Axially symmetric mass distributions
At this point, it is convenient to adopt standard spherical coordinates,
, aligned along the
-axis. These coordinates are related to
regular Cartesian coordinates as follows (see Section A.8):
Consider an axially symmetric mass distribution; that is, a
that is independent of the azimuthal angle,
. We would expect
such a mass distribution to generated an axially symmetric gravitational
potential,
. Hence, without loss of generality, we can
set
when evaluating
from Equation (3.10). In fact,
given that
in spherical coordinates, this equation yields
 |
(3.27) |
with the right-hand side evaluated at
. However, because
is independent of
, Equation (3.27)
can also be written
 |
(3.28) |
where
denotes an average over the azimuthal angle.
Now,
 |
(3.29) |
and
 |
(3.30) |
where (at
)
 |
(3.31) |
Hence,
 |
(3.32) |
Suppose that
. In this case, we can expand
as a convergent power series in
, to give
![$\displaystyle \vert{\bf r}'-{\bf r}\vert^{-1}= \frac{1}{r}\left[
1 + \left(\fra...
...rac{r'}{r}\right)^2(3\,F^{\,2}-1)
+ {\cal O}\left(\frac{r'}{r}\right)^3\right].$](img404.png) |
(3.33) |
Let us now average this expression over the azimuthal angle,
. Because
,
, and
, it is easily seen that
Hence,
Now, the well-known Legendre polynomials,
, are defined (Abramowitz and Stegun 1965b) as
![$\displaystyle P_n(x) = \frac{1}{2^n\,n!}\,\frac{d^{\,n}}{dx^{\,n}}\!\left[(x^{\,2}-1)^n\right],$](img417.png) |
(3.37) |
for
.
It follows that
and so on.
The Legendre polynomials are mutually
orthogonal:
 |
(3.42) |
(Abramowitz and Stegun 1965b).
Here,
is 1 if
, and 0 otherwise. The Legendre polynomials also form a complete set; any function
of
that is well behaved in the interval
can be represented as a weighted sum of the
. Likewise,
any function of
that is well behaved in the interval
can
be represented as a weighted
sum of the
.
A comparison of Equation (3.36) and Equations (3.38)–(3.40) makes it reasonably clear that, when
, the complete expansion
of
is
 |
(3.43) |
Similarly, when
, we can expand in powers of
to obtain
 |
(3.44) |
It follows from Equations (3.28), (3.43), and (3.44)
that
 |
(3.45) |
where
Given that the
form a complete set, we can always
write
 |
(3.47) |
This expression can be inverted, with the aid of Equation (3.42), to
give
 |
(3.48) |
Hence, Equation (3.46) reduces to
 |
(3.49) |
Thus, we now have a general expression for the gravitational potential,
,
generated by an axially symmetric mass distribution,
.