Difference between revisions of "User:Tohline/Appendix/Mathematics/ToroidalSynopsis01"

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</table>
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The three-dimensional differential volume element is,
<span id="DiffVolElement">The three-dimensional differential volume element is,</span>
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
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</div>
</div>


==Exploration==
Also, given that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\cosh\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\biggl[ e^\eta + e^{-\eta} \biggr]</math>
  </td>
  <td align="center">&nbsp; &nbsp; and, &nbsp; &nbsp;</td>
  <td align="right">
<math>~\sinh\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\biggl[ e^\eta - e^{-\eta} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\coth\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ e^\eta + e^{-\eta} \biggr]\biggl[ e^\eta - e^{-\eta} \biggr]^{-1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{r_1}{r_2} +  \frac{r_2}{r_1} \biggr]\biggl[  \frac{r_1}{r_2} - \frac{r_2}{r_1} \biggr]^{-1}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="1">
<math>~\biggl[ \frac{r_1^2 + r_2^2}{r_1 r_2}  \biggr]\biggl[  \frac{r_1^2 - r_2^2}{r_1 r_2} \biggr]^{-1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="1">
<math>~\biggl[ \frac{r_1^2 + r_2^2}{r_1^2 - r_2^2}  \biggr]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
\frac{  \varpi^2 + a^2 + (z - Z_0)^2 }{ 2a\varpi } \, .
</math>
  </td>
</tr>
</table>
</div>
 
==Arguments of Q and K==


Want to explore argument of <math>~Q_{-1 / 2}(\Chi)</math>, namely,
Want to explore argument of <math>~Q_{-1 / 2}(\Chi)</math>, namely,
<div align="center">
<div align="center">
<math>
<math>
\Chi \equiv \frac{(\varpi^')^2 + \varpi^2 + (Z^' - Z)^2}{2\varpi^'  \varpi} .
\Chi \equiv \frac{(\varpi^')^2 + \varpi^2 + (z^' - z)^2}{2\varpi^'  \varpi} .
</math>
</div>
Therefore,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~2\varpi \biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\varpi^')^2 + \varpi^2 + (z^' - z)^2 -
[\varpi^2 + a^2 + (z - Z_0)^2 ]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\varpi^')^2 - a^2 +  [ (z^')^2 - 2z^' z + z^2]-
[z^2 - 2zZ_0 + Z_0^2]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\varpi^')^2 - a^2 +  (z^')^2- Z_0^2  +2z(Z_0 - z^' )
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~2a\biggl[ \frac{\sinh\eta }{(\cosh\eta - \cos\theta)}  \biggr]\biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\varpi^')^2 - a^2 +  (z^')^2- Z_0^2  +2(Z_0 - z^' )\biggl[ Z_0 + \frac{a \sin\theta}{(\cosh\eta - \cos\theta)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2aC_0
+2a(Z_0 - z^' )\biggl[ \frac{\sin\theta}{(\cosh\eta - \cos\theta)} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \sinh\eta \biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
C_0 (\cosh\eta - \cos\theta) +
(Z_0 - z^' ) \sin\theta
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \varpi^' \Chi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\sinh\eta} \biggl[ C_0 (\cosh\eta - \cos\theta) + (Z_0 - z^' ) \sin\theta + a\cosh\eta\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \Chi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\varpi^' \sinh\eta} \biggl[ (C_0 + a)\cosh\eta + (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
where,
<div align="center">
<math>~
C_0 \equiv
\frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 +  (z^')^2- Z_0^2  +2Z_0 (Z_0 - z^' ) \biggr]
=
\frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 +  (z^')^2 +Z_0^2 - 2Z_0 z^'  \biggr]
=
\frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 + (z^' - Z_0)^2  \biggr] \, .
</math>
</div>
 
Now, notice that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
( \varpi^')^2 + a^2 + (z^' - Z_0)^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
2a\varpi^'~\coth\eta^'
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
( \varpi^')^2 - a^2 + (z^' - Z_0)^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
2a\varpi^'~\coth\eta^' - 2a^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
C_0 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
\varpi^'~\coth\eta^' - a
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
\biggl[ \frac{a \sinh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] ~\coth\eta^' - a
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" colspan="3">
<math>~
\biggl[ \frac{a \cosh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] - a \, .
</math>
  </td>
</tr>
</table>
</div>
 
Hence,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \Chi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\cosh\eta}{\varpi^' \sinh\eta}  \biggl[ \varpi^' \coth\eta^' \biggr] 
+
\frac{1}{\sinh\eta}  \biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta^' } \biggr]  \biggl[  (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\coth\eta \cdot \coth\eta^'
+
\biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl[  (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\coth\eta \cdot \coth\eta^'
-
\biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl\{ \biggl[ \frac{a \sin\theta^'}{(\cosh\eta^' - \cos\theta^')} \biggr] \sin\theta + \biggl[ \frac{a \cosh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] \cos\theta - a\cos\theta\biggr\} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\coth\eta \cdot \coth\eta^'
-
\biggl[ \frac{1 }{ \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl\{  \sin\theta^' \sin\theta +  \cosh\eta^'  \cos\theta - (\cosh\eta^' - \cos\theta^')\cos\theta\biggr\} 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\coth\eta \cdot \coth\eta^'
-
\biggl[ \frac{\sin\theta^' \sin\theta +\cos\theta^'\cos\theta }{ \sinh\eta \cdot \sinh\eta^' } \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
[[User:Tohline/2DStructure/ToroidalCoordinates#Expression_for_the_Axisymmetric_Potential|Also]],
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \Chi +1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\sinh\eta \cdot \sinh\eta^' + \cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~\mu^2 \equiv \frac{ 2 }{\Chi +1 }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2 \sinh\eta \cdot \sinh\eta^' }{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) } \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
<span id="SummaryTable">NOTE by Tohline:  On 5 June 2018, I used Excel to test the validity of the toroidal-coordinate-based expressions that have been derived here, and summarized in the following table.</span>
 
<table border="1" align="center" cellpadding="10">
<tr>
  <th align="center" colspan="3">
Summary Table
  </th>
</tr>
<tr>
  <th align="center">
Quantity
  </th>
  <th align="center">
Raw Expression in Cylindrical Coordinates
  </th>
  <th align="center">
Expression in Terms of Toroidal Coordinates
  </th>
</tr>
<tr>
  <td align="center">
<math>~\Chi</math>
  </td>
  <td align="center">
<math>
\frac{(\varpi^')^2 + \varpi^2 + (z^' - z)^2}{2\varpi^'  \varpi}
</math>
  </td>
  <td align="center">
<math>~
\frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' }
</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~\mu^2 \equiv \frac{2}{\Chi + 1}</math>
  </td>
  <td align="center">
<math>
\frac{4\varpi^' \varpi}{(\varpi^' + \varpi)^2 + (z^' - z)^2}
</math>
  </td>
  <td align="center">
<math>~
\frac{2 \sinh\eta \cdot \sinh\eta^' }{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) }
</math>
  </td>
</tr>
</table>
 
==Potential==
 
The potential, <math>~U({\vec{r}}~')</math>, at a point <math>~{\vec{r}}~'</math> due to an arbitrary mass distribution, <math>~\rho({\vec{r}})</math>, is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~U({\vec{r}}~')</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-G \iiint \frac{\rho(\vec{r}) d^3r}{|~\vec{r} - {\vec{r}}^{~'} ~|} \, .</math>
  </td>
</tr>
 
</table>
</div>
 
===Volume Element===
 
See [[#DiffVolElement|above]].
 
===Green's Function===
Wong (1973) points out that in toroidal coordinates the Green's function is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{1}{|~\vec{r} - {\vec{r}}^{~'} ~|} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi a} \biggl[ (\cosh\eta - \cos\theta)(\cosh \eta^' - \cos\theta^') \biggr]^{1 /2 }
\sum\limits_{m,n} (-1)^m \epsilon_m \epsilon_n ~\frac{\Gamma(n-m+\tfrac{1}{2})}{\Gamma(n + m + \tfrac{1}{2})}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times \cos[m(\psi - \psi^')][\cos[n(\theta - \theta^')] ~\begin{cases}P^m_{n-1 / 2}(\cosh\eta) ~Q^m_{n-1 / 2}(\cosh\eta^') ~~~\eta^' > \eta \\P^m_{n-1 / 2}(\cosh\eta^') ~Q^m_{n-1 / 2}(\cosh\eta)~~~\eta^' < \eta
\end{cases}\, ,
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.53)
  </td>
</tr>
</table>
</div>
where, <math>~P^m_{n-1 / 2}, Q^m_{n-1 / 2}</math> are "<font color="darkgreen">Legendre functions of the first and second kind with order <math>~n - \tfrac{1}{2}</math> and degree <math>~m</math> (toroidal harmonics)</font>," and <math>~\epsilon_m</math> is the Neumann factor, that is, <math>~\epsilon_0 = 1</math> and <math>~\epsilon_m = 2</math> for all <math>~m \ge 1</math>.  According to [http://adsabs.harvard.edu/abs/1999ApJ...527...86C CT99], the Green's function written in toroidal coordinates is,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{1}{|\vec{x} - \vec{x}^{~'}|}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi \sqrt{\varpi \varpi^'}} \sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] Q_{m- 1 / 2}(\Chi)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a\pi}
\biggl[  \frac{(\cosh\eta^' - \cos\theta^')}{\sinh\eta^' } \frac{(\cosh\eta - \cos\theta)}{\sinh\eta } \biggr]^{1 / 2}
\sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] Q_{m- 1 / 2}(\Chi) \, .
</math>
  </td>
</tr>
</table>
 
