Jan 19, 2000

Notes for Lecture Notes for Lecture #2

``Proof'' of the identity (Eq. (1.31))

Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
= - 4 pd3(r-r¢).
(1)

Noting that

ó
õ


[b]1insmall sphereabout r¢  
d3r   d3(r-r¢) f(r) = f(r¢),
(2)

we see that we must show that

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) = - 4 pf(r¢).
(3)

We introduce a small radius a such that:

1
|r-r¢|
=
lim
a® 0 
1
Ö
|r-r¢|2 + a2
 
.
(4)

For a fixed value of a,

Ñ2 1
Ö
|r-r¢|2 + a2
 
= -3 a2
(|r-r¢|2 + a2)5/2
.
(5)

If the function f(r) is continuous, we can make a Tayor expansion about the point r = r¢. Jackson's text shows that it is necessary to keep only the leading term. The integral over the small sphere about r¢ can be carried out analytically, by changing to a coordinate system centered at r¢;

u = r-r¢,
(6)

so that

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) » f(r¢) ó
õ


u < R 
  d3u -3 a2
(u2 + a2)5/2
.
(7)

We note that

ó
õ


u < R 
  d3u -3 a2
(u2 + a2)5/2
= 4 p ó
õ
R

0 
  du    -3 a2 u2
(u2 + a2)5/2
= 4 p -R3
(R2 + a2)3/2
.
(8)

If the infinitesimal value a is a << R, then (R2 +a2)3/2 » R3 and the right hand side of Eq. 8 is - 4 p. Therefore, Eq. 7 becomes,

ó
õ


[b]1insmall sphereaboutr¢  
d3r   Ñ2 æ
ç
è
1
|r-r¢|
ö
÷
ø
f(r) » f(r¢) (- 4 p),
(9)

which is consistent with Eq. 3.

Examples of solutions of the one-dimensional Poisson equation

Consider the following one dimensional charge distribution:

r(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
-r0
for -a < x < 0
+r0
for 0 < x < a
0
for x > a
(10)

We want to find the electrostatic potential such that

d2 F(x)
d x2
= - r(x)
e0
,
(11)
with the boundary condition F(-¥) = 0.

In class, we showed that the solution is given by:

F(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
(r0/(2e0)) (x + a)2
for -a < x < 0
-(r0/(2e0)) (x - a)2 + (r0 a2)/e0
for 0 < x < a
(r0 a2)/e0
for x > a
.
(12)

The electrostatic field is given by:

E(x) = ì
ï
ï
í
ï
ï
î
0
for x < -a
-(r0/e0) (x + a)
for -a < x < 0
(r0/e0) (x - a)
for 0 < x < a
0
for x > a
.
(13)

The electrostatic potential can be determined by piecewise solution within each of the four regions or by use of the Green's function G(x,x¢) = x < , where,

F(x) = 1
e0
ó
õ
¥

-¥ 
G(x,x¢) r(x¢) d x¢.
(14)

In the expression for G(x,x¢) , x < should be taken as the smaller of x and x¢. It can be shown that Eq. 14 gives the identical result for F(x) as given in Eq. 12.


File translated from TEX by TTH, version 2.20.
On 19 Jan 2000, 16:58.