Chapter 10 Potentials and Fields
10.1 The Potential Formulation
10.2 Continuous Distributions
10.3 Point Charges
10.1 The Potential Formulation
10.1.1 Scalar and vector potentials
10.1.2 Gauge transformation
10.1.3 Coulomb gauge and Lorentz gauge
10.1.1 Scalar and Vector Potentials
 
E( r, t ) field formulism
 
B( r, t )
or

V
(
 r, t ) potential formulism
 
A( r, t )
Maxwell’s eqs



 



E MS


B
B  A
E 
B   J   


t 





B


ES   E
B  0
E  
0
E   V
t





  E   (  A )
B  A
t

  

A
  ( E  A)  0
E  V 
t
t
  
E  A  V
t

( r, t )
 
J( r, t ) 
0
0
0
0
10.1.1 (2)
 
E 

0

A




2
  [ V 
]
 V  (  A )  
t
0
t
0



E
  B  0 J  00
t



 
  (  A)   J    [V  A]
t
t



2
  (  A )  (  A )   A
0
0
0
2




A
V
2
( A   0  0 2 )  (  A   0  0
)   0 J
t
t
10.1.1 (3)
Ex.10.1
V0
  0 k (ct  x )2 zˆ
A   4c
0
for x  ct
for x  ct
where k is a constant, c  ( 0  0 )
1
2

Sol: E   A    0 k (ct  x )zˆ
t
2


 k 
 k
B  A   0
(ct  x )2 yˆ   0 (ct  x )yˆ
4c x
2c
 
, ( E  B  0 for x  ct )
for x  ct


E  0
B  0

0 k 
0 k
E  
(ct  x )( yˆ )  
yˆ
2 x
2

k 
k
B  
(ct  x )(zˆ )  
zˆ
2c x
2
0
0

, J  ?
10.1.1 (4)

E
 kc

zˆ
t
2

   E  0

B
k

yˆ
t
2
0
0
0

 1

E
k
 kc
J  (  B)  
  zˆ  
zˆ  0

t
2c
2
0
0
0
(  
0
at x  0,
E
x 0
z
B
x 0
y
[( 
E
B
x 0
z
x 0
y
, Ex  0
as E1  E 2
1
0
B
x 0
y

1
0
B
x 0
y


 K f  xˆ
kt
kt
)  (  )] yˆ  K f ( zˆ  xˆ ) 
Kf  kt zˆ
2
2
 yˆ
0
0
1
)
c
2
10.1.2 Gauge transformation


A
E  V 
t
V


for V  V  , ( r, t )  ( t )    0

 
How about A  A   ?



  


B    A    A for A  A   ,   ( r, t )
     0




A
A 

E   V 
  V   

t
t
t
A

  V 
 (  )
t
t


    k( t )  
A
t
t
k(t )
  V 
t
   

t
10.1.2 (2)
When 

A   A   

 Gauge transforma tion
V  V 
t 



B    A    A



A
A

E   V 
  V 
t
t
The fields are independent of the gauges.
(note: physics is independent of the coordinates.)
10.1.3 Coulomb gauge and Lorentz gauge
Potential formulation



 V  (  A )  
t


2



A
V
2
( A   0  0 2 )  (  A  0  0
)   0 J
t
t


A

E


V


t
Sources: , J
V, A


B  A

Coulomb gauge:   A
0
2

2V  


V( r, t ) 

1
( r , t )
d

4 0
R
2

A
V
2
 A  0 0 2   0 J  0 0( )
t
t
easy to solve V

difficult to solve A
10.1.3 (2)
Lorentz gauge:

V
  A   0  0
t

2


A
2
 A   0  0 2   0 J 
t


2

V

 2 V  00 2   
t
 0 
2
2
the d’Alembertion
V
2
2
2

A  0 J
f 0
inhomogeneous wave eq.

0
2
    00 2
t
2
wave equation
[Note:Since is with t 2 ,the potentials with both
 t and  t are solutions.]
10.1.3 (3)
Gauge transformation


A  A  
V  V 

t

Coulomb gauge :   A  0




2



A


A

0


A



A



If you have a
and
,

2
Find ,
     A


Then, you have a solution A and   A  0
10.1.3 (4)

V
Lorentz gauge :   A   0  0
t


V


If you have a set of A and V  , and   A   0  0
t


V
V
 2
2
  A   0  0
   A  00
    00 2
t
t
t

 2
V
2

Find  ,     0  0 2    A   0  0
t
t

Then ,you have a set if solutions A and V , and

V
  A   0  0
t
10.2 Continuous Distributions :
With the Lorentz gauge   A   0  0 V ,

2
1
V 
0
where
t
2

2
  2  00
t





A 
2
E


V

,
B  A
A  0 J
t

2
2
(

0
)
,


,
In the static case
t

1

(
r)

