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

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<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
<table border="1" align="center" width="80%" cellpadding="10"><tr><td align="left">
<div align="center">
<div align="center">
'''Tangent Plane'''
'''Tangent Plane'''<br />
(See, for example, [http://math.furman.edu/~dcs/courses/math21/ Dan Sloughter's] ([https://www.furman.edu Furman University]) 2001 Calculus III  class lecture notes &#8212; specifically [http://math.furman.edu/~dcs/courses/math21/lectures/l-15.pdf Lecture 15])
</div>
</div>
----


The two-dimensional plane that is tangent to the <math>~\lambda_1</math> = constant surface ''at this point'' is given by the expression,
The two-dimensional plane that is tangent to the <math>~\lambda_1</math> = constant surface ''at this point'' is given by the expression,
Line 5,668: Line 5,671:
==Best Thus Far==
==Best Thus Far==


===Part A===


<table border="1" cellpadding="8" align="center">
<table border="1" cellpadding="8" align="center">
Line 5,955: Line 5,959:
</table>
</table>


Now, from [[#Eureka|above]], we know that,
The scale factor is, then,
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = AB</math>
<math>~h_3^{-2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,967: Line 5,971:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
\sum_{i=1}^3 \biggl( \frac{\partial\lambda_3}{\partial x_i}\biggr)^2
+
\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
+
\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2 \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


Hence,
<tr>
<table border="0" cellpadding="5" align="center">
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{i=1}^3 \biggl\{
\biggl[ \frac{m\lambda_3}{2AB} \biggr] \biggl[
A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i}
\biggr]
\biggr\}^2
</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~AB</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,989: Line 6,003:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln x}
\biggl[ \frac{m\lambda_3}{AB} \biggr]^2
\biggl\{  
\biggl[
x(2q^4 y^2 + p^4z^2)^2
\biggr]^2
+
+
\biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln y}
\biggl[  
q^4y(2x^2 + p^4 z^2)^2
\biggr]^2
+
+
\biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln z}
\biggl[  
p^4 z(x^2 - q^4y^2)^2
\biggr]^2
\biggr\}
</math>
</math>
   </td>
   </td>
Line 6,000: Line 6,023:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ m</math>
<math>~\Rightarrow~~~h_3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 6,007: Line 6,030:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{\partial \ln \lambda_3}{\partial \ln x}
\biggl[ \frac{AB}{m\lambda_3} \biggr]
\biggl\{  
\biggl[
x(2q^4 y^2 + p^4z^2)^2
\biggr]^2
+
+
\frac{\partial \ln \lambda_3}{\partial \ln y}
\biggl[
q^4y(2x^2 + p^4 z^2)^2
\biggr]^2
+
\biggl[
p^4 z(x^2 - q^4y^2)^2
\biggr]^2
\biggr\}^{-1 / 2} \, .
</math>
  </td>
</tr>
</table>
 
===Part B (25 February 2021)===
 
Now, from [[#Eureka|above]], we know that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} = AB</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
 
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="4">
'''Example:'''  <br /><math>~(q^2, p^2) = (2, 5)</math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>
  </td>
</tr>
 
<tr>
  <td align="center"><math>~\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2</math></td>
  <td align="center"><math>~\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2</math></td>
  <td align="center"><math>~\biggl[ p^2z( x^2 - q^4y^2 )  \biggl]^2</math></td>
  <td align="center"><math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math></td>
</tr>
 
<tr>
  <td align="center">2.14037</td>
  <td align="center">1.39187</td>
  <td align="center">0.04623</td>
  <td align="center">3.57847</td>
</tr>
</table>
 
