Editing Appendix/Ramblings/EllipticCylinderCoordinates (section)

=T5 Coordinates=

==Introduction==

As has been made clear in our above review of the Elliptic Cylinder Coordinate system <math>~(\xi_1, \xi_2, \xi_3) = (d\cosh\mu, \cos\nu, z)</math>, individual curves within a family of ''confocal'' ellipses are identified by one's choice of the "radial" coordinate parameter, <math>~\mu</math>, or, alternatively, <math>~\xi_1</math>.  Specifically, while the two foci of every ellipse are positioned along the x-axis at the same points &#8212; namely, <math>~(x, y) = (\pm~d, 0)</math> &#8212; the length of the semi-major axis is given by, <math>~a = \xi_1 = d\cosh\mu</math>.

In a [[User:Tohline/Appendix/Ramblings/T3Integrals|separate chapter]] we have introduced a different orthogonal curvilinear coordinate system that we refer to as, "[[User:Tohline/Appendix/Ramblings/T3Integrals|T3 Coordinates]]."  In this coordinate system, <math>~(\lambda_1, \lambda_2, \lambda_3)</math>, individual surfaces within a family of ''concentric'' spheroids are identified by one's choice of a different "radial" coordinate parameter, <math>~\lambda_1</math>.  Here we will adopt essentially this same set of orthogonal coordinates, using <math>~\lambda_1</math> and <math>~\lambda_2</math> to describe a family of ''concentric'' ellipses that is independent of the vertical-coordinate.  We will refer to it as the &hellip;

<table border="1" cellpadding="10" align="center" width="80%">
<tr><td align="center">
'''T5 Coordinate System'''</td></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\lambda_1</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~x \cosh \zeta </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2 y^2)^{1 / 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~x (\sinh\zeta)^{1/(1-q^2)} </math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x^{q^2}}{qy} \biggr)^{1/(q^2-1)}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~z </math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\zeta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\sinh^{-1}\biggl( \frac{qy}{x} \biggr) </math>
  </td>
</tr>
</table>
and, <math>~0 < q < \infty</math> is the (fixed) parameter used to specify the eccentricity, <math>~e = [(q^2-1)^{1 / 2}/q]</math>, of every <math>~\lambda_1 = </math> constant curve within the family of ''concentric'' ellipses.

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


Checking these expressions, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2 \equiv x (\sinh\zeta)^{1/(1-q^2)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl( \frac{qy}{x} \biggr)^{1/(1-q^2)} = x \biggl( \frac{x}{qy} \biggr)^{1/(q^2-1)} 
= \biggl( \frac{x^{q^2}}{qy} \biggr)^{1/(q^2-1)} \, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\lambda_1 \equiv x\cosh\zeta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x\biggl[ 1 + \sinh^2\zeta\biggr]^{1 / 2} = x\biggl[ 1 + \biggl( \frac{qy}{x} \biggr)^2\biggr]^{1 / 2}
= ( x^2 + q^2 y^2 )^{1 / 2} \, .</math>
  </td>
</tr>
</table>
Comparing this last expression with the [[#Background|above background description of ellipses]], we see that <math>~\lambda_1 = </math> constant &#8212; for example, <math>~\lambda_0</math> &#8212; is synonymous with an ellipse having &hellip;
<ul>
<li>A semi-major axis of length, <math>~a = \lambda_0</math>;</li>
<li>An eccentricity, <math>~e \equiv (1 - b^2/a^2)^{1 / 2} = [(q^2-1)/q^2]^{1 / 2}</math>;</li>
<li>A pair of foci whose coordinate locations along the major axis are, <math>~(x, y) = (\pm~c, 0)</math>, where, <math>~c = ae</math>.</li>
</ul>

==Invert Coordinate Mapping==

Solving for <math>~x(\lambda_1, \lambda_2)</math>, we find &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( x^2 + q^2 y^2 )^{1 / 2} </math>
  </td>
</tr>

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

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

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

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

Hence,

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

<tr>
  <td align="right">
<math>~x^{2q^2} \lambda_2^{2(1-q^2)} + x^2 - \lambda_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .  </math>
  </td>
</tr>
</table>
Alternatively, solving for <math>~y(\lambda_1, \lambda_2)</math>, we find &hellip;

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

<tr>
  <td align="right">
<math>~\lambda_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~( x^2 + q^2 y^2 )^{1 / 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~x^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_1^2 - q^2 y^2 \, .</math>
  </td>
</tr>
</table>
And,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(qy)^{1/q^2}~\lambda_2^{(q^2-1)/q^2} \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(qy)^{2/q^2}~\lambda_2^{2(q^2-1)/q^2} -\lambda_1^2 + q^2 y^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>

