Editing Appendix/Ramblings/EllipticCylinderCoordinates (section)

===Setup===

Let's see if we can solve the,
<div align="center">
<font color="maroon">'''Poisson Equation'''</font>
{{ User:Tohline/Math/EQ_Poisson01 }}
</div>
obtaining an analytic expression for the gravitational potential in the case where, independent of the coordinate, <math>~z</math>,
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<tr>
  <td align="right">
<math>~\rho = \rho_c\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \biggl[ 1 - \biggl(\frac{x^2}{a^2} + \frac{y^2}{b^2} \biggr)\biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_c \biggl[ 1 - \frac{1}{a^2}\biggl(x^2 + q^2 y^2 \biggr)\biggr] \, .</math>
  </td>
</tr>
</table>
Given that the density distribution is independent of <math>~z</math>, we expect the potential to be independent of <math>~z</math> as well.  So, in terms of T5-Coordinates, the Poisson equation may be written as,

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  <td align="right">
<math>~
\frac{\lambda_2}{\lambda_1 (q^2-1)xy\ell^2} \biggl\{
\frac{\partial}{\partial \lambda_1} \biggl[ \frac{(q^2-1)xy }{\lambda_1 \lambda_2 } \cdot \frac{\partial \Phi}{\partial \lambda_1}\biggr]
+
\frac{\partial}{\partial \lambda_2} \biggl[ \frac{\lambda_1 \lambda_2}{(q^2-1) xy } \cdot \frac{\partial \Phi}{\partial \lambda_2}\biggr]
\biggr\}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~4\pi G\rho_c \biggl[1 - \frac{\lambda_1^2}{a^2} \biggr] </math>
  </td>
</tr>
</table>
If we specifically consider the case where <math>~q^2 = a^2/b^2 = 2</math>, this can be rewritten as,

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<tr>
  <td align="right">
<math>~4\pi G\rho_c \biggl[1 - \frac{\lambda_1^2}{a^2} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2}{\lambda_1 (q^2-1)xy\ell^2} \biggl\{
\frac{\partial}{\partial \lambda_1} \biggl[ \frac{(q^2-1)xy }{\lambda_1 \lambda_2 } \cdot \frac{\partial \Phi}{\partial \lambda_1}\biggr]
+
\frac{\partial}{\partial \lambda_2} \biggl[ \frac{\lambda_1 \lambda_2}{(q^2-1) xy } \cdot \frac{\partial \Phi}{\partial \lambda_2}\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2}{\lambda_1 } \cdot (\Lambda-1)^{-3 / 2} \frac{2^2}{\lambda_2^2} \cdot \frac{\lambda_2^2}{2}(\Lambda-1)\Lambda \biggl\{
\frac{\partial}{\partial \lambda_1} \biggl[ \frac{(\Lambda-1)}{2 } \cdot \frac{\partial \Phi}{\partial \lambda_1}\biggr]
+
\frac{\partial}{\partial \lambda_2} \biggl[ \frac{2}{ (\Lambda-1) } \cdot \frac{\partial \Phi}{\partial \lambda_2}\biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4 \Lambda}{(\Lambda-1 )} \biggl\{
\frac{\partial}{\partial \lambda_1} \biggl[ \frac{(\Lambda-1)}{2 } \cdot \frac{\partial \Phi}{\partial \lambda_1}\biggr]
+
\frac{\partial}{\partial \lambda_2} \biggl[ \frac{2}{ (\Lambda-1) } \cdot \frac{\partial \Phi}{\partial \lambda_2}\biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
where we have used the following expressions [[#InvertedRelations|derived above]]:
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  <td align="right">
<math>~x^2y^2</math>
  </td>
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<math>~=</math>
  </td>
  <td align="left">
<math>~
(\Lambda-1) \biggl[ \frac{\lambda_2^2}{2^2}(\Lambda - 1) \biggr]^2  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{\ell^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
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<math>~
\frac{\lambda_2^2}{2} (\Lambda-1)\Lambda = \frac{2\lambda_1^2 \Lambda}{(\Lambda+1)} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Lambda</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[1 + \frac{4\lambda_1^2}{\lambda_2^2} \biggr]^{1 / 2} 
~~\Rightarrow~~ 
\frac{1}{2}(\Lambda^2 - 1)^{1 / 2} = \frac{\lambda_1}{\lambda_2}
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{xy}{\lambda_1 \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
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<math>~
\frac{1}{2^2}\biggl[ \frac{\lambda_2}{\lambda_1}(\Lambda - 1)^{3 / 2} \biggr]  
=
\frac{1}{2}(\Lambda - 1) \, .  
</math>
  </td>
</tr>
</table>

