The Fields Of A Charged Particle In Hyperbolic Motion
The fields of a charged particle in hyperbolic motionJoel Franklina) and David J. Griffithsb)Department of Physics, Reed College, Portland, Oregon 97202(Received 13 February 2014; accepted 24 April 2014)A particle in hyperbolic motion produces electric fields that appear to terminate in mid-air,violating Gauss’s law. The resolution to this paradox has been known for sixty years but exactlywhy the naive approach fails is not so clear. VC 2014 American Association of Physics Teachers.[http://dx.doi.org/10.1119/1.4875195]1u ¼ c r ' v ¼ ðcr ' rvÞ;I. INTRODUCTIONIn special relativity, a particle of mass m subject to a constant force F undergoes “hyperbolic ffiffiffiffiffiffiffizðtÞ ¼ b2 þ ðctÞ2 ;(1)2where b % mc F. The particle flies in from infinity along(say) the z-axis, comes to rest at z(0) ¼ b, and returns to infinity; its velocity approaches 6c asymptotically as t ! 61(Fig. 1).Because information cannot travel faster than the speed oflight, the region below the main diagonal (z ¼ –ct) is ignorant of the particle’s existence—the particle is “over thehorizon.” For someone at the origin it first comes into viewat t ¼ 0. If the particle is electrically charged, its fields arenecessarily zero for all z 0, at time t ¼ 0. But the electricfield for z 0 is not zero, and as we shall see the field linesappear to terminate in mid-air at the xy-plane.1 This wouldviolate Gauss’s law; it cannot be true. Our task is to locatethe error and fix it.2In Sec. II, we calculate the electric field of a charge q inhyperbolic motion, at time t ¼ 0. A plot of the field linesshows that they do not go continuously to zero at the xyplane. In Sec. III, we explore the case of “truncated” hyperbolic motion (hyperbolic motion back to time t ¼ –t0,adjoined to constant velocity for earlier times). In this case,the field lines make a sharp turn as they approach thexy-plane, and there is no violation of Gauss’s law. In Sec.IV, we work out the potentials for a charge in hyperbolicmotion, finding once again that we must adjoin “by hand” aterm inspired by the truncated case. In Sec. V, we ask howthe naive calculations missed the extra term, and concludewith the puzzle unresolved. Appendices A and B supplysome algebraic details, and Appendix C examines the radiation from a charge in hyperbolic motion; surprisingly, the“extra” terms do not contribute.randc2 t rv ¼ ��ffiffiffiffi z ;b2 þ ðctr Þ2ðbcÞ2a ¼ ��ffiffiffiffi%3 z :b2 þ ðctr Þ(4)(5)(6)2The retarded time tr is defined in general byr ¼ cðt ' tr Þ;(7)but for the moment we’ll assume t ¼ 0 (so tr is negative). Then ��ffiffiffiffi%222ðctr Þ ¼ x þ z ' b2 þ ðctr ��ffiffiffiffiffi¼ x2 þ z2 ' 2z b2 þ ðctr Þ2 þ b2 þ ðctr Þ2 ;(8)and henceII. ELECTRIC FIELD OF A CHARGE INHYPERBOLIC MOTIONWe begin by calculating the electric field at the pointr ¼ (x, 0, z), with z 0. According to the standard formula,3Eðr; tÞ ¼#qr " 2ðc ' v2 Þu þ r ( ðu ( aÞ ;34p!0 ðr & uÞ(2) (3)wherer ¼ x x þ z ffiffiffiffiffiffi%b2 þ ðctr Þ2 z ;Am. J. Phys. 82 (8), August 2014http://aapt.org/ajpFig. 1. Hyperbolic motion.C 2014 American Association of Physics TeachersV755