As an aside, note that,
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~AB</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln x}
+
\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln y}
+
\biggl[ \frac{AB}{m} \biggr]  \frac{\partial \ln \lambda_3}{\partial \ln z}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ m</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\partial \ln \lambda_3}{\partial \ln x}
+
\frac{\partial \ln \lambda_3}{\partial \ln y}
+
+
\frac{\partial \ln \lambda_3}{\partial \ln z} \, .
\frac{\partial \ln \lambda_3}{\partial \ln z} \, .
</math>
  </td>
</tr>
</table>
We realize that this ratio of lengths may also be written in the form,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6x^2 q^4y^2 p^4 z^2
+
x^4(4q^4y^2 + p^4z^2)
+
q^8y^4(4x^2 + p^4z^2)
+
p^8z^4(x^2 + q^4y^2) \, .
</math>
  </td>
</tr>
</table>
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="5">
'''Same Example, but Different Expression:'''  <br /><math>~(q^2, p^2) = (2, 5)</math> &nbsp; &nbsp; and &nbsp; &nbsp; <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>
  </td>
</tr>
<tr>
  <td align="center"><math>~6x^2 q^4y^2 p^4 z^2</math></td>
  <td align="center"><math>~x^4(4q^4y^2 + p^4z^2)</math></td>
  <td align="center"><math>~q^8y^4(4x^2 + p^4z^2)</math></td>
  <td align="center"><math>~p^8z^4(x^2 + q^4y^2)</math></td>
  <td align="center"><math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math></td>
</tr>
<tr>
  <td align="center">0.67620</td>
  <td align="center">0.94359</td>
  <td align="center">1.87054</td>
  <td align="center">0.08813</td>
  <td align="center">3.57847</td>
</tr>
</table>
Let's try &hellip;
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial x}</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>~\frac{\partial \lambda_5}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2 y(p^4z^2 + 2x^2 ) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~p^2z( x^2 - q^4y^2 ) \, .</math>
  </td>
</tr>
</table>
This means that the relevant scale factor is,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~h_5^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ -x(2 q^4y^2 + p^4z^2 ) \biggr]^2
+
\biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggr]^2
+
\biggl[ p^2z( x^2 - q^4y^2 ) \biggr]^2
=
\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~h_5</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,
</math>
  </td>
</tr>
</table>
and the three associated direction cosines are,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\gamma_{51} = h_5 \biggl( \frac{\partial \lambda_5}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-x(2 q^4y^2 + p^4z^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\gamma_{52} = h_5 \biggl( \frac{\partial \lambda_5}{\partial y} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2 y(p^4z^2 + 2x^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\gamma_{53} = h_5 \biggl( \frac{\partial \lambda_5}{\partial z} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~p^2z( x^2 - q^4y^2 )\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr) \, .</math>
  </td>
</tr>
</table>
<span id="PartBCoordinatesT10">These direction cosines</span> exactly match what is required in order to ensure that the coordinate, <math>~\lambda_5</math>, is everywhere orthogonal to both <math>~\lambda_1</math> and <math>~\lambda_4</math>.  <font color="red">'''GREAT!'''</font>  The resulting summary table is, therefore:
<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T10 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\lambda_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
  <td align="center"><math>~\gamma_{n1}</math></td>
  <td align="center"><math>~\gamma_{n2}</math></td>
  <td align="center"><math>~\gamma_{n3}</math></td>
</tr>
<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
  <td align="center"><math>~(x) \ell_{3D}</math></td>
  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
</tr>
<tr>
  <td align="center"><math>~4</math></td>
  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
</tr>
<tr>
  <td align="center"><math>~5</math></td>
  <td align="center">---</td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}</math></td>
  <td align="center"><math>~-x(2 q^4y^2 + p^4z^2)</math></td>
  <td align="center"><math>~q^2 y(p^4z^2 + 2x^2 )</math></td>
  <td align="center"><math>~p^2z( x^2 - q^4y^2 )</math></td>
  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
</tr>
<tr>
  <td align="left" colspan="9">
<table border="0" cellpadding="8" align="center">
<tr>
  <td align="right">
<math>~\ell_{3D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
  </td>
</tr>
</table>
  </td>
</tr>
</table>
Try &hellip;
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\lambda_5</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^{-2q^4} \cdot y^{2q^2}
+
y^{q^2p^4} \cdot z^{-q^4p^2}
+
x^{-p^4} \cdot z^{p^2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{ y^{2q^2} }{ x^{2q^4} }
+
\frac{ y^{q^2p^4} }{ z^{q^4p^2} }
+
\frac{ z^{p^4} }{ x^{p^2} }
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^{2q^4 + p^2} z^{q^4p^2} }\biggl\{
[x^{p^2}] y^{2q^2} [z^{q^4p^2}]
+
[x^{2q^4 + p^2}]y^{q^2p^4}
+
[x^{2q^4}]z^{p^4+q^4p^2}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^{2q^4 + p^2} z^{q^4p^2} }\biggl\{
\mathfrak{F_5}
\biggr\} \, .
</math>
  </td>
</tr>
</table>
This gives,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{2q^4}{x}\biggl( \frac{ y^{2q^2} }{ x^{2q^4} } \biggr)
- \frac{p^4}{x}\biggl( \frac{ z^{p^2} }{ x^{p^2} } \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{x^{2q^4 + p^2 + 1}}\biggl[
2q^4y^{2q^2} x^{p^2}
+ p^4 x^{2q^4}z^{p^2}
\biggr] \, .
</math>
  </td>
</tr>
</table>
Or, given that,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x^{2q^4 + p^2 + 1} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x}{ z^{q^4p^2} }\biggl\{
\frac{\mathfrak{F_5}}{\lambda_5}
\biggr\}
\, ,
</math>
  </td>
</tr>
</table>
we can also write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{ z^{q^4p^2} }{x}\biggl\{
\frac{\lambda_5}{\mathfrak{F_5}}
\biggr\}
\biggl[ 2q^4y^{2q^2} x^{p^2} + p^4 x^{2q^4}z^{p^2} \biggr]
</math>
  </td>
</tr>
</table>
Similarly,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2q^2}{y} \biggl( \frac{ y^{2q^2} }{ x^{2q^4} } \biggr)
+
\frac{q^2p^4}{y} \biggl( \frac{ y^{q^2p^4} }{ z^{q^4p^2} } \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{ x^{2q^4} z^{q^4p^2} }
\biggl[ \frac{2q^2}{y} \biggl( y^{2q^2} z^{q^4p^2} \biggr)
+
\frac{q^2p^4}{y} \biggl( y^{q^2p^4} x^{2q^4} \biggr)\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{x^{p^2}}{ y } \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\}
\biggl[ 2q^2 \biggl( y^{2q^2} z^{q^4p^2} \biggr)
+
q^2p^4 \biggl( y^{q^2p^4} x^{2q^4} \biggr)\biggr] \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\partial \lambda_5}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{q^4p^2}{z} \biggl( \frac{ y^{q^2p^4} }{ z^{q^4p^2} } \biggr)
+
\frac{p^4}{z} \biggl( \frac{ z^{p^4} }{ x^{p^2} } \biggr)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ x^{p^2} z^{q^4p^2 }} \biggl[
\frac{p^4}{z} \biggl( z^{p^4+q^4p^2}  \biggr)
- \frac{q^4p^2}{z} \biggl( x^{p^2} y^{q^2p^4} \biggr) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x^{2q^4} }{z} \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\} \biggl[
p^4 \biggl( z^{p^4 + q^4p^2 }  \biggr)
-
q^4p^2 \biggl( x^{p^2} y^{q^2p^4}  \biggr)\biggr]
</math>
  </td>
</tr>
</table>
===Understanding the Volume Element===
Let's see if the expression for the volume element makes sense; that is, does
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~dx dy dz \, ?</math>
  </td>
</tr>
</table>
First, let's make sure that we understand how to relate the components of the Cartesian line element with the components of our T10 coordinates.
====Line Element====
MF53 claim that the following relation gives the various expressions for the scale factors; we will go ahead and incorporate the expectation that, since our coordinate system is orthogonal, the off-diagonal elements are zero.
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ds^2 = dx^2 + dy^2 + dz^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\sum_{i=1,4,5} h_i^2 d\lambda_i^2 \, .
</math>
  </td>
</tr>
</table>
Let's see.  The first term on the RHS is,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~h_1^2 d\lambda_1^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_1^2 \biggl[ 
\biggl( \frac{\partial \lambda_1}{\partial x}\biggr)dx
+
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)dy
+
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)dz
\biggr]^2
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_1^2 \biggl[ 
\biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2 dx^2
+
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 dy^2
+
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 dz^2
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dx~dy
+
\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)} dx~dz
+
\cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dy~dz
\biggr] \, ;
</math>
  </td>
</tr>
</table>
the other two terms assume easily deduced, similar forms.  When put together and after regrouping terms, we can write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\sum_{i=1,4,5} h_i^2 d\lambda_i^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2
+
h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2
+
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
\biggr] dx^2
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
+
h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2
+
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
\biggr] dy^2
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 
h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
+
h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2
+
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
\biggr] dz^2 \, .
</math>
  </td>
</tr>
</table>
Given that this summation should also equal the square of the Cartesian line element, <math>~(dx^2 + dy^2 + dz^2)</math>, we conclude that the three terms enclosed inside each of the pair of brackets must sum to unity.  Specifically, from the coefficient of <math>~dx^2</math>, we can write,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1
-
h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \, .
</math>
  </td>
</tr>
</table>
Using this relation to replace <math>~h_1^2</math> in each of the other two bracketed expressions, we find for the coefficients of <math>~dy^2</math> and <math>~dz^2</math>, respectively,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[
1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
\biggr]
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1
-
h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
\biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, ;
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl[
1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2
\biggr]
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 1
-
h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
\biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, .
</math>
  </td>
</tr>
</table>
We can use the first of these two expressions to solve for <math>~h_4^2</math> in terms of <math>~h_5^2</math>, namely,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 
- h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 
- h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~
h_4^2
\biggl[
\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 
+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
</math>
  </td>
</tr>
</table>
Analogously, the second of these two expressions gives,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
h_4^2 \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
</math>
  </td>
</tr>
</table>
Eliminating <math>~h_4</math> between the two gives the desired overall expression for <math>~h_5</math>, namely,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggr]
\biggl[
\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-
\biggl[
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 
+ h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-
h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggr]
\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
h_5^2 \biggl\{
\biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
-
\biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr]
\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-
\biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
-
\biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr]
\biggl[
\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr]
\biggl[
\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+
\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
-
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2  \biggr]
\biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
- \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{
\biggl[ \gamma_{51}^2 \gamma_{12}^2 
-
\gamma_{52}^2 \gamma_{11}^2 \biggr]
\biggl[ \gamma_{43}^2 \gamma_{11}^2
- \gamma_{41}^2 \gamma_{13}^2
\biggr]
-
\biggl[ \gamma_{51}^2 \gamma_{13}^2
-
\gamma_{53}^2 \gamma_{11}^2
\biggr]
\biggl[
\gamma_{42}^2 \gamma_{11}^2
- \gamma_{41}^2 \gamma_{12}^2
\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{h_4^2 h_1^2}\biggl\{
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggl[ \gamma_{43}^2 \gamma_{11}^2
- \gamma_{41}^2 \gamma_{13}^2
\biggr]
-
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggl[ \gamma_{43}^2\gamma_{11}^2
-~ \gamma_{41}^2 \gamma_{13}^2
\biggr]
- \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggl[
\gamma_{42}^2 \gamma_{11}^2
- \gamma_{41}^2 \gamma_{12}^2
\biggr]
+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggl[
\gamma_{42}^2 \gamma_{11}^2
- \gamma_{41}^2\gamma_{12}^2
\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{
\biggl[ \gamma_{51}^2 \gamma_{12}^2 
-
\gamma_{52}^2 \gamma_{11}^2 \biggr]
\biggl[ \gamma_{43} \gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]
\biggl[ \gamma_{43} \gamma_{11}
- \gamma_{41} \gamma_{13}
\biggr]
-
\biggl[ \gamma_{51}^2 \gamma_{13}^2
-
\gamma_{53}^2 \gamma_{11}^2
\biggr]
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41} \gamma_{12}
\biggr]
\biggl[
\gamma_{42} \gamma_{11}
- \gamma_{41} \gamma_{12}
\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{h_4^2 h_1^2}\biggl\{
\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggl[ \gamma_{43} \gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]
\biggl[ \gamma_{43} \gamma_{11}
- \gamma_{41} \gamma_{13}
\biggr]
-
\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2
\biggl[ \gamma_{43}\gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]
\biggl[ \gamma_{43}\gamma_{11}
-~ \gamma_{41} \gamma_{13}
\biggr]
- \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41} \gamma_{12}
\biggr]
\biggl[
\gamma_{42} \gamma_{11}
- \gamma_{41} \gamma_{12}
\biggr]
+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41}\gamma_{12}
\biggr]
\biggl[
\gamma_{42} \gamma_{11}
- \gamma_{41}\gamma_{12}
\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{h_1^4 h_4^2 }\biggl\{
-
\biggl[ \gamma_{51} \gamma_{12} 
+
\gamma_{52} \gamma_{11} \biggr]
\gamma_{43}
\biggl[ \gamma_{43} \gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]
\gamma_{52}
+
\biggl[ \gamma_{51} \gamma_{13}
+
\gamma_{53} \gamma_{11}
\biggr]
\gamma_{42}
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41} \gamma_{12}
\biggr] \gamma_{53}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{ h_1^4 h_4^2}\biggl\{
\gamma_{12}^2
\biggl[ \gamma_{43}\gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]\gamma_{52}
-\gamma_{11}^{2}
\biggl[ \gamma_{43} \gamma_{11}
+ \gamma_{41} \gamma_{13}
\biggr]\gamma_{52}
- \gamma_{11}^{2}
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41} \gamma_{12}
\biggr]\gamma_{53}
+ \gamma_{13}^2
\biggl[
\gamma_{42} \gamma_{11}
+ \gamma_{41}\gamma_{12}
\biggr]\gamma_{53}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{h_1^4 h_4^2 }\biggl\{
\biggl[
(- \gamma_{51} \gamma_{12} 
-
\gamma_{52} \gamma_{11} )
\gamma_{43}
+
\gamma_{12}^2
-\gamma_{11}^{2}
\biggr]
(\gamma_{43} \gamma_{11}
+ \gamma_{41} \gamma_{13} )
\gamma_{52}
+
\biggl[
(\gamma_{51} \gamma_{13}
+
\gamma_{53} \gamma_{11} )
\gamma_{42}
- \gamma_{11}^{2}
+ \gamma_{13}^2
\biggr]
(\gamma_{42} \gamma_{11}
+ \gamma_{41}\gamma_{12} )
\gamma_{53}
\biggr\}
</math>
  </td>
</tr>
</table>
&hellip; Not sure this is headed anywhere useful!
====Volume Element====
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(h_1 h_4 h_5)
\biggl[
\biggl( \frac{\partial \lambda_1}{\partial x}\biggr) dx
+ \biggl( \frac{\partial \lambda_1}{\partial y}\biggr) dy
+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr) dz
\biggr]
\biggl[
\biggl( \frac{\partial \lambda_4}{\partial x}\biggr) dx
+ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr) dy
+ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr) dz
\biggr]
\biggl[
\biggl( \frac{\partial \lambda_5}{\partial x}\biggr) dx
+ \biggl( \frac{\partial \lambda_5}{\partial y}\biggr) dy
+ \biggl( \frac{\partial \lambda_5}{\partial z}\biggr) dz
\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
\biggl( \gamma_{11}\biggr) dx
+ \biggl( \gamma_{12}\biggr) dy
+ \biggl( \gamma_{13}\biggr) dz
\biggr]
\biggl[
\biggl( \gamma_{41}\biggr) dx
+ \biggl( \gamma_{42}\biggr) dy
+ \biggl( \gamma_{43}\biggr) dz
\biggr]
\biggl[
\biggl( \gamma_{51} \biggr) dx
+ \biggl( \gamma_{52} \biggr) dy
+ \biggl( \gamma_{53} \biggr) dz
\biggr]
</math>
</math>
   </td>
   </td>