<span id="InvertedRelations">&nbsp;</span>
<table border="1" align="center" width="80%" cellpadding="10">
<tr><td align="center">'''Summary of Inverted Relations'''</td></tr>
<tr><td align="left">

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

<tr>
  <td align="right">
<math>~\lambda_2^2\biggl( \frac{x}{\lambda_2}\biggr)^{2q^2} + x^2 - \lambda_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\lambda_2^2 \biggl( \frac{qy}{\lambda_2} \biggr)^{2/q^2} + q^2 y^2 - \lambda_1^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</td></tr>
<tr><td align="left">
'''Example:''' &nbsp; &nbsp; <math>~q^2 = 2</math>

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

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^{4} \lambda_2^{-2} + x^2 - \lambda_1^2</math>
  </td>
  <td align="right"><math>~\leftarrow</math>&nbsp; &nbsp;'''Quadratic Eq.''' in x<sup>2</sup></td>
</tr>

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

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y^2 + (2^{1 / 2}\lambda_2)~y -\lambda_1^2 </math>
  </td>
  <td align="right"><math>~\leftarrow</math>&nbsp; &nbsp;'''Quadratic Eq.''' in y</td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~y </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}\biggl\{
-2^{1 / 2} \lambda_2 ~\pm ~ \biggl[2\lambda_2^2 + 8\lambda_1^2  \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
  <td align="right">&nbsp;</td>
</tr>

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

where,
<div align="center">
<math>~\Lambda \equiv \biggl[1 + \frac{4\lambda_1^2}{\lambda_2^2} \biggr]^{1 / 2} \, .</math>
</div>

Note &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{y}{x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_2}{2^{3 / 2}} \cdot \frac{2}{\lambda_2^2} = \frac{1}{\sqrt{2} \lambda_2} \, ;</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
<math>~\frac{1}{\ell^2} \equiv (x^2 +  4y^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sqrt{2} \lambda_2 y \biggl[1 + \frac{4\lambda_1^2}{\lambda_2^2} \biggr]^{1 / 2} 
= \sqrt{2} \lambda_2 y \Lambda 
= \frac{\lambda_2^2}{2} \cdot \Lambda (\Lambda - 1)
\, .
</math>
  </td>
</tr>
</table>
Note as well that, <math>~\ell^{-2} = 2\lambda_1^2 \Lambda/(\Lambda + 1) \, .</math>
</td></tr>

<tr><td align="left">
'''Example:''' &nbsp; &nbsp; <math>~q^2 = \frac{3}{2}</math>

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

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2\biggl( \frac{x}{\lambda_2}\biggr)^{3} + x^2 - \lambda_1^2</math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/Appendix/Ramblings/T1Coordinates#Second_Special_Case_.28cubic.29|Cubic Eq.]]''' in x</td>
</tr>

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2 \biggl( \frac{3 y^2}{2\lambda_2^2} \biggr)^{2/3} + \frac{3}{2} y^2 - \lambda_1^2 </math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/Appendix/Ramblings/T1Coordinates#Second_Special_Case_.28cubic.29|Cubic Eq.]]''' in y<sup>2/3</sup></td>
</tr>
</table>
</td></tr>

<tr><td align="left">
'''Example:''' &nbsp; &nbsp; <math>~q^2 = 3</math>

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

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2\biggl( \frac{x}{\lambda_2}\biggr)^{6} + x^2 - \lambda_1^2</math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/Appendix/Ramblings/T1Coordinates#Second_Special_Case_.28cubic.29|Cubic Eq.]]''' in x<sup>2</sup></td>
</tr>

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2 \biggl( \frac{3y^2}{\lambda_2^2} \biggr)^{1/3} + 3 y^2 - \lambda_1^2 </math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/Appendix/Ramblings/T1Coordinates#Second_Special_Case_.28cubic.29|Cubic Eq.]]''' in y<sup>2/3</sup></td>
</tr>
</table>
</td></tr>

<tr><td align="left">
'''Example:''' &nbsp; &nbsp; <math>~q^2 = 4</math>

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

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2\biggl( \frac{x}{\lambda_2}\biggr)^{8} + x^2 - \lambda_1^2</math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/SSC/Stability/BiPolytrope0_0Details#Roots_of_Quartic_Equation|Quartic Eq.]]''' in x<sup>2</sup></td>
</tr>

