The fields of a charged particle in hyperbolic motion Joel 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 exactly why the naive approach fails is not so clear. VC 2014 American Association of Physics Teachers. [http://dx.doi.org/10.1119/1.4875195]

1 I. INTRODUCTION u c^r v cr rv ; (4) ¼ À ¼ r ð À Þ In special relativity, a particle of mass m subject to a con- c2t stant force F undergoes “hyperbolic motion”: v r z^ ; (5) ¼ 2 2 b ctr 2 þð Þ z t b2 ct ; (1) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð Þ¼ þð Þ and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 where b mc =F. The particle flies in from infinity along bc 2 (say) the z-axis, comes to rest at z(0) b, and returns to infin- a ð Þ z^ : (6) ¼ ¼ 3 ity; its velocity approaches 6c asymptotically as t 6 2 2 ! 1 b ctr (Fig. 1). þð Þ Because information cannot travel faster than the speed of qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi light, the region below the main diagonal (z –ct) is igno- The retarded time tr is defined in general by rant of the particle’s existence—the particle¼ is “over the r c t tr ; (7) horizon.” For someone at the origin it first comes into view ¼ ð À Þ at t 0. If the particle is electrically charged, its fields are but for the moment we’ll assume t 0(sot is negative). Then ¼ r necessarily zero for all z < 0, at time t 0. But the electric ¼ 2 ¼ field for z > 0 is not zero, and as we shall see the field lines 2 2 2 2 1 ctr x z b ctr appear to terminate in mid-air at the xy-plane. This would ð Þ ¼ þ À þð Þ violate Gauss’s law; it cannot be true. Our task is to locate qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 x2 z2 2z b2 ct 2 b2 ct 2; (8) the error and fix it. ¼ þ À þð rÞ þ þð rÞ In Sec. II, we calculate the electric field of a charge q in qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi hyperbolic motion, at time t 0. A plot of the field lines and hence shows that they do not go continuously¼ to zero at the xy- plane. In Sec. III, we explore the case of “truncated” hyper- bolic motion (hyperbolic motion back to time t –t0, adjoined to constant velocity for earlier times). In this¼ case, the field lines make a sharp turn as they approach the xy-plane, and there is no violation of Gauss’s law. In Sec. IV, we work out the potentials for a charge in hyperbolic motion, finding once again that we must adjoin “by hand” a term inspired by the truncated case. In Sec. V, we ask how the naive calculations missed the extra term, and conclude with the puzzle unresolved. Appendices A and B supply some algebraic details, and Appendix C examines the radia- tion from a charge in hyperbolic motion; surprisingly, the “extra” terms do not contribute.

II. ELECTRIC FIELD OF A CHARGE IN HYPERBOLIC MOTION We begin by calculating the electric field at the point r (x, 0, z), with z > 0. According to the standard formula,3 ¼ q r 2 2 E r; t 3 c v u r u a ; (2) ð Þ¼4p0 r u ð À Þ þ Âð Â Þ ð Á Þ ÂÃ where

2 2 r x x^ z b ctr z^ ; (3) ¼ þ À þð Þ Fig. 1. Hyperbolic motion. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

