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Chapter Fourteen Radiation by Moving Charges

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Chapter Fourteen Radiation by Moving Charges Chapter Fourteen Radiation by Moving Charges Sir Joseph Larmor (1857 - 1942) September 17, 2001 Contents 1 Li¶enard-Wiechert potentials 1 2 Radiation from an Accelerated Charge; the Larmor Formula 9 2.1 Relativistic Larmor Formula . 12 2.1.1 Example: Synchrotron . 14 2.1.2 Example: Linear Acceleration . 15 3 Angular distribution of radiation 17 3.1 Example: Parallel acceleration and velocity . 17 3.2 Example: Acceleration Perpendicular to Velocity . 20 3.3 Comparison of Examples . 21 3.4 Radiation of an Ultrarelativistic Charged Particle . 22 4 Frequency Distribution of the Radiated Energy 25 4.1 Continuous Frequency Distribution . 26 4.2 Discrete Frequency Distribution . 30 1 4.3 Examples . 32 4.3.1 A Particle in Instantaneous Circular Motion . 32 4.3.2 A Particle in Circular Motion . 37 5 Thomson Scattering; Blue Sky 42 6 Cherenkov Radiation Revisited 46 7 Cherenkov Radiation; Transition Radiation 51 7.1 Cherenkov Radiation in a Dilute Collisionless Plasma . 55 8 Example Problems 57 8.1 A Relativistic Particle in a Capacitor . 57 8.2 Relativistic Electrons at SLAC . 58 We have already calculated the radiation produced by some known charge and current distribution. Now we are going to do it again. This time, however, we shall consider that the source is a single charge moving in some fairly arbitrary, possibly relativistic, fashion. Here the methods of chapter 9, e.g., multipole expansions, are impractical and there are better ways to approach the problem. 1 Li¶enard-Wiechert potentials The current and charge densities produced by a charge e in motion are ½(x; t) = e±(x x(t)) ¡ J(x; t) = ev(t)±(x x(t)) (1) ¡ if x(t) is the position of the particle at time t and v(t) dx(t)=dt x_ (t) is its ´ ´ velocity. In four-vector notation, J ¹(x; t) = ec¯¹±(x x(t)) (2) ¡ 2 where ¯¹ (1; ¯) is not a four-vector; ¯ = v=c. ´ From J and ½, one ¯nds A and ©. We can do this in an in¯nite space by making use of the retarded Green's function G(x; t; x0; t0) = ±(t t0 x x0 =c)= x x0 . ¡ ¡ j ¡ j j ¡ j From here, we can evaluate the electromagnetic ¯eld as appropriate derivatives of the potentials. All of these manipulations are straightforward. Furthermore, the integrations are relatively easy because there are many delta functions. The problem becomes interesting and unfamiliar, however, for highly relativistic particles which produce large retardation e®ects. Let us start from the integral expression for the potentials: ¹ 3 ¹ A (x; t) = d x0dt0 G(x; t; x0; t0)J (x0; t0) Z ¹ 1 3 ±(t t0 x x0 =c)J (x0; t0) = d x0dt0 ¡ ¡ j ¡ j c x x Z j ¡ 0j 3 ¹ ±(x0 x(t0))±(t t0 x x0 =c) = e d x0dt0 ¯ (t0) ¡ ¡ ¡ j ¡ j x x Z j ¡ 0j ¹ ±(t