Green s function for the wave equation Non-relativistic case January 2018 1 The wave equations In the Lorentz Gauge, the wave equations for the potentials are (Notes 1 eqns 44 and 43): 1 2 2 2 2 0 (1) and 1 2 Φ 2 2 2 Φ (2) 0 The Gauge condition is (Notes 1 eqn 42): + 1 Φ 2 0 (3) In Coulomb Gauge we have the Gauge condition: 0 (4) which leads to the wave equations (Notes 1, eqn on top of page 20, with 0) 2 Φ (5) 0 and (Notes 1, eqn 41, with 0) 1 2 2 2 2 0 1 2 Φ (6) 2 Longitudinal and transverse currents The Coulomb Gauge wave equation for (6) is awkward because it contains the scalar potential Φ. We can eliminate the potential Φ to obtain an equation for alone. First we separate the current into two pieces, called the longitudinal current and the transverse current : + where 0 (7) and 0 (8) 1
(Basically we are applying the Helmholtz theorem. We can always do this: See Lea Appendix II). From charge conservation (Notes 1 eqn 7): + 0 and using equation (5) to eliminate, this becomes: 0 2 Φ ³ 0 Φ Thus, up to an irrelevant constant, 0 Φ + But from (8) 0+ ³ 0 ³ 2 So we take 0and then 0 Φ (9) Using equation (9) in equation (6), we have: 2 1 2 µ 2 2 0 0 Φ 0 ³ 0 (10) In the Coulomb Gauge, the transverse current is the source of We can also use result (9) to express in terms of Since equation (5) is the same as in the static case, the solution is also the same (Notes 1 eqn 29). Thus ( ) 1 4 Z ( 0 ) 0 3 0 1 Z 4 (0 ) 0 3 0 ( ) 1 Z 0 4 ( 0 ) 0 3 0 (11) where we used charge conservation in the last step. We can also express in terms of as follows, using (11), + 1 Z 4 0 0 3 0 2
Let s work on the integral. We start with an "integration by parts": Z 0 ( 0 Z Ã! ) 0 3 0 0 ( 0 Z ) 0 3 0 0 1 0 3 0 Z ( 0 Z ) ˆ 0 2 0 + Z 0+ ( 0 ) 0 3 0 The surface integral is zero provided that is localized. Then à ( ) ( )+ 1 Z! 4 ( 0 ) 0 3 0 " + 1 4 + 1 4 + 1 4 à Z " à Z " à Z à Z ( 0 ) 0 3 0 ( 0 ) 0 3 0 ( 0 ) 0 3 0 1 0 3 0 (Notes 1 eqn 19)!!! Z # ( + 2 0 ) 0 3 0 Z + Z + # 2 1 0 3 0 [ 4 ( 0 )] 3 0 # ( ) 1 4 ( 0 ) 0 3 0 (12) We still have the rather unphysical result from equation (5) that Φ changes instantaneously everywhere as changes. In classical physics the potential is just a mathematical construct that we use to find the fields, and it can be shown (e.g. J prob 6.20) that is causal even though Φ is not.! 3 The Green s function With either gauge we have a wave equation of the form 2 Φ 1 2 Φ 2 2 (source) where Φ may be either the scalar potential (in Lorentz Gauge) or a Cartesian component of (In Coulomb Gauge the scalar potential is found using the methods we have already developed for the static case.) The corresponding Green s function problem is: 2 ( ; 0 0 ) 1 2 2 2 4 ( 0 ) ( 0 ) (13) where the source is now a unit event located at position 0 and happening at time 0 As usual, the primed coordinates 0 0 are considered fixed for the moment. To solve this 3