755 Am. J. Phys. 82 (8), August 2014 http://aapt.org/ajp VC 2014 American Association of Physics Teachers 755 1 2 even though the pillbox encloses no charge. Something is ct x2 z2 b2 2zb 2: (9) r ¼À2z ðÞþ þ Àð Þ obviously amiss—we appear to have lost a crucial piece of qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi the field at z 0. Putting all this together and simplifying gives ¼ III. TRUNCATED HYPERBOLIC MOTION qb2 z2 x2 b2 z^ 2xz x^ ðÞ E x; 0; z À À þð Þ 3 : (10) Suppose the acceleration does not extend all the way back ð Þ¼p0 2 2 z2 x2 b2 2zb to t but begins at time t0 ab=c (for some a > 0), ð þ þ Þ Àð Þ when¼À the1 particle was at ¼À qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi That is for z > 0, of course; for z < 0 the field is zero. In 2 z t0 b 1 a ; (14) cylindrical coordinates s; /; z , then4 ð Þ¼ þ ð Þ and its velocityp wasffiffiffiffiffiffiffiffiffiffiffiffiffi qb2 z2 s2 b2 ^z 2z s ðÞ E s; /; z À À þ 3 h z ; ac ð Þ¼p0 ð Þ 2 2 v t0 z^ ; (15) z2 s2 b2 2zb ð Þ¼Àp1 a2 ð þ þ Þ Àð Þ þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (11) prior to t0 the velocityffiffiffiffiffiffiffiffiffiffiffiffiffi was constant. In other words, replace Eq. (1) with where h z is the step function (1 if z > 0, otherwise 0). This ð Þ field is plotted in Fig. 2; the field lines are circles, centered 1 on the s axis and passing through the instantaneous position b act t < t0 ab=c p1 a2 ð À Þð ¼À Þ of the charge. z t 8 þ (16) As required by Gauss’s law, E 0 for all z > 0 ð Þ¼> <> ffiffiffiffiffiffiffiffiffiffiffiffiffi2 2 (except at the point s 0, z b, whererÁ the¼ charge is located). b ct t t0 : ¼ ¼ þð Þ ð  Þ However, E is plainly not divergenceless at the xy-plane, > qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi where the field lines terminate in mid-air. Indeed, the field At time :>t 0, for all points outside a sphere of radius ¼ immediately to the right of the z 0 plane is r ct0 ab, centered at z(t0), the field is that of a charge ¼ moving¼À at¼ constant velocity—the “flattened” Heaviside field5 qb2 1 radiating from the place q would have reached, had it contin- 0 ^ (12) E s; /; þ 2 z; p 2 ð Þ¼Àp0 s2 b2 ued on its original flight plan b= 1 a : ð þ Þ ð þ Þ 2 ffiffiffiffiffiffiffiffiffiffiffiffiffi and the flux of through a cylindrical Gaussian “pillbox” of q 1 v=c R E E : (17) Àð2 Þ2 3=2 3 radius r, centered at the origin and straddling the plane, with ¼ 4p0 1 v=c sin h R infinitesimal thickness, is ½ Àð Þ Š The left edge of the sphere is at p1 a2 a b (which is qb2 r 1 q r2 þ À 2 always positive, but goes to zero as a ). Inside the E da 2 ps ds 2 2 ; ÀÁffiffiffiffiffiffiffiffiffiffiffiffiffi!1 Á ¼Àp0 0 s2 b2 ¼À0 r b sphere, where news of the acceleration has been received, ð ð ð þ Þ þ (13) the field is given by Eq. (11) (Fig. 3). The field lines

Fig. 2. Field of a charged particle in hyperbolic motion at t 0 (particle at ¼ z b). Fig. 3. Field lines for truncated hyperbolic motion (b 1, a 12=5). ¼ ¼ ¼

756 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 756 evidently join up in a thin layer at the surface of the sphere, sum of these fields that gives Eq. (20). The true field of a representing the brief interval during which the motion charge in hyperbolic motion is evidently7 switches from uniform to hyperbolic. As alpha increases (that is, as t recedes into the more dis- qb2 z2 s2 b2 ^z 2z s 0 ðÞ E s; /; z À À þ 3 h z tant past), the radius of the sphere increases, and its left sur- ð Þ¼p0 ð Þ face flattens out against the xy-plane. Meanwhile, the z2 s2 b2 2 2zb 2 “outside” field compresses into a disk perpendicular to the ð þ þ Þ Àð Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi motion, and squeezes also onto the xy-plane. The complete q s 2 2 d z ; field lines now execute a 908 turn at z 0, as required to res- þ 2p0 s b ð Þ ¼ þ cue Gauss’s law. Indeed, for a the constant velocity (21) portion of the field approaches that!1 of a point charge moving at speed c:6 and it does not look like Fig. 2, but rather Fig. 5. As a check, let’s calculate the divergence of . Writing q s E E E z E (in an obvious notation), we have E0 s; /; z 2 d z : (18) hh d ð Þ¼2p0 s ð Þ ¼ ð Þþ Ehh z Eh h z Eh h : (22) Using the same Gaussian pillbox as before, this field yields rÁ½ ð ފ ¼ ðrÁ Þ ð Þþ Á ½rð ފ

