Editing Appendix/Ramblings/Bordeaux (section)

===Compare First Terms===

Rewriting the first term in the [https://ui.adsabs.harvard.edu/abs/2020MNRAS.494.5825H/abstract Hur&eacute;, et al. (2020)] series expression for the potential, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\Psi_0}{GM} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2}{\pi} \biggl\{ \frac{\boldsymbol{K}([k_H]_0) }{[ (\varpi_W + R_c)^2 + z_W^2]^{1 / 2}} \biggr\} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[k_H]_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\varpi_W R_c}{\Delta_0^2}  \biggr]^{1 / 2}
=
\biggl\{ \frac{4\varpi_W R_c}{ (\varpi_W + R_c)^2 + z_W^2} \biggr\}^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

For comparison, the first term in Wong's expression is,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\Phi_\mathrm{W0}}{GM} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl( \frac{2^{3} }{3\pi^3} \biggr)
\Upsilon_{W0}(\eta_0) \biggl\{
\frac{\boldsymbol{K}(k) }{ r_1 }  \biggr\}\, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a^2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
R^2 - d^2 ~~~\Rightarrow ~~~ a = R_c(1 - e^2)^{1 / 2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r_1^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \varpi + R_c (1 - e^2 )^{1 / 2}\biggr]^2 + \biggl[z - Z_0 \biggr]^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~k</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{4\varpi R_c(1-e^2)^{1 / 2}}{[\varpi + R_c(1-e^2)^{1 / 2}]^2 + [z - Z_0]^2} \biggr\}^{1 / 2} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Upsilon_{W0}(\eta_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\sinh\eta_0}{\cosh\eta_0}\biggl\{
K(k_0)\cdot K(k_0) [ \cosh\eta_0(1 - \cosh\eta_0) ] 
+ 2K(k_0)\cdot E(k_0) [ \cosh^2\eta_0  + 1 ] 
- E(k_0)\cdot E(k_0) [ \cosh\eta_0(1 + \cosh\eta_0)  ]
\biggr\}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-e^2)^{1 / 2}}{e^2} \biggl\{
- K(k_0)\cdot K(k_0) (1-e) 
+ 2K(k_0)\cdot E(k_0) (1+e^2) 
- E(k_0)\cdot E(k_0) (1+e)
\biggr\}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~k_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2}{1+1/e} \biggr]^{1 / 2}
=
\biggl[ \frac{2e}{1+e} \biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
This expression is correct for any value of the aspect ratio, <math>~e</math>.  But let's set <math>~Z_0 = 0</math> &#8212; as Hur&eacute;, et al. (2020) have done &#8212; then see how the expression simplifies for an infinitesimally thin hoop, that is, if we let <math>~e \rightarrow 0</math>.  First we note that,

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

<tr>
  <td align="right">
<math>~k\biggr|_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{4\varpi R_c}{[\varpi + R_c]^2 + z^2} \biggr\}^{1 / 2} 
\, ,
</math>
  </td>
</tr>
</table>
so in this limit the modulus of the complete elliptic integral of the first kind becomes identical to the modulus used by Hur&eacute;, et al. (2020), <math>~[k_H]_0</math>.  Next, we note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r_1\biggr|_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[( \varpi + R_c )^2 + z^2]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
As a result, we can write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\Phi_\mathrm{W0}}{GM} \biggr|_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Psi_0}{GM}  \cdot
\biggl[
\biggl( \frac{2^{2} }{3\pi^2} \biggr)
\Upsilon_{W0}(\eta_0) 
\biggr]_{e\rightarrow 0} \, .
</math>
  </td>
</tr>
</table>
Now let's evaluate the coefficient, <math>~\Upsilon_{W0}</math>, in the limit of <math>~e \rightarrow 0</math>.


<table align="center"  border="1" width="100%" cellpadding="8"><tr><td align="left">

<div align="center">
<math>~\Upsilon_{W0}</math>, in the limit of <math>~e \rightarrow 0</math>.
</div>
First, drawing from our [[Apps/Wong1973Potential#Phase_0C|separate examination of the behavior of complete elliptic integral functions]], we appreciate that,

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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k_0) \cdot  E(k_0) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 ~+~\frac{1}{2^5} ~k_0^4 
~+~\frac{1}{2^5} ~ k_0^6 
+ \mathcal{O}(k_0^{8}) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[K(k_0) \cdot  K(k_0) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{2} k_0^2
+ \frac{11}{2^5} ~k_0^4 
+ \frac{17}{2^6} ~ k_0^6 
+ \mathcal{O}(k_0^{8}) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[E(k_0) \cdot  E(k_0) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 - \frac{1}{2} ~k_0^2 ~-~ \frac{1}{2^5} ~ k_0^4 ~-~\frac{1}{2^6} ~ k_0^6
+ \mathcal{O}(k_0^{8}) \, .
</math>
  </td>
</tr>
</table>
Next, employing the [[Appendix/Ramblings/PowerSeriesExpressions#Binomial|binomial expansion]], we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~k_0^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2e(1+e)^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2e(  1 - e +e^2 - e^3 + e^4 - e^5 + \cdots ) \, ;</math>
  </td>
</tr>
</table>
and,

