Difference between revisions of "User:Tohline/SSC/BipolytropeGeneralization"

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(→‎Simplest Bipolytrope: Elaborate more on expressions for Lambda_eq)
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=Bipolytrope Generalization=
=Bipolytrope Generalization=
{{LSU_HBook_header}}
{{LSU_HBook_header}}
==Setup==


In a [[User:Tohline/SphericallySymmetricConfigurations/Virial#Free_Energy_Expression|more general context]], we have discussed a Gibbs-like free-energy function of the generic form,
On <font color="red">26 August 2014</font>, Tohline finished rewriting the chapter titled "Bipolytrope Generalization" in a very concise manner ([[User:Tohline/SSC/BipolytropeGeneralization_Version2#Bipolytrope_Generalization|go here for this Version2 chapter]]) then set this chapter aside to provide a collection of older attempts at the derivations.  While much of what follows is technically correct, it is overly detailed and cumbersome.  Because it likely also contains some misguided steps, we label it in entirety as Work in Progress.
<div align="center">
 
<math>
 
\mathfrak{G} = W_\mathrm{grav} + \mathfrak{S}_\mathrm{therm} + T_\mathrm{kin} + P_e V + \cdots
{{LSU_WorkInProgress}}
</math>
 
</div>
==Old Stuff==
Here we are interested in examining the free energy of isolated, nonrotating, spherically symmetric bipolytropes, so we can drop the term that accounts for the influence of an external pressure and we can drop the kinetic energy term.  But we need to consider separately the contributions to the reservoir of thermodynamic energy by the core and envelope.  In particular, we will assume that compressions/expansions occur adiabatically, but that the core and the envelope evolve along separate adiabats &#8212; <math>~\gamma_c</math> and <math>~\gamma_e</math>, respectively.  Guided by our associated discussion of [[User:Tohline/SphericallySymmetricConfigurations/Virial#Dependence_on_Size|spherically symmetric, polytropic configurations]], we have,
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   <td align="left">
   <td align="left">
<math>
<math>
~\frac{3}{2}\biggl[ M_\mathrm{core} \biggl( \frac{P_{i}}{\rho_{ic}} \biggr) \biggr] s_\mathrm{core}  
~\frac{3}{2}\biggl[ M_\mathrm{core} \biggl( \frac{P_{ic}}{\rho_{ic}} \biggr) \biggr] s_\mathrm{core}  
= \frac{3}{2}\biggl[ M_\mathrm{core} K_c \rho_{ic}^{\gamma_c-1} \biggr] s_\mathrm{core}  
= \frac{3}{2} \biggl[ \nu M_\mathrm{tot}  P_{ic} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{core}
= 2\pi R^3 P_{ic} \biggl[ q^3 s_\mathrm{core} \biggr]
\, ,</math>
\, ,</math>
   </td>
   </td>
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   <td align="left">
   <td align="left">
<math>
<math>
~\frac{3}{2}\biggl[ M_\mathrm{env} \biggl( \frac{P_{i}}{\rho_{ie}} \biggr) \biggr] s_\mathrm{env}  
~\frac{3}{2}\biggl[ M_\mathrm{env} \biggl( \frac{P_{ie}}{\rho_{ie}} \biggr) \biggr] s_\mathrm{env}  
= \frac{3}{2}\biggl[ M_\mathrm{env} K_e \rho_{ie}^{\gamma_e-1} \biggr] s_\mathrm{env}  
= \frac{3}{2} \biggl[ (1-\nu) M_\mathrm{tot}  P_{ie} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{env}
= 2\pi R^3 P_{ie} \biggl[ (1-q^3) s_\mathrm{env} \biggr]
\, ,</math>
\, ,</math>
   </td>
   </td>
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where the subscript "<math>i</math>" means "at the interface," and <math>~\mathfrak{f}_{WM},</math> <math>~s_\mathrm{core},</math> and <math>~s_\mathrm{env}</math> are dimensionless functions of order unity (all three functions to be determined) akin to the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Structural_Form_Factors|structural form factors]] used in our examination of isolated polytropes.   
where the subscript "<math>i</math>" means "at the interface," and <math>~\mathfrak{f}_{WM},</math> <math>~s_\mathrm{core},</math> and <math>~s_\mathrm{env}</math> are dimensionless functions of order unity (all three functions to be determined) akin to the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Structural_Form_Factors|structural form factors]] used in our examination of isolated polytropes.   


While exploring how the free-energy function varies across parameter space, we choose to hold <math>~M_\mathrm{tot}</math> and <math>~K_c</math> fixed.  By dimensional analysis, it is therefore reasonable to normalize all energies, length scales, and densities by, respectively,
While exploring how the free-energy function varies across parameter space, we choose to hold <math>~M_\mathrm{tot}</math> and <math>~K_c</math> fixed.  By dimensional analysis, it is therefore reasonable to normalize all energies, length scales, densities and pressures by, respectively,
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \, .</math>
<math>~\frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{G^{3\gamma_c} M_\mathrm{tot}^{2\gamma_c}}{K_c^4} \biggr]^{1/(3\gamma_c -4)}  \, .</math>
   </td>
   </td>
</tr>
</tr>
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<math>
<math>
- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM}
- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM}
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_c \rho_{ic}^{\gamma_c-1} }{E_\mathrm{norm}} \biggr]   
+ \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R^3 P_{ic} }{E_\mathrm{norm}} \biggr] 
+ \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{ R^3 P_{ie} }{E_\mathrm{norm}} \biggr]   
</math>
</math>
   </td>
   </td>
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   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_e \rho_{ie}^{\gamma_e-1} }{E_\mathrm{norm}} \biggr]  
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1}
+ \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr] 
\biggl[ \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr]
</math>
</math>
   </td>
   </td>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
+ \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr]
+ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr]
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}
\biggl[ \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr]
</math>
</math>
   </td>
   </td>
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   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>
~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr)
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1}  
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}  
+ \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, ,
\biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_c}
+ \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr]  
\biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_e}
</math>
</math>
   </td>
   </td>
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \nu s_\mathrm{core}  \, ,</math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} 
\biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  \, ,
</math>
   </td>
   </td>
</tr>
</tr>
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl( \frac{K_e}{K_c} \biggr)
<math>\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env}  
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}  \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}  
\biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .</math>
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} (1-\nu) s_\mathrm{env} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
As is [[User:Tohline/SSC/BipolytropeGeneralization#pressureMatch|further detailed below]], the second expression for the coefficient, <math>~\mathcal{C}</math>, ensures that the pressure at the "surface" of the core matches the pressure at the "base" of the envelope; but it should only be employed ''after an equilibrium radius'', <math>~\chi_\mathrm{eq}</math>, ''has been identified by locating an extremum in the free energy.''
  <td align="right">
 
&nbsp;
==Simplest Bipolytrope==
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} (1-\nu) s_\mathrm{env} \, .</math>
  </td>
</tr>
</table>
</div>


As is [[User:Tohline/SSC/BipolytropeGeneralization#pressureMatch|further detailed below]], the second expression for the coefficient, <math>~\mathcal{C}</math>, ensures that the pressure at the "surface" of the core matches the pressure at the "base" of the envelope; but it should only be employed ''after an equilibrium radius'', <math>~\chi_\mathrm{eq}</math>, ''has been identified by locating an extremum in the free energy.''
===Familiar Setup===
 
==Simplest Bipolytrope==
As has been shown in [[User:Tohline/SSC/BipolytropeGeneralization#.280.2C_0.29_Bipolytropes|an accompanying presentation]], for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope,  
As has been shown in [[User:Tohline/SSC/BipolytropeGeneralization#.280.2C_0.29_Bipolytropes|an accompanying presentation]], for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope,  
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</tr>
</tr>
</table>
</table>
</div>
where (see an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Gravitational_Potential_Energy|associated discussion]] or the [[User:Tohline/SSC/VirialStability#Energy_Expressions|original derivation]]),
<div align="center">
<math>
f\biggl(q, \frac{\rho_e}{\rho_c}\biggr) \equiv 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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \, ,
</math>
</div>
</div>


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</div>
</div>


Therefore, for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope, the coefficients in the normalized free-energy function are,
===Cleaner Virial Presentation===
 
In an effort to show the similarity in structure among the several energy terms, we have also found it useful to write their expressions in the following forms:
 
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A}</math>
<math>~W_\mathrm{grav}</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}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \frac{\nu^2}{q} \cdot f
~\frac{\nu^2}{5q} \cdot f
= - 4\pi P_i R^3 \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_i f \, ,</math>
= \frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) q^3 (1 - q^2 )
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr)
\biggr] \, ,</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{B}</math>
<math>~S_\mathrm{core}</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[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c} q^3 [1 + \Lambda_\mathrm{eq}] \, ,</math>
<math>~2\pi P_{ic} R^3 \biggl[ q^3 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ic} \biggr]   \, ,</math>
   </td>
   </td>
</tr>
</tr>
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<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{C}</math>
<math>~S_\mathrm{env}</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[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c} (1-q^3)
<math>~2\pi P_{ie} R^3 \biggl[ (1-q^3) + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ie} \mathfrak{F} \biggr]   \, ,</math>
\biggl\{ 1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)} \biggl[  \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)  \biggr] \biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}  </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where (see an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Gravitational_Potential_Energy|associated discussion]] or the [[User:Tohline/SSC/VirialStability#Energy_Expressions|original derivation]]),
<div align="center">
<math>
f\biggl(q, \frac{\rho_e}{\rho_c}\biggr) = 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3 - q^5 )
+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] \, ,
</math>
</div>
and where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\lambda_i</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ C^' \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, .</math>
<math>~\frac{GM_\mathrm{tot}^2}{R^4 P_i} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


The equilibrium condition is,
<div align="center">
<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} 
\, .</math>
</div>
==Example==
<table border="1" cellpadding="5" align="center">
<tr>
<tr>
  <th align="center" colspan="9">
   <td align="right">
<font size="+1">Identification of Local ''Minimum'' in Free Energy</font>
<math>~\mathfrak{F} </math>
  </th>
</tr>
<tr>
   <td align="center">
<math>~\nu</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~q</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</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>
   </td>
   </td>
   <td align="center">
</tr>
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\Lambda_\mathrm{eq}</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\chi_\mathrm{eq}</math>
<math>~
\frac{1}{\lambda_{ie}} \biggl( \frac{2^2 \cdot 5\pi}{3} \biggr) \frac{q(1-q^3)}{\nu^2} (s_\mathrm{env} -1)
\, .</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~\mathcal{A}</math>
</table>
</div>
 
<div align="center">
<table border="1" cellpadding="10" width="80%">
<tr><td align="left">
This also means that the three key terms used as shorthand notation in the above expressions for the three energy terms have the following definitions:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\mathfrak{f}_{WM}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\mathcal{B}</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\mathcal{C}</math>
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.2</math>
<math>~s_\mathrm{core}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~9^{-1/3} = 0.48075</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~12.5644</math>
<math>
1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \, ,
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~2.091312</math>
 
<tr>
   <td align="right">
<math>~s_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.366531</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.037453</math>
<math>
1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \lambda_{ie} \mathfrak{F}
\, ,</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.2090801</math>
</table>
</div>
 
</td></tr>
</table>
</div>
 
Hence, if all the interface pressures are equal &#8212; that is, if <math>~P_i = P_{ic} = P_{ie}</math> and, hence also, <math>~\lambda_{i} = \lambda_{ic} = \lambda_{ie}</math> &#8212; then the total thermal energy is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
   <td align="right">
<math>~S_\mathrm{tot} = S_\mathrm{core} + S_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.2308269</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~2.06252 \times 10^{-4}</math>
<math>
2\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F}) \biggr] \, ;
</math>
   </td>
   </td>
</tr>
</tr>
 
</table>
</div>
and the virial is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.4</math>
<math>~2S_\mathrm{tot} + W_\mathrm{grav}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~4^{-1/3} = 0.62996</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~4.21974</math>
<math>
   </td>
4\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F} - f ) \biggr]
   <td align="center">
\, .
<math>~1.56498</math>
</math>
   </td>
</tr>
</table>
</div>
 
The virial should sum to zero in equilibrium, which means,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.707989</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.0220475</math>
<math>
\biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot (f - 1- \mathfrak{F} )
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.2143496</math>
 
<tr>
   <td align="right">
<math>\Rightarrow ~~~~ \biggl[ \biggl( \frac{2^2\cdot 5\pi}{3} \biggr) \frac{q}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.5635746</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~4.4626 \times 10^{-5}</math>
<math>
f - 1- \mathfrak{F}  
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.5</math>
<math>\Rightarrow ~~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggl[ \biggl( \frac{2^3\pi}{3} \biggr) \frac{q^6}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~3^{-1/3} = 0.693361</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~2.985115</math>
<math>
\biggl[ (q^3 - q^5 )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] -
\biggl[  (-2q^2 + 3q^3 - q^5) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (-\frac{3}{5}  +3q^2 - 3q^3 + \frac{3}{5} q^5) \biggr]
</math>
   </td>
   </td>
  <td align="center">
</tr>
<math>~1.42334</math>
  </td>
  <td align="center">
<math>~0.9448663</math>
  </td>
  <td align="center">
<math>~0.0152116</math>
  </td>
  <td align="center">
<math>~0.2152641</math>
  </td>
  <td align="center">
<math>~0.791882</math>
  </td>
  <td align="center">
<math>~1.5464 \times 10^{-5}</math>
  </td>
</tr>
</table>
 
 
{{LSU_WorkInProgress}}
 
=Detailed Derivations=
Dividing the free-energy expression through by <math>~E_\mathrm{norm}</math> generates,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}^* \equiv \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 588: Line 623:
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM}
2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (1 -3q^2 + 2q^3 )   
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_c \rho_{ic}^{\gamma_c-1} }{E_\mathrm{norm}} \biggr]  
</math>
</math>
   </td>
   </td>
Line 599: Line 633:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_e \rho_{ie}^{\gamma_e-1} }{E_\mathrm{norm}} \biggr]
q^2 \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} (g^2-1)  
</math>
</math>
   </td>
   </td>
Line 610: Line 644:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>\Rightarrow~~~~ \frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 617: Line 651:
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl[ \frac{K_c G^{(3\gamma_c -4)}M_\mathrm{tot}^{2(3\gamma_c -4)}}{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{1}{R} \biggr)
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1)   \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===Shift to Central Pressure Normalization===
Let's rework the definition of <math>~\lambda_i</math> in two ways:  (1) Normalize <math>~R_\mathrm{eq}</math> to <math>~R_\mathrm{norm}</math> and normalize the pressure to <math>~P_\mathrm{norm}</math>; (2) shift the referenced pressure from the pressure at the interface <math>~(P_i)</math> to the central pressure <math>~(P_0)</math>, because it is <math>~P_0</math> that is directly related to <math>~K_c</math> and <math>~\rho_c</math>; specifically, <math>P_0 = K_c \rho_c^{\gamma_c}</math>.  Appreciating that, in equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~P_i</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~P_0 - q^2 \Pi_\mathrm{eq}
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_c^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}}  
= K_c \rho_c^{\gamma_c} - \frac{3}{2^3 \pi}  
\biggr]^{1/(3\gamma_c -4)}  \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c-1}
\biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2 \, ,</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
the left-hand-side of the last expression, above, can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_e^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}}  
\biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_e-1}
</math>
   </td>
   </td>
</tr>
</tr>
Line 660: Line 703:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\frac{R_\mathrm{eq}^4}{GM_\mathrm{tot}^2}
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)  
\biggl[ P_0 - \frac{3}{2^3 \pi}  
</math>
\biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2\biggr] </math>
   </td>
   </td>
</tr>
</tr>
Line 671: Line 714:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr)
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_c-1}
\frac{\nu}{q^3} \biggr]^2 q^2 \, .</math>
\biggl[ \frac{ K_c^{3\gamma_c -3}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}}
\biggr]^{1/(3\gamma_c -4)}  \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)}  
\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_c-1)/(3\gamma_c -4)}
\biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, the virial equilibrium condition gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>
\frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr)
\frac{\nu}{q^3} \biggr]^2 q^2
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_e-1}
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1)
\biggl[ \frac{K_c^{3\gamma_c - 3} (K_e/K_c)^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} 
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)}
\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_e-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}
</math>
</math>
   </td>
   </td>
Line 703: Line 747:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>
\Rightarrow ~~~~ \frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} 
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 710: Line 756:
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2} q^2 g^2  \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This result precisely matches [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#CentralPressure|the result obtained via the detailed force-balanced conditions]] imposed through hydrostatic equilibrium.
Adopting our new variable normalizations and realizing, in particular, that,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~R_\mathrm{norm}^4 P_\mathrm{norm}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~GM_\mathrm{tot}^2 \, ,</math>
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr]  \biggl( \frac{3}{4\pi} \biggr)^{\gamma_c-1} \biggl[ M_\mathrm{tot}^{3\gamma_c-4} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)}
\biggl[ M_\mathrm{tot}^{-2} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} 
\biggl[ M_\mathrm{tot}^{-3\gamma_c+6} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)}
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)}
\biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
the expression alternatively can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2}  
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr)
= \chi_\mathrm{eq}^4 \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr) </math>
\biggl[ M_\mathrm{tot}^{2(\gamma_e-1)-2(\gamma_c-1)}\biggr]^{1/(3\gamma_c-4)}
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)}
\biggl[ \biggl( \frac{K_c}{G} \biggr)^{(\gamma_c-1)-(\gamma_e-1)} \biggr]^{3/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}
</math>
   </td>
   </td>
</tr>
</tr>
Line 759: Line 808:
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
\chi_\mathrm{eq}^4 \biggl\{  \frac{K_c \rho_c^{\gamma_c}}{P_\mathrm{norm}} - \frac{2\pi}{3}  
+ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4P_\mathrm{norm}} \biggr) \biggr\}
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}  
</math>
</math>
   </td>
   </td>
Line 771: Line 819:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>
~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr)
\chi_\mathrm{eq}^4 \biggl\\frac{K_c }{P_\mathrm{norm}} \biggl[ \frac{\rho_c}{\bar\rho} \biggl( \frac{3M_\mathrm{tot}}{4\pi R_\mathrm{norm}^3} \biggr) \chi_\mathrm{eq}^{-3} \biggr]^{\gamma_c}
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}  
- \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \chi_\mathrm{eq}^{-4} \biggr\}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<span id="pressureMatch">We also want to ensure that envelope pressure matches the core pressure at the interface.</span>  This means,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~K_e \rho_{ie}^{\gamma_e}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 796: Line 837:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~K_c \rho_{ic}^{\gamma_c}</math>
<math>
\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \frac{K_c }{P_\mathrm{norm}} \biggl( \frac{M_\mathrm{tot}^{\gamma_c}}{R_\mathrm{norm}^{3\gamma_c}} \biggr)
- \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2
</math>
   </td>
   </td>
</tr>
</tr>
Line 802: Line 846:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 808: Line 852:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>
<math>
\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c}
- \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Normalized in this manner, the virial equilibrium (as well as the hydrostatic balance) condition gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>
\chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c}
- \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 820: Line 876:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
<math>
\rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1) 
</math>
   </td>
   </td>
</tr>
</tr>
Line 827: Line 884:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>
\Rightarrow ~~~~ \chi_\mathrm{eq}^{4-3\gamma_c}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 833: Line 892:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
<math>
\biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \biggr\}^{\gamma_c - \gamma_e}</math>
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2-\gamma_c} q^2 g^2 \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
===Free-Energy Coefficients===
Therefore, for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope, the coefficients in the normalized free-energy function are,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
<math>~\mathcal{A}</math>
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 847: Line 913:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c}
<math>
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
~\frac{\nu^2}{5q} \cdot f
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}
= \frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^\biggl[
</math>
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) q^3 (1 - q^2 )
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr)
\biggr] \, ,</math>
   </td>
   </td>
</tr>
</tr>
Line 856: Line 924:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{B}</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[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} 
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}
\biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c}  
=\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} 
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}  
</math>
</math>
   </td>
   </td>
Line 878: Line 948:
   <td align="left">
   <td align="left">
<math>
<math>
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr]
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}  
=\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr]
\chi_\mathrm{eq}^{3\gamma_c-4}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, we can write the normalized (and dimensionless) free energy as,


