Editing SSC/Virial/Polytropes/Pt2 (section)

====Example====

=====Outline=====
Let's identify an equilibrium configuration numerically, using the free-energy expression.  From [[SSCpt1/Virial#Gathering_it_all_Together|our introductory discussion]], the relevant expression is,
<div align="center">
<math>
\mathfrak{G}^* = 
-3\mathcal{A} \chi^{-1} -~ \frac{1}{(1-\gamma_g)} \mathcal{B} \chi^{3-3\gamma_g} +~ \mathcal{D}\chi^3 \, ,
</math>
</div>
where, 
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>\frac{1}{5} \cdot \biggl[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\mathfrak{f}_M} \biggr]^2 \cdot \mathfrak{f}_W \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\frac{4\pi}{3} 
\biggl[ \frac{3}{4\pi} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}}\biggr)  \frac{1}{\mathfrak{f}_M} \biggr]_\mathrm{eq}^{\gamma}
\cdot \mathfrak{f}_A 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi}{3} 
\biggl[ \biggl( \frac{P_c}{P_\mathrm{norm}} \biggr)\chi^{3\gamma} \biggr]_\mathrm{eq} 
\cdot \mathfrak{f}_A \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \, .
</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="10" align="center">
<tr><td align="left">

For later use, note that,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{5^n} \biggl( \frac{4\pi}{3} \biggr)  \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}}\biggr)^{n-1}
\cdot \frac{\mathfrak{f}_W^n}{\mathfrak{f}_A^n \mathfrak{f}_M^{1-n}} \, ;
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
5^{3(n+1)}  \biggl[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\mathfrak{f}_M} \biggr]^{-6(n+1)} 
\biggl( \frac{4\pi}{3} \biggr)^{4n-4(n+1)}
\biggl[ \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}}\biggr)  \frac{1}{\mathfrak{f}_M} \biggr]^{4(n+1)} \cdot\frac{\mathfrak{f}_A^{4n}}{\mathfrak{f}_W^{3(n+1)}}
</math>
  </td>
</tr>

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

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

Now, we could just blindly start setting values of the three leading coefficients, <math>~\mathcal{A}</math>, <math>~\mathcal{B}</math>, and <math>~\mathcal{D}</math>, then plot <math>~\mathfrak{G}^*(\chi)</math> to look for extrema.  But let's accept a little guidance from this chapter's virial analysis before choosing the coefficient values.  For embedded polytropes, we know that the structural form factors are,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_M = \frac{\bar\rho}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ - \frac{3\Theta^'}{\xi} \biggr]_{\tilde\xi} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde\mathfrak{f}_W </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3^2\cdot 5}{5-n} \biggl[ \frac{\Theta^'}{\xi} \biggr]^2_{\tilde\xi} = \biggl( \frac{5}{5-n} \biggr) \tilde\mathfrak{f}_M^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde\mathfrak{f}_A  = \frac{\bar{P}}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{3(n+1) }{(5-n)} ~\biggl( \Theta^' \biggr)^2_{\tilde\xi}  + \tilde\Theta^{n+1} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the coefficient expressions become,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{1}{(5-n)} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi}{3} 
\biggl( \frac{\bar{P}}{P_\mathrm{norm}} \biggr)\chi^{3(n+1)/n}_\mathrm{eq} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{4\pi}{3} \biggr) \frac{P_e}{P_\mathrm{norm}} \, .
</math>
  </td>
</tr>
</table>
</div>

=====Strategy=====

Generic setup:
* Choose the polytropic index, <math>~n</math>, which also sets the value of the adiabatic index, <math>~\gamma=(n+1)/n</math>.
* Fix <math>~M_\mathrm{tot}</math> and <math>~K</math>, so that the radial and pressure normalizations are fixed; specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~R_\mathrm{norm}^{n-3} = \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} </math>
  </td>
  <td align="center">
&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>~P_\mathrm{norm}^{(n-3)} = K^{4n} (G^3 M_\mathrm{tot}^2)^{-(n+1)} \, .</math>
  </td>
</tr>
</table>
</div>
* Fix <math>~M_\mathrm{limit}</math>, and let it be the normalization mass; that is, set <math>~M_\mathrm{limit} = M_\mathrm{tot}</math>.
* As a result of the above choices, the value of <math>~\mathcal{A}</math> is set, and fixed; specifically,
<div align="center">
<math>~A = \frac{1}{(5-n)} \, .</math>
</div>

