Editing Appendix/Ramblings/ConcentricEllipsoidalDaringAttack (section)

=Daring Attack=

==Background==
Building on our [[User:Tohline/Appendix/Ramblings/DirectionCosines|general introduction to ''Direction Cosines'']] in the context of orthogonal curvilinear coordinate systems, and on our previous development of the so-called [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Analysis|T6]] (concentric elliptic) coordinate system, here we take a somewhat daring attack on this problem, mixing our approach to identifying the expression for the third curvilinear coordinate.  Broadly speaking, this entire study is motivated by our [[User:Tohline/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]].

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T6 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">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center"><math>~\ell_q \ell_{3D} (xp^2z)</math></td>
  <td align="center"><math>~\ell_q \ell_{3D} (q^2 y p^2z) </math></td>
  <td align="center"><math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~\tan^{-1}\biggl( \frac{y^{1/q^2}}{x} \biggr)</math></td>
  <td align="center"><math>~\frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3}</math></td>
  <td align="center"><math>~-\frac{\sin\lambda_3 \cos\lambda_3}{x}</math></td>
  <td align="center"><math>~+\frac{\sin\lambda_3 \cos\lambda_3}{q^2y}</math></td>
  <td align="center"><math>~0</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>

<tr>
  <td align="center" colspan="9">
<table border="0" cellpadding="5" 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^4 y^2 + p^4 z^2]^{- 1/ 2 }</math>
  </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>

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

As before, let's adopt the first-coordinate expression,
<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>
</table>
but for the third-coordinate expression we will abandon the trigonometric expression and instead simply use,
<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>~\frac{y^{1/q^2}}{x} \, .</math>
  </td>
</tr>
</table>
<span id="Table1DaringAttack">This modified third-coordinate expression means that the last row of the above table changes, as follows.</span>

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Daring Attack'''</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">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center"><math>~\ell_q \ell_{3D} (xp^2z)</math></td>
  <td align="center"><math>~\ell_q \ell_{3D} (q^2 y p^2z) </math></td>
  <td align="center"><math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~\frac{y^{1/q^2}}{x} </math></td>
  <td align="center"><math>~\frac{xq^2 y \ell_q}{\lambda_3}</math></td>
  <td align="center"><math>~-\frac{\lambda_3}{x}</math></td>
  <td align="center"><math>~+\frac{\lambda_3}{q^2y}</math></td>
  <td align="center"><math>~0</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>

Notice that the direction cosine functions for the (as yet, unknown) second-coordinate function remain the same.  This is because the direction-cosine functions associated with both <math>~\lambda_1</math> and <math>~\lambda_3</math> remain unchanged, so it must be true that the cross product of the first and third unit vectors leads to the same components for the second unit vector.

==New Approach==

===Setup===

The surface of an ellipsoid with semi-major axes (a, b, c) is defined by the expression,
<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>
This is identical to our expression for <math>~\lambda_1</math> if we make the associations,
<table border="0" cellpadding="5" align="center">

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

Now, given that <math>~\lambda_3</math> does not functionally depend on <math>~z</math>, let's consider that the choice of <math>~z</math> is tightly associated with the specification of the second coordinate, <math>~\lambda_2</math>.  Specifically, let's adopt the definition,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1 - \biggl( \frac{z}{c}\biggr)^2 \, ,</math>
  </td>
</tr>
</table>
in which case, we see that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~c^2(1-\lambda_2^2) = \frac{\lambda_1^2(1-\lambda_2^2)}{p^2} \, ,</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{x}{a}\biggr)^2 + \biggl( \frac{y}{b}\biggr)^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_2^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x^2 + q^2 y^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \lambda_1^2 \lambda_2^2 \, .</math>
  </td>
</tr>
</table>

[Note that <font color="red">in the case of spherical coordinates</font> (q<sup>2</sup> = p<sup>2</sup> = 1), <math>~\lambda_1 \rightarrow r</math>, and this "second" coordinate, <math>~\lambda_2</math>, becomes <math>~\sin\theta</math>.]  Combining this last expression with the <math>~x - y</math> relationship that is provided by the definition of <math>~\lambda_3</math>, gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 \lambda_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{y^{2/q^2}}{\lambda_3^2} + q^2y^2 \, .</math>
  </td>
</tr>
</table>

In general, the exponent of <math>~2q^{-2}</math> that appears in the first term on the right-hand side of this expression prevents us from being able to analytically prescribe the function, <math>~y(\lambda_1, \lambda_2, \lambda_3)</math>.  But a solution is obtainable for selected values of <math>~q^2 > 1</math>.

