Editing Appendix/Ramblings/T1Coordinates (section)

==Relationship Between the Two (to be checked and perhaps extended by Jay Call)==
The pair of HNM83 coordinate lines that correspond to the index <math>\gamma=2</math> matches identically the T1 coordinate lines in the special case highlighted above, namely, the case of <math>q^2 = 2</math>.  At the very least, this must mean that the equilibrium model constructed by HNM82 in the case of <math>\gamma=2</math> must have equipotential surfaces that are concentric oblate spheroids having axis ratio, <math>a_1/a_3 = \sqrt{2}</math>.

This makes me wonder whether I ought to redefine the T1 coordinates to conform more to the HNM82 notation.  In particular, should I think in terms of hyperbolic rather than trigonometric functions; and should I explore the utility of the coordinate "angle" that HNM82 refer to as <math>\zeta</math>?

===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>

===Coordinate Inversion===
Can we invert these expressions to obtain <math>\varpi</math> and <math>\Zeta</math> (or, equivalently, <math>z</math>) strictly in terms of <math>\chi_1</math> and <math>\chi_2</math>?  Well, we can start by squaring both coordinates and subtracting to isolate <math>\varpi^2</math>.  Specifically,
<div align="center">
<math>
\biggl( \frac{\chi_1}{B\varpi} \biggr)^2 - \biggl( \frac{\chi_2}{A\varpi^{1-q^2}} \biggr)^2 = \cosh^2\Zeta - \sinh^2\Zeta = 1 .
</math>
</div>
Hence, the desired functional expression <math>\varpi^2(\chi_1,\chi_2)</math> comes from the root(s) of the following polynomial equation:
<div align="center">
<math>
\biggl( \frac{\chi_2}{A} \biggr)^2 (\varpi^2)^{q^2} + \varpi^2 - \biggl(\frac{\chi_1}{B}\biggr)^2 = 0 .
</math>
</div>
It seems unlikely that closed-form analytic expressions can be extracted for the root(s) of this equation for an arbitrary choice of the flattening index, <math>q</math>.  But it should be possible to obtain analytically expressible roots in the case of <math>q^2 = 2</math>, <math>q^2 = 3</math>, and <math>q^2 = 4</math>.  Once the expression for <math>\varpi^2(\chi_1,\chi_2)</math> is known, a determination of <math>z(\chi_1,\chi_2)</math> is straightforward.  Specifically,
<div align="center">
<math>
z = \frac{\varpi}{q} \sinh\Zeta = \frac{\varpi}{q} \biggl[ \frac{\chi_2}{A} ~ \varpi^{q^2-1} \biggr] = \biggl[ \frac{\chi_2}{qA} ~ \varpi^{q^2} \biggr] .
</math>
</div>
In addition, from the above polynomial expression, we know that,
<div align="center">
<math>
\chi_2^2 (\varpi^2)^{q^2} = A^2 \biggl(\frac{\chi_1^2}{B^2} -  \varpi^2 \biggr) .
</math>
</div>
Hence, quite generally we deduce,
<div align="center">
<math>
z^2 = \frac{1}{q^2} \biggl(\frac{\chi_1^2}{B^2} -  \varpi^2 \biggr) ,
</math>
</div>
which, in retrospect, could have been derived more directly from the definition of <math>\chi_1</math>.

