Editing Appendix/Ramblings/ToroidalCoordinates (section)

==Confusing and Misleading Steps==

===But Not Every Circle Will Do===
It is very important to appreciate that, although surfaces of constant <math>\Chi</math> (or, equivalently, surfaces of constant <math>\xi_1</math>) are always off-center circles, it is not the case that every off-center circle will prove to be a <math>\Chi= \mathrm{constant}</math> surface in the most relevant toroidal coordinate system.  To be more specific, suppose we want to evaluate the potential at some location <math>(R,0)</math> inside or outside of a uniform-density torus whose meridional cross-section is a circle of radius <math>r_c</math> and whose center is located on the <math>x</math>-axis at position <math>x_0</math>.  The equation describing the cross-sectional surface of this torus is,
<div align="center">
<math>
(R' - x_0)^2 - {z'}^2 = r_c^2 .
</math>
</div>
Dividing through by the square of a (as yet unspecified) scale length, <math>a</math>, gives,
<div align="center">
<math>
\biggl[ \chi^2 - \frac{x_0}{a} \biggr]^2 - \zeta^2 = \frac{r_c^2}{a^2} .
</math>
</div>
This dimensionless expression will only describe a <math>\Chi = \mathrm{constant}</math> surface in an MF53 toroidal coordinate system if, simultaneously,
<div align="center">
<math>
\Chi = \frac{x_0}{a} ~~~~~\mathrm{and}~~~~~ \Chi^2 - 1 = \frac{r_c^2}{a^2} .
</math>
</div>
That is, only if,
<div align="center">
<math>
a = (x_0^2 - r_c^2)^{1/2} .
</math>
</div>
But in the above discussion we were only able to associate the dimensionless argument of the special function in CT99's CCGF expansion with the "radial" coordinate of the MF53 toroidal coordinate system by setting <math>a = R</math>, that is, only by setting the scale length equal to the cylindrical coordinate value <math>R</math> ''at which the potential is to be evaluated''.  So the surface of our torus will only align with a <math>\xi_1 = \mathrm{constant}</math> surface in a toroidal coordinate system if,
<div align="center">
<math>
R = (x_0^2 - r_c^2)^{1/2} .
</math>
</div>
This is a very tight constraint that usually will not be satisfied.


===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>

===Older, Apparently Irrelevant Material===
If we subtract "1" from both sides of this expression, the right-hand-side (RHS) takes on a familiar form:
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_\mathrm{CT99}-1
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2R R'} \biggl[ R^2 + {R'}^2 + (z - z')^2 - 2 R R' \biggr] 
</math>
  </td>
</tr>

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

</table>

It appears as though the quantity, <math>[2R R'(\chi_\mathrm{CT99}-1)]^{1/2}</math>, is the radius <math>r_c</math> of a circle whose center is located at either <math>(R,z)</math> or <math>(R',z')</math>, that is, whose center is shifted off the origin of a cylindrical coordinate system.  I'm not yet sure how/if we can benefit from recognizing this association.

<font color="red">'''Case A:'''</font> Suppose we associate <math>R'</math> with the center of a toroidal cross-section and, at the same time, associate <math>R</math> with the inner edge of a ''particular'' toroidal cross-section that is associated with the toroidal coordinate <math>\xi_1</math>.  We know that the scale-length <math>a</math> that is associated with the chosen toroidal coordinate system must be given by the ratio,
<div align="center">
<math>
a = \frac{R'}{\chi_0} .
</math>
</div>
[NOTE:  As I'm doing this, I'm realizing that it may be wiser to associate <math>a</math> directly with the coordinate location <math>R'</math>.  But let's play this out first and see.]

Then it also will be true that,
<div align="center">
<math>
a = \frac{R}{\chi_\mathrm{inner}} = R \biggl[ \frac{\xi_1+1}{\xi_1 - 1} \biggr]^{1/2}.
</math>
</div>
Hence, we conclude that,
<div align="center">
<math>
\mathrm{Case~A:}~~~~~\frac{R}{R'} = \frac{\chi_\mathrm{inner}}{\chi_0} = \biggl[\frac{\xi_1 - 1}{\xi_1 + 1}\biggr]^{1/2} \frac{(\xi_1^2 - 1)^{1/2}}{\xi_1} = \frac{(\xi_1 - 1)}{\xi_1}
</math><br />

<math>
\Rightarrow~~~~~  \xi_1 = \biggl[1 - \frac{R}{R'} \biggr]^{-1} .
</math>
</div>

What, then, is the expression for the scale-length <math>a</math> in terms of just <math>R</math> and <math>R'</math>? Well ...
<div align="center">
<math>
\xi_1 + 1 = \frac{1}{1-(R/R')} +1 = \frac{2 - R/R'}{1-(R/R')} ,
</math>
</div>
and
<div align="center">
<math>
\xi_1 - 1 = \frac{1}{1-(R/R')} -1 = \frac{R/R'}{1-(R/R')} .
</math>
</div>
Hence,
<div align="center">
<math>
a = R \biggl[ \frac{2 - R/R'}{R/R'} \biggr]^{1/2} = ( 2RR' - R^2 )^{1/2}.
</math>
</div>



<font color="red">'''Case B:'''</font> On the other hand, if we associate <math>R'</math> directly with <math>a</math>, then we conclude,
<div align="center">
<math>
\mathrm{Case~B:}~~~~~\frac{R}{R'} = \chi_\mathrm{inner} = \biggl[\frac{\xi_1 - 1}{\xi_1 + 1}\biggr]^{1/2} .
</math><br />

<math>
\Rightarrow~~~~~  (\xi_1 + 1)\biggl(\frac{R}{R'}\biggr)^2 = \xi_1 - 1 .
</math><br />

<math>
\Rightarrow~~~~~  \xi_1 \biggl[ 1 - \biggl(\frac{R}{R'}\biggr)^2 \biggr] = \biggl[1 + \biggl(\frac{R}{R'}\biggr)^2 \biggr].
</math>
<br />

<math>
\Rightarrow~~~~~  \xi_1 = \biggl[1 + \biggl(\frac{R}{R'}\biggr)^2 \biggr]\biggl[ 1 - \biggl(\frac{R}{R'}\biggr)^2 \biggr] ^{-1}.
</math>

</div>

----