Latest revision as of 22:39, 27 May 2021

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
(See, for example, Dan Sloughter's (Furman University) 2001 Calculus III class lecture notes — specifically Lecture 15)


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{k} (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!


Speculation7

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>


On my white-board I have shown that, if

<math>~\lambda_3 \equiv \ell_{3D} \mathcal{D} \, ,</math>

then everything will work out as long as,

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

<math>~=</math>

<math>~ \biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4} \, , </math>

where,

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

<math>~\equiv</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

<math>~ x^2 [16q^{16}y^8 + 16q^{12}y^6p^4z^2 + 4q^8y^4p^8z^4 + 16q^{12}y^6p^4z^2 + 16q^8y^4p^8z^4 + 4q^4y^2p^{12}z^6 + 4q^8y^4p^8z^4 + 4q^4y^2 p^{12}z^6 + p^{16}z^8] </math>

 

 

<math>~ + q^8 y^2 [16x^8 + 16x^6p^4z^2 + 4x^4p^8z^4 + 16x^6p^4z^2 + 16x^4p^8z^4 + 4x^2p^{12}z^6 + 4x^4p^8z^4 + 4x^2p^{12}z^6 + p^{16}z^8] </math>

 

 

<math>~ + p^8z^2 [x^8 - 2x^6q^4y^2 + x^4q^8y^4 - 2x^6q^4y^2 + 4x^4q^8y^4 - 2x^2q^{12}y^6 + x^4q^8y^4 - 2x^2q^{12}y^6 + q^{16}y^8] </math>

 

<math>~=</math>

<math>~ x^2 [16q^{16}y^8 + 32q^{12}y^6p^4z^2 + 24q^8y^4p^8z^4 + 8q^4y^2p^{12}z^6 + p^{16}z^8] </math>

 

 

<math>~ + q^8 y^2 [16x^8 + 32x^6p^4z^2 + 24x^4p^8z^4 + 8x^2p^{12}z^6 + p^{16}z^8] </math>

 

 

<math>~ + p^8z^2 [x^8 - 4x^6q^4y^2 + 6x^4q^8y^4 - 4x^2q^{12}y^6 + q^{16}y^8] </math>

Let's check this out.

