Difference between revisions of "User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates"

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Line 3,755: Line 3,755:
</table>
</table>


Then we have,
<span id="Eureka">Then we have,</span>
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


Line 3,937: Line 3,937:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_2</math>
<math>~\hat{e}_1 \cdot \hat{e}_3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,992: Line 3,992:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\hat{e}_3 \cdot \hat{e}_2</math>
<math>~\hat{e}_2 \cdot \hat{e}_3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,166: Line 4,166:
</table>
</table>


Not good!  Let's try, instead &hellip;
This is overly cluttered!  Let's try, instead &hellip;
 
 
<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~A \equiv \ell_{3D}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~B \equiv \mathcal{D}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>
  </td>
</tr>
</table>
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial 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 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 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 B}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{\partial B}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{\partial B}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>
  </td>
</tr>
</table>
 
</td></tr></table>
 
 
Now, let's assume that,
<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>~\biggl( \frac{A}{B} \biggr)^{1 / 2} \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{ \partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2 (AB)^{1 / 2}} \cdot \frac{\partial A}{\partial x_i}
-
\frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_i}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_3}{2AB}
\biggl[
B \cdot \frac{\partial A}{\partial x_i}
-
A \cdot \frac{\partial B}{\partial x_i}
\biggr]
\, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \cdot \frac{\partial A}{\partial x_i}
-
(x^2 + q^4y^2 + p^4 z^2 ) \cdot \frac{\partial B}{\partial x_i} \, .
</math>
  </td>
</tr>
</table>
 
<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
Looking ahead &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{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr] \biggr\}^2
+
\biggl\{ \frac{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial y}
-
A \cdot \frac{\partial B}{\partial y}
\biggr] \biggr\}^2
+
\biggl\{ \frac{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial z}
-
A \cdot \frac{\partial B}{\partial z}
\biggr] \biggr\}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[\frac{2AB}{\lambda_3}  \biggr]^2 h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y}
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z}
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[\frac{\lambda_3}{2AB}  \biggr] h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{\biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y}
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z}
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
\biggr\}^{-1 / 2}
</math>
  </td>
</tr>
</table>
Then, for example,
 
<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>~\biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr]
\biggl\{\biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y}
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z}
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
\biggr\}^{-1 / 2}
</math>
  </td>
</tr>
</table>
 
</td></tr></table>
 
As a result, we have,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
(x^2 + q^4y^2 + p^4 z^2 ) ( 4q^4y^2 + p^4 z^2 ) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
(4x^2q^4y^2 + x^2 p^4 z^2 + 4q^8y^4 + q^4y^2 p^4z^2 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[ 
-
(4q^8y^4  + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2x (2q^4y^2  + p^4z^2)^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{x}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2q^4y\biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
(x^2 + q^4y^2 + p^4 z^2 ) (p^4 z^2 + 4x^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2q^4y\biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
( x^2p^4z^2 + 4x^4 + q^4y^2p^4z^2 + 4x^2q^4y^2 + p^8z^4 + 4x^2p^4z^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2q^4y( 4x^4 + p^8z^4 + 4x^2p^4z^2)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2q^4y( 2x^2 + p^4z^2 )^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{y} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \, ;
</math>
  </td>
</tr>
</table>
 
and,
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
(x^2 + q^4y^2 + p^4 z^2 ) (q^4 y^2 + x^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) 
-
(x^2q^4y^2 + x^4 + q^8y^4 + x^2q^4y^2 + q^4y^2p^4z^2 + x^2p^4z^2)
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
( 2x^2q^4y^2) 
-
( x^4 + q^8y^4 )
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2p^4 z \biggl[
x^4 + q^8y^4
- 2x^2q^4y^2 
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2p^4 z (x^2 - q^4y^2 )^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln \lambda_3}{\partial \ln{z}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-4 \biggl[ \biggl( \frac{p^4 z^2}{4} \biggr) (x^2 - q^4y^2 )^2 \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2
\, .
</math>
  </td>
</tr>
</table>
 
<font color="red">'''Wow! &nbsp; Really close!''' (13 November 2020)</font>
 
 
Just for fun, let's see what we get for <math>~h_3</math>.  It is given by the expression,
<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 \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">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{\lambda_3}{ABx} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \biggr\}^2
+\biggl\{ \frac{\lambda_3}{ABy} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \biggr\}^2
+\biggl\{ \frac{\lambda_3}{ABz} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \biggr\}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl[ \frac{AB}{\lambda_3}\biggr]^2 h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{1}{x^2} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^4 \biggr\}
+\biggl\{ \frac{1}{y^2} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^4 \biggr\}
+\biggl\{ \frac{1}{z^2} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^4 \biggr\}
</math>
  </td>
</tr>
</table>
 
====Fiddle Around====
 
Let &hellip;
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{L}_x</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
- \biggl[
B \cdot \frac{\partial A}{\partial x}
-
A \cdot \frac{\partial B}{\partial x}
\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{x} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8x~\mathfrak{F}_x(y,z)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathcal{L}_y</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
- \biggl[
B \cdot \frac{\partial A}{\partial y}
-
A \cdot \frac{\partial B}{\partial y}
\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
8q^4y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{y} \biggl[ 2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8y~\mathfrak{F}_y(x,z)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathcal{L}_z</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
-\biggl[
B \cdot \frac{\partial A}{\partial z}
-
A \cdot \frac{\partial B}{\partial z}
\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2p^4 z \biggl(x^2 - q^4y^2 \biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{z} \biggl[p^2 z \biggl(x^2 - q^4y^2 \biggr)\biggr]^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8z~\mathfrak{F}_z(x,y)
</math>
  </td>
</tr>
</table>
With this shorthand in place, we can write,
 
<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[ x(2 q^4y^2 + p^4z^2 ) \biggl]
+ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
+ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}}
\biggl\{
-\hat\imath \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
+ \hat\jmath \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
+ \hat{k} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
\biggr\}
\, .
</math>
  </td>
</tr>
</table>
 
We therefore also recognize that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
 
Now, if &#8212; and it is a BIG "if" &#8212; <math>~h_3 = h_0(AB)^{-1 / 2}</math>, then we have,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2x \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2y \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2z\biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
</table>
 
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_0 \lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
But if this is the correct expression for <math>~\lambda_3</math> and its three partial derivatives, then it must be true that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\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>~\Rightarrow ~~~ \biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4x^2 \biggl[ \mathfrak{F}_x \biggl]
+
4y^2 \biggl[ \mathfrak{F}_y \biggl]
+
4z^2\biggl[ \mathfrak{F}_z \biggl]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4x^2 \biggl[ \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]
+
4y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]
+
4z^2\biggl[ \frac{p^4}{4}\biggl(x^2 - q^4y^2 \biggr)^2 \biggl]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 (2q^4y^2  + p^4z^2 )^2
+
q^4 y^2( 2x^2 + p^4z^2 )^2
+
p^4 z^2 (x^2 - q^4y^2 )^2
</math>
  </td>
</tr>
</table>
 
Well &hellip; the right-hand side of this expression is identical to the right-hand side of the [[#Eureka|above expression]], where we showed that it equals <math>~(\ell_{3D}/\mathcal{D})^{-2}</math>.  That is to say, we are now showing that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = [AB]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{h_3}{h_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(AB)^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
And this is ''precisely'' what, just a few lines above, we hypothesized the functional expression for <math>~h_3</math> ought to be.  <font color="red">'''EUREKA!'''</font>
 
====Summary====
 
In summary, then &hellip;
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


Line 4,172: Line 5,255:
   <td align="right">
   <td align="right">
<math>~\lambda_3</math>
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,177: Line 5,276:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 2x^2q^4y^2</math>
<math>~
-x^2 \biggl[\biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)^2 \biggl]^{1 / 2}
+ y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]^{1 / 2}
+ z^2 \biggl[ \frac{p^4}{4} \biggl(x^2 - q^4y^2 \biggr)^2 \biggl]^{1 / 2}
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,183: Line 5,286:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial \lambda_3}{\partial x}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,189: Line 5,292:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2x( 2q^4y^2 + p^4 z^2 ) \, ,</math>
<math>~
-x^2 \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)
+ q^2y^2 \biggl( x^2 + \frac{p^4z^2}{2} \biggr)
+ \frac{p^2 z^2}{2}  \biggl(x^2 - q^4y^2 \biggr) \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial \lambda_3}{\partial y}</math>
<math>~h_3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,201: Line 5,312:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2q^4 y (p^4 z^2 + 2x^2) \, ,</math>
<math>~(AB)^{-1 / 2} </math>
   </td>
   </td>
</tr>
</tr>
Line 4,207: Line 5,318:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial \lambda_3}{\partial z}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,213: Line 5,324:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~2p^4z(q^4 y^2 + x^2 ) \, .</math>
<math>~\biggl[
(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
\biggr]^{-1 / 2} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
No!  Once again this does not work.  The direction cosines &#8212; and, hence, the components of the <math>~\hat{e}_3</math> unit vector &#8212; are not correct!


=See Also=
=See Also=

Revision as of 00:16, 15 November 2020

Concentric Ellipsoidal (T6) Coordinates

Whitworth's (1981) Isothermal Free-Energy Surface
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Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T6) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.

Orthogonal Coordinates

Primary (radial-like) Coordinate

We start by defining a "radial" coordinate whose values identify various concentric ellipsoidal shells,

<math>~\lambda_1</math>

<math>~\equiv</math>

<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, .</math>

When <math>~\lambda_1 = a</math>, we obtain the standard definition of an ellipsoidal surface, it being understood that, <math>~q^2 = a^2/b^2</math> and <math>~p^2 = a^2/c^2</math>. (We will assume that <math>~a > b > c</math>, that is, <math>~p^2 > q^2 > 1</math>.)

