Editing 2DStructure/ToroidalCoordinates (section)

===Using Toroidal Coordinates with Special Alignment===

If, instead, we use a toroidal coordinate system, we have (see p. 666 of MF53),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~dV_{3D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- [h_1 d\xi_1 ] [h_2 d\xi_2] [h_3 d\xi_3] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\biggl[ \frac{a d\xi_1}{ (\xi_1-\xi_2)\sqrt{\xi_1^2-1} } \biggr]~ 
\biggl[\frac{a d\xi_2}{ (\xi_1-\xi_2)\sqrt{1-\xi_2^2} } \biggr] ~
\biggl[ \biggl( \frac{\xi_1^2 - 1}{1 - \xi_3^2} \biggr)^{1/2} \frac{a d\xi_3}{(\xi_1-\xi_2)} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- d\sigma~
\biggl[ \frac{a(\xi_1^2 - 1)^{1/2}}{(\xi_1-\xi_2)} \frac{d(\cos\varphi)}{\sqrt{1 - \cos^2\varphi}} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>+~[\varpi d\varphi ] d\sigma \, , </math>
  </td>
</tr>
</table>
</div>
where, as employed above, a differential area element in the meridional plane of a toroidal-coordinate system is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~d\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2\biggl[ \frac{d\xi_1}{ (\xi_1-\xi_2)\sqrt{\xi_1^2-1} } \biggr]~ 
\biggl[\frac{d\xi_2}{ (\xi_1-\xi_2)\sqrt{1-\xi_2^2} } \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
As in the case of the cylindrical coordinate system, because the torus is axisymmetric, integration over the azimuthal angular coordinate in a toroidal coordinate system gives,

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

<tr>
  <td align="right">
<math>~dV_{2D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\varpi d\sigma \int_0^{2\pi} d\varphi  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi \varpi d\sigma \, .</math>
  </td>
</tr>
</table>
</div>

<table border="1" align="center" width="90%" cellpadding="8">
<tr><th align="center">[https://en.wikipedia.org/wiki/Toroidal_coordinates#Scale_factors Cross-check:  Wikipedia's Differential Volume Element]</th></tr>
<tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~dV_{3D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~[h_1 d\tau ] [h_2 d\theta] [h_3 d\varphi] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^3 \biggl[\frac{ d\tau}{\cosh\tau - \cos\theta} \biggr]  \biggl[\frac{ d\theta}{\cosh\tau - \cos\theta} \biggr] 
\biggl[\frac{\sinh\tau ~d\varphi }{\cosh\tau - \cos\theta} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~d\sigma \varpi d\varphi \, , </math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~d\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{ a}{\cosh\tau - \cos\theta} \biggr]^2 d\tau d\theta \, , 
</math>
  </td>
</tr>
</table>
and, [[#Properties|as above]], 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\varpi}{a}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\sinh\tau }{\cosh\tau - \cos\theta}  \, .
</math>
  </td>
</tr>
</table>
Hence, in agreement with the expression derived using the notation of MF53,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~dV_{2D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\varpi d\sigma \int_0^{2\pi} d\varphi  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi \varpi d\sigma \, .</math>
  </td>
</tr>
</table>

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

Now, if we choose a toroidal coordinate system whose origin is located exactly as defined in our [[#Special_Case|special case, above]], the ''radial''-coordinate integration limits should be,
<div align="center">
<math>~\xi_1|_\mathrm{min} = \frac{\varpi_t}{r_t} </math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; <math>~~\xi_1|_\mathrm{max} = \infty  \, ;</math>
</div>
and the integral over the "angular" toroidal coordinate will have the simple limits,
<div align="center">
<math>~\xi_2|_- = -1  </math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; <math>~\xi_2|_+ = +1 \, .</math>
</div>
[<font color="red">COMMENT:</font> Actually, these limits will only capture integration over either the upper hemisphere (<math>~Z</math> positive) or the lower hemisphere (<math>~Z</math> negative).  So I will probably need to double the volume expression that results from these limits.]


