Geoscience Reference
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with
ν ne
ν ni
a =
+ ν n ,
b =
+ 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 ×
+ ν e
en ) v e +
B 0 ]
( ν ei + en ) v i ,
(1.61)
e
m i E =
e
m i c [ v i ×
( ν ie + in ) v e
+ ν i
in ) v i ,
B 0 ]+(
(1.62)
where ν i = ν ie + ν in 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 =(
+ ν e
en ) v e + ω ce Sv e
( ν ei + en ) v i ,
(1.63)
e
m i E =
( ν ie + in ) v e
ω ci Sv i +(
+ ν i
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 + en ) v ( ± )
en ] v ( ± )
[
i ( ω
±
ω ce )+ ν e
=
(1.65)
e
i
e
m i E ( ± ) .
in ] v ( ± )
( ν ie + in ) v ( ± )
+[
i ( ω
ω ci )+ ν i
=
(1.66)
e
i
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