Editing SSC/FreeEnergy/PolytropesEmbedded

=Chapter Divisions=
In <font color="red">October 2023</font>, this very long chapter was subdivided in order to more effectively accommodate edits.  Here is a list of the resulting set of shorter chapters:
<ol>
  <li>[[SSC/FreeEnergy/PolytropesEmbedded/Pt1|Free-Energy Synopsis]]</li>
  <li>[[SSC/FreeEnergy/PolytropesEmbedded/Pt2|Free-Energy of Truncated Polytropes]]</li>
  <li>Free-Energy of BiPolytropes
    <ul>
<li>[[SSC/FreeEnergy/PolytropesEmbedded/Pt3A|Focus on Five-One Free-Energy Expression]]</li>
<li>[[SSC/FreeEnergy/PolytropesEmbedded/Pt3B|Focus on Zero-Zero Free-Energy Expression]]</li>
<li>[[SSC/FreeEnergy/PolytropesEmbedded/Pt3C|Overview]]</li>
    </ul>
  </li>
</ol>

=Free-Energy Synopsis=

All of the self-gravitating configurations considered below have an associated Gibbs-like free-energy that can be expressed analytically as a power-law function of the dimensionless configuration radius, <math>~x</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{G}^*_\mathrm{type}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-ax^{-1} + b x^{-3/n} + c x^{-3/j} + \mathfrak{G}_0 \, .</math>
  </td>
</tr>
</table>
</div>

==Equilibrium Radii and Critical Radii==
The first and second (partial) derivatives with respect to <math>~x</math> are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial\mathfrak{G}^*_\mathrm{type}}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ax^{-2} - \biggl(\frac{ 3b}{n}\biggr) x^{-3/n -1} -\biggl(\frac{3 c}{j}\biggr) x^{-3/j-1} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{x^2} \biggl[ a - \biggl(\frac{ 3b}{n}\biggr) x^{(n-3)/n } -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j} \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2ax^{-3} + \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{-3/n -2} + \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr)  x^{-3/j-2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^3} \biggl\{ \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{(n-3)/n} 
+ \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr)  x^{(j-3)/j}  -2a\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Equilibrium configurations are identified by setting the first derivative to zero.  This gives,

<div align="center">
<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>~a - \biggl(\frac{ 3b}{n}\biggr) x^{(n-3)/n }_\mathrm{eq} -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~x^{(n-3)/n }_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a  -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} \biggr] \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} -  \frac{a}{3c} + \frac{1}{j}\cdot  x^{(j-3)/j}_\mathrm{eq}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  0 \, .</math>
  </td>
</tr>
</table>
</div>

We conclude, as well, that ''at'' this equilibrium radius, the second (partial) derivative assumes the value,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl(\frac{ 3b}{n}\biggr) \biggl( \frac{n+3}{n}\biggr) x^{(n-3)/n} 
+ \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr)  x^{(j-3)/j}  -2a\biggr\}_\mathrm{eq} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl( \frac{n+3}{n}\biggr) \biggl[a  -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} \biggr]  
+ \biggl(\frac{3 c}{j}\biggr)\biggl( \frac{j+3}{j}\biggr)  x^{(j-3)/j}_\mathrm{eq}  -2a\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{x^3_\mathrm{eq} } \biggl\{ \biggl(\frac{3 c}{j}\biggr) \biggl[ \biggl( \frac{j+3}{j}\biggr)  
-\biggl( \frac{n+3}{n}\biggr) \biggl] x^{(j-3)/j}_\mathrm{eq} 
+ \biggl( \frac{3-n}{n}\biggr)  a\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

Hence, equilibrium configurations for which the ''second'' (as well as first) derivative of the free energy is zero are found at "critical" radii given by the expression,
<div align="center">
<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{3 c}{j}\biggr) \biggl[ \biggl( \frac{j+3}{j}\biggr)  
-\biggl( \frac{n+3}{n}\biggr) \biggl] [x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} 
+ \biggl( \frac{3-n}{n}\biggr)  a
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~[x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{j^2 a(n-3)}{3 c}\biggr] [ n(j+3) - j(n+3)  ]^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{a}{3^2c}\biggl[ \frac{j^2(n-3)}{n-j} \biggr] 
\, .
</math>
  </td>
</tr>
</table>
</div>

==Examples==
===Pressure-Truncated Polytropes===
For pressure-truncated polytropes of index <math>~n</math>, we set, <math>~j = -1</math>, in which case,

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

<tr>
  <td align="right">
<math>~\frac{ b}{nc}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c}  - x^{4}_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0  \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\biggl[ \biggl(\frac{3}{4\pi}\biggr) \frac{ M_\mathrm{tot}}{M_\mathrm{SWS}} \biggr]^{(n+1)/n}  x^{(n-3)/n }_\mathrm{eq} 
- \frac{3}{20\pi} \biggl( \frac{n+1}{n}\biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}} \biggr)^{2}  - x^{4}_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0  \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~x^{(n-3)/n }_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a  + 3cx^{4}_\mathrm{eq} \biggr] \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
and
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{a(n-3)}{3^2 c (n+1)} \biggr]^{1/4} \, .
</math>
  </td>
</tr>
</table>
</div>

====Case M====
<!-- Next segment supports PowerPoint presentation
<div align="center">
<math>( {\tilde{\mathfrak{f}}}_M, {\tilde{\mathfrak{f}}}_W, {\tilde{\mathfrak{f}}}_A)</math>
</div>

<div align="center">
<math>~\frac{d}{dx_\mathrm{eq}}\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr)= 0</math>
</div>
-->

