Editing Appendix/Ramblings/ConcentricEllipsoidalT12Coordinates

__FORCETOC__
<!-- __NOTOC__ will force TOC off -->
=Concentric Ellipsoidal (T12) Coordinates=

==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 [[User:Tohline/Appendix/Ramblings/T3Integrals|T3]] (concentric oblate-spheroidal) and [[User:Tohline/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 [[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]].

Note that, in a [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Background|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.

The 1<sup>st</sup> coordinate and its associated unit vector are as follows:
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\hat{e}_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \ell_{3D} \biggl[
\hat\imath (x)  + \hat\jmath (q^2y ) + \hat{k} (p^2 z) 
\biggr] \, ,
</math>
  </td>
</tr>
</table>
where,
<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^4y^2 + p^4 z^2)^{- 1 / 2} \, .
</math>
  </td>
</tr>
</table>

==Generalized Prescription for 2<sup>nd</sup> Coordinate==

===Default===

Let's adopt the following generalized prescription for the 2<sup>nd</sup> coordinate:
<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>~
x^a y^b z^c \, ,
</math>
  </td>
</tr>
</table>
in which case,
<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>~
\frac{1}{\mathfrak{L}}
\biggl[
\hat\imath \biggl(\frac{yz}{bc}\biggr)  
+ \hat\jmath \biggl(\frac{xz}{ac}\biggr)  
+ \hat{k} \biggl(\frac{xy}{ab}\biggr)  
\biggr] \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{L}^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a^2b^2c^2} \biggl[  
a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2
\biggr] \, .
</math>
  </td>
</tr>
</table>

<span id="OneTwoPerp">Now, to ensure that <math>~\hat{e}_2</math> is perpendicular to <math>~\hat{e}_1</math>, we need,</span>
<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>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\ell_{3D}}{\mathfrak{L}} \biggl[
\frac{xyz}{bc} + \frac{q^2xyz}{ac} + \frac{p^2xyz}{ab}
\biggr]
=
\frac{\ell_{3D} (xyz)}{\mathfrak{L}(abc)} \biggl[
a + q^2b + p^2 c
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
a + q^2b + p^2 c
\biggr]\, .
</math>
  </td>
</tr>
</table>
Henceforth, we will refer to this algebraic relation as the "<font color="red">One-Two Perpendicular Constraint</font>."

===Arctangent===

Instead, let's try,
<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>~
\tan^{-1} [x^a y^b z^c] \, .
</math>
  </td>
</tr>
</table>
Then,
<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>~
\biggl[ \frac{1}{1+\tan^2\lambda_2} \biggr]
\frac{\partial }{\partial x_i} \biggl[ x^a y^b z^c \biggr] 
=
\cos^2\lambda_2 \cdot \frac{\partial }{\partial x_i} \biggl[ x^a y^b z^c \biggr] 
\, ;
</math>
  </td>
</tr>
</table>

that is,

<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>~
\cos^2\lambda_2 \cdot 
\biggl[ x^a y^b z^c \biggr] \frac{a}{x}
=
\cos^2\lambda_2 \cdot 
\biggl[ \tan\lambda_2 \biggr] \frac{a}{x}
=
\sin\lambda_2 \cos\lambda_2  \biggl( \frac{a}{x} \biggr) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\lambda_2 \cos\lambda_2  \biggl( \frac{b}{y} \biggr) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sin\lambda_2 \cos\lambda_2  \biggl( \frac{c}{z} \biggr) \ .
</math>
  </td>
</tr>
</table>
Hence,
<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>~
\sin^2\lambda_2 \cos^2\lambda_2 \biggl[
\frac{a^2}{x^2} + \frac{b^2}{y^2} + \frac{c^2}{z^2}
\biggr]
=
\frac{\sin^2\lambda_2 \cos^2\lambda_2}{(xyz)^2} \biggl[
a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2
\biggr]
</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}{\sin\lambda_2 \cos\lambda_2} \biggl[
\frac{xyz}{ \mathfrak{L} (abc)}
\biggr] \, .
</math>
  </td>
</tr>
</table>
And the associated unit vector is,
<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>~
\frac{1}{\mathfrak{L}}
\biggl[
\hat\imath \biggl(\frac{yz}{bc}\biggr)  
+ \hat\jmath \biggl(\frac{xz}{ac}\biggr)  
+ \hat{k} \biggl(\frac{xy}{ab}\biggr)  
\biggr] \, ,
</math>
  </td>
</tr>
</table>
which is the same as our default situation.