1
2
V
(
r
)

d
 V 

4 0
R
0



   0 J( r )
2 A  0 J
A( r ) 
d

4 R
10.2 (2)
For nonstatic case, the above solutions only valid
 
when r  r   ( t  t r )c




for V( r, t ), and A( r, t ) due to  ( r , t r ) and J( r , t r ) ,
where t r is the retarded time. Because the message of


the pensence of
and J must travel a distance R  r  r  ,
the delay is R / c ; that is ,
R
tr  t 
c
(Causality)
10.2 (3)
The solutions of retarded potentials for nonstatic sources are


(
r ,tr )
1

V( r, t ) 
d

4 0
R

 
 0 J( r , t r )
A( r, t ) 
d

4
R
Proof:
1 
1
1 

V (r , t ) 
 ( )  ( )d 
4 0 
R
R 
1
1
Rˆ
 
( )   2 R   2
R  r  r
R
R
R
R

1


( r , t r  t  ) 
t r   (  R )   Rˆ
c
t r
c
c
10.2 (4)
  Rˆ
1
Rˆ 
V 

  2 d


4 0  c R
R 
1  1  Rˆ
Rˆ   Rˆ
Rˆ  
 V
    ( )        2  ()    ( 2 ) d
4 0  c  R
R  R
R 


 
t r   Rˆ
t r
c
Rˆ
Rˆ
1 
1
3 
2 1
  ( 2 )  4 ( R )
 ( )  2
(R
) 2
R
R
R
R R
R
1  1  1    1 
2
3  
 V
 
  4 ( R ) d
  2 
2
2
4 0  c R c R   c R



 ( r, t r  t )
1 2
1  ( r , t r )
 2 2(
d) 

R
0
c  t 4 0
1  2V 1 

 2
 ( r, t )
The same procedure is for proving A.
2
c t
0
2
10.2 (5)
Example 10.2
0, for
I( t )  
I 0 , for
Solution:
0V0

 0  I( t r )
A(s, t ) 
zˆ 
dz
4
R
for
for
s
t ,
c
s
t ,
c

A(s, t )  0
only


E(s, t )  B(s, t )  0
z  (ct )2  s 2
contribute s

t  0 E(s, t )  ?

t  0 B(s, t )  ?
10.2 (6)

0I0
dz
( ct )2 s 2
A(s, t )  (
zˆ )20
4
s2  z2
 I
 0 0 zˆ ln( s 2  z 2  z )
2
0
d
ln( s 2  z 2  z )
dz
( ct )2 s 2

0 I0

zˆ ln( ct  (ct )2  s 2 )  ln s
2
 0 I 0 ct  (ct )2  s 2

ln(
)zˆ
2
s


0 I0c
A
E(s, t )  

zˆ
2
2
t
2 (ct )  s


A z ˆ
B(s, t )    A  

s
0I0
s

2 ct  (ct )2  s 2
1
 2
s


1
2
2z
s2 z2
1
s2  z2  z
1
 2
s  z2

 ( 2s)
2
2 
s  (ct  (ct )  s ) ˆ

2
2
 2 (ct )  s

10.2 (7)
 s 2  ct (ct )2  s 2  (ct )2  s 2 
0I0
1


2s ct  (ct )2  s 2 
(ct )2  s 2


0I0
ct
B(s, t ) 
ˆ
2
2
2s (ct )  s
Note:
ct  (ct )2  s 2
ct
D
   2  1

s
s

c 
c1
2 
ln D 
ln D 
1

t
s 
s D  2  2  1
c 1   2  1 c
1


2
sD
s 2  1
 1
c
1
c


s ct 2
(ct )2  s 2
( ) 1
s
10.2 (8)

 
 t 
ln D 
ln D 

 ln D
s
s 
s  t
1 s 
 ct(  2 )   ln D
s c t
t
c

s (ct )2  s 2
t  ,

E0
 0 I0
B
ˆ
2s
recover the static case
10.3 Point Charges
10.3.1 Lienard-Wiechert Potentials
10.3.2 The Fields of a Moving Point Charge
10.3.1 Lienard-Wiechert potentials

Consider a point charge q moving on a trajectory W(t )
  
R  r  w( t r )
retarded position
location of the observer at time t
tr  t 
R
c
Two issues
•There is at most one point on the trajectory communicating

with r at any time t.
Suppose there are two points:


R1  c( t  t1 ) R 2  c( t  t 2 ),





R1  R 2
R 1  R 2  c( t 2  t1 )  V 
c
t 2  t1
Since q can not move at the speed of light,
there is only one point at meet.
10.3.1 (2)