More specifically, the expression that describes the [[#Case_M_Free-Energy_Surface|"Case M" free-energy surface]] is,
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^{-1} +~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^{-3/n} 
+~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \biggl(\frac{R}{R_\mathrm{norm}}\biggr)^3 
\, .
</math>
  </td>
</tr>
</table>

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


Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~n\mathcal{B} = n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>

where the structural form factors for pressure-truncated polytropes are precisely defined [[SSCpt1/Virial/FormFactors#PTtable|here]].  Therefore, the statement of virial equilibrium is,

<div align="center">
<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>~\frac{ b}{nc}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c}  - x^{4}_\mathrm{eq}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi}\biggr)c x_\mathrm{eq}^4  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \frac{ b}{n}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3}  \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot  x^{(n-3)/n }_\mathrm{eq} 
- \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot  x^{(n-3)/n }_\mathrm{eq} 
- \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}   \, .</math>
  </td>
</tr>
<!-- NEXT STEP IS FOR POWERPOINT PRESENTATION ...

<tr>
  <td align="right">
<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{20\pi} \biggl[ 5\biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot  x^{(n-3)/n }_\mathrm{eq} 
- \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  \biggr] </math>
  </td>
</tr>
-->
</table>
</div>

And we conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~3c[x_\mathrm{eq}]^4_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(n-3)}{5(n+1)}  \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) [x_\mathrm{eq}]^4_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi}  \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
- \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi}  \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \cdot [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi}  \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} 
+ \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi}  \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi} \biggl(\frac{4\pi}{3} \biggr)^{(n+1)/n}   \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n} }
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[\frac{4n}{15(n+1)}  \biggl(\frac{4\pi}{3} \biggr)^{1/n}  \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n} } \biggr]^{n/(n-3)} \, .
</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="8" width="90%"><tr><td align="left">
<font color="red"><b>ASIDE:</b></font>&nbsp; Let's see what this requires for the case of <math>~n=5</math>, where everything is specifiable analytically.  We have gathered together:
* Form factors from [[SSCpt1/Virial/FormFactors#Summary_.28n.3D5.29|here]].
* Hoerdt's equilibrium expressions from [[SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|here]].
* Conversion from Horedt's units to ours as specified [[SSCpt1/Virial#Choices_Made_by_Other_Researchers|here]].

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

<tr>
  <td align="right">
<math>~{\tilde\mathfrak{f}}_M</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
( 1 + \ell^2 )^{-3/2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~{\tilde\mathfrak{f}}_W</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  
\frac{5}{2^4} \cdot \ell^{-5}  \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~{\tilde\mathfrak{f}}_A</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} = \frac{R_\mathrm{eq}}{R_\mathrm{Horedt}} \biggl[ \frac{4\pi}{(n+1)^n} \biggr]^{1/(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ 3 \biggl[ \frac{(\xi_e^2/3)^5}{(1+\xi_e^2/3)^{6}} \biggr] \biggr\}^{-1/2}\biggl[ \frac{4\pi}{(n+1)^n} \biggr]^{1/(n-3)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(1+\ell^2)^{3}}{\ell^{5}} \biggr] \biggl[ \frac{\pi}{2^3\cdot 3^6} \biggr]^{1/2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{P_e}{P_\mathrm{norm}} = \frac{P_e}{P_\mathrm{Horedt}} \biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)/(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^3 \biggl[ \frac{(\xi_e^2/3)^3}{(1+\xi_e^2/3)^{4}} \biggr]^3 \biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)/(n-3)}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\ell^{18}}{(1+\ell^2)^{12}} \biggr] \biggl[ \frac{2 \cdot 3^4}{\pi} \biggr]^{3}</math>
  </td>
</tr>
</table>

So, the radius of the critical equilibrium state should be,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^4_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(n-3)}{3\cdot 5(n+1)}   \biggl(\frac{3}{2^2\pi}\biggr) \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr)^{-1}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2^2\cdot 3\cdot 5 \pi}  
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{18}}  \biggl[ \frac{\pi}{2 \cdot 3^4} \biggr]^{3}\biggr\} (1+\ell^2)^3
\cdot \biggl\{ \frac{5}{2^4} \cdot \ell^{-5}  \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi^2}{2^9\cdot 3^{13}}  
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{23}}  \biggr\} 
\cdot \biggl\{ \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) \biggr] \biggr\} \, ;
</math>
  </td>
</tr>
</table>
whereas, each equilibrium configuration has,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^4 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] \, .</math>
  </td>
</tr>
</table>
So the equilibrium state that marks the critical configuration must have a value of <math>~\ell</math> that satisfies the relation,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi^2}{2^9\cdot 3^{13}}  
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{23}}  \biggr\} 
\cdot \biggl\{ \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) \biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~2^3\cdot 3 \ell^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ell ( \ell^4 - \frac{8}{3}\ell^2 - 1 ) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{(1 + \ell^2)^{3}}{\ell} \biggr] \tan^{-1}(\ell ) </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \frac{80}{3}\cdot \ell^2 -\ell^4  \, .
</math>
  </td>
</tr>
</table>

The solution is:  <math>~\ell_\mathrm{crit} \approx 2.223175 \, .</math>
</td></tr></table>
</div>


In addition, we know from [[SSC/Virial/PolytropesEmbedded/SecondEffortAgain#Virial_Equilibrium_of_Adiabatic_Spheres_.28Summary.29|our dissection of Hoerdt's work on detailed force-balance models]] that, in the equilibrium state,

