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Mar 26, 2002

Notes for Lecture Notes for Lecture #22

Derivation of the Lienard-Wiechert potentials and fields

Consider a point charge q moving on a trajectory Rq(t). We can write its charge density as

r(r,t) = q d3(r-Rq(t)),
(1)
and the current density as
J(r,t) = q .
R
 
q 
 (t)d3(r-Rq(t)),
(2)
where
.
R
 
q 
 (t) º d Rq(t)
dt
 .
(3)
Evaluating the scalar and vector potentials in the Lorentz gauge,
F(r,t) = 1
4pe0
 ó
õ 		ó
õ 		 d3r¢ dt¢  r(r¢,t¢)
|r-r¢|
  d(t¢-(t-|r-r¢|/c)),
(4)
and
A(r,t) = 1
4pe0 c2
 ó
õ 		ó
õ 		 d3r¢ dt¢  J(r¢,t¢)
|r-r¢|
  d(t¢-(t-|r-r¢|/c)).
(5)
We performing the integrations over first d3r¢ and then dt¢, and make use of the fact that for any function of t¢,
ó
õ 		¥

¥ 
 f(t¢) d(t¢-(t-|r-Rq(t¢)|/c)) = f(tr)
1-
.
R
 
q 
 (tr)·(r-Rq(tr))
c|r-Rq(tr)|
,
(6)
where the ``retarded time'' is defined to be
tr º t - |r- Rq(tr)|
(7)
. We find
F(r,t) = q
4pe0
 1
R - v·R
c
,
(8)
and
A(r,t) = q
4pe0 c2
 v
R - v·R
c
,
(9)
where we have used the shorthand notation R º r- Rq(tr) and v º [(R)\dot]q(tr).

In order to find the electric and magnetic fields, we need to evaluate

E(r,t) = -ÑF(r,t) - A(r,t)
t
(10)
and
B(r,t) = Ñ×A(r,t).
(11)
The trick of evaluating these derivatives is that the retarded time (7) depends on position r and on itself. We can show the following results using the shorthand notation defined above:
Ñtr = - R
c æ
ç
è 		R- v·R
c
 ö
÷
ø
,
(12)
and
tr
t
 = R
æ
ç
è 		R- v·R
c
 ö
÷
ø
.
(13)

Evaluating the gradient of the scalar potential, we find:

-ÑF(r,t) = q
4 pe0
 1
æ
ç
è 		R- v·R
c
 ö
÷
ø 		3

 
é
ê
ê
ê
ë 		R æ
ç
è 		1- v2
c2
 ö
÷
ø 		- v
c
 æ
ç
è 		R - v·R
c
 ö
÷
ø 		+ R
.
v
 
 ·R
c2
 ù
ú
ú
ú
û 		,
(14)
and
- A(r,t)
t
 = q
4 pe0
 1
æ
ç
è 		R- v·R
c
 ö
÷
ø 		3

 
é
ê
ê
ê
ë 		vR
c
 æ
ç
ç
ç
è 		v2
c2
 - v·R
Rc
 -
.
v
 
 ·R
c2
 ö
÷
÷
÷
ø 		-
.
v
 
 R
c2
 æ
ç
è 		R- v·R
c
 ö
÷
ø 		ù
ú
ú
ú
û 		.
(15)
These results can be combined to determine the electric field:
E(r,t) = q
4 pe0
 1
æ
ç
è 		R- v·R
c
 ö
÷
ø 		3

 
é
ê
ê
ê
ë 		æ
ç
è 		R- vR
c
 ö
÷
ø 		æ
ç
è 		1 - v2
c2
 ö
÷
ø 		+ æ
ç
ç
ç
è 		R× ì
ï
í
ï
î 		æ
ç
è 		R- vR
c
 ö
÷
ø 		×
.
v
 
c2
 ü
ï
ý
ï
þ 		ö
÷
÷
÷
ø 		ù
ú
ú
ú
û 		.
(16)
We can also evaluate the curl of A to find the magnetic field:
B(r,t) = q
4 pe0 c2
 é
ê
ê
ê
ê
ê
ê
ë 		-R×v
æ
ç
è 		R- v·R
c
 ö
÷
ø 		3

 
æ
ç
ç
ç
è 		1 - v2
c2
 +
.
v
 
 ·R
c2
 ö
÷
÷
÷
ø 		-
R× .
v
 
 /c
æ
ç
è 		R- v·R
c
 ö
÷
ø 		2

 
ù
ú
ú
ú
ú
ú
ú
û 		.
(17)
One can show that the electric and magnetic fields are related according to
B(r,t) = R×E(r,t)
c R
 .
(18)

File translated from TEX by TTH, version 2.20.
On 26 Mar 2002, 17:50.