equation we first Fourier transform in time (see, eg, Lea pg 503): ( ; 0 0 ) 1 Z ( ; 0 0 ) 2 and the transformed equation (13) becomes µ 2 + 2 ( ; 0 0 ) 4 1 Z ( 0 ) ( 0 ) 2 2 4 2 ( 0 ) 0 So let ( ; 0 0 )( 0 ) 0 2 and then satisfies the equation 2 + 2 4 ( 0 ) (14) where In free space without boundaries, must be a function only of 0 and must posess spherical symmetry about the source point 1. Thus in spherical coordinates with origin at the point 0 with coordinates 0, we can write: 1 2 ³ 2 ()+2 4 (15) For 6 0 the right hand side is zero. Then the function satisfies the exponential equation, and the solution is: + 1 + 6 0 (16) Near the origin, where the delta-function contributes, the second term on the LHS of (14) is negligible compared with the first, and equation (14) becomes: 2 ' 4 ( 0 ) We recognize that this equation has the solution 1 This is consistent with equation (16) as 0 provided that + 1 Thus we have the solution ( ; 0 0 ) 1 +(1 ) 0 2 You should convince yourself that this solution is correct by differentiating and stuffing back into equation (15). 1 Note that the operator 2 + 2 is spherically symmetric. 4
Nowwedotheinversetransform: Z ( ; 0 0 1 1 ) + 0 2 2 1 2 Z { exp [ ( + 0 )] + exp [ ( + 0 )]} [0 ( )] + [0 ( + )] (17) The second term is usually rejected (take 0 and thus 1)because it predicts a response to an event occurring in the future. However, Feynman and Wheeler 2 have proposed a theory in which both terms are kept. They show that this theory can be consistent with observed causality provided that the universe is perfectly absorbing in the infinite future. (This now appears unlikely.) The time that appears in the first term is called the retarded time ret. Thus we take ( ; 0 0 ) 1 [0 ( )] 1 (0 ret ) (18) Causality (an event cannot precede its cause) requires that the symmetry of this Green s function is: ( ; 0 0 )( 0 0 ; ) (See Morse and Feshbach Ch 7 pg 834-835). and also ( ; 0 0 )0 and ( ; 0 0 )0for 0 4 The potentials Now that we have the Green s function (18), we can solve our original equations. Modifying eqn 1.44 in Jackson ("Formal" Notes eqn 7) to include time dependence, and with we get an integral over a volume in space-time rather than just space: Φ ( ) 1 Z ( 0 0 ) ( ; 0 0 ) 0 3 0 4 0 Thus, inserting (18), we have Z 1 ( 0 0 ) Φ ( ) 4 0 (0 ret ) 0 3 0 (19) Z 1 ( 0 ret ) 3 0 (20) 4 0 ( ret ) in Lorentz Gauge, and similarly: ( ) 0 4 Z ( 0 ret ) ( ret ) 3 0 (21) Notice that these equations have the same form as the static potentials (equations 1.17 and 2 Reviews of Modern Physics, 1949, 21,425 5
5.32 in Jackson, eqns 21 and 29 in Notes 1), but we must evaluate the source and the distance at the retarded time. This allows for the time for a signal to travel from the source to the observer at speed Note that ret is a function of both and 0,aswellas Similar equations hold in Coulomb Gauge. The scalar potential (Notes 1 eqn. 29) changes instantaneously as changes, and the vector potential involves the transverse current only in equation (21). In spite of this peculiarity, the fields and are causal. (See problem 6.20 which demonstrates this in a relatively simple case. This problem may be "relatively" simple, but it is not easy.) When spatial boundaries are present the analysis is more complicated. We must use the same kind of techniques that we used in the static case, expanding in eigenfunctions. However, the vector nature of makes the problem much harder. (See Chapter 9 sections 6-12.) 5 Radiation from a moving point charge 5.1 The Lienard-Wiechert potentials Here the source is a point charge with position () that is moving with velocity () The charge and current densities are () [ ()] and () [ ()] Because the source terms are delta-functions, it turns out to be easier to back up one step. Then from equation (19), we have: Φ ( ) 1 Z [ 0 ( 0 )] ( 0 ret ) 0 3 0 4 0 We do the integral over the spatial coordinates first. Then Φ () 1 Z [0 + ( 0 ) ] 4 0 ( 0 0 ) where ( 0 ) ( 0 ) To do the 0 integral, we must re-express the delta-function. Recall (Lea eqn 6.10) where ( )0 In this case: [ ()] X ( 0 ) 0 + (0 ) 1 0 ( ) ( ) (22) 6
and, since 0 ( 0 ) 1+ 1 0 1+ 1 p [ (0 0 )] [ ( 0 )] 1+ 1 [ ( 0 )] ( 0 ) 0 [ (0 )] 1 [ (0 )] ( 0 ) 1 ˆ 1 The derivative 0 is always positive, since The function is zero when 0 equals the solution of the equation ret ( ret ) A space-time diagram shows this most easily: ret is found from the intersection of the backward light cone from with the charge s world line. There is only one root. (23) Thus, evaluating the integral using (22) and (23), we get: Φ () 1 Z ( 0 ret ) ³ 0 1 4 0 ( 0 ) 1 4 0 and similarly ( ) 0 4 ³ 1 ³ 1 (24) ret (25) ret 7
These are the Lienhard-Wiechert potentials. It is convenient to use the shorthand Ã! 1 (26) so that ( ) 0 4 ret 2 Φ (27) 5.2 Calculating the fields The fields are found using the usual relations (notes 1 eqns 40 and 20) Φ and But our expressions for the potentials are in terms of and ret not and so we have to be very careful in taking the partial derivatives. We can put the origin at the instantaneous position of the charge to simplify things. Then, using spherical coordinates, Our potential may be written: Φ ( ) Ψ ( ret ) A differential change in the potential due to a change in the coordinates is Φ Φ + Φ const Ψ + Ψ ret Ψ const ret ret But ret so Φ + Φ const Ψ Ψ const ret ret + Ψ ret This must be true for any and Comparing the coefficient of on both sides, we see that the component of Φ must be modified: Φ Ψ 1 Ψ (28) const const ret ret With this result, we can calculate the fields: Φ 1 1 4 0 4 0 2 (29) where µ ˆ + ˆθ µ + ˆφ µ sin We can choose our axes with polar axis along the instantaneous direction of Then cos and ³1 cos ˆ + ˆθ ³ sin In this coordinate system ³ ˆ ˆ cos ˆ sin 8
so ˆ (30) In the non-relativistic limit, 1 to zeroth order in this becomes ˆ We are also going to need (31) Then, using (27), we have Φ + 1 Φ const tret ˆ Φ + 1 Φ const tret ˆ 1 2 (Φ) µ ˆ Φ + 1 Φ const tret Inserting our results (29), (30) and (31), we get 1 ½ µ 4 0 2 ˆ (ˆ ) 2 Combining the terms that involve the acceleration, we have µ Φ 2 ¾ 2 µ 1 4 0 2 ˆ (ˆ )(ˆ ) (1 ˆ ) + 4 0 2 (1 ˆ ) µ 4 0 2 ˆ µ + 0 4 (1 ˆ ) ˆ ˆ Taking the non-relativistic limit 3 1, (33) becomes: 1 4 0 2 ˆ [ˆ (ˆ )] 2 The first term is the usual Coulomb field which goes as 1 2. The second term depends on the acceleration : this is the radiation field. rad 0 [ˆ (ˆ )] (34) 4 This term decreases as 1 and dominates at large Note also that rad is perpendicular to ˆ (32) (33) 3 Eqn (33) is not quite correct if is not small, as we have not done a correct relativistic treatment of time. We are missing some factors of and 1 ˆ 9
Next let s calculate the magnetic field: const µ 1 const ret ˆ 0 4 µ 0 1 4 1 µ ˆ 2 µ 0 1 4 2 ˆ + ˆ µ + 2 0 4 2 ˆ 0 4 (1 ˆ) (ˆ )(1 ˆ)+(ˆ ) We can simpify this result using (32) for rad and the fact that ˆ ˆ 0 0 4 2 ˆ + 0 ˆ [(ˆ )(ˆ ) (1 ˆ )] (35) 4 (1 ˆ) B-S +ˆ rad In the limit 1 the first term, which goes as 1 2 is the usual Biot-Savart law result. The second term, which goes as 1, is the radiation field. Notice that rad ˆ rad as expected for an EM wave. In the non-relativistic limit, rad 0 ˆ (36) 4 The Poynting flux for the radiation field is: Ã 1 rad rad 1 ˆ rad! rad 0 0 2 rad ˆ 0 (37) where from equation (34): rad 0 1 sin sin 4 4 0 2 and istheanglebetween and ˆ Thus, in the non-relativistic limit 1 µ 2 1 0 4 0 2 sin 2 2 (4) 2 0 3 2 sin2 Notice that 1 2 the usual inverse square law for light. is the power radiated per unit area of wavefront, Writing in terms of solid angle, 2 Ω we find the power radiated per unit solid angle is independent of distance: Ω 2 2 2 (4) 2 0 3 sin2 (38) 10
Finally the total power radiated is Z Ω Ω 2 2 (4) 2 0 3 where as usual cos Thus 2 2 (4 0 )2 3 Z 2 Z +1 0 1 +1 µ 3 3 1 1 2 2 2 2 3 4 0 3 (39) This result is called the Larmor formula. In the relativistic case, we use (33) in (37) to get: 1 µ 0 ˆ 0 4 (1 ˆ ) ˆ ˆ 2 and Ω 1 0 2 µˆ (4) 2 (1 ˆ ) 4 ˆ 2 The denominator (1 ˆ ) 4 indicates that the radiation is beamed into a small cone around the velocity vector when 1 Note here that this derivation is not strictly correct when is not negligible, and it leads to the wrong power of (1 ˆ ) in the denominator. It does indicate qualitatively how the radiation is beamed. For the correct relativistic derivation see Jackson Ch 14 or http://www.physics.sfsu.edu/~lea/courses/grad/radgen.pdf. Example A particle of charge and mass is moving in the presence of a uniform magnetic field 0ˆ Its speed Find the power radiated. First we compute the acceleration: so (40) Only the component of perpendicular to contributes, and the motion is a circle with pointing toward the center. If the component of parallel to is not zero, the motion is a helix. k remains unchanged and does not affect the radiation if, so it will be ignored from now on. (There are important beaming effects if is NOT 1) 11
Choose coordinates as shown in the diagram above, and choose 0when the particle is on the axis. Then the angle between ˆ and is found from ˆ ˆ cos sincos ( ) where is the cyclotron frequency (eqn 40). Thus from (38), Ω 2 ³ 1 (4) 2 0 3 0 2 sin 2 cos 2 ( ) Taking the time average, we get Ω 4 ( 0 ) 2 2 (4) 2 0 3 µ 1 sin2 4 0 2 1+cos 2 2 16 2 0 3 3 where is the particle s kinetic energy 1 2 2. Radiation is maximum along the direction of and minimum in the plane perpendicular to The total power radiated is (39): 2 2 ³ 3 4 0 3 2 0 2 4 ( 0 ) 2 3 4 0 3 2 4 0 2 3 0 3 3 The radiated power is proportional to the square of the magnetic field strength. We can check the physical dimensions: 4 0 2 0 6 0 0 2 ³ 2 4 3 2 0 (2 ) 2 ³ 2 12
Since the energy of a pair of point charges is we have 2 4 0 (energy length) 2 energy length volume time 1 (energy) 2 energy time as required. Things get much more interesting as Synchrotron radiation is discussed in J Ch 14 and also in http://www.physics.sfsu.edu/~lea/courses/grad/radiation.pdf. 13