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

<tr>
  <td align="right">
<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) \biggl(\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\tilde\theta^{n+1} }{(4\pi)(n+1)( -\tilde\theta' )^{2}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 3c x_\mathrm{eq}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\tilde\theta^{n+1} }{(n+1)( -\tilde\theta' )^{2}} \biggr] 
\, .
</math>
  </td>
</tr>
</table>
</div>

This means that, for any chosen polytropic index, the critical equilibrium state is the equilibrium configuration for which (needs to be checked),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2(9-2n){\tilde\theta}^{n+1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3(n-3)\biggl[ (- {\tilde\theta}^')^2 - \frac{\tilde\theta(-{\tilde\theta}^')}{\tilde\xi}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
 
We note, as well, that by combining the Horedt expression for <math>~x_\mathrm{eq}</math> with our virial equilibrium expression, we find (needs to be checked),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_\mathrm{eq}^{n-3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3}\biggl[ \frac{3}{(n+1)\tilde\xi^2} + \frac{{\tilde\mathfrak{f}}_{W} - {\tilde\mathfrak{f}}_{M}}{5\tilde\mathfrak{f}_A} \biggr]^{n} {\tilde\mathfrak{f}}_{M}^{1-n} \, .</math>
  </td>
</tr>
</table>
</div>

====Case P====

=====First Pass=====
Alternatively, let's examine the [[#Case_P_Free-Energy_Surface|"Case P" free-energy surface]].  Drawing on [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler's presentation]], we adopt the following radius and mass normalizations:
<div align="center">
<math>M_\mathrm{SWS} = 
\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \, ,</math>
</div>
<div align="center">
<math>
R_\mathrm{SWS} = \biggl( \frac{n+1}{n} \biggr)^{1/2} G^{-1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, .
</math>
</div>
In terms of these new normalizations, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R_\mathrm{norm} \equiv \biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{1/(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{G}{K} \biggr)^{n/(n-3)} M_\mathrm{tot}^{(n-1)/(n-3)}  
R_\mathrm{SWS} \biggl( \frac{n+1}{n} \biggr)^{-1/2} G^{1/2} K_n^{-n/(n+1)} P_\mathrm{e}^{-(1-n)/[2(n+1)]}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+
M_\mathrm{SWS}^{-(n-1)/(n-3)}  \biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^{(n-1)/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}
\biggl( \frac{n+1}{n} \biggr)^{[3(n-1)-(n-3)]/[2(n-3)]} 
G^{[2n+(n-3)-3(n-1)]/[2(n-3)]}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~+ K_n^{n[2(n-1) - (n+1) - (n-3)]/[(n+1)(n-3)]} P_\mathrm{e}^{-(n-1)(3-n)/[2(n+1)(n-3)]}
P_\mathrm{e}^{(n-1)(3-n)/[2(n+1)(n-3)]} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \, .
</math>
  </td>
</tr>
</table>
</div>

and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~P_\mathrm{norm} \equiv \biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)} } \biggr]^{1/(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{K^{4n}}{G^{3(n+1)} } \biggr]^{1/(n-3)}
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)}
\biggl\{ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}\biggr\}^{-2(n+1)/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)}
\biggl( \frac{n+1}{n} \biggr)^{-3(n+1)/(n-3)}
K^{4n/(n-3)} G^{-3(n+1)/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\times~
G^{3(n+1)/(n-3)} K_n^{-4n/(n-3)}
\biggl\{ P_\mathrm{e}^{-(n-3)/[2(n+1)]}\biggr\}^{-2(n+1)/(n-3)}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~P_e 
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{-2(n+1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{-3(n+1)/(n-3)} \, .
</math>
  </td>
</tr>
</table>
</div>

<span id="FirstPassFreeEnergy">Rewriting the expression for the free energy gives,</span>

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr) 
+~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{3/n}
+~ \biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{-3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~ n\mathcal{B} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr]^{3/n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{2(n+1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{3(n+1)/(n-3)} 
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)}\biggr]^{-3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} 
+~ n\mathcal{B}   \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} 
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)} 
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, .
</math>
  </td>
</tr>
</table>
</div>


Therefore, in this case, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  
\biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)}  \, ,
</math>
  </td>
</tr>
</table>
</div>

where the structural form factors for pressure-truncated polytropes are precisely defined [[SSCpt1/Virial/FormFactors#PTtable|here]].  The statement of virial equilibrium is, therefore,

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

<tr>
  <td align="right">
<math>~x^{4}_\mathrm{eq} + \alpha  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\beta  x^{(n-3)/n }_\mathrm{eq} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha \equiv \frac{a}{3c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} 
\biggl\{ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{20\pi}  \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\beta \equiv \frac{b}{nc}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  
\biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]}
\biggl\{ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathfrak{m}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

From a previous derivation, we have,
<div align="center">
<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>~
\frac{ b}{nc}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c}  - x^{4}_\mathrm{eq}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{4\pi} \biggl( \frac{n+1}{n}  \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggl\{
\biggl(\frac{3}{4\pi} \biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  
\biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggr\} \cdot  x^{(n-3)/n }_\mathrm{eq} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{3}{4\pi} \biggl( \frac{n+1}{n}  \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)} \biggl\{
\frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}  \biggr\}
- x^{4}_\mathrm{eq}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n}  x^{(n-3)/n }_\mathrm{eq} 
- \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  - x^{4}_\mathrm{eq}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \tilde{\mathfrak{f}}_A \biggl[ \biggl(\frac{3}{4\pi} \biggr) \frac{1}{\tilde{\mathfrak{f}}_M}  
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{(n+1)/n}  x^{(n-3)/n }_\mathrm{eq} 
- \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggl[ \biggl(\frac{3}{4\pi} \biggr) \frac{1}{\tilde{\mathfrak{f}}_M}  
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{2}  - x^{4}_\mathrm{eq}
</math>
  </td>
</tr>
</table>
</div>

which, thankfully, matches!  We conclude as well that the transition from stable to dynamically unstable configurations occurs at,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \, .
</math>
  </td>
</tr>