The first term gives q=0, for the point charge q at z b; as q r q for the second term, ¼ E0 da 2 2pr d z dz : (19) Á ¼ 2p0 r ð Þ ð Þ ¼ 0 ð ð @h h z z^ d z z^ ; (23) This is appropriate, of course—had the particle continued at r ð Þ¼@z ¼ ð Þ its original velocity (c), it would now be inside the box (at the origin). so Awkwardly, however, this is not what was needed to can- E h z rÁ½Šh ð Þ cel the flux from the hyperbolic part of the field [Eq. (13)]. 2 2 2 2 For that purpose the field on the xy-plane should have been q qb z s b ð À À Þ 3=2 d z ¼ 0 þ p0 ð Þ q s z2 s2 b2 2 2zb 2 E s; /; z 2 2 d z : (20) ð þ þ Þ Àð Þ ð Þ¼2p0 s b ð Þ 2 hi þ q qb 1 2 d z : (24) It must be that the “connecting” field in the spherical shell ¼ 0 À p0 s2 b2 ð Þ ð þ Þ (the field produced during the transition from uniform to Meanwhile, hyperbolic motion), which (in the limit) coincides with the xy- plane and which we have ignored, accounts for the difference, q s as suggested in Fig. 4. The net field in the xy-plane consists of Ed 2 2 d z rÁ ¼ 2p0 rÁ s b ð Þ two parts: the field E0 due to the portion of the motion at con- þ stant velocity, given (in the limit a ) by Eq. (18), and the q 1 @ s2 !1 connecting field that joins it to the hyperbolic part. It is the 2 2 d z (25) ¼ 2p0 s @s s b ð Þ þ so

Fig. 4. Truncated hyperbolic motion for large a, showing the “connecting” field. Fig. 5. Field of a particle in hyperbolic motion (corrected).

757 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 757 q b2 q 1 v Ed d z : (26) V r; t ; A r; t 2 V r; t ; rÁ ¼  2 2 2 ð Þ ð Þ¼4p0 r r v =c ð Þ¼c ð Þ p 0 s b ½ Àð Á Þ Š ð þ Þ (28) This is just right to cancel the extra term in Ehh z , and Gauss’s law is sustained: rÁ½Šð Þ where r and v are evaluated at the retarded time tr. For the q point r (s, z), E : (27) ¼ rÁ ¼ 0 1 T À T s2 z2 b2 T2 r ¼ 2 T2 z2 ½ ð þ þ À Þ IV. POTENTIAL FORMULATION ð À Þ 2 A. Lienard-Wiechert potentials z 4b2 T2 z2 s2 z2 b2 T2 (29) À ð À Þþð þ þ À Þ Š The truncated hyperbolic problem guided us to the “extra” qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 (delta-function) term in Eq. (21),butitdoesnotexplainhowwe (T ct and Tr ctr). This is for z > T; as we approach missed that term in the first place. Did it perhaps get lost in going the horizon z  T , the retarded timeÀ goes to , and ð !À Þ À1 from the potentials to the fields? Let’s work out the Lienard- for z < T there is no solution with T > Tr. Wiechert potentials,8 and calculate the field more carefully: The scalarÀ potential is

q 1 z s2 z2 b2 T2 V 2 2 T ð þ þ À Þ h T z ; (30) ¼ 4p0 T z À 2 ð þ Þ ð À Þ 2 4b2 T2 z2 s2 z2 b2 T2 3 ð À Þþð þ þ À Þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 and the vector potential is

q 1 T s2 z2 b2 T2 A 2 2 z ð þ þ À Þ h T z ^z: (31) ¼ 4p0c T z À 2 ð þ Þ ð À Þ 2 4b2 T2 z2 s2 z2 b2 T2 3 ð À Þþð þ þ À Þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 The electric field is