1ctr ¼ x2 þ z2 þ b2 Þ2 ' ð2zbÞ2 :(9)Putting all this together and simplifying givesEðx; 0; zÞ ¼ qb2 ðz2 ' x2 ' b2 Þ z þ ð2xzÞ x ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%3 :p!0ðz2 þ x2 þ b2 Þ2 ' ð2zbÞ2III. TRUNCATED HYPERBOLIC MOTION(10)That is for z 0, of course; for z 0 the field is zero. Incylindrical coordinates ðs; /; zÞ, then4Eðs; /; zÞ ¼ðz2 ' s2 ' b2 Þ z þ 2z sqb2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%3 hðzÞ;p!0ðz2 þ s2 þ b2 Þ2 ' ð2zbÞ2(11)where hðzÞ is the step function (1 if z 0, otherwise 0). Thisfield is plotted in Fig. 2; the field lines are circles, centeredon the s axis and passing through the instantaneous positionof the charge.As required by Gauss’s law, r & E ¼ 0 for all z 0(except at the point s ¼ 0, z ¼ b, where the charge is located).However, E is plainly not divergenceless at the xy-plane,where the field lines terminate in mid-air. Indeed, the fieldimmediately to the right of the z ¼ 0 plane isEðs; /; 0þ Þ ¼ 'qb21 z ;p!0 ðs2 þ b2 Þ2(12)and the flux of E through a cylindrical Gaussian “pillbox” ofradius r, centered at the origin and straddling the plane, withinfinitesimal thickness, is 2 %ððqb2 r1qrE & da ¼ '2ps ds ¼ ';p!0 0 ðs2 þ b2 Þ2!0 r 2 þ b2(13)Fig. 2. Field of a charged particle in hyperbolic motion at t ¼ 0 (particle atz ¼ b).756Am. J. Phys., Vol. 82, No. 8, August 2014even though the pillbox encloses no charge. Something isobviously amiss—we appear to have lost a crucial piece ofthe field at z ¼ 0.Suppose the acceleration does not extend all the way backto t ¼ '1 but begins at time t0 ¼ 'ab c (for some a 0),when the particle was 0 Þ ¼ b 1 þ a2 ;and its velocity wasacvðt0 Þ ¼ ' pffiffiffiffiffiffiffiffiffiffiffiffiffi z ;1 þ a2(15)prior to t0 the velocity was constant. In other words, replaceEq. (1) with81 pffiffiffiffiffiffiffiffiffiffiffiffiffi2 ðb ' actÞ ðt t0 ¼ 'ab cÞ1þa(16)zðtÞ �ffiffiffiffi 2: b2 þ ðctÞðt ) t0 Þ:At time t ¼ 0, for all points outside a sphere of radiusr ¼ 'ct0 ¼ ab, centered at z(t0), the field is that of a chargemoving at constant velocity—the “flattened” Heaviside field5radiating from the place q would phavereached, had it continffiffiffiffiffiffiffiffiffiffiffiffiffiued on its original flight plan ðb 1 þ a2 Þ:q1 ' ðv cÞ2R:(17)4p!0 ½1 ' ðv cÞ2 sin2 h 3 2 R3'pffiffiffiffiffiffiffiffiffiffiffiffiffi(The left edge of the sphere is at1 þ a2 ' a b (which isalways positive, but goes to zero as a ! 1). Inside thesphere, where news of the acceleration has been received,the field is given by Eq. (11) (Fig. 3). The field linesE¼Fig. 3. Field lines for truncated hyperbolic motion (b ¼ 1, a ¼ 12 5).Joel Franklin and David J. Griffiths756
