Editing Appendix/Ramblings/ToroidalCoordinates (section)

===Multipole Moment in Toroidal Coordinates===
While it might not be interesting or useful to impose this constraint ''in general'', it will likely be instructive to evaluate the potential at the location where this constraint is satisfied.  That is, we want to evaluate the potential inside a uniform density, circular-cross-section torus at the location,
<div align="center">
<math>
\Phi[(x_0^2 - r_c^2)^{1/2},0] = - \frac{2G}{(x_0^2 - r_c^2)^{1/4}} q_0 .
</math>
</div>

Since in this case the argument of <math>~Q_{-1/2}</math> can be expressed in terms of the "radial" toroidal coordinate, it is reasonable to write the relevant moment of the mass distribution, <math>~q_0</math>, entirely in terms of toroidal coordinates.  Specifically,
<div align="center">
<math>
~q_0 = a^{5/2} \int\int \biggl[ \frac{({\xi_1'}^2 - 1)^{1/2}}{\xi_1' - \xi_2'} \biggr]^{1/2} \rho(\xi_1',\xi_2') Q_{-1/2}(\xi_1') \biggl[ \frac{d\xi_1'}{(\xi_1' - \xi_2')({\xi_1'}^2 - 1)^{1/2}} \biggr] \biggl[ \frac{d\xi_2'}{(\xi_1' - \xi_2')(1-{\xi_2'}^2)^{1/2}} \biggr] .
</math>
</div>

Now suppose that the density distribution is only a function of the ''radial'' coordinate, that is, suppose that <math>~\rho = \rho(\xi_1')</math>.  Then the integral can be written as,
<div align="center">
<math>
~q_0 = a^{5/2} \int \rho(\xi_1') Q_{-1/2}(\xi_1')\biggl[ \frac{d\xi_1'}{({\xi_1'}^2 - 1)^{1/4}} \biggr] \int  \biggl[ \frac{d\xi_2'}{(\xi_1' - \xi_2')^{5/2}(1-{\xi_2'}^2)^{1/2}} \biggr] .
</math>
</div>
Presumably the integral over <math>~d\xi_2'</math> can be completed in closed form if the density distribution fills out the entire circular cross-section, that is, if the limits on integration are <math>~-1</math> to <math>~+1</math>.  Alternatively, write <math>~\xi_2'</math> in terms of <math>~\sin\theta</math> and integrate from <math>~-\tfrac{\pi}{2}</math> to <math>~\tfrac{\pi}{2}</math>. Let's do this.

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

<tr>
  <td align="right">
<math>~\frac{q_0}{a^{5/2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int\limits_0^{\xi_1^'} \rho(\xi_1') Q_{-1/2}(\xi_1')\biggl[ \frac{d\xi_1'}{({\xi_1'}^2 - 1)^{1/4}} \biggr] 
\int\limits_{-1}^{1}  \biggl[ \frac{d\xi_2'}{(\xi_1' - \xi_2')^{5/2}(1-{\xi_2'}^2)^{1/2}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int\limits_0^{\xi_1^'} \rho(\xi_1') Q_{-1/2}(\xi_1')\biggl[ \frac{d\xi_1'}{({\xi_1'}^2 - 1)^{1/4}} \biggr] 
\int\limits_{-\pi/2}^{\pi/2}  \biggl[ \frac{d\theta}{(\xi_1' - \sin\theta)^{5/2}} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Now, using WolframAlpha's online integrator, we find &hellip;
<div align="center">
[[File:TorusIntegration.png|450px|WolframAlpha Integration]]
</div>

Hence &#8212; continuing to substitute <math>~a</math> for <math>~\xi_1^'</math> &#8212; the definite integral gives,

