Green s function for the wave equation

Transcription

1 Green s funtion for the wave equation Non relativisti ase 1 The wave equations In the Lorentz Gauge, the wave equations for the potentials in Lorentz Gauge Gaussian units are: r 2 A 1 2 A 2 t = 4π 2 j (1) The Gauge ondition is: 2 The Green s funtion r = 4πρ (2) t2 r A + 1 t = 0 (3) For both potentials we have a wave equation of the form r = 4π (soure) t2 where an be either the salar potential or a artesian omponent of A. The orresponding Green s funtion problem is: r 2 G ( x,t; x 0,t 0 ) 1 2 G 2 t = 4πδ ( x 2 x0 ) δ(t t 0 ) where the soure is now a unit event loated at x = x 0 happening at t = t 0. To solve this equation we first Fourier transform in time: G ( x,t; x 0,t 0 ) = p 1 Z 1 G ( x,ω; x 0,t 0 )e iωt dω 2π the equation beomes µr 2 + ω2 G( x,ω; x 0,t 0 ) = 4πp 1 Z 1 δ ( x x 0 )δ (t t 0 )e iωt dt 2π 2 = 2 p 2πδ ( x x 0 ) e iωt0 1

2 So let G( x,ω; x 0,t 0 ) = g ( x, x 0 ) e iωt0 / p 2π then g satisfies the equation r 2 + k 2 g = 4πδ ( x x 0 ) Now in free spae without boundaries, g must be a funtion only of R = j x x 0 j must posess spherial symmetry about the soure point. Thus in spherial oordinates, we an write: 1 d 2 ³ R dr 2 (Rg) + k2 g = 4πδ R (4) For R 6= 0, the right h side is zero. Then the funtion Rg satisfies the exponential equation, the solution is: Rg = Ae ikr + Be ikr g = 1 Ae ikr + Be ikr R (5) Near the origin, where the delta funtion ontributes, the seond term on the LHS is negligible ompared with the first, the eequation beomes: we know that this has solution r 2 g = 4πδ ( x x 0 ) g = 1 R This is onsistent with equation 5 provided that A + B = 1 (You should onvine yourself that this solution is orret by differentiating stuffing bak into equation 4.) Thus we have the solution G ( x,ω; x 0,t 0 ) = p 1 Ae ikr + Be ikr e iωt0 2πR Now we do the inverse transform: Z G( x,t; x 0,t ) = p p Ae ikr + Be ikr e iωt0 e iωt dω 2π 2πR = 1 2π Z 1 Aexp(iω (R/ + t 0 t)) + B exp (iω ( R/+ t 0 t)) dω = Aδ (t 0 (t R/)) + Bδ (t 0 (t + R/)) (6) The seond term is usually rejeted beause it predits a response to an event ourring in the future. However, Feynman Wheeler have proposed a theory in whih both terms are kept. They show that this theory an be onsistent with observed ausality provided that the universe is perfetly absorbing in the infinite future. The time t R/ that appears in the first term is alled the retarded time t ret. The symmetry of this Green s funtion is: G ( x,t; x 0,t 0 ) = G( x 0, t 0 ; x, t) (See Morse Feshbah Ch 7 pg ). Causality requires: G ( x,; x 0,t 0 ) = 0 2

3 G ( x,t; x 0,t 0 ) = 0 for t < t 0 3 The potentials Now that we have the Green s funtion, we an solve our original equations. Z ρ( x 0,t 0 ) ( x,t) = R δ (t0 t ret )dt 0 d 3 x 0 (7) Z ρ( x 0,t ret ) = d 3 x 0 (8) R in Lorentz Gauge, similarly: A( x,t) = 1 Z j ( x 0,t ret ) d 3 x 0 (9) R Notie that these equations have the same form as the stati potentials (equations in Jakson). 4 Radiation from a moving point harge (non relativisti ase) 4.1 The Lienard Wiehert potentials Our soure is a point harge moving with veloity v (t). Then the harge urrent densities are ρ( x,t) = qδ( x r (t)) j ( x,t) = q vδ( x r (t)) Then from equation 7, we have: Z qδ( x 0 r (t 0 )) ( x,t) = δ (t 0 t ret )dt 0 d 3 x 0 R We do the integral over the spatial oordinates first. Z ( x,t) = q δ (t0 + R(t 0 )/ t) R(t 0 dt 0 ) where R(t 0 ) = j x r (t 0 )j. Now to do the t 0 integral, we must reexpress the delta funtion. Reall: δ (f (x)) = X 1 jf 0 (x i )j δ (x x i) 3

4 where f (x i ) = 0. In this ase:, sine v = d r/dt, f 0 (t 0 ) = 1+ 1 f (t 0 ) = t 0 + R(t0 ) t dr dt 0 = 1 1 ( x r(t 0 )) j x r(t 0 )j d dt 0 ( x r (t0 )) = 1 v ( x r(t0 )) j x r (t 0 )j = 1 v R R The funtion f is zero when t 0 = t ret = t R/. Thus, evaluating the integral, we get: q ( x,t) = ³ R 1 v R (10) R t ret And similarly q v A ( x,t) = ³ R 1 v R (11) R t ret These are the Lienhard Wiehert potentials. It is onvenient to use the shorth à r v = R 1 v! R = R v R (12) R 4.2 Calulating the fields In Lorentz Gauge, the fields are found using E = r 1 A t B = r A But our expressions for the potentials are in terms of x t ret, not x t, so we have to be very areful in taking the partial derivatives. We an put the origin at the instantaneous position of the harge to simplify things. Then R = r. Our potential may be written: ( x,t) ª ( x,t ret ) Then d = r d x + onst t t dt rª d x + ª dt ret onst tret t ret But dt ret = dt dr/, so r d x + onst t t dt rª d x ª dr onst tret t ret + ª dt t ret 4

5 Thus the r omponent of r must be modified: = ª 1 ª (13) r onst t r onst tret t ret Now we an alulate the fields: r = r q = q rr r v r 2 v v rr v = µ r v r ^r + ^µ µ r v r + ^Á µ r v r r r θ r sinθ φ We an hoose our axes with polar axis along the instantaneous diretion of v. Then r v = rv os θ, rr v = ³1 v os θ ^r + ^θ ³r v sinθ r In the non relativisti limit, v/ 1, to zeroth order in v/, this is We are also going to need Then we have r v t rr v = ^r = r a E = ~r + 1 onst tret t ^r 1 A t q ^r q µ ^r q v rv 2 r 2 v t r v = q µ rv 2 ^r 1+ q a + q v µ r v rv 2 r again taking the non relativisti limit, this beomes: E = q r 2 ^r µ 1+ q a r The first term is the usual Coulomb field. The other two terms depend on a : these are the radiation field. E rad = q [(^r a) ^r a] r = q (^r (^r a)) (14) r 5

6 4.3 Radiated power The Poynting vetor is S = E 4π rad B rad = E 4π rad ³^n E rad = = 4π = 4π E2 rad^n q 2 4πR 2 à q R ^n Thus the power radiated per unit solid angle is: dp d = R2 S ^n à = q2 4π ^n " ^n d #! 2 β ^n dt " ^n ^n d 2 β # ^n dt ^n d β dt q 2 4π 3 a2 sin 2 θ (15) Thus is the Larmor formula, where a = dv/dt is the aeleration θ is the angle between a ^n. The total power radiated in the non relativisti ase is: P = Z Z dp +1 d d = dµ +1 µµ µ3 = q2 2 3a2 3 Z 2π 0 2! dφ q2 4π 3a2 1 µ 2 = 2q2 3 3 a2 (16) 6