<!-- FOR POWERPOINT SYNOPSIS
<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \frac{a}{3c}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(n-3)}{20\pi n}  \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{n}{n-3} \biggr)^{(1-n)} 
\biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr)^{(n+1)} 
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}  \biggr)^{2n} \, .
</math>
  </td>
</tr>
-->
</table>
</div>

When combined with the statement of virial equilibrium ''at'' this critical point, we find,

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

<tr>
  <td align="right">
<math>~
\biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr]  + 1\biggr\}\frac{ \alpha }{\beta} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
[x_\mathrm{eq}]^{(n-3)/n }_\mathrm{crit} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \biggr\}^{(n-3)/(4n) } 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[ \frac{4n}{3 (n+1)} \biggr]^{4n} \biggl( \frac{ \alpha }{\beta} \biggr)^{4n}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{(n-3)}{3 (n+1)} \biggr]^{(n-3)}  \alpha^{(n-3)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl[ \frac{3 n}{(n-3)} \biggl( \frac{n+1}{n}\biggr) \biggr]^{(3-n)} \biggl[ \frac{3 }{4} \biggl( \frac{n+1}{n}\biggr) \biggr]^{4n}   
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha^{3(n+1)} \beta^{-4n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2}  \biggr\}^{3(n+1)} 
\biggl\{ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n}  \biggr\}^{-4n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{\mathfrak{f}}_A^{-4n} 
\biggl[ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggr]^{3(n+1)} \mathfrak{m}^{2(n+1)}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~
\mathfrak{m}^{2(n+1)}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr)\biggr]^{-3(n+1)} 
\biggl[ \frac{3 n}{(n-3)} \biggl( \frac{n+1}{n}\biggr) \biggr]^{(3-n)} \biggl[ \frac{3\tilde{\mathfrak{f}}_A }{4} \biggl( \frac{n+1}{n}\biggr) \biggr]^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl( \frac{3\cdot 5}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{3(n+1)} 
\biggl[ \frac{3 n}{(n-3)}  \biggr]^{(3-n)} \biggl[ \frac{3\tilde{\mathfrak{f}}_A }{4}  \biggr]^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3^2 \cdot 5 n}{ 4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W}  \biggr]^{(3-n)} 
\biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W}  \biggr]^{4n} \, .
</math>
  </td>
</tr>
</table>
</div>

This also means that the critical radius is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ \frac{3^2\cdot 5 n}{4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W}\biggr]^{-1}   \mathfrak{m}^{2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{3^2\cdot 5 n}{4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W} \biggr]^{-(n+1)} 
\biggl[ \frac{3^2 \cdot 5 n}{ 4\pi(n-3)} \cdot \frac{1}{\tilde{\mathfrak{f}}_W}  \biggr]^{(3-n)} 
\biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W}  \biggr]^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi(n-3)}{3^2\cdot 5 n} \cdot \tilde{\mathfrak{f}}_W \biggr]^{2(n-1)} 
\biggl[ \biggl( \frac{3^2\cdot 5}{2^4\pi}\biggr) \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_W}  \biggr]^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n}{(n-3)} \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr) \biggr]^{(1-n)} 
\biggl[ \biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr) \frac{\tilde{\mathfrak{f}}_A}{4}  \biggr]^{2n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{n}{n-3} \biggr)^{(1-n)} 
\biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr)^{(n+1)} 
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}  \biggr)^{2n} \, .
</math>
  </td>
</tr>
</table>
</div>


<!-- THERE IS A MISTAKE IN THIS OMITTED SUBSECTION

From an earlier derivation, we obtained,


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

<tr>
  <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] \frac{a}{c}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] 
\biggl[ \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}   \biggr]
\biggl[ \frac{3}{4\pi} \biggl( \frac{n+1}{n} \biggr)^{-3/(n-3)} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(n-5)/(n-3)}  \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(n-3)}{20\pi n}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} 
= \tilde{\mathfrak{f}}_W \cdot \frac{(n-3)}{15 n} \biggl(\frac{4\pi}{3}\biggr) \biggl[\biggl(\frac{3}{4\pi}\biggr)\frac{1}{\tilde{\mathfrak{f}}_M}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]^{2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[\biggl(\frac{3}{4\pi}\biggr)\frac{1}{\tilde{\mathfrak{f}}_M}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2}  [x_\mathrm{eq}]_\mathrm{crit}^2</math>
  </td>
</tr>
</table>
</div>

