View source for Temp001

===Based on Detailed Force Balance===
====The Steady-State Condition====

As has been pointed out in our [[PGE/RotatingFrame#Euler_Equation_.28rotating_frame.29|introductory discussion of the Principal Governing Equations]], quite generally we can write the

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


where, <math>~{\vec{\Omega}}_f </math> specifies the time-invariant rotation frequency of the frame and the orientation of the vector about which the frame spins. The condition for detailed force balance in a steady-state configuration is obtained by setting <math>~[\partial \vec{v}/\partial t]_\mathrm{rot} = 0</math>.  If we furthermore make the substitution, <math>~\nabla H = \nabla P/\rho</math>, where <math>~H</math> is enthalpy &#8212; an equation of state relation that is appropriate for a [[SR#Barotropic_Structure|barytropic system]] &#8212; we obtain,

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

<tr>
  <td align="right">
<math>~({\vec{v}}_\mathrm{rot} \cdot \nabla ){\vec{v}}_\mathrm{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \nabla\biggl[ H + \Phi_\mathrm{grav} - \frac{1}{2} \Omega_f^2(x^2 + y^2)\biggr] - 2{\vec{\Omega}}_f \times {\vec{v}}_\mathrm{rot} \, .
</math>
  </td>
</tr>
</table>

{{ Ou2006 }}, p. 550, &sect;2, Eq. (4)

====Adopted Velocity Flow-Field====

As {{ Ou2006 }} has pointed out [text that is taken directly from that publication appears here in an orange-colored font], <font color="orange">the velocity field of a Riemann S-type ellipsoid as viewed from a frame rotating with angular velocity <math>~{\vec{\Omega}}_f = \boldsymbol{\hat{k}} \Omega_f</math> takes the following form:</font>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~{\vec{v}}_\mathrm{rot}</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>

{{ LL96 }}, p. 700, &sect;2, Eq. (1)<br />
{{ Ou2006 }}, p. 550, &sect;2, Eq. (3)
</div>

<font color="orange">where <math>~\lambda</math> is a constant that determines the magnitude of the internal motion of the fluid, and the origin of the x-y coordinate system is at the center of the ellipsoid.  This velocity field <math>~{\vec{v}}_\mathrm{rot}</math> is designed so that velocity vectors everywhere are always aligned with elliptical stream lines by demanding that they be tangent to the</font> equi-''effective''-potential <font color="orange"> contours, which are concentric ellipses.</font>

<table border="1" align="center" cellpadding="10" width="65%"><tr><td align="left">
Plugging Ou's expression for  <math>~{\vec{v}}_\mathrm{rot}</math> into the expression on the left-hand side of the steady-state Euler equation, we see that for Riemann S-type ellipsoids,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~({\vec{v}}_\mathrm{rot} \cdot \nabla){\vec{v}}_\mathrm{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \lambda\biggl(\frac{a}{b}\biggr)y \frac{\partial}{\partial x} - \lambda\biggl(\frac{b}{a}\biggr)x \frac{\partial}{\partial y} \biggr] 
\biggl[\boldsymbol{\hat{\imath}}\lambda\biggl(\frac{a}{b}\biggr)y - \boldsymbol{\hat{\jmath}} \lambda\biggl( \frac{b}{a}\biggr)x \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \boldsymbol{\hat{\jmath}} \biggl[ \lambda\biggl(\frac{a}{b}\biggr)y\biggr] \frac{\partial}{\partial x}  
\biggl[ \lambda\biggl( \frac{b}{a}\biggr)x \biggr]
- \boldsymbol{\hat{\imath}}
\biggl[ \lambda\biggl(\frac{b}{a}\biggr)x\biggr] \frac{\partial}{\partial y}  
\biggl[\lambda\biggl(\frac{a}{b}\biggr)y \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\lambda^2\biggl[ \boldsymbol{\hat{\imath}} x + \boldsymbol{\hat{\jmath}} y \biggr] 
=
-\nabla\biggl[\frac{1}{2}\lambda^2(x^2 + y^2)  \biggr] \, .
</math>
  </td>
</tr>
</table>

