Geoscience Reference
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point
x
1
. For example, from (4.62) we get
ξ
y
∝
1
/
[
x
−
(
x
1
+
iδ
)]
and
b
y
∝
1
/
[
x
−
(
x
1
+
iδ
)]
.
The direction for bypassing the singular point with small losses is determined
by the sign of the Alfven velocity derivative. If
c
A
(
x
) increases with
x
, then
δ>
0 and the singular point is bypassed from below. But, if
c
A
(
x
) decreases
with
x
, then
δ<
0 and the singular point is bypassed from above. Phases of
ξ
y
,E
x
and
b
y
change rapidly through +
π
with increasing
c
A
(
x
) while passing
the resonance point in the positive direction of
x
and in the opposite case of
the decreasing of
c
A
(
x
)
,
the phase change is
−
π.
Energy Losses
Let us find the relation between energy losses on a resonance magnetic surface,
and the amplitude of the longitudinal magnetic field on this surface. Consider
the most interesting case of low losses in the medium. It is important that as
small as we wish conductivity results in finite energy dissipation at a resonance
field-line. Energy losses can be found by calculating the difference between the
Pointing vector (
S
) components normal to the resonance surface to its left and
right.
We shall find here the energy dissipation on the resonance surface by
direct integrating of Joule losses. Consider a cylinder crossed by the resonance
surface and the cross-section of a unit area. Integrate the averaged Joule losses
over the cylinder. Losses caused by the longitudinal conductivity vanish at
σ
→∞
. The averaged density of Joule dissipation caused by the transversal
conductivity is given by
1
2
(
j
⊥
·
)=
σ
2
E
∗
⊥
E
∗
⊥
(
E
⊥
·
)
.
The
E
y
component having a logarithmic singularity at point
x
1
does not
contribute to losses at
σ
⊥
→
0 . Therefore, the Joule dissipation at small
δ
is
ω
c
|
2
+
∞
+
∞
∆S
=
1
2
=
σ
2
k
y
k
A
(
x
1
)
dx
2
dx
b
0
|
σ
⊥
|
E
x
|
x
1
)
2
+
δ
2
.
(
x
−
−∞
−∞
The scale of plasma density variation at
x
=
x
1
can be estimated as
a
(
x
1
)=
ρ
(
x
1
)
ρ
(
x
1
)
,
where the prime denotes the derivation with respect to
x.
Substituting the
value of the integral and
k
2
||
a
k
A
(
x
1
)
a
k
A
(
x
1
)
≈
=
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