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and we find a new tensor given by
R
R
−
1
σ
=
R
·
σ
·
For the simple case when geomagnetic and geographic north coincide and the
dip angle is
I
R
corresponds to a rotation counterclockwise by the angle
I
about
the
x
axis and
,
⎛
⎞
1
0
0
⎝
⎠
R
=
0
cos
I
−
sin
I
0
+
sin
I
cos
I
R
to
E
determines
J
in the geographic coordinate system.
Following Forbes (1981), we denote the eastward direction as the
Application of
σ
λ
axis, the
northward as the
θ
axis, and the vertical as the
z
-axis. Then
⎛
⎞
σ
λλ
σ
λθ
σ
λ
z
⎝
⎠
R
σ
=
σ
θλ
σ
θθ
σ
θ
z
(3.15)
σ
z
λ
σ
z
θ
σ
zz
with
σ
λλ
=
σ
P
σ
λθ
=−
σ
θλ
=−
σ
H
sin
I
σ
λ
z
=−
σ
z
λ
=+
σ
H
cos
I
σ
θθ
=
σ
P
sin
2
I
(3.16)
+
σ
0
cos
2
I
σ
θ
z
=
σ
z
θ
=
(σ
0
−
σ
P
)
sin
I
cos
I
σ
zz
=
σ
P
cos
2
I
+
σ
0
sin
2
I
Note in this form that if we set
I
=
0, we recover a matrix identical to
σ
in (3.6).
Taking the divergence of
J
and substituting
E
=−∇
φ
, we have the dynamo
equation
R
∇·
σ
·
(
−∇
φ
+
×
)
=
U
B
0
where the quantities are all measured in the earth-fixed frame and expressed in
the rotated coordinate system. This may be written as,
R
R
∇·
σ
·∇
φ
=∇·
σ
·
(
U
×
B
)
(3.17)
R
as known functions, this equation is a complicated par-
tial differential equation with nonconstant coefficients. A common simplifying
Even taking
U
and
σ
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