A vector, <math>~\bold{\hat{n}}</math>, that is normal to the <math>~\lambda_1</math> = constant surface is given by the gradient of the function,

<math>~F(x, y, z)</math>

<math>~\equiv</math>

<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} - \lambda_1 \, .</math>

In Cartesian coordinates, this means,

<math>~\bold{\hat{n}}(x, y, z)</math>

<math>~=</math>

<math>~ \hat\imath \biggl( \frac{\partial F}{\partial x} \biggr) + \hat\jmath \biggl( \frac{\partial F}{\partial y} \biggr) + \hat{k} \biggl( \frac{\partial F}{\partial z} \biggr) </math>

 

<math>~=</math>

<math>~ \hat\imath \biggl[ x(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr] + \hat\jmath \biggl[ q^2y(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr] + \hat{k}\biggl[ p^2 z(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr] </math>

 

<math>~=</math>

<math>~ \hat\imath \biggl( \frac{x}{\lambda_1} \biggr) + \hat\jmath \biggl( \frac{q^2y}{\lambda_1} \biggr) + \hat{k}\biggl(\frac{p^2 z}{\lambda_1} \biggr) \, , </math>

where it is understood that this expression is only to be evaluated at points, <math>~(x, y, z)</math>, that lie on the selected <math>~\lambda_1</math> surface — that is, at points for which the function, <math>~F(x,y,z) = 0</math>. The length of this normal vector is given by the expression,

<math>~[ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2}</math>

<math>~=</math>

<math>~ \biggl[ \biggl( \frac{\partial F}{\partial x} \biggr)^2 + \biggl( \frac{\partial F}{\partial y} \biggr)^2 + \biggl( \frac{\partial F}{\partial z} \biggr)^2 \biggr]^{1 / 2} </math>

 

<math>~=</math>

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

 

<math>~=</math>

<math>~ \frac{1}{\lambda_1 \ell_{3D}} </math>

where,

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~\biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]^{- 1 / 2} \, .</math>

It is therefore clear that the properly normalized normal unit vector that should be associated with any <math>~\lambda_1</math> = constant ellipsoidal surface is,

<math>~\hat{e}_1 </math>

<math>~\equiv</math>

<math>~ \frac{ \bold\hat{n} }{ [ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2} } = \hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, . </math>

From our accompanying discussion of direction cosines, it is clear, as well, that the scale factor associated with the <math>~\lambda_1</math> coordinate is,

<math>~h_1^2</math>

<math>~=</math>

<math>~\lambda_1^2 \ell_{3D}^2 \, .</math>

We can also fill in the top line of our direction-cosines table, namely,

Direction Cosines for T6 Coordinates
<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>

<math>~n</math> <math>~i = x, y, z</math>
<math>~1</math>  

<math>~x\ell_{3D}</math>
 

<math>~q^2 y \ell_{3D}</math> <math>~p^2 z \ell_{3D}</math>
<math>~2</math>

 
---
 

 
---
 

 
---
 

<math>~3</math>

 
---
 

 
---
 

 
---
 

Other Coordinate Pair in the Tangent Plane

Let's focus on a particular point on the <math>~\lambda_1</math> = constant surface, <math>~(x_0, y_0, z_0)</math>, that necessarily satisfies the function, <math>~F(x_0, y_0, z_0) = 0</math>. We have already derived the expression for the unit vector that is normal to the ellipsoidal surface at this point, namely,

<math>~\hat{e}_1 </math>

<math>~\equiv</math>

<math>~ \hat\imath (x_0 \ell_{3D}) + \hat\jmath (q^2y_0 \ell_{3D}) + \hat\jmath (p^2 z_0 \ell_{3D}) \, , </math>

where, for this specific point on the surface,

<math>~\ell_{3D}</math>

<math>~=</math>

<math>~\biggl[ x_0^2 + q^4y_0^2 + p^4 z_0^2 \biggr]^{- 1 / 2} \, .</math>


Tangent Plane

The two-dimensional plane that is tangent to the <math>~\lambda_1</math> = constant surface at this point is given by the expression,

<math>~0</math>

<math>~=</math>

<math>~ (x - x_0) \biggl[ \frac{\partial \lambda_1}{\partial x} \biggr]_0 + (y - y_0) \biggl[\frac{\partial \lambda_1}{\partial y} \biggr]_0 + (z - z_0) \biggl[\frac{\partial \lambda_1}{\partial z} \biggr]_0 </math>

 

<math>~=</math>

<math>~ (x - x_0) \biggl[ \frac{\partial F}{\partial x} \biggr]_0 + (y - y_0) \biggl[\frac{\partial F}{\partial y} \biggr]_0 + (z - z_0) \biggl[ \frac{\partial F}{\partial z} \biggr]_0 </math>

 

<math>~=</math>

<math>~ (x - x_0) \biggl( \frac{x}{\lambda_1}\biggr)_0 + (y - y_0)\biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + (z - z_0)\biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0 </math>

<math>~\Rightarrow~~~ x \biggl( \frac{x}{\lambda_1}\biggr)_0 + y \biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + z \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0 </math>

<math>~=</math>

<math>~ x_0 \biggl( \frac{x}{\lambda_1}\biggr)_0 + y_0 \biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + z_0 \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0 </math>

<math>~\Rightarrow~~~ x x_0 + q^2 y y_0 + p^2 z z_0 </math>

<math>~=</math>

<math>~ x_0^2 + q^2 y_0^2 + p^2 z_0^2 </math>

<math>~\Rightarrow~~~ x x_0 + q^2 y y_0 + p^2 z z_0 </math>

<math>~=</math>

<math>~ (\lambda_1^2)_0 \, . </math>

Fix the value of <math>~\lambda_1</math>. This means that the relevant ellipsoidal surface is defined by the expression,

<math>~\lambda_1^2</math>

<math>~=</math>

<math>~x^2 + q^2y^2 + p^2z^2 \, .</math>

If <math>~z = 0</math>, the semi-major axis of the relevant x-y ellipse is <math>~\lambda_1</math>, and the square of the semi-minor axis is <math>~\lambda_1^2/q^2</math>. At any other value, <math>~z = z_0 < c</math>, the square of the semi-major axis of the relevant x-y ellipse is, <math>~(\lambda_1^2 - p^2z_0^2)</math> and the square of the corresponding semi-minor axis is, <math>~(\lambda_1^2 - p^2z_0^2)/q^2</math>. Now, for any chosen <math>~x_0^2 \le (\lambda_1^2 - p^2z_0^2)</math>, the y-coordinate of the point on the <math>~\lambda_1</math> surface is given by the expression,

<math>~y_0^2</math>

<math>~=</math>

<math>~\frac{1}{q^2}\biggl[ \lambda_1^2 - p^2 z_0 -x_0^2 \biggr] \, .</math>

The slope of the line that lies in the z = z0 plane and that is tangent to the ellipsoidal surface at <math>~(x_0, y_0)</math> is,

<math>~m \equiv \frac{dy}{dx}\biggr|_{z_0}</math>

<math>~=</math>

<math>~- \frac{x_0}{q^2y_0}</math>

Speculation1

Building on our experience developing T3 Coordinates and, more recently, T5 Coordinates, let's define the two "angles,"

<math>~\Zeta</math>

<math>~\equiv</math>

<math>~\sinh^{-1}\biggl(\frac{qy}{x} \biggr)</math>

      and,      

<math>~\Upsilon</math>

<math>~\equiv</math>

<math>~\sinh^{-1}\biggl(\frac{pz}{x} \biggr) \, ,</math>

in which case we can write,

<math>~\lambda_1^2</math>

<math>~=</math>

<math>~x^2(\cosh^2\Zeta + \sinh^2\Upsilon)\, .</math>

We speculate that the other two orthogonal coordinates may be defined by the expressions,

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~x \biggl[ \sinh\Zeta \biggr]^{1/(1-q^2)} = x \biggl[ \frac{qy}{x}\biggr]^{1/(1-q^2)} = x \biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)} = \biggl[ \frac{x^{q^2}}{qy}\biggr]^{1/(q^2-1)} \, ,</math>

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~x \biggl[ \sinh\Upsilon \biggr]^{1/(1-p^2)} = x \biggl[ \frac{pz}{x}\biggr]^{1/(1-p^2)} = x \biggl[ \frac{x}{pz}\biggr]^{1/(p^2-1)} = \biggl[ \frac{x^{p^2}}{pz}\biggr]^{1/(p^2-1)} \, .</math>

Some relevant partial derivatives are,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~\biggl[ \frac{1}{qy}\biggr]^{1/(q^2-1)} \biggl[ \frac{q^2}{q^2-1} \biggr]x^{1/(q^2-1)} = \biggl[ \frac{q^2}{q^2-1} \biggr]\biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)} = \biggl[ \frac{q^2}{q^2-1} \biggr]\frac{\lambda_2}{x} \, ; </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~\biggl[ \frac{x^{q^2}}{q}\biggr]^{1/(q^2-1)} \biggl[ \frac{1}{1-q^2} \biggr] y^{q^2/(1-q^2)} = - \biggl[ \frac{1}{q^2-1} \biggr] \frac{\lambda_2}{y} \, ;</math>

<math>~\frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ \biggl[ \frac{p^2}{p^2-1} \biggr]\frac{\lambda_3}{x} \, ; </math>