<tr>
  <td align="right">
<math>~0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_2^2 \biggl( \frac{2y}{\lambda_2} \biggr)^{1 / 2} + 4 y^2 - \lambda_1^2 </math>
  </td>
  <td align="left"> &nbsp; &nbsp; &nbsp; &nbsp;<math>~\leftarrow</math>&nbsp; &nbsp;'''[[User:Tohline/SSC/Stability/BiPolytrope0_0Details#Roots_of_Quartic_Equation|Quartic Eq.]]''' in y<sup>1/2</sup></td>
</tr>
</table>

</td></tr></table>
&nbsp;

==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]],

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

<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>

==Scale Factors, Direction Cosines &amp; Unit Vectors==

From our [[User:Tohline/Appendix/Ramblings/DirectionCosines#Usage|accompanying generic discussion of direction cosines]], we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_1}{\partial y } \biggr)^2 + \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)^2\biggr]^{-1}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\lambda_1^2 \ell^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y } \biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2\biggr\}^{-1}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{(q^2-1)xy \ell}{\lambda_2}\biggr]^2\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_3^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y } \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2\biggr\}^{-1} = 1 \, ;</math>
  </td>
</tr>
</table>
where,
<div align="center">
<math>~\ell \equiv (x^2 + q^4 y^2)^{- 1 / 2} \, .</math>
</div>


<table border="1" cellpadding="8" align="center" width="60%">
<tr>
  <td align="center" colspan="4">
'''Direction Cosines for T5 Coordinates'''
<br />
<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="center" width="10%"><math>~n</math></td>
  <td align="center" colspan="3"><math>~i = x, y, z</math>
</tr>
<tr>
  <td align="center"><math>~1</math></td>
  <td align="center">&nbsp;<br />
<math>~x\ell</math><br />&nbsp;</td>
  <td align="center"><math>~q^2 y \ell</math></td>
  <td align="center"><math>~0</math></td>
</tr>
<tr>
  <td align="center"><math>~2</math></td>
  <td align="center">
&nbsp;<br />
<math>~q^2 y \ell </math>
<br /></td>
  <td align="center">
<math>~ - x\ell
</math></td>
  <td align="center"><math>~0</math></td>
</tr>
<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">&nbsp;<br /><math>~0</math><br />&nbsp;</td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~1</math></td>
</tr>
</table>

The unit vectors are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_n</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \gamma_{n1} + \hat\jmath \gamma_{n2} + \hat{k}\gamma_{n3} \, ,
</math>
  </td>
</tr>
</table>
that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath (x\ell) + \hat\jmath (q^2 y \ell)  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath (q^2 y\ell) - \hat\jmath (x \ell)  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{k}  \, .
</math>
  </td>
</tr>
</table>
Notice that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
q^2 xy\ell^2 - q^2 xy\ell^2 = 0 \, ,
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2\ell^2 + q^4 y^2 \ell^2 
= \ell^2 (x^2 + q^4 y^2) = 1 \, .
</math>
  </td>
</tr>
</table>
These are both desired orthogonality conditions.  Alternatively,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\hat\imath
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \gamma_{11} + \hat{e}_2 \gamma_{21} + \hat{e}_3 \gamma_{31} 
=
\hat{e}_1 (x\ell) + \hat{e}_2 (q^2 y\ell)  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\hat\jmath
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \gamma_{12} + \hat{e}_2 \gamma_{22} + \hat{e}_3 \gamma_{32} 
=
\hat{e}_1 (q^2y\ell) - \hat{e}_2 (x\ell) \,  ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\hat{k}
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \gamma_{13} + \hat{e}_2 \gamma_{23} + \hat{e}_3 \gamma_{33} = \hat{e}_3 \, .
</math>
  </td>
</tr>
</table>

==Spatial Operators==

<table border="1" align="center" cellpadding="10" width="80%">
<tr>
  <td align="center">'''Summary Reminder'''</td>
</tr>
<tr><td align="left">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~h_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\lambda_1^2 \ell^2 \, ;
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~h_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{(q^2-1)xy \ell}{\lambda_2}\biggr]^2\, ;
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
   <td align="right">
<math>~h_3^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 \, ;</math>
  </td>
</tr>
</table>

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

In T5 Coordinates, a couple of relevant operators are:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla F</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_1 \biggl[\frac{1}{h_1} \frac{\partial F}{\partial \lambda_1} \biggr]
+
\hat{e}_2 \biggl[\frac{1}{h_2} \frac{\partial F}{\partial \lambda_2} \biggr]
+
\hat{e}_3 \biggl[\frac{1}{h_3} \frac{\partial F}{\partial \lambda_3} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_1 \biggl( \frac{1}{\lambda_1 \ell} \biggr) \frac{\partial F}{\partial \lambda_1} 
+
\hat{e}_2 \biggl[\frac{\lambda_2}{(q^2-1)xy\ell} \biggr] \frac{\partial F}{\partial \lambda_2} 
+
\hat{e}_3 \frac{\partial F}{\partial \lambda_3} \, .
</math>
  </td>
</tr>
</table>


