Jan 23, 2002
Notes for Lecture
Notes for Lecture #3
[0][0]
Form of Green's function solutions to the Poisson equation
According to Eq. 1.35 of your text for any two three-dimensional functions
f(r) and y(r),
|
|
ó
õ
|
Vol
|
( f(r) Ñ2 y(r) - y(r) Ñ2 f(r) ) d3r = | ó
(ç)
õ
|
Surf
|
(f(r) Ñy(r) - y(r) Ñf(r) ) · |
^
r
|
d2r, |
| (1) |
where [^(r)] denotes a unit vector normal to the integration surface.
We can choose to evaluate this expression with f(r) = F(r)
(the electrostatic potential) and y(r) = G(r,r¢), and also make
use of the identities:
and
|
Ñ2 G(r,r¢) = - 4 pd(r- r¢). |
| (3) |
Then, the Green's identity (1) becomes
|
-4 p |
ó
õ
|
Vol
|
|
æ
ç
è
|
F(r) d(r- r¢) - G(r,r¢) |
r(r)
4 pe0
|
|
ö
÷
ø
|
d3r = | ó
(ç)
õ
|
Surf
|
{F(r) ÑG(r,r¢) - G(r,r¢) ÑF(r) } · |
^
r
|
d2r. |
| (4) |
This expression can be further evaluated. If the arbitrary position, r¢
is included in the integration volume, then the equation (4) becomes
|
F(r¢) = |
ó
õ
|
Vol
|
G(r,r¢) |
r(r)
4 pe0
|
d3r + |
1
4p
|
| ó
(ç)
õ
|
Surf
|
{ G(r,r¢) ÑF(r) - F(r) ÑG(r,r¢) } · |
^
r
|
d2r. |
| (5) |
This expression is the same as Eq. 1.42 of your text if we switch the variables
r¢Û r and also use the fact that Green's function
is symmetric in its arguments: G(r,r¢) º G(r¢,r).
Mean value theorem for solutions to the Laplace equation
Consider an electrostatic field F(r) in a
charge-free region so that it satisfies the Laplace equation:
The ``mean value theorem'' value theorem (problem 1.10 of your textbook)
states that the value of
F(r) at the arbitrary (charge-free) point
r is equal to the average of
F(r¢) over the surface of any sphere
centered on the point r (see Jackson problem #1.10).
One way to prove this theorem is the following. Consider a point
r¢ = r + u, where u will
describe a sphere of radius R about the fixed point
r. We can make a Taylor series expansion of the
electrostatic potential F(r¢) about the
fixed point r:
|
F(r + u ) = F(r) + u·Ñ F(r) + |
1
2!
|
(u·Ñ)2 F(r) + |
1
3!
|
(u·Ñ)3 F(r) + |
1
4!
|
(u·Ñ)4 F(r) + ¼. |
| (7) |
According to the premise of the theorem, we want to integrate both
sides of the equation 7 over a sphere of radius R in
the variable u:
|
|
ó
õ
|
sphere
|
dSu = R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu). |
| (8) |
We note that
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) 1 = 4 pR2, |
| (9) |
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) u ·Ñ = 0, |
| (10) |
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) (u ·Ñ)2 = |
4 pR4
3
|
Ñ2, |
| (11) |
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) (u ·Ñ)3 = 0, |
| (12) |
and
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) (u ·Ñ)4 = |
4 pR6
5
|
Ñ4. |
| (13) |
Since Ñ2 F(r) = 0, the only non-zero term of
the average it thus the first term:
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) F(r + u) = 4 pR2 F(r), |
| (14) |
or
|
F(r) = |
1
4 pR2
|
R2 |
ó
õ
|
2p
0
|
dfu |
ó
õ
|
+1
-1
|
dcos(qu) F(r + u). |
| (15) |
Since this result is independent of the radius R, we see that
we have proven the theorem.
File translated from TEX by TTH, version 2.20.
On 23 Jan 2002, 12:15.