<math>~\frac{\partial \lambda_3}{\partial z}</math>

<math>~=</math>

<math>~ - \biggl[ \frac{1}{p^2-1} \biggr] \frac{\lambda_3}{z} \, .</math>

And the associated scale factors are,

<math>~h_2^2</math>

<math>~=</math>

<math>~ \biggl\{ \biggl[ \biggl( \frac{q^2}{q^2-1} \biggr)\frac{\lambda_2}{x} \biggr]^2 + \biggl[ - \biggl( \frac{1}{q^2-1} \biggr) \frac{\lambda_2}{y} \biggr]^2 \biggr\}^{-1} </math>

 

<math>~=</math>

<math>~ \biggl\{ \biggl( \frac{q^2}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{x^2} + \biggl( \frac{1}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{y^2} \biggr\}^{-1} </math>

 

<math>~=</math>

<math>~ \biggl\{x^2 + q^4 y^2 \biggr\}^{-1} \biggl[ \frac{(q^2 - 1)^2x^2 y^2}{\lambda_2^2} \biggr] \, ; </math>

<math>~h_3^2</math>

<math>~=</math>

<math>~ \biggl\{x^2 + p^4 z^2 \biggr\}^{-1} \biggl[ \frac{(p^2 - 1)^2x^2 z^2}{\lambda_3^2} \biggr] \, . </math>

We can now fill in the rest of our direction-cosines table, namely,

Direction Cosines for T6 Coordinates
<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>

<math>~n</math> <math>~i = x, y, z</math>
<math>~1</math>  

<math>~x\ell_{3D}</math>
 

<math>~q^2 y \ell_{3D}</math> <math>~p^2 z \ell_{3D}</math>
<math>~2</math>

<math>~q^2 y \ell_q </math>

<math>~-x\ell_q</math>

<math>~0</math>

<math>~3</math>

<math>~p^2 z \ell_p</math>

<math>~0</math>

<math>~-x\ell_p</math>

Hence,

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat\imath \gamma_{21} + \hat\jmath \gamma_{22} +\hat{k} \gamma_{23} = \hat\imath (q^2y\ell_q) - \hat\jmath (x\ell_q) \, ; </math>

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~ \hat\imath \gamma_{31} + \hat\jmath \gamma_{32} +\hat{k} \gamma_{33} = \hat\imath (p^2z\ell_p) -\hat{k} (x\ell_p) \, . </math>

Check:

<math>~\hat{e}_2 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ (q^2y\ell_q)^2 + (x\ell_q)^2 = 1 \, ; </math>

<math>~\hat{e}_3 \cdot \hat{e}_3</math>

<math>~=</math>

<math>~ (p^2z\ell_p)^2 + (x\ell_p)^2 = 1 \, ; </math>

<math>~\hat{e}_2 \cdot \hat{e}_3</math>

<math>~=</math>

<math>~ (q^2y\ell_q)(p^2z\ell_p) \ne 0 \, . </math>

Speculation2

Try,

<math>~\lambda_2</math>

<math>~=</math>

<math>~ \frac{x}{y^{1/q^2} z^{1/p^2}} \, , </math>

in which case,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{\lambda_2}{x} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{x}{z^{1/p^2}} \biggl(-\frac{1}{q^2}\biggr) y^{-1/q^2 - 1} = -\frac{\lambda_2}{q^2 y} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ -\frac{\lambda_2}{p^2 z} \, . </math>

The associated scale factor is, then,

<math>~h_2^2</math>

<math>~=</math>

<math>~\biggl[ \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 \biggr]^{-1} </math>

 

<math>~=</math>

<math>~\biggl[ \biggl( \frac{ \lambda_2}{x} \biggr)^2 + \biggl( -\frac{\lambda_2}{q^2y} \biggr)^2 + \biggl( - \frac{\lambda_2}{p^2z} \biggr)^2 \biggr]^{-1} </math>

Speculation3

Try,

<math>~\lambda_2</math>

<math>~=</math>

<math>~ \frac{(x+p^2 z)^{1 / 2}}{y^{1/q^2} } \, , </math>

in which case,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{2y^{1/q^2}}\biggl(x + p^2z\biggr)^{- 1 / 2} = \frac{\lambda_2}{2(x + p^2z) } \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ -\frac{\lambda_2}{q^2y} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~

\, . </math>

Speculation4

Development

Here we stick with the primary (radial-like) coordinate as defined above; for example,

<math>~\lambda_1</math>

<math>~\equiv</math>

<math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ,</math>

<math>~h_1</math>

<math>~=</math>

<math>~\lambda_1 \ell_{3D} \, ,</math>

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~[ x^2 + q^4y^2 + p^4 z^2 ]^{- 1 / 2} \, ,</math>

<math>~\hat{e}_1 </math>

<math>~=</math>

<math>~ \hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, . </math>

Note that, <math>~\hat{e}_1 \cdot \hat{e}_1 = 1</math>, which means that this is, indeed, a properly normalized unit vector.

Then, drawing from our earliest discussions of "T1 Coordinates", we'll try defining the second coordinate as,

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~ \tan^{-1} u \, , </math>       where,

<math>~u</math>

<math>~\equiv</math>

<math>\frac{y^{1/q^2}}{x} \, .</math>

The relevant partial derivatives are,

<math>~\frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{1 + u^2} \biggl[ - \frac{y^{1/q^2}}{x^2} \biggr] = - \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{x} = - \frac{\sin\lambda_3 \cos\lambda_3}{x} \, , </math>

<math>~\frac{\partial \lambda_3}{\partial y}</math>

<math>~=</math>

<math>~ \frac{1}{1 + u^2} \biggl[ \frac{y^{(1/q^2-1)}}{q^2x} \biggr] = \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{q^2y} = \frac{\sin\lambda_3 \cos\lambda_3}{q^2y} \, , </math>

which means that,

<math>~h_3^2</math>

<math>~=</math>

<math>~ \biggl[ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2 \biggr]^{-1} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{u}{1 + u^2}\biggr]^{-2} \biggl[ \frac{1}{x^2} + \frac{1}{q^4y^2} \biggr]^{-1} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{1 + u^2}{u}\biggr]^{2} \biggl[ \frac{x^2 + q^4y^2}{x^2q^4y^2} \biggr]^{-1} </math>

<math>~\Rightarrow~~~h_3</math>

<math>~=</math>

<math>~ \biggl[ \frac{1 + u^2}{u}\biggr]xq^2 y \ell_q = \frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3} \, , </math>       where,

<math>~\ell_q</math>

<math>~\equiv</math>

<math>~[x^2 + q^4 y^2]^{-1 / 2} \, .</math>

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

Direction Cosines for T6 Coordinates
<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>

<math>~n</math> <math>~i = x, y, z</math>
<math>~1</math>  

<math>~x\ell_{3D}</math>
 

<math>~q^2 y \ell_{3D}</math> <math>~p^2 z \ell_{3D}</math>
<math>~2</math>

 
---
 

 
---
 

 
---
 

<math>~3</math>

<math>~-q^2 y \ell_q</math>

<math>~x \ell_q</math>

<math>~0</math>

which means that the associated unit vector is,

<math>~\hat{e}_3 </math>

<math>~=</math>

<math>~ -\hat\imath (q^2 y \ell_{q}) + \hat\jmath (x \ell_{q}) \, . </math>

Note that, <math>~\hat{e}_3 \cdot \hat{e}_3 = 1</math>, which means that this also is a properly normalized unit vector. Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other. Let's see …

<math>~\hat{e}_3 \cdot \hat{e}_1</math>

<math>~=</math>

<math>~ (- q^2y \ell_q)x\ell_{3D} + (x\ell_q) q^2y\ell_{3D} = 0 \, . </math>

Q.E.D.

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

<math>~\hat{e}_2 \equiv \hat{e}_3 \times \hat{e}_1</math>

<math>~=</math>

<math>~ \hat\imath \biggl[ (e_3)_2 (e_1)_3 - (e_3)_3(e_1)_2 \biggr] + \hat\jmath \biggl[ (e_3)_3 (e_1)_1 - (e_3)_1(e_1)_3 \biggr] + \hat{k} \biggl[ (e_3)_1 (e_1)_2 - (e_3)_2(e_1)_1 \biggr] </math>

 

<math>~=</math>

<math>~ \hat\imath \biggl[ (x \ell_q) (p^2 z \ell_{3D}) - 0 \biggr] + \hat\jmath \biggl[ 0 - (-q^2y \ell_q)(p^2z \ell_{3D}) \biggr] + \hat{k} \biggl[ (-q^2y \ell_q) (q^2 y \ell_{3D}) - (x\ell_q)(x\ell_{3D}) \biggr] </math>

 

<math>~=</math>

<math>~\ell_q \ell_{3D}\biggl[ \hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} ( x^2 + q^4 y^2 ) \biggr] </math>

 

<math>~=</math>

<math>~\ell_q \ell_{3D}\biggl[ \hat\imath ( xp^2 z ) + \hat\jmath ( q^2y p^2z ) - \hat{k} \biggl( \frac{1}{\ell_q^2} \biggr) \biggr] \, . </math>

Note that,

<math>~\hat{e}_3 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~\ell_q^2 \ell_{3D} \biggl[ (- q^2y )x p^2 z + (x) q^2y p^2 z \biggr] = 0 \, ; </math>

and,

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ (x\ell_{3D})xp^2z \ell_q \ell_{3D} + (q^2y \ell_{3D}) q^2yp^2 z \ell_q \ell_{3D} - (x^2 + q^4 y^2)\ell_q \ell_{3D} (p^2 z \ell_{3D} ) </math>

 

<math>~=</math>

<math>~ \ell_q \ell_{3D}^2 \biggl[ x^2p^2z + (q^4y^2 ) p^2 z - (x^2 + q^4 y^2) (p^2 z ) \biggr] = 0 \, . </math>

We conclude, therefore, that <math>~\hat{e}_2</math> is perpendicular to both of the other unit vectors. Hooray!


