Green s funtion for the wave equation Non-relativisti ase January 2019 1 The wave equations In the Lorentz Gauge, the wave equations for the potentials are (Notes 1 eqns 43 and 44): 1 2 A 2 2 2 A = µ 0 j (1) and 1 2 Φ 2 2 2 Φ = ρ (2) ε 0 The Gauge ondition is (Notes 1 eqn 42): A + 1 Φ 2 =0 (3) In Coulomb Gauge we have the Gauge ondition: A =0 (4) whih leads to the wave equations (Notes 1, eqn on top of page 20, with A =0) 2 Φ = ρ (5) ε 0 and (Notes 1, eqn 41, with A =0) 1 2 A 2 2 2 A = µ 0 j 1 2 Φ (6) 2 Longitudinal and transverse urrents The Coulomb Gauge wave equation for A (6) is awkward beause it ontains the salar potential Φ. We an eliminate the potential Φ to obtain an equation for A alone. First we separate the urrent into two piees, alled the longitudinal urrent J and the transverse urrent J t : J = J + J t where J t =0 (7) and J =0 (8) 1
(Basially we are applying the Helmholtz theorem. We an always do this: See Lea Appendix II). From harge onservation (Notes 1 eqn 7): J + ρ =0 and using equation (5) to eliminate ρ, this beomes: J = ε0 2 Φ = ε 0 Φ Thus, up to an irrelevant onstant, J = ε 0 Φ + u Butfrom(8), J is the gradient of a salar, so we take u =0, and then J = ε 0 Φ (9) Using equation (9) in equation (6), we have: 2 A 1 2 A 2 2 = µ 0 J ε 0 Φ = µ 0 J J = µ 0 J t (10) In the Coulomb Gauge, the transverse urrent J t is the soure of A. We an also use result (9) to express J l in terms of J. Sine equation (5) is the same as in the stati ase, the solution is also the same (Notes 1 eqn 29). Thus J (x, t) = 1 4π ρ (x,t) x x d3 x = 1 4π ρ (x,t) x x d 3 x J (x, t) = 1 4π J (x ) x x d 3 x (11) where we used harge onservation in the last step. We an also express J t in terms of J as follows, using (11), J t = J J = J + 1 4π J x x d3 x Let s work on the integral. We start with an "integration by parts": J (x ) x x d 3 x = J (x ) x x d 3 x J J (x ) ˆn = S x x d2 x + J (x = 0+ ) x x d3 x 2 1 x x d3 x J 1 x x d3 x (Notes 1 eqn 19)
The surfae integral is zero provided that J is loalized. Then J t (x, t) = J (x, t)+ 1 4π J (x,t) x x d3 x = J + 1 4π = J + 1 4π = J + 1 4π J (x,t) x x d3 x J (x,t) x x d3 x J (x,t) x x d3 x + 2 J (x,t) x x d3 x + + J 2 1 x x d3 x J [ 4πδ (x x )] d 3 x J t (x, t) = 1 4π J (x,t) x x d3 x (12) We still have the rather unphysial result from equation (5) that Φ hanges instantaneously everywhere as ρ hanges. In lassial physis the potential is just a mathematial onstrut that we use to find the fields, and it an be shown (e.g. J prob 6.20) that E is ausal even though Φ is not. 3 The Green s funtion Witheithergaugewehaveawaveequationoftheform 2 Φ 1 2 Φ 2 2 =(soure) where Φ may be either the salar potential (in Lorentz Gauge) or a Cartesian omponent of A. (In Coulomb Gauge the salar potential is found using the methods we have already developed for the stati ase.) The orresponding Green s funtion problem is: 2 G (x, t; x,t ) 1 2 G 2 2 = 4πδ (x x ) δ (t t ) (13) where the soure is now a unit event loated at position x = x and happening at time t = t. As usual, the primed oordinates x,t, are onsidered fixed for the moment. To solve this equation we first Fourier transform in time (see, eg, Lea pg 503): G (x, t; x,t )= 1 G (x, ω; x,t ) e iωt dω 2π and the transformed equation (13) beomes 2 + ω2 G (x, ω; x,t ) = 4π 1 δ (x x ) δ (t t ) e iωt dt 2π 2 = 4π 2π δ (x x ) e iωt 3