t0 x x(t0) =c) = e dt0 ¯ (t0) ¡ ¡ j ¡ j (3) x x(t ) Z j ¡ 0 j This form reects the retarded nature of the problem. The particle is not seen by the observer at its present location x(t), but it appears to be at x(t') since light traveling from the particle x at x(t') arrives at the observer at time t. O n(t') e x (t') x(t) 3 The evaluation of this integral is not totally simple because the argument of the ±- function is not a simple function of the time t0. In general when one faces an integral of this form, one invokes the rule dg dt0 f(t0)±[g(t0)] = f(t0) (4) ,¯dt0 ¯ Z ¯ ¯t0 ¯ ¯ ¯ ¯ ¯ 1 ¯ where t0 is the zero (there may be more than one) of g , i.e., g(t0) = 0. Applying this to the present case, we have 1=2 g(t0) = t0 + x x(t0) =c t = t0 + [(x x(t0)) (x x(t0))] =c t; (5) j ¡ j ¡ ¡ ¢ ¡ ¡ so dg 1 dx(t0) (x x(t0)) v(t0) (x x(t0)) = 1 + 2 ¡ = 1 ¢ ¡ 1 ¯(t0) n(t0): (6) dt 2c á dt ! ¢ x x(t ) ¡ c x x(t ) ´ ¡ ¢ 0 0 j ¡ 0 j j ¡ 0 j The unit vector n points from the point x(t0) on the particle's path toward the ¯eld point x; it, and ¯, must be evaluated at a time t0 which is earlier than t, the time at which the ¯eld is evaluated, by some amount which is determined by solving the equation t0 + x x(t0) =c = t: (7) j ¡ j Combining Eqs. (3), (4), and (6) we ¯nd that the potentials are given simply by ¯¹ e¯¹ A¹(x; t) = e = (8) " x x(t ) (1 ¯ n)# " R· # j ¡ 0 j ¡ ¢ ret ret where R x x(t0) , · = 1 ¯ n, and the subscript ret means that the quantity ´ j ¡ j ¡ ¢ in brackets [...] must be evaluated at the retarded time t0 determined from Eq. (7). Our potentials, Eq. (8), are known as the Li¶enard-Wieckert potentials. Probably their most signi¯cant feature is the fact that they vary inversely as 1 ¯ n or ·; this ¡ ¢ factor can be very close to zero for n ¯ if ¯ is close to one, i.e., for highly relativistic k 1 More generally ±[g(x)] = ±(x x )= g0(x ) where x are the simple zeroes of g(x). If j ¡ j j j j j g0(xj) = 0 (a complex zero), thenP ±[g(x)] makes no sense. 4 particles, meaning that there is a strong maximum in the potentials produced by a relativistic particle in the direction of the particle's velocity (at some retarded time). We wish next to ¯nd the electromagnetic ¯eld. One may do this in a variety of ways. One is simply carefully to take derivatives of the Li¶enard-Wieckert potentials, a procedure followed in, e.g., Landau and Lifshitz' book, The Classical Theory of Fields. Another, much more elegant, is to ¯nd the ¯elds in the instantaneous rest frame of the particle and to Lorentz-transform them to the frame interest. Another, not very elegant at all, and about to be employed here, is to go back to the integrals, Eq. (3), for the potentials and take derivatives of these expressions. Consider just the vector potential, ¯(t0)±(t0 + R(t0)=c t) A(x; t) = e dt0 ¡ : (9) Z R(t0) If we wish to ¯nd