<math>~\mathrm{RHS}~\equiv \biggl(\frac{\mathcal{D}}{\ell_{3D}} \biggr)^2 \frac{1}{\ell_{3D}^4}</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 )^3</math>

 

<math>~=</math>

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

 

<math>~=</math>

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

 

 

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

 

<math>~=</math>

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

 

 

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

Best Thus Far

Part A

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>

 

<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>

Try …

<math>~\lambda_3</math>

<math>~\equiv</math>

<math>~ \biggl(\frac{B}{A}\biggr)^{m/2} = (D \ell_{3D})^m </math>

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

<math>~\equiv</math>

<math>~ \frac{1}{2}\biggl\{ A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i} \biggr\} </math>

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

<math>~\equiv</math>

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

In this case we find,

<math>~\frac{x}{2}\biggl\{~~\biggr\}_x</math>

<math>~=</math>

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

<math>~\frac{y}{2}\biggl\{~~\biggr\}_y</math>

<math>~=</math>

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

<math>~\frac{z}{2}\biggl\{~~\biggr\}_z</math>

<math>~=</math>

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

The scale factor is, then,

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

<math>~=</math>

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

 

<math>~=</math>

<math>~ \sum_{i=1}^3 \biggl\{ \biggl[ \frac{m\lambda_3}{2AB} \biggr] \biggl[ A \cdot \frac{\partial B}{\partial x_i} - B \cdot \frac{\partial A}{\partial x_i} \biggr] \biggr\}^2 </math>

 

<math>~=</math>

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

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

<math>~=</math>

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

Part B (25 February 2021)

Now, from above, we know that,

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

Example:
<math>~(q^2, p^2) = (2, 5)</math>     and     <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>

<math>~\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2</math> <math>~\biggl[ q^2 y(p^4z^2 + 2x^2 ) \biggl]^2</math> <math>~\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2</math> <math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math>
2.14037 1.39187 0.04623 3.57847

As an aside, note that,

<math>~AB</math>

<math>~=</math>

<math>~ \biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln x} + \biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln y} + \biggl[ \frac{AB}{m} \biggr] \frac{\partial \ln \lambda_3}{\partial \ln z} </math>

<math>~\Rightarrow ~~~ m</math>

<math>~=</math>

<math>~ \frac{\partial \ln \lambda_3}{\partial \ln x} + \frac{\partial \ln \lambda_3}{\partial \ln y} + \frac{\partial \ln \lambda_3}{\partial \ln z} \, . </math>

We realize that this ratio of lengths may also be written in the form,

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

<math>~=</math>

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

Same Example, but Different Expression:
<math>~(q^2, p^2) = (2, 5)</math>     and     <math>~(x, y, z) = (0.7, \sqrt{0.23}, 0.1)~~\Rightarrow~~ \lambda_1 = 1.0</math>

<math>~6x^2 q^4y^2 p^4 z^2</math> <math>~x^4(4q^4y^2 + p^4z^2)</math> <math>~q^8y^4(4x^2 + p^4z^2)</math> <math>~p^8z^4(x^2 + q^4y^2)</math> <math>~\biggl( \frac{\mathcal{D}}{\ell_{3D}} \biggr)^{2}</math>
0.67620 0.94359 1.87054 0.08813 3.57847


Let's try …

<math>~\frac{\partial \lambda_5}{\partial x}</math>

<math>~=</math>

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

<math>~\frac{\partial \lambda_5}{\partial y}</math>

<math>~=</math>

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

<math>~\frac{\partial \lambda_5}{\partial z}</math>

<math>~=</math>

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

This means that the relevant scale factor is,

<math>~h_5^{-2}</math>

<math>~=</math>

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

<math>~\Rightarrow~~~h_5</math>

<math>~=</math>

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

and the three associated direction cosines are,

<math>~\gamma_{51} = h_5 \biggl( \frac{\partial \lambda_5}{\partial x} \biggr)</math>

<math>~=</math>

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

<math>~\gamma_{52} = h_5 \biggl( \frac{\partial \lambda_5}{\partial y} \biggr)</math>

<math>~=</math>

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

<math>~\gamma_{53} = h_5 \biggl( \frac{\partial \lambda_5}{\partial z} \biggr)</math>

<math>~=</math>

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

These direction cosines exactly match what is required in order to ensure that the coordinate, <math>~\lambda_5</math>, is everywhere orthogonal to both <math>~\lambda_1</math> and <math>~\lambda_4</math>. GREAT! The resulting summary table is, therefore:

Direction Cosine Components for T10 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>~4</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>~5</math> --- <math>~\frac{\ell_{3D}}{\mathcal{D}}</math> <math>~-x(2 q^4y^2 + p^4z^2)</math> <math>~q^2 y(p^4z^2 + 2x^2 )</math> <math>~p^2z( x^2 - q^4y^2 )</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>


Try …

<math>~\lambda_5</math>

<math>~=</math>

<math>~ x^{-2q^4} \cdot y^{2q^2} + y^{q^2p^4} \cdot z^{-q^4p^2} + x^{-p^4} \cdot z^{p^2} </math>

 

<math>~=</math>

<math>~ \frac{ y^{2q^2} }{ x^{2q^4} } + \frac{ y^{q^2p^4} }{ z^{q^4p^2} } + \frac{ z^{p^4} }{ x^{p^2} } </math>

 

<math>~=</math>

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

 

<math>~=</math>

<math>~ \frac{1}{x^{2q^4 + p^2} z^{q^4p^2} }\biggl\{ \mathfrak{F_5} \biggr\} \, . </math>

This gives,

<math>~\frac{\partial \lambda_5}{\partial x}</math>

<math>~=</math>

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

 

<math>~=</math>

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

Or, given that,

<math>~x^{2q^4 + p^2 + 1} </math>

<math>~=</math>

<math>~ \frac{x}{ z^{q^4p^2} }\biggl\{ \frac{\mathfrak{F_5}}{\lambda_5} \biggr\} \, , </math>

we can also write,

<math>~\frac{\partial \lambda_5}{\partial x}</math>

<math>~=</math>

<math>~ - \frac{ z^{q^4p^2} }{x}\biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\}

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

Similarly,

<math>~\frac{\partial \lambda_5}{\partial y}</math>

<math>~=</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

<math>~\frac{x^{p^2}}{ y } \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\} \biggl[ 2q^2 \biggl( y^{2q^2} z^{q^4p^2} \biggr) + q^2p^4 \biggl( y^{q^2p^4} x^{2q^4} \biggr)\biggr] \, ; </math>

<math>~\frac{\partial \lambda_5}{\partial z}</math>

<math>~=</math>

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

 

<math>~=</math>

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

 

<math>~=</math>

<math>~ \frac{x^{2q^4} }{z} \biggl\{ \frac{\lambda_5}{\mathfrak{F_5}} \biggr\} \biggl[ p^4 \biggl( z^{p^4 + q^4p^2 } \biggr) - q^4p^2 \biggl( x^{p^2} y^{q^2p^4} \biggr)\biggr] </math>

Understanding the Volume Element

Let's see if the expression for the volume element makes sense; that is, does

<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>

<math>~=</math>

<math>~dx dy dz \, ?</math>

First, let's make sure that we understand how to relate the components of the Cartesian line element with the components of our T10 coordinates.

Line Element

MF53 claim that the following relation gives the various expressions for the scale factors; we will go ahead and incorporate the expectation that, since our coordinate system is orthogonal, the off-diagonal elements are zero.