<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 898: Line 965:
   <td align="left">
   <td align="left">
<math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c}
+ \biggl\{ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} + \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{ \rho_{ic} }{ \rho_{ie} } \biggr)\biggr\}  
- \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
+ \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr\}
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} \, .
\chi_\mathrm{eq}^{3\gamma_c-4}  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<tr>
<div align="center">
   <td align="right">
<math>
&nbsp;
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .
</math>
</div>
 
==Free Energy and Its Derivatives==
 
Now, the free energy can be written as,
<div align="center">
<table border="0">
 
<tr>
   <td align="right">
<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>~U_\mathrm{tot} + W</math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} +
\chi_\mathrm{eq}^{3\gamma_c-4} \biggl[\frac{3}{2^2\cdot 5\pi} - \frac{3}{2^3\pi} \biggr]\frac{\nu^2}{q^4} \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 938: Line 993:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} + W</math>
<math>
\biggl\{ \nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} -
\chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3^2}{2^3\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggl( \frac{4\pi }{3} \biggr) q^3 \biggr\}
=
\nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} -
\chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3}{10} \biggr) \frac{\nu^2}{q}  
</math>
   </td>
   </td>
</tr>
</tr>
Line 947: Line 1,008:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{C}</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{2}{3(\gamma_c - 1)} \biggr] C_\mathrm{core} R^{3-3\gamma_c}  
<math>\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env}  
+ \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \, .</math>
\biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .</math>
   </td>
   </td>
</tr>
</tr>
Line 960: Line 1,021:
</div>
</div>


The first derivative of the free energy with respect to radius is, then,
Note that, because <math>~P_{ie} = P_{ic}</math> in equilibrium, the ratio of coefficients,
 
<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>~\frac{d\mathfrak{G}}{dR}</math>
<math>~\frac{\mathcal{C}}{\mathcal{B}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -2 C_\mathrm{core} R^{2-3\gamma_c} -2 C_\mathrm{env} R^{2-3\gamma_e} + A R^{-2} \, .</math>
<math>~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl\{\frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr\}</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And the second derivative is,
<div align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{d^2\mathfrak{G}}{dR^2}</math>
<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \biggl( \frac{\mathcal{C}}{\mathcal{B}} \biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -2 (2-3\gamma_c) C_\mathrm{core} R^{1-3\gamma_c} -2 (2-3\gamma_e) C_\mathrm{env} R^{1-3\gamma_e} - 2A R^{-3} \, .</math>
<math>
\frac{  20\pi q(1-q^3) \lambda_i^{-1} + 3\nu^2 \mathfrak{F}   }{ 20\pi q^4 \lambda_i^{-1} + 3\nu^} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
The equilibrium condition is,
<div align="center">
<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} 
\, ,</math>
</div>
where,
<div align="center">
<math>
\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c-\gamma_e)}  \, .
</math>
</div>
===More General Derivation of Free-Energy Coefficients B and C===
Keep in mind that, generally,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>GM_\mathrm{tot}^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) C_\mathrm{core} R^{3-3\gamma_c} +
<math>~R_\mathrm{norm}^4 P_\mathrm{norm} = E_\mathrm{norm} R_\mathrm{norm} \, ;</math>
(3\gamma_e-2) C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \biggr]</math>
   </td>
   </td>
</tr>
</tr>
Line 1,011: Line 1,085:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\lambda_i}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) S_\mathrm{core} +
<math>
(3\gamma_e-2) S_\mathrm{env} +W \biggr] \, .</math>
\frac{R^4 P_i}{GM_\mathrm{tot}^2} = \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr) \chi^4
</math>
&nbsp;&nbsp;&hellip; and, note that &hellip; &nbsp;&nbsp;
<math>
\frac{1}{\Lambda} = \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \frac{1}{\lambda_i} \cdot \frac{1}{q^2 \sigma^2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
==Equilibrium==
The radius, <math>~R_\mathrm{eq}</math>, of the equilibrium configuration(s) is determined by setting the first derivative of the free energy to zero.  Hence,
<div align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~0 </math>
<math>~\frac{\Pi}{P_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2 C_\mathrm{core} R_\mathrm{eq}^{2-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{2-3\gamma_e} - A R_\mathrm{eq}^{-2} </math>
<math>
\frac{3}{2^3\pi} \biggl( \frac{GM_\mathrm{tot}^2}{P_\mathrm{norm} R^4} \biggr) \frac{\nu^2}{q^6}
=\biggl( \frac{2\pi}{3} \biggr) \sigma^2 \chi^{-4} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,044: Line 1,118:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{K_c \rho_c^{\gamma_c} }{P_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 C_\mathrm{core} R_\mathrm{eq}^{3-3\gamma_c} +
<math>
2 C_\mathrm{env} R_\mathrm{eq}^{3-3\gamma_e} - A R_\mathrm{eq}^{-1} \biggr]</math>
\frac{K_c}{P_\mathrm{norm}} \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c}
= \frac{K_c M_\mathrm{tot}^{\gamma_c} }{R_\mathrm{norm}^{3\gamma_c} P_\mathrm{norm}} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \chi^{-3\gamma_c} = \sigma^{\gamma_c} \chi^{-3\gamma_c} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
where we have introduced the notation,
<div align="center">
<math>
\sigma \equiv \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \, .
</math>
</div>


So, the free-energy coefficient,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{B}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 S_\mathrm{core} +
<math>
2 S_\mathrm{env} +W \biggr]</math>
\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} 
\biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c}  
=\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core}  
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,070: Line 1,161:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~ 2 S_\mathrm{tot} + W </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, .</math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr]_\mathrm{eq}
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4}
=\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}}  + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr]_\mathrm{eq}
\chi_\mathrm{eq}^{3\gamma_c-4}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This is the familiar statement of virial equilibrium.  From it we should always be able to derive the radius of equilibrium configurations.


==Stability==
To assess the relative stability of an equilibrium configuration, we need to determine the sign of the second derivative of the free energy, evaluated at the equilibrium radius.  If the sign of the second derivative is positive, the system is dynamically stable; if the sign is negative, he system is dynamically unstable.  Using the above statement of virial equilibrium, that is, setting,
<div align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2 S_\mathrm{tot} + W</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, ,</math>
<math>
  </td>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \chi_\mathrm{eq}^4 \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr)_\mathrm{eq} +
</tr>
\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} \, .
 
</math>
<tr>
  <td align="right">
<math>\Rightarrow ~~~~ S_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~- S_\mathrm{core} - \frac{W}{2} \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
we obtain,


And the free-energy coefficient,
<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>~\frac{d^2\mathfrak{G}}{dR^2}\biggr|_\mathrm{eq}</math>
<math>~\mathcal{C}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~\equiv</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ (3\gamma_c-2) S_\mathrm{core}  
<math>
+W - (3\gamma_e-2)\biggl( S_\mathrm{core} + \frac{W}{2}\biggr) \biggr]_\mathrm{eq} </math>
\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env}  
\biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  
= \biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env}  
\biggl[ \frac{1 }{\lambda_{ie}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,136: Line 1,220:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ 3(\gamma_c-\gamma_e) S_\mathrm{core}  
<math>
+ \biggl(2 - \frac{3}{2}\gamma_e\biggr)W \biggr]_\mathrm{eq} </math>
\biggl( \frac{4\pi }{3} \biggr) (1-q^3) \biggl\{ \frac{1 }{\lambda_{ie}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)}
\cdot \mathfrak{F}  \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,149: Line 1,235:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ (\gamma_c-\gamma_e) S_\mathrm{core}  
<math>
+ \frac{1}{2}\biggl(\frac{4}{3} - \gamma_e\biggr)W \biggr]_\mathrm{eq} </math>
\biggl( \frac{4\pi }{3} \biggr)
\biggl\{ (1-q^3) \chi_\mathrm{eq}^4 \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr)_\mathrm{eq}+ \biggl( \frac{2\pi}{3} \biggr) \sigma^2  
\biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<div align="center" id="DerivationTable">
<table border="1" align="center" cellpadding="10">
<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
OLD DERIVATION
<div align="center">
<math>P_{ic} = K_c \rho_c^{\gamma_c}</math>
</div>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
NEW DERIVATION
<div align="center">
<math>P_{ic} = P_0 - q^2\Pi = K_c \rho_c^{\gamma_c} - q^2\Pi</math>
</div>
   </td>
   </td>
   <td align="left">
</tr>
<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ -\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) -
 
(\gamma_e-\gamma_c) S_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>
 
<tr>
   <td align="center" colspan="2">
&hellip; therefore &hellip;
   </td>
   </td>
</tr>
</tr>
</table>
</div>
So, if when evaluated at the equilibrium state, the expression inside of the square brackets of this last expression is negative, the equilibrium configuration will be dynamically unstable.  We have chosen to write the expression in this particular final form because we generally will be interested in bipolytropes for which the adiabatic exponent of the envelope is greater than <math>~4/3</math> and the adiabatic exponent of the core is less than or equal to <math>~4/3</math> &#8212; that is, <math>~\gamma_e > 4/3 \ge \gamma_c</math>.  Hence, because the gravitational potential energy, <math>~W</math>, is intrinsically negative, the system will be dynamically unstable only if the second term (involving <math>~S_\mathrm{core}</math>) is greater in magnitude than the first term (involving <math>~W</math>).


It is worth noting that, instead of drawing upon <math>~S_\mathrm{core}</math> and <math>~W</math> to define the stability condition, we could have used an appropriate combination of <math>~S_\mathrm{env}</math> and <math>~W</math>, or the <math>~S_\mathrm{core}</math> and <math>~S_\mathrm{env}</math> pair.  Also, for example, because the virial equilibrium condition is <math>~S_\mathrm{tot} = -W/2</math>, it is easy to see that the following inequality also equivalently defines stability:
 
<div align="center">
<tr>
<table border="0">
  <td align="left">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ S_\mathrm{tot}\biggl( \gamma_e - \frac{4}{3}\biggr) -
<math>~\mathcal{B}_\mathrm{OLD}</math>
(\gamma_e-\gamma_c) S_\mathrm{core} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~>~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 0 \, .</math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} +
\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


=Examples=
==(0, 0) Bipolytropes==
===Review===
In an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|accompanying discussion]] we have derived analytic expressions describing the equilibrium structure and the stability of bipolytropes in which both the core and the envelope have uniform densities, that is, bipolytropes with <math>~(n_c, n_e) = (0, 0)</math>.  From this work, we find that integrals over the mass and pressure distributions give:
<div align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~ \frac{W}{R_\mathrm{eq}^3 P_i} = - \frac{A}{R_\mathrm{eq}^4 P_i} </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
&nbsp;<math>- ~ \frac{3}{5} \biggl[ \frac{GM_\mathrm{tot}^2}{R^4P_i} \biggr] \biggl( \frac{\nu^2}{q} \biggr) f </math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} +
\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr
</math>
   </td>
   </td>
</tr>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{B}_\mathrm{NEW}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
&nbsp;<math>- ~4\pi q^3 \Lambda f \, ,</math>
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c}
-  \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2
+ \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,225: Line 1,329:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{S_\mathrm{core}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{core}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
~2\pi q^3 (1 + \Lambda) \, ,</math>
\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} 
-  \biggl( \frac{2\pi}{5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr]
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
</table>
  <td align="right">
<math>~\frac{S_\mathrm{env}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{env}</math>
   </td>
   </td>
  <td align="center">
</tr>
<math>~=~</math>
 
  </td>
 
   <td align="left">
 
<math>
<tr>
2\pi \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
   <td align="center" colspan="2">
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,</math>
&hellip; and, enforcing in equilibrium <math>~P_{ie} = P_{ic}</math> &hellip;
   </td>
   </td>
</tr>
</tr>


</table>
 
</div>
<tr>
where,
  <td align="left">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Lambda</math>  
<math>~\mathcal{C}_\mathrm{OLD}</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{3}{2^2 \cdot 5\pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} \, ,</math>
\biggl( \frac{4\pi }{3} \biggr)
\biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2
\biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,271: Line 1,377:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~f(q,\rho_e/\rho_c)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
<math>
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] </math>
\biggl( \frac{4\pi }{3} \biggr)  
\biggl[ (1-q^3) \sigma^{\gamma_c} + \biggl( \frac{2\pi}{3} \biggr) \sigma^2  
\biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr) \chi_\mathrm{eq}^{3\gamma_e-4}\biggr]  
</math>
   </td>
   </td>
</tr>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{C}_\mathrm{NEW}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
<math>
+ \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \, ,</math>
\biggl( \frac{4\pi }{3} \biggr)  
\biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2  \sigma^2 \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2
\biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,297: Line 1,413:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~g^2(q,\rho_e/\rho_c)</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>1  +
<math>
\biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
\biggl( \frac{4\pi }{3} \biggr)  
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>
\biggl\{ (1-q^3) \sigma^{\gamma_c} \chi_\mathrm{eq}^{3\gamma_e - 3\gamma_c}
+ \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[  \biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr)
- q^2 (1-q^3)  \biggr] \chi_\mathrm{eq}^{3\gamma_e-4} \biggr\}
</math>
  </td>
</tr>
 
</table>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
  <td align="right">
   <td align="center" colspan="2">
&nbsp;
&hellip; and, also &hellip;
  </td>
   <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>1 + \biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q)
+ \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 )  \biggr] \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


===Renormalize===
 
Let's renormalize these energy terms in order to more readily relate them to the [[User:Tohline/SSC/BipolytropeGeneralization#Setup|generalized expressions derived above]].
<tr>
<div align="center">
  <td align="left">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~R^3 P_i</math>
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,339: Line 1,454:
   <td align="left">
   <td align="left">
<math>
<math>
~R^3 K_c \biggl(\frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi}\biggr) \frac{M_\mathrm{tot}}{R^3} \biggr]^{\gamma_c}
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c}  
</math>
</math>
   </td>
   </td>
Line 1,346: Line 1,461:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,352: Line 1,467:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}  
<math>
K_c M_\mathrm{tot}^{\gamma_c} R_\mathrm{norm}^{3-3\gamma_c}</math>
\frac{1}{q^2 } \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \sigma^{\gamma_c - 2} \chi_\mathrm{eq}^{4 - 3\gamma_c}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,365: Line 1,481:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}  
<math>
\biggl\{ K_c^{3\gamma_c -4} M_\mathrm{tot}^{\gamma_c(3\gamma_c -4)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{3-3\gamma_c} \biggr\}^{1/(3\gamma_c -4)}</math>
\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c}
</math>
   </td>
   </td>
</tr>
</tr>
 
</table>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
<table border="0" cellpadding="5" align="center">
\biggl\{ \frac{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} }{K_c} \biggr\}^{1/(3\gamma_c -4)}</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,391: Line 1,499:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
<math>
E_\mathrm{norm} \, .</math>
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\biggl[ \frac{GM_\mathrm{tot}^2}{R} \biggr]^{3\gamma_c -4}</math>
<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,411: Line 1,514:
   <td align="left">
   <td align="left">
<math>
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} G^{3\gamma_c -4} M_\mathrm{tot}^{6\gamma_c -8}
\frac{1}{q^2 \sigma^2} \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \biggl[
\biggl( \frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma_c}
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 \biggr]
</math>
</math>
   </td>
   </td>
Line 1,426: Line 1,529:
   <td align="left">
   <td align="left">
<math>
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} \biggl[ \frac{ G^{3\gamma_c -3} M_\mathrm{tot}^{5\gamma_c -6} }{K_c} \biggr]
\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c- \frac{5}{2}
</math>
</math>
   </td>
   </td>
</tr>
</tr>


</table>
  </td>
</tr>
</table>
</div>
===Extrema===
Extrema in the free energy occur when,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathcal{A}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,439: Line 1,553:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{C} \chi_\mathrm{eq}^{4-3\gamma_e} \, .</math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} E_\mathrm{norm}^{3\gamma_c-4} \, .
  </td>
</math>
</tr>
</table>
</div>
 
Also, as stated above, because <math>~P_{ie} = P_{ic}</math> in equilibrium, the ratio of coefficients,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\mathcal{C}}{\mathcal{B}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
When put together, these two relations imply,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\mathcal{A}</math>
\Rightarrow ~~~~\frac{GM_\mathrm{tot}^2}{R^4 P_i} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,454: Line 1,588:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \chi_\mathrm{eq}^{4-3\gamma_c} \mathcal{B}
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}  
\biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] </math>
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3\gamma_c - 3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,469: Line 1,601:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .</math>
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,477: Line 1,607:
</div>
</div>


Hence,
But the definition of <math>~\mathcal{B}</math> gives,
<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>~\Lambda</math>  
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c}</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{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c}  
\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core}  
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}  
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \, .
\, .</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
 
Hence, extrema occur when,
Given that <math>~\rho_{ic}/\bar\rho = \nu/q^3</math> for the <math>~(n_c, n_e) = (0, 0)</math> bipolytrope,  we can finally write,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,504: Line 1,632:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{R^3 P_i}{E_\mathrm{norm}}</math>
<math>~\mathcal{A}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,510: Line 1,638:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}  
<math>~\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} 
\, ,</math>
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] </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>~\Lambda</math>  
<math>\Rightarrow ~~~~ \biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot f</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{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c}
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq}   \biggl[ q^3 s_\mathrm{core} (1-q^3) s_\mathrm{env} \biggr] </math>
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} =
\frac{1}{5} \frac{\nu}{q} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}
\, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence the renormalized gravitational potential energy becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
\frac{W_\mathrm{grav}}{E_\mathrm{norm}}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,552: Line 1,664:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f
\frac{q^3}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q}  
\, ;</math>
+ \frac{(1-q^3)}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \mathfrak{F}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and the two, renormalized contributions to the thermal energy become,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]</math>
<math>\Rightarrow ~~~~\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)}
\biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot (f - 1 - \mathfrak{F})
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}
</math>
\, ,</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
\frac{4\pi}{3(\gamma_e-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \biggl( \frac{K_e}{K_c} \biggr)  
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2 - 1 ) \, .
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(3\gamma_c-3\gamma_e)/(3\gamma_c-4)}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_e}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\times \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\, ,</math>
  </td>
</tr>
</table>
</table>
</div>
</div>


Finally, then, we can state that,
In what follows, keep in mind that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,618: Line 1,707:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{f}_{WM}</math>
<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
<math>
\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c}
= R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \, ;
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,630: Line 1,722:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~s_\mathrm{core}</math>
<math>~K_c \rho_c^{\gamma_c}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
1 + \Lambda \, ,
K_c \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c}
= K_c \sigma^{\gamma_c} M_\mathrm{tot}^{\gamma_c} R^{-3\gamma_c} \, ;
</math>
</math>
   </td>
   </td>
Line 1,644: Line 1,737:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~(1-q^3) s_\mathrm{env}</math>
<math>~\Pi</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>
(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \frac{\nu^2}{q^6}
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
= \frac{2\pi}{3} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \sigma^2 \, .
\, .</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,659: Line 1,752:
</div>
</div>


===Virial Equilibrium and Stability Evaluation===
 
With these expressions in hand, we can deduce the equilibrium radius and relativity stability of <math>~(n_c, n_e) = (0, 0)</math> bipolytropes using the generalized expressions provided above.  For example, from the statement of virial equilibrium <math>~(2S_\mathrm{tot} = - W )</math> we obtain,
 
