Editing Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates (section)

==Search for the Third Coordinate==

===Cross Product of First Two Unit Vectors===

The cross-product of these two unit vectors should give the third, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 = \hat{e}_1 \times \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[ {e}_{1y}{e}_{2z} - {e}_{1z}{e}_{2y} \biggr]
+
\hat\jmath~\biggl[ {e}_{1z}{e}_{2x} - {e}_{1x}{e}_{2z} \biggr]
+
\hat{k}~\biggl[ {e}_{1x}{e}_{2y} - {e}_{1y}{e}_{2x} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[  \frac{x p^2z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggr]
+
\hat\jmath~\biggl[ \frac{q^2 y p^2z \ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
-
\hat{k}~\biggl[ \frac{x^2\ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}}  ~+~ \frac{q^4 y^2\ell_{3D} }{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} \biggl\{~
\hat\imath~( x p^2z )
+
\hat\jmath~(q^2 y p^2z  )
-
\hat{k}~(x^2  + q^4y^2)
~\biggr\} \, .
</math>
  </td>
</tr>
</table>
Inserting these component expressions into the last row of the T8 Direction Cosine table gives &hellip;

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T8 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\lambda_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
  <td align="center"><math>~\gamma_{n1}</math></td>
  <td align="center"><math>~\gamma_{n2}</math></td>
  <td align="center"><math>~\gamma_{n3}</math></td>
</tr>

<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
  <td align="center"><math>~(x) \ell_{3D}</math></td>
  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~\frac{x}{  y^{1/q^2}}</math></td>
  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math></td>
  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
  <td align="center"><math>~-\frac{\lambda_2}{q^2 y}</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~0</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center"><math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
</tr>

<tr>
  <td align="left" colspan="9">
<table border="0" cellpadding="8" align="center">

<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>

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


===Associated h<sub>3</sub> Scale Factor===
<table border="1" align="right" cellpadding="10"><tr><td align="center">
[[File:EUREKA 21Jan2021 sm.png|350px|Whiteboard EUREKA moment]]</td></tr></table>
After working through various scenarios on my whiteboard today (21 January 2021), I propose that,
<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{xp^2z}{(x^2 + q^4y^2)} \, ;</math>
  </td>
<td align="center>&nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)} \, ;</math>
  </td>
<td align="center>&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-1 \, .</math>
  </td>
</tr>
</table>

This 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>~\sum_{i=1}^3 \biggl( \frac{\partial \lambda_3}{\partial x_i}\biggr)^2</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{p^4z^2(x^2 + q^4y^2)}{(x^2 + q^4y^2)^2} 
+
1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + q^4y^2 +p^4z^2)}{(x^2 + q^4y^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\ell_{3D}^2 (x^2 + q^4y^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ell_{3D} (x^2 + q^4y^2)^{1 / 2} \, . 
</math>
  </td>
</tr>
</table>
This seems to work well because, when combined with the three separate expressions for <math>~\partial \lambda_3/\partial x_i</math>,  this single expression for <math>~h_3</math> generates all three components of the third unit vector, that is, all three direction cosines, <math>~\gamma_{3i}</math>.  All of the elements of this new "EUREKA moment" result have been entered into the following table.


<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T8 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\lambda_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
  <td align="center"><math>~\gamma_{n1}</math></td>
  <td align="center"><math>~\gamma_{n2}</math></td>
  <td align="center"><math>~\gamma_{n3}</math></td>
</tr>

<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
  <td align="center"><math>~(x) \ell_{3D}</math></td>
  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~\frac{x}{  y^{1/q^2}}</math></td>
  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y }{(x^2 + q^4y^2)^{1 / 2}}\biggr] </math></td>
  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
  <td align="center"><math>~-\frac{\lambda_2}{q^2 y}</math></td>
  <td align="center"><math>~0</math></td>
  <td align="center"><math>~\frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~- \frac{x }{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~0</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">---</td>
  <td align="center"><math>~\ell_{3D}(x^2 + q^4 y^2)^{1 / 2}</math></td>
  <td align="center"><math>~\frac{xp^2z}{(x^2 + q^4y^2)} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2z}{(x^2 + q^4y^2)}</math></td>
  <td align="center"><math>~-1</math></td>
  <td align="center"><math>~\frac{x p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~\frac{q^2 y p^2 z\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
  <td align="center"><math>~-\frac{(x^2 + q^4 y^2)\ell_{3D}}{(x^2 + q^4y^2)^{1 / 2}} </math></td>
</tr>