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

<tr>
  <td align="right">
<math>~\int\limits_{-\pi/2}^{\pi/2}  \biggl[ \frac{d\theta}{(a - \sin\theta)^{5/2}} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{2}{3(a^2-1)^2 (a-\sin\theta)^{3/2}}
\biggl[
\cos\theta ( -5a^2 + 4a\sin\theta + 1) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ (a+1)(a-1)^2 \biggl( \frac{a-\sin\theta}{a-1} \biggr)^{3/2} F\biggl(\frac{\pi - 2\theta}{4} \biggr| \frac{-2}{a-1}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 4a(a-1)(\sin\theta-a) \biggl( \frac{a-\sin\theta}{a-1} \biggr)^{1/2} E\biggl(\frac{\pi - 2\theta}{4} \biggr| \frac{-2}{a-1}  \biggr)
\biggr] \biggr\}_{-\tfrac{\pi}{2}}^{\tfrac{\pi}{2}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{2}{3(a^2-1)^2 (a-1)^{3/2}}
\biggl[(a+1)(a-1)^2 \biggl( \frac{a-1}{a-1} \biggr)^{3/2} F\biggl(0\biggr| \frac{-2}{a-1}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 4a(a-1)(1-a) \biggl( \frac{a-1}{a-1} \biggr)^{1/2} E\biggl(0 \biggr| \frac{-2}{a-1}  \biggr)
\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-\biggl\{ \frac{2}{3(a^2-1)^2 (a+1)^{3/2}}
\biggl[(a+1)(a-1)^2 \biggl( \frac{a+1}{a-1} \biggr)^{3/2} F\biggl(\frac{\pi}{2}\biggr| \frac{-2}{a-1}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 4a(a-1)(1+a) \biggl( \frac{a+1}{a-1} \biggr)^{1/2} E\biggl(\frac{\pi}{2} \biggr| \frac{-2}{a-1}  \biggr)
\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \frac{2}{3(a-1)^2 (a+1)^2 (a-1)^{3/2}}\biggl[(a+1)(a-1)^2 F\biggl(0\biggr| \frac{2}{1-a}  \biggr)
+ 4a(a-1)(1-a) E\biggl(0 \biggr| \frac{2}{1-a}  \biggr)
\biggr] \biggr\}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-\biggl\{ \frac{2}{3(a-1)^2 (a+1)^2 (a+1)^{3/2}}
\biggl[(a+1)(a-1)^2 \biggl( \frac{a+1}{a-1} \biggr)^{3/2} F\biggl(\frac{\pi}{2}\biggr| \frac{2}{1-a}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 4a(a-1)(1+a) \biggl( \frac{a+1}{a-1} \biggr)^{1/2} E\biggl(\frac{\pi}{2} \biggr| \frac{2}{1-a}    \biggr)
\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3 (a+1) (a-1)^{3/2}} ~F\biggl(0\biggr| \frac{2}{1-a}  \biggr)
- \frac{8a }{3 (a+1)^2 (a-1)^{3/2}}~ E\biggl(0 \biggr| \frac{2}{1-a}  \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~-\frac{ 2 }{3 (a+1) (a-1)^{3/2}} ~ F\biggl(\frac{\pi}{2}\biggr| \frac{2}{1-a}  \biggr)
+ \frac{8a }{3 (a+1)^{2} (a-1)^{3/2}} ~ E\biggl(\frac{\pi}{2} \biggr| \frac{2}{1-a}    \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3 (a+1) (a-1)^{3/2}} ~\biggl\{ \biggl[ F\biggl(0\biggr| \frac{2}{1-a}  \biggr) -  F\biggl(\frac{\pi}{2}\biggr| \frac{2}{1-a}  \biggr) \biggr]
- \frac{4a }{(a+1) }~ \biggl[ E\biggl(0 \biggr| \frac{2}{1-a}  \biggr) - E\biggl(\frac{\pi}{2} \biggr| \frac{2}{1-a}    \biggr) \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Now, according to (for example) [http://dlmf.nist.gov/19.6#ii NIST's ''Digital Library of Mathematical Functions''],
<div align="center">
<math>~F(0,k) = 0</math> &nbsp; and &nbsp; <math>F(\tfrac{\pi}{2},k) = K(k) \, ,</math>
</div>
where, <math>~K(k)</math> is the complete elliptic integral of the first kind.  Also, according to [http://dlmf.nist.gov/19.6#iii NIST's ''Digital Library of Mathematical Functions''],
<div align="center">
<math>~E(0,k) = 0</math> &nbsp; and &nbsp; <math>E(\tfrac{\pi}{2},k) = E(k) \, ,</math>
</div>
where, <math>~E(k)</math> is the complete elliptic integral of the second kind.  Hence we deduce that,

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

<tr>
  <td align="right">
<math>~\int\limits_{-\pi/2}^{\pi/2}  \biggl[ \frac{d\theta}{(\xi_1' - \sin\theta)^{5/2}} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{3 ({\xi_2'}^2-1) (\xi_1'-1)^{1/2}} ~\biggl[ 
\frac{4\xi_1' }{(\xi_1'+1) } \cdot  E\biggl(\frac{2}{1-\xi_1'}    \biggr)  -  K\biggl(\frac{2}{1-\xi_1'}  \biggr) 
\biggr] \, ,
</math>
  </td>
</tr>

</table>
</div>
which implies,

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

<tr>
  <td align="right">
<math>~\frac{3q_0}{2a^{5/2}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int\limits_0^{\xi_1^'} \rho(\xi_1') Q_{-1/2}(\xi_1')\biggl[ \frac{d\xi_1'}{({\xi_1'}^2 - 1)^{5/4} (\xi_1'-1)^{1/2}} \biggr] ~\biggl[ 
\frac{4\xi_1' }{(\xi_1'+1) } \cdot  E\biggl(\frac{2}{1-\xi_1'}    \biggr)  -  K\biggl(\frac{2}{1-\xi_1'}  \biggr) 
\biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>

Next, we might as well also insert the [http://dlmf.nist.gov/14.5#v NIST relation],
<div align="center">
<math>Q_{-1/2}(\cos\theta) = K\biggl[\cos(\tfrac{1}{2}\theta)\biggr]\, .</math>
</div>