Editing VE/RiemannEllipsoids (section)

==Riemann S-Type Ellipsoids==
In this case, we assume that <math>~\vec{\Omega}</math> and <math>~\vec\zeta</math> are aligned with each other and, as well, are aligned with the <math>~z</math>-axis; that is to say, in addition to setting <math>~(\Omega_1, \zeta_1) = (0, 0)</math> we also set <math>~(\Omega_2, \zeta_2) = (0, 0)</math>.  So, there are only three unknowns &#8212; <math>~\Pi, (\Omega_3, \zeta_3)</math> &#8212; and they can be determined by ignoring off-axis expressions and simultaneously solving the ''diagonal element'' expressions  displayed in our above [[#SummaryTable|''Summary Table'']].  Furthermore, two of the three diagonal-element expressions can be simplified because we are setting <math>~(\Omega_2, \zeta_2) = (0, 0)</math>.  The three relevant equilibrium constraints are:


<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">Indices</td>
  <td align="center" rowspan="2">2<sup>nd</sup>-Order TVE Expressions that are Relevant to Riemann S-Type Ellipsoids</td>
</tr>
<tr>
  <td align="center" width="5%"><math>~i</math></td>
  <td align="center" width="5%"><math>~j</math></td>
</tr>
<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~1</math></td>
  <td align="left">

<table align="left" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3\cdot 5}{2^2\pi a b c\rho} \biggr] \Pi
+\biggl\{ 
\Omega_3^2
+ 2  \biggl[ \frac{b^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 
~-~(2\pi G\rho) A_1 
\biggr\} a^2 
+ \biggl[ \frac{a^2}{a^2 + b^2}\biggr]^2 \zeta_3^2 b^2
</math>
  </td>
</tr>
</table>

  </td>
</tr>

<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~2</math></td>
  <td align="left">

<table align="left" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3\cdot 5}{2^2\pi a b c \rho} \biggr]\Pi
+ \biggl[ \frac{b^2}{b^2+a^2}\biggr]^2 \zeta_3^2 a^2
+ \biggl\{
\Omega_3^2  
+ 2 \biggl[ \frac{a^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3  
~-~( 2\pi G \rho) A_2 
\biggr\}b^2 
</math>
  </td>
</tr>
</table>

  </td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center"><math>~3</math></td>
  <td align="left">

<table align="left" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3\cdot 5}{2^2\pi abc\rho} \biggr]\Pi
- (2\pi G \rho)A_3 c^2 
</math>
  </td>
</tr>
</table>

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


The <math>~(i, j) = (3, 3)</math> component expression immediately identifies the value of one of the unknowns, namely,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Pi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2^3\pi^2}{3\cdot 5} \biggr) G \rho^2A_3 a b c^3 \, .
</math>
  </td>
</tr>
</table>

From the remaining pair of diagonal-element expressions, we therefore have,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~
0 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
a^2 \Omega_3^2
+ 2  \biggl[ \frac{b^2a^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 
+ \biggl[ \frac{a^2}{a^2 + b^2}\biggr]^2 \zeta_3^2 b^2
~+~(2\pi G\rho)(A_3 c^2 - A_1  a^2 ) \, ,
</math>
  </td>
</tr>
</table>

and,
<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~
0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{b^2}{b^2+a^2}\biggr]^2 \zeta_3^2 a^2
+ 
b^2 \Omega_3^2  
+ 2 \biggl[ \frac{a^2b^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3  
~+~( 2\pi G \rho)(A_3 c^2 - A_2 b^2) \, .
</math>
  </td>
</tr>
</table>
Multiplying the first of these two expressions through by <math>~b^2</math> and the second through by <math>~a^2</math>, then subtracting the second from the first gives,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
b^2\biggl\{ 
2  \biggl[ \frac{b^2a^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 
+ \biggl[ \frac{a^2}{a^2 + b^2}\biggr]^2 \zeta_3^2 b^2
~+~(2\pi G\rho)(A_3 c^2 - A_1  a^2 ) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
-~
a^2\biggl\{
\biggl[ \frac{b^2}{b^2+a^2}\biggr]^2 \zeta_3^2 a^2
+ 2 \biggl[ \frac{a^2b^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3  
~+~( 2\pi G \rho)(A_3 c^2 - A_2 b^2) 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ 
2  \biggl[ \frac{b^4 a^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 
~+~(2\pi G\rho)(A_3 c^2 - A_1  a^2 )b^2 \biggr\}
~-~
\biggl\{
2 \biggl[ \frac{a^4 b^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3  
~+~( 2\pi G \rho)(A_3 c^2 - A_2 b^2) a^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[ \frac{b^2 a^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi G\rho \biggl[ \frac{(A_3 c^2 - A_2 b^2) a^2 ~-~(A_3 c^2 - A_1  a^2 )b^2}{ b^2 - a^2} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi G\rho \biggl[ \frac{(A_1 - A_2)a^2b^2}{ b^2 - a^2} - A_3 c^2\biggr] \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 7, &sect;48, Eq. (30)</font> ]</td></tr>
</table>
Note that &#8212; as EFE has done and as we have recorded in a [[ThreeDimensionalConfigurations/JacobiEllipsoids#Equilibrium_Conditions_for_Jacobi_Ellipsoids|related discussion]] &#8212; the first term on the right-hand-side of this last expression can be expressed more compactly in terms of the coefficient, <math>~A_{12}</math>.

Alternatively, dividing the first expression through by <math>~a^2</math> and the second by <math>~b^2</math>, then adding the pair of expressions gives,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~
0 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Omega_3^2
+ 2  \biggl[ \frac{b^2}{b^2+a^2}\biggr] \Omega_3 \zeta_3 
+ \biggl[ \frac{a^2b^2}{(a^2 + b^2)^2}\biggr] \zeta_3^2 
~+~(2\pi G\rho)(A_3 c^2 - A_1  a^2 )\frac{1}{a^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+~
\biggl[ \frac{a^2 b^2}{(b^2+a^2)^2}\biggr] \zeta_3^2 
+ 
\Omega_3^2  
+ 2 \biggl[ \frac{a^2}{a^2 + b^2}\biggr] \Omega_3 \zeta_3  
~+~( 2\pi G \rho)(A_3 c^2 - A_2 b^2) \frac{1}{b^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2\Omega_3^2 + 2   \Omega_3 \zeta_3 
+ 2\biggl[ \frac{a^2b^2}{(a^2 + b^2)^2}\biggr] \zeta_3^2 
~+~2\pi G\rho \biggl[ \frac{A_3 c^2 - A_1  a^2 }{a^2} + \frac{A_3c^2 - A_2 b^2}{b^2}\biggr] \, .
</math>
  </td>
</tr>
</table>

If we divide through by 2, then replace the product, <math>~\Omega_3\zeta_3</math>, in this expression by the relation derived immediately above, we have,

<table align="center" border=0 cellpadding="3">
<tr>
  <td align="right">
<math>~
\Omega_3^2  
+ \biggl[ \frac{a^2b^2}{(a^2 + b^2)^2}\biggr] \zeta_3^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
~-~\pi G\rho \biggl[ \frac{b^2 (A_3 c^2 - A_1  a^2) + a^2(A_3c^2 - A_2 b^2 ) }{a^2b^2} \biggr] 
~-~   
\pi G\rho \biggl[ \frac{(A_1 - A_2)a^2b^2 - A_3 c^2(b^2 - a^2)}{ b^2 - a^2} \biggr]\biggl[ \frac{b^2+a^2}{b^2 a^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\pi G\rho}{ a^2b^2(a^2-b^2) } 
\biggl\{ [ b^2 (A_3 c^2 - A_1  a^2) + a^2(A_3c^2 - A_2 b^2 )](b^2-a^2) 
~+~   
[ (A_1 - A_2)a^2b^2 - A_3 c^2(b^2 - a^2) ](b^2+a^2) 
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\pi G\rho}{ (a^2-b^2) } 
\biggl[ 
A_1   a^2 
- A_2  b^2 
\biggr] \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 7, &sect;48, Eq. (29)</font> ]</td></tr>
</table>

<span id="fDefined">It has become customary to characterize each Riemann S-Type ellipsoid by the value of its equilibrium frequency ratio, </span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\zeta_3}{\Omega_3} \, ,</math>
  </td>
</tr>
</table>
in which case the relevant pair of constraint equations becomes,

<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="center">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[ \frac{b^2 a^2}{b^2+a^2}\biggr] f \Omega_3^2   
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\pi G\rho \biggl[ \frac{(A_1 - A_2)a^2b^2}{ b^2 - a^2} - A_3 c^2\biggr] \, ;
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 7, &sect;48, Eq. (34)</font> ]</td></tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\Omega_3^2 \biggl\{1
+ \biggl[ \frac{a^2b^2}{(a^2 + b^2)^2}\biggr] f^2 \biggr\}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2\pi G\rho}{ (a^2-b^2) } 
\biggl[ 
A_1   a^2 
- A_2  b^2 
\biggr] \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[User:Tohline/Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 7, &sect;48, Eq. (33)</font> ]</td></tr>
</table>

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

These two equations can straightforwardly be combined to generate a quadratic equation for the frequency ratio, <math>~f</math>.  Then, once the value of <math>~f</math> has been determined, either expression can be used to determine the corresponding equilibrium value for <math>~\Omega_3</math> in the unit of <math>~(\pi G \rho)^{1 / 2}</math>.  The fact that the value of <math>~f</math> is determined from the solution of  a quadratic equation underscores the realization that, for a given specification of the ellipsoidal geometry <math>~(a, b, c)</math>, if an equilibrium exists &#8212; ''i.e.,'' if the solution for <math>~f</math> is real rather than imaginary &#8212; then two equally valid, and usually different (''i.e.,'' non-degenerate), values of <math>~f</math> will be realized.  This means that two different underlying flows &#8212; one ''direct'' and the other ''adjoint'' &#8212; will sustain the shape of the ellipsoidal configuration, as viewed from a frame that is rotating about the <math>~z</math>-axis with frequency, <math>~\Omega_3</math>.