Editing ThreeDimensionalConfigurations/Challenges (section)

===Ou's (2006) Detailed Force Balance===

In a separate [[ThreeDimensionalConfigurations/RiemannStype#Based_on_Detailed_Force_Balance|accompanying chapter]], we have described in detail how [https://ui.adsabs.harvard.edu/abs/2006ApJ...639..549O/abstract Ou(2006)] used, essentially, the HSCF technique to solve the detailed force-balance equations.  Beginning with the,

<div align="center">
<font color="#770000">'''Eulerian Representation'''</font><br />
of the Euler Equation <br />
<font color="#770000">'''as viewed from a Rotating Reference Frame'''</font>

<math>\biggl[\frac{\partial\vec{v}}{\partial t}\biggr]_{rot} + ({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}= - \frac{1}{\rho} \nabla P - \nabla \Phi_\mathrm{grav} 
- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x}) - 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} \, ,</math> </div>


it can be shown that, for the velocity fields associated with all Riemann S-type ellipsoids,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~({\vec{v}}_{rot}\cdot \nabla) {\vec{v}}_{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\nabla \biggl[ \frac{1}{2} \lambda^2(x^2 + y^2) \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~- {\vec{\Omega}}_f \times ({\vec{\Omega}}_f \times \vec{x})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
+\nabla\biggl[\frac{1}{2} \Omega_f^2 (x^2 + y^2)  \biggr] \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~- 2{\vec{\Omega}}_f \times {\vec{v}}_{rot} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \nabla\biggl[ \Omega_f \lambda\biggl( \frac{b}{a} x^2 + \frac{a}{b}y^2 \biggr)  \biggr] \, .
</math>
  </td>
</tr>
</table>

<font color="orange">Hence, within</font> each steady-state <font color="orange"> configuration the following Bernoulli's function must be uniform in space:</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
H + \Phi_\mathrm{grav} - \frac{1}{2} \Omega_f^2(x^2 + y^2) 
- \frac{1}{2} \lambda^2(x^2 + y^2)  
+ \Omega_f \lambda \biggl(\frac{b}{a}x^2 + \frac{a}{b}y^2 \biggr)  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
C_B \, ,
</math>
  </td>
</tr>
</table>

[https://ui.adsabs.harvard.edu/abs/2006ApJ...639..549O/abstract Ou(2006)], p. 550, &sect;2, Eq. (6)
</div>

<font color="orange">where <math>~C_B</math> is a constant.</font>  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>~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>~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>~\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>

This third expression can be used to replace the left-hand-side of the first and second expressions.  Then via some additional algebraic manipulation, the first and second expressions can be combined to provide the desired solutions for the parameter pair, <math>~(\Omega_f, \lambda)</math>, namely,
<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>
[https://ui.adsabs.harvard.edu/abs/2006ApJ...639..549O/abstract Ou(2006)], 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>