<div align="center">
<div align="center">
<table border="0">
<table border="1" align="center" cellpadding="10">
<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~q^3 (1 + \Lambda) + (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
OLD DERIVATION
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) </math>
<div align="center">
<math>P_{i} = K_c \rho_c^{\gamma_c}</math>
 
<math>\Rightarrow ~~~~ K_c  = P_{i} \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c} </math></div>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
NEW DERIVATION
<div align="center">
<math>P_0 = K_c \rho_c^{\gamma_c} </math>
 
<math>\Rightarrow ~~~~ K_c = P_0 \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c}  </math>
</div>
   </td>
   </td>
   <td align="left">
</tr>
<math>~q^3 \Lambda \biggl[ 1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
 
+ \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \biggr] </math>
 
<tr>
   <td align="center" colspan="2">
&hellip; hence, as derived in the above table &hellip;
   </td>
   </td>
</tr>
</tr>
</table>
</div>


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


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~ \frac{1}{\Lambda}</math>
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3)
<math>
+ \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5)
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c}  
- \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
</math>
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
   </td>
   </td>
</tr>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3 + 2 -3q +q^3)
<math>
+ \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5 +3 - 15q^2+15q^3 -3q^5) 
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
  </td>
</tr>
<tr>
  <td align="center" colspan="2">
&hellip; which, when combined with the condition that identifies extrema, gives &hellip;
  </td>
</tr>
<tr>
  <td align="left">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\chi_\mathrm{eq}^{4 - 3\gamma_c}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{5}{2}\biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q)
<math>
+ \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 ) \biggr]
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )
</math>
</math>
   </td>
   </td>
Line 1,728: Line 1,849:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{5}{2}(g^2-1)  
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )
</math>
</math>
   </td>
   </td>
Line 1,741: Line 1,863:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~ \biggl[ \frac{P_i}{GM_\mathrm{tot}^2} \biggr] R_\mathrm{eq}^4</math>
<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4}  P_i }{GM_\mathrm{tot}^{2} } </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}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \, .
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2-1)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
  </td>
 
  <td align="left">
Or, given the above renormalization, this expression can be written as,
<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>
<math>~\sigma^{\gamma_c}\chi_\mathrm{eq}^{4 - 3\gamma_c} \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2</math>
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c } \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
</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}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1)  
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^2 q^2 (g^2 - 1 )
</math>
</math>
   </td>
   </td>
Line 1,775: Line 1,896:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>
\Rightarrow ~~~~ \frac{R}{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\{ \biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1)
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggr\}^{1/(4-3\gamma_c)} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
And the condition for dynamical stability is,
<div align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>-\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) -
<math>\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>
(\gamma_e-\gamma_c) S_\mathrm{core} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; <math>~>~</math>&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, .</math>
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,812: Line 1,924:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow ~~~~ 2\pi q^3 \Lambda \biggl[ \biggl( \gamma_e - \frac{4}{3}\biggr) f -
<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4}  P_0 }{GM_\mathrm{tot}^{2} } </math>
(\gamma_e-\gamma_c) \biggl( 1 + \frac{1}{\Lambda}\biggr) \biggr] </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; <math>~>~</math>&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~0 \, .</math>
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 g^2
</math>
  </td>
</tr>
 
</table>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center" colspan="2">
<math>~\biggl( \gamma_e - \frac{4}{3} \biggr)f - (\gamma_e - \gamma_c) \biggl[1 + \frac{5}{2}(g^2-1) \biggr]</math>
These are consistent results because they result in the detailed force-balance relation, <math>P_0 - P_i = q^2 \Pi_\mathrm{eq} \, .</math>
  </td>
  <td align="center">
&nbsp; <math>~>~</math>&nbsp;
  </td>
  <td align="left">
<math>~0 \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>


==(5, 1) Bipolytropes==
==Examples==
In another [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|accompanying discussion]] we have derived analytic expressions describing the equilibrium structure of bipolytropes with <math>~(n_c, n_e) = (5, 1)</math>.  Can we perform a similar stability analysis of these configurations?
Work in progress!
 
=Best of the Best=
 
==One Derivation of Free Energy==
 
<div align="center">
<table border="0" cellpadding="5" align="center">


<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="11">
<font size="+1">Identification of Local ''Extrema'' in Free Energy</font>
  </th>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~\mathfrak{G}^*</math>
<math>~\nu</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~q</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~ \frac{\rho_e}{\rho_c} </math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
+ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>  
<math>~\Lambda_\mathrm{eq}</math>
~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .
</math>
   </td>
   </td>
</tr>
   <td align="center">
</table>
<math>~\chi_\mathrm{eq}</math>
</div>
 
==Another Derivation of Free Energy==
Hence the renormalized gravitational potential energy becomes,
<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>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\mathcal{A}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~\mathcal{B}</math>
- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f
\, ;</math>
   </td>
   </td>
</tr>
   <td align="center">
</table>
<math>~\mathcal{C}</math>
</div>
and the two, renormalized contributions to the thermal energy become,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
MIN/MAX
  </td>
  <td align="left">
<math>
\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~0.2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~9^{-1/3} = 0.48075</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.5</math>
\frac{\nu (1 + \Lambda) }{(\gamma_c-1)}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} \chi^{3-3\gamma_c}
\, ,</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~12.5644</math>
<tr>
   <td align="right">
<math>~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~2.091312</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.366531</math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{R^3 P_{ie}}{ E_\mathrm{norm} } \biggr]
\biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~0.037453</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~0.2090801</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.2308269</math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
R^3 K_e \rho_{ie}^{\gamma_e}
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~2.06252 \times 10^{-4}</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
MIN
  </td>
  <td align="left">
<math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
R^3 K_e \rho_\mathrm{norm}^{\gamma_e} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e} \biggl( \frac{\bar\rho}{\rho_\mathrm{norm}} \biggr)^{\gamma_e}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~0.4</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~4^{-1/3} = 0.62996</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.5</math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
  </td>
( \rho_\mathrm{norm} R_\mathrm{norm}^3) K_e \rho_\mathrm{norm}^{\gamma_e-1} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e}
  <td align="center">
</math>
<math>~4.21974</math>
  </td>
  <td align="center">
<math>~1.56498</math>
  </td>
  <td align="center">
<math>~0.707989</math>
  </td>
  <td align="center">
<math>~0.0220475</math>
  </td>
  <td align="center">
<math>~0.2143496</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~0.5635746</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~4.4626 \times 10^{-5}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
MIN
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr) \frac{K_e}{E_\mathrm{norm}} \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1}
\biggl[ \frac{G^3 M_\mathrm{tot}^2 }{K_c^3}\biggr]^{(\gamma_e-1)/(3\gamma_c-4)}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~0.473473</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~0.681838</math>
  </td>
  <td align="center">
<math>~0.516107</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
&nbsp;
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr]  \frac{K_c M_\mathrm{tot}}{E_\mathrm{norm}}
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<tr>
   <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~0.462927</math>
  </td>
  <td align="center">
<math>~0.08255</math>
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<font color="red">MAX</font>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] 
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
\biggl[ \frac{K_c^{3\gamma_c-4} M_\mathrm{tot}^{3\gamma_c-4}}{ G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} K_c^{-1}} \biggr]^{1/(3\gamma_c-4)}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~0.5</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~3^{-1/3} = 0.693361</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.5</math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] 
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
\biggl[ \frac{K_c^{3}}{G^{3} M_\mathrm{tot}^{2}} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)}
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~2.985115</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~1.42334</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~0.9448663</math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c-4)} \biggl[ \mathrm{BigTerm}\biggr]  \chi^{3-3\gamma_e}
</math>
   </td>
   </td>
</tr>
   <td align="center">
</table>
<math>~0.0152116</math>
</div>
 
Finally, then, we can state that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\mathfrak{f}_{WM}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~0.2152641</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
<math>~0.791882</math>
  </td>
  <td align="center">
<math>~1.5464 \times 10^{-5}</math>
  </td>
  <td align="center">
MIN
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~s_\mathrm{core}</math>
<math>~0.559839</math>
  </td>
  <td align="center">
<math>~0.729581</math>
  </td>
  <td align="center">
<math>~0.499188</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~0.75089</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~0.032196</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
"
1 + \Lambda \, ,
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
"
<tr>
   <td align="right">
<math>~(1-q^3) s_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
"
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<font color="red">MAX</font>
(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


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


<table border="1" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
  <th align="center" colspan="13">
<math>~\Lambda</math>  
<font size="+1">Free Energy Extrema when: &nbsp;&nbsp;&nbsp;  </font><math>~~~~~\frac{\rho_e}{\rho_c} = \frac{1}{2} ~~~~\Rightarrow~~~~ q^3 = \frac{\nu}{2-\nu}</math>
  </th>
</tr>
<tr>
   <td align="center">
<math>~\nu</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv~</math>
<math>~q</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~ \frac{\rho_e}{\rho_c} </math>
\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c}
\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}
= \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\chi_\mathrm{eq}^{3\gamma_c - 4}
\, .</math>
   </td>
   </td>
</tr>
   <td align="center">
</table>
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
</div>
 
We also want to ensure that envelope pressure matches the core pressure at the interface.  This means,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~K_e \rho_{ie}^{\gamma_e}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~K_c \rho_{ic}^{\gamma_c}</math>
<math>~\frac{1}{\lambda_{i}}\biggr|_\mathrm{eq}</math>
   </td>
   </td>
</tr>
  <td align="center">
 
<math>~\chi_\mathrm{eq}</math>
<tr>
  </td>
   <td align="right">
   <td align="center">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>
<math>~\mathcal{A}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\mathcal{B}_\mathrm{NEW}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>
<math>~\mathcal{C}_\mathrm{NEW}</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~G^*</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<font color="red">Stability</font>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
MIN/MAX
\rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
<math>~0.5</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\biggl( \frac{1}{3} \biggr)^{1/3}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
<math>~0.5</math>
\biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \biggr\}^{\gamma_c - \gamma_e}</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~2.985115</math>
<tr>
   <td align="right">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~1.423340</math>
  </td>
  <td align="center">
<math>~0.05466039</math>
  </td>
  <td align="center">
<math>~0.3152983</math>
  </td>
  <td align="center">
<math>~0.21526406</math>
  </td>
  <td align="center">
<math>~0.23552725</math>
  </td>
  <td align="center">
<math>~6.643899 \times 10^{-3}</math>
  </td>
  <td align="center">
<math>~+ 0.5176146</math>
  </td>
  <td align="center">
<math>~+0.429245</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c}
MIN
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
&nbsp;
&hellip;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&hellip;
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
&hellip;
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}
</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<tr>
   <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
&nbsp;
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
  </td>
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}
  <td align="center">
</math>
<math>~0.6674</math>
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
<math>~+0.55572115</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<font color="red">MAX</font>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<div align="center">
<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .
</math>
</div>


==Summary==
===Understanding Free-Energy Behavior===
'''<font color="#997700">Step 1:</font>''' Pick values for the separate coefficients, <math>\mathcal{A}, \mathcal{B},</math> and <math>\mathcal{C},</math> of the three terms in the normalized free-energy expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~\mathfrak{G}^*</math>
<math>~0.6</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~\biggl( \frac{3}{7} \biggr)^{1/3}</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} </math>
<math>~0.5</math>
   </td>
   </td>
</tr>
  <td align="center">
</table>
<math>~2.2507129</math>
</div>
  </td>
then plot the function, <math>\mathfrak{G}^*(\chi)</math>, and identify the value(s) of <math>~\chi_\mathrm{eq}</math> at which the function has an extremum (or multiple extrema).
   <td align="center">
 
<math>~1.31282895</math>
'''<font color="#997700">Step 2:</font>''' Note that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\mathcal{A}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~0.04160318</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\frac{\nu^2}{5q} \biggl\{ 
<math>~0.3411545</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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]
\biggr\}</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~0.21493717</math>
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~0.26165939</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ 
<math>~5.208750 \times 10^{-3}</math>
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr)
\biggr]</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<math>~+ 0.73532249</math>
<tr>
   <td align="right">
<math>~\mathcal{B}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~+0.0935217</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}  \biggl[ 
MIN
1+\Lambda_\mathrm{eq}
\biggr]</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="center">
<math>~\mathcal{C}</math>
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~(1-\nu)\biggl( \frac{K_e}{K_c} \biggr)^*
&nbsp;
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1}  \biggl\{ 
1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)}\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\biggr\}</math>
   </td>
   </td>
</tr>
   <td align="center">
 
<tr>
   <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~0.431745</math>
  </td>
  <td align="center">
"
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}
"
\biggl\{ 
\frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}</math>
   </td>
   </td>
</tr>
  <td align="center">
</table>
"
</div>
  </td>
where (see, for example, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Expression_for_Free_Energy|in the context of its original definition]]),
   <td align="center">
<div align="center">
<math>~+0.7367797</math>
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\Lambda_\mathrm{eq} \equiv
\frac{3}{2^2\pi \cdot 5} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4}
</math>  
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<font color="red">MAX</font>
\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\chi_\mathrm{eq}^{3\gamma_c - 4}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and, where,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="center">
<math>\biggl( \frac{K_e}{K_c} \biggr)^* \equiv \frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
<math>~0.7</math>
</math>
  </td>
  <td align="center">
<math>~\biggl( \frac{7}{13} \biggr)^{1/3}</math>
  </td>
  <td align="center">
<math>~0.5</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~1.7707809</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>
<math>~1.2209446</math>
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{1-\nu}{1-q^3} \biggr]^{-\gamma_e}
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c}
\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, .
</math>
   </td>
   </td>
</tr>
  <td align="center">
 
<math>~0.029500</math>
</table>
  </td>
</div>
  <td align="center">
 
<math>~0.3589388</math>
Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<div align="center">
<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~
\frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, .
</math>
</div>
 
'''<font color="#997700">Step 3:</font>'''  An analytic evaluation tells us that the following ''should'' happen.  Using the numerically derived value for <math>~\chi_\mathrm{eq}</math>, define,
<div align="center">
<math>~\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \, .</math>
</div>
We should then discover that,
<div align="center">
<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} =
\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\, .</math>
</div>
 
===Check It===
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{B} + \mathcal{C}^'</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~0.21330744</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~
<math>~0.28172532</math>
\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \biggl[  1+\Lambda_\mathrm{eq} \biggr]
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  \biggr\}
</math>
   </td>
   </td>
</tr>
  <td align="center">
</table>
<math>~3.389793 \times 10^{-3}</math>
</div>
  </td>
 
   <td align="center">
<div align="center">
<math>~+ 0.8953395</math>
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>\Rightarrow~~~~\mathcal{A} \biggl[ \Lambda_\mathrm{eq} \cdot 5\biggl( \frac{q}{\nu^2} \biggr)  \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~- 0.0767108</math>
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~
<font color="red">MAX</font>
1+\Lambda_\mathrm{eq}
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="center">
<math>
&hellip;
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{  1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
  </td>
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]   \biggr\}
   <td align="center">
</math>
&hellip;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&hellip;
   </td>
   </td>
   <td align="left">
   <td align="center">
<math>~
&nbsp;
1+\Lambda_\mathrm{eq}
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
   </td>
   </td>
</tr>
  <td align="center">
</table>
&nbsp;
</div>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~0.270615</math>
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
<math>~+0.89227216</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
MIN
  </td>
</tr>


<tr>
  <th align="center" colspan="13">
System should be <font color="red">''stable''</font> (with free energy ''minimum'') if: &nbsp;&nbsp;&nbsp; <math>~~~~~\frac{( \gamma_e - \frac{4}{3})}{(\gamma_e - \gamma_c)}  f - \biggl[1 + \frac{5}{2} (g^2-1) \biggr] ~>~ 0</math>
  </th>
</tr>
</table>
==Solution Strategy==
For a given set of free-energy coefficients, <math>~\mathcal{A}, \mathcal{B},</math> and <math>~\mathcal{C}</math>, along with a choice of the two adiabatic exponents <math>~(\gamma_c, \gamma_e)</math>, here's how to determine all of the physical parameters that are detailed in the above example table.
* '''<font color="darkgreen">Step 1:</font>'''  Guess a value of <math>~0 < q < 1</math>.
* '''<font color="darkgreen">Step 2:</font>'''  Given the pair of parameter values, <math>~(\mathcal{A}, q)</math>, determine the interface-density ratio, <math>~\rho_e/\rho_c</math>, by finding the appropriate root of the expression that defines the function, <math>~\mathcal{A}(q, \rho_e/\rho_c)</math>.  This can be straightforwardly accomplished because, [[User:Tohline/SSC/BipolytropeGeneralization#Explain_Logic|as demonstrated below]], the relevant expression can be written as a quadratic function of <math>~(\rho_e/\rho_c)</math>.
* '''<font color="darkgreen">Step 3:</font>'''  Given the pair of parameter values, <math>~(q, \rho_e/\rho_c)</math>, determine the value of the core-to-total mass ratio, <math>~\nu</math>, from the expression that was obtained from an integration over the mass, namely,
<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>
<math>~\frac{1}{\nu}</math>
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]  \biggr\}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,507: Line 2,504:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{q^3} - 1 \biggr) \, .</math>
\frac{1}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,516: Line 2,510:
</div>
</div>


* '''<font color="darkgreen">Step 4:</font>'''  Given the value of <math>~\mathcal{B}</math> along with the pair of parameter values, <math>~(q, \nu)</math>, the above expression that defines <math>~\mathcal{B}</math> can be solved to give the relevant value of the dimensionless parameter, <math>~\Lambda_\mathrm{eq}.</math>
* '''<font color="darkgreen">Step 5:</font>'''  The value of <math>~\mathcal{C}^'</math> &#8212; the coefficient that appears on the right-hand-side of the above expression that defines <math>~\mathcal{C}</math> &#8212; can be determined, given the values of parameter triplet, <math>~(q, \nu, \Lambda_\mathrm{eq})</math>.
* '''<font color="darkgreen">Step 6:</font>'''  Given the value of <math>~\mathcal{C}</math> and the just-determined value of the coefficient <math>~\mathcal{C}^'</math>, the normalized equilibrium radius, <math>~\chi_\mathrm{eq},</math> that corresponds to the value of <math>~q</math> that was ''guessed'' in Step #1 can be determined from the above definition of <math>~\mathcal{C}</math>, specifically,
<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>\Rightarrow~~~~
<math>~\chi_\mathrm{eq}\biggr|_\mathrm{guess} </math>
\frac{1}{\Lambda_\mathrm{eq}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,529: Line 2,525:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)} \, .</math>
q^3 \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]  \biggr\}
- \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