which, when combined with the statement of virial equilibrium ''at'' this critical point, implies,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde{\mathfrak{f}}_A \biggl\{ \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2}  [x_\mathrm{eq}]_\mathrm{crit}^2\biggr\}^{(n+1)/n}  [x_\mathrm{eq}]_\mathrm{crit}^{(n-3)/n }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
[x_\mathrm{eq}]_\mathrm{crit}^4
+ \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl\{ \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{1/2}  [x_\mathrm{eq}]_\mathrm{crit}^2\biggr\}^{2}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \tilde{\mathfrak{f}}_A  \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{(n+1)/(2n)}  
[x_\mathrm{eq}]_\mathrm{crit}^{(3n-1)/n }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
[x_\mathrm{eq}]_\mathrm{crit}^4 \biggl\{ 1
+ \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{2} \biggr\}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ [x_\mathrm{eq}]_\mathrm{crit}^{(n+1)/n}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\tilde{\mathfrak{f}}_A  \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{(n+1)/(2n)}  
\biggl\{ 1+ \frac{1}{5} \biggl(\frac{4\pi}{3}\biggr) \cdot \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \biggl[ \frac{1}{\tilde{\mathfrak{f}}_W} \cdot \frac{15 n}{(n-3)} \biggl(\frac{3}{4\pi}\biggr)\biggr]^{2} \biggr\}^{-1} 
</math>
  </td>
</tr>
</table>
</div>

END OMITTED SUBSECTION -->

The following parallel derivation was done independently.  [<font color="red">Note that a factor of 2n/(n-1) appears to correct a mistake made during the original derivation.</font>]  Beginning with the virial expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\beta  x^{(n-3)/n }_\mathrm{eq}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\alpha + x^{4}_\mathrm{eq}
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~
\biggl(\frac{3}{4\pi} \biggr)^{(n+1)/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n}  [x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  
+ \frac{(n-3)}{20\pi n}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{(n-1)}{10\pi n} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl[ \frac{2n}{(n-1)}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{2(n-1)}{15 n} \biggl(\frac{4\pi}{3} \biggr)^{1/n}  \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A\tilde{\mathfrak{f}}_M^{(n-1)/n}} \cdot
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/n}  \biggl[ \frac{2n}{(n-1)}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n-3) }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr)  \frac{\tilde{\mathfrak{f}}_W^n}{\tilde{\mathfrak{f}}_A^n \tilde{\mathfrak{f}}_M^{(n-1)}} 
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)}  \biggl[ \frac{2n}{(n-1)}\biggr]^n
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr)  \frac{\tilde{\mathfrak{f}}_W^n}{\tilde{\mathfrak{f}}_A^n \tilde{\mathfrak{f}}_M^{(n-1)}} 
\biggl\{ \biggl[ \frac{20\pi n}{(n-3)} \biggr]^{(n-1)/2}  \biggl( \frac{\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W} \biggr)^{(n-1)/2} [x_\mathrm{eq} ]^{2(n-1) }_\mathrm{crit} \biggr\}\biggl[ \frac{2n}{(n-1)}\biggr]^n
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\biggl[\frac{2(n-1)}{15 n} \biggr]^n \biggl(\frac{4\pi}{3} \biggr)  \frac{\tilde{\mathfrak{f}}_W^{(n+1)/2}}{\tilde{\mathfrak{f}}_A^n } 
\biggl\{ \biggl[ \frac{20\pi n}{(n-3)} \biggr]^{(n-1)/2}   [x_\mathrm{eq} ]^{2(n-1) }_\mathrm{crit} \biggr\}\biggl[ \frac{2n}{(n-1)}\biggr]^n
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl(\frac{3}{4\pi} \biggr) 
\biggl[\frac{15 n}{2(n-1)} \biggr]^n
\biggl[ \frac{(n-3)}{20\pi n} \biggr]^{(n-1)/2}   \frac{\tilde{\mathfrak{f}}_A^n }{\tilde{\mathfrak{f}}_W^{(n+1)/2}} \biggl[ \frac{(n-1)}{2n} \biggr]^n
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~  \biggl(\frac{3}{4\pi} \biggr) 
\biggl[\frac{15 }{2^2} \biggr]^n
\biggl[ \frac{(n-3)}{20\pi n} \biggr]^{(n-1)/2}   \frac{\tilde{\mathfrak{f}}_A^n }{\tilde{\mathfrak{f}}_W^{(n+1)/2}} 
</math>
  </td>
</tr>
</table>
</div>



Also from [[SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler's work]] we know that the equilibrium mass and radius are,
<div align="center">
<table border="0" cellpadding="3">

<tr>
  <td align="right">
<math>
~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl[ {\tilde\theta}_n^{(n-3)/2} {\tilde\xi}^2 (-{\tilde\theta}^')
\biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl[ \tilde\xi {\tilde\theta}_n^{(n-1)/2} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Additional details in support of an associated PowerPoint presentation can be found [[SSC/FreeEnergy/PowerPoint|here]].

====Reconcile====


<div align="center">

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

<tr>
  <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n-3)}{20\pi (n+1)} \biggr] \biggl(\frac{n+1}{n}\biggr)
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} 
\frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} 
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>~\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{20\pi}  \biggl( \frac{n-3}{n+1} \biggr) \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
  </td>
</tr>
</table>