So let G (x, ω; x,t )=g(x, x ) e iωt / 2π and then g satisfies the equation 2 + k 2 g = 4πδ (x x ) (14) where k = ω/. In free spae without boundaries, g must be a funtion only of R = x x and must posess spherial symmetry about the soure point 1. Thus in spherial oordinates with origin at the point P with oordinates x, we an write: 1 d 2 R dr 2 (Rg)+k2 g = 4πδ R (15) For R = 0, the right hand side is zero. Then the funtion Rg satisfies the exponential equation, and the solution is: Rg = Ae ikr + Be ikr g = 1 Ae ikr + Be ikr R = 0 (16) R Near the origin, where the delta-funtion ontributes, the seond term on the LHS of (14) is negligible ompared with the first, and equation (14) beomes: 2 g 4πδ (x x ) We reognize that this equation has the solution g = 1 R This is onsistent with equation (16) as R 0, provided that A + B =1 Thus we have the solution G (x, ω; x,t )= 1 Ae ikr +(1 A) e ikr e iωt 2πR You should onvine yourself that this solution is orret by differentiating and stuffing bak into equation (15). Now we do the inverse transform: G (x, t; x,t ) = = 1 2π 1 2πR 1 2πR Ae ikr + Be ikr e iωt e iωt dω {A exp [iω (R/ + t t)] + B exp [iω ( R/ + t t)]} dω = A R δ [t (t R/)] + B R δ [t (t + R/)] (17) The seond term is usually rejeted (take B 0 and thus A =1)beause it predits a response to an event ourring in the future. However, Feynman and Wheeler 2 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. (This now appears unlikely.) The time t R/ that appears in the first term is alled 1 Note that the operator 2 + k 2 is spherially symmetri. 2 Reviews of Modern Physis, 1949, 21,425 4
the retarded time t ret. Thus we take G (x, t; x,t )= 1 R δ [t (t R/)] = 1 R δ (t t ret ) (18) Causality (an event annot preede its ause) requires that the symmetry of this Green s funtion is: G (x, t; x,t )=G(x, t ; x, t) (See Morse and Feshbah Ch 7 pg 834-835). and also G (x, ; x,t )=0 and G (x, t; x,t )=0for t<t 4 The potentials Now that we have the Green s funtion (18), we an solve our original equations. Modifying eqn 1.44 in Jakson ("Formal" Notes eqn 7) to inlude time dependene, and with S, we get an integral over a volume in spae-time rather than just spae: Φ (x, t) = 1 ρ (x,t ) G (x, t; x,t ) dt d 3 x 4πε 0 Thus, inserting (18), we have 1 ρ (x,t ) Φ (x, t) = 4πε 0 R δ (t t ret ) dt d 3 x (19) 1 ρ (x,t ret ) = d 3 x (20) 4πε 0 R (t ret ) in Lorentz Gauge, and similarly: A (x, t) = µ 0 4π j (x,t ret ) R (t ret ) d 3 x (21) Notie that these equations have the same form as the stati potentials (equations 1.17 and 5.32 in Jakson, eqns 21 and 29 in Notes 1), but we must evaluate the soure and the distane R at the retarded time. This allows for the time for a signal to travel from the soure to the observer at speed. Note that t ret is a funtion of both x and x,aswellast. Similar equations hold in Coulomb Gauge. The salar potential (Notes 1 eqn. 29) hanges instantaneously as ρ hanges, and the vetor potential involves the transverse urrent only in equation (21). In spite of this peuliarity, the fields E and B are ausal. (See problem 6.20 whih demonstrates this in a relatively simple ase. This problem may be "relatively" simple, but it is not easy.) When spatial boundaries are present the analysis is more ompliated. We must use the same kind of tehniques that we used in the stati ase, expanding G in eigenfuntions. However, the vetor nature of A makes the problem muh harder. (See Chapter 9 setions 6-12.) 5