===Examine the Case: q<sup>2</sup> = 2===

If we set <math>~q^2 = 2</math>, then this last combined expression becomes a quadratic equation for <math>~y</math>.  Specifically, we find,
<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>~
2y^2 + \frac{y}{\lambda_3^2} - \lambda_1^2 \lambda_2^2
</math>
  </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\{
-\frac{1}{\lambda_3^2} \pm \biggl[ \frac{1}{\lambda_3^4} + 8 \lambda_1^2 \lambda_2^2 \biggr]^{1 / 2}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} \, .
</math>
  </td>
</tr>
</table>
(Note that, for reasons of simplicity <font color="red">for the time being</font>, in this last expression we have retained only the "positive" solution.)  Again, calling upon the <math>~x - y</math> relationship that is provided through the definition of <math>~\lambda_3</math>, we find (when q<sup>2</sup> = 2),
<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{y}{\lambda_3^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^4} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pm 
\frac{1}{2\lambda_3^{2}} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\}^{1 / 2} \, .
</math>
  </td>
</tr>
</table>


<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
<div align="center">Summary (q<sup>2</sup> = 2)</div>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~z(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\} 
=
\frac{(\Lambda - 1)}{4\lambda_3^2}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\lambda_3^{2}} \biggl\{
\biggl[ 1 + 8 \lambda_1^2 \lambda_2^2 \lambda_3^4\biggr]^{1 / 2} - 1
\biggr\}^{1 / 2} 
=
\frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}}
\, .
</math>
  </td>
</tr>
</table>
</td></tr></table>

For convenience, we have defined,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Lambda^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~1 + 8\lambda_1^2 \lambda_2^2 \lambda_3^4 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \lambda_1^2 \lambda_2^2 \lambda_3^4 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{8}\biggl( \Lambda^2 - 1\biggr) \, .</math>
  </td>
</tr>
</table>

<table border="1" width="90%" align="center" cellpadding="8">
<tr>
<td align="center" bgcolor="pink">
'''Test Example'''
</td>
</tr>
<tr>
<td align="left">
<math>~q^2 = 2, p^2=3.15, (x, y, z) = (0.4, 0.63581, 0.1)</math><p></p>
<math>~(\lambda_1, \lambda_2, \lambda_3) = (1, 0.98412, 1.99344)</math><p></p>
<math>~\ell_{3D} = 0.730058, ~~ \ell_q = 0.750164</math><p></p>
<math>~h_1 = 0.730058</math>
</td>
</tr>

<tr>
<td align="center">
<math>~\Lambda^2-1 = 122.34879 ~~~\Rightarrow ~~~ \Lambda = 11.10625</math><p></p>
</td>
</tr>

<tr><td align="left">
Do we get the correct values of <math>~(x, y, z)</math> &nbsp;?

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

<tr>
  <td align="right">
<math>~z(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_1(1-\lambda_2^2)^{1 / 2}}{p} = 0.1000000 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~y(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda - 1)}{4\lambda_3^2} = 0.635807\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x(\lambda_1, \lambda_2, \lambda_3)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda - 1)^{1 / 2}}{2\lambda_3^{2}} = 0.400000
\, .
</math>
  </td>
</tr>
</table>
</td></tr>

<tr><td align="left">
Evaluate a few partial derivatives &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\lambda_2^2)^{1 / 2}}{p} = 0.1\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]
= 0.693054
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
= 0.218008
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
\biggl(\frac{\partial x}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2
\biggr]^{1 / 2} 
=
0.733383
\, .
</math>
  </td>
</tr>
</table>
This matches the numerical value for <math>~h_1</math> as determined [[#h1evaluated|below]], but it does not match the numerical value obtained previously (0.730058) for <math>~h_1</math>.  The most likely piece that needs adjustment is the partial of "z" with respect to &lambda;<sub>1</sub>.  It needs to be &hellip;
<div align="center"><math>~\frac{\partial z}{\partial \lambda_1} = \biggl[ h_1^2 - \biggl( \frac{\partial x}{\partial \lambda_1} \biggr)^2 - \biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2 \biggr]^{1 / 2} = 0.071647</math>.</div>

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

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)
=
(0.730058)^2 \biggl[ \frac{p^2z}{\lambda_1} \biggr]
</math>
  </td>
</tr>
</table>

