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:
For a fixed value of a,
Ñ2 |
1
|
= |
-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¢;
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:
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
|
|
| |
| |
-(r0/(2e0)) (x - a)2 + (r0 a2)/e0 |
| |
| |
|
. |
| (12) |
The electrostatic field is given by:
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.