Editing Apps/Wong1973Potential (section)

====Attempt #1====
Now, drawing upon the Key Relation,
<div align="center">
{{ Math/EQ_PminusHalf01 }}
</div>
we can alternatively write,

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

<tr>
  <td align="right">
<math>\biggl(\frac{a}{GM} \biggr) \Phi_{\mathrm{W}0} (\eta,\theta) \biggr|_\mathrm{Exterior} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-\frac{2 D_0\cdot C_0}{\pi} 
\biggl[ \frac{(\cosh\eta - \cos\theta)^{1 / 2}}{\cosh(\eta/2)}\biggr]~K\biggl( \tanh \frac{\eta}{2} \biggr) \, .
</math>
  </td>
</tr>
</table>

Postponing for the moment an evaluation of the leading, constant coefficient, we see that this expression matches equation (2.66) of {{ Wong73 }}.  Immediately following his equation (2.66), Wong goes on to state that, <font color="darkgreen">"This can be shown to be identical to the result for a thin ring obtained in a simple integration without using the toroidal coordinates &hellip;"</font>  

Our own initial attempt to demonstrate the correctness of this assertion proved to be unsuccessful.  Our second attempt ([[#Attempt_.232|immediately below]]) proved to be successful.  In this second attempt, we started from an alternate expression for the toroidal function, <math>P_{-1 / 2}(z)</math>, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>P_{-1 / 2}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sqrt{2}}{\pi}~ (\sinh\eta)^{-1 / 2} Q_{-1 / 2}(\coth\eta) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sqrt{2}}{\pi}~ (\sinh\eta)^{-1 / 2} kK(k) \, ,
</math>
  </td>
</tr>
</table>
where, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>k</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{2}{\coth\eta + 1}\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

In order to obtain this last expression, we have drawn upon the Key Relation,
{{ Math/EQ_QminusHalf01 }}

For closure, it is important that we demonstrate that the expression for <math>P_{-1 / 2}(z)</math> that we initially drew from eq. (8.13.2) of [https://books.google.com/books?id=MtU8uP7XMvoC&printsec=frontcover&dq=Abramowitz+and+stegun&hl=en&sa=X&ved=0ahUKEwialra5xNbaAhWKna0KHcLAASAQ6AEILDAA#v=onepage&q=Abramowitz%20and%20stegun&f=false Abramowitz &amp; Stegun (1995)] &#8212; and that was used by Wong &#8212; is, indeed, identical to this alternate, "[[#Attempt_.232|Attempt #2]]" expression for  <math>~P_{-1 / 2}(z)</math>.  

<table border="1" cellpadding="10" align="center" width="85%"><tr><td align="left">
Following advice that we have received from Howard Cohl (private communication), we can demonstrate this most efficiently by employing the [https://dlmf.nist.gov/19.8#ii ''Descending Landen Transformation'' for the complete elliptic integral of the first kind], that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>K(k)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> 
(1 + k_1)K(k_1) \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>k_1</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math> 
\frac{1-\sqrt{1-k^2}}{1+\sqrt{1-k^2}} \, .
</math>
  </td>
</tr>
</table>
Given our just-stated definition of the parameter, <math>k</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\sqrt{1-k^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ 1 - \frac{2}{\coth\eta + 1} \biggr]^{1 / 2} = \biggl[ \frac{\coth\eta - 1 }{\coth\eta + 1} \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 1 \pm \sqrt{1-k^2}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 \pm  \biggl[ \frac{\coth\eta - 1 }{\coth\eta + 1} \biggr]^{1 / 2} = \frac{ [\coth\eta + 1]^{1 / 2} \pm  [\coth\eta - 1]^{1 / 2}  }{ [\coth\eta + 1]^{1 / 2} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~k_1 \equiv \frac{1 - \sqrt{1-k^2}}{1 + \sqrt{1-k^2}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ [\coth\eta + 1]^{1 / 2} -  [\coth\eta - 1]^{1 / 2}  }{ [\coth\eta + 1]^{1 / 2} +  [\coth\eta - 1]^{1 / 2}  }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~k_1^2 </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ [\coth\eta + 1] -2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2} +  [\coth\eta - 1]  }{ [\coth\eta + 1] + 2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2} +  [\coth\eta - 1] }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ 2\coth\eta -2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2}  }{ 2\coth\eta + 2[\coth\eta + 1]^{1 / 2}[\coth\eta - 1]^{1 / 2}  }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ \coth\eta - [\coth^2\eta - 1]^{1 / 2}  }{ \coth\eta + [\coth^2\eta - 1]^{1 / 2}  }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{ \coth\eta/[\coth^2\eta - 1]^{1 / 2}  - 1 }{ \coth\eta/[\coth^2\eta - 1]^{1 / 2}  +  1 }
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{\cosh\eta - 1}{\cosh\eta + 1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\tanh^2\biggl(\frac{\eta}{2}\biggr) \, .</math>
  </td>
</tr>
</table>

Note as well that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>1 + k_1</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1+ \frac{1 - \sqrt{1-k^2}}{1 + \sqrt{1-k^2}} = \frac{2}{1 + \sqrt{1-k^2}} \, .</math>
  </td>
</tr>
</table>

Hence, the "second attempt" expression for <math>P_{1 / 2}(z)</math> becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>P_{-1 / 2}(\cosh\eta)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\sqrt{2}}{\pi}~ (\sinh\eta)^{-1 / 2} \biggl[ \frac{2k}{1 + \sqrt{1-k^2}} \biggr] K\biggl(\tanh \frac{\eta}{2} \biggr) 
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\pi}{2} \cdot \cosh\frac{\eta}{2}  \biggr]^{-1} K\biggl(\tanh \frac{\eta}{2} \biggr) \, .
</math>
  </td>
</tr>
</table>
This is, indeed, identical to the "Attempt #1" expression for  <math>P_{-1 / 2}(z)</math> that was used by Wong.

'''Q.E.D.'''

</td></tr></table>
<!--
Drawing upon our [[2DStructure/ToroidalGreenFunction#Basic_Elements_of_a_Toroidal_Coordinate_System|accompanying notes]] to switch back to cylindrical coordinates, we recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\sinh\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^\eta - e^{-\eta} = \frac{r_1}{r_2} - \frac{r_2}{r_1}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ 4a\varpi }{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} } \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2\cosh\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
e^\eta + e^{-\eta} = \frac{r_1}{r_2} + \frac{r_2}{r_1}
</math>
  </td>
</tr>

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

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

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \cosh\eta + 1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ [ \varpi^2  + a^2 + z^2 ] +  \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2}}{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} } 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ [ \varpi^2  + a^2 + z^2 ] + \{ \varpi^4 - 2a^2\varpi^2 + a^4 + z^2[2\varpi^2 + 2a^2] + z^4 \}^{1 / 2} }{ \{ [(\varpi - a)^2 + z^2]\cdot [(\varpi + a)^2 + z^2] \}^{1 / 2} } 
</math>
  </td>
</tr>
</table>

in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\tanh\biggl(\frac{\eta}{2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sinh\eta}{1+\cosh\eta}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4a\varpi}{[ \varpi^2  + a^2 + z^2 ] + \{ \varpi^4 - 2a^2\varpi^2 + a^4 + z^2[2\varpi^2 + 2a^2] + z^4 \}^{1 / 2}}
</math>
  </td>
</tr>
</table>
-->