evidently join up in a thin layer at the surface of the sphere,representing the brief interval during which the motionswitches from uniform to hyperbolic.As alpha increases (that is, as t0 recedes into the more distant past), the radius of the sphere increases, and its left surface flattens out against the xy-plane. Meanwhile, the“outside” field compresses into a disk perpendicular to themotion, and squeezes also onto the xy-plane. The completefield lines now execute a 908 turn at z ¼ 0, as required to rescue Gauss’s law. Indeed, for a ! 1 the constant velocityportion of the field approaches that of a point charge movingat speed c:6E0 ðs; /; zÞ ¼q sdðzÞ:2p!0 s2(18)Using the same Gaussian pillbox as before, this field yieldsððq rqð2prÞdðzÞ dz ¼ :(19)E0 & da ¼2p!0 r 2!0This is appropriate, of course—had the particle continued atits original velocity (c), it would now be inside the box (atthe origin).Awkwardly, however, this is not what was needed to cancel the flux from the hyperbolic part of the field [Eq. (13)].For that purpose the field on the xy-plane should have beenqsEðs; /; zÞ ¼dðzÞ:2p!0 s2 þ b2sum of these fields that gives Eq. (20). The true field of acharge in hyperbolic motion is evidently7Eðs; /; zÞ ¼þqsdðzÞ;2p!0 s2 þ b2(21)and it does not look like Fig. 2, but rather Fig. 5.As a check, let’s calculate the divergence of E. WritingE ¼ Eh hðzÞ þ Ed (in an obvious notation), we haver & ½Eh hðzÞ ¼ ðr & Eh ÞhðzÞ þ Eh & ½rðhÞ :(22)The first term gives q !0 , for the point charge q at z ¼ b; asfor the second term,rhðzÞ ¼@h z ¼ dðzÞ z ;@z(23)sor & ½Eh hðzÞ ¼(20)It must be that the “connecting” field in the spherical shell(the field produced during the transition from uniform tohyperbolic motion), which (in the limit) coincides with the xyplane and which we have ignored, accounts for the difference,as suggested in Fig. 4. The net field in the xy-plane consists oftwo parts: the field E0 due to the portion of the motion at constant velocity, given (in the limit a ! 1) by Eq. (18), and theconnecting field that joins it to the hyperbolic part. It is theðz2 ' s2 ' b2 Þ z þ 2z sqb2 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%3 hðzÞp!0ðz2 þ s2 þ b2 Þ2 ' ð2zbÞ2¼q qb2þh!0 p!0ðz2 ' s2 ' b2 Þðz2 þ s2 þ b2 Þ2 ' ð2zbÞ2q qb21'dðzÞ:!0 p!0 ðs2 þ b2 Þ2i3 2 dðzÞ(24)Meanwhile,)*qsr& 2dðzÞ2p!0s þ b2) 2*q 1@s¼dðzÞ2p!0 s @s s2 þ b2r & Ed ¼(25)soFig. 4. Truncated hyperbolic motion for large a, showing the “connecting”field.757Am. J. Phys., Vol. 82, No. 8, August 2014Fig. 5. Field of a particle in hyperbolic motion (corrected).Joel Franklin and David J. Griffiths757
r & Ed ¼qb2dðzÞ:p!0 ðs2 þ b2 Þ2(26)This is just right to cancel the extra term in r & ½Eh hðzÞ , andGauss’s law is sustained:qr&E¼ :(27)!0IV. POTENTIAL FORMULATIONA. Li!enard-Wiechert potentialsThe truncated hyperbolic problem guided us to the “extra”(delta-function) term in Eq. (21), but it does not explain how wemissed that term in the first place. Did it perhaps get lost in goingfrom the potentials to the fields? Let’s work out the Li!enardWiechert potentials,8 and calculate the field more carefully:V¼Vðr; tÞ ¼q1;4p!0 ½r ' ðr & vÞ c Aðr; tÞ ¼vVðr; tÞ;c2(28)where r and v are evaluated at the retarded time tr. For thepoint r ¼ (s, z),Tr ¼'1½Tðs2 þ z2 þ b2 ' T 2 Þ2ðT 2 ' z2 �ffiffiffiffiffi' z 4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2 (29)(T % ct and Tr % ctr ).9 This is for z 'T; as we approachthe horizon ðz ! 