Editing AxisymmetricConfigurations/PoissonEq (section)

==Intermediate==

<table border="1" width="95%" align="center" cellpadding="10"><tr><td align="left">
Much of the material enclosed in this ''text box'' has been drawn from &sect;3.1 of Jackson (1975).

In spherical coordinates, <math>~(r,\theta,\phi)</math>, the Poisson equation takes the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~4\pi G \rho(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{r} \frac{\partial^2}{\partial r^2} (r\Phi) 
+ \frac{1}{r^2 \sin\theta} \frac{\partial}{\partial\theta} \biggl( \sin\theta \frac{\partial\Phi}{\partial\theta}\biggr)
+ \frac{1}{r^2 \sin^2\theta} \frac{\partial^2\Phi}{\partial\phi^2} \, .
</math>
  </td>
</tr>
</table>
As it turns out, the potential &#8212; which, generally, is a function of all three coordinates &#8212; can be written as the product of three functions, each of which is a function of only one coordinate.  To demonstrate this, specifically suppose that,
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<tr>
  <td align="right">
<math>~\Phi(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{U( r )}{r} P(\theta)Q(\phi) \, .
</math>
  </td>
</tr>
</table>
Then the differential form of the Poisson equation can be rewritten as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~4\pi G \rho(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{PQ}{r} \frac{d^2U}{dr^2} 
+ \frac{UQ}{r^3\sin\theta} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP}{d\theta}\biggr) 
+ \frac{UP}{r^3\sin^2\theta} \frac{d^2Q}{d\phi^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[ \frac{r^3 \sin^2\theta}{ UPQ} \biggr] 4\pi G \rho(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{r^3 \sin^2\theta}{ UPQ} \biggr]  \biggl\{
\frac{PQ}{r} \frac{d^2U}{dr^2} 
+ \frac{UQ}{r^3\sin\theta} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP}{d\theta}\biggr) 
+ \frac{UP}{r^3\sin^2\theta} \frac{d^2Q}{d\phi^2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{r^2\sin^2\theta}{U} \biggr] \frac{d^2U}{dr^2} 
+ \frac{\sin\theta}{P} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP}{d\theta}\biggr) 
+ \frac{1}{Q} \frac{d^2Q}{d\phi^2} \, .
</math>
  </td>
</tr>
</table>
Setting, <math>~Q = e^{\pm im\phi}</math>, gives,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \biggl[ \frac{r^3 \sin^2\theta}{ UP e^{\pm i m\phi }} \biggr] 4\pi G \rho(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{r^2\sin^2\theta}{U} \biggr] \frac{d^2U}{dr^2} 
+ \frac{\sin\theta}{P} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP}{d\theta}\biggr) 
-m^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[ \frac{r^3 }{ UP e^{\pm i m\phi}} \biggr] 4\pi G \rho(r,\theta,\phi)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{r^2 }{U} \biggr) \frac{d^2U}{dr^2} 
+ \biggl[ \frac{1}{P\sin\theta} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP}{d\theta}\biggr) 
- \frac{m^2}{\sin^2\theta} \biggr] \, .
</math>
  </td>
</tr>
</table>

Now, if for each selected pair of the integer indexes, <math>~(\ell, m)</math>, the function <math>~P</math> is identified as the associated Legendre function, <math>~P_\ell^m</math>, as defined above in Table 2, then the set of terms inside the square brackets on the right-hand-side of this last expression can be set equal to <math>~[-\ell(\ell + 1)]</math> because every associated Legendre function satisfies the differential equation,

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<tr>
  <td align="right">
<math>~ \frac{1}{P_\ell^m \sin\theta} \frac{d}{d\theta}\biggl(\sin\theta \frac{dP_\ell^m}{d\theta}\biggr) + \ell(\ell+1) - \frac{m^2}{\sin^2\theta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>

Rewriting our adopted ''separable'' expression for the potential as the sum over all possible <math>~(\ell,m)</math>-component solutions, that is,
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<tr>
  <td align="right">
<math>~\Phi(r,\theta,\phi) = \sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell [\Phi(r,\theta,\phi)]_{\ell m}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{\ell=0}^\infty \sum_{m=-\ell}^\ell  \frac{U_{\ell m}( r )}{r} P_\ell^m(\theta)Q_m(\phi) \, ,
</math>
  </td>
</tr>
</table>

a solution to the Poisson equation then imposes the requirement that, for each <math>~(\ell,m)</math> pair,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl( \frac{r^2 }{U_{\ell m}} \biggr) \frac{d^2U_{\ell m} }{dr^2} - \ell(\ell+1)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ \frac{r^3 }{ U_{\ell m} P_\ell^m  e^{\pm i m\phi}} \biggr] 4\pi G [\rho(r,\theta\phi)]_{\ell m} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\frac{1}{r} \frac{d^2U_{\ell m} }{dr^2} - \biggl[ \frac{\ell(\ell+1)}{r^2} \biggr]\frac{U_{\ell m} }{r}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ \frac{1}{ P_\ell^m  e^{\pm i m\phi}} \biggr] 4\pi G [\rho(r,\theta\phi)]_{\ell m}  \, .</math>
  </td>
</tr>
</table>

</td></tr></table>