Geology Reference
In-Depth Information
Taking the divergence,
2
C
·
R
2
C
R
R
3
.
·
∇
=−
(1.309)
R
Then, taking the constant vector to be
F
0
8π(λ
+
2μ)
,
(1.310)
we see that
1
8π(λ
+
F
0
·
R
L
=
(1.311)
2μ)
R
is a particular solution of (1.286) for the scalar displacement potential. For gauge
∇·
A
=
0, equation (1.287) gives, for the
i
th component of the vector displacement
potential,
ξ
ijk
F
0
j
(
x
k
−
ξ
k
)
R
3
1
4π
2
A
i
μ
∇
=−
(
A
F
)
i
=−
,
(1.312)
using expression (1.305) for
A
F
. Next, taking the constant vector
C
to be the
vectors (
e
3
F
02
−
e
2
F
03
), (
e
1
F
03
−
e
3
F
01
)and(
e
2
F
01
−
e
1
F
02
), in turn, we find that
2
(
e
3
F
02
−
e
2
F
03
)
·
R
2
(
e
3
F
02
−
e
2
F
03
)
·
R
∇
=−
,
R
3
R
2
(
e
1
F
03
−
e
3
F
01
)
·
R
2
(
e
1
F
03
−
e
3
F
01
)
·
R
∇
=−
,
(1.313)
R
R
3
2
(
e
2
F
01
−
e
1
F
02
)
·
R
2
(
e
2
F
01
−
e
1
F
02
)
·
R
∇
=−
,
R
R
3
with (
e
1
,
e
2
,
e
3
) representing the Cartesian unit vectors. Combining these results,
we obtain
2
F
0
×
R
2
F
0
×
R
R
3
.
∇
=−
(1.314)
R
We are then led to
1
8πμ
F
0
×
R
A
=
(1.315)
R
as a particular solution of (1.287). Since
F
0
×
R
(
F
0
×
R
)
·
R
∇·
=−
=
0,
(1.316)
R
R
3
the assumed gauge,
∇·
A
=
0, is verified.
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