Editing ThreeDimensionalConfigurations/Challenges (section)

==Riemann S-type Ellipsoids==

Usually, the density, <math>~\rho</math>, and the pair of axis ratios, <math>~b/a</math> and <math>~c/a</math>, are specified.  Then, the Poisson equation is solved to obtain <math>~\Phi_\mathrm{grav}</math> in terms of <math>~A_1</math>, <math>~A_2</math>, and <math>~A_3</math>.  The aim, then, is to determine the value of the central enthalpy, <math>~H_0</math> &#8212; alternatively, the thermal energy density, <math>~\Pi</math> &#8212; and the two parameters, <math>~\Omega_f</math> and <math>~\lambda</math>, that determine the magnitude of the velocity flow-field.  Keep in mind that, as viewed from a frame of reference that is spinning with the ellipsoid (at angular frequency, <math>~\Omega_f</math>), the adopted (rotating-frame) velocity field is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda \biggl[ \boldsymbol{\hat\imath} \biggl( \frac{a}{b}\biggr) y - \boldsymbol{\hat\jmath} \biggl( \frac{b}{a} \biggr) x \biggr] \, .</math>
  </td>
</tr>
</table>
Hence, the inertial-frame velocity is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{u} + \bold{\hat{e}}_\varphi \Omega_f \varpi \, .</math>
  </td>
</tr>
</table>

While we will fundamentally rely on the <math>~(\Omega_f, \lambda)</math> parameter pair to define the velocity flow-field, in discussions of Riemann S-type ellipsoids it is customary to also refer to the following two additional parameters:  The (rotating-frame) vorticity,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\boldsymbol{\zeta} \equiv \boldsymbol{\nabla \times}\bold{u}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\boldsymbol{\hat\imath} \biggl[ \frac{\partial u_z}{\partial y} - \frac{\partial u_y}{\partial z} \biggr]
+ \boldsymbol{\hat\jmath} \biggl[ \frac{\partial u_x}{\partial z} - \frac{\partial u_z}{\partial x} \biggr]
+ \bold{\hat{k}} \biggl[ \frac{\partial u_y}{\partial x} - \frac{\partial u_x}{\partial y} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{k}} \biggl[ - \lambda \biggl(\frac{b}{a} + \frac{a}{b}\biggr) \biggr] \, ;</math>
  </td>
</tr>
</table>
and the dimensionless frequency ratio,
<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}{\Omega_f} \, .</math>
  </td>
</tr>
</table>

===2<sup>nd</sup>-Order TVE Expressions===

As we have discussed in detail in an [[VE/RiemannEllipsoids#Riemann_S-Type_Ellipsoids|accompanying chapter]], the three diagonal elements of the <math>~(3 \times 3)</math> 2<sup>nd</sup>-order tensor virial equation are sufficient to determine the equilibrium values of <math>~\Pi</math>, <math>~\Omega_3</math>, and <math>~\zeta_3</math>.

<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> element gives <math>~\Pi</math> directly in terms of known parameters.  The <math>~(1, 1)</math> and <math>~(2, 2)</math> elements can then be combined in a couple of different ways to obtain  a coupled set of expressions that define <math>~\Omega_3</math> and <math>~f \equiv \zeta_3/\Omega_3</math>.


<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">[ [[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">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 7, &sect;48, Eq. (33)</font> ]</td></tr>
</table>

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


===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>

===Hybrid Scheme===

In a separate chapter we have detailed the [[Appendix/Ramblings/HybridSchemeImplications#Hybrid_Scheme|hybrid scheme]].  For steady-state configurations, the three components of the combined Euler + Continuity equations give,

<table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left">
<div align="center">'''Hybrid Scheme Summary for ''Steady-State'' Configurations'''</div>

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right"><math>~\boldsymbol{\hat{k}:}</math></td>
  <td align="right">
<math>~
\bold\nabla \cdot (\rho v_z \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{k}} \cdot (\rho \bold{a}) \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\bold{\hat{e}_\varpi:}</math></td>
  <td align="right">
<math>~
\bold\nabla \cdot (\rho v_\varpi \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{e}}_\varpi \cdot (\rho \bold{a}) + \frac{v_\varphi^2}{\varpi} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right"><math>~\bold{\hat{e}_\varphi:}</math></td>
  <td align="right">
<math>~ 
\bold\nabla \cdot (\rho \varpi v_\varphi \bold{u})
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\bold{\hat{e}}_\varphi \cdot (\rho \varpi \bold{a}) \, .</math>
  </td>
</tr>
</table>

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

In this context, the vector acceleration that drives the fluid flow is, simply,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\bold{a}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\nabla(H + \Phi_\mathrm{grav} ) \, .</math>
  </td>
</tr>
</table>

Then, for the velocity flow-patterns in Riemann S-type ellipsoids, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho v_z \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;(because <math>~v_z = 0</math>);
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho v_\varpi \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda^2}{\varpi^3} \biggl[\frac{a}{b} - \frac{b}{a}  \biggr] \biggl\{
y^4  \biggl(\frac{a}{b}\biggr) 
- 
x^4 \biggl(\frac{b}{a}\biggr) 
\biggr\}\rho \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nabla \cdot (\rho \varpi v_\varphi \bold{u})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \lambda xy \Omega_f \biggl[\frac{a}{b} - \frac{b}{a} \biggr]\rho  \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\varpi v_\varphi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-
\biggl[ \lambda \biggl(\frac{b}{a}\biggr) - \Omega_f\biggr]x^2
-
\biggl[ \lambda \biggl(\frac{a}{b}\biggr) - \Omega_f\biggr]y^2
\, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Vertical Component:'''</font> &nbsp;  Given that <math>~\bold{\hat{k}}\cdot (\rho \bold{a}) = 0</math>, we deduce that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~H_0  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\pi G \rho c^2 A_3  \, . </math>
  </td>
</tr>
</table>

<font color="red">'''Azimuthal Component:'''</font> &nbsp; Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ 
- a b \lambda \Omega_f 
</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}  - c^2 A_3 
\biggr] \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Radial Component:'''</font> &nbsp; After inserting the "azimuthal component" relation and marching through a fair amount of algebraic manipulation, we find that Irrespective of the <math>~(x, y, z)</math> location of each fluid element, this component requires,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{2\pi G \rho }{(a^2 - b^2) }
\biggl[  A_1 a^2 - A_2 b^2  \biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \lambda^2 + \Omega_f^2\biggr] \, .
</math>
  </td>
</tr>
</table>