Difference between revisions of "User:Tohline/SSC/FreeEnergy/PolytropesEmbedded"

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(→‎Case P: Insert link to new chapter that supports PowerPoint presentation)
(→‎Summary: Finished proving that all four expressions are correct for case of n=5 as well!)
 
(48 intermediate revisions by the same user not shown)
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where the structural form factors for pressure-truncated polytropes are precisely defined [[User:Tohline/SSC/Virial/FormFactors#PTtable|here]].  We immediately conclude that,
where the structural form factors for pressure-truncated polytropes are precisely defined [[User:Tohline/SSC/Virial/FormFactors#PTtable|here]].  Therefore, the statement of virial equilibrium is,
 
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~3c[x_\mathrm{eq}]^4_\mathrm{crit} </math>
<math>~</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{(n-3)}{5(n+1)} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}
<math>~\frac{ b}{nc}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c}  - x^{4}_\mathrm{eq}</math>
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math>
<math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi}\biggr)c x_\mathrm{eq}^</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>
<math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \frac{ b}{n}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3} \biggr]</math>
   </td>
   </td>
</tr>
</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 [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|here]].
* Hoerdt's equilibrium expressions from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|here]].
* Conversion from Horedt's units to ours as specified [[User:Tohline/SphericallySymmetricConfigurations/Virial#Choices_Made_by_Other_Researchers|here]].
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~{\tilde\mathfrak{f}}_M</math>
<math>~\Rightarrow ~~~ \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4  </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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}
( 1 + \ell^2 )^{-3/2}   
- \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  \biggr]</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~{\tilde\mathfrak{f}}_W</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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{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]
- \frac{3}{20\pi} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  \, .</math>
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~{\tilde\mathfrak{f}}_A</math>
<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4  </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ]
- \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] </math>
</math>
   </td>
   </td>
</tr>
</tr>
-->
</table>
</div>
And we conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <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>
<math>~3c[x_\mathrm{eq}]^4_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>
<math>~\frac{(n-3)}{5(n+1)} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) [x_\mathrm{eq}]^4_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>
<math>~\frac{1}{20\pi}  \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
   </td>
   </td>
</tr>
</tr>
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<tr>
   <td align="right">
   <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>
<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>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>
<math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>
   </td>
   </td>
</tr>
</tr>
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<tr>
   <td align="right">
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&nbsp;
<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>
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\ell^{18}}{(1+\ell^2)^{12}} \biggr] \biggl[ \frac{2 \cdot 3^4}{\pi} \biggr]^{3}</math>
<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>
   </td>
</tr>
</tr>
</table>
So, the radius of the critical equilibrium state should be,
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^4_\mathrm{crit} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>~\frac{1}{20\pi} \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  
</math>
</math>
   </td>
   </td>
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<tr>
   <td align="right">
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&nbsp;
<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{2^2\cdot 3\cdot 5 \pi}
<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} }
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{18}}  \biggl[ \frac{\pi}{2 \cdot 3^4} \biggr]^{3}\biggr\} (1+\ell^2)^3
</math>
\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>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi^2}{2^9\cdot 3^{13}}   
<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)} \, .
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
whereas, each equilibrium configuration has,
</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 [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|here]].
* Hoerdt's equilibrium expressions from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|here]].
* Conversion from Horedt's units to ours as specified [[User:Tohline/SphericallySymmetricConfigurations/Virial#Choices_Made_by_Other_Researchers|here]].
 
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^4 </math>
<math>~{\tilde\mathfrak{f}}_M</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] \, .</math>
<math>~
( 1 + \ell^2 )^{-3/2}
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] </math>
<math>~{\tilde\mathfrak{f}}_W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi^2}{2^9\cdot 3^{13}} 
<math>~
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{23}\biggr\}
\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]
\cdot \biggl\{ \biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) \biggr] \biggr\}
</math>
</math>
   </td>
   </td>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~2^3\cdot 3 \ell^3</math>
<math>~{\tilde\mathfrak{f}}_A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   <td align="left">
   <td align="left">
<math>~
<math>~
\ell ( \ell^4 - \frac{8}{3}\ell^2 - 1 ) + (1 + \ell^2)^{3}\tan^{-1}(\ell )
\frac{3}{2^3} \ell^{-3}  [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2}
</math>
</math>
   </td>
   </td>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{(1 + \ell^2)^{3}}{\ell} \biggr] \tan^{-1}(\ell ) </math>
<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>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1 + \frac{80}{3}\cdot \ell^2 -\ell^\, .
<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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
The solution is:  <math>~\ell_\mathrm{crit} \approx 2.223175 \, .</math>
</td></tr></table>
</div>
In addition, we know from [[User:Tohline/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,
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<table border="0" cellpadding="5">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) \biggl(\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl[ \frac{(1+\ell^2)^{3}}{\ell^{5}} \biggr] \biggl[ \frac{\pi}{2^3\cdot 3^6} \biggr]^{1/2}</math>
\biggl[ \frac{\tilde\theta^{n+1} }{(4\pi)(n+1)( -\tilde\theta' )^{2}} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ 3c x_\mathrm{eq}^4</math>
<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>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<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>
\biggl[ \frac{\tilde\theta^{n+1} }{(n+1)( -\tilde\theta' )^{2}} \biggr]
\, .
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~2(9-2n){\tilde\theta}^{n+1}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{\ell^{18}}{(1+\ell^2)^{12}} \biggr] \biggl[ \frac{2 \cdot 3^4}{\pi} \biggr]^{3}</math>
3(n-3)\biggl[ (- {\tilde\theta}^')^2 - \frac{\tilde\theta(-{\tilde\theta}^')}{\tilde\xi}\biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
 
So, the radius of the critical equilibrium state should be,
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">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x_\mathrm{eq}^{n-3}</math>
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^4_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
</table>
</div>
====Case P====
Alternatively, let's examine the [[#Case_P_Free-Energy_Surface|"Case P" free-energy surface]].  Drawing on [[User:Tohline/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>
<tr>
   <td align="right">
   <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>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{2^2\cdot 3\cdot 5 \pi} 
\biggl( \frac{G}{K} \biggr)^{n/(n-3)} M_\mathrm{tot}^{(n-1)/(n-3)} 
\biggl\{\frac{(1+\ell^2)^{12}}{\ell^{18}}  \biggl[ \frac{\pi}{2 \cdot 3^4} \biggr]^{3}\biggr\} (1+\ell^2)^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)]}
\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>
</math>
   </td>
   </td>
Line 730: Line 684:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~+
<math>~\frac{\pi^2}{2^9\cdot 3^{13}}   
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)}
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
whereas, each equilibrium configuration has,
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^4 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 747: Line 705:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] \, .</math>
\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>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <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)]}
<math>~\frac{\pi^2}{2^9\cdot 3^{13}} 
P_\mathrm{e}^{(n-1)(3-n)/[2(n+1)(n-3)]}  
\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>
</math>
   </td>
   </td>
Line 770: Line 729:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~2^3\cdot 3 \ell^3</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 776: Line 735:
   </td>
   </td>
   <td align="left">
   <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>~
\ell ( \ell^4 - \frac{8}{3}\ell^2 - 1 ) + (1 + \ell^2)^{3}\tan^{-1}(\ell )
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <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>
<math>~\Rightarrow ~~~\biggl[ \frac{(1 + \ell^2)^{3}}{\ell} \biggr] \tan^{-1}(\ell ) </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 795: Line 749:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 + \frac{80}{3}\cdot \ell^2 -\ell^\, .
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
The solution is:  <math>~\ell_\mathrm{crit} \approx 2.223175 \, .</math>
</td></tr></table>
</div>


<tr>
 
  <td align="right">
In addition, we know from [[User:Tohline/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,
&nbsp;
 
  </td>
<div align="center">
  <td align="center">
<table border="0" cellpadding="5">
<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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) \biggl(\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\times~
<math>
G^{3(n+1)/(n-3)} K_n^{-4n/(n-3)}
\biggl[ \frac{\tilde\theta^{n+1} }{(4\pi)(n+1)( -\tilde\theta' )^{2}} \biggr]
\biggl\{ P_\mathrm{e}^{-(n-3)/[2(n+1)]}\biggr\}^{-2(n+1)/(n-3)}
</math>
</math>
   </td>
   </td>
Line 836: Line 781:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ 3c x_\mathrm{eq}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 842: Line 787:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~P_e
<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)} \, .
\biggl[ \frac{\tilde\theta^{n+1} }{(n+1)( -\tilde\theta' )^{2}} \biggr]
\, .
</math>
</math>
   </td>
   </td>
Line 850: Line 796:
</div>
</div>


Rewriting the expression for the free energy gives,
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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
<math>~2(9-2n){\tilde\theta}^{n+1}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 863: Line 809:
   <td align="left">
   <td align="left">
<math>~
<math>~
-3\mathcal{A} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)
3(n-3)\biggl[ (- {\tilde\theta}^')^2 - \frac{\tilde\theta(-{\tilde\theta}^')}{\tilde\xi}\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>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x_\mathrm{eq}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 878: Line 828:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
-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>
   </td>
</tr>
</tr>
</table>
</div>
====Case P====
=====First Pass=====
Alternatively, let's examine the [[#Case_P_Free-Energy_Surface|"Case P" free-energy surface]].  Drawing on [[User:Tohline/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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~R_\mathrm{norm} \equiv \biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{1/(n-3)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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}
\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>
</math>
   </td>
   </td>
Line 906: Line 874:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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)}  
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)}
\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>
</math>
   </td>
   </td>
Line 921: Line 888:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~R_\mathrm{SWS} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}
-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}  
\biggl( \frac{n+1}{n} \biggr)^{[3(n-1)-(n-3)]/[2(n-3)]}  
+~ n\mathcal{B}  \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
G^{[2n+(n-3)-3(n-1)]/[2(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>
</math>
   </td>
   </td>
Line 937: Line 903:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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)]}
+~ \biggl( \frac{4\pi}{3} \biggr) \biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}  
P_\mathrm{e}^{(n-1)(3-n)/[2(n+1)(n-3)]}
\biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr)^{(5-n)/(n-3)}  
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Therefore, in this case, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 960: Line 917:
   </td>
   </td>
   <td align="left">
   <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>~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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b</math>
<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>
   <td align="center">
   <td align="center">
Line 973: Line 936:
   </td>
   </td>
   <td align="left">
   <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}
<math>~
\biggl( \frac{n+1}{n} \biggr)^{3/(n-3)}
\biggl[ \frac{K^{4n}}{G^{3(n+1)} } \biggr]^{1/(n-3)}
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(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>
</math>
   </td>
   </td>
Line 982: Line 946:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~c</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 988: Line 952:
   </td>
   </td>
   <td align="left">
   <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>~
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


where the structural form factors for pressure-truncated polytropes are precisely defined [[User:Tohline/SSC/Virial/FormFactors#PTtable|here]].  We immediately conclude that,
<tr>
<!-- Supports PowerPoint Synopsis
  <td align="right">
<div align="center">
&nbsp;
<table border="0" cellpadding="5" align="center">
  </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>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl(\frac{4\pi}{3}\biggr)^{-1/n} \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{3(n-1)/[n(n-3)]} x_\mathrm{eq}^{(n-3)/n}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,008: Line 983:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~P_e
\frac{1}{5}\biggl(\frac{n+1}{n}\biggr) \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/(n-3)}
\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)} \, .
+ \frac{4\pi}{3} \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(5-n)/(n-3)} x_\mathrm{eq}^{4}
</math>
</math>
   </td>
   </td>
Line 1,016: Line 990:
</table>
</table>
</div>
</div>
-->
 
<span id="FirstPassFreeEnergy">Rewriting the expression for the free energy gives,</span>


<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,029: Line 1,003:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] \frac{a}{c}
<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>
</math>
   </td>
   </td>
Line 1,042: Line 1,019:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr]
<math>~
\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]
-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]  
\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>
</math>
   </td>
   </td>
Line 1,054: Line 1,030:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{(n-3)}{20\pi (n+1)} \biggr] \biggl(\frac{n+1}{n}\biggr)
<math>~
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  
+~ 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}
\frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


 
<tr>
Also from [[User:Tohline/SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler's work]] we know that the equilibrium mass and radius are,
  <td align="right">
<div align="center">
&nbsp;
<table border="0" cellpadding="3">
  </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>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl[ {\tilde\theta}_n^{(n-3)/2} {\tilde\xi}^2 (-{\tilde\theta}^')
-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}
\biggr] \, ,
+~ 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>
</math>
   </td>
   </td>
Line 1,090: Line 1,072:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl[ \tilde\xi {\tilde\theta}_n^{(n-1)/2} \biggr] \, .
+~ \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>
</math>
   </td>
   </td>
Line 1,105: Line 1,087:
</table>
</table>
</div>
</div>
Additional details in support of an associated PowerPoint presentation can be found [[User:Tohline/SSC/FreeEnergy/PowerPoint|here]].
====Reconcile====




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


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
<math>~a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,123: Line 1,101:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{(n-3)}{20\pi (n+1)} \biggr] \biggl(\frac{n+1}{n}\biggr)
<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)} \, ,
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2}  
\frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


<table border="0" cellpadding="5" align="center">
<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>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math>
<math>~c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,141: Line 1,130:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where the structural form factors for pressure-truncated polytropes are precisely defined [[User:Tohline/SSC/Virial/FormFactors#PTtable|here]].  The statement of virial equilibrium is, therefore,


</div>
Taking the ratio, the RHS is,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,153: Line 1,144:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math>
<math>~x^{4}_\mathrm{eq} + \alpha  </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,159: Line 1,150:
   </td>
   </td>
   <td align="left">
   <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)}
<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]^{-2}
\beta  x^{(n-3)/n }_\mathrm{eq} \, ,
\biggl( \frac{n+1}{n}\biggr)</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\alpha \equiv \frac{a}{3c}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,173: Line 1,169:
   </td>
   </td>
   <td align="left">
   <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)}
<math>~
\biggl[  G^{3} K_n^{-4n/(n+1)} P_\mathrm{e}^{(n-3)/(n+1)} \biggr]</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>
   </td>
</tr>
</tr>
Line 1,186: Line 1,184:
   </td>
   </td>
   <td align="left">
   <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)}
<math>~
\biggl[ K_n^{[(n+1)+(n-3)]/[(n+1)(n-3)] } \biggr]^{-4n} P_\mathrm{e}^{[(n+1)+  (n-3)]/(n+1)} </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>
   </td>
</tr>
</tr>
Line 1,199: Line 1,198:
   </td>
   </td>
   <td align="left">
   <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)}  
<math>~
K_n^{-8(n-1)/[(n+1)(n-3)] }  P_\mathrm{e}^{2(n-1)/(n+1)} \, ;</math>
\biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


while the LHS is,
<tr>
 
  <td align="right">
<div align="center">
<math>~\beta \equiv \frac{b}{nc}</math>
<table border="0" cellpadding="5" align="center">
  </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>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,220: Line 1,230:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{4/(n-3)}
\tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, ,
\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>
</math>
   </td>
   </td>
Line 1,228: Line 1,237:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{m}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}  
<math>~
M_\mathrm{tot}^{4(n-1)/(n-3)}
\biggl(\frac{3}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr) \, .
G^{[6(n-1)]/(n-3)}  
K^{-8n(n-1)/[(n-3)(n+1)] }  P_\mathrm{e}^{2(n-1)/(n+1)} \, .
</math>
</math>
   </td>
   </td>
Line 1,244: Line 1,251:
</div>
</div>


Q.E.D.
From a previous derivation, we have,
 
<!--
<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>
-->
 
By inspection, in the specific case of <math>~n=5</math> (see above), this critical configuration appears to coincide with one of the [[User:Tohline/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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,264: Line 1,257:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{n-3}{n-1} \biggr]_{n=5}</math>
<math>~0 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,270: Line 1,263:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\tilde\xi \tilde\theta^{n}}{(-\tilde\theta^')}\biggr]_{n=5}</math>
<math>~
\frac{ b}{nc}\cdot  x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c}  - x^{4}_\mathrm{eq}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,276: Line 1,271:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\frac{1}{2} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,282: Line 1,277:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
Line 1,291: Line 1,290:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3(1 + \ell^2)^{-1} </math>
<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>
   </td>
</tr>
</tr>
Line 1,300: Line 1,303:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~ \ell </math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,306: Line 1,309:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~5^{1/2} \, .</math>
<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>
   </td>
</tr>
</tr>
Line 1,312: Line 1,335:
</div>
</div>


Hence, according to Kimura, the turning point associated with the configuration with the largest equilibrium radius, corresponds to the equilibrium configuration having,
which, thankfully, matches!  We conclude as well that the transition from stable to dynamically unstable configurations occurs at,
<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!
 
===Five-One Bipolytropes===
For analytically prescribed, "five-one" bipolytropes, <math>~n = 5</math> and <math>~j =1</math>, in which case,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,326: Line 1,341:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x^{2/5 }_\mathrm{eq}</math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,332: Line 1,347:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{5}{ 3b}\biggr) \biggl[a  -3 c x^{-2}_\mathrm{eq} \biggr] \, ;</math>
<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \, .
</math>
   </td>
   </td>
</tr>
</tr>


<!-- FOR POWERPOINT SYNOPSIS
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
and
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
&nbsp;
<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \frac{a}{3c}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,350: Line 1,368:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,356: Line 1,374:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{18 c}{a }\biggr]^{1/2} \, .
<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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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>
<tr>
   <td align="right">
   <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>
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,380: Line 1,388:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{\ell_i^2} \biggl[
<math>~
\Chi^{-3/5} (5 \mathfrak{L}_i)
\biggl( \frac{n}{n-3} \biggr)^{(1-n)}  
+\Chi^{-3} (4\mathfrak{K}_i)
\biggl( \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W} \biggr)^{(n+1)}  
-\Chi^{-1} (3\mathfrak{L}_i +12\mathfrak{K}_i ) \biggr] \, .
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}  \biggr)^{2n} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
-->
</td></tr>
</table>
</table>
</div>
</div>


Hence, we have,
When combined with the statement of virial equilibrium ''at'' this critical point, we find,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,398: Line 1,406:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a</math>
<math>~
\biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr]  + 1\biggr\}\frac{ \alpha }{\beta}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i) \, ,
<math>~
[x_\mathrm{eq}]^{(n-3)/n }_\mathrm{crit}  
</math>
</math>
   </td>
   </td>
Line 1,411: Line 1,422:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 5 \mathfrak{L}_i \chi_\mathrm{eq}^{3/5} \, ,
<math>~
\biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \biggr\}^{(n-3)/(4n) }  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~c</math>
<math>~\Rightarrow~~~
\biggl[ \frac{4n}{3 (n+1)} \biggr]^{4n} \biggl( \frac{ \alpha }{\beta} \biggr)^{4n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 4 \mathfrak{K}_i \chi_\mathrm{eq}^{3}  \, ,
<math>~
\biggl[ \frac{(n-3)}{3 (n+1)} \biggr]^{(n-3)}  \alpha^{(n-3)}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[\chi_\mathrm{eq}]_\mathrm{crit}</math>
<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>
   <td align="center">
   <td align="center">
Line 1,448: Line 1,460:
   </td>
   </td>
   <td align="left">
   <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>
<math>~
\alpha^{3(n+1)} \beta^{-4n}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,460: Line 1,474:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
Line 1,466: Line 1,483:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~\biggl[ \frac{24 \mathfrak{K}_i  }{  (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]_\mathrm{crit}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,472: Line 1,489:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1  </math>
<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>
   </td>
</tr>
</tr>
Line 1,478: Line 1,497:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{L}_i  }{ \mathfrak{K}_i } \biggr]_\mathrm{crit}</math>
<math>~\Rightarrow ~~~
\mathfrak{m}^{2(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,484: Line 1,505:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~20 \, .  </math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
<span id="FiveOneRadius">Also, from our [[#Summary51|detailed force balance derivations]], we know that,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,502: Line 1,520:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
</table>
</div>


===Zero-Zero Bipolytropes===
<tr>
 
  <td align="right">
====General Form====
&nbsp;
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,
  </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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,518: Line 1,550:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,524: Line 1,556:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</tr>
Line 1,533: Line 1,566:
   </td>
   </td>
   <td align="center">
   <td align="center">
and
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
&nbsp;
<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>
   </td>
</tr>
</tr>
Line 1,542: Line 1,576:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,548: Line 1,582:
   </td>
   </td>
   <td align="left">
   <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>~\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>
</math>
   </td>
   </td>
Line 1,561: Line 1,597:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, .
<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>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}^*_{00} \equiv 5 \biggl(\frac{q}{\nu^2}\biggr) \chi_\mathrm{eq}  
<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math>
\biggl[\frac{\mathfrak{G}_{00}}{E_\mathrm{norm}} \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,586: Line 1,611:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{5}{2q^3} \biggl[
<math>~
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \biggr] \, .
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</div>
Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>~
3f \chi_\mathrm{eq}  \, ,
\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>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math>
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr] \frac{a}{c}
n_c \chi_\mathrm{eq}^{3/n_c}  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]   \, ,
</math>
</math>
   </td>
   </td>
Line 1,630: Line 1,660:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2  </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr)
<math>~\biggl[ \frac{(n-3)}{3^2 (n+1)} \biggr]
\biggl[\frac{2}{5} q^3  f - A_2\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>
</math>
   </td>
   </td>
Line 1,644: Line 1,675:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,650: Line 1,681:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
n_e \chi_\mathrm{eq}^{3/n_e}  \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, ,
\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>
</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>
   </td>
</tr>
</tr>
Line 1,658: Line 1,704:
</div>
</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,
which, when combined with the statement of virial equilibrium ''at'' this critical point, implies,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
<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>
   <td align="center">
   <td align="center">
Line 1,671: Line 1,715:
   </td>
   </td>
   <td align="left">
   <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>~  
[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>
</math>
   </td>
   </td>
Line 1,678: Line 1,724:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
<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>
   <td align="center">
   <td align="center">
Line 1,684: Line 1,731:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math>
<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>
   </td>
</tr>
</tr>
Line 1,690: Line 1,740:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~ [x_\mathrm{eq}]_\mathrm{crit}^{(n+1)/n} 
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,696: Line 1,747:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math>
<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>
   </td>
</tr>
</tr>
Line 1,702: Line 1,756:
</div>
</div>


END OMITTED SUBSECTION -->


From our [[#Equilibrium_Radius_2|associated detailed-force-balance derivation]], we know that the associated equilibrium radius is,
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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,709: Line 1,764:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}</math>
<math>~\beta  x^{(n-3)/n }_\mathrm{eq} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,716: Line 1,771:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c}
\alpha + x^{4}_\mathrm{eq}
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{n_c} \biggr\}^{1/(n_c-3)}
\, .
</math>
</math>
   </td>
   </td>
Line 1,725: Line 1,778:
</div>
</div>


<!--  Coefficient mistake, I think!
We have deduced that the system is unstable if,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,732: Line 1,783:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>
<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>
   <td align="center">
   <td align="center">
<math>~< </math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  
\frac{A_2}{C_2}
\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{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, .
+ \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>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,763: Line 1,807:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{-3} \nu^{4} q^{-2}  
\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]
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5} \biggr\}^{1/2}
</math>
</math>
   </td>
   </td>
Line 1,772: Line 1,815:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,778: Line 1,823:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~  
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\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
\frac{1}{\sqrt{3}} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5/2} \, ;
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n-1)/n} \biggl[ \frac{2n}{(n-1)}\biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
whereas, rewriting the [[#FiveOneRadius|above relation]] gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}\biggr|_{51}</math>
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n-3) }_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,798: Line 1,840:
   </td>
   </td>
   <td align="left">
   <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>
<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>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,817: Line 1,855:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{5}{6} \, .</math>
<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>
   </td>
</tr>
</tr>
<!--
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~q^3 (f - 1-\mathfrak{F} )</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,829: Line 1,870:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{5}{6} \cdot f - 1</math>
<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>
   </td>
</tr>
</tr>
Line 1,835: Line 1,879:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]_\mathrm{crit} </math>
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,841: Line 1,887:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 1 + \frac{2}{5}\biggl(\frac{5}{6} \cdot f - 1\biggr)</math>
<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>
   </td>
</tr>
</tr>
Line 1,847: Line 1,896:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~
[x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,853: Line 1,904:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{f}{3} + \frac{3}{5}</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
Also from [[User:Tohline/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>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ [\chi_\mathrm{eq}]_\mathrm{crit}</math>
<math>
~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl[ {\tilde\theta}_n^{(n-3)/2} {\tilde\xi}^2 (-{\tilde\theta}^')
\frac{1}{\sqrt{3}} \biggl[\frac{f}{3}  + \frac{3}{5}\biggr]^{5/2} \, .
\biggr] \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
-->
</table>
</div>
=Free-Energy of Truncated Polytropes=
<!-- Equation for PowerPoint slide presentation
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial\mathfrak{G}|_{K,M,P_e}}{\partial R} = 0</math>
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\partial^2\mathfrak{G}|_{K,M,P_e}}{\partial R^2} \biggr]_\mathrm{eq} = 0</math>
<math>
\biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl[ \tilde\xi {\tilde\theta}_n^{(n-1)/2} \biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
-->


In this case, the Gibbs-like free energy is given by the sum of three separate energies,
Additional details in support of an associated PowerPoint presentation can be found [[User:Tohline/SSC/FreeEnergy/PowerPoint|here]].
 
====Reconcile====
 
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}</math>
<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,907: Line 1,971:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm} + P_eV</math>
<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>
   </td>
</tr>
</tr>
<!-- HIDE INTERMEDIATE EXPRESSION ...
</table>
 
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,919: Line 1,989:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
- \biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}
+ \biggl[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
+ \frac{4\pi}{3} \cdot P_e R^3
</math>
   </td>
   </td>
</tr>
</tr>
END EXPRESSION HIDING -->
</table>
 
</div>
Taking the ratio, the RHS is,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,935: Line 2,007:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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)}  
- 3\mathcal{A} \biggl[\frac{GM^2}{R} \biggr] + n\mathcal{B} \biggl[ \frac{KM^{(n+1)/n}}{R^{3/n}} \biggr] + \frac{4\pi}{3} \cdot P_e R^3 \, ,</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]^{-2}
\biggl( \frac{n+1}{n}\biggr)</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


where the constants,
<div align="center">
<table border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, ,</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
and, as [[User:Tohline/SSC/Virial/FormFactors#PTtable|derived elsewhere]],


<div align="center" id="PTtable">
<table border="1" align="center" cellpadding="5">
<tr>
<th align="center" colspan="1">
Structural Form Factors for <font color="red">Pressure-Truncated</font> Polytropes <math>~(n \ne 5)</math>
</th>
</tr>
<tr>
<td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\tilde\mathfrak{f}_M</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,980: Line 2,034:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl( - \frac{3\tilde\theta^'}{\tilde\xi} \biggr) </math>
<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>
   </td>
</tr>
</tr>
Line 1,986: Line 2,041:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\tilde\mathfrak{f}_W</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>\frac{3\cdot 5}{(5-n)\tilde\xi^2}  
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}  M_\mathrm{tot}^{4(n-1)/(n-3)} G^{[6(n-1)]/(n-3)}  
\biggl[\tilde\theta^{n+1} + 3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr]
K_n^{-8(n-1)/[(n+1)(n-3)] } P_\mathrm{e}^{2(n-1)/(n+1)} \, ;</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
while the LHS is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>
\tilde\mathfrak{f}_A
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,008: Line 2,067:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{(5-n)} \biggl\{ 6\tilde\theta^{n+1} (n+1)
<math>~
\biggl[3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr] \biggr\}
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td>
</tr>
<tr>
  <td align="left" colspan="1">
As [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|we have shown separately]], for the singular case of <math>~n = 5</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_M</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,031: Line 2,082:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl( \frac{n+1}{n} \biggr)^{-2}
( 1 + \ell^2 )^{-3/2}
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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Q.E.D.
Now, given that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_W</math>
<math>~M_\mathrm{SWS}^{-4(n-1)/(n-3)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,045: Line 2,106:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
\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>
   </td>
</tr>
</tr>
Line 2,054: Line 2,112:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_A</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,060: Line 2,118:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
\frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] 
</math>
  </td>
</tr>
</table>
</div>
where, <math>~\ell \equiv \tilde\xi/\sqrt{3} </math>
   </td>
   </td>
</tr>
</tr>
Line 2,073: Line 2,124:
</div>
</div>


 
we have,
In general, then, the warped free-energy surface drapes across a four-dimensional parameter "plane" such that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,080: Line 2,130:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}</math>
<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,086: Line 2,136:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\mathfrak{G}(R, K, M, P_e) \, .</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
In order to effectively visualize the structure of this free-energy surface, we will reduce the parameter space from four to two, in two separate ways:  First, we will hold constant the parameter pair, <math>~(K,M)</math>; giving a nod to [http://adsabs.harvard.edu/abs/1981PASJ...33..299K Kimura's (1981b)] nomenclature, we will refer to the resulting function, <math>~\mathfrak{G}_{K,M}(R,P_e)</math>, as a "Case M" free-energy surface because the mass is being held constant.  Second, we will hold constant the parameter pair, <math>~(K,P_e)</math>, and examine the resulting "Case P" free-energy surface, <math>~\mathfrak{G}_{K,P_e}(R,M)</math>.
==Case M Free-Energy Surface==
It is useful to rewrite the free-energy function in terms of dimensionless parameters.  Here we need to pick normalizations for energy, radius, and pressure that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~M</math>.  We have chosen to use,
<!-- Equation for use in PowerPoint presentation
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
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   <td align="right">
   <td align="right">
<math>~x</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{R}{R_0}</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
-->
<div align="center">
<table border="0" cellpadding="3">


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


which, as is detailed in an [[User:Tohline/SphericallySymmetricConfigurations/Virial#Choices_Made_by_Other_Researchers|accompanying discussion]], are similar but not identical to the normalizations used by [http://adsabs.harvard.edu/abs/1970MNRAS.151...81H Horedt (1970)] and by [http://adsabs.harvard.edu/abs/1981MNRAS.195..967W Whitworth (1981)].  The self-consistent energy normalization is,


By inspection, in the specific case of <math>~n=5</math> (see above), this critical configuration appears to coincide with one of the [[User:Tohline/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">
<div align="center">
<table border="0" cellpadding="3">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~E_\mathrm{norm}</math>
<math>~\biggl[ \frac{n-3}{n-1} \biggr]_{n=5}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~P_\mathrm{norm} R^3_\mathrm{norm} \, .</math>
<math>~\biggl[ \frac{\tilde\xi \tilde\theta^{n}}{(-\tilde\theta^')}\biggr]_{n=5}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
As we have [[User:Tohline/SphericallySymmetricConfigurations/Virial#Gathering_it_all_Together|demonstrated elsewhere]], after implementing these normalizations, the expression that describes the "Case M" free-energy surface is,
<div align="center">
<math>
\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} =
-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>
</div>
Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case M" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.
<!-- Supports PowerPoint summary
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{P_e}{P_\mathrm{norm}}</math>
<math>~\Rightarrow ~~~\frac{1}{2} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,189: Line 2,200:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{20\pi x_\mathrm{eq}^4} \biggl[ 15\biggl(\frac{3}{4\pi}\biggr)^{1/n} x_\mathrm{eq}^{(n-3)/n} - 3\biggr]</math>
<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>
   </td>
</tr>
</tr>
Line 2,195: Line 2,206:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{crit}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,201: Line 2,212:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{4n}{15(n+1)} \biggl( \frac{4\pi}{3} \biggr)^{1/n} \biggr]^{n/(n-3)}</math>
<math>~3(1 + \ell^2)^{-1} </math>
   </td>
   </td>
</tr>
</tr>
Line 2,207: Line 2,218:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[x_\mathrm{eq}]_\mathrm{turn}</math>
<math>~\Rightarrow~~~ \ell </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,213: Line 2,224:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{4n}{15(n+1)} \biggl( \frac{4\pi}{3} \biggr)^{1/n} \biggr]^{n/(n-3)}</math>
<math>~5^{1/2} \, .</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{P_\mathrm{max}}{P_\mathrm{norm}}</math>
<math>~\frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,225: Line 2,252:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \biggl[ \frac{15(n+1)}{4n}\biggl(\frac{3}{4\pi}\biggr)^{1/n} \biggr]^{4n/(n-3)}</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
-->
==Case P Free-Energy Surface==
Again, it is useful to rewrite the free-energy function in terms of dimensionless parameters.  But here we need to pick normalizations for energy, radius, and mass that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~P_e</math>.  As is [[User:Tohline/SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|detailed in an accompanying discussion]], we have chosen to use the normalizations defined by [http://adsabs.harvard.edu/abs/1983ApJ...268..165S Stahler (1983)], namely,
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~R_\mathrm{SWS}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, ,</math>
<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>
   </td>
</tr>
</tr>
Line 2,253: Line 2,278:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~M_\mathrm{SWS}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}  \, .</math>
<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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>


The self-consistent energy normalization is,
Defining a new normalization energy,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="3">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~E_\mathrm{SWS} \equiv \biggl( \frac{n}{n+1} \biggr) \frac{GM_\mathrm{SWS}^2}{R_\mathrm{SWS}}</math>
<math>~E_\mathrm{SWS}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
After implementing these normalizations &#8212; see our [[User:Tohline/SSC/Virial/PolytropesEmbeddedOutline#Our_Case_M_Analysis|accompanying analysis]] for details &#8212; the expression that describes the "Case P" free-energy surface is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}_{K,P_e}^* \equiv \frac{\mathfrak{G}_{K,P_e}}{E_\mathrm{SWS}}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<!-- HIDE LONG RE-DERIVATION ...
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl\{\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \biggr\}^{-1}
\biggl(\frac{n+1}{n}\biggr)^{3/2} K^{3n/(n+1)} G^{-3/2} P_e^{(5-n)/[2(n+1)]} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
we can write,


<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{SWS}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\times \biggl\{- \biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}
-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}
+ \biggl[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}\biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
+~ n\mathcal{B}  
+ \frac{4\pi}{3} \cdot P_e R^3  
\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}
\biggr\}
+\frac{4\pi}{3} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^3 \, ,
</math>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,330: Line 2,360:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
\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
</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>
   </td>
</tr>
</tr>
Line 2,338: Line 2,368:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ n\biggl(\frac{3}{4\pi}\biggr)^{1/n}  
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}
\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[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
= \biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n}
- \frac{4\pi}{3} \cdot P_e R^3
\biggr\}
</math>
</math>
   </td>
   </td>
Line 2,355: Line 2,383:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,361: Line 2,389:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~ \frac{4\pi}{3} \, ,
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, as above,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{m}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2 \frac{R_\mathrm{SWS}}{R} \biggl[G M_\mathrm{SWS}^2 R_\mathrm{SWS}^{-1} \biggr]
   </td>
</math>
   </td>
</tr>
</tr>
</table>
</div>
Now, if we define the pair of parameters,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\alpha</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{a}{3c}</math>
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
\biggl[K M_\mathrm{SWS}^{(n+1)/n} R_\mathrm{SWS}^{-3/n}  \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,398: Line 2,431:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\beta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{b}{nc} \, ,</math>
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ P_e R_\mathrm{SWS}^3 \biggr]
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
then the statement of virial equilibrium is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~x_\mathrm{eq}^4 + \alpha</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,419: Line 2,454:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~\beta x_\mathrm{eq}^{(n-3)/n} \, ,</math>
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and the boundary between dynamical stability and instability occurs at,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \, .</math>
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) G \biggl[\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^2 \biggl[\biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]}  \biggr]^{-1}
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~[x_\mathrm{eq}]^{(n-3)/n}_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{n-3}{3(n+1)} + 1\biggr] \frac{\alpha}{\beta} </math>
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr]
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
K \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}  \biggr]^{(n+1)/n} \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]}  \biggr]^{-3/n} 
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,458: Line 2,494:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~
\biggl\{ \biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \biggr\}^{(n-3)/(4n)} 
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 P_e \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \biggr]^3
\biggl[ \frac{ 4n }{3(n+1)}\biggr]  \frac{\alpha}{\beta}  
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}_{K,P_e}^*</math>
<math>~\Rightarrow ~~~
\beta \alpha^{-3(n+1)/(4n)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,485: Line 2,518:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
\biggl[ \frac{ 4n }{3(n+1)}\biggr] \biggl[ \frac{n-3}{n} \cdot \frac{n}{3(n+1)} \biggr]^{(3-n)/(4n)}  
</math>
</math>
   </td>
   </td>
Line 2,496: Line 2,529:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
4 \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)/(4n)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)/(4n)}  
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) G \biggl[\biggl( \frac{n+1}{nG} \biggr)^{3} K^{4n/(n+1)} P_\mathrm{e}^{(3-n)/[(n+1)]} \biggr]
\biggl[\biggl( \frac{n+1}{nG} \biggr)^{-1/2} K^{-n/(n+1)} P_\mathrm{e}^{(n-1)/[2(n+1)]} \biggr]
</math>
</math>
   </td>
   </td>
Line 2,509: Line 2,540:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~
\biggl( \frac{\beta}{4}\biggr)^{4n} \alpha^{-3(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}\biggr]  
\biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)}  
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
K \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3(n+1)/(2n)} K^{2} P_\mathrm{e}^{(3-n)/(2n)} \biggr]
\biggl[ \biggl( \frac{n+1}{nG} \biggr)^{-3/2n} K^{-3/(n+1)} P_\mathrm{e}^{3(n-1)/[2n(n+1)]} \biggr] 
</math>
</math>
   </td>
   </td>
Line 2,526: Line 2,556:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~
\biggl( \frac{\beta}{4}\biggr)^{4n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 P_e \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{3n/(n+1)} P_\mathrm{e}^{3(1-n)/[2(n+1)]} \biggr]
\biggl[ \frac{ n\alpha }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3}  \, .
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
======Case M======
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, ,
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
</math>
</math>
   </td>
   </td>
Line 2,555: Line 2,591:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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}}  \, ,
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) \biggl[\biggl( \frac{n+1}{n} \biggr)^{5/2} G^{-3/2} K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]}  \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, .
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr]
</math>
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
\biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]}  \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\alpha</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \biggr]
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,601: Line 2,633:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\beta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,607: Line 2,639:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<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>
\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr)
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So the dynamical stability conditions are:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggl( \frac{n}{n-3} \biggr) [x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
+ \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr]  
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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>
   <td align="center">
   <td align="center">
Line 2,638: Line 2,677:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~-
<math>~
\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl[ \frac{ n}{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3}
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1}
\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>
</math>
   </td>
   </td>
Line 2,647: Line 2,687:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}} \biggr]
\biggl[\biggl(\frac{ n}{n+1}\biggr) \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5} \biggr]^{3(n+1)}  
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  
\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}
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3  
</math>
</math>
   </td>
   </td>
Line 2,663: Line 2,704:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~\biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
END SUPPRESSION OF LONG DERIVATION -->
   <td align="left">
   <td align="left">
<math>~- 3 \mathcal{A} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2 \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1}
<math>~
+ n\mathcal{B} \biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n}
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \, .
\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>
</math>
   </td>
   </td>
Line 2,679: Line 2,721:
</div>
</div>


Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case P" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.
Together, then,
 