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

Direction Cosines for T6 Coordinates
<math>~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)</math>

<math>~n</math> <math>~i = x, y, z</math>
<math>~1</math>  

<math>~x\ell_{3D}</math>
 

<math>~q^2 y \ell_{3D}</math> <math>~p^2 z \ell_{3D}</math>
<math>~2</math>

<math>~x p^2 z\ell_q \ell_{3D}</math>

<math>~q^2y p^2 z\ell_q \ell_{3D}</math>

<math>~-(x^2 + q^4y^2)\ell_q \ell_{3D} = - \ell_{3D}/\ell_q</math>

<math>~3</math>

<math>~-q^2 y \ell_q</math>

<math>~x \ell_q</math>

<math>~0</math>

Analysis

Let's break down each direction cosine into its components.

Direction Cosine Components for T6 Coordinates
<math>~n</math> <math>~\lambda_n</math> <math>~h_n</math> <math>~\frac{\partial \lambda_n}{\partial x}</math> <math>~\frac{\partial \lambda_n}{\partial y}</math> <math>~\frac{\partial \lambda_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\lambda_1 \ell_{3D}</math> <math>~\frac{x}{\lambda_1}</math> <math>~\frac{q^2 y}{\lambda_1}</math> <math>~\frac{p^2 z}{\lambda_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> --- --- --- --- --- <math>~\ell_q \ell_{3D} (xp^2z)</math> <math>~\ell_q \ell_{3D} (q^2 y p^2z) </math> <math>~- (x^2 + q^4y^2)\ell_q \ell_{3D}</math>
<math>~3</math> <math>~\tan^{-1}\biggl( \frac{y^{1/q^2}}{x} \biggr)</math> <math>~\frac{xq^2 y \ell_q}{\sin\lambda_3 \cos\lambda_3}</math> <math>~-\frac{\sin\lambda_3 \cos\lambda_3}{x}</math> <math>~+\frac{\sin\lambda_3 \cos\lambda_3}{q^2y}</math> <math>~0</math> <math>~-q^2 y \ell_q</math> <math>~x\ell_q</math> <math>~0</math>

Try,

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~ \tan^{-1} w \, , </math>       where,

<math>~w</math>

<math>~\equiv</math>

<math>\frac{(x^2 + q^2y^2)^{1 / 2}}{z^{1/p^2}} ~~~\Rightarrow~~~\frac{1}{z^{1 / p^2} } = \frac{w}{(x^2 + q^2 y^2)^{1 / 2}} \, .</math>

The relevant partial derivatives are,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)^{1 / 2}~z^{1/p^2}} \biggr] = \frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr] \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{1}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)^{1 / 2}~z^{1/p^2}} \biggr] = \frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr] \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ \frac{w}{1 + w^2} \biggl[- \frac{1}{p^2 z} \biggr] \, , </math>

which means that,

<math>~h_2^{-2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{x^2}{(x^2 + q^2y^2)^2} \biggr] + \biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{q^4 y^2}{(x^2 + q^2y^2)^2} \biggr] + \biggl[ \biggl( \frac{w}{1 + w^2}\biggr)^2 \frac{1}{p^4 z^2} \biggr] </math>

 

<math>~=</math>

<math>~\biggl( \frac{w}{1 + w^2}\biggr)^2 \biggl[ \frac{(x^2 + q^4y^2)(p^4 z^2) + (x^2 + q^2y^2)^2}{(x^2 + q^2y^2)^2~p^4 z^2} \biggr] </math>

<math>~\Rightarrow~~~ h_2</math>

<math>~=</math>

<math>~ \biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}} \biggr\} \, , </math>

where,

<math>~\mathcal{D}</math>

<math>~\equiv</math>

<math>~[(x^2 + q^4y^2)(p^4 z^2) + (x^2 + q^2y^2)^2]^{1 / 2} \, .</math>

Hence, the trio of associated direction cosines are,

<math>~\gamma_{21} = h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)</math>

<math>~=</math>

<math>~ \biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}} \biggr\}\frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr] = \biggl\{ \frac{x~p^2 z}{ \mathcal{D}} \biggr\} \, , </math>

<math>~\gamma_{22} = h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)</math>

<math>~=</math>

<math>~ \biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}} \biggr\} \frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr] = \biggl\{ \frac{q^2 y~p^2 z}{ \mathcal{D}} \biggr\} \, , </math>

<math>~\gamma_{23} = h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)</math>

<math>~=</math>

<math>~ \biggl( \frac{1 + w^2}{w}\biggr) \biggl\{ \frac{(x^2 + q^2y^2)~p^2 z}{ \mathcal{D}} \biggr\}\frac{w}{1 + w^2} \biggl[- \frac{1}{p^2 z} \biggr] = \biggl\{- \frac{(x^2 + q^2y^2)}{ \mathcal{D}} \biggr\} \, . </math>

VERY close!

Let's examine the function, <math>~\mathcal{D}^2</math>.

<math>~\frac{1}{\ell_{3D}^2 \ell_d^2}</math>

<math>~=</math>

<math>~ (x^2 + q^4 y^2)(x^2 + q^4 y + p^4 z) = (x^2 + q^4 y^2)p^4 z + (x^2 + q^4 y^2)^2 \, . </math>

Eureka (NOT!)

Try,

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~ \tan^{-1} w \, , </math>       where,

<math>~w</math>

<math>~\equiv</math>

<math>\frac{(x^2 + q^2y^2)^{1 / 2}}{p^2 z} ~~~\Rightarrow~~~\frac{1}{p^2 z } = \frac{w}{(x^2 + q^2 y^2)^{1 / 2}} \, .</math>

The relevant partial derivatives are,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)^{1 / 2}~p^2 z} \biggr] = \frac{w}{1 + w^2} \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr] \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{1}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)^{1 / 2}~p^2z} \biggr] = \frac{w}{1 + w^2} \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr] \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ \frac{1}{1 + w^2} \biggl[- \frac{(x^2 + q^2y^2)^{1 / 2}}{~p^2z^2} \biggr] = \frac{w}{1 + w^2} \biggl[- \frac{1}{z} \biggr] \, , </math>

which means that,

<math>~h_2^{-2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl[ \frac{w}{1 + w^2} \biggr]^2 \biggl\{ \biggl[ \frac{x}{(x^2 + q^2y^2)} \biggr]^2 + \biggl[ \frac{q^2y}{(x^2 + q^2y^2)} \biggr]^2 + \biggl[ - \frac{1}{z} \biggr]^2 \biggr\} </math>

 

<math>~=</math>

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

Speculation5

Spherical Coordinates

<math>~r\cos\theta</math>

<math>~=</math>

<math>~z \, ,</math>

<math>~r\sin\theta</math>

<math>~=</math>

<math>~(x^2 + y^2)^{1 / 2} \, ,</math>

<math>~\tan\varphi</math>

<math>~=</math>

<math>~\frac{y}{x} \, .</math>

Use λ1 Instead of r

Here, as above, we define,

<math>~\lambda_1^2</math>

<math>~\equiv</math>

<math>~x^2 + q^2 y^2 + p^2 z^2 </math>

Using this expression to eliminate "x" (in favor of λ1) in each of the three spherical-coordinate definitions, we obtain,

<math>~r^2 \equiv x^2 + y^2 + z^2</math>

<math>~=</math>

<math>~\lambda_1^2 - y^2(q^2-1) - z^2(p^2-1) \, ;</math>

<math>~\tan^2\theta \equiv \frac{x^2 + y^2}{z^2}</math>

<math>~=</math>

<math>~\frac{1}{z^2}\biggl[ \lambda_1^2 -y^2(q^2-1) -p^2z^2 \biggr] \, ;</math>

<math>~\frac{1}{\tan^2\varphi} \equiv \frac{x^2}{y^2}</math>

<math>~=</math>

<math>~ \frac{\lambda_1^2 - p^2z^2}{y^2} - q^2 \, . </math>

After a bit of additional algebraic manipulation, we find that,

<math>\frac{z^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ \frac{ (1+\tan^2\varphi)}{\mathcal{D}^2} \, , </math>

<math>~\frac{y^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ \biggl[\frac{ \mathcal{D}^2 \tan^2\varphi - p^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) \mathcal{D}^2} \biggr] \, , </math>

<math>~\frac{x^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ 1 - q^2\biggl(\frac{y^2}{\lambda_1^2} \biggr) - p^2\biggl(\frac{z^2}{\lambda_1^2}\biggr) \, , </math>

where,

<math>~\mathcal{D}^2</math>

<math>~\equiv</math>

<math>~ \biggl[ (1 + q^2\tan^2\varphi)(p^2 + \tan^2\theta) - p^2(q^2-1)\tan^2\varphi \biggr] \, . </math>

As a check, let's set <math>~q^2 = p^2 = 1</math>, which should reduce to the normal spherical coordinate system.