</div>
Taking the ratio, the RHS is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~P_e M_\mathrm{tot}^2 \biggl[ \frac{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)} }{K^{4n}} \biggr]^{1/(n-3)} 
\biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^{-2}
\biggl( \frac{n+1}{n}\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}P_e M_\mathrm{tot}^2 \biggl[ G^{3} M_\mathrm{tot}^{2} \biggr]^{(n+1)/(n-3)} K_n^{-4n/(n-3)}
\biggl[  G^{3} K_n^{-4n/(n+1)} P_\mathrm{e}^{(n-3)/(n+1)} \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}  \biggl[ G^{3} M_\mathrm{tot}^{2} \biggr]^{[(n-3)+(n+1)]/(n-3)} 
\biggl[ K_n^{[(n+1)+(n-3)]/[(n+1)(n-3)] } \biggr]^{-4n} P_\mathrm{e}^{[(n+1)+  (n-3)]/(n+1)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}  M_\mathrm{tot}^{4(n-1)/(n-3)} G^{[6(n-1)]/(n-3)} 
K_n^{-8(n-1)/[(n+1)(n-3)] }  P_\mathrm{e}^{2(n-1)/(n+1)} \, ;</math>
  </td>
</tr>
</table>
</div>

while the LHS is,

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

<tr>
  <td align="right">
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{4/(n-3)} 
\biggl\{\biggl( \frac{n+1}{n} \biggr)^{1/2} G^{-1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]}\biggr\}^{-4}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2} 
M_\mathrm{tot}^{4(n-1)/(n-3)} 
G^{[6(n-1)]/(n-3)} 
K^{-8n(n-1)/[(n-3)(n+1)] }  P_\mathrm{e}^{2(n-1)/(n+1)} \, .
</math>
  </td>
</tr>
</table>
</div>

Q.E.D.

Now, given that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_\mathrm{SWS}^{-4(n-1)/(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}\biggr]^{-4(n-1)/(n-3)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-6(n-1)/(n-3)} G^{6(n-1)/(n-3)} K_n^{-8n(n-1)/[(n+1)(n-3)]} P_\mathrm{e}^{2(n-1)/(n+1)} </math>
  </td>
</tr>
</table>
</div>

we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2} 
\biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{4(n-1)/(n-3)} 
\biggl( \frac{n+1}{n} \biggr)^{6(n-1)/(n-3)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{4(n-1)/(n-3)} 
\biggl( \frac{n+1}{n} \biggr)^{4n/(n-3)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{n-3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{n-1} 
\biggl( \frac{n+1}{n} \biggr)^{n} 
</math>
  </td>
</tr>
</table>
</div>


By inspection, in the specific case of <math>~n=5</math> (see above), this critical configuration appears to coincide with one of the [[SSC/Structure/PolytropesEmbedded#Other_Limits|"turning points" identified by Kimura]].  Specifically, it appears to coincide with the "extremal in r<sub>1</sub>" along an M<sub>1</sub> sequence, which satisfies the condition,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{n-3}{n-1} \biggr]_{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\tilde\xi \tilde\theta^{n}}{(-\tilde\theta^')}\biggr]_{n=5}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1/2}\ell \biggl[ (1 + \ell^2)^{-1/2} \biggr]^5 \biggl[ \frac{\ell}{3^{1/2}} (1+\ell^2 )^{-3/2} \biggr]^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3(1 + \ell^2)^{-1} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \ell </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~5^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>

Hence, according to Kimura, the turning point associated with the configuration with the largest equilibrium radius, corresponds to the equilibrium configuration having,
<div align="center">
<math>~\ell |_\mathrm{R_{max}} = \sqrt{5} \approx 2.2360680 \, .</math>
</div>
This is, indeed, very close to &#8212; but decidedly different from &#8212; the value of <math>~\ell_\mathrm{crit}</math> determined, above!


====Streamlined====

Let's copy the expression for the [[#FirstPassFreeEnergy|"Case P" free energy derived above]], then factor out a common term:


<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl( \frac{n+1}{n} \biggr)^{n/(n-3)} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} 
+~ n\mathcal{B}   \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)} 
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)} 
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(5-n)/(n-3)} \biggl(\frac{n+1}{n}\biggr)^{3/(n-3)} \biggl\{
-3\mathcal{A} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} 
+~ n\mathcal{B}   
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}
+\frac{4\pi}{3}  \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggr\}
</math>
  </td>
</tr>
</table>
</div>

Defining a new normalization energy,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~E_\mathrm{SWS}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~E_\mathrm{norm} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(5-n)/(n-3)} \biggl(\frac{n+1}{n}\biggr)^{3/(n-3)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n+1}{n}\biggr)^{3/2} K^{3n/(n+1)} G^{-3/2} P_e^{(5-n)/[2(n+1)]} \, ,
</math>
  </td>
</tr>
</table>
</div>
we can write,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{SWS}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-3\mathcal{A} \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} 
+~ n\mathcal{B}   
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}
+\frac{4\pi}{3}  \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, ,
</math>
  </td>
</tr>
</table>
</div>
in which case the coefficients of the generic free-energy expression are,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{5} \cdot \frac{ \tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl(\frac{n+1}{n}\biggr)\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^2
= \frac{3}{5} \cdot \biggl( \frac{4\pi }{3}\biggr)^2 \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ n\biggl(\frac{3}{4\pi}\biggr)^{1/n} 
\frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n}
= \biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{4\pi}{3} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, as above,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{m}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>

Now, if we define the pair of parameters,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{a}{3c}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{b}{nc} \, ,</math>
  </td>
</tr>
</table>
</div>
then the statement of virial equilibrium is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~x_\mathrm{eq}^4 + \alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\beta x_\mathrm{eq}^{(n-3)/n} \, ,</math>
  </td>
</tr>
</table>
</div>
and the boundary between dynamical stability and instability occurs at,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \, .</math>
  </td>
</tr>
</table>
</div>
Combining these last two expressions means that the boundary between dynamical stability and instability is associated with the parameter condition,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]^{(n-3)/n}_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n-3}{3(n+1)} + 1\biggr]  \frac{\alpha}{\beta} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\biggl\{ \biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \biggr\}^{(n-3)/(4n)}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ 4n }{3(n+1)}\biggr]  \frac{\alpha}{\beta} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\beta \alpha^{-3(n+1)/(4n)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ 4n }{3(n+1)}\biggr] \biggl[ \frac{n-3}{n} \cdot \frac{n}{3(n+1)} \biggr]^{(3-n)/(4n)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4 \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)/(4n)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)/(4n)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\biggl( \frac{\beta}{4}\biggr)^{4n} \alpha^{-3(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)}   
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 
\biggl( \frac{\beta}{4}\biggr)^{4n} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ n\alpha }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3}   \, .
</math>
  </td>
</tr>
</table>
</div>