So, integration over the remaining two (meridional-plane) coordinates gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi \int_\Sigma \varpi d\sigma </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\int\limits_{-1}^{1} \frac{(\xi_1^2 - 1)^{1/2}}{(\xi_1-\xi_2)} \biggl[ \frac{1}{ (\xi_1-\xi_2)\sqrt{\xi_1^2-1} } \biggr]~ 
\biggl[\frac{1}{ (\xi_1-\xi_2)\sqrt{1-\xi_2^2} } \biggr] d\xi_2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\int\limits_{-1}^{1}  \biggl[\frac{d\xi_2}{ (\xi_1-\xi_2)^3\sqrt{1-\xi_2^2} } \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
If, following MF53, we make the substitution, 
<div align="center">
<math>~\xi_2 ~\rightarrow \cos\zeta</math>
&nbsp; &nbsp; &nbsp; <math>~\Rightarrow </math>&nbsp; &nbsp; &nbsp; 
<math>~d\xi_2 ~\rightarrow -\sin\zeta d\zeta \, ,</math>
</div>
we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\int\limits_{0}^{\pi}  \biggl[\frac{d\zeta}{ (\xi_1-\cos\zeta)^3 } \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\biggl\{
\frac{\sin\zeta [ 4\xi_1^2 - 3\xi_1 \cos\zeta - 1]}{(\xi_1^2-1)^2 (\xi_1 - \cos\zeta)^2}
- \frac{2(2\xi_1^2 + 1)\tanh^{-1}\biggl[\frac{(\xi_1 + 1)\tan(\zeta/2)}{\sqrt{1-\xi_1^2}}  \biggr]}{(1-\xi_1^2)^{5/2}}
\biggr\}_{0}^{\pi} \, ,
</math>
  </td>
</tr>

<!-- COMMENT OUT MANY LINES ...
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\biggl\{
\frac{\sin\zeta [ 4\xi_1^2 - 3\xi_1 \cos\zeta - 1]}{(\xi_1^2-1)^2 (\xi_1 - \cos\zeta)^2}
\biggr\}_{0}^{\pi/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1 \biggl[ \frac{ 4\xi_1^2 - 1}{(\xi_1^2-1)^2 \xi_1^2 } \biggr]
=
2\pi a^3 \int\limits_{\cosh^{-1}(\varpi_t/r_t)}^\infty dx \biggl[ \frac{ 4\cosh^2x - 1}{\sinh^3 x \cosh^2 x } \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}\pi a^3 \biggl\{
-3\sinh^{-2}(x/2) - 3 \cosh^{-2}(x/2) + \frac{8}{\cosh(x)} - 4\ln[\sinh(x/2)] + 4\ln[\cosh(x/2)]
\biggr\}_{\cosh^{-1}(\varpi_t/r_t)}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}\pi a^3 \biggl\{
-3\biggl[\frac{1}{\sinh^{2}(x/2)} + \frac{1}{\cosh^{2}(x/2)}\biggr] + \frac{8}{\cosh(x)} + \ln[\coth^4(x/2)]
\biggr\}_{\cosh^{-1}(\varpi_t/r_t)}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4}\pi a^3 \biggl\{
-3\biggl[\frac{2}{(\cosh x - 1)} + \frac{2}{(\cosh x + 1)}\biggr] + \frac{8}{\cosh(x)} + \ln\biggl[\frac{\cosh x + 1}{\cosh x - 1} \biggr]^2
\biggr\}_{\cosh^{-1}(\varpi_t/r_t)}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\pi a^3 \biggl\{
-3\biggl[\frac{1}{(\xi_1 - 1)} + \frac{1}{(\xi_1 + 1)}\biggr] + \frac{4}{\xi_1} + \ln\biggl[\frac{1 + 1/\xi_1}{1 - 1/\xi_1} \biggr]
\biggr\}_{\varpi_t/r_t}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\pi a^3 \biggl\{
-3\biggl[\frac{2\xi_1}{(\xi_1^2 - 1)} \biggr] + \frac{4}{\xi_1} + \ln\biggl[\frac{1 + 1/\xi_1}{1 - 1/\xi_1} \biggr]
\biggr\}_{\varpi_t/r_t}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \biggl\{
\frac{-(2+\xi_1^2)}{\xi_1(\xi_1^2 - 1)}  + \ln\biggl[\frac{1 + 1/\xi_1}{1 - 1/\xi_1} \biggr]^{1/2}
\biggr\}_{\varpi_t/r_t}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \biggl\{
\frac{2+(\varpi_t/r_t)^2}{\varpi_t/r_t[(\varpi_t/r_t)^2 - 1]}  - \ln\biggl[\frac{\varpi_t + r_t}{\varpi_t - r_t} \biggr]^{1/2}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi \biggl\{ r_t \biggl[\biggl( \frac{\varpi_t}{r_t} \biggr)^2 - 1  \biggr]^{1/2} \biggr\}^3 \biggl\{
\frac{r_t(2r_t^2+\varpi_t^2)}{\varpi_t(\varpi_t^2 - r_t^2)}  - \ln\biggl[\frac{\varpi_t + r_t}{\varpi_t - r_t} \biggr]^{1/2}
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi (\varpi_t^2 - r_t^2)^{3/2} \biggl\{
\frac{r_t(2r_t^2+\varpi_t^2)}{\varpi_t(\varpi_t^2 - r_t^2)}  - \ln\biggl[\frac{\varpi_t + r_t}{\varpi_t - r_t} \biggr]^{1/2}
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