=Free-Energy of Bipolytropes=
 
In this case, the Gibbs-like free energy is given by the sum of four separate energies,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,689: Line 2,727:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}</math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n-3)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,695: Line 2,733:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{core} + \biggl[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{env} \, .
\biggl[ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \frac{n}{n-3} \biggr]^{-(n-3)} </math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
In addition to specifying (generally) separate polytropic indexes for the core, <math>~n_c</math>, and envelope, <math>~n_e</math>, and an envelope-to-core mean molecular weight ratio, <math>~\mu_e/\mu_c</math>, we will assume that the system is fully defined via specification of the following five physical parameters:
* Total mass, <math>~M_\mathrm{tot}</math>;
* Total radius, <math>~R</math>;
* Interface radius, <math>~R_i</math>, and associated dimensionless interface marker, <math>~q \equiv R_i/R</math>;
* Core mass, <math>~M_c</math>, and associated dimensionless mass fraction, <math>~\nu \equiv M_c/M_\mathrm{tot}</math>;
* Polytropic constant in the core, <math>~K_c</math>.
In general, the warped free-energy surface drapes across a five-dimensional parameter "plane" such that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,723: Line 2,746:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\mathfrak{G}(R, K_c, M_\mathrm{tot}, q, \nu) \, .</math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
==Order of Magnitude Derivation==
Let's begin by providing very rough, approximate expressions for each of these four terms, assuming that <math>~n_c = 5</math> and <math>~n_e = 1</math>. 
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot} M_c}{(R_i/2)} \biggr]  
<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)}  
= - 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr] \, ;</math>
\biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{-4n}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,749: Line 2,770:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
<math>~\Rightarrow ~~~[x_\mathrm{eq}]_\mathrm{crit}^{(n-3)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot} M_e}{(R_i+R)/2} \biggr]
<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)}  \, .
= - 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ;</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
======Case P======
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{core} = U_\mathrm{int}\biggr|_\mathrm{core} </math>
<math>~a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\mathfrak{b}_c \cdot n_cK_c M_c ({\bar\rho}_c)^{1/n_c}
<math>~
= 5\mathfrak{b}_c \cdot K_c M_\mathrm{tot}\nu \biggl[ \frac{3M_c}{4\pi R_i^3} \biggr]^{1/5}  
\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>
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,776: Line 2,802:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,782: Line 2,808:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5} 
<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>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\mathfrak{b}_e \cdot n_eK_e M_\mathrm{env} ({\bar\rho}_e)^{1/n_e}
= \mathfrak{b}_e \cdot K_e M_\mathrm{tot}(1-\nu) \biggl[ \frac{3M_\mathrm{env}}{4\pi (R^3-R_i^3)} \biggr]
</math>
</math>
   </td>
   </td>
Line 2,803: Line 2,815:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,809: Line 2,821:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ \frac{4\pi}{3} \, ,
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) K_e [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1} \, .
</math>
</math>
   </td>
   </td>
Line 2,816: Line 2,827:
</table>
</table>
</div>
</div>
In writing this last expression, it has been necessary to (temporarily) introduce a sixth physical parameter, namely, the polytropic constant that characterizes the envelope material, <math>~K_e</math>.  But this constant can be expressed in terms of <math>~K_c</math> via a relation that ensures continuity of pressure across the interface while taking into account the drop in mean molecular weight across the interface, that is,
where, as above,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,822: Line 2,833:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~K_e ({\bar\rho}_e)^{(n_e+1)/n_e}</math>
<math>~\mathfrak{m}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~K_c ({\bar\rho}_c)^{(n_c+1)/n_c}</math>
<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M}  \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ K_e \biggl[\biggl( \frac{\mu_e}{\mu_c} \biggr) {\bar\rho}_c\biggr]^{2}</math>
<math>~\alpha</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~K_c ({\bar\rho}_c)^{6/5}</math>
<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>
   </td>
</tr>
</tr>
Line 2,846: Line 2,864:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ \frac{K_e}{K_c} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2}</math>
<math>~\beta</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} \, .</math>
<math>~
\tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence, the fourth energy term may be rewritten in the form,
So the dynamical stability conditions are:
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,864: Line 2,883:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
   </td>
K_c\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1}  
</math>
   </td>
</tr>
</tr>


Line 2,885: Line 2,901:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr) \mathfrak{m}^{2} </math>
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
and,
Putting all the terms together gives,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,900: Line 2,912:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}</math>
<math>~  
\biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \mathfrak{m}^{4(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr]
\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}
- 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr]
+ \mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5}
</math>
</math>
   </td>
   </td>
Line 2,916: Line 2,928:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~ \Rightarrow~~~
  </td>
\mathfrak{m}^{2(n+1)}
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)}  
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,938: Line 2,937:
   <td align="left">
   <td align="left">
<math>~
<math>~
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr]  
\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} \, .
+ \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{(\nu M_\mathrm{tot})^{2}}{ qR} \biggr]^{3/5}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Together, then,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,953: Line 2,957:
   <td align="left">
   <td align="left">
<math>~
<math>~
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr] \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
\biggl[ \frac{n-3}{n} \biggr]^{(n+1)} \biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr)^{(n+1)}
+ \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} K_c \biggl[\frac{M_\mathrm{tot}^{2}}{ R} \biggr]^{3/5}\biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
\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>
</math>
   </td>
   </td>
Line 2,968: Line 2,972:
   <td align="left">
   <td align="left">
<math>~
<math>~
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]  
\biggl[ \frac{n-3}{n} \biggr]^{2(n-1)}  
+ \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl[\frac{R_\mathrm{norm}}{ R} \biggr]^{3/5} \, ,
\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>
</math>
   </td>
   </td>
Line 2,976: Line 2,980:
</div>
</div>


where,
======Compare======
 
Let's see if the two cases, in fact, provide the same answer.
<!--
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 2,982: Line 2,989:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A}_\mathrm{biP}</math>
<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>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr)  + \mathfrak{a}_e  \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ,</math>
<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>
   </td>
</tr>
</tr>
Line 2,994: Line 3,005:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{B}_\mathrm{biP}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{3}{2^2\pi} \biggr)^{1/5} \biggl[5\mathfrak{b}_c 
<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}  
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr] \, .</math>
\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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
-->
==Equilibrium Radius==
===Order of Magnitude Estimate===
This means that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,015: Line 3,025:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial\mathfrak{G}}{\partial R}</math>
<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>
   <td align="center">
   <td align="center">
Line 3,022: Line 3,032:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ 2 \mathcal{A}_\mathrm{biP}\biggl[ \frac{GM_\mathrm{tot}^2 }{R^2} \biggr]
\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}
- \frac{3}{5} \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{\nu^{2}}{ q} \biggr]^{3/5} M_\mathrm{tot}^{6/5} R^{-8/5}
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, because equilibrium radii are identified by setting <math>~\partial\mathfrak{G}/\partial R = 0</math>, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,043: Line 3,046:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl[\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}\biggr]^{5/2} \biggl(\frac{ q} {\nu^{2}}\biggr)^{3/2} \, .
<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>
</math>
   </td>
   </td>
Line 3,050: Line 3,055:
</div>
</div>


===Reconcile With Known Analytic Expression===
===Five-One Bipolytropes===
From our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#The_Core_2|earlier derivations]], it appears as though,
For analytically prescribed, "five-one" bipolytropes, <math>~n = 5</math> and <math>~j =1</math>, in which case,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,057: Line 3,063:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq} \equiv \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
<math>~x^{2/5 }_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,063: Line 3,069:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{3^8}{2^5\pi} \biggr)^{-1/2} 
<math>~\biggl(\frac{5}{ 3b}\biggr) \biggl[a  -3 c x^{-2}_\mathrm{eq} \biggr] \, ;</math>
\biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,074: Line 3,078:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
and
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl(\frac{q}{\nu^2} \biggr)^{3/2}
&nbsp;
\biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2}  \biggl(\frac{\nu^2}{q} \biggr)^{5/2}
\frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This implies that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}</math>
<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{18 c}{a }\biggr]^{1/2} \, .
\biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2}  \biggl(\frac{\nu^2}{q} \biggr)^{5/2}
\frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr]^{2/5}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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>
   <td align="center">
   <td align="center">
Line 3,113: Line 3,117:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl(\frac{\nu^2}{q} \biggr)
<math>~ \frac{1}{\ell_i^2} \biggl[
\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/5}   \frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2}
\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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</div>
Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr)  + \mathfrak{a}_e  \biggl(\frac{1-\nu}{1+q}\biggr)  \biggr] </math>
<math>~a</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{2^2\cdot 5}\biggl(\frac{\nu^2}{q} \biggr)
<math>~ 3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i) \, ,
\frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c 
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr]
</math>
</math>
   </td>
   </td>
Line 3,136: Line 3,148:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c  + \mathfrak{a}_e \cdot  \frac{q(1-\nu)}{\nu(1+q)}  \biggr] </math>
<math>~b</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\nu}{2^2\cdot 5}
<math>~ 5 \mathfrak{L}_i \chi_\mathrm{eq}^{3/5} \, ,
\frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c 
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
==Focus on Five-One Free-Energy Expression==
===Approximate Expressions===
Let's plug this equilibrium radius back into each term of the free-energy expression.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{core}</math>
<math>~c</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- 2\mathfrak{a}_c \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{\nu}{q}\biggr) \biggr] </math>
<math>~ 4 \mathfrak{K}_i \chi_\mathrm{eq}^{3} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~[\chi_\mathrm{eq}]_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,179: Line 3,185:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- 2\mathfrak{a}_c \biggl(\frac{\nu}{q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>
<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>
   </td>
</tr>
</tr>
Line 3,185: Line 3,191:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{env}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- 2\mathfrak{a}_e \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] </math>
<math>~[\chi_\mathrm{eq}]_\mathrm{crit}\biggl[ \frac{24 \mathfrak{K}_i  }{ (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2}  </math>
   </td>
   </td>
</tr>
</tr>
Line 3,197: Line 3,203:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~\biggl[ \frac{24 \mathfrak{K}_i  }{  (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,203: Line 3,209:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- 2\mathfrak{a}_e \biggl(\frac{1-\nu}{1+q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>
<math>~1 </math>
   </td>
   </td>
</tr>
</tr>
Line 3,209: Line 3,215:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_c-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{core} </math>
<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{L}_i  }{ \mathfrak{K}_i } \biggr]_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
<math>~20 \, . </math>
K_c (M_\mathrm{tot}\nu)^{6/5} (R_\mathrm{eq}q)^{-3/5}  
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<span id="FiveOneRadius">Also, from our [[#Summary51|detailed force balance derivations]], we know that,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,229: Line 3,239:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<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>
\biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5}
\biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5}
\, ;</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===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>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{S_\mathrm{env}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_e-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{env} </math>
<math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\approx</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3}{2}\biggr]
<math>~\frac{n_c}{ 3b} \biggl[a  -\biggl(\frac{3 c}{n_e}\biggr) x^{(n_e-3)/n_e}_\mathrm{eq} \biggr] \, ;</math>
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
\biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} K_c M_\mathrm{tot}^{6/5}R_\mathrm{eq}^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)}
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,256: Line 3,270:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
and
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3}{2}\biggr]
&nbsp;
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)}
\biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===From Detailed Force-Balance Models===
In the following derivations, we will use the expression,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,284: Line 3,285:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{\mu_e}{\mu_c} \biggr)^3 \biggl( \frac{\pi}{2^3} \biggr)^{1/2} \frac{1}{A^2\eta_s}
<math>~\biggl\{\frac{n_e^2(n_c-3)}{3[ n_c (n_e+3) - n_e(n_c+3)  ]}\biggr\} \frac{a}{c}  
= \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>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Keep in mind, as well &#8212; as derived in an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Background|accompanying discussion]] &#8212; that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,303: Line 3,298:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ (m_3^2 \ell_i^3) (1 + \ell_i^2)^{-1/2} [1 + (1-m_3)^2 \ell_i^2]^{-1/2}  \biggl[ m_3\ell_i + (1+\ell_i^2) \biggl(\frac{\pi}{2} + \tan^{-1} \Lambda_i \biggr) \biggr]^{-1} \, ,</math>
<math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
<div align="center">
<math>m_3 \equiv 3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \, .</math>
</div>


From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
 
<div align="center">
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">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~q</math>
<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>
   <td align="center">
   <td align="center">
Line 3,325: Line 3,323:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\eta_i}{\eta_s}
<math>~ \frac{5}{2q^3} \biggl[
= \eta_i \biggl\{\frac{\pi}{2} + \eta_i + \tan^{-1}\biggl[ \frac{1}{\eta_i} - \ell_i \biggr] \biggr\}^{-1}</math>
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</div>
Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
  <td align="left">
<td align="left">
<math>~
<math>
\eta_i \biggl\{\eta_i + \cot^{-1}\biggl[ \ell_i - \frac{1}{\eta_i} \biggr] \biggr\}^{-1} \, ,
3f \chi_\mathrm{eq}  \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\eta_i</math>
<math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~m_3 \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
<math>
n_c \chi_\mathrm{eq}^{3/n_c}  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]   \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's also define the following shorthand notation:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{L}_i</math>
<math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2  </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,375: Line 3,373:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ;</math>
<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>
   </td>
</tr>
</tr>
Line 3,381: Line 3,381:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{K}_i</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + \frac{\Lambda_i}{\eta_i} \, .</math>
<math>~  
n_e \chi_\mathrm{eq}^{3/n_e} \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,393: Line 3,395:
</div>
</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">


====Gravitational Potential Energy of the Core====
<tr>
 
Pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
<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}  
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] \, .</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,415: Line 3,415:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ -\chi_\mathrm{eq} \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math>
\biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]
\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>
   </td>
</tr>
</tr>
Line 3,434: Line 3,430:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math>
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)  + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,446: Line 3,439:
</div>
</div>


Out of equilibrium, then, we should expect,
 
From our [[#Equilibrium_Radius_2|associated detailed-force-balance derivation]], we know that the associated equilibrium radius is,
<div align="center">
<div align="center">
<table border="0" cellpadding="4">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{W_\mathrm{core}}{E_\mathrm{norm}} </math>
<math>~\chi_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \chi^{-1}
<math>~
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5}
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c}  
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)   + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{n_c} \biggr\}^{1/(n_c-3)}
\, .
</math>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~< </math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
- \chi^{-1}
\frac{A_2}{C_2}
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}
= \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, .
\biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr]
\, ,
</math>
</math>
   </td>
   </td>
Line 3,483: Line 3,483:
</table>
</table>
</div>
</div>
which, in comparison with our above approximate expression, implies,
 
-->
 
====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">
<div align="center">
<table border="0" cellpadding="4">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{a}_c  </math>
<math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\biggl( \frac{3}{2^5} \biggr) \frac{\nu}{\ell_i^5}
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{-3} \nu^{4} q^{-2}  
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)   + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]  \, .
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5} \biggr\}^{1/2}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


====Thermal Energy of the Core====
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}   + \tan^{-1}(\ell_i) \biggr] </math>
<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>
   </td>
</tr>
</tr>
</table>
</div>
whereas, rewriting the [[#FiveOneRadius|above relation]] gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ \chi_\mathrm{eq}^{3} \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]^5_\mathrm{eq}</math>
<math>~\chi_\mathrm{eq}\biggr|_{51}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<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>
\frac{1}{2^5} \biggl( \frac{3^8}{2^5\pi} \biggr)^{5/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}   + \tan^{-1}(\ell_i) \biggr]^5
\biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3}
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11}
<math>~ \frac{5}{6} \, .</math>
\biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]^5  
\biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
 
<!--
</table>
</div>
 
Out of equilibrium, we should then expect,
 
<div align="center">
<table border="0" cellpadding="4">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}}</math>
<math>~\Rightarrow~~~q^3 (f - 1-\mathfrak{F} )</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~ \frac{5}{6} \cdot f - 1</math>
\biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1}
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i
\, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
In comparison with our above approximate expression, we therefore have,
<div align="center">
<table border="0" cellpadding="4">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~\Rightarrow~~~\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]_\mathrm{crit} </math>
\biggl[ \biggl(\frac{3}{2\cdot 5}\biggr)\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} 
\biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggr]^5</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11}  
<math>~ 1 + \frac{2}{5}\biggl(\frac{5}{6} \cdot f - 1\biggr)</math>
\biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}   + \tan^{-1}(\ell_i) \biggr]^5
\biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,603: Line 3,584:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~~
&nbsp;
\mathfrak{b}_c 
  </td>
</math>
  <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>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{ 3 }{2^3\ell_i^{3}(1+\ell_i^2)^{6/5}}
<math>~
\biggl[ \ell_i (\ell_i^4 - 1 )  + (1+\ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] \, .
\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>
</math>
   </td>
   </td>
</tr>
</tr>
 
-->
</table>
</table>
</div>
</div>


 
=Free-Energy of Truncated Polytropes=
====Gravitational Potential Energy of the Envelope====
<!-- Equation for PowerPoint slide presentation
 
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]] and appreciating, in particular, that (see, for example, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Equilibrium_Condition|our notes on equilibrium conditions]]),
<div align="center">
<div align="center">
<table border="0" cellpadding="3">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~A</math>
<math>~\frac{\partial\mathfrak{G}|_{K,M,P_e}}{\partial R} = 0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\eta_i}{\sin(\eta_i - B)} \, ,</math>
<math>~\biggl[ \frac{\partial^2\mathfrak{G}|_{K,M,P_e}}{\partial R^2} \biggr]_\mathrm{eq} = 0</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
-->
In this case, the Gibbs-like free energy is given by the sum of three separate energies,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~(\eta_s - B)</math>
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\pi \, ,</math>
<math>~W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm} + P_eV</math>
   </td>
   </td>
</tr>
</tr>
 
<!-- HIDE INTERMEDIATE EXPRESSION ...
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\eta_i - B</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\, ,</math>
<math>~
- \biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}
+ \biggl[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
+ \frac{4\pi}{3} \cdot P_e R^3
</math>
   </td>
   </td>
</tr>
</tr>
</table>
END EXPRESSION HIDING -->
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \sin(\eta_i -B) = (1+\Lambda_i^2)^{-1/2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\sin[2(\eta_i-B)] = 2\Lambda_i(1 + \Lambda_i^2)^{-1} \ ,</math>
<math>~
- 3\mathcal{A} \biggl[\frac{GM^2}{R} \biggr] + n\mathcal{B} \biggl[ \frac{KM^{(n+1)/n}}{R^{3/n}} \biggr] + \frac{4\pi}{3} \cdot P_e R^3 \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
we have,


where the constants,
<div align="center">
<div align="center">
<table border="0" cellpadding="4">
<table border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
&nbsp; &nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2
<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, ,</math>
\biggl\{ \biggl[6(\eta_s-B) - 3\sin[2(\eta_s - B)] -4\eta_s\sin^2(\eta_s-B) + 4B\biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, as [[User:Tohline/SSC/Virial/FormFactors#PTtable|derived elsewhere]],
<div align="center" id="PTtable">
<table border="1" align="center" cellpadding="5">
<tr>
<th align="center" colspan="1">
Structural Form Factors for <font color="red">Pressure-Truncated</font> Polytropes <math>~(n \ne 5)</math>
</th>
</tr>
<tr>
<td align="center">


<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\tilde\mathfrak{f}_M</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) + 4B \biggr]\biggr\}
<math>~ \biggl( - \frac{3\tilde\theta^'}{\tilde\xi} \biggr) </math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,713: Line 3,723:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>\tilde\mathfrak{f}_W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
  <td align="left">
<td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2
<math>\frac{3\cdot 5}{(5-n)\tilde\xi^2}  
\biggl\{ 6\pi - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) \biggr]\biggr\}
\biggl[\tilde\theta^{n+1} + 3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr]  
</math>
</math>
   </td>
   </td>
Line 3,727: Line 3,737:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
\tilde\mathfrak{f}_A
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2(1+\Lambda_i^2)
<math>~\frac{1}{(5-n)} \biggl\{ 6\tilde\theta^{n+1} (n+1)
\biggl\{ 6\pi - 6\biggl[\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\biggr] + 6\biggl[ \frac{\Lambda_i}{(1 + \Lambda_i^2)}  \biggr]
\biggl[3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr] \biggr\}
+ 4\eta_i \biggl[ \frac{1}{(1+\Lambda_i^2)} \biggr] \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td>
</tr>
<tr>
  <td align="left" colspan="1">
As [[User:Tohline/SSC/Virial/FormFactors#Summary_.28n.3D5.29|we have shown separately]], for the singular case of <math>~n = 5</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{f}_M</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
<math>~   
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\} \, .
( 1 + \ell^2 )^{-3/2}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, in equilibrium we can write,


<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~-\chi_\mathrm{eq}\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
<math>~\mathfrak{f}_W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
<math>~
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\}
\frac{5}{2^4} \cdot \ell^{-5}   
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}
\biggl[ \ell \biggl( \ell^4 - \frac{8}{3}\ell^2 - 1 \biggr)(1 + \ell^2)^{-3} + \tan^{-1}(\ell ) \biggr]
</math>
</math>
   </td>
   </td>
Line 3,777: Line 3,791:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{f}_A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{3}{2^2} \biggl(\frac{\eta_i}{m_3}\biggr)^3
<math>~
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\}
\frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2}
\frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{\ell_i^5}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, <math>~\ell \equiv \tilde\xi/\sqrt{3} </math>
  </td>
</tr>
</table>
</div>
In general, then, the warped free-energy surface drapes across a four-dimensional parameter "plane" such that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
<math>~\mathfrak{G}(R, K, M, P_e) \, .</math>
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,806: Line 3,829:
</div>
</div>


And out of equilibrium,
In order to effectively visualize the structure of this free-energy surface, we will reduce the parameter space from four to two, in two separate ways:  First, we will hold constant the parameter pair, <math>~(K,M)</math>; giving a nod to [http://adsabs.harvard.edu/abs/1981PASJ...33..299K Kimura's (1981b)] nomenclature, we will refer to the resulting function, <math>~\mathfrak{G}_{K,M}(R,P_e)</math>, as a "Case M" free-energy surface because the mass is being held constant.  Second, we will hold constant the parameter pair, <math>~(K,P_e)</math>, and examine the resulting "Case P" free-energy surface, <math>~\mathfrak{G}_{K,P_e}(R,M)</math>.


==Virial Equilibrium and Dynamical Stability==
The first (partial) derivative of <math>~\mathfrak{G}</math> with respect to <math>~R</math> is,
<div align="center">
<div align="center">
<table border="0" cellpadding="4">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{W_\mathrm{env}}{E_\mathrm{norm}}</math>
<math>~\frac{\partial \mathfrak{G}}{\partial R}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
-\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
\frac{1}{R} \biggl[
\biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr]  
3\mathcal{A} GM^2 R^{-1} - 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 4\pi P_e R^3
\, .
\biggr] \, ;
</math>
</math>
   </td>
   </td>
Line 3,827: Line 3,853:
</table>
</table>
</div>
</div>
and the second (partial) derivative is,
<div align="center">
<table border="0" cellpadding="5" align="center">


This, in turn, implies that both in and out of equilibrium,
<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{a}_e </math>
<math>~\frac{\partial^2 \mathfrak{G}}{\partial R^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{3}{2^3} \biggl[\frac{\nu^2(1+q)}{q(1-\nu)} \biggr] \frac{1}{\ell_i^2}
<math>~
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, .
\frac{1}{R^2} \biggl[
-6\mathcal{A} GM^2 R^{-1} + \biggl(\frac{n+3}{n}\biggr) 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi P_e R^3
\biggr] \, .
</math>
</math>
   </td>
   </td>
Line 3,847: Line 3,874:
</table>
</table>
</div>
</div>
The virial equilibrium radius is identified by setting the first derivative to zero. This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">


====Thermal Energy of the Envelope====
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~3\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2
<math>~
\biggl\{  \biggl[6(\eta_s - B) - 3\sin[2(\eta_s-B)] \biggr] - \biggl[6(\eta_i - B) - 3\sin[2(\eta_i-B)] \biggr] \biggr\}</math>
3\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi P_e R_\mathrm{eq}^3 \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


This expression can be usefully rewritten in the following forms:
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr>
  <td colspan="2" align="center">Virial Equilibrium Condition</td>
</tr>
<tr>
  <td align="center">Case 1:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~3(n+3)\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2
<math>~
\biggl\{ 6\pi - 6(\eta_i - B) + 3\sin[2(\eta_i-B)]  \biggr\}</math>
3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi (n+3) P_e R_\mathrm{eq}^3
</math>
  </td>
</tr>
</table>
   </td>
   </td>
</tr>
</tr>


<tr>
  <td align="center">Case 2:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 (1 + \Lambda_i^2)  
<math>~ 8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n</math>
\biggl\{ 6\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + 6\biggl[\Lambda_i(1 + \Lambda_i^2)^{-1} \biggr] \biggr\}</math>
  </td>
</tr>
</table>
   </td>
   </td>
</tr>
</tr>


<tr>
  <td align="center">Case 3:</td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
8\pi nP_e R_\mathrm{eq}^3
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 
<math>~6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>
\biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} \, .</math>
  </td>
</tr>
</table>
   </td>
   </td>
</tr>
</tr>
Line 3,908: Line 3,965:
</div>
</div>


So, in equilibrium we can write,
Dynamical stability is determined by the sign of the second derivative expression ''evaluated at the equilibrium radius''; setting the second derivative to zero identifies the transition from stable to unstable configurations.  The criterion is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}^{3}\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^
<math>~\biggl[
\biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\}
-6n\mathcal{A} GM^2 R^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi nP_e R^3\biggr]_{R_\mathrm{eq}}
\biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3}
</math>
  </td>
</tr>
</table>
</div>
 
===Case 1 Stability Criterion===
Using the "Case 1" virial expression to define the equilibrium radius means that the stability criterion 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>~
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi (n+3) P_e R_\mathrm{eq}^3 + 8\pi nP_e R_\mathrm{eq}^3
</math>
</math>
   </td>
   </td>
Line 3,932: Line 4,009:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{3^2\pi^2}{2^{12}} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3 
<math>~
\biggl\{  \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\}
\mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n+3)- 6n ] + 4\pi P_e R_\mathrm{eq}^3 [(n+3)   + 2n]
\biggl[\frac{(1+\ell_i^2)^9}{3^9\ell_i^{15}}\biggr]
</math>
</math>
   </td>
   </td>
Line 3,944: Line 4,020:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~
4\pi P_e R_\mathrm{eq}^3 [3(n+1) ]
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr)  \biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]
<math>~
\biggl\{  \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\} \, .
\mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n-3)]  
</math>
</math>
   </td>
   </td>
</tr>
</tr>


</table>
</div>
And, out of equilibrium,
<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~\Rightarrow ~~~
4\pi P_e R_\mathrm{eq}^4 (n+1)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~
<math>~
\chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr)  
\mathcal{A} GM^2 (n-3)
\biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K}
\, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>


====Combined in Equilibrium====
===Case 2 Stability Criterion===
 
Using the "Case 2" virial expression to define the equilibrium radius means that the stability criterion is,
Notice that, in combination,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 3,990: Line 4,059:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{2S_\mathrm{env} + W_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 3,996: Line 4,065:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3  
<math>~
8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n+ 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 8\pi nP_e R_\mathrm{eq}^3
</math>
</math>
   </td>
   </td>
Line 4,009: Line 4,079:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3}  
<math>~
\biggl[3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i \biggl( 1 + \ell_i^2 \biggr)^{-1}\biggr]^3
16\pi nP_e R_\mathrm{eq}^3 - [3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}
</math>
</math>
   </td>
   </td>
Line 4,017: Line 4,087:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~
16\pi nP_e R_\mathrm{eq}^3 
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,023: Line 4,095:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl( \frac{2\cdot 3^6}{\pi} \biggr)^{1/2}
<math>~
\biggl[\frac{\ell_i^3}{( 1 + \ell_i^2)^3}\biggr] \, .
[3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Also, from above,
<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[  \frac{2S_\mathrm{core}+W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~\Rightarrow~~~
16\pi nP_e R_\mathrm{eq}^{3(n+1)/n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
<math>~
\biggl[ \ell_i \biggl(- \frac{8}{3} \ell_i^2 \biggr) (1 + \ell_i^2)^{-3}   \biggr] </math>
[3(n-3)]\mathcal{B}KM^{(n+1)/n}
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ + \biggl( \frac{2\cdot 3^6}{\pi } \biggr)^{1/2}
\biggl[ \frac{\ell_i^3}{(1 + \ell_i^2)^{3}}  \biggr] \, .</math>
  </td>
</tr>
</table>
</table>
</div>
</div>


So, in equilibrium, these terms from the core and envelope sum to zero, as they should.
===Case 3 Stability Criterion===
 
Using the "Case 3" virial expression to define the equilibrium radius means that the stability criterion is,
====Out of Equilibrium====
And now, in combination ''out'' of equilibrium,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 4,073: Line 4,126:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,079: Line 4,132:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} 
-6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}  
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\}
+\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{2n_c}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2n_e}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
\, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, quite generally ''out'' of equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,103: Line 4,146:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
-\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}   
-12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}   
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\}
+ [6n +3(n+3)] \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  
-\frac{3}{5}\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{10}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
-3\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's see what the value of this derivative is if the dimensionless radius, <math>~\chi</math>, is set to the value that has been determined, via a detailed force-balanced analysis, to be the equilibrium radius, namely, <math>~\chi = \chi_\mathrm{eq}</math>.  In this case, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl\{\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] \biggr\}_\mathrm{\chi \rightarrow \chi_\mathrm{eq}}</math>
<math>~\Rightarrow~~~
9(n+1) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,125: Line 4,163:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\chi_\mathrm{eq}^{-1}\biggl\{
<math>~
\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}   
12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1}   
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+2\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+2\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
But, according to the virial theorem &#8212; and, as we have just demonstrated &#8212; the four terms inside the curly braces sum to zero.  So this demonstrates that the derivative of our out-of-equilibrium free-energy expression does go to zero at the equilibrium radius, as it should!
===Summary51===
In summary, the desired ''out'' of equilibrium free-energy expression is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
<math>~\Rightarrow~~~
R_\mathrm{eq}^{n-3}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,151: Line 4,179:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\frac{W_\mathrm{core}}{E_\mathrm{norm}}  + \frac{W_\mathrm{env}}{E_\mathrm{norm}}
\biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1}  
+\biggl(\frac{2n_c}{3}\biggr)\frac{S_\mathrm{core}}{E_\mathrm{norm}}
+\biggl(\frac{2n_e}{3}\biggr)\frac{S_\mathrm{env}}{E_\mathrm{norm}}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===Case M===
Now, in our discussion of "Case M" sequence analyses, the configuration's radius is normalized to,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~R_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~[ G^n K^{-n} M^{n-1} ]^{1/(n-3)} \, .</math>
- \chi^{-1}
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}
\biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr]
-\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
\biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Our "Case 3" stability criterion directly relates.  We conclude that the transition from stability to dynamical instability occurs when,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~
\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+
\biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n
\biggl(\frac{2\cdot 5}{3}\biggr) \biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1}
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i
+\biggl(\frac{2}{3}\biggr)
\chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) 
\biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K}
</math>
</math>
   </td>
   </td>
Line 4,198: Line 4,228:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow~~~
\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{(n-3)/n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,204: Line 4,236:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
-
\frac{4n}{15(n+1) } \biggl(\frac{4\pi}{3}\biggr)^{1/n}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n}}  
\biggl( \frac{3}{2^4} \biggr) \biggl[\chi^{-1}\frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}\biggr]
\biggl[ \mathfrak{L}_i + 4\mathfrak{K}_i \biggr] 
+
\biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl(\frac{3\cdot 5}{2^3}\biggr) \biggl[ \chi^{-1}  
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5}  
\mathfrak{L}_i
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also in the "Case M" discussions, the external pressure is normalized to,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~P_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~[ G^{-3(n+1)} K^{4n} M^{-2(n+1)} ]^{1/(n-3)} \, .</math>
+ \biggl( \frac{\pi}{2^{5}\cdot 3^6} \biggr)
\biggl[\chi^{-1}\biggl(\frac{\nu^2}{q} \biggr) 
\frac{(1+\ell_i^2)^2}{\ell_i^{4}}\biggr]^3\mathfrak{K} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,234: Line 4,261:
</div>
</div>


Or, in terms of the ratio,  
If we raise the "Case 1" stability criterion expression to the <math>~(n-3)</math> power, then divide it by the "Case 3" stability criterion expression raised to the fourth power, we find,
 
<div align="center">
<div align="center">
<math>\Chi \equiv \frac{\chi}{\chi_\mathrm{eq}} \, ,</math>
<table border="0" cellpadding="5" align="center">
</div>
and pulling from the above expressions,


<div align="center">
<table border="0" cellpadding="4">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~\Rightarrow ~~~
[P_e]_\mathrm{crit}^{n-3}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
<math>~
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3+ \tan^{-1}(\ell_i) \biggr] </math>
\biggl[\frac{\mathcal{A} GM^2 (n-3)}{4\pi (n+1)}\biggr]^{n-3}\biggl\{ \biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1}  \biggr\}^{-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,260: Line 4,287:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}
<math>~
\biggl[ \mathfrak{L}_i - \frac{8}{3}\biggr]
\biggl[\frac{\mathcal{A}  (n-3)}{4\pi (n+1)}\biggr]^{n-3} G^{n-3} M^{2(n-3)} \biggl[\frac{3(n+1) \mathcal{B}}{4n\mathcal{A}} \biggr]^{4n} \biggl(\frac{K}{G}\biggr)^{4n} M^{4(1-n)}
</math>
</math>
   </td>
   </td>
Line 4,271: Line 4,298:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
<math>~
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\}  
\biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n} M^{-2(n+1)} G^{-3(n+1)}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~
\biggl[\frac{P_e}{P_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -\biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}
<math>~
\biggl[4\mathfrak{K}_i + \frac{8}{3} \biggr]
\biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n}  
</math>
</math>
   </td>
   </td>
Line 4,298: Line 4,328:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}   + \tan^{-1}(\ell_i) \biggr] </math>
<math>~
\biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \biggl[ \frac{ 5\tilde{\mathfrak{f}}_M^2 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)}  
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[ \frac{ \tilde{\mathfrak{f}}_A }{ \tilde{\mathfrak{f}}_M^{(n+1)/n} } \biggr]^{4n}
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,313: Line 4,346:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}\mathfrak{L}_i  </math>
<math>~
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n}  
\biggl[ \frac{ 5 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)} \tilde{\mathfrak{f}}_M^{2(n+1)} \tilde{\mathfrak{f}}_A^{4n}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===Case P===
Flipping around this expression for <math>~[P_e]_\mathrm{crit}</math>, we also can write,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
<math>~
[M]_\mathrm{crit}^{2(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^
<math>~
\biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} </math>
\biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n}   G^{-3(n+1)} P_e^{3-n} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Now, in our "Case P" discussions we normalized the mass to
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~M_\mathrm{SWS}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3(4\mathfrak{K}_i) \, ,
<math>~
\biggl(\frac{n+1}{n}\biggr)^{3/2} G^{-3/2} K^{2n/(n+1)} P_e^{(3-n)/[2(n+1)]} \, .
</math>
</math>
   </td>
   </td>
Line 4,347: Line 4,401:
</table>
</table>
</div>
</div>
Hence, we have,


we have the streamlined,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 4,354: Line 4,408:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{2^5\pi}{3^6} \biggr)^{1/2} \biggl[ \frac{(1+\ell_i^2)}{\ell_i} \biggr]^{3}  \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,360: Line 4,415:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
+\Chi^{-3/5} (5 \mathfrak{L}_i)
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \biggl(\frac{n+1}{n}\biggr)^{-3(n+1)}  
+\Chi^{-3} (4\mathfrak{K}_i)
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
or, better yet,
<div align="center" id="BiPolytropeFreeEnergy">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with</font> <math>~(n_c, n_e) = (5, 1)</math>
</th>
</tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2^4\biggl( \frac{q\ell_i^2}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,391: Line 4,429:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\Chi^{-3/5} (5 \mathfrak{L}_i)
\biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \, ,
+\Chi^{-3} (4\mathfrak{K}_i)
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</table>
</div>
</div>


 
where the constants,  
where,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{L}_i</math>
<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp; &nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>
<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So we can furthermore conclude that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{K}_i</math>
<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\Lambda_i}{\eta_i} + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>
<math>~
\biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \biggl\{ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr\}^{-3(n+1)}  
\biggl\{ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggr\}^{4n}
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,435: Line 4,477:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Lambda_i</math>
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\eta_i</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,453: Line 4,483:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
<math>~\biggl(\frac{3}{4\pi} \biggr)^{4}
\biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n} \biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]^{(n+1)} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,459: Line 4,491:
</div>
</div>


From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
 
Our expression for <math>~[M]_\mathrm{crit}^{2(n+1)}</math> can also be combined with the "Case 2 stability criterion" to eliminate the mass entirely, giving,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 4,465: Line 4,498:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{q}</math>
<math>~
\biggl\{ 16\pi nP_e R_\mathrm{eq}^{3(n+1)/n}  \biggr\}^{2n}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,471: Line 4,506:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\eta_s}{\eta_i}
<math>~
= 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>
\biggl\{ [3(n-3)]\mathcal{B}K\biggr\}^{2n}  
\biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n}  G^{-3(n+1)} P_e^{3-n}
</math>
   </td>
   </td>
</tr>
</tr>
Line 4,478: Line 4,515:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\nu</math>
<math>~\Rightarrow~~~
R_\mathrm{eq}^{6(n+1)}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,485: Line 4,524:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, .
\biggl[ \frac{ 3(n-3)}{16\pi n} \biggr]^{2n}
\biggl[\frac{  (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{6n} K^{6n}  G^{-3(n+1)} P_e^{3(1-n)}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">Radial Derivatives</td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>
<math>~\Rightarrow~~~
R_\mathrm{eq}^{2(n+1)}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,508: Line 4,541:
   <td align="left">
   <td align="left">
<math>~
<math>~
-\Chi^{-8/5} (3 \mathfrak{L}_i)
\biggl\{ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{2n} \biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl[\frac{(n+1) }{n} \biggr]^{4n+(3-n)} \biggl(\frac{3 }{4} \biggr)^{6n} \biggr\}^{1/3}
-\Chi^{-4} (12\mathfrak{K}_i)
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n}  G^{-(n+1)} P_e^{(1-n)
+\Chi^{-2} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
</math>
   </td>
   </td>
Line 4,517: Line 4,549:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\partial^2 \mathfrak{G}^*}{\partial \Chi^2}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,523: Line 4,555:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3}{5}\biggl[  
<math>~
\Chi^{-13/5} (8\mathfrak{L}_i)
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl[\frac{(n+1) }{n} \biggr]^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n}
+\Chi^{-5} (80\mathfrak{K}_i)
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n}   G^{-(n+1)} P_e^{(1-n)}  \, .
-\Chi^{-1} (10\mathfrak{L}_i  +40\mathfrak{K}_i )\biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</table>
</div>
</div>


Consistent with our [[User:Tohline/SSC/BipolytropeGeneralization#Free_Energy_and_Its_Derivatives|generic discussion of the stability of bipolytropes]] and the ''specific'' discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|the stability of bipolytropes having]] <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,
Finally, recognizing that in our "Case P" discussions we normalized the radius to
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi = \chi_\mathrm{eq}</math>
<math>~R_\mathrm{SWS}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <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>
<math>~
\biggl(\frac{n+1}{n}\biggr)^{1/2} G^{-1/2} K^{n/(n+1)} P_e^{(1-n)/[2(n+1)]} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
we have,
Furthermore, the equilibrium configuration is unstable whenever <math>~\partial^2 \mathfrak{G}/\partial \chi^2 < 0</math>, that is, it is unstable whenever,
 
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<tr>
   <td align="right">
   <td align="right">
<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>
<math>~
[R_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~></math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~20 \, .</math>
<math>~
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{n+1 }{n} \biggr)^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n}
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n} \biggl\{ R_\mathrm{SWS}\biggl(\frac{n+1 }{n} \biggr)^{-1/2} \biggr\}^{2(n+1)} 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
[[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|Table 1 of an accompanying chapter]] &#8212; and the red-dashed curve in the figure adjacent to that table &#8212; identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.
==Focus on Zero-Zero Free-Energy Expression==
Here, we will draw heavily from the following accompanying chapters:
*  [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Step_7:__Surface_Boundary_Condition|Analytic Detailed Force Balance Models]]
*  [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Free_Energy_of_BiPolytrope_with|Free-Energy Analysis]]
===From Detailed Force-Balance Models===
====Equilibrium Radius====
=====First View=====
In an [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Virial_Theorem|accompanying chapter]] we find,
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<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~\Rightarrow~~~
\frac{P_0 R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>
\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
<math>~
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
\biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{3 }{4} \biggr)^{2n}
\mathcal{A}^{-(n+1)} \mathcal{B}^{2n}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


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


Here, we prefer to normalize the equilibrium radius to <math>~R_\mathrm{norm}</math>.  So, let's replace the central pressure with its expression in terms of <math>~K_c</math>.  Specifically,
==Case M Free-Energy Surface==
 