@A qb2 2z s s2 z2 b2 T2 z^ (32) E V Àð À þ þ Þ 3=2 h T z ¼Àr À @t ¼ p0 ð þ Þ 8 4b2 T2 z2 s2 z2 b2 T2 2 9 < ð À Þþð þ þ À Þ = hi : ; which reduces to Eq. (21)—without the extra term—when t 0. Notice that the derivatives of the theta function contribute nothing (we use an overbar to denote the potentials shorn of their¼ h’s):

2 2 2 2 " " q 1 T z s z b T d T z V cAz d T z 2 2 T z ð þ Þð þ þ À Þ À ð þ Þð þ Þ¼À4p0 ð þ Þ T z ð þ ÞÀ 2 ð À Þ 2 4b2 T2 z2 s2 z2 b2 T2 3 ð À Þþð þ þ À Þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 q d T z s2 b2 ð þ Þ 1 ð 2 þ 2Þ 0: (33) ¼À4p0 T z À s b ¼ ð À Þ "#ð þ Þ

Evidently there is something wrong with the Lienard- q s VI0 0; AI0 2 h z T ; (34) Wiechert potentials themselves; they too are missing a ¼ ¼À2p0c s ð þ Þ critical term. To fix them, we play the same game as before: truncate the hyperbolic motion. We might as well go straight to the limit, with the truncation receding to ; we need q s À1 VII0 ln d z T ; the potentials of a point charge moving at speed c. There are ¼À2p0 b ð þ Þ two candidates in the literature10 [which differ by a gauge  (35) q s transformation, though both satisfy the Lorenz condition, AII0 ln d z T ^z @V=@t c2 A ]: ¼ 2p0c b ð þ Þ ¼À ðrÁ Þ 

758 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 758 (in the second case b could actually be any constant with the q s2 b2 VII ln þ2 d z T ; dimensions of length, but we might as well use a parameter ¼À4p0 b ð þ Þ that is already on the table).  (37) q s2 b2 We also need the “connecting” potentials; our experience AII ln þ2 d z T z^ : with the fields [going from Eq. (18) to Eq. (20)] suggests the ¼ 4p0c b ð þ Þ following ansatz It is easy to check, in either case, that we recover the correct “extra” term in the field [Eq. (21)]. However, we prefer VII and A because they preserve the Lorenz gauge.11 q s II 0 (36) The correct potentials for a point charge in hyperbolic VI ; AI 2 2 h z T ; 12 ¼ ¼À2p0c s b ð þ Þ motion are thus ð þ Þ

q 1 z s2 z2 b2 T2 s2 b2 V T T z ln z T ; (38) 8 2 2 2 ð þ þ À Þ 3h þ2 d 9 ¼ 4p0 T z À 2 2 2 2 2 2 2 2 ð þ ÞÀ b ð þ Þ <> ð À Þ 4b T z s z b T  => 6 ð À Þþð þ þ À Þ 7 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 :> ;> q 1 T s2 z2 b2 T2 s2 b2 A z T z ln z T z^ : (39) 8 2 2 2 ð þ þ À Þ 3h þ2 d 9 ¼ 4p0c T z À 2 2 2 2 2 2 2 2 ð þ Þþ b ð þ Þ <> ð À Þ 4b T z s z b T  => 6 ð À Þþð þ þ À Þ 7 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 :> ;>