Things to note:
<ol>
  <li>The argument of <math>~Q_{m - 1 / 2}</math> in the CT99 expression is very different from the argument of <math>~Q^m_{n - 1 / 2}</math> (or <math>~P^m_{n - 1 / 2}</math>) in Wong's expression.</li>
  <li>In both expressions, <math>~m</math> is the integer multiplying the azimuthal angle, <math>~\psi</math>, but in the CT99 expression this index serves as the ''subscript'' index of the function, <math>~Q</math>, whereas in Wong's expression it serves as the ''superscript'' index of both functions, <math>~Q</math> and <math>~P</math>.  In this context, note that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\sqrt{\frac{\pi}{2}} ~\Gamma(m-n+\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{n}_{m - \frac{1}{2}} (\coth\eta) \, .
</math>
  </td>
</tr>
</table>
  </li>
  <li>Wong's expression contains not only a summation over the index, <math>~m</math>, but also an explicit summation over the index, <math>~n</math>, which multiplies the "polar" angle, <math>~\theta</math>; no such additional summation appears in the CT99 expression, indicating that the summation over <math>~n</math> has implicitly already been completed.  In this context, note that the [[User:Tohline/Appendix/Mathematics/ToroidalConfusion#Joel.27s_Additional_Manipulations|summation expression]] gives,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right)
+
2\sum_{n=1}^{\infty}
Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left[ n (\theta - \theta^') \right]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ e^{\mu\pi i} \Gamma\left(\mu+ \tfrac{1}{2} \right) \biggl[
\dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu
}}{\left\{ \cosh\xi -\cos\left[ n (\theta - \theta^') \right] \right\}^{\mu+(1/2)}}\biggr]
\, ;
</math>
  </td>
</tr>
</table>
or, specifically for the case of <math>~\mu = 0</math>,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
\sum_{n=0}^{\infty} \epsilon_n
Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left[ n(\theta - \theta^') \right]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\dfrac{ \pi/\sqrt{2} }{\left[ \cosh\xi-\cos(\theta - \theta^') \right]^{\frac{1}{2}}  } \, .
</math>
  </td>
</tr>
</table>
  </li>
  <li>Next thought &hellip;</li>
</ol>
 
==New Insight==
 
===Identical Green's Function Expressions===
 
[https://authors.library.caltech.edu/43491/ Caltech's electronic version] of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions''; in particular, &sect;3.11, p. 169 of Volume I gives,
{{ User:Tohline/Math/EQ_Toroidal01 }}
 
If we make the association, <math>~t \leftrightarrow \coth\eta</math>, then we also have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{1}{\sinh\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{t^2 - 1} \, ,</math>
  </td>
</tr>
</table>
</div>
 
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~ \Chi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' }
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
t t^' - (t^2-1)^{1 / 2}(t^{'2}-1)^{1 / 2}\cos(\theta^' - \theta) \, .
</math>
  </td>
</tr>
</table>
</div>
 
Put together, then, these expressions mean,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
Q_{m-1 / 2}(\coth\eta) P_{m - 1 / 2}(\coth\eta^') + 2\sum_{n=1}^\infty (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)]
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)] \, .
</math>
  </td>
</tr>
</table>
</div>
</div>


