Editing Appendix/Ramblings/MacSphCriticalPoints (section)

==Model01==

<table border="1" align="center" cellpadding="8" width="80%">
<tr>
  <td align="center">Model</td>
  <td align="center"><math>\frac{b}{a}</math></td>
  <td align="center"><math>\frac{c}{a}</math></td>
  <td align="center"><math>e</math></td>
  <td align="center"><math>\frac{\Omega^2}{\pi G\rho}</math></td>
  <td align="center"><math>\frac{L}{(GM^3\bar{a})^{1 / 2}}</math></td>
  <td align="center"><math>\frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math></td>
</tr>
<tr>
  <td align="center">01</td>
  <td align="center"><math>1</math></td>
  <td align="center"><math>\frac{\sqrt{3}}{2} = 0.866025404</math></td>
  <td align="center"><math>\sin(\pi/6) = 0.5</math></td>
  <td align="center"><math>\biggl[\frac{10\pi}{ 3^{1 / 2}} - 18\biggr]</math><br />&nbsp;<br /><math>= 0.137993640</math></td>
  <td align="center"><math>\frac{1}{5}\biggl( \frac{2^{14}}{3}\biggr)^{1 / 12} \biggl[ 5\pi - 3^{5/2}  \biggr]^{1 / 2}</math><br />&nbsp;<br /><math> = 0.141633637</math></td>
  <td align="center"><math>\biggl[\frac{3}{\pi}\biggl( 2\pi - 3^{3/2}\biggr) - 1 \biggr]</math><br />&nbsp;<br /><math>=0.038039942</math></td>
</tr>

<tr>
  <td align="left" colspan="7">

Model01 is a Maclaurin spheroid for which <math>(a, b, c) = (1, 1, \sqrt{3}/2)</math>.  This specific model has been chosen because &hellip;
<ul>
<li>It is axisymmetric &#8212; hence, it can, in principle, be modeled with a two-dimensional (rather than 3D) code;</li>
<li>It is rotating rather slowly, so it is only mildly distorted from a sphere &#8212; in the EFE Diagram it lies between a spherically symmetric model (upper-right corner) and the point at which the Jacobi/Dedekind sequence bifurcates from the Maclaurin-spheroid sequence (see our discussion of Model03, below);</li>
<li>Its eccentricity, <math>e</math>, is precisely <math>1/2</math> and is among the models for which the numerical value of various physical parameters can be found in Chapter 5, Table I (p. 78) of EFE &#8212; this provides a cross-check for our model;</li>
<li>Because <math>\sin^{-1}e = \pi/6</math> precisely, the expression for <math>\Omega^2</math> can be written in algebraic form.</li>
</ul>
  </td>
</tr>
</table>

For this chosen value of <math>c/a</math>, we appreciate that the eccentricity is, 
<table border="0" align="center" cellpadding="8">
<tr>
 <td align="right"><math>e</math></td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\biggl[1 - \frac{c^2}{a^2}\biggr]^{1 / 2}
=
\biggl[1 - \frac{3}{4}\biggr]^{1 / 2}
=
\frac{1}{2} \, ,
</math>
  </td>
</tr>
</table>
in which case,
<table border="0" align="center" cellpadding="8">
<tr>
 <td align="right"><math>\frac{\sin^{-1}e}{e}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{(\pi/6)}{(1/2)} = \frac{\pi}{3}\, .
</math>
  </td>
</tr>
</table>
Hence, from our [[Apps/MaclaurinSpheroidSequence#MaclaurinFrequency|accompanying discussion of the Maclaurin spheroid sequence]],

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

<tr>
  <td align="right">
<math>
\frac{\Omega^2}{\pi G \rho }
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2}{e^2}(3-2e^2)(1-e^2)^{1 / 2} \cdot \frac{\sin^{-1}e}{e} 
- \frac{6(1-e^2)}{e^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2}{(1/2)^2}\biggl[ 3-2\biggl(\frac{1}{2}\biggr)^2 \biggr]\frac{3^{1 / 2}}{2} \cdot \frac{\pi}{3} 
- \frac{6(3/4)}{(1/2)^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
2^3\biggl[ \frac{5}{2} \biggr]\frac{\pi}{2\cdot 3^{1 / 2}} 
- 18
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{10\pi}{ 3^{1 / 2}} - 18 \, .
</math>
  </td>
</tr>
</table>

The [[Apps/MaclaurinSpheroidSequence#Corresponding_Total_Angular_Momentum|corresponding total angular momentum]] is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{L}{(GM^3\bar{a})^{1 / 2}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{6^{1 / 2}}{5} \biggl[  A_1 - A_3 (1-e^2) \biggr]^{1 / 2}(1 - e^2)^{-1 / 3} \, ,</math>
  </td>
</tr>
</table>
where,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>
A_1
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2}  \biggr](1-e^2)^{1/2} 
= 0.627598728\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
A_3
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{1/2} 
= 0.744802542\, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ A_1 - A_3 (1-e^2) 
</math>
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{e^2} \biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2}  \biggr](1-e^2)^{1/2}
-
\frac{2}{e^2} \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)^{3/2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{(1-e^2)^{1/2}}{e^2} \biggl\{ 
\biggl[\frac{\sin^{-1}e}{e} - (1-e^2)^{1/2}  \biggr]
-
2 \biggl[(1-e^2)^{-1/2} -\frac{\sin^{-1}e}{e} \biggr](1-e^2)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{(1-e^2)^{1/2}}{e^2} \biggl\{ 
3 \biggl[ \frac{\sin^{-1}e}{e}\biggr] - 3(1-e^2)^{1/2}  
-
2 e \sin^{-1}e 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{\sqrt{3}/2}{(1/2)^2} \biggl\{ 
\pi - \frac{3\sqrt{3}}{2}  
-
\frac{\pi}{6} 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\frac{1}{\sqrt{3}} \biggl[ 5\pi - 3^{5/2}  \biggr]  = 0.068996821 \, .
</math>
  </td>
</tr>
</table>
Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{L}{(GM^3\bar{a})^{1 / 2}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{6^{1 / 2}}{5} \biggl\{  3^{-1 / 2} \biggl[ 5\pi - 3^{5/2}  \biggr]\biggr\}^{1 / 2}\biggl(\frac{4}{3} \biggr)^{1 / 3}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{2^{1 / 2} \cdot 3^{1 / 2}}{3^{1 / 4} \cdot 5}  \biggl[ 5\pi - 3^{5/2}  \biggr]^{1 / 2}\cdot \frac{2^{2 / 3}}{3^{1 / 3}}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>  
\frac{1}{5}\biggl( \frac{2^{14}}{3}\biggr)^{1 / 12} \biggl[ 5\pi - 3^{5/2}  \biggr]^{1 / 2} 
= 0.141633637\, .
</math>
  </td>
</tr>
</table>

Finally, the [[Apps/MaclaurinSpheroidSequence#Alternate_Sequence_Diagrams|ratio of rotational to gravitational potential energy]] is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tau \equiv \frac{T_\mathrm{rot}}{|W_\mathrm{grav}|}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2e^2}\biggl[ 1 - \frac{e(1-e^2)^{1 / 2}}{\sin^{-1} e}\biggr] - 1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{3}{\pi}\biggl[ 2\pi - 3^{3/2}\biggr] - 1
= 0.038039942 \, .</math>
  </td>
</tr>
</table>