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Next: Potential due to uniform Up: Newtonian gravity Previous: Gravitational potential energy


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):

    $\displaystyle x$ $\displaystyle = r\,\sin\theta\,\cos\phi,$ (3.24)
    $\displaystyle y$ $\displaystyle =r\,\sin\theta\,\sin\phi,$ (3.25)
and   $\displaystyle z$ $\displaystyle = r\,\cos\theta.$ (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

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

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

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

Now,

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

and

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

where (at $ \phi =0$ )

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

Hence,

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

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

    $\displaystyle \langle F\rangle$ $\displaystyle =\cos\theta\,\cos\theta',$ (3.34)
and   $\displaystyle \langle F^{\,2}\rangle$ $\displaystyle = \frac{1}{2}\,\sin^2\theta\,\sin^2\theta' + \cos^2\theta\,\cos^2\theta'$    
      $\displaystyle = \frac{1}{3}+ \frac{2}{3}\left(\frac{3}{2}\,\cos^2\theta-\frac{1}{2}\right)\left(\frac{3}{2}\,\cos^2\theta'-\frac{1}{2}\right).$     (3.35)

Hence,

$\displaystyle \left\langle \vert{\bf r}'-{\bf r}\vert^{-1}\right\rangle$ $\displaystyle = \frac{1}{r}\left[ 1 + \left(\frac{r'}{r}\right)\cos\theta\,\cos\theta' \right.$    
  $\displaystyle \phantom{=}+\left.\left(\frac{r'}{r}\right)^2\left(\frac{3}{2}\co...
...}\cos^2\theta'-\frac{1}{2}\right) + {\cal O}\left(\frac{r'}{r}\right)^3\right].$ (3.36)

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

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

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

$\displaystyle P_0(x)$ $\displaystyle = 1,$ (3.38)
$\displaystyle P_1(x)$ $\displaystyle = x,$ (3.39)
$\displaystyle P_2(x)$ $\displaystyle = \frac{1}{2}\,(3\,x^{\,2}-1),$ (3.40)
$\displaystyle P_3(x)$ $\displaystyle =\frac{1}{2}\,(5\,x^{\,3}-x),$ (3.41)

and so on. The Legendre polynomials are mutually orthogonal:

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

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

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

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

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

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

where

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

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

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

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

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

Hence, Equation (3.46) reduces to

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


next up previous
Next: Potential due to uniform Up: Newtonian gravity Previous: Gravitational potential energy
Richard Fitzpatrick 2016-03-31