Now,
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<math>~\frac{\partial(\Lambda-1)}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl( \frac{8\lambda_1}{\lambda_2^2} \biggr)
=
\frac{4}{\lambda_1 \Lambda} \biggl( \frac{\lambda_1^2}{\lambda_2^2} \biggr)
=
\frac{(\Lambda^2-1)}{\lambda_1 \Lambda} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial(\Lambda-1)^{-1}}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{(\Lambda-1)^2} \biggl[ \frac{1}{2\Lambda} \biggr] \biggl(- \frac{8\lambda_1^2}{\lambda_2^3}  \biggr)
=
\frac{4}{(\Lambda-1)^2} \biggl[ \frac{1}{\lambda_2 \Lambda} \biggr] \biggl(\frac{\lambda_1^2}{\lambda_2^2}  \biggr)
=
\frac{\Lambda + 1}{(\Lambda-1)} \biggl[ \frac{1}{\lambda_2 \Lambda} \biggr]  \, .
</math>
  </td>
</tr>
</table>

Hence,

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  <td align="right">
<math>~4\pi G\rho_c \biggl[1 - \frac{\lambda_1^2}{a^2} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4 \Lambda}{(\Lambda-1 )} \biggl\{
\frac{(\Lambda-1)}{2 }\biggl[ \frac{\partial^2 \Phi}{\partial \lambda_1^2}\biggr]
+
\frac{\partial \Phi}{\partial \lambda_1} \cdot \frac{\partial}{\partial \lambda_1} \biggl[ \frac{(\Lambda-1)}{2 } \biggr]
+
\frac{2}{ (\Lambda-1) } \biggl[  \frac{\partial^2 \Phi}{\partial \lambda_2^2}\biggr]
+
\frac{\partial \Phi}{\partial \lambda_2} \cdot \frac{\partial}{\partial \lambda_2} \biggl[ \frac{2}{ (\Lambda-1) } \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{4 \Lambda}{(\Lambda-1 )} \biggl\{
\frac{(\Lambda-1)}{2 }\biggl[ \frac{\partial^2 \Phi}{\partial \lambda_1^2}\biggr]
+
\biggl[ \frac{(\Lambda^2-1)}{2 \lambda_1\Lambda} \biggr] \frac{\partial \Phi}{\partial \lambda_1} 
+
\frac{2}{ (\Lambda-1) } \biggl[  \frac{\partial^2 \Phi}{\partial \lambda_2^2}\biggr]
+
\biggl[ \frac{2(\Lambda+1)}{ (\Lambda-1)\lambda_2\Lambda } \biggr] \frac{\partial \Phi}{\partial \lambda_2} 
\biggr\}
</math>
  </td>
</tr>

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  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
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<math>~
2\Lambda \biggl[ \frac{\partial^2 \Phi}{\partial \lambda_1^2}\biggr]
+
\biggl[ \frac{2(\Lambda + 1)}{ \lambda_1} \biggr] \frac{\partial \Phi}{\partial \lambda_1} 
+
\frac{8 \Lambda }{ (\Lambda-1)^2 } \biggl[  \frac{\partial^2 \Phi}{\partial \lambda_2^2}\biggr]
+
\biggl[ \frac{8(\Lambda+1)}{ (\Lambda-1)^2\lambda_2 } \biggr] \frac{\partial \Phi}{\partial \lambda_2} 
</math>
  </td>
</tr>

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  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\lambda_1} \biggl\{ \Lambda \lambda_1 \biggl[ \frac{\partial^2 \Phi}{\partial \lambda_1^2}\biggr]
+
(\Lambda + 1)  \frac{\partial \Phi}{\partial \lambda_1} \biggr\}
+
\frac{8}{\lambda_2 (\Lambda-1)^2}\biggl\{ 
\Lambda \lambda_2  \biggl[  \frac{\partial^2 \Phi}{\partial \lambda_2^2}\biggr]
+
(\Lambda+1) \frac{\partial \Phi}{\partial \lambda_2} 
\biggr\} \, .
</math>
  </td>
</tr>
</table>