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

Next, let's examine all nine partial derivatives, noting at the start that,

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{\partial\Lambda}{\partial \lambda_1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[16\lambda_1 \lambda_2^2 \lambda_3^4 \biggr] 
=
\frac{(\Lambda^2-1)}{\lambda_1 \Lambda} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial \lambda_2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[16\lambda_1^2 \lambda_2 \lambda_3^4 \biggr] 
=
\frac{(\Lambda^2-1)}{\lambda_2 \Lambda} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial\Lambda}{\partial \lambda_3} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\Lambda}\biggl[32\lambda_1^2 \lambda_2^2 \lambda_3^3 \biggr] 
=
\frac{2(\Lambda^2-1)}{\lambda_3 \Lambda} 
\, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(1-\lambda_2^2)^{1 / 2}}{p} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{\lambda_1 \lambda_2}{p(1 - \lambda_2^2)^{1 / 2}} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_1}
=
\biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_1 \Lambda} \biggr]
=
\biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_1 \Lambda} \biggr]
=
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda} \biggr]
\, ,
</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}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_2}
=
\biggl[ \frac{(\Lambda^2-1)}{4\lambda_3^2\lambda_2 \Lambda} \biggr]
=
\biggl[ \frac{8\lambda_1^2 \lambda_2^2 \lambda_3^4}{4\lambda_3^2\lambda_2 \Lambda} \biggr]
=
\biggl[ \frac{2\lambda_1^2 \lambda_2 \lambda_3^2}{\Lambda} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^2} \cdot \frac{\partial \Lambda}{\partial \lambda_3}
- \frac{(\Lambda - 1)}{2\lambda_3^3}
=
\frac{1}{4\lambda_3^2} \cdot \biggl[ \frac{2(\Lambda^2 - 1)}{\lambda_3\Lambda} \biggr]
- \frac{\Lambda(\Lambda - 1)}{2\lambda_3^3 \Lambda}
=
\biggl[ \frac{(\Lambda^2 - 1) - \Lambda(\Lambda-1)}{2\lambda_3^3\Lambda} \biggr]
=
\frac{(\Lambda - 1) }{2\lambda_3^3\Lambda}
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_1}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{(\Lambda^2 - 1)}{\lambda_1 \Lambda} \biggr]
=
\frac{(\Lambda^2 - 1)}{4 \lambda_1 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} 
=
\frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_2}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{(\Lambda^2 - 1)}{\lambda_2 \Lambda} \biggr]
=
\frac{(\Lambda^2 - 1)}{4 \lambda_2 \lambda_3^{2} \Lambda (\Lambda-1)^{1 / 2}} 
=
\frac{2\lambda_1^2 \lambda_2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} 
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \cdot \frac{\partial \Lambda}{\partial \lambda_3}
- \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}}
=
\frac{1}{4\lambda_3^{2}(\Lambda-1)^{1 / 2}} \biggl[  \frac{2(\Lambda^2 - 1)}{\lambda_3 \Lambda} \biggr]
- \frac{(\Lambda-1)^{1 / 2}}{\lambda_3^{3}}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(\Lambda^2 - 1) -2\Lambda (\Lambda - 1) }{2\lambda_3^{3} \Lambda (\Lambda-1)^{1 / 2}} 
=
- \frac{ (\Lambda - 1)^{3 / 2}  }{2\lambda_3^{3} \Lambda} 
\, .
</math>
  </td>
</tr>
</table>

What about the derived scale-factors?
<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(\frac{\partial x}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial y}{\partial \lambda_1}\biggr)^2
+ \biggl(\frac{\partial z}{\partial \lambda_1}\biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^{2}}{\Lambda (\Lambda-1)^{1 / 2}} \biggr]^2
+ \biggl[ \frac{2\lambda_1 \lambda_2^2 \lambda_3^2}{\Lambda}  \biggr]^2
+ \biggl[ \frac{(1-\lambda_2^2)^{1 / 2}}{p}  \biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^{4 }}{\Lambda^2 (\Lambda-1)}  \biggr]
+ \biggl[ \frac{4\lambda_1^2 \lambda_2^4 \lambda_3^4}{\Lambda^2}  \biggr]
+ \biggl[ \frac{(1-\lambda_2^2)}{p^2}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{p^2 \Lambda^2(\Lambda - 1)} \biggl[
4 p^2 \lambda_1^2 \lambda_2^4 \lambda_3^4 \Lambda 
+ (1-\lambda_2^2) \Lambda^2(\Lambda - 1)  
\biggr] \, .
</math>
  </td>
</tr>
</table>