<math>~\lambda_1^2</math>

<math>~\rightarrow</math>

<math>~ r^2 \, , </math>

      and,      

<math>~\mathcal{D}^2</math>

<math>~\rightarrow</math>

<math>~ \biggl[ (1 + \tan^2\varphi)(1 + \tan^2\theta) \biggr] \, . </math>

<math>~\Rightarrow ~~~ \frac{z^2}{\lambda_1^2}</math>

<math>~\rightarrow</math>

<math>~ \frac{1}{1+\tan^2\theta} = \cos^2\theta = \frac{z^2}{r^2} \, ; </math>

<math>~\frac{y^2}{\lambda_1^2}</math>

<math>~\rightarrow</math>

<math>~ \biggl[\frac{(1 + \tan^2\varphi)(1 + \tan^2\theta) \tan^2\varphi - \tan^2\varphi (1+\tan^2\varphi)}{(1+\tan^2\varphi) (1 + \tan^2\varphi)(1 + \tan^2\theta)} \biggr] </math>

 

<math>~=</math>

<math>~\frac{\tan^2\varphi}{(1 + \tan^2\varphi)} \biggl[\frac{\tan^2\theta }{ (1 + \tan^2\theta)} \biggr] = \sin^2\theta \sin^2\varphi = \frac{y^2}{r^2} \,; </math>

<math>~\frac{x^2}{\lambda_1^2}</math>

<math>~\rightarrow</math>

<math>~ 1 - \biggl(\frac{y^2}{\lambda_1^2} \biggr) - \biggl(\frac{z^2}{\lambda_1^2}\biggr) \, , </math>

 

<math>~\rightarrow</math>

<math>~ 1 - \sin^2\theta \sin^2\varphi - \cos^2\theta = - \sin^2\theta \sin^2\varphi + \sin^2\theta = \sin^2\theta \cos^2\varphi = \frac{x^2}{r^2} \, . </math>

Relationship To T3 Coordinates

If we set, <math>~q = 1</math>, but continue to assume that <math>~p > 1</math>, we expect to see a representation that resembles our previously discussed, T3 Coordinates. Note, for example, that the new "radial" coordinate is,

<math>~\lambda_1^2</math>

<math>~\rightarrow</math>

<math>~ (\varpi^2 + p^2z^2) \, , </math>

      and,      

<math>~\mathcal{D}^2</math>

<math>~\rightarrow</math>

<math>~ \biggl[ (1 + \tan^2\varphi)(p^2 + \tan^2\theta) \biggr] \, . </math>

<math>~\Rightarrow ~~~ \frac{p^2z^2}{\lambda_1^2}</math>

<math>~\rightarrow</math>

<math>~\frac{ p^2}{(p^2 + \tan^2\theta)} = \frac{ 1}{(1 + p^{-2} \tan^2\theta)}\, ,</math>

<math>~\frac{\varpi^2}{\lambda_1^2} = \frac{x^2}{\lambda_1^2} + \frac{y^2}{\lambda_1^2}</math>

<math>~\rightarrow</math>

<math>~1 - p^2 \biggl( \frac{z^2}{\lambda_1^2}\biggr) = \biggl[1 - \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr] \, .</math>

We also see that,

<math>~\frac{\varpi^2}{p^2z^2}</math>

<math>~\rightarrow</math>

<math>~ (1 + p^{-2}\tan^2\theta)\biggl[1 - \frac{ 1}{(1 + p^{-2}\tan^2\theta)} \biggr] = p^{-2}\tan^2\theta \, . </math>


Again Consider Full 3D Ellipsoid

Let's try to replace everywhere, <math>~[\varpi/(pz)]^2 = p^{-2}\tan^2\theta</math> with <math>~\lambda_2</math>. This gives,

<math>~\frac{\mathcal{D}^2}{p^2}</math>

<math>~\equiv</math>

<math>~ \biggl[ (1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi \biggr] \, . </math>

which means that,

<math>\frac{p^2 z^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ \frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]} </math>

 

<math>~=</math>

<math>~ \frac{ (1+\tan^2\varphi)/\tan^2\varphi}{[q^2 \lambda_2 (1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1)]} = \frac{1/\sin^2\varphi}{[q^2\lambda_2 Q^2 - (q^2-1) ]} \, , </math>

<math>~\frac{q^2y^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } - \frac{ q^2 \tan^2\varphi (1+\tan^2\varphi)}{(1+q^2\tan^2\varphi) [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] } </math>

 

<math>~=</math>

<math>~ \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1 - \frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] } \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1 - \frac{ (1+\tan^2\varphi)/\tan^2\varphi}{ [q^2\lambda_2(1 + q^2\tan^2\varphi)/(q^2\tan^2\varphi) - (q^2-1) ] } \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{ 1}{Q^2 } \biggl\{1 - \frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] } \biggr\} = \frac{1}{Q^2[~~]} \biggl[ [~~] - \frac{1}{\sin^2\varphi} \biggr] \, , </math>

<math>~\frac{x^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~ 1 - \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } \biggl\{1 - \frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] } \biggr\} </math>

 

 

<math>~ - \frac{ (1+\tan^2\varphi)}{[(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi]} </math>

 

<math>~=</math>

<math>~ 1 - \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) } - \biggl\{\frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] } \biggr\} </math>

 

<math>~=</math>

<math>~ 1 - \frac{ 1 }{ Q^2 } - \biggl\{\frac{ 1/\sin^2\varphi}{ [q^2\lambda_2 Q^2 - (q^2-1) ] } \biggr\} = \frac{1}{Q^2 [~~] } \biggl\{ Q^2[~~] - [~~] - \frac{Q^2}{\sin^2\varphi} \biggr\} \, , </math>

<math>~\frac{x^2 + q^2y^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~1 - \biggl[1 + \frac{ q^2 \tan^2\varphi }{(1+q^2\tan^2\varphi) }\biggr] \biggl\{ \frac{ (1+\tan^2\varphi)}{ [(1 + q^2\tan^2\varphi)\lambda_2 - (q^2-1)\tan^2\varphi] } \biggr\} \, . </math>

Now, notice that,

<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>

<math>~=</math>

<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, ,</math>

and,

<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>

<math>~=</math>

<math>~1 - \frac{1}{[~~]\sin^2\varphi} \, .</math>

Hence,

<math>~\frac{x^2}{\lambda_1^2} + \frac{1}{Q^2}</math>

<math>~=</math>

<math>~\frac{q^2 y^2 Q^2}{\lambda_1^2}</math>

<math>~\Rightarrow~~~ 0</math>

<math>~=</math>

<math>~Q^4 \biggl( \frac{q^2 y^2}{\lambda_1^2} \biggr) - Q^2 \biggl( \frac{x^2}{\lambda_1^2} \biggr) - 1</math>

 

<math>~=</math>

<math>~Q^4 - Q^2 \biggl( \frac{x^2}{q^2 y^2} \biggr) - \biggl( \frac{\lambda_1^2}{q^2 y^2} \biggr) \, ,</math>

where,

<math>~Q^2</math>

<math>~\equiv</math>

<math>~\frac{1 + q^2\tan^2\varphi}{q^2\tan^2\varphi} \, .</math>

Solving the quadratic equation, we have,

<math>~Q^2</math>

<math>~=</math>

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

 

<math>~=</math>

<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1 \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2} \biggr\} \, .</math>

Tentative Summary

<math>~\lambda_1</math>

<math>~\equiv</math>

<math>~(x^2 + q^2y^2 + p^2 z^2)^{1 / 2} \, ,</math>

<math>~\lambda_2</math>

<math>~\equiv</math>

<math>~ \frac{(x^2 + y^2)^{1 / 2}}{pz} \, ,</math>

<math>~\lambda_3 = Q^2</math>

<math>~\equiv</math>

<math>~\biggl( \frac{x^2}{2q^2 y^2} \biggr) \biggl\{ 1 \pm \biggl[ 1 + 4\biggl( \frac{\lambda_1^2 q^2 y^2}{x^4} \biggr) \biggr]^{1 / 2} \biggr\} \, .</math>

Partial Derivatives & Scale Factors

First Coordinate

<math>~\frac{\partial \lambda_1}{\partial x}</math>

<math>~=</math>

<math>~\frac{x}{\lambda_1} \, ,</math>

<math>~\frac{\partial \lambda_1}{\partial y}</math>

<math>~=</math>

<math>~\frac{q^2 y}{\lambda_1} \, ,</math>

<math>~\frac{\partial \lambda_1}{\partial z}</math>

<math>~=</math>

<math>~\frac{p^2 z}{\lambda_1} \, .</math>

<math>~h_1^{-2}</math>

<math>~=</math>

<math>~ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial \lambda_1}{\partial z} \biggr)^2 </math>

<math>~=</math>

<math>~ \biggl( \frac{x}{\lambda_1} \biggr)^2 + \biggl( \frac{q^2 y}{\lambda_1} \biggr)^2 + \biggl( \frac{p^2 z}{\lambda_1} \biggr)^2 \, .</math>

<math>~\Rightarrow ~~~ h_1</math>

<math>~=</math>

<math>~ \lambda_1 \ell_{3D} \, , </math>

where,

<math>\ell_{3D} \equiv (x^2 + q^4y^2 + p^4z^2)^{-1 / 2} \, .</math>

As a result, the associated unit vector is,

<math>~\hat{e}_1</math>

<math>~=</math>

<math>~ \hat{\imath} h_1 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr) + \hat{\jmath} h_1 \biggl( \frac{\partial \lambda_1}{\partial y} \biggr) + \hat{k} h_1 \biggl( \frac{\partial \lambda_1}{\partial z} \biggr) </math>

 

<math>~=</math>

<math>~ \hat{\imath} x \ell_{3D} + \hat{\jmath} q^2 y\ell_{3D} + \hat{k} p^2 z \ell_{3D} \, . </math>

Notice that,

<math>~\hat{e}_1 \cdot \hat{e}_1</math>

<math>~=</math>

<math>~ (x^2 + q^4 y^2 + p^4 z^2) \ell_{3D}^2 = 1 \, . </math>


Second Coordinate (1st Try)

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr] \, ,</math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ - \frac{(x^2 + y^2)^{1 / 2}}{pz^2} \, .</math>

<math>~h_2^{-2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl\{ \frac{1}{pz} \biggl[ \frac{x}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2 + \biggl\{ \frac{1}{pz} \biggl[ \frac{y}{(x^2 + y^2)^{1 / 2}} \biggr] \biggr\}^2 + \biggl\{ \frac{(x^2 + y^2)^{1 / 2}}{pz^2} \biggr\}^2 </math>

 

<math>~=</math>

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

 

<math>~=</math>

<math>~ \frac{1}{p^2 z^2} + \frac{(x^2 + y^2)}{p^2 z^4} = \frac{(x^2 + y^2 + z^2)}{p^2 z^4} </math>

<math>~\Rightarrow~~~h_2</math>

<math>~=</math>

<math>~ \frac{p z^2}{r } </math>

As a result, the associated unit vector is,

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr) + \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr) + \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr) </math>

 

<math>~=</math>

<math>~ \hat{\imath} \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr] + \hat{\jmath} \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr] - \hat{k} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr] \, . </math>

Notice that,

<math>~\hat{e}_2 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ \biggl[ \frac{x^2 z^2}{r^2(x^2 + y^2)} \biggr] + \biggl[ \frac{y^2 z^2}{r^2(x^2 + y^2)} \biggr] + \biggl[ \frac{(x^2 + y^2)}{r^2} \biggr] = 1 \, . </math>

Let's check to see if this "second" unit vector is orthogonal to the "first."