How did the standard Lienard-Wiechert construction miss and we need q r0; tr , where (for t 0) the extra (delta function) terms? Was it perhaps in the deri- ð Þ ¼ vation of the Lienard-Wiechert potentials from the retarded 2 2 2 ct r r0 x x y y z z : potentials? À r ¼j À j¼ ð À 0Þ þð À 0Þ þð À 0Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(42) B. Retarded potentials Thus

Let’s take a further step back, then, and examine the 3 q r0; t qd x0 x^ y0 y^ z0 z^ retarded potential13 ð rÞ¼ þ þ  2 2 2 1 q r0; tr 3 b2 x x y y z z ^z : V s; z ð Þ d r0: (40) À þð À 0Þ þð À 0Þ þð À 0Þ ð Þ¼4p0 r qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ð (43) In this case Because of the delta function, the denominator r ð ¼ r r0 in Eq. (40) comes outside the integral—with r0, now,j À atjÞ the retarded point (where the argument of the delta- q r; t qd3 r b2 ct 2 z^ (41) ð Þ¼ À þð Þ function vanishes). What remains is qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

3 2 2 2 2 q r0; t d r0 q d x0 d y0 d z0 b x x y y z z dx0dy0dz0 ð rÞ ¼ ð Þ ð Þ À þð À 0Þ þð À 0Þ þð À 0Þ ð ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 q d z0 b s z z dz0 q d f z0 dz0 Q; (44) ¼ À þ þð À 0Þ ¼ ð Þ  ð ð qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ÀÁ

2 2 2 where s x y and 2 2 2 ¼ þ z0 b s z z0 ; (46) ¼ þ þð À Þ or qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 2 f z0 z0 b s z z0 : (45) 1 2 2 2 ð Þ À þ þð À Þ z0 s z b : (47) qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2z ðÞþ þ

The argument of the delta function vanishes when z0 z0, Note that z is non-negative, so there is no solution when ¼ 0 given by f(z0) 0: z < 0. Now ¼

759 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 759 16 17 df z z0 hyperbolic case? Zangwill offers a careful, step-by-step 1 ð À Þ ; (48) dz ¼ þ 2 derivation of the retarded potentials; one of those steps must 0 b2 s2 z z þ þð À 0Þ fail, but we have been unsuccessful in identifying the guilty qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi party. And although it is easy to construct configurations for so which the retarded potentials break down, we know of no other case for which the field formula [Eq. (2)] fails. 2 z z0 z 2z f 0 z0 1 ð À Þ 2 2 2 ; (49) ð Þ¼ þ z0 ¼ z0 ¼ s z b ACKNOWLEDGMENT þ þ and hence The authors thank Colin LaMont for introducing us to this problem.18 1 d f z0 d z0 z0 ð Þ ¼ f z ð À Þ 0 0 APPENDIX A: RETARDED TIME FOR POINTS ON ÀÁj ð Þj s2 z2 b2 THE xy-PLANE þ þ d z0 z0 : (50) 2 2 ¼ z ð À Þ The retarded time for a point in the xy-plane, at time t, is Thus given by

2 2 2 2 2 2 2 2 2 2 2 s z b c t tr x y z tr x y b c tr ; Q q þ 2þ h z : (51) ð À Þ¼ þ þ ð Þ ¼ þ þ þ ¼ 2z ð Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(A1) The retarded potential is or

1 Q 2 2 2 2 2 2 2 2 2 2 V ; (52) c t 2c ttr c tr x y b c tr ; (A2) ¼ 4p0 r À þ ¼ þ þ þ so and from Eq. (29) (with t 0) ¼ c2t2 x2 y2 b2 t2 a2 1 2 2 2 2 2 tr À À À À ; (A3) r ctr Tr s z b 2bz ; (53) 2c2t 2t ¼À ¼À ¼ 2z ð þ þ Þ Àð Þ ¼ ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi so where 2 2 2 2 2 2 x y b q s z b a2 þ þ : (A4) V ð þ þ Þ h z ; (54)  c2 ¼ 4  2 2 ð Þ p 0 z s2 z2 b2 2bz ð þ þ Þ Àð Þ In Fig. 6, t is plotted (as a function of t) for a 1. It is clear qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r ¼ and we recover Eq. (30) (for t 0). Still no sign of the extra that tr < t for all positive t, but tr > t for all negative t. The term in Eq. (38); evidently the¼ retarded potentials themselves latter is no good, of course, but we do get an acceptable solu- are incorrect, in this case. tion for all t 0. For t 0 the retarded time is (minus) infin- ity, regardless of the values¼ of x and y.

