ELECTRODYNAMICS—lecture notes
second semester 2005
Ora Entin-Wohlman
References:
1. J. D. Jackson, “Classical Electrodynamics”, Wiley.
2. G. B. Arfken, "Mathematical methods for Physicists", Academic Press.
Summary of vector analysis
1. Gradient,



F
F
F
F  xˆ
 yˆ
 zˆ
x
y
z
,
 

a. F (r ) is perpendicular to the surface of constant F that includes the point r .


r dF
b. The gradient of F (r ) : F 
r dr
2. Divergence,
a.
 
 V
  Vx Vy Vz



  ,  V 
x
y
z
is the flux per unit volume through an infinitesimal closed surface,
 
lim 1 
 V 
V  nˆda
v  0 v S
where  is the volume, S is the surface, and n̂ is a unit vector normal to the surface.
3. Curl (rotor),
a. The
n̂
xˆ






V

 ,
x
Vx
yˆ

y
Vy
zˆ

z
Vz
component of the rotor is related to its circulation by
 
lim 1  
nˆ    V 
V  d
S  0 S C
where
n̂
is normal to the plane where C lies.
 
 
 
F
(r
)
V

r
F
(r
)


V

0
,

V  0 .
Exercise: Given the vector
, find
so that
Solution:
F (r )  const. / r 3
4. Div.grad=Laplacian,
5. Rot  grad =0,
 ,
2
2F 2F 2F
 F 2  2  2
x
y
z
2
 
  F  0
2
  
   V  0
6. Div.rot=0,
7. Rot 

 
  

2
rot,   (  V )  (  V )   V
8. Gauss’s theorem

 
ˆ
V

n
da

d
r

  V
S
V
Important consequence:

  1
  r  4
 dr    r   dr   r 3   0

  
Exercise: Show that when V  A  P, where A is a constant vector, Gauss’s theorem gives

 
ˆ
n

P
da

d
r

   P . Hint: Use
S
V
 
 
 
        
ˆn  A  P  nˆ  P  A  nˆ  P  A and   A  P  P    A  A    P .

Solution: When A is a constant vector, then
  
  
  A P  A    P .
Gauss's theorem gives
 



  
 
ˆ
ˆ
d
r


A

P

n

A

P


A

n

P


A

d
r



  P.
S
S
V
The last equality gives the desired identity.
9. Stoke’s theorem
 
 
V

d




  V  nˆda
S

 
Exercise: Show that when V  AF , where A is a constant vector,
Stoke’s theorem gives


ˆ
n


Fda

Fd

  . Hint: use
S
3



 

  ( AF )  F  A  (F )  A .

Solution: When A is a constant vector, then




  ( AF )  (F )  A .
Stoke's theorem gives
 
 
 



 



ˆ
ˆ
ˆ
 FA  d      AF  nda   F  A  n  A   n  F  A   Fd .
S
S
S
The last equality gives the desired identity.
The delta-function
Dirac delta-function obeys the following in one-dimension:
 ( x  a)  0, x  a
a
 dx ( x  a)  1
a
and has the properties
 dxf ( x) ( x  a)  f (a)
  f ( x)   
i
1
df
dx
 dxf ( x) ' ( x)   f ' (a)
 ( x  xi )
xi
Delta functions of trigonometric functions (n and m are integers):
2
 d cos m cos n  
n ,m
0
2
 d sin m sin n  
n ,m
0
4
2
 d cos m sin n  0
0
We also use:
 n 
 n'   a
x  sin 
x    nn'
a

 a
 2
a
 dx sin 
0
a/2
1
1
dx cos(x / a) cos(' x / a)    '

a a / 2
2

Exercise: Calculate
 dxf ( x) (ax  b) 

1
b
f ( ) .
|a|
a
Solution: When a>0, we write y=ax+b and obtain

1
1
b
 y b
dyf

(
y
)

f
(

)


a   a 
a
a
When a<0, we write y=-|a|x+b and obtain

b y
1
1
b
1  b

dyf



(
y
)

f
(
)

f  
 |a| 
| a | 
|a| |a| |a|  a.


Both results are obtained by using
5
 (ax  b) 
1
b
 (x  ) ,
|a|
a
with no need for the change of variables.
In more than one dimension, the delta-function becomes a product. In Cartesian coordinates


 (r  R)   ( x  X ) ( y  Y ) ( z  Z )
In spherical coordinates
 
 (r  r ' ) 
Exercise: Calculate
1
 (r  r ' ) (cos   cos ' ) (   ' )
r2

 
dxdyF
(
r
)

(
A
r
 B),

where A is a two by two matrix,

B
and

r
are
two-dimensional vectors.
Solution: Written explicitly, we have

dxdyF
(
r
) ( A11 x  A12 y  Bx ) ( A21 x  A22 y  B y ) .

We now change the variables of the integration:
x'  A11 x  A12 y
y '  A21 x  A22 y ,
1
( A22 x' A12 y ' )
det A
.
1
y
( A21 x' A11 y ' )
det A
x
and
This can be put in matrix notations as
It remains to note that that
 
 
Ar  r ' and A1r '  r .
x
x'
dxdy 
y
x'
x
dx' dy '
y '
dx' dy ' 
y
| det A | ,
y '
6
to obtain the integral in the form

1
1
1 
1
dx' dy ' F ( A r ' ) ( x'  Bx ) ( y ' By ) 
F ( A B).
| det A | 
| det A |
Examples:
1. Gaussian representation of a one-dimensional delta-function, the limit   0 of
  x 2 
 ( x) 
exp    
    
 
1
2. Gaussian representation of a three-dimensional delta-function, the limit   0 of
 1 
 (r )   
 2 
3/ 2


 1
2
2
2 
exp

x

y

z
 2 2

3
1
3. Other one-dimensional delta-functions: (a) Lorentzian--the limit   0 of
 ( x) 
(b) of
1
1
;
 1  ( x /  ) 2
sin( x /  )
1
 ( x) 

x
2
1/ 
ixt
dte

1 / 
(c) and of
 ( x) 
1

e ( x /  ) , x  0
4. From Gauss’s theorem,
 (r )  
1 2 1 
  
4
r
7
8