Editing SSC/Virial/Polytropes (section)

=====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,
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<math>~R_\mathrm{norm}^{n-3} = \biggl( \frac{G}{K} \biggr)^n M_\mathrm{tot}^{n-1} </math>
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&nbsp;&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;&nbsp;
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<math>~P_\mathrm{norm}^{(n-3)} = K^{4n} (G^3 M_\mathrm{tot}^2)^{-(n+1)} \, .</math>
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* 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,
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<math>~A = \frac{1}{(5-n)} \, .</math>
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'''<font color="red">Case I:</font>'''  
* Fix <math>~\mathcal{D}</math>, which fixes the external pressure; specifically,
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<math>~P_e = \frac{3}{4\pi} \cdot \mathcal{D} P_\mathrm{norm} \, .</math>
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* 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,
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<math>~ \bar{P} = \biggl(\frac{3}{4\pi} \biggr) P_\mathrm{norm}~\mathcal{B} ~\chi^{-3(n+1)/n}_\mathrm{eq}  \, .</math>
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* 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,
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<math>~R_\mathrm{eq} </math>
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<math>~=</math>
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<math>~a_n \xi_e \, ,</math>&nbsp; &nbsp; &nbsp; &nbsp; where,
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<math>~a_\mathrm{n} </math>
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<math>~\equiv</math>
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<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>
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</tr>
</table>
</div>

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

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&nbsp;
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<math>~=</math>
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<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>
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</tr>

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&nbsp;
  </td>
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<math>~=</math>
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<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>
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</table>
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Hence,
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<math>~\xi_e^2</math>
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<math>~=</math>
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<math>~\biggl( \frac{R_\mathrm{eq}}{a_n} \biggr)^2 = \chi_\mathrm{eq}^2 \biggl( \frac{a_n}{R_\mathrm{norm}} \biggr)^{-2}</math>
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&nbsp;
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<math>~=</math>
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<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>
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&nbsp;
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<math>~=</math>
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<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>
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</table>
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But, from above, we also know that,
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<math>~\frac{3}{4\pi}\biggl( \frac{M_\mathrm{limit}}{M_\mathrm{tot}} \biggr) \frac{1}{\tilde\mathfrak{f}_M} </math>
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<math>~=</math>
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<math>~\biggl[ \frac{3\mathcal{B}}{4\pi} \cdot \frac{1}{\tilde\mathfrak{f}_A} \biggr]^{n/(n+1)} \, ,</math>
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where,
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<math>\tilde\mathfrak{f}_A  = \frac{\bar{P}}{P_c}</math>
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<math>~=</math>
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<math>
\frac{3(n+1) }{(5-n)} ~\biggl( \Theta^' \biggr)^2_{\tilde\xi}  + \tilde\Theta^{n+1} \, ,
</math>
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</table>
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Hence we can write,
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<math>~\xi_e^2</math>
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<math>~=</math>
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<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>
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</table>
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or,

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<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>
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<math>~=</math>
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<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.