Geology Reference
In-Depth Information
depends only from the distance
r
. This equation
has the form:
f
(
r
)
D
g
(™,¥). If we fix latitude and
longitude, the right-hand side of (
4.77
) does not
change. Therefore, function
f
must be a constant:
f
(
r
)
D
k
. For reasons that will be clear later, it is
convenient to set
k
D
n
(
n
C
1). In this instance,
taking the left-hand side of (
4.77
)gives:
Then,
sin ™
dT
d™
d
2
L
d
¥
2
1
T sin ™
d
d™
1
Lsin
2
™
C
D
n.n
C
1/
where we have substituted total derivatives for
single variable functions
T
and
L
with respect to
™ and ¥. Finally,
r
2
dR
dr
1
R
d
dr
D
n.n
C
1/
sin ™
dT
d™
sin ™
T
d
d™
C
n.n
C
1/sin
2
™
This
is
an
ordinary
differential
equation,
which can be rewritten as follows:
d
2
L
d¥
2
1
L
D
(4.81)
r
2
d
2
R
dr
2
C
2r
dR
dr
n.n
C
1/R.r/
D
0 (4.78)
This equation has the form
f
(™)
D
g
(¥). If
we fix a longitude ¥, then the right-hand side
of (
4.81
) does not change. Therefore, function
f
must be a constant:
f
(™)
D
k
. Setting
k
D
m
2
and
considering the right-hand side of (
4.81
)gives:
The solutions of (
4.78
) have the form:
r
n
r
.nC1/
R.r/
D
(4.79)
d
2
L
d¥
2
D
m
2
L.¥/
Therefore, we have two classes of solutions
of Laplace's equations, one characterized by in-
creasing values of the radial component for in-
creasing distances from the Earth's center, and
one that decreases with the distance from the
Earth:
(4.82)
This is another ordinary differential equation,
whose solutions have the form:
sin m¥
cosm¥
L.¥/
D
(4.83)
V
e
.r;™;¥/
D
r
n
Y.™;¥/
V
i
.r;™;¥/
D
r
.nC1/
Y.™;¥/
(4.80)
V.r;™;¥/
D
Finally, let us consider the left-hand side of
(
4.81
). This equation is more complicate:
The first class of solutions clearly corresponds
to external sources for the geomagnetic poten-
tial (magnetosphere and ionosphere), whereas
the second class arises from the internal sources
(crustal and core). Now let us focus on the right-
hand side of (
4.77
). We perform further separa-
tion of the variables by assuming that the compo-
nent depending only from colatitude and longi-
tude has the form:
Y
(™,¥)
D
T
(™)
L
(¥). Equating
right-hand side of (
4.77
)to
n
(
n
C
1) gives:
sin ™
dT
d™
sin ™
T
d
d™
C
n.n
C
1/sin
2
™
D
m
2
If we perform the substitution:
x
D
cos™ in this
equation, in a few steps we obtain:
1
x
2
d
2
T
dx
2
2x
dT
dx
n.n
C
1/
1
x
2
T.x/
D
0 (4.84)
m
2
C
sin ™
@Y
@™
@
2
Y
@¥
2
1
Y sin ™
@
@™
1
Y sin
2
™
C
This is the well-known
Legendre equation
,
whose solutions are the
Legendre associate poly-
nomials
. In particular, it can be shown that the
D
n.n
C
1/