5 Radiation from a moving point harge 5.1 The Lienard-Wiehert potentials Here the soure is a point harge q with position r (t) that is moving with veloity v (t). The harge and urrent densities are ρ (x,t) =qδ [x r (t)] and j (x,t) =qvδ [x r (t)] Beause the soure terms are delta-funtions, it turns out to be easier to bak up one step. Then from equation (19), we have: Φ (x, t) = 1 qδ [x r (t )] δ (t t ret ) dt d 3 x 4πε 0 R We do the integral over the spatial oordinates first. Then Φ (x,t) = 1 q δ [t + R (t ) / t] 4πε 0 R (t dt ) where R (t )= x r (t ). To do the t integral, we must re-express the delta-funtion. Reall (Lea eqn 6.10) where f (x i )=0. In this ase: δ [f (x)] = i 1 f (x i ) δ (x x i) (22) f (t )=t + R (t ) t and, sine v = dr/dt, f (t ) = 1+ 1 dr dt =1+ 1 d [x r (t dt )] [x r (t )] = 1+ 1 [x r (t )] x r (t ) d dt [x r (t )] = 1 v [x r (t )] x r (t ) =1 v R R =1 v ˆR The derivative f is always positive, sine v<. The funtion f is zero when t equals the solution of the equation t ret = t R (t ret ) /. A spae-time diagram shows this most easily: t ret is found from the intersetion of the bakward light one from P with the harge s world line. There is only one root. (23) 6
Thus, evaluating the integral using (22) and (23), we get: Φ (x,t) = 1 δ (t t ret ) q dt = 1 q 4πε 0 R (t v R ) 1 4πε 0 v R (24) R R 1 R t ret and similarly A (x, t) = µ 0 qv 4π R 1 v R (25) R t ret These are the Lienhard-Wiehert potentials. It is onvenient to use the shorthand v R v R r v = R 1 = R (26) R so that A (x, t) = µ 0 qv 4π = v tret 2 Φ (27) r v 7
5.2 Calulating the fields The fields are found using the usual relations (notes 1 eqns 40 and 20) E = Φ A and B = A But our expressions for the potentials are in terms of x and t ret, not x and 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, using spherial oordinates, R = r. Our potential may be written: Φ (x, t) Ψ (x,t ret ) A differential hange in the potential due to a hange in the oordinates is dφ = Φ dx + Φ onst t dt Ψ dx + Ψ dt ret = dψ onst tret ret But dt ret = dt dr/, so Φ dx + Φ onst t dt Ψ dx Ψ dr onst tret ret + Ψ dt ret This must be true for any dx and dt. Comparing the oeffiient of dr on both sides, we see that the r omponent of Φ must be modified: Φ r = Ψ onst t r 1 Ψ (28) onst tret ret With this result, we an alulate the fields: Φ = 1 q = 1 q r 4πε 0 r v 4πε 0 rv 2 v (29) where r v = r v r ˆr + ˆθ r v r + ˆφ r v r r r θ r sin θ φ We an hoose our axes with polar axis along the instantaneous diretion of v. Then r v = rv os θ, and r v = 1 v os θ ˆr + ˆθ r v r sin θ In this oordinate system v = vẑ = v ˆr os θ ˆθ sin θ so r v =ˆr v (30) In the non-relativisti limit, v/ 1, to zeroth order in v/, this beomes r v =ˆr. We are also going to need r v = r a (31) 8
Then, using (27), we have E = Φ + 1 Φ onst tret ˆr A = Φ + 1 Φ onst tret ˆr 1 2 (vφ) = Φ + 1 Φ ˆr v Φ onst tret 2 v Inserting our results (29), (30) and (31), we get E = 1 q 4πε 0 rv 2 ˆr v (ˆr v/) q r a rv 2 q a r v 2 Combining the terms that involve the aeleration, we have 1 q E = 4πε 0 rv 2 ˆr v q (ˆr v/)(ˆr a) a (1 ˆr v/) + 4πε 0 r v 2 (1 ˆr v/) q = 4πε 0 rv 2 ˆr v µ + 0 q 4πr v (1 ˆr v/) ˆr ˆr v a Taking the non-relativisti limit 3 v/ 1, (33) beomes: E = 1 q 4πε 0 r 2 ˆr + q [ˆr (ˆr a)] 2 r The first term is the usual Coulomb field whih goes as 1/r 2. The seond term depends on the aeleration a: this is the radiation field. E rad = µ 0 q [ˆr (ˆr a)] (34) 4π r This term dereases as 1/r and dominates at large r. Note also that E rad is perpendiular to ˆr. Next let s alulate the magneti field: B = A onst t = 1 onst tret ˆr µ 0 qv 4π r v B = µ 0 1 4π q v 1 a r v ˆr v r v r v rv 2 = µ 0 1 4π q rv 2 ˆr v v ˆr a + v r a r v rv 2 = µ 0q 4πrv 2 v ˆr µ 0 q/ ˆr a (ˆr a)(1 v ˆr/)+(ˆr v) 4π r v (1 v ˆr/) (32) (33) 3 Eqn (33) is not quite orret if v/ is not small, as we have not done a orret relativisti treatment of time. We are missing some fators of γ and 1 β ˆr. 9