Written in terms of Cartesian coordinates, this becomes,

<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>~\frac{ 8 \lambda_1^2 \lambda_2^2 \lambda_3^4 (\lambda_2^2 ) }{2 \Lambda (\Lambda - 1)} 
+ \frac{(1-\lambda_2^2)}{p^2}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\lambda_1^2} \biggl[
\frac{ (\Lambda+1)(x^2 + 2y^2) }{2 \Lambda } 
+ z^2
\biggr] \, .
</math>
  </td>
</tr>
</table>
Note that,
<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>~4x^2\lambda_3^4 = 4x^2 \biggl( \frac{y^2}{x^4} \biggr) = 4\biggl( \frac{y^2}{x^2} \biggr) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \Lambda </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x^2 + 4y^2}{x^2} \, .</math>
  </td>
</tr>
</table>

<span id="h1evaluated">Hence, the scale factor becomes,</span>

<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>~ \frac{1}{2 \lambda_1^2} \biggl[
(x^2 + 2y^2)  
+ \frac{ x^2(x^2 + 2y^2) }{(x^2 + 4y^2) } 
+ 2z^2
\biggr] 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2 \lambda_1^2(x^2 + 4y^2) } \biggl[
(2x^4 + 8x^2y^2 +8y^4)
+ 2z^2(x^2 + 4y^2) 
\biggr] 
= \frac{1.911525}{3.554} = 0.537852
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
0.733384
\, .
</math>
  </td>
</tr>
</table>


----


Compare this expression with the one derived earlier, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_1^2 \biggr|_{q^2 = 2} = \biggl[\lambda_1^2 \ell_{3D}^2 \biggr]_{q^2 = 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + 2y^2 + p^2z^2)}{x^2 + 4y^2 + p^4z^2} \, .
</math>
  </td>
</tr>
</table>
Well &hellip; first we recognize that, when q<sup>2</sup> = 2,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1^2 = x^2 + 2y^2 + p^2z^2 \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="center">
<math>~\lambda_2^2 = \frac{\lambda_1^2 - p^2 z^2}{\lambda_1^2}
=
\frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \, ,
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="left">
<math>~\lambda_3^2 = \frac{y}{x^2} \, .</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(\Lambda^2 - 1) = 8\lambda_1^2 \lambda_2^2 \lambda_3^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
8(x^2 + 2y^2 + p^2z^2)\biggl[ \frac{x^2 + 2y^2}{x^2 + 2y^2 + p^2z^2} \biggr]\frac{y^2}{x^4}
=
\biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \Lambda^2  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \biggl[ \frac{8y^2(x^2 + 2y^2)}{x^4 } \biggr]
=
\frac{1}{x^4}\biggl[x^4 + 8x^2y^2 + 16y^4  \biggr]
=
\frac{1}{x^4}\biggl[x^2 + 4y^4  \biggr]^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (\Lambda+1)  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x^2 + 4y^4)}{x^2} + 1
=
\frac{2x^2 + 4y^4}{x^2} \, ,
</math>
  </td>
</tr>
</table>
which means,

<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>~\frac{1}{8\lambda_1^2 \Lambda^2} \biggl\{
4\lambda_1^2 \lambda_2^2 \lambda_3 (\Lambda+1)
+ 4\lambda_1^2 \lambda_2^2 (\Lambda^2 - 1)
+ 8z^2\Lambda^2
\biggr\}
</math>
  </td>
</tr>
</table>

==Think Again==

===Firm Relations===

In addition to the functions that are specified in our above [[#Table1DaringAttack|Daring Attack Table]], we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)
=
\biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{x}{\lambda_1}
=
x \lambda_1 \ell_{3D}^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)
=
\biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{q^2y}{\lambda_1}
=
q^2 y \lambda_1 \ell_{3D}^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)
=
\biggl(\lambda_1 \ell_{3D} \biggr)^2 \frac{p^2z}{\lambda_1}
=
p^2 z \lambda_1 \ell_{3D}^2 \, .
</math>
  </td>
</tr>
</table>