'TÞ, the retarded time goes to '1, andfor z 'T there is no solution with T Tr .The scalar potential is23q1zðs2 þ z2 þ b2 ' T 2 ffiffiffiffiffiqT'45hðT þ zÞ;4p!0 ðT 2 ' z2 Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2(30)and the vector potential is23q1Tðs2 þ z2 þ b2 ' T 2 ÞA¼4z ' ��ffiffiffiffi5hðT þ zÞ z :4p!0 c ðT 2 ' z2 Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2(31)The electric field is98 @A qb2 2z s ' ðs2 ' z2 þ b2 þ T 2 Þ z¼E ¼ 'rV 'hi3 2 hðT þ zÞp!0 :@t;4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2(32)which reduces to Eq. (21)—without the extra term—when t ¼ 0. Notice that the derivatives of the theta function contributenothing (we use an overbar to denote the potentials shorn of their h’s):23q1ðT þ zÞðs2 þ z2 þ b2 ' T 2 Þ""'dðT þ zÞðV þ cA z Þ ¼ 'dðT þ zÞ 24ðT þ zÞ ' ��ffiffiffiffi54p!0ðT ' z2 Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2"#q dðT þ zÞðs2 þ b2 Þ¼ 0:1' 2¼'4p!0 ðT ' zÞðs þ b2 ÞEvidently there is something wrong with the Li!enardWiechert potentials themselves; they too are missing acritical term. To fix them, we play the same game as before:truncate the hyperbolic motion. We might as well go straightto the limit, with the truncation receding to '1; we needthe potentials of a point charge moving at speed c. There aretwo candidates in the literature10 [which differ by a gaugetransformation, though both satisfy the Lorenz condition,@V @t ¼ 'c2 ðr & AÞ]:758Am. J. Phys., Vol. 82, No. 8, August 2014VI0 ¼ 0;A0I ¼ 'q shðz þ TÞ;2p!0 c s2 %qslndðz þ TÞ;2p!0b %qslndðz þ TÞ zA0II ¼2p!0 cbVII0 ¼ 'Joel Franklin and David J. Griffiths(33)(34)(35)758
%qs2 þ b 2lndðz þ TÞ;b24p!0 %(37)qs2 þ b2lndðz þ TÞ z :AII ¼4p!0 cb2It is easy to check, in either case, that we recover the correct“extra” term in the field [Eq. (21)]. However, we prefer VIIand AII because they preserve the Lorenz gauge.11The correct potentials for a point charge in hyperbolicmotion are thus12(in the second case b could actually be any constant with thedimensions of length, but we might as well use a parameterthat is already on the table).We also need the “connecting” potentials; our experiencewith the fields [going from Eq. (18) to Eq. (20)] suggests thefollowing ansatzVI ¼ 0;AI ¼ 'qshðz þ TÞ;2p!0 c ðs2 þ b2 ÞVII ¼ '(36)9823 % 222222q1zðs þ z þ b ' T Þs �ffiffiffiffiffiffi;hðTþzÞ'lnT'dðzþTÞV¼4522 b24p!0 ;: ðT ' z Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2A¼8 222232 2q1Tðs þ z þ b ' T Þs þb67z ' ��ffiffiffiffi5hðT þ zÞþ ln2 ' z2 Þ 4b24p!0 c ðT:4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ22%dðz þ TÞ9 ;(38) z :(39)How did the standard Li!enard-Wiechert construction missthe extra (delta function) terms? Was it perhaps in the derivation of the Li!enard-Wiechert potentials from the retardedpotentials?and we need qðr0 ; tr Þ, where (for t ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffi'ctr ¼ jr ' r0 j ¼ ðx ' x0 Þ2 þ ðy ' y0 Þ2 þ ðz ' z0 Þ2 :B. Retarded potentialsThusLet’s take a further step