Also, from [[User:Tohline/Appendix/Mathematics/ToroidalConfusion#QPrelation|our derived <math>~Q-P</math> relation]],
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 192: Line 958:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~ 2\varpi^' \varpi \Chi</math>
<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 199: Line 965:
   <td align="left">
   <td align="left">
<math>~
<math>~
(\varpi^')^2 + \varpi^2 + (Z^' - Z)^2
\sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{m - \frac{1}{2}} (\coth\eta)
</math>
</math>
   </td>
   </td>
Line 206: Line 972:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~ 2\Chi \biggl[ \frac{a \sinh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr]\biggl[ \frac{a \sinh\eta }{(\cosh\eta - \cos\theta)} \biggr]</math>
<math>~\Rightarrow ~~~
P^{-n}_{m - \frac{1}{2}} (\coth\eta)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 213: Line 980:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{a \sinh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr]^2 + \biggl[ \frac{a \sinh\eta }{(\cosh\eta - \cos\theta)} \biggr]^2
\sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta)  
+ \biggl[ \frac{a \sin\theta^'}{(\cosh\eta^' - \cos\theta^')} - \frac{a \sin\theta}{(\cosh\eta - \cos\theta)} \biggr]^2
\, .
</math>
</math>
   </td>
   </td>
Line 220: Line 987:
</table>
</table>
</div>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta)
\biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\}
\cos[n(\theta^' - \theta)]
</math>
  </td>
</tr>
</table>
</div>
Next, we pull from the [[User:Tohline/Appendix/Mathematics/ToroidalConfusion#Gil|accompanying discussion of the Gil et al. (2000) expression]],
{{ User:Tohline/Math/EQ_Toroidal02 }}
Identifying  <math>~x</math> with <math>~\cosh\eta</math>, in which case we have <math>~\lambda = \coth\eta</math>, and, switching index notation, <math>~n \leftrightarrow m</math>, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{m-1 / 2}^n (\coth\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^m
\frac{\pi^{3/2}}{\sqrt{2} \Gamma(m-n+\frac{1}{2})} (\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta) \, ,
</math>
  </td>
</tr>
</table>
</div>
where, this last step also incorporates the [[User:Tohline/Appendix/Mathematics/ToroidalConfusion#Proper_Interpretation_of_DLMF_Expression|"Euler reflection formula for gamma functions"]], namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\Gamma(m-n+\tfrac{1}{2})} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\Gamma(n-m+\frac{1}{2}) }{\pi (-1)^{m+n}} \, .</math>
  </td>
</tr>
</table>
</div>
So we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
Q_{m - 1 / 2}(\Chi)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{n=0}^\infty \epsilon_n (-1)^n
\biggl\{(-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta)\biggr\}
\biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\}
\cos[n(\theta^' - \theta)]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\sqrt{\sinh\eta^'} \sqrt{\sinh\eta}
\sum_{n=0}^\infty \epsilon_n  (-1)^m
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})}
P_{n-1 / 2}^m(\cosh\eta)
Q^m_{n-\frac{1}{2}}(\cosh\eta^')
\cos[n(\theta^' - \theta)] \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the CT99 Green's function may be rewritten as,
<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right">
<math>~ \frac{1}{|\vec{x} - \vec{x}^{~'}|}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a\pi}
[ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2}
\sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')]
\sum_{n=0}^\infty \epsilon_n  (-1)^m
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})}
P_{n-1 / 2}^m(\cosh\eta)
Q^m_{n-\frac{1}{2}}(\cosh\eta^')
\cos[n(\theta^' - \theta)]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a\pi}
[ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2}
\sum_{m=0}^{\infty} \sum_{n=0}^\infty \epsilon_m\epsilon_n  (-1)^m 
\frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})}
\cos[m(\psi - \psi^')] \cos[n(\theta^' - \theta)]
P_{n-1 / 2}^m(\cosh\eta)
Q^m_{n-\frac{1}{2}}(\cosh\eta^') \, .
</math>
  </td>
</tr>
</table>
Let's compare this with Wong's (1973) Green's function, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{|~\vec{r} - {\vec{r}}^{~'} ~|} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\pi a} \biggl[ (\cosh\eta - \cos\theta)(\cosh \eta^' - \cos\theta^') \biggr]^{1 /2 }
\sum\limits_{m,n} (-1)^m \epsilon_m \epsilon_n ~\frac{\Gamma(n-m+\tfrac{1}{2})}{\Gamma(n + m + \tfrac{1}{2})}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
\times \cos[m(\psi - \psi^')][\cos[n(\theta - \theta^')] ~\begin{cases}P^m_{n-1 / 2}(\cosh\eta) ~Q^m_{n-1 / 2}(\cosh\eta^') ~~~\eta^' > \eta \\P^m_{n-1 / 2}(\cosh\eta^') ~Q^m_{n-1 / 2}(\cosh\eta)~~~\eta^' < \eta
\end{cases}\, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.53)
  </td>
</tr>
</table>
</div>
[June 10, 2018]  <font color="red"><b>Amazing! </b></font> The two expressions match precisely!
===Integral Over Polar Angle===
On p. 293 of his article, [http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)] references [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' and states, "It can be shown that &hellip;"
<div align="center">
<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int^\pi_{-\pi} d\theta (\cosh\eta - \cos\theta)^{-\tfrac{5}{2}} \cos[n(\theta - \theta^')]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(8\sqrt{2}/3) Q^2_{n - \frac{1}{2}} (\cosh\eta) \cos (n\theta^')/\sinh^2\eta \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.56)
  </td>
</tr>
</table>
</td></tr></table>
</div>
Let's see if we can replicate this integration result.  (We tried using WolframAlpha's integration tool, but were unsuccessful.)  We presume that Wong initially took the following steps to simplify the left-hand-side of this integral expression:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\int_{-\pi}^{\pi} \frac{\cos[n(\theta - \theta^')] d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} </math>
  </td>
  <td align="right">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\cos(n\theta^') \int_{-\pi}^{\pi} \frac{ \cos(n\theta) ~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}}
+ \sin(n\theta^') \cancelto{0}{ \int_{-\pi}^{\pi} \frac{ \sin(n\theta)  d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} }
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \cos(n\theta^') \int_{0}^{\pi} \frac{ \cos(n\theta)~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} \, .
</math>
  </td>
</tr>
</table>
</div>
That is to say, given that the limits of the integration are <math>~-\pi</math> to <math>~+\pi</math>: &nbsp; The second integral on the right-hand-side will go to zero because the numerator of its integrand &#8212; ''i.e.,'' <math>~\sin(n\theta)</math> &#8212; is an odd function; and, with regard to the first integral on the right-hand-side, the lower integration limit can be set to zero and the result doubled because the numerator of its integrand &#8212; ''i.e.,'' <math>~\cos(n\theta)</math> &#8212; is an even function.
Now, examining Wong's reference to [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'', we find:
<ul>
  <li>
Equation (5) in &sect;3.7, p. 155 of Volume I gives,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_\nu^\mu(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^{i \mu \pi} ~2^{-\nu - 1} \frac{\Gamma(\nu + \mu + 1) }{\Gamma(\nu + 1) } (z^2 - 1)^{-\mu/2} \int_0^\pi (z+\cos t)^{\mu - \nu - 1} (\sin t)^{2\nu + 1} dt \, .
</math>
  </td>
</tr>
</table>
This is valid for,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathrm{Re} ~\nu > -1</math>&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; and &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\mathrm{Re} (\nu + \mu + 1) > 0 \, .</math>
  </td>
</tr>
</table>
  </li>
  <li>
Equation (10) in &sect;3.7, p. 156 of Volume I gives,
{{ User:Tohline/Math/EQ_Toroidal03 }}
  </li>
</ul>
Focusing in on this second integral definition of the Legendre function, <math>~Q^\mu_\nu</math>, let's set  <math>~z = \cosh\eta</math>, <math>~t = \theta</math>, <math>~\mu = 2</math>, and, <math>~\nu = n - \tfrac{1}{2}</math>, where <math>~n</math> is zero or a positive integer.  in this case we have,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^2 (\cosh\eta)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi)^{-\frac{1}{2}} (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~\biggl\{
\int_0^\pi (\cosh\eta - \cos \theta)^{-\frac{5}{2}} \cos(n\theta) ~d\theta
- \cancelto{0}{\cos[(n-\tfrac{1}{2})\pi] }~~\int_0^\infty (\cosh\eta + \cosh \theta)^{- \frac{5}{2}} e^{-n\theta} ~d\theta
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
where the prefactor of the second term &#8212; that is, <math>~\cos[(n-\tfrac{1}{2})\pi] </math> &#8212; goes to zero for all allowable values of the integer, <math>~n</math>.  Hence,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2\cos(n\theta^')
\int_0^\pi \frac{ \cos(n\theta)~d\theta }{ (\cosh\eta - \cos \theta)^{\frac{5}{2}} }
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{ 2(2\pi)^{\frac{1}{2}} Q_{n - \frac{1}{2}}^2 (\cosh\eta) \cos(n\theta^')}{ (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~ }
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{ 2^3 \sqrt{2} }{ 3 } \biggr] \frac{  Q_{n - \frac{1}{2}}^2(\cosh\eta) \cos(n\theta^')}{ \sinh^2\eta  }
\, .
</math>
  </td>
</tr>
</table>
where we have set,
<div align="center">
<math>~
\Gamma(\tfrac{5}{2}) = \Gamma(\tfrac{1}{2} + 2) = \frac{ \sqrt{\pi} \cdot 4! }{4^2 \cdot 2!}
= \frac{\sqrt{\pi} \cdot 2^3\cdot 3}{  2^5 } = \frac{3 \sqrt{\pi}}{2^2} \, .
</math>
</div>
The right-hand-side of this last expression exactly matches the result published by Wong (1973) and [[#Integral_Over_Polar_Angle|rewritten inside the box, above]]. 
Q.E.D.
===Integral Over Radial Coordinate===
On p. 294 of his article, [http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)] references [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' and states that,
<div align="center">
<table border="1" cellpadding="8" width="80%" align="center"><tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int^x dt \Biggl[ \frac{Q^2_{n - \frac{1}{2}}(t) X_{n-\frac{1}{2}}(t)}{t^2 - 1} \Biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl[
(n-\tfrac{3}{2}) X_{n-\frac{1}{2}}(x)~ Q^2_{n+\frac{1}{2}}(x)
- (n+\tfrac{1}{2}) X_{n+\frac{1}{2}}(x) ~Q^2_{n-\frac{1}{2}}(x)
\biggr] \, .
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[http://adsabs.harvard.edu/abs/1973AnPhy..77..279W Wong (1973)], Eq. (2.58)
  </td>
</tr>
</table>
</td></tr></table>
</div>
Let's see if we can replicate this integration result.  Let's start with the "Key Equation",
{{ User:Tohline/Math/EQ_Toroidal05 }}
<!--numbered equation under &sect;3.12 (p. 169) of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi (1953)], which is prefaced by the statement, "If <math>~w_\nu^\mu(z)</math> and <math>w_\sigma^\rho(z)</math> denote any solutions of Legendre's differential equation &hellip; with the parameters <math>~\nu, \mu</math> and <math>~\sigma, \rho</math>, respectively, then it follows &hellip; that,"
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int_a^b\biggl[(\nu - \sigma)(\nu + \sigma + 1) + (\rho^2 - \mu^2)(1 - z^2)^{-1} \biggr] w_\nu^\mu ~w_\sigma^\rho ~dz
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
z(\nu-\sigma) w_\nu^\mu ~w_\sigma^\rho + (\sigma+\rho) w_\nu^\mu ~ w_{\sigma-1}^\rho - (\nu + \mu) w_{\nu - 1}^\mu ~w_\sigma^\rho
\biggr]_a^b \, .
</math>
  </td>
</tr>
</table>
-->
In order to match the left-hand side of Wong's expression, we should adopt the associations: &nbsp; <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math>, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, <math>~\rho \rightarrow 0</math>, and <math>~\sigma \rightarrow ( n - \tfrac{1}{2})</math>.  In which case, [https://authors.library.caltech.edu/43491/1/Volume%201.pdf Erd&eacute;lyi's (1953)] expression becomes,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int_a^b\biggl[ -4(1 - t^2)^{-1} \biggr] Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) ~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
(n - \tfrac{1}{2} ) Q_{n - \frac{1}{2}}^2(t) ~ X_{n - \frac{3}{2}}(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t)
\biggr]_a^b \, .
</math>
  </td>
</tr>
</table>
This is very similar to, but does not appear to match, Wong's expression. 
<font color="maroon">'''Reconciliation Attempt #1:'''</font> &nbsp; Keeping in mind that,
{{ User:Tohline/Math/EQ_Toroidal04 }}
which means, after making the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 0</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, that,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2n t X_{n-\frac{1}{2}}(t) - (n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~(n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>
  </td>
</tr>
</table>
the integral can be rewritten as,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int_a^b\biggl[  \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
\biggl[ 2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) \biggr]  Q_{n - \frac{1}{2}}^2(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t)
\biggr\}_a^b
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
\biggl[ 2n t  Q_{n - \frac{1}{2}}^2(t)
- (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) \biggr] X_{n - \frac{1}{2}}(t)
-  (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t)
\biggr\}_a^b
</math>
  </td>
</tr>
</table>
Returning to the same ''recurrence'' "Key Equation," but this time adopting the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, we can write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2n t Q^2_{n - \frac{1}{2}}(t) - (n + \tfrac{3}{2}) Q^2_{n - \frac{3}{2}} (t) \, ,
</math>
  </td>
</tr>
</table>
in which case the integral becomes,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\int_a^b\biggl[  \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4} \biggl\{
(n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t) ~ X_{n - \frac{1}{2}}(t)
-  (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t)
\biggr\}_a^b \, .
</math>
  </td>
</tr>
</table>
Hooray!  This ''does'' indeed match Wong's relation (2.58)!
===Evaluating Q<sup>2</sup><sub>&nu;</sub>===
How do we evaluate an "order 2" associated Legendre function, such as, <math>~Q_\nu^2</math> ?
Start by recognizing that, from our identified set of "Key Equations,"
{{ User:Tohline/Math/EQ_Toroidal07 }}
Hence, after adopting the association, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, we have, when <math>~\mu = 0</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^{1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-\tfrac{1}{2}) (z^2-1)^{-\frac{1}{2}} [z Q_{n - \frac{1}{2}}(z) - Q_{n - \frac{3}{2}}(z)]
</math>
  </td>
  <td allign="center">&nbsp; &nbsp; &hellip; &nbsp; &nbsp;</td>
  <td align="left">
for <math>~n \ge 1 \, ,</math>
  </td>
</tr>
</table>
and, when <math>~\mu = 1</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{n - \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2-1)^{-\frac{1}{2}} \{ (n-\tfrac{3}{2}) z Q^1_{n - \frac{1}{2}}(z) - (n+\tfrac{1}{2})Q^1_{n - \frac{3}{2}}(z)\}
</math>
  </td>
  <td allign="center">&nbsp; &nbsp; &hellip; &nbsp; &nbsp;</td>
  <td align="left">
for <math>~n \ge 1 \, .</math>
  </td>
</tr>
</table>
All we are missing, then, is expressions for the index, <math>~n=0</math>, that is, we need independent expressions for <math>~Q^1_{-\frac{1}{2}}</math> and for <math>~Q^2_{-\frac{1}{2}}</math>.
From [https://dlmf.nist.gov/14.6#E2 DLMF] &sect;14.6.2, or from  &sect;3.6.1, eq. (7) (p. 149) of [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' we find,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathsf{Q}^{m}_{\nu}\left(x\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(-1)^{m}\left(1-x^{2}\right)^{m/2}\frac{{\mathrm{d}}^{m}\mathsf{Q}_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>
  </td>
</tr>
</table>
or, from [https://dlmf.nist.gov/14.6#E4 DLMF] &sect;14.6.4, and from &sect;3.6.1, eq. (5) (p. 148) of  [https://authors.library.caltech.edu/43491/1/Volume%201.pdf A. Erd&eacute;lyi's (1953)] ''Higher Transcendental Functions'' we have,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^{m}_{\nu}\left(x\right)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\left(x^{2}-1\right)^{m/2}\frac{{\mathrm{d}}^{m}Q_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>
  </td>
</tr>
</table>
Leaning on the latter of the two possible expressions, we therefore have,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1)^{1 / 2} \frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, .
</math>
  </td>
</tr>
</table>
Therefore, starting from the "Key Equation",
{{ User:Tohline/Math/EQ_QminusHalf01 }}
we associate <math>~k^2 \leftrightarrow 2/(z+1)</math>, which implies,
<div align="center">
<math>~\frac{dk}{dz} = - [2(z+1)^3]^{-1 / 2} \, .</math>
</div>
Hence, drawing on the [https://dlmf.nist.gov/19.4#i DLMF's &sect;19.4 expressions for the derivatives of complete elliptic integrals] and appreciating that, <math>~(k')^2 \equiv (1-k^2)</math>, we find,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dk}{dz} \cdot \frac{d}{dk} \biggl[ k K(k) \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{dk}{dz} \cdot \biggl[ K(k) + \frac{E(k) - (k')^2 K(k) }{(k')^2} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} \biggl[ \frac{E(k)  }{(k')^2} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} E(k)  \biggl[ \frac{z+1}{z-1} \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl[\frac{1}{2(z+1)(z-1)^2} \biggr]^{1 / 2} E(k) \, .</math>
  </td>
</tr>
</table>
As a result,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (z^2 - 1)^{1 / 2} \biggl[\frac{1}{2(z^2-1)(z-1)} \biggr]^{1 / 2} E(k)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E\biggl( \sqrt{ \frac{2}{z+1} } \biggr)  \, .
</math>
  </td>
</tr>
</table>
With the exception of the leading negative sign, this appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [<b>[[User:Tohline/Appendix/References#MF53|<font color="red">MF53</font>]]</b>].
Now, let's evaluate the second derivative &hellip;
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 2^{-\frac{1}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}E(k) </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 2^{-\frac{1}{2}} \biggl\{
-\frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k)
- (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-1} \frac{dk}{dz} \cdot \frac{dE(k)}{dk}
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} \biggl\{
\frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k)
+ (z+1)^{-\frac{1}{2}}(z-1)^{-1} [2(z+1)^3]^{-1 / 2} \biggl[ \frac{E(k) - K(k)}{k} \biggr]
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{
\frac{1}{2} (z-1) E(k)
+ (z+1)  E(k)
+ (z+1)(z-1) [2(z+1)^3]^{-1 / 2} \biggl[ E(k) - K(k)\biggr]\biggl[ \frac{z+1}{2} \biggr]^{\frac{1}{2}}
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{
\frac{1}{2} (z-1) E(k)
+ (z+1)  E(k)
+ \frac{1}{2} (z-1)  \biggl[ E(k) - K(k)\biggr]
\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{
4 z E(k) - (z-1) K(k)
\biggr\} \, .</math>
  </td>
</tr>
</table>
Hence, we have,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}\biggl\{
4 z E(k) - (z-1) K(k)
\biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>
This ''also'' appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [<b>[[User:Tohline/Appendix/References#MF53|<font color="red">MF53</font>]]</b>].
<div align="center" id="Q1Q2Summary">
<table border="1" align="center" cellpadding="8">
<tr>
  <th align="center">Summary</th>
</tr>
<tr>
  <td align="left">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q^1_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-  \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k) 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~Q^2_{-\frac{1}{2}}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}}
</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
where: &nbsp; <math>~k = \sqrt{ \frac{2}{z+1}} \, .</math>
  </td>
</tr>
</table>
<table align="center" cellpadding="5" border="1">
<tr>
  <td align="center"><math>~| ~Q^1_{-\frac{1}{2}}(z) ~|</math></td>
  <td align="center"><math>~Q^2_{-\frac{1}{2}}(z)</math></td>
</tr>
<tr>
  <td align="center">
[[File:ABSQ1minusHalf.png|250px|P0minusHalf]]
  </td>
  <td align="center">
[[File:Q2minusHalf.png|250px|P0plusHalf]]
  </td>
</tr>
<tr>
  <td align="left" colspan="2">
See [[User:Tohline/Appendix/Equation_templates#Caption|relevant caption]].
  </td>
</tr>
</table>
  </td>
</tr>
</table>
</div>
Let's push forward a bit more; specifically, let's find the expressions that are relevant when <math>~n = +1</math>.  When <math>~\mu = 0</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{1}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z Q_{+\frac{1}{2}}(z) - Q_{ - \frac{1}{2}}(z) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z^2 k~K ( k ) ~-~ [2z^2(z+1)]^{1 / 2} E( k ) ~-~ k K(k)\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}  \biggl\{ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl[ \frac{2z^2(z+1)}{z^2-1} \biggr]^{1 / 2} E( k ) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}  \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr] \, .
</math>
  </td>
</tr>
</table>
And, when <math>~\mu = 1</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z Q^1_{+\frac{1}{2}}(z) ~+~3Q^1_{- \frac{1}{2}}(z) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ \frac{z}{2} \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr]
~-~3 \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k)\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z (z^2 - 1)^{1 / 2} k~K ( k ) 
~-~(z^2+3) \biggl( \frac{2}{z-1} \biggr)^{1 / 2} E(k)\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\frac{1}{2^2}\biggl\{ z k~K ( k ) 
~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, .
</math>
  </td>
</tr>
</table>