We an simpify this result using (32) for E rad and the fat that ˆr ˆr 0. B = µ 0 q 4πrv 2 v ˆr + µ 0 q ˆr [(ˆr v/)(ˆr a) a (1 ˆr v/)] (35) 4π r v (1 v ˆr/) = B B-S +ˆr E rad / In the limit v/ 1, the first term, whih goes as 1/r 2, is the usual Biot-Savart law result. The seond term, whih goes as 1/r, is the radiation field. Notie that B rad =ˆr E rad / as expeted for an EM wave in free spae. In the non-relativisti limit, B rad = µ 0 q a ˆr (36) 4π r The Poynting flux for the radiation field is: S = 1 E rad B rad = 1 ˆr E rad E rad µ 0 µ 0 = E2 rad ˆr µ 0 (37) where from equation (34): E rad = µ 0 q 1 q a sin θ = a sin θ 4π r 4πε 0 r2 and θ is the angle between a and ˆr. Thus, in the non-relativisti limit S = 1 2 1 q µ 0 4πε 0 r 2 a sin θ q 2 a 2 = (4π) 2 ε 0 3 r 2 sin2 θ Notie that S 1/r 2, the usual inverse square law for light. S is the power radiated per unit area of wavefront, S = dp/da. Writing da in terms of solid angle, da = r 2 dω, we find the power radiated per unit solid angle is independent of distane: dp dω = r2 S = q2 a 2 (4π) 2 ε 0 3 sin2 θ (38) Finally the total power radiated is dp P = dω dω = q2 a 2 2π +1 1 µ 2 (4π) 2 dφdµ ε 0 3 0 1 where as usual µ =osθ. Thus P = q 2 a 2 µ (4πε 0 )2 3 µ3 +1 3 1 P = 2 q 2 a 2 3 4πε 0 3 (39) This result is alled the Larmor formula. In the relativisti ase, we use (33) in (37) to get: S = 1 µ 0 ˆr µ 0 q 4πr v (1 ˆr v/) ˆr ˆr v 2 a 10
and dp dω = 1 µ 0 q 2 ˆr (4π) 2 (1 ˆr v/) 4 ˆr v 2 a The denominator (1 ˆr v/) 4 indiates that the radiation is beamed into a small one around the veloity vetor when v/ 1. Note here that this derivation is not stritly orret when v/ is not negligible, and it leads to the wrong power of (1 ˆr v/) in the denominator. It does indiate qualitatively how the radiation is beamed. For the orret relativisti derivation see Jakson Ch 14 or http://www.physis.sfsu.edu/~lea/ourses/grad/radgen.pdf. Example A partile of harge q and mass m is moving in the presene of a uniform magneti field B = B 0 ẑ. Its speed v. Find the power radiated. First we ompute the aeleration: F = ma = qv B so a = q m v B = q B m v = ω v (40) Only the omponent of v perpendiular to B ontributes, and the motion is a irle with a pointing toward the enter. If the omponent of v parallel to B is not zero, the motion is a helix. v remains unhanged and does not affet the radiation if v,soitwillbeignored from now on. (There are important beaming effets if v/ is NOT 1.) Choose oordinates as shown in the diagram above, and hoose t =0when the partile is on the x axis. Then the angle χ between ˆr and a is found from ˆr â =osχ =sinθos (ωt φ) 11
where ω = qb m is the ylotron frequeny (eqn 40). Thus from (38), dp dω = q 2 q 2 1 (4π) 2 ε 0 3 m v B 0 sin 2 θ os 2 (ωt φ) Taking the time average, we get < dp dω > = q4 (v B 0 ) 2 m 2 (4π) 2 1 sin2 θ = q4 B0K 2 1+os 2 ε 0 3 2 16π 2 ε 0 m 3 3 θ where K is the partile s kineti energy 1 2 mv2. Radiation is maximum along the diretion of B, and minimum in the plane perpendiular to B. The total power radiated is (39): P = 2 q 2 q 2 3 4πε 0 3 m v 2 q 4 (v B 0 ) 2 B 0 = 3 4πε 0 3 m 2 = q4 B0K 2 3πε 0 3 m 3 The radiated power is proportional to the square of the magneti field strength. We an hek the physial dimensions: P = q4 µ 0 B0 2 v 2 q 4 u B v 2 = 6πε 0 µ 0 m 2 3πε 2 0 (m2 ) 2 Sine the energy of a pair of point harges is U = q2 4πε 0 d we have P =(energy length) 2 length energy 1 time volume (energy) 2 = energy time as required. Things get muh more interesting as v. Synhrotron radiation is disussed in J Ch 14 and also in http://www.physis.sfsu.edu/~lea/ourses/grad/radiation.pdf. 12