<math>~ds^2 = dx^2 + dy^2 + dz^2</math>

<math>~=</math>

<math>~ \sum_{i=1,4,5} h_i^2 d\lambda_i^2 \, . </math>

Let's see. The first term on the RHS is,

<math>~h_1^2 d\lambda_1^2</math>

<math>~=</math>

<math>~ h_1^2 \biggl[ \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)dx + \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)dy + \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)dz \biggr]^2 </math>

 

<math>~=</math>

<math>~ h_1^2 \biggl[ \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2 dx^2 + \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 dy^2 + \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 dz^2 </math>

 

 

<math>~ + \cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dx~dy + \cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial z}\biggr)} dx~dz + \cancel{2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)} dy~dz \biggr] \, ; </math>

the other two terms assume easily deduced, similar forms. When put together and after regrouping terms, we can write,

<math>~ \sum_{i=1,4,5} h_i^2 d\lambda_i^2 </math>

<math>~=</math>

<math>~ \biggl[ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2 + h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggr] dx^2 </math>

 

 

<math>~ + \biggl[ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 + h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggr] dy^2 </math>

 

 

<math>~ + \biggl[ h_1^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 + h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2 \biggr] dz^2 \, . </math>

Given that this summation should also equal the square of the Cartesian line element, <math>~(dx^2 + dy^2 + dz^2)</math>, we conclude that the three terms enclosed inside each of the pair of brackets must sum to unity. Specifically, from the coefficient of <math>~dx^2</math>, we can write,

<math>~h_1^2 \biggl( \frac{\partial \lambda_1}{\partial x}\biggr)^2 </math>

<math>~=</math>

<math>~ 1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \, . </math>

Using this relation to replace <math>~h_1^2</math> in each of the other two bracketed expressions, we find for the coefficients of <math>~dy^2</math> and <math>~dz^2</math>, respectively,

<math>~\biggl[ 1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggr] \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 </math>

<math>~=</math>

<math>~ \biggl[ 1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, ; </math>

<math>~\biggl[ 1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggr] \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 </math>

<math>~=</math>

<math>~ \biggl[ 1 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2 \biggr]\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \, . </math>

We can use the first of these two expressions to solve for <math>~h_4^2</math> in terms of <math>~h_5^2</math>, namely,

<math>~\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 - h_4^2\biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 </math>

<math>~=</math>

<math>~ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - h_4^2\biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} </math>

<math>~\Rightarrow ~~~ h_4^2 \biggl[ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggr] </math>

<math>~=</math>

<math>~ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} </math>

Analogously, the second of these two expressions gives,

<math>~ h_4^2 \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr]

</math>

<math>~=</math>

<math>~ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} </math>

Eliminating <math>~h_4</math> between the two gives the desired overall expression for <math>~h_5</math>, namely,

<math>~0 </math>

<math>~=</math>

<math>~\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggr] </math>

 

 

<math>~ - \biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 + h_5^2 \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 - h_5^2 \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr] </math>

 

<math>~=</math>

<math>~ h_5^2 \biggl\{ \biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 - \biggl( \frac{\partial \lambda_5}{\partial y}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr] </math>

 

 

<math>~ - \biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 - \biggl( \frac{\partial \lambda_5}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggr] \biggr\} </math>

 

 

<math>~ -\biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial y}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2\biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggr] </math>

 

 

<math>~ + \biggl[ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial z}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} - \biggl( \frac{\partial \lambda_4}{\partial x}\biggr)^2 \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggr] </math>

 

<math>~=</math>

<math>~ \frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{ \biggl[ \gamma_{51}^2 \gamma_{12}^2 - \gamma_{52}^2 \gamma_{11}^2 \biggr] \biggl[ \gamma_{43}^2 \gamma_{11}^2 - \gamma_{41}^2 \gamma_{13}^2 \biggr] - \biggl[ \gamma_{51}^2 \gamma_{13}^2 - \gamma_{53}^2 \gamma_{11}^2 \biggr] \biggl[ \gamma_{42}^2 \gamma_{11}^2 - \gamma_{41}^2 \gamma_{12}^2 \biggr] \biggr\} </math>

 

 

<math>~ + \frac{1}{h_4^2 h_1^2}\biggl\{ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggl[ \gamma_{43}^2 \gamma_{11}^2 - \gamma_{41}^2 \gamma_{13}^2 \biggr] - \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggl[ \gamma_{43}^2\gamma_{11}^2 -~ \gamma_{41}^2 \gamma_{13}^2 \biggr] - \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggl[ \gamma_{42}^2 \gamma_{11}^2 - \gamma_{41}^2 \gamma_{12}^2 \biggr] + \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggl[ \gamma_{42}^2 \gamma_{11}^2 - \gamma_{41}^2\gamma_{12}^2 \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{h_5^2}{h_1^4 h_4^2 h_5^2}\biggl\{ \biggl[ \gamma_{51}^2 \gamma_{12}^2 - \gamma_{52}^2 \gamma_{11}^2 \biggr]

\biggl[ \gamma_{43} \gamma_{11} + \gamma_{41} \gamma_{13} \biggr] \biggl[ \gamma_{43} \gamma_{11} - \gamma_{41} \gamma_{13} \biggr]

- \biggl[ \gamma_{51}^2 \gamma_{13}^2 - \gamma_{53}^2 \gamma_{11}^2 \biggr]

\biggl[ \gamma_{42} \gamma_{11} + \gamma_{41} \gamma_{12} \biggr] \biggl[ \gamma_{42} \gamma_{11} - \gamma_{41} \gamma_{12} \biggr] \biggr\} </math>

 

 

<math>~ + \frac{1}{h_4^2 h_1^2}\biggl\{ \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2}

\biggl[ \gamma_{43} \gamma_{11} + \gamma_{41} \gamma_{13} \biggr] \biggl[ \gamma_{43} \gamma_{11} - \gamma_{41} \gamma_{13} \biggr]

- \biggl( \frac{\partial \lambda_1}{\partial y}\biggr)^2 \biggl[ \gamma_{43}\gamma_{11} + \gamma_{41} \gamma_{13} \biggr] \biggl[ \gamma_{43}\gamma_{11} -~ \gamma_{41} \gamma_{13} \biggr]

- \biggl( \frac{\partial \lambda_1}{\partial x} \biggr)^{2} \biggl[ \gamma_{42} \gamma_{11} + \gamma_{41} \gamma_{12} \biggr] \biggl[ \gamma_{42} \gamma_{11} - \gamma_{41} \gamma_{12} \biggr]


+ \biggl( \frac{\partial \lambda_1}{\partial z}\biggr)^2 \biggl[ \gamma_{42} \gamma_{11} + \gamma_{41}\gamma_{12} \biggr] \biggl[ \gamma_{42} \gamma_{11} - \gamma_{41}\gamma_{12} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{h_1^4 h_4^2 }\biggl\{ - \biggl[ \gamma_{51} \gamma_{12} + \gamma_{52} \gamma_{11} \biggr] \gamma_{43} \biggl[ \gamma_{43} \gamma_{11} + \gamma_{41} \gamma_{13} \biggr] \gamma_{52}

+ \biggl[ \gamma_{51} \gamma_{13} + \gamma_{53} \gamma_{11} \biggr] \gamma_{42} \biggl[ \gamma_{42} \gamma_{11} + \gamma_{41} \gamma_{12} \biggr] \gamma_{53}

\biggr\} </math>

 

 

<math>~ + \frac{1}{ h_1^4 h_4^2}\biggl\{ \gamma_{12}^2 \biggl[ \gamma_{43}\gamma_{11} + \gamma_{41} \gamma_{13} \biggr]\gamma_{52} -\gamma_{11}^{2} \biggl[ \gamma_{43} \gamma_{11} + \gamma_{41} \gamma_{13} \biggr]\gamma_{52}


- \gamma_{11}^{2} \biggl[ \gamma_{42} \gamma_{11} + \gamma_{41} \gamma_{12} \biggr]\gamma_{53}


+ \gamma_{13}^2 \biggl[ \gamma_{42} \gamma_{11} + \gamma_{41}\gamma_{12} \biggr]\gamma_{53} \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{1}{h_1^4 h_4^2 }\biggl\{ \biggl[ (- \gamma_{51} \gamma_{12} - \gamma_{52} \gamma_{11} ) \gamma_{43} + \gamma_{12}^2

-\gamma_{11}^{2} \biggr] (\gamma_{43} \gamma_{11} + \gamma_{41} \gamma_{13} ) \gamma_{52} + \biggl[ (\gamma_{51} \gamma_{13} + \gamma_{53} \gamma_{11} ) \gamma_{42}

- \gamma_{11}^{2}

+ \gamma_{13}^2 \biggr] (\gamma_{42} \gamma_{11} + \gamma_{41}\gamma_{12} ) \gamma_{53} \biggr\} </math>

… Not sure this is headed anywhere useful!