Check &hellip;
<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( \frac{\partial x}{\partial \lambda_1} \biggr)^2
+
\biggl( \frac{\partial y}{\partial \lambda_1} \biggr)^2
+
\biggl( \frac{\partial z}{\partial \lambda_1} \biggr)^2
=
\lambda_1^2 \ell_{3D}^4 \biggl[
x^2 + q^4 y^2 + p^4z^2
\biggr]
=
\lambda_1^2 \ell_{3D}^2 \, .
</math>
&nbsp; &nbsp; &nbsp; <font color="red">(Yes!)</font>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_3^2 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)
=
\biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( - \frac{\lambda_3}{x} \biggr)
=
- q^4 y^2 \ell_q^2 \biggl( \frac{x}{\lambda_3} \biggr) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_3^2 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)
=
\biggl[ \frac{xq^2y \ell_q}{\lambda_3} \biggr]^2 \biggl( + \frac{\lambda_3}{q^2y} \biggr)
=
x^2 \ell_q^2 \biggl( \frac{q^2y} {\lambda_3}\biggr) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_3^2 \biggl( \frac{\partial \lambda_3}{\partial z} \biggr)
=
0 \, .
</math>
  </td>
</tr>
</table>

Check &hellip;
<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( \frac{\partial x}{\partial \lambda_3} \biggr)^2
+
\biggl( \frac{\partial y}{\partial \lambda_3} \biggr)^2
+
\biggl( \frac{\partial z}{\partial \lambda_3} \biggr)^2
=
\frac{\ell_q^4}{\lambda_3^2} \biggl[x^2 q^8 y^4 + x^4 q^4y^2  \biggr]
=
\frac{x^2 q^4 y^2\ell_q^4}{\lambda_3^2} \biggl[q^4 y^2 + x^2  \biggr]
=
\frac{x^2 q^4 y^2\ell_q^2}{\lambda_3^2} \,  .
</math>
&nbsp; &nbsp; &nbsp; <font color="red">(Yes!)</font>
  </td>
</tr>
</table>

And, last &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \gamma_{21}
=
h_2 \ell_q \ell_{3D} (xp^2z) \, ,
</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>~
h_2 \gamma_{22}
=
h_2 \ell_q \ell_{3D} (q^2 y p^2 z) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \gamma_{23}
=
- h_2 \ell_q \ell_{3D}(x^2 + q^4y^2) \, .
</math>
  </td>
</tr>
</table>

===Speculation===
====First====
From the direction-cosine expressions for <math>~\partial\lambda_2/\partial x_i</math> that have been summarized in our above [[#Table1DaringAttack|Daring Attack Table]], it seems reasonable to suggest that,

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

<tr>
  <td align="right">
<math>~h_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\ell_q \ell_{3D})^2 
=
\biggl[ (x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) \biggr]^{-1}
\, ,
</math>
  </td>
</tr>
</table>
in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~xp^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2yp^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(x^2 + q^4y^2) \, ;</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\ell_q \ell_{3D})^2 xp^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial y}{\partial \lambda_2} = h_2^2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\ell_q \ell_{3D})^2 q^2yp^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_2} = h_2^2 \biggl(\frac{\partial \lambda_2}{\partial z} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(\ell_q \ell_{3D})^2 (x^2 + q^4y^2) \, .</math>
  </td>
</tr>
</table>

====Second====
Alternatively, after examining the direction-cosine expressions for <math>~\partial x_i/\partial \lambda_2</math> that we have just provided, one might suggest that,

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

<tr>
  <td align="right">
<math>~h_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(\ell_q \ell_{3D})^{-2} 
=
(x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) 
=
p^4z^2(x^2 + q^4y^2) + (x^2 + q^4y^2)^2
\, ,
</math>
  </td>
</tr>
</table>
in which case, the expressions provided for <math>~\partial \lambda_2/\partial x_i</math> and  <math>~\partial x_i/\partial \lambda_2</math> must be swapped relative to our ''First'' speculation.

====Third====
Noticing that <math>~h_1^2</math> is proportional to <math>~\lambda_1^2</math> and that <math>~h_3^2</math> is inversely proportional to <math>~\lambda_3^2</math>, let's consider both as possible behaviors for the 2<sup>nd</sup> scale factor.  Let's try the first of these behaviors.  Specifically, what if we assume &hellip;

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{xp^2 z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2y p^2z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .</math>
  </td>
</tr>
</table>
Then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_2^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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
=
[\lambda_2 \ell_q \ell_{3D} ]^{-2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\lambda_2 \ell_q \ell_{3D} \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
Primary implication:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{21} = h_2 \biggl(\frac{\partial \lambda_2}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(xp^2 z) \ell_q \ell_{3D} \ ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\gamma_{22} = h_2 \biggl(\frac{\partial \lambda_2}{\partial y} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(q^2 y p^2 z) \ell_q \ell_{3D} \ ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\gamma_{23} = h_2 \biggl(\frac{\partial \lambda_2}{\partial z} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(x^2 + q^4 y^2) \ell_q \ell_{3D} \ .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
These perfectly match the direction-cosine expressions (<math>~\gamma_{2i}</math> for i = 1, 3)<br />that have been summarized in our above [[#Table1DaringAttack|Daring Attack Table]].
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \gamma_{21}
=
\lambda_2 \ell_q^2 \ell_{3D}^2 (xp^2z) \, ,
</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>~
h_2 \gamma_{22}
=
\lambda_2 \ell_q^2 \ell_{3D}^2 (q^2 y p^2 z) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial z}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \gamma_{23}
=
- \lambda_2 \ell_q^2 \ell_{3D}^2(x^2 + q^4y^2) \, .
</math>
  </td>
</tr>
</table>
</td></tr></table>


Now, what specifically is the function, <math>~\lambda_2(x, y, z)</math> ?  Start by rewriting the three partial derivatives as,

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

<tr>
  <td align="right">
<math>~\frac{1}{2} \frac{\partial (\lambda_2^2)}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~xp^2 z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2y p^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{1}{2} \frac{\partial (\lambda_2)^2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-(x^2 + q^4y^2) \, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\lambda_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2)p^2z \, .</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2^2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2xp^2z \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; and, &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2^2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^2 yp^2z \, .</math>&nbsp; &nbsp; &nbsp; <font color="red">Great!</font>
  </td>
</tr>
</table>

But this cannot be the correct expression for <math>~\lambda_2^2</math> because,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2^2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^2y^2)p^2 \, ,</math>
  </td>
</tr>
</table>
which does not match the desired partial derivative with respect to <math>~z</math>.

====Fourth====

Alternatively, if we assume &hellip;

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_2}{xp^2 z} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_2}{q^2y p^2z} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{\lambda_2}{(x^2 + q^4y^2)} \, ,</math>
  </td>
</tr>
</table>
then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_2^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\lambda_2}{xp^2 z} \biggr)^2
+ \biggl( \frac{\lambda_2}{q^2y p^2z} \biggr)^2
+\biggl( \frac{\lambda_2}{x^2 + q^4y^2} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (h_2 \lambda_2)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2  (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4}{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ h_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\lambda_2} \biggl[
\frac{x^2 q^4y^2p^8z^4(x^2 + q^4y^2)^2}{ q^4y^2p^4z^2 (x^2 + q^4y^2)^2 + x^2p^4z^2  (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^8z^4}
\biggr]^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\lambda_2} \biggl\{
\frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ q^4y^2 (x^2 + q^4y^2)^2 + x^2  (x^2 + q^4y^2)^2 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\lambda_2} \biggl\{
\frac{x q^2y p^2z(x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}}
\biggr\}
</math>
  </td>
</tr>
</table>
Let's check for consistency with one of the direction-cosines.
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{21} = h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{
\frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{ \gamma_{21} }{\ell_q(xp^2z) }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(x^2 + q^4y^2)^{1 / 2}}{xp^2z} \biggl\{
\frac{q^2y (x^2 + q^4y^2)}{ [ (x^2 + q^4y^2)^3 + x^2 q^4y^2 p^4z^2 ]^{1 / 2}}
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2y}{xp^2z} 
\biggl[1 + \frac{x^2q^4y^2p^4z^2}{(x^2 + q^4y^2)^3}  \biggr]^{-1 / 2} \, .
</math>
  </td>
</tr>
</table>
This does not match the term in the expression for <math>~\gamma_{21}</math> &#8212; namely, <math>~\ell_{3D}</math> &#8212; that is expected from the original tabulation.

====Better Organized====

From our above [[#Table1DaringAttack|Daring Attack Table]], we appreciate that the three direction cosines associated with the (as yet unknown) second curvilinear coordinate are,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{21}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D} (xp^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\gamma_{22}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\ell_q \ell_{3D} (q^2 y p^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\gamma_{23}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\ell_q \ell_{3D} (x^2 + q^4 y^2) \, .</math>
  </td>
</tr>
</table>

It is easy to see that the desired ''orthogonality'' relationship,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sum_{i=1}^3 (\gamma_{2i})^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 \, ,</math>
  </td>
</tr>
</table>
is satisfied because,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(xp^2z)^2 + (q^2y p^2z)^2 + (x^2 + q^4y^2)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2)(x^2 + q^4y^2 + p^4z^2) = ( \ell_q \ell_{3D} )^{-2} \, .</math>
  </td>
</tr>
</table>