===First Special Case (quadratic)===
When <math>q^2 = 2</math>, the above polynomial becomes a quadratic equation and therefore its roots can be determined analytically.  Setting,
<div align="center">
<math>
x \equiv \varpi^2 ,
</math>
</div>
the relevant quadratic equation is,
<div align="center">
<math>
\biggl( \frac{\chi_2}{A} \biggr)^2 x^2 +  x - \biggl( \frac{\chi_1}{B} \biggr)^2 = 0 .
</math>
</div>
The root of this function that provides positive values for <math>x</math> (''i.e.,'' for <math>\varpi^2</math>) is,
<div align="center">
<math>
x = \frac{A^2}{2\chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] ,
</math>
</div>
where,
<div align="center">
<math>
\Chi_\mathrm{quad} \equiv \frac{2\chi_1 \chi_2}{AB} = \sinh (2\Zeta).
</math>
</div>
This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 2</math>.  The desired expression for <math>z(\chi_1,\chi_2)</math> is,
<div align="center">
<math>
z^2 =  \frac{\chi_1^2}{q^2 B^2}- \frac{A^2}{2q^2 \chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] = \frac{A^2}{4 \chi_2^2}\biggl[ \frac{1}{2} \Chi_\mathrm{quad}^2 +1 -  \sqrt{1 + \Chi_\mathrm{quad}^2}\biggr] .
</math>
</div>

We note as well that in this special case, where <math>q^2 = 2</math>, 
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
4 L^2 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\varpi^2 \biggl[1 + 2 \sinh^2 \Zeta  \biggr] = \varpi^2 \cosh(2\Zeta) = \varpi^2 \biggl[ 1+\sinh^2(2\Zeta) \biggr]^{1/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{A^2}{2\chi_2^2} \biggl[ \sqrt{1 + \Chi_\mathrm{quad}^2} - 1 \biggr] \sqrt{ 1 + \Chi_\mathrm{quad}^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{A^2}{2\chi_2^2} \biggl[ \Chi_\mathrm{quad}^2 + 1 - \sqrt{ 1 + \Chi_\mathrm{quad}^2} \biggr] .
</math>
  </td>
</tr>
</table>
Note that this expression is very similar, but not identical, to the expression just derived for <math>z^2</math>.


===Second Special Case (cubic)===
When <math>q^2 = 3</math>, the above polynomial becomes a cubic equation and therefore its roots also can be determined analytically.  Again setting,
<div align="center">
<math>
x \equiv \varpi^2 ,
</math>
</div>
the relevant cubic equation is,
<div align="center">
<math>
x^3 +  \biggl( \frac{A}{\chi_2} \biggr)^2 x - \biggl( \frac{A \chi_1}{B \chi_2} \biggr)^2 = 0 .
</math>
</div>


<table align="center" border="5" cellpadding="20" bordercolor="purple" width="75%">
<tr><td>
A cubic equation of the form,
<div align="center">
<math>
x^3 + ax + b = 0 ,
</math>
</div>
has the following real root:
<div align="center">
<math>
x = \alpha + \beta ,
</math>
</div>
where,
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\alpha^3
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
-\frac{b}{2} \biggl\{ 1 - \biggl[ 1 + \biggl( \frac{2}{b} \biggr)^2 \biggl( \frac{a}{3} \biggr)^3 \biggr]^{1/2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\beta^3
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
-\frac{b}{2} \biggl\{ 1 + \biggl[ 1 + \biggl( \frac{2}{b} \biggr)^2 \biggl( \frac{a}{3} \biggr)^3 \biggr]^{1/2} \biggr\}
</math>
  </td>
</tr>

</table>

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


The root of this function that provides positive values for <math>x</math> (''i.e.,'' for <math>\varpi^2</math>) is,
<div align="center">
<math>
x = \biggl( \frac{A^2 \chi_1^2}{2 B^2\chi_2^2} \biggr)^{1/3} \biggl\{ \biggl[ 1 + \sqrt{1 + \Chi_\mathrm{cube}^2} \biggr]^{1/3} + \biggl[ 1 - \sqrt{1 + \Chi_\mathrm{cube}^2} \biggr]^{1/3} \biggr\} ,
</math>
</div>
where,
<div align="center">
<math>
\Chi_\mathrm{cube}^2 \equiv \frac{2^2 B^4 A^2}{3^3 \chi_1^4 \chi_2^2} .
</math>
</div>
This is the desired inversion which gives <math>\varpi(\chi_1,\chi_2)</math> in the case <math>q^2 = 3</math>.