It is useful to rewrite the free-energy function in terms of dimensionless parameters.  Here we need to pick normalizations for energy, radius, and pressure that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~M</math>.  We have chosen to use,
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<math>~P_0</math>
<math>~x</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{R}{R_0}</math>
K_c \rho_c^{\gamma_c} = K_c \biggl[ \frac{3M_\mathrm{core}}{4\pi R_i^3} \biggr]^{\gamma_c}
= K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c}
~~~\Rightarrow~~~ \frac{P_0}{P_\mathrm{norm}} = \biggl[ \frac{3}{4\pi}\biggl(\frac{\nu}{q^3}\biggr) \frac{1}{\chi_\mathrm{eq}^3}\biggr]^{(n_c+1)/n_c}  
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
-->
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c}
<math>~R_\mathrm{norm}</math>
\frac{R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} \, ,</math>
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~R_\mathrm{eq}^{(n_c-3)/n_c}
<math>~P_\mathrm{norm}</math>
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} \, ,</math>
\biggl(\frac{G}{K_c}\biggr) M_\mathrm{tot}^{(n_c-1)/n_c} \biggl[ \frac{3\nu }{4\pi q^3 } \biggr]^{-(n_c+1)/n_c}
\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
</table>
   <td align="right">
</div>
<math>~\Rightarrow~~~\chi_\mathrm{eq}^{(n_c-3)/n_c} \equiv \biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^{(n_c-3)/n_c}</math>
 
which, as is detailed in an [[User:Tohline/SphericallySymmetricConfigurations/Virial#Choices_Made_by_Other_Researchers|accompanying discussion]], are similar but not identical to the normalizations used by [http://adsabs.harvard.edu/abs/1970MNRAS.151...81H Horedt (1970)] and by [http://adsabs.harvard.edu/abs/1981MNRAS.195..967W Whitworth (1981)].  The self-consistent energy normalization is,
 
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
   <td align="right">
<math>~E_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~P_\mathrm{norm} R^3_\mathrm{norm} \, .</math>
\frac{1}{2}\biggl(\frac{4\pi}{3} \biggr)^{1/n_c}
\biggl( \frac{\nu}{q^3}\biggr)^{(n_c-1)/n_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>


Or, in terms of <math>~\gamma_c</math>,
As we have [[User:Tohline/SphericallySymmetricConfigurations/Virial#Gathering_it_all_Together|demonstrated elsewhere]], after implementing these normalizations, the expression that describes the "Case M" free-energy surface is,
<div align="center">
<math>
\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} =
-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>
</div>
 
Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case M" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.
 
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   <td align="right">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>
<math>~\frac{P_e}{P_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\frac{1}{20\pi x_\mathrm{eq}^4} \biggl[ 15\biggl(\frac{3}{4\pi}\biggr)^{1/n} x_\mathrm{eq}^{(n-3)/n} - 3\biggr]</math>
\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
=====Second View=====
Alternatively, from our derivation and discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#CentralPressure|analytic detailed force-balance models]],


<div align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~[x_\mathrm{eq}]_\mathrm{crit}</math>
\biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr] P_0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; <math>~=</math>&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} \, ,</math>
<math>~\biggl[ \frac{4n}{15(n+1)} \biggl( \frac{4\pi}{3} \biggr)^{1/n} \biggr]^{n/(n-3)}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


where,
<tr>
<div align="center">
  <td align="right">
<table border="0" cellpadding="5" align="center">
<math>~[x_\mathrm{eq}]_\mathrm{turn}</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} \biggr]^{n/(n-3)}</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~[g(\nu,q)]^2</math>
<math>~\frac{P_\mathrm{max}}{P_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \biggl[ \frac{15(n+1)}{4n}\biggl(\frac{3}{4\pi}\biggr)^{1/n} \biggr]^{4n/(n-3)}</math>
1  + \biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
-->
==Case P Free-Energy Surface==
Again, it is useful to rewrite the free-energy function in terms of dimensionless parameters.  But here we need to pick normalizations for energy, radius, and mass that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~P_e</math>.  As is [[User:Tohline/SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|detailed in an accompanying discussion]], we have chosen to use the normalizations defined by [http://adsabs.harvard.edu/abs/1983ApJ...268..165S Stahler (1983)], namely,


In order to show that this expression is the same as the other one, [[#First_View_2|above]], we need to show that,
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<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="left">
   <td align="right">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
<math>~R_\mathrm{SWS}</math>
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} </math>
<math>~\biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, ,</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~
<math>~M_\mathrm{SWS}</math>
f - 1-\mathfrak{F}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{5}{2q^3} \biggl[g^2-1\biggr]</math>
<math>~\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}  \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
The self-consistent energy normalization is,
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~E_\mathrm{SWS} \equiv \biggl( \frac{n}{n+1} \biggr) \frac{GM_\mathrm{SWS}^2}{R_\mathrm{SWS}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,831: Line 4,856:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{5}{2q^3}  \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
<math>~
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>
\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
After implementing these normalizations &#8212; see our [[User:Tohline/SSC/Virial/PolytropesEmbeddedOutline#Our_Case_M_Analysis|accompanying analysis]] for details &#8212; the expression that describes the "Case P" free-energy surface is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}_{K,P_e}^* \equiv \frac{\mathfrak{G}_{K,P_e}}{E_\mathrm{SWS}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<!-- HIDE LONG RE-DERIVATION ...
   <td align="left">
   <td align="left">
<math>~\frac{5}{2q^5} \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl\{ 2 ( q^2 - q^3 )  
<math>~
+ \frac{\rho_e}{\rho_0}\biggl[ 1 - 3q^2+ 2q^3 \biggr] \biggr\} \, .</math>
\biggl\{\biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \biggr\}^{-1}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Let's see &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
&nbsp;
f - 1-\mathfrak{F}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c}
\times \biggl\{- \biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}  
\biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] -
+ \biggl[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}} \biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[  (-2q^2 + 3q^3 - q^5) +
+ \frac{4\pi}{3} \cdot P_e R^3  
\frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,883: Line 4,907:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~-
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr)  
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}
- \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) \biggr]
</math>
</math>
   </td>
   </td>
Line 4,899: Line 4,922:
   <td align="left">
   <td align="left">
<math>~
<math>~
- \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \frac{GM^2}{R}  
+\biggl( \frac{\rho_e}{\rho_c}  \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]
- \biggl[\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \frac{nKM^{(n+1)/n}}{R^{3/n}}
- \frac{4\pi}{3} \cdot P_e R^3
\biggr\}
</math>
</math>
   </td>
   </td>
Line 4,913: Line 4,938:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~-
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl\{ (q^3- q^5 )  
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
+ (2q^2 - 3q^3 + q^5) \biggr\}
</math>
</math>
   </td>
   </td>
Line 4,929: Line 4,953:
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 3 (1 -5q^2 + 5q^3 - q^5) \biggr]
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2 \frac{R_\mathrm{SWS}}{R} \biggl[G M_\mathrm{SWS}^2 R_\mathrm{SWS}^{-1}  \biggr]
+\frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 2 - 2q^5 + 5\biggl( q^5-q^3\biggr)\biggr]  
</math>
</math>
   </td>
   </td>
Line 4,940: Line 4,963:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3- q^5 ) + (2q^2 - 3q^3 + q^5) \biggr]
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr] \biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
+ \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 3 (1 -5q^2 + 5q^3 - q^5)+2 - 2q^5 + 5( q^5-q^3) \biggr]
\biggl[K M_\mathrm{SWS}^{(n+1)/n} R_\mathrm{SWS}^{-3/n} \biggr]
</math>
</math>
   </td>
   </td>
Line 4,955: Line 4,978:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ 2q^2 - 2q^3 \biggr]
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ P_e R_\mathrm{SWS}^3 \biggr]
+ \frac{5}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[  1 - 3q^2 + 2q^3  \biggr] \, .
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Q.E.D.
Hence, the equilibrium radius can also be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 4,980: Line 4,996:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~-
\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c} q^2 g^2 \, ;
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
or, in terms of the polytropic index,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}^{n_c-3} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \, .
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) G \biggl[\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]}  \biggr]^2 \biggl[\biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \biggr]^{-1}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
====Gravitational Potential Energy====
Also from our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Gravitational_Potential_Energy|accompanying discussion]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}  </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>~
- \Chi^{-1} \biggl( \frac{3}{5}\biggr) \biggl(\frac{\nu}{q^3} \biggr)^2 q^5
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}} \biggr]
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{-1/(n_c-3)f(\nu,q)
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
K \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \biggr]^{(n+1)/n} \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \biggr]^{-3/n}   
</math>
</math>
   </td>
   </td>
Line 5,035: Line 5,038:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ 2^{n_c-(n_c-3)} \biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} b_\xi^{n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~=</math>
<math>~
  </td>
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 P_e \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \biggr]^3
<td align="left">
\biggr\}
<math>
- \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ \biggl(\frac{6}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, .
</math>
</math>
   </td>
   </td>
Line 5,062: Line 5,050:
</div>
</div>


====Internal Energy Components====


=====First View=====
Before writing out the expressions for the internal energy of the core and of the envelope, we [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Virial_Theorem|note from our separate detailed derivation]] that, in either case,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 5,072: Line 5,056:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\frac{P_i \chi^{3\gamma}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma}</math>
<math>~\mathfrak{G}_{K,P_e}^*</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,078: Line 5,062:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\biggl(\frac{P_i }{P_0} \biggr) \biggl(\frac{P_0 }{P_\mathrm{norm}} \biggr)\chi^{3}\biggr]_\mathrm{eq} \biggl[\frac{\chi}{\chi_\mathrm{eq}}\biggr]^{3-3\gamma}</math>
<math>~-
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,087: Line 5,073:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma} \, ,</math>
<math>~
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) G \biggl[\biggl( \frac{n+1}{nG} \biggr)^{3} K^{4n/(n+1)} P_\mathrm{e}^{(3-n)/[(n+1)]}  \biggr]
\biggl[\biggl( \frac{n+1}{nG} \biggr)^{-1/2} K^{-n/(n+1)} P_\mathrm{e}^{(n-1)/[2(n+1)]} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where, in equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl(\frac{P_i }{P_0} \biggr)_\mathrm{eq}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1 - b_\xi q^2</math>
<math>~
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr]
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
K \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3(n+1)/(2n)} K^{2} P_\mathrm{e}^{(3-n)/(2n)}  \biggr]
\biggl[ \biggl( \frac{n+1}{nG} \biggr)^{-3/2n} K^{-3/(n+1)} P_\mathrm{e}^{3(n-1)/[2n(n+1)]}  \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,114: Line 5,103:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b_\xi q^2</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{\frac{2}{5}q^3 f + \biggl[1 - \frac{2}{5} q^3( 1+\mathfrak{F} ) \biggr]\biggr\}^{-1} </math>
<math>~
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 P_e \biggl[ \biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{3n/(n+1)} P_\mathrm{e}^{3(1-n)/[2(n+1)]} \biggr]
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,132: Line 5,124:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr]^{-1} </math>
<math>~-
\biggl( \frac{n}{n+1} \biggr)^{3/2} G^{3/2}K^{-3n/(n+1)} P_\mathrm{e}^{(n-5)/[2(n+1)]}  
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, copying from our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#InternalEnergies|accompanying detailed derivation]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{4\pi/3 }{({\gamma_c}-1)} 
\times \biggl\{\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_c}
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr) \biggl[\biggl( \frac{n+1}{n} \biggr)^{5/2} G^{-3/2} K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]\biggr]
\biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\}
</math>
</math>
   </td>
   </td>
Line 5,163: Line 5,150:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{1 }{({\gamma_c}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
- \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}} \biggr]
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_c}  
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
\biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2} K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]}  \biggr]
</math>
</math>
   </td>
   </td>
Line 5,176: Line 5,163:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{4\pi/3 }{({\gamma_e}-1)}
- \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \biggl[ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \biggr]
\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_e}
\biggr\}
\biggl\{ (1-q^3) + b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr] \biggr\}  
</math>
</math>
   </td>
   </td>
Line 5,199: Line 5,184:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~-
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr] \biggl( \frac{n+1}{n} \biggr) \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr]
\biggl(\frac{R_\mathrm{SWS}}{R}\biggr)
\Chi^{3-3\gamma_e} \biggl(\frac{P_i }{P_0} \biggr)
\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr]  \biggr\}
</math>
</math>
   </td>
   </td>
Line 5,214: Line 5,196:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
+ \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}} \biggr]
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R_\mathrm{SWS}}{R}\biggr)^{3/n}
\Chi^{3-3\gamma_e}  
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3
\biggl\{ (1-b_\xi q^2)(1-q^3) +  b_\xi  \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr]  \biggr\}
</math>
</math>
   </td>
   </td>
Line 5,234: Line 5,215:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~-
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl[\frac{3}{5}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggl( \frac{n+1}{n} \biggr)\biggr] \biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr
\biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1}
\Chi^{3-3\gamma_e} (1-q^3)
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Furthermore,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
&nbsp;
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{3}{4\pi}\biggr)^{\gamma_c - 1}
+ \biggl[n\biggl(\frac{3}{4\pi}\biggr)^{1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{\mathfrak{f}}}_M^{(n+1)/n}}  \biggr]
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}
\biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  
\biggl\{\chi_\mathrm{eq}^{4-3\gamma_c}\biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3
</math>
</math>
   </td>
   </td>
Line 5,274: Line 5,245:
<math>~=</math>
<math>~=</math>
   </td>
   </td>
END SUPPRESSION OF LONG DERIVATION -->
   <td align="left">
   <td align="left">
<math>~
<math>~- 3 \mathcal{A} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2 \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1}
\biggl(\frac{3}{4\pi}\biggr)^{\gamma_c - 1}
+ n\mathcal{B} \biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n}  
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}
+ \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \, .
\biggl\{\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case P" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.
==Summary==
<div align="center">
<table border="1" align="center" cellpadding="5">
<tr>
  <th align="center">&nbsp;</th>
  <th align="center">DFB Equilibrium</th>
  <th align="center">Onset of Dynamical Instability</th>
</tr>
<tr>
  <th align="center" rowspan="2"><font size="+1">Case M:</font></th>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,294: Line 5,281:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{3}{4\pi}\biggr)^{(\gamma_c - 1)/(4-3\gamma_c)}
\biggl[ \frac{4\pi}{(n+1)^n} \biggr] {\tilde\xi}^{(n-3)} (-{\tilde\xi}^2 \tilde{\theta^'})^{(1-n)}
\biggl( \frac{\nu}{q^3} \biggr)^{(6-5\gamma_c)(4-3\gamma_c)}
\biggl\{\frac{q^2}{2}
\biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,311: Line 5,300:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{3}{4\pi}\biggr)^{1/(n_c-3)}
\biggl[ \frac{4n}{15(n+1) }\biggr]^\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{\nu}{q^3} \biggr)^{(n_c-5)(n_c-3)}
\biggl\{\frac{q^2}{2}
\biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{-3/(n_c-3)}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


  </td>
</tr>
<tr>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,328: Line 5,321:
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}  
\biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)} {\tilde\theta}^{(n+1)(n-3)} (-{\tilde\xi}^2 \tilde{\theta^'})^{2(n+1)}
\, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


Hence, we have,
  </td>
<div align="center">
  <td align="center">
 
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,348: Line 5,339:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
n_c  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)}  
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{-3/n_c}  
\biggl[ \frac{ 5^3 \tilde{\mathfrak{f}}_M^2}{ \tilde{\mathfrak{f}}_W^3 } \biggr]^{n+1} \biggl( \frac{3\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
  </td>
</tr>
<tr>
  <th align="center" rowspan="2"><font size="+1">Case P:</font></th>
  <td align="center">


<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,364: Line 5,362:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
n_c 
\biggl(\frac{n}{4\pi}\biggr) {\tilde\xi}^2 {\tilde\theta}^{n-1}
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}  
\Chi^{-3/n_c}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


  </td>
  <td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]_\mathrm{crit}^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,381: Line 5,381:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
n_e
\biggl[ \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr]
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}
\biggl[ \frac{n-3}{n} \biggr]^{(n-1)/(n+1)}
\Chi^{-3/n_e} (1-q^3)
\biggl[\frac{ \tilde{\mathfrak{f}}_A}{4} \biggr]^{2n/(n+1)}    
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2   \biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


  </td>
</tr>
<tr>
  <td align="center">


=====Second View=====
In our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#PiDefinition|accompanying discussion of energies associated with detailed force balance models]], we used the notation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Pi</math>
<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]^{2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
\biggl(\frac{3}{2^3\pi}\biggr) \frac{GM_\mathrm{tot}^2}{R^4} \biggl(\frac{\nu}{q^3}\biggr)^2
\biggl( \frac{n^3}{4\pi}\biggr) {\tilde\theta}^{n-3} (-{\tilde\xi}^2 {\tilde{\theta^'}})^2
= P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2 \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
 
which allows us to rewrite the [[#Second_View|above quoted relationship]] between the central pressure and the radius of the bipolytrope as,
  </td>
<div align="center">
  <td align="center">
<math>~P_0 = \Pi (qg)^2 \, .</math>
 
</div>
We [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Virial_Equilibrium|also showed]] that, in equilibrium, the relationship between the central pressure and the interface pressure is,
<div align="center">
<math>~P_0 =P_i + \Pi_\mathrm{eq} q^2 \, .</math>
</div>
This means that, in equilibrium, the ratio of the interface pressure to the central pressure is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>
<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,435: Line 5,423:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0}
<math>~\biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]\biggl(\frac{3}{4\pi} \biggr)^{4/(n+1)}
= 1- \frac{1}{g^2} \, ,
\biggl[\frac{  (n-3)}{4\pi n}\biggr]^{(n-3)/(n+1)} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n/(n+1)}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
or given that (see [[#Second_View|above]]),


<div align="center">
  </td>
</tr>
<tr>
  <td align="left" colspan="3">
In all four cases, the expression on right intersects (is equal to) the expression on the left when the following condition applies:
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
  <td align="left">
For <math>~n \ne 5</math>: &nbsp; &nbsp; &nbsp;
  </td>
   <td align="right">
   <td align="right">
<math>~
<math>~
\frac{5}{2q^3} \biggl[g^2-1\biggr]
2(9-2n){\tilde\theta}^{n+1}
</math>
</math>
   </td>
   </td>
Line 5,458: Line 5,451:
   <td align="left">
   <td align="left">
<math>~
<math>~
f - 1-\mathfrak{F}  
3(n-3) \biggl[ (-{\tilde\theta}^')^2 - \biggl( -\frac{\tilde\theta {\tilde\theta}^'}{\tilde\xi} \biggr)\biggr] \, ;
</math>
</math>
   </td>
   </td>
Line 5,464: Line 5,457:


<tr>
<tr>
  <td align="left">
For <math>~n = 5</math>: &nbsp; &nbsp; &nbsp;
  </td>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~~
<math>~
g^2
\biggl[\frac{2^4\cdot 5}{3}\biggr] \ell^3
</math>
</math>
   </td>
   </td>
Line 5,473: Line 5,469:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} ) \, ,
<math>~
(1+\ell^2)^3 \tan^{-1}\ell + \ell(\ell^4-1) \, .
</math>
</math>
  </td>
</tr>
</table>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
we have,
 
 
If (for <math>n\ne 5</math>) we adopt the shorthand notation,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 5,485: Line 5,488:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>
<math>~\Upsilon</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0}
<math>~[3 (-{\tilde\theta}^')^2 - {\tilde\mathfrak{f}}_M \tilde\theta] =
= 1- \biggl[ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} ) \biggr]^{-1} \, .
3\biggl[ (-{\tilde\theta}^')^2 - \biggl( -\frac{\tilde\theta {\tilde\theta}^'}{\tilde\xi} \biggr)\biggr] \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
This is exactly the pressure-ratio expression presented in our "first view" and unveils the notation association,
  <td align="center" colspan="3">
<div align="center">
and
<table border="0">
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~b_\xi q^2</math>
<math>~\tau</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\leftrightarrow~</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~{\tilde\theta}^{n+1} \, ,
\frac{1}{g^2} \, .
</math>
</math>
   </td>
   </td>
Line 5,518: Line 5,521:
</div>
</div>


From [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Thermal_Energy_Content|our separate derivation]], we have, in equilibrium,
then the critical condition becomes,
<div align="center">
<div align="center">
<table border="0">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}_\mathrm{core} = \biggl(\frac{2n_c}{3}\biggr) S_\mathrm{core}</math>
<math>~(n-3)\Upsilon</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\biggl(\frac{2n_c}{3}\biggr) \biggl( \frac{4\pi}{5} \biggr) R_\mathrm{eq}^3 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr)_\mathrm{eq} </math>
<math>~2(9-2n)\tau \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and ''at'' the critical state, the expressions for the structural form-factors become,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~{\tilde\mathfrak{f}}_A</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\biggl( \frac{ q^5n_c}{5} \biggr) R_\mathrm{eq}^3  \biggl( \frac{2^3\pi}{3} \biggr) \Pi_\mathrm{eq} \biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>
<math>~\frac{1}{(5-n)} \biggl[6\tau + (n+1)\Upsilon  \biggr] </math>
   </td>
   </td>
</tr>
</tr>
Line 5,550: Line 5,560:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\biggl( \frac{ n_c}{5} \biggr) \biggl[ R_\mathrm{norm}^3 
<math>~\frac{1}{(5-n)} \biggl\{ 6 + (n+1)\biggl[ \frac{2(9-2n)}{n-3} \biggr\biggr\}\tau </math>
P_\mathrm{norm} \biggr] \chi_\mathrm{eq}^{-1} \biggl(\frac{\nu^2}{q}\biggr)
\biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{P_0} \biggr)_\mathrm{eq}\biggl( \frac{P_0}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>
   </td>
   </td>
</tr>
</tr>
Line 5,561: Line 5,569:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{ n_c}{5} \biggr) \biggl(\frac{\nu^2}{q}\biggr) 
<math>~\frac{1}{(5-n)} \biggl[ \frac{6(n-3) + 2(9-2n)(n+1)}{n-3} \biggr] \tau </math>
\biggl[\frac{5}{2q^2} \biggl( 1-\frac{1}{g^2} \biggr)\biggl( q^2g^2\biggr) + 1 \biggr] \chi_\mathrm{eq}^{-1} </math>
   </td>
   </td>
</tr>
</tr>
Line 5,577: Line 5,584:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{ n_c}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr) 
<math>~\frac{1}{(5-n)} \biggl[ \frac{4n(5-n)}{n-3} \biggr\tau </math>
\biggl[ g^2-\frac{3}{5}  \biggr]
\biggl\{\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c}
\biggr\}^{-1/(n_c-3)}</math>
   </td>
   </td>
</tr>
</tr>
Line 5,593: Line 5,596:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2}  \biggr]
<math>~\frac{4n\tau}{(n-3)} \, ;</math>
\biggl( \frac{ 1}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr)  g^2
\biggl\{2^{n_c}\biggl(\frac{3}{4\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} (q g)^{-2n_c}
\biggr\}^{1/(n_c-3)}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~{\tilde\mathfrak{f}}_W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2}  \biggr]
<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl[\tau + \Upsilon  \biggr]</math>
\biggl\{2^{n_c}\cdot 2^{(3-n_c)}\biggl(\frac{3}{4\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} \biggl(\frac{\nu}{q^3}\biggr)^{2(n_c-3)} q^{5(n_c-3)} q^{-2n_c} g^{-2n_c} g^{2(n_c-3)}
\biggr\}^{1/(n_c-3)}</math>
   </td>
   </td>
</tr>
</tr>
Line 5,624: Line 5,625:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2} \biggr]
<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl\{1 + \biggl[ \frac{2(9-2n)}{n-3} \biggr] \biggr\}\tau</math>
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15} g^{-6}
\biggr\}^{1/(n_c-3)} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,


<div align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2 \biggr]
<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl[ \frac{3(5-n)}{n-3} \biggr\tau</math>
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15}  b_\xi^3 q^{6}  
\biggr\}^{1/(n_c-3)} </math>
   </td>
   </td>
</tr>
</tr>
Line 5,659: Line 5,649:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~n_c q^3 \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2 \biggr]
<math>~\frac{3^2\cdot 5 \tau}{(n-3) {\tilde\xi}^2} </math>
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^3 \biggr\}^{1/(n_c-3)} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression.  Also from our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Thermal_Energy_Content|previous derivation]], we can write,
<div align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}_\mathrm{env} = \biggl(\frac{2n_e}{3}\biggr) S_\mathrm{env}</math>
<math>~\Rightarrow~~~
\frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr)
<math>~
R_\mathrm{eq}^3 \Pi_\mathrm{eq} \biggl\{ (1-q^3) \biggl(\frac{P_i  }{\Pi}\biggr)_\mathrm{eq} 
\biggl[\frac{(n-3) {\tilde\xi}^2}{3^2\tau} \biggr]^{3} \biggl(-\frac{3 {\tilde\theta}^'}{\tilde\xi} \biggr)^2
+ \biggl( \frac{\rho_e}{\rho_0} \biggr)\biggl[ (-2q^2 + 3q^3 - q^5 )
= 3^2\biggl[\frac{(n-3) {\tilde\xi}^2}{3^2\tau} \biggr]^{3} \biggl(-\frac{{\tilde\xi}^2 {\tilde\theta}^'}{ {\tilde\xi}^3 } \biggr)^2
  + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_0} \biggr) ( -1 + 5q^2 -5q^3 +  q^5 )\biggr]\biggr\}
</math>
</math>
   </td>
   </td>
Line 5,693: Line 5,678:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr)
<math>~
R_\mathrm{eq}^3 \biggl[  P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2\biggr]_\mathrm{eq}
\biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr] (-{\tilde\xi}^2 {\tilde\theta}^' )^2 \, .
\biggl\{ (1-q^3) q^2(g^2-1)  + \biggl(\frac{2}{5}\biggr) q^5 \mathfrak{F} \biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Hence (1),
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \biggl[ P_\mathrm{norm} R_\mathrm{norm}^3 \biggr] \frac{n_e}{2}
<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr]
\biggl(\frac{\nu^2}{q^4}\biggr)(1-q^3)
\biggl[ \frac{4n}{15}\biggr]^n  \biggl(\frac{1}{3}\biggr)  
\biggl\{ (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}  
\biggl[ \frac{3^2\cdot 5 }{4n {\tilde\xi}^2}   \biggr]^n
\chi^{-1}_\mathrm{eq}  
\tilde{\mathfrak{f}}_M^{1-n}
</math>
</math>
   </td>
   </td>
Line 5,721: Line 5,710:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e (1-q^3)
<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr
\biggl\{ (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \frac{q^2}{2}\biggl(\frac{\nu}{q^3}\biggr)^2
\biggl[ \frac{1}{ {\tilde\xi}^{2n}}   \biggr]
\biggl[\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\biggl( \frac{-{\tilde\xi}^2{\tilde\theta}^'}{{\tilde\xi}^3} \biggr)^{1-n}
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c}\biggr]^{-1/(n_c-3)}  
</math>
</math>
   </td>
   </td>
Line 5,740: Line 5,728:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e (1-q^3)
<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr
\biggl\{ (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}  
{\tilde\xi}^{n-3}
\biggl[2^{[n_c-(n_c-3)]} \biggl(\frac{3}{4\pi} \biggr)
(-{\tilde\xi}^2{\tilde\theta}^')^{1-n}
\biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} q^{2(n_c-3)-2n_c} g^{-2n_c} \biggr]^{1/(n_c-3)}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Q.E.D.


<tr>
 
And (2),
<div align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e (1-q^3)
<math>~
\biggl\{  (g^2-1) + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}  
\biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{  (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)}  
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl[ \frac{ 5^3 \tilde{\mathfrak{f}}_M^2}{ \tilde{\mathfrak{f}}_W^3 } \biggr]^{n+1} \biggl( \frac{3\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n}  
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} g^{-2n_c} \biggr]^{1/(n_c-3)} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And, finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e (1-q^3) (b_\xi q^2)^{-1}
<math>~3^{4} (4\pi)^{-(n+1)}
\biggl\{  1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\}
\biggl(\frac{ n-3}{n}\biggr)^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)}
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr]^{n+1} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} (b_\xi q^2)^{n_c} \biggr]^{1/(n_c-3)}  
\biggl[ \frac{3n\tau}{n-3}  \biggr]^{4n}  
</math>
</math>
   </td>
   </td>
Line 5,794: Line 5,782:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e (1-q^3)  
<math>~3^{4} (4\pi)^{-(n+1)}
\biggl\{ 1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\}  
\biggl(\frac{ n-3}{n}\biggr)^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)} n^{3(n+1)}
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl[ \frac{n-3}{n} \biggr]^{3(n+1)} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6-2(n_c-3)+2n_c} b_\xi^{3-n_c+n_c} \biggr]^{1/(n_c-3)}  
\biggl[ \frac{n}{n-3} \biggr]^{4n} \tau^{4n-3(n+1)} 3^{4n-4(n+1)}
</math>
</math>
   </td>
   </td>
Line 5,810: Line 5,798:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ n_e\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{3} \biggr]^{1/(n_c-3)} (1-q^3)
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{n+1}
\biggl\{ 1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\} \, ,
(-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)}
\tau^{n-3}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression.
====Summary00====
In summary, the desired ''out'' of equilibrium free-energy expression is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
<math>~\Rightarrow ~~~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,836: Line 5,816:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{n+1}
A_0\Chi^{-3/n_c} + B_0\Chi^{-3/n_e} - C_0\Chi^{-1}
(-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} {\tilde\theta}^{(n+1)(n-3)} \, .
</math>
</math>
   </td>
   </td>
Line 5,843: Line 5,823:
</table>
</table>
</div>
</div>
where,
Q.E.D.
 


And (3),
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~A_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{core}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>
<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^{2(n+1)}_\mathrm{crit}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,856: Line 5,837:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{n_c}{b_\xi} 
\biggl[ \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr]^{n+1}
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{1/(n_c-3)}  
\biggl[ \frac{n-3}{n} \biggr]^{(n-1) }  
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
\biggl[\frac{ \tilde{\mathfrak{f}}_A}{4} \biggr]^{2n } 
</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 5,866: Line 5,847:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~B_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{env}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,872: Line 5,853:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{n_e}{b_\xi}
\biggl[ \frac{3^2\cdot 5}{4\pi }\biggr]^{n+1}
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c} \biggr]^{1/(n_c-3)} (1-q^3)
\biggl[ \frac{n-3}{n} \biggr]^{(n-1) }  
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} \, ,
\biggl[ \frac{ n\tau}{n-3} \biggr]^{2n }   \biggl[ \frac{(n-3){\tilde\xi}^2}{3^2\cdot 5 \tau} \biggr]^{n+1}
</math>
</math>
   </td>
   </td>
Line 5,882: Line 5,863:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~C_0 \equiv \biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{eq} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>~
\biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ \frac{1}{4\pi }\biggr]^{n+1}
\biggl[ \biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, .
\biggl[ \frac{ n}{n-3} \biggr]^{n+1 }  \biggl[ (n-3){\tilde\xi}^2\biggr]^{n+1} \tau^{n-1}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Or, in a more compact form,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}^* \equiv \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{-1/(n_c-3)}  
<math>~\Rightarrow ~~~ \biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^{2}_\mathrm{crit}</math>
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,910: Line 5,884:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
n_c A_1\Chi^{-3/n_c} + n_e B_1\Chi^{-3/n_e} - 3C_1\Chi^{-1}
\biggl( \frac{n}{4\pi }\biggr) {\tilde\xi}^2 {\tilde\theta}^{n-1}
</math>
</math>
   </td>
   </td>
Line 5,917: Line 5,891:
</table>
</table>
</div>
</div>
where,
Q.E.D.


And (4),
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~A_1 </math>
<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2(n+1)}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]^{n+1} \biggl(\frac{3}{4\pi} \biggr)^{4}
\frac{1}{b_\xi} (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
\biggl[\frac{  (n-3)}{4\pi n}\biggr]^{(n-3)} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n}
</math>
</math>
   </td>
   </td>
Line 5,938: Line 5,913:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~B_1 </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl\{ \biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr] (-{\tilde\xi}^2 {\tilde\theta}^' )^2 \biggr\}^{n+1}  3^4(4\pi)^{-(n+1)}
\frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} \, ,
\biggl(\frac{ n-3}{n}\biggr)^{(n-3)}  
\biggl[\frac{3n\tau }{n-3} \biggr]^{4n}  
</math>
</math>
   </td>
   </td>
Line 5,952: Line 5,928:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~C_1 </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>~ \biggl[ \frac{n^3}{4\pi}\biggr]^{n+1}(-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)}
\biggl( \frac{2}{5}\biggr) q^5 f  \, .
\tau^{n-3}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Let's examine the behavior of the first radial derivative.
<tr>
<div align="center">
   <td align="right">
<table border="0" cellpadding="5" align="center">
<math>~\Rightarrow~~~
 
\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2}
<tr>
</math>
   <td align="right">
<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,978: Line 5,950:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3}{\Chi} \biggl[ - A_1\Chi^{-3/n_c} - B_1\Chi^{-3/n_e} + C_1\Chi^{-1} \biggr] \, .</math>
<math>~ \biggl[ \frac{n^3}{4\pi}\biggr](-{\tilde\xi}^2 {\tilde\theta}^' )^{2}  
{\tilde\theta}^{n-3}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Q.E.D.


Let's see whether the sum of terms inside the square brackets is zero at the derived equilibrium radius, that is, when <math>~\Chi = 1</math> and, hence, when
=Free-Energy of Bipolytropes=
 
In this case, the Gibbs-like free energy is given by the sum of four separate energies,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 5,990: Line 5,967:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi = \chi_\mathrm{eq}</math>
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 5,996: Line 5,973:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{1/(n_c-3)}  
\biggl[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{core} + \biggl[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{env} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
In addition to specifying (generally) separate polytropic indexes for the core, <math>~n_c</math>, and envelope, <math>~n_e</math>, and an envelope-to-core mean molecular weight ratio, <math>~\mu_e/\mu_c</math>, we will assume that the system is fully defined via specification of the following five physical parameters:
* Total mass, <math>~M_\mathrm{tot}</math>;
* Total radius, <math>~R</math>;
* Interface radius, <math>~R_i</math>, and associated dimensionless interface marker, <math>~q \equiv R_i/R</math>;
* Core mass, <math>~M_c</math>, and associated dimensionless mass fraction, <math>~\nu \equiv M_c/M_\mathrm{tot}</math>;
* Polytropic constant in the core, <math>~K_c</math>.
In general, the warped free-energy surface drapes across a five-dimensional parameter "plane" such that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 6,010: Line 6,001:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~\mathfrak{G}(R, K_c, M_\mathrm{tot}, q, \nu) \, .</math>
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} b_\xi^{-n_c} \biggr]^{1/(n_c-3)}
\, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,019: Line 6,007:
</div>
</div>


==Order of Magnitude Derivation==
Let's begin by providing very rough, approximate expressions for each of these four terms, assuming that <math>~n_c = 5</math> and <math>~n_e = 1</math>. 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 6,024: Line 6,014:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~
<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>
C_1 - A_1 - B_1
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- \mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot} M_c}{(R_i/2)} \biggr]  
\biggl( \frac{2}{5}\biggr) q^5 f
= - 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr] \, ;</math>
- \frac{1}{b_\xi}  (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
- \frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,042: Line 6,027:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- \mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot} M_e}{(R_i+R)/2} \biggr]  
\biggl( \frac{2}{5}\biggr) q^5 f
= - 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ;</math>
- \frac{1}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2   \biggr\}
+ \frac{q^3}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
- \frac{q^3}{b_\xi}  \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,059: Line 6,040:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{core} = U_\mathrm{int}\biggr|_\mathrm{core} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\mathfrak{b}_c \cdot n_cK_c M_c ({\bar\rho}_c)^{1/n_c}
\biggl( \frac{2}{5}\biggr) q^5 f - \frac{1}{b_\xi}  
= 5\mathfrak{b}_c \cdot K_c M_\mathrm{tot}\nu \biggl[ \frac{3M_c}{4\pi R_i^3} \biggr]^{1/5}  
+ \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^2 
+ \frac{q^3}{b_\xi}
- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^
- \frac{q^3}{b_\xi} + \biggl( \frac{3}{5} \biggr) q^5
</math>
</math>
   </td>
   </td>
Line 6,083: Line 6,060:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^2\biggl\{
<math>~\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5}
\biggl( \frac{2}{5}\biggr) q^3 f - \frac{1}{b_\xi q^2}
\, ;</math>
+ \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]  (1-q^3)
+ \biggl( \frac{3}{5} \biggr) q^3
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,094: Line 6,067:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^2\biggl\{
<math>~\mathfrak{b}_e \cdot n_eK_e M_\mathrm{env} ({\bar\rho}_e)^{1/n_e}
\biggl( \frac{2}{5}\biggr) q^3 f - \biggl[ 1+\frac{2}{5} q^3(f-1-\mathfrak{F}) \biggr]
= \mathfrak{b}_e \cdot K_e M_\mathrm{tot}(1-\nu) \biggl[ \frac{3M_\mathrm{env}}{4\pi (R^3-R_i^3)} \biggr]
+ \biggl[ (1-q^3) - \frac{2}{5} q^3 \mathfrak{F} \biggr]
+ \biggl( \frac{3}{5} \biggr) q^3
\biggr\}
</math>
</math>
   </td>
   </td>
Line 6,117: Line 6,087:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^2\biggl\{0\biggr\} \, .
<math>~
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) K_e [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1} \, .
</math>
</math>
   </td>
   </td>
Line 6,123: Line 6,094:
</table>
</table>
</div>
</div>
Q.E.D.
In writing this last expression, it has been necessary to (temporarily) introduce a sixth physical parameter, namely, the polytropic constant that characterizes the envelope material, <math>~K_e</math>. But this constant can be expressed in terms of <math>~K_c</math> via a relation that ensures continuity of pressure across the interface while taking into account the drop in mean molecular weight across the interface, that is,
 
Even slightly better:
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 6,131: Line 6,100:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{q^2}\biggl[ \biggl(\frac{\pi}{2\cdot 3}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(5-n_c)} b_\xi^{-n_c}\biggr]^{1/(n_c-3)}
<math>~K_e ({\bar\rho}_e)^{(n_e+1)/n_e}</math>
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~K_c ({\bar\rho}_c)^{(n_c+1)/n_c}</math>
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
or, better yet,
<div align="center" id="BiPolytropeFreeEnergy">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with ''Structural'' </font> <math>~(n_c, n_e) = (0, 0)</math>
</th>
</tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2\biggl(\frac{q^2}{\nu}\biggr)^2 \chi_\mathrm{eq}
<math>~\Rightarrow ~~~~ K_e \biggl[\biggl( \frac{\mu_e}{\mu_c} \biggr) {\bar\rho}_c\biggr]^{2}</math>
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~K_c ({\bar\rho}_c)^{6/5}</math>
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</div>
where, keeping in mind that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{(b_\xi q^2)}</math>
<math>~\Rightarrow ~~~~ \frac{K_e}{K_c} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>
<math>~\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
we have,
Hence, the fourth energy term may be rewritten in the form,


<div align="center">
<div align="center">
Line 6,201: Line 6,142:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~A_2 \equiv \frac{A_1}{q^2} </math>
<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{q^3}{(b_\xi q^2)} \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]  
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
K_c\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1}
</math>
</math>
   </td>
   </td>
Line 6,221: Line 6,163:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
q^3  \biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr]  - \biggl( \frac{3}{5} \biggr) \biggr\}
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Putting all the terms together gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{2}{5}q^3  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]   \, ,
- 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr]
- 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr]  
+ \mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5}
</math>
</math>
   </td>
   </td>
Line 6,243: Line 6,194:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~B_2 \equiv \frac{B_1}{q^2} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{1}{(b_\xi q^2)} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
+ \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)}  
</math>
</math>
   </td>
   </td>
Line 6,263: Line 6,215:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
(1-q^3)\biggl\{ \frac{1}{(b_\xi q^2)} -1  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\}  
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr]
+ \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{(\nu M_\mathrm{tot})^{2}}{ qR} \biggr]^{3/5}
</math>
</math>
   </td>
   </td>
Line 6,271: Line 6,224:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~~ \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 6,277: Line 6,230:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
(1-q^3)\biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] - 1 + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\}  
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr] \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
+ \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} K_c \biggl[\frac{M_\mathrm{tot}^{2}}{ R} \biggr]^{3/5}\biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
</math>
</math>
   </td>
   </td>
Line 6,291: Line 6,245:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{2}{5} q^3 \biggl\{ (1-q^3) (f - 1-\mathfrak{F} )  + \mathfrak{F} \biggr\}
- 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]
+ \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl[\frac{R_\mathrm{norm}}{ R} \biggr]^{3/5} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{A}_\mathrm{biP}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr)  + \mathfrak{a}_e  \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ,</math>
\frac{2}{5} q^3 \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,313: Line 6,272:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{B}_\mathrm{biP}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl( \frac{3}{2^2\pi} \biggr)^{1/5} \biggl[5\mathfrak{b}_c 
\frac{2}{5} q^3 f - A_2 \, ,
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr] \, .</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
==Equilibrium Radius==
===Order of Magnitude Estimate===
This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~C_2 \equiv \frac{C_1}{q^2} </math>
<math>~\frac{\partial\mathfrak{G}}{\partial R}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
<td align="left">
  <td align="left">
<math>
<math>~
\frac{2}{5} q^3 \, .
+ 2 \mathcal{A}_\mathrm{biP}\biggl[ \frac{GM_\mathrm{tot}^2 }{R^2} \biggr]
- \frac{3}{5} \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{\nu^{2}}{ q} \biggr]^{3/5} M_\mathrm{tot}^{6/5} R^{-8/5}
\, .
</math>
</math>
   </td>
   </td>
Line 6,341: Line 6,309:
</div>
</div>


As before, the equilibrium system is dynamically unstable if <math>~\partial^2 \mathfrak{G}/\partial \Chi^2 < 0</math>.  We have deduced that the system is unstable if,
Hence, because equilibrium radii are identified by setting <math>~\partial\mathfrak{G}/\partial R = 0</math>, we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 6,347: Line 6,315:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>
<math>~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~< </math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl[\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}\biggr]^{5/2} \biggl(\frac{ q} {\nu^{2}}\biggr)^{3/2} \, .
\frac{A_2}{C_2}
= \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, .
</math>
</math>
   </td>
   </td>
Line 6,362: Line 6,328:
</div>
</div>


 
===Reconcile With Known Analytic Expression===
==Overview==
From our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5_1#The_Core_2|earlier derivations]], it appears as though,
 
<div align="center">
===BiPolytrope51===
 
====Key Analytic Expressions====
 
<div align="center" id="FreeEnergy51">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with</font> <math>~(n_c, n_e) = (5, 1)</math>
</th>
</tr>
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <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>
<math>~\chi_\mathrm{eq} \equiv \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 6,388: Line 6,341:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{1}{\ell_i^2} \biggl[
<math>~\biggl( \frac{3^8}{2^5\pi} \biggr)^{-1/2}
\Chi^{-3/5} (5 \mathfrak{L}_i)
\biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}  
+\Chi^{-3} (4\mathfrak{K}_i)
</math>
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i ) \biggr]
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl(\frac{q}{\nu^2} \biggr)^{3/2}
\biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2} \biggl(\frac{\nu^2}{q} \biggr)^{5/2}  
\frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr] \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</td></tr>
</table>
</table>
</div>
</div>
where,


This implies that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 6,406: Line 6,370:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{L}_i</math>
<math>~\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>
<math>~
\biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2}  \biggl(\frac{\nu^2}{q} \biggr)^{5/2}  
\frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr]^{2/5}
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,418: Line 6,385:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{K}_i</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\Lambda_i}{\eta_i}  + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>
<math>~\biggl(\frac{\nu^2}{q} \biggr)
\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/5}   \frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2}
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,430: Line 6,399:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Lambda_i</math>
<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr)  + \mathfrak{a}_e  \biggl(\frac{1-\nu}{1+q}\biggr)  \biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>
<math>~\frac{1}{2^2\cdot 5}\biggl(\frac{\nu^2}{q} \biggr)
\frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c 
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,442: Line 6,414:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\eta_i</math>
<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c  + \mathfrak{a}_e \cdot  \frac{q(1-\nu)}{\nu(1+q)}  \biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
<math>~\frac{\nu}{2^2\cdot 5}
\frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c 
+ \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr]  
</math>
   </td>
   </td>
</tr>
</tr>
Line 6,454: Line 6,429:
</div>
</div>


From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
==Focus on Five-One Free-Energy Expression==
 
===Approximate Expressions===
 
Let's plug this equilibrium radius back into each term of the free-energy expression.
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 6,460: Line 6,439:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{q}</math>
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{core}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\eta_s}{\eta_i}
<math>~- 2\mathfrak{a}_c \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{\nu}{q}\biggr) \biggr] </math>
= 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 6,473: Line 6,451:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\nu</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 6,479: Line 6,457:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~- 2\mathfrak{a}_c \biggl(\frac{\nu}{q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>
\frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Consistent with our [[User:Tohline/SSC/BipolytropeGeneralization#Free_Energy_and_Its_Derivatives|generic discussion of the stability of bipolytropes]] and the ''specific'' discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|the stability of bipolytropes having]] <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}^*_{51}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi = \chi_\mathrm{eq}</math>
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <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>
<math>~- 2\mathfrak{a}_e \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] </math>
   </td>
   </td>
</tr>
</tr>
Line 6,511: Line 6,481:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c} \biggl[\frac{(1+\ell_i^2)^{6/5}}{(3\ell_i^2)} \biggr]^{n_c} \biggr\}^{1/(n_c-3)} \, ,</math>
<math>~- 2\mathfrak{a}_e \biggl(\frac{1-\nu}{1+q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where the last expression has been cast into a form that more clearly highlights overlap with the expression, below, for the equilibrium radius for zero-zero bipolytropes.  Furthermore, the equilibrium configuration is unstable whenever,
<div align="center">
<math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_{51}}{\partial \chi^2}\biggr]_{\Chi=1} < 0 \, ,</math>
</div>
that is, it is unstable whenever,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>
<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_c-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{core} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~></math>
<math>~\approx</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~20 \, .</math>
<math>~\biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2}
K_c (M_\mathrm{tot}\nu)^{6/5} (R_\mathrm{eq}q)^{-3/5} 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} 
\biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5}
\, ;</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{S_\mathrm{env}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_e-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3}{2}\biggr]
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
\biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} K_c M_\mathrm{tot}^{6/5}R_\mathrm{eq}^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3}{2}\biggr]
\mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2}
\biggl[ \frac{q^3}{\nu} \biggr]^{4/5}  \frac{(1-\nu)^2}{(1-q^3)}
\biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5} \, .
</math>
  </td>
</tr>
 
</table>
</div>
</div>
[[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|Table 1 of an accompanying chapter]] &#8212; and the red-dashed curve in the figure adjacent to that table &#8212; identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.


====Behavior of Equilibrium Sequence====
===From Detailed Force-Balance Models===
 
In the following derivations, we will use the expression,
<div align="center">
<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{\mu_e}{\mu_c} \biggr)^3 \biggl( \frac{\pi}{2^3} \biggr)^{1/2} \frac{1}{A^2\eta_s}
= \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>
</div>
Keep in mind, as well &#8212; as derived in an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Background|accompanying discussion]] &#8212; that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (m_3^2 \ell_i^3) (1 + \ell_i^2)^{-1/2} [1 + (1-m_3)^2 \ell_i^2]^{-1/2}  \biggl[ m_3\ell_i + (1+\ell_i^2) \biggl(\frac{\pi}{2} + \tan^{-1} \Lambda_i \biggr) \biggr]^{-1} \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>m_3 \equiv 3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \, .</math>
</div>
From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~q</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta_i}{\eta_s}
