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where
s
is a distance measured in the direction parallel to
k
. The perturbation
vector
k
is confined to the
x
-
z
plane, while
∇
n
is in the
y
direction. Because both
elements
σ
P
of the tensor conductivity are proportional to the density, it
follows that they will depend on
s
in a similar manner,
σ
0
and
)
=
σ
0
+
δσ
0
e
i
(
ks
−
ω
t
)
σ
0
(
s
,
t
)
=
σ
P
+
δσ
P
e
i
(
ks
−
ω
t
)
σ
P
(
s
,
t
Our goal is to determine the perturbation electric field
δ
E
k
that is parallel to
k
, since we are dealing with an electrostatic wave,
E
k
e
i
(
ks
−
ω
t
)
E
k
(
s
,
t
)
=
δ
We now solve for
δ
E
k
in terms of the initial perturbation in
δ
n
by setting
∇·
J
=
0. In the
x
-
z
plane,
J
=
σ
0
E
z
ˆ
a
z
+
σ
P
E
x
ˆ
a
x
For now, we take the zero-order perpendicular electric field to be zero. Since
the zero-order parallel electric field is upward in the case illustrated by Fig. 10.15,
we have
)
)
(
(
σ
0
+
δσ
0
e
i
k
x
x
+
k
z
z
−
ω
t
e
i
k
x
x
+
k
z
z
−
ω
t
J
=
−
E
||
+
δ
E
k
sin
θ
a
z
ˆ
)
)
(
(
σ
p
+
δσ
p
e
i
k
x
x
+
k
z
z
−
ω
t
e
i
k
x
x
+
k
z
z
−
ω
t
+
δ
E
k
cos
θ
a
x
ˆ
Evaluating
∇·
J
, dropping second-order terms, and setting the result equal to
zero yields
−
θ
e
i
(
)
=
k
x
x
+
k
z
z
−
ω
t
ik
z
δσ
0
E
||
+
ik
z
σ
0
δ
E
k
sin
θ
+
ik
x
δσ
p
δ
E
k
cos
0
Substituting
k
z
=
k
sin
θ
and
k
x
=
k
cos
θ
and solving for
δ
E
k
,
θ
σ
0
sin
2
θ
+
σ
p
cos
2
δ
E
k
=
E
||
δσ
0
sin
θ
which may be written
δσ
0
σ
0
sin
θ
δ
E
k
+
E
θ
+
σ
p
/σ
0
cos
2
(10.7)
||
sin
2
θ
Our goal is not to carry out a full algebraic analysis of the process, which is
quite messy, but rather to gain some physical insight. To proceed, however, we
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