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The Fields of a Charged Particle in Hyperbolic Motion Joel Franklina) and David J

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The Fields of a Charged Particle in Hyperbolic Motion Joel Franklina) and David J 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 Âà 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 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 Þ ¼ ðrÁ Þ ð Þþ Á ½rð Þ 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 rÁ½h ð Þ cel the flux from the hyperbolic part of the field [Eq.
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