Geoscience Reference
In-Depth Information
with
ν
ne
ν
ni
a
=
iω
+
ν
n
,
b
=
iω
+
ν
n
,
n
=
ν
ne
+
ν
ni
,
−
−
m
e
m
n
N
N
n
ν
en
,
m
i
m
n
N
N
n
ν
in
.
ν
ne
=
ni
=
Substituting (1.60) into (1.44) and (1.45), we obtain
e
m
e
E
=(
e
m
e
c
[
v
e
×
−
−
iω
+
ν
e
−
aν
en
)
v
e
+
B
0
]
−
(
ν
ei
+
bν
en
)
v
i
,
(1.61)
e
m
i
E
=
e
m
i
c
[
v
i
×
−
(
ν
ie
+
aν
in
)
v
e
−
−
iω
+
ν
i
−
bν
in
)
v
i
,
B
0
]+(
(1.62)
where
ν
i
=
ν
ie
+
ν
in
,ν
ie
=(
m
e
/m
i
)
ν
ei
. Equations (1.61) and (1.62) in
Cartesian coordinates
with the
z
-axis pointing along
B
0
(
B
0
=
B
0
z
,
for definiteness,
B
0
>
0), become
{
x, y, z
}
e
m
e
E
=(
−
−
iω
+
ν
e
−
aν
en
)
v
e
+
ω
ce
Sv
e
−
(
ν
ei
+
bν
en
)
v
i
,
(1.63)
e
m
i
E
=
−
(
ν
ie
+
aν
in
)
v
e
−
ω
ci
Sv
i
+(
−
iω
+
ν
i
−
bν
in
)
v
i
,
(1.64)
where the electron and ion cyclotron frequencies are
eB
0
m
e
c
,
ω
ci
=
eB
0
ω
ce
=
m
i
c
,
and
⎛
⎞
⎛
⎞
v
e,i
=
v
e,i⊥
v
e,i
=
010
v
e,i x
v
e,i y
v
e,i z
⎝
⎠
,
⎝
⎠
.
S
=
−
100
001
The easiest way to solve (1.63) and (1.64) is turn to the basis of circle
polarizations
v
(
±
)
e
=
v
ex
±
iv
ey
,
e
=
v
ez
,
v
(
±
)
i
=
v
ix
±
iv
iy
,
i
=
v
iz
,
E
(
±
)
=
E
x
±
iE
y
,
E
=
E
z
.
Then (1.63) and (1.64) for the transversal components yield
e
m
e
E
(
±
)
,
(
ν
ei
+
bν
en
)
v
(
±
)
aν
en
]
v
(
±
)
[
−
i
(
ω
±
ω
ce
)+
ν
e
−
−
=
−
(1.65)
e
i
e
m
i
E
(
±
)
.
bν
in
]
v
(
±
)
(
ν
ie
+
aν
in
)
v
(
±
)
−
+[
−
i
(
ω
∓
ω
ci
)+
ν
i
−
=
(1.66)
e
i
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