==Necessary 3<sup>rd</sup> Coordinate==

The unit vector associated with the 3<sup>rd</sup> coordinate is obtained from the cross product of the first two unit vectors.  That is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_1 \times \hat{e}_2</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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>~\frac{\ell_{3D}}{\mathfrak{L}} \biggl\{
\hat\imath \biggl[ (q^2y) \biggl( \frac{xy}{ab} \biggr) - (p^2z) \biggl( \frac{xz}{ac} \biggr) \biggr]
+
\hat\jmath \biggl[ (p^2z) \biggl( \frac{yz}{bc} \biggr) - (x)\biggl( \frac{xy}{ab} \biggr) \biggr]
+
\hat{k} \biggl[ (x) \biggl( \frac{xz}{ac} \biggr) - (q^2y) \biggl( \frac{yz}{bc} \biggr) \biggr]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathfrak{L}(abc)} \biggl\{
\hat\imath \biggl[ (cq^2y^2)  - (b p^2z^2) \biggr]x
+
\hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y
+
\hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z
\biggr\}
</math>
  </td>
</tr>
</table>

==Old Examples==

===T6 Coordinates===
In the set that we have [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Analysis|elsewhere referenced as T6 coordinates]], we chose:  a = - 1, b = q<sup>-2</sup>, c = 0.  We note, first, that this set of parameter values satisfies the [[#OneTwoPerp|above-defined]] <font color="red">One-Two Perpendicular Constraint</font>.  In this case, our ''generalized prescription for the 2<sup>nd</sup> coordinate'' generates a unit vector of the form,
<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>~
\frac{1}{\mathfrak{L}_{T6} (abc)}
\biggl[
\hat\imath (ayz)  
+ \hat\jmath (bxz)  
+ \hat{k} (cxy)  
\biggr] 
=
\frac{z}{q^2 \mathfrak{L}_{T6} (abc)}
\biggl[
-\hat\imath (q^2y)  
+ \hat\jmath (x)  
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
-\hat\imath (q^2y)  
+ \hat\jmath (x)  
\biggr] \ell_q \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ell_q^{-2} \equiv \biggl[ \frac{q^4\mathfrak{L}_{T6}^2(abc)^2}{z^2}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^4}{z^2}\biggl[  
(yz)^2 + b^2(xz)^2 
\biggr] 
=
\biggl[ x^2 + q^4y^2 \biggr] \, .
</math>
  </td>
</tr>
</table>

And it implies a unit vector for the 3<sup>rd</sup> coordinate of the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathfrak{L}_{T6} (abc)} \biggl\{
\hat\imath \biggl[ (cq^2y^2)  - (b p^2z^2) \biggr]x
+
\hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y
+
\hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z
\biggr\}
</math>
  </td>
</tr>

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

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

===T10 Coordinates===
In the set that we have [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#PartBCoordinatesT10|elsewhere referenced as T10 coordinates]], we chose:  a = 1, b = q<sup>-2</sup>, c = - 2p<sup>-2</sup>.  We note, first, that this set of parameter values satisfies the [[#OneTwoPerp|above-defined]] <font color="red">One-Two Perpendicular Constraint</font>. In this case, our ''generalized prescription for the 2<sup>nd</sup> coordinate'' generates a unit vector of the form,
<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>~
\frac{1}{\mathfrak{L}_{T10} (abc)}
\biggl[
\hat\imath (ayz)  
+ \hat\jmath (bxz)  
+ \hat{k} (cxy)  
\biggr] 
=
\frac{1}{q^2 p^2 \mathfrak{L}_{T10} (abc)}
\biggl[
\hat\imath (q^2y p^2z)  
+ \hat\jmath ( x p^2 z )  
- \hat{k} ( 2xq^2y)  
\biggr] 
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(abc)^2\mathfrak{L}^2_{T10}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[  
a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2
\biggr]
=
 \biggl[  
y^2z^2 + \frac{x^2 z^2}{q^4} + \frac{4x^2 y^2}{p^4}
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\mathcal{D}^2 \equiv q^4p^4(abc)^2\mathfrak{L}^2_{T10}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
 \biggl[  
q^4y^2 p^4z^2 + x^2 p^4z^2 + 4x^2 q^4y^2 
\biggr] \, .
</math>
  </td>
</tr>
</table>


