Editing Appendix/Ramblings/ConcentricEllipsoidalT8Coordinates (section)

=Concentric Ellipsoidal (T8) Coordinates=

==Background==
Building on our [[Appendix/Ramblings/DirectionCosines|general introduction to ''Direction Cosines'']] in the context of orthogonal curvilinear coordinate systems, and on our previous development of [[Appendix/Ramblings/T3Integrals|T3]] (concentric oblate-spheroidal) and [[Appendix/Ramblings/EllipticCylinderCoordinates#T5_Coordinates|T5]] (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T8) coordinate system.  This is motivated by our [[ThreeDimensionalConfigurations/Challenges#Trial_.232|desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids]].

Note that, in a [[Appendix/Ramblings/ConcentricEllipsoidalCoordinates|separate but closely related discussion]], we made attempts to define this coordinate system, numbering the trials up through "T7."  In this "T7" effort, we were able to define a set of three, mutually orthogonal unit vectors that should work to define a fully three-dimensional, concentric  ellipsoidal coordinate system.  But we were unable to figure out what coordinate function, <math>~\lambda_3(x, y, z)</math>, was associated with the third unit vector.  In addition, we found the <math>~\lambda_2</math> coordinate to be rather strange in that it was not oriented in a manner that resembled the classic spherical coordinate system.  Here we begin by redefining the <math>~\lambda_2</math> coordinate such that its associated <math>~\hat{e}_3</math> unit vector lies parallel to the x-y plane.

==Realigning the Second Coordinate==

The first coordinate remains the same as before, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 + q^2 y^2 + p^2 z^2 \, .</math>
  </td>
</tr>
</table>
This may be rewritten as,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~a = \lambda_1 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="center">
<math>~b = \frac{\lambda_1}{q} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="left">
<math>~c = \frac{\lambda_1}{p} \, .</math>
  </td>
</tr>
</table>

By specifying the value of <math>~z = z_0 < c</math>, as well as the value of <math>~\lambda_1</math>, we are picking a plane that lies parallel to, but a distance <math>~z_0</math> above, the equatorial plane.  The elliptical curve that defines the intersection of the <math>~\lambda_1</math>-constant surface with this plane is defined by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 - p^2z_0^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2 + q^2 y^2  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{x}{a_{2D}}\biggr)^2 + \biggl( \frac{y}{b_{2D}}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_{2D} = \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp;&nbsp; &nbsp;</td>
  <td align="left">
<math>~b_{2D} = \frac{1}{q} \biggl(\lambda_1^2 - p^2z_0^2 \biggr)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

At each point along this elliptic curve, the line that is tangent to the curve has a slope that can be determined by simply differentiating the equation that describes the curve, that is,
<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>~\frac{2x dx}{a_{2D}^2} + \frac{2y dy}{b_{2D}^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{dy}{dx}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{2x}{a_{2D}^2} \cdot \frac{b_{2D}^2}{2y} = - \frac{x}{q^2y} \, .</math>
  </td>
</tr>

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

The unit vector that lies tangent to any point on this elliptical curve will be described by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~ \biggl\{ \frac{\Delta x}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\}
+
\hat\jmath~ \biggl\{ \frac{\Delta y}{[ (\Delta x)^2 + (\Delta y)^2 ]^{1 / 2}} \biggr\} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~ \biggl\{ \frac{1}{[ 1 + x^2/(q^4y^2) ]^{1 / 2}} \biggr\}
-
\hat\jmath~ \biggl\{ \frac{x/(q^2y)}{[ 1 + 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>~
\hat\imath~ \biggl\{ \frac{q^2y}{[ x^2 + q^4y^2 ]^{1 / 2}} \biggr\}
-
\hat\jmath~ \biggl\{ \frac{x}{[ x^2 + q^4y^2  ]^{1 / 2}} \biggr\} \, .</math>
  </td>
</tr>
</table>
As we have discovered, the coordinate that gives rise to this unit vector is,
<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>~\frac{x}{y^{1/q^2}} \, .</math>
  </td>
</tr>
</table>

Other properties that result from this definition of <math>~\lambda_2</math> are presented in 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">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</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>

The associated unit vector is, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath~\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] - \hat\jmath~\biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] \, .
</math>
  </td>
</tr>
</table>

It is easy to see that <math>~\hat{e}_2 \cdot \hat{e}_2 = 1</math>.  We also see 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>~
x \ell_{3D}\biggl[ \frac{q^2 y }{(x^2 + q^4y^2)^{1 / 2}}  \biggr]
-
q^2y \ell_{3D} \biggl[ \frac{x }{(x^2 + q^4y^2)^{1 / 2}}  \biggr] = 0 \, ,
</math>
  </td>
</tr>
</table>
so it is clear that these first two unit vectors are orthogonal to one another.  

