Editing ThreeDimensionalConfigurations/RiemannStype (section)

====Implied Parameter Values====

So, at the surface of the ellipsoid (where the enthalpy ''H = 0'') on each of its three principal axes, the equilibrium conditions demanded by the expression for detailed force balance become, respectively:
<ol type="I">
<li>On the x-axis, where (x, y, z) = (a, 0, 0):
<br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_B</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi G \rho \biggl[ I_\mathrm{BT} a^2 - A_1 a^2   \biggr]
- \frac{1}{2} \Omega_f^2(a^2 ) 
- \frac{1}{2} \lambda^2(a^2)  
+ \Omega_f \lambda \biggl(\frac{b}{a}\cdot a^2  \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~2\biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_1  - \Omega_f^2 - \lambda^2  + 2\Omega_f \lambda \biggl(\frac{b}{a}  \biggr)  
</math>
  </td>
</tr>
</table>
</li>
<li>On the y-axis, where (x, y, z) = (0, b, 0):
<br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_B</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi G \rho \biggl[ I_\mathrm{BT} a^2 - A_2 b^2 \biggr]
- \frac{1}{2} \Omega_f^2(b^2) 
- \frac{1}{2} \lambda^2(b^2)  
+ \Omega_f \lambda \biggl(\frac{a}{b}\cdot b^2 \biggr)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2\biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_2 \biggl( \frac{b^2}{a^2}\biggr) 
- \Omega_f^2 \biggl( \frac{b^2}{a^2} \biggr) 
- \lambda^2\biggl( \frac{b^2}{a^2} \biggr) 
+ 2\Omega_f \lambda \biggl(\frac{b}{a}\biggr)  
</math>
  </td>
</tr>
</table>
</li>
<li>On the z-axis, where (x, y, z) = (0, 0, c):
<br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~C_B</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\pi G \rho \biggl[ I_\mathrm{BT} a^2 - A_3 c^2  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2 \biggl[ \frac{C_B}{a^2} + (\pi G\rho)I_\mathrm{BT}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr)
</math>
  </td>
</tr>
</table>
</li>
</ol>

Using the result from "III" to replace the left-hand side of both relation "I" and relation "II", we find that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_1  - \Omega_f^2 - \lambda^2  + 2\Omega_f \lambda \biggl(\frac{b}{a}  \biggr)  \, ,
</math>
  </td>
</tr>
</table>

and,

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

<tr>
  <td align="right">
<math>~(2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_2 \biggl( \frac{b^2}{a^2}\biggr) 
- \Omega_f^2 \biggl( \frac{b^2}{a^2} \biggr) 
- \lambda^2\biggl( \frac{b^2}{a^2} \biggr) 
+ 2\Omega_f \lambda \biggl(\frac{b}{a}\biggr)  \, .
</math>
  </td>
</tr>
</table>
</div>

Multiplying the first of these two expressions by <math>~(b/a)^2</math> then subtracting it from the second gives,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr) - \biggl(\frac{b}{a}\biggr)^2 (2\pi G \rho) A_3 \biggl( \frac{c^2}{a^2}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) A_2 \biggl( \frac{b^2}{a^2}\biggr) 
+ 2\Omega_f \lambda \biggl(\frac{b}{a}\biggr) 
- \biggl(\frac{b}{a}\biggr)^2 \biggl[ (2\pi G \rho) A_1    + 2\Omega_f \lambda \biggl(\frac{b}{a}  \biggr)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{(\pi G \rho)c^2}{ab} \biggl[  A_3 a^2  -   A_3 b^2  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\pi G \rho) (A_2 - A_1) a b 
+ \Omega_f \lambda a^2  
- ab \biggl[ \Omega_f \lambda \biggl(\frac{b}{a}  \biggr)  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(\pi G \rho) (A_2 - A_1) a b 
+ \Omega_f \lambda ( a^2 -  b^2 )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\frac{\Omega_f \lambda}{\pi G \rho} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a b ( a^2 -  b^2 )}\biggl[ A_3 ( a^2  -  b^2 )c^2  -  (A_2 - A_1) a^2 b^2 \biggr] \, .
</math>
  </td>
</tr>
</table>

Alternatively, just subtracting the first expression from the second gives,

<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>~
(2\pi G \rho) A_2 \biggl( \frac{b^2}{a^2}\biggr) 
- \Omega_f^2 \biggl( \frac{b^2}{a^2} \biggr) 
- \lambda^2\biggl( \frac{b^2}{a^2} \biggr) 
-
\biggl[ (2\pi G \rho) A_1  - \Omega_f^2 - \lambda^2   \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(2\pi G \rho) \biggl[ A_2 \biggl( \frac{b^2}{a^2}\biggr) -  A_1  \biggr]
+ \Omega_f^2 \biggl[1 - \frac{b^2}{a^2}  \biggr] + \lambda^2  \biggl[1 - \frac{b^2}{a^2}  \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\Omega_f^2 + \lambda^2}{\pi G \rho}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] \, .
</math>
  </td>
</tr>
</table>

We can eliminate <math>~\lambda</math> between these last two expressions as follows:  From the first of the two, we have

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

<tr>
  <td align="right">
<math>~
\lambda 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\Omega_f} \biggl\{ \frac{\pi G \rho}{a b ( a^2 -  b^2 )}\biggl[ A_3 ( a^2  -  b^2 )c^2  -  (A_2 - A_1) a^2 b^2 \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
Hence, the second gives,

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

<tr>
  <td align="right">
<math>~\Omega_f^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 (\pi G \rho)\biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] - \lambda^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 (\pi G \rho)\biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] -   
\frac{1}{\Omega_f^2} \biggl\{ \frac{\pi G \rho}{a b ( a^2 -  b^2 )}\biggl[ A_3 ( a^2  -  b^2 )c^2  -  (A_2 - A_1) a^2 b^2 \biggr] \biggr\}^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{\Omega_f^4}{(\pi G \rho)^2} 
- \frac{2\Omega_f^2}{(\pi G \rho)} \biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] +   
\biggl\{ \frac{1}{a b ( a^2 -  b^2 )}\biggl[ A_3 ( a^2  -  b^2 )c^2  -  (A_2 - A_1) a^2 b^2 \biggr] \biggr\}^2 \, .
</math>
  </td>
</tr>
</table>
This is a quadratic equation whose solution gives <math>~\Omega_f^2/(\pi G \rho)</math> and, in turn,  <math>~\lambda^2/(\pi G \rho)</math>.  Specifically for ''Direct'' configurations, we find that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\Omega_f^2}{(\pi G \rho)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \biggl[M + \sqrt{ M^2 - 4N^2} \biggr] \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{\lambda^2}{(\pi G \rho)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \biggl[M - \sqrt{ M^2 - 4N^2} \biggr] \, ,</math>
  </td>
</tr>
</table>
{{ Ou2006 }}, p. 551, &sect;2, Eqs. (15) &amp; (16)
</div>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
2\biggl[ A_1 - A_2 \biggl( \frac{b^2}{a^2}\biggr) \biggr]\biggl[ \frac{a^2}{a^2 - b^2} \biggr] \, ,</math> &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>~N</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{1}{a b ( a^2 -  b^2 )}\biggl[ A_3 ( a^2  -  b^2 )c^2  -  (A_2 - A_1) a^2 b^2 \biggr] \, .
</math>
  </td>
</tr>
</table>


<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="11"><b>TEST (part 3)</b></td>
</tr>
<tr>
  <td align="center" rowspan="1"><math>~\frac{b}{a}</math></td>
  <td align="center" rowspan="1"><math>~\frac{c}{a}</math></td>
  <td align="center" rowspan="1"><math>~A_1</math></td>
  <td align="center" rowspan="1"><math>~A_2</math></td>
  <td align="center" rowspan="1"><math>~A_3</math></td>
  <td align="center" rowspan="1"><math>~M</math></td>
  <td align="center" rowspan="1"><math>~N</math></td>
  <td align="center" rowspan="1"><math>~\frac{\Omega_f^2}{\pi G \rho}</math></td>
  <td align="center" rowspan="1"><math>~\frac{\lambda^2}{\pi G \rho}</math></td>
  <td align="center" rowspan="1"><math>~\frac{\Omega_f}{\sqrt{G \rho}}</math></td>
  <td align="center" rowspan="1"><math>~\frac{\lambda}{\sqrt{G \rho}}</math></td>
</tr>
<tr>
  <td align="center">0.9</td>
  <td align="center">0.641</td>
  <td align="center">0.521450273</td>
  <td align="center">0.595131012</td>
  <td align="center">0.883418715</td>
  <td align="center">0.414682903</td>
  <td align="center">0.054301271</td>
  <td align="center">0.407446048</td>
  <td align="center">0.007236855</td>
  <td align="center">1.131383892</td>
  <td align="center">0.150782130</td>
</tr>
</table>


The numerical values listed in the last two columns of this "part 3" test match the values listed [[#TestPart2|above in "part 2" of our test]] for, respectively, <math>~\omega</math> and <math>~\omega^\dagger</math>.