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Radiation by Moving Charges Radiation by Moving Charges May 19, 20101 1J.D.Jackson, "Classical Electrodynamics", 3rd Edition, Chapter 14 Radiation by Moving Charges Li´enard - Wiechert Potentials I The Li´enard-Wiechert potential describes the electromagnetic effect of a moving charge. I Built directly from Maxwell's equations, this potential describes the complete, relativistically correct, time-varying electromagnetic field for a point-charge in arbitrary motion. I These classical equations harmonize with the 20th century development of special relativity, but are not corrected for quantum-mechanical effects. I Electromagnetic radiation in the form of waves are a natural result of the solutions to these equations. I These equations were developed in part by Emil Wiechert around 1898 and continued into the early 1900s. Radiation by Moving Charges Li´enard - Wiechert Potentials We will study potentials and fields produced by a point charge, for which 0 a trajectory ~x0(t ) has been defined a priori. It is obvious that when a charge q is radiating is giving away momentum and energy, and possibly angular momentum and this emission affects the trajectory. This will be studied later. For the moment, we assume that the particle is moving with a velocity much smaller than c. The density of the moving charge is given by 0 0 0 0 ρ(~x ; t ) = qδ(~x − ~x0[t ]) (1) and since in general the current density ~J is ρ~v, we also have d~x ~J(~x 0; t0) = q~vδ(~x 0 − ~x [t0]) ; where ~v(t0) = 0 (2) 0 dt0 In the Lorentz gauge ( r~ · A~ + (1=c)@Φ=@t = 0) the potential satisfy the wave equations (??) and (??) whose solutions are the retarded functions Z ρ(~x 0; t − j~x − ~x 0j=c) Φ(~x; t) = d 3x 0 (3) j~x − ~x 0j Radiation by Moving Charges 1 Z ~J(~x 0; t − j~x − ~x 0j=c) A~ (~x; t) = d 3x 0 (4) c j~x − ~x 0j It is not difficult to see that these retarded potentials take into account the finite propagation speed of the EM disturbances since an effect measured at ~x and t was produced at the position of the source at time j~x − ~x (~t)j ~t = t − 0 (5) c Thus, using our expressions for ρ and ~J from eqns (1) and (2) and putting β~ ≡ ~v=c, Z δ(~x 0 − ~x [t − j~x − ~x 0j=c]) Φ(~x; t) = q 0 d 3x 0 (6) j~x − ~x 0j Z β~(t − j~x − ~x 0j=c)δ(~x 0 − ~x [t − j~x − ~x 0j=c]) A~ (~x; t) = q 0 d 3x 0 (7) j~x − ~x 0j Radiation by Moving Charges Note that for a given space-time point( ~x; t), there exists only one point on the whole trajectory, the retarded coordinate ~x coresponding to the retarded time ~t defined in (5) which produces a contribution ~ ~x = ~x0(~t) = ~x0 (t − j~x − ~x0j=c) (8) Let us also define the vector ~ 0 0 R(t ) = ~x − ~x0(t ) (9) in the direction ~n ≡ R~ =R. Then Z δ(~x 0 − ~x [t − R(t0)=c]) Φ(~x; t) = q 0 d 3x 0 (10) R(t0) Z β~(t − R(t0)=c)δ(~x 0 − ~x [t − R(t0)=c]) A~ (~x; t) = q 0 d 3x 0 (11) R(t0) Because the integration variable ~x 0 appears in R(t0) we transform it by introducing a new parameter ~r ∗, where ∗ 0 0 ~x = ~x − ~x0 [t − R(t )=c] (12) Radiation by Moving Charges The volume elements d 3x ∗ and d 3x 0 are related by the Jacobian transformation h i d 3x ∗ = Jd 3x 0 ; where J ≡ 1 − ~n(t0) · β~(t0) (13) is the Jacobian (how?). With the new integration variable, the integrals for the potential transform to Z δ(~x ∗) d 3x ∗ Φ(~x; t) = q (14) ∗ ~ j~x − ~x − ~x0(~t)j(1 − ~n · β) and Z β~(~t) δ(~x ∗) d 3x ∗ A~ (~x; t) = q (15) ∗ ~ j~x − ~x − ~x0(~t)j(1 − ~n · β) which can be evaluated trivially, since the argument of the Dirac delta function restricts ~x ∗ to a single value " # " # q q Φ(~x; t) = = (16) (1 − ~n · β~)j~x − ~xj (1 − ~n · β~)R ~t ~t " # " # qβ~ qβ~ A~ (~x; t) = = (17) (1 − ~n · β~)j~x − ~xj (1 − ~n · β~)R ~t ~t Radiation by Moving Charges Li´enard - Wiechert potentials " # " # q q Φ(~x; t) = = (18) (1 − ~n · β~)j~x − ~xj (1 − ~n · β~)R ~t ~t " # " # qβ~ qβ~ A~ (~x; t) = = (19) (1 − ~n · β~)j~x − ~xj (1 − ~n · β~)R ~t ~t These are