= \eta_i \biggl\{\frac{\pi}{2} + \eta_i + \tan^{-1}\biggl[ \frac{1}{\eta_i} - \ell_i \biggr]  \biggr\}^{-1}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\eta_i \biggl\{\eta_i + \cot^{-1}\biggl[ \ell_i - \frac{1}{\eta_i} \biggr]  \biggr\}^{-1} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\eta_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m_3 \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Let's also define the following shorthand notation:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{L}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ;</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathfrak{K}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + \frac{\Lambda_i}{\eta_i} \, .</math>
  </td>
</tr>
</table>
</div>
====Gravitational Potential Energy of the Core====
Pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[  \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr] \, .</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ -\chi_\mathrm{eq} \biggl[  \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]
\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>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)  + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]
</math>
  </td>
</tr>
</table>
</div>
Out of equilibrium, then, we should expect,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\frac{W_\mathrm{core}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ - \chi^{-1}
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)  + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
- \chi^{-1}
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}
\biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr] 
\, ,
</math>
  </td>
</tr>
</table>
</div>
which, in comparison with our above approximate expression, implies,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\mathfrak{a}_c  </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{2^5} \biggr) \frac{\nu}{\ell_i^5}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr)  + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>
====Thermal Energy of the Core====
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \chi_\mathrm{eq}^{3} \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]^5_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^5} \biggl( \frac{3^8}{2^5\pi} \biggr)^{5/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]^5
\biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11}
\biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]^5
\biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Out of equilibrium, we should then expect,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1}
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i
\, .
</math>
  </td>
</tr>
</table>
</div>
In comparison with our above approximate expression, we therefore have,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~
\biggl[ \biggl(\frac{3}{2\cdot 5}\biggr)\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} 
\biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggr]^5</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11}
\biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr]^5
\biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\mathfrak{b}_c 
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\frac{ 3 }{2^3\ell_i^{3}(1+\ell_i^2)^{6/5}}
\biggl[ \ell_i (\ell_i^4 - 1 )  + (1+\ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
====Gravitational Potential Energy of the Envelope====
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]] and appreciating, in particular, that (see, for example, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Equilibrium_Condition|our notes on equilibrium conditions]]),
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="right">
<math>~A</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\frac{\eta_i}{\sin(\eta_i - B)} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~(\eta_s - B)</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\pi \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\eta_i - B</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\, ,</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \sin(\eta_i -B) = (1+\Lambda_i^2)^{-1/2}</math>
  </td>
  <td align="center">
&nbsp; &nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>~\sin[2(\eta_i-B)] = 2\Lambda_i(1 + \Lambda_i^2)^{-1} \ ,</math>
  </td>
</tr>
</table>
</div>
we have,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2
\biggl\{ \biggl[6(\eta_s-B) - 3\sin[2(\eta_s - B)] -4\eta_s\sin^2(\eta_s-B) + 4B\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) + 4B \biggr]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2
\biggl\{ 6\pi - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) \biggr]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2(1+\Lambda_i^2)
\biggl\{ 6\pi - 6\biggl[\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\biggr] + 6\biggl[ \frac{\Lambda_i}{(1 + \Lambda_i^2)}  \biggr]
+ 4\eta_i \biggl[ \frac{1}{(1+\Lambda_i^2)} \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
So, in equilibrium we can write,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~-\chi_\mathrm{eq}\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\}
\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>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2^2} \biggl(\frac{\eta_i}{m_3}\biggr)^3
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\}
\frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{\ell_i^5}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
And out of equilibrium,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\frac{W_\mathrm{env}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
-\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
\biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr]
\, .
</math>
  </td>
</tr>
</table>
</div>
This, in turn, implies that both in and out of equilibrium,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\mathfrak{a}_e </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2^3} \biggl[\frac{\nu^2(1+q)}{q(1-\nu)} \biggr] \frac{1}{\ell_i^2}
\biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
====Thermal Energy of the Envelope====
Again, pulling from [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|our detailed derivations]],
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2
\biggl\{  \biggl[6(\eta_s - B) - 3\sin[2(\eta_s-B)] \biggr] - \biggl[6(\eta_i - B) - 3\sin[2(\eta_i-B)] \biggr] \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2
\biggl\{ 6\pi  - 6(\eta_i - B) + 3\sin[2(\eta_i-B)]  \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 (1 + \Lambda_i^2) 
\biggl\{ 6\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + 6\biggl[\Lambda_i(1 + \Lambda_i^2)^{-1} \biggr] \biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 
\biggl\{  (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
So, in equilibrium we can write,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}^{3}\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 
\biggl\{  (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\}
\biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{3^2\pi^2}{2^{12}} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3 
\biggl\{  \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\}
\biggl[\frac{(1+\ell_i^2)^9}{3^9\ell_i^{15}}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr)  \biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]
\biggl\{  \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
And, out of equilibrium,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~
\chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) 
\biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K}
\, .
</math>
  </td>
</tr>
</table>
</div>
====Combined in Equilibrium====
Notice that, in combination,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[\frac{2S_\mathrm{env} + W_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3}
\biggl[3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i \biggl( 1 + \ell_i^2 \biggr)^{-1}\biggr]^3
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \biggl( \frac{2\cdot 3^6}{\pi} \biggr)^{1/2} 
\biggl[\frac{\ell_i^3}{( 1 + \ell_i^2)^3}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Also, from above,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[  \frac{2S_\mathrm{core}+W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
\biggl[ \ell_i \biggl(- \frac{8}{3} \ell_i^2 \biggr) (1 + \ell_i^2)^{-3}  \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ + \biggl( \frac{2\cdot 3^6}{\pi } \biggr)^{1/2}
\biggl[ \frac{\ell_i^3}{(1 + \ell_i^2)^{3}}  \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
So, in equilibrium, these terms from the core and envelope sum to zero, as they should.
====Out of Equilibrium====
And now, in combination ''out'' of equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} 
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\}
+\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{2n_c}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2n_e}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
\, .
</math>
  </td>
</tr>
</table>
</div>
Hence, quite generally ''out'' of equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} 
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\}
-\frac{3}{5}\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{10}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
-3\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \, .
</math>
  </td>
</tr>
</table>
</div>
Let's see what the value of this derivative is if the dimensionless radius, <math>~\chi</math>, is set to the value that has been determined, via a detailed force-balanced analysis, to be the equilibrium radius, namely, <math>~\chi = \chi_\mathrm{eq}</math>.  In this case, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl\{\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] \biggr\}_\mathrm{\chi \rightarrow \chi_\mathrm{eq}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -\chi_\mathrm{eq}^{-1}\biggl\{
\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} 
+ \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+2\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}
+2\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
But, according to the virial theorem &#8212; and, as we have just demonstrated &#8212; the four terms inside the curly braces sum to zero.  So this demonstrates that the derivative of our out-of-equilibrium free-energy expression does go to zero at the equilibrium radius, as it should!
===Summary51===
In summary, the desired ''out'' of equilibrium free-energy expression is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{W_\mathrm{core}}{E_\mathrm{norm}}  + \frac{W_\mathrm{env}}{E_\mathrm{norm}}
+\biggl(\frac{2n_c}{3}\biggr)\frac{S_\mathrm{core}}{E_\mathrm{norm}}
+\biggl(\frac{2n_e}{3}\biggr)\frac{S_\mathrm{env}}{E_\mathrm{norm}}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \chi^{-1}
\biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}
\biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr] 
-\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2}
\biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~ 
+
\biggl(\frac{2\cdot 5}{3}\biggr) \biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1}
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i
+\biggl(\frac{2}{3}\biggr)
\chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) 
\biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-
\biggl( \frac{3}{2^4} \biggr) \biggl[\chi^{-1}\frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}\biggr]
\biggl[ \mathfrak{L}_i + 4\mathfrak{K}_i \biggr] 
+
\biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl(\frac{3\cdot 5}{2^3}\biggr) \biggl[ \chi^{-1}
\biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5}
\mathfrak{L}_i
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl( \frac{\pi}{2^{5}\cdot 3^6} \biggr)
\biggl[\chi^{-1}\biggl(\frac{\nu^2}{q} \biggr) 
\frac{(1+\ell_i^2)^2}{\ell_i^{4}}\biggr]^3\mathfrak{K} \, .
</math>
  </td>
</tr>
</table>
</div>
Or, in terms of the ratio,
<div align="center">
<math>\Chi \equiv \frac{\chi}{\chi_\mathrm{eq}} \, ,</math>
</div>
and pulling from the above expressions,
<div align="center">
<table border="0" cellpadding="4">
<tr>
  <td align="right">
<math>~\biggl[  \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2}
\biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}
\biggl[ \mathfrak{L}_i - \frac{8}{3}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2
\biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ -\biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2}  \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}
\biggl[4\mathfrak{K}_i + \frac{8}{3} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3}  + \tan^{-1}(\ell_i) \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}\mathfrak{L}_i  </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 
\biggl\{  (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ ~\frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}  (4\mathfrak{K}_i) \, ,
</math>
  </td>
</tr>
</table>
</div>
we have the streamlined,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{2^5\pi}{3^6} \biggr)^{1/2} \biggl[ \frac{(1+\ell_i^2)}{\ell_i} \biggr]^{3}  \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
+\Chi^{-3/5} (5 \mathfrak{L}_i)
+\Chi^{-3} (4\mathfrak{K}_i)
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
  </td>
</tr>
</table>
</div>
or, better yet,
<div align="center" id="BiPolytropeFreeEnergy">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with</font> <math>~(n_c, n_e) = (5, 1)</math>
</th>
</tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2^4\biggl( \frac{q\ell_i^2}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Chi^{-3/5} (5 \mathfrak{L}_i)
+\Chi^{-3} (4\mathfrak{K}_i)
-\Chi^{-1} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{L}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathfrak{K}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\Lambda_i}{\eta_i}  + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Lambda_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\eta_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{q}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta_s}{\eta_i}
= 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, .
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">Radial Derivatives</td></tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\Chi^{-8/5} (3 \mathfrak{L}_i)
-\Chi^{-4} (12\mathfrak{K}_i)
+\Chi^{-2} (3\mathfrak{L}_i  +12\mathfrak{K}_i )
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\partial^2 \mathfrak{G}^*}{\partial \Chi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{5}\biggl[
\Chi^{-13/5} (8\mathfrak{L}_i)
+\Chi^{-5} (80\mathfrak{K}_i)
-\Chi^{-1} (10\mathfrak{L}_i  +40\mathfrak{K}_i )\biggr]
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
Consistent with our [[User:Tohline/SSC/BipolytropeGeneralization#Free_Energy_and_Its_Derivatives|generic discussion of the stability of bipolytropes]] and the ''specific'' discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|the stability of bipolytropes having]] <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi = \chi_\mathrm{eq}</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>
</div>
Furthermore, the equilibrium configuration is unstable whenever <math>~\partial^2 \mathfrak{G}/\partial \chi^2 < 0</math>, that is, it is unstable whenever,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>
  </td>
  <td align="center">
<math>~></math>
  </td>
  <td align="left">
<math>~20 \, .</math>
  </td>
</tr>
</table>
</div>
[[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|Table 1 of an accompanying chapter]] &#8212; and the red-dashed curve in the figure adjacent to that table &#8212; identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.
==Focus on Zero-Zero Free-Energy Expression==
Here, we will draw heavily from the following accompanying chapters:
*  [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Step_7:__Surface_Boundary_Condition|Analytic Detailed Force Balance Models]]
*  [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Free_Energy_of_BiPolytrope_with|Free-Energy Analysis]]
===From Detailed Force-Balance Models===
====Equilibrium Radius====
=====First View=====
In an [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Virial_Theorem|accompanying chapter]] we find,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{P_0 R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~f</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
<td align="left">
<math>
1+
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} 
\biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathfrak{F} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[  (-2q^2 + 3q^3 - q^5) +
\frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{\rho_e}{\rho_c} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{q^3(1-\nu)}{\nu(1-q^3)} \, .
</math>
  </td>
</tr>
</table>
</div>
Here, we prefer to normalize the equilibrium radius to <math>~R_\mathrm{norm}</math>.  So, let's replace the central pressure with its expression in terms of <math>~K_c</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~P_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
K_c \rho_c^{\gamma_c} = K_c \biggl[ \frac{3M_\mathrm{core}}{4\pi R_i^3} \biggr]^{\gamma_c}
= K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c}
~~~\Rightarrow~~~ \frac{P_0}{P_\mathrm{norm}} = \biggl[ \frac{3}{4\pi}\biggl(\frac{\nu}{q^3}\biggr) \frac{1}{\chi_\mathrm{eq}^3}\biggr]^{(n_c+1)/n_c}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c}
\frac{R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~R_\mathrm{eq}^{(n_c-3)/n_c}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{G}{K_c}\biggr) M_\mathrm{tot}^{(n_c-1)/n_c} \biggl[ \frac{3\nu }{4\pi q^3 } \biggr]^{-(n_c+1)/n_c}
\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\chi_\mathrm{eq}^{(n_c-3)/n_c} \equiv \biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^{(n_c-3)/n_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl(\frac{4\pi}{3} \biggr)^{1/n_c}
\biggl( \frac{\nu}{q^3}\biggr)^{(n_c-1)/n_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Or, in terms of <math>~\gamma_c</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
=====Second View=====
Alternatively, from our derivation and discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#CentralPressure|analytic detailed force-balance models]],
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>
\biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr] P_0</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp;
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~[g(\nu,q)]^2</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
1  + \biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]  \, .
</math>
  </td>
</tr>
</table>
</div>
In order to show that this expression is the same as the other one, [[#First_View_2|above]], we need to show that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~
f - 1-\mathfrak{F}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{5}{2q^3} \biggl[g^2-1\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{5}{2q^3}  \biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{5}{2q^5} \biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl\{ 2 ( q^2 - q^3 )
+ \frac{\rho_e}{\rho_0}\biggl[  1 - 3q^2+ 2q^3 \biggr] \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>
Let's see &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
f - 1-\mathfrak{F}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} 
\biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] -
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[  (-2q^2 + 3q^3 - q^5) +
\frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr)
- \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[  (-2q^2 + 3q^3 - q^5) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]
+\biggl( \frac{\rho_e}{\rho_c}  \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl\{ (q^3- q^5 )
+ (2q^2 - 3q^3 + q^5) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 3 (1 -5q^2 + 5q^3 - q^5) \biggr]
+\frac{1}{2} \biggl( \frac{\rho_e}{\rho_c}  \biggr)^2 \frac{1}{q^5} \biggl[ 2 - 2q^5 + 5\biggl( q^5-q^3\biggr)\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3- q^5 ) + (2q^2 - 3q^3 + q^5) \biggr]
+ \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[  3 (1 -5q^2 + 5q^3 - q^5)+2 - 2q^5 + 5( q^5-q^3)  \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ 2q^2 - 2q^3  \biggr]
+ \frac{5}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[  1 - 3q^2 + 2q^3  \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Q.E.D.
Hence, the equilibrium radius can also be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c} q^2 g^2 \, ;
</math>
  </td>
</tr>
</table>
</div>
or, in terms of the polytropic index,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}^{n_c-3} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \, .
</math>
  </td>
</tr>
</table>
</div>
====Gravitational Potential Energy====
Also from our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Gravitational_Potential_Energy|accompanying discussion]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \Chi^{-1} \biggl( \frac{3}{5}\biggr) \biggl(\frac{\nu}{q^3} \biggr)^2 q^5
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{-1/(n_c-3)}  f(\nu,q)
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ 2^{n_c-(n_c-3)} \biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} b_\xi^{n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
- \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ \biggl(\frac{6}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, .
</math>
  </td>
</tr>
</table>
</div>
====Internal Energy Components====
=====First View=====
Before writing out the expressions for the internal energy of the core and of the envelope, we [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#Virial_Theorem|note from our separate detailed derivation]] that, in either case,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[\frac{P_i \chi^{3\gamma}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\biggl(\frac{P_i }{P_0} \biggr) \biggl(\frac{P_0 }{P_\mathrm{norm}} \biggr)\chi^{3}\biggr]_\mathrm{eq} \biggl[\frac{\chi}{\chi_\mathrm{eq}}\biggr]^{3-3\gamma}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma} \, ,</math>
  </td>
</tr>
</table>
</div>
where, in equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl(\frac{P_i }{P_0} \biggr)_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - b_\xi q^2</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~b_\xi q^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{2}{5}q^3 f + \biggl[1 - \frac{2}{5} q^3( 1+\mathfrak{F} ) \biggr]\biggr\}^{-1} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr]^{-1} </math>
  </td>
</tr>
</table>
</div>
So, copying from our [[User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy0_0#InternalEnergies|accompanying detailed derivation]], we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi/3 }{({\gamma_c}-1)} 
\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_c}
\biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1 }{({\gamma_c}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_c}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi/3 }{({\gamma_e}-1)} 
\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_e}
\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] 
\Chi^{3-3\gamma_e} \biggl(\frac{P_i }{P_0} \biggr)
\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] 
\Chi^{3-3\gamma_e}
\biggl\{ (1-b_\xi q^2)(1-q^3) +  b_\xi  \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr]  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{({\gamma_e}-1)}  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] 
\Chi^{3-3\gamma_e} (1-q^3)
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Furthermore,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{4\pi}\biggr)^{\gamma_c - 1}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}
\biggl\{\chi_\mathrm{eq}^{4-3\gamma_c}\biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
</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)^{\gamma_c - 1}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}
\biggl\{\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c}
\biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
</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)^{(\gamma_c - 1)/(4-3\gamma_c)}
\biggl( \frac{\nu}{q^3} \biggr)^{(6-5\gamma_c)(4-3\gamma_c)}
\biggl\{\frac{q^2}{2}
\biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)}
</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)^{1/(n_c-3)}
\biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)(n_c-3)}
\biggl\{\frac{q^2}{2}
\biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{-3/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}
\, .
</math>
  </td>
</tr>
</table>
</div>
Hence, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
n_c  \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c}
\biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{-3/n_c}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
n_c 
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}
\Chi^{-3/n_c}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
n_e
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)}
\Chi^{-3/n_e} (1-q^3)
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
=====Second View=====
In our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#PiDefinition|accompanying discussion of energies associated with detailed force balance models]], we used the notation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\Pi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3}{2^3\pi}\biggr) \frac{GM_\mathrm{tot}^2}{R^4} \biggl(\frac{\nu}{q^3}\biggr)^2
= P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2 \, ,
</math>
  </td>
</tr>
</table>
</div>
which allows us to rewrite the [[#Second_View|above quoted relationship]] between the central pressure and the radius of the bipolytrope as,
<div align="center">
<math>~P_0 = \Pi (qg)^2 \, .</math>
</div>
We [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Virial_Equilibrium|also showed]] that, in equilibrium, the relationship between the central pressure and the interface pressure is,
<div align="center">
<math>~P_0 =P_i + \Pi_\mathrm{eq} q^2 \, .</math>
</div>
This means that, in equilibrium, the ratio of the interface pressure to the central pressure is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0}
= 1- \frac{1}{g^2} \, ,
</math>
  </td>
</tr>
</table>
</div>
or given that (see [[#Second_View|above]]),
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
\frac{5}{2q^3} \biggl[g^2-1\biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
f - 1-\mathfrak{F}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~~
g^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} ) \, ,
</math>
  </td>
</tr>
</table>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0}
= 1- \biggl[ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} )  \biggr]^{-1} \, .
</math>
  </td>
</tr>
</table>
</div>
This is exactly the pressure-ratio expression presented in our "first view" and unveils the notation association,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~b_\xi q^2</math>
  </td>
  <td align="center">
<math>~\leftrightarrow~</math>
  </td>
  <td align="left">
<math>
\frac{1}{g^2} \, .
</math>
  </td>
</tr>
</table>
</div>
From [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Thermal_Energy_Content|our separate derivation]], we have, in equilibrium,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~\mathfrak{G}_\mathrm{core} = \biggl(\frac{2n_c}{3}\biggr) S_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl(\frac{2n_c}{3}\biggr) \biggl( \frac{4\pi}{5} \biggr) R_\mathrm{eq}^3 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr)_\mathrm{eq} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl( \frac{ q^5n_c}{5} \biggr) R_\mathrm{eq}^3  \biggl( \frac{2^3\pi}{3} \biggr) \Pi_\mathrm{eq} \biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl( \frac{ n_c}{5} \biggr) \biggl[ R_\mathrm{norm}^3 
P_\mathrm{norm} \biggr] \chi_\mathrm{eq}^{-1} \biggl(\frac{\nu^2}{q}\biggr) 
\biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{P_0} \biggr)_\mathrm{eq}\biggl( \frac{P_0}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{ n_c}{5} \biggr) \biggl(\frac{\nu^2}{q}\biggr) 
\biggl[\frac{5}{2q^2} \biggl( 1-\frac{1}{g^2} \biggr)\biggl( q^2g^2\biggr) + 1 \biggr] \chi_\mathrm{eq}^{-1} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{ n_c}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr) 
\biggl[ g^2-\frac{3}{5}  \biggr]
\biggl\{\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c}
\biggr\}^{-1/(n_c-3)}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2}  \biggr]
\biggl( \frac{ 1}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr)  g^2
\biggl\{2^{n_c}\biggl(\frac{3}{4\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} (q g)^{-2n_c}
\biggr\}^{1/(n_c-3)}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2}  \biggr]
\biggl\{2^{n_c}\cdot 2^{(3-n_c)}\biggl(\frac{3}{4\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} \biggl(\frac{\nu}{q^3}\biggr)^{2(n_c-3)} q^{5(n_c-3)} q^{-2n_c} g^{-2n_c} g^{2(n_c-3)}
\biggr\}^{1/(n_c-3)}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2}  \biggr]
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15}  g^{-6}
\biggr\}^{1/(n_c-3)} \, .</math>
  </td>
</tr>
</table>
</div>
Finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2  \biggr]
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15}  b_\xi^3 q^{6}
\biggr\}^{1/(n_c-3)} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~n_c q^3 \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2  \biggr]
\biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^3 \biggr\}^{1/(n_c-3)} \, ,</math>
  </td>
</tr>
</table>
</div>
which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression.  Also from our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Thermal_Energy_Content|previous derivation]], we can write,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~\mathfrak{G}_\mathrm{env} = \biggl(\frac{2n_e}{3}\biggr) S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr)
R_\mathrm{eq}^3 \Pi_\mathrm{eq} \biggl\{ (1-q^3) \biggl(\frac{P_i  }{\Pi}\biggr)_\mathrm{eq} 
+ \biggl( \frac{\rho_e}{\rho_0} \biggr)\biggl[ (-2q^2 + 3q^3 - q^5 )
  + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_0} \biggr) ( -1 + 5q^2 -5q^3 +  q^5 )\biggr]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr)
R_\mathrm{eq}^3 \biggl[  P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2\biggr]_\mathrm{eq}
\biggl\{ (1-q^3) q^2(g^2-1)  + \biggl(\frac{2}{5}\biggr) q^5 \mathfrak{F} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \biggl[  P_\mathrm{norm} R_\mathrm{norm}^3 \biggr] \frac{n_e}{2}
\biggl(\frac{\nu^2}{q^4}\biggr)(1-q^3)
\biggl\{  (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}
\chi^{-1}_\mathrm{eq}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e (1-q^3)
\biggl\{  (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \frac{q^2}{2}\biggl(\frac{\nu}{q^3}\biggr)^2
\biggl[\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c}\biggr]^{-1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e (1-q^3)
\biggl\{  (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}
\biggl[2^{[n_c-(n_c-3)]} \biggl(\frac{3}{4\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} q^{2(n_c-3)-2n_c} g^{-2n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e (1-q^3)
\biggl\{  (g^2-1)  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\}
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} g^{-2n_c} \biggr]^{1/(n_c-3)} \, .
</math>
  </td>
</tr>
</table>
</div>
And, finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e (1-q^3) (b_\xi q^2)^{-1}
\biggl\{  1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\}
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} (b_\xi q^2)^{n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e (1-q^3)
\biggl\{  1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\}
\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)
\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6-2(n_c-3)+2n_c}  b_\xi^{3-n_c+n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ n_e\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{3} \biggr]^{1/(n_c-3)} (1-q^3)
\biggl\{  1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression.
====Summary00====
In summary, the desired ''out'' of equilibrium free-energy expression is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
A_0\Chi^{-3/n_c} + B_0\Chi^{-3/n_e} - C_0\Chi^{-1}
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~A_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{core}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{n_c}{b_\xi} 
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{1/(n_c-3)}
q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~B_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{env}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{n_e}{b_\xi}
\biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c} \biggr]^{1/(n_c-3)} (1-q^3)
\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~C_0 \equiv \biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
<td align="left">
<math>
\biggl( \frac{6}{5}\biggr) q^5 f
\biggl[ \biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, .
</math>
  </td>
</tr>
</table>
</div>
Or, in a more compact form,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}^* \equiv \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{-1/(n_c-3)}
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n_c A_1\Chi^{-3/n_c} + n_e B_1\Chi^{-3/n_e} - 3C_1\Chi^{-1}
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~A_1 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\frac{1}{b_\xi}  (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~B_1 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~C_1 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
<td align="left">
<math>
\biggl( \frac{2}{5}\biggr) q^5 f  \, .
</math>
  </td>
</tr>
</table>
</div>
Let's examine the behavior of the first radial derivative.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3}{\Chi} \biggl[ - A_1\Chi^{-3/n_c} - B_1\Chi^{-3/n_e} + C_1\Chi^{-1} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Let's see whether the sum of terms inside the square brackets is zero at the derived equilibrium radius, that is, when <math>~\Chi = 1</math> and, hence, when
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi = \chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{1/(n_c-3)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} b_\xi^{-n_c} \biggr]^{1/(n_c-3)}
\, .
</math>
  </td>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
C_1 - A_1 - B_1
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{5}\biggr) q^5 f
- \frac{1}{b_\xi}  (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
- \frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{5}\biggr) q^5 f
- \frac{1}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
+ \frac{q^3}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\}
- \frac{q^3}{b_\xi}  \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{2}{5}\biggr) q^5 f - \frac{1}{b_\xi}
+ \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^2 
+ \frac{q^3}{b_\xi}
- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^5 
- \frac{q^3}{b_\xi} + \biggl( \frac{3}{5} \biggr) q^5
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2\biggl\{
\biggl( \frac{2}{5}\biggr) q^3 f - \frac{1}{b_\xi q^2}
+ \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]  (1-q^3)
+ \biggl( \frac{3}{5} \biggr) q^3
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2\biggl\{
\biggl( \frac{2}{5}\biggr) q^3 f - \biggl[ 1+\frac{2}{5} q^3(f-1-\mathfrak{F}) \biggr]
+ \biggl[  (1-q^3) - \frac{2}{5} q^3 \mathfrak{F} \biggr]
+ \biggl( \frac{3}{5} \biggr) q^3
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^2\biggl\{0\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
Q.E.D.
Even slightly better:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{q^2}\biggl[ \biggl(\frac{\pi}{2\cdot 3}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(5-n_c)} b_\xi^{-n_c}\biggr]^{1/(n_c-3)}
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \, ,
</math>
  </td>
</tr>
</table>
</div>
or, better yet,
<div align="center" id="BiPolytropeFreeEnergy">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with ''Structural'' </font> <math>~(n_c, n_e) = (0, 0)</math>
</th>
</tr>
<tr><td align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2\biggl(\frac{q^2}{\nu}\biggr)^2 \chi_\mathrm{eq}
\biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1}
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
where, keeping in mind that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{(b_\xi q^2)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>
  </td>
</tr>
</table>
</div>
we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~A_2 \equiv \frac{A_1}{q^2} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\frac{q^3}{(b_\xi q^2)}  \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^3  \biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr]  - \biggl( \frac{3}{5} \biggr) \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5}q^3  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]  \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~B_2 \equiv \frac{B_1}{q^2} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\frac{1}{(b_\xi q^2)} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2  \biggr\} 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(1-q^3)\biggl\{ \frac{1}{(b_\xi q^2)} -1  + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
(1-q^3)\biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] - 1 + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5} q^3 \biggl\{ (1-q^3) (f - 1-\mathfrak{F} )  + \mathfrak{F} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5} q^3 \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{5} q^3  f - A_2 \, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~C_2 \equiv \frac{C_1}{q^2} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
<td align="left">
<math>
\frac{2}{5} q^3 f  \, .
</math>
  </td>
</tr>
</table>
</div>
As before, the equilibrium system is dynamically unstable if <math>~\partial^2 \mathfrak{G}/\partial \Chi^2 < 0</math>.  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>
==Overview==
===BiPolytrope51===
====Key Analytic Expressions====
<div align="center" id="FreeEnergy51">
<table border="1" cellpadding="5" align="center">
<tr>
<th align="center">
<font size="+1">Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with</font> <math>~(n_c, n_e) = (5, 1)</math>
</th>
</tr>
<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>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{L}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\mathfrak{K}_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\Lambda_i}{\eta_i}  + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Lambda_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\eta_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
From the [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Parameter_Values|accompanying Table 1 parameter values]], we also can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{q}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta_s}{\eta_i}
= 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, .
</math>
  </td>
</tr>
</table>
</div>
Consistent with our [[User:Tohline/SSC/BipolytropeGeneralization#Free_Energy_and_Its_Derivatives|generic discussion of the stability of bipolytropes]] and the ''specific'' discussion of [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|the stability of bipolytropes having]] <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}^*_{51}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi = \chi_\mathrm{eq}</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>
<tr>
  <td align="right">
&nbsp;
  </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[\frac{(1+\ell_i^2)^{6/5}}{(3\ell_i^2)} \biggr]^{n_c} \biggr\}^{1/(n_c-3)} \, ,</math>
  </td>
</tr>
</table>
</div>
where the last expression has been cast into a form that more clearly highlights overlap with the expression, below, for the equilibrium radius for zero-zero bipolytropes.  Furthermore, the equilibrium configuration is unstable whenever,
<div align="center">
<math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_{51}}{\partial \chi^2}\biggr]_{\Chi=1} < 0 \, ,</math>
</div>
that is, it is unstable whenever,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>
  </td>
  <td align="center">
<math>~></math>
  </td>
  <td align="left">
<math>~20 \, .</math>
  </td>
</tr>
</table>
</div>
[[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Stability_Condition|Table 1 of an accompanying chapter]] &#8212; and the red-dashed curve in the figure adjacent to that table &#8212; identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.
====Behavior of Equilibrium Sequence====
Here we reprint Figure 1 from an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Model_Sequences|accompanying chapter wherein the structure of five-one bipolytropes has been derived]]. It displays detailed force-balance sequences in the <math>~q - \nu</math> plane for a variety of choices of the ratio of mean-molecular-weights, <math>~\mu_e/\mu_c</math>, as labeled.
[[File:PlotSequencesBest02.png|450px|center|Five-One Bipolytropic Equilibrium Sequences for Various ratios of the mean molecular weight]]
=====Limiting Values=====
Each sequence begins <math>~(\ell_i = 0)</math> at the origin, that is, at <math>~(q,\nu) = (0,0)</math>.  As <math>~\ell_i \rightarrow \infty</math>, however, the sequences terminate at different coordinate locations, depending on the value of <math>~m_3 \equiv 3(\mu_e/\mu_c)</math>.  In deriving the various limits, it will be useful to note that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\eta_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1 + \ell_i^2)}{m_3 \ell_i} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Lambda_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(1+\ell_i^2)}{m_3\ell_i}-\ell_i</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{m_3\ell_i} + \biggl[\frac{(1 -m_3)}{m_3} \biggr]\ell_i </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{m_3\ell_i} \biggl[ 1 - (m_3-1) \ell_i^2\biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{ (m_3-1) \ell_i}{m_3} \biggl[ 1 - \frac{1}{(m_3-1) \ell_i^2}\biggr] \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~1 + \Lambda_i^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \frac{1}{m_3^2\ell_i^2}\biggl[1 + (1 -m_3) \ell_i^2 \biggr]^2</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{m_3^2\ell_i^2}\biggl\{ m_3^2\ell_i^2 + \biggl[1 + (1 -m_3) \ell_i^2 \biggr]^2\biggr\}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{m_3^2\ell_i^2}\biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (1 -m_3)^2 \ell_i^4 \biggr\}</math>
  </td>
</tr>
</table>
</div>
Examining the three relevant parameter regimes, we see that:
* For <math>~\mu_e/\mu_c < \tfrac{1}{3}</math>, that is, <math>~m_3 < 1</math> &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx </math>
  </td>
  <td align="left">
<math>~\tan^{-1} \biggl[\frac{(1 -m_3)}{m_3} \biggr]\ell_i</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx </math>
  </td>
  <td align="left">
<math>~\frac{\pi}{2} -  \biggl[\frac{m_3}{(1 -m_3)\ell_i} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
1 + \frac{(1 + \ell_i^2)}{m_3 \ell_i}\biggl[\pi - \frac{m_3}{(1 -m_3)\ell_i} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{m_3 + \pi \ell_i}{m_3}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{1}{1 + (\pi \ell_i/m_3)} \rightarrow 0 \, .
</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>~\biggl(\frac{\nu}{q}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_i^2}{1 + \Lambda_i^2}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m_3^2\ell_i^4 \biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (1 -m_3)^2 \ell_i^4 \biggr\}^{-1}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m_3^2\biggl\{ \ell_i^{-4} + (2 -2m_3 + m_3^2) \ell_i^{-2} + (1 -m_3)^2 \biggr\}^{-1}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\nu}{q}\biggr|_{\ell_i\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{m_3}{1-m_3} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{m_3}{1-m_3}\biggr]\frac{1}{1 + (\pi \ell_i/m_3)} \rightarrow 0 \, . </math>
  </td>
</tr>
</table>
</div>
*  For <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, that is, <math>~m_3 = 1</math> &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tan^{-1} \Lambda_i </math>
  </td>
  <td align="center">
<math>~= </math>
  </td>
  <td align="left">
<math>~\tan^{-1} \biggl(\frac{1}{\ell_i}\biggr)</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx </math>
  </td>
  <td align="left">
<math>~\frac{1}{\ell_i}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
1 + \frac{(1 + \ell_i^2)}{\ell_i}\biggl[\frac{\pi}{2} + \frac{1}{\ell_i }\biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2}\biggr)\ell_i</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{2}{\pi \ell_i} \rightarrow 0</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>~\biggl(\frac{\nu}{q}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_i}{(1 + 1/\ell_i^2)^{1/2}}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\ell_i \biggl(\frac{2}{\pi \ell_i} \biggr) = \frac{2}{\pi} \approx 0.63662</math>
  </td>
</tr>
</table>
</div>
*  For <math>~\mu_e/\mu_c > \tfrac{1}{3}</math>, that is, <math>~m_3 > 1</math> &hellip;
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx </math>
  </td>
  <td align="left">
<math>~\tan^{-1} \biggl[-\biggl(\frac{m_3-1}{m_3} \biggr)\ell_i\biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx </math>
  </td>
  <td align="left">
<math>~-\frac{\pi}{2} +  \biggl[\frac{m_3}{(m_3-1)\ell_i} \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
1 + \frac{(1 + \ell_i^2)}{m_3 \ell_i}\biggl[ \frac{m_3}{(m_3-1)\ell_i} \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1 + \frac{(1 + 1/\ell_i^2)}{(m_3-1) }
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
1 + \frac{1}{(m_3-1) } = \frac{m_3}{(m_3-1)}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(m_3-1)}{m_3}
</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>~\biggl(\frac{\nu}{q}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_i^2}{1 + \Lambda_i^2}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m_3^2\ell_i^4 \biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (m_3-1)^2 \ell_i^4 \biggr\}^{-1}</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~m_3^2\biggl\{ \ell_i^{-4} + (2 -2m_3 + m_3^2) \ell_i^{-2} + (m_3-1)^2 \biggr\}^{-1}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\nu}{q}\biggr|_{\ell_i\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~\frac{m_3}{m_3-1} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i\rightarrow \infty}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{m_3}{m_3-1} \biggl[\frac{m_3 - 1}{m_3}\biggr] \rightarrow 1 \, . </math>
  </td>
</tr>
</table>
</div>
Summarizing:
* For <math>~\mu_e/\mu_c < \tfrac{1}{3}</math>, that is, <math>~m_3 < 1</math> &nbsp;  &nbsp;  &nbsp; &hellip; &nbsp;  &nbsp;  &nbsp;  <math>~(q,\nu)_{\ell_i \rightarrow \infty} = (0, 0) \, .</math>
*  For <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, that is, <math>~m_3 = 1</math> &nbsp;  &nbsp;  &nbsp; &hellip; &nbsp;  &nbsp;  &nbsp;  <math>~(q,\nu)_{\ell_i \rightarrow \infty} = (0, \tfrac{2}{\pi}) \, .</math>
*  For <math>~\mu_e/\mu_c > \tfrac{1}{3}</math>, that is, <math>~m_3 > 1</math> &nbsp;  &nbsp;  &nbsp; &hellip; &nbsp;  &nbsp;  &nbsp;  <math>~(q,\nu)_{\ell_i \rightarrow \infty} = [(m_3-1)/m_3, 1] \, .</math>
=====Turning Points=====
Let's identify the location of two turning points along the <math>~\nu(q)</math> sequence &#8212; one defines <math>~q_\mathrm{max}</math> and the other identifies <math>~\nu_\mathrm{max}</math>.  They occur, respectively, where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\ln q}{d\ln \ell_i} = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{d\ln \nu}{d\ln \ell_i} = 0 \, .</math>
  </td>
</tr>
</table>
</div>
In deriving these expressions, we will use the relations,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\eta_i}{d\ell_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{m_3 (1-\ell_i^2)}{(1+\ell_i^2)^2} \, ,</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\frac{d\Lambda_i}{d\ell_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{1}{m_3\ell_i^2} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~m_3 \equiv 3\biggl(\frac{\mu_e}{\mu_c}\biggr) \, .</math>
</div>
Given that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~q </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{ 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \biggr\}^{-1} \, ,</math>
  </td>
</tr>
</table>
</div>
we find,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\ln q}{d\ln \ell_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_i}{q} \cdot ( -q^2) \frac{d}{d\ell_i} \biggl\{
\frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-q\ell_i \biggl\{-\frac{1}{\eta_i^2}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]\frac{d\eta_i}{d\ell_i}
+ \frac{1}{\eta_i(1+\Lambda_i^2)} \frac{d\Lambda_i}{d\ell_i}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q\ell_i \biggl\{\frac{(1-\ell_i^2)}{m_3 \ell_i^2}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{(1+\ell_i^2)}{m_3^2 \ell_i^3(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{q}{m_3^2}{\ell_i^2}\biggl\{m_3 \ell_i (1-\ell_i^2) \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
And, given that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\nu </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, .
</math>
  </td>
</tr>
</table>
</div>
we find,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\ln \nu}{d\ln \ell_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_i}{\nu} \biggl\{
\frac{q}{(1+\Lambda_i^2)^{1/2}} + \frac{q}{(1+\Lambda_i^2)^{1/2}} \frac{d\ln q}{d\ln \ell_i}
- \frac{\ell_i q \Lambda_i }{(1+\Lambda_i^2)^{3/2}} \frac{d\Lambda_i}{d\ell_i}
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q \ell_i}{ \nu(1+\Lambda_i^2)^{1/2}}\biggl\{
1 + \frac{d\ln q}{d\ln \ell_i}
+ \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
In summary, then, the <math>~q_\mathrm{max}</math> turning point occurs where,
<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>~ m_3 \ell_i (1-\ell_i^2) \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
\, ;
</math>
  </td>
</tr>
-->
<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ (1+\Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{(1+\ell_i^2)}{m_3 \ell_i (1-\ell_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
\, ;
</math>
  </td>
</tr>
</table>
</div>
and the <math>~\nu_\mathrm{max}</math> turning point occurs where,
<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>~
1 + \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] 
+ \frac{q \ell_i^3  (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{q\ell_i^2}{m_3^2}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1
+ \frac{q \ell_i^3  (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \biggl[ \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} 
+ \frac{q\ell_i^2}{m_3^2}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggr] \cdot \biggl[ 1 - \ell_i^2(1-m_3) \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1
+ \frac{q \ell_i^3  (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
+ \frac{1}{m_3 \ell_i} \biggl[ \frac{\Lambda_i }{(1+\Lambda_i^2)} 
+ \frac{q\ell_i^3}{m_3}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggr] \cdot \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
<table border="0" width="80%" cellpadding="5" align="center"><tr><td align="left">
<font color="red"><b>NOTE:</b></font>&nbsp; As we show [[#Limiting_Values|above]], for the special case of <math>~m_3 = 1</math> &#8212; that is, when <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, precisely &#8212; the equilibrium sequence (as <math>~\ell_i \rightarrow \infty</math>) intersects the <math>~q = 0</math> axis at precisely the value, <math>~\nu = 2/\pi</math>.  As is illustrated graphically in [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Model_Sequences|Figure 1 of an accompanying chapter]], no <math>~\nu_\mathrm{max}</math> turning point exists for values of  <math>~m_3 > 1</math>. 
</td></tr>
</table>
For the record, we repeat, as well, that the transition from stable to dynamically unstable configurations occurs along the sequence when,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~20 \biggl\{ \frac{\Lambda_i}{\eta_i}  + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{20(1+\Lambda_i^2)(1+\ell_i^2)}{m_3\ell_i} \biggl\{ \frac{\Lambda_i}{(1+\Lambda_i^2)}  + \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ m_3 \ell_i (\ell_i^4-1) + m_3(1+\ell_i^2)^3\cdot \tan^{-1}\ell_i </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~20\ell_i^2 (1+\Lambda_i^2)(1+\ell_i^2) \biggl\{ \frac{\Lambda_i}{(1+\Lambda_i^2)}  + \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{m_3 \ell_i (\ell_i^4-1) + m_3(1+\ell_i^2)^3\cdot \tan^{-1}\ell_i }{ 20\ell_i^2 (1+\ell_i^2)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\Lambda_i  + (1+\Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr]  \, .</math>
  </td>
</tr>
</table>
</div>
In order to clarify what equilibrium sequences do not have any turning points, let's examine how the <math>~q_\mathrm{max}</math> turning-point expression behaves as <math>~\ell_i \rightarrow \infty</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ m_3 \ell_i (\ell_i^2-1) \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2)}{ m_3 \ell_i (\ell_i^2-1)  } \biggl[ 1 + \ell_i^2(m_3-1) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{ 1 + \frac{1}{m_3^2\ell_i^2}\biggl[ (m_3-1) \ell_i^2-1 \biggr]^2 \biggr\} \biggl\{ \frac{\pi}{2}
+ \biggl[ -\frac{\pi}{2} - \frac{1}{\Lambda_i} + \frac{1 }{3\Lambda_i^3} + \mathcal{O}(\Lambda_i^{-5} )\biggr] \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2) \ell_i^2(m_3-1)}{ m_3 \ell_i (\ell_i^2-1)  } \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl\{ 1 + \frac{(m_3-1)^2 \ell_i^2}{m_3^2}\biggl[ 1 - \frac{1}{ (m_3-1) \ell_i^2 } \biggr]^2 \biggr\} \cdot 
\frac{1}{(-\Lambda_i)} \biggl[  1 - \frac{1 }{3\Lambda_i^2} + \cancelto{0}{\mathcal{O}(\Lambda_i^{-4} )}\biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2)}{ \ell_i (\ell_i^2-1)  } \cdot \frac{m_3}{(m_3-1)} \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ 1 - \frac{2}{ (m_3-1) \ell_i^2 } + \frac{m_3^2}{(m_3-1)^2 \ell_i^2}  +  \frac{1}{ (m_3-1)^2 \ell_i^4 } \biggr] \cdot  \frac{m_3}{(m_3-1)\ell_i} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-1}
\biggl\{  1 - \frac{m_3^2}{3(m_3-1)^2\ell_i^2} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-2} \biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~ \Rightarrow ~~~\biggl(1 + \frac{1}{\ell_i^2}  \biggr)  \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]\biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1 - \frac{1}{\ell_i^2}  \biggr) \biggl[ 1 - \frac{2}{ (m_3-1) \ell_i^2 } + \frac{m_3^2}{(m_3-1)^2 \ell_i^2}  +  \frac{1}{ (m_3-1)^2 \ell_i^4 } \biggr]
\biggl\{  1 - \frac{m_3^2}{3(m_3-1)^2\ell_i^2} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-2} \biggr\}
</math>
  </td>
</tr>
</table>
</div>
The leading-order term is unity on both sides of this expression, so they cancel; let's see what results from keeping terms <math>~\propto \ell_i^{-2}</math>.
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{1}{\ell_i^2} \biggl[ 1 + \frac{1}{(m_3-1)}  - \frac{1}{(m_3-1)}\biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\ell_i^2} \biggl[- 1 - \frac{2}{ (m_3-1) } + \frac{m_3^2}{(m_3-1)^2 }  - \frac{m_3^2}{3(m_3-1)^2}  \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ 2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \frac{2}{ (m_3-1) } + \frac{2m_3^2}{3(m_3-1)^2 } 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ 6(m_3-1)^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 6(m_3-1) + 2m_3^2 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ 6m_3^2-12m_3 + 6 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - 6m_3+6 + 2m_3^2 
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow~~~ m_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{3}{2} \, . 
</math>
  </td>
</tr>
</table>
</div>
We therefore conclude that the <math>~q_\mathrm{max}</math> turning point does not appear along any sequence for which,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~m_3</math>
  </td>
  <td align="center">
<math>~></math>
  </td>
  <td align="left">
<math>~\frac{3}{2}</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\mu_e}{\mu_c}</math>
  </td>
  <td align="center">
<math>~></math>
  </td>
  <td align="left">
<math>~\frac{1}{2}\, .</math>
  </td>
</tr>
</table>
</div>
<!--
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ \frac{1}{\ell_i^2} \biggl[ 2 + \frac{1}{\alpha m_3} \biggr]</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ \frac{1}{\ell_i^2}\biggl[
\frac{1}{\alpha^2 }  - \frac{2}{ m_3 \alpha  } -\frac{1}{3\alpha^2 } \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[ 2\alpha^2 m_3 + \alpha \biggr]</math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~ \biggl[
\frac{2m_3}{3}  - 2\alpha  \biggr]
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ 2\alpha^2 m_3 </math>
  </td>
  <td align="center">
<math>~\approx</math>
  </td>
  <td align="left">
<math>~
\frac{2m_3}{3}  - 3\alpha 
</math>
  </td>
</tr>
</table>
</div>
-->
<div align="center">
<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="2">Five-One Bipolytrope Equilibrium Sequences in <math>~q - \nu</math> Plane</td>
</tr>
<tr>
  <td align="center" width="50%">
Full Sequences for Various <math>~\frac{\mu_e}{\mu_c}</math>
  </td>
  <td align="center" width="50%">
Magnified View with Turning Points and Stability Transition-Points Identified
  </td>
</tr>
<tr>
  <td align="center" colspan="2">
[[File:Qvsnu51combined.png|750 px|Five-One Sequences]]
  </td>
</tr>
</table>
</div>