V. WHAT WENT WRONG? Straightforward application of the standard formulas for the field [Eq. (2)], the Lienard-Weichert potentials [Eq. (28)], and the retarded potential [Eq. (40)], yield incorrect results (inconsistent with Maxwell’s equations) in the case of a charged particle in hyperbolic motion—they all miss an essential delta-function contribution. How did this happen? Bondi and Gold14 write,

“The failure of the method of retarded potentials to give the correct field is hardly surprising. The solution of the wave equation by retarded potentials is valid only if the contributions due to distant regions fall off sufficiently rapidly with distance.” Fulton and Rohrlich15 write, “The Lienard/Wiechert potentials are not valid in the present case at T z 0, because their derivation assumes that the sourceþ ¼ is not at infinity.” But where, exactly, do the standard derivations make these assumptions, and how can they be generalized to cover the Fig. 6. Graph of the retarded time, as a function of t,forpointsinthexy-plane.

760 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 760 APPENDIX B: POTENTIALS r v 2 2 Tr d r Á T Tr z b Tr The vector from the (retarded) position of the charge to  À c ¼ð À ÞÀ À þ b2 T2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ r the point r (s, z), is zT ¼ T r : pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(B3) 2 2 ¼ À b Tr 2 2 þ r s z b Tr z^ : (B1) ¼ þ À þ It pays to usep Eq.ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi(B2) to eliminate the radical: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi The retarded time is given by 2z b2 T2 s2 z2 b2 T2 2TT ; (B4) þ r ¼ð þ þ À Þþ r 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi T T r s2 z b2 T2 : (B2) so À r ¼ ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiþ À þ r qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2z Tr Squaring twice and solving the resulting quadratic yields Eq. d T : (B5) ¼ À s2 z2 b2 T2 2TT (29).19 ð þ þ À Þþ r Referring back to Eqs. (3) and (5), the denominator in Eq. Putting in Eq. (29), and simplifying, (28) is

T2 z2 s2 z2 b2 T2 2 4b2 T2 z2 d ð À Þ ð þ þ À Þ þ ð À Þ ; (B6) ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 T s2 z2 b2 T2 4b2 T2 z2 z s2 z2 b2 T2 ð þ þ À Þ þ ð À Þ À ð þ þ À Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi so q 1 q 1 z s2 z2 b2 T2 V 2 2 T ð þ þ À Þ (B7) ¼ 4p0 d ¼ 4p0 T z À 2 ð À Þ 2 4b2 T2 z2 s2 z2 b2 T2 3 ð À Þþð þ þ À Þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 [Eq. (30)]. Meanwhile, the vector potential [Eq. (28)] is

v Tr 2zTr q A 2 V V z^ 2 2 2 2 z^ ¼ c ¼ c b2 T2 ¼ c s z b T 2TTr 4p0d þ r ½ð þ þ À Þþ Š  2 q pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi zTr 2 2 2 2 2 z^ ¼ 4p0c T s z b T 2TT 2z T (B8) ½ð þ þ À Þþ rŠÀ r q 1 s2 z2 b2 T2 2 2 z T ð þ þ À Þ ^z ¼ 4p0c T z À 2 ð À Þ 2 4b2 T2 z2 s2 z2 b2 T2 3 ð À Þþð þ þ À Þ 4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5 [Eq. (31)].