=See Also=
=See Also=
* [https://authors.library.caltech.edu/43491/1/Volume%201.pdf Arthur Erd&eacute;lyi (1953, New York:  McGraw-Hill)] &#8212; ''Higher Transcendental Functions'', especially Volume I, Chapter 3 (pp. 120 - 174)
* [http://adsabs.harvard.edu/abs/1940QJMat..11..222C T. G. Cowling (1940, Quart. J. Math. Oxford Ser., 11, 222 - 224)] &#8212; ''On Certain Expansions Involving Products of Legendre Functions''




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Latest revision as of 03:48, 26 September 2018

Synopsis of Toroidal Coordinate Approach

Whitworth's (1981) Isothermal Free-Energy Surface
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Basics

Here we attempt to bring together — in as succinct a manner as possible — our approach and C.-Y. Wong's (1973) approach to determining the gravitational potential of an axisymmetric, uniform-density torus that has a major radius, <math>~R</math>, and a minor, cross-sectional radius, <math>~d</math>. The relevant toroidal coordinate system is one based on an anchor ring of major radius,

<math>~a^2 \equiv R^2 - d^2 \, .</math>

If the meridional-plane location of the anchor ring — as written in cylindrical coordinates — is, <math>~(\varpi, z) = (a,Z_0)</math>, then the preferred toroidal-coordinate system has meridional-plane coordinates, <math>~(\eta, \theta)</math>, defined such that,

<math>~\eta</math>

<math>~=</math>

<math>~\ln\biggl(\frac{r_1}{r_2} \biggr) \, ,</math>

    and,    

<math>~\cos\theta</math>

<math>~=</math>

<math>~\frac{(r_1^2 + r_2^2 - 4a^2)}{2r_1 r_2} \, ,</math>

where,

<math>~r_1^2 </math>

<math>~\equiv</math>

<math>~(\varpi + a)^2 + (z-Z_0)^2 \, ,</math>

    and,    

<math>~r_2^2 </math>

<math>~\equiv</math>

<math>~(\varpi - a)^2 + (z-Z_0)^2 \, ,</math>

and <math>~\theta</math> has the same sign as <math>~(z-Z_0)</math>. Mapping the other direction, we have,

<math>~\varpi</math>

<math>~=</math>

<math>~\frac{a \sinh\eta }{(\cosh\eta - \cos\theta)} \, ,</math>

    and,    

<math>~z-Z_0</math>

<math>~=</math>

<math>~\frac{a \sin\theta}{(\cosh\eta - \cos\theta)} \, .</math>

The three-dimensional differential volume element is,

<math>~d^3 r</math>

<math>~=</math>

<math>\varpi d\varpi ~dz ~d\psi</math>

<math>~=</math>

<math>~\biggl[ \frac{a^3\sinh\eta}{(\cosh\eta - \cos\theta)^3} \biggr] d\eta~ d\theta~ d\psi \, .</math>

Note that, if <math>~\eta_0</math> identifies the surface of the uniform-density torus, then,

<math>~\cosh\eta_0</math>

<math>~=</math>

<math>~\frac{R}{d} \, ,</math>

     

<math>~\sinh\eta_0</math>

<math>~=</math>

<math>~\frac{a}{d} \, ,</math>

    and,    

<math>~\coth\eta_0</math>

<math>~=</math>

<math>~\frac{R}{a} \, ;</math>

and when the integral over the volume element is completed — that is, over all <math>~\psi</math>, over all <math>~\theta</math>, and over the "radial" interval, <math>~\eta_0 \le \eta \le \infty</math> — the resulting volume is,

<math>~V</math>

<math>~=</math>

<math>~\frac{2\pi^2 \cosh\eta_0}{\sinh^3\eta_0}</math>

<math>~=</math>

<math>~2\pi^2 Rd^2 \, .</math>

Also, given that,

<math>~\cosh\eta</math>

<math>~=</math>

<math>~\frac{1}{2}\biggl[ e^\eta + e^{-\eta} \biggr]</math>

    and,    

<math>~\sinh\eta</math>

<math>~=</math>

<math>~\frac{1}{2}\biggl[ e^\eta - e^{-\eta} \biggr] \, ,</math>

we have,

<math>~\coth\eta</math>

<math>~=</math>

<math>~\biggl[ e^\eta + e^{-\eta} \biggr]\biggl[ e^\eta - e^{-\eta} \biggr]^{-1}</math>

<math>~=</math>

<math>~\biggl[ \frac{r_1}{r_2} + \frac{r_2}{r_1} \biggr]\biggl[ \frac{r_1}{r_2} - \frac{r_2}{r_1} \biggr]^{-1}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{r_1^2 + r_2^2}{r_1 r_2} \biggr]\biggl[ \frac{r_1^2 - r_2^2}{r_1 r_2} \biggr]^{-1}</math>

<math>~=</math>

<math>~\biggl[ \frac{r_1^2 + r_2^2}{r_1^2 - r_2^2} \biggr]</math>

 

<math>~=</math>

<math>~ \frac{ \varpi^2 + a^2 + (z - Z_0)^2 }{ 2a\varpi } \, . </math>

Arguments of Q and K

Want to explore argument of <math>~Q_{-1 / 2}(\Chi)</math>, namely,

<math> \Chi \equiv \frac{(\varpi^')^2 + \varpi^2 + (z^' - z)^2}{2\varpi^' \varpi} . </math>

