A discontinuity in the electromagnetic field of a uniformly accelerated charge
aa r X i v : . [ phy s i c s . c l a ss - ph ] J un A discontinuity in the electromagnetic field of auniformly accelerated charge
Ashok K. Singal
Astronomy and Astrophysics Division, Physical Research Laboratory, Navrangpura,Ahmedabad - 380 009, IndiaE-mail: [email protected]
30 June 2020
Abstract.
The electric field of a uniformly accelerated charge shows a plane ofdiscontinuity, where the field extending only on one side of the plane, terminatesabruptly on the plane with a finite value. This indicates a non-zero divergence ofthe electric field in a source-free region, implying a violation of Gauss law. In orderto make the field compliant with Maxwell’s equations everywhere, an additional fieldcomponent, proportional to a δ -function at the plane of discontinuity, is required. Sucha “ δ -field” might be the electromagnetic field of the charge, moving with a uniformvelocity approaching c , the speed of light, prior to the imposition of acceleration atinfinity. However, some attempts to derive this δ -field for such a case, have not beenentirely successful. Some of the claims of the derivation involve elaborate calculationswith some not-so-obvious mathematical approximations. Since the result to be derivedis already known from the constraint of its compliance with Maxwell’s equations, andthe derivation involves the familiar text-book expressions for the field of a uniformlymoving charge, one would expect an easy, simple approach, to lead to the correctresult. Here, starting from the electromagnetic field of a uniformly accelerated chargein the instantaneous rest frame, in terms of the position and motion of the chargeat the retarded time, we derive this δ -field, consistent with Maxwell’s equations, in afairly simple manner. This is followed by a calculation of the energy in the δ -field, in ananalytical manner without making any approximation, where we show that this energyis exactly the one that would be lost by the charge because of the radiation reactionon the charge, proportional to its rate of change of acceleration, that was imposed onit at a distant past. discontinuity in the electromagnetic field of a uniformly accelerated charge
1. Introduction
Electromagnetic field of a charge undergoing a uniform proper acceleration was derivedfirst by Born [1], and later by many more authors [3, 2, 4]. From Born’s solution, Paulinoticed that at a specific instant, when the charge was instantly stationary, the magneticfield was zero everywhere and from that he concluded that there can be no radiationfrom a uniformly accelerated charge [5].Bondi and Gold [6] first drew attention to an anomaly apparent in Born’s solutionthat the electric field component along the direction of motion terminates with a non-zero value at a plane normal to the direction of motion; the discontinuity of electricfield implying a breakdown of one of Maxwell’s equations, viz. Gauss law. Such adiscontinuity, would normally be accompanying a surface charge density at the planeof discontinuity [7]. However, as no such charge density is pre-specified in the presentcase, one could be led to the conclusion that the Maxwell’s equations are incompatiblewith the existence of a single charge, uniformly accelerated at all times.Bondi and Gold [6] proposed that a consistency with Maxwell’s equations could berestored if owing to some process there were a certain δ -function field at that planeof discontinuity. One of the scenario suggested for this purpose was that prior tosome instant in distant past, say t = − τ , the charge moved with a uniform velocityand after t = − τ , with a constant acceleration. Then in the limit τ → ∞ , one mayobtain the required “ δ -field.” Though they themselves [6] did not further pursue thisline of thinking, but afterwards in literature this seems to have become the standardexplanation [8, 9, 10, 11] for the δ -field, required to make total field conform to Gausslaw. However, one of the exasperating aspect of this is that a simple derivation of the δ -field from the electromagnetic field of a charge that moved with a uniform velocitybefore t = − τ with τ → ∞ in the limit, has not been always successful. Since the resultto be obtained is already known from the constraint of its compliance with Maxwell’sequations, and the derivation involves the familiar, text-book expressions for the field ofa uniformly moving charge, one would expect a simple, an almost ingenuous approach,to lead to a correct result. This two stage motion has been termed ‘truncated hyperbolicmotion’ and a recent attempt to derive the δ -field from the truncated motion, using asimple method, remained unsuccessful [12, 13]. The δ -field was derived by Cross [14],but the derivation involved a