Volume Element

<math>~(h_1 h_4 h_5) d\lambda_1 d\lambda_4 d\lambda_5</math>

<math>~=</math>

<math>~ (h_1 h_4 h_5) \biggl[ \biggl( \frac{\partial \lambda_1}{\partial x}\biggr) dx + \biggl( \frac{\partial \lambda_1}{\partial y}\biggr) dy + \biggl( \frac{\partial \lambda_1}{\partial z}\biggr) dz \biggr] \biggl[ \biggl( \frac{\partial \lambda_4}{\partial x}\biggr) dx + \biggl( \frac{\partial \lambda_4}{\partial y}\biggr) dy + \biggl( \frac{\partial \lambda_4}{\partial z}\biggr) dz \biggr] \biggl[ \biggl( \frac{\partial \lambda_5}{\partial x}\biggr) dx + \biggl( \frac{\partial \lambda_5}{\partial y}\biggr) dy + \biggl( \frac{\partial \lambda_5}{\partial z}\biggr) dz \biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \biggl( \gamma_{11}\biggr) dx + \biggl( \gamma_{12}\biggr) dy + \biggl( \gamma_{13}\biggr) dz \biggr] \biggl[ \biggl( \gamma_{41}\biggr) dx + \biggl( \gamma_{42}\biggr) dy + \biggl( \gamma_{43}\biggr) dz \biggr] \biggl[ \biggl( \gamma_{51} \biggr) dx + \biggl( \gamma_{52} \biggr) dy + \biggl( \gamma_{53} \biggr) dz \biggr] </math>

COLLADA

Here we try to use the 3D-visualization capabilities of COLLADA to test whether or not the three coordinates associated with the T6 Coordinate system are indeed orthogonal to one another. We begin by making a copy of the Inertial17.dae text file, which we obtain from an accompanying discussion. When viewed with the Mac's Preview application, this group of COLLADA-based instructions displays a purple ellipsoid with axis ratios, (b/a, c/a) = (0.41, 0.385). This means that we are dealing with an ellipsoid for which,

<math>~q \equiv \frac{a}{b}</math>

<math>~=</math>

<math>~2.44</math>

      and,      

<math>~p \equiv \frac{a}{c}</math>

<math>~=</math>

<math>~2.60 \, .</math>

<math>~\lambda_1</math>

<math>~\equiv</math>

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

<math>~\lambda_2</math>

<math>~=</math>

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

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

<math>~\equiv</math>

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

First Trial

First Trial
(specified variable values have bgcolor="pink")
x y z <math>~\lambda_1</math> <math>~\ell_{3D}</math> <math>~\mathcal{D}</math>
0.5 0.35493 0.00000 1 0.46052 2.11310

Unit Vectors

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

<math>~=</math>

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

 

<math>~=</math>

<math>~ \hat\imath (0.23026) + \hat\jmath (0.97313) \, ; </math>

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

<math>~=</math>

<math>~ \frac{x_0 q^2 y_0 p^2 z_0}{\mathcal{D}} \biggl\{ \hat{\imath} \biggl( \frac{1}{x_0} \biggr) + \hat{\jmath} \biggl( \frac{1}{q^2 y_0} \biggr) + \hat{k} \biggl( -\frac{2}{p^2 z_0} \biggr) \biggr\} </math>

 

<math>~=</math>

<math>~-\hat{k} ~\biggl( \frac{2x_0 q^2 y_0 }{\mathcal{D}} \biggr) </math>

 

<math>~=</math>

<math>~-\hat{k} ~\biggl( 1 \biggr) \ ; </math>

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

<math>~=</math>

<math>~\frac{\ell_{3D}}{\mathcal{D}} \biggl\{ -\hat\imath \biggl[ x_0(2 q^4y_0^2 + p^4z_0^2 ) \biggl] + \hat\jmath \biggl[ q^2 y_0(p^4z_0^2 + 2x_0^2 ) \biggl] + \hat{k} \biggl[ p^2z_0( x_0^2 - q^4y_0^2 ) \biggl] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{2q^2 x_0 y_0\ell_{3D}}{\mathcal{D}} \biggl\{ -\hat\imath ( q^2y_0 ) + \hat\jmath (x_0) \biggr\} = \biggl(1\biggr)\ell_{3D}\biggl\{ -\hat\imath ( q^2y_0 ) + \hat\jmath (x_0) \biggr\} </math>

 

<math>~=</math>

<math>~ -\hat\imath (0.97313 ) + \hat\jmath (0.23026) \, . </math>

Tangent Plane

From our above derivation, the plane that is tangent to the ellipsoid's surface at <math>~(x_0, y_0, z_0)</math> is given by the expression,

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

<math>~=</math>

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

For this First Trial, we have (for all values of <math>~z</math>, given that <math>~z_0 = 0</math>) …

<math>~ (0.5)x + (2.11310)y </math>

<math>~=</math>

<math>~ 1 </math>

<math>~\Rightarrow ~~~ y </math>

<math>~=</math>

<math>~ \frac{(1 - 0.5x)}{2.11310} \, . </math>

So let's plot a segment of the tangent plane whose four corners are given by the coordinates,

Corner x y z
A x_0 - 0.25 = +0.25 0.41408 -0.25
B x_0 + 0.25 = +0.75 0.29577 -0.25
C x_0 - 0.25 = +0.25 0.41408 +0.25
D x_0 + 0.25 = +0.75 0.29577 +0.25

Now, in order to give some thickness to this tangent-plane, let's adjust the four corner locations by a distance of <math>~\pm 0.1</math> in the <math>~\hat{e}_1</math> direction.

Eight Corners of Tangent Plane

Corner 1: Shift surface-point location <math>~(x_0, y_0, z_0)</math> by <math>~(+\Delta e_1)</math> in the <math>~\hat{e}_1</math> direction, by <math>~(+\Delta e_2)</math> in the <math>~\hat{e}_2</math> direction, and by by <math>~(+\Delta e_3)</math> in the <math>~\hat{e}_3</math> direction. This gives …

<math>~x_1</math>

<math>~=</math>

<math>~x_0 + (\Delta e_1)0.23026 - (\Delta e_2)0.97313</math>

Second Trial

Second Trial … <math>~(q = 2.44, p = 2.60)</math>
[specified variable values have bgcolor="pink"]
x_0 y_0 z_0 <math>~\lambda_1</math> <math>~\ell_{3D}</math> <math>~\mathcal{D}</math>
0.5 0.35493 0.00000 1 0.46052 2.11310

Generic Unit Vector Expressions

Let's adopt the notation,

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

<math>~=</math>

<math>~\hat\imath ~[e_{ix}] + \hat\jmath ~[e_{iy}] + \hat{k} ~[e_{iz}]</math>

      for,       <math>~i = 1,3 \, .</math>

Then, for the T6 Coordinate system, we have,

<math>~e_{1x}</math>

<math>~\equiv</math>

<math>~x_0 \ell_{3D} \, ;</math>

     

<math>~e_{1y}</math>

<math>~\equiv</math>

<math>~q^2y_0 \ell_{3D} \, ;</math>

     