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

<tr>
  <td align="right">
<math>~\nabla^2 F</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{h_1 h_2 h_3} \biggl[
\frac{\partial}{\partial \lambda_1} \biggl( \frac{h_2 h_3}{h_1} \cdot \frac{\partial F}{\partial \lambda_1}\biggr)
+
\frac{\partial}{\partial \lambda_2} \biggl( \frac{h_3 h_1}{h_2} \cdot \frac{\partial F}{\partial \lambda_2}\biggr)
+
\frac{\partial}{\partial \lambda_3} \biggl( \frac{h_1 h_2}{h_3} \cdot \frac{\partial F}{\partial \lambda_3}\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 (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 F}{\partial \lambda_1}\biggr]
+
\frac{\partial}{\partial \lambda_2} \biggl[ \frac{\lambda_1 \lambda_2}{(q^2-1) xy } \cdot \frac{\partial F}{\partial \lambda_2}\biggr]
+
\frac{\partial}{\partial \lambda_3} \biggl[ \frac{\lambda_1  (q^2-1) xy \ell^2 }{\lambda_2} \cdot \frac{\partial F}{\partial \lambda_3}\biggr]
\biggr\}
</math>
  </td>
</tr>
</table>

And if <math>~F</math> is a function only of <math>~\lambda_1</math>, then,

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

<tr>
  <td align="right">
<math>~\nabla^2 F</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 F}{\partial \lambda_1}\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 (q^2-1)xy\ell^2} \biggl\{
\biggl[ \frac{(q^2-1)xy }{\lambda_1 \lambda_2 } \biggr] \biggl[ \frac{\partial^2 F}{\partial \lambda_1^2}\biggr]
+
\frac{\partial F}{\partial \lambda_1} \cdot \frac{\partial}{\partial \lambda_1} \biggl[ \frac{(q^2-1)xy }{\lambda_1 \lambda_2 } \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{\lambda_1^2  \ell^2} \biggr] 
\biggl[ \frac{\partial^2 F}{\partial \lambda_1^2}\biggr]
+
\biggl[ \frac{1}{\lambda_1 xy\ell^2} \biggr]
\frac{\partial F}{\partial \lambda_1} \cdot \frac{\partial}{\partial \lambda_1} \biggl[ \frac{xy }{\lambda_1  } \biggr] \, .
</math>
  </td>
</tr>
</table>
In order to complete this evaluation, we need a couple of "complementary partial derivatives."  Referencing the [[#ComplementaryDerivatives|relation provided above]], we find,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{xy}{\lambda_1^2} 
+
\frac{y}{\lambda_1} \biggl[h_1^2 \frac{\partial \lambda_1}{\partial x} \biggr]
+
\frac{x}{\lambda_1} \biggl[h_1^2 \frac{\partial \lambda_1}{\partial y} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{xy}{\lambda_1^2} 
+
\frac{h_1^2}{\lambda_1} \biggl[\frac{xy}{\lambda_1}  
+
\frac{q^2 x y}{\lambda_1}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{xy}{\lambda_1^2} \biggl[\lambda_1^2 \ell^2 (1 + q^2)  - 1 \biggr] 
\, .
</math>
  </td>
</tr>
</table>
<span id="Laplacian">Hence,</span>

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

<tr>
  <td align="right">
<math>~\nabla^2 F</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{\lambda_1^2  \ell^2} \biggr] 
\biggl[ \frac{\partial^2 F}{\partial \lambda_1^2}\biggr]
+
\frac{xy}{\lambda_1^2} \biggl[\lambda_1^2 \ell^2 (1 + q^2)  - 1 \biggr]
\biggl[ \frac{1}{\lambda_1 xy\ell^2} \biggr]
\frac{\partial F}{\partial \lambda_1}  
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{\lambda_1^2  \ell^2} \biggr] 
\biggl[ \frac{\partial^2 F}{\partial \lambda_1^2}\biggr]
-
\biggl[ \frac{1}{\lambda_1^3 \ell^2} \biggr]
\frac{\partial F}{\partial \lambda_1}  
+
\biggl[ \frac{(1 + q^2)}{\lambda_1 } \biggr]
\frac{\partial F}{\partial \lambda_1}  \, .
</math>
  </td>
</tr>
</table>