(
•  r , t r )d  q
the point chage
q


 ( r , t r )d  1  Rˆ  V / c
due to Doppler –shift effect as the point charge is considered as
an extended charge.
Proof.
consider the extended charge has a length L as a train
(a) moving directly to the observer
time for the light to arrive the observer.
L  x L  L x


c
v
c
E
L 
F
L
1 v / c
10.3.1 (3)
(b)moving with an angle  to the observer
L cos   x L  L x


c
v
c
L
L 
1  v cos  / c
 The apparent volume

  ( r , t r )d 
q

1  Rˆ  v / c
 


Rˆ  v
1
c
actual volume
10.3.1 (4)

V( r, t ) 

1 ( r , t r )
1
q

d




4 0
R
4 0 R (1  Rˆ  v )
c

V( r, t ) 
1
qc
 
4 0 Rc  R  v



 
 0 ( r , t r )v( t r )
0 v


A( r, t ) 
d



(

 r , t r ) d
4
R
4 R


 
0
qcv
v

A( r, t ) 
   2 V( r, t )
4 ( Rc  R  v) c
Lienard-Wiechert Potentials for a moving point charge
10.3.1 (5)
Example 10.3
q

v  const
Solution:
let


V( r, t )  ?
 
A( r, t )  ?
w ( t  0)  0


w ( t )  vt
 
R  r  vt r  c( t  t r )
 
2
2
r 2  2 r  vt r  v 2 t r  c 2 ( t 2  2 t t r  t r )
 
 
(c 2 t  r  v )  (c 2 t  r  v )2  (c 2  v 2 )( r 2  c 2 t 2 )
tr 
c2  v2

r
r
tr  t   t 
c
c
 choose  sign

consider v  0,


retarded 


1
10.3.1 (6)
R  c( t  t r )
 
r  vt r
ˆ
R
c( t  t r )

  
 v r  vt r 
Rˆ  v
 R (1 
)  c( t  t r )1  

c
c
c
(
t

t
)

r 
 
v  r v2
 c( t  t r ) 
 tr
c
c
1 1
 

(c 2 t  r  v )2  (c 2  v 2 )( r 2  c 2 t 2 )
c
1
qc
1
qc

V( r, t ) 
 
ˆ
Rv
4 0
4 0 (c 2 t  r  v )2  (c 2  v 2 )( r 2  c 2 t 2 )
Rc(1 
)
c


 
v 
0
qcv
A( r, t )  2 V( r, t )
  2
2
2
2
2
2 2
1
4

c
(
c
t

r

v
)

(
c

v
)(
r

c
t )



0 0
2
c
10.3.2 The Fields of a Moving Point Charge
Lienard-Weichert potentials:

1
qc

v


 
A
(
r
,
t
)

V
(
r, t )
2
4 0 ( Rc  R  v)
c




A
B  A
E  V 
t
  
R  r  w( t r ), R  c( t  t r ), and

V( r, t ) 
Math., Math., and Math,…. are in the following:
 
qc
1
V 
  2 ( Rc  R  v)
4 0 ( Rc  R  v)
 
vw
(t r )
10.3.2 (2)
=
R  ct r
 

 

 
 
(R  v)  (R  )v  ( v  )R  R  (  v)  v  (  R )

t r ˆ  


ˆ
( R   )v  R i  i v j ( t r ) j  R i
vj
j  a( R  t r )
t r
i

 



 
( v  )R  ( v  )r  ( v  )w
a  v



( v  ) r  vi  i rjz ˆj  vi ij ˆj  v
w j t r


ˆ
ˆj  v ( v  t r )
( v   )w  v i  i w j ( t r ) j  v i
t r i
v j t r


ˆ
  v   i v j ( t r )k 
kˆ  a j i t r kˆ  a  t r
t r i





  R    r    w( t r )  (v  t r )  v  t r
   
 
  
 
(R  v)  a(R  t r )  v  v( v  t r )  R  (a  t r )  v  ( v  t r )
  0
0
 
a(R  t r )  t r (R  a )
10.3.2 (4)
 
qc
1
V 
  2 ( Rc  R  v)
4 0 ( Rc  R  v)


 
qc
1

2
2


c

t

v

(
R

a

v
)t r
  2
r
4 0 ( Rc  R  v)


1
qc
R 


2
2

v

(
c

v

R

a
)
 
  2
4 0 ( Rc  R  v) 
Rc  R  v 


  
  
1
qc
2
2

  3 ( Rc  R  v)v  (c  v  R  a )R
4 0 ( Rc  R  v)



  
A Prob.10.17 1
qc
R
a
R 2



2
)  ( c  v  R  a )v 
  3 ( Rc  R  v)(  v 
t
4 0 ( Rc  R  v) 
c
c

10.3.2 (3)
 
 
 
 0  v  ( v  t r )  v( v  t r )  t r ( v  v)
 
 v( v  t r )  v 2t r
 
 