<tr>
  <td align="left" colspan="9">
<table border="0" cellpadding="8" align="center">

<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>

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

===What is the Third Coordinate Function, &lambda;<sub>3</sub>===

The remaining [https://en.wikipedia.org/wiki/The_$64,000_Question $64,000 question] is, "What is the actual expression for <math>~\lambda_3(x, y, z)</math>&nbsp; ? &nbsp;"  

Notice that the (partial) derivatives of <math>~\lambda_3</math> with respect to <math>~x</math> and <math>~y</math> may be rewritten, respectively, in the form
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{q^2 y}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^2 y}{x(1 + \eta^2)}
=
\frac{\eta}{(1 + \eta^2)} \, ,
</math>
&nbsp; &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>~\biggl( \frac{x}{p^2z} \biggr) \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^2y}{x(1 + \eta^2)}
=
\frac{\eta}{(1 + \eta^2)} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{q^2y}{x} ~~~~\Rightarrow~~ \frac{\partial \ln\eta}{\partial \ln x} = -1 \, , ~~~~\frac{\partial \ln\eta}{\partial \ln y} = +1 \, . 
</math>
  </td>
</tr>
</table>

Then, after searching through the [[Appendix/References#CRC|CRC Mathematical Handbook's]] pages of familiar derivative expressions, we appreciate that

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

<tr>
  <td align="right">
<math>~\frac{d}{dx_i} \biggl[\frac{1}{\cosh\gamma}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{\tanh\gamma}{\cosh\gamma} \biggr] \frac{d\gamma}{dx_i} \, .</math>
  </td>
</tr>
</table>

Hence, it will be useful to adopt the mapping,

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

<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\sinh \gamma \, ,</math>
  </td>
</tr>
</table>
because the right-hand side of both partial-derivative expressions becomes,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\eta}{(1+\eta^2)}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\frac{\sinh\gamma}{\cosh^2\gamma} = \frac{\tanh\gamma}{\cosh\gamma} \, .</math>
  </td>
</tr>
</table>

====Guess A====
In particular, this suggests that we set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{A}{\cosh\gamma} \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\sinh^{-1}\eta = \pm \cosh^{-1}[\eta^2 + 1]^{1 / 2} \, .</math>
  </td>
</tr>
</table>
In other words,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A[\eta^2 + 1]^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
Let's check the first and second partial derivatives.
<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{A}{2} \biggl[ \frac{2\eta}{(\eta^2 + 1)^{3 / 2}} \biggr] \frac{\partial \eta}{\partial x}
</math>
  </td>
</tr>
</table>

====Guess B====

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

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\ln(1+\eta^2) \, .</math>
  </td>
</tr>
</table>
Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\lambda_3}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\eta}{(1+\eta^2)} \, ,
</math>
  </td>
</tr>
</table>
in which case we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>
  </td>
</tr>
</table>
which means,
<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{\eta}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
=
- \frac{x}{(x^2 + q^4y^2)} \biggl[ \frac{q^4 y^2}{x^2} \biggr] \, .
</math>
  </td>
</tr>
</table>

====Guess C====

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

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \ln(1+\eta^{-2}) \, .</math>
  </td>
</tr>
</table>
Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{d\lambda_3}{d\eta}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-
\frac{\eta^{-3}}{(1+\eta^{-2})} 
=
-\frac{ \eta^{-1}}{(1+\eta^{2})} 
</math>
  </td>
</tr>
</table>
in which case we find,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial x_i} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{d\lambda_3}{d\eta} \cdot \frac{\partial\eta}{\partial x_i}</math>
  </td>
</tr>
</table>
which means,
<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{\eta^{-1}}{(1+\eta^2)}  \biggl[ - \frac{q^2 y}{x^2} \biggr]
=
\frac{x}{(x^2 + q^4y^2)}  \, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\lambda_3}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\eta^{-1}}{(1+\eta^2)}  \biggl[ \frac{q^2 }{x} \biggr]
=
-\frac{x^2}{y(x^2+ q^4y^2)}  
</math>
  </td>
</tr>
</table>