'''<font color="red">Case I:</font>'''  
* Fix <math>~\mathcal{D}</math>, which fixes the external pressure; specifically,
<div align="center">
<math>~P_e = \frac{3}{4\pi} \cdot \mathcal{D} P_\mathrm{norm} \, .</math>
</div> 
* Choose a variety of values of the remaining coefficient, <math>~\mathcal{B}</math>; then, for each value, plot <math>\mathfrak{G}^*(\chi)</math> and locate one or more extrema along with the value of <math>~\chi_\mathrm{eq}</math> that is associated with each free energy extremum.  This identifies the equilibrium value of the mean pressure inside the pressure-truncated polytrope via the expression,
<div align="center">
<math>~ \bar{P} = \biggl(\frac{3}{4\pi} \biggr) P_\mathrm{norm}~\mathcal{B} ~\chi^{-3(n+1)/n}_\mathrm{eq}  \, .</math>
</div>
* In order to check whether we've identified the correct value of <math>~\chi_\mathrm{eq}</math>, we have to relate it to the radial coordinate, <math>~\xi_e</math>, used in the analytic solution of the Lane-Emden equation.  As has been explained in our [[SSC/Structure/Polytropes#Polytropic_Spheres|discussion of detailed force-balanced models of polytropes]], generically,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~R_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a_n \xi_e \, ,</math>&nbsp; &nbsp; &nbsp; &nbsp; where,
  </td>
</tr>

<tr>
  <td align="right">
<math>~a_\mathrm{n} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{1}{4\pi G}~ \biggl( \frac{H_c}{\rho_c} \biggr)\biggr]^{1/2}  
= 
\biggl[ \frac{(n+1)K}{4\pi G} \cdot \rho_c^{(1-n)/n}\biggr]^{1/2}  
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{a_n}{R_\mathrm{norm}} \biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n+1) K }{4\pi G} \biggr] \rho_c^{(1-n)/n} \cdot \frac{1}{R_\mathrm{norm}^2}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n+1) K }{4\pi G} \biggr] \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{(1-n)/n} 
\biggl( \frac{3M_\mathrm{limit}}{4\pi R_\mathrm{eq}^3} \biggr)^{(1-n)/n} 
\cdot \frac{1}{R_\mathrm{norm}^2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{(n+1) K }{4\pi G} \biggr] \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{(1-n)/n} 
\biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{(1-n)/n} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^{(1-n)/n} \chi_\mathrm{eq}^{3(n-1)/n}
\cdot R_\mathrm{norm}^{(n-3)/n}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>~\xi_e^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{R_\mathrm{eq}}{a_n} \biggr)^2 = \chi_\mathrm{eq}^2 \biggl( \frac{a_n}{R_\mathrm{norm}} \biggr)^{-2}</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi }{(n+1) }
\biggl[ \frac{3}{4\pi}\biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\tilde\mathfrak{f}_M} \biggr]^{(n-1)/n} 
\chi_\mathrm{eq}^{(3-n)/n} \, .
</math>
  </td>
</tr>
</table>
</div>
But, from above, we also know that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{3}{4\pi}\biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\tilde\mathfrak{f}_M} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{3\mathcal{B}}{4\pi} \cdot \frac{1}{\tilde\mathfrak{f}_A} \biggr]^{n/(n+1)} \, ,</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\tilde\mathfrak{f}_A  = \frac{\bar{P}}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{3(n+1) }{(5-n)} ~\biggl( \Theta^' \biggr)^2_{\tilde\xi}  + \tilde\Theta^{n+1} \, ,
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~\xi_e^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi }{(n+1) }
\biggl[ \frac{3\mathcal{B}}{4\pi} \cdot \frac{1}{\tilde\mathfrak{f}_A} \biggr]^{(n-1)/(n+1)} 
\chi_\mathrm{eq}^{(3-n)/n} \, ,
</math>
  </td>
</tr>
</table>
</div>
or,