Therefore,

<math>~2\varpi \biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>

<math>~=</math>

<math>~ (\varpi^')^2 + \varpi^2 + (z^' - z)^2 - [\varpi^2 + a^2 + (z - Z_0)^2 ] </math>

 

<math>~=</math>

<math>~ (\varpi^')^2 - a^2 + [ (z^')^2 - 2z^' z + z^2]- [z^2 - 2zZ_0 + Z_0^2] </math>

 

<math>~=</math>

<math>~ (\varpi^')^2 - a^2 + (z^')^2- Z_0^2 +2z(Z_0 - z^' ) </math>

<math>~\Rightarrow ~~~2a\biggl[ \frac{\sinh\eta }{(\cosh\eta - \cos\theta)} \biggr]\biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>

<math>~=</math>

<math>~ (\varpi^')^2 - a^2 + (z^')^2- Z_0^2 +2(Z_0 - z^' )\biggl[ Z_0 + \frac{a \sin\theta}{(\cosh\eta - \cos\theta)} \biggr] </math>

 

<math>~=</math>

<math>~ 2aC_0 +2a(Z_0 - z^' )\biggl[ \frac{\sin\theta}{(\cosh\eta - \cos\theta)} \biggr] </math>

<math>~\Rightarrow ~~~ \sinh\eta \biggl[ \varpi^' \Chi - a\coth\eta\biggr]</math>

<math>~=</math>

<math>~ C_0 (\cosh\eta - \cos\theta) + (Z_0 - z^' ) \sin\theta </math>

<math>~\Rightarrow ~~~ \varpi^' \Chi </math>

<math>~=</math>

<math>~ \frac{1}{\sinh\eta} \biggl[ C_0 (\cosh\eta - \cos\theta) + (Z_0 - z^' ) \sin\theta + a\cosh\eta\biggr] </math>

<math>~\Rightarrow ~~~ \Chi </math>

<math>~=</math>

<math>~ \frac{1}{\varpi^' \sinh\eta} \biggl[ (C_0 + a)\cosh\eta + (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr] </math>

where,

<math>~ C_0 \equiv \frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 + (z^')^2- Z_0^2 +2Z_0 (Z_0 - z^' ) \biggr] = \frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 + (z^')^2 +Z_0^2 - 2Z_0 z^' \biggr] = \frac{1}{2a}\biggl[ (\varpi^')^2 - a^2 + (z^' - Z_0)^2 \biggr] \, . </math>

Now, notice that,

<math>~ ( \varpi^')^2 + a^2 + (z^' - Z_0)^2 </math>

<math>~=</math>

<math>~ 2a\varpi^'~\coth\eta^' </math>

<math>~\Rightarrow ~~~ ( \varpi^')^2 - a^2 + (z^' - Z_0)^2 </math>

<math>~=</math>

<math>~ 2a\varpi^'~\coth\eta^' - 2a^2 </math>

<math>~\Rightarrow ~~~ C_0 </math>

<math>~=</math>

<math>~ \varpi^'~\coth\eta^' - a </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{a \sinh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] ~\coth\eta^' - a </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{a \cosh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] - a \, . </math>

Hence,

<math>~ \Chi </math>

<math>~=</math>

<math>~ \frac{\cosh\eta}{\varpi^' \sinh\eta} \biggl[ \varpi^' \coth\eta^' \biggr] + \frac{1}{\sinh\eta} \biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta^' } \biggr] \biggl[ (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr] </math>

 

<math>~=</math>

<math>~ \coth\eta \cdot \coth\eta^' +

\biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl[  (Z_0 - z^' ) \sin\theta - C_0 \cos\theta \biggr]  

</math>

 

<math>~=</math>

<math>~ \coth\eta \cdot \coth\eta^' -

\biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{a \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl\{ \biggl[ \frac{a \sin\theta^'}{(\cosh\eta^' - \cos\theta^')} \biggr] \sin\theta + \biggl[ \frac{a \cosh\eta^' }{(\cosh\eta^' - \cos\theta^')} \biggr] \cos\theta - a\cos\theta\biggr\}  

</math>

 

<math>~=</math>

<math>~ \coth\eta \cdot \coth\eta^' -

\biggl[ \frac{1 }{ \sinh\eta \cdot \sinh\eta^' } \biggr]  \biggl\{  \sin\theta^' \sin\theta +  \cosh\eta^'  \cos\theta - (\cosh\eta^' - \cos\theta^')\cos\theta\biggr\}  

</math>

 

<math>~=</math>

<math>~ \coth\eta \cdot \coth\eta^' -

\biggl[ \frac{\sin\theta^' \sin\theta +\cos\theta^'\cos\theta }{ \sinh\eta \cdot \sinh\eta^' } \biggr] 

</math>

 

<math>~=</math>

<math>~ \biggl[ \frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr] \, . </math>

Also,

<math>~ \Chi +1 </math>

<math>~=</math>

<math>~ \biggl[ \frac{\sinh\eta \cdot \sinh\eta^' + \cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } \biggr] </math>

<math>~ \Rightarrow ~~~\mu^2 \equiv \frac{ 2 }{\Chi +1 }</math>

<math>~=</math>

<math>~ \biggl[ \frac{2 \sinh\eta \cdot \sinh\eta^' }{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) } \biggr] \, . </math>

NOTE by Tohline: On 5 June 2018, I used Excel to test the validity of the toroidal-coordinate-based expressions that have been derived here, and summarized in the following table.

Summary Table

Quantity

Raw Expression in Cylindrical Coordinates

Expression in Terms of Toroidal Coordinates

<math>~\Chi</math>

<math>

\frac{(\varpi^')^2 + \varpi^2 + (z^' - z)^2}{2\varpi^'  \varpi} 

</math>

<math>~ \frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } </math>

<math>~\mu^2 \equiv \frac{2}{\Chi + 1}</math>

<math> \frac{4\varpi^' \varpi}{(\varpi^' + \varpi)^2 + (z^' - z)^2} </math>

<math>~ \frac{2 \sinh\eta \cdot \sinh\eta^' }{ \cosh(\eta^' + \eta) - \cos(\theta^' - \theta) } </math>

Potential

The potential, <math>~U({\vec{r}}~')</math>, at a point <math>~{\vec{r}}~'</math> due to an arbitrary mass distribution, <math>~\rho({\vec{r}})</math>, is,

<math>~U({\vec{r}}~')</math>

<math>~=</math>

<math>~-G \iiint \frac{\rho(\vec{r}) d^3r}{|~\vec{r} - {\vec{r}}^{~'} ~|} \, .</math>

Volume Element

See above.

Green's Function

Wong (1973) points out that in toroidal coordinates the Green's function is,

<math>~\frac{1}{|~\vec{r} - {\vec{r}}^{~'} ~|} </math>

<math>~=</math>

<math>~ \frac{1}{\pi a} \biggl[ (\cosh\eta - \cos\theta)(\cosh \eta^' - \cos\theta^') \biggr]^{1 /2 } \sum\limits_{m,n} (-1)^m \epsilon_m \epsilon_n ~\frac{\Gamma(n-m+\tfrac{1}{2})}{\Gamma(n + m + \tfrac{1}{2})} </math>

 

 

<math>~ \times \cos[m(\psi - \psi^')][\cos[n(\theta - \theta^')] ~\begin{cases}P^m_{n-1 / 2}(\cosh\eta) ~Q^m_{n-1 / 2}(\cosh\eta^') ~~~\eta^' > \eta \\P^m_{n-1 / 2}(\cosh\eta^') ~Q^m_{n-1 / 2}(\cosh\eta)~~~\eta^' < \eta \end{cases}\, , </math>

Wong (1973), Eq. (2.53)

where, <math>~P^m_{n-1 / 2}, Q^m_{n-1 / 2}</math> are "Legendre functions of the first and second kind with order <math>~n - \tfrac{1}{2}</math> and degree <math>~m</math> (toroidal harmonics)," and <math>~\epsilon_m</math> is the Neumann factor, that is, <math>~\epsilon_0 = 1</math> and <math>~\epsilon_m = 2</math> for all <math>~m \ge 1</math>. According to CT99, the Green's function written in toroidal coordinates is,

<math>~ \frac{1}{|\vec{x} - \vec{x}^{~'}|}</math>

<math>~=</math>

<math>~ \frac{1}{\pi \sqrt{\varpi \varpi^'}} \sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] Q_{m- 1 / 2}(\Chi) </math>

 

<math>~=</math>

<math>~ \frac{1}{a\pi} \biggl[ \frac{(\cosh\eta^' - \cos\theta^')}{\sinh\eta^' } \frac{(\cosh\eta - \cos\theta)}{\sinh\eta } \biggr]^{1 / 2} \sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] Q_{m- 1 / 2}(\Chi) \, . </math>

Things to note:

  1. The argument of <math>~Q_{m - 1 / 2}</math> in the CT99 expression is very different from the argument of <math>~Q^m_{n - 1 / 2}</math> (or <math>~P^m_{n - 1 / 2}</math>) in Wong's expression.
  2. In both expressions, <math>~m</math> is the integer multiplying the azimuthal angle, <math>~\psi</math>, but in the CT99 expression this index serves as the subscript index of the function, <math>~Q</math>, whereas in Wong's expression it serves as the superscript index of both functions, <math>~Q</math> and <math>~P</math>. In this context, note that,

    <math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>

    <math>~=</math>

    <math>~(-1)^m \sqrt{\frac{\pi}{2}} ~\Gamma(m-n+\tfrac{1}{2}) \biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{n}_{m - \frac{1}{2}} (\coth\eta) \, . </math>

  3. Wong's expression contains not only a summation over the index, <math>~m</math>, but also an explicit summation over the index, <math>~n</math>, which multiplies the "polar" angle, <math>~\theta</math>; no such additional summation appears in the CT99 expression, indicating that the summation over <math>~n</math> has implicitly already been completed. In this context, note that the summation expression gives,