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ x\ell_{3D} \biggl[ \frac{xz}{r(x^2 + y^2)^{1 / 2}} \biggr] + q^2 y\ell_{3D} \biggl[ \frac{yz}{r(x^2 + y^2)^{1 / 2}} \biggr] - p^2 z \ell_{3D} \biggl[ \frac{(x^2 + y^2)^{1 / 2}}{r} \biggr] </math>

 

<math>~=</math>

<math>~ \ell_{3D} \biggl\{ \biggl[ \frac{x^2z}{r(x^2 + y^2)^{1 / 2}} \biggr] + \biggl[ \frac{q^2 y^2 z}{r(x^2 + y^2)^{1 / 2}} \biggr] - \biggl[ \frac{p^2 z(x^2 + y^2)^{1 / 2}}{r} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{z\ell_{3D}}{r (x^2 + y^2)^{1 / 2}} \biggl\{ \biggl[ x^2\biggr] + \biggl[ q^2 y^2 \biggr] - \biggl[ p^2 (x^2 + y^2) \biggr] \biggr\} </math>

 

<math>~\ne</math>

<math>~ 0 \ . </math>


Second Coordinate (2nd Try)

Let's try,

<math>~\lambda_2</math>

<math>~=</math>

<math>~ \biggl[\frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz} \biggr] \, , </math>

<math>~\Rightarrow~~~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{x}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{x}{p^2 z^2 \lambda_2} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{q^2 y}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } = \frac{q^2y}{p^2 z^2 \lambda_2} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ \frac{\mathfrak{f}\cdot p^2z}{pz(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2} } - \frac{(x^2 + q^2y^2 + \mathfrak{f}\cdot p^2 z^2)^{1 / 2}}{pz^2} = \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z \biggr) - \frac{\lambda_2 }{z} = \frac{1}{p^2z^2 \lambda_2 } \biggl( \mathfrak{f}\cdot p^2z - p^2z \lambda_2^2 \biggr) \, . </math>

Hence,

<math>~h_2^{-2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl[ \frac{x}{p^2 z^2 \lambda_2} \biggr]^2 + \biggl[ \frac{q^2y}{p^2 z^2 \lambda_2} \biggr]^2 + \biggl[ \frac{ \mathfrak{f} }{z \lambda_2 } - \frac{\lambda_2 }{z}\biggr]^2 </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{x^2 + q^4 y^2}{p^4 z^4 \lambda_2^2} \biggr] + \biggl[ \frac{1}{z\lambda_2}\biggl( \mathfrak{f} - \lambda_2^2 \biggr) \biggr]^2 </math>

 

<math>~=</math>

<math>~ \frac{1}{p^4 z^4 \lambda_2^2} \biggl[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 \biggr] </math>

<math>~\Rightarrow ~~~ h_2</math>

<math>~=</math>

<math>~ \frac{p^2 z^2 \lambda_2}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \, . </math>

So, the associated unit vector is,

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr) + \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr) + \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr) </math>

 

<math>~=</math>

<math>~ \hat{\imath} \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} + \hat{\jmath} \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} + \hat{k} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} \, . </math>

Checking orthogonality …

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ x\ell_{3D} \biggl\{ \frac{x}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} + q^2 y\ell_{3D} \biggl\{ \frac{q^2 y}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} + p^2 z \ell_{3D} \biggl\{ \frac{p^2z(\mathfrak{f}-\lambda_2^2)}{[ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, . </math>

 

<math>~=</math>

<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, . </math>

If <math>~\mathfrak{f} = 0</math>, we have …

<math>~p^2 z (\mathfrak{f} - \lambda_2^2) </math>

    <math>~~~\rightarrow ~~~ </math>   

<math>~ \biggl[- p^2 z \lambda_2^2\biggr]_{\mathfrak{f} = 0} = - p^2 z\biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f} \cdot }p^2 z^2 )^{1 / 2}}{pz} \biggr]^2 = - \frac{(x^2 + q^2y^2 )}{z} \, ,</math>

which, in turn, means …

<math>~[ x^2 + q^4 y^2 + p^4 z^2 (\cancelto{0}{\mathfrak{f}} - \lambda_2^2)^2 ]^{1 / 2}</math>

<math>~=</math>

<math>~ [ x^2 + q^4 y^2 + p^4 z^2 \lambda_2^4 ]^{1 / 2} </math>

 

<math>~=</math>

<math>~ \biggl\{ x^2 + q^4 y^2 + p^4 z^2 \biggl[\frac{(x^2 + q^2y^2 + \cancelto{0}{\mathfrak{f}\cdot} p^2 z^2)^{1 / 2}}{pz} \biggr]^4 \biggr\}^{1 / 2} </math>

 

<math>~=</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

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

and,

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~\frac{\ell_{3D}}{ [ x^2 + q^4 y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)^2 ]^{1 / 2} } \biggl\{ x^2 + q^4y^2 + p^4 z^2 (\mathfrak{f} - \lambda_2^2)\biggr\} \, . </math>

Speculation6

Determine λ2

This is very similar to the above, Speculation2. Try,

<math>~\lambda_2</math>

<math>~=</math>

<math>~ \frac{x y^{1/q^2}}{ z^{2/p^2}} \, , </math>

in which case,

<math>~\frac{\partial \lambda_2}{\partial x}</math>

<math>~=</math>

<math>~ \frac{\lambda_2}{x} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial y}</math>

<math>~=</math>

<math>~ \frac{x}{z^{2/p^2}} \biggl(\frac{1}{q^2}\biggr) y^{1/q^2 - 1} = \frac{\lambda_2}{q^2 y} \, , </math>

<math>~\frac{\partial \lambda_2}{\partial z}</math>

<math>~=</math>

<math>~ -\frac{2\lambda_2}{p^2 z} \, . </math>

The associated scale factor is, then,

<math>~h_2</math>

<math>~=</math>

<math>~\biggl[ \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 \biggr]^{-1 / 2} </math>

 

<math>~=</math>

<math>~\biggl[ \biggl( \frac{ \lambda_2}{x} \biggr)^2 + \biggl( \frac{\lambda_2}{q^2y} \biggr)^2 + \biggl( - \frac{2\lambda_2}{p^2z} \biggr)^2 \biggr]^{-1 / 2} </math>

 

<math>~=</math>

<math>~\frac{1}{\lambda_2}\biggl[ \frac{ 1}{x^2} + \frac{1}{q^4y^2} + \frac{4}{p^4z^2} \biggr]^{-1 / 2} </math>

 

<math>~=</math>

<math>~\frac{1}{\lambda_2}\biggl[ \frac{ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2}{x^2 q^4 y^2 p^4 z^2} \biggr]^{-1 / 2} </math>

 

<math>~=</math>

<math>~ \frac{1}{\lambda_2}\biggl[ \frac{x q^2 y p^2 z}{ \mathcal{D}} \biggr] \, . </math>

where,

<math>~\mathcal{D}</math>

<math>~\equiv</math>

<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>

The associated unit vector is, then,

<math>~\hat{e}_2</math>

<math>~=</math>

<math>~ \hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr) + \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr) + \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr) </math>

 

<math>~=</math>

<math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl\{ \hat{\imath} \biggl( \frac{1}{x} \biggr) + \hat{\jmath} \biggl( \frac{1}{q^2 y} \biggr) + \hat{k} \biggl( -\frac{2}{p^2 z} \biggr) \biggr\} \ . </math>

Recalling that the unit vector associated with the "first" coordinate is,

<math>~\hat{e}_1 </math>

<math>~\equiv</math>

<math>~ \hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat\jmath (p^2 z \ell_{3D}) \, , </math>

where,

<math>~\ell_{3D}</math>

<math>~=</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>

let's check to see whether the "second" unit vector is orthogonal to the "first."

<math>~\hat{e}_1 \cdot \hat{e}_2</math>

<math>~=</math>

<math>~ \frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl[ 1 + 1 - 2 \biggr] = 0 \, . </math>

Hooray!

Direction Cosines for Third Unit Vector

Now, what is the unit vector, <math>~\hat{e}_3</math>, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?