<math>~e_{1z}</math>

<math>~\equiv</math>

<math>~p^2 z_0 \ell_{3D} \, ;</math>

<math>~e_{2x}</math>

<math>~\equiv</math>

<math>~ \frac{q^2 y_0 p^2 z_0}{\mathcal{D}}\, ;</math>

     

<math>~e_{2y}</math>

<math>~\equiv</math>

<math>~\frac{x_0 p^2 z_0}{\mathcal{D}} \, ;</math>

     

<math>~e_{2z}</math>

<math>~\equiv</math>

<math>~- \frac{2x_0 q^2 y_0}{\mathcal{D}}\, ;</math>

<math>~e_{3x}</math>

<math>~\equiv</math>

<math>~-x_0(2 q^4y_0^2 + p^4z_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, ;</math>

     

<math>~e_{3y}</math>

<math>~\equiv</math>

<math>~q^2 y_0(p^4z_0^2 + 2x_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, ;</math>

     

<math>~e_{3z}</math>

<math>~\equiv</math>

<math>~p^2z_0( x_0^2 - q^4y_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, .</math>


Second Trial
  x y z
<math>~e_1</math> 0.23026 0.97313 0.0
<math>~e_2</math> 0.0 0.0 -1.0
<math>~e_3</math> - 0.97313 0.23026 0.0


What are the coordinates of the eight corners of a thin tangent-plane? Let's say that we want the plane to extend …

  • From <math>~(-\Delta_1)</math> to <math>~(+\Delta_1)</math> in the <math>~\hat{e}_1</math> direction … here we set <math>~\Delta_1 = 0.05</math>;
  • From <math>~(-\Delta_2)</math> to <math>~(+\Delta_2)</math> in the <math>~\hat{e}_2</math> direction … here we set <math>~\Delta_2 = 0.25</math>;
  • From <math>~(-\Delta_3)</math> to <math>~(+\Delta_3)</math> in the <math>~\hat{e}_3</math> direction … here we set <math>~\Delta_3 = 0.25</math>.

<math>~\Delta_x</math>

<math>~=</math>

<math>~\Delta_1 e_{1x} + \Delta_2 e_{2x} + \Delta_3 e_{3x} = -0.23177 \, ;</math>

<math>~\Delta_y</math>

<math>~=</math>

<math>~\Delta_1 e_{1y} + \Delta_2 e_{2y} + \Delta_3 e_{3y} = +0.10622 \, ;</math>

<math>~\Delta_z</math>

<math>~=</math>

<math>~\Delta_1 e_{1z} + \Delta_2 e_{2z} + \Delta_3 e_{3z} = -0.25000 \, .</math>

Tangent Plane Schematic
vertex x y z   x y z
0 <math>~x_0 - |\Delta_x|</math> <math>~y_0 - |\Delta_y|</math> <math>~z_0 - |\Delta_z|</math> 0.26823 0.24871 -0.25
1 <math>~x_0 - |\Delta_x|</math> <math>~y_0 + |\Delta_y|</math> <math>~z_0 - |\Delta_z|</math> 0.26823 0.46115 -0.25
2 <math>~x_0 - |\Delta_x|</math> <math>~y_0 - |\Delta_y|</math> <math>~z_0 + |\Delta_z|</math> 0.26823 0.24871 0.25
3 <math>~x_0 - |\Delta_x|</math> <math>~y_0 + |\Delta_y|</math> <math>~z_0 + |\Delta_z|</math> 0.26823 0.46115 0.25
4 <math>~x_0 + |\Delta_x|</math> <math>~y_0 - |\Delta_y|</math> <math>~z_0 - |\Delta_z|</math> 0.73177 0.24871 -0.25
5 <math>~x_0 + |\Delta_x|</math> <math>~y_0 + |\Delta_y|</math> <math>~z_0 - |\Delta_z|</math> 0.73177 0.46115 -0.25
6 <math>~x_0 + |\Delta_x|</math> <math>~y_0 - |\Delta_y|</math> <math>~z_0 + |\Delta_z|</math> 0.73177 0.24871 0.25
7 <math>~x_0 + |\Delta_x|</math> <math>~y_0 + |\Delta_y|</math> <math>~z_0 + |\Delta_z|</math> 0.73177 0.46115 0.25

In the figure on the left, vertex 0 is hidden from view behind the (yellow) solid rectangle.

Third Trial

GoodPlane01

Third Trial … <math>~(q = 2.44, p = 2.60)</math>
[specified variable values have bgcolor="pink"]
x_0 y_0 z_0 <math>~\lambda_1</math> <math>~\ell_{3D}</math> <math>~\mathcal{D}</math>
0.8 0.24600 0.00000 1 0.59959 2.34146

Again, for the T6 Coordinate system, we have,

<math>~e_{1x}</math>

<math>~\equiv</math>

<math>~x_0 \ell_{3D} \, ;</math>

     

<math>~e_{1y}</math>

<math>~\equiv</math>

<math>~q^2y_0 \ell_{3D} \, ;</math>

     

<math>~e_{1z}</math>

<math>~\equiv</math>

<math>~p^2 z_0 \ell_{3D} \, ;</math>

<math>~e_{2x}</math>

<math>~\equiv</math>

<math>~ \frac{q^2 y_0 p^2 z_0}{\mathcal{D}}\, ;</math>

     

<math>~e_{2y}</math>

<math>~\equiv</math>

<math>~\frac{x_0 p^2 z_0}{\mathcal{D}} \, ;</math>

     

<math>~e_{2z}</math>

<math>~\equiv</math>

<math>~- \frac{2x_0 q^2 y_0}{\mathcal{D}}\, ;</math>

<math>~e_{3x}</math>

<math>~\equiv</math>

<math>~-x_0(2 q^4y_0^2 + p^4z_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, ;</math>

     

<math>~e_{3y}</math>

<math>~\equiv</math>

<math>~q^2 y_0(p^4z_0^2 + 2x_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, ;</math>

     

<math>~e_{3z}</math>

<math>~\equiv</math>

<math>~p^2z_0( x_0^2 - q^4y_0^2 )\frac{\ell_{3D}}{\mathcal{D}} \, .</math>


Third Trial
  x y z <math>~\Delta_\mathrm{TP}</math>
<math>~e_1</math> 0.47967 0.87745 0.0 0.02
<math>~e_2</math> 0.0 0.0 -1.0 0.25
<math>~e_3</math> - 0.87753 0.47952 0.0 0.25

In constructing the Tangent-Plane (TP) for a 3D COLLADA display, we first move from the point that is on the surface of the ellipsoid, <math>~\vec{x}_0 = (x_0, y_0, z_0) = (0.8, 0.246, 0.0)</math>, to

vertex
"m"
<math>~\vec{P}_m</math> Components
<math>~x_m = \hat\imath \cdot \vec{P}_m</math> <math>~y_m = \hat\jmath \cdot \vec{P}_m</math> <math>~z_m = \hat{k} \cdot \vec{P}_m</math>
0 <math>~\vec{x}_0 - \Delta_1\hat{e}_1 - \Delta_2\hat{e}_2 - \Delta_3\hat{e}_3</math>

<math>~ x_0 - \Delta_1 e_{1x} - \Delta_2 e_{2x} - \Delta_3 e_{3x} </math>

<math>~ y_0 - \Delta_1 e_{1y} - \Delta_2 e_{2y} - \Delta_3 e_{3y} </math>

<math>~ z_0 - \Delta_1 e_{1z} - \Delta_2 e_{2z} - \Delta_3 e_{3z} </math>

   

0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979

0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847

- 0.25 (-1.0) = + 0.25

1 <math>~\vec{x}_0 - \Delta_1\hat{e}_1 + \Delta_2\hat{e}_2 - \Delta_3\hat{e}_3</math>

