Editing ThreeDimensionalConfigurations/RiemannStype (section)

===Based on Virial Equilibrium===

Pulling from Chapter 7 &#8212; specifically, &sect;48 &#8212; of [<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], we understand that the semi-axis ratios, <math>~(\tfrac{b}{a}, \tfrac{c}{a})</math> associated with Riemann S-type ellipsoids are given by the roots of the equation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl[  \frac{a^2 b^2}{a^2 + b^2} \biggr] f \biggl( \frac{\Omega^2}{\pi G \rho} \biggr)
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a^2 b^2 A_{12} - c^2 A_3 \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;48, Eq. (34)</font> ]</td></tr>
</table>
</div>
and the associated value of the square of the equilibrium configuration's angular velocity is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ 1 + \frac{a^2 b^2 \cdot f^2}{(a^2 + b^2)^2}  \biggr] \frac{\Omega^2}{\pi G \rho}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2B_{12} \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;48, Eq. (33)</font> ]</td></tr>
</table>
</div>
<span id="A12B12">where,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~A_{12}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~-\frac{A_1-A_2}{(a^2 - b^2)} \, ,</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (107)</font> ]</td></tr>

<tr>
  <td align="right">
<math>~B_{12}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~A_2 - a^2A_{12} \, .</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;21, Eq. (105)</font> ]<br />See also the ''note'' immediately following &sect;21, Eq. (127)</td></tr>
</table>
</div>

<table border="1" align="center" width="80%" cellpadding="8"><tr><td align="left">
In a [[VE/RiemannEllipsoids#Riemann_S-Type_Ellipsoids|subsection of a separate chapter]], we show in detail how the above set of equilibrium conditions arises from the,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3">
<font color="#770000">'''2<sup>nd</sup>-Order Tensor Virial Equation (TVE)'''</font>
  </td>
</tr>
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \mathfrak{T}_{ij} + \mathfrak{W}_{ij} + \delta_{ij}\Pi 
+ \Omega^2 I_{ij} - \Omega_i\Omega_k I_{kj} + 2\epsilon_{ilm}\Omega_m \int_V \rho u_lx_j dx \, .
</math>
  </td>
</tr>
</table>

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

(Notice that if we set <math>~f \rightarrow 0</math>, this pair of expressions simplifies to the pair we have provided in a separate discussion of the [[ThreeDimensionalConfigurations/JacobiEllipsoids#Equilibrium_Conditions_for_Jacobi_Ellipsoids|equilibrium conditions for Jacobi ellipsoids]].)  Following Chandrasekhar's lead and eliminating <math>~\Omega^2</math> between these two expressions, we obtain,
<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>~
\biggl[ \frac{a^2 b^2}{(a^2 + b^2)^2} \biggr] f^2 
+ \biggl[ \frac{2a^2 b^2 B_{12}}{c^2 A_3 - a^2 b^2 A_{12}} \biggr]\frac{f}{a^2 + b^2} + 1 \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">[ [[Appendix/References#EFE|EFE]], <font color="#00CC00">&sect;48, Eq. (35)</font> ]</td></tr>
</table>
For a given <math>~f</math>, this last expression determines the ratios of the axes of the ellipsoids that are compatible with equilibrium; and the value of <math>~\Omega^2</math>, that is to be associated with a particular solution of this last expression, then follows from either one of the first two expressions.  For convenience of evaluation and for greater clarity, let's rewrite this last (quadratic) equation in the form,
<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>~
\alpha f^2 + \beta f + 1 \, .
</math>
  </td>
</tr>
</table>

<div align="center" id="TestPart2">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="7"><b>TEST (part 2)</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^2 A_{12}</math></td>
  <td align="center" rowspan="1"><math>B_{12}</math></td>
  <td align="center" rowspan="1"><math>\alpha \equiv \frac{(b/a)^2}{[ 1 + (b/a)^2]^2}  </math></td>
  <td align="center" rowspan="1"><math>\beta \equiv \biggl[ \frac{2  B_{12}}{(c/b)^2 A_3 - a^2 A_{12}} \biggr]\frac{1}{1 + (b/a)^2}</math></td>
  <td align="center" rowspan="1"><math>C_\mathrm{LL96} \equiv \frac{\beta}{2\alpha^{1 / 2}}</math></td>
</tr>
<tr>
  <td align="center" rowspan="2">0.9</td>
  <td align="center" rowspan="2">0.641</td>
  <td align="center">0.387793362</td>
  <td align="center">0.207337649</td>
  <td align="center">0.247245200</td>
  <td align="center">3.797483556</td>
  <td align="center">3.818580687</td>
</tr>
<tr>
  <td align="center">0.38779 33613 22405 96490</td>
  <td align="center">0.20733 7650610469 45704</td>
  <td align="left" colspan="3"><math>~~\Leftarrow~~</math> (January, 2022) [[Appendix/Ramblings/ForCohlHoward#Evaluation_of_Index_Symbols|Howard Cohl's Mathematica evaluation]] with 20-digit precision</td>
</tr>
</table>
</div>