==Example (q<sup>2</sup> = 2) Poisson Equation==

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

<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,

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

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

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

<tr>
  <td align="right">
<math>~x^2y^2</math>
  </td>
  <td align="center">
<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>
  <td align="left">
<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>
  <td align="left">
<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,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<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,

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

<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{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>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>

<tr>
  <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>

===Trials===

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

<tr>
  <td align="right">
<math>~\Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \lambda_1^\alpha + B\lambda_2^\beta </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{\partial \Phi}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A\alpha \lambda_1^{\alpha-1}  \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~ \frac{\partial \Phi}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~B\beta  \lambda_2^{\beta - 1} \, .</math>
  </td>
</tr>
</table>
In this case we find,

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

<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{2}{\lambda_1} \biggl\{ \Lambda \lambda_1 \frac{\partial}{\partial \lambda_1}\biggl[ \frac{\partial \Phi}{\partial \lambda_1}\biggr]
+
(\Lambda + 1)  \biggl[ \frac{\partial \Phi}{\partial \lambda_1} \biggr] \biggr\}
+
\frac{8}{\lambda_2 (\Lambda-1)^2}\biggl\{ 
\Lambda \lambda_2  \frac{\partial}{\partial \lambda_2}\biggl[  \frac{\partial \Phi}{\partial \lambda_2}\biggr]
+
(\Lambda+1) \biggl[\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{2A}{\lambda_1} \biggl\{ \Lambda \lambda_1 \frac{\partial}{\partial \lambda_1}\biggl[ \alpha \lambda_1^{\alpha-1}  \biggr]
+
(\Lambda + 1)  \biggl[ \alpha \lambda_1^{\alpha-1}  \biggr] \biggr\}
+
\frac{8B}{\lambda_2 (\Lambda-1)^2}\biggl\{ 
\Lambda \lambda_2  \frac{\partial}{\partial \lambda_2}\biggl[ \beta  \lambda_2^{\beta - 1} \biggr]
+
(\Lambda+1) \biggl[ \beta  \lambda_2^{\beta - 1} \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2A\alpha }{\lambda_1} \biggl\{ \Lambda (\alpha-1)  \lambda_1^{\alpha-1} 
+
(\Lambda + 1) \lambda_1^{\alpha-1} \biggr\}
+
\frac{8B\beta }{\lambda_2 (\Lambda-1)^2}\biggl\{ 
\Lambda(\beta-1) \lambda_2^{\beta - 1}
+
(\Lambda+1) \lambda_2^{\beta - 1}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2A\alpha \lambda_1^{\alpha-2} \biggl\{ \Lambda (\alpha-1)  
+
(\Lambda + 1)  \biggr\}
+
\frac{8B\beta \lambda_2^{\beta-2}}{ (\Lambda-1)^2}\biggl\{ 
\Lambda(\beta-1)
+
(\Lambda+1) 
\biggr\}
</math>
  </td>
</tr>

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

If <math>~\alpha = 4</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{8B\beta \lambda_2^{\beta-2}}{ (\Lambda-1)^2} \biggl[ \beta \Lambda +1 \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] -8A\lambda_1^2
- 32A\lambda_1^{2} \Lambda
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
8B\beta \lambda_2^{\beta-2} \biggl[ \beta \Lambda +1 \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 4\pi G\rho_c 
- 32A\lambda_1^{2} \Lambda
- \lambda_1^2 \biggl[\frac{4\pi G \rho_c}{a^2} + 8A  \biggr] \biggr\}
(\Lambda^2 - 2\Lambda + 1) \, .
</math>
  </td>
</tr>
</table>
If, then, <math>~8Aa^2 = -4\pi G\rho_c</math>,

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
8B\beta \lambda_2^{\beta-2} \biggl[ \beta \Lambda +1 \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\pi G\rho_c\biggl\{ 1 
+ \biggl[ \frac{4\lambda_1^{2}}{a^2} \biggr] \Lambda
\biggr\}
(\Lambda^2 - 2\Lambda + 1) \, .
</math>
  </td>
</tr>
</table>
But, we also know that, <math>\lambda_1^2 = \lambda_2^2(\Lambda^2-1)/4</math>, so &hellip;

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

<tr>
  <td align="right">
<math>~
8a^2 B\beta \lambda_2^{\beta-2} \biggl[ \beta \Lambda +1 \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\pi G\rho_c \biggl\{ a^2 
+  \lambda_2^2 (\Lambda^2-1)\Lambda 
\biggr\}
(\Lambda^2 - 2\Lambda + 1) \, .
</math>
  </td>
</tr>
</table>


<font color="green">'''(25 October 2020) I give up &hellip; for now.'''</font>