rather complicated-looking evaluation of asymptotic termsin the Li´enard-Wiechert potentials. Continuous electric field lines for truncated motionwere found recently, and the electric field δ -function derived in the limit τ → ∞ [15].Our derivation here arrives at similar results by a different method where we shallcompare the two field expressions in terms of motion at retarded time, and show thatway why the two fields differ in their transverse components; in the case of the uniformlyaccelerated charge, the transverse component is zero as the acceleration field gets neatlycancelled by the transverse component of the velocity field in the instantaneous restframe of the charge. discontinuity in the electromagnetic field of a uniformly accelerated charge δ -field. For this,we shall first derive the electromagnetic field of a uniformly accelerated charge inthe instantaneous rest frame, in terms of the position and motion of the charge atthe retarded time. From that, using an algebraic transformation, the field in termsof “real-time” position of the charge will be arrived at. This procedure allows us abetter understanding of the interrelation between the field of a uniformly acceleratedcharge and that of a uniformly moving charge, especially in the region of their commonboundary. We may state that a computation of the total electromagnetic energy in thefield of a uniformly accelerated charge and its comparison with the total energy in theelectromagnetic field of a charge moving with a uniform velocity, but with a magnitudeequal to the “present speed” of the uniformly accelerated charge, has shown that inboth cases the two quantities are exactly the same [16, 17]. Similar is the conclusionarrived at for the net Poynting flux through a given spherical surface centered on theretarded time position of the charge, from a comparison between the two cases [17].Here we will not only derive the δ -field, arising from the uniform motion of thecharge before t = − τ , with v → c for τ → ∞ , but shall follow it with a computation ofthe total energy in the field, in an exact analytical manner, showing that the energy in δ -field is the one lost by the moving charge due to radiation reaction at time t = − τ inthe limit τ → ∞ , when there was a rate of change of acceleration of the charge.
2. Electromagnetic field of a charge undergoing a uniform properacceleration
From a uniformly accelerated motion we mean a motion with a constant properacceleration. We may take it to be a one-dimensional motion since an appropriateLorentz transformation can always be employed so as to reduce the velocity componentnormal to the acceleration vector to zero.Electromagnetic field of a charge e , moving with v k ˙ v , can be written for any giventime t as [18, 19, 20, 21] E = " e ( n − v /c ) γ r (1 − n · v /c ) + e n × ( n × ˙ v ) rc (1 − n · v /c ) t r (1)where quantities within the square brackets are to be evaluated at the correspondingretarded time t r = t − r/c . The acceleration field (the second term within the squarebrackets), transverse to n and falling with distance as 1 /r , is usually assumed to beresponsible for radiation from a charge.Using the vector identity v = n ( v . n ) − n × { n × v } , we can rewrite the velocityfield (first term within the square brackets), in terms of radial (along n ) and transversecomponents (normal to n ) as [16, 17] " e ( n − v /c ) γ r (1 − n · v /c ) t r = " e n γ r (1 − n · v /c ) + e n × ( n × γ v ) γ r c (1 − n · v /c ) t r . (2)This is also the expression for the electric field of a charge moving with a uniformvelocity v . Thus the total electric field of a uniformly moving charge or the velocity discontinuity in the electromagnetic field of a uniformly accelerated charge a = γ ˙ v , the “ present” velocity v o at time t is related to its value v r at the retardedtime t r = t − r/c as γ v = γ r v r + ( t − t r ) a = " γ v + rγ ˙ v c t r , (3)where γ , γ r are the corresponding Lorentz factors.Substituting Eq. (2) in Eq. (1), and utilizing Eq. (3), we find that the net electricfield of a uniformly accelerated charge is E = " e n γ r (1 − n · v /c ) + e n × ( n × γ v /c ) γ r (1 − n · v /c ) t r . (4)The magnetic field can be computed from the electric field [18, 19, 20, 21], which in thecase of a uniformly accelerated charge, turns out from Eq. (4), to be B = [ n ] × E = " − e n × γ v /cγ r (1 − n · v /c ) t r . (5)In the instantaneous rest-frame, where, by definition, the present velocity v = 0,we have then a simple expression for the total electric field of a uniformly acceleratedcharge as E = " e n γ r (1 − n · v /c ) t r , (6)where there is only a radial electric field with respect to the charge position at theretarded time. There is of course, a nil magnetic field, B = 0, in the instantaneousrest-frame.