<math>~ x_0 - \Delta_1 e_{1x} + \Delta_2 e_{2x} - \Delta_3 e_{3x} </math>

<math>~ y_0 - \Delta_1 e_{1y} + \Delta_2 e_{2y} - \Delta_3 e_{3y} </math>

<math>~ z_0 - \Delta_1 e_{1z} + \Delta_2 e_{2z} - \Delta_3 e_{3z} </math>

   

0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979

0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847

+ 0.25 (-1.0) = - 0.25

2 <math>~\vec{x}_0 - \Delta_1\hat{e}_1 - \Delta_2\hat{e}_2 + \Delta_3\hat{e}_3</math>

<math>~ x_0 - \Delta_1 e_{1x} - \Delta_2 e_{2x} + \Delta_3 e_{3x} </math>

<math>~ y_0 - \Delta_1 e_{1y} - \Delta_2 e_{2y} + \Delta_3 e_{3y} </math>

<math>~ z_0 - \Delta_1 e_{1z} - \Delta_2 e_{2z} + \Delta_3 e_{3z} </math>

   

0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.56307

0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823

- 0.25 (-1.0) = + 0.25

3 <math>~\vec{x}_0 - \Delta_1\hat{e}_1 + \Delta_2\hat{e}_2 + \Delta_3\hat{e}_3</math>

<math>~ x_0 - \Delta_1 e_{1x} + \Delta_2 e_{2x} + \Delta_3 e_{3x} </math>

<math>~ y_0 - \Delta_1 e_{1y} + \Delta_2 e_{2y} + \Delta_3 e_{3y} </math>

<math>~ z_0 - \Delta_1 e_{1z} + \Delta_2 e_{2z} + \Delta_3 e_{3z} </math>

   

0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.57103

0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823

+ 0.25 (-1.0) = - 0.25

4  

0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290

0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436

- 0.25 (-1.0) = + 0.25

5  

0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290

0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436

+ 0.25 (-1.0) = - 0.25

6  

0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021

0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333

- 0.25 (-1.0) = + 0.25

7  

0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021

0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333

+ 0.25 (-1.0) = - 0.25


Tangent Plane Schematic Vertex Locations via Excel
<math>~x_0 = 0.8, z_0 = 0.0, y_0 = 0.246, \lambda_1 = 1.0</math>
Tangent Plane Schematic
<math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math>

GoodPlane02

<math>~x_0 = 0.075, z_0 = 0.0, y_0 = 0.4089, \lambda_1 = 1.0</math>
Tangent Plane Schematic
<math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math>

GoodPlane03

<math>~x_0 = 0.25, z_0 = 0.20, y_0 = 0.33501, \lambda_1 = 1.0</math>
Tangent Plane Schematic Tangent Plane Schematic
<math>~\Delta_1 = 0.02, \Delta_2 = 0.25, \Delta_3 = 0.25</math> <math>~\Delta_1 = 0.02, \Delta_2 = 0.10, \Delta_3 = 0.25</math>

CAPTION:   The image on the right differs from the image on the left in only one way — <math>~\Delta_2</math> = 0.1 instead of 0.25. It illustrates more clearly that the <math>~\hat{e}_3</math> (longest) coordinate axis is not parallel to the z-axis when <math>~z_0 \ne 0.</math>

GoodPlane04

<math>~x_0 = 0.25, z_0 = 1/3, y_0 = 0.1777, \lambda_1 = 1.0</math>
Tangent Plane Schematic
<math>~\Delta_1 = 0.02, \Delta_2 = 0.10, \Delta_3 = 0.25</math>

Further Exploration

Let's set: <math>~x_0 = 0.25, y_0 = 0.33501, z_0 = 0.2 ~~~\Rightarrow ~~~\lambda_1 = 1.00000, \lambda_2 = 0.33521 \, .</math>


<math>~q \equiv \frac{a}{b}</math>

<math>~=</math>

<math>~2.43972</math>

      and,      

<math>~p \equiv \frac{a}{c}</math>

<math>~=</math>

<math>~2.5974 \, .</math>

<math>~\lambda_1</math>

<math>~\equiv</math>

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

<math>~\lambda_2</math>

<math>~=</math>

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

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

<math>~\equiv</math>

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

Next, let's examine the curve that results from varying <math>~z</math> while <math>~\lambda_1</math> and <math>~\lambda_2</math> are held fixed. From the expression for <math>~\lambda_2</math> we appreciate that,

<math>~x</math>

<math>~=</math>

<math>~\frac{\lambda_2 z^{2/p^2}}{y^{1/q^2}} \, ;</math>

and from the expression for <math>~\lambda_1</math> we have,

<math>~x^2</math>

<math>~=</math>

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

Hence, the relationship between <math>~y</math> and <math>~z</math> is,

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

<math>~=</math>

<math>~\lambda_1^2 - q^2y^2 - p^2z^2</math>

<math>~\Rightarrow ~~~ \lambda_2^2 z^{4/p^2} </math>

<math>~=</math>

<math>~y^{2/q^2} \biggl[ \lambda_1^2 - q^2y^2 - p^2z^2\biggr] \, .</math>


Alternatively,

<math>~y</math>

<math>~=</math>

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

Hence, the relationship between <math>~x</math> and <math>~z</math> is,

<math>~x^2</math>

<math>~=</math>

<math>~\lambda_1^2 - p^2z^2 - q^2\biggl[ \frac{\lambda_2 z^{2/p^2}}{x}\biggr]^{2q^2} </math>

<math>~\Rightarrow~~~ x^{2(q^2+1)}</math>

<math>~=</math>

<math>~x^{2q^2} \biggl[\lambda_1^2 - p^2z^2\biggr] - q^2\biggl[ \lambda_2 z^{2/p^2}\biggr]^{2q^2} </math>

<math>~\Rightarrow~~~ x^{2}</math>

<math>~=</math>

<math>~\biggl\{ x^{2q^2} \biggl[\lambda_1^2 - p^2z^2\biggr] - q^2\biggl[ \lambda_2 z^{2/p^2}\biggr]^{2q^2} \biggr\}^{1/(q^2+1)}</math>

Here are some example values …

<math>~\lambda_1 = 1</math>    and,     <math>~\lambda_2 = 0.33521</math>
<math>~z</math> 1st Solution   2nd Solution lambda_3 coordinate
<math>~y_1</math> <math>~x_1</math> <math>~y_2</math> <math>~x_2</math>
0.01 0.407825695 0.0995168 - -
0.03 0.40481851 0.138 - -
0.04 0.40309223 0.1503934 - -
0.08 0.393779065 0.1854283 - -
0.12 0.37990705 0.2103761 - -
0.16 0.36067787 0.23111 1.04123×10-4 0.9095546
0.2 0.33500747 0.2500033 2.23778×10-4 0.85448
0.22 0.31923525 0.2592611 3.36653 ×10-4 0.82065
0.24 0.30106924 0.2686685 5.2327 ×10-4 0.78192
0.26 0.2799962 0.2784963 8.53243 ×10-4 0.73752
0.28 0.25521147 0.2891526 1.491545 ×10-3 0.68634
0.3 0.22530908 0.3013752 2.89262 ×10-3 0.62671
0.32 0.1873233 0.3168808 6.6223 ×10-3 0.55579
0.34 0.13149897 0.3423994 2.09221 ×10-2 0.46637
0.343 0.1191543 0.3490285 0.026458 0.4496
0.344 0.1145 0.3517 0.02880 0.4435
0.345 0.1093972 0.354688 0.03155965 0.4371186
0.3485 0.0847372 0.3713588 0.0480478 0.4085204

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