Editing 2DStructure/ToroidalCoordinates (section)

===Connection With the Physical Problem===
[[#OffCenterCircle|Earlier]], we stated that the off-center circle with purple perimeter displayed in Figure 1 is prescribed by the algebraic expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(R_0 - \varpi)^2 + (Z_0 - Z)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_0^2 \, .</math>
  </td>
</tr>
</table>
</div>
Let's plug in the "toroidal-coordinate" expressions for each parameter that appears on the left-hand side of this relation and see whether, after simplification, it reduces to the right-hand side. 
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(R_0 - \varpi)^2 + (Z_0 - Z)^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{a\xi_1}{(\xi_1^2 - 1)^{1/2}}  - \frac{a(\xi_1^2 - 1)^{1/2}}{\xi_1 - \xi_2} \biggr]^2 
+ \biggl[\pm~\frac{a(1-\xi_2^2)^{1/2}}{\xi_1 - \xi_2}\biggr]^2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2}{(\xi_1^2 - 1)} \biggl\{
\biggl[ \xi_1  - \frac{(\xi_1^2 - 1)}{\xi_1 - \xi_2} \biggr]^2 
+ \biggl[\frac{(1-\xi_2^2)^{1/2}(\xi_1^2 - 1)^{1/2} }{\xi_1 - \xi_2}\biggr]^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2}{(\xi_1^2 - 1)(\xi_1-\xi_2)^2} \biggl\{
\biggl[ \xi_1(\xi_1-\xi_2)  - (\xi_1^2 - 1) \biggr]^2 
+ (1-\xi_2^2)(\xi_1^2 - 1)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2}{(\xi_1^2 - 1)(\xi_1-\xi_2)^2} \biggl[
( 1-\xi_1\xi_2 )^2 + (\xi_1^2 - 1 -\xi_1^2\xi_2^2 + \xi_2^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2}{(\xi_1^2 - 1)(\xi_1-\xi_2)^2} \biggl[
\xi_1^2  -2\xi_1\xi_2 +  \xi_2^2
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2}{(\xi_1^2 - 1)} \, .
</math>
  </td>
</tr>
</table>
</div>
This, indeed, equals the right-hand side of the relation, which is, <math>~r_0^2</math>.  It all nicely checks out!

Next, taking a hint from the EUREKA! moment recorded in [[Appendix/Ramblings/ToroidalCoordinates#Relating_CCGF_Expansion_to_Toroidal_Coordinates|our accompanying notes]], let's rewrite the function <math>~\Chi</math> in terms of toroidal rather than cylindrical coordinates, where <math>~\Chi</math> is the argument of the special function, <math>~Q_{-1/2}</math>, that appears in the [[2DStructure/ToroidalCoordinates#Expression_for_the_Axisymmetric_Potential|above definition of <math>~q_0</math>]].  More specifically, let's assume that the coordinate location at which the gravitational potential is to be evaluated, <math>~(R_*, Z_*)</math>, is taken to be the cylindrical-coordinate location of the origin of the toroidal coordinate system, <math>~(a, Z_0)</math>.  Given this association, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Chi </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{R_*^2 + \varpi^2 + (Z_* - Z)^2}{2R_* \varpi}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a^2 + \varpi^2 + (Z_0 - Z)^2}{2a \varpi}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1 + (\varpi/a)^2 + [(Z_0 - Z)/a]^2}{2(\varpi/a) }</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\Chi^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 2 \biggl( \frac{\varpi}{a} \biggr) \biggr]^{-2}
\biggl\{ 1 + \biggl(\frac{\varpi}{a} \biggr)^2 + \biggl[\frac{(Z_0 - Z)}{a} \biggr]^2 \biggr\}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{2(\xi_1^2 - 1)^{1/2}}{\xi_1 - \xi_2} \biggr]^{-2}
\biggl\{ 1 + \biggl[\frac{(\xi_1^2 - 1)^{1/2}}{\xi_1 - \xi_2} \biggr]^2 
+ \biggl[\pm~\frac{(1-\xi_2^2)^{1/2}}{\xi_1 - \xi_2}  \biggr]^2 \biggr\}^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(\xi_1 - \xi_2)^2}{4(\xi_1^2 - 1)} 
\biggl[ 1 + \frac{(\xi_1^2 - 1)}{(\xi_1 - \xi_2)^2} 
+ \frac{(1-\xi_2^2)}{(\xi_1 - \xi_2)^2}  \biggr]^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4(\xi_1^2 - 1)(\xi_1 - \xi_2)^2} 
\biggl[ (\xi_1 - \xi_2)^2 + (\xi_1^2 - 1)
+ (1-\xi_2^2)  \biggr]^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{[ 2\xi_1(\xi_1 - \xi_2 ) ]^2}{4(\xi_1^2 - 1)(\xi_1 - \xi_2)^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\xi_1^2}{\xi_1^2 - 1} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, when the function, <math>~q_0</math>, is rewritten in terms of the elliptic integral of the first kind, <math>~K(\mu)</math>, the modulus of <math>~K</math> can be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2[1+\Chi]^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\biggl[1+\frac{\xi_1}{(\xi_1^2 - 1)^{1/2}} \biggr]^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2(\xi_1^2 - 1)^{1/2}}{(\xi_1^2 - 1)^{1/2}+\xi_1} \, .</math>
  </td>
</tr>
</table>
</div>

This is the key result motivating the use of a toroidal coordinate system to evaluate the gravitational potential:  When expressed in an appropriately defined toroidal coordinate system, the modulus of the special function is a function of one, rather than two, spatial coordinates.  This gives some hope that the integral over the second (angular) coordinate, <math>~\xi_2</math>, can be completed analytically, giving rise to an expression for the gravitational potential whose evaluation only requires numerical integration over a single (radial) coordinate, <math>~\xi_1</math>.