3. Electromagnetic field in the instantaneous rest frame of a uniformlyaccelerated charge in terms of its “present” position
As is well known, the only non-trivial case of uniform acceleration is that of a hyperbolicmotion [22, 23, 24, 25]. We assume that a ≡ γ ˙ v is the uniform acceleration along the z-axis. Let the charge initially moving along the − z direction from z = ∞ at time t = −∞ ,is getting constantly decelerated till, say, at time t = 0 it comes to rest momentarily at z = z , and then onwards moves with an increasing velocity along the + z direction. Themotion of the charge for t > a , starting from z at t = 0 [22, 23, 24, 25, 26, 27]. Due to thecylindrical symmetry of the system, it is convenient to employ cylindrical coordinates( ρ, φ, z ). Without any loss of generality, we can choose the origin of the coordinatesystem, z = 0, at z − c /a . discontinuity in the electromagnetic field of a uniformly accelerated charge Figure 1.
A charge, moving with a uniform proper acceleration, a = γ ˙ v , alongthe z-axis, and coming from z = ∞ at t = −∞ with an initial velocity along − z direction, comes to rest momentarily at z = z at time t = 0. We want to determinethe electromagnetic field at time t = 0, at a point P ( ρ, z ), which is at a distance r from the charge position z r at the corresponding retarded time t r = − r/c , when it wasmoving with a velocity v r along − z direction. Let z t be the position of the charge at a time t , where it has the velocity v t = at/γ t ,with γ t = [1 + ( at/c ) ] / = [1 + ( ct/z ) ] / . Then we have z t = z + Z t at [1 + ( at/c ) ] / d t = z + c a (cid:20)n at/c ) o / − (cid:21) = h z + c t i / = z γ t . (7)The choice of origin, so that z = c /a , makes the relation between charge position andtime a rather convenient one.We can express the electromagnetic field of a uniformly accelerated charge, notnecessarily in terms of motion of the charge at retarded time as in Eqs. (4), (5) or (6),instead in terms of the “real-time” value of the charge position. Here we shall, startingfrom Eq. (6), determine the electromagnetic field in the instantaneous rest frame of theuniformly accelerated charge, where v = 0 at t = 0.If P ( ρ, z ) be the field point, where we wish to determine the electromagnetic fieldat a time t = 0, in terms of the “real-time” position of the charge z when its velocity v = 0, all we require is the quantity rγ r (1 − v r cos θ/c ) in Eq. (6) to be expressed interms of ρ , z and z . (For the field expression in a more general case, i.e. at t = 0 whenthe uniformly accelerated charge may be moving with some finite velocity, we beginfrom Eqs. (4) and (5), see the Appendix.)Let z r = z γ r be the position of the charge at the corresponding retarded time t r = − r/c , where it had the velocity v r = − arcγ r = − crz γ r (8) discontinuity in the electromagnetic field of a uniformly accelerated charge − z direction (Fig. 1). Then wehave z = z γ = z " (cid:18) ct r z (cid:19) = z + r . (9)From Fig. 1, we have the relations z r = z − r cos θ, ρ = r sin θ , (10)where θ is the angle with respect to the z-axis. This gives us z + ρ = z + r − rz cos θ . (11)From Eqs. (9) and (11), we get2 rz cos θ = − ( z + ρ − z ) . (12)After squaring, and using Eq. (10), we can obtain r = [( z + ρ − z ) + 4 z ρ ] / z . (13)Also γ r (1 − v r cos θ/c ) = z r z + r cos θz = zz . (14)From Eqs. (13) and (14), we get rγ r (1 − v r cos θ/c ) = [( z + ρ − z ) + 4 z ρ ] / z = ξ z . (15)where ξ = [( z + ρ − z ) + 4 z ρ ] / . (16)Substituting Eqs. (15) in Eq. (6), we get the electromagnetic field at P ( ρ, z ) at time t = 0 as E = 4 ez n ξ , (17)Using (12) and (13), the field components are then written as E z = 4 ez cos θξ = − ez ( z + ρ − z ) ξ E ρ = 4 ez ξ ρr = 8 ez ρzξ , (18)the remaining field components being zero.The above field expressions are in the instantaneous rest frame of the uniformlyaccelerated charge, and agree with those derived earlier [12, 15]. For the more generalcase, when the uniformly accelerated charge, at t = 0, would be located elsewhere andmoving with a finite velocity, the derivation of field expressions, though on similar lines,is a bit more involved and is given in the Appendix (Eq. (A.13)). discontinuity in the electromagnetic field of a uniformly accelerated charge
4. A discontinuity in electromagnetic field of a uniformly acceleratedcharge at the z = 0 plane From causality arguments, field of a uniformly accelerated charge, beginning from aninfinite past, are to be found only in the region z > z →
0, from Eq. (18), wecan write the field as E ρ = 0 E z = − ez ( z + ρ ) . (19)Since the field of the uniformly accelerated charge at t = 0 does not extend intoregion z ≤
0, the field component E z in Eq. (19) terminates with a finite value as the z = 0 plane is approached and there is thus a discontinuity in electromagnetic field ofa uniformly accelerated charge at the z = 0 plane. Of course this can happen if, at the z = 0 plane, there is a surface charge density σ = 14 π E z | z → = − ez π ( z + ρ ) (20)amounting to a total charge − e . There is of course no surface charge density at the z = 0 plane pre-specified in our case, where the only charge in the picture is the onewith uniform acceleration. Therefore this discontinuity at the z = 0 plane implies aviolation of one of Maxwell’s equations, viz. Gauss law.However, compatibility with Maxwell’s equations could be restored if at the z = 0plane, there were additional field proportional to the δ -function [6] E ρ = 2 eρz + ρ δ ( z ) . (21)To verify, the contribution of the “ δ -field” (Eq. (21)) to ∇ · E at the z = 0 plane is1 ρ ∂ ( ρE ρ ) ∂ρ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) z=0 = 4 ez ( z + ρ ) δ ( z ) (22)which neatly cancels the surface charge density inferred in (Eq. (20)), and we thenobtain ∇ · E = 0 at the z = 0 plane.Of course, there will be an accompanying magnetic field, B = v i × E /c , which inthe limit | v i | → c , is B φ = − eρz + ρ δ ( z ) . (23)We want to derive δ -field (Eq. (21)) from the charge that moved with a uniformvelocity prior to some instant in distant past, and afterwards has been moving witha constant acceleration. Let us assume that the uniform acceleration was imposedupon the charge starting from a time t i = − τ onwards and that until then the charge,in continuation with the ensuing accelerated motion specifications, was moving witha constant velocity v i = − c τ / ( z + c τ ) / , with a corresponding Lorentz factor γ i = [1 + ( cτ /z ) ] / . In that case, at time t = 0 while field in the region r < cτ (i.e., discontinuity in the electromagnetic field of a uniformly accelerated charge Figure 2.