* '''<font color="darkgreen">Step 7:</font>'''  But, independent of this ''guessed'' value of <math>~\chi_\mathrm{eq},</math> the condition for virial equilibrium &#8212; which identifies extrema in the free-energy function &#8212; gives the following expression for the normalized equilibrium radius:
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow~~~~
<math>~\chi_\mathrm{eq}\biggr|_\mathrm{virial} </math>
\frac{2}{\Lambda_\mathrm{eq}}  
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,548: Line 2,542:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~\biggl[ \frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} \biggr]^{1/(4 - 3\gamma_c)} \, .</math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) ( q - q^3 )
+ \frac{2}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (1 - q^5 ) - \frac{5}{2}( q^3 - q^5 ) \biggr]
- 5 \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3)  
- \frac{3}{q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
* '''<font color="darkgreen">Step 8:</font>'''  If <math>~\chi_\mathrm{eq}|_\mathrm{guess} \ne \chi_\mathrm{eq}|_\mathrm{virial}</math>, return to Step #1 and guess a different value of <math>~q</math>.  Repeat Steps #1 through #7 until the two independently derived values of the normalized radius match, to a desired level of precision.
* '''<font color="darkgreen">Keep in mind:</font>'''  (A) A graphical representation of the free-energy function, <math>~\mathfrak{G}(\chi)</math>, can also be used to identify the "correct" value of <math>~\chi_\mathrm{eq}</math> and, ultimately, the above-described iteration loop should converge on this value.  (B) The free-energy function may exhibit more than one (or, actually, no) extrema, in which case more than one (or no) value of <math>~q</math> should lead to convergence of the above-described iteration loop.
{{LSU_WorkInProgress}}
=Detailed Derivations=
Dividing the free-energy expression through by <math>~E_\mathrm{norm}</math> generates,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}^* \equiv \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,566: Line 2,569:
   <td align="left">
   <td align="left">
<math>
<math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ ( q - q^3 ) - (-2 + 3q - q^3) \biggr]
- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM}
+ \frac{1}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_c \rho_{ic}^{\gamma_c-1} }{E_\mathrm{norm}} \biggr]
\biggl[ 2(1 - q^5 ) - 5( q^3 - q^5 ) - 3(-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
</math>
   </td>
   </td>
Line 2,578: Line 2,580:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>  
10\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 1-q \biggr]
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_e \rho_{ie}^{\gamma_e-1} }{E_\mathrm{norm}}  \biggr]  
+ \frac{5}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2
\biggl[ 1  - 3q^2 + 2q^3 \biggr]
</math>
</math>
   </td>
   </td>
Line 2,591: Line 2,591:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>\Rightarrow~~~~
&nbsp;
\frac{1}{\Lambda_\mathrm{eq}} \biggl[ \frac{2q^2}{5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggr]
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,600: Line 2,598:
   <td align="left">
   <td align="left">
<math>
<math>
2q^2 (1-q )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) ( 1 - 3q^2 + 2q^3 )
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl[ \frac{K_c G^{(3\gamma_c -4)}M_\mathrm{tot}^{2(3\gamma_c -4)}}{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{1}{R} \biggr)  
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Fortunately, this precisely matches our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#LambdaDeff|earlier derivation]], which states that,
<div align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{1}{\Lambda}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; <math>~=</math>&nbsp;  
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>\frac{5}{2}(g^2-1) =
<math>
\frac{5}{2}\biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) +
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_c^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}}  
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \, .</math>
\biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c-1}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
=Playing With One Example=
By setting,
<table border="1" align="center" cellpadding="5">


<tr>
<tr>
   <td align="center" colspan="3">
  <td align="right">
<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>
&nbsp;
  </td>
   <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>  
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_e^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}}
\biggr]^{1/(3\gamma_c -4)}  \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_e-1}
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
  <th align="center">
   <td align="right">
<math>~\mathcal{A}</math>
&nbsp;
  </th>
  <th align="center">
<math>~\mathcal{B}</math>
  </th>
  <th align="center">
<math>~\mathcal{C}</math>
  </th>
</tr>
 
<tr>
   <td align="center">
2.5
   </td>
   </td>
   <td align="center">
   <td align="center">
1.0
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
2.0
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)
</math>
   </td>
   </td>
</tr>
</tr>
</table>
a plot of <math>~\mathfrak{G}^*</math> versus <math>~\chi</math> exhibits the following, two extrema:
<table border="1" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
extremum
&nbsp;
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\mathfrak{G}^*</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)}</math>
<math>
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_c-1}
\biggl[ \frac{ K_c^{3\gamma_c -3} }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}}
\biggr]^{1/(3\gamma_c -4)}  \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)}  
\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_c-1)/(3\gamma_c -4)}
\biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~\mathcal{C}^'</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\frac{\mathcal{A}}{\mathcal{B} +\mathcal{C}^'}</math>
<math>  
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_e-1}
\biggl[ \frac{K_c^{3\gamma_c - 3} (K_e/K_c)^{3\gamma_c -4}  }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} 
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)}
\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_e-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}  
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,692: Line 2,684:
<tr>
<tr>
   <td align="right">
   <td align="right">
MIN
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>1.1824</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>-0.611367</math>
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr)
</math>
   </td>
   </td>
   <td align="center">
</tr>
&nbsp;<math>~~~~\Rightarrow~~~~</math>&nbsp;
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>0.66891</math>
&nbsp;
  </td>
  <td align="left">
<math>
+ \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr]  \biggl( \frac{3}{4\pi} \biggr)^{\gamma_c-1} \biggl[ M_\mathrm{tot}^{3\gamma_c-4} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)}
\biggl[ M_\mathrm{tot}^{-2} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} 
\biggl[ M_\mathrm{tot}^{-3\gamma_c+6} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)}
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)}
\biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1}
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>1.3378</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>1.0693</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>1.0694</math>
<math>  
~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ M_\mathrm{tot}^{2(\gamma_e-1)-2(\gamma_c-1)}\biggr]^{1/(3\gamma_c-4)}
\biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)}
\biggl[ \biggl( \frac{K_c}{G} \biggr)^{(\gamma_c-1)-(\gamma_e-1)} \biggr]^{3/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1}
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,719: Line 2,733:
<tr>
<tr>
   <td align="right">
   <td align="right">
MAX
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>9.6722</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>+0.508104</math>
<math>
  </td>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
  <td align="center">
+ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
<math>~\Rightarrow</math>
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}
  </td>
</math>
  <td align="center">
<math>0.004313</math>
  </td>
  <td align="center">
<math>0.008625</math>
  </td>
  <td align="center">
<math>2.4786</math>
  </td>
  <td align="center">
<math>2.4786</math>
   </td>
   </td>
</tr>
</tr>
</table>
The last two columns of this table confirm the internal consistency of the relationships presented in '''<font color="#997700">Step 3</font>''', above.  But what does this mean in terms of the values of <math>~\nu</math>, <math>~q</math>, and the related ratio of densities at the interface, <math>~\rho_e/\rho_c</math>?
Let's assume that what we're trying to display and examine is the behavior of the free-energy ''surface'' for a fixed value of the ratio of densities at the interface.  Once the value of <math>~\rho_e/\rho_c</math> has been specified, it is clear that the value of <math>~q</math> (and, hence, also <math>~\nu</math>) is set because <math>~\mathcal{A}</math> has also been specified.  But our specification of <math>~\mathcal{B}</math> along with <math>~\rho_e/\rho_c</math> also forces a particular value of <math>~q</math>.  It is unlikely that these two values of <math>~q</math> will be the same.  In reality, once <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> have both been specified, they force a particular <math>~(\nu, q)</math> pair.  How do we (easily) figure out what this pair is?


Let's begin by rewriting the expressions for <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> in terms of just <math>~q</math> and the ratio, <math>~\rho_e/\rho_c</math>, keeping in mind that, for the case of a uniform-density core (of density, <math>~\rho_c</math>) and a uniform-density envelope (of density, <math>~\rho_e</math>),
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\rho_e}{\rho_c}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, ,</math>
<math>  
~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}  \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
hence,
 
<span id="pressureMatch">We also want to ensure that envelope pressure matches the core pressure at the interface.</span>  This means,
<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>~\nu</math>
<math>~K_e \rho_{ie}^{\gamma_e}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,775: Line 2,777:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ 1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{(1-q^3)}{q^3} \biggr]^{-1}</math>  
<math>~K_c \rho_{ic}^{\gamma_c}</math>
   </td>
   </td>
  <td align="center">&nbsp;&nbsp; and &nbsp; &nbsp;</td>
</tr>
 
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{q^3}{\nu}</math>
<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,785: Line 2,789:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr] \, .</math>  
<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Putting the expression for <math>~\mathcal{A}</math> in the desired form is simple because <math>~\nu</math> only appears as a leading factor.  Specifically, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{A}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,803: Line 2,801:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}  
\biggl\{  1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2} - 1 \biggr)
\rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]
\biggr\}</math>
   </td>
   </td>
</tr>
</tr>
Line 2,818: Line 2,814:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi }{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\biggl\{ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2
\biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \biggr\}^{\gamma_c - \gamma_e}</math>
\biggl[1-\frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr]
\biggr\} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
The expression for <math>~\mathcal{B}</math> can be written in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{B}</math>
<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 2,839: Line 2,828:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c}  
1+\frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\chi_\mathrm{eq}^{3\gamma_c - 4}  
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}  
\biggr\}</math>
</math>
   </td>
   </td>
</tr>
</tr>
Line 2,854: Line 2,843:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\nu \biggl[ \biggl( \frac{4\pi}{3} \biggr)\frac{q^3}{\nu} \biggr]^{1-\gamma_c}  
<math>
+\frac{\pi q^5}{5} \biggl( \frac{\nu^2}{q^6} \biggr) \chi_\mathrm{eq}^{3\gamma_c - 4}  
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}
</math>
</math>
   </td>
   </td>
Line 2,868: Line 2,858:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \frac{q^3}{\nu} \biggr]^{-\gamma_c}  
<math>
+\frac{\pi q^5}{5} \biggl( \frac{q^3}{\nu} \biggr)^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4}  
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence, we can write the normalized (and dimensionless) free energy as,


<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 2,882: Line 2,878:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3  \biggr]^{-\gamma_c} 
<math>
+\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}  
+ \biggl\{ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} + \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{ \rho_{ic} }{ \rho_{ie} } \biggr)\biggr\}  
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}  
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} \, .
</math>
</math>
   </td>
   </td>
Line 2,889: Line 2,888:
</table>
</table>
</div>
</div>
Generally speaking, the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, which appears in the expression for <math>~\mathcal{B}</math>, is not known ahead of time.  Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, <math>~\mathcal{A}</math>, <math>~\mathcal{B}</math>, and <math>~\mathcal{C}</math>, and determine the equilibrium radius by looking for extrema in the free-energy function.  And because <math>~\chi_\mathrm{eq}</math> is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, <math>~(\nu, q)</math>, give self-consistent values for the coefficient pair, <math>~(\mathcal{A}, \mathcal{B})</math>.


Because we are using a uniform density core and uniform density envelope as our base model, however, we ''do'' know the analytic solution for <math>~\chi_\mathrm{eq}</math>.  As stated above, it is,
Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<div align="center">
<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .
</math>
</div>
 
==Free Energy and Its Derivatives==
 
Now, the free energy can be written as,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</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>~\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} </math>
<math>~U_\mathrm{tot} + W</math>
   </td>
   </td>
</tr>
</tr>
Line 2,912: Line 2,919:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi q^2}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}\biggl( \frac{\nu}{q^3} \biggr)
<math>~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} + W</math>
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]  
\cdot \biggl[ \frac{q^3}{\nu} \biggr]^{\gamma_c-1} </math>
   </td>
   </td>
</tr>
</tr>
Line 2,927: Line 2,931:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\pi}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}
<math>~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] C_\mathrm{core} R^{3-3\gamma_c}  
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) ( q^2-q^3 ) +
+ \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \, .</math>
\frac{\rho_e}{\rho_c} (1 - q^2) \biggr]  
\cdot \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3\biggr]^{\gamma_c-2} </math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Combining this expression with the one for <math>~\mathcal{B}</math> gives us the desired result &#8212; although, strictly speaking, it is cheating!  We can now methodically choose <math>~(\nu, q)</math> pairs and map them into the corresponding values of <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.  And, via an analogous "cheat," the choice of <math>~(\nu, q)</math> also gives us the self-consistent value of <math>~\mathcal{C}</math>.  In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.


=Second Example=
The first derivative of the free energy with respect to radius is, then,
==Explain Logic==
 
[[Image:FreeEnergyExample.jpg|right|400px]]  The figure presented here, on the right, shows a plot of the free energy, as a function of the dimensionless radius, <math>~\mathfrak{G}^*(\chi)</math>, where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathfrak{G}^*</math>
<math>~\frac{d\mathfrak{G}}{dR}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,</math>
<math>~ -2 C_\mathrm{core} R^{2-3\gamma_c} -2 C_\mathrm{env} R^{2-3\gamma_e} + A R^{-2} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
and, where we have used the parameter values,
</div>
<table border="1" align="center" cellpadding="5">
 
<tr>
And the second derivative is,
  <td align="center" colspan="3">
<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>
  </td>
</tr>


<tr>
<div align="center">
  <th align="center">
<table border="0">
<math>~\mathcal{A}</math>
  </th>
  <th align="center">
<math>~\mathcal{B}</math>
  </th>
  <th align="center">
<math>~\mathcal{C}</math>
  </th>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
0.201707
<math>~\frac{d^2\mathfrak{G}}{dR^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
0.0896
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
0.002484
<math>~ -2 (2-3\gamma_c) C_\mathrm{core} R^{1-3\gamma_c} -2 (2-3\gamma_e) C_\mathrm{env} R^{1-3\gamma_e} - 2A R^{-3} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
Directly from this plot we deduce that this free-energy function exhibits a minimum at <math>~\chi_\mathrm{eq} = 0.1235</math> and that, at this equilibrium radius, the configuration has a free-energy value, <math>~\mathfrak{G}^*(\chi_\mathrm{eq} ) = -2.0097</math>.  Via the steps described below, we demonstrate that this identified equilibrium radius is appropriate for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope (with the just-specified core and envelope adiabatic indexes) that has the following physical properties:
* Fractional core mass, <math>~\nu = 0.1</math>;
* Core-envelope interface located at <math>~r_i/R = q = 0.435</math>;
* Density jump at the core-envelope interface, <math>~\rho_e/\rho_c = 0.8</math>.
'''<font color="red">Step 1:</font>'''  Because the ratio, <math>~q^3/\nu</math>, is a linear function of the density ratio, <math>~\rho_e/\rho_c</math>, the full definition of the free-energy coefficient, <math>~\mathcal{A}</math>, can be restructured into a quadratic equation that gives the density ratio for any choice of the parameter pair, <math>~(q, \mathcal{A})</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~5 \biggl( \frac{q^3}{\nu} \biggr)^2 \mathcal{A}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>  
<math>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) C_\mathrm{core} R^{3-3\gamma_c} +
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
(3\gamma_e-2) C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \biggr]</math>
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr)
</math>
   </td>
   </td>
</tr>
</tr>
Line 3,015: Line 2,992:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~~ 5\mathcal{A} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^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>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) S_\mathrm{core} +
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
(3\gamma_e-2) S_\mathrm{env} +W \biggr] \, .</math>
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
and this can be written in the form,
 
==Equilibrium==
The radius, <math>~R_\mathrm{eq}</math>, of the equilibrium configuration(s) is determined by setting the first derivative of the free energy to zero.  Hence,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math> \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 a + \biggl( \frac{\rho_e}{\rho_c} \biggr) b + c</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>  
<math>~ 2 C_\mathrm{core} R_\mathrm{eq}^{2-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{2-3\gamma_e} - A R_\mathrm{eq}^{-2} </math>
0 \, ,
</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>~a</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\mathcal{A} (1-q^3)^2 - 1 + \frac{5}{2}q^3 - \frac{3}{2}q^5 \, ,</math>
<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 C_\mathrm{core} R_\mathrm{eq}^{3-3\gamma_c} +  
2 C_\mathrm{env} R_\mathrm{eq}^{3-3\gamma_e} - A R_\mathrm{eq}^{-1} \biggr]</math>
   </td>
   </td>
</tr>
</tr>
Line 3,066: Line 3,038:
<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>~10\mathcal{A} q^3(1-q^3) - \frac{5}{2}q^3 (1-q^2) \, ,</math>
<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 S_\mathrm{core} +
2 S_\mathrm{env} +W \biggr]</math>
   </td>
   </td>
</tr>
</tr>
Line 3,078: Line 3,051:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~c</math>
<math>\Rightarrow ~~~~ 2 S_\mathrm{tot} + W </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~5\mathcal{A} q^6 - q^5 \, .</math>
<math>~0 \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence,
This is the familiar statement of virial equilibrium.  From it we should always be able to derive the radius of equilibrium configurations.
 
==Stability==
 
To assess the relative stability of an equilibrium configuration, we need to determine the sign of the second derivative of the free energy, evaluated at the equilibrium radius.  If the sign of the second derivative is positive, the system is dynamically stable; if the sign is negative, he system is dynamically unstable.  Using the above statement of virial equilibrium, that is, setting,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{\rho_e}{\rho_c}</math>
<math>~2 S_\mathrm{tot} + W</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=~</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{2a} \biggl[\pm ( b^2 - 4ac)^{1/2} - b \biggr] \, .</math>
<math>~0 \, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
(For our physical problem it appears as though only the positive root is relevant.)  For the purposes of this example, we set <math>~\mathcal{A} = 0.2017</math> and examined a range of values of <math>~q</math> to find a physically interesting value for the density ratio.  We picked:
<table border="1" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~\mathcal{A}</math>
<math>\Rightarrow ~~~~ S_\mathrm{env} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~q</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
&nbsp;
<math>~- S_\mathrm{core} - \frac{W}{2} \, ,</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~a</math>
</table>
</div>
we obtain,
 
<div align="center">
<table border="0">
 
<tr>
   <td align="right">
<math>~\frac{d^2\mathfrak{G}}{dR^2}\biggr|_\mathrm{eq}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~b</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~c</math>
<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ (3\gamma_c-2) S_\mathrm{core}
+W - (3\gamma_e-2)\biggl( S_\mathrm{core} + \frac{W}{2}\biggr) \biggr]_\mathrm{eq} </math>
   </td>
   </td>
   <td align="center">
</tr>
 
<tr>
   <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\frac{\rho_e}{\rho_c}</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
&nbsp;
<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ 3(\gamma_c-\gamma_e) S_\mathrm{core}
  </td>
+ \biggl(2 - \frac{3}{2}\gamma_e\biggr)W \biggr]_\mathrm{eq} </math>
  <td align="center">
<math>~\nu</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.2017</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.435</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>\Rightarrow</math>
<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ (\gamma_c-\gamma_e) S_\mathrm{core}
+ \frac{1}{2}\biggl(\frac{4}{3} - \gamma_e\biggr)W \biggr]_\mathrm{eq} </math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.03173</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~-0.01448</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~-0.008743</math>
<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ -\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) -
(\gamma_e-\gamma_c) S_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>\Rightarrow</math>
</table>
</div>
So, if when evaluated at the equilibrium state, the expression inside of the square brackets of this last expression is negative, the equilibrium configuration will be dynamically unstable.  We have chosen to write the expression in this particular final form because we generally will be interested in bipolytropes for which the adiabatic exponent of the envelope is greater than <math>~4/3</math> and the adiabatic exponent of the core is less than or equal to <math>~4/3</math> &#8212; that is, <math>~\gamma_e > 4/3 \ge \gamma_c</math>.  Hence, because the gravitational potential energy, <math>~W</math>, is intrinsically negative, the system will be dynamically unstable only if the second term (involving <math>~S_\mathrm{core}</math>) is greater in magnitude than the first term (involving <math>~W</math>).
 
