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ϑ
r
r
r
S
r
r |
Figure 1.1 Small sphere of radius =
, surrounding the source point
r , as the source point and field point, r , approach coincidence.
R
=|
r
For r r , the gradient of 1/ R is
1
R
1
|
r
R 3 .
r
=∇
=−
(1.170)
r |
r
Taking the divergence yields
2 1
R
2 1
|
=∇
r
r |
2 x 1
x 3 2
x 1 2
2 x 2
x 2 2
2 x 3
+
+
3
2
3
R 3 =
=
0.
(1.171)
R 5
Now suppose the source point is close to the field point so that R is the very small
quantity . Surround the source point with a sphere of radius , as illustrated in
Figure 1.1.
Now take the Laplacian of expression (1.168). As the source point and field point
approach coincidence, the right side approaches
u ( r )
2 1
R
d
1
∇·
V .
(1.172)
By the divergence theorem of Gauss, the volume integral throughout the small
sphere is the integral over its surface of the outward normal component of
(1/ R ),
3 . The surface integral is then
3
2
equal to
4π. Thus, the volume
integral (1.168) is the general solution of the Poisson equation (1.163).
An arbitrary vector field u can then be broken into a lamellar part, determined
within a constant by the scalar potential φ given by (1.168), and a solenoidal part
/
/
×
=−
 
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