A charge coming from infinity ( z = ∞ at t = −∞ ) with an initial uniformvelocity v i , with a corresponding Lorentz factor γ i , moving along − z direction, isapplied a constant proper acceleration, a = γ ˙ v along the z-axis, at time t i = − τ when it is at a location z i , and thereafter comes to rest ( v = 0) momentarily, at z = z at time t = 0. The charge would have reached its extrapolated “present”position z µ at time t = 0, if it had continued to move with its initial uniform velocity v i . At time t = 0, the electromagnetic field at all points within Σ, boundary of a sphereof radius r = cτ = z γ i | v i | /c , centered on the charge position z i , is that of a uniformlyaccelerated charge, with only a radial component along ˆr . However, at all pointsoutside the sphere, the field is that of a charge at location z µ , moving with a uniformvelocity v i , having both a radial component along ˆr and a transverse component along ˆ θ . The spherical boundary Σ intersects the z-axis at a point z σ . The box shows z σ and z ρ both expanded upto first order terms in 1 /γ i . The plot in the figure is for γ i = 2 . | v i | ≈ . c . in the region inside of the spherical surface Σ of radius r centred at z i = ( z + c τ ) / ,position of the charge at time t i = − τ ) is that of a uniformly accelerated charge, thefield in the region r > cτ = z γ i | v i | /c is that of a uniformly moving charge.The electric field just inside the boundary Σ, being that of the charge moving witha uniformly acceleration, from Eq. (6), can be written as E a = " e ˆr γ r (1 − n · v i /c ) t i , (24)while, the electric field just outside the boundary Σ, being that of the charge movingwith a uniform velocity v i , from Eq. (2) can be written as E v = " e ˆr γ r (1 − n · v i /c ) t i + " ev i sin θ ˆ θγ r c (1 − n · v i /c ) t i . (25) discontinuity in the electromagnetic field of a uniformly accelerated charge Figure 3.
The distribution of electric field lines of a charge moving with a uniformvelocity v and a corresponding Lorentz factor γ . The field line distributions, centeredon the “present” position of the moving charge, are shown for different Lorentz factors,with γ increasing in steps of 3.16 (= √ γ → ∞ , with v → c , the electric fieldincreasingly resembles a δ -field. The radial components (along ˆr = n , Fig. 2) on the boundary Σ are exactly the samein Eqs. (24) and (25), implying a continuity in the radial components. We see that for afinite τ , there is no discontinuity in the radial component across the spherical boundaryΣ, even though the electric field of the uniformly accelerated charge terminates witha finite value at the boundary Σ. The difference in the electric fields at the boundaryΣ lies only in the transverse components; in the case of the uniformly moving charge,there is a finite transverse component ( ∝ v i sin θ ) along ˆ θ (Fig. 2), while in the case ofthe uniformly accelerated charge, the transverse component is zero as the accelerationfield, responsible for radiation in a usual case, gets neatly cancelled by the transversecomponent of the velocity field at t = 0, when the instantaneous velocity of the uniformlyaccelerated charge becomes zero.The uniformly moving charge has also a magnetic field ∝ v i sin θ , B = " ev i sin θ ˆ φγ r c (1 − n · v i /c ) t i . (26)The electric field due to the charge moving with uniform velocity v i before t i = − τ ,can be written equivalently, in an alternative form, with respect to the extrapolated“present” position z µ of the charge. This is the position the charge would occupy if therewere no acceleration