureinLondon,pointsoutthattheideaofchargedparticlesinatomscanbeconsistent with Maxwell's andFaraday’s ideas, helping topave thewayforourmodem picture ofparticles andficlds interacting instead ofthinking about everything asadisturbance oftheaether, aswaspopular after Maxwell. 1881-AlbertMichelsonandEdwinMorleyattempttomeasurethemotionoftheearththroughtheaetherbysin interferometry. They find norelative velocity. Michelson interprets thisresult assupporting Stokes hypothesis inwhich theactherintheneighborhood oftheearthmovesattheearth'svelocity. 1883 -Fitz Gerald proposes testing Maxwell's theory byusing oscillating currents inwhat wewould now calla magnetic dipole antenna (loop ofwire). Heperforms theanalysis anddiscovers thatvery high frequencies arerequired to make thetest.Later thatyearheproposes obtaining therequired highfrequencies bydischarging acapacitor intoacircuit. 1883-5 -Horace Lamb and Oliver ‘Heaviside analyze the interaction ofoscillating electromagnetic fields with conductors anddiscover theeffect ofskindepth. 1884 -John Poynting shows thatMaxwell's equations predict thatenergy flows through empty space with theenergy fluxgiven byExB/4x. Healso investigates energy flow inFaraday fashion byassigning energy tomoving tubes of electric andmagnetic flux. 1884 -Heinrich Hertz asserts that_E made bycharges and_made byachanging magnetic field areidentical. Workingfromdynamical ideasbasedonthisassumption andsomeofMaxwell's equations, Hertzisabletoderivetherestofthem. 1887 -Svante Arrhenius deduces that indilute solutions electrolytes arecompletely dissociated into positive and negative ions. 1887-Hertzfindsthatultraviolet lightfallingonthenegativeelectrodeinasparkgapfacilitatesconduction bythe%®@inthegap. | 12 OS oe~ =e 1888 -R.T.Glazebrook revives oneofCauchy's wave theories andcombines itwith Stokes anisotropic aether inertia theory togetagreement with theexperiments ofStokes in1867. Oss +Hertzdiscovers thatoscillating sparkscanbeproducedinanopensecondary circuitifthefrequency ofthe primary isresonant withthesecondary. Heusesthisradiator toshow thatelectrical signals arepropagated along wires and through theairatabout thesame speed, both about thespeed oflight. Healsoshows thathiselectric radiations, when passed through aslitinascreen, exhibit diffraction effects. Polarization effects using agrating ofparallel metal wires are alsoobserved. [inother words, radio waves dothesame things thatlight waves do] 1888 -Roentgen shows thatwhen anuncharged dielectric ismoved atright angles toamagnetic field isproduced. 7? 1889 -Hertz gives thetheory ofradiation from hisoscillating spark gap. 1889 -Oliver Heaviside finds thecorrect form fortheelectric andmagnetic fields ofamovingchargedparticle,validfor allspeeds v<e. 1889 -J.J.Thomson shows thatCanton's effect (1762) inwhich aredhotpoker canneutralize theelectrification ofa small charged body isduetoelectron emission causing theairbetween thepoker andthebody tobecome conducting. 1890 -Fitz Gerald uses theretarded potentials ofL.Lorenz,tocalculateelectricdipoleradiation fromHertz'sradiator. 1892 -Oliver Lodge performs experiments onthepropagation oflight near rapidly moving steel disks totestStokes hypothesis thatmoving matter drags theaether with it.Nosuch effect isobserved. eo”-HendrikAntonLorentz,presentshiselectrontheoryofelectrified matter andtheaether. This theory combines Maxwell's equations, with thesource terms pandJ,with theLorentz force lawfortheacceleration ofcharged particles: ma=qE+q vxB.Lorentz's acther issimply space endowed with certain dynamical properties. Lorentz. gives themodern theory ofdielectrics involving DandP,andalsoincludes theeffect ofmagnetized matter. Healsogives what wenow calltheDrude-Lorentz harmonic oscillator model oftheindex ofrefraction. ButLorentz's theory hasa“stationary ether", which conflicts with thenegative Michelson-Morley result. 1892-GeorgeFitz.Geraldproposes lengthcontraction asawaytoreconcile Lorentz’s theoryandthenullresultsonthemotionoftheearththroughtheaether.Attheendofthis year Lorentz. endorses thisidea. 1894 -J.J,Thomson measures thespeed ofcathode rays andshows thatthey travel much more slowly than thespeed oflight.Theaethermodelofcathode rays begins todie. 1894 -Philip Lenard studies thepenetration ofcathode raysthrough matter. 1895-Pierre Curie experimentally discovers Curie's lawforparamagnetism andalsoshows thatthereisnotemperature effect fordiamagnetism. 1895 -Lorentz, inhis“Search foratheory ofelectrical andoptical effects inmoving bodies" gives theLorentz transformation tofirst order inv/c. The transformed time variable hecalls *"local time! 1895 -Wilhelm Roentgen discovers X-rays produced bybremsstrahlung incathode raytubes. @.,..~ArthurShuster,EmilWiechert,andGeorgeStokesproposethatX-raysareaetherwavesofexceedinglysmall wavelength. [sonow wehave light waves, radio waves, andX-rays allbeing thesame thing] 13 1896 -J.J. Thomson discovers thatmaterials through which X-rays pass arerendered conducting. 1896-HenriBecquerel discoversthatsomesortofnaturalradiationfromuraniumsaltscanexposeaphotographic =) ‘wrapped inthick black paper. 1896 -P,Zeeman discovers thesplitting ofatomic linespectra byamagnetic field. 1896 -Lorentz givesanelectrontheoryoftheZeemaneffect. 1897 -J.J.Thomson argues thatcathode rays must becharged particles smaller insizethan atoms (EmilWiechert made thesame suggestion independently inthissame year). Inresponse Fitz Gerald suggests that “we aredealing with free electrons inthese cathode rays." 1897 -W.Wien discovers that positively-charged moving particles canalso bemade (the so-called canal rays ofE. Goldstein) andthatthey have amuch smaller q/mratio than cathode rays. 1897 -J.J. Thomson deflects cathode rays bycrossed electric andmagnetic fields andmeasures e/m. 1898 -Marie andPierre Curie separate from pitchblende two highly radioactive elements which they name polonium and radium, 1899 -Emest Rutherford discovers that therays from uranium come intwo types, which hecalls alpha and beta radiation, 1900-MaricandPierreCurieshowthatbetaraysandcathoderaysareidentical. e@ The period 1900-1925 1900. -Emil Wiechert shows that simply replacing thedistributed charge from Lorentz’s theory with thecharge ofa movingpointparticlegivesincorrect results,InsteadtheLienard-Wiechertretardedpotentials mustbeused. 1900 -Joseph Larmor obtains thesecond order corrections totheLorentz. Transformation, 1901-R.Blondlot performs experiments thatshowthatLorentz’stheoryinwhichthereisnomovingaethergivesthe correct result incases where thehypothesis ofamovingaethergivesthewrongresult. 1902-LordRayleigh performs experiments totestwhethertheFitzGeraldcontraction iscapableofcausingdouble refraction inmoving transparent substances. Nosuch effect isfound. 1903 -TheHagen-Rubens connections between theconduetivity ofmetals andtheir optical properties areexperimentally established. 1903 -Lorentz gives thefamous square root formulas fortheLorentz. transformation giving theeffect toallorders in vie. 1904-Lorentzgiveshiselectron-collision theoryofelectricalconduction e 4 ee o_— 1905 -H.A.Wilson performs experiments similar tothose ofBlondlot; again, Lorentz's theory isfound togive the correct result, @...;-AlbertEinsteincompletesLorentz'sworkonspace-timetransformations andrelativityisborn.[thereisnoether, sothere isnospecial frame inwhich theether isatrest,soallframes areequivalent andrelative ] Index Aepinus, Francis Ulrich Theodore -1759 Airy, George -1846 Ampere, Andre Marie -1820, 1825 Arago,Francois -1811,1816 Arthenius, Svante -1887 Bartoli, A.-1876 Becquerel, Henri -1896 e Biot, Jean-Baptiste -1812, 1820 Blondlot, R.-1901 Boltzmann, Ludwig -1879 Boussinesq, Joseph -1867 Bradley, James -1728 Brewster, David-1815,1816 Cabeo, Niccolo -1620 Canton -1762 Carlisle, Anthony -1800 eu Vincenzo -1630Cauchy, Augustine Louis -1828, 1839 Cavendish, Henry -1775 Coulomb, Charles Augustin -1785 Crookes, William -1879 Curie, Marie -1895, 1900 Curie, Pierre -1895, 1898, 1900 Davy, Humphrey -1807, 1821 Descartes, Rene-1638,1644 Doppler, Christian -1842 duFay, Charles Francois -1733 Einstein, Albert -1905 Faraday, Michael-1812,1821,1831,1833,1834,1837,1838,1841,1845,1846,1854,1855 Fermat, Pierre de-1657 Fitz Gerald, George -1881, 1883, 1890, 1892 Fizeau, A.-1849 Foucault, Jean -1850 Franklin, Benjamin -1747 Fresnel,Augustin -1817,1825,1827 Galileo -1638 Galvani, Luigi -1780 Gauss,KarlFriedrich-1813 @‘ites,William-1600Glazebrook, R.T.-1888 Gray, Stephen -1729 1s Green, George -1828, 1837, 1839 Grimaldi, Francesco Maria -1665 Hagen -1903Hall,Edwin-1879 @ Heaviside, Oliver -1883, 1889 Helmholtz, Hermann von -1847, 1870, 1881 Henry, Joseph -1832, 1842 Herschel, John -1820 Hertz, Heinrich -1884, 1887, 1888, 1889 Hittorf -1869 Hooke, Robert -1667 Huygens, Christiaan -1678 Jacquiér, Francis -1742 Joule, James Prescott -1841 Kerr, John -1875 Kirchoff, Gustav-1848,1857 Lagrange, Joseph Louis -1764, 1777 Lamb, Horace -1883 Laplace, Pierre Simon -1782, 1808, 1813 Larmor, Joseph -1900 LeSeur,Thomas -1742 Lenard, Philip -1894 Lenz,Emil -1834 Lodge, Oliver -1892 Lorentz,HendrikAnton-1892,1895,1896,1903,1904 }Lorenz, Ludwig -1867 MacCullagh, James -1835, 1839 Magnus -900BC Malus, Etienne Louis -1808 Mascart, E.-1872 Maxwell, James Clerk -1855, 1861, 1864, 1868, 1869, 1873 Mayer, Julius Robert -1842 Michell, John -1750 Michelson, Albert -1881 Morley, Edwin -1881 Musschenbroek, Pieter van -1745 Navier, Claude Louis Marie Henri -1827 ‘Neumann, Franz -1835, 1842, 1845 Newton, {sac -1671, 1717 Nicholson, William -1800 Nollet, Abbe Jean-Antoine -1749 O'Brien, Matthew -1842 Oersted, Hans Christian -1820 Ohm,Georg Simon -1826 Ostrogradsky -1831 Peltier, Jean Charles -1834 Peregrinus, Peytrus-1269 e@Poisson, Simeon Denis -1812, 1824, 1828 Poynting, John -1884 16 uo Priestly, Joseph -1766 Rayleigh, Lord -1902 or. Bernhard -1861,1867 Roemer, Olaf -1675 Roentgen, Wilhelm -1888, 1895 Rowland, Henry -1876, 1880 Rubens -1903 Rutherford, Emest -1899 Savart, Felix -1820 Savery, F.-1827 Seebeck, ThomasJohann-1822 Shuster, Arthur -1896 Stefan, J.-1879 Stokes, George Gabriel -1825, 1849, 1850, 1867, 1896 Sulzer, Johann -1752 Thales ofMiletos -600 BC . ‘Thomson, J.J.-1881, 1889, 1894, 1896, 1897 ‘Thomson, William (Lord Kelvin) -1839, 1846, 1850, 1851, 1853, 1854 Volta, Alessandro -1793 Watson,William-1748 |Weber, Wilhelm -1846, 1847 Wiechert, Emil -1896, 1900 Wien, W. -1897 Wilson,H.A.-1905 ow.‘Thomas -1801, 1803 ‘Zeeman, Pieter -1896 Next: Review Sheet Up:NoTitle Previous: Homework Assignments Ross Spencer ‘Tue Apr 1310:47:17 MDT 1999 7 John David Jackson - Classical Electrodynamics 3rd Ed. eJackson Chapter 1Notes PhL 1.28.03 Chapter 1:Introduction toElectrostatics This isapretty heavy-duty introduction! Itsetsuptheentire mechanism forcomputing potential with theDirichlet orNeumann boundary conditions using Green's Functions. 1.1 Coulomb's Law. 1.2 Electric Field. Ofapointcharge,thenmanypointcharges,thenintegraloverp.Mentionofthe "statcoulomb"withmksputinappendix. Reviewofdelta function technology. 1.3Gauss's Law. Proves thisforapoint charge intheintegral sense, result (1.11) 1.4Differential Form ofGauss's Law. Jackson justquotes "the divergence theorem" without comment. Atleast Portis proves it.Iremember being hitflatinthefacebythisstrange newtheorem. SoVeE =xp. 1.5The Potential. Defines ifforgeneral pbysaying E=-V@andthen $iswhat itis.Since Ehasthis form, weareguaranteed byvector identity tohave VxE=0. Relation towork. 6aslineintegral ofE. 1.6Surface charge candtheDipole Layer. Shows how tocompute itusing pillbox andget4no = eAEyom,Thedipolelayerisdefinedastwoolayersseparatedbydsuchthatod=Dasd—>0,allfunctionsofx.SoyoudealonlywiththeD(x)result.Now=~fDdQandAp=4xDacrossthelayer. 1.7. Poisson and Laplace Equations. Poisson haspontheright, Laplace does not. Finds thatGreen's isURsee(1.31). 1.8Green's Theorem Applied to4.Inthedivergence theorem setA=§Vy andyou getthe"first identity". Swap thefunctions, subtract, andyougetthe"second identity" weusually justcall"Green's Theorem" (1.35). Sowehave justderived thisfrom thedivergence theorem. Now set =I/Rand6=6 Note thatVy=413while V*}=4zp,andvoilal, outpopstheunderpinning equation ofelectrostatics (1.36). Ittells you how tofind thepotential given somepandsomeboundaryconditionsonaclosed surface. Interpretation isvery important andvery tricky here! Ifpointxisoutsidetheclosedsurface,thederivation showsthattheLHSof(1.36) iszero andsothe twoterms ontheright must exactly cancel. Inthiscase, youaregetting nostatement atallabout what isinside oroutside thesurface. Ifthesurface isjustamathematical onearound some charge distribution p,allyoucansayisthatthattherighttermsurfacetermwillcanceltheptermontheRHS.Nowthefirst term istheinfinite-surface solution forthepotential thatwearefamiliar with, duetocharges p.The second term canbeinterpreted asarising from amysterious distribution ofchargec anddipole layer D onthesurface thatmakes apotential which exactly cancels thefirstterm. Suppose weréally could put thisdistribution ofrealcharge and Donsuch asurface, gluing itinto free space somehow. Then ifwe went outside (now with aninfinite surface beyond us),wewould addthepotentials from theoriginalp e@term andthen from thereal surface oandDterms, these would cancel, andwewould conclude that6=0 1 everywhereoutsidethesurface.Wecanthenreversethissituationbysayingthefollowing.Supposewe ehave acavity ingrounded metal which matches oursurface. Weknow that§=0everywhere outside the surface! Then wecanconclude thattheoandDinthiscasereally exist ontheinside surface ofthecavity andthese realthings arewhat cancels thep-generated potential! {Iamnotquite sure what theDterm means, although Iknow what Dmeans; theoterm means realsurface charge. } Ifpointxisinsidetheclosedsurface,wegetanequationforwhichistheusualptermplusthe surface terms, Ifwehave amathematical surface only, which supports nocharge orD,then thesurface term willgive exactly 0(assuming there arenometal surfaces anywhere). Butifweareinametal cavity andwehave real¢andDonthemetal surface arranged asdescribed above sothey cancel ourppotential ontheoutside, then thisequation tellsushowtofirdthecompleted inside thecavity, andinthiscase the second term will NOT bezero. Wehave aseeming paradox here. Wewant thesurface terms tobe0ifwehave xontheinside ofa purely mathematical surface, butwewant thesurface terms tobenon-zero andexactly cancel thep term ifweareontheoutside ofthesurface, because inthiscase thetheorem says theLHS =0.Sohow canthe same surface terms be0inone case andnon-zero intheother case! ‘The. answer isthat thesurface term integral really isdifferent depending onwhether xisinside oroutside. Ididnotunderstand thisuntil Idid thefollowing example case, andthen itbecame clear. Example: Apointchargeqinthecenterofasphereofradius a.Inthiscase, r"istheintegration variable thatwanders overthesurface andwecanreplace da’witha’dQ’. Wechoose thez'axisalong the direction ofvector r.Wethenreplace da’with2a’dx’where x'=cos6'. Here aresome intermediate steps incomputing thesurface integrals: e /an'=d/ér Ra+7-2arcos! =A+Bx' Asa?+?B=-2ar and @/én'(1/R) =-(a-rx'YR° atthesurface Weknow with apurely math surface that =q/r’atthesurface, so 8/6n'() =-q/a? atthesurface Sothecomplete surfacebusinessisthis: (1/41)2a?Sort Cala?YR.-(q/a)l-(a=rx'YR?]} where ofcourse R=R(x’), andintegrate -1to1.There arethree terms tointegrate here. Ifweplug this integral symbolically intoMaple, wegetapretty bigmesswhich involves\fA¥B and«/A-B. Weknow thatA>B,sothatisnottheproblem. This iswhere thesituation a>rora>rmakes ahuge difference. We have: [ANB =[1 =sign(a-r)*(a-t) =Ja] //critical thingissign(a-r) | AB =V@Hy =(@H) 1noconfusion here 2 Whenr<a(meaningourpointxisinsidethesurface),theentiresurfaceintegrationgives0,showingus ethatouranswer isjust =q/rfrom thepterm. Ontheother hand, when r>a(meaning weareonthe outside), thenthebigmess comes outbeing -q/tandexactly cancels thepterm! Themain point istounderstand that, although thesurface integral "looks thesame" inboth thex inside andxoutside cases, itisnotthesame! Note: Regarding theintegrand VA+Bx', ithasabranch cuttotheleftof-1which liesoutside our integration range. Weareworking ontherealanalytic sheet where wetake the+square root, theother sheethasthe-square root,wenevergothere. When wehavetodealwiththeendpoint value/(a-r)’, we must therefore take thepositive square rootwhich is|a-r|. There isnomoretoitthanthat! Thissamesituation occurs inamoretrivialsituation. Imagine computing thepotential insideasphere ofuniformsurfacechargeo.Wedoexactlyasaboveandwegetfdx’/Randwegetananswerthat looks like {la-r| -(atr)}/2ar. Ifr>a, wegetaI/ranswer, andifr<awegetaconstant I/aanswer --the potential isconstant inside thesphere. Atexactly 0both answers arethesame. Now wecome toanother main point. Ingeneral, forxinside, weshould just regard (1.36) asa complicated "integral equation" thatwehave tosolve ford. Butwelookattheequation andwonder: isit possible that wecantake some forced values for$and4’onthesurface, andthenusetheequation to compute $everywhere inside (also knowing pofcourse)? Ifso,thatwould beapowerful tool, andthisis thesubjectofthenextsection. 1.9.Comments onSolutions intheDirichlet, Neumann andCauchy cases. Itturns outthatifyoutry @ toforceboth$and¢'onaclosedsurface(Cauchy), youhave"overspecified” theproblemandthereisnosolution generally speaking. However, ifyouspecify either}(Dirichlet, theusualcase)or6'(Neumann)ontheclosedsurface, theninfactthisdoescompletely determine @inside,butyouhavetodosomework tofind thesolution, which wewill dobelow. Atableonpage17summarizes things,and]putnowsomeoldernotesrighthere. Instatics, youhave thewave-equation with k=0(A=«)with pasthedriving source [known as Poisson], asin(1.28), andthegeneral solution is(1.17) where 1/RistheGreen's function orpropagator (think exp(ikR)/R withk=0). This isthe"particular solution", andyoucanaddsolutions toV*=0[ known asLaplace ],asin(1.36), inorder tofindacomplete solution thatmatches séme boundary conditions. Thetwoboundary surface terms canbeinterpreted asfrom aneffective surface charge and "dipole layer", page 15.[scediscussion above! ] Intheboundary conditions, youareeitherspecifying thenormalEfieldatthesurface (Neumann), oryouspecify@itself(Dirichlet). Ifyouspecifyeitheroneonaclosedboundary, yougetauniqueand correct solution, butifyoutrytospecify both (Cauchy) ,itis"too much” andtheonly solution is0. Specifying anyofthese three onanopen (partial) surface is“notenough" togetasolution. This isshown inthefirst column ofthetable onpage 17. Notice thesecond column which applies to"hyperbolic" ODE's likethewave equation with k#0.In thiscase, specifying anything onaclosed surface istoomuch, andtheonly chance youhave isdoing Cauchy (ie,specifying 4and¢')onanopen partial surface. InKirchhoff scalar diffraction theory (field y)applied toanaperture, youareineffect trying tospecifyyand@y/énonaclosedsurface(inthehole, youareassuming youhave theincident field unaltered, onthescreen youassume 0,onthegreat sphere youassume0).ThusyouaretryingtodoCauchyonthewaveequation,andthatisknowntoonlygivea t) Osolution. Thefixisthatintheaperture, thingsarenotquiteunaltered! 3 ‘ThethirdcolumnappliestoODE'sliketheheatequationwhichIhaveneverreallystudied!Jackson r} doesnotprovethesethings,butrefersustofamoushistoricsourceslikeMorse&Feshbach, andSommerfeld himself. 1.10. How tosolve for$with boundary conditions using theGmethod. Thebasic problem isthat you would like toeliminate one ortheother ofthetwo surface terms mentioned above. Earlier weused Green's Theorem with y=1/Rand$=4.Ifweinstead usey=G(x.x)), weget(1.42). Ifwecould somehow solvethisforG(x,x),obviously wecanfind4.Ifwecansomehow arrangeforGtobothsolve thePoisson equation AND tobe0onasurface, then wewill have eliminated thefirst surface term, and wecanthengetouranswer byknowing only$onthesurface (that is,wedon't have topreknow 4',nor dowewanttoknowitifweknowthere).SothisismethodtosolveaDirichlet problem! Ontheother hand, suppose wecould find aGwith G'=0 onthesurface. That would appear to eliminate thesecond term in(1.42). HOWEVER, thisleads toaninconsistency! Itiseasy toshow that thesurface integral ofG'mustbe-4x,soyoucanneverhaveG'=0.ThefixistosetG’=-4n/Area =a constant. Ifyoudothis,thenthesecond surface term in(1.42) does notquite vanish andyouendupwith anextratermI/Area* ("[email protected]"extrafirsttermyou seein(1.46) .Ifweagree toonly doNeumann problems working inthespace between afinite surface andaninfinite one, which together make upoursurface S,then thisaverage 4term vanishes. This is tricky, though, because youhave toremember thatyour true "interior" region isnow "exterior" tothe inner surface. Finally,Jacksonin(1.40)writesageneralformfortheGreen'sfunctionwhichsolves(1.39).theusual URistheparticularsolution,butFcanbeinprincipleanysolutionofLaplace!Ourgameisgoingtobe r) findingF(x,x’)thatgivesusstufflikeG=0onsomesurface.ThetermFcanbedirectlyassociated with theinduced surface charges inaproblem with conductors. 1.11Electrostatic potential energyandenergyoftheelectricfield.Jacksonstartsbydoingthesimple not-equal sum between asetofcharges toaddupthepotential energy asin(1.50). Forcontinuous this appears as1.52 andhenotes thatnow wehave included thediagonal orself-energy stuff. But1.52 gives 1.53 where you integrate p},and then you fiddle easily toshow 1.54 which weknow well, that the density is[E}*/8x.Heshows withanexample what must always betrue,thattheselfenergy terms will make suretheenergy density cannever benegative, though thenon-self-energy component caninfactbe negative. Hementions thatyoucanusethevirtual displacement ideatoseehow energy changes andthus doaforce computation. Wethen have 13problems. Iseem tohave done 7ofthem during thecourse! 4 / JacksonChapter2Notes PhL 1.28.03 ©notice that Jackson isin3rd edition 1998 at$94 Amazon. Itseems tomehehasadded alotmore applications, inPortis style, andhasgotten ridoftheentire multipole expansion chapter 16andreplaced it with some stuff about radiation damping. Maybe hedecided themultipole wasjusttoohard! Otherwise, the3rdedition hasthesame chapters andbasic contents asmyIstedition. Chapter 2:Green's Function solutions forMetal Sphere, andother things Images isreally theGreen's thing indisguise. Why charges can't leave acharged sphere. Inversion. Then orthogonal functions andseparation ofvariables. 2.1Method ofImages. The idea here isthatyou considerapointchargenearsomemetalconductor, andyouknow thatinthiscase, $(x)=qG(x,x') where x’isthelocation ofthecharge. Think ofimage charges ascreating theF(x,x’) discussed earlier. Youthencarefully select theimage orimages tomake G == 0onyour conductor surface. Ifyoucanfindasetofimage charges thatdothetrick, then youhave foundAsolutionfor}inavolumebounded byasurfaceonwhich6isknown.Sinceweknowthatin thiscase (known onaclosed surface, Dirichlet) thesolution isunique, then thepoint charge +image charge configuration must beTHE solution totheproblem. Alessfancy waytothink ofimage charges is thatyouselect them tomake field lines perpendicular tothemetal surface sothere isnotangential Efield. Theusual situation isapoint charge above aninfinite plane, thentheimage hasopposite charge andisan equal distance behind thesurface. Asusual, thefield implied intheimage problem “ontheother side of e theboundary"isfictitiousandhasnobearingonthephysicalproblem. 2.2Point Charge near Grounded Metal Sphere .Inever realized until justnowthesignificance ofthis problem and how itrelates tosomethingIrecentlywaswonderingabout.Hereisthelogicflow.First, Jacksonputsapoint charge qdistance yfrom thecenter ofagrounded sphere ofradius a,andputs it outside soy>a.Heshows thatyoucanget§=G=0onthesphere byassuming oneimage charge inside thesphere whose sizeis-q(a/y) andwhose distance from sphere center isa’/y.Thepotential solution is thengivenbythefirst2termsof(2.8),andyoucanthengoandcompute whatever youwant.Inthis solution, thegrounded sphere brings infrom ground atotal charge ofq'=-q(a/y), asGauss's lawtells you.Again,wesecthattheimagechargeissimulating whatthesurfacechargedistribution reallydoes, 2.3Point Charge near charged, insulated metal sphere. Suppose you addtotheabove solution a spherewithtotalchargeQi=Q-[-q(a/y)].ThetotalsolutionthenhasQonthesphereandsolution(2.8) with thethird term. This isthen thesolution yougetbyputting apoint charge outside acharged insulated sphere. Butwhy isthisinteresting? Thepotential isthesumofthree terms, andifyoudoV},yougetthe électric field (F(x)) atallpoints inspace, anditisafunction ofa,y,andQ.However, ifyou want to know theforce onthepoint charge q,youonly consider thelasttwoterms in(2.8)and setx=y, andthis then gives (2.9). Thepoint charge feels aforce from theQatsphere center, andfrom theimage charge . which liesunder thesphere surface below thepoint charge. Thedistance between thepoint charge andthe imagechargeisgivenbyy-a’/y.Asthepointchargeapproaches thesurface, theimagecharge approaches fromtheothersideandit'sforcedominates. Theimagechargealwayshasasignopposite q.If | Qandqhaveoppositesigns,thenqisattractedbythecentralQandbytheimagecharge,Themore |e@ interestingcaseiswhenQandqhavethesamesign.‘ThenqisrepelledbythecentralQ,butattractedby 1 theimagecharge.Thetwoforcesareequalwhenqliesabout(a/2)/q/@abovethespheresurfaceinthe @ case Q>>q, Foranelectron over amacroscopic sphere ofsome reasonable charge, thisisavery small distance. Asqapproaches thesurface, theimage charge wins out(thereforming surface charge, thatisto say)andtheforce becomes infinite right atthesurface. Sohereisthepoint: anindividual electron ona charged sphere isheldonbyaninfinite restoring force ifittriestoleave. Nodoubt ifyouwork witha non-idealized metal surface, youwillfindtheforce isfinite, andafinite energy cangetanelectron over theequilibrium hump andthisiscalled thework function. Nottoolong agoImade abogus argument about why electrons didnotleave acharged metal object inwhich Ireferred totheeVchemical idea. I thought theelectron washeldbyitschemical binding stability energy. That might betrue, butthisimage charge force seems much more significant. This whole discussion could have been tried with regard toacharge over aflatmetal surface, but there itdoes notwork soeasily. You could imagine aninsulated large round plate thatisgrounded and then yougettheimage charge solution assuming very large radius. Butthen youthink ofthatplate as having charge Qinaddition andyoucompute ofthis,things have logdivergence, notvery nice. You have toputinacutoff radius R,butIthink thewhole thing canbemade towork. Forradius Rofthe plate, Ithink thisisthepotential thepoint charge seesfrom theimage andtheplate: $(y)=-a/y) +(QtqV@R?) [RF -y] IfTset$'(y)=0withlargeQ,Ifindthaty=RA/xq/2Q istheequilibrium point.Yes,itisquiteugly.The problem here isthatdistances domatter, youcannot justignore faraway things andsaythatlocally the surfaceisflatandwewillusethe"flatsolution".Jacksonsolutiondivergesfora->ooaswell.Sothere e@sphere isavery instructive case indeed thatshows clearly theprinciple involved. Inother cases, thesame general idea will probably betrue,butmuchhardertocompute. 2.4Point charge near ametal sphere offixed potential. This isidentical totheprevious problem except youreplace Q+aq/y with Vainthethird term. Wearejust superposing another sphere onthe originalproblem. Wehavenowchangedtheydependence ofeverything alittlebydoingthis. 2.5MetalSphereinauniform Efield.Usesimagechargesat4R>+tosimulatetheEfieldand findsthatthesphereactsasadipoleobject.Sothisexampleuses2imagechargesandnopointchargeto |start with, Resulting oonthesphereisaverysimplecos@formsuggesting dipole. 2.6.Method ofInversion. Equation (2.17) with (2.18) makes thebasic symmetry theorem. You cantake theknown solution ofoneproblem and"invert it"toapply toanother problem. Anexample isthatyou cansolve thesphere above aplane inthisway asinFigure 2.10, There aretricky points here, Ihave not studied it.Ithink thisisaspecial caseofthemore general conformal mapping ideas thatJackson gave us notes onandprobably hasadded tohislater editions? 2.7Green's funetion foraSphere. Weoutlined theidea inthelastchapter: findGthatvanishes ona sphere, then usethatin(1.42) tofind§givensomevalueof}onthesphere.WealreadyknowtherightG from oursphere image solution, itis(2.22) which becomes (2.23) inpolar. Thederivative is(2.24). Since there isnopintheclosed region (which, bytheway, isexterior here tothesphere), only oneterm ofthe @ threein(1.42)surviveanditisshownin(2.25). 2 2.8Applicationtotwohemispheresat+Vand-V.Jameverythingin,get(2.27)which,Iamhappyto t) see,"cannotbeintegrated inclosedform".Hecangettheresulto1!-axis,andhecanpower-series expandthings andhegetssomething thatlooks likemultipole, butthatideahasnotbeen setupyet. : 2.9.Orthogonal Funetions andExpansions, Covers thebasic ideaofexpanding f{x)interms ofsome orthogonal functions $,(x), theorthonormality condition, thecompleteness condition. Fourier Series and Integral arestated. Iam pretty solid onthisstuff. 2.10 Separation ofVariables. Theproblem isacube with 5faces at@=0andsome V(x,y) onthetop surface. Thesolution istousecartesians, separate variables, findsome basis functions thatsatisfy the3 zero conditions, then force thelastonebysumming with coefficients asin(2.63) withcoeffs inthenext equation. Thecomment made isthatyouoften endup"making your own" custom setofortho-functions foraparticular classofproblems. 11problems, Ihave done 4itwould appear. 3 wo. Jackson Chapter3Notes PhL 1.28.03 ® Chapter 3:Solving spherical andcylindrical electrostatic problems. This chapter talks about thetechnology needed forworking inthe(r,0,4) and(p,2,) worlds. Ineach world, theradial equation implies certain special functions (Legendre andBessel). The final section treats theflatmetal charged disk asanexample ofCauchy mixed boundary conditions. 3.1Laplace Equation in(1,0,). Assume (3.2) asseparated form, find thethree separated ODE equations. >iseasy andcauses quantized integral parameter m.The1/@separation constant iswritten ‘{(t+1) anditislater shown why/mustbeaninteger.Sonowwehavethethreeequations,andwehave solved oneofthem. Theradial equation istrivial andthesolution ist’or1/r*!asin(3.8). Soonlythe@ equation requires work, andthisisLegendre. 3.2Legendre Land. The6equation iswritten with x=cos0 ,buthere only inthecase m=0, Assume +x power series, getrecursion forcoefficients, discover thattogetconvergence atx=+1youneed truncation, andthatintummakes/beintegral. Thesolutions forgiven/arethefamousP(x).The second kind solutions don't falloutherebecause thepower series assumes convergence atx=0. TheP{x) form anortho set,alltheproperties aregiven andsome arederived indetail. Page 59shows the expansion andcoefficient formulas, theexpansion andprojection asIliketocallthem. 3.3Azimuthal Symmetry problems in(r,9,9). This makes m=0 sowethen have theP(x) andthe absolutelymostgeneralsolutionis(3.33),|agree.Thismakesiteasytosolvenearaspherewithan C) arbitraryV(8)potentialonit,specialize to3.34andyouaredoneasin3.35,atleastasaseries.Wenow dothe+Vhemispheres forasecond time andget(3.37), amore organized answer than lasttime. Anice trick isthis. Suppose youknow 6(2) only onthesymmetry axis where expansion reduces to 3.38 since P|(1) =1,theforward direction. Then usethatdatatogetthecoefficients, andaddback theP, andyouhave your fullanswer! Wefound theaxissolution exactly fortheV problem, sowecandothat here asathird way tosolve this same problem! Next, 1/Risexpanded inP,(cosy) asshown in(3.41) andaproof isgiven. Recall ther<notation trick! Hisproof isjusttolook onthezaxis andgetaseries there, then addP,back in. Another sample problem: find6neararingofcharge, page 63.Again, weknow iteasily ontheaxis. Expand andanswer is(3.46) ontheaxis still, then addP,(cos@) and3.48 isallyours. Iliked itlasttime, and Istill like it. 3.4ThePla(x) andYjq(0,6) Worlds. Youneedthisfornoazimuthal symmetry. Alljustthebasic nuts andbolts. Oursame most general expansion isnow written as(3.61). Jackson istrying tothrottle therate atwhichcomplexity buildsup,nothittingthepoorstudentswithmrightoffthebat. 3.5.The YimAddition Theorem. ‘Thething isshown in(3.62). The important point isthatthislet's you separate thevariables ofthetwo vectors which aremaking thecosy angle. When you then stick this expansion intoour1/RLegendre expansion, yougettheglorious (3.70), which provides separation inall three variables, atthecost ofamessy double sum. Still, itletsyou dosomething when you might e otherwisebeblocked.Youcanbetthisresultwillplayaroleinthemultipolefieldexpansion! 1 3.6Cylindrical SymmetryandBesselWorld.Westartover,separateinz,9,9.Wegettwoconstantsas e beforek?andv*,The4andzequations arebothtrivialgivingexposasin3.76implying v=m=integer. ‘Thepequation then ends upasBessel's Equation where thekconstant isabsorbed intotheargument Jukp), sox=kpinthisworld. Wegetvarious Jproperties. Forintegral vweneed thesecond kinders No(kp) which aretheNeumann's. TheHankels arealsodefined (third kind). Various properties ofthings aregiven, very useful andhere allinoneplace. Now when youtrytomake orthonormal functions over some range 0aforp,youendupusing J,(kp) with k=zeros/a. This isneeded tomake parts goaway in theproof ororthogonality. This iscalled Fourier-Bessel series, Ihave never used itinanyofmydoings. Jackson finally mentions some other versions ofBessel series with names Kapteyn, Schlomilch, and Watson's Bessel book iswhere tolook forthisstuff. Weassumed non-phase expo forz,andthisledto softBessels intheother direction. Ifweassume softexpo phasor form inz,then wegetthehard other kind ofBessels, theIandKfunctions, so-called "modified Bessel functions". Hard means they blow up insome direction, soft means sine like. 3.7Cylinder/Bessel problems. Cylinder with 6=0 everywhere butprescribed toafunction ononeend. Then anapplication oftheFourer-Bessel series. Skip thisstuff, usewhen youneed it!Jdon't think we willbedoing ourscattering problem incylindricals, butifso,come back here. 3.8Doing ageneral Green's analysis inr,0,4. WedidtheGreen's deal inDirichlet forasphere (find G such thatG=0 onasphere) andtheimage charge trick made short work ofit.Butsuppose youhave some other spherically symmetrical situation, such asconcentric spheres. Thegeneral form ofGis(3.118) with (3.119), that is,adouble harmonic expansion with some radial green's function g(r,t’) that wehave to figureoutforourgeometry. Forthesinglespherethatfunctionisshownin3.114,whichwecanthinkof e asthe1/Rfactor (initsfancy expansion) plus Fastheexpansion oftheimage charge inasimilar fashion. Fortheconcentric spheres case, theresulting gisthemuch messier (3.122), andthefullGreen's isshown in3.125. Forthiscase, youwould need aninfinite setofimage charges togetthesame result, Jackson claims. Remember, you aretrying tomake G=0 onboth spheres atthesame time! This section also gives some useful delta function stuff inthespherical coordinates, page 79.This isapretty tough rowto hoe! Each sample problem isawhole paper youcould write. 3.9Spherical Problems. WedidSpherical setup work above, then wedidcylindrical setup work, and then cylinder problems, butwenever didspherical problems inthegeneral case, sohere wearenow. This isgetting closer tomymain interest theme right now which isapplying multipole toplane wave scattering, butatthemoment weareback inElectrostatic Land andweareabout todosomething here along these lines. Here,wearefirstreminded ofthe general Dirichlet 2-term result (3.126). Thefirstterm isthep term ifyouhave any, thesecond term isintegrating aprescribed} over your surface with dG/dn asafactor. In 3.127 wecompute this dG/dn forsphere(b), then 3.128 gives thenon-p term ofthe3.126. Sointhe following twoexamples, weuseagrounded sphereandpractice doingthepterm.Inthegeneral case,you | cansuperposethissolutionwithsolutiontoaspherewithsomegeneralonitssurface.Forexample,you |could doaring-charge inside thenow-infamous #Vsphere. Example1:putaringofchargeinsideagroundedmetalsphere.Inthiscaseofcoursethesurface |integral is0,andwearegoing topractice doing thepGintegral term. This pisgiven with delta functions which inturnkilloffthedV’integration intheptermintegral (6isjust2x)giving 3.130. Note thatone .e@ deltaisforcingcos6'=0sincechargeringisinthatplane,hencethenoriginalYiq(6',9')withinGgets 2 pinnedtoYin(0,-)andm=0duetosymmetryandthatiswhyyouseeP\(0)sittingin3.130.TheotherP,is t) ofcoursefromtheotherYfactorinG.So3.130isthesolutiontodurproblem, asumover£. Example 2:putadiameter linecharge inside agrounded metal sphere. Wehave anewpwithits deltas,andwetakenoteoftheI/Fintheseforms. Theresult is3.133. Since inthiscaseonlyhasdelta in0',westillhave ther'-integration todoandthefinal result is3.136. Jackson notes thatthisthing actually diverges onthezaxis(itshould, thatiswhere thecharge is),BUT manages tobe0onthesphere, meeting the$=0requirement there (grounded sphere), 3.10 Green's for (p,z,6). Write theusual equating defining G.Expand both sides. TheRHS isobvious, theLHS expansion ofGshown in3.140 isjustified because wesaythatthecylindrical solutions had exp(ikx) exp(imd), where wehavenowusedtheI,Ksignofk?.Wehavethus"separated variables" inG andhave g(p.p') tofigure out.Inatypical tour-de-force, Jackson shows thatthesolution totheradialp equation isg(p,p') =4xIn(kp.) Ka(kp.). Recall thatIistheonethatisbest-behaved atp=0andno surprise thatitisconnected with p<.Sowecanjamthisinto3.140 togetthecylinder Green's G.Butwe know thatthisis1/Raswell, sowegetthe3.148 expansion of1/R.This result ispretty messy! Jackson then shows thatifyoucompute In(1/R) from thismessy thing, yougettherelatively simple expansion shown as3.152. Very nice. Noexamples however. 3.11 The Eigenfunction Method. This isaquantum-mechanics Schroedinger-equation approach. Take your ODE asin3.153 andfind some eigensolutions asin3.154 which have some desired boundary conditions, such asvanishing atthewalls ofaboxthatyouwant tostudy. Thegreens must atonce beof thegeneralform3.157expandingontotheeigenfunctionbasis.Goofaroundandendupwith3.160.One r) pointnotquitemadeclearlyisthatyoumighthave4=0in3.153(asintheLaplaceequation), butyouwillstillofcoursehavenon-zero eigenvalues in3.154.Inparticular, forthatmentioned boxincartesians, youreigenfunctions are3.166 and2pasin3.165, although there isno2.=k?intheLaplace, That iswhy Weseejust2,"inthedenominator ofthefinalGforthissituation in3.167, callthisexample 1.For example 2,wetakethesame A=0Laplace equation, butinstead ofaboxwedoinfinite space. Inthiscase wegetthat2,=k?which isnowgoing tobeacontinuous eigenvalue, and3.160 becomes 3.164. Thisis aninteresting waytointerpretthisniceformfor1/R.Example3istodothethingincylindricals, andwe then get G=I/Rexpressedas3.168. Comments onthismethod? The idea istodecide ahead oftime onsome boundary conditions, then solve foreigenfunctions byinsisting onthese boundary conditions foralltheeigenfunctions, andby inserting some Ayintoyour original equation, which really comes from separating thevariables, even if’ =0inthefullequation forG.‘That is,wegetaseparate "wave equation" foreach variable-separated function, The fulloriginal equation still has4=0. Weknow what each equation looks like from our earlier work, andonly oneofthethree variables willbetough. Notice from page 48thatthesum ofthe effective k*valueshastobezero,soiftwoofthemarephasortype,thethirdmustbehardexpotype. ‘You cannot have allthree bephasor type because they would then adduptoanegative value, not0. 3.12 AnExample ofmixed (Cauchy) boundary conditions. The problem isacharged, insulated metal disk,soundssimpleenough!Wetakeasourclosedsurfaceaninfinitehemisphere ononesideofthedisk. Weknow that6=constant onthedisk and§=0ontheinfinite half-sphere, butwedon't know 6inthe diskplaneoutsidethedisk!However,wedoknow8$/6zoutsidethedisk,thealong-axisEfield,andthat rymust bezero bysymmetry. Sohere wehave aCauchy situation that really was notmentioned in 3 Jackson's table. Wereally dohave aclosed surface, andwehave}prescribedonsomeofit,and¢!onthe r) restofit. OK, next since weknow there'is no dependence wespecialize thecylindrical expansion toget 3.170, butwedon't yetknow what f{k)is.Wequickly getthetwo"integral equations” forf(k)shown in 3.173 from themixed boundary conditions, Jackson isforced tojusttellusthatthesolution is3.176 without 2derivation, butofcourse wecanverify thatitworks. This gives 3.177 forthefullsolution. The integral there canbedone togive thefinal form 3.178, anice compact closed-form solution tothis problem with nospecial functions required! Jackson computesooneachsideofthedisk(sameonboth sides ofcourse) andfinds aspike attheedge asyouwould expect. Itisasifthere were some kind of centrifugal force pushing thecharge outtotheedge tomaintain noradial Efield. Jackson finally notes thattheproblem canbeexactly solved inelliptical coordinates, something hetreats does inthisbook. Fourteen problems, ofwhich Ihave done 5.Stupidly, Ididnotkeep myproblem sets, isthatpossible? 