==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>

==Inverting Coordinate Relations==

===In a Plane Perpendicular to the Z-Axis===
====General Case====
At a fixed value of <math>~z = z_0</math>, let's invert the <math>~\lambda_1(x, y)</math> and <math>~\lambda_2(x, y)</math> relations to obtain expressions for <math>~x(\lambda_1, \lambda_2)</math> and <math>~y(\lambda_1, \lambda_2)</math>.  Perhaps this will help us determine what the third coordinate expression should be.  

We start with,
<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^2y^2 + p^2 z_0^2)^{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>~\frac{x}{y^{1/q^2}} \, .</math>
  </td>
</tr>
</table>
This means that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2(\ln x - \ln\lambda_2 ) \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~q^2 y^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\lambda_1^2 - p^2z_0^2) - x^2</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~2\ln y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} \, .</math>
  </td>
</tr>
</table>
Together, this gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ln\biggl\{ \frac{1}{q^2} \biggl[ (\lambda_1^2 - p^2z_0^2) - x^2 \biggr] \biggr\} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^2(\ln x - \ln\lambda_2 ) </math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~(\lambda_1^2 - p^2z_0^2) - x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ x^2 +q^2\biggl[\frac{x}{\lambda_2} \biggr]^{2q^2} +  (p^2z_0^2 - \lambda_1^2) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>

====What if Axisymmetric (q<sup>2</sup> = 1)====

In an axisymmetric configuration, <math>~q^2 = 1</math> and <math>~(\lambda_1^2 - p^2z_0^2) = \varpi^2</math>, so this general expression for <math>~x</math> becomes,
<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^2 + \biggl[\frac{x}{\lambda_2} \biggr]^{2} -\varpi^2 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\varpi \biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
Given that, for axisymmetric systems,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\varpi \cos\varphi \, ,</math>
  </td>
</tr>
</table>
we conclude that when <math>~q^2 = 1</math>,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{\lambda_2^2}{1 + \lambda_2^2}\biggr]^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cos\varphi </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\lambda_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\cot^2\varphi </math>
  </td>
</tr>
</table>

====What if q<sup>2</sup> = 2====
For example, if we choose <math>~q^2 = 2</math>, we have a quadratic expression for <math>~x^2</math>, namely,

<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^2 + 2\biggl[\frac{x}{\lambda_2} \biggr]^{4} +  (p^2z_0^2 - \lambda_1^2) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^{4} + \frac{1}{2} \lambda_2^4~ x^2 +  \frac{1}{2}\lambda_2^4(p^2z_0^2 - \lambda_1^2) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{\lambda_2^4}{2} \pm \biggl[ \frac{\lambda_2^8}{4} - 2\lambda_2^4(p^2z_0^2 - \lambda_1^2) \biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{2} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>
Given that, for <math>~q^2 = 2</math>, one of the two defining expression means, <math>~\lambda_2 = x/\sqrt{y}</math>, we also have,

<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}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>

===New 2<sup>nd</sup> Coordinate===

Apparently it will be cleaner to define a new "2<sup>nd</sup> coordinate," <math>~\kappa_2</math>, such that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\kappa_2^{2q^2}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
q^2\biggl[\frac{1}{\lambda_2} \biggr]^{2q^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\kappa_2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
q^{1/q^2}\biggl[\frac{1}{\lambda_2} \biggr] 
= 
q^{1/q^2}\biggl[\frac{y^{1/q^2}}{x} \biggr]
= 
\frac{(qy)^{1/q^2}}{x}  \, .
</math>
  </td>
</tr>
</table>
(With this new definition, <math>~\kappa_2 \sim \tan\varphi</math>; it is exactly this when <math>~q^2 = 1</math>.)  Then we can rewrite the last expression from the [[#General_Case|above general case]] as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2 +(\kappa_2 x)^{2q^2} -  (\lambda_1^2 - p^2z_0^2 ) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
When <math>~q=1</math> (the axisymmetric case), this gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x^2(1+\kappa_2^2)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (\lambda_1^2 - p^2z_0^2 ) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{x^2}{(\lambda_1^2 - p^2z_0^2 ) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{(1+\kappa_2^2)} \, ,</math>
  </td>
</tr>
</table>
which means that <math>~\kappa_2 = \tan\varphi</math>.  And, for the case of <math>~q^2 = 2</math>, after making the substitution, 
<div align="center"><math>~\lambda_2 \rightarrow (q)^{1/q^2}\kappa_2^{-1} = \frac{2^{1 / 4}}{\kappa_2} \, ,</math></div>
we find,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2^4}{4} \biggl\{ \biggl[ 1 - \frac{8(p^2z_0^2 - \lambda_1^2)}{\lambda_2^4} \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, ,
</math>
  </td>
</tr>
</table>
and,

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

<tr>
  <td align="right">
<math>~y = \biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr) x^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{\kappa_2^2}{2^{1 / 2}} \biggr)
\frac{1}{2\kappa_2^4} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^{3 / 2}\kappa_2^2} \biggl\{ \biggl[ 1 + 4\kappa_2^4(\lambda_1^2 - p^2z_0^2 ) \biggr]^{1 / 2} - 1 \biggr\} \, .
</math>
  </td>
</tr>
</table>