Editing Appendix/Ramblings/T1Coordinates (section)

===T2 Coordinates===

====Trial Definition====
If I define the dimensionless angle,
<div align="center">
<math>
\Zeta \equiv \sinh^{-1}\biggl( \frac{qz}{\varpi} \biggr) ,
</math>
</div>
which is similar but not identical to the HNM82 angle, <math>\zeta</math>, the two key "T2" coordinates can be written as,
<div align="center">
<math>
\chi_1 \equiv B \varpi \cosh\Zeta ~~~~~\mathrm{and} ~~~~~ \chi_2 \equiv \frac{A \sinh\Zeta}{ \varpi^{q^2-1}}  .
</math>
</div>
(To maintain some flexibility, I have inserted two, as yet unspecified, coefficients, <math>B</math> and <math>A</math>, into the expression for <math>\chi_1</math> and <math>\chi_2</math>, respectively.) Let's check to see how these two coordinates are related to the original T1 coordinates:
<div align="center">
<math>
\chi_1 = B \varpi \biggl[ 1+\sinh^2\Zeta \biggr]^{1/2} = B \varpi \biggl[ 1+\biggl( \frac{qz}{\varpi} \biggr)^2 \biggr]^{1/2} = B ( \varpi^2 + q^2z^2 )^{1/2} = qB \xi_1 ;
</math>
</div> 
and,
<div align="center">
<math>
\chi_2 = \frac{A}{ \varpi^{q^2-1}} \biggl( \frac{qz}{\varpi} \biggr) = \frac{Aqz}{\varpi^{q^2}} = Aq \biggl[\frac{1}{\tan\xi_2} \biggr]^{q^2} .
</math>
</div> 
While the relationship between <math>\chi_2</math> and <math>\xi_2</math> looks rather convoluted, there is at least a rational reason (provided by HNM82) for specifying the form of <math>\chi_2</math>, whereas the original specification of <math>\xi_2</math> was chosen rather arbitrarily in an effort to look like spherical coordinates.

We should probably address as well the structure of the awkward function that appears in the denominator of the T1-coordinate scale factors, namely,
<div align="center">
<math>
L \equiv \biggl[ z^2 + \frac{\varpi^2}{q^4} \biggr]^{1/2} .
</math>
</div>
In terms of our newly introduced hyperbolic functions, this may be written as,
<div align="center">
<math>
q^4 L^2 = \varpi^2 + q^4 z^2 = \varpi^2 \biggl[1 + q^2 \sinh^2 \Zeta  \biggr] .
</math>
</div>

====Scale Factors====
To determine the scale factors and position vector, we need to first derive a variety of partial derivatives.  

<table align="center" border="1" cellpadding="5">
<tr>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial x}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial y}
</math>
  </td>
  <td align="center">
<math>
\frac{\partial}{\partial z}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\chi_1</math>
  </td>
  <td align="center">
<math>
\biggl(\frac{B^2}{\chi_1}\biggr) x
</math>
  </td>
  <td align="center">
<math>
\biggl(\frac{B^2}{\chi_1}\biggr) y
</math>
  </td>
  <td align="center">
<math>
\biggl(\frac{B^2}{\chi_1}\biggr) q^2 z
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\chi_2</math>
  </td>
  <td align="center">
<math>
- \biggl( \frac{q^3 A z}{\varpi^{q^2+2}} \biggr) x
</math>
  </td>
  <td align="center">
<math>
- \biggl( \frac{q^3 A z}{\varpi^{q^2+2}} \biggr) y
</math>
  </td>
  <td align="center">
<math>
\frac{qA}{\varpi^{q^2}}
</math>
  </td>
</tr>

<tr>
  <td align="center">
<math>\chi_3</math>
  </td>
  <td align="center">
<math>
- \biggl( \frac{1}{\varpi^{2}} \biggr) y
</math>
  </td>
  <td align="center">
<math>
+ \biggl( \frac{1}{\varpi^{2}} \biggr) x
</math>
  </td>
  <td align="center">
<math>
0
</math>
  </td>
</tr>
</table>

The scale factors are, therefore:

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>h_1^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\chi_1}{\partial x} \biggr)^2 + \biggl( \frac{\partial\chi_1}{\partial y} \biggr)^2 + \biggl( \frac{\partial\chi_1}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{B^2}{\chi_1}\biggr)^2 (\varpi^2 + q^4 z^2) \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{\chi_1^2}{B^4 (\varpi^2 + q^4 z^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>h_2^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\chi_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial\chi_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial\chi_2}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{q A}{\varpi^{q^2+1}}\biggr)^2 (q^4 z^2 + \varpi^2) \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{z^2 \varpi^2 }{\chi_2^2 (\varpi^2 + q^4 z^2)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>h_3^2</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{\partial\chi_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial\chi_3}{\partial y} \biggr)^2 + \biggl( \frac{\partial\chi_3}{\partial z} \biggr)^2 \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{1}{\varpi^2}\biggr)^2 (x^2 + y^2) \biggr]^{-1}
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\varpi^2 
</math>
  </td>
</tr>
</table>

====Position Vector====

When written in terms of the T2-coordinate unit vectors, the Cartesian unit vectors are:

<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\hat{i}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl( h_1 \frac{\partial\chi_1}{\partial x} \biggr) 
+ \hat{e}_2 \biggl( h_2 \frac{\partial\chi_2}{\partial x} \biggr)
+ \hat{e}_3 \biggl( h_3 \frac{\partial\chi_3}{\partial x} \biggr)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{x}{(\varpi^2 + q^4 z^2)^{1/2}} \biggl[ \hat{e}_1 - \hat{e}_2\biggl(\frac{q^2 z}{\varpi}\biggr) \biggr] - \hat{e}_3 \biggl(\frac{y}{\varpi}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\hat{j}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl( h_1 \frac{\partial\chi_1}{\partial y} \biggr) 
+ \hat{e}_2 \biggl( h_2 \frac{\partial\chi_2}{\partial y} \biggr)
+ \hat{e}_3 \biggl( h_3 \frac{\partial\chi_3}{\partial y} \biggr)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{y}{(\varpi^2 + q^4 z^2)^{1/2}} \biggl[ \hat{e}_1 - \hat{e}_2\biggl(\frac{q^2 z}{\varpi}\biggr) \biggr] + \hat{e}_3 \biggl(\frac{x}{\varpi}\biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\hat{k}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 \biggl( h_1 \frac{\partial\chi_1}{\partial z} \biggr) 
+ \hat{e}_2 \biggl( h_2 \frac{\partial\chi_2}{\partial z} \biggr)
+ \hat{e}_3 \biggl( h_3 \frac{\partial\chi_3}{\partial z} \biggr)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\varpi^2 + q^4 z^2)^{1/2}} \biggl[ \hat{e}_1 (q^2 z) + \hat{e}_2 \varpi \biggr]
</math>
  </td>
</tr>
</table>

Therefore, the position vector is,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>\vec{x}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{i}x + \hat{j}y + \hat{k}z
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\hat{e}_1 (h_1 \chi_1) + (1 - q^2) \hat{e}_2 (h_2 \chi_2)
</math>
  </td>
</tr>
</table>