4 Jackson Chapter4Notes Ph 1.29.03 e Chapter 4:TheScalar Multipole Expansion, andDielectrics Itisalittle oddthatthese twosubjects aremixed inonechapter. Itistrue, however, thatdielectric is reallybasedononeofthe multipole moments (dipole), soitisgood forthereader toatleast know what thesemoments are!Jacksontreatstheusualsetofdielectric problems, butalsodelves intomodels fory andforthemicroscopic which hecalls y.This latter istreated forboth polar andnon-polar substances. This isavery good anduseful chapter. 4.1TheMultipole Expansion. Finally wearegetting where wewant tobe.Wealready know from 3.61 what themost general form ofapotential isinspherical coordinates. You canpick whatever coefficients youwant here, andtheresult willformally satisfy theLaplace Equation. Notice in3.61 that wehave thetwo characteristic radial functions r‘andr*', Now, weimagine some pdistribution thatisallcontained within some mathematical bounding sphere. Then weknow thatoutside thatsphere, wecannot haverterms inthepotential because theywill diverge at«©,soweonlyhave ther!terms, sowewrite 4.1asshown. This equation isTHE MULTIPOLE EXPANSIONofthe(scalar)potential.Thatis,youareexpandingitontoeigenfunctions of theLaplace equation inspherical coordinates. Weknow thisisacomplete expansion, nothing isleftout, anditisjustaquestionoffinding thecoefficients which areherecalled qe,.Nosurprise, these aregiven in4.3andyouseethatyouarejustintegrating your friendly Yqqwith r*against pover thesphere that containsalltheaction.For£=0wegetthecharge.For£=1wegettheelectricdipoleasshown,although e@thepthings appear asraising andlowering operator typethings duetothewaysphericals work. Asforthe £=2,theQjarewhat yougetfrom acertain cartesian Taylor expansion ofshown in4.10. Togetthis result,youfirstexpand1/Ras(1/r)[1-rer/r*+ete]whereR=|r-r'| andwetreatrassmallandjust doaregular Taylor series.Theninsertthisinto=fp/RdVandyouhaveit Sothatisreally allthere istoit!‘Theideaisthentothink ofthespecificé,m terms intheexpansion bythemselves, callthese things }m,.Bach oneimplies itsownelectric fieldEyqasin4.11. For£=1weget theusual dipole result shown in4.13, notquoted yetinthistext. 4.2.MultipoleExpansionofInteractionEnergy.Thepotentialenergyofachargeqinanexternal |potential @isjustq®. This would bethework needed tobring inqfrom infinity tosome location r.The work tobring intwocharges would be(q1+42). Note thatthisdoes notinclude the"self energy", the work needed toassemble qlandq2near each other without @present, which iscertainly non-zero. I thinkthenthatfp@dVistheenergyofsome charges justduetotheexternal field, andifyouwanted the totalenergy youwould havetocompute ¢p+) dVwhere @wasthepotential justduetopwithno external ®.Jackson isthinking ofthisfp®dVasthe“interaction energy" between anexistingpand some externally generated ,asortofinteraction Hamiltonian orLagrangian type thing. Ifthisintegral is broken intomultipole terms, youget4.17 which shows how each moment ofpinteracts with the®field. e Asthemoments goup,thederivatives [email protected],youcouldruleoutaquadrupole interaction if 1 youknewthatyourextemalEfieldhadnogradient.Notethatthisisnotthecaseinaplanewaveinthez t) direction, butwereallyhaven'tdonewavesyet,solet'sholdoffonthisidea.Page 102thengives anuclear physics application. Anucleus hasaquantum stateandisexpected to have aqzymoment. Anucleus inacrystal isexpected toseeamacroscopic} which hasfirstandsecond order gradients, sothere isexpected tobeaninteraction between thecrystal field andthenuclear Q moment. InQMweknow thatthisinteraction willsplit theotherwise degenerate mlevels, akin tothe Zeeman effect forelectrons asIdimly recall. SoRFtechniques canthen beused tomeasure theQ moments inthisway,givenaknowledge ofthelocalEfield. Jackson's lastgasp inthissection istocomment ontheinteraction between anexternal dipole field andanacted-upon dipole field, weget4.19. Iamnotsurewhy wecareabout thisbutOK, maybe hewill need itlater forsomething. Hemight have saidsomething about thedipoles ofwater molecules here, but hedid not. 4.3Dielectric Materials. This iscertainly achange intopic within thischapter, butitisalsosomething I ‘want toknow about, sinceIamthinkingoffloatersasdielectricspheres. Thisdiscussion isVERY carefully constructed. Westart with some microscopic Maxwell equations 4.20 where ¢andp'arethedetailed microfield andcharge distributions, atotal mess. ‘Weknow thatwe have togetintoaverage fields towork atamacroscopic scale. Wearegoing toaverage themicro field andthemicro passhown in4.22. Thestarting point is4.23, standard Efrom p,asuperposition ofpoint charges ifyoulike.Wethendoourmultipole expansion ofthe¢field.Thisjustmeansdoanexpansion of1/Randputitin4.23. Wedidthisin4.10, thecartesian Taylor expansion through quadrupole. Jackson leaves ¥on1/R, there isnotreason toevaluate itright now. Here, hedrops quadrupole andabove, with ‘theclaimthatsincefieldvariationsarelargerthanatomicsize(5000Alargerthan1A),youwon'tbe e@ doing much quadrupole action. This claim isnotsupported very well, butitmust betrue, because you don'tseeanyonetalkingaboutsomekindofQpolarization tensoronafootingwithPasbeingimportant inadielectric. Sothings aresimplified alot,weonly have two terms asin4.26. Here Jackson is summing overdistinct charges ¢.Hereplaces thosewithafakeintegration in4.28byusingdeltafunction pandx,wherenow1isamicroscopic levelseaoflittle dipoles asin4.27, just asIwould do this. Thesubscript "mol" isjusttoremind usthatthese arenotthemacroscopicp andP.Jackson likes to think of"molecules" instead of"atoms" which isfine. Wenow take 4.28 andapply theaveraging operation over avolume asshown in4.22. Hetreats thepterm, andsays thepterm works exactly the same way. In4.30 theaveraging isapplied andwemove ittotheright, where ofcourse itjustgives the "average charge" inthevolume duetothedelta, Butthen youcanwrite thisasthemolecule density N times theaverage charge permolecule in4.31, alljustwords, nothing tricky ishappening here. Thenet result isthatin4.28 wegettoreplace protwith <N><e> asshown in4.33. Similarly, wereplace ting,with <N><p>, sotheaverage polarization inavolume isNtimes theaverage polarization permolecule. Then thebigstep iscoming. After all,wewant toknow what Vee isgoing tobe[don't confuse Jackson's choice of€asmicrofield with¢asdielectric constant soontocome!]Applying VgivesusaV?ineach term applied to1/Randweknow that gives delta --THIS iswhy Jackson just lefttheVoperatorssitting. ‘The final result isthevery simple 4.34. Jackson interprets the"extra" term ontheright asaneffective polarization charge density that isgetting crunched around asinfigure 4.2, affecting theEfield divergence. Butthenext step istorewrite as4.35 andidentify thepieces asIhavemarked. Andsoweend ‘upwiththefinalfactsthatP=NpandphewritesasNe+Pex.Allthroughout, this<e>thinghasbeen presenttoaccountfortheaveragecharge,butnormailythisisgoingtobeexactly0,andthenyouonly @‘want toworry about the“external toatoms" charge, thefreecharge thatcanmove around andthat isPex, 2 SothisIthinkisaveryclearexplanationofwhatPis,andwhatDis,andwhythereisnoQorhigher e@term toworry about! Iwillbeinterested toseeifwereally ignore QinMieScattering theory. Finally,wegetthetraditional restatement ofthetwoMaxwell equations forE,andthedivergence oneshowsthatDseesonlythefreecharge,sothatitwillbeDaomathatwillseethefreesurfacechargec.So thisisgoing tobetheonebigdifference ataboundary! Avery good section, 4.4Dielectric Boundary Conditions. Here thefactthatP=7Bisquotedasanexperimental factthat hasvery little error inmost substances, ie,there isnotensor aspect toit,seegraph ofsome evidence on thisonpage 109. Theformal boundary conditions areasin4.45. Iremember being confused about thisonafinalexaminsomesimpleproblemtheygave,ouch,itreallyisstupidlysimple. 45Dielectric boundary problems. Ourfirstproblem isthetraditional point charge near aplane interface. Earlier wemight havedone thiswithaconductor, butJackson skipped thattraditional problem andwent right tothesphere. Looking from theright, thesolution istohave apoint charge attheimage point ontheleftwithaspecial magic charge q'.Looking from theleft,weseeasingle charge attheq location butwith magic charge q”.Theboundary conditions areasin4.48, andthese arethen used tofind thesizesofq'andq".Youhavetodifferentiate thepotentials onbothsidesintherightdirections and apply yourBC's. Theanswer isin4.51, sothecomplete problem issolved, There isapaionthe interface asin4.53thatyoucompute with apillbox, knowing P|andP2onthetwosides. Oursecond problem isanother traditional onethathasavery simple answer. Thedielectric sphere in theconstant zdirection Efield. Imight imagine thatifthisE-field were inaplane wave, itmight somehowinteractwiththeoveralldipoleappearancethatthespherehaslookingfromtheoutside,aswe e@shall soon seeandcause some dipote radiation. Tosolve thisproblem, weassume appropriate expansions for$onthetwosides inourspherical coords, asin4.54, Weapply thedielectric BC's atthesurface (well suited certainly tospherical coordinates!) anddosome shuffle toconclude thatonly theA:,ByandC; coefficients arenon-zero. ‘Thefinalanswer isin4.60andisveryinteresting, Inside, wehave aperfectly parallel Efieldthatisalong theoriginal fieldbutreduced by3/(c+2). Outside, wehave ouroriginal field plusthefield ofwhat appears tobeapoint-dipole located atsphere center. This dipole points along the original field andhasthestrength shown in4.62. ,soPisasin4.63inside thesphere. Finally, we compute theGpai. Thethird problem isaspherical cavity inadielectric, Thetrick hereistolookatthesecond equation in4.56 andmove theetotheother side, which islikedoing >I/e.Theanswer isthenthesame except thedipole direction isnow reversed against theoriginal field andhasadifferent magnitude asshown in 4.66. Sol thinkIgetthebasicideahereandIthinkIcansolveanyproblemofthis type. Here isafourth problem wewill need right away. Imagine acavity inaninfinite dielectric. What is theEfieldatthecenter ofthehole? Weknow howPsitsinthedielectric, andweknow thegpa,charge sitsontheboundary, andtheEfield sees thischarge aswell asfreecharge. Thecharge isPcos0) andthez component ofthefielditmakesatthecentermakesanothercos®,soweget4.68wherethe1?factorsof thedAandthefield 1/1?cancel andweget1/3x° atIand-1which is2/3,then 2nfrom azimuth andwe getouranswer 4nP/3. 4.6,TheClausius-Mossotti Modelfor.Anotherexcellentsection.First,weask:inadielectricwith euniform macroscopic (average) field E,what field does aparticular molecule actually see? Insection 4.3, 3 wedidnotaskthisquestion.TherewejustcomputedEasanaveragevalueofthemicroEandcomputed t) thedivergence ofthisaverageE.Wedidnotaskthequestionthatisaskedhere.Theanswerisquiteamazing Ithink. Theanswer isthataparticular molecules sees E'=E+(4n/3) P,where Eisthe macroscopic field wealways use. Iffrom some other model youcancompute that p=yE; onthe microscale foryourmolecules, then P=Np =NyE; andyoucombine togetP/Ny =E+(4n/3) P,andyou canthensolve forx=P/Eandyouget4.74. Theideaisthatthisrelates themacroscopicx,parameterto themicroscopic yparameter. Theresult isClausius-Mossotti andisOKforgases, roughly trueforliquids andgases. Ithink forwater theresult isoffby50% orso.Comment thatthestatice forwater is80, obviously nottrueatoptical frequencies! So,howdowearrive atthefactthatamolecule seesE'=E+(4n/3) P?TheEpartisbecause we assume itispresent from some source, soitistheextra correction term thatisinteresting. Ifyoucarve outaspherical cavity around your testmolecule, thedistance Pbulk creates the(4/3) Pfield atthe center. Forthispart, wearenotsurprised tofindthatthisextra field isinthesame direction asEsince E ismaking it.Thisresult isconsistent withoursolution ofthecavity problem inthelastsection, where the field inacavity isstronger than thefield intheenclosing dielectric, The real mystery isthis: obviously youshould now addtheeffect ofacarvedoutsphere.Jackson shows that foracubic lattice, theeffect isexactly 0,just because youareinthemidst ofasymmetric environment. (Wehave already added theexternal field E,sodon't want toaddithereagain) Well, you must ask,what happens ifwetakeourlittle cavity tobehuge, itwillalways give0.Suppose ourentire dielectric were alarger sphere. Iguess theanswer isthateventually you getsome dielectric surface charge todeal with. Jackson then argues thatthislocal contribution isroughly zero inanynormal material. e 4.7Modelsfory.Weskippedthissectioninthecourse,Ireaditnow. First, wegetamodel foryfornon-polar objects (molecules) just from atoyelectron model ofa harmonic oscillator andwegety=e’/may? where@pisourresonant HOfrequency whichwewouldsetto hu=optical levels. Foramixture ofmolecule types, weget4.78. Jackson argues thatthiseffect canonly make dielectric constants ontheorder of1.00XX andhegives examples ofair,helium, etc.That is,his examples arenon-polar gases. Jackson wonders whether thermal agitation affects thisresult. Intuition says nosince themolecules oughttoself-polarize regardless ofjiggling around,theEfieldisnotmuchdifferent innon-relativistic action. Jackson proves thisisinfacttrueusing astatmech trick thatIunfortunately have forgotten but know atonetime. Youweight things withtheBoltzmann factor incomputing athermal average. Sothe answer here really isitmakes nodifference. ‘Themore significant situation iswhen youhave fixed dipole moments asinwater orCOgas.Here theHamiltonian termis-peEandwhenbeBoltzmanize thisonewegetthaty=p’/(3kT).Obviously thermalfights thelineup,andlarger kTrelative topeEenergy isgoing tocostyouandreducey.Sadly, Jackson does notquote ¢values forpolar substances, butIthink water atDCfrequencies has©=80,so thiscompletely swamps thenon-polar effect. So,avery clear andtothepoint section! 4.8Electrostatic Energy inaDielectric. 1have only skimmed thissection today, itisnotofdirect interest tomeright now. Thequestion isthis: is4.86 stilltrueinterms ofmacroscopic quantities ifyou e assemblechargesinadielectricenvironment? Jacksonconcludesthatyesitis,providedthedielectricis 4 "linear"so4.91isOK.ThetotalenergydensityisEeD/8nasshownin4.92.Heconcludesthatthe r) dielectric holdsanextraenergydensityduetoPshownin4,97,andcomments onthe1/2. ‘Thenext subject here iswhat happens when you"move in"adielectric object intosome fixed fields, orbetween some battery-fed electrodes heldatconstant potential. Since there areenergy changes, wecan expect forces onthedielectric object andthese arediscussed. Maybe youcanlevitate something inthisway.Thatis,anEfieldwithgradientoughttohaveaforceonaneutralchunkofdielectric. We know this istrue forafixed dipole moment, etcetc.Havetokeepitfromrotating. Nine problems, Ididonly 3inthecourse itwould appear. 5 « eJackson Chapter 6Notes PhL 1.31.03 Chapter 6:Faraday, Maxwell, Potentials, Gauges, Grecn's Solutions, Conservation Laws This isagrab-bag chapter thatconstructs theknown important results for"time-varying fields". Earlier in thebook wedidthings inelectrostatics andmagnetostatics, here those things areupdated. 6.1Faraday's Law. Inthissection, Jackson reviews the1831 discovery ofFaraday thatifyouchange themagnetic fluxsurface integral onacurrent loop, avoltage isinduced around thatloop which causes a current toflow inthatloop according toOhm's Law. This voltage isusually written€ andishistorically called the“electromotive force". Itis,however, thesame kind ofpotential (voltage) wehad in electrostatics, anditisasifabattery were inseries with theloop, although thebattery cannot belocalized anywhere intheloop. Wenowknow thisvoltage isduetotheMaxwell equation VxE=-B/c. Jackson notes thattheterm Lenz's Law isused todeseribe thesign oftheinduced voltage --itopposes thechange |inlux. Inthislittlesection,JacksonfirstassumesanunknownconstantkinFaraday'slaw,asifVxE=-kB. Hethen considers aloop inmotion anduses Galilean invariance toshow thatinthemoving loop, charges areseeing afield E'=E+k(vxB). Butfrom Biot-Savart (force onacurrent inaBfield) weconclude that k= Ie, Recall thatBiot-Savart was developed inChapter 5,astheforce onatestcurrent similar to Coulomb's lawoftheforce onatestcharge. [Ihave aseparate note written uponthissection.] Bythe way,thisforceonaparticleisvalidallthewayuptothespeedoflightforv. e So,theendresult hereisVxE=-B/c. Jackson doesnotclaim to"derive" thisequation. Hejust found what the constant mustbetobeconsistent withBiot-Savart. 6.2Energy inaBfield. Ihave spent about 3hours trying toclean upJackson's logic inthissection, but tonoavail. Itried todraw areasonable tiling picture andsoon. Myconclusion isthat this isabogus presentation, something Jackson normally does notdo! Hestarts with theidea that ifyou have awire with acurrent flowing init,andyou increase theflux through thewire bymaking achange 6Binthe magnetic field inthevicinity, then some work 5W=I5V=(I/c)d(flux) must bedone bythecurrent in the wire tomake this new increased Bfield. Jackson then tries totile his wire circuit somehow with a mesh ofloops, andthen hetries toapply this5Widea tooneofthelittle tiling loops. Itisvery unclear whether thetiling loops aresupposed toreplace thewire, orwhether there issupposed tobecurrent densityJinallspaceinsidethestartingwire,orwhat!Heassumesthatthetilingwireshaveacross sectional area Acand thatJdV=IdéwheredV=Aodé.SothefactthathehasJbeingalongthetiling loopsuggeststhatheistryingtoreplacetheoriginalwirewithameshofwires.Butdoesthismeshthen form athinopen surface with some small thickness thatspans theoriginal wire loop? Ifso,then the integral in6.12 would bejustover thevolume ofthismesh, notover thevolume ofallspace. Then the conclusion would bethatthetotal work done is6.15 integrated over thevolume ofthismesh, Butthis seems wrong, because weknow itshould beover thevolume ofallspace where wehave fields from our wire. Thewhole presentation isvery, very ugly andIamsurethatJacksonregretteditandhasreplacedit with some better argument inhis2ndand3rdeditions. Therefore,IamabandoningthisentiresectionofJackson'sbook,includingthetrailingcommentson t) workdoneinamagnetization etc.‘Theresultisofcoursewellknown. 1 e 6.3Maxwell's Contribution. Jacksonclaimsthatin1865,Maxwell wasstaringattheequations 6.22thatseem toexplain allelectrical phenomena tothatdate, ButMaxwell sawsomething wrong. Since diveurl=0, thesecond equation implied thatdivJ=0,inviolation ofcontinuity. Heknew thisshould be dp/dt. Hegotthisfixed upbyadding 1/dAD/ét totheRHS, andsince divD =4np,thiswasexactly what wasneeded. Besides making theequations consistent with continuity, thislittle addition allowed oneto show thateither field satisfies thewave equation, andthisallowed waves assolutions totheequations, andthisledtoMaxwell's idea thatprobably light wasthese waves, traveling through theEMether. Maxell noticed thatthevelocity implied byhisequations waspretty close tothemeasured speed oflight. ‘This wasahuge breakthrough andthatiswhy hegetshisname onthese equations, although heonly really added thedisplacement current term. : 6.4 Vector and Scalar Potentials. (1)Conjecture thatanyBcanbewritten asB=VxAwhere Aissome "vector potential”. Atleast we knowthatthisformguarantees thatVeB=0sinceingeneraldivcurl=0foranything, (2)Instatics weknew thatVxE=0soweconjectured thatE=-V6since curlgrad=0foranything. We wereabletoalsosatisfyVeE=0ifweinsistedthatV?)=0. (3)Withtimedependence, wehaveVxE=-(1/c) Binstead of0.WithB=VxAthismeans wehave thatVx(E+A/c)=0sonowweshouldchooseE+A/c=-Vé@sincethatformsatisfiesthisequation e@since curlgrad =0.Soweendupwith: B=VxA and E=- Ak-Vo Wehave now replaced 6quantities with 4quantities, perhaps animprovement. Atthispoint, wehave satisfied twoofMaxwell's equations: VeB =0andFaraday. ‘Theother twoMaxwell's equations are shown topofpage 180.Theabove equations canboth bewritten asFP"?=3'A° -2°A", sotheEandB fields arereally part ofatensor. Seepage 379. 6.5A The Lorentz Gauge (4)Ifweassume thecondition @,A" =0(known astheLorentz. Condition) onthepotential (wheredy= fe@=-A/e andforproper scaling, andx°=ct),thenthetwoequations decouple intothetwoshown on page bottom, which wecancombine as(7A! =4xJ",Soatthispoint wehave "three" equations, those two onthebottom, and theLorentz. condition, (5)Consider thegauge transformation A'=A+VAalong with@'=6- A/e,or A= AP+A, Notice thatsuch atransformation does notalter theEandBfields shown in6.29 and6.31. Thus thepotentials A"andA"areequivalent solutions toaproblem that give thesame observable field results. There appearstobeawhole"family"ofsolutions A"thatareequivalent. Inthetensornotationfromabovewe 2 hadthatF*”=@"A®-d°A"sowhenyoudoA"=AY+A,yougetF'™=F"becausethetwosecond r) derivative termsobviously canceleachother. (©)Consider nowour"three equations" asnoted in(4)above. Suppose ourinitial pickforA!gives 2," =f#0. Then 0,A"=6,A" +0,0"A=f +C?A. ThenallweneeddoispickAsuchthatCRA~-f[asort ofPoisson equation in4D]andwehayeachieved 2," =0.Sothere isasubset ofourinitial "family" of solutionsthatgivethesamephysicalfields,whichsubsetmeetstheLorentzCondition.Thissmaller |family ofsolutions iscalled theLorentz Gauge. (6A) IfweareintheLorentz Gauge, then AF=a,{GA”- PAN}=AG,A°)-CPAM=-PRAM=-3,2" sothatwecanwrite 3," =O?A*=(4n/c)J". (1)Once weare"intheLorentz gauge", wecanstilldofurther transformations A"=A"+3A’ where CPA’=0[asortofLaplace equation in4DJ,andweremain" intheLorentz Gauge. These further transformations move uswithin thesmaller family ofsolutions. 6.5B The Coulomb Gauge (aka theTransverse Gauge) (8)Instead ofrequiring thecondition 6,A" =0,wehererequire thatVeA =0. Thefirstthing youseeon thetopofpage180isthatelectrostatics aswestudieditjustcontinues! Wesolvefordintermsofpjust e aswealways did,which iswhat 6.45 says. Butnow the"other" equation 6.33 hasajunk term initas shown 6.46. Ittums outthatyoucanpartition Jasshown in6.48. Toprove this(Ididitinproblem 6.6), youfirst show thattheexpressions shown in6.49and6.50really doadduptoJ,thisisanon-trivial piece ofwork requiring twovector identities. Secondly, younote thattherequired curl anddivconditions onthetwo pieces aremetbecause curlgrad=0anddivcurl =0.Again, thisisahighly non-trivial decomposition. Atthispoint therestiseasy. That "junk term" Imentioned isjust4nJ,/eandyouarrive attheverysimpleresult6.52whichsaysthat,inthisgauge,AisdrivenonlybyJ,.OfcoursefiguringoutwhatJ, actually isrequires doing thatmessy integration 6.50, butitcanbedone. Sointhisgauge, youhave again obtained decoupling oftheAand@equations. Specifically, isasinelectrostatics with 6.45, driven instantaneously bythecharge. Then youhave only 6.52 tosolve with 6.50 asthesource. ‘This tooisan instantaneous integral. Application ofthisgauge:nosources(nop’andnoJ)impliesthat=0andalsoJ,=0,soyourentire problem isthehomo wave equation forAwith nosources and6.53. 6.6The Green's function forthewave equation. Iremember doing thisindetail, andthere areseveral results tonote. Theoriginal equation is6.54 with a driving term shown anda4xsitting there too—ascalar time-dependent wave equation. TheGdefinition also hasthe-4masin6.55, andthen 6.56 says how wewould usethisGifweknew what itwas tothen solvethefullproblem.ThesolutionforGinmomentumspaceis6.59,andwhenwe4-Fourierthatback e@intoreal4-space wegetthefamous result which is6.64 which hasthetime-retarding delta function times 3 UR.SoifwelookathowweuseGtogetafullsolution,wehave6.65,andyouseethefamousI/R eappearing there. Ifyoudothetime integral, yougettheeven moré famous solution which is6.66. So:the solutionofthetime-dependent waveequationdrivenbyfunction-4xfisthespaceintegraloffatretarded time, divided byR. 6.7Solving thetime-dependent scalar wave equation with andwithout sources fusing theGreen's Function Method: thePoissoninitial-time solution, andtheKirchhoff Huygensintegral. Back onpage 18inelectrostatics, wehadshowed that1/RwastheGreen's Function oftheLaplace, and wecommented there thatyoucould then addtoyour formal 1/RGreen's function anysolution F(x,x') of thehomogeneous Laplace equation, andthatthisextra freedom allows ustomatch boundary conditions, such asmaking G=0onsome closed surface. This concept wassimply adding homo solutions tothe "particular" solution. Here wehave asimilar situation, Wehave found that6(time)/R istheformal Green's function, and ‘weought tobeable toaddtoitanother function Fwhich solves thehomogeneous wave equation. We nowhave ahyperbolic ODE sotheonly waytohave awinisspecifying thesolutionyandyy’onafinite ‘open surface. Inour4Dequation, sotospeak, an“open surface" isusually taken tobeallof3Dspace plus theboundary t=to,some initial time. Sotheidea istospecifyy andy’(normal derivative) ona closed 3Dsurface atthisinitial time. Remember thatthiswas"too thuch” intheelectrostatic case, butit iswhat we need here. But Jackson does not"add something totheGreen's Function" theway hedidback inthat electrostatics chapter. Instead, hesoft ofstarts from scratch. Using Green's Theorem, and aninfinite volumewithboundingsurfaceSatinfinity,wegettheresult6.70wherewecanjunkthatlastsurface r) integraltermbecausewejustassumeydropsoffthereIguess.Thereisobviously somecondition to think about, butJackson hashishands fullwithother details. Attime to=0,suppose wehavey andy" described bysome functions FandDasshow in6.71. Then ifyoupark yourselfattheoriginbysettingx=0,youcangetresult 6.73. Thefirst term isourfamiliar particular solution from 6.66. Added tothis, weseeasurface term atradius ct.Wehave found asolution totheproblem thatmeets theboundary conditions 6.71 anditisgood atanytime t.It'sonly weakness isthatitisonly atx=0,butIimagine the references show thegeneral result. Notice thatyouhave tointegrate theboundary value stuff over the entireclosed2Dsurface,all4steradians. Thisfancy6.73solutioniscalledthePoisson's solution. It seems thatthesurface integral samples ashell ofboth boundary value functions, Themore time passes t, thefarther outisthat shell being sampled. Iamnotsure exactly how onewould usethis, Ineed tosee some sample problems, butIthinkitisbeyondthisbookbutwehavesomereferences. Again, theboundary conditions FandDhave tobeknown ONLY attime t=0,that isthe"open surface"idea.Afterthat,ittakesoffonitsownaccording to6.73.Noticethatatalatertime,itissamplingthe surface functionsastheywereatt=0,butofcourse ithastogoouttodistance 1’=cttofind those values. Going back to6.70, suppose wecan drop theinitial value parts terms, and suppose there areno sources f,then wehave only thesurface term which isthelastterm in6.70 which wedropped earlier. So wehave then 6.74, Bymanipulation, Jackson converts thisto6.76 which gives thefieldyinsidea volumeintermsofasurfaceareaintegralaroundthevolumeofyandvariousderivativesofsame(space andtime). This isallbeing done inthetime domain, sowehave the"retarded" label ontheguts. This is theKirchhoffdiffraetion-typeintegralformula,butitlooksquiteforeigntomebecauseTamusedto eseeing inwith sine time dependence only, where wehave just two terms andnotthree, andwhere we have exp(ikR)/R andnotjust 1/R. Iknow thatwhen youundo theretardation intoatime integral with the 4 deltashowingandputinthesinetimedependence,thatishowyoupickuptheexp(ikR)factor.How t) threetermsbecome twotermsisprettyhazy,Iwouldhavetodotheconversion tokspaceindetail.//Well ifwesneak apeak ahead atpage 281, wehave arepeat ofourresult 6.76, andJackson takes this right tokspace andnowIamremindedthattherereallyare3terms,thoughweusuallydroponeofthem. This then serves asJackson's derivation ofthe Kirchhoff scalar formula, When hedoes his vector Kirchhoff, hestarts from scratch again, 6.8Poynting's Theorem (1884): Conservation ofEnergy Thisisaverygoodandquickderivation. Howcanwe"dowork"onasystemofEandBfields? Think ofapoint charge inmotion v.Theforce thatanE,Bsystem offields exerts onthischarge isF= QE+(1/c)vxB.TheworkdonebythefieldsonthechargeisdW=Fedx={qE+(I/c)vxB}dx. ‘Thework perunittime isthengiven asdW/dt ={qE+(1/c) vxB}ev=qE y. Theimportant point is thatthemagnetic field cannever dowork onaparticle because themagnetic force isalways atright angles tothevelocity vector oftheparticle! Now write (dW/dt)dV =Eevq&(r-a dV=Ee Sav. This isthework done persecond inasmall volume dVbythefields onacurrent. Sothisthen takes usto Jackson's starting point: powerappliedbyfieldstosources=fHeJdV. r)JacksonthengoofsaroundwithMaxwell'sequationsandmanagestorewritetheRHSaboveastwoterms,shown intheRHS of6.81. This 6.81 istheconservation ofenergy. Itsaysthat, intime interval dt, decrease instored energy = energy lostbyfields tosources +energy lostthrough bounding surface Indifferential form wehave 6.82. Ineither case weinterpret thequantity (c/4x) ExHas therateatwhich power flows outperunitarea, andthisiscalled S,Poynting's vector. Itisenergy flow perunit area, We have notdone anytime averages here, this isallinstantaneous. You could add anarbitrary curl of something toSandnotchange anything, butnooneever does that. Jackson hasassumed thatthemedia arelinear, otherwise youhave extra energy storage andlossmechanisms such ashysteresis, 6.9 Conservation ofLinear Momentum Wetake asimilar starting point withF=gE+(1/c)vxBastheforceoffieldsonacharge.Butthisis dp/dt, themomentum change put onto thecharge. Integrate this asin6.89 over dVtogetthetotal momentum transferred from fields tocharges. Think ofthisasanincrease inPe. Now thegame isto fiddle with theRHS of6.89 using thetwoMaxwells shown in6.90. After purely mathematical shufilings, weendupwith 6.100 asarestatement ofourequation. Now onereason fortheextra complication we encounter here isthatthething being conserved, linear momentum, isavector, whereas intheprevious section itwas thescalar: energy. ‘The resulting equation like 6.100 will beavector equation. Inorder to e maintainanelegantnotation,thedyadicnotationisused,butyoucanreallyjustregard6,100asthethree 5 componentequationsandforeachcomponentiwehavethedivergenceofTee;=Tjwhichwemay @regard asavector intheindex j.Buttheresult really istensor innfture, nogetting around that. Sowhat does 6.100 say? Thefirst issue istheidentification ofthelinear momentum Pp oftheE andBfields according to6.94. Jackson hasnotmentioned photons, butIwill, inorder toconfirm this result. Write: energypassingthruwallpieceofareadAintimedt=(c/4n)ExBdAdt=SdAdt #photonspassingthruwallpieceofareadAintimedt=(e/4zt)ExBdAdt/hao momentum perphoton =ha/e total mom passing thruwall piece ofarea dAintime dt= {(c/4n) ExB dAdt/he}*(Aca/e)=AP volume density ofmomentum P=AP/4V =AP/(dA(celt)] ={(cl4n) ExBdAét/he}*(Reale) /[dA(edt)] ={(c/4n) EXB}*(1/0?) =(1/4nc) ExB=Sic? Thisconfirms thefactor of1/c?thathegets.Onefactor comes from momentum perphoton, theothercomesfromthethicknessofthe volume. Solooking attheLHS of6.93 or6.100, weseethatthelefthand siderepresents therateofincrease inmechanical momentum ofthecharges plus therateofincrease inmomentum stored inthefield within ourvolume, Momentum isthus "coming infrom somewhere" andadding toboth these terms. Thus, in e 6.100,wehavetointerpret thevectorquantity neTasamomentum fluxonthesurfacethatisfeedingmomentum inthrough theboundary! Justaswemade adistinction between stored energy andenergy flux through theboundary, here wemake adistinction between stored momentum andmomentum flux through theboundary. What symbols arewegoingtouse: usenergy density $=energy flux P=momentum density F=momentum flux(theMaxwell stresstensor) ‘Now imagine thatatourwall wehave acomplete absorber onthe“right side". Then intimedt we knowthati«¥dAmomentum sinksintotheabsorber, sothepressure onthesurface willbe pressure onsurface=fi ©T Forexample, ifthe surface isinthez-constant plane, wehave 2 Pressure=2©T=Ty=avector Pressures =Ts3=(1/4z)[EsE3+ByBs-(1/2)(E’+B’)] e Thiswouldbethenormalpressureonthewall,pushingitback--wecallthisradiation pressure. However, forgeneral fields, there aregoing toalso betangential pressure components. Simple case isa 6 planewavehittinganabsorberatanangle.Weexpecttangentialpressurepushingthewallsideways,in r) additiontothenormalpressurecompénent.Application: putablack ballonaspindle andshine alaser atitatangrazing angle. Weexpect the laser tokeep theballrotating againstthelossoffriction, andwecould compute theforce ontheballusing thisTthing! {Hey guess what: Ithink thatiswhat aradiometer is!One invacuum should rotate away from thewhite ormirrored sides. Butones withgasgotheother direction duetoathermal effect. ] Question: what aretheabsolute intensities ofEfields inlight ofvarious kinds encountered every day? Think ofsunlight asakilowatt persquare meter, integrated from IRthrough UVIsuppose. Assume allatanaverage 2ofS000A. Conversion sayswehavethenS=10°erg/sec /em?. Thistellsus thatE*=10°erg/cm’, sothatE=10°dyne/esu. Thus, fieldstorage isontheorder ofmilli-ergs/cc and radiation pressure ismilli-dynes/em?, Suppose aradiometervaneweighs1gramandis1cm?inarea.The linear acceleration onthiscould be10°cm/sec” soin|second itisnotgoing veryfast.Toobad,websites saythatradiation pressure isnotenough andthatheat convection does it!Thetoyradiometer spins the wrong Way from aradiation pressure point ofview. 6.10 Macroscopic Equations Jackson hasalready come upwith expressions foruandSand#interms offields E,D,B,H. Inthis, section, hethinks again about themacro fields being now spatial andtemporal averages ofmicro fields (including thistime@andA).Oneimportantresultis6.112whichremindsusexactlyabouttheeffective chargeduetothepresenceofadielectric(usuallyonaboundary),aswellastheeffectivecurrentarising from magnetization Mandalsofrom P.ThisdP/dt term seems newtome,Naturally itwasnotpresent in thestaticequations!Jackson'smainpointinthissectionisthatheisdoingeverythingfromscratch,using e@only Maxwell's equations. [Sodoes this mean IcanuseJ=dP/dt tocompute adielectric radiation situation? Iwasalways sortofstumped onhow tostart inthecalculation ofAwhenyoucouldnotdothe little parts trick]. So,ifwestartwiththeenergyconservation lawintermsofthe DandHfields andstart stripping out thewrappings onthese fields, thelawappears as6.117. The point isthat wehave now identified the extra terms duetoMandP,andthese areshowing upasadders tothecurrent asin6.112. Thepoint is thatthefield notonly does work onfieecurrent Jintheform JeB, butitalso does work (atleast reactive work) onMandPexactly asshown inthisequation! You basically have inductance andcapacitive work staring you intheface here. Since these terms arereactive, andrepresent temporary storage andnot power loss, Jackson argues thatitisnatural toincorporate them intheDandHfield presentation. Problems. Interestingly, theones Ihave done orlooked atbelow arethesame ones (apart from 6)that JacksonhadusdowhenItooktheclass. Problem 6.6. See hand-written solution, shows that you can separate currents asclaimed forthe Coulomb Gauge. Problem 6.8: statement ofthemain chapter conserved quantities Wehave already derived theexpressions foru.Keep inmind theasymmetrical way inwhich DandH e aredefined: 7 D=cE VoD=4npice t) H=(1B VxH*(4n/c) Jee Sothefirstenergy formula hereforuhasED+BHinside. ‘TheSwearehappy with. The thing heiscalling ghere isourmomentum density#,andseemsthattheresulthereisDxB which isaslightly unexpected combination. Ifyoutrace things through onpage 192, this isinfactthe result. ‘Wecouldgotraceoutthederivation ofthetensor(IhavesomepencilmarksinthebookfromwhenI once didthis), andtheresult isthatyoualways have DEandHBateach quadratic location. SoIamnot really doing thisproblem, Iamjustreviewing theresults thatitstates. Problem 6.9: Angular Momentum oftheFieldsanditsConservation Law Tam notdoing thisproblem, juststudying itsresults. Ifwelook atangular momentum instead oflinear momentum, here arethethingswefind: L=rxP =ex (1/4nc)DxB] since 9=(/4nc)DxB Forangular momentum (avector!) weagain have toaskwhat isthetensor thatdescribes thedescribes the e2 flowofangularmomentumthroughaboundaryandtheanswerisM=Txr. eNow weknow thatinquantum mechanics wehave L=exp =(i)rxV p=@i)V ‘These aredifferential operators. One should notconfuse these operators with the£and #density functions foranEMfield. You cannot takeLsandapply ittosomething likeEorBorAor6andexpect tomagically getthesame thing multiplied by: 1andclaim thatyouhave located aphoton andthisisits spinrelative tothezaxis. Ontheother hand, youshould findthistobethecase when youapply Lstoa multipole field Yq, thatis,youarediagonal with eigenvalues m.Seenext couple ofproblems, Problem 6.11. Classical model foraphoton group TheEfieldshown hassomekindofunspecified slow(relative to4)transverse cutofffunction (could useaGaussian, asinaGaussian Beam),andhascircularpolarization, soyoumighttrytothinkofthisas some kind ofsemi-localized photon, When you setVE =0, you getexactly 0from thefunction shown. ‘The@,picks upikfrom theexponent, andthezfactors then cancel thed,and@,factors. Theform forB comes from theVxE Maxwell equation whereweagreetodropsecondderivatives ofEo. ‘Whatdoesthissetoffieldslooklike?TheBisalwaysperptoEduetothe90degree rotation ofthei factor. Thewave isgoing totheright. Inone, theBleads by90degrees, andintheother itlags bysame.‘Thereisaz.component duetothetransverse confinement ofthebeam,duetothedivergence requirement. 8g e@ Problem6.12,Continuation of6.11. ForL3,Iguess wemight compute wuand£foraninfinite cross section disk across thebeam, offinite thickness, Wewould count ontheEyfunction toconverge this integration. Wewould dothis forboth w and£andthen take theratio. There aresome things thatpuzzle mehere, however. What istitemeaningoftheobjectEeBhere?Weknowthetransverse partisgoingtobe0,butthisisnotobviouswiththe— well Iguess itjust works, yes,just doit,fine. Clearly thezdirection cannot converge sodon't tryto ‘integrate that way. Now let's assume wegettheresult shown, Ihave notdone this calculation, Ifwe interpret this plane wave assome numberNofphotonsallinthesamestate,thenw/N=oandLIN=+h andtheratio isthen exactly asshown, sothatwould beourinterpretation. Asforthefinal question, ifa field hascylsymmetry, then notransverse direction inspace ispicked outsoyou cannot have any transverse vector quantity benon-zero (isthisright..2) 9 eProofof(6.5)Jackson PhL 1.22.03 Imagine acontour thatismoving inadirection v.Imagine weareinterested intheareaintegral over a surface bounded bythiscontour ofsome vector field A, I=fAGedA cw) Ifwewant thetotal derivative ofI,wegetonecontribution from dA/dt, butwegetanother onebecause thecontour moves. The first term isobvious. The second term can bewritten like this: ( fA@edA -SAGedA ]/dt C(tdt) co Ifyoudraw apicture ofthecontour moving indirection v,youseethatthevalues ofAthatarepickedup attime trdt were thevalues thatAhadattbutatadistance shifted byvdt. This isjustbecause theentire contour anditsattached surface hasshifted byvdt.Thus wecanrewrite theabove as: | [fAcivdtyedA -fAQedA ]/ét=f(veVA@edA cH oO) 0) e wherethis6Afirsttermexpansion usedhereisobviousifyoujustwriteoutthecomponents. A(rta) =AG) +@*V)AQ +. ‘Now wehave this little vector identity, VxAxv=(veV)A- v(VeA) wherevisaconstantsoallgradients onitvanish,killingtwooftheusualtermsinthisidentity.Now,ifA happens tobeB,themagnetic field, weknow thatVeB =0soweignore thesecond term. Sowerewrite our term asfollows: . =SVx(AxvjedA cH) ButStokes letsusrelate anarea integral likethiswith acur!tolineintegral =f(Axvpedl ct) e@ andthisthenistherightmostterminJackson'sequation(6.5)onpage172. 1 Cidedom 6,6 2-2~08 — 24 |Gah= a(sree) Ways SPSe)a5,] .C90.JsshowactmyonR,andSG)cacanst mech. Sse] -1-9) O= “VAIO =WHET), 2°ODTheanew Wawed laneds sino het O+9=0- Wontmddoster: [a7 |We) veSede]=o.YowuaeSiafad: _f 9(SO)=Se9-F(RI +kSe) Dur permesaigunee -=\SoHtst =0 ae, e Chapa 8 Jackson, Chapter 8:"WaveGuides andResonant Cavitites.” . - @8.1Fictes atsurface ofondWithinaGdnductor. - . 