Now, as we attempt to determine the functional form of the second curvilinear coordinate, <math>~\lambda_2(x, y, z)</math>, a seemingly useful intermediate step is to determine the functional form of each of the three partial derivatives of this key coordinate function, namely, <math>~\partial \lambda_2/\partial x_i</math>, for i = 1, 3.  Here, we will accomplish this intermediate step by ''guessing'' the functional form of the second scale factor, <math>~h_2(x, y, z)</math>, then applying the relation,

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\gamma_{2i}}{h_2} \, .</math>
  </td>
</tr>
</table>
Notice that, without violating the above-state ''orthogonality'' relationship, we can adopt virtually any functional form for <math>~h_2(x, y, z)</math> and deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A(x, y, z) (xp^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A(x, y, z) (q^2 y p^2z) \, ,</math>
  </td>
  <td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-A(x, y, z) (x^2 + q^4 y^2) \, ,</math>
  </td>
</tr>
</table>
as long as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A(x, y, z)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{ \ell_q \ell_{3D} }{h_2} \, .
</math>
  </td>
</tr>
</table>

This key, leading coefficient function is unity &#8212; and, hence, is independent of position &#8212; if, as in our [[#First|''First'' speculation]] above, we ''guess'' that <math>~h_2^2 = (\ell_q \ell_{3D})^2</math>.  If, as in our [[#Second|''Second'' speculation]] above, we ''guess'' that <math>~h_2^2 = (\ell_q \ell_{3D})^{-2}</math>, we find that, <math>~A = (\ell_q \ell_{3D})^2</math>.  Our above [[#Third|''Third'' speculation]] is replicated if we ''guess'' that <math>~h_2^2 = (\lambda_2 \ell_q \ell_{3D})^2</math>; we immediately see that, in this ''Third'' case,

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{xp^2 z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^2y p^2z}{\lambda_2} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{(x^2 + q^4y^2)}{\lambda_2} \, .</math>
  </td>
</tr>
</table>

==Study the Functional Forms==

We know the functional forms of two of the desired curvilinear coordinates, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_1(x, y, z)</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>~\lambda_3(x, y, z)</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{y^{1/q^2}}{x} \, ,</math>
  </td>
</tr>
</table>
but we do not yet have a valid expression for the 2<sup>nd</sup> coordinate, <math>~\lambda_2(x, y, z)</math>.  Nevertheless, let's see if we can ''guess'' the functional forms for <math>~x_i(\lambda_1, \lambda_2, \lambda_3)</math>, by inverting the two known curvilinear-coordinate functions.  As a starting point, let's impose the following mappings:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left" colspan="3">
<math>~\frac{y^{1/q^2}}{\lambda_3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~
\frac{1}{p}\biggl[
\lambda_1^2 - q^2y^2 - x^2
\biggr]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~
\frac{1}{p}\biggl[ \lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2} \biggr]^{1 / 2} 
=
\frac{1}{p\lambda_3}\biggl[ \lambda_1^2 \lambda_3^2 - (qy\lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2} 
\, .
</math>
  </td>
</tr>
</table>
This means, for example, that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ell_q^{-2}</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~
q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} 
=
\lambda_3^{-2} \biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr]
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\ell_{3D}^{-2}</math>
  </td>
  <td align="center">
<math>~~\rightarrow~~</math>
  </td>
  <td align="left">
<math>~
q^4y^2 + \frac{y^{2/q^2}}{\lambda_3^2} +
p^2\biggl[
\lambda_1^2 - q^2y^2 - \frac{y^{2/q^2}}{\lambda_3^2}
\biggr]
=
\lambda_3^{-2} \biggl[
(q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2
\biggr]
\, .
</math>
  </td>
</tr>
</table>

===Derivatives of x===
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x \lambda_1 \ell_{3D}^2 
=
y^{1/q^2} \lambda_1 \lambda_3 \biggl[
(q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2
\biggr]^{-1} \, ,
 </math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\frac{\partial x}{\partial \lambda_2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \ell_q \ell_{3D} (xp^2z)
=
\biggl[ h_2 \ell_q \ell_{3D}\biggr] \biggl(y^{1/q^2} \biggr)
\frac{p}{\lambda_3^2}\biggl[ \lambda_1^2 \lambda_3^2 - (qy\lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_2 \biggl( y^{1/q^2} \biggr) p \biggl[ \lambda_1^2 \lambda_3^2 - (qy \lambda_3)^2 - y^{2/q^2} \biggr]^{1 / 2}
\biggl[ q^2(qy\lambda_3)^2 + y^{2/q^2} \biggr]^{-1 / 2}
\biggl[
(q^2-p^2)(qy \lambda_3)^2 + (1-p^2)y^{2/q^2} + p^2\lambda_1^2\lambda_3^2
\biggr]^{-1 / 2}
</math>
  </td>
</tr>
</table>