The pair of solutions is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_\pm</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2\alpha}\biggr\{
- \beta \pm \biggl[ \beta^2 - 4\alpha \biggr]^{1 / 2}
\biggr\} \, ;
</math>
  </td>
</tr>
</table>
and the corresponding values of the angular velocity (in units of [G &rho;]<sup>½</sup>) are provided by the expression,

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

<tr>
  <td align="right">
<math>\omega_\pm \equiv \frac{\Omega}{\sqrt{G\rho}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{ 2B_{12} \biggl[ 1 + \alpha f_\pm^2 \biggr]^{-1} \biggr\}^{1 / 2} \cdot \sqrt{\pi} \, .
</math>
  </td>
</tr>
</table>

As an aid in determining both values of the parameter, <math>~f</math>, we note as well that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f_+ \cdot f_-</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\alpha} = \biggl[ \frac{a^2 + b^2}{ab}\biggr]^2 \, .
</math>
  </td>
</tr>
</table>

<div align="center" id="TestPart21">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="12"><b>TEST (part 2.1)</b> &nbsp; &nbsp; <math>(b/a, c/a) = (0.9, 0.641)</math></td>
</tr>
<tr>
  <td align="center" rowspan="2">EFE Derivation</td>
  <td align="center"><math>f_+</math></td>
  <td align="center"><math>\frac{\omega_+}{\sqrt\pi}</math></td>
  <td align="center"><math>\frac{\zeta_+}{\sqrt{\pi G\rho}} = f_+\cdot\frac{\omega_+}{\sqrt\pi}</math></td>
  <td align="center"><math>x_+ = \alpha^{1 / 2}f_+</math></td>
  <td align="center"><math>\lambda_+ = -\alpha^{1 / 2}\frac{\zeta_+}{\sqrt{\pi G\rho}}</math></td>
  <td align="center" rowspan="4" bgcolor="lightgray" width="1%">&nbsp;</td>
  <td align="center"><math>f_-</math></td>
  <td align="center"><math>\frac{\omega_-}{\sqrt\pi}</math></td>
  <td align="center"><math>\frac{\zeta_-}{\sqrt{\pi G\rho}} = f_-\cdot\frac{\omega_-}{\sqrt\pi}</math></td>
  <td align="center"><math>x_- = \alpha^{1 / 2}f_-</math></td>
  <td align="center"><math>\lambda_- = -\alpha^{1 / 2}\frac{\zeta_-}{\sqrt{\pi G\rho}}</math></td>
</tr>
<tr>
  <td align="right" bgcolor="yellow">-0.268009</td>
  <td align="right">0.638310</td>
  <td align="right">-0.1710727</td>
  <td align="right">-0.133264</td>
  <td align="right">0.0850638</td>

  <td align="right" bgcolor="lightblue">-15.09117</td>
  <td align="right" bgcolor="lightblue">0.0850638</td>
  <td align="right" bgcolor="lightblue">-1.28371</td>
  <td align="right" bgcolor="lightblue">-7.50390</td>
  <td align="right" bgcolor="lightblue">0.638310</td>
</tr>
<tr>
  <td align="center" rowspan="2">Adjoint</td>
  <td align="center"><math>f^\dagger_+ = \frac{1}{\alpha f_+}</math></td>
  <td align="center"><math>\frac{\omega^\dagger_+}{\sqrt\pi} = \biggl| \frac{x_+ \omega_+}{\sqrt\pi} \biggr|</math></td>
  <td align="center"><math>\frac{\zeta^\dagger_+}{\sqrt{\pi G\rho}} = f^\dagger_+\cdot\frac{\omega^\dagger_+}{\sqrt\pi}</math></td>
  <td align="center"><math>x^\dagger_+ = \frac{1}{x_+}</math></td>
  <td align="center"><math>\lambda^\dagger_+ = -\alpha^{1 / 2}\frac{\zeta^\dagger_+}{\sqrt{\pi G\rho}}</math></td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
</tr>
<tr>
  <td align="right" bgcolor="lightblue">-15.09117</td>
  <td align="right" bgcolor="lightblue">0.085064</td>
  <td align="right" bgcolor="lightblue">-1.28371</td>
  <td align="right" bgcolor="lightblue">-7.50390</td>
  <td align="right" bgcolor="lightblue">0.638310</td>