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

<tr>
  <td align="right">
<math>~k_0^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4e^2(1+e)^{-2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4e^2(  1 - 2e + 3e^2 - 4e^3 + 5e^4 - 6e^5 + \cdots ) \, .</math>
  </td>
</tr>
</table>
Hence, we have,

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

<tr>
  <td align="right">
<math>~\frac{2^2}{\pi^2} \biggl[ \frac{e^2}{(1 - e^2)^{1 / 2}} \biggr] \Upsilon_{W0}(\eta_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (1-e) \biggl[ 1 + \frac{1}{2} k_0^2
+ \frac{11}{2^5} ~k_0^4 
+ \mathcal{O}(k_0^{6}) 
 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 (1+e^2) \biggl[ 1 ~+~\frac{1}{2^5} ~k_0^4 
+ \mathcal{O}(k_0^{6}) 
 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (1+e) \biggl[ 1 - \frac{1}{2} ~k_0^2 ~-~ \frac{1}{2^5} ~ k_0^4 ~
+ \mathcal{O}(k_0^{6}) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (1-e) \biggl[ 1 + e (  1 - e +e^2 - e^3 + e^4 - e^5 + \cdots )
+ \frac{11}{2^3} \cdot ~e^2(  1 - 2e + 3e^2 - 4e^3 + 5e^4 - 6e^5 + \cdots ) 
+ \mathcal{O}(e^{3}) 
 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ 2 (1+e^2) \biggl[ 1 ~+~\frac{1}{2^3} \cdot~e^2(  1 - 2e + 3e^2 - 4e^3 + 5e^4 - 6e^5 + \cdots ) 
+ \mathcal{O}(e^{3}) 
 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- (1+e) \biggl[ 1 - e ~(  1 - e +e^2 - e^3 + e^4 - e^5 + \cdots ) ~
-~ \frac{1}{2^3} \cdot~ e^2(  1 - 2e + 3e^2 - 4e^3 + 5e^4 - 6e^5 + \cdots ) ~
+ \mathcal{O}(e^{3}) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- (1-e) \biggl[ 1 + e (  1 - e  )
+ \frac{11}{2^3} \cdot ~e^2 
 \biggr]
+ 2 (1+e^2) \biggl[ 1 ~+~\frac{1}{2^3} \cdot~e^2 
 \biggr]
- (1+e) \biggl[ 1 - e ~(  1 - e   ) ~
-~ \frac{1}{2^3} \cdot~ e^2 ~
\biggr]
+ \mathcal{O}(e^{3}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[ 1 + e (  1 - e  )
+ \frac{11}{2^3} \cdot ~e^2 
 \biggr]
+e \biggl[ 1 + e 
\biggr]
+ 2 \biggl[ 1 ~+~\frac{1}{2^3} \cdot~e^2 
 \biggr]
+ 2 e^2 
- \biggl[ 1 - e ~(  1 - e   ) ~
-~ \frac{1}{2^3} \cdot~ e^2 ~
\biggr]
- e \biggl[ 1 - e 
\biggr]
+ \mathcal{O}(e^{3}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-1 - e + e^2
- \frac{11}{2^3} \cdot ~e^2 
+ e + e^2
+ 2 ~+~\frac{1}{2^2} \cdot~e^2 
+ 2 e^2 
-1 + e - e^2  ~
+~ \frac{1}{2^3} \cdot~ e^2 ~
- e +e^2 
+ \mathcal{O}(e^{3}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{e^2}{2^3}
\biggl[ 2^5
- 11~ 
~+~3 \biggr]~
+ \mathcal{O}(e^{3}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3e^2
+ \mathcal{O}(e^{3}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{2^2}{3\pi^2} \Upsilon_{W0}(\eta_0)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[1 + \mathcal{O}(e^{1})]\cdot (1 - e^2)^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[ \frac{2^2}{3\pi^2} \Upsilon_{W0}(\eta_0) \biggr]_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 \, .</math>
  </td>
</tr>
</table>

</td></tr></table>

Given that, 

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \biggl[ \frac{2^2}{3\pi^2} \Upsilon_{W0}(\eta_0) \biggr]_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 \, ,</math>
  </td>
</tr>
</table>
we conclude that,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\Phi_\mathrm{W0}}{GM} \biggr|_{e\rightarrow 0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Psi_0}{GM}  \, ,
</math>
  </td>
</tr>
</table>
that is, we conclude that <math>~\Psi_0</math> matches <math>~\Phi_{W0}</math> in the limit of, <math>~e\rightarrow 0</math>.