====Graphical Depiction of Free-Energy Surface====
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Latest revision as of 20:43, 23 September 2016

Free-Energy Synopsis

Whitworth's (1981) Isothermal Free-Energy Surface
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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,

<math>~\mathfrak{G}^*_\mathrm{type}</math>

<math>~=</math>

<math>~-ax^{-1} + b x^{-3/n} + c x^{-3/j} + \mathfrak{G}_0 \, .</math>

Equilibrium Radii and Critical Radii

The first and second (partial) derivatives with respect to <math>~x</math> are, respectively,

<math>~\frac{\partial\mathfrak{G}^*_\mathrm{type}}{\partial x}</math>

<math>~=</math>

<math>~ax^{-2} - \biggl(\frac{ 3b}{n}\biggr) x^{-3/n -1} -\biggl(\frac{3 c}{j}\biggr) x^{-3/j-1} </math>

 

<math>~=</math>

<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>

<math>~\frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

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

<math>~0</math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~x^{(n-3)/n }_\mathrm{eq}</math>

<math>~=</math>

<math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a -\biggl(\frac{3 c}{j}\biggr) x^{(j-3)/j}_\mathrm{eq} \biggr] \, .</math>

<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>

<math>~=</math>

<math>~ 0 \, .</math>

We conclude, as well, that at this equilibrium radius, the second (partial) derivative assumes the value,

<math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_\mathrm{type}}{\partial x^2} \biggr]_\mathrm{eq}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

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,

<math>~0</math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~[x_\mathrm{eq}^{(j-3)/j}]_\mathrm{crit} </math>

<math>~=</math>

<math>~\biggl[ \frac{j^2 a(n-3)}{3 c}\biggr] [ n(j+3) - j(n+3) ]^{-1} </math>

 

<math>~=</math>

<math>~ \frac{a}{3^2c}\biggl[ \frac{j^2(n-3)}{n-j} \biggr] \, . </math>

Examples

Pressure-Truncated Polytropes

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

<math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq}</math>

<math>~=</math>

<math>~0 \, ;</math>

<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>

<math>~=</math>

<math>~0 \, ;</math>

<math>~x^{(n-3)/n }_\mathrm{eq}</math>

<math>~=</math>

<math>~\biggl(\frac{n}{ 3b}\biggr) \biggl[a + 3cx^{4}_\mathrm{eq} \biggr] \, ;</math>

 

and

 

<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>

<math>~=</math>

<math>~\biggl[ \frac{a(n-3)}{3^2 c (n+1)} \biggr]^{1/4} \, . </math>

Case M

More specifically, the expression that describes the "Case M" free-energy surface is,

<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>

<math>~=</math>

<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>


Hence, we have,

<math>~a</math>

<math>~\equiv</math>

<math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, , </math>

<math>~b</math>

<math>~\equiv</math>

<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>

<math>~c</math>

<math>~\equiv</math>

<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, , </math>

where the structural form factors for pressure-truncated polytropes are precisely defined here. Therefore, the statement of virial equilibrium is,

<math>~0 </math>

<math>~=</math>

<math>~\frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq}</math>

<math>~\Rightarrow ~~~ \biggl(\frac{3}{4\pi}\biggr)c x_\mathrm{eq}^4 </math>

<math>~=</math>

<math>~\biggl(\frac{3}{4\pi}\biggr) \biggl[ \frac{ b}{n}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3} \biggr]</math>

<math>~\Rightarrow ~~~ \biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) x_\mathrm{eq}^4 </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

And we conclude that,

<math>~3c[x_\mathrm{eq}]^4_\mathrm{crit} </math>

<math>~=</math>

<math>~\frac{(n-3)}{5(n+1)} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>

<math>~\Rightarrow ~~~ \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) [x_\mathrm{eq}]^4_\mathrm{crit}</math>

<math>~=</math>

<math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>

<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>

<math>~=</math>

<math>~\frac{1}{20\pi} \biggl( \frac{n-3}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>

<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>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\frac{1}{20\pi} \biggl( \frac{4n}{n+1} \biggr) \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} </math>

<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ [ x_\mathrm{eq} ]_\mathrm{crit} </math>

<math>~=</math>

<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>

ASIDE:  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 here.
  • Hoerdt's equilibrium expressions from here.
  • Conversion from Horedt's units to ours as specified here.

<math>~{\tilde\mathfrak{f}}_M</math>

<math>~=</math>

<math>~ ( 1 + \ell^2 )^{-3/2} </math>

<math>~{\tilde\mathfrak{f}}_W</math>

<math>~=</math>

<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>

<math>~{\tilde\mathfrak{f}}_A</math>

<math>~=</math>

<math>~ \frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] </math>

<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>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl[ \frac{(1+\ell^2)^{3}}{\ell^{5}} \biggr] \biggl[ \frac{\pi}{2^3\cdot 3^6} \biggr]^{1/2}</math>

<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>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl[ \frac{\ell^{18}}{(1+\ell^2)^{12}} \biggr] \biggl[ \frac{2 \cdot 3^4}{\pi} \biggr]^{3}</math>

So, the radius of the critical equilibrium state should be,

<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^4_\mathrm{crit} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

whereas, each equilibrium configuration has,

<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^4 </math>

<math>~=</math>

<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] \, .</math>

So the equilibrium state that marks the critical configuration must have a value of <math>~\ell</math> that satisfies the relation,

<math>~\frac{\pi^2}{2^6\cdot 3^{12}} \biggl[ \frac{(1+\ell^2)^{12}}{\ell^{20}} \biggr] </math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~2^3\cdot 3 \ell^3</math>

<math>~=</math>

<math>~ \ell ( \ell^4 - \frac{8}{3}\ell^2 - 1 ) + (1 + \ell^2)^{3}\tan^{-1}(\ell ) </math>

<math>~\Rightarrow ~~~\biggl[ \frac{(1 + \ell^2)^{3}}{\ell} \biggr] \tan^{-1}(\ell ) </math>

<math>~=</math>

<math>~1 + \frac{80}{3}\cdot \ell^2 -\ell^4 \, . </math>

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


In addition, we know from our dissection of Hoerdt's work on detailed force-balance models that, in the equilibrium state,

<math>~\biggl(\frac{P_e}{P_\mathrm{norm}}\biggr) \biggl(\frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4</math>

<math>~=</math>

<math> \biggl[ \frac{\tilde\theta^{n+1} }{(4\pi)(n+1)( -\tilde\theta' )^{2}} \biggr] </math>

<math>~\Rightarrow ~~~ 3c x_\mathrm{eq}^4</math>

<math>~=</math>

<math> \biggl[ \frac{\tilde\theta^{n+1} }{(n+1)( -\tilde\theta' )^{2}} \biggr] \, . </math>

This means that, for any chosen polytropic index, the critical equilibrium state is the equilibrium configuration for which (needs to be checked),

<math>~2(9-2n){\tilde\theta}^{n+1}</math>

<math>~=</math>

<math>~ 3(n-3)\biggl[ (- {\tilde\theta}^')^2 - \frac{\tilde\theta(-{\tilde\theta}^')}{\tilde\xi}\biggr] \, . </math>

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),

<math>~x_\mathrm{eq}^{n-3}</math>

<math>~=</math>

<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>

Case P

First Pass

Alternatively, let's examine the "Case P" free-energy surface. Drawing on Stahler's presentation, we adopt the following radius and mass normalizations:

<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>

<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>

In terms of these new normalizations, we have,

<math>~R_\mathrm{norm} \equiv \biggl[\biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{(n-1)} \biggr]^{1/(n-3)}</math>

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<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>

and,

<math>~P_\mathrm{norm} \equiv \biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)} } \biggr]^{1/(n-3)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<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>

Rewriting the expression for the free energy gives,

<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

 

<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>

 

 

<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>

 

<math>~=</math>

<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>

 

 

<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>


Therefore, in this case, we have,

<math>~a</math>

<math>~=</math>

<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>

<math>~b</math>

<math>~=</math>

<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>

<math>~c</math>

<math>~=</math>

<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>

where the structural form factors for pressure-truncated polytropes are precisely defined here. The statement of virial equilibrium is, therefore,

<math>~x^{4}_\mathrm{eq} + \alpha </math>

<math>~=</math>

<math>~ \beta x^{(n-3)/n }_\mathrm{eq} \, , </math>

where,

<math>~\alpha \equiv \frac{a}{3c}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \biggl( \frac{4\pi}{3\cdot 5}\biggr) \tilde{\mathfrak{f}}_W \biggl( \frac{n+1}{n} \biggr) \mathfrak{m}^{2} \, , </math>

<math>~\beta \equiv \frac{b}{nc}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, , </math>

<math>~\mathfrak{m}</math>

<math>~\equiv</math>

<math>~ \biggl(\frac{3}{4\pi}\biggr) \frac{1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot} }{M_\mathrm{SWS}} \biggr) \, . </math>

From a previous derivation, we have,

<math>~0 </math>

<math>~=</math>

<math>~ \frac{ b}{nc}\cdot x^{(n-3)/n }_\mathrm{eq} - \frac{a}{3c} - x^{4}_\mathrm{eq} </math>

 

<math>~=</math>

<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>

 

 

<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>

<math>~0</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

which, thankfully, matches! We conclude as well that the transition from stable to dynamically unstable configurations occurs at,

<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>

<math>~=</math>

<math>~\biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \, . </math>

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

<math>~ \biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] + 1\biggr\}\frac{ \alpha }{\beta} </math>

<math>~=</math>

<math>~ [x_\mathrm{eq}]^{(n-3)/n }_\mathrm{crit} </math>

 

<math>~=</math>

<math>~ \biggl\{ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr] \alpha \biggr\}^{(n-3)/(4n) } </math>

<math>~\Rightarrow~~~ \biggl[ \frac{4n}{3 (n+1)} \biggr]^{4n} \biggl( \frac{ \alpha }{\beta} \biggr)^{4n} </math>

<math>~=</math>

<math>~ \biggl[ \frac{(n-3)}{3 (n+1)} \biggr]^{(n-3)} \alpha^{(n-3)} </math>

<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>

<math>~=</math>

<math>~ \alpha^{3(n+1)} \beta^{-4n} </math>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ \mathfrak{m}^{2(n+1)} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

This also means that the critical radius is,

<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~ [x_\mathrm{eq}]_\mathrm{crit}^{2(n+1)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>


The following parallel derivation was done independently. [Note that a factor of 2n/(n-1) appears to correct a mistake made during the original derivation.] Beginning with the virial expression,

<math>~\beta x^{(n-3)/n }_\mathrm{eq} </math>

<math>~=</math>

<math>~ \alpha + x^{4}_\mathrm{eq} </math>

<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>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n-3)/n }_\mathrm{crit} </math>

<math>~=</math>

<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>

<math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n-3) }_\mathrm{crit} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit} </math>

<math>~=</math>

<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>

<math>~\Rightarrow~~~ [x_\mathrm{eq} ]^{(n+1) }_\mathrm{crit} </math>

<math>~=</math>

<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>


Also from Stahler's work we know that the equilibrium mass and radius are,

<math> ~\frac{M_\mathrm{tot}}{M_\mathrm{SWS}} </math>

<math>~=~</math>

<math> \biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl[ {\tilde\theta}_n^{(n-3)/2} {\tilde\xi}^2 (-{\tilde\theta}^') \biggr] \, , </math>

<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} </math>

<math>~=~</math>

<math> \biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl[ \tilde\xi {\tilde\theta}_n^{(n-1)/2} \biggr] \, . </math>

Additional details in support of an associated PowerPoint presentation can be found here.

Reconcile

<math>~\biggl[\frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^4_\mathrm{crit} </math>

<math>~=</math>

<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>

<math>~\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4_\mathrm{crit}</math>

<math>~=</math>

<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>

Taking the ratio, the RHS is,

<math>~\biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{2} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

while the LHS is,

<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

Q.E.D.

Now, given that,

<math>~M_\mathrm{SWS}^{-4(n-1)/(n-3)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

we have,

<math>~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{4}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~\biggl( \frac{R_\mathrm{norm}}{R_\mathrm{SWS}} \biggr)^{n-3}</math>

<math>~=</math>

<math>~ \biggl(\frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr)^{n-1} \biggl( \frac{n+1}{n} \biggr)^{n} </math>


By inspection, in the specific case of <math>~n=5</math> (see above), this critical configuration appears to coincide with one of the "turning points" identified by Kimura. Specifically, it appears to coincide with the "extremal in r1" along an M1 sequence, which satisfies the condition,

<math>~\biggl[ \frac{n-3}{n-1} \biggr]_{n=5}</math>

<math>~=</math>

<math>~\biggl[ \frac{\tilde\xi \tilde\theta^{n}}{(-\tilde\theta^')}\biggr]_{n=5}</math>

<math>~\Rightarrow ~~~\frac{1}{2} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~3(1 + \ell^2)^{-1} </math>

<math>~\Rightarrow~~~ \ell </math>

<math>~=</math>

<math>~5^{1/2} \, .</math>

Hence, according to Kimura, the turning point associated with the configuration with the largest equilibrium radius, corresponds to the equilibrium configuration having,

<math>~\ell |_\mathrm{R_{max}} = \sqrt{5} \approx 2.2360680 \, .</math>

This is, indeed, very close to — but decidedly different from — the value of <math>~\ell_\mathrm{crit}</math> determined, above!


Streamlined

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


<math>~\frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} </math>

<math>~=</math>

<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>

 

 

<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>

 

<math>~=</math>

<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>

Defining a new normalization energy,

<math>~E_\mathrm{SWS}</math>

<math>~\equiv</math>

<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>

 

<math>~=</math>

<math>~ \biggl(\frac{n+1}{n}\biggr)^{3/2} K^{3n/(n+1)} G^{-3/2} P_e^{(5-n)/[2(n+1)]} \, , </math>

we can write,

<math>~\mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{SWS}} </math>

<math>~=</math>

<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>

in which case the coefficients of the generic free-energy expression are,

<math>~a</math>

<math>~=</math>

<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>

<math>~b</math>

<math>~=</math>

<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>

<math>~c</math>

<math>~=</math>

<math>~ \frac{4\pi}{3} \, , </math>

where, as above,

<math>~\mathfrak{m}</math>

<math>~\equiv</math>

<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>

Now, if we define the pair of parameters,

<math>~\alpha</math>

<math>~\equiv</math>

<math>~\frac{a}{3c}</math>

<math>~\beta</math>

<math>~\equiv</math>

<math>~\frac{b}{nc} \, ,</math>

then the statement of virial equilibrium is,

<math>~x_\mathrm{eq}^4 + \alpha</math>

<math>~=</math>

<math>~\beta x_\mathrm{eq}^{(n-3)/n} \, ,</math>

and the boundary between dynamical stability and instability occurs at,

<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>

<math>~=</math>

<math>~\biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \, .</math>

Combining these last two expressions means that the boundary between dynamical stability and instability is associated with the parameter condition,

<math>~[x_\mathrm{eq}]^{(n-3)/n}_\mathrm{crit} </math>

<math>~=</math>

<math>~\biggl[ \frac{n-3}{3(n+1)} + 1\biggr] \frac{\alpha}{\beta} </math>

<math>~\Rightarrow ~~~ \biggl\{ \biggl[ \frac{n-3}{3(n+1)} \biggr]\alpha \biggr\}^{(n-3)/(4n)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ 4n }{3(n+1)}\biggr] \frac{\alpha}{\beta} </math>

<math>~\Rightarrow ~~~ \beta \alpha^{-3(n+1)/(4n)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ 4n }{3(n+1)}\biggr] \biggl[ \frac{n-3}{n} \cdot \frac{n}{3(n+1)} \biggr]^{(3-n)/(4n)} </math>

 

<math>~=</math>

<math>~ 4 \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)/(4n)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)/(4n)} </math>

<math>~\Rightarrow ~~~ \biggl( \frac{\beta}{4}\biggr)^{4n} \alpha^{-3(n+1)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ n }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{(3-n)} </math>

<math>~\Rightarrow ~~~ \biggl( \frac{\beta}{4}\biggr)^{4n} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ n\alpha }{3(n+1)}\biggr]^{3(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{n-3} \, . </math>

Case M

<math>~a</math>

<math>~\equiv</math>

<math>~3\mathcal{A} = \frac{3}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}\, , </math>

<math>~b</math>

<math>~\equiv</math>

<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>

<math>~c</math>

<math>~\equiv</math>

<math>~\frac{4\pi}{3}\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \, . </math>

Hence,

<math>~\alpha</math>

<math>~=</math>

<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>

<math>~\beta</math>

<math>~=</math>

<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>

So the dynamical stability conditions are:

<math>~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggl( \frac{n}{n-3} \biggr) [x_\mathrm{eq}]_\mathrm{crit}^4</math>

<math>~=</math>

<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>

and,

<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>

<math>~=</math>

<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>

<math>~\Rightarrow~~~ \biggl(\frac{\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} </math>

<math>~=</math>

<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>

<math>~\Rightarrow~~~\biggl[ \frac{n}{n-3} \biggl( \frac{P_e}{P_\mathrm{norm}} \biggr)\biggr]^{n-3} </math>

<math>~=</math>

<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>

Together, then,

<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n-3)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

 

<math>~=</math>

<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>

<math>~\Rightarrow ~~~[x_\mathrm{eq}]_\mathrm{crit}^{(n-3)}</math>

<math>~=</math>

<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>

Case P

<math>~a</math>

<math>~=</math>

<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>

<math>~b</math>

<math>~=</math>

<math>~\biggl( \frac{4\pi n}{3}\biggr) \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} </math>

<math>~c</math>

<math>~=</math>

<math>~ \frac{4\pi}{3} \, , </math>

where, as above,

<math>~\mathfrak{m}</math>

<math>~\equiv</math>

<math>~\biggl( \frac{3}{4\pi} \biggr) \frac{ 1}{\tilde{\mathfrak{f}}_M} \biggl( \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr) \, .</math>

Hence,

<math>~\alpha</math>

<math>~=</math>

<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>

<math>~\beta</math>

<math>~=</math>

<math>~ \tilde{\mathfrak{f}}_A \mathfrak{m}^{(n+1)/n} \, . </math>

So the dynamical stability conditions are:

<math>~[x_\mathrm{eq}]_\mathrm{crit}^4</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl[ \frac{n-3}{n} \biggr]\biggl( \frac{4\pi \tilde{\mathfrak{f}}_W}{3^2\cdot 5}\biggr) \mathfrak{m}^{2} </math>

and,

<math>~ \biggl( \frac{\tilde{\mathfrak{f}}_A }{4}\biggr)^{4n} \mathfrak{m}^{4(n+1)} </math>

<math>~=</math>

<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>

<math>~ \Rightarrow~~~ \mathfrak{m}^{2(n+1)} </math>

<math>~=</math>

<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>

Together, then,

<math>~[x_\mathrm{eq}]_\mathrm{crit}^{4(n+1)}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

Compare

Let's see if the two cases, in fact, provide the same answer.

<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>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

Five-One Bipolytropes

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

<math>~x^{2/5 }_\mathrm{eq}</math>

<math>~=</math>

<math>~\biggl(\frac{5}{ 3b}\biggr) \biggl[a -3 c x^{-2}_\mathrm{eq} \biggr] \, ;</math>

 

and

 

<math>~[x_\mathrm{eq}]_\mathrm{crit} </math>

<math>~=</math>

<math>~\biggl[ \frac{18 c}{a }\biggr]^{1/2} \, . </math>


More specifically, the expression that describes the free-energy surface is,

<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>

<math>~=</math>

<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>

Hence, we have,

<math>~a</math>

<math>~\equiv</math>

<math>~ 3\chi_\mathrm{eq}(\mathfrak{L}_i + 4\mathfrak{K}_i) \, , </math>

<math>~b</math>

<math>~\equiv</math>

<math>~ 5 \mathfrak{L}_i \chi_\mathrm{eq}^{3/5} \, , </math>

<math>~c</math>

<math>~\equiv</math>

<math>~ 4 \mathfrak{K}_i \chi_\mathrm{eq}^{3} \, , </math>

and conclude that,

<math>~[\chi_\mathrm{eq}]_\mathrm{crit}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~[\chi_\mathrm{eq}]_\mathrm{crit}\biggl[ \frac{24 \mathfrak{K}_i }{ (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]^{1/2} </math>

<math>~\Rightarrow~~~\biggl[ \frac{24 \mathfrak{K}_i }{ (\mathfrak{L}_i + 4\mathfrak{K}_i)} \biggr]_\mathrm{crit}</math>

<math>~=</math>

<math>~1 </math>

<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{L}_i }{ \mathfrak{K}_i } \biggr]_\mathrm{crit}</math>

<math>~=</math>

<math>~20 \, . </math>

Also, from our detailed force balance derivations, we know that,

<math>~\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>

<math>~=</math>

<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>

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,

<math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math>

<math>~=</math>

<math>~\frac{n_c}{ 3b} \biggl[a -\biggl(\frac{3 c}{n_e}\biggr) x^{(n_e-3)/n_e}_\mathrm{eq} \biggr] \, ;</math>

 

and

 

<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, . </math>


And here, the expression that describes the free-energy surface is,

<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>

<math>~=</math>

<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>

Hence, we have,

<math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math>

<math>~=</math>

<math> 3f \chi_\mathrm{eq} \, , </math>

<math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math>

<math>~\equiv</math>

<math> n_c \chi_\mathrm{eq}^{3/n_c} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>

<math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2 </math>

<math>~\equiv</math>

<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>

 

<math>~=</math>

<math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, , </math>

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

<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>

<math>~=</math>

<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>

<math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>

<math>~=</math>

<math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math>

 

<math>~=</math>

<math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math>


From our associated detailed-force-balance derivation, we know that the associated equilibrium radius is,

<math>~\chi_\mathrm{eq}</math>

<math>~=</math>

<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>


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 comparable expression shown above for the "Five-One" Bipolytrope. Here we have,

<math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<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>

whereas, rewriting the above relation gives,

<math>~\chi_\mathrm{eq}\biggr|_{51}</math>

<math>~=</math>

<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>

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

<math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>

<math>~=</math>

<math>~ \frac{5}{6} \, .</math>

Free-Energy of Truncated Polytropes

In this case, the Gibbs-like free energy is given by the sum of three separate energies,

<math>~\mathfrak{G}</math>

<math>~=</math>

<math>~W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm} + P_eV</math>

 

<math>~=</math>

<math>~ - 3\mathcal{A} \biggl[\frac{GM^2}{R} \biggr] + n\mathcal{B} \biggl[ \frac{KM^{(n+1)/n}}{R^{3/n}} \biggr] + \frac{4\pi}{3} \cdot P_e R^3 \, ,</math>

where the constants,

<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>

      and     

<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, ,</math>

and, as derived elsewhere,

Structural Form Factors for Pressure-Truncated Polytropes <math>~(n \ne 5)</math>

<math>~\tilde\mathfrak{f}_M</math>

<math>~=</math>

<math>~ \biggl( - \frac{3\tilde\theta^'}{\tilde\xi} \biggr) </math>

<math>\tilde\mathfrak{f}_W</math>

<math>~=</math>

<math>\frac{3\cdot 5}{(5-n)\tilde\xi^2} \biggl[\tilde\theta^{n+1} + 3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr] </math>

<math>~ \tilde\mathfrak{f}_A </math>

<math>~=</math>

<math>~\frac{1}{(5-n)} \biggl\{ 6\tilde\theta^{n+1} + (n+1) \biggl[3 (\tilde\theta^')^2 - \tilde\mathfrak{f}_M \tilde\theta \biggr] \biggr\} </math>

As we have shown separately, for the singular case of <math>~n = 5</math>,

<math>~\mathfrak{f}_M</math>

<math>~=</math>

<math>~ ( 1 + \ell^2 )^{-3/2} </math>

<math>~\mathfrak{f}_W</math>

<math>~=</math>

<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>

<math>~\mathfrak{f}_A</math>

<math>~=</math>

<math>~ \frac{3}{2^3} \ell^{-3} [ \tan^{-1}(\ell ) + \ell (\ell^2-1) (1+\ell^2)^{-2} ] </math>

where, <math>~\ell \equiv \tilde\xi/\sqrt{3} </math>


In general, then, the warped free-energy surface drapes across a four-dimensional parameter "plane" such that,

<math>~\mathfrak{G}</math>

<math>~=</math>

<math>~\mathfrak{G}(R, K, M, P_e) \, .</math>

In order to effectively visualize the structure of this free-energy surface, we will reduce the parameter space from four to two, in two separate ways: First, we will hold constant the parameter pair, <math>~(K,M)</math>; giving a nod to Kimura's (1981b) nomenclature, we will refer to the resulting function, <math>~\mathfrak{G}_{K,M}(R,P_e)</math>, as a "Case M" free-energy surface because the mass is being held constant. Second, we will hold constant the parameter pair, <math>~(K,P_e)</math>, and examine the resulting "Case P" free-energy surface, <math>~\mathfrak{G}_{K,P_e}(R,M)</math>.