APPENDIX C: RADIATION From the potentials [Eqs. (38) and (39)], we obtain the fields:20

qb2 z2 s2 b2 T2 z^ 2z s q s E s; ; z; t ðÞ z T z T ; / À À À þ 3=2 h 2 2 d (C1) ð Þ¼p0 ð þ Þþ 2p0 s b ð þ Þ 8 z2 s2 b2 T2 2 4b2 z2 T2 9 þ < ð þ þ À Þ À ð À Þ = hi : ;

qb2 2Ts q s 1 /^ ^ B s; /; z; t 3=2 h z T 2 2 d z T r E : ð Þ¼ p0c ð þ ÞÀ 2p0c s b ð þ Þ ¼ c ð Â Þ 8 z2 s2 b2 T2 2 4b2 z2 T2 þ 9 < ð þ þ À Þ À ð À Þ = hi (C2) : ;

761 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 761 The power radiated is21 1 r v P lim 1 Á E2r2 r E 2 sinh dh d/ : ¼ r l c À rc ½ Àð Á Þ Š !10 ð (C6)

Using the relevant fields [Eq. (C1)], we find

2 2 2 r Tr cq cq P lim þ 2 2 : (C7) ¼ r r 6p0b ¼ 6p0b !1()

Perhaps surprisingly, it is constant (independent of Tr, and hence the same for all points on the trajectory),22 and it agrees with the Lienard formula23 (for collinear v and a) 2 q 6 2 P 3 c a : (C8) ¼ 6p0c