back, then, and examine theretarded potential13ð1qðr0 ; tr Þ 3 0Vðs; zÞ ¼d r:(40)r4p!0In this case qðr; tÞ ¼ qd3 r ��ffiffiffiffi %b2 þ ðctÞ2 z(41)(42) þ y0 y þ z0 zqðr0 ; tr Þ ¼ qd3 x0 x'(43)Because of the delta function, the denominator ðr ¼jr ' r0 jÞ in Eq. (40) comes outside the integral—with r0 ,now, at the retarded point (where the argument of the deltafunction vanishes). What remains is �ffiffiffiffiffiffi%000qðr ; tr Þ d r ¼ q dðx Þdðy Þd z ' b2 þ ðx ' x0 Þ2 þ ðy ' y0 Þ2 þ ðz ' z0 Þ2 dx0 dy0 dz0ð ��ffiffiffiffi%'(0¼ q d z ' b2 þ s2 þ ðz ' z0 Þ2 dz0 ¼ q d f ðz0 Þ dz0 % Q;03 0where s2 ¼ x2 þ y2 andf ðz0 Þ % z0 �ffiffiffiffib2 þ s2 þ ðz ' z0 Þ2 :(45)The argument of the delta function vanishes when z0 ¼ z0 ,given by f(z0) ¼ �ffiffiffiffiffiffiffi %b2 þ ðx ' x0 Þ2 þ ðy ' y0 Þ2 þ ðz ' z0 Þ2 z :Am. J. Phys., Vol. 82, No. 8, August 2014(44)z0 ��ffiffiffiffib2 þ s2 þ ðz ' z0 Þ2 ;(46)z0 ¼1 2ðs þ z2 þ b2 Þ:2z(47)orNote that z0 is non-negative, so there is no solution whenz 0. NowJoel Franklin and David J. Griffiths759
dfðz ' z0 ��ffiffiffiffi;¼1þdz0b2 þ s2 þ ðz ' z0 Þ2sof 0 ðz0 Þ ¼ 1 þðz ' z0 Þz2z2¼ ¼ 2;z0z0 s þ z2 þ b2(48)(49)'(d f ðz0 Þ ¼(50)Thus %s2 þ z2 þ b2Q¼qhðzÞ:2z2(51)and from Eq. (29) (with t ¼ s2 þ z2 þ b2 Þ2 ' ð2bzÞ2 ;r ¼ 'ctr ¼ 'Tr ¼2zThe retarded time for a point in the xy-plane, at time t, isgiven �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffifficðt ' tr Þ ¼ x2 þ y2 þ zðtr Þ2 ¼ x2 þ y2 þ b2 þ c2 t2r ;orc2 t2 ' 2c2 ttr þ c2 t2r ¼ x2 þ y2 þ b2 þ c2 t2r ;(52)(A2)so(53)tr ¼c2 t2 ' x2 ' y2 ' b2 t2 ' a2¼;2c2 t2t(A3)x2 þ y2 þ b2:c2(A4)wheresoqðs2 þ z2 þ b2 ��ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi hðzÞ;V¼4p!0z ðs2 þ z2 þ b2 Þ2 ' ð2bzÞ2APPENDIX A: RETARDED TIME FOR POINTS ONTHE xy-PLANE(A1)The retarded potential is1 Q;V¼4p!0 rACKNOWLEDGMENTThe authors thank Colin LaMont for introducing us to thisproblem.18and hence1dðz0 ' z0 Þjf 0 ðz0 Þj %s2 þ z2 þ b2dðz0 ' z0 Þ:¼2z2hyperbolic case?16 Zangwill17 offers a careful, step-by-stepderivation of the retarded potentials; one of those steps mustfail, but we have been unsuccessful in identifying the guiltyparty. And although it is easy to construct configurations forwhich the retarded potentials break down, we know of noother case for which the field formula [Eq. (2)] fails.(54)and we recover Eq. (30) (for t ¼ 0). Still no sign of the extraterm in Eq. (38); evidently the retarded potentials themselvesare incorrect, in this case.a2 %In Fig. 6, tr is plotted (as a function of t) for a ¼ 1. It is clearthat tr t for all positive t, but tr t for all negative t. Thelatter is no good, of course, but we do get an acceptable solution for all t ) 0. For t ¼ 0 the retarded time is (minus) infinity, regardless of the values of x and y.V. WHAT WENT WRONG?Straightforward application of the standard formulas forthe field [Eq. (2)], the Li!enard-Weichert potentials [Eq.