( R  v)  v  ( R  a  v 2 )t r
 
 ct r  R  ( R  R ) 
=
 
1
  ( R  R )
2 RR
 

1 
 ( R  )R  R  (  R )
R
 
R  v(R  t r ) v  t
r
 
1   
 R  v( R  t r )  R  ( v  t r )
R

 

 
1 
v(R  t r )  t r (R  v)
 R  ( R  v)t r 
R
R
t r 
 
Rc  R  v
=
10.3.2 (5)

 
  
  
A
1
qc
2
2
E( r, t )  V   
  3 ( Rc  R  v)v  (c  v  R  a )R
t
4 0 ( Rc  R  v)
   Ra R 2
  
2
 ( Rc  R  v)(  v 
)  ( c  v  R  a )v 
c
c

 

1
qR

 
2
2
ˆ

  3 (c  v  R  a )( cR  v)  R  (cRˆ  v)a ]
4 0 ( Rc  R  v)


define
 
 


ˆ
u  cR  v  R  u  Rc  R  v
 
E( r, t ) 


1
qR
     
2
2 
  3 (c  v )u  u( R  a )  a( R  u )
4 0 ( R  u )
 
  
1
qR
2
2 
E( r, t ) 
  3 (c  v )u  R  ( u  a )
4 0 ( R  u )


generalized Coulomb field

if a  0, v  0,
 
E( r, t ) 
1 q ˆ
R
2
4 0 R


u  cRˆ  v
radiation field or acceleration
field dominates at large R
Electrostatic field
10.3.2 (6)
 v
A 2V


1
1
c

 

B  A


(
V
v
)

V
(


v
)  v  (V)
2
2
c
c
 
aR


   v  a  t r 
 
Rc  R  v

  
  
1
qc
2
2
V 
  3 ( Rc  R  v)v  (c  v  R  a )R
4 0 ( Rc  R  v)

 
 1  1
   
qc
aR
1
qc
2
2
B 2 
 
 
  3 ( c  v  R  a )v  R 
c  4 0 ( Rc  R  v) ( Rc  R  v) 4 0 ( Rc  R  v)

10.3.2 (7)



1 q
1
       2
2

  3 R   a( R  u )  v( R  a )  v(c  v )
c 4 0 ( R  u )


  2
       
1ˆ  1
qR
2
 R
  3 ( R  v)(c  v )  ( R  v)(R  a )  a( R  u ) 
c
 4 0 ( R  u )



1ˆ  1
qR  2
      
2
 R
  3 u( c  v )  u ( R  a )  a ( R  u ) 
c
 4 0 ( R  u )



   
1ˆ  1
qR  2
2
 R
  3 u( c  v )  R  ( u  a ) 
c
 4 0 ( R  u )

1ˆ  
 R  E( r, t )
c
 
1ˆ  
B( r, t )  R  E( r, t )
c
10.3.2 (8)

The force on a test charge Q with velocity V due to a moving
charge q with velocity v is

 
F  Q(E  VB)


   
qQ
R  2
  V ˆ
2 
2
2 

  3 (c  v )u  R  ( u  a )   R  (c  v )u  R  ( u  a )  
4 0 ( R  u ) 
c

 
  
Where R, u, v, and a are all evaluated at t r

10.3.2 (9)
Example 10.4
Solution:
q
 
wvt

a0

v  const
t0

w
at
 
E( r, t )  ?
 
B( r, t )  ?
origin
q (c 2  v 2 )R 
  3 u
4 0 ( R  u )


 



Ru  cR  Rv  c( r  Vt r )  c( t  t r )v  c( r  vt )
 
E( r, t ) 


1
 
 Ex.10.3
  2
2
2
2
2
2 2 2
R  u  Rc  R  v
(c t  r  v)  (c  v )(r  c t )
Prob.10.14
Rc 1  v sin  / c

cR
2
 
E( r, t ) 
2
q
(c 2  v 2 )
1 3
4 0


2
2
2
Rc(1  v sin  / c) 


  
R  r  vt
10.3.2 (10)
 
E( r, t ) 
q
1  v2 / c2
Rˆ
4 0 (1  v 2 sin2  / c 2 ) 3 2 R 2
 
1 ˆ  
B( r, t )  R  E( r, t )
c
 
1   
B( r, t )  2 v  E( r, t )
c

E point to pˆ
by coincidenc e

 r  vtr (r  vt )  (t  tr )v P v
R

 
R
R
R c
when v 2  c 2 ,
 
1 q ˆ
E (r , t ) 
R,
2
4 0 P
Coulomb`s law

p
 
 q 
B(r , t )  0 2 (v  Pˆ )
4 P
“Biot-savart Law for a point charge.”