Geoscience Reference
In-Depth Information
Integrating (15.17) over the thin current layer of the thickness
l
, we find
dV
dz
=0
,
z
1
dA
x
dz
+
ilΩζ
1
A
x
1
−
lζ
1
V
1
=0
.
(15.18)
z
1
where
denotes the discontinuity of functions at the boundary
z
=
z
1
and
ζ
1
,A
x
1
,V
1
=
ζ, A
x
,V
(
x
=
x
1
). The velocity
V
on both sides is approxi-
mately equal, the difference being of the order of
Ωl
{}
0
.
1.
Let the boundary coincide with the plane
z
=
z
1
and let the speed and
potential in the rest of space obey the equations
∼
∂
2
V
∂ z
2
−
2
∂V
∂ z
+
Ω
2
V
=0
,
∂
2
A
x
∂ z
2
=0
.
Let us consider magnetic perturbations caused by an upward propagating
acoustic wave. This is
V
=
V
0
exp [(1 +
k
)
z
]
,
k
=
1
−
Ω
2
,
Im
k>
0
.
(15.19)
Assume that the ground is a perfect conductor. Then
A
x
=
(2
z
1
−
z
)
d,
for
z
1
< z,
(15.20)
dz,
for 0
< z<z
1
.
Substituting (15.20) into the second relationship of (15.18), we can express
the integration constant
d
in terms of the gas velocity on the boundary
z
=
z
1
:
ζ
1
l
iΩζ
1
lz
1
−
2
V
(
z
1
)
.
d
=
(15.21)
A uniform gas flux, i.e.
Ω
= 0 produces a magnetic perturbation
b
y
=
4
πσ
1
vHB
0
/c
2
.
Plane Acoustic Pulse
Let the initial speed of gas displacement be
dδ
(
t
)
dt
V
(
t
)=
V
0
t
0
,
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