(28)], and the retarded potential [Eq. (40)], yield incorrectresults (inconsistent with Maxwell’s equations) in the case ofa charged particle in hyperbolic motion—they all miss anessential delta-function contribution. How did this happen?Bondi and Gold14 write,“The failure of the method of retarded potentials togive the correct field is hardly surprising. Thesolution of the wave equation by retarded potentialsis valid only if the contributions due to distantregions fall off sufficiently rapidly with distance.”Fulton and Rohrlich15 write,“The Li!enard/Wiechert potentials are not valid in thepresent case at T þ z ¼ 0, because their derivationassumes that the source is not at infinity.”But where, exactly, do the standard derivations make theseassumptions, and how can they be generalized to cover the760Am. J. Phys., Vol. 82, No. 8, August 2014Fig. 6. Graph of the retarded time, as a function of t, for points in the xy-plane.Joel Franklin and David J. Griffiths760
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%Trd %r'¼ ðT ' Tr Þ ' z ' b2 þ Tr2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2cb þ Tr2zTr¼ T ' pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :(B3)b2 þ Tr2The vector from the (retarded) position of the charge tothe point r ¼ (s, z), is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%r ¼ s þ z ' b2 þ Tr2 z :(B1)The retarded time is given ��ffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi%22T ' Tr ¼ r ¼ s þ z ' b2 þ Tr2 :It pays to use Eq. (B2) to eliminate the �ffi2z b2 þ Tr2 ¼ ðs2 þ z2 þ b2 ' T 2 Þ þ 2TTr ;(B4)so(B2)d¼T'Squaring twice and solving the resulting quadratic yields Eq.(29).19Referring back to Eqs. (3) and (5), the denominator in Eq.(28) isso r&vAPPENDIX B: POTENTIALSðs2þz22z2 Tr:þ b2 ' T 2 Þ þ 2TTr(B5)Putting in Eq. (29), and ��ffiffiffiffiffiffiffiffiðT 2 ' z2 Þ ðs2 þ z2 þ b2 ' T 2 Þ2 þ 4b2 ðT 2 ' z2 Þ;d ¼ ��ffiffiffiffiT ðs2 þ z2 þ b2 ' T 2 Þ2 þ 4b2 ðT 2 ' z2 Þ ' zðs2 þ z2 þ b2 ' T 2 Þ(B6)23q 1q1zðs2 þ z2 þ b2 ' T 2 Þ¼V¼4T ' ��ffiffiffiffi54p!0 d 4p!0 ðT 2 ' z2 Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2(B7)[Eq. (30)].Meanwhile, the vector potential [Eq. (28)] is �ffip zVz¼V¼c½ðs2 þ z2 þ b2 ' T 2 Þ þ 2TTr 4p!0 dc2c b2 þ Tr2 %q2zTr z¼2224p!0 c T½ðs þ z þ b ' T 2 Þ þ 2TTr ' 2z2 Tr23q1ðs2 þ z2 þ b2 ' T 2 Þ¼4z ' T ��ffiffiffiffi5 z4p!0 c ðT 2 ' z2 Þ4b2 ðT 2 ' z2 Þ þ ðs2 þ z2 þ b2 ' T 2 Þ2A¼(B8)[Eq. (31)].APPENDIX C: RADIATIONFrom the potentials [Eqs. (38) and (39)], we obtain the fields:208qb2 Eðs; /; z; tÞ ¼hp!0 :ðz2 ' s2 ' b2 ' T 2 Þ z þ 2z sðz2 þ s2 þ b2 '8 qb2Bðs; /; z; tÞ ¼h: p!0 c761T 2 Þ2qsdðz þ TÞ;i3 2 hðz þ TÞ þ2 þ b22p!s0;' 4b2 ðz2 ' T 2 Þ2Tsðz2 þ s2 þ b2 ' T 2 Þ2 ' 4b2 ðz2 ' T 2 ÞAm. J. Phys., Vol. 82, No. 8, August 20149 9 qs1dðz þ TÞ /¼ ð r ( EÞ:i3 2 hðz þ TÞ '222p!0 c s þ bc;Joel Franklin and David J. Griffiths(C1)(C2)761
