● wave properties 洛仑兹规范下的达朗贝尔方程是两个波动方程, 因此由它们求出的(A2q)及(E,B)均为波动 形式,反映了电磁场的波动性。 o highly symmetric and independent to each other Get one, get 2nd for free Solution of d'alembert eq under lorenz gauge indicates that EM interaction takes time To study radiation we use lorenz gauge
洛仑兹规范下的达朗贝尔方程是两个波动方程， 因此由它们求出的 及 均为波动 形式，反映了电磁场的波动性。 (A, ) (E, B) l wave properties l highly symmetric and independent to each other Get one, get 2nd for free. Solution of d’alembert eq under Lorenz gauge indicates that EM interaction takes time. To study radiation, we use Lorenz gauge
37.2 Retarded potential 1. Solve d'Alembert equation ssume pr, t)is known. We first solve point charge problem, then use superposition to get general solution
§7.2 Retarded potential Assume is known. We first solve point charge problem，then use superposition to get general solution (x,t) = 1. Solve d’Alembert equation
Assume point charge at origin, p(r, t =g(t8(x) symmetry indicates o=p(r, t)is independent of 0, p, so dAlembert eq for scalar potential is 1oqQ()δ(r) 2 2 2 大 asr≠0 1o=0 2 u(rt let (r, t)= 0
Assume point charge at origin， ， symmetry indicates is independent of , so d’Alembert eq for scalar potential is (x,t) Q(t) (x) = = (r,t) , as r 0 2 2 2 2 1 1 ( ) 0 r r r r c t − = 2 2 2 2 2 0 1 1 ( ) ( ) ( ) Q t r r r c t r r − = − ＊ r u r t r t ( , ) ( , ) = 2 2 2 2 2 1 0 u u r c t − = let
The general solution for 1d wave equation is (r≠0) (=ft-)+8(+)a( g(+ C Outward spherical wave For radiation gt+ c Inward wave g(t+ Qt Compared with static potential we have: o(r, t) )c、cr 兀60 O(xlt-r If point charge is placed at x p(x, t) 容易证明上述解的形式满足波动方程式
r c r f (t − ) r c r g(t + ) Outward spherical wave Inward wave The general solution for 1D wave equation is ( , ) ( ) ( ) c r g t c r u r t = f t − + + r c r g t r c r f t r t ( ) ( ) ( , ) + + − = (r 0) Compared with static potential , we have： r c r Q t r t 0 4 ( ) ( , ) − = ( + ) = 0 c r g t For radiation If point charge is placed at x r c r Q x t x t 4 0 ( , ) ( , ) − = 容易证明上述解的形式满足波动方程＊式
For contineous charge distribution p(',t (x,t) V 4IEor Since A satisfies identical equation as so the solution 4丌V
For contineous charge distribution 0 ( , ) ( , ) V 4 r x t c x t dV r − = dV r c r J x t A x t V − = ( , ) 4 ( , ) 0 Since satisfies identical equation as , so the solution： A