Virial Equilibrium and Dynamical Stability

The first (partial) derivative of <math>~\mathfrak{G}</math> with respect to <math>~R</math> is,

<math>~\frac{\partial \mathfrak{G}}{\partial R}</math>

<math>~=</math>

<math>~ \frac{1}{R} \biggl[ 3\mathcal{A} GM^2 R^{-1} - 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 4\pi P_e R^3 \biggr] \, ; </math>

and the second (partial) derivative is,

<math>~\frac{\partial^2 \mathfrak{G}}{\partial R^2}</math>

<math>~=</math>

<math>~ \frac{1}{R^2} \biggl[ -6\mathcal{A} GM^2 R^{-1} + \biggl(\frac{n+3}{n}\biggr) 3\mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi P_e R^3 \biggr] \, . </math>

The virial equilibrium radius is identified by setting the first derivative to zero. This means that,

<math>~3\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>

<math>~=</math>

<math>~ 3\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi P_e R_\mathrm{eq}^3 \, . </math>

This expression can be usefully rewritten in the following forms:

Virial Equilibrium Condition
Case 1:

<math>~3(n+3)\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}</math>

<math>~=</math>

<math>~ 3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi (n+3) P_e R_\mathrm{eq}^3 </math>

Case 2:

<math>~ -6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>

<math>~=</math>

<math>~ 8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} </math>

Case 3:

<math>~ 8\pi nP_e R_\mathrm{eq}^3 </math>

<math>~=</math>

<math>~6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>

Dynamical stability is determined by the sign of the second derivative expression evaluated at the equilibrium radius; setting the second derivative to zero identifies the transition from stable to unstable configurations. The criterion is,

<math>~0</math>

<math>~=</math>

<math>~\biggl[ -6n\mathcal{A} GM^2 R^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R^{-3/n} + 8\pi nP_e R^3\biggr]_{R_\mathrm{eq}} </math>

Case 1 Stability Criterion

Using the "Case 1" virial expression to define the equilibrium radius means that the stability criterion is,

<math>~0</math>

<math>~=</math>

<math>~ -6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3)\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 4\pi (n+3) P_e R_\mathrm{eq}^3 + 8\pi nP_e R_\mathrm{eq}^3 </math>

 

<math>~=</math>

<math>~ \mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n+3)- 6n ] + 4\pi P_e R_\mathrm{eq}^3 [(n+3) + 2n] </math>

<math>~\Rightarrow ~~~ 4\pi P_e R_\mathrm{eq}^3 [3(n+1) ] </math>

<math>~=</math>

<math>~ \mathcal{A} GM^2 R_\mathrm{eq}^{-1} [3(n-3)] </math>

<math>~\Rightarrow ~~~ 4\pi P_e R_\mathrm{eq}^4 (n+1) </math>

<math>~=</math>

<math>~ \mathcal{A} GM^2 (n-3) </math>

Case 2 Stability Criterion

Using the "Case 2" virial expression to define the equilibrium radius means that the stability criterion is,

<math>~0</math>

<math>~=</math>

<math>~

8\pi nP_e R_\mathrm{eq}^3 - 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n}  + 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 8\pi nP_e R_\mathrm{eq}^3

</math>

 

<math>~=</math>

<math>~ 16\pi nP_e R_\mathrm{eq}^3 - [3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} </math>

<math>~\Rightarrow~~~ 16\pi nP_e R_\mathrm{eq}^3 </math>

<math>~=</math>

<math>~ [3(n-3)]\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} </math>

<math>~\Rightarrow~~~ 16\pi nP_e R_\mathrm{eq}^{3(n+1)/n} </math>

<math>~=</math>

<math>~ [3(n-3)]\mathcal{B}KM^{(n+1)/n} </math>

Case 3 Stability Criterion

Using the "Case 3" virial expression to define the equilibrium radius means that the stability criterion is,

<math>~0</math>

<math>~=</math>

<math>~ -6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + 3(n+3) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} + 6n\mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} - 6n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>

 

<math>~=</math>

<math>~ -12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} + [6n +3(n+3)] \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} </math>

<math>~\Rightarrow~~~ 9(n+1) \mathcal{B}KM^{(n+1)/n} R_\mathrm{eq}^{-3/n} </math>

<math>~=</math>

<math>~ 12n\mathcal{A} GM^2 R_\mathrm{eq}^{-1} </math>

<math>~\Rightarrow~~~ R_\mathrm{eq}^{n-3} </math>

<math>~=</math>

<math>~ \biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1} </math>

Case M

Now, in our discussion of "Case M" sequence analyses, the configuration's radius is normalized to,

<math>~R_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~[ G^n K^{-n} M^{n-1} ]^{1/(n-3)} \, .</math>

Our "Case 3" stability criterion directly relates. We conclude that the transition from stability to dynamical instability occurs when,

<math>~ \biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3} </math>

<math>~=</math>

<math>~ \biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n </math>

<math>~\Rightarrow~~~ \biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]_\mathrm{crit}^{(n-3)/n} </math>

<math>~=</math>

<math>~ \frac{4n}{15(n+1) } \biggl(\frac{4\pi}{3}\biggr)^{1/n}\cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_A \tilde{\mathfrak{f}}_M^{(n-1)/n}} </math>

Also in the "Case M" discussions, the external pressure is normalized to,

<math>~P_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~[ G^{-3(n+1)} K^{4n} M^{-2(n+1)} ]^{1/(n-3)} \, .</math>

If we raise the "Case 1" stability criterion expression to the <math>~(n-3)</math> power, then divide it by the "Case 3" stability criterion expression raised to the fourth power, we find,

<math>~\Rightarrow ~~~ [P_e]_\mathrm{crit}^{n-3} </math>

<math>~=</math>

<math>~ \biggl[\frac{\mathcal{A} GM^2 (n-3)}{4\pi (n+1)}\biggr]^{n-3}\biggl\{ \biggl[\frac{4n\mathcal{A}}{3(n+1) \mathcal{B}} \biggr]^n \biggl(\frac{G}{K}\biggr)^n M^{n-1} \biggr\}^{-4} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{\mathcal{A} (n-3)}{4\pi (n+1)}\biggr]^{n-3} G^{n-3} M^{2(n-3)} \biggl[\frac{3(n+1) \mathcal{B}}{4n\mathcal{A}} \biggr]^{4n} \biggl(\frac{K}{G}\biggr)^{4n} M^{4(1-n)} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n} M^{-2(n+1)} G^{-3(n+1)} </math>

<math>~\Rightarrow ~~~ \biggl[\frac{P_e}{P_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3} </math>

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \biggl[ \frac{ 5\tilde{\mathfrak{f}}_M^2 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)} \biggl( \frac{3}{4\pi}\biggr)^4 \biggl[ \frac{ \tilde{\mathfrak{f}}_A }{ \tilde{\mathfrak{f}}_M^{(n+1)/n} } \biggr]^{4n} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \biggl[ \frac{ 5 }{ \tilde{\mathfrak{f}}_W } \biggr]^{3(n+1)} \tilde{\mathfrak{f}}_M^{2(n+1)} \tilde{\mathfrak{f}}_A^{4n} </math>

Case P

Flipping around this expression for <math>~[P_e]_\mathrm{crit}</math>, we also can write,

<math>~ [M]_\mathrm{crit}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n} G^{-3(n+1)} P_e^{3-n} \, . </math>

Now, in our "Case P" discussions we normalized the mass to

<math>~M_\mathrm{SWS}</math>

<math>~\equiv</math>

<math>~ \biggl(\frac{n+1}{n}\biggr)^{3/2} G^{-3/2} K^{2n/(n+1)} P_e^{(3-n)/[2(n+1)]} \, . </math>

Hence, we have,

<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \biggl(\frac{n+1}{n}\biggr)^{-3(n+1)} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} \, , </math>

where the constants,

<math>~\mathcal{A} \equiv \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2}</math>

      and     

<math>\mathcal{B} \equiv \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \, .</math>

So we can furthermore conclude that,

<math>~\biggl[\frac{M}{M_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3 }{4} \biggr)^{4n} \biggl\{ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr\}^{-3(n+1)} \biggl\{ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggr\}^{4n} </math>

 

<math>~=</math>

<math>~\biggl(\frac{3}{4\pi} \biggr)^{4} \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n} \biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]^{(n+1)} \, . </math>


Our expression for <math>~[M]_\mathrm{crit}^{2(n+1)}</math> can also be combined with the "Case 2 stability criterion" to eliminate the mass entirely, giving,

<math>~ \biggl\{ 16\pi nP_e R_\mathrm{eq}^{3(n+1)/n} \biggr\}^{2n} </math>

<math>~=</math>

<math>~ \biggl\{ [3(n-3)]\mathcal{B}K\biggr\}^{2n} \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{4n} K^{4n} G^{-3(n+1)} P_e^{3-n} </math>

<math>~\Rightarrow~~~ R_\mathrm{eq}^{6(n+1)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ 3(n-3)}{16\pi n} \biggr]^{2n} \biggl[\frac{ (n-3)}{4\pi (n+1)}\biggr]^{n-3} \biggl[\frac{3(n+1) }{4n} \biggr]^{4n} \mathcal{A}^{-3(n+1)} \mathcal{B}^{6n} K^{6n} G^{-3(n+1)} P_e^{3(1-n)} </math>

<math>~\Rightarrow~~~ R_\mathrm{eq}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl\{ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{2n} \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl[\frac{(n+1) }{n} \biggr]^{4n+(3-n)} \biggl(\frac{3 }{4} \biggr)^{6n} \biggr\}^{1/3} \mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n} G^{-(n+1)} P_e^{(1-n)} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl[\frac{(n+1) }{n} \biggr]^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n} \mathcal{A}^{-(n+1)} \mathcal{B}^{2n} K^{2n} G^{-(n+1)} P_e^{(1-n)} \, . </math>

Finally, recognizing that in our "Case P" discussions we normalized the radius to

<math>~R_\mathrm{SWS}</math>

<math>~\equiv</math>

<math>~ \biggl(\frac{n+1}{n}\biggr)^{1/2} G^{-1/2} K^{n/(n+1)} P_e^{(1-n)/[2(n+1)]} \, , </math>

we have,

<math>~ [R_\mathrm{eq}]_\mathrm{crit}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{n+1 }{n} \biggr)^{(n+1)} \biggl(\frac{3 }{4} \biggr)^{2n} \mathcal{A}^{-(n+1)} \mathcal{B}^{2n} \biggl\{ R_\mathrm{SWS}\biggl(\frac{n+1 }{n} \biggr)^{-1/2} \biggr\}^{2(n+1)} </math>

<math>~\Rightarrow~~~ \biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]_\mathrm{crit}^{2(n+1)} </math>

<math>~=</math>

<math>~ \biggl[ \frac{ (n-3)}{4\pi n} \biggr]^{(n-1)} \biggl(\frac{3 }{4} \biggr)^{2n} \mathcal{A}^{-(n+1)} \mathcal{B}^{2n} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{n}{n-3} \biggr]^{(1-n)} (4\pi)^{1-n}\biggl(\frac{3 }{4} \biggr)^{2n} \biggl[ \frac{1}{5} \cdot \frac{\tilde{\mathfrak{f}}_W}{\tilde{\mathfrak{f}}_M^2} \biggr]^{-(n+1)} \biggl[ \biggl(\frac{4\pi}{3} \biggr)^{-1/n} \frac{\tilde{\mathfrak{f}}_A}{\tilde{\mathfrak{f}}_M^{(n+1)/n}} \biggr]^{2n} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{n}{n-3} \biggr]^{(1-n)} (4\pi)^{1-n -2} 3^{2n+2} 4^{-2n} \biggl[ \frac{5\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W}\biggr]^{(n+1)} \biggl[ \frac{\tilde{\mathfrak{f}}_A^{2n}}{\tilde{\mathfrak{f}}_M^{2(n+1)}} \biggr] </math>

 

<math>~=</math>

<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>

Case M Free-Energy Surface

It is useful to rewrite the free-energy function in terms of dimensionless parameters. Here we need to pick normalizations for energy, radius, and pressure that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~M</math>. We have chosen to use,

<math>~R_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~\biggl[ \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} \biggr]^{1/(n-3)} \, ,</math>

<math>~P_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~\biggl[ \frac{K^{4n}}{G^{3(n+1)} M_\mathrm{tot}^{2(n+1)}} \biggr]^{1/(n-3)} \, ,</math>

which, as is detailed in an accompanying discussion, are similar but not identical to the normalizations used by Horedt (1970) and by Whitworth (1981). The self-consistent energy normalization is,

<math>~E_\mathrm{norm}</math>

<math>~\equiv</math>

<math>~P_\mathrm{norm} R^3_\mathrm{norm} \, .</math>

As we have demonstrated elsewhere, after implementing these normalizations, the expression that describes the "Case M" free-energy surface is,

<math> \mathfrak{G}_{K,M}^* \equiv \frac{\mathfrak{G}_{K,M}}{E_\mathrm{norm}} = -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>

Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case M" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.


Case P Free-Energy Surface

Again, it is useful to rewrite the free-energy function in terms of dimensionless parameters. But here we need to pick normalizations for energy, radius, and mass that are expressed in terms of the gravitational constant, <math>~G</math>, and the two fixed parameters, <math>~K</math> and <math>~P_e</math>. As is detailed in an accompanying discussion, we have chosen to use the normalizations defined by Stahler (1983), namely,

<math>~R_\mathrm{SWS}</math>

<math>~\equiv</math>

<math>~\biggl( \frac{n+1}{nG} \biggr)^{1/2} K^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, ,</math>

<math>~M_\mathrm{SWS}</math>

<math>~\equiv</math>

<math>~\biggl( \frac{n+1}{nG} \biggr)^{3/2} K^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \, .</math>

The self-consistent energy normalization is,

<math>~E_\mathrm{SWS} \equiv \biggl( \frac{n}{n+1} \biggr) \frac{GM_\mathrm{SWS}^2}{R_\mathrm{SWS}}</math>

<math>~=</math>

<math>~ \biggl( \frac{n+1}{n} \biggr)^{3/2} G^{-3/2}K^{3n/(n+1)} P_\mathrm{e}^{(5-n)/[2(n+1)]} \, .</math>

After implementing these normalizations — see our accompanying analysis for details — the expression that describes the "Case P" free-energy surface is,

<math>~\mathfrak{G}_{K,P_e}^* \equiv \frac{\mathfrak{G}_{K,P_e}}{E_\mathrm{SWS}}</math>

<math>~=</math>

<math>~- 3 \mathcal{A} \biggl( \frac{n+1}{n} \biggr)\biggl( \frac{M}{M_\mathrm{SWS}}\biggr)^2 \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-1} + n\mathcal{B} \biggl(\frac{M}{M_\mathrm{SWS}}\biggr)^{(n+1)/n} \biggl(\frac{R}{R_\mathrm{SWS}}\biggr)^{-3/n} + \frac{4\pi}{3} \cdot \biggl( \frac{R}{R_\mathrm{SWS}}\biggr)^3 \, . </math>

Given the polytropic index, <math>~n</math>, we expect to obtain a different "Case P" free-energy surface for each choice of the dimensionless truncation radius, <math>~\tilde\xi</math>; this choice will imply corresponding values for <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math> and, hence also, corresponding (constant) values of the coefficients, <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.

Summary

  DFB Equilibrium Onset of Dynamical Instability
Case M:

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]^{n-3}</math>

<math>~=</math>

<math>~ \biggl[ \frac{4\pi}{(n+1)^n} \biggr] {\tilde\xi}^{(n-3)} (-{\tilde\xi}^2 \tilde{\theta^'})^{(1-n)} </math>

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3}</math>

<math>~=</math>

<math>~ \biggl[ \frac{4n}{15(n+1) }\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)}} </math>

<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]^{n-3}</math>

<math>~=</math>

<math>~ \biggl[ \frac{(n+1)^3}{4\pi} \biggr]^{(n+1)} {\tilde\theta}^{(n+1)(n-3)} (-{\tilde\xi}^2 \tilde{\theta^'})^{2(n+1)} </math>

<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>

<math>~=</math>

<math>~ \biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)} \biggl[ \frac{ 5^3 \tilde{\mathfrak{f}}_M^2}{ \tilde{\mathfrak{f}}_W^3 } \biggr]^{n+1} \biggl( \frac{3\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} </math>

Case P:

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^2</math>

<math>~=</math>

<math>~ \biggl(\frac{n}{4\pi}\biggr) {\tilde\xi}^2 {\tilde\theta}^{n-1} </math>

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]_\mathrm{crit}^2</math>

<math>~=</math>

<math>~ \biggl[ \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr] \biggl[ \frac{n-3}{n} \biggr]^{(n-1)/(n+1)} \biggl[\frac{ \tilde{\mathfrak{f}}_A}{4} \biggr]^{2n/(n+1)} </math>

<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]^{2}</math>

<math>~=</math>

<math>~ \biggl( \frac{n^3}{4\pi}\biggr) {\tilde\theta}^{n-3} (-{\tilde\xi}^2 {\tilde{\theta^'}})^2 </math>

<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2}</math>

<math>~=</math>

<math>~\biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]\biggl(\frac{3}{4\pi} \biggr)^{4/(n+1)} \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{(n-3)/(n+1)} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n/(n+1)} </math>

In all four cases, the expression on right intersects (is equal to) the expression on the left when the following condition applies:

For <math>~n \ne 5</math>:      

<math>~ 2(9-2n){\tilde\theta}^{n+1} </math>

<math>~=</math>

<math>~ 3(n-3) \biggl[ (-{\tilde\theta}^')^2 - \biggl( -\frac{\tilde\theta {\tilde\theta}^'}{\tilde\xi} \biggr)\biggr] \, ; </math>

For <math>~n = 5</math>:      

<math>~ \biggl[\frac{2^4\cdot 5}{3}\biggr] \ell^3 </math>

<math>~=</math>

<math>~ (1+\ell^2)^3 \tan^{-1}\ell + \ell(\ell^4-1) \, . </math>


If (for <math>n\ne 5</math>) we adopt the shorthand notation,

<math>~\Upsilon</math>

<math>~\equiv</math>

<math>~[3 (-{\tilde\theta}^')^2 - {\tilde\mathfrak{f}}_M \tilde\theta] = 3\biggl[ (-{\tilde\theta}^')^2 - \biggl( -\frac{\tilde\theta {\tilde\theta}^'}{\tilde\xi} \biggr)\biggr] \, , </math>

and

<math>~\tau</math>

<math>~\equiv</math>

<math>~{\tilde\theta}^{n+1} \, , </math>

then the critical condition becomes,

<math>~(n-3)\Upsilon</math>

<math>~=</math>

<math>~2(9-2n)\tau \, ,</math>

and at the critical state, the expressions for the structural form-factors become,

<math>~{\tilde\mathfrak{f}}_A</math>

<math>~=</math>

<math>~\frac{1}{(5-n)} \biggl[6\tau + (n+1)\Upsilon \biggr] </math>

 

<math>~=</math>

<math>~\frac{1}{(5-n)} \biggl\{ 6 + (n+1)\biggl[ \frac{2(9-2n)}{n-3} \biggr] \biggr\}\tau </math>

 

<math>~=</math>

<math>~\frac{1}{(5-n)} \biggl[ \frac{6(n-3) + 2(9-2n)(n+1)}{n-3} \biggr] \tau </math>

 

<math>~=</math>

<math>~\frac{1}{(5-n)} \biggl[ \frac{4n(5-n)}{n-3} \biggr] \tau </math>

 

<math>~=</math>

<math>~\frac{4n\tau}{(n-3)} \, ;</math>

<math>~{\tilde\mathfrak{f}}_W</math>

<math>~=</math>

<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl[\tau + \Upsilon \biggr]</math>

 

<math>~=</math>

<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl\{1 + \biggl[ \frac{2(9-2n)}{n-3} \biggr] \biggr\}\tau</math>

 

<math>~=</math>

<math>~\frac{3\cdot 5}{(5-n) {\tilde\xi}^2} \biggl[ \frac{3(5-n)}{n-3} \biggr] \tau</math>

 

<math>~=</math>

<math>~\frac{3^2\cdot 5 \tau}{(n-3) {\tilde\xi}^2} </math>

<math>~\Rightarrow~~~ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} </math>

<math>~=</math>

<math>~ \biggl[\frac{(n-3) {\tilde\xi}^2}{3^2\tau} \biggr]^{3} \biggl(-\frac{3 {\tilde\theta}^'}{\tilde\xi} \biggr)^2 = 3^2\biggl[\frac{(n-3) {\tilde\xi}^2}{3^2\tau} \biggr]^{3} \biggl(-\frac{{\tilde\xi}^2 {\tilde\theta}^'}{ {\tilde\xi}^3 } \biggr)^2 </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr] (-{\tilde\xi}^2 {\tilde\theta}^' )^2 \, . </math>

Hence (1),

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr]_\mathrm{crit}^{n-3}</math>

<math>~=</math>

<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr] \biggl[ \frac{4n}{15}\biggr]^n \biggl(\frac{1}{3}\biggr) \biggl[ \frac{3^2\cdot 5 }{4n {\tilde\xi}^2} \biggr]^n \tilde{\mathfrak{f}}_M^{1-n} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr] \biggl[ \frac{1}{ {\tilde\xi}^{2n}} \biggr] \biggl( \frac{-{\tilde\xi}^2{\tilde\theta}^'}{{\tilde\xi}^3} \biggr)^{1-n} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{4\pi}{(n+1)^n }\biggr] {\tilde\xi}^{n-3} (-{\tilde\xi}^2{\tilde\theta}^')^{1-n} </math>

Q.E.D.


And (2),

<math>~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>

<math>~=</math>

<math>~ \biggl( \frac{3}{4\pi}\biggr)^4 \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)} \biggl[ \frac{ 5^3 \tilde{\mathfrak{f}}_M^2}{ \tilde{\mathfrak{f}}_W^3 } \biggr]^{n+1} \biggl( \frac{3\tilde{\mathfrak{f}}_A}{4}\biggr)^{4n} </math>

 

<math>~=</math>

<math>~3^{4} (4\pi)^{-(n+1)} \biggl(\frac{ n-3}{n}\biggr)^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)} \biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr]^{n+1} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} \biggl[ \frac{3n\tau}{n-3} \biggr]^{4n} </math>

 

<math>~=</math>

<math>~3^{4} (4\pi)^{-(n+1)} \biggl(\frac{ n-3}{n}\biggr)^{n-3} \biggl(\frac{n+1}{n} \biggr)^{3(n+1)} n^{3(n+1)} \biggl[ \frac{n-3}{n} \biggr]^{3(n+1)} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} \biggl[ \frac{n}{n-3} \biggr]^{4n} \tau^{4n-3(n+1)} 3^{4n-4(n+1)} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{n+1} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} \tau^{n-3} </math>

<math>~\Rightarrow ~~~\biggl[ \frac{P_\mathrm{e}}{P_\mathrm{norm}}\biggr]_\mathrm{crit}^{n-3}</math>

<math>~=</math>

<math>~\biggl[ \frac{(n+1)^3}{4\pi}\biggr]^{n+1} (-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} {\tilde\theta}^{(n+1)(n-3)} \, . </math>

Q.E.D.


And (3),

<math>~\biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^{2(n+1)}_\mathrm{crit}</math>

<math>~=</math>

<math>~ \biggl[ \frac{3^2\cdot 5}{4\pi \tilde{\mathfrak{f}}_W}\biggr]^{n+1} \biggl[ \frac{n-3}{n} \biggr]^{(n-1) } \biggl[\frac{ \tilde{\mathfrak{f}}_A}{4} \biggr]^{2n } </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{3^2\cdot 5}{4\pi }\biggr]^{n+1} \biggl[ \frac{n-3}{n} \biggr]^{(n-1) } \biggl[ \frac{ n\tau}{n-3} \biggr]^{2n } \biggl[ \frac{(n-3){\tilde\xi}^2}{3^2\cdot 5 \tau} \biggr]^{n+1} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{1}{4\pi }\biggr]^{n+1} \biggl[ \frac{ n}{n-3} \biggr]^{n+1 } \biggl[ (n-3){\tilde\xi}^2\biggr]^{n+1} \tau^{n-1} </math>

<math>~\Rightarrow ~~~ \biggl[ \frac{R_\mathrm{eq}}{R_\mathrm{SWS}} \biggr]^{2}_\mathrm{crit}</math>

<math>~=</math>

<math>~ \biggl( \frac{n}{4\pi }\biggr) {\tilde\xi}^2 {\tilde\theta}^{n-1} </math>

Q.E.D.

And (4),

<math>~\biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2(n+1)}</math>

<math>~=</math>

<math>~\biggl[ \frac{5^3\tilde{\mathfrak{f}}_M^2}{\tilde{\mathfrak{f}}_W^3} \biggr]^{n+1} \biggl(\frac{3}{4\pi} \biggr)^{4} \biggl[\frac{ (n-3)}{4\pi n}\biggr]^{(n-3)} \biggl(\frac{3\tilde{\mathfrak{f}}_A }{4} \biggr)^{4n} </math>

 

<math>~=</math>

<math>~\biggl\{ \biggl[ \frac{(n-3)^3}{3^4\tau^3} \biggr] (-{\tilde\xi}^2 {\tilde\theta}^' )^2 \biggr\}^{n+1} 3^4(4\pi)^{-(n+1)} \biggl(\frac{ n-3}{n}\biggr)^{(n-3)} \biggl[\frac{3n\tau }{n-3} \biggr]^{4n} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{n^3}{4\pi}\biggr]^{n+1}(-{\tilde\xi}^2 {\tilde\theta}^' )^{2(n+1)} \tau^{n-3} </math>

<math>~\Rightarrow~~~ \biggl[ \frac{M_\mathrm{tot}}{M_\mathrm{SWS}}\biggr]_\mathrm{crit}^{2} </math>

<math>~=</math>

<math>~ \biggl[ \frac{n^3}{4\pi}\biggr](-{\tilde\xi}^2 {\tilde\theta}^' )^{2} {\tilde\theta}^{n-3} </math>

Q.E.D.

Free-Energy of Bipolytropes

In this case, the Gibbs-like free energy is given by the sum of four separate energies,

<math>~\mathfrak{G}</math>

<math>~=</math>

<math>~ \biggl[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{core} + \biggl[W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm}\biggr]_\mathrm{env} \, . </math>

In addition to specifying (generally) separate polytropic indexes for the core, <math>~n_c</math>, and envelope, <math>~n_e</math>, and an envelope-to-core mean molecular weight ratio, <math>~\mu_e/\mu_c</math>, we will assume that the system is fully defined via specification of the following five physical parameters:

  • Total mass, <math>~M_\mathrm{tot}</math>;
  • Total radius, <math>~R</math>;
  • Interface radius, <math>~R_i</math>, and associated dimensionless interface marker, <math>~q \equiv R_i/R</math>;
  • Core mass, <math>~M_c</math>, and associated dimensionless mass fraction, <math>~\nu \equiv M_c/M_\mathrm{tot}</math>;
  • Polytropic constant in the core, <math>~K_c</math>.

In general, the warped free-energy surface drapes across a five-dimensional parameter "plane" such that,

<math>~\mathfrak{G}</math>

<math>~=</math>

<math>~\mathfrak{G}(R, K_c, M_\mathrm{tot}, q, \nu) \, .</math>

Order of Magnitude Derivation

Let's begin by providing very rough, approximate expressions for each of these four terms, assuming that <math>~n_c = 5</math> and <math>~n_e = 1</math>.

<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>

<math>~\approx</math>

<math>~- \mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot} M_c}{(R_i/2)} \biggr] = - 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr] \, ;</math>

<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>

<math>~\approx</math>

<math>~- \mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot} M_e}{(R_i+R)/2} \biggr] = - 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ;</math>

<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{core} = U_\mathrm{int}\biggr|_\mathrm{core} </math>

<math>~\approx</math>

<math>~\mathfrak{b}_c \cdot n_cK_c M_c ({\bar\rho}_c)^{1/n_c} = 5\mathfrak{b}_c \cdot K_c M_\mathrm{tot}\nu \biggl[ \frac{3M_c}{4\pi R_i^3} \biggr]^{1/5} </math>

 

<math>~=</math>

<math>~\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5} \, ;</math>

<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>

<math>~\approx</math>

<math>~\mathfrak{b}_e \cdot n_eK_e M_\mathrm{env} ({\bar\rho}_e)^{1/n_e} = \mathfrak{b}_e \cdot K_e M_\mathrm{tot}(1-\nu) \biggl[ \frac{3M_\mathrm{env}}{4\pi (R^3-R_i^3)} \biggr] </math>

 

<math>~=</math>

<math>~ \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) K_e [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1} \, . </math>

In writing this last expression, it has been necessary to (temporarily) introduce a sixth physical parameter, namely, the polytropic constant that characterizes the envelope material, <math>~K_e</math>. But this constant can be expressed in terms of <math>~K_c</math> via a relation that ensures continuity of pressure across the interface while taking into account the drop in mean molecular weight across the interface, that is,

<math>~K_e ({\bar\rho}_e)^{(n_e+1)/n_e}</math>

<math>~\approx</math>

<math>~K_c ({\bar\rho}_c)^{(n_c+1)/n_c}</math>

<math>~\Rightarrow ~~~~ K_e \biggl[\biggl( \frac{\mu_e}{\mu_c} \biggr) {\bar\rho}_c\biggr]^{2}</math>

<math>~\approx</math>

<math>~K_c ({\bar\rho}_c)^{6/5}</math>

<math>~\Rightarrow ~~~~ \frac{K_e}{K_c} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2}</math>

<math>~\approx</math>

<math>~\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} \, .</math>

Hence, the fourth energy term may be rewritten in the form,

<math>~\mathfrak{S}_\mathrm{therm}\biggr|_\mathrm{env} = U_\mathrm{int}\biggr|_\mathrm{env} </math>

<math>~\approx</math>

<math>~ \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr) \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} K_c\biggl[ \frac{3M_\mathrm{tot}\nu}{4\pi (Rq)^3} \biggr]^{-4/5} [M_\mathrm{tot}(1-\nu)]^2 [R^3(1-q^3)]^{-1} </math>

 

<math>~=</math>

<math>~ \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)} \, . </math>

Putting all the terms together gives,

<math>~\mathfrak{G}</math>

<math>~\approx</math>

<math>~ - 2\mathfrak{a}_c \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{\nu}{q}\biggr) \biggr] - 2\mathfrak{a}_e \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] + \mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} K_c (M_\mathrm{tot}\nu)^{6/5} (Rq)^{-3/5} </math>

 

 

<math>~ + \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} K_c M_\mathrm{tot}^{6/5}R^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)} </math>

 

<math>~=</math>

<math>~ - 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr] + \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{(\nu M_\mathrm{tot})^{2}}{ qR} \biggr]^{3/5} </math>

<math>~\Rightarrow ~~~~ \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>

<math>~=</math>

<math>~ - 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{GM_\mathrm{tot}^2 }{R} \biggr] \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} + \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} K_c \biggl[\frac{M_\mathrm{tot}^{2}}{ R} \biggr]^{3/5}\biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} </math>

 

<math>~=</math>

<math>~ - 2 \mathcal{A}_\mathrm{biP} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr] + \mathcal{B}_\mathrm{biP} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl[\frac{R_\mathrm{norm}}{ R} \biggr]^{3/5} \, , </math>

where,

<math>~\mathcal{A}_\mathrm{biP}</math>

<math>~\equiv</math>

<math>~\biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr) + \mathfrak{a}_e \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] \, ,</math>

<math>~\mathcal{B}_\mathrm{biP}</math>

<math>~\equiv</math>

<math>~\biggl( \frac{3}{2^2\pi} \biggr)^{1/5} \biggl[5\mathfrak{b}_c + \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr] \, .</math>

Equilibrium Radius

Order of Magnitude Estimate

This means that,

<math>~\frac{\partial\mathfrak{G}}{\partial R}</math>

<math>~=</math>

<math>~ + 2 \mathcal{A}_\mathrm{biP}\biggl[ \frac{GM_\mathrm{tot}^2 }{R^2} \biggr] - \frac{3}{5} \mathcal{B}_\mathrm{biP} K_c \biggl[\frac{\nu^{2}}{ q} \biggr]^{3/5} M_\mathrm{tot}^{6/5} R^{-8/5} \, . </math>

Hence, because equilibrium radii are identified by setting <math>~\partial\mathfrak{G}/\partial R = 0</math>, we have,

<math>~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>

<math>~=</math>

<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl[\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}\biggr]^{5/2} \biggl(\frac{ q} {\nu^{2}}\biggr)^{3/2} \, . </math>

Reconcile With Known Analytic Expression

From our earlier derivations, it appears as though,

<math>~\chi_\mathrm{eq} \equiv \frac{R_\mathrm{eq}}{R_\mathrm{norm}}</math>

<math>~=</math>

<math>~\biggl( \frac{3^8}{2^5\pi} \biggr)^{-1/2} \biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3} </math>

 

<math>~=</math>

<math>~\biggl(\frac{2\cdot 5}{3}\biggr)^{5/2} \biggl(\frac{q}{\nu^2} \biggr)^{3/2} \biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2} \biggl(\frac{\nu^2}{q} \biggr)^{5/2} \frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr] \, . </math>

This implies that,

<math>~\frac{\mathcal{A}_\mathrm{biP} }{\mathcal{B}_\mathrm{biP}}</math>

<math>~\approx</math>

<math>~ \biggl[\biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/2} \biggl(\frac{\nu^2}{q} \biggr)^{5/2} \frac{(1 + \ell_i^2)^3}{\ell_i^5} \biggr]^{2/5} </math>

 

<math>~=</math>

<math>~\biggl(\frac{\nu^2}{q} \biggr) \biggl( \frac{\pi}{2^8 \cdot 3 \cdot 5^5} \biggr)^{1/5} \frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} </math>

<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c\biggl(\frac{\nu}{q}\biggr) + \mathfrak{a}_e \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] </math>

<math>~\approx</math>

<math>~\frac{1}{2^2\cdot 5}\biggl(\frac{\nu^2}{q} \biggr) \frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c + \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr] </math>

<math>~\Rightarrow ~~~~ \biggl[ \mathfrak{a}_c + \mathfrak{a}_e \cdot \frac{q(1-\nu)}{\nu(1+q)} \biggr] </math>

<math>~\approx</math>

<math>~\frac{\nu}{2^2\cdot 5} \frac{(1 + \ell_i^2)^{6/5}}{\ell_i^2} \biggl[5\mathfrak{b}_c + \mathfrak{b}_e \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \frac{q^3(1-\nu)^2}{\nu^2(1-q^3)} \biggr] </math>

Focus on Five-One Free-Energy Expression

Approximate Expressions

Let's plug this equilibrium radius back into each term of the free-energy expression.

<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{core}</math>

<math>~\approx</math>

<math>~- 2\mathfrak{a}_c \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{\nu}{q}\biggr) \biggr] </math>

 

<math>~=</math>

<math>~- 2\mathfrak{a}_c \biggl(\frac{\nu}{q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>

<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}}\biggr|_\mathrm{env}</math>

<math>~\approx</math>

<math>~- 2\mathfrak{a}_e \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} \biggl[ \frac{GM_\mathrm{tot}^2 }{R_\mathrm{eq}} \biggl(\frac{1-\nu}{1+q}\biggr) \biggr] </math>

 

<math>~=</math>

<math>~- 2\mathfrak{a}_e \biggl(\frac{1-\nu}{1+q}\biggr) \biggl[ \frac{R_\mathrm{norm} }{R_\mathrm{eq}} \biggr] \, ;</math>

<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_c-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{core} </math>

<math>~\approx</math>

<math>~\biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} K_c (M_\mathrm{tot}\nu)^{6/5} (R_\mathrm{eq}q)^{-3/5} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{3}{2\cdot 5}\biggr]\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5} \, ;</math>

<math>~\frac{S_\mathrm{env}}{E_\mathrm{norm}} = \biggl[\frac{3(\gamma_e-1)}{2}\biggr] \frac{U_\mathrm{int}}{E_\mathrm{norm}}\biggr|_\mathrm{env} </math>

<math>~\approx</math>

<math>~\biggl[\frac{3}{2}\biggr] \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl(\frac{G^3}{K_c^5}\biggr)^{1/2} K_c M_\mathrm{tot}^{6/5}R_\mathrm{eq}^{-3/5}\biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)} </math>

 

<math>~=</math>

<math>~\biggl[\frac{3}{2}\biggr] \mathfrak{b}_e \biggl( \frac{3}{2^2\pi } \biggr)^{1/5} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-2} \biggl[ \frac{q^3}{\nu} \biggr]^{4/5} \frac{(1-\nu)^2}{(1-q^3)} \biggl(\frac{R_\mathrm{norm}}{R_\mathrm{eq}}\biggr)^{3/5} \, . </math>

From Detailed Force-Balance Models

In the following derivations, we will use the expression,

<math>~\chi_\mathrm{eq} \equiv \frac{ R_\mathrm{eq}}{R_\mathrm{norm}}</math>

<math>~=</math>

<math>~\biggl( \frac{\mu_e}{\mu_c} \biggr)^3 \biggl( \frac{\pi}{2^3} \biggr)^{1/2} \frac{1}{A^2\eta_s} = \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>

Keep in mind, as well — as derived in an accompanying discussion — that,

<math>~\nu \equiv \frac{M_\mathrm{core}}{M_\mathrm{tot}} </math>

<math>~=</math>

<math>~ (m_3^2 \ell_i^3) (1 + \ell_i^2)^{-1/2} [1 + (1-m_3)^2 \ell_i^2]^{-1/2} \biggl[ m_3\ell_i + (1+\ell_i^2) \biggl(\frac{\pi}{2} + \tan^{-1} \Lambda_i \biggr) \biggr]^{-1} \, ,</math>

where,

<math>m_3 \equiv 3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \, .</math>

From the accompanying Table 1 parameter values, we also can write,

<math>~q</math>

<math>~=</math>

<math>~\frac{\eta_i}{\eta_s} = \eta_i \biggl\{\frac{\pi}{2} + \eta_i + \tan^{-1}\biggl[ \frac{1}{\eta_i} - \ell_i \biggr] \biggr\}^{-1}</math>

 

<math>~=</math>

<math>~ \eta_i \biggl\{\eta_i + \cot^{-1}\biggl[ \ell_i - \frac{1}{\eta_i} \biggr] \biggr\}^{-1} \, , </math>

where,

<math>~\eta_i</math>

<math>~=</math>

<math>~m_3 \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>

Let's also define the following shorthand notation:

<math>~\mathfrak{L}_i</math>

<math>~\equiv</math>

<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ;</math>

<math>~\mathfrak{K}_i</math>

<math>~\equiv</math>

<math>~\frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] + \frac{\Lambda_i}{\eta_i} \, .</math>


Gravitational Potential Energy of the Core

Pulling from our detailed derivations,

<math>~\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] \, .</math>

<math>~\Rightarrow ~~~~ -\chi_\mathrm{eq} \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] \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>

 

<math>~=~</math>

<math>~ \biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] </math>

Out of equilibrium, then, we should expect,

<math>~\frac{W_\mathrm{core}}{E_\mathrm{norm}} </math>

<math>~=~</math>

<math>~ - \chi^{-1} \biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^5} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] </math>

 

<math>~=~</math>

<math>~ - \chi^{-1} \biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2} \biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr] \, , </math>

which, in comparison with our above approximate expression, implies,

<math>~\mathfrak{a}_c </math>

<math>~=~</math>

<math>~ \biggl( \frac{3}{2^5} \biggr) \frac{\nu}{\ell_i^5} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) + (1 + \ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] \, . </math>

Thermal Energy of the Core

Again, pulling from our detailed derivations,

<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] </math>

<math>~\Rightarrow ~~~~ \chi_\mathrm{eq}^{3} \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]^5_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \frac{1}{2^5} \biggl( \frac{3^8}{2^5\pi} \biggr)^{5/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr]^5 \biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3} </math>

 

<math>~=~</math>

<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11} \biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr]^5 \biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr] \, . </math>

Out of equilibrium, we should then expect,

<math>~\frac{S_\mathrm{core}}{E_\mathrm{norm}}</math>

<math>~=~</math>

<math>~ \biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1} \biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i \, . </math>

In comparison with our above approximate expression, we therefore have,

<math>~ \biggl[ \biggl(\frac{3}{2\cdot 5}\biggr)\mathfrak{b}_c \biggl( \frac{3\cdot 5^5}{2^2\pi} \biggr)^{1/5} \biggl(\frac{\nu^2}{q}\biggr)^{3/5} \biggr]^5</math>

<math>~=~</math>

<math>~ \frac{1}{\pi}\biggl(\frac{3}{2^{2}}\biggr)^{11} \biggl(\frac{\nu^2}{q}\biggr)^{3} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr]^5 \biggl[\frac{(1+\ell_i^2)^9}{\ell_i^{15}}\biggr] </math>

<math>~\Rightarrow~~~~ \mathfrak{b}_c </math>

<math>~=~</math>

<math>~\frac{ 3 }{2^3\ell_i^{3}(1+\ell_i^2)^{6/5}} \biggl[ \ell_i (\ell_i^4 - 1 ) + (1+\ell_i^2)^{3}\tan^{-1}(\ell_i) \biggr] \, . </math>


Gravitational Potential Energy of the Envelope

Again, pulling from our detailed derivations and appreciating, in particular, that (see, for example, our notes on equilibrium conditions),

<math>~A</math>

<math>~=~</math>

<math>~\frac{\eta_i}{\sin(\eta_i - B)} \, ,</math>

<math>~(\eta_s - B)</math>

<math>~=~</math>

<math>~\pi \, ,</math>

<math>~\eta_i - B</math>

<math>~=~</math>

<math>~\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\, ,</math>

<math>~\Rightarrow ~~~ \sin(\eta_i -B) = (1+\Lambda_i^2)^{-1/2}</math>

     and    

<math>~\sin[2(\eta_i-B)] = 2\Lambda_i(1 + \Lambda_i^2)^{-1} \ ,</math>

we have,

<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2 \biggl\{ \biggl[6(\eta_s-B) - 3\sin[2(\eta_s - B)] -4\eta_s\sin^2(\eta_s-B) + 4B\biggr] </math>

 

 

<math>~ - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) + 4B \biggr]\biggr\} </math>

 

<math>~=~</math>

<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2 \biggl\{ 6\pi - \biggl[6(\eta_i-B) - 3\sin[2(\eta_i - B)] -4\eta_i\sin^2(\eta_i-B) \biggr]\biggr\} </math>

 

<math>~=~</math>

<math>~ -\biggl( \frac{1}{2^3\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2(1+\Lambda_i^2) \biggl\{ 6\pi - 6\biggl[\frac{\pi}{2} - \tan^{-1}(\Lambda_i)\biggr] + 6\biggl[ \frac{\Lambda_i}{(1 + \Lambda_i^2)} \biggr] + 4\eta_i \biggl[ \frac{1}{(1+\Lambda_i^2)} \biggr] \biggr\} </math>

 

<math>~=~</math>

<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\} \, . </math>

So, in equilibrium we can write,

<math>~-\chi_\mathrm{eq}\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\} \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>

 

<math>~=~</math>

<math>~ \frac{3}{2^2} \biggl(\frac{\eta_i}{m_3}\biggr)^3 \biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{\ell_i^5} </math>

 

<math>~=~</math>

<math>~ \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2} \biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, . </math>

And out of equilibrium,

<math>~\frac{W_\mathrm{env}}{E_\mathrm{norm}}</math>

<math>~=~</math>

<math>~ -\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2} \biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr] \, . </math>

This, in turn, implies that both in and out of equilibrium,

<math>~\mathfrak{a}_e </math>

<math>~=~</math>

<math>~ \frac{3}{2^3} \biggl[\frac{\nu^2(1+q)}{q(1-\nu)} \biggr] \frac{1}{\ell_i^2} \biggl\{ \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \frac{\Lambda_i}{\eta_i} + \frac{2}{3} \biggr\} \, . </math>

Thermal Energy of the Envelope

Again, pulling from our detailed derivations,

<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2 \biggl\{ \biggl[6(\eta_s - B) - 3\sin[2(\eta_s-B)] \biggr] - \biggl[6(\eta_i - B) - 3\sin[2(\eta_i-B)] \biggr] \biggr\}</math>

 

<math>~=~</math>

<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[\frac{\eta_i}{\sin(\eta_i - B)} \biggr]^2 \biggl\{ 6\pi - 6(\eta_i - B) + 3\sin[2(\eta_i-B)] \biggr\}</math>

 

<math>~=~</math>

<math>~ ~ \biggl( \frac{1}{2^5\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 (1 + \Lambda_i^2) \biggl\{ 6\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + 6\biggl[\Lambda_i(1 + \Lambda_i^2)^{-1} \biggr] \biggr\}</math>

 

<math>~=~</math>

<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} \, .</math>

So, in equilibrium we can write,

<math>~\chi_\mathrm{eq}^{3}\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} \biggl[\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{(1+\ell_i^2)^3}{3^3\ell_i^5}\biggr]^{3} </math>

 

<math>~=~</math>

<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{3^2\pi^2}{2^{12}} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3 \biggl\{ \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\} \biggl[\frac{(1+\ell_i^2)^9}{3^9\ell_i^{15}}\biggr] </math>

 

<math>~=~</math>

<math>~ ~\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) \biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr] \biggl\{ \frac{(1 + \Lambda_i^2)}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \frac{\Lambda_i}{\eta_i} \biggr\} \, . </math>

And, out of equilibrium,

<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ ~ \chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) \biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K} \, . </math>

Combined in Equilibrium

Notice that, in combination,

<math>~\biggl[\frac{2S_\mathrm{env} + W_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=</math>

<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^3 </math>

 

<math>~=</math>

<math>~ - \frac{2}{3}\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \biggl[3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i \biggl( 1 + \ell_i^2 \biggr)^{-1}\biggr]^3 </math>

 

<math>~=</math>

<math>~ - \biggl( \frac{2\cdot 3^6}{\pi} \biggr)^{1/2} \biggl[\frac{\ell_i^3}{( 1 + \ell_i^2)^3}\biggr] \, . </math>

Also, from above,

<math>~\biggl[ \frac{2S_\mathrm{core}+W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \ell_i \biggl(- \frac{8}{3} \ell_i^2 \biggr) (1 + \ell_i^2)^{-3} \biggr] </math>

 

<math>~=~</math>

<math>~ + \biggl( \frac{2\cdot 3^6}{\pi } \biggr)^{1/2} \biggl[ \frac{\ell_i^3}{(1 + \ell_i^2)^{3}} \biggr] \, .</math>

So, in equilibrium, these terms from the core and envelope sum to zero, as they should.

