Editing Appendix/Ramblings/ConcentricEllipsoidalCoordinates (section)

====Development====

Here we stick with the [[#Primary_.28radial-like.29_Coordinate|primary (radial-like) coordinate as defined above]]; for example,
<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^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda_1 \ell_{3D} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\ell_{3D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[ x^2 + q^4y^2 + p^4 z^2 ]^{- 1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\hat{e}_1 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, .
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that, <math>~\hat{e}_1 \cdot \hat{e}_1 = 1</math>, which means that this is, indeed, a properly normalized ''unit'' vector.
</td></tr></table>

Then, drawing from our [https://www.phys.lsu.edu/astro/H_Book.current/Appendices/Mathematics/operators.tohline1.pdf earliest discussions of "T1 Coordinates"], we'll try defining the ''second'' coordinate as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\tan^{-1} u \, ,
</math>
&nbsp; &nbsp; &nbsp; where,
  </td>
</tr>

<tr>
  <td align="right">
<math>~u</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>\frac{y^{1/q^2}}{x} \, .</math>
  </td>
</tr>
</table>
The relevant partial derivatives are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{1 + u^2} \biggl[ - \frac{y^{1/q^2}}{x^2} \biggr]
=
- \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{x} 
=
- \frac{\sin\lambda_3 \cos\lambda_3}{x} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{1 + u^2} \biggl[ \frac{y^{(1/q^2-1)}}{q^2x} \biggr]
=
\biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{q^2y} 
=
\frac{\sin\lambda_3 \cos\lambda_3}{q^2y} \, ,
</math>
  </td>
</tr>
</table>
which means that,
<table border="0" cellpadding="5" align="center">

<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 \biggr]^{-1}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1 + u^2}{u}\biggr]xq^2 y \ell_q = \frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3} \, ,
</math>
&nbsp; &nbsp; &nbsp; where, 
  </td>
</tr>

<tr>
  <td align="right">
<math>~\ell_q</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~[x^2 + q^4 y^2]^{-1 / 2} \, .</math>
  </td>
</tr>
</table>

The third row of direction cosines can now be filled in to give,

<table border="1" cellpadding="8" align="center" width="60%">
<tr>
  <td align="center" colspan="4">
'''Direction Cosines for T6 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_{3D}</math><br />&nbsp;
  <td align="center"><math>~q^2 y \ell_{3D}</math>
  <td align="center"><math>~p^2 z \ell_{3D}</math>
</tr>
<tr>
  <td align="center"><math>~2</math></td>
  <td align="center">
&nbsp;<br />
---
<br />&nbsp;
  <td align="center">
&nbsp;<br />
---
<br />&nbsp;
  <td align="center">
&nbsp;<br />
---
<br />&nbsp;
  </td>
</tr>
<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">
<math>~-q^2 y \ell_q</math>
  </td>
  <td align="center">
<math>~x \ell_q</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
</tr>
</table>
which means that the associated unit vector is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\hat\imath (q^2 y \ell_{q}) + \hat\jmath (x \ell_{q})  \, .
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that, <math>~\hat{e}_3 \cdot \hat{e}_3 = 1</math>, which means that this also is a properly normalized ''unit'' vector.  Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other.  Let's see &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 \cdot \hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(- q^2y \ell_q)x\ell_{3D} + (x\ell_q) q^2y\ell_{3D} = 0 \, .
</math>
  </td>
</tr>
</table>
Q.E.D.
</td></tr></table>

Now, even though we have not yet determined the proper expression for the "second" orthogonal coordinate, <math>~\lambda_2</math>, we should be able to obtain an expression for its associated unit vector from the cross product of the "third" and "first" unit vectors.  Specifically we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2 \equiv \hat{e}_3 \times \hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \biggl[ (e_3)_2 (e_1)_3 - (e_3)_3(e_1)_2 \biggr]
+
\hat\jmath \biggl[ (e_3)_3 (e_1)_1 - (e_3)_1(e_1)_3 \biggr]
+
\hat{k} \biggl[ (e_3)_1 (e_1)_2 - (e_3)_2(e_1)_1 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \biggl[ (x \ell_q) (p^2 z \ell_{3D}) - 0 \biggr]
+
\hat\jmath \biggl[ 0 - (-q^2y \ell_q)(p^2z \ell_{3D}) \biggr]
+
\hat{k} \biggl[ (-q^2y \ell_q) (q^2 y \ell_{3D}) - (x\ell_q)(x\ell_{3D}) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D}\biggl[
\hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} ( x^2 + q^4 y^2 )
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D}\biggl[
\hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} \biggl( \frac{1}{\ell_q^2}  \biggr) 
\biggr] \, .
</math>
  </td>
</tr>
</table>

<table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left">
Note that, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 \cdot \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q^2 \ell_{3D} \biggl[
(- q^2y )x p^2 z  
+ (x) q^2y p^2 z  
\biggr] = 0 \, ;
</math>
  </td>
</tr>
</table>
and,
<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>~
(x\ell_{3D})xp^2z \ell_q \ell_{3D}
+ (q^2y \ell_{3D}) q^2yp^2 z \ell_q \ell_{3D}
- (x^2 + q^4 y^2)\ell_q \ell_{3D} (p^2 z \ell_{3D} )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \ell_q \ell_{3D}^2 \biggl[
x^2p^2z 
+ (q^4y^2 ) p^2 z 
- (x^2 + q^4 y^2) (p^2 z )
\biggr] = 0 \, .
</math>
  </td>
</tr>
</table>
We conclude, therefore, that <math>~\hat{e}_2</math> is perpendicular to both of the other unit vectors. <font color="red">'''Hooray!'''</font>
</td></tr></table>


Filling in the second row of the direction cosines table gives,

<table border="1" cellpadding="8" align="center" width="60%">
<tr>
  <td align="center" colspan="4">
'''Direction Cosines for T6 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_{3D}</math><br />&nbsp;
  <td align="center"><math>~q^2 y \ell_{3D}</math>
  <td align="center"><math>~p^2 z \ell_{3D}</math>
</tr>
<tr>
  <td align="center"><math>~2</math></td>
  <td align="center">
<math>~x p^2 z\ell_q \ell_{3D}</math>
  <td align="center">
<math>~q^2y p^2 z\ell_q \ell_{3D}</math>
  <td align="center">
<math>~-(x^2 + q^4y^2)\ell_q \ell_{3D} = - \ell_{3D}/\ell_q</math>
  </td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">
<math>~-q^2 y \ell_q</math>
  </td>
  <td align="center">
<math>~x \ell_q</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
</tr>
</table>