Axially symmetric mass distributions

At this point, it is convenient to adopt standard spherical coordinates, $r,\, \theta,\, \phi$, aligned along the $z$-axis. These coordinates are related to regular Cartesian coordinates as follows (see Section A.8):

$$x = r\,\sin\theta\,\cos\phi,$$ (3.24)

$$y =r\,\sin\theta\,\sin\phi,$$ (3.25)

$$z = r\,\cos\theta.$$ (3.26)

Consider an axially symmetric mass distribution; that is, a $\rho({\bf r})$ that is independent of the azimuthal angle, $\phi$. We would expect such a mass distribution to generated an axially symmetric gravitational potential, ${\mit\Phi}(r,\theta)$. Hence, without loss of generality, we can set $\phi =0$ when evaluating ${\mit\Phi}({\bf r})$ from Equation (3.10). In fact, given that $d^{\,3}{\bf r}' = r'^{\,2}\,\sin\theta'\,dr'\,d\theta'\,d\phi'$ in spherical coordinates, this equation yields

$${\mit\Phi}(r,\theta) = - G\int_0^\infty\int_0^\pi\int_0^{2\pi} \f... ...o(r',\theta')\,\sin\theta'}{\vert{\bf r}-{\bf r}'\vert}\,d\phi'\,d\theta'\,dr',$$ (3.27)

with the right-hand side evaluated at $\phi =0$. However, because $\rho(r',\theta')$ is independent of $\phi'$, Equation (3.27) can also be written

$${\mit\Phi}(r,\theta) = - 2\pi\,G\int_0^\infty\int_0^\pi r'^{\,2}\... ...a')\,\sin\theta'\,\langle\vert{\bf r}-{\bf r}'\vert^{-1}\rangle\,d\theta'\,dr',$$ (3.28)

where $\langle\cdots\rangle\equiv \oint (\cdots)\,d\phi'/2\pi$ denotes an average over the azimuthal angle.

Now,

$$\vert{\bf r}'-{\bf r}\vert^{-1} = (r^{\,2}-2\,{\bf r}\cdot{\bf r}' + r'^{\,2})^{-1/2},$$ (3.29)

and

$${\bf r}\cdot{\bf r}' = r\,r'\,F,$$ (3.30)

where (at $\phi =0$)

$$F = \sin\theta\,\sin\theta'\,\cos\phi' + \cos\theta\,\cos\theta'.$$ (3.31)

Hence,

$$\vert{\bf r}'-{\bf r}\vert^{-1} = (r^{\,2}-2\,r\,r'\,F + r'^{\,2})^{-1/2}.$$ (3.32)

Suppose that $r > r'$. In this case, we can expand $\vert{\bf r}'-{\bf r}\vert^{-1}$ as a convergent power series in $r'/r$, to give

$$\vert{\bf r}'-{\bf r}\vert^{-1}= \frac{1}{r}\left[ 1 + \left(\fra... ...rac{r'}{r}\right)^2(3\,F^{\,2}-1) + {\cal O}\left(\frac{r'}{r}\right)^3\right].$$ (3.33)

Let us now average this expression over the azimuthal angle, $\phi'$. Because $\langle 1\rangle =1$, $\langle\cos\phi'\rangle = 0$, and $\langle \cos^2\phi'\rangle = 1/2$, it is easily seen that

$$\langle F\rangle =\cos\theta\,\cos\theta',$$ (3.34)

$$\langle F^{\,2}\rangle = \frac{1}{2}\,\sin^2\theta\,\sin^2\theta' + \cos^2\theta\,\cos^2\theta'$$

$$= \frac{1}{3}+ \frac{2}{3}\left(\frac{3}{2}\,\cos^2\theta-\frac{1}{2}\right)\left(\frac{3}{2}\,\cos^2\theta'-\frac{1}{2}\right).$$ (3.35)

Hence,

$$\left\langle \vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \frac{1}{r}\left[ 1 + \left(\frac{r'}{r}\right)\cos\theta\,\cos\theta' \right.$$

$$\phantom{=}+\left.\left(\frac{r'}{r}\right)^2\left(\frac{3}{2}\co... ...}\cos^2\theta'-\frac{1}{2}\right) + {\cal O}\left(\frac{r'}{r}\right)^3\right].$$ (3.36)