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

<tr>
  <td align="right">
<math>~\xi_e^2 \biggl[ \frac{3(n+1) }{(5-n)} ~\biggl( \Theta^' \biggr)^2_{\tilde\xi}  + \tilde\Theta^{n+1} \biggr]^{(n-1)/(n+1)}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi }{(n+1) }
\biggl[ \frac{3\mathcal{B}}{4\pi} \biggr]^{(n-1)/(n+1)} 
\chi_\mathrm{eq}^{(3-n)/n} \, .
</math>
  </td>
</tr>
</table>
</div>
This last expression may be useful because the numerical value of the right-hand-side will be known once an extremum of a free-energy plot has been identified, while the function on the left-hand side can be evaluated separately, from knowledge of the internal structure of detailed force-balanced, ''isolated'' polytropes.

=====Strategy2=====
* Pick the desired polytropic index, <math>~n</math>, and a radial coordinate within the isolated polytropic model, <math>~\tilde\xi \leq \xi_1</math>, that will serve as the truncated edge of the embedded polytrope.
* Knowledge of the isolated polytrope's internal structure will give the value of the Lane-Emden function, <math>~\tilde\theta</math>, and its radial derivative, <math>~{\tilde\theta'}</math>, at this truncated edge of the structure.
* According to {{ Horedt70 }} &#8212; see our [[SSC/Structure/PolytropesEmbedded#Horedt.27s_Presentation|accompanying discussion of detailed force-balanced models]] &#8212; the physical radius and external pressure that corresponds to this choice of the truncated edge is given by the expressions,
<div align="center">
<table border="0" cellpadding="3">

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

<tr>
  <td align="right">
<math>
~\frac{P_e}{P_\mathrm{norm}} = p_a \biggl( \frac{P_\mathrm{Horedt}}{P_\mathrm{norm}} \biggr)
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} \biggl[ \frac{(n+1)^3}{4\pi} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^{-2} \biggr]^{(n+1)/(n-3)}  \, .
</math>
  </td>
</tr>
</table>
</div>

* Using the chosen value of <math>~\tilde\xi</math> and its associated function values, <math>~\tilde\theta</math> and <math>~\tilde\theta^'</math>, determine the values of the three relevant structural form factors via the following analytic relations:

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_M = \frac{\bar\rho}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \biggl[ - \frac{3 \tilde\theta^'}{\tilde\xi} \biggr] \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde\mathfrak{f}_W </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3^2\cdot 5}{5-n} \biggl[ \frac{\tilde\theta^'}{\tilde\xi} \biggr]^2 = \biggl( \frac{5}{5-n} \biggr) \tilde\mathfrak{f}_M^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\tilde\mathfrak{f}_A  = \frac{\bar{P}}{P_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{3(n+1) }{(5-n)} ~( \tilde\theta^')^2  + \tilde\theta^{n+1} \, .
</math>
  </td>
</tr>
</table>
</div>
* Using these values of the structural form factors, determine the values of the three free-energy coefficients (set <math>~M_\mathrm{limi}/M_\mathrm{tot} = 1</math> for the time being) via the expressions:
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>~\mathcal{A}_\mathrm{mod} \equiv (5-n)\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{(5-n)}{5} \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^{2}  \frac{\tilde\mathfrak{f}_W}{\tilde\mathfrak{f}_M^2} 
= \biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr)^{2}  \, ,</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>~\mathcal{D}_\mathrm{mod} \equiv (5-n) \mathcal{D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{4\pi}{3} \biggr) (5-n) \frac{P_e}{P_\mathrm{norm}} \, .
</math>
  </td>
</tr>
</table>
</div>
* Plot the following free-energy function and see if the value of <math>~\chi</math> associated with the extremum is equal to the dimensionless equilibrium radius, <math>~R_\mathrm{eq}/R_\mathrm{norm}</math>, as predicted by the {{ Horedt70 }} expression, above:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(5-n)\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3\mathcal{A}_\mathrm{mod} \chi^{-1} -~ \frac{1}{(1-\gamma_g)} \mathcal{B}_\mathrm{mod} \chi^{3-3\gamma_g} +~ \mathcal{D}_\mathrm{mod}\chi^3</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-3\mathcal{A}_\mathrm{mod} \chi^{-1} +n\mathcal{B}_\mathrm{mod} \chi^{-3/n} +~ \mathcal{D}_\mathrm{mod}\chi^3 \, .</math>
  </td>
</tr>
</table>
</div>
* Virial equilbrium &#8212; that is, an extremum in the free energy function &#8212; occurs when <math>~\partial\mathfrak{G}^*/\partial\chi = 0</math>, that is, where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{B}_\mathrm{mod} \chi_\mathrm{eq}^{(n-3)/n} - \mathcal{D}_\mathrm{mod}\chi_\mathrm{eq}^4</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{A}_\mathrm{mod} \, .</math>
  </td>
</tr>
</table>
</div>