1,Suppose.you,have2perfect, gonductor.on one.sidg.ofaplanesurface.Inelectrostatics, |youknowthat'B +0inside.the conductor. Sincethereisnocurrent, flowing inside, the conductor (statics), amagnetic fig}d goes.right through the.baundary. with nochange. (Ze, noeurface current either). Thus ifyouput,a magnet npan a,piece ofmetal, the, Bfield goes through with no-change, atthe surface. ; ‘The situatiqn isalittle different, fortime-varying fields. Ifslowly- varying, ' | youstillget=0Ansideconductor (sogecondition forthisapproxofcourserequires @large enough; we,agsume@®).,Alsoyoucannothaveatime-varying H-fieldinsidea | perfect conductor because, itwould: induce circular curents, ie, 1twould induce an |Bfield, but, you are not allowed tohave. anBfigld. .So4fyoushine sayaplane. wave- om@perfect conductor, there will beacertainsurface chargedensity endcurrent density s0.astoexactly cancel,EandBinsidethe | conductor. Justoutside thesurface, youknowthat,y=0.fromdivil-0, endyouknow that By=0from curlf =... Figure 8.1shows. thesituation. . .Ifconductor.hasfiniteconductivity,Yougetcertaini,andEB,asin(8.9)and r)(8.10)whichshowsthatHynowpenetrates toskindepthintotheconductor, andthet = there isactually some small Ey ash penetrating, though B,< Hy.Thus, there is also some Eyoutside thesurface, sosome Pointing energy isflowing into theconductor, obvious ohmic losses. - ., - Asfrequency-increases, skin, depth, decreases and ohmic loss increases. , ao ‘Now heassumés ‘perfect conduction agéin, dnd auniform cross section onyour ~ netallic guide. Inside issome material with pebutnow something like air, You write your Maxwells equations and insist that solution propagates down the guide with some kand frequency W. This reduces waveequations to(8.19); notice that both the &and jtcomponents ofEand Bobey this equation; notice that wehave only specified the2andtdependence oftheE,andB,fields. Ttturns out the a“babis" for describing any wave consists ofthe TEand TMmodes. ATE mode isoneinwhich E,=0idneitcally, andB,satisfies acertain boundary condition around thesurace ofsomecross section, namely, 4“™{ B,hastohit eachboundarywithzeroslope.TheTMmodesarethose'where B,=oidentically and @5,v0.wnonely B,=0wherever itbitetheboundary. (Notethatonesyouknow E,andB,forsomemode, youcancompute theB,andBytrivially from(8.19). ) Obviously for Tmmodes, the electric field problem islike aparticle inatwo dimensional box, orbetter yet, atimpani drum surface. Clearly there aregoing to beeigen-values ofk.. r Foragiven geometry, yousolve the2-dim BYproblem andyougetsome eigenvalues “~~ called %,.These thendetermine youreigen-k vectors kyasin(8.37). Defining the cutoff frequency obviously (for given mode) youofcourse getk=k(S), ie,adisperions relation. Youlearn that foragiven frequency d\,there areonly afinite number ofuseable (ie, kreal) kyvalues, ie, afinite number ofmodes. Analogous toparticle inpoetntial well where there are only afinite number ofbound states. Usuaily, the lowest mode ofallisthebottom TEmode. Soifyou choose geometry nicely, you can arrange for this mode tobethe only "activated" or"proagating™’ mode atagivenworking frequency. Foreachmode,thereisaf¥equency below‘which that mode wili not go. That isthe cutoff frequency ofthat mode. Inaddition tothe TEand TMmodes ofawaveguide, there isaspecial mode in whichB,=0andB,=0identically. Te,thisissortofaTEandT™mode,orTEM. |Ie,this mode propagates completely ‘trangversely, asifitwere inaninfinite medium. However, theelectric transverse field must satisfy anelectrostatic BVproblem, so this mode cannot goindide ahoBlow vavity. This TEM thing isthe principle way energy goes down acoax line orparallel transmission line. Inwaveguides, onthe other hand, ~youaretalking theTEandTMstuffbecause TEMdoesnotgo.atall. e@ S.bRectangular Waveguide. . Here your BVproblem has eigenvalues indicated bymand nintheusual way. Modes are this called TE, and1™,,. First index refers tofirst ‘dimension, Ifthet dimension islargest, thenTEqisthelowestmode(modewiththelowestcutofffrequency). Bychoosing a4b, youcanmake only this mode go;ifaebthen youarestuck with - two degenerate modes, maybe harder tocontrol atentrance and exit ofwav eguide. | . e JacksonVectorKirchhoff PhL 1.26.03 ®‘Thetraditional scalar form is(9.67) with both terms inthere (unlike Sommerfeld). Gisshown, theusual thing, Since eachcomponent ofEandBsatisfies thesamewave equation asy,weget(9.68) andthe same forB.Jackson then launches intoawhirlwind ofvector identities with farsurface integrals vanishing andwhen thedust settles, youhave (9.75) and(9.76). Ihave nodoubt hehasdone thisall correctly, myjobistomake sureIknowhowtointerpretandusetheresult. Atthispoint, itHielps tolookback atpage236where hediscusses "what happens” atthesurface ofa conductor.Youseethatn¢Eat2surfaceisameasureofthesurfacechargegthere.AndnxBisameasureofthesurfacecurrentK(different dimensions fromJ).Inotherwords,thepresenceofgallows orsupports anormal Efield, while thepresence ofasurface current Kallows atangential B.Theother twoitems arezero: tangential Eandnormal B.Inside themetal allfour field pieces are0.These conclusions arise trivially from thedivE =5andcurlB =Jequations applied toapillbox andloop straddling thesfirface. Ofcourse things don't change instantancously atthesurface, sowehave picture on page 239 showing that there issome skin thickness toallthis activity just below themathematical surface. Now back topage 285andthevector Kformulas. Thesurface Sinthese formulas isacomplete bounding surface intheGreen sense. Ifyouassume radiation field types, youcanlimit $toyour finite surface S1,some closed boundary located atsome localized spot inthemiddle ofyour volume. Sothisis what (9.77) issaying: hehastaken (9.75) andincluded only theinterior surface $1,andhaschanged thedefinition ofnso‘itpointsoutofthe interior ballandintoour"region ofinterest". Jackson goes onto"interpret" theterms in(9.75) or(9.77). Since these aresurface integrals, for those parts ofthesurface thatare"metal", wecaninterpret ngEasasurface chargegthatmustbethere e from ourgeneral surface rules. And nxBisasurface current. Butthen wehave tointerpret theother things likenxE andngB asmagical "magnetic"«andK.Interpretation isfree,whattheformulasaysis what counts. So,equations (9.77) anditscounterpart forBaretheequations touseifyouaredoing diffraction from some 3Dlocalized object (such asasphere, notnecessarily metal), Youhave tointegrate 9.77 over theentire surface ofthat object. Next, Jackson wants toflat-adapt thisgeneral formula foruseintheusual holes-in-metal applications. Heimagines onpage 286thattheinternal surface S1ismade intoapancake. Thetotal internal surface in thispicture isboth sides, sohehasnow redefined thesymbol S1sothatS1(in integral) =$1+SI’, You have tointegrate over both SIand SI’. This isdone in(9.68) butthetwo surface sides aresoclose that thenormal vector geometry isthesame, sothings canbewritten asin(9.78). This isjustafolding ofthe twosides intoanintegral over oneside, nothing new hasbeen done yet. Now look atthethree difference terms in(9.78) which areintegrated only over theright-side surface SI.Think oftheactual pancake right surface asmetal with holes, Thefirstterm will benon-zero only in theholes, since itcontains nxE,Jackson wants tomake theother twoterms vanish everywhere onSl.If hecanmake (9.79) betrue, then heachieves thatgoal. And (9.79) willbetrueandvarious required field conditions willalsobetrueifhemakes (9.80) betrue. These equations arebasically assumptions madeforfieldsE!andB'onthejeg?sideofthepancake,sortoflike doing image charges. Infact, wearereally doing Sommerfeld’s image charge method here, andwewillendupwith hisresult! Jackson imagines that currents andcharges runaround inside thepancake however they must inorder togenerate theE’andB’ fields heisassuming. Thepancake hasametal sheet only ontheright, Iguess, orifnot,then heimagines surfacechargesandsurfacecurrentsontheleftsurfaceofthemetal.Theideaisthatthesefieldsimagined e ontheleftside will affect things ontheleftside, butwedon't care about theleftside. The fields there are 1 to Justconstruedtomakethecomputationontherightsidebesimpler.(Jacksoncouldhaveclarifiedthis r) important pointIthinkbetterthanhédoes.)Theendresultisthen(9.81)whichyouintegrate onlyovertheholes onS1.Sothisisthevector version oftheKirchhoff formulafor flatsurfaces that wemost commonly use! Heswitches theyoperator toget(9.82) which seems easier towork with, since G= e(iKRYR. Hepoints outthatthisresult isEXACT ifyoureally know thetangential Efield inyour apertures. Now, let'slookcarefully at(9.81). Since G=exp(ikR)/4gR where R=|r-r,wecancompute: VG=(k7/4q) ™*[ /kR-1MKRP]R where yR=R Now wemake theusual assumption thatwearemany 2,away atourobservation point, soKR>>1,of R>>p,, soweget vGe ikRG Notethaty'G=-7G,sowepickupaminus signhere!Next, let'sput =2without giving upanything. Wethen getthisexact result, apart from theapproximation made above incomputing yG: EmDikldg¢dA'(CMIRY) (2)(Ey2EpRV(ez)) R=(ex,y-y,22) eTheintegration coordinate isr=(x'y,z’) andwemightaswellset2'=0rightnowtoget: E@)=Vig,SdA!RY) {2E(r)-2ER) } R=(ex;yy,2) where E,isthesangential fieldintheapertures, E,=(EB, 0) andtheintegral isonly overtheapertures. Thesecond term issmaller thanthefirstbyfactors like(x-x’Vz andthisterm canbeneglected intheusual diffraction situation where wearefarfrom asmall diffractor, thisisthez>> dbusiness. Ifweselect ourcoordinate system ofintegration such that2’=0,wethen get: Ete)=Vin,aA"EA)(CMR)AR) andthis isexactly theSommerfeld Iformula that Ilike, page 49ofGoodman. Notice that thefield direction just passes through. Reminder: thisfinal form isonly correct forz.>> d.Ifyouwant tolook closer in(oratlarge angles away from thezaxis), thenusetheprevious result above! Azcomponent of theresulting field should beexpected, since theradiated EisperptothelineR. Ofcourseonceyouhavecomputed Easabove,youcangetBfromthecurlequation, e B=(/k) yxE Masshownas(9.5). 2 TERevealsToskseeShopssushiropa»Posiatbynnage) RT a)Chinfunsuintinaginbak,|omeSeoSEES)ReRL .oeeeA Aoucrus.you.ovine, Shboll.yorivaafiA eeSASS NS eS rea ae a eeee pe ee ~ABs SSE Jee5GeRUD]BESEo)/R ~-—t “ees see ~~Dadegatepbedpink,aglace32Ale)|somoRelZ-RAM).Thor - mop abet paotie GSfhxe2)A@——__——---— es spe Fd Lt eefp 4 ©.be,ReSab oh Nenation peDEE_.we eeeBale ifsorlhingSonggaadeAintwd.gibiACE),Dainokgusitied Bodailedas,aiid, hs{2412|cuba,Ramflongre iadknachannediga 4,tudeodes, pe tes FaSerne ENeT OTR ae ee1 LitvenksWiedwikgrits 1gaa=.te-ke[EABL a a -ae Re A aatiasinhabale1!Redhae a{Seba ee ~ah dsWs : Se nNOS OeENiacanecompulakin dugatBandReeaeSeeOOReh fp 2ec etai2Peep OT 2ea=|Sertet2STEPSg paga iingadedt fo.Se ae a eeSe 2Suhel -5- te ee . eee ee | GiseiBakRoET(A h rood '¥ 5 rood ijtahgevesplerer ae eran see ee oP ep “a a a|geht ali-- a 7 —_TOF RTS oe TO .ro Guleyole quesBRC deat, Beata 2SG <LonerSomatfadbl =nGousnalayad Aahcape ath aman _ TroinpctundnaOueitaDaa 1ate i =tO nana oe iee—_:Bong:re“shone pda,ERtalE&Stelade.wena ae bo=aulgee poo.oe a +fee .ba _ oyhaeilNoss"SlodwtxTawanSHFsega 7 twee t eee aee — ++elyhee}yy miwip)J cern BOmagsiUpee,LkRenaatinpapanmttandiniene |4 —+}-arReape ="4---—+onae 4Bs -gg --—- geeeee en|-Mrannecads aalewlansk.Dspiagtifiandaco) b-—.doaWeer—f4t 2+ -4-f-- 4 po SaAP ELEPSEER oe DP |)BERR] nce is”eSCEke. kdehyfae)ekeSsSg.7afin tsjttaay(|sina amintararantt leak—no +Meordata GaspolenedoemnDey t|Woynshin nara oe_os+YyarinSatadl,iyodeaee 8nego uy LBs Eevee) epES eRe aSs) eemB - oop fd fo pe ae -- oe bar sfhackionin fasedaaboan oyBfiand.SARNCA)jomnchJagugasithbosap =bady.godegtmmiendeta. Vporsmatd SaleofWiryinase gagenbe aban ~SEED Aeeto=Ko fy -- —-————!Deedee gadis i~ateetal oktYe,earhateonaete ||oe4;Sancti4 +W—— tCI 1. rn oea we ee tot. ¢-lip te pode -4TL.ba . tus. y )Daf.pedabadShe shhaet solo oil ——2Rileyalas |tyBgeoten oemR aeeeOnaTLTF aeSaSetgfate208(165)|agon SeesbyeSoke"Tanwar____"___yaa__ Senneee eei ealaepeethAedoolgleagJna 2eeeee ee = sa Asah peesere asmock tbbeye:oan Phan oe i ™ ' — 1 i poe ore a poe tfip seh eiegpalsi a ‘ amekie. aloTe Ra=gtMaeEDEL. BQ, * : LRDEGrinQpeatercd KYbeKo Fiduiadono;age)=for.fea oo1flueBoa26alifeisha LEK, me0ya&Ser VE q OwPMime SHi: % > Foxoutdonenduin oe ~~ € - eJacksonChapter16Notes PHL 1.31.03+2.5.03 Chapter 16: Multipole Fields. This chapter shows how tosolve Maxwell's equations infullglory inspherical coordinates. Theusuale™ time dependence isassumed. The general idea isthatyouexpand thefields ineach region ofyour problem infullgenerality (which means all£mvalues andboth electric andmagnetic modes), then you require that field boundary conditions bemetattheboundaries. Several examples aregiven, and comments aremade regarding both atomic andnuclear emissions. When sources arepresent, itisshown how tocompute allthemoments. Inthegeneral case there are4such moments ineach émharmonic. As anexample ofaproblem with "sources", thecenter-fed antenna istreated with theusual King Smile assumption about thecurrent shape. Theresult iscompared with afulltreatment inChapter 9thatdidall themultipoles atonce, butagain intheKing Smile limit, Thesecond example isscattering from a conducting sphere where wegetheavily involved intheformalism. Ihave done athird example of scattering from adielectric sphere inaseparate document, asomewhat harder problem that the conducting sphere. ‘Thereason thatthemultipole expansion canhelpsolve problems istheusual one. Therotation group has"good quantum numbers" ¢andmsince each represents aconserved quantity. Ifthecorresponding operators commute with theHamiltonian, thesolution canbediagonalized intotheindividual partial waves émandthere isnointeraction between different partial waves. Sotheproblem issimplified intoa setofsmaller problems, oneineach partial wave. Within apartial wave, thegeneral form ofasolutionis equite limited, there areonly tworadial function possibilities. Ifyoudosomething likescatter offacube, theboundary conditions willbefunctions ofspherical angles,andthiswillmixadjacent harmonics of different ém,Ithink, thiscorresponding tothatfactthatthe"potential" thatthiscube represents isno longer spherically symmetrical. Forproblems with radiating sources, theexpansion isuseful inlong limit because theterms drop offquickly. 16.1 Solutions ofthescalar wave equation in(r,0,9). Back onpage 54where wehadtheLaplace ‘equation (wave equation withk’=0),weassumed U(n)/r astheradial form andgot(3.7) astheradial equation which ledtoréandr*"asthesolutions. Weended upwithanexpansion asin3.61onpage67. Here, wehavethatextra k?term, sotheradial functions aregoing tobedifferent. Theexpansion ofa sealar wave equation solution isgiven in16.4andtheradial functions turnouttobethespherical Bessel functions ofvarious types. Sooneway toshow theexpansion isin16.16 using theHankels. Naturally, Jackson takes thisopportunity totellusallabout these Besse! functions. Note thattheHankels arejust thoselinearcombinations ofthe jandnfunctions which match outgoing orincoming spherical waves at large argument. Now,theGreen'sfunction inthepresence ofk’+0isel“*/R asin16.18, Ofcourse wewanttoexpand this intheusual double-spherical-harmonic type formula and that turns outtobe16.22 which wecan compare totheLaplace version in3.70 page 69.Wenow have non-trivial radial functions inthisthing, they areofcourse thenew spherical Bessel functions. Forr,wehave theHankel which dies as¢/x, and forr.wehave thejtype which iswell-behaved attheorigin. Wearenowtopofpage542.Here,welookagainattheODEfortheYqufunctions.Jacksondefines tavector ofoperators which include 0/0) and4/20. Hecalls thisvector bythenameL,andidentifies this 1 thingsquaredastheLHSofthat ODE! Iknow thatthisisjustaspecific representation ofthese operators,andthattheyhaveproperties thatarerepresentation-independent. WegettheraisingandlowerL® operators asshown. Page 543 shows some commutators: TLL =0 (p,L]=0 Then 16.30 shows usthatwereally already know about L?, Remember from quantum wave mechanics thatp=(A/i)V and L=rx p=(Wi) rxV. These arethe representations ofpandLinthe"x,y,z"world,andyoucouldcompute themomentum sayofastateby doing this: <ylply> =<w|x><x|p|x’><x'ly> andthething inthemiddle is8(x-x’)(f/i)V (1think). Here in E&M Jackson isusing the same operators with f=1 just because they are convenient forhis manipulations which aregoing toinvolve thesame Yq,thatappear inspherical coordinates inQM. Inthis chapter wearenot"interpreting" Lverymuch, justusing itasatool.Ifyouwrite V?insphericals, you get16.30whereL?takescareoftheangular stuff.So,Iamhappywithallthisstuffonp542-543. 16.2 Multipole Expansion oftheFields. First offthebat,Jackson writes two3-equation setswhich are claimed tobeequivalent toMaxwell's 4equations. Ineach case, thefourth istruebecause divcurl=0. Solet'sgowith aBexpansion, since either willdo.Since Basavector solves thewave equation, weget 16.35. BUT, thisexpansion aswritten only solves thewave equation member ofour3-equation set.We have toimpose thatVeB =0onthecoefficients. This turns outtobeahuge mess, andJackson takes us through thegory details, andwhen thedust settles, wereplace 16.35 with asum with some as-yet-not shown coefficients over asetofbasic fields By,given by16.42. These have theform ofanyradial solutiontimesnotYmn(0,6)butLYqn(0,9). In16.45Jacksonnormalizes thesethingsandcallsthem t Xu(0,9). Heselects Xasaletter close toY,andXisavector. Now here isamajor point. Notice thatr*Bj,= 0because reL=0inanoperator sense from 16.27. (From mechanics, theangular momentum Lisperpendicular tobothrandp,sinceL=rxp.)Thismeans thatallthesolutions By,in16.42are“transverse magnetic" orTM.TheBfieldisexactly perpendicular tor. Note thatthere isingeneral noreason why thismust betrueinageneral solution, although itisina plane wave solution ofMaxwell's. Butthissetofsolutions does have theTMproperty. And ofcourse if youstartwithEinstead, yougettheE,,shownin15.44andthesewillallbeTEsolutions. Luckily thesetwosetsofsolutions formacomplete set!TheseEyeandB;marecalled"themultipole fields". Now TM isbetter called "electric" because weshall soon seethat thedriving source ofsuch fields is electric charge, whereas the"magnetic" TEsolutions aredriven bymagnetic charge. Recall right now from electric dipole radiation thatEstays parallel topnomatter where yougo,sothiscannot beaTE mode. Itturns outtobeaTMmodeandthedipoleBfieldisgoingtobefullytransverse. ‘SonowwehavetheGrand Finaleofthissection which isexpansion 16.47. Notice thatBandE shown have allpossible TM andTEcomponents each. The f,andg,arenotspecific radial functions, they aresome arbitrary linear combinations ofthetwo radial basis functions. Notice that thesame coefficients appear ineach equation likeagbecausethatisthewaythesolutions areproportioned, justdo it!Similarly, whatever thefunction f,is,itmust bethesame function inboth equations, andthesame for Be This discussion implies thatyoucanandshould study theindividual modes bythemselves. Thebig advantagehereisthatwecanapplysphericalsystemboundaryconditionsmostconveniently inthisform! e 2 1633PropertiesoftheMultipoleFields.Thefirstorderofbusinessistostudythesefieldsatverysmall rdandvery large radius r, ‘ as “ First thesmall rsituation. Forelectric, weget15.61 which replicates ourmultipole expansion equation 4.1fromelectrostatics, and16.48goeswithit.TheBfieldhereiskrtimessmallerinsizethat theEfield, another reason wemight callthis"electric" instead ofTM. Forthe"magnetic" modes, opposite istrue, andyoucanalways reverse things with 16.52, Now forlarge rwith ourusual outgoing boundary condition. Inthiscase theradial function hastobe theHankel-1 which hastheright large-r form. (Note thatHankel 2isthecomplex conjugate and therefore goes asexp(-ikr)/r which isincoming, notoutgoing) Fortheelectric wegetBasin16.53 asthe Jarge-r limit, andthen Eistheusual far-field Bxnresult, sointhefarzone BandEareequal insize. Note Well: allmultipoles have solutions which goas1/randthuscanexist inthelarge-r limit, there is nothing special about dipole. ‘Now westayinthislarge-r limit andaskthequestions: howmuch energy, andhowmuch angular momentum iscarted offbyagiven multipole field? Onpage 548Jackson decides tolook attheelectric multipoles {Isuspect theconclusions arethesame forthemagnetics, butIdon'tthinkhecommentson this}andhecomputes theanswers tothetwoquestions. Hegives theelectric multipole itsusual coefficient agas shown in16.57. Wearegoing toassume thatthisisallwehave atthemoment. Recall from page 200thattheinstantaneous linear momentum density is?=wsS/c? =S/v? where S =(c/4n) ExH..Also recall from thatpage thattheinstantaneous angular momentum density is=rx9. Ifweuseourtime average theorem (seeChap 7notes), thenweadd1/2andput*onthesecond field, so wethenhave P=(1/8nc) ExH*inthissense, and£=r x#=(1/8nc) rx(ExH*). Instead ofusing the symbol £,Jackson usessymbol mforthisin16.61. {Hedoes notwant toconfuse £with L} So,howmuchUandhowmuch£iscontainedinaparticularmultipolefield?Thetimeaveraged e energy dUcontained inashell ofthickness drisgiven by16.60. Thetimeaveraged£inthesameshellis given in16.64, Inboth cases, theangle stuff isexact, andweuselarge-r (radiation zone) fortheradial stuff. Yes, only £5does notvanish because oftheraising/lowering business. Wegetthefamous result then that£/U=mi/hw and£;=£2= 0.Wearesaying thatanelectric multipole field éminthe radiation zone only hasangular momentum about thepolar zaxis andtheamount is"asif"thefield were made of"photons" having energy Awandhaving L3=mfi.This issome kindofspecial £mphoton! Onpage549Jackson triestoexplain whywasdon'tgetan&(¢+i) formfor|M]thewaywedoin quantum mechanics, buthisdiscussion evades me,heisreally quoting from another source. This isone example ofwhyengineers probably don't likeJackson's book much, because ofthis"real physics stuff", Now, let's talkabout how this"ém photon" isproduced byquantum state transitions ofmatter, perhaps from atoms. Suppose thequantum states are[JM> and|M'>. Then theselection rules arethese: J’must becontained within theseries J@£=J-e upto J+é M=M+m //Jacksonhastheothersignonm,Idon'tseewhy!forparity conservation, must have Parity(\S'M>) =Parity(\JM>) *(-1)! forelectric, (-1)"" formagnetic Thefirsttworules arejustconservation ofangular momentum . Thisprocessofemittingan£mphotoniscalleda"multipoletransition", Idon'thaveanyideaofwhat isreally going oninsuch atransition. Ingeneral itmust involve multiple truephotons, since each photon onlyreallyhasj=1.DoIhaveabookwherethissubjectisdiscussed? OnthewebIsceaguysayingthat ®inspherical geometry, youreally dohave émphotons. Ihave toimagine thisasacoherent state of 3 multiple photons -amulti-photon state thathasquantum numbers ém.Still, various people dospeak of theémthing as"aphoton" Iwillhave toputthissubject offtoanotherday.Ithinkthereareinteresting e implications. Although Ihave many physics books, none ofthem seem toaddress thissubject, butIhave notlooked indepth. (Question: inpositronium wehave sl,s2andLcombining tomake J.Ina2-photon state, isthere something like L?) Thebottom lineformeregarding multipoles isthis: probably inadielectric material likewater, selection rules areprobably going tocause alltransitions other than electric dipole(electric, =1)tobe very weak. Ifamaterial canemit €=5, thatisfine, andthiswillradiate justasfaras€=1, Butthis probably does nothappen innormal materials. 16.4Angular Distribution oftheMultipole Fields. Recall thatwehadthegeneral multipole expansion forBandEshowing inGrand Finale 16.47. Atthattime, wewere supposed tothink ofthef,type function asanarbitrary lincomb asin16.43 with twomore arbitrary coefficients thatdepend oné, Since wehave another coefficient layer ontopofthis(theagforexample),wecoulddecidetonormalizethese lower level coefficients sothesum ofsquares is1.Inanyevent, inthefarzone where wehave only hl, wenow give itaunity lower level coefficient. Doing this, wethen arrive at16,69 forthelarge-r form of thefullfancyexpansion! Inthislimit,thetwofieldshavethesimplenxrelationship, andnowweseethe all-important upper level coefficients exposed inthislimit. Gettingthepowerperanglefunction 16.70isatrivial step. Forasingle pole term weget16.71 in termsofthe fancy Xharmonics. Ifyouexpand thatoutinterms oftheYfunctions, yougetamixture or raising, lowering, andnon-changing Loperators, soweexpect toseeamixture ofadjacent mvalues. The result isstated in16.72 without proof, Jackson's table onpage 551thenshows theexact results oftheX objectforZe1and€=2,Thedipoleinterpretation isthis:m=Otermwouldcomefromaz-aligned dipole @ofsizep=a/k’asIlearn bycomparing to9.23. Ifyouhadanx-aligned, youwould getamixof m=41 inthe usual combination. Weseetheplots forthe£=1,2 multiple cases (allofthem). Jackson's comment isagood oneonpage 553: Ifyouhave some atoms radiating thermally, say,youwould expect anincoherent sumofallthem multipoles foragiven é,and16.73 shows thattheresult isisotropic asitofcourse hastobe.But, ifwe stimulate atomswithaspecialfield(planewave,eg),thiswillnotbethecase,Isuspect.Heconcludes by integrating thedifferential power over solid angle togetvery simple results. 16.5 Sources ofMultipole Radiation: theMoments. Digression onmagnetization M.Look back atpage 151inthemagnetics areawhich Iskipped onthis reading. Equation 5,77 shows thenotion ofamagnetization densityM=n<m>inexactlythewaywe talked about P=n<p> onpage 118. Asinthepcase, you canhave already-existing mdipoles ina material likeironwhich justgetlined up,oryoucaninduce your ownmdipoles.Interestingly, Jackson does nothave alittle section on"models forM"inhisbook, theway hedoes forP.Portis does discuss thisonpage 243, however. Roughly setting M=Bandstudying asimple model foranelectron, hegets result (24) forx.This is"diamagnetism". Electrons tend tospinaround inaBfield incircles andmake moments m,soindiamagnetism wearecreating these moments. Inparamagnetism these orbits already exist andwethermally linethem up,giving 38.Spin gets intotheactaswell, soyoucanseethatthisisa complicated subject. Now, forourcurrent purposes, themain point tonote about Misthat itspresence in amaterial makes aneffective current density whic isJy =©VxMason page 152. Thisis analogousto pp=-VeP asonpage 112.Recall thatyoucansaythatVeE =4x(Pise +Pe)butVeD =4npiee where Dis 4 defined exactly soitonlyseesfreecharge. Insimilar fashion youdefine HsothatVxB=(4ndc)(Suet @ Jy)butVxH=(4n/c) Since- . So,aswestart outhere, Jackson seems tobesettingp=Pree+Pr,buthehasJ=Jeeandhebreaksout theMstuffseparately,callingitscriptM.Next,JacksonhastodefineanewfieldvariableE!asin16.79 toincorporate thefreeJjusttosimplify themanipulations thatarecoming. Thefirstorder ofbusiness istonow rewrite our"wave equation sets" byadding thesources, andthis isdone in16,80 and81.Theresult isanincredible mess! Youwonder whypeople don't dotheexpansion inthevector potential Awhere thesources aresimple, butIguess thereason isthatwedon't know how to putboundary conditions onAtofindasolution. SotheBwave equation isdriven byV xJandadouble curlonMwhich seems pretty clear. TheE’equation issymmetrical inthatitisdriven byV xManda double curiofJ.Hey, where didpgo? Now weseethemotivation forE'!Itisdefined exactly sothat ‘VeE' =0.TheJterm isadded toE’andthen VeJisknown interms ofpfromcontinuity. Sotheanswer isthatthewave equations don't have pbecause they come from theMaxwell curls only. Then thisEtrick gets ridofpasjustmentioned. Fine. ‘Now things aregoing togetVERY serious! First, thenew BandEequations match theformer B andEequations outside thesources, sothere weknow theFinale 16.47 applies forBandE’(another reason todoE'asdone). Soletstrysomething like16.82 with asymptotic limits 16.83 tomake things consistent with ourgeneral notation inthefarzone, That is,this16.83 issetting thescale onour functions fandg.Jackson then swirls around tocome upwithODE's forfandgwhich aredriven by complex source terms, seefirsttwoequations onpage 555. TheODE leftsides areexactly what wehad earlier wayback in16.5page 539(thespherical Bessel operator, sotospeak). Wealready computed the e Green'sfunction forthisoperator onpage54116.21,andnowwearegoingtouseit!ItisjustaproductoftwoBessel functions, theinner onej,theouter oneh.So,creating ashorthand notation Keforthe messy right sideof16.84, weinstantly have theGreen's solution 16.86 forf.Taking thelarge rlimit, we arethen able toidentify theagcoefficient, andthis isthen stated in16.88. Same foraybelow that Obviously, these things are"moments" ofthesource situation thatdrive themultipole fields. Thetworesults forthese moments arethen “simplified” andpresented again onpage 556. Each one isnow anintegral ofsource stuff against aYopfunction, thewhole thing volume-integrated over the source region, andwenotice thatourfriend phasreappeared inthefirstequation. Let's pause tothink about this. Ifwestimulate awater sphere with aplane wave, wecanprobably decompose theplane wave into multipole components (soon!). Wecanthen think ofthese assomehow inducing aJ,pandMinthewater droplet --andthatstage ofcourse involves the"physics" ofthematter known aswater; pethaps there isnosignificant M.Those induced sources then inturn imply some multipole moments according totheintegrals shown onpage 556,andthese thengiveradiated multipole fields forBandE'Weknow what each moment field E’orBlooks likeforlarge r,andintheory wecan compute thedetails atanyr,butthen wehave todothedetailed integrals asin16.86. This isbecause we need toknow thefandgfunctions which appear inassumed 16.82. Thegood news isthatwehave a general solution totheentire problem interms ofsome integrations. Bytheway, tocarry outthe induction program mentioned here, wehave toassume theradiated fields arelocally weak compared to theincoming field, something Iwanttoseeaproofofhereatsomepoint. Now inthelarge 2limit (meaning kd>>I where source hassized),things simplify because wecan take limits oftheBessel functions. Weendupwith four kinds ofmoments called Q,Q’,MandM’as detailedonpage556-7.WerecognizeQasourelectrostatic moments! Wehaveag~Q+Q'andthis @second Q'thing canarise only ifyouhave M,andinthatcase, Q’<<Q.Sothemain idea isthatagis 5 mainly driven byp,andthat iswhy wecalled theTM modes "electric". Wealso have ay~M +M;,whereMisdrivenbythefreecurrentJ,andM'comesonlyifthereisM.MandM'areusuallythesame @ sizesoyouhave toworry about both ofthem ifMispresent. Now weseewhy heused script M—-to avoid confusion with these final long-2. moments! Sowow, again, wehave acomplete solution ofthings inaform geared todistant radiation problems from alocalized source that issmall relative to2. 16.6 Radiation from Atoms and Nuclei. Another section very important tome. For both atoms and nuclei, wearegoing tousethesame format "ball park" estimates fortheelectric andmagnetic sources Rule 1:forEMradiation byeither atoms ornuclei, weareusually inthelonga. limit kd>>1.Certainly foratoms and visible light this isthecase, compare 1AtoS0Q0A. Inour simple atomic model, our estimate forkdiska~(Zzr/137)*. Fornuclear, kacover toolarge arange forustosaysomething about it. Tosaythatwehave apthat isgoing todrive aparticular multiple electric field istosay16.99. Then intheintegral 16.94 onlyoneQmisgenerated. Thedimensions require some kindofe/a?factor, where e isthechargeofthe"thing”thatisradiating (electron inatom,protoninnucleus), whileaisthesizeofthe thing youareinterested (atomic ornuclear radius). Jackson throws inafactor of3here probably justto make things look nicer down theroad, nocomment isgiven. Maybe each atom has3radiating electrons? Ithinkitisjusttomakethingssimplerinourlatercomparisons. SoIagreethat16.99isreasonableforp. Now 16.101 istrickier tounderstand. Here, hecombines theMand M'integrands together. We dimly recall how magnetic dipoles areenhanced byg-factors, sowewill allow such adimensionless factor.ThebasicunitisgoingtobetheBohrmageton foratoms.InLiveseypage163weseethatthisisetheclassical magnetic dipolemomentofacirculating electron m=ef/2m,usuallycalledyis. Theelectron spin onitsown hasasimilar moment, give ortake afactor of2.Forthenucleus,theappropriate massis m,andagain moments come from both “orbital” and"spin" presence inthenucleus. So,in16.101 wehave thegfactor, wehavea1/a’forourvolume dimension factor, andwehave an appropriate magneton factor. The I/rthere must beduetoourhaving VeM andnotM.Note thatin spherical coordinates, thedivergence ofaconstant vectorA=A®isVeA=A(2/t).Alllthedivergenceterms have aI/rinthem duetothe?drvolume factor, soIamhappy tohave aI/r,andweneed itas welljusttohave thedimensions beright. Finally, wehave ourusual Y_yforthisassumed multipole. Using these crude estimates forthedriving factors ofourmoment integrals, wegetthethree simple results shown onpage 558forQ,Q'andM+M’. Right offthebat,wegetamajor result: Rule 2;Foreither atoms ornuclei, Q’<<< Qduetorelation 16.103, soforget Q'(the "induced electric") Atthesame time, weuse16.98 totalk about ourtransition rates forradiation between two quantum states, That is,wearesimply dividing thetotal power (ascomputed in16.97 foramultipole) bythe photon energy which hasdimensions I/time. Weassociated 1/¢with therateoftransition measured in number ofphotons/sec. Using thisgeneral formula andinserting ourelectric result 16.100 weget16.104 forour transition rate. ‘Aside: {Inanswer toanearlier question Ihad,youcanaddjt¢BintoaHamiltonian andgetquantum splitlevelsduetothisthingandIpresumethesewouldbeMItransitions. } e 6 Iht Dt aonhile Now,in16.105wearecomparingtherateofelectric,versusmagneticforthesameorderé,andthenin @ 16.106 wecompare order£toorderé+1,whichcomparison isthesameforeitherelectricormagnetic. Atomic Radiation. Inthiscase, thefactor in16.105 isroughly (Zy/137)* which isaverysmall number unless youaretalking X-rays from adeep inner shell ofaveryheavyatom.Forwaterthisisgoingtobe onthe orderof107to10%.Sowegetthisrule: Rule3;Inatomic radiation oforder £,magnetic isweaker thanelectric by10?to10°[(Zeq /137)°]. So ‘youwillseethemagnetic only iftheelectric issomehow forbidden byselection rules. Rule4;Ineitheratomicofnuclearradiation ofeithertype(MorE),thetransition rateinorder£+1is smaller thaninorder£by~(kd)*~(Zyp/137), asin16.108. Therefore, within amultipole family (Mor E),thelowest allowed £dominates. Usually thisiselectric dipole fortheEfamily. So,when electric dipole isfully allowed, youdon't have tothink about higher order Qmoments, assuming theyare "stimulated" inasimilar strength inaninduced radiation situation. ‘Thus, even ifwefindthataplane wave haslarge higher Qmoments beyond é=1,theyarenotgoing tomatter when wescatter lightoff water drops. Nowsomenuclear-only comments, startingbottomofpage559. From the16.104 estimate, youcanplotlifetime versus @(ora=E)asshown, andyougetadifferent lineforeachorder£duetothepowerlawyouseesittingin16.104,whichis@**soonloglogplotwe e getlogt=-(2¢+1) logE+constant(e). Onepoint tobemade hereisthis: forafixed energy release E( draw avertical lineintheplot), youfindthatthehigher order£transitionshavelongertimesandlower transitionrates. Jackson claims thatifyoustudy aboatload ofnuclear EMtransitions, they fallinbands close tothelines shown inthefigure. ‘Nowfornuclear,whataboutratio16.105forEversusMofthesameorder£?Ingeneral,electricare stronger by25-120 times except for£=1where theElisinhibited byacharacteristic ofnuclei (perhapsnaivelyyoudon'tseemuchapmomentinaballofprotons),soElandMIareaboutthesameinthisone case, Now aslightly different reading onratio 16.106. Foragiven parity, ifthelowest order ofMisé,then thelowest order inthEfamily forthesame transition willbe£+1,so16.110 isaratioyoureally care about.Innuclei,youmightfindthatQ2is5%ofM1asbeingtypical.SoifthelowestallowedisM,then |youmight have some noticeable Einthenext higher order. However, ifthelowest allowed isE,,then 16.111 shows thatthenextparity allowed transition M,s;iscompletely negligible. 16.7 Thevery famous center fedantenna. Total antenna length isd.DoNOT assume kd>>1.Study | thiscase very carefully please. There isnoM,soignore those terms in16.91 and 16.92. Inside the antennawires,rxJ=0.Thiskillsofftheonlyremaining termintheayequation, s0amajorfastresult: (1)there arenomagnetic modes atall,soallmodes aregoing tobeelectrics suchaselectric dipole E] ‘Nowforcurrent,JacksonforthemomentjustassumessomeunknownI(1)thatvanishesattheendsofthe e antenna wires. Hecanthenwrite Jasin16.113 andthenpasin16.114, using delta functions. Since Jis 1 ee inthefdirection, andsincewehavereJintheagequation, theintegrand hasnodependence, whichmeansthat: @ (2)only them=0 electric modes survive. Weknew thisanyway from symmetry. Now, theintegrand hasno@dependence either, apart from theYm function. Since theeven €Yo functions areodd, they arekilled offaswell, so (3)only theé=oddmodes survive. Sowenow want tocompute upthesurviving modes which are£oddandm=0using16.118,andnowwe need amode! forthecurrent I(1). The wave equation suggests exp(ikz) type dependence along thewires, asifwehadaplane wave inspace along thezdirection. Ifyou made aparallel wire transmission line, youwould expecttoseeexp(ikz)alongtheline,butthisisanothersubjectJacksondoesnottreatinhis book (well maybe hedoes). You could regard theconductors asmere boundary conditions andthewave travels along them atthefullspeed oflight with exp(ikz) intheideal case, When there areohmic and radiative losses, youhave tomodify thisresult. SoJackson issaying letstrytoignore radiation loss and ignore ohmic lossandputintheidealized current shape which then is16.119 since ithastovanish atthe ends. Boom, youcannow gettheresult asin16.1201 Alldone! This isthegeneral result (with our assumptions asstated) regardless oftherelationship between dand2. Jackson next considers thehalf-wave andfull-wave antennas asexamples. Fortheshorter antenna, thetable shows that theElismost ofit,and theE3isabout 5% and theESis0.1%, wearetalking amplitudehere,Forthelongerantenna,thingsconvergemoreslowly.TheE3isnow33%ofEllandtherESis3%inamplitude. Things getmessier when theyarelarge compared to2,keep thatinmind! Hegoes ontocompute thepower distribution forthesum ofE1+E3forthese twoantennas. The math isdone onpage 565with some mildly messy numerical results. ‘Now back inChapter 9wesolved thearbitrary length center-fed antenna with thesame assumptions ashere(same current assumed, same ignoring ofradiative andohmic losses), butthere weused thefull power far-zone result (9.8) sowecomputed allorders atonce! Here wearejustdoing El+E3,soitis interesting tocompare theresults. Jackson didwrite down thehalfandfullwave “exact” results onpage 279, sowecancompare ourapproximation here tothat. Theplots onpage 566areforthispurpose. ‘Aside: Irecall how myHarvard independent study guyKing used topooh-pooh books likeJackson's forassuming thecurrent isnotaffected bytheradiation. "Real" antenna people likeKing take thisinto consideration.. Recall also thatIhaveawholefatTransmission LineTheorybookbythissameKing. 16.8 Expand aplane wave insphericals. The result wewant is16.127 or16.128, andJackson derives these quickly byjusttaking thelarge rlimit oftheexp(ikRYR result wealready found in16.22. Hegetsheretousehis"sumrule"foundcarlier.(Itisalwaysnicetogetfutureuseoutofanobscureresult!) Look at16.129. Forsure all£orders appear here, although only with m=0. The z-axis here canbe thought ofasthekaxisinwhich case [email protected],soyou cannot sayaplane wave isadipole Eloranything likethat! Infact, aplane wave isgoing totryto induce allYomodes intoanobject thatwillradiate. Weshould seethishappen intheconducting sphere example tofollow. Some dayIwilltrytographically addtheterms toseehowaplane wave arises here! ‘Now Jackson starts over without clearly explaining hismotivation, butIamnow aware ofit.Inthissection,Jacksonsays:"Thinkabout16.128butnowaddaunitvectortoitsoyouhavearealfield=@ 8 insteadofjustascalarfunction.Infact,takethetraditionallinearcombinations ofthoseunitvectorsthat @ define thecircular polarizations. Welikethese because theyaregoing toallow ustouseLsoperators in ourcalculation which willgreatly simplify things. Now using this, let'sderive result (16.139). Weare going toneed thisresult inthenextsection intheproblem ofscattering from aconducting sphere! " Sidetrack andConfusion: Itriedtoshow thethese circular polarizations somehow diagonalizeL, =-id with =1eigenvalues, butconcluded thatinfact, foragiven tandz,circ polplane wave hasno dependence, soL,=0.Infact, if|compute Le™*"IfindthatdL=Le™"dv=kel* sind§dVfora littlevolume dVatlocation 1,0,4 relative tok.If]integrate thisaround the@ringatr,8,wegetzero.So thentheintegral overtheentire plane wave isalso0.Sothetotal "angular momentum" ofaplanewaveis 0.BUT,IhaveherenotreallycomputedtheLofafield,onlyofascalarfunctione™*" ,andLcanonly beapplied toascalar function. Still, onemight talkaboutL Ewhere wereally mean LE; andthen itis true. Inaplane wave, wehave photons traveling inthekdirection, sotheringintegral justdescribed really iszero. Idon't seehow thecircpolchanges thissituation, However: ifwelook intheproblems onpage 200, weseethatJackson hasinfactpondered this question with amedium ¢,u.Intheabove paragraph, Iwasthinking ofLastheoperator L=(1/i) rxV. However, onpage 200weshow thataEMfield hasanangular momentum density£=rx9wherethis thing#isthelinear-momentum densityofthefield,givenby?=S/v?whereSisthePoynting vector, andv=velocity =cA/pe ,soP=peS/c®.Inproblems 6.11and6.12,which havenowmoreorless done, weshow thatalocalized plane wave (Gaussian beam like) withcircpoldoes infacthave az component ofangular momentum thatis:+1 when scaled intheusual wayNA/fo. Somyconfusion was toconfusepwith@andLwith£Also,inthefollowingwhenweexpandatrueplanewavein e sphericals, wewill again find that foreach multipole componentéofsuchaplaneway,westillgetaz component ofangular momentum thatis+1. Here theunits arem=+1.There arenocomponents ofa plane wave that have m>1. Resume: Sohowdoes Jackson derive 16.139? Hestarts byassuming theexpansion 16.131 foraplane wave ofcircular polarization. This istheGrand Finale 16.47 where wehave assumed jforboth radial functions. Thereason fordoing thisisthatanything elseblows upat0,butplane waves donotblow up there. From thisexpansion, wecanextract thecoefficients asin16.133,4. Now in16.133 forEweput oursimple form16.130, theplanewaveforme"*.Thisisthepointatwhich ourchoice ofcircpolgives ustheL.operators in16.135, resulting in16.136. Atthispoint weshove inourearlier-expansion 16.127 foreandoutcome simple results 16.137,8. Thus, wenowknow thecoefficients inouroriginal expansion 16.131 andtheresult is16.139. Notice thatthemsumwent away because allthecoefficients only exist form=1orm=-1, depending onwhich circpolyouaretalking about. Ifwethink ofourplane wave asphotons, although westillgetasum over all£(recall theém photons discussed earlier), weonly get\m/=1, suggesting thattheplane wave does have acomponent of, angular momentum about thezaxisthatisinfact: 1.This makes usthink ofaphoton with twostates of Jongitudina} polarization, butweknow photons only have transverse polarization, sonotquite clear how youfitthephoton picture inwith ourplane wave expansion! Defer thistoarainy day. 16.9 The Conducting Sphere Scattering Problem! e Thisisabigproblem, andisprobably theclosestthinginJacksontomydielectric waterspherescatteringproblem, sopayclose attention! Seedetails ofthesolution attheendofthisdocument! Weassume 9 oo separate forms forincoming andoutgoing fields. Fortheincoming fields, weuseourplane waveexpanded intheveryfancy16.139thatwejustpondered intheprevious section.Fortheoutgoing weuse @ 16.141 where NOW (unlike intheplane wave expansion inthelastsection) weputHankel-1 everywhere because italone hastheright asymptotic behavior e™/r. Now, why doweonly have m=+1inexpansion 16.141? Jackson isuncharacteristically silent onthis question. Wesuspect ithastodowith conservation of2s.Iwill create myown argument. Suppose we localize theincoming plane wave toavery long butfinite packet, byslightly mixing frequencies. Then wecantalkabout "before" and"after" thescattering process. Thesphere might absorb some energy inthe process,butmaybeitisnotgoingtoabsorbanyangularmomentum. Ifweknewthistobetrue,thenwemightarguethattheoutgoing wavemusthavethesameLsastheincoming. Butstillthisseemsvagueto me. Let's seeifFizpatrick hasanything tosayabout this. This isaguyontheweb who hasmade lotsof Jackson-like notes. Here isthetoplevel forhisnotes from oneclass: hitp://farside.ph.utexas.edu/~rfitzp/teaching/jk1 /lectures/lectures.html This istheHTML presentation ofhisnotes. Hehasanother web page http://farside.ph.utexas.edw/—~rfitzp/teaching htm! which shows ALL hiscourse, andalso leads toPDF versions ofhisnotes. These aremuch better since the equation numbers arecorrect. Ihave downloaded hismultipole notes. Hisnotes follow extremely closely toJackson, same notation, headmits thatJackson ishismain source.Hemakesthiscomment: “aspherically symmetric scatterer cannotcoupledifferent mcomponents”. eWellthatisnotmuchofaproof?IjustaskedJimaboutthis. Ithink therealanswer isthis: youcould assume allthemterms inthescattered wave andthen you would have coefficients a(¢,m) intheoutgoing equation. Butwhen youthen tried tohave thetotal fields satisfy theboundary conditions atthesphere, youwould findthattheother mwaves have tobezero or youcannot make thematch. Inother words, what youwould getissomething likethis: FeB=0=> Lflémsma)Xyu =F(E,t=a) Xeu andyou would then saythat since each X¢q hasitsown characteristic angular dependence, thematch requires that alltheother mcoefficients vanish. This would betrue foranyboundary conditions that were "spherically symmetric" because thesame general argument above applies. SoOK. Tnow accept 16.141 asthegeneral form ofthescattered wave. The boundary conditions are extremely elegant, being 16.142 where8=.Theform16.143isveryhelpfulinarrivingattheboundary conditions 16.144,alongwiththefactthat*Xjq=0. NoticethatthefeB=0equation determines the @coefficients, andthefxE=0determines theBones. Ifyouwrite these coefficients only using hankels, youseethateach ratio in16.145 isjustaphasor asshown in16.146, sothismakes foravery simple formula forthecoefficients 16.148. Thesolution totheentire problem isinthese little phase shifts called 8,and8,which arefunctions ofk,aand£.The large andsmall klimits areshown in16.149, 150. Noticethatand6areassociated withthemagnetic (TE)fieldsin16.141.Thisisthecoefficient calledayin16.47. t 10 Sothefinalresult forthescattered wave is,statedin16.151wheretheoandBaredoneasphasors @ times sines. Keepinmind: thisistheexactresult forthisproblemi, evenverycloseintothesphere, sowe really DOhave thii,problem solved. Intheradiation zone thelimit forBisshown in16.152. Since the phases inapartial Wavearenotlikelytobeequal, thepolarization ineachpartial waveoftheresulting wave iselliptical. Jackson goes ontowrite down thedifferential cross section asin16.155 andheremarks that itis “rather complicated". Inthedielectric sphere problem, Jimthought youneed 40partial waves before you actually seetherainbow stuff. Wecanofcourse dothelarge andsmall klimits (relative tosphere radius a).Theexact total cross section however canbedone (asisoften thecase, recall unitarity) andyoufind thateachpartialwavecontributes as(2¢+1)timesthesumofthesinesofthephaseshifts.ThelongA(smallk)limitisdominated by£=1.Notice,bytheway,that£0playsnoroleinthis problem. You seethat£=0ismissing from 16.139, theincident wave. Infact, Xep=LYo~ 1xVYoo=0 because Yooisjust aconstant, Soanyway, the&=1dominates andisstated in16,157 which isrestated in 16.160 andisthesame foreither circpolincident state. Notice thatboth the and8terms contribute to this result, soitisamix ofElandM1.Theplotshowsthatmostofthescatteringisbackwardinthis limit. The front/back asymmetry arises from thecos@ term which arises from thecross terms which come frominterference betweentheElandM1.Finally, thetotalcross section inthislimitshows theRayleigh's Lawk‘dependence. Iagree thatwe have seen thisarise inourcurrent problem, butitisnotobvious tomethatthiswould becase forany scattering thathasmainly £=1. Nodoubt youcanprove ittobethecase, andweknow that itisthecase insimple dipole radiation models aswehave made earlier, Forexample, onpage 271weknow thatSthe Poynting willbeproportional tok‘inanydipole problem, soIguess thatclinches it! e Inanyevent,theworkdoneinthissectionwasfirstdonebyMieandDebyein1908-1909. References: Born andWolf give thegeneral sphere solution, allowing ittohave dielectric and conducting properties! Imight gocheck thisoutsomewhere. //Ijust web-ordered 7thedition for$64 hardback, Ihope itstillhasthatsolution Jackson mentions. (herefsa1959 Principles ofOptics edition). Problems: Thedielectric sphere problem isassigned asproblem 16.12, thelastone, Idon’t seeasolution tothismessy problem ontheweb, Icanimagine why, itisamess. Inotice thatourfriend Fitzpatrick entitles thissection ofhis"version ofJackson” asMiescattering, In doing so,henotes thatyoucanwrite outeverything interms ofthecoefficients without ever using the boundary conditions, then apply them atthelastminute. Hethen does thesame conductor asJackson, so hedoes notattempt thegeneral problem! Appendix A.Details ofJackson's Boundary Conditions fortheConducting Sphere Thefirstthing tounderstand isthis: thethree vectors (f,Xm>#XXm)aremutually perpendicular. How doweknow this? Well, Xem =ot¢(LYm)wherea=(IA/2(E+1) ).Weknowforsurethatf©Xu=0 because weknow f¢L=0.ThissaysthatXmliesinaplane perpendicular totheunitvector? .The third vector heregives 0when dotted with?orXemsinceA*(AxB)=0,soitalsoliesinthatsame @plane withXjabutisperptoXa,Ofthese three vectors, weareonlyclaiming that hasunitlength. The ul vector Xeq isavector offunctions of6and4soislikelynottobeofunitlengthforallsuchangles.YoucanandshouldthinkofXpqandfxXpqasthetwoperpendicular "transverse" vectors,asifyouwere e thinking about polarization ofaplanewaveinthefdirection. Thesecond thing tounderstand isJackson's 16.143. Ihave proven thisbyhand foranarbitrary radial function f(r),thefunction doesn't have tobespherical Bessel's theway Jackson might imply. The important thing torealize isthat this equation isshowing that thequantity Vx[f(r)Xm]canbe decomposed intoaradial piece (thefirstterm), andatransverse piece thatisinthefxXmdirection. Now about theboundary conditions. The twoshown weunderstand: there canbenotangential E field, andthere canbenonormal Bfield. What about theother twopossible boundary conditions here? What about normal E?Theanswer is(pill boxstraddles surface) thatsurface charge will arrange itself however itneeds toinorder tomake normal Ebewhat oursolution makes itcome outbeing! Thesame thingistruefortangential B,thinkofaloopandB=0insidetheconductor. SurfacecurrentsKwillflow however they must tomake tangential Bcome asitcomes out. Sothisiswhy wedonottryto"impose" twomore boundary conditions. Here then arethefields asJackson arranges them: Bie=(U2)EHARE) [lke)Kear#2,Vx{iAkit)Xess}] Biine= (1/2)E iff4n(2E+1) [F2ijdkir)Xa~2i/kiVx{je(kit)Xess}J Eye (1/2)Eitnf4nQe+1) [aOHAkit)XessBelkVx{hMKit) Xess}J BY(1/2)Eityf4n(2e+1) [FiBeOHdKie)Xess~its(AfkerVx{h(Kit)Xeas}] r) Ourboundary conditions arethese: PeBit Bre) =O #1 PxBlinc+E's)=0 #2 Forcondition #1,notice thatsince ?¢X;_=0,weonly pick offthesecond term ineach Bfield expansion. Since thesecond termitselfisthenexpanded asshown in16.143, andsincef#(#xXm)=0, allthatsurvives isthefirst term in16.143 foreach field. Ifweignore theoverall constant factors andthe Yonthen wegetthis: >2iMkyjkr)~ite(OVHOAir)=0 #1 andthisisjustwhat Jackson shows in16.144, Now consider condition #2;more terms survive. Outofeach field wewillnowhave an¥xX;qterm andanXuterm. Thefirsttermineachfieldexpansion obviously contributes tothe#xXmamount, but what about thesecond term. Ifweexpand thesecond term using 16.143, thefirst term yields nothing since fxf=0. Weareleftwithacontribution fromthesecond termwhich looks likef xxXm,.But this iseasily shown tobe-Xm_- Now, since thetwo terms arealong different (and mutually perp) 12 a —_—e transversedirectionvectors,thecoefficientsofthetwotermsmustseparatelymeettheboundary @ condition! Thus,wearriveattwocoviditions fromcondition #2: $xX;qterms: 2jdkin) +oe(2)HO(kyr) =0 ~Xom /tterms: +2MkyAfrjar) 1+Be(OM Afrh(ki)] =0 Now itislucky thatthefirst condition here replicates thecondition wealready gotfrom BC#1.The second condition isthen asJackson states in16.144b. | Notethatwehave2equations in2unknowns (ignoring +ete). Appendix B. Changes inthepresence ofand ¢ Inorder toeven beabletoattack thedielectric sphere problem (which Ido-in aseparate document), you havetoknowhowto"interpret" alloftheChapter16formulas inthecasethatyouhaveauniforme and h.[thinkIhavefoundagoodwaytodoit. Section 16.2onderiving theGrand Finale formula. Youmight think youcould justreplace BwithH andEwithDandhavethepestuffclearedoutofMaxwell's equations, butthatisnotthecase.Lookingat page 178,theproblem isthattheFaraday's lawwillthen saythatcurl D=-1/e@,H_ *jis.Soweneed another plan, Gobacktothestartingpoint16.31andaddteasshownbasedon7.1page202whichshowsthefour @Maxwell's inthiscase, notice there isasingle insertion point. Goahead anddothetime derivatives and get16.32. Wearegoing toenduprescalingtheBfield,sothedivequationswon'tchange.Wecareabout thecurl equations. These are: VxE=ikB k=@=9+t=KB B=1A\pe VxB=-ikyeE =-ikB? B Jue=p" Now define theusual modified k(call itk’)asshown above right. Then weget VxE =ik'BB ‘Vxk =ik(BB) VxE =ik VxB =-ik'B"E Vx(6B) =-ik'E VxB' =-ik'E Therefore, wehaveshownthatwecanrewrite16.32,ourstartingpoint,bymakingtworeplacements: =L={ 1)replacekwithk’ K=k/B B=1Ape n=eye 2)replaceBwithB’ BI=BB ‘Once wehave done this,allfourequations lookexactly thesame. This isthemain point!!! Thetriplet equation groups page 543. These areasshown, except make thetworeplacements shown @ above. 13 ‘ . Themultipole fields page 545. These areasshown, except make thetworeplacements shown above. A NoticethatallBesselfunctions willnowhave(k't)asargument, e Grand Finale 16.47 page 546. Putink’forkeverywhere, andthefirstequation isforB',These arejust thesame changes made inthesteps above. Section 16.3 equations. Equations like 16.48 and16.51 arejust limits with constants notshown, sowe arenotbothered bynotseeing 1/ein16.51. Theswap rule 16.52 ofcourse should have ourB’inplace of B.Iwould putaB'in16.53 andtherefore alsoin16.56. Same in16.57 along with replacing kwith k’ there. Now in16.58 wehave tomake adifferent change! The true formula foruhasEDandBH. Sowe would then write thisaseEE+(I/) BB. IfwenowrescaletoourB=BYB,weget€EE+(1/)B?B'B’ =e[EE+BB). Intheradiation limit being done here, wehave |E|=[B'|from modified 16.56. Then we canwrite theenergy u=2e (BY. Therefore, 16.60 willhaveanextra ¢outfront. Then in16.61 wereally should have Hthere andH=B/u. Butthen putinB=BYB, andthen 16.61 willhave 1/(yB) outfront. Using 16.57 with itsactual k'andB'then gives 16.62 with both B"inside, and with (1/1) outfront. Thus, theresult 16.65 will have this 1/uoutfront. The ratio in16.66 will then say that M/U(modified) =M/U (shown) *(1/1) *(/e). Sotheratio isthen nolonger asshown andIdonot know how tointerpret themodified result. Section16.4ondistribution. ‘ThetruetimeaveragedpowerwillbeExH=(1/y)ExB=(1/s8)BxB'= VEEX,Thus,weshouldhavethissquarerootfactoroutinfrontof16.70. Step1:Let'sgothroughthegeneralmultipole expansion andaddthepresenceofuniform and¢.Start with 7.1page 202which shows thefourMaxwell's inthiscase, notice there isasingle insertion point. So, Thave marked uphow thetwo3-equation setsonpage $43arealtered. One alteration isthat you have tothink ofkas@/vinsteadofw/c,butonlyinplacesthatgottheirk from thewave equation! This means inside anyBessel function likej(kt), themultiplying kinthere must bethek=olyversion ofk.However, theexternal factors ofkareunchanged, namely, the1/kthat youseeinthemultipole fields on$45except youhave toaddonetefactorasshownforEintheelectric ‘mode case only. Looking then atthegeneral expansion 16.47, wehave toaddsame pefactor thatweadded into 16.42. Note thatthisisNOT just arescaling oftheagcoefficient because thetwoexpansions arecoupled. Onpage 548Iadded new factors asshown. Next, how isthehuge 16.139 going tobealtered? Section 16.5 onmoments. Looking at16.77, Jackson isstarting with thefullequations including H,and hehasleftthepolarization charge grouped inwith p.SoIhave nomodifications tomake tothisentire section. Section 16.7 oncenter-fed antenna. Everything goes exactly asstated. However, when wegettothe radiatedpowerformula16.121,weshouldaddthesameséfactoroutfrontasonpage550described above,andinallotherpowerformulasinthissection. 9 4 ‘ Section16.8onPlaneWaveExpansions. Firstconsider16.130.Asusual,k=k’intheexponent.The @second equation there should bemodified according to'7.13 onpaie 204which saysBy=" Eo.Thus the second equation iscorrect ifwejustreplace BbyB'.Sowejustmake theusual 2changes in16.130. Asfor16.131, itisjust acase oftheGrand Finale 16.47, andweknow tothink ofB'ontheleftofthe second, andallkarek’.Luckily forus,allthesolution forthecoefficients isunchanged andweendup with 16.139 with justourusual 2changes. Section 16.9 ontheConducting Sphere. Asusual, replace Bwith B'everywhere andkwith k’.Wedo thisintheassumed form oftheoutgoing waves aswellasin16.139. Theboundary condition canbetaken asneB' =0sonothing changes throughout. ‘The final results 16.151 and16.152 arethesame except interpret kandBask’ andB'. ‘Nowwecometothepowerformulas. Thereshouldbeaffactoroutfrontin16.155forthesame reason asdiscussed above (really itisExH =(1/1) ExB =(1/Bq) ExB' ). 15 From (busyureter Ty = a ©: Kil=Figs+Vim SodolceLHS41bondexpand orefoltre wAguyK|=Fol(eee)(HH wek =V(obee)xLK)-+(rhe)VeAH]=O+ =Se[rhea SeO=Serer fe)=LBTekaaX. 1txX [email protected],SoweVor aad doshow Dud Wd,omSethwachaXGyCoedegif- [obec,camedLeff)sureeafacharmvolsihe 9x0 Ym)=AAU) SeYim®(randbeShed ayy(20) Yim =A)Ge42)Vow—“GgAx@xsye @ ; n \EGRET Cidon 2\Y EeimicoY =Gyn |YIN NE AMG ©NewwaeolladelLemnadoceefut “Pe e COT[PxDaal=CDFTen amdKArmaings tostnvDenJwot COPTVen=e EY)2 Swice webund LsAL) omYim: Buk[6,20soy aI, —=.BukWYon,=O,$ Dim=Xone Pah Caryyn,=Got[1Yn]=GaTYin)F Qe.e 6 rms VxLltxWes]=tV$e.8) @Pood:weeVx(ABdarksdsgat Lis=PVF~GEYG-F)AGE9)—(FB) 0)® ® @ Vw©iadeonemnr, a0wantde shay O1@+O=0. ©=-—WH).% @:Noleak TE= St3h +$1o96 %=OEEHONE =@No=BE+HteBE%36 e & Cse$ ~+[O9.5+G4), >|<+494) ©=NeW) =(LWF) seeVel)=Fe Thnfee O10. “OP Ferrel =o Ge e ©Yadkoow (1620)Pror) 049556 © eiss aan jell GAMO7a [Yaktse(1695)|— 4 =ky)AVia)Et),EisD5Ak (2) @Do aparts dowwe 35 =—keAVaj[eter) JeanAk =4keSAVAnSeLT‘ =+k,SavAeUxLeta] (2) eyWarVF.duetsop = ~s ¥ 5FHL OR=TGuey)Ea}+3099wit0 % ByLockwowAerm ©whedy Var) =[dejac]A oeLL=AXV WerowAjaceex —yPts 2x9 Axe AxGXT) =(ABYR(ARV e~(aet)2= 8S “ro =0D Dreufrre , . £ “s@ VL =+8fd. O4sGey Sv) amd Soour rseutt ofMiepodk a) . ‘ Lus=keavAe[RAO.j1) BH+jecaval «) (c)VowwaeSeritity (1641) soDot “SVE =AVN(IraeTao =AE Bawe\ewad Dak te -e_OgHe e vw aoe--Xaay; 0hat > us=k,wwAston Psishaje+04)52| = Conlodne Meche YorJoma to9 ~ a a. Saf.uleH); kos LUS=eSAVA®|(Ad493,Gea)V6LOOlayFE](5)Yo -% Heim (2)=Hera A:[Sa] =HeJAVA SpA =bySavo:[acg] =HhoSavFae +AIST there Bay=AGale)4459 [Jaco] Rawat : e |Fiom(5)=FeSavSAS HE Nes Se=HHay=HOA#2 (RWB) =(FAYO) Soogaun L 3 4;’(Mom(0)=Ferav|OA) +RA)F GrantOriewie(3)doqik Lisakeay[AIS+DPSMEclef fheRUSA(1840) *Paeraee obdre(R'9)ondhecns ® e B(n)\=Ase) +18.(yates FW=42-Ga] 49,249-Ljab} ile) =a=3- Thom By=ag!19,C4) =Agtiltg'+Bag!)=ting!digs Tramteakcogent29:oot84)e{BitesMpg—SUS)oe =fring’uleadff>+445) ~GinLg"32d Meow wasdreBE(16.7) daa &seluad bya=Aclbr) SoOho Igisagcomet =ee- ts-103)=]5grey| - e 4§=~in]i+| Sowewow hon tgs=Ha)EGA) NCHAERSHEEA Orenoubrme uns=+k,\av[(9-A)ag+eaa.g]-(F) [22xsity]et |~adaav[FA]Egseal~( ANER Fs]|e @(rdPasataPort! theomraet RUS4(4) bP) ge=nk)gS ~BeeVo]S59.)PRSg| +g; =aeSat92x(ry){ose5)AV eivwredaneowalsorveinfic)Maxustterns oe(1891). Vector Spherical Harmonics Cantetor. Paoro8 e X=BVYon KY=V.E=2.%=0Y=AVbm, RXK=0Z=2VYeens Jackson Y= = 23 _ >vaan REVbeTaney Liaw MondWald MsHOVX(RY=—LaiQdx Vr—(Xtike) N=GER =LHR IA](TInt) eA=tYxNa A=Yom orQinmennits within UM =F0=0 GHik)rso 3(V+EM=0 and(VAE)N=O TM(elechic) Bene=Aeon)MewEan=LNB eiAe(Gon)Now ~ => — TE(magnate) Ean=Amr) Mane Bom=LV e =iA (hm)Nim e Physics 712—Handout 3—Feb22,1995 ‘Vector Spherical Harmonics This handout summarizes thevector spherical harmonic formalism dis- cussed inlectures. This treatment wasprepared byProfessor Carleton De- ‘Tarandistaken from lecture notes byProfessor Richard Price andTracy Steelhammer. Orthogonal Expansion Spherical harmonics provide anextremely useful orthonormal basis forexpansion ofscalar functions onasphere. Wewould liketofindasimilar orthonormal basis forvector functions on sphere. ‘These functions arevector spherical harmonics (VSH’s). There areavariety ofchoices forbasis sets, butweshall choose asetthatisclosely related to that ofJackson, Chapter 16. Define thevector harmonics L Xem(9,46) =rxV¥in(8,¢) a) TC Yem(8,¢) =£¥im(9,4) (2) 1Zem(8,8)=rV¥in(8,8) ®Y @These vector functions formacomplete basis forexpanding avector function ofthespherical polar angles @andHfIneexpressionfotKydiesfrom Jackson’s byafactor ofi/&{E+1).“HeusesL=3(rxV)MSeeJackson,Eq (16.25-30) foradiscussion ofproperties ofL.]Thenormalization ofthese .harmonics iseasily computed: /XiXemd®=L416 bmi ww |YimsYimnd® =bebagi (8)Yaes [PimLind =C+1bmm OK [Kem-Zimnd® =0. (nw Theharmonics arenotjustorthogonal inthesense ofanintegral overthe sphere, butthey arealsoorthogonal everywhere inthesense ofthevector dotproduct: Xime¥en =0 ew Yim Zim =0. (9) e TO EaOtoe— | 1 t Ru(Ads —4 Anarbitrary (well-behaved) vector fieldA(r,0,) canbeexpanded interms ofthese VSH’s as mn 3 as 2(78,8)=DARKem+ALYMem+AP)Zem]- (10) a ‘Theradial coefficient functions arefound from theorthogonality properties: Qa) = —1 + .420=gayJ%in(0,8)--A(%6,6)40 (1) AR)=f¥in(0,4)-A(r8,6)40 a2yby Mey = —t s :ADU)=qrpayfPhnl6s):Alo8,90. (13), Calculus Exercises with Vector Spherical Harmonic Components Weplan toexpress Maxwell’s equations interms ofthecomponents ofthe VSH’s, butfirstwemust work outtheexpression forthedivergence andcurl oftheharmonics.Inthenextsectionwepresentresults.Togetafeelingfor ethemethod ofderiving them, wegive asample calculation here. Consider thedivergence ofthe(r)component term: VARA ¥em]=VAAL/2)2¥em(8, 8) (14)dlr) ye (7)~ =[EAR mer+342/1¥in(8,d)~ (15) Ld)aar) =FECA inOT (6) Note that theterm inr-VYm(0,) vanishes since thegradient ofYim,has noradial component. Some other useful identities may befound inJackson, Chapter 16,including these: (fxV)¥in =6+ D¥imv (17) Vm=—EtDy, ~ (18) re(rxV) =OV (19)Ve(rxV) =ao” (20) Div and Curl ofthe Vector Spherical Harmonics Aswewill see what isparticularly useful about theVSH isthat thedivand curl can be as aeyByes eee _ r) KXBLS GF.(Rx4)=VITetPW -010=0 7 D Bir =OrBe tovty=0+e[Le] =Ay vv me Son 208, Y x(Ms EEG e 72 ae se aay VvZAG) =VBteFSH=LAT Bos 3s, e XKUx(PxBD =-2PKTH wre| expressed interms ofthesame £mcomponents. There isnomixing between components ofdifferent £m. Taking A(r,0,¢) asexpanded above, wehave, after some calculation, 1d)aym,_€E+)) .) VemDiagal)—avin Ya NotethatthetermsinXimdonotcontribute tothedivergence” Theex- pression forthecurl: VxA=Bis VxA=DIBQ2 Xen+BOVim+BO(r)Zem), (22) im where ag=de cen om214) a 4BR=Trae) (24) y @214am, 14 ym BE=~2Ald+le(ral). (25) e Parity oftheVector Spherical Harmonics Underspatialinversion(0,9) becomes (n~@,6+x) andYem(4,¢) becomes Yim(a —6,6+7) = (-)'¥im(8,4). Since vectors like#andVreverse signunder parity, the parity transformations oftheVSH can besummarized asfollows: oe Xen(t- 06+) =(-)Kem(84) (26) Yen(t- 06+) =-(-)'¥em(0,4) (27) Zem(a~ %d4R) =—(-)'Zem(0,4). (28) Since Maxwell’s equations areform-invariant under parity transformations, theparity transformation properties oftheVSH canbeofhelp inreducing thenumber ofcomponents weneed toconsider insome applications. Illustration: Electrostatic Multipole What follows isnottherecom- mended way tosolve anelectrostatics problem, butitillustrates theway theVSH’s work. Westart with therequirement that VxE=0.Interms oftheVSHcomponents ofB:B{"),o{),andBl)thevanishing ofthecurl implies that B®=0 (23)“ (r) dBQ=FE), (30) andfrom V-E =4xpandtheexpression forthedivergence interms ofthe VSH components weget ld (re+),SPH) AY20)=trom @nw wherepémistheusualspherical multipole momentofthechargedensityp.Putting together these twoequations gives theradial Poisson equation: Tda2.ayke+1), 20) 4 atsr(r’strz, Jr Ep)=48pm. 32) D(C)+EI) GAaeGelimJaen) =4 (32) Notice thattheVSHcomponent rE?)appears inthisequation where we would expect tofindthepotential multipole component ~dym. This result canbeexplained bynoting thatifweinsert themultipole expansion forthe scalar potential ¢into theexpression —V¢ wefind ; a 3 x—V(bemYem) =-——(dtm)Yem| IZem a S~0.6 A (emYen)=-(Gem)Ye , (33) e socomparing thisresultwiththeexpansion ofBintermsofVSH’swesee that EL)=~bim/* (34) the asexpected. Wealsoseethatthe(r)and(1)components arecorrectly pe . correlated according totherequirement thatVxE=0,andthatthe(2) 2component ismissing, asalsorequired. a Illustration: Magnetostatic Multipole Wearegiven alocalized cur- rent distribution Jandwant toexpress themagnetic field Boutside the ~ source asamultipole field expansion. Inanalogy with themultipole expan- sionfortheelectric fieldwewrite B=—V@outside thesource, where the ot magnetic scalarpotential hasthemultipole expansion ones. o“Bs yo,Ss Bt bem ern Aa xv P=Dain Roane Sotheproblem istofindanexpression forthemagnetic multipole moments Himinterms ofJ.(Recall thatthesolution tothisproblem wasnotgiven inJackson, Chapter 5.) Zo Wewant tosolve Ampére’s Law VxB= 4n/cJ andV-B=0. Our strategy istoexpand both Band Jinterms ofVSH and use theabove soy| oR)——, ee weTT ¥ OAteowhe, oatok 4G) +6=BE ©Atsnkfit—-*=7S ®MStay =MSVAy Tre 40 ere lyWeey=sin’[Siseny| Kevanies own Aamgs 0+ aeo=wrx524) LY>Vin=13248" Beag-4Rex bog eo aA toa expression forthecurltogetasetofequations fortheVSH components. Thevanishing divergence ofboth BandJwillgiveusextra conditions todetermine allofthecomponents. ofFirst, let’s write theVSH components ofBoutsidethesourceinterms ofthemultipole moments imbycalculating thegradientofthemagnetic scalarpotential explicitly. Weget )_4(+1) > A :Bim=gyqperaHem (36)ina. B®= 1 (37) q inBee person Boa=o 7 (38) outside the source. FromAmpére’sLawandtheexpressionforthecurlwehave J 4 4) cies - xip=dae v wow Daeyp214630) (40)“a ews c td 4 4; logyld ow 2=TB+EBM). (41) icethatthefirsttwoequations correlate the(r)and(1)components of ~ J adrequired-fom V-J =0.Notice thatoutside thecurrent distribution BE)=0,PromV-B=0wehave Adzp) E41)pay_ 4 al”Bin)- Ben,=0 (#2) Wesolvethisequation forB{”)andplugintoEq(41)anddividebyrtoget a +B, i@OrL241). ain)_4aDe(8)+eol sqlBQ)LEDGai)=Zeerie (43)~~ ThisisagajdtheradialPoissonequation.“£his timetHeroleofthepoten-tialisplyfed byrB{{)."Using theusualradialGreen’s function togetthe solution wefind outside thesource (0={FMEA 1 1D)ergyrBe)=(fas yan, (44) ~~ 5 59°ESUayltay7de Y=oA = < . Comparing with Eq(36), weget, a=) fptexs. 42, Bem=aeey|XnI)rtdrd (45) or,interms ofspherical harmonics explicitly, =! fateoye). So Bem=gery[MexV8)sav. (46) Thisistheexpression wewanted.Se€+modM}y Tacksnr1696,sup[et Maxwell’s Equations Letuswrite theharmonic formofMaxwell’s equa- tions interms ofVSH components. Theequations wewant torewrite are V-E =4xp (47) VB=0 git: (48)VxE =KB (49) e VxB=-ikE+5 (50) Interms ofVSH components thefirsttwoequations read Ldarye+ ve Bag) =trom kM aaa(ul)Ld) apiy Meth . V6 SH -49-Jy=0 62) andthesecond twoMaxwell equations read £641 epp(t)HDge=ikBy) p(3) id ; . ote —pRreQ) =ska) @(4) | 1oe ld .. “PE+T(E) =ike ad(55)& MEDD 2tes IQ4(66) | Teak) =e+ «on | oot law, ld ; 4n vFP+TORR) =ike+ZIP. 68) e ofp=ade olEeebe(=br,EL 6 Theseeightequations dividengatlyintotwoindependent setsoffourequa-tions,onesetinvolvingoyBe,BE),andBG)andtheotherinvolvingonlyE{2,Bf),andB{").“The waveequation forB!2)isobtained bysubsti- tuting Eqs(56)and(57)intoEq(55)toeliminate 5!and£{").Asimilar manipulation yields awaveequation forE{2)withtheresult: “ ¢ 1@ ert dry electric3+22;Frys -SDia =9-Zedy60)Wh 1@ ‘ TONEgret +(eEee 2alta. (60)WVgree 24yor rar + c—, aSsis ‘Theseareinhomogeneous radialHelmholtz waveequations foraeEQ)°The firstequationyields“electric” multipole wavesandthesecond ue[ike“magnetic” multipole waves.Theseequations canbesolvedusingtheradial pe"]“ge0Green’s functions fortheHelmholtz waveequation, discussed inJackson, Chapter 16,andsummarized inthesection after thenext below. Resonances inaConducting Spherical Cavity Wenowconsiderthe e homogeneous solutions ofthewave equations fortheinterior ofahollow spherical perfectly conducting shell ofradius a.Wewant solutions that are regular attheoriginandsatisfytheperfect conductor boundary conditions:#-B=0and#xE= 0)‘Looking attheform oftheVSH terms, wesee ~ that these conditions require BE)(a)=BM(a) =BM(a)=0. (61)4 NNForzerocurrent,theequations(59)and(60)arejusttheequationsforthe esphericalBesselfiions’Therequirement thatthesolutionberegularat rothe orighgives thesolutions ¥ ou BQ=waidlér) (62)Een=ehnidkr). (63) Theboundary condition onthesurface permits twotypesofsolutions, just asinawaveguide: electric multipole, or“TM”, or“even” withBo)= 0everywhere andmagnetic multipole, or“TE”,or“odd”withEf)=0everywhere. FoythedlpetricmultipolesolutionsBY)=0everywhere impliesphat EQ)=Bo©0everywhere, ascanbeseenfromEgs58)and(56)."Theboundary condition B{)(a) =0together withEq.(57)"requires that a J qlrBe (r)]leae =0 (64) atthesurface fortheelectric multipole solutions. Interms ofthespherical Besselfunction, werequirethat we a.glidelene=0. (65)Sda Thiscondition canbesatisfied onlyforspecial values ofk=w/canddeter- mines theeigenfrequencies. Forthemagnetic multipole solutiony’ therequirement thatEO=0everywhere implies similarly thatBC)“=Bl?)“=0everywhere, andthe boundary condition B{2\(a) =0requires that(W 7 e je{ko)=0. (00)/a Radiation from aLocalized Source Equations (59)and(60)maybe solved togiveelectric andmagnetic dipole radiation fields outside alocalized . source. Therelevant Green’s function satisifies theequation @2d4K+). A1, je.29 tptieeeEarp =—kae- ey)er(6) Thesolution thatisregular attheorigin andhasoutgoing-wave boundary conditions asr—ooisgiven byJackson inhisEq(16.21): adryr!)=thie(br<)A9 (br). (68)({6-ti] Thus wemay solve Eq.(59)forthe electric multipole field: yewmayek(treemuralAg ”) Anik Lim_ty 5ker!de!, OoBy BIN)=SEAMery[BUD=galIDVideUoPa.(68)4 3ysComparingtheintegrandwiththeexpressionfofthecurlintermsofVSH, oa Eq.(25),werecognize the(2)-component pfthecurlofJ.Thecontribu. tionofthismultipole tothemagnetic aj(4electric fields canbefound by tay e@Aade oa)USalalg? 10)awd(U1 =)eho) THBEe\¢ }pers(10)amd(ui) Sy . =m2coeffeine4beVxT Seexpanse v — x iY .=. XVTQ Xederm=Fat ——, e aac) Corletnn. comparing with Eqs(10)and(50). Wemay therefore rewrite thissolution outside thesource as - o a! atk Ber)=az(t,m)hQ(kr)Xem(8,8) =TA—(70) yenE(r)=;VxB (71)A(50) ———DASoN . an where) Taga*e*petTAUXSpeer (x8) \anil as(tsm)=ee[idee eXE(0)- VI). (TGGY ‘ erDe Yroaa Anotherexpression foraz(¢,m)canbeobtainedbyintegrating bypartsand on using thecurrent conservation condition relating thecharge density tothe % current density: iwp=V-J. After some computation weget qu Ark?. aq. ik [ee an(tsm)=~Zr|YinONoe)Elrdelbr+Se-aeyjderyer. em( e+ a ¢ ak (73) Asimilaranalysisofthewaveequationforthemagneticmultipolefields, . @Eq(60), canbedone with theresult outside thesource: Ele) =am(,m)Al(br)Xem(0,4) (74)¢ Bir)=zxE (75) qe where au(t,m)=—42[idler %n(@',8)-Te!). (18) iM\e, =e+teI em a ‘These results maybecompared withJackson (16.46), (16.89), and(16.91). (Thenormalization ofXmisdifferent here).Ifthesource hasdefinite parity, then there arerestrictions onthemul- tipole moments thatcanbegenerated. Consider Eq(78). Ifthecurrent density isoddunder parity, thenthecharge density must beeven, andonly even ¢electric multipole moments aregenerated. Thus thelowest possible electric multipole moment is£=2,orelectric quadrupole. Likewise, inview ofEq(76)suchasource generates onlyodd£magnetic multipole moments. Thelowest termismagnetic dipole. Conversely, ifthecurrent density is evenunder parity, sothatthecharge density isodd,thenwegetonlyodd electric multipole moments (electric dipole andhigher) andeven¢magnetic multipole moments (magnetic quadrupole andhigher). These requirements e@ ae”Scanaren CsA isa&X= Ge aS5Vato are summarized inthe table below. Ifthe source does not have definite parity, then both types ofmultipole fields canbegenerated. ParityofJ electricCy =(-) magnetic -C¥ [OF Radiation from aSmall Source Ifthesize ofalocalized source dis considerably Jessthan thewavelength oftheradiation, ie.kd<1,the expressions fortheelectric andmagnetic dipole coefficients canbesimplified further. Thesimplification makes useofthesmall argument approximation forthespherical Bessel function: . (aryeil)=OPED (77) Forasmallsource thearguments oftheBessel functions intheexpressionsforap(¢,m) anday4(é,m) aresmall intheregion where theintegrand is non-vanishing. Theéthelectricdipolemomentcanbeseentobeofthe r) order(kd)‘.Sincethecurrent density istypically oftheorderkedp(r),themagnetic multipole moments areoftheorder (&d)‘*1. Thelowest non- vanishing moment isthedominant oneineach series. Amagnetic dipole moment isexpected tobeofthesame sizeasanelectric quadrupole moment. . Intheexpression fortheelectric multipole coefficient ag(£,m), thedominant termisthefirstoneinvolving thecharge density, since theJ-rtermisusually smaller bytwopowers ofkd.When thespherical Bessel function isreplaced byasimple power ofrtheintegralsforthemultipolemomentsthenbecome proportional totheexpressions forthestatic multipole moments: ankt#? | ax(tim)=areayatn (78) dnktt? tym) =ATR autem)=Terpyihin (79) where thestatic electric multipole moment is dem=fYentto(a)er (60) andthestatic magnetic multipole moment isgiven byEq(45). Jo Relationship toCartesian Multipole Moments Forlow-order mul- tipole moments itisoften convenient toexpress the’fields interms ofthe Cartesian multipole moments, instead of£mmoments. The static electric dipole, magnetic dipole, andelectric quadrupole moments aregiven respec- tively by p=/re(r)d°r (81) 1aex[rxseer (82) Gao=fBrarp~1bap)pla)dPr. (83) Toillustrate howwerewrite theexpressions forthecorresponding low-order multipole fields interms ofthese Cartesiaty’moments, consider theelectric dipole fields obtained bysubstituting (78)into(70). 1 age4nk B=DyPamhl (er)(exV)¥im(6, 4) (84)met e EsaxB. (85) ‘Thesumover minvolves only ¢imYim- Using thehelpful identity SS ¥E,8)Vin(0,8) =8-83yin(6',S')¥im(8, 8)=Ff, (86) where fhasspherical polar angles 6’,¢’andrhasangles6,4,weget 4J Dam¥in(0, 4)=por. (87) 3 Substituting thisresult intoEq(84)andusing theexplicit expression for hy(kr) gives ke= Pex p= a-2B= M(Exp)—(1-z) (88)fhe ik B=@xp)xi+He-p)-plG-Be* (80) Asimilar treatment forthemagnetic dipole fields gives theanalogous expression B=-¥exp 0-2) 90*FxAO~Fe ($0) | tl . eftr . 1ik,inp B=Pexa) xt +ne-m)— aS-He” (on) ‘Toconvert theexpression fortheelectric quadrupole fields into Cartesian moments, weusetheidentity ar 2 £0. oftFDtem¥an(9) =DoFaQaaho- (02)met rt The fields are then written interms ofthe vector Qa(t)=Y>Qapie. (93) The result is 4 BeEnlMcenyex ae) (94) i E={VxB. (95) Inthe radiation zone thefields have theform ieikt t) B=we, xQ(?) (96) ike eltB==eexQixF (7) Forhigher multipole moments theexpressions interms ofCartesian mo- . ments isclumsy, and the£mmoments arepreferred. Exercises bee. 1.Derive thedivandcurlexpressions (21), (23), (24), (25). 2.Derive thewaveequations (59)’and (60). 3.Given aconducting spherical resonant cavity ofradius a,write the complete expression fortheelectric andmagnetic fields corresponding tothelowest electric dipole andlowest magnetic dipole resonance, specifying theresonant frequencies inboth cases. 4,Prove theidentities (86) and(92). 5.Derive (90)and(91). 6.Derive (94). Iv VectorSphericalHarmonics PhL 2.8.03 eThese aremyextended notes onthe 1995 Physics 712 handout onthesubject ofVSH which was assembled byCarleton from notes ofRichard and ‘Tracy. Lots ofcalculations arerecorded inmy appendix. Iplan tosummarize thissétofnotes inanother document, leaving outallthedetails, and trying tofocus onthemain points. 1.The Scalar Function ydefines anaxis rotation, Consider thethree vectors asfollows: rxVy ty ry where yisanycomplex function y(@,9). Notice that allthree vectors aredimensionless, These three vectors aremutually perpendicular: reVy =roy=0 since yisnotafunction ofr re(rx Vy) =0 since Ae(AxB) =0foranyvectors Vy«(rx Vy) =0 same reason e Whatarethesevectors?Theradialoneisjustofmagnitudeypointingradiallyoutward,thatispretty clear. TherVyvector points inadirection tangent tothesphere's surface inthedirection inwhichyhas itsmaximum increase. And rxVyisalsotangent also andisperp totheother twovectors. Now let'strytoexpand anarbitrary complex function A(0,6) asfollows: A=arVy +Bf ytyrx Vy where a,Bandyarecomplex functions [email protected] general expansion ofA?Iwill argue thatitis,andhere arethecomponents: a=Ae(rVy) B=(Ae fw yaAe(rx Dy) Allwearedoing ischanging from x,y,z. tox,y’! where thenewcoordinate system isafunction ofour pilot function y. 2.Specific choice foryasY,~(0,) andsome reasons forthischoice. Wecould, forexample, sety=Yon(8,4), then weget: r) reVY¥m=0 1 re(rxVYen)=0 t) V¥me(rxV Ym)=0 Let's now give these three vector functions Carleton's names, Xon=1XVYn Yon=FYon Zr=19Ym Inpassing, wenote thatJackson also uses anXfunction, buthisisscaled differently, see(16.45). XplACKSON-—L_ Xm= LYen iL=rxVinfer) eer!) ‘TheiherecomesfromJackson's inclusion ofaniinhisdefinition ofL,makingitconsistent withtheL thatweuseinquantummechanics. Forourpurposes here,however, thisfactorcausesonlytrouble.Imighthavecalledthesemoresuggestive names,likeMo,RumandGay(angularMomentum, Radial, Gradient) .One problem isthatyoucannever remember which letter goes with which function. In retrospect, Ithink Ilikethesimple X,Y,Z choice. Now,foreachchoiceofvaluesém,wehavealittleperpendicular vectortriadwhereonevectoris radial andtheother twoaretangential tothesurface ofthesphere. Notice, however, thattheorientation ofthetwotangential vectors relative tothesphere isgoing tobedependent onthechoice ofand m,whereas theradialvectorisalwaysradial. @ Why areweinterested inthese vectors with y=Yqq?Well, firstoff,inourappendix attheendof thispaper wepainstakingly calculate many things, andindoing sowelearn many things. One factwe findisthatapplication ofanyofthestandardoperators likecurl,div,grad,lapl,r©andrxtothethreebasisfunctions produces eitheralinearcombination ofthesamebasisfunctions (whichreinforces our idea ofthem spanning the3-space), orproduces amultiple ofYm inthecase ofanoperation that produces ascalar result likediv.Inthese proofs, wesometimes encounter theV?operator andwereplaceitwiththeoperator -1/r”L?whenitactsonafunction of@and$,whereiLisrxV.Itisatthispointthat wemake useofthespecific properties oftheYmfunctions, thatL?Yon=£(C+1) Yn. Itisthisfactthat keeps everything closed within thelittle ¢mmanifold. Forexample, you never encounter anL,type operator thattries toraise youtoYma+1 ,forexample. Let's look more closely atthisclaim that nothing canpush usoutoftheémspace.Nomatterhow much vector stuff weapply ontheleft, when wework through allthevector identities, these aretheonly dimensionless vector things wecan"end upwith": FYem Y 1VYmm Z xVYimx Fee V¥m) 0 PYVXVY mn 0PPV aBVVg==PL?Yaw==C+)PYes=-£(2+DY ri) 2 wee e@ Theonlythingswehaveto"playwith"arethevectorrorf,thescalarr,andtheVoperator. Nowforanypilotfunction (6,4)inplaceofY,thefirst6itemswouldstillbetrue,andourspacewouldbeclosed under these operations. Itisthelastitem that iscritical. With thegeneric function wewould endup with?V’y="?g,some newfunction, andwecould thenendupwithfgasanewvector. Asanexample, weknow (byexplicit caleinourappendix) thatV*applied toanyofthevector basis functions justyields alinear combination ofthose basis functions. Wealsoknow thatanypurely-r operator canonly shuffle thepowers ofraroundinsomeway,withoutaffecting theYq,--certainly we willnotshiftanyémbyapplying combinations ofd,andr.Therefore, itistruethattheL?operator when acting onanyofthevector basis functions willgiveatworst alinear combination ofsame. Since [L?,L J=0, wecanseethatL?X=&(¢+1) X.For¥andZ1don'tthinkthisistrue,butL?willcertainly yielda linear combination ofthevector basis functions asperourdiscussion above. Inparticular, thevector operators inMaxwell's Equations won't take you outofém, soeach £m manifold represents a"separate problem", asisthegoalinanydiagonalization. Ofcourse theunderlying factisthatMaxwell's equations areinvariant under SO(3) andthisiswhy L?Lsdiagonalization is possible. TheHamiltonian orLagrangian would besomething likeF“F., +A",andisclearly SO(3) invariant, nottomention SO(3,1) invariant. Sothese comments Ithink provide alittle extra motivation fortheunderpinning ofthe vector multipole expansion. Before going on,weshould notethatJackson's "grand finale multipole expansion ofEorB"16.47 hastwovectorterms,butaccording to16.143, thesecond termitselfbreaks intotwootherterms,andyouthenendupwiththree vector terms ineach equation, andthese areprecisely thevector basis functions we areheretalkingabout.Jacksonhoweverdoesnotreallydiscussthefactthatthesethreefunctionsspan @ thespace; heisjustsilent about thatandgrinds outtheresults, 3.Orthogonality Proofs Let's think oforthogonality notjustinterms ofdotproducts, butinterms ofcomplex dotproducts with theleftitemcomplex conjugated, andintegrated overallsolidangledQ.Aspecial "innerproduct". YwithY.Orthogonality oftheY'sisobvious fromthey's,since fe#=1.Thatis,weknow thatour choice ofy=Yax(0,9) precisely havethisorthonormality, namely (stated as(5)inthepaper) SfIOPYen",0) ©FYen'(0.9)=Sf82Yen"C,8) Yon‘0.9)=BesBurm Xwith X.Osthogonality oftheX'srequires thefollowing work: (there isprobably aneasier way, see below) L281, HL, 1, Define thefollowing e@ esktif => Slee,=2andsy'ees=0 3 bebel e ‘You then find that L=(1/2)[eLteL] +2L, and then (Lye@ )#0) {Lo L.+L! L.)Fly Ly where Ihave leftspace foryoutowrite inY¢q andYminthespaces. Thetwo raising ortwo lowering operators dothesame change tothetwoYfunctions, sothefinal delta form isnotchanged, butyoupick upthesquare oftheraising algebra coefficients Jackson shows onpage 542. Then added tothelast squared coefficient m?youget£(é+1), asCarleton shows inhis(4).So, SfS2LYe'O,8) #LYe8,9)=CEL)BeBan Noticethatthisproofdoesnotworkwithjustanyoldfunctions! Hereisanotherproofthatismoreto thepoint: SdQ.LY¢m'(0,9) #LYew(8,9)=SdQ<E'm'|£)O>*<84|L|ém>=<é'm![Le£|ém> e =<l'im'[£?|fm>=(641)<Cm'|fen>=(041)BeyBare where £istheabstract operator intheusual Hilbert space wedeal with inthese matters. Again, itisthe Casimir operator ofthegroup andallthatstuff. Zwith Z.Finally, what about orthogonality oftheZ's? Igotstuck onthatforawhile, butthen Irealized this theorem se2wenno aslong as£(0,6) isfinite at@=+n,andisperiodic in6.Seeproofonscratch.Youbasicallytreatthe divergence astwo terms andineach term you have aperfect differential with respect tooneofthetwo angle variables andthentheassumptions given make theresult zero. Using asimple vector identity, a corollary tothistheoremis fire Vg)=-sdQev'e) 4 OO mm Aat dla justasifyouweredoingapartsintegrationandthrowing:away theparts.WeknowfromJacksonp543 e that-V?g=(I/t’)L?gandifgisaYmthenL?becorhes é(¢+1).Thisthenprovesorthogonality ofthe Z's,equation (6), PsaoVYew'(8,9) *VYen(8,6)=€(E+1)8,¢Baten Ywith ZandYwith X. These function setsareorthogonal simply because thevectors areorthogonal from ourearlier observations, namely, that reV¥m=0 =>YwithZ=0re(rxV Ym)=0 =>YwithX=0 Zwith X.This istheoneremaining combination, anditisnotsrivially truebecause, although VY¥me(rxV Ym)=0 we have Vm @(rxV Ym)#0 Hereisaproofthatwiththeangularintegralincluded,ZisorthogonaltoX,Startwith eVe[f Lg]=Vfelg +fVeLg /justavector identity, thinking f=Yu" Integration over solid angle gives 0forthéleftfactor from ourtheorem above, assuming fandgarefinite ete.Thefirsttermontherightistheonewewanttoshowis0(itis~Z*©X),sowewillhaveproven (7)ifwecanshow thatthesecond term ontheright iszero. But VeLg=Ve (rxVg)=Vge(Vxr)-re(Vx Vg)=0+0=0 since(Vxr)=0andVx(Vg)=0,QED. 4.The Expansion Theorem Wecancombine alltheorthogonality conditions proven above involving theX,Y,Z into asingle statement. Todothis, let's redefine thevector functions asfollows Khem=keXen=kerxVYon=ikeLYoo Mim =Ym =fYou Wea =eZn=ketVem e wherewehavedefined 5 kp1Afee+1) e Then intheprevious section wehave proven thefollowing fact: S82Xea(0,0)%* Xnl8,9) =BreBurnBin orthogonality Furthermore, wehave shown that thefunctions X",,(8,6) asvectors in3-space form atriad of perpendicular vectors which, ifnormalized tounity, would just besome rotation ofthecanonical unit vectors, with thespecific rotation being afunction of£,m,©and4.Therefore, interms of3-space, we know that thevectors X"p.(8,6) clearly "span" the3-space. Any vector could bewritten asalinear combination ofthesethreevectors. ‘Now, wewant tomake thisconjecture: anyvector function of6and¢canbeexpanded asfollows: A) =Loan Aeon X"ea(0,4) @ ‘Theradialprojection ofthis equation is, FeA@¢) =AO9)= ZinsAon(F©X"rm(0,4)) =ZanAont¥m(8.4) andweknow from scalar theory thatthisisinfacta"complete" expansion forthescalar functionA,(@,4). Soatleast theconjecture seems possibly true. Theorthogonality makes usthink itistrue! Atleast within‘eachpartialwave,wecanseethatwearespanningthefull3-vectorspacewithourthreebasisfunctions. r) Ifweassume theabove expansion istrue, wecanusetheorthonormality condition asfollows. Start with theexpansion, A) =Ean Acne X°1n(0,4) Peruse theorthonormality, S42XaG,)**Xml.)=BeeBainBrin Apply integral operator toboth sides SIX"co(0,6)**AO,8)=ZeonAtmaf42X"rui(8,9)*4 X'en(0,0) =EmmaAon86ninBrin=Arms And there isourprojection, Nes=f42X"cur(0,9)*AO,9) Ihaveprovennothingofcourse,butitdoesseemreasonable, Hereistheimpliedcompleteness condition: e 6 a Ta aoale e@ ZonXn(,03][X"en(O'9) Jr=Bir<BEIOS>=Bir(-2 which would becalled anaddition theorem fortheX's. Ifwecould prove thisdirectly, 1think our expansion conjecture would then bejustified. However, Ithink theproof would bevery difficult. The usualproofistoshowcompleteness someotherway,thentheabove hastobetrue! Ina different notation wemight say, <ném|i0$> <i0{n'£m'> =<ném|n'ém’> orthogonality <iG6|ném><ném|'i'0'g'> =<i04)'70'9> completeness Finally, ifA=A(¢,0,9), treatrasaparameter beingcarriedalong,andwethenget AG0,8) =Eoan Avan0)X"on(0,4) Aem(t)= f42X"a(0,0)** A(,0,6) This isreally amultipole orpartial wave expansion ofavectorfield,andwearegoingtoapplyittothe electricandmagnetic fields soon, What wearedoing issimilar toJackson Chapter 16butmaybe-a little more organized. Weshall findthatwecandealwith each partial wave separately andthereby simplify our e attackoncertainproblems. 5.Apply divergence totheexpansion Inthesenotes,Imistakenly usedthesymbol tinstead ofrasthelabelforthecoefficient oftheradial function Y.Ithen stayed consistent with thatuse. Thetwotransverse vectors ZandXgetlabels 1and2. SinceIwanttocheckthingsagainst Carleton's writeup, let'susehisformofthe expansion A= Em [A2X +A-¥ +A: Z] VeoA =Lm [Ve(ArX) +V0(A,¥) +V0(AZ) ]=Em[1+24+3] From ourappendix wehave VeXm =0 PeXm =0 VeYen=(2/t) Yon #8Yen=Yen Ve Zen=-&e+1)(Wt)Yon FeZn=0 Soweget 7 1=V0(A.X) =VAreX +A. V0X =0,Ar(feX)+ Ao(VOX)=,Ar(FoX)+Ar(VeX)=9,A2(0)+A2(0)=0 e 2=Ve(Ar¥) =OAr(PO¥) +Ay(VY) =(Ac)YentAc(2/t)Yom =UP(PFAe)Ym 3=Ve(AiZ) =OAr(FeZ)+ Ai(V©Z)=,Ai(0)+Ar(E41)(It)Yon) =~As(241)(It)Yon Sohereistheresult A= Dm [AX +A¥+AiZ] VeA=Lom[Ve(ArX) +Ve(A,¥) +V(AZ) J =Zam{WP aC? Ac)=ArSCHL)(1/t)}Yom. whichagreeswith(21) ‘Now, suppose someone imposes thisrequirement, Vea =4np Then thisreally mustbesatisfiedineachpartialwave,sowefindthat e {POC Ay)~ArCH)(I/t)}=420m ‘Thus, thisdivergence condition (perhaps V¢B=0foramagneticfield,orV#E=pforEfield)putsno restriction atallonthefunction Az,butrequires theother two functions toberelated bythis little DE. ‘This begins toshow how wemight apply Maxwell's equations andend upwith constraints onthe multipole coefficients ofthefieldswithineachpartialwave. 5.Apply cur! totheexpansion SinceIwanttocheckthingsagainstCarleton's writeup,let'susehisformoftheexpansion A= 3m [AX +AY¥Y AZ] VxA=Deo[Vx(ArX) +Vx(ALY)+X(AiZ)]=Zim[1+2+3] From ourappendix wehave VxXm=~[ltt] Yon-(Wt)Zen BxXm =-Zin VxVem=>(It)Xen #xYm=0 r | 8 |: eVxZmn=(1h)Xen PxZm=Xun Soweget 1=Vx(AaX)=VAaxX +A2VxX =0,Ao(PxX)+A2(VXX) =OyAa(-Zam) +Ar(-[E(E+1)/F ]You-(1/t)Zon) =[8p Aa+(I/t) ]Zag ~Az€(C+1)/t Ym =(UO,(Fz)Zim=AnC041Yow 2=Vx(Ac¥) =O,Ac(PXY)+A(V XY)=(Ac)OFACI)Xan =-AdI/t) Xen 3=Vx(AiZ) =,Ar(FxZ)+As(VXZ)=0, At(Kon)As(I/t)Xen =(8,Ar+Ad/F]Xen=(IG(7At)Xen Sohere isthe result A= 2m [AX +A¥ +AiZ] VxA =Bm [Vx(A2X) +Vx(Ar¥) #Vx(AiZ) ] e =Zan{(11)(Aa)Zin-E+IMEAYog+[(U)(@(6Ar)>Ad]Xm} 1 1 2 andamazingly enough, thisexactly agrees with (22)-(25), ‘Now, suppose someone imposes thisrequirement, VxA=0 ‘Then thisreally must besatisfied ineach partial wave, andforeach component! Soweget ~(Uie)0, (Aa) =0 =>nothing since Az=0from nextequation ~&(e+l)irAn=0 =>Ar=0 (YG,(6Ai)Aye=0 >A(rA=A, Sowith this condition, ineach partial wave wehave knocked outtheA,component! And theother two components arerelated bythissimple differential equation! 9 6.Doing Electrostatics with VSH Wehaveourtwoequationsalreadynoted,repeatthemhere: e Vek =4np which implies (148,17 B,)-Ey£241)(1A)}=4Don (a) verifies (31) and VxE=0 which implies -Met irEy=0 >E=0 (Ut)(@,(¢ EiEr=0 >a(rE)=E, (b) ThetwoDE'sbothinvolvedE;andE,.Ifwecombinetheseweget(32)asasingleDEforEj,andifwe then define =-1E,then equation (31) reads asfollows: OF}+2lt)A-CEO=-AnDom e@ Compare thistoJackson 16.5withk?=0andasource added, andyouseethatwehave discovered our friend thePoisson equation where the"potential" seems tobe¢=-rE;.Sothink ofthisassaying thatwe solve thePoisson equation for6,andthenwehaveE,=~$/r, E,=-d,6,E,=0.ThenourEfieldmustbe, EB=2m [EY +B) Z]=Lm LAOY -orZ)= =Eon[dom®Yon+donVYon} where now wehave shown thesubscripts on which wewere toolazy toaddbefore. Dowerecognize this solution 2GotoJackson (4.1) onthemultipole expansion for},andassumewehavefoundsome solution (68,9) =Emm(4n/(2E+1)) QnUe" Yen=Eenen(t) Yen(8.4) where Sent) =(4/(26+1)) GonVe" Then the electric field should be E=-Vo =-Len VIGen(F) ¥on(8.6) ]=~Ean€VIdantT)]Yen(6.4)+en(t)VYen(8-9)}e 10 -_ 7” Oe e ==Dan{Aben(t)FYou(8.9) +bent 1VYon(0.4) } =-Emm {Abma(0) ¥+buat Z} which agrees with ourVSH result. 7.Doing Magnetostatics with VSH Before looking atMaxwell's here, (35)isjusttheelectrostatic expansion restated formagnetic quantities, including themagnetic potential. IfwesetB=-V weinstantly endupwith (36) through (38), no problem here. Now, let's look atMaxwell's equations. Wehave ourtwo equations already noted, repeat them here: VeB=0 which implies {UP8,(7B,)-Bi&(é+1)(1)}=0 see(42) and @ VxB=4nleS which implies . ~(11)8, (FB) =Ande Sy /Z (40) =Mtl) By=4nle Jy 1Y G9) (1de)(@;(tBy)-Bur=4adeJp IX(41) ThetwoDE's both involved B;andB,.Ihave combined these toget(43)forB*driven byJ. ‘Now derivation of(44) requires abitofdigression! Jackson (3.120) shows thedefinition oftheGreen's functionforthisequation andtheformofthesolution. Iabbreviate asfollows: op[g] =-4n/r* 8(r-1') 3.120) ‘Then ifyouhave oplF] =s the solution will be e u a FO)=-fate)3(°)(7/4)de? t)0 asyou know byapplying optoboth sides. Weshow thesource function saslocalized. Forroutside this region, wehaver>1’andweknow thatg(r,r) =B'(ryr""". Similarly, ifwelookatverysmall rnear=0, we conclude that forr<1’wehaveg(c,’)=A(e)e’.Fromtherequiredsymmetry ofg,thistellsusthat geyr)=Ch rf JLi/nf'] =4n/(2¢+1) refi Wedetermine theconstant according topage 80andwefind that C=4n/(2é+1) asshown above. The solution ofourposed problem isthen a F(r)=~1(2é+1) *1!*sore dr 0 ‘Now inourparticular problem wehave s(r)=4z/e &(¢+1) Jx(r) andF=rB,andthisgives (44) 1! Sothepoint isthatwedidourVHS thing forthismagnetic problem andweended upwith an equation (43) forB,driven byJzandwehave solved itin(44)forB,Comparing (44)to(36)gives a Hon£16fTam(e)vede e 0 butofcourse wearestillin£mpartial waves here, Wecanuse(11)toreplace Jzm(?) with theprojection integral, andthisthen gives (45) which canthen berewritten easily as(46). This result (46) claims totellusthemagnetic moment ofacurrent distribution. InJackson this appears in(16.95), buttheforms don't exactly match, Here ishowwefixthings up.Consider: Vel Vn rxS] HexdeV[Vil+1¥mVe(exd)=142 T=rxS©VC)YonHEXerV¥m=OFTVVggXS Soweget VelYmPxF]=f¥meV0(ext)trV¥merxd In(16.96 a)thedivergence integral will vanish, soweget 0=Jackson +Carleton. The only disagreement isa-sign (both have minus signs). Itisalittle hazy how Jackson defined hismoments, perhaps thisminus arises because ayyhasaminus sign in16.47. Ignoring thisminor detail, Carleton's exercise here wastoconsider magnetostatics, useapotential to define themoments, andthen solve fortheBfield using theVSH anddoacomparison toendupwith a classicresultforthemoments. e 12 @ 8.Maxwell's Equations withVSH. WelistoffthefourfullycoupledstandardMaxwell equations withsources,andthenwewritedownall8 ‘VSH equations. These follow trivially from mydivandcurl results above, allarecorrect. Ithen did exactly what Carleton said, andgotthetwodriven wave equations shown in(59) and(60). Butwehold now todoanother exercise section. 9.Fields inconducting cavity. (1)Duplicate stufffrom above, butsettoBandreplace Vwith? since wewantf«B=0 B= Dm [BX +B,¥ +B)Z] feB= Sm[f¢(B2X) +f©(B.Y) +?e(BiZ) ]=ImBs Ym =>B.=0 using VeXm =0 FeX =0 V©Yem =(2/8)Yom £©Yen=Yon e VeLem=-£(l+1)(1/2)Vem PeZm=0 (2)Duplicate stufffrom above, butsettoEandreplace Vwith? since wewantfxE=0 E= 2m [EX +E,Y +E\Z] FxE= Lm[Px(E2X) +Px(B,Y) +8x(E1Z) }=Zam[-E2 Zon+E:Xm] >EL=E =0 using VxXm =[ECCI] Yen-(Wt)Zan #xXXm =-Zin VxYim=~(It)Xen Px¥m =0 VxLom(It)Xen FxZog=Xm So,theboundary conditions foreither aspherical conducting cavity oraspherical conducting sphere are very simple: . FeB=0 =>B,(a)=0 PxE=0 =>E,(a) =E\(a) =0 13 where these fields areevaluated atthesurface ofthesphere, r=a.Certainly asimple result! Nowiftherearenosources,wearejustlookingforstand-alone solutions thatcanexist,theyreferto ethese asresonances. They willofcourse have surface currents andcharges, butthese arenotthe"free" sources implied bytheequations. Now wearegoing to"look forsolutions" ofthe8equations (51)-(58) subject totheboundary conditions (61), everything issourceless. Recall thatE,andB,refer toRADIAL components ofthese fields. Foreach field, weknow thattheother twocomponents areTRANSVERSE. SoNOW finally we areseeing themeaning oftheindices 1,2andr--better latethan never. 1and2arethetwo transverse directions that weassociate with the functions Zand X. Ourequations divide intotheo.group andtheBgroup asshown. Suppose inthe@group welook forsolutions with E,=0.These would be"transverse electric" or TE,sincethereisnoradialEinasolutionofthissort.Inthiscase,lookingatthecegroup,wegetEy=0 from (51)andthenBy=0from (55).Then equation (56)and(57)areboth 0=0andaddnothing new. ‘Thesurviving fields areE,andB,andB,which areallintheBgroup. SotheTEsolution meansthatall three fields inthe group arenulleverywhere. ‘TE=magnetic: a:E,=E)=B)=0 B:B,,By,E,allactive TM=electric: B:B,=B,=E,=0 a:E,,Ej,Ballactive ForTE,tofindthenon-zero Bfields,weseethatE,mustsolve(60),andforthecavitycase,ithastobeajdkr).Noticethatofthe three conditions (61), themiddle oneisintheo:group, theother twoareintheB. SoforourTMsolution inthecavity which involvesBfieldsonly,thecondition E,(a)=0impliesj4ka)=0whichgivesthekmodeeigenvalues fromthezerosofthisfunction.ButwealsohavethatB,(a)=0toe think about, butequation (53) shows thatthisissatisfied atonce ifEx(a) =0. ‘Suppose welook forTMsolutions with thewhole agroup being zero. Again, just assuming B,=0 forces Ey=0from (53) andthen By=0from $4.Inthiscase, wetake intheogroup thatByisaj,(kr) for thesame reason, finite attheorigin. But now theBCisthat E)(a) =0and from (57) this puts the condition that 3;(rB2(a))=0=0,(rjkr)" asclaimed in(65). Sofrom this little exercise, weseethepower oftheentire formalism here. We have thetwo decoupled equation setswhich wecanthink ofcompletely independently, thisiswhy wehave thefamous ‘TEandTM "modes" inallthese problems, Inever quite understood thatbefore. InJackson page 543, he produced two3-equation setsbutthen hehadtodosome arm-wavingonthenextpage. Solving theconducting cavity problem waseasy! Theboundary conditions were trivial, wethought abouteachofthetwomodes, andforeachmodewereabletowritedown"byinspection" thesolution for theprimary active field, either B,orBy.These aretheXorLycomponents,reallyEna(t)keepinmind. These aretheprimary EandMfields inJackson's 16.47 grand finale. Now ineither case, once youhave theprimary field, theonly other existing fields areoftheother type. Forexample, inTEyoufindthatEy isyour Bessel function, andthen theonly other existing fields areB;andB,.These aretrivial towrite down using (53) and(54). And similarly fortheother case. Soagain, once yougointopartial waves, this spherically symmetric problem istrivial tosolve, andyou dogeteigenvalues fork,This isavery excellent problem toconsider. 14 : 10.RadiationfromaLocalizedSource e@Sonow weputthesources back inandwetrytosolve (59), (60) using theGreen's Function method. From Jackson weknow that theGreen's is(68), andthisgives solution (69). You canalways verify a result likethisbyapplying opf]toboth sides which Ijustdid.Imodified (69)with anupper endpoint on theintegral, butnow seenotnecessary. Now youhave torecognize theJ-function there asbeing the2-coefficient (the Xcoefficient) inthe expansion youcouldgetforVxJ.Youcouldthenusetheinversion formulaforthiscoefficient asIhave shown after equation (69). Inserting thisinversion into(69) then gives theresult (70) with (72). And (71) istrivial from (50). Stop! We have now derived thestandard Jackson formula foranelectric orTM partial wave multipole field! Wearereminded thatitiscalled electric since pappears inthecoefficient. Ijustspent about 4hours deriving Jackson's identity (16.90). Ittook $pages, andImade about 5 mistakes along theway, each ofwhich hadtobeidentified andcorrected. Very painful, retain innotes. SoIamnow happy with Carleton’s (72) and(73). You have torealize thatboth Carleton's Xandhisaz arescaled relative toJackson. Now look back at(59) and (60). Wejust used theGreen's method tosolve (59) fortheTM modes andourresult was (70)-(73). Solving (60) instead fortheTEmode issimpler since theJstructure ismore straightforward, Inanalogy with (69)wegetthesquare current bracket replaced with [-ikrJ?](which cancels thefirstI/r'factor) ,andofcourse B?isreplaced withE*.Therefore, ourresult willbe(72)with curlJreplaced by-ikJ,andthisgivesprecisely (76). 1justreadthetrailingpartofthehandoutondipolesandquadrupoleandlong2limitsandCartesian emoments andallthat, itisfine, butthisisnotwhat interests meinthishandout, sononotes Done!NowIwouldliketowritea"summary" ofwhatallhappened here!Forestfortrees.Amazingly ittook me18pages ofnotes todigest thislittle handout which isonly 12pages! Appendix 1:Horrendous operations ontheBasis Functions reXg ~POXm ~FOL Ym~re(rxV)Yn=O PeXm=08Yen=F°FYen=FYon PeXm=~1eZm= ~6VYmo~tVeYon=0because Yimisnotafunction ofr PXX=KerxXen=kerXXVYom=Ke[VY an)f=VYan]=~ke?VYon =ekyrZam==tXv rxX=rxYn=PYm=O e 1XXm=kyPXZag=keX(tVYon)=ketFXVYom=KetXe=FX 15 VeXtm=kyVeXm=k,V0(rxVYm)=ke(VYe(VxX8)-re(VXV¥m)]=0 e VeXm=VeYen=VO(PYm)=[V¥me+¥en(Vet )]=Yon(2/t) =(2h)You VeXekeV#Zan=keV8(fVYon)=ke[Vi#VYostt(VOVYod]=FkeV?Yen==ke(E+)(U8)Yen==(Wtke)Yon VxXlaq=keVXXoq=keVx(xVYu)=?//Thisisaharderonetodo,towit:f=0,4) Vx(rx VA)=r(VeVE)-VE(Ver) +(VieV)r-(reV)VF =1+24+344 1=rV'f anyf 2=-3Vf anyf 3=Af dn=Of; =Of=VE anyf 4a-ndOF=-n BOF =-3(nA+Gn\(af) =-8,(eV) +Of=0+VE= VEtrueonlyforf=(6,4) SoVx(rx Vf)=rV'F-VEF andforf=Yomget=-[£E+1)/t] YenF-VYon$0repeat: VxXmkyVXXankVx(rxVYom)=ke{-[6241 Yon®-VYom} aRel -(C41Vr]Yon-(1/t)Zon}==[k+l] Xe=(It)om VxXm=VXYom=VX(FYan)=VYonXP+Yon(VX)=(it)xVYon, e (Lt) Xen=-1/kt)Xo VxXm=VXKyZen=hyVXV¥on)=Ke[VEXV8+(VXVYos)= ke[8xVYen+0]=Kt)xVYon=KALt)Xem=(1/4)Xb ‘Now letsengage thecur! ofcur! battle! VxVXXbom=VX(-[kel(lt] Vt]Xm-(l/r)Xe) +ke) VxL(t)Xan]=Vx[(U/t)Mew)==kLCC) -2] (1)=Vx[(1/2) Xen] =Vt) xXe+(Lt) VxXe =e) {PxX) Ht)(VxXo} =CUP) (0 }+A) {-(kat)Xe}=Cla?) Xho [2]=Vx[(1/r)X°eu)=V(t)xXm+(It)VXXan =GUC)FxXan} (1A)(VXXem}=(VP) Xam}+(1){t)Xm} “9 ° ® 16 meee ve e =>VXVXgg==kel(C+)[1]=[2]=~ke€(0+1)(Lika?) KbELH Xp => VXVXXl =ACHP Xen Vx(VxX im)=Vx{=(Ilka)Xen}=CUK)Vx{(1/t)Xen}=CUK) (CUP) {BxXe} +(1) VXXm} ] =(Uk) (C1?) (=Pag}+(Ut)(=tel(C41Vt]Xen=(Wt)Xe}] =Uke) [(M){~[kef(€+1/t ]Xm}] =+P Xan Jinteresting! Vx{VxXm} =Vx{(lt) Xm} = =(CUP) (FxX'm)} +(It)(VxXom}] =[(/t){-[ke(¢+1)/t] Xm}]fromprevious calculation=kyf(t? Xm //breaking therulesetbytheprevious two! Summarize these double-curl results: Vx(Vx Xe) =ACHP Xlom Vx{VXXen}SLEYI?Xe Vx{Vx Xm} ==kpAEH Xm e@ Nowlet'strysomeLaplacians: VWXm=VEV0Xe}{VXVXX em} =V{0} -{&(+1)/P Xen} =-6+) Xm VX m=V{VeXm p={VXVX Xm} =V¢(2Mt) Yom}-{€(E+1Y? Xow } =2{V(C1)You]}=4+LPXP, | =2{CUP)PYen+(1/1)VYon}-(2+? Xow 2{CUP) Xam (UP) Xam/ke}=AEE Xn ) =(2ke?)om~[ECE#N)#2]1?Xam /notverynice VXim=VEVOXmn}(VXVXXm} SV{=Ky(E41) (1)Ym}=(>Ke(E41 Xm } kyE41)CV[1Yon)}+KeCEH Xe, sake QE+L) {(1A?)FYen+(It)VYom}+ky(C41 Xm ==keROFL) (CU) XmHIP)Xm/Ke}kyCLEP Xp SDK LEMS Xm =EY PXom 7 4 Iamgetting tired ofallthisstuff. Wondering ifMaple candoit?No,butIfound apackage thatIcan addtomyMaple,looksinteresting... Well,theproblem isrelatedto"inert"versus"doit”.Idon'treally e want thething tobecalculated, Iwant ittoremain insymbolic form, likeVY. Back tohand work! TheLaplacians look suspicious, {bettheyshould allcome outthesame, butthere issomuch algebra here Ihave norealway tocheck things, andearly errors propagate! Atleast Ihave awritten derivation record soIshould beable torepair things atsome point. Let's summarize some ofthese result inatable: Xow=kXen=kerxVYm=ikeLYenXm=2xVYon Xm =¥en=PYn Yon=FYon Wom =kyZn=kerVYon Zen=1V¥on eX 0 FeXm =0 BeX=Yn #6Yen=YonFeX'm=0 FeZm=0 FxX ==Xm FXXmq =-Zm PxXa =0 £x¥m=0 fxXen=Xho #XxZon=Xn e PexxX'_=-P XXA=Xow FxPxXen=-Xew PxP xXm=0 xP xYpq=0 PxPxXm=PoxX=-Xm PxPxZim5+Zem Vex a0 VeXm =0 VeXmn=(2/t) Yon VeYom=(2/t)Yow VeX= kyMEF)(It)Yom VeZoo==(641)(It)Yom VxXow~[keEFLMt]Kem=(Ut)Xe VxXm==[Lett] Yen~(It)Zon VxXm=-(kt)X'om VxYom==(1/2)Xen VxXm=(It)Xion VXZen=(It)Xen Ve(rxVA=0foranyf 18 I iee Summary ofCarelton's VSH Handout Notes PhL 2.13.03 The Basis Functions andExpansion Theorem Definethesethreevectorfunctions onthesurfaceofasphere6,6whichhasradiusr: Xm=xVYon 2 =Xan VE(E+1) Yen=FYon rt =Xe Zn=1VYon 1 =XmVeer) where Yqistheusual spherical harmonic asinJackson. These vectors aremutually perpendicular, with the¥function being radial and ZandXbeing transverse tothesphere. InanE&M problem, one is familiar with thetwo transverse directions, and |and2areexactlythosedirections. Withintheémmanifold(partialwave)ofthemultipoleexpansionofavectorfunctionA(@,6),the three vectors above form atrue setofbasis functions, andweendupwith thefollowing expansion theorem including theprojection formula AGO) =ZonnAennX"m(8,$) Ana=fdaX"mn'(8,9)* A@,6) . expansion projection t)‘Theseimplythefollowingorthogonality andcompleteness relations: sinX"em(0,0)**X"mn(09) =SeeSernSen orthogonality Zeon [Xen (8,0)] [X"m(8'.0") Ji==Six8(Q-Q') completeness Inthedetailed notes, weprove orthogonality, andthefactthatthethree vectors areperpendicular toeach other, butwehave tosimply accept asbeing awfully reasonable thefact that thebasis functions are complete, sotheexpansion isvalid foranyreasonable vector function. Icould notfind atrivial proof of thisfact, butthere probably isone. The above result isthen trivially extended tovector functions of1,0,@byadding rasaride-along parameter, andwethen get, A(0,) =ZeusAcma(t)X"em(8,) Aam()= fdaX"em(8,6)A(8,6) Inourapplications wewill often think ofAasbeingE,BorJ.Theaboveisthegeneralization ofthe partialwaveexpansion ofascalarfunction,whichweknowworkslikethis: 1 e A(9,6)=ZonAea(t)¥en(,$) Aem(t)=f42Yen*(8,6)A(6,0.4) S92Yew*(0,6) Yen(0,0) =BecBrion orthogonality LmYoo* (0,9) Yen(0'$') =5(Q -2’) completeness Wearefamiliar with theuseofthismultipole expansion inelectrostatic potential theory forthepotential Some Useful Properties oftheBasis Functions VeXm =0 BPeXm =0 VeYin=(2/t)Yen PX =Yon Ve Lon=-(641)(Vt)Yon PeZm=0 VxXom ==(EEE Yem=(Ut)Zim FXXm=~Ze Vx¥emq=-(It)Xen TxYm =0 VxZm= (1)Xm PxLen=Xen Itturns outthatalloperations youcanthink oftoapply tothebasis functions, ortoscalar functions ofr etimes thebasis functions, always keep you within theémmanifold. The reason forthis fact issummarized bythefollowing exhaustion ofallpossible dimensionless vectors youcanmake outofthebuilding blocks rand V, FYm Y 1VYm Z rxVYenx fre Vm) t) PYXVY mn 0PPVCV nmPeV?Yon==PL?Yon=~C(E+1)PYin==CHI). ‘The lastitem istheonly butcrucial place where theproperties oftheYmxarerequired, and noother functions willdo!TheYm»work because they arecoordinate-space representations ofangular momentum states {¢m>which diagonalize L?andLs,which ispossible because theHamiltonian isarotational scalar. From now on,wewillwrite theexpansion formula using Carleton's X,Y,Z form, towit, A= Ym [A2X +A,¥ +A:Z] inashorthandnotationwherethethreecoefficientsreallyhaveémlabelsandarefunctionsorr,andof r)course the basis vectors also have these labels. Note that |and 2indicate the transverse vector coefficient | functionsofr,whiletistheradialvectorcoefficientfunction. 2 e@ Diagonalizing Maxell’sequationsinvolvesapplyingVandVxappliedtotheaboveexpansionfortheE andBfields. Using theproperties noted above, onecanshow these essential facts: VeA =Sm [Ve(ArX) +V*(Ar¥) +V*(AiZ) ] =Em{1478,(7A.)~Ar£(E+1)(1/t)}Yom Theresult isofcourse ascalar, andtheAgcoefficient does notappear intheresult, indicating thatno constraint isputonA,byadivergence condition onemight findinMaxell's equations. Similarly wehave VxA=Em[Vx(ArX) +Vx(A,¥)+Vx(AiZ)J =Lem{-{(1t)8¢ (Az) Zen~26+1MA2Ven +[(1/t (Ge(6Ar) Adt] Xm } 1 + 2 This result ismore complicated and allthree coefficients areinvolved. We have shown thecorrect expansion labels fortheresulting coefficient combinations. Atthispoint inthehandout, Carleton does twowarm-up exercises. Electrostaties Exercise. Use V«E=4zp andVxE=0andseewhat happens, Theresult isthatE=0 e andwegettwocoupledfirstorderDEsforE,andE,whichwecancombine togetthePoissonradialequation forthequantity (rE')which wecaninterpret asthepotential -.‘Thisequation isdriven by source term4zPm. WefindinfactthatEley=-don/rwhere thedo,aretheusual scalar multipole coefficients used inelectrostatics multipole expansion. Magnetostatics Exercise. Use V¢B=0andVxB=4r/cJandseewhat happens. TheVxBequation results inthree equations (39)-(41), andtheV©Bequation gives afourth (42). Bycombining twoof these, weobtain again Poisson's radial equation, butitisnow forthequantity (rB") andtheequation is driven byaF*term.Carleton provides in(44)theGreen's Function solution forB*.Viaaseparate path, hewrites theconventional scalar multipole expansion formagnetic potential 6interms ofmoments ton Then B=-V@isexpanded inVSH giving equations (36)-(38). Comparing ourGreen's solution with (36) gives atraditional formula forthemagnetic moment asanintegral over thecurrent distribution J. After these warm-up exercises, wegoforthewhole nine yards. Onpage 6wewrite outthefour Maxwell equations (with sources), andapplying theabove divandcurl expansions, weendupwith 8little equations forthe6VSH component functions. Now comes thekeyfact. ‘These 8equations can be grouped intotwosetsof4equations which Ihave called theagroup andtheBgroup. Each groupinvolvesonly3ofthe6unknown functions, sowehavereallynwocompletely independent solutionsets. TE=magnetic: a:E,=E,=By=0 B:B,,By,Epallactive e TM-=electric: BB, =B,=E)=0 a:E,,Ey,Byallactive 3 ThenameTEarisesbecauseE,=0inthissolutionset--thereisnoradialEfield.WeknowfromJackson t) thattheconventional namesmagnetic andelectricariselateronwhenweseewhattheradiation solutionslook like. Ingeneral, yousolve theTEandTM problems independently, andthen ofcourse you could superpose solutions together. Notice thisfact: ineach active set,yousolve fora#2field, andthatisALL ofthatkindoffield. The other twofields inthesetareoftheOTHER type. SoinTEtheentire Efield is simply E*,forexample, andtheother twofields areBtype. Think now about what wehave done! Wetakethefourvery complicated Maxwell vector equations with nasty curls and vector components, and wereduce them inthepartial wave expansion to2 independent setsof4scalar equations forthree field components. Ineach set,wecanshow that the#2 field component function must solve theradial wave equation driven byanappropriate current mixture, as shown in(59) and(60). Full Maxwell Exercise: thecavity resonator. Imagine nosources andaconducting metal cavity. What E&M modes (resonances) canexist inthere? TheBC's aretheusual? ¢B=0andrxE=Oatr=a, which boildown totheextremely simple (61). Since nosources, solutions oftheradial wave equation are spherical Bessel's likejkr). TheTEandTMsolutions arejust multiples ofjkr) asin(62,3). However, theBC's cause thesolution values ofktobequantized, andthese arethen thecavity normal modes. The conditions forTMandTEaredifferent, (65) and(66), somodes have different eigenfrequencies. This is afairly complex problem toeven think about, buttheVSH method provides analmost instant solution! Ofcourse thegeometry isoptimal forVSH work! Bytheway, these same BC's apply toJackson's scattering from aconducting sphere. Inthatproblem, wehave extra BC's inthefullcoordinate space ofanincomingplanewaveandoutgoingsphericalwave. InowrealizethatJackson16.139isnothingmorethantheexpansionofaplanewavetravelinginthez e@direction onto theVSH basis functions. Heuses circular polarizations. Irecall atthetime how completely obscure this seemed! You can seehow Jackson prefers tousethecombination Vxf;(r)Xm to incorporate the¥andZcomponents oftheexpansion,andhehasevenprovidedalittleorthogonality for this combination in16.132 which isunbelievably obtuse! Carleton has shown how this can allbe understood inasystematic fashion. Icould resolve thisproblem using theVSH from thestart, project the plane wave onto thebasis functions from scratch, butofcourse Iwould endupwith thesame 16.139 with threetermsinplaceofJackson's two. Radiation from alocalized source. Here wejust take those two#2component wave equations anddo theGreen's function solutions, asdone inJackson. TheB*solution isshown in(69) (thisisofcourse the particular solution, towhich wecanaddsolutions ofthesourceless problem, likethose cavity modes, in ordertomatchsomeboundary conditions.) ‘Now onpage 9wecome tothenormal "statement" ofmultipole fields. ForTM, weknow that theB field hasonly theBycomponent, and that iswhat (70) says. Here weusetheHankel 1function because weareimplicitly matching thespherical outgoing boundary condition, since ourtopic is"radiation from a localized source". Jackson in6.42 isslightly more general, notspecifying thetype ofradial function. Away from thesource, yougettheBfield from theBfield intrivial fashion justfrom theusual Maxwell. Byinstalling ourspecific Green's solution (69) forBz,wegetaspecific result fortheradiation field in terms ofnewly minted coefficient agasshown. Itisanintegral over thecurl ofJinthesource times other stuff asin(72), andthisistheresult thattook me5pages ofnotes toconvert intoform (73) where yousee that aTMmode isdriven bycharge pifitispresent, hence thename “electric”. 4 Atthispoint,declarationoftheformoftheTEmodefollowsatoncewith(76),justlookingatthe r) different sourceintheradialequation. Inthiscase,wedon'thavetodotheelaborate conversion, weseethatayisanintegral ofJasshown. Sofine, wehave recovered Jackson's multipole expansion results inamuch more organized approach, with much lessmystery. Ipresume Jackson hasheld firm inhis2ndand3rdeditions. Comment onWave Equations. Weknow that components ofEandBinfree space solve thewave equation, letthere belight! Inthepreseice ofsources pandJ,thesewaveequationsarenotsonice.Here isaquick shot, (V?+)E=4nVp-4nik/e J and=ikB=VxE (VW?+k?)B=-4nlc VT and ikE=-VxB +4n/cJ3 This iswhy weintroduce thevector potential A. Now intheVSH approach, wefind that ineach mode, the#2component satisfies aradial wave equation. The B2solution fortheTMisdriven bythe#2VSH expansion coefficient ofVxJ,inanalogy with thesecond line above, giving usresult (72). However, after thetransformation to(73) using continuity, wefind that infactpisreally themain driver oftheTMmode B,contrary towhat appears above, where wedon't even seepinthesecond line, Similarly, theEpsolution forTEisdriven bythe#2coefficient ofJ,asshownin(76).Thisdimly reflectsthefirstlineabove,butagainwemarvelattheabsenceofpinaffectingE,directly.Ofcoursein e@light ofcontinuity, thetwosources pandJarealways connected andallthese comments arenotvery productive Iguess. Idon't think thevector basis functions satisfy anywave equations, though Isaw someone claim this once. See other notes ontheNandMfunctions. 5 MieScatteringandtheMandNfunctions PhL 2.14.03 ® These started outbeing notes onmylittle MieTheory 4-page downloaded paper which happens tostate allthecoefficients informulas that look like mine, Review. Wehave seen how thefunctions X,¥andZspan the4mmanifold. Wehave alsoseen how one solves anE&M problem separately inthetwo"modes" known asTMandTE.Each mode hasitsown private setofequations andoperates independently. Ineither Jackson orCarleton’s notes wehave this basic idea: TM: Ben=ag(ém) f(r)Xem(8,9) Emm=(Uk) VxBrn TE: Emm=au(lm) gdr)Xon(9,) Bry=Ci/k) VxEm ‘That istosay,foreach ém,wecanhave thetwoindependent solutions noted above. ‘The quantities f(r) andg(r) each represent alinear combination ofspherical Bessel functions. The twocoefficients ofthislinear combination aredependent on¢andmust bedetermined from boundary conditions. Usually they arenormalized sothesum oftheir squares is1,butthisisnotnecessary, since theag(ém) typecoefficient canpickuptheslack. Theasymptotic behaviors ofj(kr) andh(kr) are especiallyusefulinmeetingtheseboundaryconditions,thefirstbeingfiniteatr=0,thesecondhaving e outgoing spherical wave behavior. ‘Thenormalization ofthefunction Xqq(®,6) varies with author, butalways hasthisform Xen(0,$) =C(ém)*rxVYen(8,6) Ceém= +1 Carleton 1 Mie — ihackson “fecen) ‘When wewrite ZandY,wealways mean Carleton's normalization forthese twobasis functions. Itwould seemthattheXvector basisfunction, oneofthetwotransverse ones,issomehow more "fundamental" than theother two, since itprovides the"main field" ineach ofthemodes. The"other" field thatgoes with each mode isfound from theVxoperation asshown, andwecanwrite out: Vx(FA) Xem(8,4)] =-C(Em) {(£'dr) +1/1)Zem(8,0) +A(E+1) F/T Veu(8,0) 3} Thus, aswewell know, the“other field” inagiven (TM orTE,transverse) mode hascomponents inboth theZ(other transverse) andtheY(radial) direction. Wecanrewrite theabove vector expansion as follows: @ Vx[F(2)Xem(0,)]=£10)FXXen(8,O) =CCm)€CE+1)freFYon(0,4) 1 wherewehavekeptitgeneralsothatanybody'sXcanbeusedhere.Thisappearsas16.143inJackson r where hisC(4,m) isasshown above. Todothis, weused thefact(from ourappendix inother paper) that Z=- xXcatwon =-(1/C(ém)) xX ‘Nowwhenwewritethefullexpansion overallthemodes,wegetthisresult: B= 2m [ az(ém) £7)Xem(8,6) + Uk)Vx{a(ém) gt) Xem(8,9)} J E=Em[(Wk) Vx{as(ém) fr)Xen(8,9)} + am(lm) gr) Xem(8,6) J andthisappears asthe"grand finale" 16.47 inJackson, which Inow sceisnotallthatgrand. Itisjustthe statement ofwhat wearesaying allalong intheVSH expansion. The game istodetermine thetwo coefficients ap(€m)anday(£m)aswellasthespherical Besselimpliedlinearcombination coefficients for each ém. Notice that intheform above, there isnoimmediate need toexpand thesecondary terms in either way shown above, butweknow wecandoit. TheMiebasisfunctions MandN.IcallthemMiebecauseIsuspecttheyappeared inhis1908paperin thisform. Inmylittle 4-page downloaded Mie Theory paper weseethefollowing definition: Men= Vx[FAt) Yom(8,9) =VXE(FUE) Yen(8,6)] e Warning: The Yim(0,9) which appear inthisMie paper may have different normalization than weare used to,butwewill absorb anydifference intoourlinear combination ofspherical Bessel coefficients, whicharealreadyimpliedwhenwewritethegenericfr) From thefirst form, wecanquickly simplify asfollows: Vx[fat) Yon(0,9) ¥]=V[£40)You(8,6)] XP becausethesecondtermhasVxrwhichis0.Inthenextstep,canseethatthe V[£46)Yen(8,8)] =£0)VYn(8,0) +£10) PYen(8,6) sothesecond termgives nothing sincefxr=0.Soweendupwith Mea(t,8,9) =Vx[f4t) Yom(0,6) F]=-£40) rxVYom(0,9) =-£40)Xen(@,0VC(Em) _//ouroldfriend Thus,wecanwriteourexactsamemultipole fieldsas TM: Ben=~az(£m)/C(é,m)Men(t,8,9) e Emm=(Uk) VxB 2 | e TE: Eye==ay(Em)/C(E,m)Men(0,6) Byy=(ilk) VXB Atthispoint, why notdefine Nom(1s0.6) =(1/k)VxMew(£,8,4) =(1/k)Vx¥x[fi(0)Yon(0,6) ‘Then wecanbeeven more compact andsay TM: Ben=[-ap(ém)/C(ém)] Myae(1,9,9) Een=i[-az(ém)/C(é,m)] Nowe(r,6,9) TE: Em= [-ay(ém)/C(é,m)] Meaw(t,0.9) Bem=~i[-ane(ém)/C(£,m)]} Nemna(t,6,9) Here Ihave added E(electric) andM(magnetic) labels onthe MandNfunctions tomake usremember thatcach onecanhave adifferent spherical Bessel function, That is,wemight write Mone(+,9,0) =Vx[ft) Yon(8,6) F] Monm(?,8,9) =Vx[841 Yom(8,6) e@ Nowlet'srescaleourcoefficients intheobviousmannercp(£m)=-az(¢m)/C(4,m)togetthisfinalform ‘TM: Bm= Ce(ém) Mene(t,9,9) =ceMeme Eem= ice(ém) Noue(t,0,6) =ice Nowe TE: Eoa= em(2m) Menm(1,8,6)= ooMeant Bam=icu(Cm)Nomra(t,0,9) =iCreNemns where wearetrying toachieve agoal ofvery compact notation. Atthispoint wecanwrite ourfull expansion as B= Ym CeMme -iceNene J E=2m [iceNeme +¢1Menm | which Ihave toadmit ispretty compact. Now thepaperIhavehasadifferentnotationforthings: them me even e => aspecial meaning notdiscussed yet!(electric) odd0=>aspecialmeaningnotdiscussedyet!(magnetic,secbelow) e@ na =£,themainindexonthespherical harmonics : 3 ) a = mytheminor indexm =n,theindexofrefraction (theycallthisthe“opticalconstant") t)ka =x,knownas“thesizeparameter" ‘The superscript ontheNandMfunctions relates tothechoice ofBessel function, The functions have label 3(which would beaHankel 1)andfield hassubscript s.The"internal" wave (inside thedielectric) haslabel1(which wouldbej,)andsubscript t(transmitted), whiletheplanewaveexpression alsohasthe internal 1label andsubscript i.Thethree labels arecorrect. You canseethatthecoefficients in(1.4) and (5) aredefined inapeculiar manner, perhaps asMiedidthings. Status: So,atthispoint Iamhappy about theexpansions (1.4) forthescattered wave, and(1.5) forthe transmitted orinternal wave. Mybigproblem is(1.6) fortheplane wave incident. Iknow from Jackson thatthesimplest forms fortheplane wave arewith circular polarizations, andwehave E,=E, (ane) [jAKt)Xear E(W/K)Vx{jak)Xe} Be=E, (Vane) [Fijak) Xear -(WK)VX{jdke)Xe}] Foreach partial wave é,weknow thatthem=+1areinvolved. The formula (1.6) does notshow this detail. What ismost annoying isthattheauthor simple makes nocomment whatsoever about polarization, asifitdid not even exist. Aside: Whatwould happen ifweaveraged overthetwopolarizations fortheintensity? Wecould thinkaboutthePoyntingvectortimeaveragedwhichwouldbe e <S> =c/8n ExxBy” Because ofthe*,therewillbenon-vanishing crossterms. Rewrite as . BL=E, (fame) [—jdkt)Xess #(UM) (Cen) {(£40) +Ut)Zs(0.6) +CE) FEV Yeu.) 3] Bs=E, ()V4n(Qe+1) [Fijer) Xa =GK){+Cai) (£14) +1A)Z'a0) +ACI) Fee Yew04) 391 ItistheX-Z cross terms that survive asinCarleton (7),along with thediagonal terms. Iseenotrivial argument thatsaysyoucansomehow replace X,.1 with Xjoinanykind ofaverage sense. SoIhavenow(momentarily)" lostthefaith"inthispaperduetotheunsupported expansion (1.6). This makes menotunderstand thequoted Mie coefficients, although they dolook very much liketheones 1got.Ididfindthatthecoefficients infactdidnotdependonthepolarization state,something that seemed strange atthetime. Soforthemoment, |willnotregard thisasaconfirmation source formy coefficients, butwill instead learn some generalities about Mie scattering! Some new realizations. For given index and aparameters, there will bevalues ofkforwhich the coefficients blowup--poles!Ididnotthinkthiscouldhappen. Perhapsthenumerators compensate so@things don't goinfinite, butthese poles atleast tellyouwhere there will belotsoflarge field action! In 4 experimentswithalaser,youletdropletsevaporatesoradiusadecreases,andthenyoucanpassthrough e resonances. Nodoubtthesearetheresonances Ofthedielectric spherejustaswehadinthecaseofthe conducting cavity! The Qvalues quoted arefinite, even forlossless dielectrics, sothenumerators must beproviding cancellation. (?)Author states thatyoucangenerate fields neartheobject thatareamillion times stronger than theincident field, duetotheresonance effects! This paper isnodoubt correct, butitisnottheright place tolearn thissubject. Ididmanage toleam a fewthings from it,however, such astheNandMbasis functions, Recall that there arewhole BOOKS Justonthesubjectofthiskindofscattering!Inoticeonthewebdiscussion ofsandppolarization states! Fact: Icannot findarealMiederivation ontheweb! Ithink itwillbeinmynewbook arriving soon. Otherwise Iwillhave togotoMarriott dodigsomething uponthissubject! //Ididjustthat! Ididfind onelittle paper thathadsimilar MandNnotation, andasIwasjustgiving upandstoring PDF documents away, Ilooked inthekrugal oneandfound how thatplane wave expansion works! Plane Wave Expansion Explained. Seedetails inAppendix A,butthisisthebasic idea. Write down Jackson's exact result 16.139. ‘Then taketheappropriate linear combination ofhisresults thatcorresponds toE=(E,+E.)/2 =%e™.Hisplane wave formulas thenread, EE, (12M) Vam(2er1) [jkr [Xe Xe]+(UK) Vxjkr) [Xeon Xe} ] Bed, (1/2)(i)V4mQe+1) [-ijkr) [Xe-Neal -WK)Vx{jkr) [Xpa+ Xa)}] e Thenmakethefollowing definitions foro=odd,ande=even: Mare =jdkt) rxVyore where Wor P,\(cos®) sing Mae=jdkt)rxViyere where WiePr'(cos®) cos (note thatsing is“odd” in4)with corresponding formulas fortheNfunctions Note =(I/k) VxMore =(Ik) Vx[idkt) FxVere J Nae #(I/k) VXMae =(IK) Vxfidkt) rxViyere J Thevector form ofMisthesame asabove, butwehave something other thanYmasthepilot function. It isinfactthesum oftwoYm's ofopposite mvalues, allowing ustomake thisreplacement inJackson's formula, [|2eH |[YeniA)+Yel@O)] =Zin[Gay Wow 2641 Ye+1(0,0) -Y,1(0,6)]=2hap Vel LY¥1(6,9)- ¥1.1€6,0)] aneery Y Then Jackson's plane wave formulas shown above boildown tothis: 3 eeelke,y2H ay_extE,=fe LiFe Mote>ENI e =Belts yye2H )=;x) B=yel= Yi“enh(MQ, otNo,] IntheBequation, thefirstterm isa(-i)from theJackson formula, butthen wedivide byibecause the difference ofY'sdoesnothavetheithatthesumdidhave,andwechangeotoeaswell.Forthesecond term, weget-itimes theNfunction thatmatches thefirstterm intheEequation. //Tnow have ageneral rulefordoing Bfrom Einanyexpansion: cross mult coeff's by-iandsteal thenature, This result agrees with (1.6) inallrespects except one: inthederivation above, weclearly seethatthem- index should have thevalue 1,asshown inthesubscript labels above. Butin(1.6), thissubscript appears asthegeneral m(which they happen tocall£).This isanerror in(1.6)! IntheKrugal PDF paper, the above formula appears exactly asIhavestateditasequation2.44.Thesuperscript (1)impliesthatweare using thej,radial functions. Recall thatJackson's "first term” intheEequation 16.139 isthe"magnetic" orTEterm, andthisis stilltruewhen wecombine thetwohelicity solutions. Thus, wecanassociate "magnetic" with "odd", and ofcourse "electric" with "even". That isanice label tohave onthe Mand Nfunctions. Soifthe first term isodd, wealways expect thesecond term tobeodd. AsideonKrugel: |stumbled ontothecontentsandfirstthreechapterson-lineofthefollowing book: E.Krugel,ThePhysicsofInterstellar Dust,12/02 $135Amazon e Heistheonewho cleared uptheabove plane wave mystery. Krugel isattheMax Planck Institute (MP1) forRadio Astronomy inBonn. Inotice thatthePDF documents cannot beprinted, norcananything be selected forcutandpaste! Attempt tosetupthescattering problem intheNandMworld. First, lets look back atourearlier derivation where Iwrote down thefour boundary conditions. Mytwo curlconditions implied matching oftangential field components. TherxBcondition isproblematical for aconductor because there really isasurface current. However, Ifyou useHinstead, Iguess Hdoes not seethesurface current, theway Ddoes notseesurface charge. SoIguess Iaccept Krugel's two rx boundary conditions inthegeneral case ofarbitrary isotropic 1ands,which iswhat heistalking about. Krugel thennotes thateachofthese rxboundary conditions implie 2equations ifyouusethe and6 directions fortangentialness. Hesays youthen getfour equations forthefour unknown coefficients and youcanthen solve. When Ididthisproblem, Iwasindeed finding redundancy with myboundary conditions. SoIguess theanswer isthatthetworxconditions areallyouneed! Here isKrugel's setup oftheproblem: ; 204 “ck FE,=2,E.(M9),-in,] EmTap //incident planewave 6 elie alti alin e E,=2E.(-bM®, +a,N°,] incidentplanewave,(3)meansh{kr) Ey=Z.E, [eeMO), -id,nofal //incident planewave,(1)meansjkr) Ipresume theexpressions fortheHfields arethesame astheabove, except Iexpect toseetheeven and oddindices switched. This isbecause the"first term” inJackson's Bfields arealways going tobethe “other kind” from what youseeintheEfield. What isthegeneral ruleforconverting from anEequation toaBequation? Ididitfortheplane wave above: (1)negate thefirst term's sign, andreplace oddwith even (2)keep thesecond term's sign andreplace even with odd, Soapplying these rules totheabove, = -M® 2iN =toh H,=2,H,[-M@, -in®,]Heay 3) H,=EHe[ MG, +a,NO,J eHy=2H[=oMQ,-idND] Now theboundary conditions are Hg=H+H,» rxH=0 Hig =Hy+Hyp Bo =En +E rxE=0 By =En+Ey This wilbecontinued inanother paper! Appendix A.Trying togetasimple plane wave expansion interms oftheNandMfunctions Attempt #1(not successful) Let's just trythisasanexercise intheVSH X,Y,Z that wethink weknow about. Westart with | E=c™% e Ourfullexpansions arewritten, 7 | B=Im[ a(Ein)£46)Xen(8,4) +ilk)ay(Zim)Vx{20)Xea(0.4)} e E=Ln (Wk) an(ém) Vx{f40)Xm(B,9)} + alm) Bt) Xem(,6) J buttrytoshorten itbywriting, B=lal apfX +Gikyam {Vx(gX)} ] E=Em[(ik)azr {Vx(fX) }+ amgX ] Usethefactthat(again incompact notation justsoweknow where factors aresitting! ) Vx(FX)=-C{(f+ MDZ+e+)fir¥] andwethenhave B= Xm[agfX +Gilk)aw {-C[(g'+ 1)Z+e+)gr¥]} J E=Dm[(Wkae{-CL(f +l)Z+MEH)frY]}+ayBX. ] which wethen group byterm as B= 2m{ {af} X+{(Wk)Came(l+l) gr}¥+{W@K)C am(g'+)}Z e B= ml {ang} X-{(ik)CagMet)fir}¥-(WK)C a(f'+1)}Z J Inthelower equation, weswap theEandMlabels, thefandgfunctions, AND thesecondary terms have anextra minus sign. ‘Now let'strytocompute thethree projections asin(11) -(13). Start with Egy=[Ve(+1)] f32X aleE=[UCKE+)] fAQXmek whereinthelastformwemeanCarleton's X.FortheEfieldweshalluse Exe!%=kEy(i)fAnQl*) jkr)Ye0(0,6) 11Jackson16.129 Installing thisgives, B= [Ue(e+l)1ZeCotfax@er) jee)f40Y¥¢0(0,6)[Xew'(8,0) ©2] Atthispoint, weneed togooffandoanoverdue piece ofwork, which istorelate theCartesian unit, vectorstotheX,Y,Zvectors.AsshowninAppendixBwehave r) 8 EEE EEE - 42 Ym oy’ e (Xemn"@.8) ©81=[86“SQ-CoCFS) I which makes forafairly unpleasant angular integral, A ; OY OY" Brow=[WE(C+1)]Zp(i)Vf4n(2E#1) jkr)S92Yeo8.9-S—E™-(CoCrras J Butnow what dowedo? Notice thatwecannot conclude thatm=0 duetotheother factors present. This shows thewisdom ofJackson's approach using theraising andlowering operators onpage 568. Although Icannot trivially dotheintegral shown, Ithink Icanconclude thatwhatever itis,itisafunction. of6andm.Isuspect theintegral willvanish unless m=-1,0,or+1based onmyintuitive angular momentum additionnotions.Alsoitwillvanishunjess&=¢-1,£or£+1Butforthemomentlet'swrite: Bray=[1ME(41) ]De(i) 4e(20+1) jet) Kem = Presumably theother twocomponents have similar results. Then wehave toputthese sums over’ into ouroriginal sumoverfm,Theresult isamess andIhave nodirection really astowhere Iamgoing with this, solet's can itfornow. e Attempt#2(successful) Let's here piggyback onJackson's calculations! Weknow that: E.=@ #if)e™ gives theresult Ex=E,(i)V4n2e+l) [ jdkt)Xia£(Wk)Vx{jkr)Xe}J BeaE,(I)Y4n(2er1) CFijer)Xper-WK)Vx{jkXe}] Solet's use this E=(,+E)2 =£e*B=(B,+B.y2 =f /Isee16.130andfactthatB=2xE Then our results will be | B=E,(V2)V4mer1) [jkr [XeXea]+(Wk)Vx{jkr)[Xeo-Xa}] eB=Y, (U2) ()'Y4nQe+l) [=14) (Xeo1-Xea] -Wk)Vx(jer) [XetXa]}] 9 Thave confirmed thissecond equation several times, every signandifactor isimportant! ‘NowIamhavingtroublerelatingthisformto(1.6)inmylittleMiewebpaper. t BUT WAIT!!! Think about what these things look like: [Xero+Xa]=Cem)*xV[LY41(8,9)+¥,-1(8,0)] Look atJackson 3.53 and 3.54 =a[2H ps 8 Y,+1(0,0)=an) Pe'(cos6) e' cay yeaa. (eK preoeayei Y¢.1(8,0)=C1)!Yess(0,6) ale) Pe'(cos8) ¢° Therefore, 26H wet _[ae , CY¥e118,9)+¥e18,0)] =“Gallery P,(cos8) {e*-e*}=2iFee P/'(cos6)sind Ifweleave offtheconstant, mykrugel PDF claims that Wore=Pr'(cos6) sing e ‘And then the"odd" Mfunction isthecurl ofthisthing! That iswhy itis "odd", perhaps. That istosay Mote =jkr) rxVote Ifwekeep track oftheconstants jdkr) [Xoo +Xe] =jAkeyCEm) *rxVLY,41(8,9) +¥e-1(0,9)] Z=jke)C(E,m) 21fatehrxV{P/(cos0) sind} “i 26H =21 | jyexVver Yuen ane) i Qe . =ti JO janx¥yor Veen \)ace* " pet 1 2FSeyMow ‘ThenthefirstterminJackson's sumbecomes: e 10 e E=E,(12Mi)V4nQQeel) [jk [XenXe]J . 2041=E,()'— [Mone+ OFey Mow Now recall that theNfunction isdefined as N=(Ik) VxM =(1/k)Vx{ike)rxVy} ‘You can see that the second term has the difference ofthetwo Yfunctions Jdkt) [Xeni Xa] =jkerCEm) *rxV[¥e(8,9) -¥e-1(6,9)] =jdkC(m) *rx7(2|2A1. p2c0s6)cose} ‘ m ane eon~Pett3 - :=(2/i) anACDMar Mar=jdkr) rxVere WerePr'(cos®)cosh The second term inthe Jackson sum isthen @ xcmFe509 FKe1-Ked =AVHC219>PEE Hag j o>Xe andean Md ert 224 |= —— (iNanerty “NW whichhasthesameformasourMresultabove,sothefinalresultisthis: 4 aye Bet .E=fe8 =EEOFayMate-iNae] And sowehave precisely produced equation (1.6) now at5ofmidnight 2/14/03. Thus; wemay conclude thattheyareconsidering fixed polarization intheXdirection! ‘Thebigmystery isfinally resolved. Now what about thecorresponding Bequation? Comments. You getthedefinite feeling that wearecombining aspin I"thing" (polarization Vector, say) with £mspatial functions, soweshould bedoing some kind ofj=@1somewhere. The underlying group theory isbeing masked inourefforts sofar,butIthink Ihave found asource that discusses this subject. Ithink theidea might bethisconcerning theplane wave expansion: Ifyou pre-combine into values ofj,then youneed only have asum over j.Iknow thisisavague comment, butitmight explain |e whythepolarization isnotappearing intheplanewaveexpansion Ikeeplookingat. IL IdidaMarriott library triptoday onthissubject, wasreminded thatyoucannot park before 6PM. I found alittle paper which Tcopied abitof,anditrepeats thefunny plane wave expansion interms ofNe andMfunctions, However, thispaper gives alotmore detail onwhat these functions look like . Appendix A.Relating Cartesian unit vectors toX,Y,Z Xm =FxVYou Ym=?Yim Lm =VYom Itispossible todothis, buttheresults arenotvery nice. Forexample ©XVVin=RL-SoYo-(CoCy Se)¥y.J+§[ChYo-(CoSy1So)¥y]+2[UrYo] PYm= R[SC,Y]+ $1S.SyY]+ 21CY] FVYem= £[CoCy¥e-SYSo Yy]+—F[CoChYe +CySoYy]+ 2[-SeYo] where Ypmeans the 6derivative ofY,etc. Asnoted elsewhere, this must somehow bethe same asa certain rotation acting onourunitvectors. 12 —_ eMore ontheMandNFunctions PhL 2.16.03 1.About the Mfunetion Inthepreviouspaperonthistopic,Imadethisidentification forthe“firstterm"multipole fields, Mon(,0,9) =Vx[£(0)You8]=~£0)FxVYen==£40)Xea/C(m) where Xea(0,9) =C(E,m) *rxVYen(0,9) C(am)= +1 Carleton <1 Mie(not sure what thismeans now) “iTen Jackson Sometimes, such asinKrugel, wefound thatY,.,wasreplace with alinear combination ofYmWithin the same égroup. eLet's now prove afewtheorems concerning theMfunctions Theorem 1:IfM =Vx(¢y)where ciseither aconstant vector orr,then M=- exVy. ProofVx(cy)=Vyxc+wV xe=Vyxe= -exVy Theorem 2:Let M=Vx(cy). Ifciseither aconstant vector orr,andifysatisfies thewave equation (V?+k?) y=0, thensodoesM: (V?+k?)M=0 Proof: Ifweapply V?ontoM,itmoves through ontothe¢y,andthenwecanapply ouridentity Vicy) =cVy +yVe +2Vee Vy where we think ofacomponent of¢atatimehere,butwriteitasavector.Nowifeisaconstant,thelast two terms vanish andwegetourresult, VMeVxV (cy) =Vx(eVy)= -KVx(ey) =-KM Itturns outhat thisstill works when e+reven though risnotaconstant vector. Consider, r} Very)=rV’ytyVor+2VreVy=-K(ry)+0+20H 1 where wehaveusedthefactthatV’r=0since 8,,,= 68:=0.Obviously since rislinear, itcannot x) haveaLaplacian, Inthesecondterm,wenotethatVr=I,theidentitymatrix,whichwemightwritein theformTandthenwewouldsaythat‘T»Vy=Vy(ordoascomponents!) .Soatthispointwehave VMsVxV (ry) =VxEk (ry) +2Vy]=-PM+2VxVy =-KM QED Theorem 3:Lot M=Vx(ry)= -rxVy. [fysatisfies thewave-like equation Vy=f{r)y, thenM satisfies thesame equation, VM=f(r)M..{Thistheorem isnottrueifwereplace rwithconstant ¢.} Proof! VM=VxV?(ry)= Vx(rV7y)= Vx(0) y)r]=VRy)xr+(FO)y)Vx =Velt)y)xr=[f0)Vy+yVfi)[xr=xf()Vy+0=f(0)rxVy=f)MbyTheorem 1 Example ofTheorem 3.Consider Xu=1xVYm Sothaty=-Yim. Weknow thatV?Yin=-&(2+1)i7 Yon Sothatf(r)=-&(£+1)/r°. ThusweknowthatV?Xim=~&(£+1)/" Xym.Weprovedthisfactingorydetailinour"vectorharmonies.doc” paper.Wemightthinkofthisas"thewaveequationwithinapartial wave", loosely speaking. Summary ofwhat wehave sofar: *DefineM=Vx(ry).Then:©M=-rx Vy,whichisoneofourusualVSHbasisfunctionsX,uptoaconstant .eae *Ifysatisfiesawaveorwave-like equation,then sodoes M. Theorem 4.Suppose wedefine M=f(r)My=£46)Vx(ry)=-f(1)rxVy, where M,=~rxVyand where: Q)y=¥O.6); (2)wsatisfies thewave-like equation V*y=g(r)y where g(r)=-£(é+1)/1"; (3)£0) isalinear combination ofspherical Bessel functions such asjkr), where kistheconstant appearing intheBessel equation. Then (V?+k*) M=0. Corollary: Ifisanylinearcombination ofspherical harmonics Y.withinthe£manifold, theorem4is true, ProofofTheorem 4:FromTheorem 2weknowthatV?M,=g(t)My.Soconsider, VM=V?[fd)Mo]=[V?£0]Me+fn)V?M,+2Vf) ©VMy ={WP a[7a £4] +8£4)}Mo+2£149) 3,M, e Nowcomesasomewhat unobvious factthatIgotcaughton.Consider: 2 t) -M,=rxVy=rx(Itve6+Ir1Syeh) =yod-1/S0yy6. SoM, isnoteven afunction ofr,andtherefore d,M,=0. Thus weget: VM= {lra[rdfAn]-ee+L)/r fr)}My={-kK?f{r)}}M,=-12M QED Since fr)isaBessel function asnoted, wegetthesimplification of{...}shown ontheright above. Comment: Theabovetheoremshowsaninteresting result.Weknowthatf1)y;satisfiesthescalar wave equation (where f(r)isaspherical Bessel ofparameterk,andisalinearcombination ofspherical harmonics within é). Ifweareinterestedinsolvingthevectorwaveequation,thenM=-f(r)rxVy works. Ithastheextra nice property that VeM=0. Observation: Themultipole field "main terms" forE andBareofexactly thisform. Usually wehave yi asastraight You,butitisOKtoinstead usesomething likey, =C[Ym Yen] ifthatisconvenient. For theexpansion ofan&polarized planewave,thisdoesturnouttobeconvenient, andm=1.Wearealso satisfying V¢E,B=0withthissolution, e 2.AbouttheNfunction Now let's look atNdefinedby: N=VxM where Msatisfies thewave equation. Then Nmust alsosatisfy thewave equation, since VN=V?VxM=Vx(V?M)=Vx(-kM)= -KVxM= -EN Notice also that VeN =0. Usually, amultiplicative constant isused outinfront, N=(1/k) VxM since thisremoves afactor of/kinthe"second terms" inthemultipole expansions forEandB,Wehave justshown, then, thatthese "second terms" satisfy both thedivergence condition andwave equation, so they represent asecond solution tothewave equation mentioned earlier. So,wecannow extend our earlier comment: Comment:Weknowthatf(1)y;satisfiesthesealarwaveequation(wheref(t)isasphericalBesselof r) parameter k,andw;isalinearcombination ofspherical harmonics within£),Ifweareinterested in solving theveetor wave equation, then weknow oftwosolutions: (1)M=-£4)rxVye,andwealsoknowthatVeM=0. Q) N=(1/k) VxM= =(1k) Vx(£(6)rxVyc]andV*N=0, ‘Theseareexactly thefirstandsecond terms weseeinJackson's multipole expansion formulas, apart from overall constants. ‘We can think ofMasM=constantxf(r)*X_whereXisoneofCarleton's threeVSH's.Weknow thatwecanexplode Nintoalinearcombination oftheYandZVSHbasisvectors. OnlyXofthese three basis vectors satisfies thewave equation! However, theNlinearcombination ofYandZalso solves it! One other fact. With Nscaled bythe(1/k) factor asshown VxN=kM Sowereally have this symmetrical situation: N=(/Q)VxM M=(1K) VxN ‘Also weknow that(from inspection ofM) @ reM=0 soMisthe"transverse" solution OfcourseweknowthissinceM~XandweknowthatXandZ.atetransverse, Yradial.Notethatsince Nis amixture ofYandZ,ithasbothtransverse andradialpieces. 4 ‘Yeftoy Spherical Harmonic --fromMathWorld URLOOS 'Page 1of3 — WineanAd. WOLFRAM Page“ mathworld.wolfram.com: MBRESEARCH ——provucrs —senvices soLUMONs |etSOURCELNORARYY) NEWSONUNESTOREOURCompan AWOLFRAMWesRESOURCE ERIC¥iTEIN'S Algebra +Vector Algebra +yore oF Calculus andAnalysis »Special Functions »Spherical Harmonics +MATHEMATICS . — “Vector Spherical Harmonic™ stmnch.————_.. | Thespherical harmonics canbegeneralized tovectorspherical harmonics byQ lookingforascalarfunction yandaconstant vectoresuchthat 'SScalar Tunction y constant vector v , M=Vx(ey)=a(Vxe)+(Ve)xe~ € - Ls ome2(We)xe=-ex Ww =>Da -EOe () ©Coleulus andAnalysis so‘DiscreteMathematicsioFoundcionsofHtbematiés “v.m=o. Huedanycurt. Q) ohne Nowusethevectoridentities, asak JoribebilyundStites... VM=RAkVA yruay” Vit=—Ox(@xR)oo =-Ux(-7 0)PS sid AVI=Vx(CVH)Y =Ux(eSH)Gy amagenca. moe9 WM~BYx(ev)=Vx(ex), (/ @ABOUT THIS SITE— ° K AUTHOR'S NOTE 50oraas / ‘pwrnew M+PM= ayaK (5)‘orinboientay” * VIM+BM=Vx[e(V'o+kh,‘oneaconmaisuton . ; . ;2yovTueauesncox, andMsatisfies thevector Helmholtz differential equationif y,satisfies the‘oeMAiLCOMMENTS .a ~~ewwwenerntps scalarHelmholtzdifferentialequation 4 Remeron: Vuh+ky=0. (6) —o Construct another vector function. 9ORDERBOOKHERE naZxM oK royomeBOOKFROM a whichalsosatisfiesthevectorHelmholtzdifferentialequationsince e@ vnLivw xM)thyx(VM) | (8) |hitp/imathworld.wolfram.com/VectorSphericalHarmonic tl 218/03 Yector Spherical Harmonic --fromMathWorld Page2of3 4“A ityx(-k°M) =-kVxM=-#N, which gives onsen nnW ) Wehavetheadgitonaidentity 41 _ as (26 VxNn%ZVx(7xM)=Pye =}Vxew)=>Kv¥(4ekk“*)7AtaVaca: HEM =kM (10) Inthisformalism, yiscalled thegenerating function ande iscalled thepilot vector. Thechoice ofgenerating function isdetermined bythesymmetry of thescalarequation,i.e.,itischosentosolvethedesiredscalardifferential oe"equation’ IfMistakenas ytBae,bak qjsmt&conshourh11 ae(et M=Vx(r¥), aoe) a2e@U7* whereyistheradiusvector,thenMisasolutiontothevectorwaveequation inspherical coordinates, Ifwewant vector solutions which aretangential to theradius vector, osm Morar: (Vpxe)=(Vuexr)=0, (12) so exr=0 (13) andwemay take c=r (14) (Arfken 1985, pp.707-711; Bohren andHuffman 1983, p.88). Anumber ofconventions areinuse. Hill (1954) defines t+1 1 on”, VP=->yt+———— "16e Pe-Vorei' '*7pnaen 0+iM/(U+ D+1)sinoyj"d (15) http://mathworld.wolfram.com/VectorSphericalHarmonic.htm| 218/03 Vector Spherical Harmonic --from MathWorld Page 3of3 ft 1oy, iM. + wr = {ht +—8+ e 'ai! A+1)86 aisano? 16) M “a ioy", xp=-_—1_yn- 4_O75 P=-Titehend Viger oF” ay Morse andFeshbach (1953) define vector harmonics called B,C,andP using rather complicated expressions. References Arfken, G."Vector Spherical Harmonics." §12.11 inMathematical Methods forPhysicists, 3rded.Orlando, FL:Academic Press, pp.707-711, 1985. Blatt, J.M.andWeisskopf, V."Vector Spherical Harmonics." Appendix B,§1in Theoretical Nuclear Physics. New York: Wiley, pp.796-799, 1952. Bohren, C.F.andHuffman, D.R.Absorption andScattering ofLight bySmall Particles. New York: Wiley, 1983, e@ Hill,E.H."TheTheoryofVectorSphericalHarmonics." Amer.J.Phys.22,211-214,1954. Jackson, J.D.Classical Electrodynamics, 2nded.NewYork:Wiley,pp.744-755, 1975. Morse, P.M.andFeshbach, H.Methods ofTheoretical Physics, Part II.New York: McGraw-Hill, pp.1898-1901, 1953. Author: Eric W.Weisstein ©1999 CRC Press LLC, ©1999-2003 Wolfram Research, Inc. Rolated Wolfram Research Products Include: BWMathematica GOCalculationCenter 3CalculusWIZ hitp://mathworld.wolfram.comy/ VectorSphericalHarmonic.html 2/8/03 NOTES sos EffectiveAreaofanAntenna PhL- 2.24.03 |e Therearemanyideasherethatarenewtome,soIhavetriedtonumbereachofthestepstokeepthem from blurring intoeach other. Ihadguidance from: | http://farside.ph.utexas.edw/~rfitzp/teaching/jk1/lectures/node83.html 1,Radiation resistance R,,q ofaHertzianDipoleantenna,FromJackson,weknowthepowerand | pattern ofa"Hertzian Dipole" antenna, acenter-fed antenna which isd/2.inlength oneach side, and : where d<<, That power is,from 9.29, PSI,*(kd)'/12¢ =1,7/2 [24 07/120 *(WA)] =1,7/2 *Ras Here,I,isapeakcurrent, soI,?/2=PinsandRragmustthenbetheradiation resistance ofthis antenna, of course hereincgsunits. Secondly, weknow thepower dipole radiation pattern issin’®. | 2,Converting Rag tomks units. 1Coulomb =c(egsy/10esu so. amp=o(cgs)/10 statamp e 1volt=1/300statvolt1ohm=1volt/1amp=1/300statvolt/[c(cgs)/10 statamp]=1/[30c(cgs) ]statohms =>R(ohms) =30¢(cgs) R(statohms) ‘Thus, fortheHertzian antenna wehave Ryas(Statohms) =[27/3¢*(d/A)"] Reg(ohms) =[20x2*(d/A)*] =197(a/A)? 3.Model forareceiving antenna. When anyantenna isconfigured forreception, itismodeled asan induced voltage inseries with Ras, andyour best betforpower intoaload istohave Riss =Rae«Ifthe antennaemfsourcegenerates V,peakvoltage,halfofthatwillthenbeontheload,sotheloadreceives power of Prous(1/2)(Vol2)?(Read=(18)Vo"/Res where thefirstfactor of1/2comes because wewant totalkrmspower, notpeak power. Inanymatched antenna,therefore, fullyhalfthereceived powerisre-radiated intoRus! 1 oor 4.Pia FortheHertzian Dipole antenna. Weknow thatV.=Eo(d/2) [ortwice thisforpeak topeak], justfromdefinition ofthefield,whereE,isthepeakelectricfieldamplitude. Thus,fromtheprevious eparagraph, | Phoad=(1/8)Vo?/Reng=(1/32)Bo”d?/Resa Ontheother hand, from 2paragraphs agoweknow thatforthisantenna, inmks units, | Raa(ohms)=202*(/A)? Therefore, thepower delivered tothematched load ofashortdipoleantennais . Prcaa=1/132*20*n?] *E,?*2? which isafundamental result, saying power received isproportional to?andtoE,”. Wesuspect that theE,?proportionality willholdforanyantenna, notjustashort dipole. 5.Effective area concept. Foranyantenna, wecanimagine an"effective antenna ofareaA"thatscoops up100%ofthePoynting fluxitreceivesandconvertsthat100%toPias.Wewouldthensay Pras =A *S 6.PoyntingfluxinmksunitsandZo.Incgsunits,foraplanewaveinfreespaceweknowthatthetime® averaged Sisgiven by S=(c/8x)ExH* =>S=(c/8n)EB,” Converting thistomks asonpage 619Jackson requires collie Ep>rane, Ey |sowewould expect toget | S=(18mVice,) 4ne.Be=(Afnse E22=(E22]/Z, =>S=[E221/Z | whereZ=376.7=120nohms=1/(¢,¢). 7,Area ofaHertzian Dipoleantenna. Combining theprevioustwoitems,wefindthat Prod =A*S=A* [E,/2 ]/Zy | beingthedefinition ofthiseffective areaA.Butforthematched Hertzian Dipoleweknowthat e 2 pe Reta Mies nial t Phosa=1/[32*20%n7] *B22? from which wemay find theeffective area A*[EJ/2 ]/Z, =1/[32*20*2"] *B? *2? A=1/[32*20*n?] *A?*2.Z,=1/[32*20%n"] *02*240m=12/,32*n]*A?=(3/8n)A? => A(short dipole) =(3/8)2? 8.Gain oftheHertzian dipole antenna. Now theHertzian dipole isnotanisotropic radiator, buthasa sin’@ dependence fordP/dQ.. Let's compare adipdle radiator toanisotropic one: P/Q =D sin’ => P=D*8x/3 éP/4Q.=1 =>PHI*dn Ifwewant both these radiators toradiate thesame total power, wemust have D*8n/3 =I*4n 9=>#23 =I e Nowifweconsiderthedipoleinitspeakradiatingdirection, wehavedP/dQ.=D,sothedipoleantenna therefore hasaGAIN over thesame-total-power isotropic antenna ofG=3/2. 9.Thehypothetical isotropic antenna. Imagineatransmitter sendingasignaltoourHertzian antenna from apoint inthepeak ofthebeam ofourreceiving antenna. Wehave Piag =A *Sinourreceiver load. Now, suppose wereplace ourreceiving antenna with ahypothetical isotropic one. From reciprocity, ‘weknow thatthisantenna forthesame incoming Sisgoing tohave lessPraag byexactly thepower gain factor ofthe Hertz. antenna. Sowewould write Ping =A*S 1]Hertz antenna Plow =A'*S isotropic antenna Piost/ Plana =G= gain Therefore, AJA'= Phoaa/ Png =G Soyouimagine thatyoustart with anisotropic receiving antenna ofarea Aoandtheantenna's gain makes theeffective area larger by A=G*Ao, 3 10.Effective area AOofthehypothetical isotropic antenna. So,atthispoint wecomputed theAfora HertzianshortdipoletobeA=(3/87)A2,andweknowthatthegainforthisantennaisG=3/2.Ifwet)replaced thisantenna with 2hypothetical isotropic receiving antenna, thathypothetical antenna would have asmaller effective areaA,andwouldhaveasmallerpowerdeliveredtothereceivingantennaload. We would have Ay=(3/8m)A7G =(3/8n)A?*2/3 =A?/4n=n? K=N2n 11,Interaction between twoantennas. Imagine now atransmitting antenna driven byPrnt allofwhich isdelivered intothetransmitting radiation resistance. Thefluxfrom thisantenna atdistance risgoing to beGxthefluxfromanisotropicradiatorofthe same power, so S=G,* Panic! (nr?) Thepower delivered totheload inareceiving antenna will be Plat=Ar*S=G,Ay*{G,*Panit/(40?)} =G,GyPanis(27/4)/(Ant?)=GeGLM4retJ?Pamit=[Ge97/4m1](G,22/4]PeniA? =[AAUR] Paani! ThisistheFriisTransmission Formula, symmetric inbothdirections ofcourse. @ 12.Conclusions about Area ‘Therefore wearrive atthese conclusions: (1)Theeffective areaAoofahypothetical (andnon-existent) isotropic antennaisaxX?whichistheareaofacircleofradiusX.‘Themeaningofthisareaisthat,ifyouactuallyhadsuchareceiving antenna,and putitintotheflux Sofsome transmitted plane wave, andifyoumatched thisantenna toitsload, the power delivered totheload would beS*Ao. (2)Anyrealantenna hasaneffective area A=GAp=GxX?,where Gisthedirectivity orgainfactor in presumably thebest direction, which isassumed tobetheway itwould beused. Again, thepower delivered toamatched load would beS*A. 4 bt rn eo Fe Pointchargesascurrent:usingdeltafunetions Phi. 1.28.03 ©Consider apoint charge qmoving atsome slow constant velocity v.What isJandwhat isp? P(r) =q8(r=a) point charge located atposition a Weknow thisisright because weintegrate overasurrounding volumetogetq.Next,let'sguessforJ: Ser)=qv8(r- a) Toverify thisconjecture, consider (visaconstant vector) VeI(r) =(veV)8(r- a) Ontheother hand, wecanseethat Ap(t)=9&B(r-a)=qGa)©VaB(r-a)=qv6CV,)8(r-a) =-9(veV)B(e-a) Since thissatisfies VeJ=~Gip(r), weknow itmust betheright form forJ.Rememberthatcontinuity . justsaysthatifthere isanetoutflow ofcurrent from avolume, theenclosed charge must decrease. e Exampleinthezdirection Letv=v2Z. ThenJ(r)=qv &(r-a)2. ThenVeI(r)=8,J,= qv,8(r-a),where2,actsonlyonthez factor ofthedelta function. Now imagine analigned boxcontaining thecharge. Integrate thisdivergence h overtheboxandwegetqvfd3,6(z-a.)=onlytheparts=qv[8(h-a.)-(2)].Thisisonlynon- 0 zero when thecharge ispassing through theupper orlower face ofthebox. Ontheother hand, ifwe integrate thecurrent density over thebox, wegetQ(a,) =40(a,) O(h- a,).‘That is,itisnon-zero only ifa, isinthebox.Ifweapplyatimederivative tothisQ,weuse0,8(x)=6(x)toget 8,Qax) =qA[8(a.)B(h- a.)]=qGa,/6t Axe[0(a.)6(h- a,)] —//chain rule =qV Gaz[6(a.)O(h- a.)] Wrecognizev :=qv{0(a,)[-8(h-a,)]+8(a,)Oh-a,)} 1productrule =qv{OCh)[-5(hea)]+5(@,) O(h-0) } /Luse deltas =av {[-3(h-a]+3(a)}=-qv[5(h-a,) -8a.)] Sothingdoworkout.ItjustseemsalittleoddtosaythatVeJ(r)=qv0,8(r~a)inthiscase,and generally e p(t)=48(r-a)Ur)=qvd(r-a) VeS(r)=~dplat=qvV&(r-a) i Application toamechanicaldipoleradiator © Imagine charges qand-qseparated bydistance acos(«t). Inother words, +qislocated at qisat:a(t)=(d/2) exp(-iot) 2 -qisat:-a(t)=-(d/2)exp(-iot) 2 P(r) =(+q) 5(r- a)+-q)B(r+ a)=q[5(r- a)-5(r+a)) Thevelocityofthe+qchargeis2,a(t)=-ioa(t),whilethatof-qistheopposite. Sowehave ‘Thecurrent density is I(r)=+q)- ioalt))8(r-a)+(-qX+ iea(t))8(r+a)=-ieqa(t)[5(r-a)+B(r+a)] ‘Now useJackson (9.13) tocompute then=0term ofthevector potential! Note thatthetwocurrents arein thesame direction! Weget: co & ot . eo AG)=Farsik>[qa]=-ik=[24(42)exp(-iot)2}=-ik—(qd2}exp(-iot) Thething (qd2}isusuallycalledp,thedipolemoment, p=qd2 AG)=ik=p agreeing with(9.16) ‘Wecould also compute pbyintegrating : p=ferdr -fq[8(@r-a) -&(r+a)] rd'r=qa-(q¢-a))=2qa=qd2 Note thatthings aredefined with andwithout thetime phase invarious places. 2 Dds =a =whe. @ ploy P=KE ” tT“elachicsusceptibuily * D-e€ E=[+UTKe Tawdiclechue constant! kewyee=WY ye ML.oe eG Ife mn=Ve wev=S- BB, nade =Niirke e al —_ \ || Yo \odlhe-o8) J(RPok in 2 id Pongobaponbni(k) 90)SG) meee °wal=X. e Syy.se[Salapet =wa. .(me) 6S Ques ok=[Piah.dat. Avs & sint2h _eostaklel<9Wy.|eaeaes| iddk=\fal4\—Leastat=2(l-cos@4) r) Sore =zC\—cese), Sosk=2(\-cose) ~)a| =You =Y= eo Tj [j=Soke!) _coke’) bd e Some Questions PhL 1.30.03 1,What istheenergy density inaplane wave? Instatics, page21Jackson shows that|E/'/8n istheenergy density ofanEfield. The||isthere only because Eisavector, notbecause something iscomplex-valued. The understanding here isthatEisa real-valued vector. When wedoplane waves, weusually sayE=Eoexp(ikz) 2.Wereally mean thereal partofthisthingwhich isBacos(kz) 2.Thepeakenergy density ofsuchaplane wavestored onlyinthe Efield would thenbeEy'/8. Theaverage however istheaverage ofcosine squared which is1/2,so <Ug> =Es'/16n. Butinaplane wave, theBfieldhasthesame energy density, weget<U>=Ey/8nas thetotal average energy density. Notice that theaverage energy density ofaplane wave isindependent offrequency @.Higher frequency plane waves have afewer number ofphotons thateach have ahigher energy quantum. 2.What isthe momentum ofaplanewave? (a)Photon hasenergy hvandmomentum psuchthatE”-p’=p,p"=mc?=0sop=hu/e=hp.This would then bep=hay/2nc =fk.That sounds familiar. ButE&M classical theory does notknow about h, soquestion remains unanswered. td (b)Usethefactthat,forphotons,p=E/cwhereEisenergy.ThenaverageenergydensityUinavolume isE,'/8n infreespace, so:dEnergy =(Bo'/8x) dVanddp=(Eq’/8nc) dV.Sotheaverage momentum density isP=(E9'/8nc). Foraunitamplitude plane wave, wehaveP=(1/8nc). ‘Thenumber ofphotons perunitvolume isn=(E/AV)/E, =(Eq?/82) /Ao=(Ex?/8n)/Ao =n. Thenumber ofphotons inavolume dzdAisndzdA, they allbounce offourmirror intime At,soforce F=Ap/At. We have Ap=2np dAdz, andAt=dz/c (thetime forthisvolume tohitthewall completely), soF=2npcdA orPressure =2npe = (B:/4n). Soweareintheendabletoshow these things about momentum without reference to2photon, except weneed theidea that p=E/eandthen P=U/e. Iamafraid thiscanonly come from quantum theory. Notice that themomentum ofanindividual photonAkseemstobeirrelevantsoourmomentum calculation. The momentum ofaplanewaveisnotproportional tok.Inaplanewave,asweincreasek,weincrease theenergy ofaphoton, butatthesamé time wedecrease thenumber ofphotons, sothe momentum porPforagivenvatueofEpremainsunchanged. Summary foraplanewaveoftheformE=Eyexp(ikz) 2: average energy density =(Fy/8n) =U average momentum density =(Ep’/8nc) =P=U/e average energy flow perunitareaperunittime=AU/At=(Eq’/8z)dz/dz/e=(E,'/8n)*c=Ute | Thisisalsocalled"energyflow"or"power/area” andsymbolSofPoyntingisusuallyused. | eSome people refer topower/area asintensity andusesymbol I. 1 r)averagemomentum flowperunitareperunittime=AP/At=(E,'/8nc)dz/dz/e=(E;2/8n)=U =P*e This iscalled "momentum flow" butthere isnoterm like"power" todescribe thisthing. average pressure onamirror =F/dA =(Ap/At)/dA =2*PdV/(dA dz/c) =24(P*c) =2*U=(Eq’/4n) Comment: Ifyouhaveanabsorber insteadofamirror,radiation pressureishalftheabove. 3.Problematic bogus calculation ofPressure Average energy density inaplane wave isE,’/8x, seeabove. Imagine plane wave bouncing offamirror. Push themirror dzintothefield, stored fieldenergy decreases bydzE,'/8n perunitareajustbecause there islessfield present. And thisreduction occurs inboth theincoming andtheoutgoing beams? Or shouldyoufirstcomputethetotalcoherentEandBfieldstogetthetruestoredenergyinthebeamspace? IfIdothis,|findthatE=2Eysin(kz)sin(@t) =B.ThenIget<E’>=(2Eo)"*1/2=2Ex?andthesamefor<B’>. Then Iwould claim thattheaverage energy density inthevolume totheleftofthemirror isthe sumofthese which would be4Ey”,thenweaddthe8toget<u>= 4Ey'/8x=Ey'/2.Ifthisisright, thenwewould claim thatafterthemirror hasmoved dz,thefieldhasloststored energy (Eo"/2x)dzper unit area, Now, thepushing hand does work onthemirror which does work onthefield, soyou would think thefield energy would increase, notdecrease! This simple argument gives twice theright answer forpressure with thewrong sign, since pressure =F/4A =dU/(d2dA)! Something isnotkosher with this argument. e Idea: while wearepushing onthemirror fortime dtanditismoving atspeed v,thereflected beam is Doppler up-shifted. The frequency ofthereflected beam isupshifted infrequency byamount L/(1-2v/c) ~142v/e during thetime ofmovement. This ought tohave something todowith theproblem! After spending about 2fullhours onthisparadox, Ihave nosolution toit.SoIdon't know how to compute radiation pressureusingachange-in-energy-stored argument. 1sentanemailtoJim,Ibethe will have theobvious answer thatIamunable toseetoday. 4,What about Poynting's Vector? StartoffwithS=kExBwherekistobedetermined. ThisisforrealfieldsEandB.IfwewriteS=kExBinesuunits, then|S|=k|Einstantaneous. Then |Slpex =kEo?and<|S|> =kE9'/2. Butfrom the section above thismust bethismust beU*c=(Eo’/8z)*c =(Ec’/2)*(c/4n), sok=c/4z. Therefore S=(c/4n) ExB instantaneous energy flow => —_{[Slreac =(c/4n) Ey?=(Ex'/4n)*e <S>=(c/8n) ExB=(Ec/8n)*e How doweadapt thisforcomplex EandB?Try$'=j ExB*where jistobedetermined. Forplane wave inesuunits wehave |S'|=jEx?because thephasors cancel. Clearly S'sodefined cannot refertoan instantaneous value atsome zbecause there isnofunction ofzleft.Consider: 2 e |S'|=jEo?=(/k)kEg?=(j/k)[Sipeate=(2i/K)<]S|> Ifweselect j=k,then S'refers toapeak value, andifwepick j=k/2,then S'isanaverage value. Ithink Jackson onpage 205wants tobethinking average flow, sohewould say S'=(c/8x) ExB* average energy flow Hfasweseein7.14 |5.How dowecome toassociated pwith V,and inwhat context(s) ? Inwave theory, consider aplane wave exp(ikx). The operator (#/i)V yields Akwhich weknow isthe momentum ofaphotonintheplanewave(see2(a)above).Sowewouldassociate p=(A/i)V.Sothis association, which isthesame intheSaxon QM book, requires quantum theory. Onpage 542ofJackson, weseem himusep=(1/i)¥ asashorthand, sohecanusL=rxpasanother shorthand. Forhim, these areoperators closely related tothequantum operators, butthey arehave theft scaledout.IfweapplyJackson's poperatortoexp(ikx), wegetjustk,themomentum ofaphotoninunitsoff.Ishouldthinkofthese things asmere shorthands inJackson. Butifwetalkabout individual photons, they have thatinterpretation. e 6.Whatisthepotential ofaplanewave,both$andA? Letébeanarbitrary unitvectortransverse tokofthe plane wave. Then trythisform: A=(1/ik) exp(iker) & Then B=Vx A=(1/ik){VLexpliker)] x&}+exp(iker) Vxé, Thesecond term is0since &isafixed vector. Then V[exp(iker)] =ikexp(iker) sowehave B=Vx A=(1/ik) {ikexp(iker)x €}=exp(iker) Rx& Sothis looks like areasonable Bfield foraplanewaveinthekdirection.GiventhisBandthisk,wethen know that E=exp(iker) & Ax(Rx& =k because then ExB isinthe kdirection. Now weknow that e@ E=-V6-(1/e),A=-V6-(1/o)(-i@)(I/ik) exp(iker) 8=-V6+expliker) &=-Vo+E 3 t)‘Thus,ifwetake$=constantinspace,wegettherightanswer.Summary: Ifwestartwiththesepotentials: (Ao) =(eE ik,60) weobtain these plane-wave fields: Ese" & Boe™ Rxd where Eispolarized inthe&direction. Noticethat VeA=(1ik) &oV(e™) =e RoBi) Heaglar= "(Ve Since these don't addupto0,wearenotinaLorentz. gauge here (see6.36). Nor areweintheCoulomb gauge. This isoneofaninfinite possible setsofpotentials thatgive these fields. 7.Innon-relativistic classical mechanics, howdoEandBfieldsinteractwithaparticle? e (a)TheEandBfieldsputaforceonaparticlewhichisF=qE+(v/c)xB.Thefirsttermisnormalinelectrostatics. The second term represents anEfield thatappears inamoving frame ofreference. Ithink thisforce lawcomes entirely from Maxwell's four equations. (b)Astatic charged particle makes anEfield which isE=-V) where}isthepotentialmadebythe particle, and wefind 6bysolving thePoisson equation (©)Iftheparticle makes acurrent Jbecause itismoving, then wegetanAthatisacertain integral ofJ, andthen B=VxA andEthen comes from aMaxwell equation. (@)Ingeneral, wehave J*driving anODE forAwhich encapsulates (b)and(c)above. However, weknow that thefields affect thecharges andthecharges affect thefields, sonotobvious what general solutions exist. Weoften talkabout aparticle insome "external fields" which areimposed from theoutside, Wealsotalkabout particles moving andmaking fields, forexample, some “fixed currents" in anantenna. Wealsotalkaboutparticleswithoutfieldsandfieldswithoutparticles. 8,Innon-relativistic quantum mechanics, howdoEandBfields interact with 2particle? Ithinkthisisgoingtohavesomething todowithAeJandmaybe§p. 4 Forelectricfield,weshouldbeabletoaddaHamiltonianterm-eEzforfieldinzdirection.Maybeadd t) something like-evxBformagnetic field.Iwillfindsomebookswherethisisdone. 9.Inrelativistic classical mechanics, howdoEandBfieldsinteract withaparticle? Jackson talksaboutsomeofthisinhisbook.Atleasthowfastmoving particles generate fields.1forgethowtheforceequation changes (ifitdoesatall)forafast-moving particle 10.Inrelativistic quantum mechanics, howdoEandBfieldsinteract withaparticle? This isperhaps Dirac electron theory with fields notyetquantized. 11.Inrelativistic quantum mechanics andquantum fieldtheory, howdoEandBfieldsinteract with aparticle? ‘Thegeneral ideaisyouwrite aLagrangian andaddtheinteraction term eA"J,which couples thephoton field totheelectric current, represented bytheelectron Dirac spinor business. 5 VOL. 79,NO.7 JOURNAL OFGEOPHYSICAL RESEARCH MARCH 1,1974 @ TheAverage Auroral ZoneElectric Field F.S, Mozen ano P.Lucu Physics Department andSpace Sciences Laboratory University ofCalifornia, Berkeley, California 94720 outhundredandseventyighthousofestrioldataobainedbythofightof32ballone between =5and L=8.2 havebeen averaged todetermine mean properties oftheauroral . zone electric field. When itismapped into theequatorial plane, theaverage EXB flow hasa sunward component atallloeal mes and producas aconvection pattern consistent with thatexpectedforflowaround«rotatingobstacle.TheaverageauroralzoneelectricfoldstrengthIncreasesbyafactorofabout2asKpincreasesfrom0to6,butthelarger-scale convectionpatter,isnot greatly affected bythelevelofmagneticactivity.Theaverageelectricldvariationwith p's small compared tothet caused bylocal turbulent fluctuations, andsolocal field strengths fanbolargo during qulet periods orsmall during magnetic activity. ‘The average electric eld Iagaitude isenhanced byafactorofabout2whentheinterplanetary magoeticfekdhasasouth ‘ward component; thisfinding provides further evidence infavor ofthegeneration oftheauroral Zone electric field bymagaetie Held reconnection. “The ycomponent ofthointerplanctary mag netic field causes thenorchera auroral zone electri field strength tobelarger than average teatlocaldawn,when8,>0,andlargerthanaveragenearlocaldisk,whenB,<0.Thesebehaviorsareunderstood intermsofamodelinwhichthetwo-cellconvection patternisrotatedaboutthefun-earth Tinebyanamount that depends onthesign and roaguitude ofB,.This model and the dala suggest either that important parallel potential drope exist along auroral zone magnetie field lines orthat simple eonjugney between thetwohemispheres does notexist along such field lines. Auroral ionospheric andmagnetospheric electric fidlds timeandageneral trend intheaverage fieldasafunction havo been mossured byavariety oftechniques (see,Mozer ofloval time, Whereas thedata arecharacterized asmuch [1973] forareview ofthvarious types ofmeasurements), bytheir spread asbytheir mean diurnal pattern, thistur- andmuchinformationoncertainaspectsoftheirmorphol-bulencewillgenerallybeeliminatedfromfurtherdiseussion t)ogyhasbeenobtained [Moser, 1971b;Cauffman andinthepresentpaperbyaveraging overthedatacollectedatGumnett, 1972; Douprik etol,1972; Haerendel, 1972; anygiven local time. Properties ofthisturbulent behavior Melluain, 1972]. However, relatively fewattempts have have been described elsewhere [Mozer, 1971a]. een made todetermine the diurnal, latitude, magnetioactivity,andotherdependencies oftheaverageauroralzoneAviorLanceSoateMaanerosrnenic EuxotnicFrio electric eld, probably because sufficiently large bodies of fyFiguro2thehourly&thedataof lctriofield, “ ¢hourlyaveragesof ofFigure1are directlyinterpretable electricfielddatahavenotyetbeenplottedwitherrorbarsthatrepresent thestandard deviationsobtained. Since oneofthelarger bodies ofsuchdatahas ofthemeans. The.solid curvesofthisfigurearesmoothfits resultedfromtheflightof32balloons between Z=54totheexperimental datadrawnwiththeconstraint thatthe andL=82(Moser andSerlin, 1969; Mozer andManka, 24.hour average ofthewestward fieldcomponent bezeroin 1071], the478hours ofdatethereby collected havebeen order tosatisfy VXE=0.Inagreement withexpectation analyzed todetermine thedependence oftheaverage theelectric eldpatterne ofFigure 2imply asunward ExBauroralzoneelectricfieldontheseparameters. Wheress convection, sincethedominantfildcomponent issouthvardthiebodyofdatahasbeensufficienttoelucidatecertainneardawnandnorthward neardusk.Thesedataarealsoingeneral relationships that willbediscussed below, itisexcellent agreement withaverage flows deduced bybariumimportant tonotethatfurtherdetailedanslyseswillprob-cloudmeasurements {Haerendel, 1972],groundmagnetometer ablyrequirethousands ofhoursofdatainordertoaverage observations [Akasofu, 1965],andradarincoherent. back- tDetteroverlocal,turbulent fluctuations ofthesefields, soatter[Doupnik efal.,1972].Itisfurthernotedthatthe D southward fieldcomponent istypically afactorofabout3 ¢ are larger thanthewestward component, inagreement withsll ‘Table 1summarizes thelocations andquantities ofdata oftheabove techniques. collected from exch ofthoseven sites inwhich balloon ‘Todediice large-scale magnetospheric convection patterns measurements ofauroral zone ionospheric electric fields from theaverage ionospheric data ofFigure 2,itiseon- havo been made. One-hour averages ofthedata from each venient tomap theionospheric electric fields tothemag- ofthose fights are presented inanonrotating frame of netic equatorial plane.Ontheassumptionthatthevariation reference inFigure 1.Because ofthescientific interest in ofthepotential drop along neighboring magnetic fieldlines nighttime measurements, data collection hasbeen greatest issmall compared with thedifference inpotential between @]} eT cesakeofdatebtaned byanyiocheUns,horatosoftheSonnptre totheentail time isthe31hours acquired at0200 LIT,andthesmallest electri fieldcomponents aregiven bygeometric factors that amount isthe6hours ofdata obtained at1900 LT, depend onthemagnetic field configuration [AMozer, 1970] Figure 1shows both aspread inthedata atanylocal andaroplotted inFigure 8foranempirical magnetic field model (Fairfield, 1968] having magnetic feldlines thatenter Copyright ©1974bytheAmerican Geophysical Unio. theionosphere atthree latitudes. Because theequatorial 1001 002 Morea ano Lucnn: Avsnaoe Auronat, Zowe Exacratc Fre. TABLE 1.Summary ofData 0 NumberofHoursof Site LValue Flights Date —_* e Or er Sef |Churehilt b2 12 160 238o arseYellowknife 82 3 62 aaUraniumCity cat 4 6 £8-20| Great Whale 66 1 2 a‘Thompson 64 8 126 ad-College 54 3 at wD .MoMurray 54 1 “ see 7 —Zi a } magnetic fieldstrength varies rapidly withradius nearlocal Ee N midnight inthefiedmodel (Figure 3,top),theratioofthe 98-20]ionospheric totheequatorial electricfieldcomponents, which Z 4arogiven inthebottom plots ofFigure 3,depond critically ~+0 onthelatitude oftheionospheric endofthemagnetic field a CAL TIME”line.Thusiftheequatorialelectricfieldstrengthisroughly aan tierindependent ofpositionnearmidnightoveraregionofafew|Fig.2.HourlyaveragesofthedatapresentedinFigure1i ‘Theerror barssrestandard devintions ofthemeans, andthe arthradiinear10Zs,theionospheric fieldshouldvaryoacurvesareempiricalfitstothedatasalisfyingtheconatrait rapidlywithlatitude, especially pearmidnight. ‘imposedbyVE=0thatthe24-haveragewestward fieldbe Todetermine whether such arapid variation ofiono- xero. spheric electric field strength with latitude ispresent in theexperimental data, thecurves ofFigure 1have been theequatorial plane. The fields obtained bythis mapping averaged over twoJatitude intervals andplotted inFig- andgiven bythevectors inFigure 5might beinerror by uro4.The higher-latitude data ofthis figure come from factors oftheorder of2near midnight, owing touncer- thefirstfoursiteslisted inTable 1,anddata during periods tainties inthemagnetio fieldgeometry. Nevertheless, cortain ofextreme magnetic activity have been deleted from these general features ofthe deduced equatorial electric field averages, Thedifferences between thehigher- andthelower- pattern must stillbevalid. Thefielddirection isgenerally latitude data aregenerally small near midnight andarenot from dawn todusk, corresponding tosunward convection, statistically significant atothertimes(nearlocalnoonthereandthefieldmagnitude istypicallyabout1mV/m.Theearaonly two lower-latitude measurements). Thus cither field issmall near 2200 and 1400 LT, corresponding to thomagnetic field model overestimates theradial gradient times when theaverage ionospheric fieldisnearly zero, oftheequatorial magnetic field atlocal midnight near 10 Re,otthere isarapid radial increase ofequatorial electric —_—120, ——o field strength that nearly compensates forthelarge mag- —+notiefieldgradient,Thereisindependent evidencethatthe©yagformer interpretation ismore likely tobecorret (Sugiura,1972). 23Fortheabove reasons the67°curves ofFigure 3have ©40} been used tomap themean electric fields ofFigure 2into ° 2002 80 \ BeEota0l4 160ae a$8 VR | << 2 sg74 aa | a230 2 ry 80! ah 2 an 5 « 2«0NN ‘A es 2: BANOO. od 2, 33 ope ay #300 Ed Li NEOcoCANYS a oe ORO OE 8-40AY ol oe 4a -80) t)Trae me co)semmuonsSl,ret ig.3.(Top) Equatorial magnetic fieldstrengths asa LocalTIME ~function oflocaltimeonfeldlinesthatintersect theiono- Fig. 1.One-hour averages oflectric field components sphere atthreo diferent latitudes. (Middle andBottom) The measured in nonrotating frame ofreference on82balloons ratios ofionospheric toequatorial electrio field componentsflownintheauroralzone. onthesamethreemagneticfieldlines. | ‘MozenanoLose:AvanicsAvronalZowsEscratcFie 1003, 50 g >825) y) aes Ser < Y)98-2s| = i)ae a y) a ms we YIN 8825) Bm 77,AN N Fig.4Hourly averages oftheelectri Seldcomponents AZ INNS BOBTheasured atthe, higher- andlowerlatitade sites. i Kp Equipotentials oftheelectricfieldpatternarepresented Fig.6.Histogramofthoamount,ofclectricfielddata inFigure5asdeducodfromthemeasuredvectorsandthehisinedas&tunetionofXp,shadedareasdefiningregionsassumptions thattheelectricfiedisuniformacrossthe om atailatatailwarddistanceof15Re,thattheplasmapause te weardawnan aisanclectricequipotental, andthatthemagnetopause is1fetthatthefieldsneasdawnandduskarelargerthanrot.Sinestheexperimental dataalonedonotprovide Sversselittsthepotential dropacrossapasaoe; equipotential plasmspause toabouthalfofthatexisting it enough information fordrawing equipotential patterns; theelectric fieldwereuniform everywhere everywhere,assumptions otherthanthesewillproduce oeeeeenennnthesbow fonscombine differentpatterns, However, theexperimental datadoplace 1experiment 1aboveassumptions com! tosuggest anequatorial potential dropofabout60kVand important constraints onsuch patterns. The average total " P . "anEXB potential flowinagreement withearlier theories potential difference inside 82,cannot greatly exceed 30EV.There arenallyintheflewneat1460antaonLip, G22FeviewbyAzjord(1909)andwithexpectations basedon flowaroundarotatingimperviousobject.Theequipotential 0600 plotsofFigure5alsogenerallyagreewithaverageequatorialfield patterns deduced from plasma measurements [Mcllwain, i pinwe 1972,andrevision, privatecommunication, 1973].Vawwarion oprate Avensax Eurovero Frau Wirt ‘Macxenic Acrivery i Figure 6presents ahistogram ofthenumber ofhours of _ GOWclectricfielddatacollectedasafunctionoftbeplanetary— KindexKp.Thedatahavebeendivided intohigh-and Asony low-Kp intervals asillustrated inFigure 6such that roughly = 40% ofthedata liesineach interval, The hourly averages ‘aony oftheelectric fieldcomponents forthese high- andlow-Kp intervals aro plotted inFigure 7.Within the statistical L q 2 accuracy ofthis figure theelectric field isroughly twice asEAT Fpew IOKlargeduringperiodsofenhanced KpasitiswhenKpissmall, but the large-scale convection electric field pattern ZW isnotstrongly dependent onthelevel ofmagnetic activity. 4 a owEe50 oh ow Eg? x 4as ses°Sayey a82 -$8 -2 2 8Qerasaaan 1800RES 0 A ps Fig.5.Hourly. averaged clectric fieldvectors plotted in 198 theequatorial plane ofanonrotating frame ofreference, as ~S0L 4 iz = Zs viewed from above thenorth pole. Thesolid curves are LOCAL” TIMEelectric equipotentials deduced from the experimental data fontheassumption that. theelectric fcld isuniform across Fig. 7.Hourly averages ofelectric field components measured thetailat«tailward distance of15Ryfrom theearth. ‘uring periods ofhigh andlowKp,aadefined inFigure 6. 1004 MozmxanoLucu:AvensonAvaorat,ZoweRuscrsioFre.o {09 varies bylessthan=10% withlocaltimeoverthisinterval. E |,01000600 LT Although there isasmall dependence offieldsirength on> Kp,asillustrated bythelinear least squares fittothedata,>eo= . themajor variations ofthefied arenotexplained inthis - : on : way. Inconclusion theaverage electric field strength varies g soe . byafactor ofroughly 2with magnetic activity, butthisZoo] .- isa aan variation islessthanthatcaused bylocal turbulent varia-irs wt : tions, 249 pidy! 7 pyapane Devexoaxce oP7HAvenace Exsctaic Fitooxnite ..Etay we =Poot. Tvrmertaenary Macneric Frew gol: = One-hour averages oftheelectric fielddatshavebeen a divided into two subsets depending onthe sign ofthe w northward component oftheinterplanetary magnetic field, B,,andthedataineachsubsethavebeenaveragedto o 2 ry ry %produce thediurnalplotsofFigures9and10.ThetimeKe delays Atofthesefigures aretheintervals between mea- win 8. setter i tu a cleouie tug“Ue of,the interplanetary magnetic field and. thei.8.Scatterdiagramof1-haveragesofallelectric ionospherie electricfield.IfattentionisfocusedonthelocalFae aTee aecareastsauaree timeinterval of0000-1000 because differences between thefittothe data twocurves arestatistically most significant during these times, then forany time delay between —4and +4hours ‘Thedependence ofelectric field.strength oninagnetic itisobserved thattheaverage auroral zoneelectric fieldis activity isfurther elucidated bythescatter plotofI1-h Toughly twiceaslargewhentheinterplanetary magnetic fieldaverages oftheelectric fieldstrength versus Kppresented hasasouthward component asitiswhen themagnetic field inFigure 8Data from thelocaltimeinterval of0100-0600 isnorthward. Thisresult isconsistent withearlier electric have been selected forthisstudy both because large fieldmeasurements [Mozer, 19716; Hacrendel, 1972] and amount ofdataexiste andbecause theaverage fieldstrgngth offers strong evidence thattheeonvection electric fieldis @FBr<o x!BPO @:<0 x:B>0 pe Ehad ° SDobttenncsecconese >eee a 3| gq“ COPS eseeeeysene tpt Paasa ©7 e go dOP percept eed Bgpte’Seeeeity, -, e“E g- a Pte F = iets, {iB° arf) 2) Sd ' =~«0 40 Ot ategatergy4 ensills ats01 | oaee620a8 a D LOCAL TIME LocAL TIMEFig.9.HowiyoveragesofthewashedeseGoldFig10,HowdyavengeoftewouthvadelitmeeeQD ‘measured when the northward component ofthe interplane- sured when the northward component ofthe interplanetary lary magnetic fiold was positive and when itwas negative. magnetic field was positive and when itwas negative. The ‘The quantity Atrefer tothetime delay between themea- quantity Alrefers tothetime delay between themeasure- surement ofthe interplanetary. magnetic field and thét of ment ofthe interplanetary magnetic field and that ofthe theionospheric electro field, ionospheric electric field. ‘Mozen ano Lucw: Avenmot Avnonat, Zone Buseraie Fret 1005 produced bymagnetic field reconnection [Dungey, 1961]. ©2820 -x:8,<o Itisalsonotedthattheshapeoftheconvection pattern is ofapproximatelyindependentofB,andisgivenbythecurves biyeee eofFigure5eventhoughtheaverage magnitude ofthe 9 a refconvectionisrelatedtothiesignofB,.Thattheshapesof «opthecurves ofFigures 9and10arenotstrongly dependent, «0‘ontheassumedtimedelayisprobablyaccountedforbythe patah> statistical uncertainties inthecurves andbytheobserva- ° ca retentntdco tinesellofte“erseneay et +magnetio fieldisapparently several hours. a‘Thodependenceoftheaverageauroralzoneelectrifield zcafontheycomponent oftheinterplanetary magnetic fieldis Q “4 “gael ‘given inFigures 11and12,Foranytime delay lessthar! SG-40) about 8hours thesouthward component ofelectric field is a largest inthepostmidnight timeinterval whenB,>0. 2basThecurvesofFigure12alsosuggestwithlessstatistical oPBeaters adsignificance that thomagnitude ofthesouthward field com- Q ponent islargest neardusk when B,<0.These results a“| maybeunderstood interms ofareconnection model (W. GS40 tates Gonzalez andF.S.Mozer, unpublished manuscript, 1973) oe) alias 4 inwhich thelineofreconnection depends onthedirection i ry oftheinterplanetary magnetic fied. Forapurely south- ~o ‘wardmagneticfieldthereeonneetionlineisintheequatorial @+aeSeer cael28 Ce Zo S 40 ” ee “pi g osaa ae N« eo aris " LOCAL TIME 3re measuredwhontheycomponentoftheinterplanetarymag-a netic field was either positive ornegative. Teng rents §o«of plane;andthereconnection linerotatesoutofthisplane: 1sthointerplanetary fieldrotates about thesun-earthean ling.Sincethepostreconnection flowisinitiallyperpendic- SOP epee ulartothereconnection line,thetwo-cell magnetosphericaof convectionpatterntendstorotateaboutthesun-earthline astheinterplanetary magnetic field direction rotates. ‘This,j o4ae rotationcausesalargerreturnflowatnfixednorthernhem-a eeere isphere sub-polareaplatitudeatdawn,whenB,>0,and g—atdusk, when B,<0.This rotation ofthetwo-cell eonvee- q 5 tionpattern hasbeenobserved inpolar capclectrie fieldasscoe data(Heppner,1972;Mozeretal.,1974]aswellasintheaes southward electric fieldcomponent ofFigure 12,Ttisalso bo consistentwithgroundobservationsofmagneticficldvafin- 40tionsinthepolarcap[Svalgaard,1968;Mangurov,1969; 40]Borthelier and Guerin, 1972}. That thesign B,haslittleof: aes effectonthewestwardfieldcomponent ofFigure11isunder- cee stoodfromthefatsthatthemajoreffectisexpectednear r) 40) Joealdawnandduskandthatthesearetimeswhentheaver- os8620ee agowestwardelectricfieldiszero.LOCAL TIME ‘Since thehypothesized rotation ofthetwo-cell convection 11,Hourly averages ofthewe fieid Pattern occurring when B,>0causes anenhanced electriceiee eeeereereacer eldFeldneardawnstafixedlatitude inthenorthern hemi- netic fieldwaseither positive ornegative. sphere andareduced fied atthesame time andfatitude in 1006 ‘MozsnapLucu:AvinaceAunoaatZoneEuzormeFini thesouthern hemisphere, theionospheric electric fields in Hamberger, 8.M.,andJ.Jancarik, Dependence ofanomalous thodawn-dusk magnetic meridional planeatequalnorthern sonduetivity ofplasma ontheturbulent epectrum, Physandsouthern hemisphere auroralzonelatitudes arenot ,,/*ev.Lett26,990,1970.Hoppnor, J.P.,Polarnpelectrio fielddistributions related equal. Figure 12suggests thatthemagnitude ofthisdiffer- “{othe’interplanetary magnetic fielddirection, J.Geophys. ‘enco isareasonable fraction ofthelarger fied strength. This Res, 77,4877, 1972.resultmaybeduetoalackofconjugacy atauroralzoneKelley,M.C,F.8,Mozer,andU,V.Fahleson,Plectrio latitudesneardawnandduskormaybeduetoparallel fieldsinthenighttinganddaytimesuroralsone,J-Geo- potentialdropsthatarecomparable tothe~60-kVaverage Jcalloy,M."C.,F.8.Moser,G,Haerendel, U.V.Fahleson, potential drop across themagnetosphere, There isconsider- "andA.Kavadas, Rocket deelectric field’ measurements in ableexperimental (Moser andFahleson, 1970; Kelley-et al, 8.1000-garuma magnetie substorm, BotTrans. AGU, 62,320,“ 197la, b;Mozer etal.1973] andtheoretical [Kindel and 1071b. snstabitit Kennel, 1971;Coroniti andKennel, 1972]evidence that*pésh5,MuandCXKennel,Topsidecurrentinstabilities, parallelpotential dropsoftensofkilovolts existinthemag-Mansurov, S.M.,Newevidence ofarelationship betweennetosphere. Such results axe consistent with many labora- magnetic fields inspace and onearth, Geomagn.' Aeron, 0, toryexperiments [Hamberger andJancark,1970)and6221900. thevisiityofth one fllwain, C.K,Plasma convection inthevicinity of the plasmatheoretical calculations [Buneman, 1958]. MoCcedchitinBevih'eMemnetuephine Prosetes, Acknowledgments. Thisworkwassupported bytheAtmo- _Stlited by,B.MMeCormac, p.268,D.Reidel, Dordrecht, spheric Sciences Section oftheNational Science Foundation Netherlands, 1972 ; ,undergrantsGA1317,GA11259,GA28207,GA17328,andMoser,F.8,Wleetricfieldmappingintheionosphere attheGA33112X, bythoOfficeofNavalResearch underconiracts ,,¢austoritl plane,Planet, SpaceSoi18,250,1970.1N00OL¢-67-A0112-0068 andNOO014-69-A-02000-01016, andbyMoser,F.SPowerspectra ofthemagnetospheric electricNASAundergrantNAS6-0502. old,J.Geophys.Res,76,3651,19T1a.Mona, FS.Originhdfc! oflessee during.is “+latedmagnetospheric substorms, J.Geophys.Res,76,755, ‘TheBatorthanksJ.R.Doupnile andA.Pedersen fortheir Igri. ™ sscistance inevaluating thispaper. Motor, F.S.,Analysis oftechniques formeasuring deand actosphere, iRev,1h, ‘Ruvenanors aleeSedeinthemagnstonphere, Space$4.Revy1, ‘Akusofu, S-L, Dynamic morphology ofaur feSei, Morer, F.8,andU.V.Fahleson, Parallel andperpendicularBeena8160ne MODbaloey ofsuroran, Spacs electricfieldsinanaurora,Planet.SpaceSei,18,1563,1970. Axford, ‘W.I,Magnetospheric convection, Rev. Geophys. Mozer, F.8,andR.H.’Manka, Magnetospheric electric "Space Phys, 379,1000, field’ properties dedueed ‘from simultaneous balloon Sights,Berthelier, A.,andC.Guerin,Influence ofthepolarityof,J:Geophys. Res,76,1607,1971.he.interplanetary. maguctiotivity.athighlattedes,Mozer,F.'S,andR.Serlin,Magnetospheric electricfeldr) Rep. GRI/NT/87, 11pp,Groupe deRech. Tonos, Centre measurements withballoons, J.Geophys, Res. 74,4780, 1960.patdeIsReth,'Se,SentMaur,Franc,1072 Mover,F8.FHBogothand8.ThuruianRelationbe; oneman,©.Tnstebility, turbulence, and’conductivity inteenfonospherie electricfieldsandenergetic. traps current-carrying plasma, Phys. Rev. Lett, 1,8,1958. precipitating electrons, J.Geophys. Ret, 78,690,1073, Cauffman, D.P,andD.A,Guruett, Satellite measurements Mozer, F.S.,W.D.Gonsales, F.H.Bogott, M.C.Kelley, ofhigh’Intitude convection electric fields, Space SoiRev, andS.Schuts, High-latitude electric fields andthethree13,389,1072, dimensional interaction betweentheinterplanetary andtheCoroniti,F.V.,andC.F.Kennel,Polarization oftheauroral. lervestrilmagneticfields,J.Geophys.Res.70,66,1074‘lectrojet,J.Geophys.Ren,77,2895,1972. Sugiura,M,Theringcurrentandassociated ‘phenomens,Dovpnik, J.R.,P.M. Banks, M.J.Baron, C.L.Reno, and inCritical Problems ofMagnetospheric Physics, edited by ‘1.Pettceks, ‘Direct meesurement ofplasiaa drift velocities E-R.Dyer, p.195,National Academy ofSciences, Wash- athighmagnetic latitudes, J.Geophys. Ree, 77,4268, 1072, ington, D.C, 1972.Dungoy, J.W.,Interplanetary magnetic fieldand’the auroral Svalgaard, J.»Sector gtructure oftheinterplanetary magnetic ones’ Phye, Rev.Lett, 6,47,190% fieldanddaily variations ofthegeomagnetic feld athigh Faitfeld, D.Hl,Average magnetic field configuration ofthe ltlitudes, Geophye. Pap. R-6, Dan. Meteorol. Inst, Copen- ‘outer magnetosphere, J.Geophys. Res, 78,7820, 1968. Inagen, Denmark, 1968. Hacrendel, G., Plasma drifts inthe auroral ionosphere de- rived from’ barium releases, inBarth's Afagnetospheric Processes, edited by.B.M.MoCormac, p.246, D.Reidel, Received May 29,1973; Dordrecht, Netherlands, 1972. accepted November 7,1973.) \ ' . zghe DIPOLE FIELD s. Note:Ispentawholehalfdaythinkingaboutthesethings;Idonotwantto ever think about them again. - pose. ni potential: V=ptcos(theta)/r#*2 V=Esat field: B,,=2pcos(theta)/r*#3 Enreese Beneta7Pesin(these)/ra#3 Se=ewe.a equipotential lines: +?2[¥cos(theta) AsLese ; %fieldlines: retrgp?(theta) AeLite). Note: the potential Visnot aharmonic function intwo dimensions, so complex methods are useless. ave B()| Boose+1‘he& Bly) = (4-34 : (Re) = seal=t)ds Fsin’o * L=eséCe) =sacar). . a wr. 5dik=SateSCErtdecH) HORM Ze C1Teieace =BERGE RaneROEeae Pialrera eeePS a REREEEins MondoattSiWg, AERAow weakBiter Lu |CSSA seteemcan cree2heeee) aittt,+ 1 >Ba aca oteen AGREE AssShem co BRE |Cabrera:counts fluxquantatofind:aDiracmonopole .pumcrseTakeasuperconducting loop,placeitinring.Suchadetector isinsensitive toany’encouragement fromtheexperi-Peer gatcrs.0n ultralow magnetic-field device, and themonopole velocity, mass, electric mental side.”Rabameiccmonitor thecurrentwithaSupercon- chargeormagneticdipolemoment. Butpracticallyallprevioussearches Bete27-ducting QuantumInterference Device Asimilarapproachwasusedin1970formonopoles—in cosraicrays,aparti- eS&formany‘months. Thatwastheexperi- byPhilippe Eberhard, RonaldRossandcleaccelerators, andinmatter—would *Peat <<ment that Blas Cabrera. ofStanford Luis Alvarez (Berkeley) and Robert nothave detected anextremely mas- Begesst xeUniversity didinhissearchforaniov-Watt(SLA).Theycirculatedmoonsivemonopole. Currentlypopular SER“sy,ingmagnetic monopole. OnValen- dust(collected bytheApollo11astro-grandunifiedtheoriespredicttheexis. Reeeere;tine’sDayhe-found asingle:eventnauts,NealArmstrong, EdwinAldrin tenceofmonopoles, whichareindeed = theSo{faiconsistent withoneDiracunitofmag-andMichael Collins) through asuper- extremely massive—10-* gramsor ~Bgee5:0neticcharge.. Whenrumorsof.theob--conducting solenoid andlookedforthe10'*.GeV.BS =):Servation spread, theorganizers oftheemf.generated. Cabrera’s paperwas In1975P.Buford Price, Edward<A¥°i,-ThirdWorkshoponGrandUnification.published.inthe17May.issueofPhys-Shirk(Berkeley),ZackOsborneand iggphonedCabrerainBlacksburg, Virgin*.”ical.ReviewLetters....:+vem.LawrencePinsky(UniversityofHous- iia(where.he wasgiving’a colloquium), <2;Theexistence ofmagnetic monopoles ton)reported findingincosmicraysa¥;7and askedhimto,address. the’Work-’«wasproposed byPaulDiracin1931inmonopolewithtwicetheDiracstrength agoshops,the:next:morning“(16Aprill.,‘anattempttoexplain‘the existenceofandamassatleast200protonmasses 4 “aieAftetalong;wee-hours-of-the-morning the’smallest electric charge, e.Dirac’s (puysics Topay;October 1975,page17). igpearive:to:ChapelHill,N:.C; Cabreraquantization condition=said thattheThe’Pricegrouplaterwithdrewits \felectrified. theworkshop.:participants,, strength ofasinglemagnetic pole,g,claimtohavefoundamonopoie. .Aislacareflow-key éccountofhiswouldborestrictedbythe-relation Sabrar’swayadetect‘amonopoleis eteabossible‘monopole:discovery:U}>>HEME eeegtymt to,sethequantjuxinasupercon- WetOnehoted-theorist remarked,;"We° Fit. BARE ove.|ductor.Theuxquantumgy=he/2eEaeametoscofl-and stayed-to praise.”It.In1948Diracshowed thatif'amagnetic_givesadirectmeasurement ofmagnet- ‘ ESijwas: a=erysimpressive--glitch.-- One.chargeexists,theintegrityofquantum“fccharge. ‘ixshouldn't beconvinced byoneevent.” mechanics, andinparticular thequan-_Cabrera’s apparatus isafour-turn,S- EURBut it’schout.as impressive asone tization ofangular momentum, isvio- cm-diameter loopmade ofniobium,po- capeventcanbe.”feat Sry2vslated.unless thereisasmallest electric sitionedwithitsexisvertical; theloop f22422Byusingasuperconductingring;the~charges) fransx isconnectedbytwistedpairleadstothe FzgCabrera experiment candetectamov-Inremarksprepared. fortheMono-superconducting inputcoil.ofasquy SfsSingmagneticcharge:solely:from.the’pole.MeetingheldinTriestelastDe-magnetometer. IfasingleDiraccliargeéz-long-range electromagnetic interac- cember, Diracsaid,“Iaminclined now_pesses-.through theloop;onewould828ionsbetweenthemagneticchargeandtobelievethatmonopolesdonotexist,expect.an,8,changein.theGut FENthemacroscopic quantum stateoftheSo-many yearshavegonebywithout through thesuperconducting circuit, Fee 2sgfens Seog ate! helees oe | pss Eien gentsaia Seer8" EOSast Uae6 sage Se TRACE scan RM SymoneITcpaleeaten oFageae=oF T)_egFsenatate monopoleevent:found byBiasCabreraSanit“fedetctoroo,anewouldexpecanSfchangeinthefoxrough*,|epee iteee ‘chargepassesithrough theApthetsuperconducting circu. aERM RANT Gt ra SE a a2”rrgpaiatasiee ESEPHYSICSTODAY/JUNE198217 ./ / >consistingofthedetectionloopandthe well-defined stablelevelsforatleast% I ‘squtDinputcoil(afactorof2from P ° fonehourbeforeandafter.Onlyssix 4ng=249andof4fromthenumberof“& ||eventswererecordedduringthe70%of“7turnsinthepick-up loop). fi1' "therunning timewhenthelabwas“% q‘Thesquipandlooparemounted id| TAunoccupied. \OO:220cm-diameter, I-meter-long iti iySourcesoferror.Discussingpossible q ylindricel superconducting shield “*neaip) ; ‘if sources ofspurious detector response, ~closedatthebottom.Surrounding the I P\\id Cabrerasaysline-voltage fluctuations” superconducting shieldisasinglemu- Hl ${||causedbytwopoweroutagesandac- metalcylinderthatreducestheEarth's =,{| 1\lH companying transients didnotcause fieldtoafewmilligauss. Thecombined, iy 1ildetectable offsets.Radiofrequency in-~shielding provides 180db‘isolation ai if-Cslitrstion —terference fromthemotorbrushes ofa fromexternal magnetic-field changes .,Supepidhsing. 4ool heatgunfailedtoproduceanyoffsets _andanambientfieldof5x10~*gauss... °°?i! CAS fit whenoperatedclosetothedetector.Afluxquantum inthissizecylinder 1 ——2| ‘Thecritical current oftheloopisnot“4‘wouldproduceafieldofabout9%10-* Hi reached forcurrents 1000times--4|gauss.So,Cabreratoldus,thecylinder ! B greater. Noseismicdisturbance oc-hasafewunpaired vortices. Cabrera ; curred on14February.hadusedasimilarultralowfielddevice i ‘| Couldanenergeticcosmicrayhitting q forhisPhDresearch(whichheearned li.‘ fthewirecauseapieceofittogo in1974under William Fairbank's di- Hit)|“Sigeemteow |) normal? Cabrera saysthatsuchreysrection). |]coulddepositasmuchas1GeV/emin[BlasCabrerais-athird-generation (|' {Jtraversing thewire,raisingthelocalphysicist. Hisfather,Nicolas, isasol- | i wiretemperature byabout0.01K.Butid-state physicist who spent 23years at a5-Kchange would beneeded toreach theUniversity ofVirginia(rherehis J thecriticaltemperature. -SonearnedhisBS),andisnow-stthewonopole-search apparatus. Theloopis_Cabrerahasintentionally generated ‘Autonomous University ofMadrid Moneteswnompmcstotesccionesne, mechanically induced offsetsbyhitting“ThefatherofNicolas,alsonamedBlas,tometer.Thecombinedshieldingprovidesthedetectorwithascrewdriver handle, specialized inmagnetism andwasan 180-<b isolation romexternal fedchanges forexample, Outof25attempts, two | organizerofthe1930SolvayCongress, andanambientfieldof5x10-*gauss. produced offsetsupto6¢,,butthe| alongwithNielsBohr,MarieCurieand signalwasnotcleanenoughtomimica .1Albert Einstein. Heissometimes cred- monopole event, Cabrera told us. 3] ~itedwith founding modern experimen- three orfour expansions, thefield in- There was always anovershoot and talphysics inSpain.) sidetheapparatus isdown to5x10~* settling-in period lasting several hours. qTomakethesuperconducting shield,gauss.ThenCabreralowershisappa- Theonlyreasonable occurrence thatbreratakesanexpandableleadfoil,ratusintotheultralowfieldregionwouldmimicamonopolepassage,Ca- 60microns thick,andcoolsittoliquid- through anairlock. (Stanford hasfourbrerafeels,isaspontaneous internal i>helium temperature’ whileitisaccor- suchultralow fielddevices operating; stress-release mechanism. Iftheturnsdion-pleated intoalong, thin ribbon. twoofthem areused forthegyro intheloopshifted with respecttoeach Whenthefoilcontainer becomes super- relativity experiment.) TheDewarisother,theinductance wouldchangeconducting, ithastrapped theambient maintained continuously atliquid-heli- andproduce acurrent change. A1% qfield. Then Cabrerausesamechanical umtemperature. shiftininductance couldmimicthe plungertoexpand thefoilintoacylin- Thefour-turn loopisaflipcoilthatevent. i der, still keeping itatliquid-helium canberotated 180" intheplane ofthe Cabrera says hissignal-to-noise level i temperature. During theexpansion, coil(toallowcalibration oftheabsolute approaches 15:1forthepassageofa4 the magnetic field that occupied the magnetic field). Cabrera had been us- single Dirac charge. The 1970 experi- 4 volume ofcylinder before expansion is ingthe facility foranexperiment to ment ofAlvarez and hiscollaborators * expelled from thevolume byinduced measure h/m,. Inhis monopole was notsensitive toasingle Dirac q ‘supercurrents onthesurface, leavinga search, heleftthecoilinafixed posi- charge inasingle pass. .Sothey circu- lower magnetic field inside the foil tion with itsaxis coincident tothe lated their moon-rock samples several container. Although some trapped shield axis. timesthrough thesuperconducting coil |fieldremains inside theshield because Bymid-May, Cabrera hadbeenmoni- toincrease thesignal-to-noise ratio. 4 ‘ofthefolded configuration, theinside toring thedccurrent inthesupercon- Cabrera toldusthat theadvantage of qofthecylinder hasafieldafactorof10-ductingloopcontinuously forover200thistypeofexperiment, compared to 100 lower than thefield outside. days and hadseen orily-one candidate other monopole searches, isthatapar- ‘ThenCabrera takesasecondfolded event—recorded at1:53PMonSunday, ticlewithanyvelocity ormasscarrying shieldribbon,placedinsideacooling 14February, whenthelabwasunoccu- amagnetic chargeaffectstheexperi- tube, which can bethought ofasapied. Cabrera found theeventanhourmentinthesameway.Thecouplingis vacuumjacketwithalong,glasstube.andahalflater,whenhereadhisstriponlythrough long-range electromag- ‘The tube allows him tocontrol the chart recorder. netic fields. Although hispresent sys- temperature ofthe ioner shield:while Inhis Phys. Rev. Letter, Cabrera tem, with itslowbandwidth, isunable‘theoutershieldiskeptsuperconduct- reportsthatthe-event isconsistent _todiscerntheexpectedmonopoleveloc- . ing. Heletswarm helium: gesflow. with thepassage ofasingle Dirac ity(about 10° c),some squip systemsthroughtheglasstube,thenslowlychargewithinacombined uncertainty builtforresearchhavebeenmadewithreducestheflowsothattheidnnerof+5%.Itisthelargesteventofanysufficientbandwidth. Ciscooled through itstransition kind intherecord. Hefound atotal of Improved experiment. Cabrera, Mi- ‘temperature. Becausetheinnershield27eventsexceeding athreshold of0.2chaelTaber,SusanFelch,Robert isinthelower magnetic field provided 45,which remain after excluding Gardnerand John Bourgarebuildinga "by thepreviously expanded shield,aknowndisturbances such.astransfers newdetectorwiththreemutuallyorth-stilllower fieldistrapped. He-contin-. .ofliquid helium.and nitrogen. Hede- ogonal loops wound onaspherical Py-uesthisbootstrap process,sothatafter‘fines‘an-event asa,sharpoffsetwithrexbulb. Unliketheflipcoil,thethree- q Sig. 18 PHYSICS TODAY /JUNE 1982 ow OAFLSE : BAe eee e ey atte ee | 7 — rr EE: . a Selooparrangement ishighlystableChicago)pointedoutthatifmonopoles looklikeamagneticmonopole?” AfeweyMechanically. Forhypothetical mono- aredistributed throughout thegalaxy,. months agoAlvarez calculated, assusmSty,Polespassing through theloops,about theywouldleachenergy outoftheingthosemeasurements are‘correct,£22;70%ofthetrajectories willintersect atgalacticmagnetic fieldandsoondes-thatthemonopoles intheSunmustp ik.leasttwooftheloops,providing coinci- troythefield.Somearguethatmono- havemasses greater than10"GeVtoji:denceinformation, Thetotalareaispolesareconcentrated locally,butit’skeepthemtogetherbygravitational Bi:@}tentimesthatofthefipcoil.Andificult toproduce amechanism forattractioninspiteoftheirmutualre- Zp.additignalfactoroffiveineffectivethisconcentration. AttheGUTwork:pulsion. yaydetecting areaisgainedbecause:Ca-shopinApril,Glashow.mentioned anPerhapsmonopolescollect atthecen- ee breraalsowillbeabletodetectnear-ideadeveloped byhim,SavasDimopou- teroftheEarth,too,Alvarezspect. gowmisstrajectories thatdonotpasslos,EdwardPurcell’(Harvard) andlates.Youcouldn'thopetoputamono. BE through because theywould cause aFrank Wilczek (University ofCalifor- poleonatableandexperiment withit. iqeldchangewithintheultralow-field nia,SantaBarbara). Theyspeculate AsPaulFrampton University ofNorth "38shield.CabreraexpectsthatthegroupthatthemonopolefluxarrivingonCarolina)pointsout,“Becavesofite aig)Willbetakingdatawiththenewequip-Earthoriginates intheSun.Alvarez enormous mass,thethingwouldbe“ment beginning thismonth. hascalledattention tomeasurements unimpressed byatable.Earth’sgravi-Cabrera stresses thattheeventheoftheSun’smagnetic fieldforthelasttational pullcouldbegreater thanthe “+hasreported is“notyetadiscovery. five“quiet periods” thatonceledJohn electromagnetic forcebetweenamono. Itsaninteresting event.We'rework- WilcoxStanford) towritea1972paper poleandatypicalatom.Sothemono. "inghardwithnewapparatus toseecalled,“WhydoestheSunsometimes polewouldgostraightthrough"cnt definitively ifthere aresuch particles. 1stough tobeveryextravagant with onlyonedatapoint.Butit’salsodiffi- rT 1 " galt 9makeigoaway.We'recaughtHunting neutron—antineutron oscillation x ona knife edge.” Theory.In1974Gerard'tHooft(Uni-WhenMurrayGell-Mann andAbra-IncontrasttotheSU(6)grandunifica-YersityofUtrecht)andA.M.Polyakov hamPaispredicted in1955thattheK°tion,theMarshak-Mohapatra theoryGandau Institute forTheoretical Phy- meson should exhibit anoscillating predicts neutron oscillation onatimesics,Moscow) independently showed probability formetamorphosis intoitsscalethatmaywellbeaccessible to >thatifyouhaveanon-Abelian gaugeantiparticle, theK°,theycitedtheneu-experiment, Protondecay,ontheoth-groupthatissemisimple endthatspon-tronasacotnterexample. Theconser-erhand,isnotpermitted intheMar. *taneously undergoes abreakdown ofvationofbaryonnumber,theypointedshak-Mohapatra partialunification +>Seale, suchtheories havesolutions cor out,would prevent theneutron-anti- scheme, :responding toDiracmagnetic mono-neutronanalogofneutral-kaon oscilla Onecouldcomplicate thespontane- *poles.Theyfurthershowedthatthetion.Butnowadays, withgrandunifiedous-symmetry-breaking mechanism in }»mass ofthe:monopole would bethe theories oftheelementary particles theSU(6) theorytogetneutronoscilla- pr” characteristicscaleofsymmetry break:verymuchinfavor,allbetsbasedontionsatanobservable level.Butsuch ingdividedbythefine-structure con- baryon-number conservation areoff.Adepartures from“minimal SU(G)”stant.ThatsameyearHowardGeorgi, prodigious experimental effort(puvsics woulddeprivethetheoryofmuchofitsHelenQuinnandStevenWeinberg (alltopay,January1980,page17)atteststo simpleelegance, Georgicontends. “IthenstHarvard) showed thatforathewidespread expectation thatproton would begreatly surprised ifwefindlarge class ofgrand unified theories, decay willbeseenwithalifetime ofneutron oscillation,” hetoldus.“Butif theunification scale—the regionwhereabout10°years, wedo,itwillbemostinstructive.”strong,weakandelectromagnetic cou.ButthesameSU(6)grandunified Aesthetic prejudices notwithstand-plings become equal—is about 10 theory thatpredicts thisfiniteproton ing,Lay-nam Chang (Virginia Poly-Gey. Allthegrand unified theory lifetime doesnot(initssimplest form) technic Institute) andNgee-pong °predictions formonopole massareinpermitneutronoscillation. TheSU()Chang(CityCollegeofNewYork)have ‘theballpark ofahundred times the unification ofquarks andleptons inaproposed® amodification ofSU(6)thet unificationscale,thatis10'*GeV.singlegauge-theoretic framework, pro-wouldpermitbothneutronoscillation. Ifanymonopolesexist now,presuma- posedbyHowardGeorgiandSheldonandprotondecay’atobservablelevels blytheywouldhavebeenproduced inGlashow (bothatHarvard) in1974,byextending thetheory's minima! theveryearlyUniverse, about10-% replaces baryon conservation bythesymmetry-breaking mechanism toin- secaftertheBigBang,atatimewhenconservation ofB—Z,thedifferenceGlueHiggsbosonswithmascesaround theunified interactions breakapart between baryon andleptonnumber, 10*or10°GeV. intostrong,weakandelectromagnetic. thusforbidding nz+f,theneutron. Becauseneutron oscillation plays 4Cabrera citesanobservational upper oscillation transition. suchacentral roleinchoosing among *"boundonthemassdensityofmono-‘Thereare,however,rivalunification competing unification theories, anume +“poles tobegivenbythelocal“missing theories. Robert Marshak (Virginia berofexperimental attempta tolook¢mass.” Thislimitisintherange 0.03- Polytechnic Institute) andRabindra forthisexotic phenomenon arenowin233,005 solar masses/cubic parsec. One -Mohapatra (CityCollege ofNewYork) various stages ofplanning, andone“3"canassumethatbecausethemonopoles haverecentlyputforward’a“partialgrouphasrecentlyannounced prelimi- {Garesomassivetheirvelocitieswouldbe(asdistinguished fromgrand)unifica-.naryresults.AttheInternational Con. Snogreaterthanabout300km/sec(astiontheory”basedonaleft-right sym-ferenceonBaryonNonconservation,Alvarezsays,justsauntering byanmetricelectroweak schemewhich, heldinBombay inJanuary, Milla <7atom60thatlittleornoionization whenjoinedtotheusualcolorSU(@)Baldo-Ceolin (University ofPadua)re7§takesplace). ThenCabrera estimates group ofquantum chromodynamics, ported thefirstresults obtained bya-syathat thenumberofmonopolespassingyieldsacolorSU(4) groupstructure CERN,Laue-Langevin, Padua,Ruth- ‘§through theEarth's surface would bemathematically similar toanearlier erford,.Sussexcollaboration usingthe EH-ggtX10"" emé*sec™'ster~!. Such aproposal ofJogesh Pati(Universityofcold-neutron facilitiesoftheLaue-Lan- Buxwouldyictd15eventsperyearMaryiand)antAbdusSalamCntaraa.geviafastsatGosoble,Theetre: &Ziehrough bisloop. tional Centre forTheoretical Physics, nullfinding—that thecharacteristic<n1969EugeneParkerUniversity ofTrieste,andImperial College,London). neutronoscillation timeisgreaterthan es - “ageHe PHYSICS TODAY/JUNE1682 19 4 UNITS SummaryofE&MUnitsHistory PhL 1.24.03, e Details areinaseparate document, herewejustwant togetthesummary ideas, ‘There arethree equations ofinterest F=kI*qQi? oo) /force between twocharges (electrostatics) dF/dx =k2*2i/D @ /force between twoparallel wires (magnetism) kl=chek2 QB) /seeotherdocforhowthisisshown. P=VI CO) /power =potential *current Thefactthatconstants KIandk2must beconnected isglear since thecharge appears inbothequations (1)and(2).Ifwedouble theunitofcharge inbothequations, wehave toreduce bothk1andk2by4,so theyareproportional. Thefactthattheconstant in(3)isc*comes from theplace where electricity and magnetism meet --waves inspace, Now inthebeginning there wasonly electrostatics, before 1800 say,andtheegs-esu system wasused where k1=1. This defined areasonable working unitofcharge, theesu=statcoulomb. Noonecared about equation (2)because itwasnotknown yet. e ‘Afer1800,batteries andcurrents inwiresappeared, and(4)appeared onthescenejustfromworkwith DC currents. Soon, inthe1825 time frame, equation (2)appeared, andpeople whoworked with currents andwires and magnetic fields andsuch used thecgs-emu system wherein k2= 1.This made kl=c?sotheunitof charge here, theabcoulomb, wasridiculously large forelectrostatic work, butfineformagnetic work. ‘This system ofcourse hasanabvolt aswell, which turns outtobequite small, Since batteries typically putout1.5e8abvolts, itwas convenient torescale anddefine a"volt" such that batteries thenputout1.5volts, sovolt/abvolt =10°.Current wasstillinabeoulombs/sec =abamperes. Next, people starting liking themkssystem with itsjoules andwatts andmeters andkg,allbeing closer to thehuman scale ofthings, somore appropriate forbuilding things (engineering). So,instead ofusing a mixed system ofvolts andabamperes, anewsystem wasdeveloped called "thepractical units". Inthisnewsystem, planners wanted tousethenormal mksunits andthevolt. Inequation (4)above, since Pwas watts andVwas volts, acurrent unit was implied which was 1/10 ofanabampere. This issimply because joule/erg =1¢7andvolt/abvolt =1e8, sodividing gives 1/10. This new unit istheampere, and thecorresponding charge isthecoulomb. Given thischarge unit, andcomparing thatunitsaytotheesu, it wasthen easy tofigure outwhat theright values arefork1andk2forthisnewsystem. Itiseasy toshow withunitconversions thatk2=107,written asjia/4x tomake H=jpBinfreespace: @ dF(dyne)~ 1*i(abamp)? =>—_nt/dyne =1/k2*(amp/abamp)* 1e5=1/k2*le-2. 1 _ dF(nt) ~k2i(amp)* e Thismadek1=107c°(mks)=9x10”writtenas1/(4m¢0)sothatD=eEinfreespace.Thefactthen thatk1/k2 =sqrt(1/(}1o¢0)) =c*isthenpretty much ano-brainer. SothisistheSIormksa orpractical system. Weendupwith volts, amps, watts, Coulombs, Teslas volts/meter, Webers andallthatstuff. Everyone ishappy, except thatBandEhave different units which isuglysince theyarebothreally partofthesame F"”,andwehavethese twoconstants floating around everywhere toannoy us.Atheorist would rather havé cfloating around inequations which atleastmeans something andcanbesetto1byrescaling time,compared to¢splitintotwopiecesjpand¢)eachof which mean nothing atall. Oneother comment. ItistruethatEisstrong andBweak innormal lifeasv/e,andthisismost clearly seeninthecgs-esu unitswhere themagnetic equation hask2=I/c*quashing magnetic effects. Thereason weendupwith EandBindifferent units inmksa isthatthismakes them similar insize forhuman life. Thenatural units (je,same forEandB)make Emuch larger than Bforhumans. However, ifwedid things with large velocities likec/2inournormal lives, EandBwould besimilar insize. This would be asifournatural time scale wasnsec instead ofsec. Imagine pieces ofourequipment thatmove o/2 relativetocachother,wewouldhavesimilarEandBfieldsfloatingaroundinthenaturalunits. 2 7 UnitsusedinE&M PhL 1.22.03 1.About Units inGeneral Define versus Use. Some equations define anewunit, other equations justusetheunitinsome manner in the solution ofaproblem.Forexample,E=mv'/2andB=mghbothuseanenergyunitontheleft,but neitheristhedefining equation foranerg.Perhaps E=FxD isthedefining equation here, Although E=my*? istrueforwhatitmeans, sincewearenotdefining theergbythisequation, wewould never say 1erg=1/2gm-cm/sec* thewaywemight say1foot=12inches. Butwewould usethegeneral notion thaterg=gm-cm/sec? interms of"dimensions" tocheck workwedo,seenextitem. Athirdtypeof equation isused torescale'a unit, such asL(in) =12L(ft) andforsuch anequation, itisalways correct to saythat |ft= 12in.Seebelow. ‘Two ways ofhandling units inanequation. Consider: F(nt) =m(kg) a(m/sec’) F=36 kg*10m/sec” =360nt F=mkg *am/sec? =m*ant Intheequation ontheleft,m(kg) means "mass measured inkilograms". Inthisform, m(kg) isjusta ‘umber, ithasnounits! Wehave already handled theunits with theparentheses. Thetwonumbers onthe right when multiplied give thenumber ontheleft. When wesolve aproblem, however, weoften write thingsasdoneonthetopright.Themethodshownonthebottomrightseemsconfusingtome-writing t) unitslikethisseemsbettersuitedtodoingnumerical calculations andmakingsureyougetnoerrors.With numbers, youwillcancel andcombine units asrequired andthatactsasacheck onyour work. Butfrom a theoretical point ofview, Ifind thenotation onthetoplefttobemuch clearer and unambiguous. Sometimes equations have adifferent form ifoneusesdifferent units, soF=mawould notbe"enough" if thatwere thecase forthat equation, which itisnot. ButfortheCoulomb force law, itisthecase! Defining aUnit byRescaling. Consider: 1foot=12inches Linch) =12L¢foot) The thing ontheleftisthedefinition ofafoot, iftheinch isalready defined. Notice how thelabels are reversed intheright equation. Converting Equations toDifferent Units. Theabove isanexample ofarescaling ofsome unit. Here is mygeneral notation fordoing this: L{inch) =L(foot) *{foot/inch } J1( right over left" } or 1foot=|inch*{foot/inch } 1("loftoverright"} The symbol {foot/inch }ismeant asanormal dimensions fraction. You putfoot =12inch onthetop, 2canceltheinch,andendupwith{...