It is worth noting that, instead of drawing upon <math>~S_\mathrm{core}</math> and <math>~W</math> to define the stability condition, we could have used an appropriate combination of <math>~S_\mathrm{env}</math> and <math>~W</math>, or the <math>~S_\mathrm{core}</math> and <math>~S_\mathrm{env}</math> pair.  Also, for example, because the virial equilibrium condition is <math>~S_\mathrm{tot} = -W/2</math>, it is easy to see that the following inequality also equivalently defines stability:
<div align="center">
<table border="0">
 
<tr>
   <td align="right">
<math>~ S_\mathrm{tot}\biggl( \gamma_e - \frac{4}{3}\biggr) -
(\gamma_e-\gamma_c) S_\mathrm{core} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.80068</math>
<math>~>~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>\Rightarrow</math>
<math>~ 0 \, .</math>
  </td>
  <td align="center">
<math>~0.10074</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


=Examples=


'''<font color="red">Step 2:</font>''' Next, we chose the parameter pair,
==(0, 0) Bipolytropes==
===Review===
In an [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|accompanying discussion]] we have derived analytic expressions describing the equilibrium structure and the stability of bipolytropes in which both the core and the envelope have uniform densities, that is, bipolytropes with <math>~(n_c, n_e) = (0, 0)</math>. From this work, we find that integrals over the mass and pressure distributions give:
<div align="center">
<div align="center">
<math>
<table border="0">
~\biggl(q, \frac{\rho_e}{\rho_c} \biggr) = (0.43500, 0.80000)
</math>
</div>
and determined the following parameter values from the known analytic solution:
<table border="1" cellpadding="5" align="center">
<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~\nu</math>
<math>~ \frac{W}{R_\mathrm{eq}^3 P_i} = - \frac{A}{R_\mathrm{eq}^4 P_i} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
&nbsp;<math>- ~ \frac{3}{5} \biggl[ \frac{GM_\mathrm{tot}^2}{R^4P_i} \biggr] \biggl( \frac{\nu^2}{q} \biggr) f </math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~\Lambda_\mathrm{eq}</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\chi_\mathrm{eq}</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\mathcal{A}</math>
&nbsp;<math>- ~4\pi q^3 \Lambda f \, ,</math>
  </td>
  <td align="center">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.100816</math>
<math>~\frac{S_\mathrm{core}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{core}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~43.16365</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~3.923017</math>
<math>
~2\pi q^3 (1 + \Lambda) \, ,</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.13684</math>
 
<tr>
   <td align="right">
<math>~\frac{S_\mathrm{env}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{env}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.12349</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.201707</math>
<math>
  </td>
2\pi \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
  <td align="center">
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,</math>
<math>~0.089625</math>
  </td>
  <td align="center">
<math>~0.002484</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where,
<div align="center">
<table border="0">


==Construction Multiple Curves to Define a Free-Energy Surface==
<tr>
Okay.  Now that we have the hang of this, let's construct a sequence of curves that represent physical evolution at a fixed interface-density ratio, <math>~\rho_e/\rho_c</math>, but for steadily increasing core-to-total mass ratio, <math>~\nu</math>.  Specifically, we choose,
  <td align="right">
<div align="center">
<math>~\Lambda</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
<math>
~\frac{\rho_e}{\rho_c} = \frac{1}{2} \, .
\frac{3}{2^2 \cdot 5\pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} \, ,</math>
</math>
   </td>
</div>
From the known analytic solution, here are parameters defining several different equilibrium models:
<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="9">
<font size="+1">Identification of Local ''Minimum'' in Free Energy</font>
   </th>
</tr>
</tr>
<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~\nu</math>
<math>~f(q,\rho_e/\rho_c)</math>  
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~q</math>
<math>~\equiv~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] </math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\Lambda_\mathrm{eq}</math>
<math>~=~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\chi_\mathrm{eq}</math>
<math>1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
+ \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \, ,</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~\mathcal{A}</math>
 
<tr>
   <td align="right">
<math>~g^2(q,\rho_e/\rho_c)</math>  
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\mathcal{B}</math>
<math>~\equiv~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\mathcal{C}</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>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.2</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~9^{-1/3} = 0.48075</math>
<math>~\equiv~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~12.5644</math>
<math>1 + \biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q)
+ \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 )  \biggr] \, ,
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~2.091312</math>
</table>
</div>
 
===Renormalize===
Let's renormalize these energy terms in order to more readily relate them to the [[User:Tohline/SSC/BipolytropeGeneralization#Setup|generalized expressions derived above]].
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~R^3 P_i</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.366531</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.037453</math>
<math>
~R^3 K_c \biggl(\frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi}\biggr) \frac{M_\mathrm{tot}}{R^3} \biggr]^{\gamma_c}
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.2090801</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.2308269</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~2.06252 \times 10^{-4}</math>
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
K_c M_\mathrm{tot}^{\gamma_c} R_\mathrm{norm}^{3-3\gamma_c}</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.4</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~4^{-1/3} = 0.62996</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~4.21974</math>
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
\biggl\{ K_c^{3\gamma_c -4} M_\mathrm{tot}^{\gamma_c(3\gamma_c -4)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{3-3\gamma_c} \biggr\}^{1/(3\gamma_c -4)}</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~1.56498</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.707989</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.0220475</math>
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
\biggl\{ \frac{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} }{K_c} \biggr\}^{1/(3\gamma_c -4)}</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.2143496</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.5635746</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~4.4626 \times 10^{-5}</math>
<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
E_\mathrm{norm} \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Also,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.5</math>
<math>~\biggl[ \frac{GM_\mathrm{tot}^2}{R} \biggr]^{3\gamma_c -4}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~3^{-1/3} = 0.693361</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~2.985115</math>
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} G^{3\gamma_c -4} M_\mathrm{tot}^{6\gamma_c -8}
\biggl( \frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma_c}
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~1.42334</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.9448663</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~0.0152116</math>
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} \biggl[ \frac{ G^{3\gamma_c -3} M_\mathrm{tot}^{5\gamma_c -6} }{K_c} \biggr]
</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~0.2152641</math>
 
<tr>
   <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0.791882</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~1.5464 \times 10^{-5}</math>
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} E_\mathrm{norm}^{3\gamma_c-4} \, .
</math>
   </td>
   </td>
</tr>
</tr>


<tr>
<tr>
   <td align="left" colspan="5">
   <td align="right">
Here we are examining the behavior of the free-energy function for bipolytropic models having <math>~(n_c, n_e) = (0, 0)</math>, <math>~(\gamma_c, \gamma_e) = (6/5, 2)</math>, and a density ratio at the core-envelope interface of <math>~\rho_e/\rho_c = 1/2</math>.  The figure shown here, on the right, displays the three separate free-energy curves, <math>~\mathfrak{G}^*(\chi)</math> &#8212; where, <math>~\chi \equiv R/R_\mathrm{norm}</math> is the normalized configuration radius &#8212; that correspond to the three values of <math>~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math> given in the first column of the above table.  Along each curve, the local free-energy minimum corresponds to the the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, recorded in column 6 of the above table.
<math>
\Rightarrow ~~~~\frac{GM_\mathrm{tot}^2}{R^4 P_i} </math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="center" colspan="4">
   <td align="left">
[[Image:ThreeFreeEnergyCurves.png|center|300px]]
<math>
~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3\gamma_c - 3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c}
</math>
   </td>
   </td>
</tr>
</tr>
</table>


Each of the free-energy curves shown above has been entirely defined by our specification of the three coefficients in the free-energy function, <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math>.  In each case, the values of these three coefficients was judiciously chosen to ''produce'' a curve with a local minimum at the correct value of <math>~\chi_\mathrm{eq}</math> corresponding to an equilibrium configuration having the desired <math>~(\nu, \rho_e/\rho_c)</math> model parameters.  Upon plotting these three curves, we noticed that two of the curves &#8212; curves for <math>~\nu = 0.4</math> and <math>~\nu = 0.5</math> &#8212; also display a local ''maximum''.  Presumably, these maxima also identify equilibrium configurations, albeit unstable ones.  From a careful inspection of the plotted curves, we have identified the value of <math>~\chi_\mathrm{eq}</math> that corresponds to the two newly discovered (unstable) equilibrium models; these values are recorded in the table that immediately follows this paragraph.  By construction, we also know what values of <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math> are associated with these two identified equilibria; these values also have been recorded in the table.  But it is not immediately obvious what the values are of the <math>~(\nu, \rho_e/\rho_c)</math> model parameters that correspond to these two equilibrium models.
<tr>
 
   <td align="right">
<table border="1" cellpadding="5" align="center">
&nbsp;
<tr>
   <td align="center" colspan="9">
Subsequently Identified Local Energy ''Maxima''
   </td>
   </td>
</tr>
<tr>
   <td align="center">
   <td align="center">
<math>~\chi_\mathrm{eq}</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\mathfrak{G}^*</math>
<math>
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} \, .
</math>
   </td>
   </td>
  <td align="center">
</tr>
<math>~\biggl( \frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} \biggr) = \chi_\mathrm{eq}^{4-3\gamma_c}</math>
</table>
</td>
</div>
   <td align="center">
 
<math>~\therefore</math>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\Lambda</math>  
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\therefore</math>
<math>~\equiv~</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\therefore</math>
<math>
\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}
\, .</math>
   </td>
   </td>
   <td align="center">
</tr>
<math>~\mathcal{A}</math>
</table>
</div>
 
Given that <math>~\rho_{ic}/\bar\rho = \nu/q^3</math> for the <math>~(n_c, n_e) = (0, 0)</math> bipolytrope,  we can finally write,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
<math>~\frac{R^3 P_i}{E_\mathrm{norm}}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\mathcal{B}</math>
<math>~=</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>~\mathcal{C}</math>
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}
\, ,</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="center">
   <td align="right">
<math>~0.08255</math>
<math>~\Lambda</math>  
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~+ 4.87562</math>
<math>~\equiv~</math>
   </td>
   </td>
   <td align="center">
  <td align="left">
<math>~0.368715</math>
<math>
\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} =
\frac{1}{5} \frac{\nu}{q} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}
\, .</math>
  </td>
</tr>
</table>
</div>
Hence the renormalized gravitational potential energy becomes,
<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>
- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f
\, ;</math>
  </td>
</tr>
</table>
</div>
and the two, renormalized contributions to the thermal energy become,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0</math>
<math>~=~</math>
   </td>
  </td>
   <td align="center">
  <td align="left">
<math>~0</math>
<math>
\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}
\, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{4\pi}{3(\gamma_e-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(3\gamma_c-3\gamma_e)/(3\gamma_c-4)}
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_e}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
\times \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\, ,</math>
  </td>
</tr>
 
</table>
</div>
 
Finally, then, we can state that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathfrak{f}_{WM}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~s_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
1 + \Lambda \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~(1-q^3) s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\, .</math>
  </td>
</tr>
</table>
</div>
 
===Virial Equilibrium and Stability Evaluation===
With these expressions in hand, we can deduce the equilibrium radius and relativity stability of <math>~(n_c, n_e) = (0, 0)</math> bipolytropes using the generalized expressions provided above.  For example, from the statement of virial equilibrium <math>~(2S_\mathrm{tot} = - W )</math> we obtain,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~q^3 (1 + \Lambda) + (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)  </math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~q^3 \Lambda \biggl[ 1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
+ \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \biggr] </math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0">
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{1}{\Lambda}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3)
+ \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5)
- \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-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) (q-q^3 + 2 -3q +q^3)
+ \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5 +3 - 15q^2+15q^3 -3q^5) 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\frac{5}{2}\biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q)
+ \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 )  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\frac{5}{2}(g^2-1)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \biggl[ \frac{P_i}{GM_\mathrm{tot}^2} \biggr] R_\mathrm{eq}^4</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \, .
</math>
  </td>
</tr>
</table>
</div>
 
Or, given the above renormalization, this expression can be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>
\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c } \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \frac{R}{R_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\biggl\{ \biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1)
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggr\}^{1/(4-3\gamma_c)} \, .
</math>
  </td>
</tr>
</table>
</div>
 
 
And the condition for dynamical stability is,
<div align="center">
<table border="0">
 
<tr>
  <td align="right">
<math>-\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) -
(\gamma_e-\gamma_c) S_\mathrm{core} </math>
  </td>
  <td align="center">
&nbsp; <math>~>~</math>&nbsp;
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~ 2\pi q^3 \Lambda \biggl[ \biggl( \gamma_e - \frac{4}{3}\biggr) f -
(\gamma_e-\gamma_c) \biggl( 1 + \frac{1}{\Lambda}\biggr) \biggr] </math>
  </td>
  <td align="center">
&nbsp; <math>~>~</math>&nbsp;
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\biggl( \gamma_e - \frac{4}{3} \biggr)f - (\gamma_e - \gamma_c) \biggl[1 + \frac{5}{2}(g^2-1) \biggr]</math>
  </td>
  <td align="center">
&nbsp; <math>~>~</math>&nbsp;
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>
 
==(5, 1) Bipolytropes==
In another [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|accompanying discussion]] we have derived analytic expressions describing the equilibrium structure of bipolytropes with <math>~(n_c, n_e) = (5, 1)</math>.  Can we perform a similar stability analysis of these configurations?
Work in progress!
 
=Best of the Best=
 
==One Derivation of Free Energy==
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
- \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1}
+ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}
\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>
~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}  \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, .
</math>
  </td>
</tr>
</table>
</div>
 
==Another Derivation of Free Energy==
Hence the renormalized gravitational potential energy becomes,
<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>
- \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f
\, ;</math>
  </td>
</tr>
</table>
</div>
and the two, renormalized contributions to the thermal energy become,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{U_\mathrm{core}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_c-1)} \biggl[ \frac{S_\mathrm{core}}{ E_\mathrm{norm} } \biggr]</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{\nu (1 + \Lambda) }{(\gamma_c-1)}
\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} \chi^{3-3\gamma_c}
\, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\frac{U_\mathrm{env}}{ E_\mathrm{norm} } = \frac{2}{3(\gamma_e-1)} \biggl[ \frac{S_\mathrm{env}}{ E_\mathrm{norm} } \biggr]</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{R^3 P_{ie}}{ E_\mathrm{norm} } \biggr]
\biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-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{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
R^3 K_e \rho_{ie}^{\gamma_e}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
R^3 K_e \rho_\mathrm{norm}^{\gamma_e} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e} \biggl( \frac{\bar\rho}{\rho_\mathrm{norm}} \biggr)^{\gamma_e}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr]
( \rho_\mathrm{norm} R_\mathrm{norm}^3) K_e \rho_\mathrm{norm}^{\gamma_e-1} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr) \frac{K_e}{E_\mathrm{norm}} \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1}
\biggl[ \frac{G^3 M_\mathrm{tot}^2 }{K_c^3}\biggr]^{(\gamma_e-1)/(3\gamma_c-4)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr]  \frac{K_c M_\mathrm{tot}}{E_\mathrm{norm}}
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] 
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
\biggl[ \frac{K_c^{3\gamma_c-4} M_\mathrm{tot}^{3\gamma_c-4}}{ G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} K_c^{-1}} \biggr]^{1/(3\gamma_c-4)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e}
\biggl[ \mathrm{BigTerm}\biggr] 
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)}
\biggl[ \frac{K_c^{3}}{G^{3} M_\mathrm{tot}^{2}} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr)
\biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c-4)} \biggl[ \mathrm{BigTerm}\biggr]  \chi^{3-3\gamma_e}
</math>
  </td>
</tr>
</table>
</div>
 
Finally, then, we can state that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathfrak{f}_{WM}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~s_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
1 + \Lambda \, ,
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~(1-q^3) s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
(1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\, .</math>
  </td>
</tr>
</table>
</div>
 
Note,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Lambda</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c}
\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4}
= \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\chi_\mathrm{eq}^{3\gamma_c - 4}
\, .</math>
  </td>
</tr>
</table>
</div>
 
We also want to ensure that envelope pressure matches the core pressure at the interface.  This means,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~K_e \rho_{ie}^{\gamma_e}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~K_c \rho_{ic}^{\gamma_c}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)}  \biggr\}^{\gamma_c - \gamma_e}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c}
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e}
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1}
\biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)}
</math>
  </td>
</tr>
</table>
</div>
 
Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<div align="center">
<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, .
</math>
</div>
 
==Summary==
===Understanding Free-Energy Behavior===
'''<font color="#997700">Step 1:</font>''' Pick values for the separate coefficients, <math>\mathcal{A}, \mathcal{B},</math> and <math>\mathcal{C},</math> of the three terms in the normalized free-energy expression,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} </math>
  </td>
</tr>
</table>
</div>
then plot the function, <math>\mathfrak{G}^*(\chi)</math>, and identify the value(s) of <math>~\chi_\mathrm{eq}</math> at which the function has an extremum (or multiple extrema).
 
'''<font color="#997700">Step 2:</font>''' Note that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\nu^2}{5q} \biggl\{ 
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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ 
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr)
\biggr]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}  \biggl[ 
1+\Lambda_\mathrm{eq}
\biggr]</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\mathcal{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(1-\nu)\biggl( \frac{K_e}{K_c} \biggr)^*
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1}  \biggl\{ 
1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)}\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}
\biggl\{ 
\frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]
\biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}</math>
  </td>
</tr>
</table>
</div>
where (see, for example, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#Expression_for_Free_Energy|in the context of its original definition]]),
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\Lambda_\mathrm{eq} \equiv
\frac{3}{2^2\pi \cdot 5} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\chi_\mathrm{eq}^{3\gamma_c - 4}
</math>
  </td>
</tr>
</table>
</div>
and, where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\biggl( \frac{K_e}{K_c} \biggr)^* \equiv \frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{1-\nu}{1-q^3} \biggr]^{-\gamma_e}
\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c}
\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, .
</math>
  </td>
</tr>
 
</table>
</div>
 
Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and
<div align="center">
<math>
\frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~
\frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, .
</math>
</div>
 
'''<font color="#997700">Step 3:</font>'''  An analytic evaluation tells us that the following ''should'' happen.  Using the numerically derived value for <math>~\chi_\mathrm{eq}</math>, define,
<div align="center">
<math>~\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \, .</math>
</div>
We should then discover that,
<div align="center">
<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} =
\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\, .</math>
</div>
 
===Check It===
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{B} + \mathcal{C}^'</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \biggl[  1+\Lambda_\mathrm{eq} \biggr]
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr]  \biggr\}
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\Rightarrow~~~~\mathcal{A} \biggl[ \Lambda_\mathrm{eq} \cdot 5\biggl( \frac{q}{\nu^2} \biggr)  \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1+\Lambda_\mathrm{eq}
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{  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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]  \biggr\}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
1+\Lambda_\mathrm{eq}
+ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>
\Rightarrow~~~~\Lambda_\mathrm{eq}  \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]  \biggr\}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
  </td>
</tr>
</table>
</div>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>\Rightarrow~~~~
\frac{1}{\Lambda_\mathrm{eq}}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^3 \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]  \biggr\}
- \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) +
\frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow~~~~
\frac{2}{\Lambda_\mathrm{eq}}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) ( q - q^3 )
+ \frac{2}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (1 - q^5 ) - \frac{5}{2}( q^3 - q^5 ) \biggr] 
- 5 \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3)
- \frac{3}{q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
5\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ ( q - q^3 ) - (-2 + 3q - q^3) \biggr]
+ \frac{1}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2
\biggl[ 2(1 - q^5 ) - 5( q^3 - q^5 ) - 3(-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>
10\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 1-q \biggr]
+ \frac{5}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2
\biggl[ 1  - 3q^2 + 2q^3  \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>\Rightarrow~~~~
\frac{1}{\Lambda_\mathrm{eq}} \biggl[ \frac{2q^2}{5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
2q^2 (1-q )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) ( 1  - 3q^2 + 2q^3 )
</math>
  </td>
</tr>
</table>
</div>
Fortunately, this precisely matches our [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#LambdaDeff|earlier derivation]], which states that,
<div align="center">
<table border="0">
 
<tr>
  <td align="right">
<math>~\frac{1}{\Lambda}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp;
  </td>
  <td align="left">
<math>\frac{5}{2}(g^2-1) =
\frac{5}{2}\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>
 
=Playing With One Example=
By setting,
<table border="1" align="center" cellpadding="5">
 
<tr>
  <td align="center" colspan="3">
<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>
  </td>
</tr>
 
<tr>
  <th align="center">
<math>~\mathcal{A}</math>
  </th>
  <th align="center">
<math>~\mathcal{B}</math>
  </th>
  <th align="center">
<math>~\mathcal{C}</math>
  </th>
</tr>
 
<tr>
  <td align="center">
2.5
  </td>
  <td align="center">
1.0
  </td>
  <td align="center">
2.0
  </td>
</tr>
</table>
a plot of <math>~\mathfrak{G}^*</math> versus <math>~\chi</math> exhibits the following, two extrema:
<table border="1" cellpadding="5" align="center">
 
<tr>
  <td align="right">
extremum
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\mathfrak{G}^*</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}^'</math>
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
  </td>
  <td align="center">
<math>~\frac{\mathcal{A}}{\mathcal{B} +\mathcal{C}^'}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
MIN
  </td>
  <td align="center">
<math>1.1824</math>
  </td>
  <td align="center">
<math>-0.611367</math>
   </td>
   <td align="center">
&nbsp;<math>~~~~\Rightarrow~~~~</math>&nbsp;
  </td>
  <td align="center">
<math>0.66891</math>
  </td>
  <td align="center">
<math>1.3378</math>
  </td>
  <td align="center">
<math>1.0693</math>
  </td>
  <td align="center">
<math>1.0694</math>
  </td>
</tr>
 
<tr>
  <td align="right">
MAX
  </td>
  <td align="center">
<math>9.6722</math>
  </td>
  <td align="center">
<math>+0.508104</math>
  </td>
  <td align="center">
<math>~\Rightarrow</math>
  </td>
  <td align="center">
<math>0.004313</math>
  </td>
  <td align="center">
<math>0.008625</math>
  </td>
  <td align="center">
<math>2.4786</math>
  </td>
  <td align="center">
<math>2.4786</math>
  </td>
</tr>
</table>
The last two columns of this table confirm the internal consistency of the relationships presented in '''<font color="#997700">Step 3</font>''', above.  But what does this mean in terms of the values of <math>~\nu</math>, <math>~q</math>, and the related ratio of densities at the interface, <math>~\rho_e/\rho_c</math>?
 