At first glance, this does not appear to give the correct expression for the volume of a torus.

END COMMENTED-OUT    -->
</table>
</div>
where the expression obtained after integrating over <math>~\zeta</math> was obtained from WolframAlpha's online integrator.  

[<font color="red">COMMENT:</font> This result is problematic because it was derived without enforcing the condition, <math>~\xi_1^2 > 1</math>.  Notice, in particular, that the last term includes a couple of square-roots of expressions that will naturally be negative.]

Carrying out this same integration (specifying wider integration limits based on the toroidal-coordinate specification [https://en.wikipedia.org/wiki/Toroidal_coordinates#Definition described in Wikipedia]) via multiple steps using the integral tables published by [http://www.mathtable.com/gr/ Gradshteyn &amp; Ryzhik], I have obtained,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\int\limits_{-\pi}^{\pi}  \biggl[\frac{d\zeta}{ (\xi_1-\cos\zeta)^3 } \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\biggl\{
\frac{\sin\zeta [ 4\xi_1^2 - 3\xi_1 \cos\zeta - 1]}{(\xi_1^2-1)^2 (\xi_1 - \cos\zeta)^2}
+ \biggl[ \frac{2(2\xi_1^2 + 1)}{(\xi_1^2-1)^{5/2}}\biggr] \tan^{-1}\biggl[\tan\biggl(\frac{\zeta}{2}\biggr) \cdot \frac{(\xi_1 + 1)^{1/2}}{(\xi_1 - 1)^{1/2}}  \biggr]
\biggr\}_{-\pi}^{\pi} \, .
</math>
  </td>
</tr>

</table>
</div>

This expression makes more sense; at least the arguments of the square-roots are all positive.  Now, evaluating the limits:  (1) The first term inside the curly braces goes to zero at both limits; and (2) the argument of the arctangent is <math>~\pm \infty.</math>  Hence, the result of taking the arctangent is <math>~+\tfrac{\pi}{2}</math>, at the upper limit, and is <math>~-\tfrac{\pi}{2}</math> at the lower limit.  Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \int\limits_{\varpi_t/r_t}^\infty d\xi_1
\biggl\{\biggl[ \frac{2(2\xi_1^2 + 1)}{(\xi_1^2-1)^{5/2}}\biggr] \biggl(\frac{\pi}{2} + \frac{\pi}{2}\biggr)  
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 a^3 \int\limits_{\varpi_t/r_t}^\infty 
\biggl[ \frac{(2\xi_1^2 + 1)}{(\xi_1^2-1)^{5/2}}\biggr] d\xi_1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 a^3 \biggl[ - \frac{\xi_1}{(\xi_1^2 - 1)^{3/2}} \biggr]_{\varpi_t/r_t}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 a^3 \biggl[ \frac{\varpi_t r_t^2}{(\varpi_t^2 - r_t)^{3/2}} \biggr]\, .
</math>
  </td>
</tr>
</table>
</div>

Now, in the special case we are considering here,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r_t \biggl[\biggl( \frac{\varpi_t}{r_t} \biggr)^2 - 1  \biggr]^{1/2} 
= [\varpi_t^2 - r_t^2]^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi^2 \varpi_t r_t^2 \, ,
</math>
  </td>
</tr>
</table>
</div>
which is the answer we were expecting for the volume of the (pink) torus.