the Li´enard - Wiechert potentials. It is worth noticing the presence of the term (1 − ~n · β~), which clearly arises from the fact that the velocity of the EM waves is finite, so the retardation effects must be taken into account in determining the fields. Radiation by Moving Charges Special Note about the shrinkage factor (1 − ~n · β~) Consider a thin cylinder moving along the x-axis with velocity v. To calculate the field at x when the ends of the cylinder are at( x1; x2), we need to know the location of the retarded points~x1 and~x2 x − x~ v x − x~ v 1 1 = and 2 2 = (20) x − x~1 c x − x~2 c by setting L~ ≡ x~2 − x~1 and L ≡ x2 − x1 and subtracting we get v L L~ − L = L~ ! L~ = (21) c 1 − v=c That is, the effective length L~ and the natural length L differ by the factor (1 − ~x · β~)−1 = (1 − v=c)−1 because the source is moving relative to the observer and its velocity must be taken into account when calculating the retardation effects. Radiation by Moving Charges Li´enard - Wiechert potentials : radiation fields The next step after calculating the potentials is to calculate the fields via the relations 1 @A~ B~ = r~ × A~ and E~ = − − r~ Φ (22) c @t and we write the Li´enard - Wiechert potentials in the equivalent form Z δ(t0 − t − R(t0)=c) Φ(~x; t) = q dt0 (23) R(t0) Z β~(t0)δ(t0 − t + R(t0)=c]) A~ (~x; t) = q dt0 (24) R(t0) 0 0 where R(t ) ≡ j~x − ~x0(t )j. This can be verified by using the following property of the Dirac delta function (how?) Z X g(x) g(x)δ[f (x)]dx = (25) jdf =dxj i f (xi )=0 which holds for regular functions g(x) and f (x) of the integration variable x where xi are the zeros of f (x). • The advantage in pursuing this path is that the derivatives in eqn (22) can be carried out before the integration over the delta function. Radiation by Moving Charges This procedure simplifies the evaluation of the fields considerably since, we do not need to keep track of the retarded time until the last step. We get for the electric field Z δ(t0 − t + R(t0)=c) E~ (~x; t)= −q r~ dt0 R(t0) q @ Z β~(t0)δ(t0 − t + R(t0)=c) − dt (26) c @t R(t0) Thus, differentiating the integrand in the first term, we get (HOW?) Z ~n R(t0) ~n R(t0) E~ (~x; t)= q δ t0 − t + − δ0 t0 − t + dt0 R2 c cR c q @ Z β~(t0)δ(t0 − t + R(t0)=c) − (27) c @t R(t0) But (HOW?) R(t0) @ R(t0) δ0 t0 − t + = − δ t0 − t + (28) c @t c Z ~n R(t0) q @ Z (~n − β~) R(t0) E~ (~x; t)= q δ t0 − t + dt0+ δ t0 − t + dt0 R2 c c @t cR(t0) c (29) Radiation by Moving Charges We evaluate the integrals using the Dirac delta function expressed in equation (25). But we need to know the derivatives of the delta function's arguments with respect to t0. Using the chain rule of differentiation d R(t0) t0 − t + = 1 − ~n · β~ (30) dt0 c ~t with which we get the result (HOW?): ( ) ( ) ~n q @ ~n − β~ E~ (~r; t) = q + (31) (1 − ~n · β~)R2 c @t (1 − ~n · β~)R ~t ~t Since @R @R @t0 @t0 @t0 @~t 1 = = −~n·~v = c 1 − ) = @t @t0 @t @t @t @t (1 − ~n · β~) (32) Thus ( ) ( ) @ ~n − β~ 1 @ ~n − β~ = (33) @t (1 − ~n · β~)R (1 − ~n · β~) @~t (1 − ~n · β~)R2 ~t ~t Radiation by Moving Charges By using the additional pieces _ ~ Rj~t = −c ~n · β (34) ~t _ c h ~ ~i ~nj~t = ~n(~n · β) − β (35) R ~t d _ 1 − ~n · β~ = − ~n · β~ + β~ · ~n_ (36) d~t ~t and we finally get 8 9 h ~ ~_i <>(~n − β~)(1 − β2) ~n × (~n − β) × β => E~ (~r; t) = q + (37) ~ ~ 3 2 ~ ~ 3 :> (1 − n · β) R c(1 − n · β) R ;> ~t A similar procedure for B~ shows that B~ (~r; t) = r~ × A~ = ~n(~t) × E~ (38) Radiation by Moving Charges Some observations I When the particle is at rest and unaccelerated with respect to us, the field reduces simply to Coulomb's law q~n=R2. whatever corrections are introduced the do not alter the empirical law. I We also see a clear separation into the near field (which falls off as 1=R2) and the radiation field (which falls off as1 =R) I Unless the particle is accelerated (β_ 6= 0), the field falls off rapidly at large distances.
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