Now, the well-known Legendre polynomials, $P_n(x)$, are defined (Abramowitz and Stegun 1965b) as

$$P_n(x) = \frac{1}{2^n\,n!}\,\frac{d^{\,n}}{dx^{\,n}}\!\left[(x^{\,2}-1)^n\right],$$ (3.37)

for $n=0,\infty$. It follows that

$$P_0(x) = 1,$$ (3.38)

$$P_1(x) = x,$$ (3.39)

$$P_2(x) = \frac{1}{2}\,(3\,x^{\,2}-1),$$ (3.40)

$$P_3(x) =\frac{1}{2}\,(5\,x^{\,3}-x),$$ (3.41)

and so on. The Legendre polynomials are mutually orthogonal:

$$\int_{-1}^1 P_n(x)\,P_m(x)\,dx = \int_0^\pi P_n(\cos\theta)\,P_m(\cos\theta)\,\sin\theta\,d\theta = \frac{\delta_{nm}}{n+1/2}$$ (3.42)

(Abramowitz and Stegun 1965b). Here, $\delta_{nm}$ is 1 if $n=m$, and 0 otherwise. The Legendre polynomials also form a complete set; any function of $x$ that is well behaved in the interval $-1\leq x\leq 1$ can be represented as a weighted sum of the $P_n(x)$. Likewise, any function of $\theta$ that is well behaved in the interval $0\leq\theta\leq \pi$ can be represented as a weighted sum of the $P_n(\cos\theta)$.

A comparison of Equation (3.36) and Equations (3.38)–(3.40) makes it reasonably clear that, when $r > r'$, the complete expansion of $\langle\vert{\bf r}'-{\bf r}\vert^{-1}\rangle$ is

$$\left\langle\vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \frac{... ...sum_{n=0,\infty} \left(\frac{r'}{r}\right)^n P_n(\cos\theta)\,P_n(\cos\theta').$$ (3.43)

Similarly, when $r < r'$, we can expand in powers of $r/r'$ to obtain

$$\left\langle\vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \frac{... ...sum_{n=0,\infty} \left(\frac{r}{r'}\right)^n P_n(\cos\theta)\,P_n(\cos\theta').$$ (3.44)

It follows from Equations (3.28), (3.43), and (3.44) that

$${\mit\Phi}(r,\theta) = \sum_{n=0,\infty} {\mit\Phi}_n(r)\,P_n(\cos\theta),$$ (3.45)

where

$${\mit\Phi}_n(r) = -\frac{2\pi\,G}{r^{\,n+1}}\int_0^r \int_0^\pi r'^{\,n+2}\, \rho(r',\theta')\,P_n(\cos\theta')\,\sin\theta'\,d\theta'\,dr'$$

$$\phantom{=}-2\pi\,G\,r^{\,n}\int_r^\infty \int_0^\pi r'^{\,1-n}\, \rho(r',\theta')\,P_n(\cos\theta')\,\sin\theta'\,d\theta'\,dr'.$$ (3.46)

Given that the $P_n(\cos\theta)$ form a complete set, we can always write

$$\rho(r,\theta) = \sum_{n=0,\infty} \rho_n(r)\,P_n(\cos\theta).$$ (3.47)

This expression can be inverted, with the aid of Equation (3.42), to give

$$\rho_n(r) = (n+1/2)\int_0^\pi\rho(r,\theta)\,P_n(\cos\theta)\,\sin\theta\,d\theta.$$ (3.48)

Hence, Equation (3.46) reduces to

$${\mit\Phi}_n(r) = -\frac{2\pi\,G}{(n+1/2)\,r^{\,n+1}}\int_0^r r'^... ...)\,dr'-\frac{2\pi\,G\,r^{\,n}}{n+1/2}\int_r^\infty r'^{\,1-n}\,\rho_n(r')\,dr'.$$ (3.49)

Thus, we now have a general expression for the gravitational potential, ${\mit\Phi}(r,\theta)$, generated by an axially symmetric mass distribution, $\rho(r,\theta)$.