  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
  <td align="center">&nbsp;</td>
</tr>
<tr>
  <td align="left" colspan="12">
Given the choice of <math>(b/a, c/a)</math> FILL OUT THIS TABLE AS FOLLOWS:  
<ol>
<li>From the pair of solutions to the quadratic equation for <math>f</math>, determine numerical values for <math>f_+</math> and <math>f_-</math>.</li>
<li>From the relation that gives <math>\Omega</math> in terms of <math>f</math>, determine numerical values for <math>\omega_+/\sqrt{\pi}</math> and <math>\omega_-/\sqrt{\pi}</math>.</li>
<li>Knowing <math>f_\pm</math> and <math>\omega_\pm</math>, determine the values of <math>\zeta_\pm</math>, <math>x_\pm</math> and <math>\lambda_\pm</math> via the relations,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\zeta</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> f \cdot \Omega</math> &nbsp; &nbsp; <math>\Rightarrow</math> &nbsp; &nbsp; <math>\frac{\zeta}{\sqrt{\pi G \rho}} = f\cdot \frac{\omega}{\sqrt{\pi}} \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], §48, p. 133, Eq. (31)
  </td>
</tr>

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> \alpha^{1 / 2}f \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], §48, p. 134, Eq. (40)
  </td>
</tr>

<tr>
  <td align="right">
<math>\frac{\zeta}{\sqrt{\pi G\rho}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math> -\lambda \biggl[\frac{a}{b} + \frac{b}{a}\biggr]</math>  &nbsp; &nbsp; <math>\Rightarrow</math> &nbsp; &nbsp; <math> \lambda = - \alpha^{1 / 2}\frac{\zeta}{\sqrt{\pi G\rho}} \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
{{ LL96 }}, &sect;2, p. 700, immediately following Eq. (1)
  </td>
</tr>
</table>

.</li>
<li>Determine the adjoint of <math>f_+</math>, <math>\omega_+</math>, and <math>x_+</math> using the indicated expressions, namely,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>f^\dagger_+</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(\alpha f_+)^{-1} \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], §48, p. 135, Eq. (44)
  </td>
</tr>

<tr>
  <td align="right">
<math>\omega_+^\dagger</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>|x_+ \omega_+| \, ,</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], §48, p. 135, Eq. (46)
  </td>
</tr>

<tr>
  <td align="right">
<math>x^\dagger_+</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>(x_+)^{-1} \, .</math>
  </td>
</tr>
<tr>
  <td align="center" colspan="3">
[<b>[[Appendix/References#EFE|<font color="red">EFE</font>]]</b>], §48, p. 135, immediately preceding Eq. (47)
  </td>
</tr>
</table>
</li>
<li>Determine the adjoint of <math>\zeta_+</math> from knowledge of the adjoint values, <math>f_+^\dagger</math> and <math>\omega_+^\dagger</math>, via the indicated expressions.</li>
<li>Then, as indicated, determine the adjoint of <math>\lambda_+</math> from knowledge of the value of <math>\zeta_+^\dagger</math>.</li>
</ol>
  </td>
</tr>
</table>
</div>

<font color="red">NOTE:&nbsp; The model that has been identified as the "negative" solution, <math>f_-</math>, of the governing quadratic relation is precisely synonymous with the model that has been identified as the "adjoint" of the "positive" solution, <math>f_+^\dagger</math>.  It should come as no surprise that the model labeled as <math>f_-^\dagger</math> is the model labeled as <math>f_+</math>.</font>