a)Electronic mail: [email protected] b)Electronic mail: griffi[email protected] 1The entire xy-plane first “sees” the charge at time t 0. If this seems sur- prising, refer to Appendix A. ¼ 2The fields of a point charge in hyperbolic motion were first considered by M. Born, “Die Theorie des starren Elektrons in der Kinematik des Relativit€atsprinzips,” Ann. Phys. 30, 1–56 (1909). For the early history of the problem see W. Pauli, Theory of Relativity (reprint by Dover, New York, 1981), Section 32(c). For a comprehensive history see E. Eriksen and Ø. Grøn, “Electrodynamics of Hyperbolically Accelerated Charges. I. The Electromagnetic Field of a Charged Particle with Hyperbolic Motion,” Ann. Phys. 286, 320–342 (2000). See also S. Lyle, Uniformly Accelerating Charged Particles: A Threat to the Equivalence Principle (Springer, Berlin, 2008). Fig. 7. Radiation from the charge at time zero (for b 1), showing the 3See, for example, D. J. Griffiths, Introduction to Electrodynamics, 4th ed. spherical surface and the delta-fields at: (a) T 0, (b) T ¼1/2, (c) T 1. ¼ ¼ ¼ (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. To calculate the power radiated by the charge at a time tr 5 (when it is located at the point z t b2 T2), we inte- E. M. Purcell and D. J. Morin, Electricity and Magnetism, 3rd ed. ð rÞ¼ À r (Cambridge U.P., Cambridge, UK, 2013), Section 5.6. grate the Poynting vector 6 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi J. M. Aguirregabiria, A. Hernandez, and M. Rivas, “d-function converging 1 sequences,” Am. J. Phys. 70, 180–185 (2002), Eq. (50). The fields of a S E B (C3) massless point charge are considered also in J. D. Jackson, Classical ¼ l 0 ð Â Þ Electrodynamics, 3rd ed. (Wiley, New York, 1999), Prob. 11.18, W. Thirring Classical Mathematical Physics: Dynamical Systems and Field over a sphere of radius r T Tr centered at z(tr), and take Theories (Springer-Verlag, New York, 1997), pp. 367–368 and M. V. the limit as r with ¼T Àct held constant. (That is, we !1 r ¼ r Kozyulin and Z. K. Silagadze, “Light bending by a Coulomb field and the track the energy as it flows outward at the speed of light; Aichelburg-Sexl ultraboost,” Eur. J. Phys. 32, 1357–1365 (2011), Eq. (20). “radiation” is the portion that makes it “all the way to infin- 7H. Bondi and T. Gold, “The field of a uniformly accelerated charge, with ity.”) We need the fields, then, at later and later times, as the special reference to the problem of gravitational acceleration,” Proc. R. sphere expands. Now, the delta-function term is confined to the Soc. London Ser. A 229, 416–424 (1955). D. G. Boulware, “Radiation from a uniformly accelerated charge,” Ann. Phys. , 169–188 (1980) plane z –T,whichrecedesfartherandfarthertotheleftas 124 ¼ obtained Eq. (21) using the truncated hyperbolic model. The latter was time goes on (Fig. 7), and the expanding sphere never catches also explored by W. Thirring A Course in Mathematical Physics: up. Curiously, then, the delta-function term does not contribute Classical Field Theory, 2nd ed. (Springer-Verlag, New York, 1992), p. 78. to the power radiated by the charge at any (finite) point on its See also Lyle, ref. 2, Section 15.9. trajectory. By the same token, thesphericalsurfaceisalwaysin 8Reference 3, Eqs. (10.46) and (10.47). 9 the region where z T > 0, so we can drop the theta functions. See Appendix B for details of these calculations. Equation (29) reduces to Now, þ Eq. (9), of course, when t 0 and s x. 10R. Jackiw, D. Kabat, and¼ M. Ortiz,¼ “Electromagnetic fields of a massless particle and the eikonal,” Phys. Lett. B 277, 148–152 (1992). 1 1 11 S E B E ^r E Potentials (30) and (31) satisfy the Lorenz gauge condition, as does Eq. ¼ l 0 ð  ޼l 0c ½ Âð  ފ (37), but Eq. (36) does not. 12 1 The delta-function terms in the potentials were first obtained by T. Fulton E2^r ^r E E ; (C4) and F. Rohrlich, “Classical radiation from a uniformly accelerated ¼ l 0c ½ Àð Á Þ Š charge,” Ann. Phys. 9, 499–517 (1960). 13Reference 3, Eq. (10.19). Since all we’re doing is searching for a missing and on the surface of the sphere da r2 sinh dh d/ ^r, giving term, we may as well concentrate on V, and set t 0. ¼ 14Reference 7, quoted in Eriksen and Grøn, Ref. 2.¼ 15Reference 12, quoted in Eriksen and Grøn, Ref. 2. 1 2 S da E2r2 r E sinh dh d/: (C5) 16Lyle (Ref. 2, page 216) thinks it “likely” that the extra terms could in fact Á ¼ l 0c ½ Àð Á Þ Š be obtained at the level of the Lienard-Wiechert potentials “if we were

762 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 762 more careful about the step function,” but he offers no justification for this 21The factor 1 r v=rc accounts for the fact that the rate at which energy conjecture. leaves a (moving)ð À Á chargeÞ is not the same as the rate at which it (later) 17A. Zangwill, Modern Electrodynamics (Cambridge U.P., Cambridge, crosses a patch of area on the sphere. See Ref. 3, page 485. 2013), Sec. 20.3. 22The fact that a charged particle in hyperbolic motion radiates has interest- 18C. LaMont, “Relativistic direct interaction electrodynamics: Theory and ing implications for the equivalence principle—in fact, it is this aspect of computation,” Reed College senior thesis, 2011. the problem that has attracted the attention of most of the authors cited 19 The sign of the radical is enforced by the condition T > Tr. here. Incidentally, the particle experiences no radiation reaction force— 20Any reader with lingering doubts is invited to check that these fields sat- see R. Peierls, Surprises in Theoretical Physics (Princeton U.P., Princeton, isfy all of Maxwell’s equations. Note the critical role of the delta functions NJ, 1979, Chapter 8.) in Gauss’s law and the Ampe`re-Maxwell law. 23Reference 3, Eq. (11.73).

763 Am. J. Phys., Vol. 82, No. 8, August 2014 Joel Franklin and David J. Griffiths 763