Let's assume that what we're trying to display and examine is the behavior of the free-energy ''surface'' for a fixed value of the ratio of densities at the interface.  Once the value of <math>~\rho_e/\rho_c</math> has been specified, it is clear that the value of <math>~q</math> (and, hence, also <math>~\nu</math>) is set because <math>~\mathcal{A}</math> has also been specified.  But our specification of <math>~\mathcal{B}</math> along with <math>~\rho_e/\rho_c</math> also forces a particular value of <math>~q</math>.  It is unlikely that these two values of <math>~q</math> will be the same.  In reality, once <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> have both been specified, they force a particular <math>~(\nu, q)</math> pair.  How do we (easily) figure out what this pair is?
 
Let's begin by rewriting the expressions for <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> in terms of just <math>~q</math> and the ratio, <math>~\rho_e/\rho_c</math>, keeping in mind that, for the case of a uniform-density core (of density, <math>~\rho_c</math>) and a uniform-density envelope (of density, <math>~\rho_e</math>),
<div align="center">
<table border="0" cellpadding="5" align="center">
<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>
hence,
<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>~\biggl[ 1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{(1-q^3)}{q^3} \biggr]^{-1}</math>
  </td>
  <td align="center">&nbsp;&nbsp; and &nbsp; &nbsp;</td>
  <td align="right">
<math>~\frac{q^3}{\nu}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr] \, .</math>
  </td>
 
</tr>
</table>
</div>
Putting the expression for <math>~\mathcal{A}</math> in the desired form is simple because <math>~\nu</math> only appears as a leading factor.  Specifically, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
\biggl\{  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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr]
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi }{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2}
\biggl\{  q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2
\biggl[1-\frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr]
\biggr\}  \, .</math>
  </td>
</tr>
 
</table>
</div>
The expression for <math>~\mathcal{B}</math> can be written in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1}  \biggl\{ 
1+\frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c}
\chi_\mathrm{eq}^{3\gamma_c - 4}
\biggr\}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\nu \biggl[ \biggl( \frac{4\pi}{3} \biggr)\frac{q^3}{\nu} \biggr]^{1-\gamma_c} 
+\frac{\pi q^5}{5} \biggl( \frac{\nu^2}{q^6} \biggr) \chi_\mathrm{eq}^{3\gamma_c - 4}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \frac{q^3}{\nu} \biggr]^{-\gamma_c} 
+\frac{\pi q^5}{5} \biggl( \frac{q^3}{\nu} \biggr)^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3  \biggr]^{-\gamma_c} 
+\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .
</math>
  </td>
</tr>
</table>
</div>
Generally speaking, the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, which appears in the expression for <math>~\mathcal{B}</math>, is not known ahead of time.  Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, <math>~\mathcal{A}</math>, <math>~\mathcal{B}</math>, and <math>~\mathcal{C}</math>, and determine the equilibrium radius by looking for extrema in the free-energy function.  And because <math>~\chi_\mathrm{eq}</math> is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, <math>~(\nu, q)</math>, give self-consistent values for the coefficient pair, <math>~(\mathcal{A}, \mathcal{B})</math>.
 
Because we are using a uniform density core and uniform density envelope as our base model, however, we ''do'' know the analytic solution for <math>~\chi_\mathrm{eq}</math>.  As stated above, it is,
<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}{\Lambda_\mathrm{eq}} \cdot \frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi q^2}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}\biggl( \frac{\nu}{q^3} \biggr)
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) +
\frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]
\cdot \biggl[ \frac{q^3}{\nu} \biggr]^{\gamma_c-1} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\pi}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}
\biggl(\frac{\rho_e}{\rho_c}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) ( q^2-q^3 ) +
\frac{\rho_e}{\rho_c} (1 - q^2) \biggr]
\cdot \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3\biggr]^{\gamma_c-2} </math>
  </td>
</tr>
</table>
</div>
Combining this expression with the one for <math>~\mathcal{B}</math> gives us the desired result &#8212; although, strictly speaking, it is cheating!  We can now methodically choose <math>~(\nu, q)</math> pairs and map them into the corresponding values of <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>.  And, via an analogous "cheat," the choice of <math>~(\nu, q)</math> also gives us the self-consistent value of <math>~\mathcal{C}</math>.  In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.
 
=Second Example=
==Explain Logic==
[[Image:FreeEnergyExample.jpg|right|400px]]  The figure presented here, on the right, shows a plot of the free energy, as a function of the dimensionless radius, <math>~\mathfrak{G}^*(\chi)</math>, where,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,</math>
  </td>
</tr>
</table>
and, where we have used the parameter values,
<table border="1" align="center" cellpadding="5">
<tr>
  <td align="center" colspan="3">
<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>
  </td>
</tr>
 
<tr>
  <th align="center">
<math>~\mathcal{A}</math>
  </th>
  <th align="center">
<math>~\mathcal{B}</math>
  </th>
  <th align="center">
<math>~\mathcal{C}</math>
  </th>
</tr>
 
<tr>
  <td align="center">
0.201707
  </td>
  <td align="center">
0.0896
  </td>
  <td align="center">
0.002484
  </td>
</tr>
</table>
Directly from this plot we deduce that this free-energy function exhibits a minimum at <math>~\chi_\mathrm{eq} = 0.1235</math> and that, at this equilibrium radius, the configuration has a free-energy value, <math>~\mathfrak{G}^*(\chi_\mathrm{eq} ) = -2.0097</math>.  Via the steps described below, we demonstrate that this identified equilibrium radius is appropriate for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope (with the just-specified core and envelope adiabatic indexes) that has the following physical properties:
* Fractional core mass, <math>~\nu = 0.1</math>;
* Core-envelope interface located at <math>~r_i/R = q = 0.435</math>;
* Density jump at the core-envelope interface, <math>~\rho_e/\rho_c = 0.8</math>.
 
 
'''<font color="red">Step 1:</font>'''  Because the ratio, <math>~q^3/\nu</math>, is a linear function of the density ratio, <math>~\rho_e/\rho_c</math>, the full definition of the free-energy coefficient, <math>~\mathcal{A}</math>, can be restructured into a quadratic equation that gives the density ratio for any choice of the parameter pair, <math>~(q, \mathcal{A})</math>.  Specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~5 \biggl( \frac{q^3}{\nu} \biggr)^2 \mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ 5\mathcal{A} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \, ,
</math>
  </td>
</tr>
</table>
</div>
and this can be written in the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math> \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 a + \biggl( \frac{\rho_e}{\rho_c} \biggr) b + c</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
0 \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~a</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~5\mathcal{A} (1-q^3)^2 - 1 + \frac{5}{2}q^3 - \frac{3}{2}q^5 \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~b</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~10\mathcal{A} q^3(1-q^3) - \frac{5}{2}q^3 (1-q^2) \, ,</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~c</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~5\mathcal{A} q^6 - q^5 \, .</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\rho_e}{\rho_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2a} \biggl[\pm ( b^2 - 4ac)^{1/2} - b \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
(For our physical problem it appears as though only the positive root is relevant.)  For the purposes of this example, we set <math>~\mathcal{A} = 0.2017</math> and examined a range of values of <math>~q</math> to find a physically interesting value for the density ratio.  We picked:
<table border="1" cellpadding="5" align="center">
 
<tr>
  <td align="center">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~q</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~a</math>
  </td>
  <td align="center">
<math>~b</math>
  </td>
  <td align="center">
<math>~c</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~\frac{\rho_e}{\rho_c}</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~\nu</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.2017</math>
  </td>
  <td align="center">
<math>~0.435</math>
  </td>
  <td align="center">
<math>\Rightarrow</math>
  </td>
  <td align="center">
<math>~0.03173</math>
  </td>
  <td align="center">
<math>~-0.01448</math>
  </td>
  <td align="center">
<math>~-0.008743</math>
  </td>
  <td align="center">
<math>\Rightarrow</math>
  </td>
  <td align="center">
<math>~0.80068</math>
  </td>
  <td align="center">
<math>\Rightarrow</math>
  </td>
  <td align="center">
<math>~0.10074</math>
  </td>
</tr>
</table>
 
 
'''<font color="red">Step 2:</font>'''  Next, we chose the parameter pair,
<div align="center">
<math>
~\biggl(q, \frac{\rho_e}{\rho_c} \biggr) = (0.43500, 0.80000)
</math>
</div>
and determined the following parameter values from the known analytic solution:
<table border="1" cellpadding="5" align="center">
<tr>
  <td align="center">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
  </td>
  <td align="center">
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
  </td>
  <td align="center">
<math>~\Lambda_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.100816</math>
  </td>
  <td align="center">
<math>~43.16365</math>
  </td>
  <td align="center">
<math>~3.923017</math>
  </td>
  <td align="center">
<math>~0.13684</math>
  </td>
  <td align="center">
<math>~0.12349</math>
  </td>
  <td align="center">
<math>~0.201707</math>
  </td>
  <td align="center">
<math>~0.089625</math>
  </td>
  <td align="center">
<math>~0.002484</math>
  </td>
</tr>
</table>
 
==Construction Multiple Curves to Define a Free-Energy Surface==
Okay.  Now that we have the hang of this, let's construct a sequence of curves that represent physical evolution at a fixed interface-density ratio, <math>~\rho_e/\rho_c</math>, but for steadily increasing core-to-total mass ratio, <math>~\nu</math>.  Specifically, we choose,
<div align="center">
<math>
~\frac{\rho_e}{\rho_c} = \frac{1}{2} \, .
</math>
</div>
From the known analytic solution, here are parameters defining several different equilibrium models:
<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="9">
<font size="+1">Identification of Local ''Minimum'' in Free Energy</font>
  </th>
</tr>
<tr>
  <td align="center">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~q</math>
  </td>
  <td align="center">
<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
  </td>
  <td align="center">
<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>
  </td>
  <td align="center">
<math>~\Lambda_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.2</math>
  </td>
  <td align="center">
<math>~9^{-1/3} = 0.48075</math>
  </td>
  <td align="center">
<math>~12.5644</math>
  </td>
  <td align="center">
<math>~2.091312</math>
  </td>
  <td align="center">
<math>~0.366531</math>
  </td>
  <td align="center">
<math>~0.037453</math>
  </td>
  <td align="center">
<math>~0.2090801</math>
  </td>
  <td align="center">
<math>~0.2308269</math>
  </td>
  <td align="center">
<math>~2.06252 \times 10^{-4}</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.4</math>
  </td>
  <td align="center">
<math>~4^{-1/3} = 0.62996</math>
  </td>
  <td align="center">
<math>~4.21974</math>
  </td>
  <td align="center">
<math>~1.56498</math>
  </td>
  <td align="center">
<math>~0.707989</math>
  </td>
  <td align="center">
<math>~0.0220475</math>
  </td>
  <td align="center">
<math>~0.2143496</math>
  </td>
  <td align="center">
<math>~0.5635746</math>
  </td>
  <td align="center">
<math>~4.4626 \times 10^{-5}</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.5</math>
  </td>
  <td align="center">
<math>~3^{-1/3} = 0.693361</math>
  </td>
  <td align="center">
<math>~2.985115</math>
  </td>
  <td align="center">
<math>~1.42334</math>
  </td>
  <td align="center">
<math>~0.9448663</math>
  </td>
  <td align="center">
<math>~0.0152116</math>
  </td>
  <td align="center">
<math>~0.2152641</math>
  </td>
  <td align="center">
<math>~0.791882</math>
  </td>
  <td align="center">
<math>~1.5464 \times 10^{-5}</math>
  </td>
</tr>
 
<tr>
  <td align="left" colspan="5">
Here we are examining the behavior of the free-energy function for bipolytropic models having <math>~(n_c, n_e) = (0, 0)</math>, <math>~(\gamma_c, \gamma_e) = (6/5, 2)</math>, and a density ratio at the core-envelope interface of <math>~\rho_e/\rho_c = 1/2</math>.  The figure shown here, on the right, displays the three separate free-energy curves, <math>~\mathfrak{G}^*(\chi)</math> &#8212; where, <math>~\chi \equiv R/R_\mathrm{norm}</math> is the normalized configuration radius &#8212; that correspond to the three values of <math>~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math> given in the first column of the above table.  Along each curve, the local free-energy minimum corresponds to the the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, recorded in column 6 of the above table.
  </td>
  <td align="center" colspan="4">
[[Image:ThreeFreeEnergyCurves.png|center|300px]]
  </td>
</tr>
</table>
 
Each of the free-energy curves shown above has been entirely defined by our specification of the three coefficients in the free-energy function, <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math>.  In each case, the values of these three coefficients was judiciously chosen to ''produce'' a curve with a local minimum at the correct value of <math>~\chi_\mathrm{eq}</math> corresponding to an equilibrium configuration having the desired <math>~(\nu, \rho_e/\rho_c)</math> model parameters.  Upon plotting these three curves, we noticed that two of the curves &#8212; curves for <math>~\nu = 0.4</math> and <math>~\nu = 0.5</math> &#8212; also display a local ''maximum''.  Presumably, these maxima also identify equilibrium configurations, albeit unstable ones.  From a careful inspection of the plotted curves, we have identified the value of <math>~\chi_\mathrm{eq}</math> that corresponds to the two newly discovered (unstable) equilibrium models; these values are recorded in the table that immediately follows this paragraph.  By construction, we also know what values of <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math> are associated with these two identified equilibria; these values also have been recorded in the table.  But it is not immediately obvious what the values are of the <math>~(\nu, \rho_e/\rho_c)</math> model parameters that correspond to these two equilibrium models.
 
<table border="1" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="9">
Subsequently Identified Local Energy ''Maxima''
  </td>
</tr>
<tr>
  <td align="center">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\mathfrak{G}^*</math>
  </td>
  <td align="center">
<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>
</td>
  <td align="center">
<math>~\therefore</math>
  </td>
  <td align="center">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}^' = \mathcal{A} \chi_\mathrm{eq}^{3\gamma_c-4} - \mathcal{B}</math>
  </td>
  <td align="center">
<math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)}</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.08255</math>
  </td>
  <td align="center">
<math>~+ 4.87562</math>
  </td>
  <td align="center">
<math>~0.368715</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~0.2143496</math>
  </td>
  <td align="center">
<math>~0.5635746</math>
  </td>
  <td align="center">
<math>~4.4626 \times 10^{-5}</math>
  </td>
  <td align="center">
<math>~1.7768 \times 10^{-2}</math>
  </td>
  <td align="center">
<math>~0.08254</math>
  </td>
</tr>
 
<tr>
  <td align="center">
<math>~0.032196</math>
  </td>
  <td align="center">
<math>~+11.5187</math>
  </td>
  <td align="center">
<math>~0.25300</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~0.2152641</math>
  </td>
  <td align="center">
<math>~0.791882</math>
  </td>
  <td align="center">
<math>~1.5464 \times 10^{-5}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~0</math>
<math>~5.8964 \times 10^{-2}</math>
   </td>
   </td>
  <td align="center">
<math>~0.2143496</math>
  </td>
  <td align="center">
<math>~0.5635746</math>
  </td>
  <td align="center">
<math>~4.4626 \times 10^{-5}</math>
  </td>
</tr>
<tr>
   <td align="center">
   <td align="center">
<math>~0.032196</math>
<math>~0.032196</math>
  </td>
  <td align="center">
<math>~+11.5187</math>
  </td>
  <td align="center">
<math>~0.25300</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~0</math>
  </td>
  <td align="center">
<math>~0.2152641</math>
  </td>
  <td align="center">
<math>~0.791882</math>
  </td>
  <td align="center">
<math>~1.5464 \times 10^{-5}</math>
   </td>
   </td>
</tr>
</tr>
Line 3,483: Line 5,464:


=Related Discussions=
=Related Discussions=
* [[User:Tohline/SSC/BipolytropeGeneralization_Version2|Newer, Version2 of ''Bipolytrope Generalization'' derivations]].
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic0_0#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|Analytic solution]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|Analytic solution]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.


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Latest revision as of 18:24, 26 August 2014

Bipolytrope Generalization

Whitworth's (1981) Isothermal Free-Energy Surface
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On 26 August 2014, Tohline finished rewriting the chapter titled "Bipolytrope Generalization" in a very concise manner (go here for this Version2 chapter) then set this chapter aside to provide a collection of older attempts at the derivations. While much of what follows is technically correct, it is overly detailed and cumbersome. Because it likely also contains some misguided steps, we label it in entirety as Work in Progress.



Work-in-progress.png

Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
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Old Stuff

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

<math>~=</math>

<math>~W_\mathrm{grav} + \mathfrak{S}_A\biggr|_\mathrm{core} + \mathfrak{S}_A\biggr|_\mathrm{env} </math>

 

<math>~=</math>

<math> ~W_\mathrm{grav} + \biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} \, . </math>

In addition to the gravitational potential energy, which is naturally written as,

<math>~W_\mathrm{grav}</math>

<math>~=</math>

<math>~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \cdot \mathfrak{f}_{WM} \, ,</math>

it seems reasonable to write the separate thermal energy contributions as,

<math>~S_\mathrm{core}</math>

<math>~=</math>

<math> ~\frac{3}{2}\biggl[ M_\mathrm{core} \biggl( \frac{P_{ic}}{\rho_{ic}} \biggr) \biggr] s_\mathrm{core} = \frac{3}{2} \biggl[ \nu M_\mathrm{tot} P_{ic} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{core} = 2\pi R^3 P_{ic} \biggl[ q^3 s_\mathrm{core} \biggr] \, ,</math>

<math>~S_\mathrm{env}</math>

<math>~=</math>

<math> ~\frac{3}{2}\biggl[ M_\mathrm{env} \biggl( \frac{P_{ie}}{\rho_{ie}} \biggr) \biggr] s_\mathrm{env} = \frac{3}{2} \biggl[ (1-\nu) M_\mathrm{tot} P_{ie} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{-1} \biggl( \frac{4\pi R^3}{3 M_\mathrm{tot}} \biggr) \biggr] s_\mathrm{env} = 2\pi R^3 P_{ie} \biggl[ (1-q^3) s_\mathrm{env} \biggr] \, ,</math>

where the subscript "<math>i</math>" means "at the interface," and <math>~\mathfrak{f}_{WM},</math> <math>~s_\mathrm{core},</math> and <math>~s_\mathrm{env}</math> are dimensionless functions of order unity (all three functions to be determined) akin to the structural form factors used in our examination of isolated polytropes.