    <math>~ Q^{\mu}_{-\frac{1}{2}}\left(\cosh\xi\right) + 2\sum_{n=1}^{\infty} Q^{\mu}_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left[ n (\theta - \theta^') \right] </math>

    <math>~=</math>

    <math>~ e^{\mu\pi i} \Gamma\left(\mu+ \tfrac{1}{2} \right) \biggl[ \dfrac{\left(\frac{1}{2}\pi\right)^{1/2}\left(\sinh\xi\right)^{\mu }}{\left\{ \cosh\xi -\cos\left[ n (\theta - \theta^') \right] \right\}^{\mu+(1/2)}}\biggr] \, ; </math>

    or, specifically for the case of <math>~\mu = 0</math>,

    <math>~ \sum_{n=0}^{\infty} \epsilon_n Q_{n-\frac{1}{2}}\left(\cosh\xi\right) \cos\left[ n(\theta - \theta^') \right] </math>

    <math>~=</math>

    <math>~ \dfrac{ \pi/\sqrt{2} }{\left[ \cosh\xi-\cos(\theta - \theta^') \right]^{\frac{1}{2}} } \, . </math>

  4. Next thought …

New Insight

Identical Green's Function Expressions

Caltech's electronic version of A. Erdélyi's (1953) Higher Transcendental Functions; in particular, §3.11, p. 169 of Volume I gives,

LSU Key.png

<math>~ Q_\nu[t t^' - (t^2-1)^{1 / 2} (t^{'2} - 1)^{1 / 2} \cos\psi] </math>

<math>~=</math>

<math>~ Q_\nu(t) P_\nu(t^') + 2\sum_{n=1}^\infty (-1)^n Q^n_\nu(t) P^{-n}_\nu(t^') \cos(n\psi) </math>

A. Erdélyi (1953):  Volume I, §3.11, p. 169, eq. (4)

Valid for:    

<math>~t, t^'</math>  real

       

<math>~1 < t^' < t</math>

       

<math>~\nu \ne -1, -2, -3, </math> …

       

<math>~\psi</math>   real

If we make the association, <math>~t \leftrightarrow \coth\eta</math>, then we also have,

<math>~\frac{1}{\sinh\eta}</math>

<math>~=</math>

<math>~\sqrt{t^2 - 1} \, ,</math>

in which case,

<math>~ \Chi </math>

<math>~=</math>

<math>~ \frac{\cosh\eta \cdot \cosh\eta^' - \cos(\theta^' - \theta) }{ \sinh\eta \cdot \sinh\eta^' } </math>

 

<math>~=</math>

<math>~ t t^' - (t^2-1)^{1 / 2}(t^{'2}-1)^{1 / 2}\cos(\theta^' - \theta) \, . </math>

Put together, then, these expressions mean,

<math>~ Q_{m - 1 / 2}(\Chi) </math>

<math>~=</math>

<math>~ Q_{m-1 / 2}(\coth\eta) P_{m - 1 / 2}(\coth\eta^') + 2\sum_{n=1}^\infty (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)] </math>

 

<math>~=</math>

<math>~ \sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta) P^{-n}_{m - 1 / 2}(\coth\eta^') \cos[n(\theta^' - \theta)] \, . </math>

Also, from our derived <math>~Q-P</math> relation,

<math>~Q^m_{n-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ \sqrt{\frac{\pi}{2}} ~\Gamma(n+m + \tfrac{1}{2}) ~(-1)^m\biggl[ \frac{1}{ \sqrt{\sinh\eta}} \biggr] P^{-n}_{m - \frac{1}{2}} (\coth\eta) </math>

<math>~\Rightarrow ~~~ P^{-n}_{m - \frac{1}{2}} (\coth\eta)</math>

<math>~=</math>

<math>~ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta) \, . </math>

we can write,

<math>~ Q_{m - 1 / 2}(\Chi) </math>

<math>~=</math>

<math>~ \sum_{n=0}^\infty \epsilon_n (-1)^n Q^n_{m - 1 / 2}(\coth\eta) \biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\} \cos[n(\theta^' - \theta)] </math>

Next, we pull from the accompanying discussion of the Gil et al. (2000) expression,

LSU Key.png

<math>~Q_{n-1 / 2}^m (\lambda)</math>

<math>~=</math>

<math>~(-1)^n \frac{\pi^{3/2}}{\sqrt{2}~ \Gamma(n-m+1 / 2)} (x^2-1)^{1 / 4} P_{m-1 / 2}^n(x) \, , </math>

Gil, Segura, & Temme (2000):  eq. (8)

where:    

<math>~\lambda \equiv x/\sqrt{x^2-1}</math>

Identifying <math>~x</math> with <math>~\cosh\eta</math>, in which case we have <math>~\lambda = \coth\eta</math>, and, switching index notation, <math>~n \leftrightarrow m</math>, gives,

<math>~Q_{m-1 / 2}^n (\coth\eta)</math>

<math>~=</math>

<math>~(-1)^m \frac{\pi^{3/2}}{\sqrt{2} \Gamma(m-n+\frac{1}{2})} (\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta) </math>

 

<math>~=</math>

<math>~ (-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta) \, , </math>

where, this last step also incorporates the "Euler reflection formula for gamma functions", namely,

<math>~\frac{1}{\Gamma(m-n+\tfrac{1}{2})} </math>

<math>~=</math>

<math>~\frac{\Gamma(n-m+\frac{1}{2}) }{\pi (-1)^{m+n}} \, .</math>

So we have,

<math>~ Q_{m - 1 / 2}(\Chi) </math>

<math>~=</math>

<math>~ \sum_{n=0}^\infty \epsilon_n (-1)^n \biggl\{(-1)^n \sqrt{ \frac{\pi}{2} } ~\Gamma(n - m + \tfrac{1}{2} )(\sinh\eta)^{1 / 2} P_{n-1 / 2}^m(\cosh\eta)\biggr\} \biggl\{ \sqrt{\frac{2}{\pi}} ~\frac{(-1)^m \sqrt{\sinh\eta^'} }{\Gamma(n+m + \tfrac{1}{2})} ~ Q^m_{n-\frac{1}{2}}(\cosh\eta^') \biggr\} \cos[n(\theta^' - \theta)] </math>

 

<math>~=</math>

<math>~\sqrt{\sinh\eta^'} \sqrt{\sinh\eta} \sum_{n=0}^\infty \epsilon_n (-1)^m \frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} P_{n-1 / 2}^m(\cosh\eta) Q^m_{n-\frac{1}{2}}(\cosh\eta^') \cos[n(\theta^' - \theta)] \, . </math>

Hence, the CT99 Green's function may be rewritten as,

<math>~ \frac{1}{|\vec{x} - \vec{x}^{~'}|}</math>

<math>~=</math>

<math>~ \frac{1}{a\pi} [ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2} \sum_{m=0}^{\infty} \epsilon_m \cos[m(\psi - \psi^')] \sum_{n=0}^\infty \epsilon_n (-1)^m \frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} P_{n-1 / 2}^m(\cosh\eta) Q^m_{n-\frac{1}{2}}(\cosh\eta^') \cos[n(\theta^' - \theta)] </math>

 

<math>~=</math>

<math>~ \frac{1}{a\pi} [ (\cosh\eta^' - \cos\theta^') (\cosh\eta - \cos\theta)]^{1 / 2} \sum_{m=0}^{\infty} \sum_{n=0}^\infty \epsilon_m\epsilon_n (-1)^m \frac{ \Gamma(n - m + \tfrac{1}{2})}{\Gamma(n+m + \tfrac{1}{2})} \cos[m(\psi - \psi^')] \cos[n(\theta^' - \theta)] P_{n-1 / 2}^m(\cosh\eta) Q^m_{n-\frac{1}{2}}(\cosh\eta^') \, . </math>

Let's compare this with Wong's (1973) Green's function, namely,

<math>~\frac{1}{|~\vec{r} - {\vec{r}}^{~'} ~|} </math>

<math>~=</math>

<math>~ \frac{1}{\pi a} \biggl[ (\cosh\eta - \cos\theta)(\cosh \eta^' - \cos\theta^') \biggr]^{1 /2 } \sum\limits_{m,n} (-1)^m \epsilon_m \epsilon_n ~\frac{\Gamma(n-m+\tfrac{1}{2})}{\Gamma(n + m + \tfrac{1}{2})} </math>

 

 

<math>~ \times \cos[m(\psi - \psi^')][\cos[n(\theta - \theta^')] ~\begin{cases}P^m_{n-1 / 2}(\cosh\eta) ~Q^m_{n-1 / 2}(\cosh\eta^') ~~~\eta^' > \eta \\P^m_{n-1 / 2}(\cosh\eta^') ~Q^m_{n-1 / 2}(\cosh\eta)~~~\eta^' < \eta \end{cases}\, . </math>

Wong (1973), Eq. (2.53)

[June 10, 2018] Amazing! The two expressions match precisely!

Integral Over Polar Angle

On p. 293 of his article, Wong (1973) references A. Erdélyi's (1953) Higher Transcendental Functions and states, "It can be shown that …"

<math>~ \int^\pi_{-\pi} d\theta (\cosh\eta - \cos\theta)^{-\tfrac{5}{2}} \cos[n(\theta - \theta^')] </math>

<math>~=</math>

<math>~ (8\sqrt{2}/3) Q^2_{n - \frac{1}{2}} (\cosh\eta) \cos (n\theta^')/\sinh^2\eta \, . </math>

Wong (1973), Eq. (2.56)

Let's see if we can replicate this integration result. (We tried using WolframAlpha's integration tool, but were unsuccessful.) We presume that Wong initially took the following steps to simplify the left-hand-side of this integral expression:

<math>~\int_{-\pi}^{\pi} \frac{\cos[n(\theta - \theta^')] d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} </math>

<math>~=</math>

<math>~ \cos(n\theta^') \int_{-\pi}^{\pi} \frac{ \cos(n\theta) ~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} + \sin(n\theta^') \cancelto{0}{ \int_{-\pi}^{\pi} \frac{ \sin(n\theta) d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} } </math>

 

<math>~=</math>

<math>~ 2 \cos(n\theta^') \int_{0}^{\pi} \frac{ \cos(n\theta)~ d\theta}{(\cosh\eta - \cos\theta)^{5 / 2}} \, . </math>

That is to say, given that the limits of the integration are <math>~-\pi</math> to <math>~+\pi</math>:   The second integral on the right-hand-side will go to zero because the numerator of its integrand — i.e., <math>~\sin(n\theta)</math> — is an odd function; and, with regard to the first integral on the right-hand-side, the lower integration limit can be set to zero and the result doubled because the numerator of its integrand — i.e., <math>~\cos(n\theta)</math> — is an even function.