<!--  COMMENT OUT
There are some differences between the last term obtained via the printed table of integrals versus via WolframAlpha's online integrator.  Eventually, I would like to reconcile these differences.  In the meantime, the integration limits appear to be a problem because (good news, perhaps) the first term involving <math>~\sin\zeta</math> goes to zero no matter how you slice it; but (bad news), the tangent function in the last term goes to <math>~\pm \infty</math>.  So I'm not sure, yet, how to assess this result.


<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tfrac{1}{2}\pi a^3 \biggl\{
\frac{2(\xi_1^2 + 2)}{\xi_1(1-\xi_1^2)} + \ln\biggl[ \frac{1+\xi_1}{1-\xi_1} \biggr]
\biggr\}_{\varpi_t/r_t}^\infty \, .
</math>
  </td>
</tr>
At first glance, this doesn't work because at both limiting values, the argument of the logarithm is negative.  Maybe I need to change to cosh function before integrating.
-->
<!-- NO LONGER INTERESTED IN THE FOLLOWING...
Let's try switching the order of the integrations.
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi a^3 
\int\limits_{-\pi}^{\pi} d\zeta \int\limits_{\varpi_t/r_t}^\infty \biggl[\frac{d\xi_1}{ (\xi_1-\cos\zeta)^3 } \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \pi a^3 \int\limits_{-\pi}^{\pi} d\zeta 
\biggl\{\frac{1}{(\xi_1-\cos\zeta)^2}
\biggr\}_{\varpi_t/r_t}^\infty
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi a^3 \int\limits_{-\pi}^{\pi}  
\biggl[\frac{d\zeta}{(\varpi_t/r_t- \cos\zeta)^2}
\biggr]
</math>
  </td>
</tr>

</table>
</div>

===Another Simple Toroidal-System Alignment===
Okay, because the struggle with our previous volume integration had to do with the limits on integration, let's try moving the origin of the toroidal-coordinate system to a position that lies outside of the (pink) torus &#8212; let's choose, <math>~0 < a < \varpi_t</math>.  For simplicity, however, let's keep the origin in the equatorial plane, that is, set <math>~Z_0 = 0</math>.  Given that <math>~\varpi_t = \tfrac{3}{4}</math> and <math>~r_t = \tfrac{1}{4}</math>, from [[#Associated_Analytic_Expressions|our above derivations]] of the "radial-coordinate" limits of integration, we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\kappa</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2  - (\varpi_t^2 - r_t^2)  = a^2 - \tfrac{1}{2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\beta_\pm</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
- \frac{\kappa}{2} \biggl[ \frac{\varpi_t \mp r_t \sqrt{C}}{(\varpi_t + r_t)(\varpi_t - r_t)} \biggr] 
= -\frac{1}{2}\biggl(a^2 - \frac{1}{2} \biggr)\biggl[ \frac{\varpi_t \mp r_t }{(\varpi_t^2 - r_t^2)} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{1}{2} - a^2\biggr)\biggl[\frac{3}{4} \mp \frac{1}{4}  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Let's try, <math>~a = \tfrac{1}{2}</math>, in which case,
<div align="center">
<math>\beta_+ = \frac{1}{8}</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\beta_- = \frac{1}{4}\, ,</math>
</div>
and,
<div align="center">
<math>\xi_1|_\mathrm{max} = 
\biggl[1-\biggl( \frac{a}{\varpi_t-\beta_+} \biggr)^2  \biggr]^{-1/2}
= \biggl[1-\biggl( \frac{4}{5} \biggr)^2  \biggr]^{-1/2}
</math>
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp; 
<math>\xi_1|_\mathrm{min} = 
\biggl[1-\biggl( \frac{a}{\varpi_t-\beta_-} \biggr)^2  \biggr]^{-1/2}
\, ,</math>
</div>

FINISHED COMMENTING-OUT LINES THAT ARE NO LONGER OF INTEREST -->