<!-- START TESTPART26
<div align="center" id="TestPart26">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="10"><b>TEST (part 2.6)</b> &nbsp; &nbsp; <math>(b/a, c/a) = (0.9, 0.641)</math></td>
</tr>
<tr>
  <td align="center" rowspan="2">EFE Derivation</td>
  <td align="center"><math>f_+</math></td>
  <td align="center"><math>\frac{\omega_+}{\sqrt\pi}</math></td>
  <td align="center"><math>\frac{1}{\alpha^{1 / 2}f_-} = \alpha^{1 / 2}f_+</math></td>
  <td align="right"><math>\biggl[ \frac{1}{\alpha^{1 / 2}f_+} \biggr]\frac{\omega_+}{\sqrt\pi}</math></td>
  <td align="center" rowspan="4" bgcolor="lightgray" width="1%">&nbsp;</td>
  <td align="center"><math>f_-</math></td>
  <td align="center"><math>\frac{\omega_-}{\sqrt\pi}</math></td>
  <td align="center"><math>\frac{1}{\alpha^{1 / 2}f_+} = \alpha^{1 / 2}f_-</math></td>
  <td align="right"><math>\biggl[ \frac{1}{\alpha^{1 / 2}f_-} \biggr]\frac{\omega_-}{\sqrt\pi}</math></td>
</tr>
<tr>
  <td align="right" bgcolor="yellow">-0.268009</td>
  <td align="right">0.638310</td>
  <td align="right">-0.133264</td>
  <td align="right">-4.78981</td>
  <td align="right" bgcolor="lightblue">-15.091171</td>
  <td align="right">0.085064</td>
  <td align="right">-7.503897</td>
  <td align="right">-0.011336</td>
</tr>
<tr>
  <td align="center" rowspan="2">{{ LL96hereafter }}<br />Eqs. (8) &amp; (9)</td>
  <td align="right"><math>\frac{1}{\alpha^{1 / 2}}\biggl[\frac{\lambda}{\omega_\mathrm{LL96}}\biggr]_+</math></td>
  <td align="center"><math>\omega_\mathrm{LL96} = \sqrt{\frac{2B_{12}}{1+x_+^2}}</math></td>
  <td align="center"><math>x_+</math></td>
  <td align="center"><math>\lambda_+ = \biggl[\frac{\omega_\mathrm{LL96}}{x}\biggr]_+</math></td>
  <td align="right"><math>\frac{1}{\alpha^{1 / 2}}\biggl[\frac{\lambda}{\omega_\mathrm{LL96}}\biggr]_-</math></td>
  <td align="center"><math>\omega_\mathrm{LL96} = \sqrt{\frac{2B_{12}}{1+x_-^2}}</math></td>
  <td align="center"><math>x_-</math></td>
  <td align="center"><math>\lambda_- = \biggl[\frac{\omega_\mathrm{LL96}}{x}\biggr]_-</math></td>
</tr>
<tr>
  <td align="right" bgcolor="lightblue">-15.091163</td>
  <td align="right">0.638310</td>
  <td align="right">-0.133264</td>
  <td align="right">-4.78982</td>
  <td align="right" bgcolor="yellow">-0.268009</td>
  <td align="right">0.085064</td>
  <td align="right">-7.503897</td>
  <td align="right">-0.011336</td>
</tr>
</table>
</div>
END TESTPART26 -->

<table border="1" width="80%" align="center" cellpadding="5"><tr><td align="left">
<br /><div align="center">Derivation by {{ LL96full }}</div>

Essentially this same derivation can be found in &sect;2 of {{ LL96 }}, hereafter {{ LL96hereafter }} &#8212; see also [[ThreeDimensionalConfigurations/Stability/RiemannEllipsoids#Lebovitz_&_Lifschitz_(1996)|our chapter discussion of this work]].  But instead of seeking solutions that are expressed in terms of the model-parameter pair <math>(\Omega_f,\zeta)</math>, and the ratio, <math>f \equiv \zeta/\Omega_f</math>, it should be recognized that {{ LL96hereafter }} use, 
[[File:CommentButton02.png|right|100px|Comment by J. E. Tohline:  LL96 do not explicitly state what sign should be assigned to the dimensionless frame-rotation frequency, &omega;.  By assuming, as has been done here, that it has the opposite sign to EFE's &Omega;, it is straightforward to show that the LL96 derivation is identical to the EFE derivation that we have presented immediately above.]]
<table border="0" cellpadding="10" align="center">

<tr>
  <td align="right">
<math>\omega_\mathrm{LL96}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-~\frac{\Omega_f}{\sqrt{\pi G \rho}} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\lambda</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \frac{\zeta}{\sqrt{\pi G \rho}} \biggl[ \frac{a_1^2 + a_2^2}{a_1 a_2} \biggr]^{-1} \, .</math>
  </td>
</tr>
</table>
So, the frequency ratio that {{ LL96hereafter }} define immediately preceding their Eq. (5),