Out of Equilibrium

And now, in combination out of equilibrium,

<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>

<math>~=</math>

<math>~ \biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} + \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\} +\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{2n_c}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} +\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2n_e}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \, . </math>

Hence, quite generally out of equilibrium,

<math>~\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] </math>

<math>~=</math>

<math>~ -\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-1} \biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} + \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}\biggr\} -\frac{3}{5}\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3/5} \biggl(\frac{10}{3}\biggr) \biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} -3\chi^{-1}\biggl(\frac{\chi}{\chi_\mathrm{eq}}\biggr)^{-3} \biggl(\frac{2}{3}\biggr)\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \, . </math>

Let's see what the value of this derivative is if the dimensionless radius, <math>~\chi</math>, is set to the value that has been determined, via a detailed force-balanced analysis, to be the equilibrium radius, namely, <math>~\chi = \chi_\mathrm{eq}</math>. In this case, we have,

<math>~\biggl\{\frac{\partial}{\partial \chi} \biggl[ \frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr] \biggr\}_\mathrm{\chi \rightarrow \chi_\mathrm{eq}}</math>

<math>~=</math>

<math>~ -\chi_\mathrm{eq}^{-1}\biggl\{ \biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} + \biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} +2\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} +2\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq} \biggr\} \, . </math>

But, according to the virial theorem — and, as we have just demonstrated — the four terms inside the curly braces sum to zero. So this demonstrates that the derivative of our out-of-equilibrium free-energy expression does go to zero at the equilibrium radius, as it should!


Summary51

In summary, the desired out of equilibrium free-energy expression is,

<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>

<math>~=</math>

<math>~ \frac{W_\mathrm{core}}{E_\mathrm{norm}} + \frac{W_\mathrm{env}}{E_\mathrm{norm}} +\biggl(\frac{2n_c}{3}\biggr)\frac{S_\mathrm{core}}{E_\mathrm{norm}} +\biggl(\frac{2n_e}{3}\biggr)\frac{S_\mathrm{env}}{E_\mathrm{norm}} </math>

 

<math>~=</math>

<math>~ - \chi^{-1} \biggl( \frac{3}{2^4} \biggr) \frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2} \biggl[ \mathfrak{L}_i - \frac{8}{3} \biggr] -\chi^{-1}\cdot \frac{3}{2^2} \biggl(\frac{\nu^2}{q} \biggr) \frac{1}{\ell_i^2} \biggl[\mathfrak{K}_i+ \frac{2}{3} \biggr] </math>

 

 

<math>~ + \biggl(\frac{2\cdot 5}{3}\biggr) \biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl[ \chi^{-1} \biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \biggl(\frac{3}{2^{2}}\biggr)^{2}\mathfrak{L}_i +\biggl(\frac{2}{3}\biggr) \chi^{-3}\biggl(\frac{\nu^2}{q} \biggr)^3 \biggl( \frac{\pi}{2^{6}\cdot 3^5} \biggr) \biggl[\frac{(1+\ell_i^2)^6}{\ell_i^{12}}\biggr]\mathfrak{K} </math>

 

<math>~=</math>

<math>~ - \biggl( \frac{3}{2^4} \biggr) \biggl[\chi^{-1}\frac{\nu^2}{q} \cdot \frac{1}{\ell_i^2}\biggr] \biggl[ \mathfrak{L}_i + 4\mathfrak{K}_i \biggr] + \biggl(\frac{3}{2^2\pi} \biggr)^{1/5}\biggl(\frac{3\cdot 5}{2^3}\biggr) \biggl[ \chi^{-1} \biggl(\frac{\nu^2}{q}\biggr) \frac{1}{(1+\ell_i^2)^{2}} \biggr]^{3/5} \mathfrak{L}_i </math>

 

 

<math>~ + \biggl( \frac{\pi}{2^{5}\cdot 3^6} \biggr) \biggl[\chi^{-1}\biggl(\frac{\nu^2}{q} \biggr) \frac{(1+\ell_i^2)^2}{\ell_i^{4}}\biggr]^3\mathfrak{K} \, . </math>

Or, in terms of the ratio,

<math>\Chi \equiv \frac{\chi}{\chi_\mathrm{eq}} \, ,</math>

and pulling from the above expressions,

<math>~\biggl[ \frac{W_\mathrm{core}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \ell_i \biggl(\ell_i^4 - \frac{8}{3} \ell_i^2 -1 \biggr) (1 + \ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] </math>

 

<math>~=~</math>

<math>~ - \biggl( \frac{3^8}{2^5\pi } \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3} \biggl[ \mathfrak{L}_i - \frac{8}{3}\biggr] </math>

<math>~\biggl[\frac{W_\mathrm{env}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ -\biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1+\Lambda_i^2)\biggl[\frac{\pi}{2}+\tan^{-1}(\Lambda_i)\biggr] + \Lambda_i + \frac{2}{3} \cdot \eta_i \biggr\} </math>

 

<math>~=~</math>

<math>~ -\biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3} \biggl[4\mathfrak{K}_i + \frac{8}{3} \biggr] </math>

<math>~\biggl[ \frac{S_\mathrm{core}}{E_\mathrm{norm}}\biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \ell_i (\ell_i^4 - 1 )(1+\ell_i^2)^{-3} + \tan^{-1}(\ell_i) \biggr] </math>

 

<math>~=~</math>

<math>~ \frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3}\mathfrak{L}_i </math>

<math>~\biggl[\frac{S_\mathrm{env}}{E_\mathrm{norm}} \biggr]_\mathrm{eq}</math>

<math>~=~</math>

<math>~ ~\frac{1}{2} \biggl( \frac{3^2}{2\pi} \biggr)^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} \eta_i^2 \biggl\{ (1 + \Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}(\Lambda_i) \biggr] + \Lambda_i \biggr\} </math>

 

<math>~=~</math>

<math>~ ~\frac{1}{2} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]^{3} (4\mathfrak{K}_i) \, , </math>

we have the streamlined,

<math>~\biggl( \frac{2^5\pi}{3^6} \biggr)^{1/2} \biggl[ \frac{(1+\ell_i^2)}{\ell_i} \biggr]^{3} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>

<math>~=</math>

<math>~ +\Chi^{-3/5} (5 \mathfrak{L}_i) +\Chi^{-3} (4\mathfrak{K}_i) -\Chi^{-1} (3\mathfrak{L}_i +12\mathfrak{K}_i ) </math>

or, better yet,

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with <math>~(n_c, n_e) = (5, 1)</math>

<math>~2^4\biggl( \frac{q\ell_i^2}{\nu^2}\biggr) \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>

<math>~=</math>

<math>~ \Chi^{-3/5} (5 \mathfrak{L}_i) +\Chi^{-3} (4\mathfrak{K}_i) -\Chi^{-1} (3\mathfrak{L}_i +12\mathfrak{K}_i ) </math>


where,

<math>~\mathfrak{L}_i</math>

<math>~\equiv</math>

<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>

<math>~\mathfrak{K}_i</math>

<math>~\equiv</math>

<math>~\frac{\Lambda_i}{\eta_i} + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>

<math>~\Lambda_i</math>

<math>~\equiv</math>

<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>

<math>~\eta_i</math>

<math>~=</math>

<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>

From the accompanying Table 1 parameter values, we also can write,

<math>~\frac{1}{q}</math>

<math>~=</math>

<math>~\frac{\eta_s}{\eta_i} = 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>

<math>~\nu</math>

<math>~=</math>

<math>~ \frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, . </math>

Radial Derivatives

<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>

<math>~=</math>

<math>~ -\Chi^{-8/5} (3 \mathfrak{L}_i) -\Chi^{-4} (12\mathfrak{K}_i) +\Chi^{-2} (3\mathfrak{L}_i +12\mathfrak{K}_i ) </math>

<math>~\frac{\partial^2 \mathfrak{G}^*}{\partial \Chi^2}</math>

<math>~=</math>

<math>~\frac{3}{5}\biggl[ \Chi^{-13/5} (8\mathfrak{L}_i) +\Chi^{-5} (80\mathfrak{K}_i) -\Chi^{-1} (10\mathfrak{L}_i +40\mathfrak{K}_i )\biggr] </math>

Consistent with our generic discussion of the stability of bipolytropes and the specific discussion of the stability of bipolytropes having <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,

<math>~\chi = \chi_\mathrm{eq}</math>

<math>~=</math>

<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>

Furthermore, the equilibrium configuration is unstable whenever <math>~\partial^2 \mathfrak{G}/\partial \chi^2 < 0</math>, that is, it is unstable whenever,

<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>

<math>~></math>

<math>~20 \, .</math>

Table 1 of an accompanying chapter — and the red-dashed curve in the figure adjacent to that table — identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.

Focus on Zero-Zero Free-Energy Expression

Here, we will draw heavily from the following accompanying chapters:


From Detailed Force-Balance Models

Equilibrium Radius

First View

In an accompanying chapter we find,

<math>~ \frac{P_0 R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>

<math>~=</math>

<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] </math>

where,

<math>~f</math>

<math>~\equiv</math>

<math> 1+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \, , </math>

<math>~\mathfrak{F} </math>

<math>~\equiv</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, , </math>

<math>~\frac{\rho_e}{\rho_c} </math>

<math>~=</math>

<math>~ \frac{q^3(1-\nu)}{\nu(1-q^3)} \, . </math>

Here, we prefer to normalize the equilibrium radius to <math>~R_\mathrm{norm}</math>. So, let's replace the central pressure with its expression in terms of <math>~K_c</math>. Specifically,

<math>~P_0</math>

<math>~=</math>

<math>~ K_c \rho_c^{\gamma_c} = K_c \biggl[ \frac{3M_\mathrm{core}}{4\pi R_i^3} \biggr]^{\gamma_c} = K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c} ~~~\Rightarrow~~~ \frac{P_0}{P_\mathrm{norm}} = \biggl[ \frac{3}{4\pi}\biggl(\frac{\nu}{q^3}\biggr) \frac{1}{\chi_\mathrm{eq}^3}\biggr]^{(n_c+1)/n_c} </math>

<math>~\Rightarrow~~~K_c \biggl[ \frac{3\nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{eq}^3} \biggr]^{(n_c+1)/n_c} \frac{R_\mathrm{eq}^4}{G M_\mathrm{tot}^2 } </math>

<math>~=</math>

<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] </math>

<math>~\Rightarrow~~~R_\mathrm{eq}^{(n_c-3)/n_c} </math>

<math>~=</math>

<math>~ \biggl(\frac{G}{K_c}\biggr) M_\mathrm{tot}^{(n_c-1)/n_c} \biggl[ \frac{3\nu }{4\pi q^3 } \biggr]^{-(n_c+1)/n_c} \biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] </math>

<math>~\Rightarrow~~~\chi_\mathrm{eq}^{(n_c-3)/n_c} \equiv \biggl[\frac{R_\mathrm{eq}}{R_\mathrm{norm}}\biggr]^{(n_c-3)/n_c}</math>

<math>~=</math>

<math>~ \frac{1}{2}\biggl(\frac{4\pi}{3} \biggr)^{1/n_c} \biggl( \frac{\nu}{q^3}\biggr)^{(n_c-1)/n_c} \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, . </math>

Or, in terms of <math>~\gamma_c</math>,

<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>

<math>~=</math>

<math>~ \frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c} \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \, . </math>

Second View

Alternatively, from our derivation and discussion of analytic detailed force-balance models,

<math> \biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr] P_0</math>

  <math>~=</math> 

<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} \, ,</math>

where,

<math>~[g(\nu,q)]^2</math>

<math>~\equiv</math>

<math> 1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \, . </math>

In order to show that this expression is the same as the other one, above, we need to show that,

<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] </math>

<math>~=</math>

<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} </math>

<math>~\Rightarrow~~~

f - 1-\mathfrak{F} 

</math>

<math>~=</math>

<math>~\frac{5}{2q^3} \biggl[g^2-1\biggr]</math>

 

<math>~=</math>

<math>~\frac{5}{2q^3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>

 

<math>~=</math>

<math>~\frac{5}{2q^5} \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl\{ 2 ( q^2 - q^3 ) + \frac{\rho_e}{\rho_0}\biggl[ 1 - 3q^2+ 2q^3 \biggr] \biggr\} \, .</math>

Let's see …

<math>~

f - 1-\mathfrak{F} 

</math>

<math>~=</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] - \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

 

<math>~=</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) - \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) \biggr] </math>

 

 

<math>~ - \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] </math>

 

<math>~=</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl\{ (q^3- q^5 ) + (2q^2 - 3q^3 + q^5) \biggr\} </math>

 

 

<math>~ + \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 3 (1 -5q^2 + 5q^3 - q^5) \biggr] +\frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 2 - 2q^5 + 5\biggl( q^5-q^3\biggr)\biggr] </math>

 

<math>~=</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3- q^5 ) + (2q^2 - 3q^3 + q^5) \biggr] + \frac{1}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \frac{1}{q^5} \biggl[ 3 (1 -5q^2 + 5q^3 - q^5)+2 - 2q^5 + 5( q^5-q^3) \biggr] </math>

 

<math>~=</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ 2q^2 - 2q^3 \biggr] + \frac{5}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 1 - 3q^2 + 2q^3 \biggr] \, . </math>

Q.E.D.

Hence, the equilibrium radius can also be written as,

<math>~\chi_\mathrm{eq}^{4-3\gamma_c} </math>

<math>~=</math>

<math>~ \frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c} q^2 g^2 \, ; </math>

or, in terms of the polytropic index,

<math>~\chi_\mathrm{eq}^{n_c-3} </math>

<math>~=</math>

<math>~ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \, . </math>

Gravitational Potential Energy

Also from our accompanying discussion, we have,

<math>~\frac{W_\mathrm{grav}}{E_\mathrm{norm}} </math>

<math>~=</math>

<math> - \Chi^{-1} \biggl( \frac{3}{5}\biggr) \biggl(\frac{\nu}{q^3} \biggr)^2 q^5 \biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{-1/(n_c-3)} f(\nu,q) </math>

 

<math>~=</math>

<math> - \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f \biggl[ 2^{n_c-(n_c-3)} \biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} b_\xi^{n_c} \biggr]^{1/(n_c-3)} </math>

 

<math>~=</math>

<math> - \Chi^{-1} \biggl( \frac{6}{5}\biggr) q^5 f \biggl[ \biggl(\frac{6}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, . </math>

Internal Energy Components

First View

Before writing out the expressions for the internal energy of the core and of the envelope, we note from our separate detailed derivation that, in either case,

<math>~\biggl[\frac{P_i \chi^{3\gamma}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma}</math>

<math>~=</math>

<math>~\biggl[\biggl(\frac{P_i }{P_0} \biggr) \biggl(\frac{P_0 }{P_\mathrm{norm}} \biggr)\chi^{3}\biggr]_\mathrm{eq} \biggl[\frac{\chi}{\chi_\mathrm{eq}}\biggr]^{3-3\gamma}</math>

 

<math>~=</math>

<math>~\biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma} \, ,</math>

where, in equilibrium,

<math>~\biggl(\frac{P_i }{P_0} \biggr)_\mathrm{eq}</math>

<math>~=</math>

<math>~1 - b_\xi q^2</math>

<math>~b_\xi q^2</math>

<math>~=</math>

<math>~\biggl\{\frac{2}{5}q^3 f + \biggl[1 - \frac{2}{5} q^3( 1+\mathfrak{F} ) \biggr]\biggr\}^{-1} </math>

 

<math>~=</math>

<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr]^{-1} </math>

So, copying from our accompanying detailed derivation, we have,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>

<math>~=</math>

<math> \frac{4\pi/3 }{({\gamma_c}-1)} \biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_c} \biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\} </math>

 

<math>~=</math>

<math> \frac{1 }{({\gamma_c}-1)} \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_c} q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, , </math>

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> \frac{4\pi/3 }{({\gamma_e}-1)} \biggl\{\biggl(\frac{P_i }{P_0} \biggr) \biggl[ \frac{3}{4\pi } \biggl( \frac{\nu}{q^3} \biggr)\biggr]^{\gamma_c}\chi^{3-3\gamma_c}\biggr\}_\mathrm{eq} \Chi^{3-3\gamma_e} \biggl\{ (1-q^3) + b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} </math>

 

<math>~=</math>

<math> \frac{1}{({\gamma_e}-1)} \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_e} \biggl(\frac{P_i }{P_0} \biggr) \biggl\{ (1-q^3) + b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} </math>

 

<math>~=</math>

<math> \frac{1}{({\gamma_e}-1)} \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_e} \biggl\{ (1-b_\xi q^2)(1-q^3) + b_\xi \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} </math>

 

<math>~=</math>

<math> \frac{1}{({\gamma_e}-1)} \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{3-3\gamma_e} (1-q^3) \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} \, . </math>

Furthermore,

<math>~\biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr]</math>

<math>~=</math>

<math>~ \biggl(\frac{3}{4\pi}\biggr)^{\gamma_c - 1} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c} \biggl\{\chi_\mathrm{eq}^{4-3\gamma_c}\biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)} </math>

 

<math>~=</math>

<math>~ \biggl(\frac{3}{4\pi}\biggr)^{\gamma_c - 1} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c} \biggl\{\frac{1}{2}\biggl(\frac{3}{4\pi} \biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3}\biggr)^{2-\gamma_c} \biggl[ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)} </math>

 

<math>~=</math>

<math>~ \biggl(\frac{3}{4\pi}\biggr)^{(\gamma_c - 1)/(4-3\gamma_c)} \biggl( \frac{\nu}{q^3} \biggr)^{(6-5\gamma_c)(4-3\gamma_c)} \biggl\{\frac{q^2}{2} \biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{(3-3\gamma_c)/(4-3\gamma_c)} </math>

 

<math>~=</math>

<math>~ \biggl(\frac{3}{4\pi}\biggr)^{1/(n_c-3)} \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)(n_c-3)} \biggl\{\frac{q^2}{2} \biggl[ 1 + \frac{2}{5} q^3( f - 1-\mathfrak{F} )\biggr] \biggr\}^{-3/(n_c-3)} </math>

 

<math>~=</math>

<math>~ \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)} \, . </math>

Hence, we have,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>

<math>~=</math>

<math> n_c \biggl[\biggl(\frac{4\pi}{3}\biggr)^{1-\gamma_c} \biggl( \frac{\nu}{q^3} \biggr)^{\gamma_c}\chi_\mathrm{eq}^{3-3\gamma_c}\biggr] \Chi^{-3/n_c} q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] </math>

 

<math>~=</math>

<math> n_c \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)} \Chi^{-3/n_c} q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, , </math>

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>

<math>~=</math>

<math> n_e \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^3\biggr]^{1/(n_c-3)} \Chi^{-3/n_e} (1-q^3) \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} \, . </math>


Second View

In our accompanying discussion of energies associated with detailed force balance models, we used the notation,

<math>~\Pi</math>

<math>~\equiv</math>

<math>~ \biggl(\frac{3}{2^3\pi}\biggr) \frac{GM_\mathrm{tot}^2}{R^4} \biggl(\frac{\nu}{q^3}\biggr)^2 = P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2 \, , </math>

which allows us to rewrite the above quoted relationship between the central pressure and the radius of the bipolytrope as,

<math>~P_0 = \Pi (qg)^2 \, .</math>

We also showed that, in equilibrium, the relationship between the central pressure and the interface pressure is,

<math>~P_0 =P_i + \Pi_\mathrm{eq} q^2 \, .</math>

This means that, in equilibrium, the ratio of the interface pressure to the central pressure is,

<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>

<math>~=</math>

<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0} = 1- \frac{1}{g^2} \, , </math>

or given that (see above),

<math>~

\frac{5}{2q^3} \biggl[g^2-1\biggr]

</math>

<math>~=</math>

<math>~ f - 1-\mathfrak{F} </math>

<math>~\Rightarrow~~~~ g^2 </math>

<math>~=</math>

<math>~ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} ) \, , </math>

we have,

<math>~\biggl(\frac{P_i}{P_0}\biggr)_\mathrm{eq}</math>

<math>~=</math>

<math>~1 - \frac{\Pi_\mathrm{eq} q^2}{P_0} = 1- \biggl[ 1+\frac{2}{5} q^3 ( f - 1-\mathfrak{F} ) \biggr]^{-1} \, . </math>

This is exactly the pressure-ratio expression presented in our "first view" and unveils the notation association,

<math>~b_\xi q^2</math>

<math>~\leftrightarrow~</math>

<math> \frac{1}{g^2} \, . </math>

From our separate derivation, we have, in equilibrium,

<math>~\mathfrak{G}_\mathrm{core} = \biggl(\frac{2n_c}{3}\biggr) S_\mathrm{core}</math>

<math>~=~</math>

<math>\biggl(\frac{2n_c}{3}\biggr) \biggl( \frac{4\pi}{5} \biggr) R_\mathrm{eq}^3 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr)_\mathrm{eq} </math>

 

<math>~=~</math>

<math>\biggl( \frac{ q^5n_c}{5} \biggr) R_\mathrm{eq}^3 \biggl( \frac{2^3\pi}{3} \biggr) \Pi_\mathrm{eq} \biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>

 

<math>~=~</math>

<math>\biggl( \frac{ n_c}{5} \biggr) \biggl[ R_\mathrm{norm}^3 P_\mathrm{norm} \biggr] \chi_\mathrm{eq}^{-1} \biggl(\frac{\nu^2}{q}\biggr) \biggl[\frac{5}{2q^2} \biggl( \frac{P_i}{P_0} \biggr)_\mathrm{eq}\biggl( \frac{P_0}{\Pi} \biggr)_\mathrm{eq} + 1 \biggr] </math>

<math>~\Rightarrow ~~~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>

<math>~=~</math>

<math>~\biggl( \frac{ n_c}{5} \biggr) \biggl(\frac{\nu^2}{q}\biggr) \biggl[\frac{5}{2q^2} \biggl( 1-\frac{1}{g^2} \biggr)\biggl( q^2g^2\biggr) + 1 \biggr] \chi_\mathrm{eq}^{-1} </math>

 

<math>~=~</math>

<math>~\biggl( \frac{ n_c}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr) \biggl[ g^2-\frac{3}{5} \biggr] \biggl\{\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr\}^{-1/(n_c-3)}</math>

 

<math>~=~</math>

<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2} \biggr] \biggl( \frac{ 1}{2} \biggr) \biggl(\frac{\nu^2}{q}\biggr) g^2 \biggl\{2^{n_c}\biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} (q g)^{-2n_c} \biggr\}^{1/(n_c-3)}</math>

 

<math>~=~</math>

<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2} \biggr] \biggl\{2^{n_c}\cdot 2^{(3-n_c)}\biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{1-n_c} \biggl(\frac{\nu}{q^3}\biggr)^{2(n_c-3)} q^{5(n_c-3)} q^{-2n_c} g^{-2n_c} g^{2(n_c-3)} \biggr\}^{1/(n_c-3)}</math>

 

<math>~=~</math>

<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) \frac{1}{g^2} \biggr] \biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15} g^{-6} \biggr\}^{1/(n_c-3)} \, .</math>

Finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,

<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{core} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>

<math>~=~</math>

<math>~n_c \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2 \biggr] \biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{3n_c-15} b_\xi^3 q^{6} \biggr\}^{1/(n_c-3)} </math>

 

<math>~=~</math>

<math>~n_c q^3 \biggl[ 1- \biggl(\frac{3}{5}\biggr) b_\xi q^2 \biggr] \biggl\{\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^3 \biggr\}^{1/(n_c-3)} \, ,</math>

which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression. Also from our previous derivation, we can write,

<math>~\mathfrak{G}_\mathrm{env} = \biggl(\frac{2n_e}{3}\biggr) S_\mathrm{env}</math>

<math>~=~</math>

<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr) R_\mathrm{eq}^3 \Pi_\mathrm{eq} \biggl\{ (1-q^3) \biggl(\frac{P_i }{\Pi}\biggr)_\mathrm{eq} + \biggl( \frac{\rho_e}{\rho_0} \biggr)\biggl[ (-2q^2 + 3q^3 - q^5 )

 + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_0} \biggr) ( -1 + 5q^2 -5q^3 +  q^5 )\biggr]\biggr\} 

</math>

 

<math>~=~</math>

<math>~ 2\pi\biggl(\frac{2n_e}{3}\biggr) R_\mathrm{eq}^3 \biggl[ P_\mathrm{norm} \chi^{-4}\biggl(\frac{3}{2^3\pi}\biggr) \biggl(\frac{\nu}{q^3}\biggr)^2\biggr]_\mathrm{eq} \biggl\{ (1-q^3) q^2(g^2-1) + \biggl(\frac{2}{5}\biggr) q^5 \mathfrak{F} \biggr\} </math>

 

<math>~=~</math>

<math>~ \biggl[ P_\mathrm{norm} R_\mathrm{norm}^3 \biggr] \frac{n_e}{2} \biggl(\frac{\nu^2}{q^4}\biggr)(1-q^3) \biggl\{ (g^2-1) + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \chi^{-1}_\mathrm{eq} </math>

<math>~\Rightarrow~~~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>

<math>~=~</math>

<math>~ n_e (1-q^3) \biggl\{ (g^2-1) + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \frac{q^2}{2}\biggl(\frac{\nu}{q^3}\biggr)^2 \biggl[\frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c}\biggr]^{-1/(n_c-3)} </math>

 

<math>~=~</math>

<math>~ n_e (1-q^3) \biggl\{ (g^2-1) + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \biggl[2^{[n_c-(n_c-3)]} \biggl(\frac{3}{4\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{(1-n_c)+2(n_c-3)} q^{2(n_c-3)-2n_c} g^{-2n_c} \biggr]^{1/(n_c-3)} </math>

 

<math>~=~</math>

<math>~ n_e (1-q^3) \biggl\{ (g^2-1) + \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr\} \biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} g^{-2n_c} \biggr]^{1/(n_c-3)} \, . </math>

And, finally, switching from the <math>~g</math> notation to the <math>~b_\xi</math> notation gives,

<math>~\biggl[ \frac{\mathfrak{G}_\mathrm{env} }{E_\mathrm{norm}}\biggr]_\mathrm{eq} </math>

<math>~=~</math>

<math>~ n_e (1-q^3) (b_\xi q^2)^{-1} \biggl\{ 1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\} \biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6} (b_\xi q^2)^{n_c} \biggr]^{1/(n_c-3)} </math>

 

<math>~=~</math>

<math>~ n_e (1-q^3) \biggl\{ 1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\} \biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} q^{-6-2(n_c-3)+2n_c} b_\xi^{3-n_c+n_c} \biggr]^{1/(n_c-3)} </math>

 

<math>~=~</math>

<math>~ n_e\biggl[\biggl(\frac{2\cdot 3}{\pi} \biggr)\biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{3} \biggr]^{1/(n_c-3)} (1-q^3) \biggl\{ 1 - \biggl[1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3} \biggr)\mathfrak{F} \biggr]b_\xi q^2\biggr\} \, , </math>

which, after setting <math>~\Chi = 1</math>, precisely matches the above, "first view" expression.


Summary00

In summary, the desired out of equilibrium free-energy expression is,

<math>~\frac{\mathfrak{G}}{E_\mathrm{norm}} </math>

<math>~=</math>

<math>~ A_0\Chi^{-3/n_c} + B_0\Chi^{-3/n_e} - C_0\Chi^{-1} </math>

where,

<math>~A_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{core}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>

<math>~=</math>

<math> \frac{n_c}{b_\xi} \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{1/(n_c-3)} q^3\biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, , </math>

<math>~B_0 \equiv \biggl( \frac{\mathfrak{S}_\mathrm{env}}{E_\mathrm{norm}} \biggr)_\mathrm{eq}</math>

<math>~=</math>

<math> \frac{n_e}{b_\xi} \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c} \biggr]^{1/(n_c-3)} (1-q^3) \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} \, , </math>

<math>~C_0 \equiv \biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{eq} </math>

<math>~=</math>

<math> \biggl( \frac{6}{5}\biggr) q^5 f \biggl[ \biggl(\frac{2\cdot 3}{\pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-5} b_\xi^{n_c} \biggr]^{1/(n_c-3)} \, . </math>

Or, in a more compact form,

<math>~\mathfrak{G}^* \equiv \biggl[ \biggl(\frac{2\cdot 3}{\pi}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(n_c-5)} b_\xi^{n_c}\biggr]^{-1/(n_c-3)} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>

<math>~=</math>

<math>~ n_c A_1\Chi^{-3/n_c} + n_e B_1\Chi^{-3/n_e} - 3C_1\Chi^{-1} </math>

where,

<math>~A_1 </math>

<math>~\equiv</math>

<math> \frac{1}{b_\xi} (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] \, , </math>

<math>~B_1 </math>

<math>~\equiv</math>

<math> \frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} \, , </math>

<math>~C_1 </math>

<math>~\equiv</math>

<math> \biggl( \frac{2}{5}\biggr) q^5 f \, . </math>

Let's examine the behavior of the first radial derivative.

<math>~\frac{\partial \mathfrak{G}^*}{\partial \Chi}</math>

<math>~=</math>

<math>~\frac{3}{\Chi} \biggl[ - A_1\Chi^{-3/n_c} - B_1\Chi^{-3/n_e} + C_1\Chi^{-1} \biggr] \, .</math>

Let's see whether the sum of terms inside the square brackets is zero at the derived equilibrium radius, that is, when <math>~\Chi = 1</math> and, hence, when

<math>~\chi = \chi_\mathrm{eq}</math>

<math>~=</math>

<math>~ \biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} (q g)^{2n_c} \biggr]^{1/(n_c-3)} </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{1}{2^{n_c}}\biggl(\frac{4\pi}{3} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^{n_c-1} b_\xi^{-n_c} \biggr]^{1/(n_c-3)} \, . </math>

<math>~ C_1 - A_1 - B_1 </math>

<math>~=</math>

<math>~ \biggl( \frac{2}{5}\biggr) q^5 f - \frac{1}{b_\xi} (q^3) \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] - \frac{1}{b_\xi} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} </math>

 

<math>~=</math>

<math>~ \biggl( \frac{2}{5}\biggr) q^5 f - \frac{1}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} + \frac{q^3}{b_\xi} \biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} - \frac{q^3}{b_\xi} \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] </math>

 

<math>~=</math>

<math>~ \biggl( \frac{2}{5}\biggr) q^5 f - \frac{1}{b_\xi} + \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^2 + \frac{q^3}{b_\xi} - \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]q^5 - \frac{q^3}{b_\xi} + \biggl( \frac{3}{5} \biggr) q^5 </math>

 

<math>~=</math>

<math>~q^2\biggl\{ \biggl( \frac{2}{5}\biggr) q^3 f - \frac{1}{b_\xi q^2} + \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr] (1-q^3) + \biggl( \frac{3}{5} \biggr) q^3 \biggr\} </math>

 

<math>~=</math>

<math>~q^2\biggl\{ \biggl( \frac{2}{5}\biggr) q^3 f - \biggl[ 1+\frac{2}{5} q^3(f-1-\mathfrak{F}) \biggr] + \biggl[ (1-q^3) - \frac{2}{5} q^3 \mathfrak{F} \biggr] + \biggl( \frac{3}{5} \biggr) q^3 \biggr\} </math>

 

<math>~=</math>

<math>~q^2\biggl\{0\biggr\} \, . </math>

Q.E.D.

Even slightly better:

<math>~\frac{1}{q^2}\biggl[ \biggl(\frac{\pi}{2\cdot 3}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{(5-n_c)} b_\xi^{-n_c}\biggr]^{1/(n_c-3)} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>

<math>~=</math>

<math>~ n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \, , </math>

or, better yet,

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with Structural <math>~(n_c, n_e) = (0, 0)</math>

<math>~2\biggl(\frac{q^2}{\nu}\biggr)^2 \chi_\mathrm{eq} \biggl[\frac{\mathfrak{G}}{E_\mathrm{norm}} \biggr]</math>

<math>~=</math>

<math>~ n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} </math>

where, keeping in mind that,

<math>~\frac{1}{(b_\xi q^2)}</math>

<math>~=</math>

<math>~\biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>

we have,

<math>~A_2 \equiv \frac{A_1}{q^2} </math>

<math>~\equiv</math>

<math> \frac{q^3}{(b_\xi q^2)} \biggl[ 1 - \biggl( \frac{3}{5} \biggr) b_\xi q^2 \biggr] </math>

 

<math>~=</math>

<math> q^3 \biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] - \biggl( \frac{3}{5} \biggr) \biggr\} </math>

 

<math>~=</math>

<math> \frac{2}{5}q^3 \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>

<math>~B_2 \equiv \frac{B_1}{q^2} </math>

<math>~\equiv</math>

<math> \frac{1}{(b_\xi q^2)} (1-q^3)\biggl\{ 1- \biggl[ 1 - \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr]b_\xi q^2 \biggr\} </math>

 

<math>~=</math>

<math> (1-q^3)\biggl\{ \frac{1}{(b_\xi q^2)} -1 + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\} </math>

 

<math>~=</math>

<math> (1-q^3)\biggl\{ \biggl[1 + \frac{2}{5}q^3 (f - 1-\mathfrak{F} ) \biggr] - 1 + \frac{2}{5} \biggl(\frac{q^3}{1-q^3}\biggr) \mathfrak{F} \biggr\} </math>

 

<math>~=</math>

<math> \frac{2}{5} q^3 \biggl\{ (1-q^3) (f - 1-\mathfrak{F} ) + \mathfrak{F} \biggr\} </math>

 

<math>~=</math>

<math> \frac{2}{5} q^3 \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]\biggr\} </math>

 

<math>~=</math>

<math> \frac{2}{5} q^3 f - A_2 \, , </math>

<math>~C_2 \equiv \frac{C_1}{q^2} </math>

<math>~\equiv</math>

<math> \frac{2}{5} q^3 f \, . </math>

As before, the equilibrium system is dynamically unstable if <math>~\partial^2 \mathfrak{G}/\partial \Chi^2 < 0</math>. We have deduced that the system is unstable if,

<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>

<math>~< </math>

<math>~ \frac{A_2}{C_2} = \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, . </math>


Overview

BiPolytrope51

Key Analytic Expressions

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with <math>~(n_c, n_e) = (5, 1)</math>

<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>

<math>~=</math>

<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>

where,

<math>~\mathfrak{L}_i</math>

<math>~\equiv</math>

<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i \, ,</math>

<math>~\mathfrak{K}_i</math>

<math>~\equiv</math>

<math>~\frac{\Lambda_i}{\eta_i} + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, ,</math>

<math>~\Lambda_i</math>

<math>~\equiv</math>

<math>~\frac{1}{\eta_i} - \ell_i \, ,</math>

<math>~\eta_i</math>

<math>~=</math>

<math>~3 \biggl( \frac{\mu_e}{\mu_c} \biggr) \biggl[\frac{\ell_i }{(1+\ell_i^2)}\biggr] \, .</math>

From the accompanying Table 1 parameter values, we also can write,

<math>~\frac{1}{q}</math>

<math>~=</math>

<math>~\frac{\eta_s}{\eta_i} = 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \, ,</math>

<math>~\nu</math>

<math>~=</math>

<math>~ \frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, . </math>

Consistent with our generic discussion of the stability of bipolytropes and the specific discussion of the stability of bipolytropes having <math>~(n_c, n_e) = (5, 1)</math>, it can straightforwardly be shown that <math>~\partial \mathfrak{G}^*_{51}/\partial \chi = 0</math> is satisfied by setting <math>~\Chi = 1</math>; that is, the equilibrium condition is,

<math>~\chi = \chi_\mathrm{eq}</math>

<math>~=</math>

<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>

 

<math>~=</math>

<math>~\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c} \biggl[\frac{(1+\ell_i^2)^{6/5}}{(3\ell_i^2)} \biggr]^{n_c} \biggr\}^{1/(n_c-3)} \, ,</math>

where the last expression has been cast into a form that more clearly highlights overlap with the expression, below, for the equilibrium radius for zero-zero bipolytropes. Furthermore, the equilibrium configuration is unstable whenever,

<math>~\biggl[ \frac{\partial^2 \mathfrak{G}^*_{51}}{\partial \chi^2}\biggr]_{\Chi=1} < 0 \, ,</math>

that is, it is unstable whenever,

<math>~\frac{ \mathfrak{L}_i}{\mathfrak{K}_i}</math>

<math>~></math>

<math>~20 \, .</math>

Table 1 of an accompanying chapter — and the red-dashed curve in the figure adjacent to that table — identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, <math>~\mu_e/\mu_c</math>.

Behavior of Equilibrium Sequence

Here we reprint Figure 1 from an accompanying chapter wherein the structure of five-one bipolytropes has been derived. It displays detailed force-balance sequences in the <math>~q - \nu</math> plane for a variety of choices of the ratio of mean-molecular-weights, <math>~\mu_e/\mu_c</math>, as labeled.