Finally, drawing on discussion in [[Appendix/Ramblings/ToroidalCoordinates#ToroidalScaleFactors|our accompanying set of notes]], we recognize that, expressed in terms of toroidal coordinates, the differential area element in the meridional plane is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~d\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a^2 \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>
  </td>
</tr>
</table>
</div>

Putting everything together, then, the (indefinite) integral expression for <math>~q_0</math>, expressed in terms of toroidal-coordinates, is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int \int \biggl[ \frac{a(\xi_1^2 - 1)^{1/2}}{\xi_1 - \xi_2} 
\biggr]^{1/2} \mu K(\mu) \rho(\xi_1, \xi_2) \biggl[ \frac{a}{(\xi_1 - \xi_2)(\xi_1^2 - 1)^{1/2}} \biggr] 
\biggl[ \frac{a}{(\xi_1 - \xi_2)(1-\xi_2^2)^{1/2}} \biggr] d\xi_1 d\xi_2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^{5/2}
\int (\xi_1^2 - 1)^{-1/4} \mu K(\mu) d\xi_1 \int  \rho(\xi_1, \xi_2) 
\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>~2^{1/2} a^{5/2}
\int [ (\xi_1^2 - 1)^{1/2}+\xi_1 ]^{-1/2} K(\mu) d\xi_1 \int  \rho(\xi_1, \xi_2) 
\biggl[ \frac{d\xi_2}{(\xi_1 - \xi_2)^{5/2}(1-\xi_2^2)^{1/2}} \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>

<!--  OLD WAY
Making the substitution, 
<div align="center">
<math>~\xi_2~ \rightarrow ~ \sin\theta</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~d\xi_2~ \rightarrow ~ \cos\theta ~d\theta</math> ,
</div>
and, 
<div align="center">
<math>~\xi_1~ \rightarrow ~ \cosh x</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~d\xi_1~ \rightarrow ~ \sinh x ~dx</math> ,
</div>
gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2^{1/2} a^{5/2}
\int\limits_{x_\mathrm{min}}^{x_\mathrm{max}} \frac{K(\mu) \sinh x ~dx}{( \sinh x+\cosh x )^{1/2}}  
\int\limits_{\theta_\mathrm{min}}^{\theta_\mathrm{max}}  \rho(\xi_1, \theta) 
\biggl[ \frac{d\theta}{(\xi_1 - \sin\theta)^{5/2}} \biggr]  \, ,
</math>
  </td>
</tr>
</table>
</div>
where, written in terms of <math>~x</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2\sinh x}{\sinh x+\cosh x}\biggr]^{1/2}
</math>
  </td>
</tr>
</table>
</div>

and where we have now explicitly introduced four parameters to set definite limits on the nested pair of integrations.
-->

Making the substitution, 
<div align="center">
<math>~\xi_2~ \rightarrow ~ \cos\theta</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~d\xi_2~ \rightarrow ~ - \sin\theta ~d\theta</math> ,
</div>
gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2^{1/2} a^{5/2}
\int\limits_{\xi_1|_\mathrm{min}}^{\xi_1|_\mathrm{max}}  \frac{K(\mu) d\xi_1}{[ (\xi_1^2 - 1)^{1/2}+\xi_1 ]^{1/2} } 
\int\limits_{\theta_\mathrm{min}}^{\theta_\mathrm{max}}  \rho(\xi_1, \theta) 
\biggl[ \frac{d\theta}{(\xi_1 - \cos\theta)^{5/2}} \biggr]  \, ,
</math>
  </td>
</tr>
</table>
</div>

where we have now explicitly introduced four parameters to set definite limits on the nested pair of integrations.  Making the additional substitution, 
<div align="center">
<math>~\xi_1~ \rightarrow ~ \cosh x</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~\Rightarrow</math> &nbsp; &nbsp; &nbsp; &nbsp;
<math>~d\xi_1~ \rightarrow ~ \sinh x ~dx</math> ,
</div>
gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2^{1/2} a^{5/2}
\int\limits_{x_\mathrm{min}}^{x_\mathrm{max}} \frac{K(\mu) \sinh x ~dx}{( \sinh x+\cosh x )^{1/2}}  
\int\limits_{\theta_\mathrm{min}}^{\theta_\mathrm{max}}  \rho(\xi_1, \theta) 
\biggl[ \frac{d\theta}{(\xi_1 - \cos\theta)^{5/2}} \biggr]  \, ,
</math>
  </td>
</tr>
</table>
</div>
where, written in terms of <math>~x</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{2\sinh x}{\sinh x+\cosh x}\biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>