======Case M======
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~n\mathcal{B} = n\biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}}  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{4\pi }{15} \biggr) \tilde{\mathfrak{f}}_W \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^2 \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde{\mathfrak{f}}_A \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n+1)/n} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-1} \, .</math>
  </td>
</tr>
</table>
</div>

So the dynamical stability conditions are:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggl( \frac{n}{n-3} \biggr) [x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr] \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^2   \, ;</math>
  </td>
</tr>
</table>
</div>

and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{4(n+1)} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-4n} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{ n}{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3}  
\biggl(\frac{4\pi \tilde{\mathfrak{f}}_W}{15} \biggr)^{3(n+1)}   
\biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{6(n+1)}  \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)^{-3(n+1)} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\biggl(\frac{ n}{n+1}\biggr)  \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{3(n+1)}   
\biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n+1)}  \biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} 
\biggl[\biggl(\frac{ n}{n+1}\biggr)  \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{-3(n+1)}   
\biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{-2(n+1)}  \, .
</math>
  </td>
</tr>
</table>
</div>

Together, then,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{n-3} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n-3)} 
\biggl[ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \frac{n}{n-3} \biggr]^{-(n-3)} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{n-3} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n-3)} 
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{-4n} 
\biggl[\biggl(\frac{ n}{n+1}\biggr)  \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{3(n+1)}   
\biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{2(n+1)}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{4n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{4(n-1)} 
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{-4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~[x_\mathrm{eq}]_\mathrm{crit}^{(n-3)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)}  \, .
</math>
  </td>
</tr>
</table>
</div>

======Case P======
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{5} \cdot \biggl( \frac{4\pi }{3}\biggr)^2 \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{4\pi}{3} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, as above,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{m}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\alpha</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{5} \cdot \biggl( \frac{4\pi }{3}\biggr) \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\beta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, .
</math>
  </td>
</tr>
</table>
</div>
So the dynamical stability conditions are:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{3(n+1)} \biggr]\biggl[ \frac{n-3}{n} \biggr]\frac{1}{5} \cdot \biggl( \frac{4\pi }{3}\biggr) \biggl(\frac{n+1}{n}\biggr)\tilde{\mathfrak{f}}_W \mathfrak{m}^{2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)  \mathfrak{m}^{2} </math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ 
\biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \mathfrak{m}^{4(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[  \biggl( \frac{4\pi }{3^2\cdot 5}\biggr)  \tilde{\mathfrak{f}}_W \mathfrak{m}^{2}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3}  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow~~~
\mathfrak{m}^{2(n+1)}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[  \biggl( \frac{4\pi }{3^2\cdot 5}\biggr)  \tilde{\mathfrak{f}}_W \biggr]^{-3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{-(n-3)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n}  \, .
</math>
  </td>
</tr>
</table>
</div>
Together, then,

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

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n-3}{n} \biggr]^{(n+1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)^{(n+1)}  
\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(n-3)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{n-3}{n} \biggr]^{2(n-1)} 
\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-2(n+1)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \, .
</math>
  </td>
</tr>
</table>
</div>

======Compare======

Let's see if the two cases, in fact, provide the same answer.
<!--
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{n-3} = \biggl[ \frac{x_\mathrm{P}}{x_\mathrm{M}} \biggr]_\mathrm{crit}^{n-3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[ \frac{n-3}{n} \biggr]^{2(n-1)} 
\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-2(n+1)} \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} 
 \biggr\}^{(n-3)/[4(n+1)]} \biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{n-3}{n}\biggr)^{2(n-1)(n-3)/[4(n+1)]} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W }{3^2\cdot 5}\biggr)^{-n -(n-3)/2} 
\biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{n+n(n-3)/(n+1)} 
\biggl( \frac{n+1}{n} \biggr)^{n} \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{1-n} 
</math>
  </td>
</tr>
</table>
</div>
-->
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}}\biggr)^{n-3} = \biggl[ \frac{x_\mathrm{P}}{x_\mathrm{M}} \biggr]_\mathrm{crit}^{n-3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)  \mathfrak{m}^{2}  \biggr\}^{(n-3)/4}
\biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)  \mathfrak{m}^{2}  \biggr\}^{(n-3)/4}
\biggl\{ \biggl[ \frac{4}{\tilde{\mathfrak{f}}_A}\biggl( \frac{n}{n+1} \biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^n \biggl(\frac{3}{4\pi \tilde{\mathfrak{f}}_M}\biggr)^{(n-1)} \biggr\}^{-1}
</math>
  </td>
</tr>
</table>
</div>

===Five-One Bipolytropes===
For analytically prescribed, "five-one" bipolytropes, <math>~n = 5</math> and <math>~j =1</math>, in which case,

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

<tr>
  <td align="right">
<math>x^{2/5 }_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{5}{ 3b}\biggr) \biggl[a  -3 c x^{-2}_\mathrm{eq} \biggr] \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
and
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>[x_\mathrm{eq}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{18 c}{a }\biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>


More specifically, [[#BiPolytrope51|the expression that describes the free-energy surface]] is,

<div align="center" id="FreeEnergy51">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>\mathfrak{G}^*_{51} \equiv 2^4\biggl( \frac{q}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}_{51}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\frac{1}{\ell_i^2} \biggl[
\Chi^{-3/5} (5 \mathfrak{L}_i)
+\Chi^{-3} (4\mathfrak{K}_i)
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>

Hence, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>a</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>b</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>5 \mathfrak{L}_i \chi_\mathrm{eq}^{3/5} \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>c</math>
  </td>
  <td align="center">
<math>\equiv</math>
  </td>
  <td align="left">
<math>4 \mathfrak{K}_i \chi_\mathrm{eq}^{3}  \, ,
</math>
  </td>
</tr>
</table>

and conclude that,

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

<tr>
  <td align="right">
<math>[\chi_\mathrm{eq}]_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{18 (4 \mathfrak{K}_i \chi_\mathrm{eq}^{3} )}{ 3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2}_\mathrm{crit}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>[\chi_\mathrm{eq}]_\mathrm{crit}\biggl[ \frac{24 \mathfrak{K}_i  }{  (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2}  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~\biggl[ \frac{24 \mathfrak{K}_i  }{  (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>1  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~\biggl[ \frac{\mathfrak{L}_i  }{ \mathfrak{K}_i } \biggr]_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>20 \, .  </math>
  </td>
</tr>
</table>
</div>

<span id="FiveOneRadius">Also, from our [[#Summary51|detailed force balance derivations]], we know that,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5} \, .</math>
  </td>
</tr>
</table>

===Zero-Zero Bipolytropes===

====General Form====
In this case, we retain full generality making the substitutions, <math>~n \rightarrow n_c</math> and <math>~j \rightarrow n_e</math>, to obtain,

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

<tr>
  <td align="right">
<math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{n_c}{ 3b} \biggl[a  -\biggl(\frac{3 c}{n_e}\biggr) x^{(n_e-3)/n_e}_\mathrm{eq} \biggr] \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
and
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{n_e^2(n_c-3)}{3[ n_c (n_e+3) - n_e(n_c+3)  ]}\biggr\} \frac{a}{c} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, .
</math>
  </td>
</tr>
</table>
</div>


And here, [[#BiPolytrope00|the expression that describes the free-energy surface]] is,

<div align="center" id="FreeEnergy00">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>~\mathfrak{G}^*_{00} \equiv 5 \biggl(\frac{q}{\nu^2}\biggr) \chi_\mathrm{eq} 
\biggl[\frac{\mathfrak{G}_{00}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{2q^3} \biggl[
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \biggr] \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
Hence, we have, 

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

<tr>
  <td align="right">
<math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
3f \chi_\mathrm{eq}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
n_c \chi_\mathrm{eq}^{3/n_c}  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]   \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2  </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) 
\biggl[\frac{2}{5} q^3  f - A_2\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n_e \chi_\mathrm{eq}^{3/n_e}  \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, , 
</math>
  </td>
</tr>
</table>
</div>

where the definitions of <math>~f</math> and <math>~\mathfrak{F}</math> are [[#BiPolytrope00|given below]].  We immediately deduce that the ''critical'' equilibrium state is identified by,

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

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{fn_e(n_c-3)}{3(n_c - n_e)}\biggr\} [\chi_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit}  \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\}^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math>
  </td>
</tr>
</table>
</div>


From our [[#Equilibrium_Radius_2|associated detailed-force-balance derivation]], we know that the associated equilibrium radius is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c} 
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{n_c} \biggr\}^{1/(n_c-3)}
\, .
</math>
  </td>
</tr>
</table>
</div>

<!--  Coefficient mistake, I think!
We have deduced that the system is unstable if,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>
  </td>
  <td align="center">
<math>~< </math>
  </td>
  <td align="left">
<math>~
\frac{A_2}{C_2}
= \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

-->

====Compare with Five-One====
It is worthwhile to set <math>~n_c = 5</math> and <math>~n_e = 1</math> in this expression and compare the result to the [[#FiveOneRadius|comparable expression shown above for the "Five-One" Bipolytrope]].  Here we have,

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

<tr>
  <td align="right">
<math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{-3} \nu^{4} q^{-2} 
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5} \biggr\}^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\frac{1}{\sqrt{3}} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5/2} \, ;
</math>
  </td>
</tr>
</table>
</div>
whereas, rewriting the [[#FiveOneRadius|above relation]] gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}\biggr|_{51}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{1}{\sqrt{3}} \biggl[\frac{(1+\ell_i^2)^{6/5}}{3\ell_i^2}\biggr]^{5/2} \, .</math>
  </td>
</tr>
</table>
</div>

And, here, we should conclude that the ''critical'' equilibrium configuration is associated with,

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

<tr>
  <td align="right">
<math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{6} \, .</math>
  </td>
</tr>
<!--
<tr>
  <td align="right">
<math>~\Rightarrow~~~q^3 (f - 1-\mathfrak{F} )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{6} \cdot f - 1</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 1 + \frac{2}{5}\biggl(\frac{5}{6} \cdot f - 1\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{f}{3}  + \frac{3}{5}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [\chi_\mathrm{eq}]_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\frac{1}{\sqrt{3}} \biggl[\frac{f}{3}  + \frac{3}{5}\biggr]^{5/2} \, .
</math>
  </td>
</tr>
-->
</table>
</div>