Five-One Bipolytropic Equilibrium Sequences for Various ratios of the mean molecular weight
Limiting Values

Each sequence begins <math>~(\ell_i = 0)</math> at the origin, that is, at <math>~(q,\nu) = (0,0)</math>. As <math>~\ell_i \rightarrow \infty</math>, however, the sequences terminate at different coordinate locations, depending on the value of <math>~m_3 \equiv 3(\mu_e/\mu_c)</math>. In deriving the various limits, it will be useful to note that,

<math>~\frac{1}{\eta_i}</math>

<math>~=</math>

<math>~\frac{(1 + \ell_i^2)}{m_3 \ell_i} \, ,</math>

<math>~\Lambda_i</math>

<math>~=</math>

<math>~\frac{(1+\ell_i^2)}{m_3\ell_i}-\ell_i</math>

 

<math>~=</math>

<math>~\frac{1}{m_3\ell_i} + \biggl[\frac{(1 -m_3)}{m_3} \biggr]\ell_i </math>

 

<math>~=</math>

<math>~\frac{1}{m_3\ell_i} \biggl[ 1 - (m_3-1) \ell_i^2\biggr] </math>

 

<math>~=</math>

<math>~- \frac{ (m_3-1) \ell_i}{m_3} \biggl[ 1 - \frac{1}{(m_3-1) \ell_i^2}\biggr] \, ,</math>

<math>~1 + \Lambda_i^2</math>

<math>~=</math>

<math>~1 + \frac{1}{m_3^2\ell_i^2}\biggl[1 + (1 -m_3) \ell_i^2 \biggr]^2</math>

 

<math>~=</math>

<math>~\frac{1}{m_3^2\ell_i^2}\biggl\{ m_3^2\ell_i^2 + \biggl[1 + (1 -m_3) \ell_i^2 \biggr]^2\biggr\}</math>

 

<math>~=</math>

<math>~\frac{1}{m_3^2\ell_i^2}\biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (1 -m_3)^2 \ell_i^4 \biggr\}</math>

Examining the three relevant parameter regimes, we see that:

  • For <math>~\mu_e/\mu_c < \tfrac{1}{3}</math>, that is, <math>~m_3 < 1</math> …

<math>~\tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx </math>

<math>~\tan^{-1} \biggl[\frac{(1 -m_3)}{m_3} \biggr]\ell_i</math>

 

<math>~\approx </math>

<math>~\frac{\pi}{2} - \biggl[\frac{m_3}{(1 -m_3)\ell_i} \biggr]</math>

<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~ 1 + \frac{(1 + \ell_i^2)}{m_3 \ell_i}\biggl[\pi - \frac{m_3}{(1 -m_3)\ell_i} \biggr] </math>

 

<math>~\approx</math>

<math>~ \frac{m_3 + \pi \ell_i}{m_3} </math>

<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~ \frac{1}{1 + (\pi \ell_i/m_3)} \rightarrow 0 \, . </math>

 

and

 

<math>~\biggl(\frac{\nu}{q}\biggr)^2</math>

<math>~=</math>

<math>~\frac{\ell_i^2}{1 + \Lambda_i^2}</math>

 

<math>~=</math>

<math>~m_3^2\ell_i^4 \biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (1 -m_3)^2 \ell_i^4 \biggr\}^{-1}</math>

 

<math>~=</math>

<math>~m_3^2\biggl\{ \ell_i^{-4} + (2 -2m_3 + m_3^2) \ell_i^{-2} + (1 -m_3)^2 \biggr\}^{-1}</math>

<math>~\Rightarrow ~~~ \frac{\nu}{q}\biggr|_{\ell_i\rightarrow \infty}</math>

<math>~\approx</math>

<math>~\frac{m_3}{1-m_3} </math>

<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i\rightarrow \infty}</math>

<math>~\approx</math>

<math>~\biggl[\frac{m_3}{1-m_3}\biggr]\frac{1}{1 + (\pi \ell_i/m_3)} \rightarrow 0 \, . </math>

  • For <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, that is, <math>~m_3 = 1</math> …

<math>~\tan^{-1} \Lambda_i </math>

<math>~= </math>

<math>~\tan^{-1} \biggl(\frac{1}{\ell_i}\biggr)</math>

<math>~\Rightarrow ~~~ \tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx </math>

<math>~\frac{1}{\ell_i}</math>

<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~ 1 + \frac{(1 + \ell_i^2)}{\ell_i}\biggl[\frac{\pi}{2} + \frac{1}{\ell_i }\biggr] </math>

 

<math>~\approx</math>

<math>~ \biggl(\frac{\pi}{2}\biggr)\ell_i</math>

<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~ \frac{2}{\pi \ell_i} \rightarrow 0</math>

 

and

 

<math>~\biggl(\frac{\nu}{q}\biggr)</math>

<math>~=</math>

<math>~\frac{\ell_i}{(1 + 1/\ell_i^2)^{1/2}}</math>

<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~\ell_i \biggl(\frac{2}{\pi \ell_i} \biggr) = \frac{2}{\pi} \approx 0.63662</math>


  • For <math>~\mu_e/\mu_c > \tfrac{1}{3}</math>, that is, <math>~m_3 > 1</math> …

<math>~\tan^{-1} \Lambda_i \biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx </math>

<math>~\tan^{-1} \biggl[-\biggl(\frac{m_3-1}{m_3} \biggr)\ell_i\biggr]</math>

 

<math>~\approx </math>

<math>~-\frac{\pi}{2} + \biggl[\frac{m_3}{(m_3-1)\ell_i} \biggr]</math>

<math>~\Rightarrow ~~~ \frac{1}{q}\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~\approx</math>

<math>~ 1 + \frac{(1 + \ell_i^2)}{m_3 \ell_i}\biggl[ \frac{m_3}{(m_3-1)\ell_i} \biggr] </math>

 

<math>~=</math>

<math>~ 1 + \frac{(1 + 1/\ell_i^2)}{(m_3-1) } </math>

 

<math>~\approx</math>

<math>~ 1 + \frac{1}{(m_3-1) } = \frac{m_3}{(m_3-1)} </math>

<math>~\Rightarrow ~~~ q\biggr|_{\ell_i \rightarrow \infty}</math>

<math>~=</math>

<math>~ \frac{(m_3-1)}{m_3} </math>

 

and

 

<math>~\biggl(\frac{\nu}{q}\biggr)^2</math>

<math>~=</math>

<math>~\frac{\ell_i^2}{1 + \Lambda_i^2}</math>

 

<math>~=</math>

<math>~m_3^2\ell_i^4 \biggl\{ 1 + (2 -2m_3 + m_3^2) \ell_i^2 + (m_3-1)^2 \ell_i^4 \biggr\}^{-1}</math>

 

<math>~=</math>

<math>~m_3^2\biggl\{ \ell_i^{-4} + (2 -2m_3 + m_3^2) \ell_i^{-2} + (m_3-1)^2 \biggr\}^{-1}</math>

<math>~\Rightarrow ~~~ \frac{\nu}{q}\biggr|_{\ell_i\rightarrow \infty}</math>

<math>~\approx</math>

<math>~\frac{m_3}{m_3-1} </math>

<math>~\Rightarrow ~~~ \nu \biggr|_{\ell_i\rightarrow \infty}</math>

<math>~=</math>

<math>~\frac{m_3}{m_3-1} \biggl[\frac{m_3 - 1}{m_3}\biggr] \rightarrow 1 \, . </math>

Summarizing:

  • For <math>~\mu_e/\mu_c < \tfrac{1}{3}</math>, that is, <math>~m_3 < 1</math>       …       <math>~(q,\nu)_{\ell_i \rightarrow \infty} = (0, 0) \, .</math>


  • For <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, that is, <math>~m_3 = 1</math>       …       <math>~(q,\nu)_{\ell_i \rightarrow \infty} = (0, \tfrac{2}{\pi}) \, .</math>


  • For <math>~\mu_e/\mu_c > \tfrac{1}{3}</math>, that is, <math>~m_3 > 1</math>       …       <math>~(q,\nu)_{\ell_i \rightarrow \infty} = [(m_3-1)/m_3, 1] \, .</math>
Turning Points

Let's identify the location of two turning points along the <math>~\nu(q)</math> sequence — one defines <math>~q_\mathrm{max}</math> and the other identifies <math>~\nu_\mathrm{max}</math>. They occur, respectively, where,

<math>~\frac{d\ln q}{d\ln \ell_i} = 0</math>

      and      

<math>~\frac{d\ln \nu}{d\ln \ell_i} = 0 \, .</math>

In deriving these expressions, we will use the relations,

<math>~\frac{d\eta_i}{d\ell_i}</math>

<math>~=</math>

<math>~\frac{m_3 (1-\ell_i^2)}{(1+\ell_i^2)^2} \, ,</math>

<math>~\frac{d\Lambda_i}{d\ell_i}</math>

<math>~=</math>

<math>~- \frac{1}{m_3\ell_i^2} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \, ,</math>

where,

<math>~m_3 \equiv 3\biggl(\frac{\mu_e}{\mu_c}\biggr) \, .</math>


Given that,

<math>~q </math>

<math>~=</math>

<math>~\biggl\{ 1 + \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \biggr\}^{-1} \, ,</math>

we find,

<math>~\frac{d\ln q}{d\ln \ell_i}</math>

<math>~=</math>

<math>~\frac{\ell_i}{q} \cdot ( -q^2) \frac{d}{d\ell_i} \biggl\{ \frac{1}{\eta_i}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~-q\ell_i \biggl\{-\frac{1}{\eta_i^2}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]\frac{d\eta_i}{d\ell_i} + \frac{1}{\eta_i(1+\Lambda_i^2)} \frac{d\Lambda_i}{d\ell_i} \biggr\} </math>

 

<math>~=</math>

<math>~q\ell_i \biggl\{\frac{(1-\ell_i^2)}{m_3 \ell_i^2}\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \frac{(1+\ell_i^2)}{m_3^2 \ell_i^3(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ \frac{q}{m_3^2}{\ell_i^2}\biggl\{m_3 \ell_i (1-\ell_i^2) \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \biggr\} \, . </math>

And, given that,

<math>~\nu </math>

<math>~=</math>

<math>~ \frac{\ell_i q}{(1+\Lambda_i^2)^{1/2}} \, . </math>

we find,

<math>~\frac{d\ln \nu}{d\ln \ell_i}</math>

<math>~=</math>

<math>~\frac{\ell_i}{\nu} \biggl\{ \frac{q}{(1+\Lambda_i^2)^{1/2}} + \frac{q}{(1+\Lambda_i^2)^{1/2}} \frac{d\ln q}{d\ln \ell_i} - \frac{\ell_i q \Lambda_i }{(1+\Lambda_i^2)^{3/2}} \frac{d\Lambda_i}{d\ell_i} \biggr\} </math>

 

<math>~=</math>

<math>~\frac{q \ell_i}{ \nu(1+\Lambda_i^2)^{1/2}}\biggl\{ 1 + \frac{d\ln q}{d\ln \ell_i} + \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \biggr\} </math>

In summary, then, the <math>~q_\mathrm{max}</math> turning point occurs where,

<math>~0</math>

<math>~=</math>

<math>~ (1+\Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \frac{(1+\ell_i^2)}{m_3 \ell_i (1-\ell_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \, ; </math>

and the <math>~\nu_\mathrm{max}</math> turning point occurs where,

<math>~0</math>

<math>~=</math>

<math>~ 1 + \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] + \frac{q \ell_i^3 (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \frac{q\ell_i^2}{m_3^2}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr] </math>

 

<math>~=</math>

<math>~ 1 + \frac{q \ell_i^3 (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \biggl[ \frac{\Lambda_i }{m_3 \ell_i (1+\Lambda_i^2)} + \frac{q\ell_i^2}{m_3^2}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggr] \cdot \biggl[ 1 - \ell_i^2(1-m_3) \biggr] </math>

 

<math>~=</math>

<math>~ 1 + \frac{q \ell_i^3 (1-\ell_i^2)}{m_3} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] + \frac{1}{m_3 \ell_i} \biggl[ \frac{\Lambda_i }{(1+\Lambda_i^2)} + \frac{q\ell_i^3}{m_3}\cdot \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggr] \cdot \biggl[ 1 - \ell_i^2(1-m_3) \biggr] \, . </math>

NOTE:  As we show above, for the special case of <math>~m_3 = 1</math> — that is, when <math>~\mu_e/\mu_c = \tfrac{1}{3}</math>, precisely — the equilibrium sequence (as <math>~\ell_i \rightarrow \infty</math>) intersects the <math>~q = 0</math> axis at precisely the value, <math>~\nu = 2/\pi</math>. As is illustrated graphically in Figure 1 of an accompanying chapter, no <math>~\nu_\mathrm{max}</math> turning point exists for values of <math>~m_3 > 1</math>.

For the record, we repeat, as well, that the transition from stable to dynamically unstable configurations occurs along the sequence when,

<math>~\frac{(\ell_i^4-1)}{\ell_i^2} + \frac{(1+\ell_i^2)^3}{\ell_i^3} \cdot \tan^{-1}\ell_i </math>

<math>~=</math>

<math>~20 \biggl\{ \frac{\Lambda_i}{\eta_i} + \frac{(1+\Lambda_i^2)}{\eta_i} \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~\frac{20(1+\Lambda_i^2)(1+\ell_i^2)}{m_3\ell_i} \biggl\{ \frac{\Lambda_i}{(1+\Lambda_i^2)} + \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>

<math>~\Rightarrow ~~~ m_3 \ell_i (\ell_i^4-1) + m_3(1+\ell_i^2)^3\cdot \tan^{-1}\ell_i </math>

<math>~=</math>

<math>~20\ell_i^2 (1+\Lambda_i^2)(1+\ell_i^2) \biggl\{ \frac{\Lambda_i}{(1+\Lambda_i^2)} + \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \biggr\} </math>

<math>~\Rightarrow ~~~ \frac{m_3 \ell_i (\ell_i^4-1) + m_3(1+\ell_i^2)^3\cdot \tan^{-1}\ell_i }{ 20\ell_i^2 (1+\ell_i^2)}</math>

<math>~=</math>

<math>~\Lambda_i + (1+\Lambda_i^2)\biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i\biggr] \, .</math>


In order to clarify what equilibrium sequences do not have any turning points, let's examine how the <math>~q_\mathrm{max}</math> turning-point expression behaves as <math>~\ell_i \rightarrow \infty</math>.

<math>~ \frac{(1+\ell_i^2)}{(1+\Lambda_i^2)} \biggl[ 1 - \ell_i^2(1-m_3) \biggr]</math>

<math>~=</math>

<math>~ m_3 \ell_i (\ell_i^2-1) \biggl[\frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr] </math>

<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2)}{ m_3 \ell_i (\ell_i^2-1) } \biggl[ 1 + \ell_i^2(m_3-1) \biggr]</math>

<math>~=</math>

<math>~ \biggl\{ 1 + \frac{1}{m_3^2\ell_i^2}\biggl[ (m_3-1) \ell_i^2-1 \biggr]^2 \biggr\} \biggl\{ \frac{\pi}{2} + \biggl[ -\frac{\pi}{2} - \frac{1}{\Lambda_i} + \frac{1 }{3\Lambda_i^3} + \mathcal{O}(\Lambda_i^{-5} )\biggr] \biggr\} </math>

<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2) \ell_i^2(m_3-1)}{ m_3 \ell_i (\ell_i^2-1) } \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]</math>

<math>~=</math>

<math>~ \biggl\{ 1 + \frac{(m_3-1)^2 \ell_i^2}{m_3^2}\biggl[ 1 - \frac{1}{ (m_3-1) \ell_i^2 } \biggr]^2 \biggr\} \cdot \frac{1}{(-\Lambda_i)} \biggl[ 1 - \frac{1 }{3\Lambda_i^2} + \cancelto{0}{\mathcal{O}(\Lambda_i^{-4} )}\biggr] </math>

<math>~ \Rightarrow ~~~ \frac{(1+\ell_i^2)}{ \ell_i (\ell_i^2-1) } \cdot \frac{m_3}{(m_3-1)} \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]</math>

<math>~=</math>

<math>~ \biggl[ 1 - \frac{2}{ (m_3-1) \ell_i^2 } + \frac{m_3^2}{(m_3-1)^2 \ell_i^2} + \frac{1}{ (m_3-1)^2 \ell_i^4 } \biggr] \cdot \frac{m_3}{(m_3-1)\ell_i} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-1} \biggl\{ 1 - \frac{m_3^2}{3(m_3-1)^2\ell_i^2} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-2} \biggr\} </math>

<math>~ \Rightarrow ~~~\biggl(1 + \frac{1}{\ell_i^2} \biggr) \biggl[ 1 + \frac{1}{\ell_i^2(m_3-1)} \biggr]\biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]</math>

<math>~=</math>

<math>~\biggl(1 - \frac{1}{\ell_i^2} \biggr) \biggl[ 1 - \frac{2}{ (m_3-1) \ell_i^2 } + \frac{m_3^2}{(m_3-1)^2 \ell_i^2} + \frac{1}{ (m_3-1)^2 \ell_i^4 } \biggr] \biggl\{ 1 - \frac{m_3^2}{3(m_3-1)^2\ell_i^2} \biggl[1 - \frac{1}{(m_3-1)\ell_i^2}\biggr]^{-2} \biggr\} </math>

The leading-order term is unity on both sides of this expression, so they cancel; let's see what results from keeping terms <math>~\propto \ell_i^{-2}</math>.

<math>~ \frac{1}{\ell_i^2} \biggl[ 1 + \frac{1}{(m_3-1)} - \frac{1}{(m_3-1)}\biggr]</math>

<math>~=</math>

<math>~ \frac{1}{\ell_i^2} \biggl[- 1 - \frac{2}{ (m_3-1) } + \frac{m_3^2}{(m_3-1)^2 } - \frac{m_3^2}{3(m_3-1)^2} \biggr] </math>

<math>~\Rightarrow~~~ 2 </math>

<math>~=</math>

<math>~ - \frac{2}{ (m_3-1) } + \frac{2m_3^2}{3(m_3-1)^2 } </math>

<math>~\Rightarrow~~~ 6(m_3-1)^2 </math>

<math>~=</math>

<math>~ - 6(m_3-1) + 2m_3^2 </math>

<math>~\Rightarrow~~~ 6m_3^2-12m_3 + 6 </math>

<math>~=</math>

<math>~ - 6m_3+6 + 2m_3^2 </math>

<math>~\Rightarrow~~~ m_3</math>

<math>~=</math>

<math>~ \frac{3}{2} \, . </math>

We therefore conclude that the <math>~q_\mathrm{max}</math> turning point does not appear along any sequence for which,

<math>~m_3</math>

<math>~></math>

<math>~\frac{3}{2}</math>

<math>~\Rightarrow ~~~ \frac{\mu_e}{\mu_c}</math>

<math>~></math>

<math>~\frac{1}{2}\, .</math>


Five-One Bipolytrope Equilibrium Sequences in <math>~q - \nu</math> Plane

Full Sequences for Various <math>~\frac{\mu_e}{\mu_c}</math>

Magnified View with Turning Points and Stability Transition-Points Identified

Five-One Sequences

Graphical Depiction of Free-Energy Surface

Figure 1:   Free-Energy Surface for <math>~(n_c,n_e) = (5,1)</math> and <math>~\frac{\mu_e}{\mu_c} = 1</math>
Free-Energy surface for 5_1 bipolytrope

Left Panel: The free energy (vertical, blue axis) is plotted as a function of the radial interface location, <math>~\xi_i</math> (red axis), and the normalized configuration radius, <math>~\Chi \equiv \chi/\chi_\mathrm{eq}</math> (green axis). Right Panel: Same as the left panel, but animated in order to highlight undulations of the surface. The value of the free energy is indicated by color as well as by the height of the warped surface — red identifies the lowest depicted energies while blue identifies the highest depicted energies; these same colors have been projected down onto the <math>~z = 0</math> plane to present a two-dimensional, color-contour plot. A multi-colored line segment drawn parallel to the <math>~\xi_i</math> axis at the value, <math>~\Chi = 1</math>, identifies the configuration's equilibrium radius for each value of the interface location. Equilibrium configurations marked in white lie at the bottom of the principal free-energy "valley" and are stable, while configurations marked in blue lie at the top of a free-energy "hill," indicating that they are unstable; the red dot identifies the location along this equilibrium sequence where the transition from stable to unstable configurations occurs.

For purposes of reproducibility, it is incumbent upon us to clarify how the values of the free energy were normalized in order to produce the free-energy surface displayed in Figure 1. The normalization steps are explicitly detailed within the fortran program, below, that generated the data for plotting purposes; here we provide a brief summary. We evaluated the normalized free energy, <math>~\mathfrak{G}^*_{51}</math>, across a <math>~200 \times 200</math> zone grid of uniform spacing, covering the following <math>~(x,y) = (\ell_i,\Chi)</math> domain:

<math>~\frac{1}{\sqrt{3}}</math>

<math>~\le \ell_i \le</math>

<math>~\frac{3}{\sqrt{3}}</math>

<math>~0.469230769</math>

<math>~\le \Chi\le</math>

<math>~2.0</math>

(With this specific definition of the y-coordinate grid, <math>~\Chi = 1</math> is associated with zone 70.) After this evaluation, a constant, <math>~E_\mathrm{fudge} = -10</math>, was added to <math>~\mathfrak{G}^*</math> in order to ensure that the free energy was negative across the entire domain. Then (inorm = 1), for each specified interface location, <math>~x = \ell_i</math>, employing the equilibrium value of the free energy,

<math>~E_0 = \mathfrak{G}^*_{51}(\ell_i, \Chi = 1) + E_\mathrm{fudge} \, ,</math>

the free energy was normalized across all values of <math>~y = \Chi</math> via the expression,

<math>~\mathrm{fe} = \frac{(\mathfrak{G}^*_{51} + E_\mathrm{fudge}) - (E_0)_i}{|E_0|_i} \, .</math>

Finally, for plotting purposes, at each grid cell vertex ("vertex") — as well as at each grid cell center ("cell") — the value of the free energy, <math>~\mathrm{fe}</math>, was renormalized as follows,

<math>~\mathrm{vertex} = \frac{\mathrm{fe} - \mathrm{min}(\mathrm{fe})}{ \mathrm{max}(\mathrm{fe}) - \mathrm{min}(\mathrm{fe})} \, .</math>

Via this last step, the minimum "vertex" energy across the entire domain was 0.0 while the maximum "vertex" energy was 1.0.


FORTRAN Program Documentation Example Evaluations

(See also associated Table 1)
Coord. Axis Coord. Name Associated Physical Quantity <math>~\frac{\mu_e}{\mu_c} = 1</math> <math>~\frac{\mu_e}{\mu_c} = 0.305</math>
x-axis bsize <math>~\ell_i \equiv \frac{\xi_i}{\sqrt{3}}</math> <math>~\frac{2.416}{\sqrt{3}} = 1.395</math> <math>~\frac{8.1938}{\sqrt{3}} = 4.7307</math> <math>~\frac{14.389}{\sqrt{3}} = 8.3076</math>
y-axis csize <math>~\Chi \equiv \frac{\chi}{\chi_\mathrm{eq}}</math> <math>~1</math> <math>~1</math> <math>~1</math>
Relevant Lines of Code  
      eta = 3.0d0*muratio*bsize/(1.0d0+bsize**2)
      Gami = 1.0d0/eta-bsize
      frakL = (bsize**4-1.0d0)/bsize**2 + &
     &        DATAN(bsize)*((1.0d0+bsize**2)/bsize)**3
      frakK = Gami/eta + ((1.0d0+Gami**2)/eta)*(pii/2.0d0+DATAN(Gami))
      E0    = ((5.0d0*frakL) + (4.0d0*frakK)&
     &        - (3.0d0*frakL+12.0d0*frakK))/bsize**2+Efudge
          fescalar(j,k) = (csize**(-0.6d0)*(5.0d0*frakL)&
     &        + csize**(-3.0d0)*(4.0d0*frakK)&
     &   - (3.0d0*frakL+12.0d0*frakK)/csize)/bsize**2 + Efudge
          if(inorm.eq.1)fescalar(j,k)=fescalar(j,k)/DABS(E0) &
     &                  - E0/DABS(E0)
Variable Represents Value calculated via the expression …
eta <math>~\eta_i</math>

<math>~3 \biggl(\frac{\mu_e}{\mu_c}\biggr)\biggl[ \frac{\ell_i}{(1+\ell_i^2)} \biggr]</math>

<math>~1.421</math> <math>~0.1851</math> <math>~0.1086</math>
Gami <math>~\Lambda_i</math> <math>~\frac{1}{\eta_i} - \ell_i</math> <math>~-0.691</math> <math>~0.6705</math> <math>~0.9033</math>
frakL <math>~\mathfrak{L}_i</math> <math>~

\frac{(\ell_i^4 - 1)}{\ell_i^2} + \biggl[ \frac{1+\ell_i^2}{\ell_i} \biggr]^3 \tan^{-1}\ell_i

</math>
<math>~10.37</math> <math>~186.80</math> <math>~937.64</math>
frakK <math>~\mathfrak{K}_i</math> <math>~

\frac{\Lambda_i}{\eta_i} + \frac{(1 + \Lambda_i^2)}{\eta_i} \biggl[ \frac{\pi}{2} + \tan^{-1}\Lambda_i \biggr]

</math>
<math>~0.518</math> <math>~20.544</math> <math>~46.882</math>
  <math>~\frac{\mathfrak{L}_i}{\mathfrak{K}_i}</math>   <math>~20</math> <math>~9.093</math> <math>~20</math>
E0 - Efudge <math>~\mathfrak{G}^*_{51}(\ell_i,\Chi=1)</math>

<math>~ \frac{1}{\ell_i^2} \biggl[ 5 \mathfrak{L}_i + 4\mathfrak{K}_i - (3\mathfrak{L}_i +12\mathfrak{K}_i ) \biggr] = \frac{2(\mathfrak{L}_i - 4\mathfrak{K}_i)}{\ell_i^2} </math>

<math>~8.525</math> <math>~9.3496</math> <math>~21.737</math>
Figure 2:   Free-Energy Surface for <math>~(n_c,n_e) = (5,1)</math> and <math>~\frac{\mu_e}{\mu_c} = 0.305</math>
Free-Energy surface for 5_1 bipolytrope

Left Panel: The free energy (vertical, blue axis) is plotted as a function of the radial interface location, <math>~\xi_i</math> (red axis), and the normalized configuration radius, <math>~\Chi \equiv \chi/\chi_\mathrm{eq}</math> (green axis). Right Panel: Same as the left panel, but animated in order to highlight undulations of the surface. The value of the free energy is indicated by color as well as by the height of the warped surface — red identifies the lowest depicted energies while blue identifies the highest depicted energies; these same colors have been projected down onto the <math>~z = 0</math> plane to present a two-dimensional, color-contour plot. A multi-colored line segment drawn parallel to the <math>~\xi_i</math> axis at the value, <math>~\Chi = 1</math>, identifies the configuration's equilibrium radius for each value of the interface location. Equilibrium configurations marked in white lie at the bottom of the principal free-energy "valley" and are stable, while configurations marked in blue lie at the top of a free-energy "hill," indicating that they are unstable; the red dot identifies the location along this equilibrium sequence where the transition from stable to unstable configurations occurs.

BiPolytrope00

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with Structural <math>~(n_c, n_e) = (0, 0)</math>

<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>

<math>~=</math>

<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>

where,

<math>~A_2 </math>

<math>~\equiv</math>

<math> \frac{2}{5}q^3 \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, , </math>

<math>~B_2 </math>

<math>~\equiv</math>

<math> \frac{2}{5} q^3 f - A_2 \, , </math>

<math>~C_2 </math>

<math>~\equiv</math>

<math> \frac{2}{5} q^3 f \, , </math>

<math>~f</math>

<math>~\equiv</math>

<math> 1+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \, , </math>

<math>~\mathfrak{F} </math>

<math>~\equiv</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, , </math>

<math>~\frac{\rho_e}{\rho_c} </math>

<math>~=</math>

<math>~ \frac{q^3(1-\nu)}{\nu(1-q^3)} \, . </math>

The associated equilibrium radius is,

<math>~\chi_\mathrm{eq}</math>

<math>~=</math>

<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>

We have deduced that the system is unstable if,

<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>

<math>~< </math>

<math>~ \frac{A_2}{C_2} = \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, . </math>

Fortran Code

This is the program that generated the free-energy data for the "five-one" bipolytrope that is displayed in the above, Figure 1 image/animation.

      PROGRAM BiPolytrope
      real*8 pii
      real*8 bmin,bmax,cmin,cmax,db,dc
      real*8 c(200),b(200),chalf(199),bhalf(199)
      real*8 bsize,csize,emin,emax
      real*8 fepoint(200,200),fescalar(199,199)
      real*8 ell(200),ellhalf(199)
      real*8 muratio,eta,Gami,frakK,frakL
      real*8 eshift,ediff
      real   xx(200),yy(200),cell(199,199),vertex(200,200)
      real   valuemin,minlog,valufudge
      real*8 q,nu,chiEq,Enorm,E0,Efudge
      integer j,k,n,nmax,inorm
  101 format(4x,'bsize',7x,'csize',8x,'xi',10x,'A',12x,'B',12x,&
     &'fM',13x,'fW',11x,'fA',/)
! 102 format(1p8d12.4)
  103 format(2i5,1p3d14.6)
  201 format(5x,'valuemin = ',1pe15.5,//)
  205 format(5x,'For Cell-center ... emin, emax = ',1p2d14.6,/)
  206 format(5x,'For Cell-vertex ... emin, emax = ',1p2d14.6,/)
  701 format(5x,1p10d10.2)  
  700 format(8x,'xi',9x,'ell',8x,'eta',8x,'Lambda',5x,'frakK',&
     &       5x,'frakL',8x,'q',5x,'nu',5x,'chiEq',8x,'E0',/)
!!!!!!!!
!!!!!!!!
      inorm=1
      pii = 4.0d0*datan(1.0d0) 
      muratio = 1.0d0
      bsize = 0.0d0
      csize = 0.0d0
      Efudge = -10.0d0
      write(*,101)
!     write(*,102)bsize,csize,xival,coefA,coefB,fM,fW,fA
!!!!!!!!!!!
!
!  In this free-energy routine, c = X = chi/chi_eq and b = xi_i
!
!!!!!!!!!!!
      nmax = 200
      bmin = 1.0d0
      bmax = 3.0d0
      db   = (bmax-bmin)/dfloat(nmax-1)
      b(1) = bmin
      ell(1) = b(1)/dsqrt(3.0d0)
!  These values of cmin and cmax ensure that X=1 occurs at zone 70
      cmin = 0.469230769d0
      cmax = 2.00d0
      dc   = (cmax-cmin)/dfloat(nmax-1)
      c(1) = cmin
      do n=2,nmax
        b(n) = b(n-1)+db
        c(n) = c(n-1)+dc
        ell(n) = b(n)/dsqrt(3.0d0)
      enddo
      do n=1,nmax-1
        bhalf(n) = 0.5d0*(b(n)+b(n+1))
        chalf(n) = 0.5d0*(c(n)+c(n+1))
        ellhalf(n) = bhalf(n)/dsqrt(3.0d0)
      enddo
!
!  BEGIN LOOP to evaluate free energy (cell center)
!
      emin = 0.0d0
      emax = 0.0d0
      write(*,700)
      do j=1,nmax-1
        bsize = ellhalf(j)
      eta = 3.0d0*muratio*bsize/(1.0d0+bsize**2)
      Gami = 1.0d0/eta-bsize
      frakL = (bsize**4-1.0d0)/bsize**2 + &
     &        DATAN(bsize)*((1.0d0+bsize**2)/bsize)**3
      frakK = Gami/eta + ((1.0d0+Gami**2)/eta)*(pii/2.0d0+DATAN(Gami))
      q = 1.0d0/(1.0d0 + (0.5d0*pii+DATAN(Gami))/eta)
      nu = bsize*q/dsqrt(1.0d0+Gami**2)
      chiEq = dsqrt(pii/8.0d0)*(nu**2/(q*bsize**2))&
     &        *((1.0d0+bsize**2)/(3.0d0*bsize))**3
      Enorm = 16.0d0*(q/nu**2)*chiEq
      E0    = ((5.0d0*frakL) + (4.0d0*frakK)&
     &        - (3.0d0*frakL+12.0d0*frakK))/bsize**2+Efudge
      write(*,701)b(j),bsize,eta,Gami,frakK,frakL,q,nu,chiEq,E0
        do k=1,nmax-1
          csize=chalf(k)
          fescalar(j,k) = (csize**(-0.6d0)*(5.0d0*frakL)&
     &        + csize**(-3.0d0)*(4.0d0*frakK)&
     &   - (3.0d0*frakL+12.0d0*frakK)/csize)/bsize**2 + Efudge
          if(inorm.eq.1)fescalar(j,k)=fescalar(j,k)/DABS(E0) &
     &                  - E0/DABS(E0)
        if(fescalar(j,k).gt.0.5d0)fescalar(j,k)=0.5d0
          if(fescalar(j,k).lt.emin)emin=fescalar(j,k)
          if(fescalar(j,k).gt.emax)emax=fescalar(j,k)
!         write(*,103)j,k,bsize,csize,fescalar(j,k)
        enddo
      enddo
      write(*,205)emin,emax
!
!  BEGIN LOOP to evaluate free energy (cell vertex)
!
      emin = 0.0d0
      emax = 0.0d0
      do j=1,nmax
        bsize = ell(j)
      eta = 3.0d0*muratio*bsize/(1.0d0+bsize**2)
      Gami = 1.0d0/eta-bsize
      frakL = (bsize**4-1.0d0)/bsize**2 + &
     &        DATAN(bsize)*((1.0d0+bsize**2)/bsize)**3
      frakK = Gami/eta + ((1.0d0+Gami**2)/eta)*(pii/2.0d0+DATAN(Gami))
      q = 1.0d0/(1.0d0 + (0.5d0*pii+DATAN(Gami))/eta)
      nu = bsize*q/dsqrt(1.0d0+Gami**2)
      chiEq = dsqrt(pii/8.0d0)*(nu**2/(q*bsize**2))&
     &        *((1.0d0+bsize**2)/(3.0d0*bsize))**3
      Enorm = 16.0d0*(q/nu**2)*chiEq
      E0    = ((5.0d0*frakL) + (4.0d0*frakK)&
     &        - (3.0d0*frakL+12.0d0*frakK))/bsize**2 + Efudge
        do k=1,nmax
          csize=c(k)
          fepoint(j,k) = (csize**(-0.6d0)*(5.0d0*frakL)&
     &        + csize**(-3.0d0)*(4.0d0*frakK)&
     &   - (3.0d0*frakL+12.0d0*frakK)/csize)/bsize**2 + Efudge
          if(inorm.eq.1)fepoint(j,k)=fepoint(j,k)/DABS(E0) &
     &                  - E0/DABS(E0)
        if(fepoint(j,k).gt.0.5d0)fepoint(j,k)=0.5d0
          if(fepoint(j,k).lt.emin)emin=fepoint(j,k)
          if(fepoint(j,k).gt.emax)emax=fepoint(j,k)
!         write(*,103)j,k,bsize,csize,fepoint(j,k)
        enddo
      enddo
      write(*,206)emin,emax
!
!  Now fill single-precision arrays for plotting.
!
      do n=1,nmax
!       xx(n)=b(n)/b(nmax)
!       yy(n)=c(n)/c(nmax)
        xx(n)=b(n)-bmin
        yy(n)=c(n)-cmin
!       xx(n)=b(n)
!       yy(n)=c(n)
      enddo
      valuemin= -emin
      valufudge = 1.0d0/(emax-emin)
      do k=1,nmax
        do j=1,nmax
          vertex(j,k)=fepoint(j,k)+valuemin
          vertex(j,k)=vertex(j,k)*valufudge
        enddo
      enddo
      do k=1,nmax-1
        do j=1,nmax-1
          cell(j,k)=fescalar(j,k)+valuemin
          cell(j,k)=cell(j,k)*valufudge
        enddo
      enddo
      call XMLwriter01(nmax,xx,yy,cell,vertex)
      stop
      END PROGRAM BiPolytrope
      Subroutine XMLwriter01(imax,x,y,cell_scalar,point_scalar)
      real x(200),y(200),z(1)
      real cell_scalar(199,199),point_scalar(200,200)
      integer imax
      integer extentX,extentY,extentZ
      integer ix0,iy0,iz0
      integer norm(200,3)
!     imax=200
      ix0=0
      iy0=0
      iz0=0
      extentX=imax-1
      extentY=imax-1
      extentZ=0
      z(1) = 0.0
! Set normal vector 1D array
      do i=1,imax
        norm(i,1)=0
        norm(i,2)=0
        norm(i,3)=1
      enddo
  201 format('<?xml version="1.0"?>')
  202 format('<VTKFile type="RectilinearGrid" version="0.1" byte_order="LittleEndian">')
  302 format('</VTKFile>')
  203 format(2x,'<RectilinearGrid WholeExtent="',6I4,'">')
  303 format(2x,'</RectilinearGrid>')
  204 format(4x,'<Piece Extent="',6I4,'">')
  304 format(4x,'</Piece>')
  205 format(6x,'<CellData Scalars="cell_scalars" Normals="magnify">')
  305 format(6x,'</CellData>')
  206 format(8x,'<DataArray type="Float32" Name="magnify" NumberOfComponents="3" format="ascii">')
  207 format(8x,'<DataArray type="Float32" Name="cell_scalars" format="ascii">')
  399 format(8x,'</DataArray>')
  208 format(6x,'<PointData Scalars="colorful" Normals="direction">')
  308 format(6x,'</PointData>')
  209 format(8x,'<DataArray type="Float32" Name="colorful" format="ascii">')
  210 format(6x,'<Coordinates>')
  310 format(6x,'</Coordinates>')
  211 format(8x,'<DataArray type="Float32" format="ascii" RangeMin="0" RangeMax="5">')
  212 format(8x,'<DataArray type="Float32" format="ascii">')
  213 format(8x,'<DataArray type="Float32" Name="direction" NumberOfComponents="3" format="ascii">')
  501 format(10f9.5)
  502 format(10f9.5)
  503 format(5x,9(1x,3I2))
  504 format(10f9.5)
  505 format(5x,10(1x,3I2))
!!!!!
!
!  Begin writing out XML tags.
!
!!!!!
      write(*,201)                              !<?xml
      write(*,202)                              !VTKFile
        write(*,203)ix0,extentX,iy0,extentY,iz0,extentZ        !  RectilinearGrid
          write(*,204)ix0,extentX,iy0,extentY,iz0,extentZ      !    Piece
            write(*,205)                        !      CellData
              write(*,207)                      !        DataArray(cell_scalars)
      do j=1,imax-1
      write(*,501)(cell_scalar(i,j),i=1,imax-1)
      enddo
              write(*,399)                      !        /DataArray
              write(*,206)                      !        DataArray(cell_scalars)
      do j=1,imax-1
      write(*,503)(norm(i,1),norm(i,2),norm(i,3),i=1,imax-1)
      enddo
              write(*,399)                      !        /DataArray
            write(*,305)                        !      /CellData
            write(*,208)                        !      PointData
              write(*,209)                      !        DataArray(points)
      write(*,502)((point_scalar(i,j),i=1,imax),j=1,imax)
              write(*,399)                      !        /DataArray
              write(*,213)                      !        DataArray(cell_scalars)
      do j=1,imax
      write(*,505)(norm(i,1),norm(i,2),norm(i,3),i=1,imax)
      enddo
              write(*,399)                      !        /DataArray
            write(*,308)                        !      /PointData
            write(*,210)                        !      Coordinates
              write(*,212)                      !        DataArray(x-direction)
        write(*,504)(x(i),i=1,imax)
              write(*,399)                      !        /DataArray
              write(*,212)                      !        DataArray(y-direction)
        write(*,504)(y(i),i=1,imax)
              write(*,399)                      !        /DataArray
              write(*,212)                      !        DataArray(z-direction)
        write(*,504)z(1)
              write(*,399)                      !        /DataArray
            write(*,310)                        !      /Coordinates
          write(*,304)                          !    /Piece
        write(*,303)                            !  /RectilinearGrid
      write(*,302)                              !/VTKFile
      return
      end

Nonstandard Examination

In our introductory remarks, above, we said the warped free-energy surface drapes across a five-dimensional parameter "plane" such that,

<math>~\mathfrak{G}_{51}</math>

<math>~=</math>

<math>~\mathfrak{G}(R, K_c, M_\mathrm{tot}, q, \nu) \, .</math>

From a more pragmatic point of view, we should have said that the "five-one" free-energy surface drapes across a five-dimensional parameter "plane" such that,

<math>~\mathfrak{G}_{51}</math>

<math>~=</math>

<math>~\mathfrak{G}(R, K_c, M_\mathrm{tot}, \ell_i, \tfrac{\mu_e}{\mu_c}) \, .</math>

In our initial, standard examination of the structure of this warped free-energy surface, we held three parameters fixed — namely, <math>~K_c, M_\mathrm{tot}, \tfrac{\mu_e}{\mu_c}</math> — and plotted <math>~\mathfrak{G}_{51}(\ell_i, \Chi\equiv R/R_\mathrm{eq})</math>. Now, let's fix <math>~\Chi = 1</math> and plot <math>~\mathfrak{G}_{51}(\ell_i, \tfrac{\mu_e}{\mu_c})</math>. The following plot shows how a portion of the <math>~(\ell_i, \mu_e/\mu_c)</math> grid maps onto the traditional <math>~(q, \nu)</math> plane. The numerical labels identify lines of constant <math>~\xi_i = \sqrt{3}\ell_i</math> — 7 (light green), 9 (pink), and 12 (orange) — and lines of constant <math>~\mu_e/\mu_c</math> — 0.330 (purple), 0.315 (dark green), and 0.305 (white/blue).

xi-ell grid drawn on q-nu grid

See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation