Editing SSC/FreeEnergy/PolytropesEmbedded/Pt1 (section)

====Case M====
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<math>( {\tilde{\mathfrak{f}}}_M, {\tilde{\mathfrak{f}}}_W, {\tilde{\mathfrak{f}}}_A)</math>
</div>

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<math>~\frac{d}{dx_\mathrm{eq}}\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr)= 0</math>
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More specifically, the expression that describes the [[#Case_M_Free-Energy_Surface|"Case M" free-energy surface]] is,
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<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>
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</td></tr>
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Hence, we have,
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<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,

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<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>
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<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,
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<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>

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<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">
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<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>