imposed on it and it had continued to move with a constant velocity v i . Noting that v i γ i = − c τ /z , at t = 0 we have z µ = z i + v i τ = z γ i − v z γ i /c = z /γ i ,with the electric field of the charge being everywhere in a radial direction ˆ R from z µ [18, 19, 20, 28]. E = e ˆ R R γ [1 − ( v /c ) sin ψ ] / = eγ i R [( γ ( z − z µ ) + ρ ] / , (27)with the magnetic field, B = v i × E /c [18, 19, 20].Figure 3 shows the distribution of electric field lines of a charge moving with a discontinuity in the electromagnetic field of a uniformly accelerated charge v and a corresponding Lorentz factor γ . The field lines, centered onthe “present” position of the moving charge, are shown for different velocities and thecorresponding Lorentz factors. As γ becomes larger, the electric field component alongthe direction of motion, becomes negligible relative to the perpendicular component,with field lines increasingly oriented perpendicular to the direction of motion. Moreoverthe field becomes negligible, except in a narrow zone along the direction of motion. inthe limit γ i → ∞ , the electric field turns into ‘ δ -field’ (Fig. 3), at the z = 0 plane(representing the surface R = cτ ) when τ → ∞ .The question of discontinuity at the boundary Σ arises only when τ → ∞ , and asa result the spherical surface Σ coincides with the z = 0 plane, and as γ i → ∞ thefield arising from the uniform motion of the charge before t = − τ become δ -field, witha negligible E z component at z = 0, then only the question of Gauss law violation at z = 0 arises.It may be tempting to calculate the field by putting z µ = z /γ i = 0 directly inEq. (27), to get the δ -field at the z = 0 plane in the limit γ i → ∞ , however as has beenshown [12, 13], it does not this way lead to the correct expression for δ -field. Instead, anappropriate limiting procedure must be followed to derive the δ -field, as shown below. δ -field From Eq. (27), We can write the field components in terms of δ ( z ) as E z = lim γ i →∞ "Z z ρ −∞ d z eγ i ( z − z µ )[( γ ( z − z µ ) + ρ ] / δ ( z ) ,E ρ = lim γ i →∞ "Z z ρ −∞ d z eγ i ρ [( γ ( z − z µ ) + ρ ] / δ ( z ) . (28)Here limits of the integral are from z = −∞ to z ρ , where z ρ is the z coordinate of thecircle of radius ρ on the spherical boundary Σ (Fig. 2).Now at time t = 0, z µ = z /γ i , which gives us E z = lim γ i →∞ " − eγ i [( γ i z ρ − z ) + ρ ] / δ ( z ) ,E ρ = eρ lim γ i →∞ " ( γ i z ρ − z )[( γ i z ρ − z ) + ρ ] / + 1 δ ( z ) . (29)We want to determine the field in the limit, z i = z γ i → ∞ with γ i → ∞ and z µ = z /γ i approach the z = 0 plane.From Fig. 2, we have ( r − ( z ρ − z σ )) + ρ = r , from which, for a fixed ρ , we getto a first order in 1 /γ i z ρ − z σ ≈ ρ / r ≈ ρ / γ i z . (30)Also, to a first order in 1 /γ i , we have z σ = γ i z (1 − v i /c ) = z /γ i (1 + v i /c ) ≈ z / γ i . (31)Thus we see that z σ ≈ z µ / z σ , z ρ , z µ , always lie at z >
0, however, inthe limit γ i →
0, they all approach the z = 0 plane. discontinuity in the electromagnetic field of a uniformly accelerated charge γ i z ρ − z in the limit γ i → ∞ . From Eqs. (30) and (31), wherewe kept terms upto first order in 1 /γ i , as those only would later survive when γ i → ∞ ,we get lim γ i →∞ [ γ i z ρ − z ] = " ρ z + z − z = 12 z h ρ − z i . (32)Substituting Eq. (32) in Eq. (29) we have E z = lim γ i →∞ " − ez γ i [ z + ρ ] δ ( z ) = 0 ,E ρ = eρ " ρ − z z + ρ + 1 δ ( z ) = 2 eρz + ρ δ ( z ) , (33)Eq. (33) is the expression for δ -field (Eq. (21)) that has been conjectured [8, 9, 10, 11, 12],and derived by others [14, 15] using differing methods.