<math>~\hat{e}_3 \equiv \hat{e}_1 \times \hat{e}_2</math>

<math>~=</math>

<math>~ \hat\imath \biggl[ ( e_{1y} )( e_{2z}) - ( e_{2y} )( e_{1z}) ) \biggl] + \hat\jmath \biggl[ ( e_{1z} )( e_{2x}) - ( e_{2z} )( e_{1x}) ) \biggl] + \hat{k} \biggl[ ( e_{1x} )( e_{2y}) - ( e_{2x} )( e_{1y}) ) \biggl] </math>

 

<math>~=</math>

<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl\{ \hat\imath \biggl[ \biggl( -\frac{2 q^2y}{p^2 z} \biggr) - \biggl( \frac{p^2z}{q^2y} \biggr) \biggl] + \hat\jmath \biggl[ \biggl( \frac{p^2z}{x} \biggr) - \biggl(-\frac{2x}{p^2z} \biggr) \biggl] + \hat{k} \biggl[ \biggl( \frac{x}{q^2y} \biggr) - \biggl( \frac{q^2y}{x} \biggr) \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl\{ -\hat\imath \biggl[ \frac{2 q^4y^2 + p^4z^2}{q^2 y p^2 z} \biggl] + \hat\jmath \biggl[ \frac{p^4z^2 + 2x^2}{xp^2 z} \biggl] + \hat{k} \biggl[ \frac{x^2 - q^4y^2}{x q^2y} \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ -\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl] + \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl] + \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl] \biggr\} \, . </math>

Is this a valid unit vector? First, note that …

<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}</math>

<math>~=</math>

<math>~ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) (x^2 + q^4y^2 + p^4 z^2 ) </math>

 

<math>~=</math>

<math>~ (x^2 q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2) + (q^8 y^4 p^4 z^2 + x^2 q^4y^2 p^4 z^2 + 4x^2q^8y^4) +(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 + 4x^2q^4y^2 p^4 z^2) </math>

 

<math>~=</math>

<math>~ 6x^2 q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4 q^4y^2) + q^8 y^4(p^4 z^2 + 4x^2) +p^8z^4(x^2 + q^4 y^2 )\, . </math>

Then we have,

<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}\hat{e}_3 \cdot \hat{e}_3</math>

<math>~=</math>

<math>~ \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2 + \biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl]^2 + \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2 </math>

 

<math>~=</math>

<math>~ x^2(4 q^8y^4 + 4q^4y^2p^4z^2 + p^8z^4 ) + q^4 y^2(p^8z^4 + 4x^2p^4z^2 + 4x^4 ) + p^4z^2( x^4 - 2x^2q^4 y^2 + q^8y^4 ) </math>

 

<math>~=</math>

<math>~ 4 x^2 q^8y^4 + 4x^2 q^4y^2p^4z^2 + x^2 p^8z^4 + q^4 y^2p^8z^4 + 4x^2q^4 y^2p^4z^2 + 4x^4q^4 y^2 + x^4p^4z^2 - 2x^2q^4 y^2p^4z^2 + q^8y^4p^4z^2 </math>

 

<math>~=</math>

<math>~ 6x^2 q^4y^2p^4z^2 + p^8z^4 (x^2 +q^4 y^2) + x^4(4q^4 y^2 + p^4z^2) + q^8 y^4(4 x^2 + p^4z^2 ) </math>

 

<math>~=</math>

<math>~ \biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} \, , </math>

which means that, <math>~\hat{e}_3\cdot \hat{e}_3 = 1</math>.    Hooray! Again (11/11/2020)!

Direction Cosine Components for T6 Coordinates
<math>~n</math> <math>~\lambda_n</math> <math>~h_n</math> <math>~\frac{\partial \lambda_n}{\partial x}</math> <math>~\frac{\partial \lambda_n}{\partial y}</math> <math>~\frac{\partial \lambda_n}{\partial z}</math> <math>~\gamma_{n1}</math> <math>~\gamma_{n2}</math> <math>~\gamma_{n3}</math>
<math>~1</math> <math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math> <math>~\lambda_1 \ell_{3D}</math> <math>~\frac{x}{\lambda_1}</math> <math>~\frac{q^2 y}{\lambda_1}</math> <math>~\frac{p^2 z}{\lambda_1}</math> <math>~(x) \ell_{3D}</math> <math>~(q^2 y)\ell_{3D}</math> <math>~(p^2z) \ell_{3D}</math>
<math>~2</math> <math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math> <math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math> <math>~\frac{\lambda_2}{x}</math> <math>~\frac{\lambda_2}{q^2 y}</math> <math>~-\frac{2\lambda_2}{p^2 z}</math> <math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math> <math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math> <math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math>
<math>~3</math> --- --- --- --- --- <math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math> <math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl]</math> <math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math>

<math>~\ell_{3D}</math>

<math>~\equiv</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>

<math>~\mathcal{D}</math>

<math>~\equiv</math>

<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>


Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.

<math>~\hat{e}_1 \cdot \hat{e}_3</math>

<math>~=</math>

<math>~\frac{\ell_{3D}^2}{\mathcal{D}} \biggl\{ -x \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl] + q^2 y \biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl] + p^2 z \biggl[ p^2z( x^2 - q^4y^2 ) \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{\ell_{3D}^2}{\mathcal{D}} \biggl\{ - (2 x^2q^4y^2 + x^2p^4z^2 ) + (q^4 y^2 p^4z^2 + 2x^2 q^4 y^2) + ( x^2p^4z^2 - q^4y^2 p^4z^2 ) \biggr\} </math>

 

<math>~=</math>

<math>~ 0 \, , </math>

and,

<math>~\hat{e}_2 \cdot \hat{e}_3</math>

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathcal{D}} \cdot \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl\{ - \biggl[ (2 q^4y^2 + p^4z^2 ) \biggl] + \biggl[ (p^4z^2 + 2x^2 ) \biggl] - \biggl[ 2( x^2 - q^4y^2 ) \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~ 0 \, . </math>

Q. E. D.

Search for Third Coordinate Expression

Let's try …

<math>~\lambda_3</math>

<math>~=</math>

<math>~\mathcal{D}^n \ell_{3D}^m </math>

 

<math>~=</math>

<math>~ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{n / 2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2} </math>

<math>~\Rightarrow ~~~ \frac{\partial \lambda_3}{\partial x_i}</math>

<math>~=</math>

<math>~ \ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{n / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr] </math>

 

 

<math>~ + \mathcal{D}^n \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr] </math>

 

<math>~=</math>

<math>~ \mathcal{D}^n \ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr] </math>

 

 

<math>~ + \mathcal{D}^n \ell_{3D}^m \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr] \, . </math>

Hence,

<math>~\frac{1}{\mathcal{D}^n \ell_{3D}^m} \cdot \frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~ \frac{n}{2\mathcal{D}^2}\frac{\partial}{\partial x} \biggl[q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr] - \frac{m \ell_{3D}^2}{2} \frac{\partial}{\partial x} \biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr] </math>

 

<math>~=</math>

<math>~x \biggl\{ \frac{n}{\mathcal{D}^2}\biggl[p^4 z^2 + 4q^4y^2 \biggr] - m \ell_{3D}^2 \biggr\} </math>

 

<math>~=</math>

<math>~x \biggl\{ \frac{n (p^4 z^2 + 4q^4y^2)}{ ( q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 ) } - \frac{m}{ ( x^2 + q^4y^2 + p^4 z^2 ) } \biggr\} </math>

This is overly cluttered! Let's try, instead …


<math>~A \equiv \ell_{3D}^{-2}</math>

<math>~=</math>

<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>

      and,      

<math>~B \equiv \mathcal{D}^2</math>

<math>~=</math>

<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>

<math>~\Rightarrow ~~~ \frac{\partial A}{\partial x}</math>

<math>~=</math>

<math>~2x \, ,</math>

<math>~\frac{\partial A}{\partial y}</math>

<math>~=</math>

<math>~2q^4 y \, ,</math>

<math>~\frac{\partial A}{\partial z}</math>

<math>~=</math>

<math>~ 2p^4 z\, ;</math>

<math>~ \frac{\partial B}{\partial x}</math>

<math>~=</math>

<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>

<math>~\frac{\partial B}{\partial y}</math>

<math>~=</math>

<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>

<math>~\frac{\partial B}{\partial z}</math>

<math>~=</math>

<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>


Now, let's assume that,

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~\biggl( \frac{A}{B} \biggr)^{1 / 2} \, ,</math>

<math>~\Rightarrow~~~ \frac{ \partial \lambda_3}{\partial x_i}</math>

<math>~=</math>

<math>~ \frac{1}{2 (AB)^{1 / 2}} \cdot \frac{\partial A}{\partial x_i} - \frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_i} </math>

 

<math>~=</math>

<math>~\frac{\lambda_3}{2AB} \biggl[ B \cdot \frac{\partial A}{\partial x_i} - A \cdot \frac{\partial B}{\partial x_i} \biggr] \, . </math>

<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>

<math>~=</math>

<math>~ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \cdot \frac{\partial A}{\partial x_i} - (x^2 + q^4y^2 + p^4 z^2 ) \cdot \frac{\partial B}{\partial x_i} \, . </math>

Looking ahead …

<math>~h_3^{-2}</math>

<math>~=</math>

<math>~ \biggl\{ \frac{\lambda_3}{2AB} \biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr] \biggr\}^2 + \biggl\{ \frac{\lambda_3}{2AB} \biggl[ B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y} \biggr] \biggr\}^2 + \biggl\{ \frac{\lambda_3}{2AB} \biggl[ B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z} \biggr] \biggr\}^2 </math>

<math>~\Rightarrow ~~~ \biggl[\frac{2AB}{\lambda_3} \biggr]^2 h_3^{-2}</math>

<math>~=</math>

<math>~ \biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z} \biggr]^2 </math>