The power radiated is21% ,ð 1r&v 2 21'½E r ' ðr & EÞ2 sinh dh d/ :P ¼ limr!1 l 0 crc(C6)Using the relevant fields [Eq. (C1)], we find( )%2r þ Trcq2cq2:¼P ¼ limr!1r6p!0 b26p!0 b2(C7)Perhaps surprisingly, it is constant (independent of Tr, andhence the same for all points on the trajectory),22 and itagrees with the Li!enard formula23 (for collinear v and a)P¼q2c6 a2 :6p!0 c3(C8)a)Electronic mail: jfrankli@reed.eduElectronic mail: griffith@reed.edu1The entire xy-plane first “sees” the charge at time t ¼ 0. If this seems surprising, refer to Appendix A.2The fields of a point charge in hyperbolic motion were first considered byM. Born, “Die Theorie des starren Elektrons in der Kinematik desRelativit atsprinzips,” Ann. Phys. 30, 1–56 (1909). For the early history ofthe problem see W. Pauli, Theory of Relativity (reprint by Dover, NewYork, 1981), Section 32(c). For a comprehensive history see E. Eriksenand Ø. Grøn, “Electrodynamics of Hyperbolically Accelerated Charges. I.The Electromagnetic Field of a Charged Particle with HyperbolicMotion,” Ann. Phys. 286, 320–342 (2000). See also S. Lyle, UniformlyAccelerating Charged Particles: A Threat to the Equivalence Principle(Springer, Berlin, 2008).3See, for example, D. J. Griffiths, Introduction to Electrodynamics, 4th ed.(Pearson, Upper Saddle River, NJ, 2013), Eq. (10.72.)4This field was first obtained by G. A. Schott, Electromagnetic Radiation(Cambridge U.P., Cambridge, UK, 1912), pp. 63–69.5E. M. Purcell and D. J. Morin, Electricity and Magnetism, 3rd ed.(Cambridge U.P., Cambridge, UK, 2013), Section 5.6.6J. M. Aguirregabiria, A. Hern!andez, and M. Rivas, “d-function convergingsequences,” Am. J. Phys. 70, 180–185 (2002), Eq. (50). The fields of amassless point charge are considered also in J. D. Jackson, ClassicalElectrodynamics, 3rd ed. (Wiley, New York, 1999), Prob. 11.18, W.Thirring Classical Mathematical Physics: Dynamical Systems and FieldTheories (Springer-Verlag, New York, 1997), pp. 367–368 and M. V.Kozyulin and Z. K. Silagadze, “Light bending by a Coulomb field and theAichelburg-Sexl ultraboost,” Eur. J. Phys. 32, 1357–1365 (2011), Eq. (20).7H. Bondi and T. Gold, “The field of a uniformly accelerated charge, withspecial reference to the problem of gravitational acceleration,” Proc. R.Soc. London Ser. A 229, 416–424 (1955). D. G. Boulware, “Radiationfrom a uniformly accelerated charge,” Ann. Phys. 124, 169–188 (1980)obtained Eq. (21) using the truncated hyperbolic model. The latter wasalso explored by W. Thirring A Course in Mathematical Physics:Classical Field Theory, 2nd ed. (Springer-Verlag, New York, 1992), p. 78.See also Lyle, ref. 2, Section 15.9.8Reference 3, Eqs. (10.46) and (10.47).9See Appendix B for details of these calculations. Equation (29) reduces toEq. (9), of course, when t ¼ 0 and s ¼ x.10R. Jackiw, D. Kabat, and M. Ortiz, “Electromagnetic fields of a masslessparticle and the eikonal,” Phys. Lett. B 277, 148–152 (1992).11Potentials (30) and (31) satisfy the Lorenz gauge condition, as does Eq.