Now, examining Wong's reference to A. Erdélyi's (1953) Higher Transcendental Functions, we find:

  • Equation (5) in §3.7, p. 155 of Volume I gives,

    <math>~Q_\nu^\mu(z)</math>

    <math>~=</math>

    <math>~ e^{i \mu \pi} ~2^{-\nu - 1} \frac{\Gamma(\nu + \mu + 1) }{\Gamma(\nu + 1) } (z^2 - 1)^{-\mu/2} \int_0^\pi (z+\cos t)^{\mu - \nu - 1} (\sin t)^{2\nu + 1} dt \, . </math>

    This is valid for,

    <math>~\mathrm{Re} ~\nu > -1</math> 

        and    

    <math>~\mathrm{Re} (\nu + \mu + 1) > 0 \, .</math>

  • Equation (10) in §3.7, p. 156 of Volume I gives,

    LSU Key.png

    <math>~Q_\nu^\mu(z)</math>

    <math>~=</math>

    <math>~ e^{i \mu \pi} ~ (2\pi)^{-\frac{1}{2}} (z^2-1)^{\mu/2} ~\Gamma(\mu + \tfrac{1}{2})~\biggl\{ \int_0^\pi (z - \cos t)^{-\mu - \frac{1}{2}} \cos[(\nu + \tfrac{1}{2})t] ~dt -\cos(\nu\pi) \int_0^\infty (z + \cosh t)^{-\mu - \frac{1}{2}} e^{-(\nu + \frac{1}{2})t} ~dt \biggr\} </math>

    A. Erdélyi (1953):  Volume I, §3.7, p. 156, eq. (10)

    Valid for:    

    <math>~\mathrm{Re} ~\nu > -\tfrac{1}{2}</math> 

        and    

    <math>~\mathrm{Re} (\nu + \mu + 1) > 0 \, .</math>

Focusing in on this second integral definition of the Legendre function, <math>~Q^\mu_\nu</math>, let's set <math>~z = \cosh\eta</math>, <math>~t = \theta</math>, <math>~\mu = 2</math>, and, <math>~\nu = n - \tfrac{1}{2}</math>, where <math>~n</math> is zero or a positive integer. in this case we have,

<math>~Q_{n - \frac{1}{2}}^2 (\cosh\eta)</math>

<math>~=</math>

<math>~ (2\pi)^{-\frac{1}{2}} (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~\biggl\{ \int_0^\pi (\cosh\eta - \cos \theta)^{-\frac{5}{2}} \cos(n\theta) ~d\theta - \cancelto{0}{\cos[(n-\tfrac{1}{2})\pi] }~~\int_0^\infty (\cosh\eta + \cosh \theta)^{- \frac{5}{2}} e^{-n\theta} ~d\theta \biggr\} \, , </math>

where the prefactor of the second term — that is, <math>~\cos[(n-\tfrac{1}{2})\pi] </math> — goes to zero for all allowable values of the integer, <math>~n</math>. Hence,

<math>~2\cos(n\theta^') \int_0^\pi \frac{ \cos(n\theta)~d\theta }{ (\cosh\eta - \cos \theta)^{\frac{5}{2}} } </math>

<math>~=</math>

<math>~\frac{ 2(2\pi)^{\frac{1}{2}} Q_{n - \frac{1}{2}}^2 (\cosh\eta) \cos(n\theta^')}{ (\cosh^2\eta-1) ~\Gamma(\tfrac{5}{2})~ } </math>

 

<math>~=</math>

<math>~\biggl[ \frac{ 2^3 \sqrt{2} }{ 3 } \biggr] \frac{ Q_{n - \frac{1}{2}}^2(\cosh\eta) \cos(n\theta^')}{ \sinh^2\eta } \, . </math>

where we have set,

<math>~ \Gamma(\tfrac{5}{2}) = \Gamma(\tfrac{1}{2} + 2) = \frac{ \sqrt{\pi} \cdot 4! }{4^2 \cdot 2!} = \frac{\sqrt{\pi} \cdot 2^3\cdot 3}{ 2^5 } = \frac{3 \sqrt{\pi}}{2^2} \, . </math>

The right-hand-side of this last expression exactly matches the result published by Wong (1973) and rewritten inside the box, above.

Q.E.D.

Integral Over Radial Coordinate

On p. 294 of his article, Wong (1973) references A. Erdélyi's (1953) Higher Transcendental Functions and states that,

<math>~ \int^x dt \Biggl[ \frac{Q^2_{n - \frac{1}{2}}(t) X_{n-\frac{1}{2}}(t)}{t^2 - 1} \Biggr] </math>

<math>~=</math>

<math>~ \frac{1}{4} \biggl[ (n-\tfrac{3}{2}) X_{n-\frac{1}{2}}(x)~ Q^2_{n+\frac{1}{2}}(x) - (n+\tfrac{1}{2}) X_{n+\frac{1}{2}}(x) ~Q^2_{n-\frac{1}{2}}(x) \biggr] \, . </math>

Wong (1973), Eq. (2.58)

Let's see if we can replicate this integration result. Let's start with the "Key Equation",

LSU Key.png

<math>~ \int_a^b\biggl[(\nu - \sigma)(\nu + \sigma + 1) + (\rho^2 - \mu^2)(1 - z^2)^{-1} \biggr] w_\nu^\mu ~w_\sigma^\rho ~dz </math>

<math>~=</math>

<math>~ \biggl[ z(\nu-\sigma) w_\nu^\mu ~w_\sigma^\rho + (\sigma+\rho) w_\nu^\mu ~ w_{\sigma-1}^\rho - (\nu + \mu) w_{\nu - 1}^\mu ~w_\sigma^\rho \biggr]_a^b </math>

A. Erdélyi (1953):  Volume I, §3.12, p. 169, eq. (1)

where, <math>~w_\nu^\mu(z)</math> and <math>~w_\sigma^\rho(z)</math> denote any solutions of Legendre's differential equation


In order to match the left-hand side of Wong's expression, we should adopt the associations:   <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math>, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, <math>~\rho \rightarrow 0</math>, and <math>~\sigma \rightarrow ( n - \tfrac{1}{2})</math>. In which case, Erdélyi's (1953) expression becomes,

<math>~ \int_a^b\biggl[ -4(1 - t^2)^{-1} \biggr] Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) ~dt </math>

<math>~=</math>

<math>~ \biggl[ (n - \tfrac{1}{2} ) Q_{n - \frac{1}{2}}^2(t) ~ X_{n - \frac{3}{2}}(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) \biggr]_a^b \, . </math>

This is very similar to, but does not appear to match, Wong's expression.

Reconciliation Attempt #1:   Keeping in mind that,

LSU Key.png

<math>~(\nu - \mu + 1)P^\mu_{\nu + 1} (z)</math>

<math>~=</math>

<math>~ (2\nu + 1)z P_\nu^\mu(z) - (\nu + \mu)P^\mu_{\nu-1}(z) </math>

Abramowitz & Stegun (1995), p. 334, eq. (8.5.3)

NOTE: <math>~Q_\nu^\mu</math>, as well as <math>~P_\nu^\mu</math>, satisfies this same recurrence relation.

which means, after making the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 0</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, that,

<math>~(n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>

<math>~=</math>

<math>~2n t X_{n-\frac{1}{2}}(t) - (n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t)</math>

<math>~\Rightarrow ~~~(n-\tfrac{1}{2})X_{n - \frac{3}{2}}(t) </math>

<math>~=</math>

<math>~2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t)</math>

the integral can be rewritten as,

<math>~ \int_a^b\biggl[ \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt </math>

<math>~=</math>

<math>~ \frac{1}{4} \biggl\{ \biggl[ 2n t X_{n-\frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) \biggr] Q_{n - \frac{1}{2}}^2(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) \biggr\}_a^b </math>

 

<math>~=</math>

<math>~ \frac{1}{4} \biggl\{ \biggl[ 2n t Q_{n - \frac{1}{2}}^2(t) - (n + \tfrac{3}{2}) Q_{n - \frac{3}{2}}^2(t) \biggr] X_{n - \frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t) \biggr\}_a^b </math>

Returning to the same recurrence "Key Equation," but this time adopting the associations, <math>~z \rightarrow t</math>, <math>~\mu \rightarrow 2</math> and <math>~\nu \rightarrow (n-\tfrac{1}{2})</math>, we can write,

<math>~(n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t)</math>

<math>~=</math>

<math>~ 2n t Q^2_{n - \frac{1}{2}}(t) - (n + \tfrac{3}{2}) Q^2_{n - \frac{3}{2}} (t) \, , </math>

in which case the integral becomes,

<math>~ \int_a^b\biggl[ \frac{Q_{n - \frac{1}{2}}^2(t) ~X_{n - \frac{1}{2}}(t) }{(t^2-1)}\biggr]~dt </math>

<math>~=</math>

<math>~ \frac{1}{4} \biggl\{ (n - \tfrac{3}{2})Q^2_{n + \frac{1}{2}}(t) ~ X_{n - \frac{1}{2}}(t) - (n+\tfrac{1}{2})X_{n+\frac{1}{2}}(t) Q_{n - \frac{1}{2}}^2(t) \biggr\}_a^b \, . </math>

Hooray! This does indeed match Wong's relation (2.58)!