B(x; t), we have to take derivatives of R with respect to various components of x. Consider, for example, df df R df f(R) = R = = n : (10) r dRr dR R dR Application of this simple rule gives (since (Ãa) = à a + à a, and r £ r £ r £ ¯(t0) = 0 due to the lack of an x dependence.) r £ ±(t0 + R=c t) B(x; t) = A(x; t) = e dt0 ¡ ¯(t0) r £ Z r à R ! £ 1 1 0 0 0 0 = e dt (n ¯) 2 ±(t + R=c t) + ± (t + R=c t) (11) Z £ ·¡R ¡ cR ¡ ¸ where the prime on the delta function denotes di®erentiation with respect to the argument. Hence ¯ n · ±0(t0 + R=c t) 0 B(x; t) = e £2 + dt ¡ (n ¯) (12) (" ·R #ret Z cR à · ! £ ) Now, from Eqs. (5) and (6) d(t0 + R=c t) ¡ = · or ·dt0 = d(t0 + R=c t) (13) dt0 ¡ 5 and so ¯ n n ¯ 0 0 0 B(x; t) = e £2 + d(t + R=c t) £ ± (t + R=c t) (" ·R #ret Z ¡ à c·R ! ¡ ) ¯ n @[(n ¯)=c·R] = e £ d(t0 + R=c t) £ ±(t0 + R=c t) (" ·R2 # ¡ ¡ @(t + R=c t) ¡ ) ret Z 0 ¡ ¯ n 1 @ n ¯ = e £2 £ (" ·R #ret ¡ "· @t0 à cR· !#ret) ¯ n 1 1 @ ¯ n = e £2 + £ : (14) (" ·R #ret c "· @t0 à ·R !#ret) The electric ¯eld can be found by similar manipulations: 1 @A(x; t) E(x; t) = ©(x; t) ¡r ¡ c @t d ±(t0 + R=c t) e ¯±0(t0 + R=c t) = e dt0 n ¡ + dt0 ¡ ¡ Z dR à R ! c Z R ±(t0 + R=c t) 1 0 0 0 = e dt n 2 ¡ + (¯ n)± (t + R=c t) Z ( R cR ¡ ¡ ) n e 1 @ ¯ n = e 2 ¡ : (15) ··R ¸ret ¡ c "· @t0 à ·R !#ret We now have expressions for E and B, but they involve time derivatives of retarded quantities. We can work out each of these derivatives. First, we consider just the derivative of n = R=R. One has @R = x_ (t0) = ¯c (16) @t0 ¡ ¡ and 1=2 @R @[(x x(t0)) (x x(t0))] 1 = ¡ ¢ ¡ = [ 2(x x(t0)) x_ (t0)] = n x_ (t0) = n ¯c @t0 @t0 2R ¡ ¡ ¢ ¡ ¢ ¡ ¢ (17) Hence, 1 dn 1 @R 1 @R 1 = 2 R = [¯ (n ¯)n] : (18) c dt0 cR @t0 ¡ cR @t0 ¡R ¡ ¢ The quantity in brackets is can be written more concisely: (n ¯)n ¯ = (n ¯)n (n n)¯ = n (n ¯); (19) ¢ ¡ ¢ ¡ ¢ £ £ 6 hence we ¯nd 1 dn 1 = n (n ¯): (20) c dt0 R £ £ Employing this useful result we have, for the electric ¯eld, n 1 n @ 1 1 @ ¯ E(x; t) = e 2 + 2 2 n (n ¯) + "·R · R £ £ c· @t0 ·R ¶ ¡ c· @t0 ÷R !#ret n(1 n ¯) 1 = e ¡2 2¢ + 2 2 [(n ¯)n ¯] " · R · R ¢ ¡ #ret n @ 1 1 @ ¯ + (21) "c· @t0 ·R ¶ ¡ c· @t0 ÷R !#ret or n ¯ n @ 1 1 @ ¯ E(x; t) = e 2¡ 2 + : (22) " · R c· @t0 ·R ¶ ¡ c· @t0 ÷R !#ret This expression may be related to that for the magnetic induction, ¯ n e 1 @ ¯ n B(x; t) = e £2 + £ " ·R #ret c "· @t0 à ·R !#ret ¯ n e 1 @ ¯ = e £2 + n " ·R #ret c "· @t0 ÷R ! £ #ret ¯ 1 +e 2 n (n ¯) "· R £ R £ £ ¶#ret ¯ n e 1 @ ¯ = e £2 + n " ·R #ret c "· @t0 ÷R ! £ #ret 1 +e 2 2 [n ((n ¯) ¯) (n ¯)(n ¯)] (23) ·· R £ ¢ ¡ £ ¢ ¸ret Next, put all terms proportional to n ¯ over the same denominator, which is ·2R2, £ and also make use of the fact that (n ¯) ¯ = 0, to ¯nd £ ¢ ¯ n e 1 @ ¯ B(x; t) = e 2£ 2 + n (24) " · R #ret c "· @t0 ÷R ! £ #ret This is our ¯nal expression.
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