Editing Apps/Wong1973Potential (section)

===Thin Ring Approximation===

Quoting from p. 295 of {{ Wong73 }}, we find the statement, <font color="orange">"For the case of a very thin ring (i.e., <math>\eta_0 \rightarrow \infty)</math>, the exterior solution has contributions mostly from the first term in the expansion of the series &hellip;"</font>  So looking again at the n = 0 contribution to the exterior potential, we have,

<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>
-D_0
[\cosh\eta - \cos\theta]^{1 / 2} ~\biggl\{
P_{-\frac{1}{2}}(\cosh\eta) [C_0(\cosh\eta_0)]
\biggr\} \, ,
</math>
  </td>
</tr>
</table>
where, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>C_0(\cosh\eta_0)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{2}~Q_{+\frac{1}{2}}(\cosh \eta_0) Q_{- \frac{1}{2}}^2(\cosh \eta_0) 
+ \frac{3}{2}~ Q_{- \frac{1}{2}}(\cosh \eta_0)~Q^2_{+ \frac{1}{2}}(\cosh \eta_0) \, .
</math>
  </td>
</tr>
</table>

Let's go ahead and look at the behavior of <math>C_0(x)</math> in the limit of <math>x = \cosh\eta_0 \rightarrow \infty</math>. Combining [http://dlmf.nist.gov//14.8.E15 DLMF eq. (14.8.15)] with [http://dlmf.nist.gov/14.8.E10 DLMF eq. (14.3.10)], we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>Q_\nu^\mu(x)\biggr|_{x \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
(-1)^\mu \biggl[ \frac{\pi^{1 / 2} \Gamma(\nu + \mu + 1) }{\Gamma(\nu + \tfrac{3}{2}) (2x)^{\nu+1}}\biggr] \, .
</math>
  </td>
</tr>
</table>
Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>C_0(x)\biggr|_{x\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[ \frac{\pi^{1 / 2} \Gamma(\tfrac{3}{2} ) }{\Gamma(2) (2x)^{3/2}}\biggr]\cdot \biggl[ \frac{\pi^{1 / 2} \Gamma(5/2) }{\Gamma(1) (2x)^{1 / 2}}\biggr]
+
\frac{3}{2}\biggl[ \frac{\pi^{1 / 2} \Gamma(\tfrac{1}{2}) }{\Gamma(1) (2x)^{1/2}}\biggr]\cdot \biggl[ \frac{\pi^{1 / 2} \Gamma(\tfrac{7}{2}) }{\Gamma(2) (2x)^{3/2}}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{\pi}{2^3 x^2} \biggl\{
\biggl[ \frac{\Gamma(\tfrac{3}{2} ) }{\Gamma(2) }\biggr]\cdot \biggl[ \frac{ \Gamma(5/2) }{\Gamma(1) }\biggr]
+
3\biggl[ \frac{ \Gamma(\tfrac{1}{2}) }{\Gamma(1) }\biggr]\cdot \biggl[ \frac{\Gamma(\tfrac{7}{2}) }{\Gamma(2) }\biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
Next, pulling from [https://en.wikipedia.org/wiki/Gamma_function#Particular_values Wikipedia's list of particular examples of gamma-function values], we obtain,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>C_0(x)\biggr|_{x\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>\frac{\pi}{2^3 x^2} \biggl\{
\biggl[ \frac{\pi^{1 / 2}}{2}\biggr]\cdot \biggl[\frac{3\pi^{1 / 2}}{2^2}\biggr]
+
3\biggl[ \pi^{1 / 2}\biggr]\cdot \biggl[ \frac{3\cdot 5 \pi^{1 / 2} }{2^3 }\biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{3 \pi^2}{2^6 x^2} \biggl\{1 + 3\cdot 5 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{3 \pi^2}{2^2} \biggr) \frac{1}{\cosh^2\eta_0} \, .
</math>
  </td>
</tr>
</table>

====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>
-->

====Attempt #2====
Now, drawing upon the Key Relation,
<div align="center">
{{ Math/EQ_Toroidal02 }}
</div>
and setting n = m = 0 while adopting the association, <math>x \rightarrow \cosh\eta</math>, we obtain,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>Q_{-1 / 2}(\coth\eta)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\pi^{3/2}}{\sqrt{2\pi}}~ (\sinh\eta)^{1 / 2} P_{-1 / 2}(\cosh\eta)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 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>
</table>
We therefore can 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{\sqrt{2} D_0\cdot C_0}{\pi} 
\biggl[ \frac{(\cosh\eta - \cos\theta)^{1 / 2}}{ (\sinh\eta)^{1 / 2}}\biggr]~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} D_0\cdot C_0}{\pi} 
\biggl[ \frac{(\cosh\eta - \cos\theta)^{1 / 2}}{ (\sinh\eta)^{1 / 2}}\biggr]~k K(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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}{ \sinh\eta + \cosh\eta}\biggr]^{1 / 2}~K(k) 
</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>k\equiv \biggl[  \frac{2}{1+\coth\eta } \biggr]^{1 / 2} = \biggl[  \frac{2\sinh\eta}{\sinh\eta+\cosh\eta } \biggr]^{1 / 2}\, .</math>
</div>

Now, converting back to cylindrical coordinates, we recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>k^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{4a\varpi}{(\varpi + a)^2 + z^2 } \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{ 2a^2}{ (\varpi + a)^2 + z^2 }  \biggr] \, . </math>
  </td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>\biggl(\frac{a}{GM} \biggr) \Phi_{\mathrm{W}0} (\varpi,z) \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{ 2a^2}{ (\varpi + a)^2 + z^2 }  \biggr]^{1 / 2}~K(k) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2^{3/2} D_0\cdot C_0}{\pi} \biggr]
 \frac{ aK(k)  }{[ (\varpi + a)^2 + z^2]^{1 / 2} }  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>
- \biggl[ \frac{2^{3/2} }{\pi} \biggr] \biggl[ \frac{2^{3/2} }{3\pi^2} \cdot \frac{\sinh^3\eta_0}{\cosh\eta_0} \biggr] \biggl[\biggl( \frac{3 \pi^2}{2^2} \biggr) \frac{1}{\cosh^2\eta_0}\biggr]
 \frac{ aK(k)  }{[ (\varpi + a)^2 + z^2]^{1 / 2} }  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>\approx</math>
  </td>
  <td align="left">
<math>- \frac{2}{\pi} \cdot 
\frac{ aK(k)  }{[ (\varpi + a)^2 + z^2]^{1 / 2} }  \, ,
</math>
  </td>
</tr>
</table>
where, in the next-to-last step, we have inserted, from above, the expression for the product of <math>D_0</math> and <math>C_0</math> in the limit of <math>\eta_0 \rightarrow \infty</math>.  This is precisely the expression for the ''Gravitational Potential in the Thin Ring Approximation, <math>\Phi_\mathrm{TR}(\varpi,z)</math>, that we have presented elsewhere.