While exploring how the free-energy function varies across parameter space, we choose to hold <math>~M_\mathrm{tot}</math> and <math>~K_c</math> fixed. By dimensional analysis, it is therefore reasonable to normalize all energies, length scales, densities and pressures by, respectively,

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

<math>~\equiv</math>

<math>~\biggl[ \frac{G^{3(\gamma_c-1)} M_\mathrm{tot}^{5\gamma_c-6}}{K_c} \biggr]^{1/(3\gamma_c -4)} \, ,</math>

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

<math>~\equiv</math>

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

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

<math>~\equiv</math>

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

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

<math>~\equiv</math>

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

As is detailed below — first, here, and via an independent derivation, here — quite generally the expression for the normalized free energy is,

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

<math>~=</math>

<math> - \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R^3 P_{ic} }{E_\mathrm{norm}} \biggr] + \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{ R^3 P_{ie} }{E_\mathrm{norm}} \biggr] </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr] \biggl[ \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr] </math>

 

 

<math> + \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{ R_\mathrm{norm}^4 P_\mathrm{norm} }{E_\mathrm{norm} R_\mathrm{norm}} \biggr] \biggl[ \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr) \chi^3 \biggr] </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \chi^{-1} + \biggl[ \frac{4\pi q^3 s_\mathrm{core} }{3(\gamma_c - 1)} \biggr] \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_c} + \biggl[ \frac{4\pi (1-q^3) s_\mathrm{env} }{3(\gamma_e - 1)} \biggr] \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3 -3\gamma_e} </math>

where we have introduced the parameter, <math>~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math>. After defining the normalized (and dimensionless) configurarion radius, <math>~\chi \equiv R/R_\mathrm{norm}</math>, we can write the normalized free energy of a bipolytrope in the following compact form:

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

<math>~=</math>

<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,</math>

where,

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~\frac{1}{5} \cdot \mathfrak{f}_{WM} \, ,</math>

<math>~\mathcal{B}</math>

<math>~\equiv</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, , </math>

<math>~\mathcal{C}</math>

<math>~\equiv</math>

<math>\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .</math>

As is further detailed below, the second expression for the coefficient, <math>~\mathcal{C}</math>, ensures that the pressure at the "surface" of the core matches the pressure at the "base" of the envelope; but it should only be employed after an equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been identified by locating an extremum in the free energy.

Simplest Bipolytrope

Familiar Setup

As has been shown in an accompanying presentation, for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope,

<math>~\mathfrak{f}_{WM}</math>

<math>~\equiv</math>

<math>~\frac{\nu^2}{q} \cdot f \, ,</math>

<math>~s_\mathrm{core}</math>

<math>~\equiv</math>

<math> 1 + \Lambda_\mathrm{eq} \, , </math>

<math>~(1-q^3) s_\mathrm{env}</math>

<math>~\equiv</math>

<math> (1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,</math>

and where (see, for example, in the context of its original definition, or another, separate derivation),

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

<math>~=~</math>

<math>

\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} 

\chi_\mathrm{eq}^{3\gamma_c - 4} </math>

 

<math>~=</math>

<math>\frac{2}{5(g^2-1)} = \biggl\{ \frac{5}{2}\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] \biggr\}^{-1} \, ,</math>

and where (see the associated discussion of relevant mass integrals),

<math> \frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, . </math>

Cleaner Virial Presentation

In an effort to show the similarity in structure among the several energy terms, we have also found it useful to write their expressions in the following forms:

<math>~W_\mathrm{grav}</math>

<math>~=</math>

<math>~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \frac{\nu^2}{q} \cdot f = - 4\pi P_i R^3 \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_i f \, ,</math>

<math>~S_\mathrm{core}</math>

<math>~=</math>

<math>~2\pi P_{ic} R^3 \biggl[ q^3 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ic} \biggr] \, ,</math>

<math>~S_\mathrm{env}</math>

<math>~=</math>

<math>~2\pi P_{ie} R^3 \biggl[ (1-q^3) + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ie} \mathfrak{F} \biggr] \, ,</math>

where (see an associated discussion or the original derivation),

<math> f\biggl(q, \frac{\rho_e}{\rho_c}\biggr) = 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] \, , </math>

and where,

<math>~\lambda_i</math>

<math>~\equiv</math>

<math>~\frac{GM_\mathrm{tot}^2}{R^4 P_i} \, ,</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>~=</math>

<math>~ \frac{1}{\lambda_{ie}} \biggl( \frac{2^2 \cdot 5\pi}{3} \biggr) \frac{q(1-q^3)}{\nu^2} (s_\mathrm{env} -1)

\, .</math>

This also means that the three key terms used as shorthand notation in the above expressions for the three energy terms have the following definitions:

<math>~\mathfrak{f}_{WM}</math>

<math>~\equiv</math>

<math>~\frac{\nu^2}{q} \cdot f \, ,</math>

<math>~s_\mathrm{core}</math>

<math>~\equiv</math>

<math> 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \, , </math>

<math>~s_\mathrm{env}</math>

<math>~\equiv</math>

<math> 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \lambda_{ie} \mathfrak{F} \, ,</math>

Hence, if all the interface pressures are equal — that is, if <math>~P_i = P_{ic} = P_{ie}</math> and, hence also, <math>~\lambda_{i} = \lambda_{ic} = \lambda_{ie}</math> — then the total thermal energy is,

<math>~S_\mathrm{tot} = S_\mathrm{core} + S_\mathrm{env}</math>

<math>~=</math>

<math> 2\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F}) \biggr] \, ; </math>

and the virial is,

<math>~2S_\mathrm{tot} + W_\mathrm{grav}</math>

<math>~=</math>

<math> 4\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F} - f ) \biggr] \, . </math>

The virial should sum to zero in equilibrium, which means,

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot (f - 1- \mathfrak{F} ) </math>

<math>\Rightarrow ~~~~ \biggl[ \biggl( \frac{2^2\cdot 5\pi}{3} \biggr) \frac{q}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>

<math>~=</math>

<math> f - 1- \mathfrak{F} </math>

<math>\Rightarrow ~~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggl[ \biggl( \frac{2^3\pi}{3} \biggr) \frac{q^6}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>

<math>~=</math>

<math> \biggl[ (q^3 - q^5 )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] - \biggl[ (-2q^2 + 3q^3 - q^5) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (-\frac{3}{5} +3q^2 - 3q^3 + \frac{3}{5} q^5) \biggr] </math>

 

<math>~=</math>

<math> 2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (1 -3q^2 + 2q^3 ) </math>

 

<math>~=</math>

<math> q^2 \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} (g^2-1) </math>

<math>\Rightarrow~~~~ \frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1) \, . </math>

Shift to Central Pressure Normalization

Let's rework the definition of <math>~\lambda_i</math> in two ways: (1) Normalize <math>~R_\mathrm{eq}</math> to <math>~R_\mathrm{norm}</math> and normalize the pressure to <math>~P_\mathrm{norm}</math>; (2) shift the referenced pressure from the pressure at the interface <math>~(P_i)</math> to the central pressure <math>~(P_0)</math>, because it is <math>~P_0</math> that is directly related to <math>~K_c</math> and <math>~\rho_c</math>; specifically, <math>P_0 = K_c \rho_c^{\gamma_c}</math>. Appreciating that, in equilibrium,

<math>~P_i</math>

<math>~=</math>

<math>~P_0 - q^2 \Pi_\mathrm{eq} = K_c \rho_c^{\gamma_c} - \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2 \, ,</math>

the left-hand-side of the last expression, above, can be rewritten as,

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~\equiv</math>

<math>~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>

 

<math>~=</math>

<math>~\frac{R_\mathrm{eq}^4}{GM_\mathrm{tot}^2} \biggl[ P_0 - \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) q^2\biggr] </math>

 

<math>~=</math>

<math>~\frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 \, .</math>

Hence, the virial equilibrium condition gives,

<math> \frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 </math>

<math>~=</math>

<math> \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1) </math>

<math> \Rightarrow ~~~~ \frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2} </math>

<math>~=</math>

<math> \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2} q^2 g^2 \, . </math>

This result precisely matches the result obtained via the detailed force-balanced conditions imposed through hydrostatic equilibrium.

Adopting our new variable normalizations and realizing, in particular, that,

<math>~R_\mathrm{norm}^4 P_\mathrm{norm}</math>

<math>~=</math>

<math>~GM_\mathrm{tot}^2 \, ,</math>

the expression alternatively can be rewritten as,

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~\equiv</math>

<math>~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} = \chi_\mathrm{eq}^4 \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr) </math>

 

<math>~=</math>

<math> \chi_\mathrm{eq}^4 \biggl\{ \frac{K_c \rho_c^{\gamma_c}}{P_\mathrm{norm}} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4P_\mathrm{norm}} \biggr) \biggr\} </math>

 

<math>~=</math>

<math> \chi_\mathrm{eq}^4 \biggl\{ \frac{K_c }{P_\mathrm{norm}} \biggl[ \frac{\rho_c}{\bar\rho} \biggl( \frac{3M_\mathrm{tot}}{4\pi R_\mathrm{norm}^3} \biggr) \chi_\mathrm{eq}^{-3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \chi_\mathrm{eq}^{-4} \biggr\} </math>

 

<math>~=</math>

<math> \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \frac{K_c }{P_\mathrm{norm}} \biggl( \frac{M_\mathrm{tot}^{\gamma_c}}{R_\mathrm{norm}^{3\gamma_c}} \biggr) - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 </math>

 

<math>~=</math>

<math> \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 \, . </math>

Normalized in this manner, the virial equilibrium (as well as the hydrostatic balance) condition gives,

<math> \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 </math>

<math>~=</math>

<math> \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1) </math>

<math> \Rightarrow ~~~~ \chi_\mathrm{eq}^{4-3\gamma_c} </math>

<math>~=</math>

<math> \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2-\gamma_c} q^2 g^2 \, . </math>

Free-Energy Coefficients

Therefore, for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope, the coefficients in the normalized free-energy function are,

<math>~\mathcal{A}</math>

<math>~=</math>

<math> ~\frac{\nu^2}{5q} \cdot f = \frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) q^3 (1 - q^2 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr) \biggr] \, ,</math>

<math>~\mathcal{B}</math>

<math>~\equiv</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c} =\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr] \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} =\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} - \frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3}\biggr]^2 q^2 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr\} \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl\{ \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} + \chi_\mathrm{eq}^{3\gamma_c-4} \biggl[\frac{3}{2^2\cdot 5\pi} - \frac{3}{2^3\pi} \biggr]\frac{\nu^2}{q^4} \biggr\} </math>

 

<math>~=</math>

<math> \biggl\{ \nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} - \chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3^2}{2^3\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggl( \frac{4\pi }{3} \biggr) q^3 \biggr\} = \nu \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} - \chi_\mathrm{eq}^{3\gamma_c-4} \biggl( \frac{3}{10} \biggr) \frac{\nu^2}{q} </math>

<math>~\mathcal{C}</math>

<math>~\equiv</math>

<math>\biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \, .</math>

Note that, because <math>~P_{ie} = P_{ic}</math> in equilibrium, the ratio of coefficients,

<math>~\frac{\mathcal{C}}{\mathcal{B}}</math>

<math>~=</math>

<math>~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl\{\frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr\}</math>

<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \biggl( \frac{\mathcal{C}}{\mathcal{B}} \biggr)</math>

<math>~=</math>

<math> \frac{ 20\pi q(1-q^3) \lambda_i^{-1} + 3\nu^2 \mathfrak{F} }{ 20\pi q^4 \lambda_i^{-1} + 3\nu^2 } \, . </math>


The equilibrium condition is,

<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} \, ,</math>

where,

<math> \mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c-\gamma_e)} \, . </math>

More General Derivation of Free-Energy Coefficients B and C

Keep in mind that, generally,

<math>GM_\mathrm{tot}^2</math>

<math>~=</math>

<math>~R_\mathrm{norm}^4 P_\mathrm{norm} = E_\mathrm{norm} R_\mathrm{norm} \, ;</math>

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

<math>~\equiv</math>

<math> \frac{R^4 P_i}{GM_\mathrm{tot}^2} = \biggl( \frac{P_i}{P_\mathrm{norm}} \biggr) \chi^4 </math>   … and, note that …    <math> \frac{1}{\Lambda} = \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \frac{1}{\lambda_i} \cdot \frac{1}{q^2 \sigma^2} \, ; </math>

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

<math>~=</math>

<math> \frac{3}{2^3\pi} \biggl( \frac{GM_\mathrm{tot}^2}{P_\mathrm{norm} R^4} \biggr) \frac{\nu^2}{q^6} =\biggl( \frac{2\pi}{3} \biggr) \sigma^2 \chi^{-4} \, ; </math>

<math>~\frac{K_c \rho_c^{\gamma_c} }{P_\mathrm{norm}}</math>

<math>~=</math>

<math> \frac{K_c}{P_\mathrm{norm}} \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c} = \frac{K_c M_\mathrm{tot}^{\gamma_c} }{R_\mathrm{norm}^{3\gamma_c} P_\mathrm{norm}} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \chi^{-3\gamma_c} = \sigma^{\gamma_c} \chi^{-3\gamma_c} \, , </math>

where we have introduced the notation,

<math> \sigma \equiv \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \, . </math>

So, the free-energy coefficient,

<math>~\mathcal{B}</math>

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{P_{ic} }{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c} =\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \biggr]_\mathrm{eq} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} =\biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \frac{1}{\lambda_{ic}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \chi_\mathrm{eq}^4 \biggl( \frac{P_{ic}}{P_\mathrm{norm}} \biggr)_\mathrm{eq} +

\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} \, .

</math>


And the free-energy coefficient,

<math>~\mathcal{C}</math>

<math>~\equiv</math>

<math> \biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} = \biggl( \frac{4\pi }{3} \biggr) (1-q^3) s_\mathrm{env} \biggl[ \frac{1 }{\lambda_{ie}} \biggr]_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) (1-q^3) \biggl\{ \frac{1 }{\lambda_{ie}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \mathfrak{F} \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \chi_\mathrm{eq}^4 \biggl( \frac{P_{ie}}{P_\mathrm{norm}} \biggr)_\mathrm{eq}+ \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} \, . </math>

OLD DERIVATION

<math>P_{ic} = K_c \rho_c^{\gamma_c}</math>

NEW DERIVATION

<math>P_{ic} = P_0 - q^2\Pi = K_c \rho_c^{\gamma_c} - q^2\Pi</math>

… therefore …

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

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} +

\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} 

</math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} +

\biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr]  

</math>

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

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 + \biggl( \frac{4\pi}{3\cdot 5} \biggr) q^2 \sigma^2 \biggr] \chi_\mathrm{eq}^{3\gamma_c-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 \biggl[ \sigma^{\gamma_c} - \biggl( \frac{2\pi}{5} \biggr) q^2 \sigma^2 \chi_\mathrm{eq}^{3\gamma_c-4} \biggr] </math>

… and, enforcing in equilibrium <math>~P_{ie} = P_{ic}</math> …

<math>~\mathcal{C}_\mathrm{OLD}</math>

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) \biggl[ (1-q^3) \sigma^{\gamma_c} + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr) \chi_\mathrm{eq}^{3\gamma_e-4}\biggr] </math>

<math>~\mathcal{C}_\mathrm{NEW}</math>

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4-3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 \biggr] + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\}_\mathrm{eq} \chi_\mathrm{eq}^{3\gamma_e-4} </math>

 

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) \biggl\{ (1-q^3) \sigma^{\gamma_c} \chi_\mathrm{eq}^{3\gamma_e - 3\gamma_c} + \biggl( \frac{2\pi}{3} \biggr) \sigma^2 \biggl[ \biggl( \frac{2}{5} q^5 \mathfrak{F} \biggr) - q^2 (1-q^3) \biggr] \chi_\mathrm{eq}^{3\gamma_e-4} \biggr\} </math>

… and, also …

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>

<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>

<math>~=</math>

<math> \frac{1}{q^2 } \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \sigma^{\gamma_c - 2} \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>

 

<math>~=</math>

<math> \biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 </math>

<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>

<math>~=</math>

<math> \frac{1}{q^2 \sigma^2} \biggl( \frac{3\cdot 5}{2^2\pi} \biggr) \biggl[ \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 \biggr] </math>

 

<math>~=</math>

<math> \biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \frac{5}{2} </math>

Extrema

Extrema in the free energy occur when,

<math>~\mathcal{A}</math>

<math>~=</math>

<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{C} \chi_\mathrm{eq}^{4-3\gamma_e} \, .</math>

Also, as stated above, because <math>~P_{ie} = P_{ic}</math> in equilibrium, the ratio of coefficients,

<math>~\frac{\mathcal{C}}{\mathcal{B}}</math>

<math>~=</math>

<math>~\chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}\biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .</math>

When put together, these two relations imply,

<math>~\mathcal{A}</math>

<math>~=</math>

<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \chi_\mathrm{eq}^{4-3\gamma_c} \mathcal{B} \biggl[ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] </math>

 

<math>~=</math>

<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] \, .</math>

But the definition of <math>~\mathcal{B}</math> gives,

<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c}</math>

<math>~=</math>

<math> \biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \, . </math>

Hence, extrema occur when,

<math>~\mathcal{A}</math>

<math>~=</math>

<math>~\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core} \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \biggl[ 1+ \frac{(1-q^3) s_\mathrm{env}}{q^3 s_\mathrm{core}} \biggr] </math>

<math>\Rightarrow ~~~~ \biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot f</math>

<math>~=</math>

<math> \biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \biggl[ q^3 s_\mathrm{core} + (1-q^3) s_\mathrm{env} \biggr] </math>

 

<math>~=</math>

<math> \frac{q^3}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} + \frac{(1-q^3)}{[\lambda_i]_\mathrm{eq}} + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \mathfrak{F} </math>

<math>\Rightarrow ~~~~\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} </math>

<math>~=</math>

<math> \biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot (f - 1 - \mathfrak{F}) </math>

 

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2 - 1 ) \, . </math>

In what follows, keep in mind that,

<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>

<math>~=</math>

<math> \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c} = R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \, ; </math>

<math>~K_c \rho_c^{\gamma_c}</math>

<math>~=</math>

<math> K_c \biggl( \frac{\rho_c}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c} = K_c \sigma^{\gamma_c} M_\mathrm{tot}^{\gamma_c} R^{-3\gamma_c} \, ; </math>

<math>~\Pi</math>

<math>~=</math>

<math> \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \frac{\nu^2}{q^6} = \frac{2\pi}{3} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \sigma^2 \, . </math>


OLD DERIVATION

<math>P_{i} = K_c \rho_c^{\gamma_c}</math>

<math>\Rightarrow ~~~~ K_c = P_{i} \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c} </math>

NEW DERIVATION

<math>P_0 = K_c \rho_c^{\gamma_c} </math>

<math>\Rightarrow ~~~~ K_c = P_0 \sigma^{-\gamma_c} M_\mathrm{tot}^{-\gamma_c} R^{+3\gamma_c} </math>

… hence, as derived in the above table …

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>

<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>

<math>~=</math>

<math> \sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 </math>

… which, when combined with the condition that identifies extrema, gives …

<math>~\chi_\mathrm{eq}^{4 - 3\gamma_c}</math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 ) </math>

<math>~\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 ) </math>

<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4} P_i }{GM_\mathrm{tot}^{2} } </math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2-1) </math>