5. Total energy in electromagnetic field including that in δ -field at time t = 0It is possible to calculate analytically the electromagnetic field energy in the case ofa charge moving with a uniform velocity or even of a charge moving with a uniformacceleration [16, 17]. We can, accordingly, calculate exactly the total energy in thefield, including that in δ -field, at time t = 0. The field in the spherical region of radius r = cτ around z i , are of a charge moving with a constant proper acceleration a , whileoutside that region the field is that of a charge moving with a uniform velocity v i (with acorresponding Lorentz factor γ i ), the latter field turning into δ -field in the limit τ → ∞ .The electromagnetic field energy is given by the volume integral E = Z E + B π d V , (34)where d V = r d r (1 − v cos θ/c ) dΩ is the volume element (Fig. 4).Using the integral Z Ω (1 − v cos θ/c ) γ (1 − v cos θ/c ) dΩ = Z π π sin θγ (1 − v cos θ/c ) d θ = 4 π , (35)the total field energy in the volume up to r = cτ , from Eqs. (6) in the instantaneousrest frame at time t = 0, is found to be E = Z cτǫ e r dr = e ǫ − e cτ . (36)Since the integral diverges for r →
0, we have restricted the lower limit of r to ǫ , whichmay represent the radius of the charged particle. For τ → ∞ , the energy in the field ofan accelerated charge, at time t = 0 (when it is instantly stationary), becomes E = e ǫ . (37)This exactly is the expression for the field energy of a charge permanently at restin an inertial frame, while in our calculations we included the contribution of the discontinuity in the electromagnetic field of a uniformly accelerated charge Figure 4.
Volume element for the computation of the field energy. Thickness of thespherical shell, shown by the dotted line, is d w = d r (1 − v cos θ/c ), along angle θ ,implying the corresponding volume element of the spherical shell as d V = r d w dΩ = r d r (1 − v cos θ/c ) dΩ, acceleration field as well, for all r . Where then is the radiation, given by Larmor’sradiation formula, emitted from such a charge throughout its accelerated motion fromtime t = − τ to t = 0?Using the integral Z Ω (1 − v cos θ/c ) sin θγ (1 − v cos θ/c ) dΩ = Z π π sin θγ (1 − v i /c cos θ ) d θ = 8 πγ , (38)from Eqs. (25) and (26), at time t = 0, the energy in electromagnetic field in the region r > cτ is found to be E = e (cid:20) γ i v i /c ) (cid:21) Z ∞ cτ d rr , (39)Using ( γ i v i ) = ( aτ ) we have E = e cτ + 2 e a τ c . (40)For τ → ∞ , the field becomes δ -field, and the energy in that is E = 2 e a τ c . (41) discontinuity in the electromagnetic field of a uniformly accelerated charge τ , at the rate, 2 e a / c , given byLarmor’s radiation formula. However, in reality, the δ -field has no causal relation withthe charge during its uniform acceleration. All fields, originating from the acceleratingcharge positions, lie in the region z > t = 0 and the radiation, if any, fromthe accelerating charge should also lie only there. In fact, because of a rate of changeof acceleration ˙ a , at time t = − τ , an event with which the δ -field has a causal relation,the charge moving with velocity v i undergoes radiation losses, at a rate ∝ − γ i v i · ˙ a [29],owing to the Lorentz-Dirac radiation reaction, thereby with a total energy loss∆ E = − e c Z − τ +0 − τ − γ i v i · ˙ a d t = 2 e c γ i | v i a | = 2 e a τ c . (42)which exactly is the electromagnetic field energy in the δ -field (Eq. (41)). Acknowledgments
I thank Jerrold Franklin for his comments and suggestions on the manuscript.
Appendix A. Electromagnetic field of a uniformly accelerated charge interms of its “real-time” motion
We want to express the electromagnetic field at some event (location and instant oftime) with respect to the motion of the charge also for the same instant. It may notbe possible to do so in a general case where the acceleration of the charge may bechanging with time. However, for a uniformly accelerated charge, it is possible to solvethe expressions for electromagnetic field, not necessarily in terms of motion of the chargeat the retarded time, instead wholly in terms of the “real-time” motion of the charge[8, 9].Let P ( ρ, z ) be the field point where we want to determine the electromagnetic fieldof a uniformly accelerated charge at some instant of time t , in terms of the “real-time”position of the charge z e and its velocity v = at/γ , implying γ v /c = ct/z (A.1)with γ = [1 + ( ct/z ) ] / . From Eq. (7), we have z e = [ z + c t ] / . Thus all werequire is the quantity rγ r (1 − v r cos θ/c ) in Eqs. (4) and (5) to be expressed in termsof ρ , z , z and t .Let z r be the position of the charge at the corresponding retarded time t r = t − r/c ,where it had the velocity v r = at r /γ r . From Eq. (7), we also have z r = [ z + c t ] / .Then, we have z = z + c t = z + c t + r − rct = z + c t + r − rct . (A.2)Now, from Fig. 5, we have the relations, z r = z − r cos θ, ρ = r sin θ , which gives us z + ρ = z + r − rz cos θ . (A.3) discontinuity in the electromagnetic field of a uniformly accelerated charge Figure A1.