===Struggling===

I have noticed that, in this last set of expressions, there are recurring terms of the form, <math>~(qy\lambda_3)</math> and <math>~(y^{2/q^2})</math>.  So, while keeping the same definition of the ccordinate, <math>~\lambda_1</math>, let's replace <math>~\lambda_2</math> and <math>~\lambda_3</math> with a pair of coordinates defined as follows:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_4 \equiv y\lambda_3 = \frac{y^{(q^2+1)/q^2}}{x} \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\lambda_5 \equiv y^{2/q^2} \, .</math>
  </td>
</tr>
</table>
This means that,
<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>~\lambda_5^{q^2/2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{y^{(q^2-1)/q^2}}{\lambda_4} 
=
\lambda_4^{-1} \lambda_5^{ (q^2-1)/2}
\, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~z^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{p^2} \biggl[\lambda_1^2 - x^2 - q^2y^2  \biggr]
=
\frac{1}{p^2 \lambda_4^2} \biggl[
\lambda_1^2 \lambda_4^2 - \lambda_5^{q^2-1} - q^2 \lambda_4^2\lambda_5^{q^2}
\biggr]
\, .</math>
  </td>
</tr>
</table>
Is this a set of orthogonal coordinates?  Well &hellip; &nbsp; &nbsp; &nbsp; <font color="red">No!</font>

==New Insight==

Following the development of our [[#Better_Organized|above, ''Better Organized'']] discussion, we reverted to several hours of pen &amp; paper derivations, primarily investigating whether it will help us to rewrite various expressions using the [<b>[[User:Tohline/Appendix/References#MF53|<font color="red">MF53</font>]]</b>] [[User:Tohline/Appendix/Mathematics/ScaleFactors#DirectionCosineRelations|Direction-Cosine Relations]].  We discovered that if we set,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_2^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[(xq^2y)\ell_q \ell_{3D}]^2 \, ,</math>
  </td>
</tr>
</table>
then,

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

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{p^2 z}{q^2y} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{p^2z}{x} \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr] \, .</math>
  </td>
</tr>
</table>

This seems to be a promising method of attack because &#8212; in all three cases, i = 1,3 &#8212; the derivative of <math>~\lambda_2</math> with respect to <math>~x_i</math> does not depend on <math>~x_i</math>.  Perhaps this simplification will help us identify the function that defines <math>~\lambda_2</math>.  This proposed prescription for <math>~h_2(x, y, z)</math> and some of its implications are reflected in the following "New Insight" table.  (Keep in mind that, although the expressions for <math>~\gamma_{21}, \gamma_{22}, ~\mathrm{and}~ \gamma_{23}</math> remain correct, the tabulated expression is a ''guess'' for <math>~h_2</math> and, hence, the tabulated expressions for all three <math>~\partial \lambda_2/\partial x_i</math> are pure speculation.)
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''New Insight'''</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">---</td>
  <td align="center"><math>~\ell_q \ell_{3D} (xq^2y)</math></td>
  <td align="center"><math>~\frac{p^2z}{q^2y}</math></td>
  <td align="center"><math>~\frac{p^2z}{x}</math></td>
  <td align="center"><math>~-\biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr]</math></td>
  <td align="center"><math>~\ell_q \ell_{3D} (xq^2y) \biggl[ \frac{p^2z}{q^2y} \biggr]</math></td>
  <td align="center"><math>~\ell_q \ell_{3D} (xq^2y) \biggl[ \frac{ p^2z}{x} \biggr] </math></td>
  <td align="center"><math>~- \ell_q \ell_{3D}  (xq^2y) \biggl[ \frac{x}{q^2y} + \frac{q^2y}{x} \biggr]</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~\frac{y^{1/q^2}}{x} </math></td>
  <td align="center"><math>~\frac{xq^2 y \ell_q}{\lambda_3}</math></td>
  <td align="center"><math>~-\frac{\lambda_3}{x}</math></td>
  <td align="center"><math>~+\frac{\lambda_3}{q^2y}</math></td>
  <td align="center"><math>~0</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>