Geology Reference
In-Depth Information
Fig. 4.24
Wavelength of
spherical harmonics as a
function of degree
Transforming this equation to local coordi-
nates gives:
Equation
4.104
is known as
Jeans relation
.
It links the wavelength of individual spherical
harmonics to the corresponding degree, indepen-
dently from the order. Figure
4.24
shows a plot of
œ as a function of
n
. We note that in order to reach
a resolution of
1,000 km it is necessary to have
n
D
40.
A 3-D view of the oscillations associated with
surface harmonics is shown in Fig.
4.25
.We
stress again the importance of the result that
any
function that satisfies Laplace's equation can
be represented by a sum of spherical harmonics
(Eq.
4.93
), which is the spherical analogue of the
classic Fourier series. We shall see soon that this
result furnishes a powerful tool for the parametric
fit to observed data and the creation of global
models of the geomagnetic field.
Let us consider again the magnetic pressure
P
m
D
B
2
/2
0
(Eq.
4.56
). This quantity has di-
mensions Pa
D
Nm
2
@
2
Y
n
@
2
Y
n
n.n
C
1/
a
2
2
Y
n
Š
Y
n
.™;¥/
(4.100)
@Ÿ
2
C
@§
2
Dr
Y
n
(™, ¥)
Therefore,
the
surface
harmonics
satisfy
the
scalar
Helmholtz's
equation
and
are
eigenvectors
of
the
two-dimensional
C
@
2
/@§
2
with eigenval-
ues
n
(
n
C
1)/
a
2
. The structure of this equation
clearly suggests a solution that is a combination
of sines and cosines. Therefore, to interpret
the harmonics
Y
n
(™, ¥)aswaves,wetrythe
following solution:
2
D
@
2
/@Ÿ
2
Laplacian
r
Y
n
.Ÿ;§/
D
e
ik
(4.101)
where ¡
D
(Ÿ,§)and
k
D
(
k
Ÿ
,
k
§
)isthe
wave
vector
. This function has wavelength œ given by:
D
Jm
3
. Therefore, it
represents an
energy density
for the magnetic
field. In fact, any magnetic field
B
is generated by
a system of macroscopic or microscopic currents.
These currents must be maintained by some emf
and require a certain amount of work. It is possi-
ble to show that the total work that is necessary
to set up the system of currents that generates a
magnetic field
B
in a region
R
is given by:
2
k
D
2
q
k
Ÿ
C
k
§
œ
D
(4.102)
Substituting the ansatz (
4.101
)into(
4.100
)
gives:
n.n
C
1/
a
2
Z
k
2
D
(4.103)
1
2
0
E
D
B .r/
B .r/dV
(4.105)
R
Therefore,
This is the total energy stored by the magnetic
field in the region
R
. Now we want to calculate
2 a
p
n.n
C
1/
Š
2 a
n
C
1=2
œ
D
(4.104)
