Geology Reference
In-Depth Information
If
F
(
r
)
0fasterthan1/
r
2
as
r
→∞
→
,
the surface integrals may be
neglected, thus
1
R
1
4π
F
(
r
)
d
V
,
L
F
=−
∇
·
(1.300)
and
1
R
1
4π
F
(
r
)
d
V
,
A
F
=
∇
×
(1.301)
where we have used the identity
∇
1
R
1
R
=−∇
.
(1.302)
1.4.6 Kelvin's problem
If the body force density is concentrated at a point, it may be represented as
F
(
r
)
=
F
0
δ(
r
−
r
0
),
(1.303)
where δ(
r
−
r
0
) is the Dirac delta distribution. In this case, the body force is
a point force
F
0
at
r
=
r
0
.If(ξ
1
,ξ
2
,ξ
3
) are the components of
r
0
,wehave
R
2
(
x
1
−
ξ
1
)
2
(
x
2
−
ξ
2
)
2
(
x
3
−
ξ
3
)
2
=
+
+
for
R
=
r
−
r
0
. The integrals (1.300) and
(1.301) then give,
1
R
1
4π
∇
1
4π
R
F
0
R
3
,
·
L
F
=−
·
F
0
=
(1.304)
and
1
R
1
4π
∇
1
4π
F
0
×
R
R
3
,
A
F
=
×
F
0
=
(1.305)
since
1
R
R
R
3
.
∇
=−
(1.306)
Taking the divergence,
F
0
×
R
3
(
F
0
×
R
)
·
R
∇·
=−
=
0,
(1.307)
R
3
R
5
verifies that
A
F
has the correct gauge,
0.
If
C
is an arbitrary constant vector, we have the gradient
∇·
A
F
=
C
·
R
C
R
−
(
C
·
R
)
R
R
3
∇
=
.
(1.308)
R
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