A charge, moving with a uniform proper acceleration, a = γ ˙ v , alongthe z-axis, and coming from z = ∞ at t = −∞ with an initial velocity along − z direction, is decelerated and comes to rest momentarily at z = z at time t = 0. Thecharge afterwards moves along the z-axis with an ever increasing velocity due to itsconstant proper acceleration. We want to determine electromagnetic field at time t , ata point P ( ρ, z ), which is at a distance r = c ( t − t r ) from the charge position z r at thecorresponding retarded time t r , when it had a velocity v r . The charge at time t is ata location z e , moving with a “present” velocity v . From Eqs. (A.2) and Eq. (A.3), we eliminate z r , to get2 rct − rz cos θ = z + c t + ρ − z . (A.4)Squaring, we get4 r ( z + c t − zct cos θ ) = ( z + c t + ρ − z ) + 4 z ρ . (A.5)Using Eq. (A.4), to eliminate the cos θ term, we can write4[ z − c t ] r + 4 rct ( z + c t + ρ − z ) − [( z + c t + ρ − z ) + 4 z ρ ] = 0 . (A.6)This quadratic equation in r has a solution r = − ct ( z + c t + ρ − z ) + z [( z + c t − ρ − z ) + 4 z ρ ] / z − c t ) . (A.7)The other possible solution of the quadratic equation is ruled out from the requirementthat r > γ r (1 − v r cos θ/c ) = z r z − at r cos θc = z r z − c ( t − r/c ) cos θz = z − ct cos θz . (A.8)From which we get[ rγ r (1 − v r cos θ/c )] = ( rz − rct cos θ ) z = r ( z + c t − zct cos θ ) − c t ρ z . (A.9) discontinuity in the electromagnetic field of a uniformly accelerated charge rγ r (1 − v r cos θ/c ) = [( z + c t − ρ − z ) + 4 z ρ ] / z = ξ z , (A.10)where ξ = [( z + c t − ρ − z ) + 4 z ρ ] / . (A.11)From Eqs. (A.10) and (A.8), we can also write r = ξ z − ct cos θ ) . (A.12)Substituting Eqs. (A.10) and (A.12) in Eq. (4), and using Eqs. (A.1) and (A.4), weget components of the electromagnetic field at field point P ( ρ, z ) at time t , as E z = 4 ez r cos θrξ − ez r sin θγ v cξ = 4 ez r cos θ ( z − ct cos θ ) − rct sin θξ = − ez ( z + c t + ρ − z ) /ξ E ρ = 4 eρz rξ + 8 eρz γ v cos θcξ = 8 ez ρ ( z − ct cos θ ) + ct cos θξ = 8 ez ρz/ξ B φ = eργ v /cγ r (1 − v r cos θ/c ) = 8 ez ρct/ξ , (A.13)the remaining field components are zero.Of course these more general expressions (Eq. (A.13)) along with (A.11)), reduceto the simpler ones derived for the instantaneous rest frame (Eq. (18)) and (16)), byputting t = 0. References [1] M. Born, “Die Theorie des starren Elektrons in der Kinematik des Relativit¨atsprinzips,” Ann.Phys. das relativit¨atprinzip , (Braunschweig, 1911).[4] G. A. Schott, Electromagnetic radiation (Univ. Press, Cambridge, 1912).[5] W. Pauli
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