Evaluating Q2ν

How do we evaluate an "order 2" associated Legendre function, such as, <math>~Q_\nu^2</math> ?

Start by recognizing that, from our identified set of "Key Equations,"

LSU Key.png

<math>~P_\nu^{\mu + 1}(z)</math>

<math>~=</math>

<math>~ (z^2-1)^{-\frac{1}{2}} \{ (\nu - \mu) z P^\mu_\nu(z) - (\nu + \mu)P^\mu_{\nu - 1}(z)\} </math>

Abramowitz & Stegun (1995), p. 333, eq. (8.5.1)

NOTE: <math>~Q_\nu^\mu</math>, as well as <math>~P_\nu^\mu</math>, satisfies this same recurrence relation.

Hence, after adopting the association, <math>~\nu \rightarrow (n - \tfrac{1}{2})</math>, we have, when <math>~\mu = 0</math>,

<math>~Q_{n - \frac{1}{2}}^{1}(z)</math>

<math>~=</math>

<math>~ (n-\tfrac{1}{2}) (z^2-1)^{-\frac{1}{2}} [z Q_{n - \frac{1}{2}}(z) - Q_{n - \frac{3}{2}}(z)] </math>

    …    

for <math>~n \ge 1 \, ,</math>

and, when <math>~\mu = 1</math>,

<math>~Q_{n - \frac{1}{2}}^{2}(z)</math>

<math>~=</math>

<math>~ (z^2-1)^{-\frac{1}{2}} \{ (n-\tfrac{3}{2}) z Q^1_{n - \frac{1}{2}}(z) - (n+\tfrac{1}{2})Q^1_{n - \frac{3}{2}}(z)\} </math>

    …    

for <math>~n \ge 1 \, .</math>

All we are missing, then, is expressions for the index, <math>~n=0</math>, that is, we need independent expressions for <math>~Q^1_{-\frac{1}{2}}</math> and for <math>~Q^2_{-\frac{1}{2}}</math>.

From DLMF §14.6.2, or from §3.6.1, eq. (7) (p. 149) of A. Erdélyi's (1953) Higher Transcendental Functions we find,

<math>~\mathsf{Q}^{m}_{\nu}\left(x\right)</math>

<math>~=</math>

<math>~(-1)^{m}\left(1-x^{2}\right)^{m/2}\frac{{\mathrm{d}}^{m}\mathsf{Q}_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>

or, from DLMF §14.6.4, and from §3.6.1, eq. (5) (p. 148) of A. Erdélyi's (1953) Higher Transcendental Functions we have,

<math>~Q^{m}_{\nu}\left(x\right)</math>

<math>~=</math>

<math>~\left(x^{2}-1\right)^{m/2}\frac{{\mathrm{d}}^{m}Q_{\nu}\left(x\right)}{{\mathrm{d}x}^{m}}</math>

Leaning on the latter of the two possible expressions, we therefore have,

<math>~Q^1_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ (z^2 - 1)^{1 / 2} \frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, ; </math>

<math>~Q^2_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ (z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] \, . </math>

Therefore, starting from the "Key Equation",

LSU Key.png

<math>~Q_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ \sqrt{ \frac{2}{z+1} } ~K\biggl( \sqrt{ \frac{2}{z+1}} \biggr) </math>

      for example …

<math>~Q_{-\frac{1}{2}}(\cosh\eta)</math>

<math>~=</math>

<math>~ 2 e^{-\eta/2} K(e^{-\eta}) </math>

Abramowitz & Stegun (1995), p. 337, eq. (8.13.3)

Abramowitz & Stegun (1995), p. 337, eq. (8.13.4)

we associate <math>~k^2 \leftrightarrow 2/(z+1)</math>, which implies,

<math>~\frac{dk}{dz} = - [2(z+1)^3]^{-1 / 2} \, .</math>

Hence, drawing on the DLMF's §19.4 expressions for the derivatives of complete elliptic integrals and appreciating that, <math>~(k')^2 \equiv (1-k^2)</math>, we find,

<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>

<math>~=</math>

<math>~\frac{dk}{dz} \cdot \frac{d}{dk} \biggl[ k K(k) \biggr]</math>

 

<math>~=</math>

<math>~\frac{dk}{dz} \cdot \biggl[ K(k) + \frac{E(k) - (k')^2 K(k) }{(k')^2} \biggr]</math>

 

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} \biggl[ \frac{E(k) }{(k')^2} \biggr]</math>

 

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z+1)^3} \biggr]^{1 / 2} E(k) \biggl[ \frac{z+1}{z-1} \biggr] </math>

 

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z+1)(z-1)^2} \biggr]^{1 / 2} E(k) \, .</math>

As a result,

<math>~Q^1_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ - (z^2 - 1)^{1 / 2} \biggl[\frac{1}{2(z^2-1)(z-1)} \biggr]^{1 / 2} E(k) </math>

 

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k) </math>

 

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E\biggl( \sqrt{ \frac{2}{z+1} } \biggr) \, . </math>

With the exception of the leading negative sign, this appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [MF53].

Now, let's evaluate the second derivative …

<math>~\frac{d}{dz} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>

<math>~=</math>

<math>~ - 2^{-\frac{1}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}E(k) </math>

<math>~\Rightarrow ~~~ \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr]</math>

<math>~=</math>

<math>~ - 2^{-\frac{1}{2}} \biggl\{ -\frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k) - (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k) + (z+1)^{-\frac{1}{2}}(z-1)^{-1} \frac{dk}{dz} \cdot \frac{dE(k)}{dk} \biggr\}</math>

 

<math>~=</math>

<math>~ 2^{-\frac{1}{2}} \biggl\{ \frac{1}{2}(z+1)^{-\frac{3}{2}}(z-1)^{-1}E(k) + (z+1)^{-\frac{1}{2}}(z-1)^{-2}E(k) + (z+1)^{-\frac{1}{2}}(z-1)^{-1} [2(z+1)^3]^{-1 / 2} \biggl[ \frac{E(k) - K(k)}{k} \biggr] \biggr\}</math>

 

<math>~=</math>

<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ \frac{1}{2} (z-1) E(k) + (z+1) E(k) + (z+1)(z-1) [2(z+1)^3]^{-1 / 2} \biggl[ E(k) - K(k)\biggr]\biggl[ \frac{z+1}{2} \biggr]^{\frac{1}{2}} \biggr\}</math>

 

<math>~=</math>

<math>~ 2^{-\frac{1}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ \frac{1}{2} (z-1) E(k) + (z+1) E(k) + \frac{1}{2} (z-1) \biggl[ E(k) - K(k)\biggr] \biggr\}</math>

 

<math>~=</math>

<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{3}{2}} (z-1)^{-2}\biggl\{ 4 z E(k) - (z-1) K(k) \biggr\} \, .</math>

Hence, we have,

<math>~Q^2_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ (z^2 - 1) \frac{d^2}{dz^2} \biggl[ Q_{-\frac{1}{2}}(z) \biggr] </math>

 

<math>~=</math>

<math>~ 2^{-\frac{3}{2}} (z+1)^{-\frac{1}{2}} (z-1)^{-1}\biggl\{ 4 z E(k) - (z-1) K(k) \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} \, . </math>

This also appears to match the tabulated values published in the bottom half of Table IX (p. 1923) of [MF53].

Summary

<math>~Q^1_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ - \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k) </math>

<math>~Q^2_{-\frac{1}{2}}(z)</math>

<math>~=</math>

<math>~ \frac{ 4 z E(k) - (z-1) K(k) }{ [2^{3} (z+1) (z-1)^{2} ]^{1 / 2}} </math>

where:   <math>~k = \sqrt{ \frac{2}{z+1}} \, .</math>

<math>~| ~Q^1_{-\frac{1}{2}}(z) ~|</math> <math>~Q^2_{-\frac{1}{2}}(z)</math>

P0minusHalf

P0plusHalf

See relevant caption.


Let's push forward a bit more; specifically, let's find the expressions that are relevant when <math>~n = +1</math>. When <math>~\mu = 0</math>,

<math>~Q_{+ \frac{1}{2}}^{1}(z)</math>

<math>~=</math>

<math>~ \frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z Q_{+\frac{1}{2}}(z) - Q_{ - \frac{1}{2}}(z) \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{2} (z^2-1)^{-\frac{1}{2}} \biggl\{ z^2 k~K ( k ) ~-~ [2z^2(z+1)]^{1 / 2} E( k ) ~-~ k K(k)\biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{2} \biggl\{ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl[ \frac{2z^2(z+1)}{z^2-1} \biggr]^{1 / 2} E( k ) \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{2} \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr] \, . </math>

And, when <math>~\mu = 1</math>,

<math>~Q_{+ \frac{1}{2}}^{2}(z)</math>

<math>~=</math>

<math>~ -~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z Q^1_{+\frac{1}{2}}(z) ~+~3Q^1_{- \frac{1}{2}}(z) \biggr\} </math>

 

<math>~=</math>

<math>~ -~\frac{1}{2}(z^2-1)^{-\frac{1}{2}} \biggl\{ \frac{z}{2} \biggl[ (z^2 - 1)^{1 / 2} k~K ( k ) ~-~ \biggl( \frac{2z^2}{z-1} \biggr)^{1 / 2} E( k ) \biggr] ~-~3 \biggl[\frac{1}{2(z-1)} \biggr]^{1 / 2} E(k)\biggr\} </math>

 

<math>~=</math>

<math>~ -~\frac{1}{2^2}(z^2-1)^{-\frac{1}{2}} \biggl\{ z (z^2 - 1)^{1 / 2} k~K ( k ) ~-~(z^2+3) \biggl( \frac{2}{z-1} \biggr)^{1 / 2} E(k)\biggr\} </math>

 

<math>~=</math>

<math>~ -~\frac{1}{2^2}\biggl\{ z k~K ( k ) ~-~(z^2+3) \biggl[ \frac{2}{(z-1)(z^2-1)} \biggr]^{1 / 2} E(k)\biggr\} \, . </math>

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

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