}=thenumber12,{foot/inch} doesnotmean"feetperinch", U because ifitdid,thatwould give1/12which isnotwhat ismeant! When adimension isonthebottom, - youhavetoreversetherule.Forexample: e v(m/sec) =v(mile/hour) *{miles/m} *{sec/hour} =v(mile/hour) *{mile/f)* {fVin}* (in/om}* {cm/m) *{sec/hour} =v(mile/hour) *{5280} *{12}* {2.54}* {1/100} *{1/3600} =.45v(m/hour) Thisprovides a100%error-free cleanwaytoconvert anyquantity fromanyunitstoanyotherunits! We willusethismethod many times inwhat follows. 2.Units used inElectricity andMagnetism; comments applicable toallSystems TheTwo Basic Equations forDefining Charge andRelated Quantities. Thefirstofthese is associated withMr.Coulomb's work done around 1788, nowknown ashislaw.Thesecond isassociated withMr.Ampere's workdonearound 1827(andotherworkofBiotandSavart andOersted around the sametime). Following Jackson's appendix, weinstall constants k1andk2intotheseequations: F=kl*qQir qa) /forcebetween twocharges <P/dx=k2*2i1/D Q forcebetweentwoparallelwires | ‘Thesecondequationisreallyanintegrationofamoreprimitiveequationwhichisthis:AF=k2*idxx(dXxn? @) e sothe factor of2appearingin(2)justcomesfromtheintegrationtogetthatresult. ‘Sometimes (1)isusedtodefine aunitofcharge, andsometimes (2)isused.Different choices ofk1 andk2yieldthedifferent"systems ofunits" used inE&M. How arekiandk2related toMaxwell's Equations? Thefirstorder ofbusiness istosayhow weare going todefine theelectric field. Every system doesthisinthesame waywithnoconstantasfollows: F=qE fromwhich weconclude that: E=ki*Q/? ‘Thatistosay,thereisnoconstant between FandqE.Theassociated Maxwell equation musthavethis form asweshow inthewords that follow: divE=4nklp feeaa= faiveav=4rk,foav When weintegrate thedivequation (Gauss's Theorem) overasphere around apoint charge, doing int(E.dA) makes 4xfromtheangle integral, sotherehastobea47putintothedivE equation RHSto cancel it.Themain point isthatthekiappearing in(1)alsoappears inthedivEequation. Insimilar fashion wecanrewrite (3)as 2 . AF=(1/o)idxxdBdB=ak2*(dXxyi? WHistisBiot-Savart (4)Weshouldthinkofthe firstequation inthemanner ofF=qE.Forelectrics,weuse"q"asatestchargeto sensetheEfield. Formagnetics, weuse"idx" asatestcurrent tosense theBfield. When equation (3)is split into twoparts thisway, allowing adefinition oftheBfield, wehave thechance toaddanew constant which wemight have called k4,butJackson calls [email protected] cancelswhen thefirsttwoequations areglued back together togive(3),itcanbeanything. [Inthe Gaussian ogssystem itissettoc,inmksto1,forexample. ] OfcoursewhenwelaterlearnthatEandBarepartofthesametensorF*,wemustchoosekl,k2 andasoEandBhadthesame dimensions, andthisisonemajor benefit oftheogs-Gaussian system. This isoflittle concern inthepractical units, SotherightequationdefinestheBfieldproducedbyapracticalcurrent,andtheleftshowstheforce ofthatfieldonatestcurrent.ButtheleftequationnowstatedthisWayalsoappliestotheforceonatestcurrent inthepresence ofafixed-magnet fieldB.Wenowknowthisisduetothesamecurrentidea,butit wasnotknown in1810(say), sothisfirstformula correlated fixed-magnet Bwithelectro-magnet Band implicitly saidthetwoB'swore thesame thing. Theinternal currents ideaofMthenquicklyfollowed. . Asnoted, thefirstequation islike F=qE, butbecause itismore complicated innature andcorrelated previously separated ideas, itgetsaname associated with it,inhonor ofBiotandSavart who, in1820, discovered thelawastheforceonawireinthepresenceofafixed-magnet. Oerstedin1819hadfoundthatacurrentdeflected acompass, butdidnotmeasureoranalyzethenatureofthatforce. Theassociated Maxwell equation isthis (Ampere's Law) e curlB=4k JBeai=fourlBda=4nak,[eda Toseetheconnection here,thesimplestcaseisthestraightwire.Integrate dBfrom(4)-rightoverthewire andyougetB=k22I/r.Compute thissame Bfrom Stokes theorem applied tothecurlequation shown above, youget2mrB ontheleftfrom thelineintegral, andyouget4xk2Iontheright, soB=k2I/r again. This shows whywehadtoputthe47andK2intothecurlBequation. So,inthissection wehave related ourconstants tothetwoMaxwell equations thathave sources (p andJ).ThedivB equation isuninteresting since itisjust0.Westillhave toworry about the4thMaxwell equation about curlE,andwehave toworry about the“other term" inthecurlBequation, known asthe "displacement current". Wedothese things right now. Asimple problem regarding theterm curl B=k4JE/at. Let's assume aconstant k4inthisequation incurrent-free space andseeifwecanconnect itwith k1or k2.Todothis, let'sstudy asimple problem. Imagine atinyring(radius a)moving toward apoint charge atspeed ¥,allflat-on andsymmetric. If|applytheaboveboldedcur!ruletothisproblem, fBea=fourtBeda =+k4d/dt( Jessa) Using thisformula, wecancompute theBfield inthetinyringtobe: @ Beklk4Qav/r ‘ 3 ee Inthisproblem, theringissosmall,thatEisconstant onthering,soelectricfluxisjustna®E=7a”kl eQU?Putting k4d/dtonthismakes -k42vir’,whiletheleftsidegives2xaB. The2nacancels onboth sides, giving theabove result. Now Itreatthesame problem bysitting ontheringandwatching Qapproach atspeed v,Ican identify QvwithIdxintheBiot-Savart Bformula above, andIgetthat Beak2Qavir where sin0from thecross product isa/r.Comparing these tworesults, wefindthat, k4=o K/L. Notice thatthisisjustwhat Jackson hasputinhis(A.8). Asimple problem regarding curlE=-k38/0t (Faraday's Law) ‘Thesimple problem needed hereistoputasquare loopattheedgeofaregionofconstantBfield,saybetween thepolesofamagnet, andhavetheloopmovetotheright,say,atspeedv.Theequation of interest here isthis Jeedl= fourlBeda =~k3d/dt( Beda) Inoursimpleproblem,thechangeoffluxontherightisByvwhereyistheheightoftheloop,sotheeRHSisthenk3Byvinmagnitude. Theelectrons inthemoving loopwirefeelaforced =(1/c)idxxdB asnoted above. Putting J=qv8(r-a) andintegrating inall3Doveratiny"wire",idx=qy,andwethen getthattheforceisF=(1/ct)qvB.Weinterpret thisasbeingduetoanelectric fieldE'=F/q=(1/o.)vB in therestframeofthewire, andthelineintegral thengives (1/«:) yvB. Thus, k3=1/o.. Jackson onpage172dealswiththisissueinamoregeneral way,andhereisasummary ofhis method. Youwrite theusual Stokes equation with theabove curlrulewith constant k3(hecalls itk there). Youthen consider thesituation ofacurrentloopmovinginthepresenceofalabBfieldofsomesort.Theelectric fieldinequation 6.4isdenoted E'toremind usthatitisthefieldinaframe moving with thewire,because acontour integration isdoneinthatframe. Thatisakeypoint. Weareinterested inthe fieldthatpushes acurrent inthewire,andthatfieldmustbethefieldthatisintherestframeofthewire. SoIthink(6.4)isevaluated inthelab,butyouusethefieldE'asjuststated. Thenin(6.5)hecomputes thefluxchange inthelabandcomes upwiththefamous twoterms --Iproved thesecond termona separate sheet. Thisletsyouwrite(6.6)andtheninterpret Easthefieldinthelab.Wehavethenproven thefieldtransform law(6.8) fornon-relativistic velocity v.Westillhaveourk3inthisformula. Now |comes themagic trick.Inthelab,anelectron inthewireisinfactacurrent oftheformqv.Weknow fromBiot-Savart thatsuchacurrent feelsaforce from theBfield, andthatforce is(I/a) qvxB.So Jackson's conclusion isthatk3=\/ct.(Inthissection heuseskfork3,butusesk3inhisappendix.) Maxwell's Equations andtheConnection toSpeed ofLight. Sofarthen,based onthediscussionsabove,wehavetheseMaxwell'sEquations: ry 4 divE=4nk1pcurlE=-(1/o)6B/ot (Faraday) ki=ceka | divB=0 curlB=4nak25+(ak2/k1)AE Ittooksome doing justtogettothispoint! Now ifyoudoVxVxE andgetthefree-space wave equation, youconclude thatk1/k2 =¢*inyoursystem ofunitswhatever itis-wehavenotpicked oneyet.Thisisbecauseplanewavesmusttravelthespeedoflight.IfyoustartwithVxVxB,youreachthesame |conclusion. Either wayyoupickupboth curlequations andyoutherefore conclude thatk1/k2 =¢. Dotheconstants kiand k2and ahave dimensions? Tfwelookbackatouroriginal twodefining equations inwhich k1andk2firstappear, andifweimagine thateach ofourconstants have some kindofdimensions, thenthedimensions inthese twoequations are onlyconsistent IFthisistrue:(k1/k2 )must havethedimensions ofm”/sec” (sayinmksa). Weknow that infactk1/k2 =c?sothisallmakes perfect sense. ThisdoesNOT telluswhat wehave todofork1andk2 separately, however. Similarly, constantcmayormaynothavedimensions,sincewecanmakeitbe anything wewant. Since aisarescaling oftheB-field, giving itdimensions willsimply change thedimensions ofthe Bfield. There arereally only tworeasonable choices tomake about dimensioning klandk2.One choice isto giveeachofthese units its"natural" dimensions such that thedimensions cancel onboth sides ofthetwo basic equations. e F(force)=k1(force-L”/charge”) *q(charge)Q(charge)/r(L)* ran) dF(force)/dx(L) =k2(force-time”/charge*)*2i(charge/time)I(charge/time)/D(L) Q Inthiscase, wecannot make anyconnection between thecharge unitandtheother units. This isagood approach totake ifneitherk1nork2isgoingtobeunity. The other reasonable choice istomake kl=Iwith nodimensions, ortomake k2=|with no dimensions. Ifthis isdone with kl,then theequation isregarded asmaking aconnection between theunit ofchargeandthebaseunitsofforceandlength.Yougetforce=charge*/L? .Ifthisisdoneinsteadwith k2,thenyougetforce =charge’/time” which isdifferent! ‘The mksa system takes thefirst approach noted above. Theesusystems setskI=|andthisdefines theesuunitofcharge suchthatdyne=esu°/em”. The,emu systems setk2=1sodyne=emu’/sec” What about Dand H? AsJackson points outonpage 617, thisallows 6more constants toappear inyour world: Deqk+AP =2'E H=(I/p) B-AM=(1/)B Thave used e'andp'because, although Jackson sayseveryone uses1.ande,IseethatBleaney onpage 19 haveD=e&9£,soIwouldthensaye'=etforBleaney, whereasPortisonp112usesjuste’=«. Since wearedefining DandHbythese equations, wearereally freetosetall4constants arbitrarily. Ofcourse PandMarezero infree space. Inthislimit,©becomes€andp.becomes }iyandthisisthe 5 a a motivation forthese notations, asifthese numbers were properties offreespace. Jackson says that’ and 4arealwaysboth1or4. e Itisdesirable tohave thedivD andthecurlH equations "look nice”. From above wehave divE =4rk1 p,soinfreespace wewould then have divD=(4g) *k1)p. The choice ofklandeyisalwayscoordinated sothatthefactorinparensiseither4xor1,for“unrationalized" or"rationalized". Inallsystems shown onpage 618,these aretheonlytwopossibilities fordivD. ‘Theother interesting equation iscurlB=(4nc.k2) J+(cck2/k1) GE. Infreespace wewould have thisresult: curlHL=(40:K2/h1o) J+(cK2/(KIpoea)) @D. Again, k2andpparechosen sothefirstfactor (infront oftheJ)iseither cor4nc.. InJackson's page618list,there aretwochoices fort (1andI/c),so weendupwith fourdifferent forms forthecurlH equation indifferent systems! Jackson does notcomment onthe7type factors onesees inP=4EandM=xH.There issome variation inhow these things aredone aswell. Bleaney saysP=xe98 butM=H, soforeach author you havetogolookatexactly howtheydothings! Isuspect thatgeophysicists areinterested inHandj and useunits different form transformer designers. ‘Why batteries areabout avolt Wearejumping ahead hereabit,butwewant toshow what the"volt" isauseful unitsize.Using the Bohr radius andtheelectric charge, wehave lel=1.6x107C p=5.3x10"'m leltp=3x10°C/m e ‘V(volts) =k;"*** Q(C)/x(m)=9x10?*3x10°=27 volts asarough scale Hydrogen levels areE=-¢”/(2rpn*)sothatV=-13.5volts/n. Inalarger atomthereisshielding sothe general formisthesame, buttheradius islarger, sosetn=4sayandwegetouter electron ofsomething likeCubeing bound byontheorder of1eV,meaning thevalence electron isatapotential of-1volt.Inabattery,youmightpullanelectronoffCutogetCu++,andyoumightputitontoZn+togetZn.Each ofthese actions willinvolve adifferent energy ontheorder of1eV,sothenetresult willbeontheorder ofaneV,sothebattery willputoutsomething ontheorder of1volt.Theionization andaffinity values areaffected bythefactthattheionswhich result arenotfree, butareinasolution, butthisdoes not change thescale ofthings. Infact,theenergy obtained byputting Cut+ insolution probably offsets its ionization potential. (seechemistry books formore detail) The potential unit determines thecurrent unit. Inanysystem, weknow thatpower P=IV. Thus, theproduct ofthecurrent andpotential units mustbe theenergy/sec unitinwhatever system youarein,Inmks wegetwatts =amperes *volts. Intheother systems wegeterg/sec =abamperes*abvolts =statamperes*statvolts. 3.Themksa System (=SIsystem =rationalized-mls). 6 Wedothisonefirstbecausemostlookupinformation isintheseunits,whicharesometimescalledthe @ "practical" units. Also, theworld hasadopted SInowandSIis“taking over". Onewrites equation (2) above as: dF(nt)/dx(m) =k2(nt-sec”/ C?)*2i(C/sec)I(C/sec)/D(m) k2(nt-sec?/ C?)=107=pigl4a andonethen defines thesizeofquantity C(theCoulomb unitofcharge) bysetting k2(nt-sec”/ C?)= 10”.Inthisunit,anexperimenter willfindthatje}=1.6e-19 C.WHY wask2selected tobethisstrange power? Because then aC/sec =ampere will beareasonable "practical unit" ofcurrent, andwill have a simple connection (1/10) with theformerly used abampere unitoftheemu system, More onthishistorical subjectlater.Infacttheampereisreallydefinedasperabove,thenCoulomb =ampere-sec. Theconstant k2isusually written asjis/4n. The4xremoves 4xfrom theMaxwell equation, andthe #9issocalled because itends up(viaalong path) intheH=ysBrelation infreespace. Inthemksa system, theconstant k2isassigned itsnatural dimensions: k2(nt-sec*/ C?)=107=po/4n k2=107nt-sec/C?_ (=10”henry/meter) Theappearance of"henry" justcomes from V=Lal, andistheway1/47 isusually expressed. Now, because k?isassigned these "natural" units, wecannot make anyconnection between theCoulomb and anyofourbase units. ‘Theunits justcancel outintheequation, aswasnoted earlier. Noticethatevenwithan"amp"ofcurrent, theforce between two"practical wires" ameter apart is tiny, about 2/10,000,000 of1nt,and think of1Ib=4.45nt,sontreallyisaneverydayunit.Sointhe magneticworld,aCoulomb/sec isnotavery"big"thing.Ifyouputthewires1cmapartandput100 e amps ineachwire,yougainafactor of10°,soforce isthen2/10=1/5nt~IIb.Thisiseasily observable byanydolt. Sointhissystem, charge isdefined bythemagnetic equation. Theelectric equation willthen read, F(nt) =k1(nt-m7/C?) *q(C)Q(C)/r(my* =9x10°*q(C)Q(C)/r(m)* since weknow from theabove work thatkl(nt-m7/C?) =k2*o(m/sec)’ =le-7*(3e8)* =9x10”.This constant klisusually expressed ask1=1/(4ne9) sowecanhave D=eBinfreespace. Aswith k2,the constant k1isgiven its"natural" dimensions: kl(nt-m?/C?)=9x 10°= 1/(4ne0) kL=9.x10?nt-m/C?_ (=9x10°meter/farad) where thelastcomes from Q=CV defining thefarad. Notice thatk1/k2 =(3*10* m/sec)’. Theelectrostatic potential isaderived concept, notadefinition. WehavedW =FdxsodW/q =F/qdx which isthen Edx.This work percharge iscalled thepotential andisthisdV =Edx=(F/q)dx. Thus, our definition ofthe volt isthis: 1volt=1nt-m/C ‘Now fortheparticular situation ofapointcharge,wecanintegrateEfromrtoinfinitytoget e ‘V(volts)=9x10”Q(C)/r(m) /=k1Qr 7 where wehave used thepoint charge forourdefinition, Now E=nt/C from itsdefinition, =volts/m here. Inelectrostatics, aCoulomb isaverylargecharge! IfyouputtwoofthemImeterapart,theforce @ between them would be9billion newtons, enough toexplode anyphysical holding object toshreds ina realhurry! Another way tosaythis(seenext paragraph) isthatanaluminum sphere ofradius 1meter holding 1Coulomb ofcharge would have apotential of9billion volts! Side question: why don’t theelectrons justflyoffachargedsphereintothevacuum?Accordingto Purcell page 50,anaverage electron atthesurface "feels" halfthefield. Perhaps wehave 10,000 volts on apractical sphere, sothefieldjustoutside is10,000 volts/meter. Butthisisamere 10volts/Angstrom. If ‘youpulloneextreme surface electron 10Aabove thesurface, you"gain" about *10° eVofenergy. But you would expect theelectron tobebound inthemetal bysomething ontheorder of1eV(see battery discussion earlier). Thus,youwouldneedperhaps|billionvoltsonyoursphere(~Coulomb) beforeelectronswouldflyoffspontaneously. Ifthereisaconductor outsidethesphere,thenthe"workfunction” toremove anelectron goes away, andyou getresistive flow. This same thing happens ifthemedium outside thesphere ionizes intoaplasma duetotheEfield (spark, lightening bolt, etc). Thissystem iscalled "rationalized mks" onlybecause wewrite kl=9x10°as1/(4ne,). Since 4k appears intheMaxwell equation, thisgets ridofthe4mfloating around. What about «.? Jackson says a=1. The place tolook forthis isinFaradays law. Bleaney page 256 shows clearly that a=1 forthat book, andthesame forPortis onp382. Bleaney claims tousethe “yationalized mks" where Portis uses the"SI units system" which incorporates rationalized mks. Another term used ismksa where astands forampere. ‘Themagnetic field formula above isdF=(1/a) idxxB which inmksa is€F=idxxB.Thus, Bhas units which arederived from other quantities B(avamp-m) similarto _E{nt/C) @ (Side note: ifyouknow about antisymmetric F"whose 6non-zero elements areBandE,itsure is uncomfortable tohave these fields have different units! ) The volt was defined asabove, and weusually useE(volt’m) incommon parlance. Itwas decided by someone that weshould have aunit ofmagnetic flux (not field) called theWeber: (Weber) =B(nt/amp-m)edA(m?) sothat 1Weber =Int-m/amp ‘Then wecanthink ofB(Weber/m?) incommon speech. The current SIunit system (see http://iserver.asa.edu.py/physicsweb/hisotry_of_units.htm) wasadoptedinaconference in1971.Atthistime,theTeslawasintroduced togivetheBfielditsownprivateunit,Weberswerenotthrownout.SinceWeberwasalreadydefinedasabove,wecanregardtheTeslaas aderived unit, orwecould formally define itthisway B(Tesla) 1Tesla=1Weber/m? =1nt/amp-m MyBleaney book is1965 sodoes notmention Tesla, butPortis is1978 anddoes. Jackson is1962-7 and also does notmention Tesla. [bytheway, Hzwasadopted bySIin1960. My1956 and1964 handbooksdon'tuseit.thinkcommonuseofHzphasedinaround1966intheUS,wascyclespersecond.] So,onemorecomment, Ahispointwehavedefinedvoltsandamperes soonemoreobviousunit @ remains: 8 e R(ohms) =V(volts)/(amps) 1obm=1volt/amp ‘There aresome other SIunits that areabitnew tome: pascal=|nt/m?forpressure siemens =amp/volt toreplace mhos forconductance Magnetization Misinteresting. Bleaney page 131istalking magnetostatics andmentions theidea ofa magnetic dipole moment m=I(current)$(area ofloop), similar totheelectric dipole p=qL. Ifyoutalk about avolume density ofmoments, thenyouhave M=Nm/V=>M(current/length) likeAmpere/m. SinceMisrelatedtoH,..wesometimesseeHwithunitsofcurrent/length, | 4,The cgs-esu-Gaussian system ofunits Inthissystem,thestartingpointistosetkl=IsothatF=qQ/r’andthisofcoursemakeselectrostatics very niceandsimple. Weregard thisequation asdefining theesuunitofcharge alsocalled astatcoulomb (Purcell calls thisanesuanddoes notmention statcoulomb). | F(dynes) =q(esu)Q(esu)/r(cm)” k1(dyne-cm*/esu”) =1butnoexplicit dimensions! : .SincekIhasnodimensions, weregardthisastheformaldefinition oftheesuintermsofbaseunits: eIdyne=1esu*/cm? Notice thatE=F/qisindynes/esuorbelowitcanbestatvolts/em. ‘Since weknow thatk1(nt-m’/C’) =9x10° fromthemksa system, weconsider: 9x10°=k1(nt-m?/C*) =k1(dyne-cm*/esu’) *{dyne/nt}* {cm/m}?* {C/esu}? =1*e-5*e-4* {Clesu}” Thus: Coulomb =3x10? esu(statcoulomb) Sothestatcoulomb ~esuisamuchsmallerthingthanaCoulomb. That|meterspherewithanesuonitwouldonlyhaveapotential ofabout3volts,muchmoremanageable! ‘Thepotential isgoing tobedV=Edx=F/qdx.Theunitwillbethestatvolt andwehave: dV(statvolts) =F(dynes)/q(esu) *dx(em) 1statvolt =1dyne-cm/esu ‘The conversion istherefore: 1statvolt =1dyne-cm/esu= 1nt-m/C *{dyne/nt} {em/m)} {C/esu} =1volt *¢-5 *e-2 *3e9 =300 volts 9 eer amet 7 Certainly wewould saythatastatamp =esu/sec, theunitofcurrent. ‘Then statohm =statvolt/statamp.TheEfieldismeasuredinstatvolts/em. r)‘Now moving tothemagnetic equation, wehave dF/dx =k2*2i1/D k2=k2(dyne-sec?/ esu®)=k1/c?=1/c?=1/(9€20) sec/em* Remember that theforce between currents was already pretty small even with amperes. Here with a current unitthat is3billion times smaller, ourforce isreallyreally tiny! Usually inequations onejust showsthe¢factorsas“c",andofcoursehereitisincgsunits.Thuswehave F(dyne)/dx(cm) =(1/e(erm/sec))* *2i(esu/sec)I(esu/sec/D(em) //force between wires Asnoted ingeneral above, interms ofunitsthisequation alsosays1dyne=1esu?/cm?. What about Band Biot-Savart? Recall that AF=(1/a) idxxdB B=0K2*(dXxyi? JacksoninhisGaussian system(Purcellalso)selectsct=sothesebecome dF=(i/c)idxxdB B=(/e)*(dXxry? This isnice inthatweassociate aI/cfactor with each current when wedothesplit here. Thefactor ofcis taken tohave itsunits with itintheGaussian system. Thus weget <F(dynes)=(I/e(cm/sec)) i(esu/sec)dx(cm) xdB(dynes/esu) e ‘Thepoint here wastoshow thatwehave B(dynes/esu) intheGauss system. Infact, thisunitiscalled a "Gauss" B(gauss) =B(dynes/esu) 1gauss =1dyne/esu Notice thatEandBboth have thesame units inthissystem, dynes/esu, although they areusually called "gauss" fortheBfield. Theterm“oersted” isusedfortheHfield, butitisthesame unitasthegauss. How aregauss andTesle's related? Iseenotrivial waytocompute this,justhavetodobrute force: AF(at)=i(C/sec)dx(m) xB(tesla) F(dynes)= (1/e(cm/sec)) i(eswsec)dx(em) xB(gauss) Divide these twoequations toget: k(t) __i(C/sec) ,dx(m), +Bitesla)_F(dynes) ~Wesw/sec) *axcemy*%™S°°) *Bigaussy (tesla) {dyne/nt}={eswC}*{em/m}*o(em/sec)ee leS=1e9 *102*3e10 *Biteslad“B(gauss)so ® 10 Btesla) . . . e Bigaussy 714 >B(gauss) =10,000B(Tesla) =>1Tesla=10,000gauss Wecould alsorelate theEfield units bycomparing volts/meter tostatvolts/em. So E(volts/m) =E(statvolt/om)* {statvolt/volt} *{m/em} =E(statvolt/em)* 300*100 =30,000 Eistatvolt/om) > 1statvolt/em =30,000 volts/m Soinmksa themagnetic field unitisbigger, buttheelectric field unitissmaller. What about magnetic flux. Ithink people used this 1maxwell =1gauss-cm? compare to: 1Weber =Tesla/m? . Divide anduseT=1e4gauss toconclude that 1Weber=10°maxwell. There arereally noother units thatpeople useinthissystem. Theinside back cover ofPurcell gives agoodsummary ofthings. 5,Theegs-emusystemofunits e Now,ontotheemusystem.Sometimes unitsherearecalled"absolute" units,hencethetermslikeabvolt.Wetake asourstarting point, dF(dyne)/dx(cm) =k2*2i(abC/sec)I(abC/sec)/D(em) k2(dyne-sec/abC*) =I(dimensionless) dF(nt/dx(m) —=k2(nt-sec”/ C?)*2i(C/sec)I(C/see)/D(m) k2(nt-sec?/ C?)=107=pole Wewrite thesecond lineonly sowecandoadivision tofindhow abC arerelated toC: | le5=1e7*{C/abC}? =>abcoulomb =10coulomb Thisfactorentirelyarisesfromthent/dyneshiftbeingoffsetbythek2shift.Ofcoursetheabamp willbeanabcoulomb/sec. Now kl=k2*c? =c?=(3e10 cm/sec)’. Thiswearegoingtohaveintheelectrostatic sideofthings, F(dynes)=o?(cm*/sec*)q(abC)Q(abCyr(em? and |E(dynes/abC) =c?(cm?/sec*) Q(abC)/r(cm)* Thepotential willbeV=Edxsothat 1abvolt=1dyne-cm/abC =Lnt-m/C*{dyne/nt}*{m/m}*{C/abC} =1volt*10%*107*107 r) =10°volt . . ul ‘NextwecometoBiot-Savart andwehavetofacetheotquestion:(k2=1) r) AF=(1/a) idxxdB dB=a (dXxry? Mist isBiot-Savart (4) Ifweset=I(asitisinmksa), thenwecould conclude thatB(Tesla) =10°B(emu). Afactor of10 comes from abcoulomb tocoulomb, andafactor of100comes from I/distance. However, Ithink noone uses anemu unit likethis, Ithink they usetheGauss from thecgs-esu system instead! This could be fixed bymaking &=1/10. Then wewould make nodistinction between thetwoogssystems forB. However, wewould than require thatdF=10idxxdB. This hastobethecase ofB=Gauss and i(abC/sec), nowayaround it!IfBisthesame inemuasesu,thensoisB-flux, themaxwell. Obviously ifwehave abamps andabvolts, wearegoing tohave abohms. 6.Some History ‘Theemusystem isthecounterpoint totheesusystem because herewearegoing totakek2=Iwithno dimensions, whereas inesuwetake kl=1.What does thissystem useforc:?Was itthehistorical system used byAmpere etal?DoIknow anysources thatusethissystem? Geophysics may have historically usedthissystem.Thenotebelowsuggeststhatabvoltswereusedpriorto1881,somaybetheemusystem with ogswasthestandard foreverything prior tothattime! Myguess isthatbefore 1881, everyone doing electric currents work wasusing these "ab"units and gsbase units, butwasjustcalling them “electromagnetic units". Surely the"ab" forabsolute isaretroactive additionfromthemoderndaytodistinguish currentampsfromhistoricalampswhicharer)today’s abamps. Theclaimmadebelow isthatin 1881abvolt wasreplaced withvoltthatwas10°times larger and more practical. ‘Why wastheampere defined as1/10ofanabamp (with theresult thattheCoulomb became 1/10of theemuunit(abcoulomb) )?Thereason justcomes from P=IVasnoted earlier, 1amp=|joule/volt =10"erg/(10°abyolt) =(1/10)abamp Sothepotential unit(volt) determines thecurrent unit(ampere) andthusthecharge unit(coulomb). Also: abohm =abvolt/abamp =10°volt/(10 amp)= 107ohm Some quotes ofinterest from theweb: "First toadopt acustomary metric system wasFrance in1799. Progress towards itsadoption elsewhere hasbeen distinctly lessrapid nowthatthose opposed toitarenolonger decapitated. Anyway, theCGS system enjoyed wide acceptance until 1881 when thethennewelectricity industry decided thatthee.m.u. abvolt wastoosmall forpractical useandsucceeded inintroducing avalue 10°times larger: ourpresent value forthevolt. Theohm andtheamp were also re-scaled. " "The First International Conference ofElectricians (Paris, 1881) adopted theBritish Association definition oftheohm andadded definitions forthevolt, ampere, coulomb, andfarad. Thus was born the“absolutepracticalsystemofelectricalunits”: r) 12 *absolute, because theunits were defined solely interms ofmechanical units (length, mass andtime) e *practical, because thesizes were much more convenient than thecgsunits. Theampere wasaderived unit, defined asthecurrent produced inaconductor with aI-ohm resistance when there wasapotential difference of1voltbetween itsends” "In1948 itwasdecided thatthecentimetre, thegram andtheerg(theCGS unitofenergy) should be replaced bythelarger metre, kilogram andjoule, This wascalled theMKSA system anditrestored coherence " More ontheorder inwhich things were done (1780 -1881) 1785 Coulomb's lawforstatic charges 1800 Volta makes batteries, allowing people tohave wires with currents inthem forthefirsttime 1820 Oersted sees current inwire deflect acompass needle 1820 Arago seescurrent inwire attract ironfilings. 1820 Biot, Savart andAmpere getforce oncurrent from Bandbetween twocurrents 1825 Ampere puts outasummarizing paper 1827 Ohm's lawinDieGalvanische Kette, Mathematisch Bearbeitet paper 1874 BAAS adopts cgs 1881 conference alsoadopts practical units includiig amp, ohm, volt e Duringallthistime,nooneknewaboutelectrons andprotons. ChargeQwasregarded assomekindof invisible fluid thatcame outofmatter, maybe twofluids. Thefluid could flow through awire, andcould beproduced with abattery. The idea thatthefluid wasmade ofindividual particles wasunknown until 1897. The connection between light waves andthischarge fluid through theMaxwell equations must have seemed very strange. That didnotstopelectric motors from being built in1831. Dalton summarized basic ideas about atoms around 1803, theywere round solid balls with interesting properties. In1897 Thomson showed thatcathode rayswere individual charged "corpuscles" much lighter than atoms, later called electrons, andthey were assumed tobepieces ofatoms. People made atomic models likethePlum Pudding Model with amixture of+and-chargesswimmingaroundintheball. Only in1911 didRutherford infer thepresence ofcharged positive atomic nucleus. Thefactthatlight wasquantized inhvwasfound separately from théphotoclectric effect (Einstein), andthiswiththe nucleus idea because thewave model forhydrogen ofBohr in1913, andsostarted theprobability idea andquantum mechanics. Wenow know thatE&M isjustquantum mechanics foramassless particle. Relativity (again Einstein) came along in1905 andsoon gotmixed with QM. 13 | RAINBOW e Rainbows A.D. Andrew G.L. Cain S.S. Crom T.D. Morley Introduction Mostofusarefamiliar withthesightofarainbow afterarainstorm. Thesunisat ourback, andweseeanarcwithviolet innermost, blending intoblue, green, yellow and orange, with aredouter band. This istheprimary rainbow, andwesecitaswell inthe spray ofawaterfall. Sometimes after arainshower wemaybefortunate enough tosee another rainbow, thesecondary rainbow, higher intheskythan theprimary rainbow. Thesequence ofcolors inthissecondary rainbow isreversed, with theredringinnermost, and the violet band outermost. td Inthisprojectyouexaminehowrainbowsareformedandseewhythesequence ofcolors inthetwobows isasdescribed above. First wemust study thepathfollowed by | lightpassingfromonemediumtoanother. A 8 4 v. 8, 2 B e@ Figure1 1 e Intraveling fromAtoB,thelightisrefracted, thatischanges direction, attheinterfacebetween thetwomedia. Theangle ofincidence, @,,andtheangle ofrefraction, 6, satisfy Snell's Law (Willebrord Snell, 1621), sin®, _sin®, yy? where¥,isthevelocityoflightinthefirstmedium, andv,isthevelocityoflight inthe second. Theratio“is 0.7508 forredlight,and0.7440 forvioletlight. Yair : Part L.The primary rainbow Thetheory ofrainbows presented herewasdeveloped byDescartes. Theprimary rainbow isformed byraysoflightwhich enter adropofwater (assumed tobespherical),arerefracted attheboundary, reflectonceinternally offthebackofthe raindrop, andareagainrefracted astheyleavethedrop,headedtowardtheobserveroftherainbow. ¥ Figure 2 Thedistance hfrom thecenter lineofthedrop totheincoming rayiscalled theimpact e parameter, andtheanglexiscalledtheangleofincidence. Theangleyiscalledthe 2 scatteringangle.Ifarayisincidentwithimpactparameter0,itisreflectedbackonits r) original path,andthescattering angleism.Astheimpactparameter increases, thescattering angle decreases, asseenintheabove figure. AsAincreases, thescattering | anglereachesaminimum ofabout 138°, andthen begins increasing. Descartes’ reasoning was that light exiting thedrop isconcentrated near the scattering angle forwhich thescattering angle isleast sensitive tochanges intheimpact parameter h.That is,itisconcentrated near thescattering angle forwhich therateof change ofscattering angle with respect tohisassmall aspossible, namely where WWLo, ah Ofcourse thishappens where thescattering angle yassumes itsminimum value, and Descartes calculated scattering angles formany values oftheimpact parameter h, thereby determining thatthereflected light isconcentrated nearthescattering angle of 138°. Theprimary rainbow isformed bylight from those drops forwhich theangle | betweentheincidentraysandtheobserver'slineofsight is180-138=42°. Exercises e 1.Fermat'sPrinciple inopticsisthatlighttravelsonthepathwhichminimizes thetime oftravel (See Project 2.).UseFermat's Principle toshow thatintraveling from AtoB (see Figure 1),light follows thepath forwhich Snell's Law sin®, _sin®, yo% holds. 2.Show that thescattering angle yisgiven bytheformula y=n+2x—4y, or went2x—4aresif"asinz]Yair This canbedone byexamining Figure 2andeither using plane geometry orbycarefully calculating theray's change ofdirection ateach refraction orreflection 3.Plotthescattering angleasafunction ofthe angle ofincidence, andcalculate the e minimumvalueofyforbothredandvioletlight.Usethistoexplainwhytherainbow 3 forms inacirculararcandwhythevioletarcistheinnermostringandtheredarcis e outermost. Part I.The secondary rainbow Thesecondary rainbow isformed byrays oflight which undergo twointernal reflections before being refracted towards theobserver. This rainbow isfainter than the primary rainbow because ateachreflection aportionofthe light istransmitted intotheair.Thepathoflightundergoing twointernalreflections isshowninFigure3. | \ Figure 3 Inthiscase, arayentering adrop atimpact parameter h=0isreflected twice, andleaves thedrop initsoriginal direction, with scattering angle y=0. Ash changes, the scattering angle increases toamaximum ofabout 130°, andthen decreases. Inthiscase,lightisconcentrated inthedirection ofthis maximum scattering angle, Exercises 4.Show thatinthecase oftwointernal reflections, thescattering angle wisgiven bythe formula y=6y-2x, or ify=6AresiiYastsinxox @ | 4 t) 5.Plotyasafunctionofx,anddeterminethemaximumvalueofyforbothredandviolet light. Explain whythesecondary rainbow appears higher intheskythanthe primary rainbow, andwhythesequence ofbands inthesecondary rainbow isreversed, with theredband innermost and violet outermost. 6.Explain whytheskyisparticularly darkintheregion between theprimary and secondary rainbows. This dark area isknown asAlexander's Dark Band. | References H.MoysesNussenveig, TheTheoryoftheRainbow,ScientificAmerican, April1977, |reprinted inLightfromtheSky,W.H.Freeman andCompany, SanFrancisco. ' A.D. Andrew, G.L. Cain, S.Crum, T.D.Morley, Calculus Projects Using Mathematica, McGraw-Hill, 1996. 5 (> restart; ‘ . {>y:=arcsin(ni*sin(x)/n2); r)[>pt=Pi+2*x-4ty; //psi(x)[>pl:=dife(p,x); //psi'(x) (>p2 :=dife(pl,x); //psi"(x) (>Nl:=sin(x)*cos(x)/(sin(p)) ;//firstfactorofdNsolution [> N2 :=abs(-pl+signum(p1)*(p1*2 +2*p2*£)*.5)/(p2);//secondfactor [> N:=N1*N2; //complete solution [>nl:=1;//Nowinsertspecificvaluesforconstants (> n2 :=1.33; [> £:=1e-4; //thisisRp/D,wewanttheeyefaraway,angularly speaking (andD>>R) >plot(N,x=0..Pi/2); 0.005: 0.004: 0.003: 0.002: @|vw: ° 02 04 a6 08 i 12 14 4 xq ‘ Page1 4 (> restart; [>¥:=aresin(n1*sin (x)/n2); e@ iac) =aresin]——— |nar) >pi=Pi +2te -4ty; //psi(x) nlsin(x)p=m+2x—4 aresin|——— L n2 >pl :=diff(p,x); //psi(x) nlcos(x pl=2-4— 2 nF? sin|no,|Sinen2? [> p2 :=diff(p1,x); //psi"(x) nlsin(x nPcos(x)? sin(xpes ei“7rs;7 n2 n?? >N1:=sin(x)*cos(x)/(sin(p)); //firstfactorofdNsolution sin(x) cos(x, nlsin(x) ein|2x—4aresin| ——— n2 >N2 :sabs(-pl+signum(p1)*(p1*2 +2*p2*£)*.5) /(p2);//secondfactor nlcos(x) . nlcos(x) N2:=|-2+4-——=—=——— +signum]2-4——————_ 'nB?sin(x)? nP?sin(x)? 2,|1-——— n2,|1-———nd? nP nlcos(x. i2-4 __ nf? sin(xna,[1eesine? nz | nlsin(x) nl?cos(x)? sin(x) \s +2] 4 -4 IfnP?sin(x)? al?sin(xY \°/ n2,|1-————|1- |n2 n? 4 nlsin(x) 4nPcos(x)?sin(x)nl?sin(x)? nF?sin(x)?2! n2,[1-2] 1-——— r) - nP Page| |[>N:=w1#N2;//completesolutionnlcos(x nlco: @N:==sin(x) cos(x)2g) sig2-4) nP?sin(x)? nP?sin(x)? n2,|1-————— n2,|1-——-——n2 n? \ nlcos(x z | nP?sin(x12,|1-——— nP nlsin(x nP cos(x)? 5 sala @, staFsins) |nPsin(x)? nb?sin(x? \°/?) 2,|1-———— |1-n2 nd |_(nlsin(x) nlsin(x) nPcos(x)? sin(x) sin]2.x—4aresin}——— ||}4—————_ _g n2nl?sin(x)? nPsin(x)? \@/2) n2,|1-—— ——|.1- n? n2? [>nl:=1;//Nowinsertspecificvaluesforconstants L nl=1 @ [rr nt33 n2:=1.33 >£:=1e-4; //thisisRp/D,wewanttheeyefaraway,angularly speaking (andD>>R) f= .0001 >plot (p,x=0..Pi/2); 344° 3 29 ~~ 28 NX 27 \N\ 26 NN25a 7 0 a2 04 06 os i 12 14 x @® > Page2 TheRainbow Phe 1.22.03 e@ ‘Look atthepencil drawing. Itshows asingle raycoming inatimpact parameter h(angle x)andemerging atanangle yrelative totheinitial direction. Nowimagine anobserving eyeviewing theemerging rayat agreatdistance. Model theeyeasacircular pupil thatcatches rays. ‘Thenumber ofrayscaptured bythe eyeisproportional todybecause thepupil hasafinite size. Since wearesofaraway, thefactthatthe location onthesphere oftheemerging rayvaries slightly asyvaries does notmatter, Intheother direction, imagine thataraycomes inatsome doazimuth (north pole isleftendof sphere) below theplane ofpaper. Thenjusttilttheentire diskdrawn inthefigure byrotating itabout the horizontal axisbyangle ¢.Theangle between theemerging rayandtheuntilted emerging rayis60= siny56asproven inourlittleappendix below. Thisisobvious when y=n/2 sosiny=1. Imagineasquare(edged)intowhichisinscribedthepupil(diameterd)ofaviewingeye(atdistance Dfromthesphere).Thesquareforces88=8y=d/D.Alllrayswithinthesesmallangleshitthesquare. Ofthese,thefractionn/4hitthediskoftheretina,butwedon'tbotherwiththisfactorinthefollowingsincewearenotlookingforanabsoluteanswer,justtheshapeofthecurve. Now, thenumber ofrayswhich passthrough anareaR?dQonthesphere (x=polar angle, 6= azimuthal angle) iso(rays/area)* R?dO.*cos(x), thislastfactor since R?dQistilted relative tothe : incoming plane wave ofrays. Wehave assumed auniform incoming plane wave raydensity o;these raysarecoming from theleftinthefigure. Wearealsoimplicitly assuming polarization perpendicular to theplane ofpaper. Theother polarization direction willhave some extra angle complications andwill probably haveaweakeroutput,butthenatureofthe result won't bedifferent. Thus wehave r) SBN(x,$)=oR?cosxsinx8x8=numberofrayshittingourdifferential area. ‘Then set6x=Sy/y'(x) and56=50/simy. Finally, use69=Sy=d/Dandtheresult is @R’sinxcosxd thingretina)=2%sinxcosxd_d__ N(numberofraysreachingretina) =< Oo) where thesecond fraction is6xwhich wehavekeptseparate forthemoment. Wehave anequation for ‘y'@) onourpencil sheet, andweknow w'(x) =0aty=137.48 °orso.Atthispoint, everything elsein theequation isfinite, sodNhasaninfinite peak here! This makes thepoint quite well thatthecritical angle completely dominates thereflected light, andthis was Descartes idea. But weknow theresult cannot really beinfinite. Tocorrect thesituation ,let's replace 8x=Sy/y'(x) with amore accurate expression. Wecanexpand (x) around thepoint xandgetthisresult, where nowwekeep anextra term: by=5x{yi+(Sx/2)* y"} Rewrite this as (w"/2) Bx? +(W1x) -By=0 A(8x)? +B(x) +C=0 solve this for8x, e dx=[-y £sqrt(yy+2y"by)]/y" 1 | @ Whichrootdowewanthere?Wewanttherootthatcausescancellation between thetwolargeterms because theresult wearelooking forisdifferentially small. Weknow thaty">0, soassume thatSy>0sotheinsideofthesquarerootislargerthan|y'|.Thenwewantthissolution: Sx=|-y'+sign(y’)*sart((y?+2y"dy)|/y" | Wehave added||becausewewant5x>0sincewearejustdoingaraydensityscaling. WesetSy=d/Dinside andthen ourformula becomes Dsiny y" Right ATthecritical point where y'=0 wecanwrite thisas ae GR’sinxcosxd d AN(oumberofraysreachingretina)=SSS ran) @) Comparing with (1)above, weseethaty’inthedenominator hasbeen replaced bysqrt(w"d/2D). This is small, butnot0,sotheresult fordNislarge butfinite, Solet'splottheabovegeneralresultfordNusingMaple!‘Theresultisveryimpressiveandhighly @ peaked asexpected. Wecouldcompute thewidthofthepeak,buttheMapleplotshowsittobeabout.04 radians with ourassumed f=d/D=.0001. This isabout 2.4degrees, very narrow, athumb width which is 2.4degrees aswell forme(1"at24"). This might agree with theobserved width ofarainbow.Probably thecolor bands overlap alot,ete. Colors arethere because theindex ofwater varies with frequency. Since visible isbelow the electronics resonance area, index (and dielectric constant) increases with increasing frequency, sored light isbent least andhasthelowest index, andwillhave asmaller angle, soredappears atthebottom oftherainbowband(ontheinsideofthebigcircle).Toseewhyarainbowisacircle,holdthepieceof paper with thefigure onitandrotate itsuch thatrayscoming from behind youallparallel cause reflection intoyour eye. Thepaper must beattached asapaddle onawheel whose axisgoes through your eye. Inthesingle bounce situation which wehave looked at,theoutput angleyrangesfrom180degrees downto137degrees andourdrawing istypical. Suppose theraydoes two integral bounces. Ihave shown asecond bounce asadotted lineinthefigure. Youcanseethatthisray.emerges onthetop,noton thebottomofthefigure.Inthiscase,redlightwithlessbendcomesoutat/argerybecausenowwis measuredaround onthetopsideofthefigure. Sothesecond rainbow hastheopposite coloring order. For nbounces, youhave tostudy thegeometry andseewhether raycomes outonthetoporthebottom. Ihave seen three rainbows inTorrey. Appendix. What istheangle between thetilted emergent rayandtheoriginal untilted emergent ray? Go outtotheeyealarge distance Daway. Make alittle coordinate system centered ontheemergence point, withxtotherightandydownandzoutofpaper.Thecoordinatesofthepointattheeyethatishitbythis r) rayare:2=Dsinydo, x=Deosy, y=Dsiny,orr=.Thecoordinates ofthepointwithnotiltarez=0, 2 |x=Deosy,y=Dsinyorr=b,Thecorrespondingraysaregivenbyrl=aaandr2=abwhereo.isa e scalingfactor.Theanglebetweentheseraysisgivenbycos@=rl-hat¢r2-hat=a-hat»b-hat.Now[al= | Dsqrt(1+(sinydo)? )whileJb]=D.Andaeb=D.Thus,cos=1/sqrt(1+(sind)? ).Butdéissmall,sowecanexpandRHStogetI-1/2((sinyd6)" .Butweknowthatforsmallangled@wehavecosd@=1-1/2d6*.Thiswegetouranswerwhichisd0=sinydd. 3