----

Alternatively, from a [[PGE/Euler#in_terms_of_the_vorticity:|separate discussion of vector identities]] we realize that,
<div align="center">
<math>
(\vec{v}\cdot\nabla)\vec{v} = \frac{1}{2}\nabla(\vec{v}\cdot \vec{v}) + \vec{\zeta}\times \vec{v} ,
</math>
</div>
where, <math>\vec\zeta \equiv \nabla\times\vec{v}</math> is the fluid vorticity.  Plugging in Ou's expression for <math>~{\vec{v}}_\mathrm{rot}</math>, we find that &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec{\zeta} = \nabla\times {\vec{v}}_\mathrm{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\boldsymbol{\hat{k}} \lambda \biggl[ \frac{b}{a} + \frac{a}{b} \biggr] \, ;</math>
  </td>
</tr>
<tr><td align="center" colspan="3">
{{ Ou2006 }}, p. 551, &sect;2, Eq. (17)
</td></tr>
<tr>
  <td align="right">
<math>~\vec{\zeta} \times {\vec{v}}_\mathrm{rot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\lambda^2\biggl[\boldsymbol{\hat{\jmath}} \biggl(1 + \frac{a^2}{b^2}\biggr)y + \boldsymbol{\hat{\imath}} \biggl(1 + \frac{b^2}{a^2}\biggr)x \biggr] \, ; </math> &nbsp; &nbsp; and,
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{2}\nabla( {\vec{v}}_\mathrm{rot} \cdot  {\vec{v}}_\mathrm{rot} )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\lambda^2\biggl[ \boldsymbol{\hat{\imath}} \biggl(\frac{b}{a}\biggr)^2x + \boldsymbol{\hat{\jmath}} \biggl(\frac{a}{b}\biggr)^2y  \biggr] \, .</math>
  </td>
</tr>
</table>

Hence, we again appreciate that, for Riemann S-type ellipsoids,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~( {\vec{v}}_\mathrm{rot} \cdot \nabla){\vec{v}}_\mathrm{rot} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\lambda^2\biggl[ \boldsymbol{\hat{\imath}} x + \boldsymbol{\hat{\jmath}} y \biggr] </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>
</table>

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


The steady-state Euler-equation specification therefore becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

{{ Ou2006 }}, p. 550, &sect;2, Eq. (5)
</div>

<font color="orange">Hence, within the 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>

{{ Ou2006 }}, p. 550, &sect;2, Eq. (6)
</div>

<font color="orange">where <math>~C_B</math> is a constant.</font>  It is customary to define an effective potential which is the sum of the gravitational potential and the system's centrifugal potential (as viewed from the rotating frame), namely,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Phi_\mathrm{eff} \equiv \Phi_\mathrm{grav} + \Psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\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>
</tr>
</table>

{{ Ou2006 }}, p. 550, &sect;2, Eq. (7)
</div>
in which case the statement of detailed force balance in Riemann S-type ellipsoids can be rewritten in the following deceptively simpler form:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~H + \Phi_\mathrm{eff}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~C_B \, .</math>
  </td>
</tr>
</table>

{{ Ou2006 }}, p. 550, &sect;2, Eq. (8)
</div>

====Evaluation of the Gravitational Potential====

Drawing from a separate discussion of the [[ThreeDimensionalConfigurations/HomogeneousEllipsoids#Gravitational_Potential|gravitational potential of homogeneous ellipsoids]], we see that for Riemann S-type ellipsoids,

<div align="center">
<math>
~\Phi_\mathrm{grav}(\vec{x}) = -\pi G \rho \biggl[ I_\mathrm{BT} a^2 - \biggl(A_1 x^2 + A_2 y^2 +A_3 z^2 \biggr) \biggr],
</math><br />

[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 3, Eq. (40)</font><sup>1,2</sup> ]<br />
[ [[Appendix/References#BT87|BT87]], <font color="#00CC00">Chapter 2, Table 2-2</font> ]
</div>

where, the normalization constant,

<table align="center" border=0 cellpadding="3">

<tr>
  <td align="right">
<math>
~I_\mathrm{BT}
</math>
  </td>
  <td align="left">
<math>
~= A_1 + A_2\biggl(\frac{b}{a}\biggr)^2+ A_3\biggl(\frac{c}{a}\biggr)^2 .
</math>
  </td>
</tr>
</table>

<div align="center">
[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">Chapter 3, Eq. (22)</font><sup>1</sup>]<br />
[ [[Appendix/References#BT87|BT87]], <font color="#00CC00">Chapter 2, Table 2-2</font> ]
</div>

====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">
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  <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>
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</table>
{{ Ou2006 }}, p. 551, &sect;2, Eqs. (15) &amp; (16)
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where,
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  <td align="right">
<math>~M</math>
  </td>
  <td align="center">
<math>~\equiv</math>
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<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,
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  <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>
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  <td align="center" colspan="11"><b>TEST (part 3)</b></td>
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<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>.