And it implies a unit vector for the 3<sup>rd</sup> coordinate of the form,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{
\hat\imath \biggl[ (cq^2y^2)  - (b p^2z^2) \biggr]x
+
\hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y
+
\hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z
\biggr\} q^2 p^2
</math>
  </td>
</tr>

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

==Develop 3<sup>rd</sup>-Coordinate Profile==

===Setup===

Reflecting back on an [[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Search_for_Third_Coordinate_Expression|earlier exploration]], let's define the two polynomials,
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>~\mathfrak{B} \equiv [\mathfrak{L}(abc)]^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, .
</math>
  </td>
</tr>
</table>


<table border="1" align="center" cellpadding="10"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{A}}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{A}}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^4 y \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{A}}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{B}}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x(b^2z^2 + c^2y^2) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{B}}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y(a^2 z^2 + c^2x^2) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{B}}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z(a^2 y^2 + b^2x^2) \, .</math>
  </td>
</tr>
</table>
</td></tr></table>

<span id="needed">Then the 3<sup>rd</sup> unit vector may be written as,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[\hat{e}_3]_\mathrm{needed}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{
\hat\imath \biggl[ (cq^2y^2)  - (b p^2z^2) \biggr]x
+
\hat\jmath \biggl[ (ap^2z^2) - (cx^2) \biggr]y
+
\hat{k} \biggl[ (bx^2) - (aq^2y^2) \biggr]z
\biggr\} \, .
</math>
  </td>
</tr>
</table>

===Guess Third Coordinate Expression===
Let's see what unit vector results if we define,
<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>~\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} \, .</math>
  </td>
</tr>
</table>

====Partial Derivatives====
<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>~
\biggl[ \frac{1}{2} \mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2}\biggr] \frac{\partial \mathfrak{A}}{\partial x_i}
- \biggl[ \frac{1}{2} \mathfrak{A}^{1 / 2} \mathfrak{B}^{- 3 / 2} \biggr] \frac{\partial \mathfrak{B}}{\partial x_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B}\cdot \frac{\partial \mathfrak{A}}{\partial x_i}
- 
\mathfrak{A}\cdot  \frac{\partial \mathfrak{B}}{\partial x_i} \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B} \biggl[ 2x \biggr]
- 
\mathfrak{A}\biggl[ 2x (b^2z^2 + c^2 y^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 
- 
(x^2 + q^4y^2 + p^4z^2)(b^2z^2 + c^2 y^2) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 
- 
x^2 (b^2z^2 + c^2 y^2) 
- 
q^4y^2 (b^2z^2 + c^2 y^2) 
- 
p^4z^2(b^2z^2 + c^2 y^2) 

\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl\{
(yz)^2[a^2 - q^4b^2 - c^2p^4] + (xz)^2[b^2 - b^2] + (xy)^2 [c^2 - c^2]
- 
c^2 q^4y^4 
- 
b^2p^4z^4 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
y^2z^2(a^2 - q^4b^2 - c^2p^4)  
- 
c^2 q^4y^4 
- 
b^2p^4z^4 
\biggr] \, ;
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B} \biggl[ 2q^4y \biggr]
- 
\mathfrak{A}\biggl[ 2y(a^2 z^2 + c^2x^2) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
q^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]
- 
(a^2 z^2 + c^2x^2) (x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
a^2q^4(yz)^2 + b^2q^4(xz)^2 + c^2q^4(xy)^2
- 
a^2 z^2 (x^2 + q^4y^2 + p^4z^2)
-
c^2x^2 (x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl\{
(yz)^2 \biggl[  a^2q^4 - a^2q^4\biggr] + (xz)^2\biggl[ b^2q^4 -a^2 - c^2p^4\biggr] + (xy)^2 \biggl[ c^2q^4 - c^2q^4 \biggr]
- 
a^2 p^4z^4
-
c^2x^4
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) 
- 
a^2 p^4z^4
-
c^2x^4
\biggr] \, ;
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{B} \biggl[ 2p^4z \biggr]
- 
\mathfrak{A}\biggl[ 2z(a^2 y^2 + b^2x^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
p^4[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2]
- 
(a^2 y^2 + b^2x^2)(x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
a^2p^4(yz)^2 + b^2p^4(xz)^2 + c^2p^4(xy)^2
- 
a^2 y^2 (x^2 + q^4y^2 + p^4z^2)
- 
b^2x^2(x^2 + q^4y^2 + p^4z^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl\{
(yz)^2 \biggl[ a^2p^4 - a^2p^4 \biggr] + (xz)^2 \biggl[ b^2p^4 - b^2p^4 \biggr] + (xy)^2 \biggl[c^2p^4 - a^2 - b^2q^4 \biggr]
- 
a^2 q^4y^4 
- 
b^2x^4
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
 (xy)^2 ( c^2p^4 - a^2 - b^2q^4 )
- 
a^2 q^4y^4 
- 
b^2x^4
\biggr] \, .
</math>
  </td>
</tr>
</table>

After completing a few squares, this last expression may be rewritten as &hellip;

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
 (xy)^2 ( c^2p^4 - a^2 - b^2q^4 )
- 
a^2 q^4y^4 
- 
b^2x^4
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
 (xy)^2 ( c^2p^4 - a^2 - b^2q^4 )
\pm 2abq^2(xy)^2 - (aq^2y^2 \pm bx^2)^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
 (xy)^2 ( c^2p^4 - a^2 - b^2q^4 
\pm 2abq^2) - (aq^2y^2 \pm bx^2)^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2z \biggl[
 (xy)^2 [ c^2p^4 -(a \pm bq^2)^2
\pm 4abq^2] - (aq^2y^2 \pm bx^2)^2
\biggr] \, .
</math>
  </td>
</tr>
</table>

Now, if we choose the ''superior'' sign throughout this expression, the above-derived <font color="red">One-Two Perpendicular Constraint</font> can be satisfied by setting, <math>~(a + bq^2) = -cp^2</math>.  The expression then becomes,

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2z \biggl[
(aq^2y^2 + bx^2)^2 - 4abq^2 (xy)^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2z (aq^2y^2 - bx^2)^2 \, .
</math>
  </td>
</tr>
</table>


----


<table border="1" cellpadding="10" align="center" width="80%"><tr><td align="left">
Alternatively, suppose 
<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>~\mathfrak{A}^{m / 2} \mathfrak{B}^{-n/ 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
m \mathfrak{B}\cdot \frac{\partial \mathfrak{A}}{\partial x_i}
- 
n \mathfrak{A}\cdot  \frac{\partial \mathfrak{B}}{\partial x_i} \, .
</math>
  </td>
</tr>
</table>
Then we have, for example,

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

<tr>
  <td align="right">
<math>~ \frac{1}{2z}\biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
m \mathfrak{B} \biggl[ p^4 \biggr]
- 
n \mathfrak{A}\biggl[ (a^2 y^2 + b^2x^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
mp^4 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
- 
n (x^2 + q^4y^2 + p^4z^2)(a^2 y^2 + b^2x^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
mp^4 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
- 
n [(x^2 + q^4y^2 + p^4z^2)a^2 y^2 + (x^2 + q^4y^2 + p^4z^2)b^2x^2]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)a^2p^4(yz)^2
+ (m-n)b^2p^4(xz)^2
+ (xy)^2[mc^2p^4 - na^2 - nb^2q^4]
- n a^2 q^4y^4 - nb^2x^4
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ]
+ (xy)^2[mc^2p^4 - n(a^2 + b^2q^4)]
- n [ (aq^2y^2 \pm bx^2)^2 \mp 2abq^2(xy)^2] \, .
</math>
  </td>
</tr>
</table>

Choosing the ''inferior'' sign then enforcing  the above-derived <font color="red">One-Two Perpendicular Constraint</font> by setting, <math>~(a + bq^2) = -cp^2</math>, gives,

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

<tr>
  <td align="right">
<math>~ \frac{1}{2z}\biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ]
+ (xy)^2[mc^2p^4 - n(a^2 + b^2q^4 + 2abq^2)]
- n (aq^2y^2 - bx^2)^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 ]
+ (xy)^2[mc^2p^4 - n(-c p^2 )^2]
- n (aq^2y^2 - bx^2)^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)p^4[ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 ]
- n (bx^2 - aq^2y^2 )^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(m-n)p^4\mathfrak{B}
- n (bx^2 - aq^2y^2 )^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
m p^4\mathfrak{B} -n [p^4\mathfrak{B}+ (bx^2 - aq^2y^2 )^2 ] \, .
</math>
  </td>
</tr>
</table>
Reflecting back on the first line of the "example" derivation, we recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~p^4\mathfrak{B}+ (bx^2 - aq^2y^2 )^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{A} (a^2 y^2 + b^2x^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~(bx^2 - aq^2y^2 )^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathfrak{A} (a^2 y^2 + b^2x^2)
-p^4\mathfrak{B}
</math>
  </td>
</tr>
</table>

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


----


Similarly, we find,

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
y^2z^2(a^2 - q^4b^2 - c^2p^4)  
- 
c^2 q^4y^4 
- 
b^2p^4z^4 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
y^2z^2(a^2 - q^4b^2 - c^2p^4)  
- 
(cq^2y^2 \pm bp^2z^2)^2 \pm 2bcq^2y^2p^2z^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(yz)^2(a^2 - q^4b^2 - c^2p^4  \pm 2bcq^2p^2)  
- 
(cq^2y^2 \pm bp^2z^2)^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(yz)^2[a^2 - (bq^2 \pm cp^2)^2  \pm 4bcq^2p^2]  
- 
(cq^2y^2 \pm bp^2z^2)^2
\biggr] \, .
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) 
- 
a^2 p^4z^4
-
c^2x^4
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
x^2z^2 ( b^2q^4 -a^2 - c^2p^4 ) 
- 
(ap^2z^2 \pm cx^2)^2 \pm 2acx^2p^2z^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
x^2z^2 ( b^2q^4 -a^2 - c^2p^4 \pm 2acp^2 ) 
- 
(ap^2z^2 \pm cx^2)^2 
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2y \biggl[
(xz)^2 [ b^2q^4 -(a\pm cp^2)^2  \pm 4acp^2 ] 
- 
(ap^2z^2 \pm cx^2)^2 
\biggr] \, .
</math>
  </td>
</tr>
</table>

So, if we again choose the ''superior'' sign throughout these expression, the above-derived <font color="red">One-Two Perpendicular Constraint</font> can be satisfied:  &nbsp; In the first by setting, <math>~(bq^2 + cp^2) = - a</math>; and in the second by setting, <math>~(a + cp^2) = - bq^2</math>.  The expressions then become, respectively,

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2x \biggl[
(cq^2y^2 + bp^2z^2)^2
- 4bcq^2 y^2 p^2 z^2  
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2x (cq^2y^2 - bp^2z^2)^2 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \biggl[ \frac{2\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 2y \biggl[
(ap^2z^2 + cx^2)^2 
- 4acx^2 p^2 z^2
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2y(ap^2z^2 - cx^2)^2 \, .
</math>
  </td>
</tr>
</table>

<table border="1" align="center" cellpadding="10" width="60%"><tr><td align="left">
<div align="center">'''Summary'''</div>
Given,
<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>~
\mathfrak{A}^{1 / 2} \mathfrak{B}^{-1 / 2} 
=
\biggl[ \frac{ (x^2 + q^4y^2 + p^4z^2) }{ a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2 } \biggr]^{1 / 2} 
</math>
  </td>
</tr>
</table>
the three relevant partial derivatives are:

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

<tr>
  <td align="right">
<math>~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- x (cq^2y^2 - bp^2z^2)^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- y(ap^2z^2 - cx^2)^2 \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-z (aq^2y^2 - bx^2)^2 \, .
</math>
  </td>
</tr>
</table>

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

====Scale Factor====
Hence, the associated scale factor is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2
+ \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2
+ \biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2
\biggl\{
\biggl[ - x (cq^2y^2 - bp^2z^2)^2  \biggr]^2
+ \biggl[ - y(ap^2z^2 - cx^2)^2 \biggr]^2
+ \biggl[ -z (aq^2y^2 - bx^2)^2 \biggr]^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\lambda_3}{\mathfrak{A}\mathfrak{B}} \biggr]^2
\biggl\{
x^2 (cq^2y^2 - bp^2z^2)^4 
+ y^2 (ap^2z^2 - cx^2)^4 
+ z^2 (aq^2y^2 - bx^2)^4 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{\mathfrak{A}\mathfrak{B}}{\lambda_3} \biggr]\frac{1}{\mathfrak{C}} \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[
x^2 (cq^2y^2 - bp^2z^2)^4 
+ y^2 (ap^2z^2 - cx^2)^4 
+ z^2 (aq^2y^2 - bx^2)^4 
\biggr]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

====Direction Cosines and Unit Vector====
And the associated triplet of direction cosines is:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{31} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- x (cq^2y^2 - bp^2z^2)^2\biggl[ \frac{1}{\mathfrak{C}} \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\gamma_{32} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- y(ap^2z^2 - cx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \gamma_{33} \equiv h_3 \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-z (aq^2y^2 - bx^2)^2 \biggl[ \frac{1}{\mathfrak{C}} \biggr] \, .
</math>
  </td>
</tr>
</table>
This means that, for our particular ''guess'' of the 3<sup>rd</sup> coordinate, the relevant unit vector is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[\hat{e}_3]_\mathrm{guess}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\mathfrak{C}} \biggl\{
- \hat\imath \biggl[ x (cq^2y^2 - bp^2z^2)^2 \biggr]
- \hat\jmath \biggl[ y(ap^2z^2 - cx^2)^2 \biggr]
- \hat{k} \biggl[ z (aq^2y^2 - bx^2)^2 \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

===Contrast===

The unit vector resulting (just derived) from our ''guess'' of the third-coordinate expression should be compared with [[#needed|the ''needed'' unit vector as described above]], namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[\hat{e}_3]_\mathrm{needed}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>~\mathfrak{A}^{-1 / 2} \mathfrak{B}^{-1 / 2} \biggl\{
\hat\imath \biggl[ x(cq^2y^2  - b p^2z^2) \biggr]
+
\hat\jmath \biggl[ y(ap^2z^2 - cx^2) \biggr]
+
\hat{k} \biggl[ z(bx^2 - aq^2y^2) \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

At first glance, apart from the difference in signs, the three terms inside the curly braces appear to be identical in these two separate unit-vector expressions.  But they are not!  Relative to the ''needed'' expression, key components of each term are '''squared''' in the ''guessed'' expression.  Very close &hellip; but no cigar!

<table border="1" cellpadding="10" align="center" width="80%"><tr><td align="left">
<div align="center">'''ASIDE'''</div>
Note that, <math>~[\hat{e}_3]_\mathrm{needed} \cdot [\hat{e}_3]_\mathrm{needed} = 1</math> implies that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B} = \biggl[ \frac{(abc)\mathfrak{L}}{\ell_{3D}} \biggr]^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ x(cq^2y^2  - b p^2z^2) \biggr]^2
+ \biggl[ y(ap^2z^2 - cx^2) \biggr]^2
+ \biggl[ z(bx^2 - aq^2y^2) \biggr]^2 \, .
</math>
  </td>
</tr>
</table>
But we also know that (see, for example, immediately below),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, .
</math>
  </td>
</tr>
</table>

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

What about the overall leading coefficient?  That is, does <math>~\mathfrak{A}\mathfrak{B} = \mathfrak{C}^2</math>&nbsp;&nbsp;? &nbsp;Well, given that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} = \ell_{3D}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4z^2)</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~\mathfrak{B} = (abc)^2\mathfrak{L}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] \, ,
</math>
  </td>
</tr>
</table>
we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{A} \mathfrak{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2 + q^4y^2 + p^4z^2)[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
+ q^4y^2 [a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
+ p^4z^2[a^2(yz)^2 + b^2(xz)^2 + c^2(xy)^2] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x^2y^2z^2[a^2 + b^2q^4 +c^2p^4]
+ x^4 [b^2 z^2 + c^2 y^2] 
+ q^4y^4 [a^2 z^2  + c^2 x^2] 
+ p^4z^4[a^2 y^2 + b^2 x^2] 
</math>
  </td>
</tr>
</table>

On the other hand,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{C}^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
x^2 (cq^2y^2 - bp^2z^2)^4 
+ y^2 (ap^2z^2 - cx^2)^4 
+ z^2 (aq^2y^2 - bx^2)^4 
 \, .
</math>
  </td>
</tr>
</table>

===Dot With 1<sup>st</sup> Unit Vector===

Is <math>~[\hat{e}_3]_\mathrm{guess}</math> orthogonal to <math>~\hat{e}_1</math>?  Let's take their dot product to see; note that, for simplicity, we will flip the sign on <math>~[\hat{e}_3]_\mathrm{guess}</math>.

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

<tr>
  <td align="right">
<math>~-\hat{e}_1 \cdot [\hat{e}_3]_\mathrm{guess} \biggl[ \frac{\mathfrak{C}}{\ell_{3D}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{  
\hat\imath (x)  + \hat\jmath (q^2y ) + \hat{k} (p^2 z) 
\biggl\} \cdot
\biggl\{
\hat\imath \biggl[ x (cq^2y^2 - bp^2z^2)^2 \biggr]
+ \hat\jmath \biggl[ y(ap^2z^2 - cx^2)^2 \biggr]
+ \hat{k} \biggl[ z (aq^2y^2 - bx^2)^2 \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
x^2 (cq^2y^2 - bp^2z^2)^2 
+ q^2y^2(ap^2z^2 - cx^2)^2 
+ p^2z^2 (aq^2y^2 - bx^2)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
x^2 (c^2q^4y^4 - 2cq^2y^2 bp^2z^2 + b^2p^4z^4) 
+ q^2y^2(a^2p^4z^4 - 2acx^2p^2z^2 + c^2x^4) 
+ p^2z^2 (a^2q^4y^4 - 2aq^2y^2 bx^2 + b^2x^4) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  -2x^2 y^2 z^2[cq^2 bp^2 + aq^2cp^2 + aq^2 bp^2 ] 
+ x^2 (c^2q^4y^4 + b^2p^4z^4) 
+ q^2y^2(a^2p^4z^4 + c^2x^4) 
+ p^2z^2 (a^2q^4y^4 + b^2x^4) \, .
</math>
  </td>
</tr>
</table>
It does not appear as though the RHS of this expression can be zero for all values of the Cartesian coordinates, (x, y, z).  Hence <math>~[\hat{e}_3]_\mathrm{guess}</math> is not orthogonal to <math>~\hat{e}_1</math>.

=See Also=

<ul>
  <li>
[[User:Tohline/Appendix/Ramblings/DirectionCosines|Direction Cosines]]
  </li>
  <li>[[User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates#Background|Trials up through T7 Coordinates]]</li>
</ul>


{{ SGFfooter }}