Note that if the coefficient, <math>~\mathcal{B}</math>, is written in terms of the normalized central pressure, the statement of virial equilibrium becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{A}_\mathrm{mod} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{4\pi}{3}\biggl( \frac{P_c}{P_\mathrm{norm}} \biggr) \chi_\mathrm{eq}^{3(n+1)/n} 
\cdot [(5-n)\tilde\mathfrak{f}_A] \biggr] \chi_\mathrm{eq}^{(n-3)/n} - \frac{4\pi}{3} (5-n)\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \chi_\mathrm{eq}^4</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3}\chi_\mathrm{eq}^4 \biggl\{\biggl( \frac{P_c}{P_\mathrm{norm}} \biggr) 
\cdot (5-n)\tilde\mathfrak{f}_A  - (5-n)\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3}\chi_\mathrm{eq}^4 \biggl\{\biggl( \frac{P_c}{P_\mathrm{norm}} \biggr) 
\cdot \biggl[ 3(n+1) (\tilde\theta^')^2 + (5-n)\tilde\theta^{n+1} \biggr]  - (5-n)\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
But, in this situation, <math>~\tilde\theta^{n+1} = P_e/P_c</math>, so virial equilibrium implies,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{3}{4\pi}\chi_\mathrm{eq}^{-4} \mathcal{A}_\mathrm{mod} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{P_c}{P_\mathrm{norm}} \biggr) 
\cdot \biggl[ 3(n+1) (\tilde\theta^')^2 + (5-n)\frac{P_e}{P_c} \biggr]  - (5-n)\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3(n+1) (\tilde\theta^')^2\biggl( \frac{P_c}{P_\mathrm{norm}} \biggr) 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\Rightarrow~~~\frac{P_c R_\mathrm{eq}^4}{GM_\mathrm{limit}^2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{4\pi(n+1) (\tilde\theta^')^2} \, .
</math>
  </td>
</tr>
</table>
</div>

=====Compare With Detailed Force Balanced Solution=====
In a [[SSC/Structure/PolytropesEmbedded#.3D_1_Polytrope|separate discussion]], we presented the detailed force-balanced model of an <math>~n=1</math> polytrope that is embedded in an external medium.  We showed that, for an applied external pressure given by,
<div align="center">
<math>\frac{P_e}{P_\mathrm{norm}} = \frac{\pi}{2} \biggl[ \frac{\sin\xi_e}{\xi_e(\sin\xi_e - \xi_e \cos\xi_e )} \biggr]^2 \, ,</math> 
</div>
the associated equilibrium radius of the pressure-confined configuration is,
<div align="center">
<math>
R_\mathrm{eq} = \xi_e a_\mathrm{n=1} = \biggl[ \frac{K}{2\pi G} \biggr]^{1/2} \xi_e \, .
</math>
</div>
Flipping this around, after we use a plot of the free-energy expression to identify the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, the corresponding dimensionless radius as used in the Lane-Emden equation should be,
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~\xi_e</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{2\pi G}{K} \biggr]^{1/2} R_\mathrm{norm} \cdot \chi_\mathrm{eq} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(2\pi)^{1/2} \chi_\mathrm{eq} \, .
</math>
  </td>
</tr>

</table>
</div>
Keep in mind that, for an [[SSC/Structure/Polytropes#.3D_1_Polytrope|isolated <math>~n=1</math> polytrope]], the (zero pressure) surface is identified by <math>~\xi_1 = \pi</math>. Hence we should expect free-energy extrema to occur at values of <math>~\chi_\mathrm{eq} \le \pi/2</math>.