<math>~\Rightarrow ~~~ \biggl[\frac{\lambda_3}{2AB} \biggr] h_3</math>

<math>~=</math>

<math>~ \biggl\{\biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z} \biggr]^2 \biggr\}^{-1 / 2} </math>

Then, for example,

<math>~\gamma_{31} \equiv h_3 \biggl(\frac{\partial \lambda_3}{\partial x} \biggr)</math>

<math>~=</math>

<math>~\biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr] \biggl\{\biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y} \biggr]^2 + \biggl[ B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z} \biggr]^2 \biggr\}^{-1 / 2} </math>

As a result, we have,

<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>

<math>~=</math>

<math>~2x \biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - (x^2 + q^4y^2 + p^4 z^2 ) ( 4q^4y^2 + p^4 z^2 ) \biggr] </math>

 

<math>~=</math>

<math>~2x \biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - (4x^2q^4y^2 + x^2 p^4 z^2 + 4q^8y^4 + q^4y^2 p^4z^2 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr] </math>

 

<math>~=</math>

<math>~2x \biggl[ - (4q^8y^4 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr] </math>

 

<math>~=</math>

<math>~-2x (2q^4y^2 + p^4z^2)^2 </math>

 

<math>~=</math>

<math>~-8x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)^2 </math>

<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{x}}</math>

<math>~=</math>

<math>~-\biggl[ 2x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2 \, ; </math>

and,

<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>

<math>~=</math>

<math>~ 2q^4y\biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - (x^2 + q^4y^2 + p^4 z^2 ) (p^4 z^2 + 4x^2) \biggr] </math>

 

<math>~=</math>

<math>~ 2q^4y\biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - ( x^2p^4z^2 + 4x^4 + q^4y^2p^4z^2 + 4x^2q^4y^2 + p^8z^4 + 4x^2p^4z^2) \biggr] </math>

 

<math>~=</math>

<math>~ -2q^4y( 4x^4 + p^8z^4 + 4x^2p^4z^2) </math>

 

<math>~=</math>

<math>~ -2q^4y( 2x^2 + p^4z^2 )^2 </math>

<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{y} }</math>

<math>~=</math>

<math>~ - \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \, ; </math>

and,

<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>

<math>~=</math>

<math>~2p^4 z \biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - (x^2 + q^4y^2 + p^4 z^2 ) (q^4 y^2 + x^2) \biggr] </math>

 

<math>~=</math>

<math>~2p^4 z \biggl[ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) - (x^2q^4y^2 + x^4 + q^8y^4 + x^2q^4y^2 + q^4y^2p^4z^2 + x^2p^4z^2) \biggr] </math>

 

<math>~=</math>

<math>~2p^4 z \biggl[ ( 2x^2q^4y^2) - ( x^4 + q^8y^4 ) \biggr] </math>

 

<math>~=</math>

<math>~-2p^4 z \biggl[ x^4 + q^8y^4 - 2x^2q^4y^2 \biggr] </math>

 

<math>~=</math>

<math>~-2p^4 z (x^2 - q^4y^2 )^2 </math>

<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln \lambda_3}{\partial \ln{z}}</math>

<math>~=</math>

<math>~-4 \biggl[ \biggl( \frac{p^4 z^2}{4} \biggr) (x^2 - q^4y^2 )^2 \biggr] </math>

 

<math>~=</math>

<math>~-\biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \, . </math>

Wow!   Really close! (13 November 2020)


Just for fun, let's see what we get for <math>~h_3</math>. It is given by the expression,

<math>~h_3^{-2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl\{ \frac{\lambda_3}{ABx} \biggl[ 2x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2 \biggr\}^2 +\biggl\{ \frac{\lambda_3}{ABy} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \biggr\}^2 +\biggl\{ \frac{\lambda_3}{ABz} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \biggr\}^2 </math>

<math>~\Rightarrow~~~ \biggl[ \frac{AB}{\lambda_3}\biggr]^2 h_3^{-2}</math>

<math>~=</math>

<math>~ \biggl\{ \frac{1}{x^2} \biggl[ 2x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)\biggr]^4 \biggr\} +\biggl\{ \frac{1}{y^2} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^4 \biggr\} +\biggl\{ \frac{1}{z^2} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^4 \biggr\} </math>

Fiddle Around

Let …

<math>~\mathcal{L}_x</math>

<math>~\equiv</math>

<math>~ - \biggl[ B \cdot \frac{\partial A}{\partial x} - A \cdot \frac{\partial B}{\partial x} \biggr] </math>

<math>~=</math>

<math>~8x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)^2 </math>

<math>~=</math>

<math>~\frac{2}{x} \biggl[ 2x \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2 </math>

<math>~=</math>

<math>~8x~\mathfrak{F}_x(y,z) </math>

<math>~\mathcal{L}_y</math>

<math>~\equiv</math>

<math>~ - \biggl[ B \cdot \frac{\partial A}{\partial y} - A \cdot \frac{\partial B}{\partial y} \biggr] </math>

<math>~=</math>

<math>~ 8q^4y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2 </math>

<math>~=</math>

<math>~ \frac{2}{y} \biggl[ 2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2 </math>

<math>~=</math>

<math>~8y~\mathfrak{F}_y(x,z) </math>

<math>~\mathcal{L}_z</math>

<math>~\equiv</math>

<math>~ -\biggl[ B \cdot \frac{\partial A}{\partial z} - A \cdot \frac{\partial B}{\partial z} \biggr] </math>

<math>~=</math>

<math>~ 2p^4 z \biggl(x^2 - q^4y^2 \biggr)^2 </math>

<math>~=</math>

<math>~ \frac{2}{z} \biggl[p^2 z \biggl(x^2 - q^4y^2 \biggr)\biggr]^2 </math>

<math>~=</math>

<math>~8z~\mathfrak{F}_z(x,y) </math>

With this shorthand in place, we can write,

<math>~\hat{e}_3</math>

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ -\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl] + \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl] + \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{(AB)^{1 / 2}} \biggl\{ -\hat\imath \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2} + \hat\jmath \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2} + \hat{k} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2} \biggr\} \, . </math>

We therefore also recognize that,

<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>

<math>~=</math>

<math>~ -\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ -\frac{1}{(AB)^{1 / 2}} \biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} \, , </math>

<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>

<math>~=</math>

<math>~ \frac{1}{(AB)^{1 / 2}} \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ \frac{1}{(AB)^{1 / 2}} \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} \, , </math>

<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>

<math>~=</math>

<math>~ \frac{1}{(AB)^{1 / 2}} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ \frac{1}{(AB)^{1 / 2}} \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} \, . </math>

Now, if — and it is a BIG "if" — <math>~h_3 = h_0(AB)^{-1 / 2}</math>, then we have,

<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>

<math>~=</math>

<math>~ -\biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ -2x \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} \, , </math>

<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>

<math>~=</math>

<math>~ \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ 2y \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} \, , </math>

<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>

<math>~=</math>

<math>~ \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} </math>

<math>~=</math>

<math>~ 2z\biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, , </math>

<math>~\Rightarrow~~~ h_0 \lambda_3</math>

<math>~=</math>

<math>~ -x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} + y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} + z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, . </math>

But if this is the correct expression for <math>~\lambda_3</math> and its three partial derivatives, then it must be true that,

<math>~h_3^{-2}</math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ \biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>

<math>~=</math>

<math>~ 4x^2 \biggl[ \mathfrak{F}_x \biggl] + 4y^2 \biggl[ \mathfrak{F}_y \biggl] + 4z^2\biggl[ \mathfrak{F}_z \biggl] </math>

 

<math>~=</math>

<math>~ 4x^2 \biggl[ \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)^2 \biggl] + 4y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl] + 4z^2\biggl[ \frac{p^4}{4}\biggl(x^2 - q^4y^2 \biggr)^2 \biggl] </math>

 

<math>~=</math>

<math>~ x^2 (2q^4y^2 + p^4z^2 )^2 + q^4 y^2( 2x^2 + p^4z^2 )^2 + p^4 z^2 (x^2 - q^4y^2 )^2 </math>

Well … the right-hand side of this expression is identical to the right-hand side of the above expression, where we showed that it equals <math>~(\ell_{3D}/\mathcal{D})^{-2}</math>. That is to say, we are now showing that,

<math>~\biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>

<math>~=</math>

<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = [AB]</math>

<math>~\Rightarrow ~~~ \frac{h_3}{h_0}</math>

<math>~=</math>

<math>~(AB)^{-1 / 2} \, .</math>

And this is precisely what, just a few lines above, we hypothesized the functional expression for <math>~h_3</math> ought to be. EUREKA!

Summary

In summary, then …

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~ -x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} + y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} + z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} </math>

 

<math>~=</math>

<math>~ -x^2 \biggl[\biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr)^2 \biggl]^{1 / 2} + y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]^{1 / 2} + z^2 \biggl[ \frac{p^4}{4} \biggl(x^2 - q^4y^2 \biggr)^2 \biggl]^{1 / 2} </math>

 

<math>~=</math>

<math>~ -x^2 \biggl(q^4y^2 + \frac{p^4z^2}{2} \biggr) + q^2y^2 \biggl( x^2 + \frac{p^4z^2}{2} \biggr) + \frac{p^2 z^2}{2} \biggl(x^2 - q^4y^2 \biggr) \, , </math>

and,

<math>~h_3</math>

<math>~=</math>

<math>~(AB)^{-1 / 2} </math>

 

<math>~=</math>

<math>~\biggl[ (x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \biggr]^{-1 / 2} \, .</math>

No! Once again this does not work. The direction cosines — and, hence, the components of the <math>~\hat{e}_3</math> unit vector — are not correct!

See Also

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation