Feb 25, 2011
RETARDED POTENTIALS
At point r =(x,y,z), integrate over charges at positions r’
 
r
 rr  rr d r 
r
r r
j  3r

1
A( r ,t)   r r d r 
c r  r 
r
 ( r ,t) 
3
(7)
(8)
where [ρ] evaluate ρ at retarded time:

 1







r
,tr

r

c


Similar for [j]
So potentials at point

r
and time t are affected by conditions at
r
1r r
pointr at a retarded time,t  r  r 
c
Given a charge and current density,


find retarded potentials and
Aby means of (7) and (8)
Then use (1) and (2) to derive E, B
Fourier transform  spectrum
Radiation from Moving Charges
The Liénard-Wiechart Potentials
Retarded potentials of single, moving charges
Charge q
moves along trajectory
velocity at time t is
charge density
current density
 
r
r
(t)
0
 
(
u
(
t)
r
)
ot




(
r
,
t
)

q
(
r

r
(
t
))
0


 

j
(
r
,
t
)

q
u
(
t
)
(
r

r
(
t
))
0
delta function
Can integrate over volume d3r to get total charge and current
 3
q(r,t)d r
   3
quj(r,t)d r
What is the scalar potential for a moving charge?
Recall




3


(
r
,t)
r
 d
r


r
where [ ] denotes
evaluation at retarded
time





 r


r

 3
 (
r
,
t
)



(
r
,
t
)

d
r
d
t
t

t







r

r
c


Substitute
 
and integrate









(
r
,
t
)

q
r

r
(
t
)
0


d
r

3
light
el
time
trav



between
rand
r

 

r
r


 q

(
r
r
(
t))
3
0




(
r
,t)
t   
t

t

drd



r
r
c




r
r

(
t
t
)
c


q
d
t


r
r
(
t)
0
Now let
then

 
R(t)rr0(t)

R(t)R(t)

vector
scalar



R
(
t
)

1



(
r
,
t
)

q
R
(
t
)
(
t

t
)
d
t

c
Now change variables again
then
R
(
t)



t
t
t
c
1



R


d
t
d
t
(
t
)
d
t
c


u(t) r0(t)

d  
 
R(t )  r r0(t)
dt

u(t)

2
2


R
(
t
)

R
(
t
)
 





2
R
(
t
)
R
(
t
)


2
R
(
t
)

u
(
t
)
Velocity
dot both sides 
Also define unit vector

 R
n
R
1
dt  dt  R(t)dt
c
 1  
 1 R(t )dt
 c

 1     
 1 n(t )  u(t )dt
 c

so



1


R
(
t
)




(
r
,
t
)

q
(
t
)
d
t



1 

1

n
(
t
)

u
(
t
)
c
This means evaluate
integral at t'‘=0,
or t‘=t(retard)
So...

(r,t)
q
(tretard
)R
(tretard
)
where
 
1
(tretard
)
(t)1cn(t)u(t)
beaming factor
or, in the bracket notation:

q
(r,t) 
R


Liénard-Wiechart
scalar potential
Similarly, one can show for the vector potential:

  qu

A

c

R


Liénard-Wiechart
vector potential
Given the potentials

q
(r,t) 
R


one can use


B



A

  qu

A

cR


1
A
E



c
t
to derive E and B.
We’ll skip the math and just talk about the result. (see Jackson §14.1)

The Result: The E, B field at point r and time t depends on
the retarded position r(ret) and retarded time t(ret) of the charge.
Let

(
u

r
)
0t
ret





u

r
(
t
)
0
ret

u


c
velocity
of
charged
particle
accelerati
on



1

n

r r
r r r
B(r,t)  n  E (r,t)


r
r2
r
r


rÝ
r r
(n  ) (1    ) q  n
 r r
E ( r ,t)  q
 
  3  ( n  )  
3 2
 R
c1 4 4R 442 4 4 4 43


1 4 44 2 4 4 43
"VELOCITY FIELD"
1
 2 Coulomb Law
R
Field of particle w/ constant velocity
"RADIATION FIELD"
1

R
Transverse field due to acceleration
Qualitative Picture:
transverse “radiation”
field propagates at velocity c