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

<tr>
  <td align="right">
<math>x</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>
\frac{\omega_\mathrm{LL96}}{\lambda} 
=
\frac{\Omega_f}{\zeta}\biggl[ \frac{a_1^2 + a_2^2}{a_1 a_2} \biggr]
=
\frac{1}{\alpha^{1 / 2}f}
\, .
</math>
  </td>
</tr>
</table>
<div align="center">[For the example, <math>(b/a, c/a) = (0.9, 0.641)</math>, parameter values: <math>(f,x) = (-15.09117122, -0.133264084), (-0.268008879, -7.503897260)</math>].</div>

When rewritten in terms of <math>x</math>, the above-derived quadratic equation for <math>f</math>, namely,
<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>~
\alpha f^2 + \beta f + 1 \, ,
</math>
  </td>
</tr>
</table>
becomes,
<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>
\alpha \biggl[ \frac{1}{\alpha^{1 / 2} x}\biggr]^2 + \beta \biggl[ \frac{1}{\alpha^{1 / 2} x}\biggr] + 1 
=
\frac{1}{x^2} + \biggl[ \frac{\beta }{\alpha^{1 / 2} x}\biggr] + 1 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 + \biggl[ \frac{\beta }{\alpha^{1 / 2} }\biggr]x + x^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
1 +2Cx + x^2 \, ,
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">{{ LL96 }}, &sect;2, Eq. (5)</td></tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>C \equiv \frac{\beta}{2\alpha^{1 / 2}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{2}\biggl[ \frac{a^2 b^2}{(a^2 + b^2)^2} \biggr]^{-1 / 2} \biggl[ \frac{2a^2 b^2 B_{12}}{c^2 A_3 - a^2 b^2 A_{12}} \biggr]\frac{1}{a^2 + b^2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{a b B_{12}}{c^2 A_3 - a^2 b^2 A_{12}} \biggr]  \, .
</math>
  </td>
</tr>
<tr><td align="center" colspan="3">{{ LL96 }}, &sect;2, Eq. (6)</td></tr>
</table>
<div align="center">[For the example, <math>(b/a, c/a) = (0.9, 0.641)</math>, we find, <math>C = 3.818580689</math>.]</div>

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

<!-- START TESTPART2.75
<div align="center" id="TestPart275">
<table border="1" align="center" cellpadding="8">
<tr>
  <td align="center" colspan="10"><b>TEST (part 2.75)</b></td>
</tr>
<tr>
  <td align="center" rowspan="2"><math>~\frac{b}{a}</math></td>
  <td align="center" rowspan="2"><math>~\frac{c}{a}</math></td>
  <td align="center" rowspan="2"><math>~a^2 A_{12}</math></td>
  <td align="center" rowspan="2"><math>~ B_{12}</math></td>
  <td align="center" rowspan="2"><math>~\alpha \equiv \frac{(b/a)^2}{[ 1 + (b/a)^2]^2}  </math></td>
  <td align="center" rowspan="2"><math>~\beta \equiv \biggl[ \frac{2  B_{12}}{(c/b)^2 A_3 - a^2 A_{12}} \biggr]\frac{1}{1 + (b/a)^2}  </math></td>
  <td align="center" rowspan="1" colspan="2">Direct</td>
  <td align="center" rowspan="1" colspan="2">Adjoint</td>
</tr>
<tr>
  <td align="center" rowspan="1"><math>~f </math></td>
  <td align="center" rowspan="1"><math>~\omega = \frac{\Omega}{\sqrt{G\rho}} </math></td>
  <td align="center" rowspan="1"><math>~f^\dagger </math></td>
  <td align="center" rowspan="1"><math>~\omega^\dagger = \frac{\Omega^\dagger}{\sqrt{G\rho}} </math></td>
</tr>
<tr>
  <td align="center" rowspan="2">0.9</td>
  <td align="center" rowspan="2">0.641</td>
  <td align="center">0.387793362</td>
  <td align="center">0.207337649</td>
  <td align="center">0.247245200</td>
  <td align="center">3.797483556</td>
  <td align="center">- 0.268008879</td>
  <td align="center">&plusmn; 1.131374734</td>
  <td align="center">-15.09117122</td>
  <td align="center">&plusmn; 0.150771618</td>
</tr>
<tr>
  <td align="center">0.38779 33613 22405 96490</td>
  <td align="center">0.20733 7650610469 45704</td>
  <td align="left" colspan="6"><math>~~\Leftarrow~~</math> (January, 2022) [[Appendix/Ramblings/ForCohlHoward#Evaluation_of_Index_Symbols|Howard Cohl's Mathematica evaluation]] with 20-digit precision</td>
</tr>
</table>
</div>
END TESTPART2.75 -->