(37), but Eq. (36) does not.12The delta-function terms in the potentials were first obtained by T. Fultonand F. Rohrlich, “Classical radiation from a uniformly acceleratedcharge,” Ann. Phys. 9, 499–517 (1960).13Reference 3, Eq. (10.19). Since all we’re doing is searching for a missingterm, we may as well concentrate on V, and set t ¼ 0.14Reference 7, quoted in Eriksen and Grøn, Ref. 2.15Reference 12, quoted in Eriksen and Grøn, Ref. 2.16Lyle (Ref. 2, page 216) thinks it “likely” that the extra terms could in factbe obtained at the level of the Li!enard-Wiechert potentials “if we wereb)Fig. 7. Radiation from the charge at time zero (for b ¼ 1), showing thespherical surface and the delta-fields at: (a) T ¼ 0, (b) T ¼ 1/2, (c) T ¼ 1.To calculate the power radiated by theffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiat a time trpcharge(when it is located at the point zðtr Þ ¼ b2 ' Tr2 ), we integrate the Poynting vectorS¼1ðE ( BÞl0(C3)over a sphere of radius r ¼ T ' Tr centered at z(tr), and takethe limit as r ! 1 with Tr ¼ ctr held constant. (That is, wetrack the energy as it flows outward at the speed of light;“radiation” is the portion that makes it “all the way to infinity.”) We need the fields, then, at later and later times, as thesphere expands. Now, the delta-function term is confined to theplane z ¼ –T, which recedes farther and farther to the left astime goes on (Fig. 7), and the expanding sphere never catchesup. Curiously, then, the delta-function term does not contributeto the power radiated by the charge at any (finite) point on itstrajectory. By the same token, the spherical surface is always inthe region where z þ T 0, so we can drop the theta functions.Now,11½E ( ð r ( EÞ ðE ( BÞ ¼l0l 0c1½E2 r ' ð r & EÞE ;¼l 0cS¼(C4)and on the surface of the sphere da ¼ r2 sinh dh d/ r , givingS & da ¼7621½E2 r2 ' ðr & EÞ2 sinh dh d/:l 0cAm. J. Phys., Vol. 82, No. 8, August 2014(C5)Joel Franklin and David J. Griffiths762
more careful about the step function,” but he offers no justification for thisconjecture.17A. Zangwill, Modern Electrodynamics (Cambridge U.P., Cambridge,2013), Sec. 20.3.18C. LaMont, “Relativistic direct interaction electrodynamics: Theory andcomputation,” Reed College senior thesis, 2011.19The sign of the radical is enforced by the condition T Tr .20Any reader with lingering doubts is invited to check that these fields satisfy all of Maxwell’s equations. Note the critical role of the delta functionsin Gauss’s law and the Ampère-Maxwell law.763Am. J. Phys., Vol. 82, No. 8, August 2014The factor ð1 ' r & v rcÞ accounts for the fact that the rate at which energyleaves a (moving) charge is not the same as the rate at which it (later)crosses a patch of area on the sphere. See Ref. 3, page 485.22The fact that a charged particle in hyperbolic motion radiates has interesting implications for the equivalence principle—in fact, it is this aspect ofthe problem that has attracted the attention of most of the authors citedhere. Incidentally, the particle experiences no radiation reaction force—see R. Peierls, Surprises in Theoretical Physics (Princeton U.P., Princeton,NJ, 1979, Chapter 8.)23Reference 3, Eq. (11.73).21Joel Franklin and David J. Griffiths763
Fig. 2. Field of a charged particle in hyperbolic motion at t¼0 (particle at z¼b). Fig. 3. Field lines for truncated hyperbolic motion (b¼1, a¼12 5). 756 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 756
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