Editing Appendix/Ramblings/EllipticCylinderCoordinates (section)

==Relevant Partial Derivatives==

Before moving forward, we need to evaluate a number of relevant partial derivatives.

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_1}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial x} \biggl[x^2 + q^2y^2\biggr]^{1 / 2} = \frac{1}{2}\biggl[x^2 + q^2y^2\biggr]^{- 1 / 2} 2x
= \frac{x}{\lambda_1} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_1}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial y} \biggl[x^2 + q^2y^2\biggr]^{1 / 2} = \frac{1}{2}\biggl[x^2 + q^2y^2\biggr]^{- 1 / 2} 2q^2 y
= \frac{q^2 y}{\lambda_1} \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial x} \biggl[x^{q^2/(q^2-1)} (q y)^{-1/(q^2-1)}\biggr] 
= \biggl[\frac{q^2}{q^2-1}\biggr] \frac{\lambda_2}{x} 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial}{\partial y} \biggl[x^{q^2/(q^2-1)} (q y)^{-1/(q^2-1)}\biggr]
= - \biggl[ \frac{1}{q^2-1} \biggr]\frac{\lambda_2}{y}  
\, .
</math>
  </td>
</tr>
</table>

<span id="ComplementaryDerivatives">We may also need the set of complementary partial derivatives</span>.  Even though we are unable to explicitly invert the coordinate mappings, once we have in hand expressions for the three scale factors (see immediately below), we can determine expressions for the set of complementary partial derivatives via the [[User:Tohline/Appendix/Ramblings/DirectionCosines#Basic_Definitions_and_Relations|generic relation]],

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<tr>
  <td align="right">
<math>~\frac{\partial x_i}{\partial\lambda_n}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~h_n^2 \cdot \frac{\partial \lambda_n}{\partial x_i} \, .</math>
  </td>
</tr>
</table>


<table border="1" align="center" width="80%" cellpadding="10">
<tr><td align="left">
'''Example:''' &nbsp; &nbsp; <math>~q^2 = 2</math>

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

<tr>
  <td align="right">
<math>~y </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_2}{2^{3 / 2}}\biggl\{
\Lambda - 1
\biggr\} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~x^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_2^2}{2} 
\biggl\{
\Lambda - 1
\biggr\} \, ,
</math> &nbsp; &nbsp; &nbsp; where, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="right">
<math>~\Lambda</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left" colspan="1">
<math>~
\biggl[1 + \frac{4\lambda_1^2}{\lambda_2^2}  \biggr]^{1 / 2}
\, .</math>
  </td>
</tr>
</table>
Noting that,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \Lambda}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\Lambda \lambda_1}\biggl[ \frac{4\lambda_1^2}{\lambda_2^2} \biggr]</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \Lambda}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{1}{\Lambda \lambda_2}\biggl[ \frac{4\lambda_1^2}{\lambda_2^2} \biggr] \, ,</math>
  </td>
</tr>
</table>
we have,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\Lambda (\Lambda - 1) - 4 \lambda_1^2/\lambda_2^2 }{2^{3 / 2}\Lambda }  
=
\frac{\Lambda^2 - \Lambda - 4 \lambda_1^2/\lambda_2^2 }{2^{3 / 2}\Lambda }  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ (1 - \Lambda) }{2^{3 / 2}\Lambda }  \, .
</math>
  </td>
</tr>
</table>

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


Let's compare by drawing from the expressions for <math>~\ell^2</math>, above, and for <math>~h_n^2</math> derived below.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ h_1^2 \cdot \frac{\partial \lambda_1}{\partial y} \biggr]_{q^2=2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \lambda_1^2 \ell^2 \biggl( \frac{q^2 y}{\lambda_1} \biggr) \biggr]_{q^2=2}
=
\biggl[ 2\lambda_1 \ell^2 y \biggr]_{q^2=2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\lambda_1 \biggl\{
\frac{1}{\sqrt{2}\lambda_2 \Lambda}
\biggr\} 
=
\frac{\sqrt{2}\lambda_1}{\lambda_2\Lambda} \, .
</math>
  </td>
</tr>
</table>
<font color="red">'''Yes!'''</font>  This, indeed matches the just-derived expression for <math>~\partial y/\partial \lambda_1</math>.  And we also have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ h_2^2 \cdot \frac{\partial \lambda_2}{\partial y} \biggr]_{q^2=2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{  - \biggl[ \frac{1}{q^2-1} \biggr]\frac{\lambda_2}{y} \biggl[ \frac{(q^2-1)xy \ell}{\lambda_2}\biggr]^2\biggr\}_{q^2=2}
=
- \biggl[ \frac{(q^2-1)x^2 y \ell^2}{\lambda_2}\biggr]_{q^2=2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1 - \Lambda )}{2^{3 / 2} \Lambda}  \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Yes, again!'''</font>