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.17):

$$x = r\,\sin\theta\,\cos\phi,$$ (869)

$$y = r\,\sin\theta\,\sin\phi,$$ (870)

$$z = r\,\cos\theta.$$ (871)

Consider an axially symmetric mass distribution: i.e., a $\rho({\bf r})$ which is independent of the azimuthal angle, $\phi$. We would expect such a mass distribution to generated an axially symmetric gravitational potential, $\Phi(r,\theta)$. Hence, without loss of generality, we can set $\phi =0$ when evaluating $\Phi$ from Equation (868). In fact, given that $d^3{\bf r}' = r'^{\,2}\,\sin\theta'\,dr'\,d\theta'\,d\phi'$ in spherical coordinates, this equation yields

$$\Phi(r,\theta) = - G\int_0^\infty\int_0^\pi\int_0^{2\pi} \fr... ...in\theta'}{\vert{\bf r}-{\bf r}'\vert}\,dr'\,d\theta'\,d\phi',$$ (872)

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

$$\Phi(r,\theta) = - 2\pi\,G\int_0^\infty\int_0^\pi r'^{\,2}\,... ...,\langle\vert{\bf r}-{\bf r}'\vert^{-1}\rangle\,dr'\,d\theta',$$ (873)

where $\langle\cdots\rangle$ denotes an average over the azimuthal angle, $\phi'$.

Now,

$$\vert{\bf r}'-{\bf r}\vert^{-1} = (r^{2}-2\,{\bf r}\cdot{\bf r}' + r'^{\,2})^{-1/2},$$ (874)

and

$${\bf r}\cdot{\bf r}' = r\,r'\,F,$$ (875)

where (at $\phi =0$)

$$F = \sin\theta\,\sin\theta'\,\cos\phi' + \cos\theta\,\cos\theta'.$$ (876)

Hence,

$$\vert{\bf r}'-{\bf r}\vert^{-1} = (r^{2}-2\,r\,r'\,F + r'^{\,2})^{-1/2}.$$ (877)

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

$$\vert{\bf r}'-{\bf r}\vert^{-1}= \frac{1}{r}\left[ 1 + \left... ...ght)^2(3\,F^2-1) + {\cal O}\left(\frac{r'}{r}\right)^3\right].$$ (878)

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

$$\langle F\rangle = \cos\theta\,\cos\theta',$$ (879)

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

$$= \frac{1}{3}+ \frac{2}{3}\left(\frac{3}{2}\,\cos^2\theta-\frac{1}{2}\right)\left(\frac{3}{2}\,\cos^2\theta'-\frac{1}{2}\right).$$ (880)

Hence,

$$\left\langle \vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \frac{1}{r}\left[ 1 + \left(\frac{r'}{r}\right)\cos\theta\,\cos\theta' \right.$$ (881)

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

Now, the well-known Legendre polynomials, $P_n(x)$, are defined

$$P_n(x) = \frac{1}{2^n\,n!}\,\frac{d^n}{dx^n}\!\left[(x^2-1)^n\right],$$ (882)

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

$$P_0(x) = 1,$$ (883)

$$P_1(x) = x,$$ (884)

$$P_2(x) = \frac{1}{2}\,(3\,x^2-1),$$ (885)

etc. The Legendre polynomials are mutually orthogonal: i.e.,

$$\int_{-1}^1 P_n(x)\,P_m(x)\,dx = \int_0^\pi P_n(\cos\theta)\... ...(\cos\theta)\,\sin\theta\,d\theta = \frac{\delta_{nm}}{n+1/2}.$$ (886)

Here, $\delta_{nm}$ is 1 if $n=m$, and 0 otherwise. The Legendre polynomials also form a complete set: i.e., any well-behaved function of $x$ can be represented as a weighted sum of the $P_n(x)$. Likewise, any well-behaved (even) function of $\theta$ can be represented as a weighted sum of the $P_n(\cos\theta)$.

A comparison of Equation (881) and Equations (883)-(885) makes it reasonably clear that, when $r > r'$, the complete expansion of $\langle\vert{\bf r}'-{\bf r}\vert^{-1}\rangle$ is

$$\left\langle\vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \... ...\left(\frac{r'}{r}\right)^n P_n(\cos\theta)\,P_n(\cos\theta').$$ (887)

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

$$\left\langle\vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle = \... ...\left(\frac{r}{r'}\right)^n P_n(\cos\theta)\,P_n(\cos\theta').$$ (888)

It follows from Equations (873), (887), and (888) that

$$\Phi(r,\theta) = \sum_{n=0,\infty} \Phi_n(r)\,P_n(\cos\theta),$$ (889)

where

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

$$-2\,\pi\,G\,r^n\int_r^\infty \int_0^\pi r'^{\,1-n}\, \rho(r',\theta')\,P_n(\cos\theta')\,\sin\theta'\,dr'\,d\theta'.$$ (890)

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

$$\rho(r,\theta) = \sum_{n=0,\infty} \rho_n(r)\,P_n(\cos\theta).$$ (891)

This expression can be inverted, with the aid of Equation (886), to give

$$\rho_n(r) = (n+1/2)\int_0^\pi\rho(r,\theta)\,P_n(\cos\theta)\,\sin\theta\,d\theta.$$ (892)

Hence, Equation (890) reduces to

$$\Phi_n(r) = -\frac{2\pi\,G}{(n+1/2)\,r^{n+1}}\int_0^r r'^{\,... ...,\pi\,G\,r^n}{n+1/2}\int_r^\infty r'^{\,1-n}\,\rho_n(r')\,dr'.$$ (893)

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