<math>~\sigma^{\gamma_c}\chi_\mathrm{eq}^{4 - 3\gamma_c} - \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2</math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^2 q^2 (g^2 - 1 ) </math>

<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2 </math>

<math>\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2 </math>

<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4} P_0 }{GM_\mathrm{tot}^{2} } </math>

<math>~=</math>

<math> \biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 g^2 </math>

These are consistent results because they result in the detailed force-balance relation, <math>P_0 - P_i = q^2 \Pi_\mathrm{eq} \, .</math>

Examples

Identification of Local Extrema in Free Energy

<math>~\nu</math>

<math>~q</math>

<math>~ \frac{\rho_e}{\rho_c} </math>

<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

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

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

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

MIN/MAX

<math>~0.2</math>

<math>~9^{-1/3} = 0.48075</math>

<math>~0.5</math>

<math>~12.5644</math>

<math>~2.091312</math>

<math>~0.366531</math>

<math>~0.037453</math>

<math>~0.2090801</math>

<math>~0.2308269</math>

<math>~2.06252 \times 10^{-4}</math>

MIN

<math>~0.4</math>

<math>~4^{-1/3} = 0.62996</math>

<math>~0.5</math>

<math>~4.21974</math>

<math>~1.56498</math>

<math>~0.707989</math>

<math>~0.0220475</math>

<math>~0.2143496</math>

<math>~0.5635746</math>

<math>~4.4626 \times 10^{-5}</math>

MIN

<math>~0.473473</math>

<math>~0.681838</math>

<math>~0.516107</math>

 

 

<math>~0.462927</math>

<math>~0.08255</math>

"

"

"

MAX

<math>~0.5</math>

<math>~3^{-1/3} = 0.693361</math>

<math>~0.5</math>

<math>~2.985115</math>

<math>~1.42334</math>

<math>~0.9448663</math>

<math>~0.0152116</math>

<math>~0.2152641</math>

<math>~0.791882</math>

<math>~1.5464 \times 10^{-5}</math>

MIN

<math>~0.559839</math>

<math>~0.729581</math>

<math>~0.499188</math>

 

 

<math>~0.75089</math>

<math>~0.032196</math>

"

"

"

MAX



Free Energy Extrema when:     <math>~~~~~\frac{\rho_e}{\rho_c} = \frac{1}{2} ~~~~\Rightarrow~~~~ q^3 = \frac{\nu}{2-\nu}</math>

<math>~\nu</math>

<math>~q</math>

<math>~ \frac{\rho_e}{\rho_c} </math>

<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

<math>~\frac{1}{\lambda_{i}}\biggr|_\mathrm{eq}</math>

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

<math>~\mathcal{A}</math>

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

<math>~\mathcal{C}_\mathrm{NEW}</math>

<math>~G^*</math>

Stability

MIN/MAX

<math>~0.5</math>

<math>~\biggl( \frac{1}{3} \biggr)^{1/3}</math>

<math>~0.5</math>

<math>~2.985115</math>

<math>~1.423340</math>

<math>~0.05466039</math>

<math>~0.3152983</math>

<math>~0.21526406</math>

<math>~0.23552725</math>

<math>~6.643899 \times 10^{-3}</math>

<math>~+ 0.5176146</math>

<math>~+0.429245</math>

MIN

 

 

 

<math>~0.6674</math>

"

"

"

<math>~+0.55572115</math>

 

MAX

<math>~0.6</math>

<math>~\biggl( \frac{3}{7} \biggr)^{1/3}</math>

<math>~0.5</math>

<math>~2.2507129</math>

<math>~1.31282895</math>

<math>~0.04160318</math>

<math>~0.3411545</math>

<math>~0.21493717</math>

<math>~0.26165939</math>

<math>~5.208750 \times 10^{-3}</math>

<math>~+ 0.73532249</math>

<math>~+0.0935217</math>

MIN

 

 

 

<math>~0.431745</math>

"

"

"

<math>~+0.7367797</math>

 

MAX

<math>~0.7</math>

<math>~\biggl( \frac{7}{13} \biggr)^{1/3}</math>

<math>~0.5</math>

<math>~1.7707809</math>

<math>~1.2209446</math>

<math>~0.029500</math>

<math>~0.3589388</math>

<math>~0.21330744</math>

<math>~0.28172532</math>

<math>~3.389793 \times 10^{-3}</math>

<math>~+ 0.8953395</math>

<math>~- 0.0767108</math>

MAX

 

 

 

<math>~0.270615</math>

"

"

"

<math>~+0.89227216</math>

 

MIN

System should be stable (with free energy minimum) if:     <math>~~~~~\frac{( \gamma_e - \frac{4}{3})}{(\gamma_e - \gamma_c)} f - \biggl[1 + \frac{5}{2} (g^2-1) \biggr] ~>~ 0</math>

Solution Strategy

For a given set of free-energy coefficients, <math>~\mathcal{A}, \mathcal{B},</math> and <math>~\mathcal{C}</math>, along with a choice of the two adiabatic exponents <math>~(\gamma_c, \gamma_e)</math>, here's how to determine all of the physical parameters that are detailed in the above example table.

  • Step 1: Guess a value of <math>~0 < q < 1</math>.
  • Step 2: Given the pair of parameter values, <math>~(\mathcal{A}, q)</math>, determine the interface-density ratio, <math>~\rho_e/\rho_c</math>, by finding the appropriate root of the expression that defines the function, <math>~\mathcal{A}(q, \rho_e/\rho_c)</math>. This can be straightforwardly accomplished because, as demonstrated below, the relevant expression can be written as a quadratic function of <math>~(\rho_e/\rho_c)</math>.
  • Step 3: Given the pair of parameter values, <math>~(q, \rho_e/\rho_c)</math>, determine the value of the core-to-total mass ratio, <math>~\nu</math>, from the expression that was obtained from an integration over the mass, namely,

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

<math>~=</math>

<math>~1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{q^3} - 1 \biggr) \, .</math>

  • Step 4: Given the value of <math>~\mathcal{B}</math> along with the pair of parameter values, <math>~(q, \nu)</math>, the above expression that defines <math>~\mathcal{B}</math> can be solved to give the relevant value of the dimensionless parameter, <math>~\Lambda_\mathrm{eq}.</math>
  • Step 5: The value of <math>~\mathcal{C}^'</math> — the coefficient that appears on the right-hand-side of the above expression that defines <math>~\mathcal{C}</math> — can be determined, given the values of parameter triplet, <math>~(q, \nu, \Lambda_\mathrm{eq})</math>.
  • Step 6: Given the value of <math>~\mathcal{C}</math> and the just-determined value of the coefficient <math>~\mathcal{C}^'</math>, the normalized equilibrium radius, <math>~\chi_\mathrm{eq},</math> that corresponds to the value of <math>~q</math> that was guessed in Step #1 can be determined from the above definition of <math>~\mathcal{C}</math>, specifically,

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

<math>~=</math>

<math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)} \, .</math>

  • Step 7: But, independent of this guessed value of <math>~\chi_\mathrm{eq},</math> the condition for virial equilibrium — which identifies extrema in the free-energy function — gives the following expression for the normalized equilibrium radius:

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

<math>~=</math>

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

  • Step 8: If <math>~\chi_\mathrm{eq}|_\mathrm{guess} \ne \chi_\mathrm{eq}|_\mathrm{virial}</math>, return to Step #1 and guess a different value of <math>~q</math>. Repeat Steps #1 through #7 until the two independently derived values of the normalized radius match, to a desired level of precision.
  • Keep in mind: (A) A graphical representation of the free-energy function, <math>~\mathfrak{G}(\chi)</math>, can also be used to identify the "correct" value of <math>~\chi_\mathrm{eq}</math> and, ultimately, the above-described iteration loop should converge on this value. (B) The free-energy function may exhibit more than one (or, actually, no) extrema, in which case more than one (or no) value of <math>~q</math> should lead to convergence of the above-described iteration loop.



Work-in-progress.png

Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
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Detailed Derivations

Dividing the free-energy expression through by <math>~E_\mathrm{norm}</math> generates,

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

<math>~=</math>

<math> - \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{E_\mathrm{norm}} \biggr) \biggl( \frac{1}{R} \biggr) \cdot \mathfrak{f}_{WM} + \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_c \rho_{ic}^{\gamma_c-1} }{E_\mathrm{norm}} \biggr] </math>

 

 

<math> ~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{M_\mathrm{tot} K_e \rho_{ie}^{\gamma_e-1} }{E_\mathrm{norm}} \biggr] </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl[ \frac{K_c G^{(3\gamma_c -4)}M_\mathrm{tot}^{2(3\gamma_c -4)}}{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{1}{R} \biggr) </math>

 

 

<math> + \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_c^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_c-1} </math>

 

 

<math> ~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl[ \frac{K_c M_\mathrm{tot}^{3\gamma_c -4} K_e^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6}} \biggr]^{1/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} \biggl[ \frac{3M_\mathrm{tot}}{4\pi R^3} \biggr]^{\gamma_e-1} </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr) </math>

 

 

<math> + \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_c-1} \biggl[ \frac{ K_c^{3\gamma_c -3} }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_c-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} </math>

 

 

<math> ~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr)^{\gamma_e-1} \biggl[ \frac{K_c^{3\gamma_c - 3} (K_e/K_c)^{3\gamma_c -4} }{G^{3\gamma_c-3} M_\mathrm{tot}^{2\gamma_c-2}} \biggr]^{1/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{-3(\gamma_e-1)/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R_\mathrm{norm}}{R} \biggr) </math>

 

 

<math> + \biggl[ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_c-1} \biggl[ M_\mathrm{tot}^{3\gamma_c-4} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)} \biggl[ M_\mathrm{tot}^{-2} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} \biggl[ M_\mathrm{tot}^{-3\gamma_c+6} \biggr]^{(\gamma_c-1)/(3\gamma_c -4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_c-1)} \biggl( \frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c-1} </math>

 

 

<math> ~+ \biggl[ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggr] \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ M_\mathrm{tot}^{2(\gamma_e-1)-2(\gamma_c-1)}\biggr]^{1/(3\gamma_c-4)} \biggl[ \frac{R_\mathrm{norm}}{R} \biggr]^{3(\gamma_e-1)} \biggl[ \biggl( \frac{K_c}{G} \biggr)^{(\gamma_c-1)-(\gamma_e-1)} \biggr]^{3/(3\gamma_c -4)} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e-1} </math>

 

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} </math>

 

 

<math> ~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, . </math>

We also want to ensure that envelope pressure matches the core pressure at the interface. This means,

<math>~K_e \rho_{ie}^{\gamma_e}</math>

<math>~=</math>

<math>~K_c \rho_{ic}^{\gamma_c}</math>

<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>

<math>~=</math>

<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)} \biggr\}^{\gamma_c - \gamma_e}</math>

<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>

<math>~=</math>

<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} </math>

 

<math>~=</math>

<math> \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c} </math>

 

<math>~=</math>

<math> \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)} </math>

Hence, we can write the normalized (and dimensionless) free energy as,

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

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \biggl\{ \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} + \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{ \rho_{ic} }{ \rho_{ie} } \biggr)\biggr\} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} \, . </math>

Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and

<math> \frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, . </math>

Free Energy and Its Derivatives

Now, the free energy can be written as,

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

<math>~=~</math>

<math>~U_\mathrm{tot} + W</math>

 

<math>~=~</math>

<math>~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] S_\mathrm{env} + W</math>

 

<math>~=~</math>

<math>~\biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] C_\mathrm{core} R^{3-3\gamma_c} + \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \, .</math>

The first derivative of the free energy with respect to radius is, then,

<math>~\frac{d\mathfrak{G}}{dR}</math>

<math>~=~</math>

<math>~ -2 C_\mathrm{core} R^{2-3\gamma_c} -2 C_\mathrm{env} R^{2-3\gamma_e} + A R^{-2} \, .</math>

And the second derivative is,

<math>~\frac{d^2\mathfrak{G}}{dR^2}</math>

<math>~=~</math>

<math>~ -2 (2-3\gamma_c) C_\mathrm{core} R^{1-3\gamma_c} -2 (2-3\gamma_e) C_\mathrm{env} R^{1-3\gamma_e} - 2A R^{-3} \, .</math>

 

<math>~=~</math>

<math>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) C_\mathrm{core} R^{3-3\gamma_c} + (3\gamma_e-2) C_\mathrm{env} R^{3-3\gamma_e} - A R^{-1} \biggr]</math>

 

<math>~=~</math>

<math>~ \frac{2}{R^2} \biggl[(3\gamma_c-2) S_\mathrm{core} + (3\gamma_e-2) S_\mathrm{env} +W \biggr] \, .</math>

Equilibrium

The radius, <math>~R_\mathrm{eq}</math>, of the equilibrium configuration(s) is determined by setting the first derivative of the free energy to zero. Hence,

<math>~0 </math>

<math>~=~</math>

<math>~ 2 C_\mathrm{core} R_\mathrm{eq}^{2-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{2-3\gamma_e} - A R_\mathrm{eq}^{-2} </math>

 

<math>~=~</math>

<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 C_\mathrm{core} R_\mathrm{eq}^{3-3\gamma_c} + 2 C_\mathrm{env} R_\mathrm{eq}^{3-3\gamma_e} - A R_\mathrm{eq}^{-1} \biggr]</math>

 

<math>~=~</math>

<math>~ R_\mathrm{eq}^{-1} \biggl[ 2 S_\mathrm{core} + 2 S_\mathrm{env} +W \biggr]</math>

<math>\Rightarrow ~~~~ 2 S_\mathrm{tot} + W </math>

<math>~=~</math>

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

This is the familiar statement of virial equilibrium. From it we should always be able to derive the radius of equilibrium configurations.

Stability

To assess the relative stability of an equilibrium configuration, we need to determine the sign of the second derivative of the free energy, evaluated at the equilibrium radius. If the sign of the second derivative is positive, the system is dynamically stable; if the sign is negative, he system is dynamically unstable. Using the above statement of virial equilibrium, that is, setting,

<math>~2 S_\mathrm{tot} + W</math>

<math>~=~</math>

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

<math>\Rightarrow ~~~~ S_\mathrm{env} </math>

<math>~=~</math>

<math>~- S_\mathrm{core} - \frac{W}{2} \, ,</math>

we obtain,

<math>~\frac{d^2\mathfrak{G}}{dR^2}\biggr|_\mathrm{eq}</math>

<math>~=~</math>

<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ (3\gamma_c-2) S_\mathrm{core} +W - (3\gamma_e-2)\biggl( S_\mathrm{core} + \frac{W}{2}\biggr) \biggr]_\mathrm{eq} </math>

 

<math>~=~</math>

<math>~ \frac{2}{R_\mathrm{eq}^2} \biggl[ 3(\gamma_c-\gamma_e) S_\mathrm{core} + \biggl(2 - \frac{3}{2}\gamma_e\biggr)W \biggr]_\mathrm{eq} </math>

 

<math>~=~</math>

<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ (\gamma_c-\gamma_e) S_\mathrm{core} + \frac{1}{2}\biggl(\frac{4}{3} - \gamma_e\biggr)W \biggr]_\mathrm{eq} </math>

 

<math>~=~</math>

<math>~ \frac{6}{R_\mathrm{eq}^2} \biggl[ -\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>

So, if when evaluated at the equilibrium state, the expression inside of the square brackets of this last expression is negative, the equilibrium configuration will be dynamically unstable. We have chosen to write the expression in this particular final form because we generally will be interested in bipolytropes for which the adiabatic exponent of the envelope is greater than <math>~4/3</math> and the adiabatic exponent of the core is less than or equal to <math>~4/3</math> — that is, <math>~\gamma_e > 4/3 \ge \gamma_c</math>. Hence, because the gravitational potential energy, <math>~W</math>, is intrinsically negative, the system will be dynamically unstable only if the second term (involving <math>~S_\mathrm{core}</math>) is greater in magnitude than the first term (involving <math>~W</math>).

It is worth noting that, instead of drawing upon <math>~S_\mathrm{core}</math> and <math>~W</math> to define the stability condition, we could have used an appropriate combination of <math>~S_\mathrm{env}</math> and <math>~W</math>, or the <math>~S_\mathrm{core}</math> and <math>~S_\mathrm{env}</math> pair. Also, for example, because the virial equilibrium condition is <math>~S_\mathrm{tot} = -W/2</math>, it is easy to see that the following inequality also equivalently defines stability:

<math>~ S_\mathrm{tot}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core} </math>

<math>~>~</math>

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

Examples

(0, 0) Bipolytropes

Review

In an accompanying discussion we have derived analytic expressions describing the equilibrium structure and the stability of bipolytropes in which both the core and the envelope have uniform densities, that is, bipolytropes with <math>~(n_c, n_e) = (0, 0)</math>. From this work, we find that integrals over the mass and pressure distributions give:

<math>~ \frac{W}{R_\mathrm{eq}^3 P_i} = - \frac{A}{R_\mathrm{eq}^4 P_i} </math>

<math>~=~</math>

 <math>- ~ \frac{3}{5} \biggl[ \frac{GM_\mathrm{tot}^2}{R^4P_i} \biggr] \biggl( \frac{\nu^2}{q} \biggr) f </math>

 

<math>~=~</math>

 <math>- ~4\pi q^3 \Lambda f \, ,</math>

<math>~\frac{S_\mathrm{core}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{core}</math>

<math>~=~</math>

<math> ~2\pi q^3 (1 + \Lambda) \, ,</math>

<math>~\frac{S_\mathrm{env}}{R_\mathrm{eq}^3 P_i} = B_\mathrm{env}</math>

<math>~=~</math>

<math> 2\pi \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,</math>

where,

<math>~\Lambda</math>

<math>~\equiv~</math>

<math>

\frac{3}{2^2 \cdot 5\pi} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} \, ,</math>

<math>~f(q,\rho_e/\rho_c)</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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] </math>

 

<math>~=~</math>

<math>1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \, ,</math>

<math>~g^2(q,\rho_e/\rho_c)</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>

 

<math>~\equiv~</math>

<math>1 + \biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q) + \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 ) \biggr] \, , </math>

Renormalize

Let's renormalize these energy terms in order to more readily relate them to the generalized expressions derived above.

<math>~R^3 P_i</math>

<math>~=</math>

<math> ~R^3 K_c \biggl(\frac{\rho_{ic}}{\bar\rho} \biggr)^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi}\biggr) \frac{M_\mathrm{tot}}{R^3} \biggr]^{\gamma_c} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} K_c M_\mathrm{tot}^{\gamma_c} R_\mathrm{norm}^{3-3\gamma_c}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggl\{ K_c^{3\gamma_c -4} M_\mathrm{tot}^{\gamma_c(3\gamma_c -4)} \biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{3-3\gamma_c} \biggr\}^{1/(3\gamma_c -4)}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggl\{ \frac{G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} }{K_c} \biggr\}^{1/(3\gamma_c -4)}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3-3\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} E_\mathrm{norm} \, .</math>

Also,

<math>~\biggl[ \frac{GM_\mathrm{tot}^2}{R} \biggr]^{3\gamma_c -4}</math>

<math>~=</math>

<math> ~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} G^{3\gamma_c -4} M_\mathrm{tot}^{6\gamma_c -8} \biggl( \frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma_c} </math>

 

<math>~=</math>

<math> ~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} \biggl[ \frac{ G^{3\gamma_c -3} M_\mathrm{tot}^{5\gamma_c -6} }{K_c} \biggr] </math>

 

<math>~=</math>

<math> ~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-(3\gamma_c-4)} E_\mathrm{norm}^{3\gamma_c-4} \, . </math>

<math> \Rightarrow ~~~~\frac{GM_\mathrm{tot}^2}{R^4 P_i} </math>

<math>~=</math>

<math> ~\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{3\gamma_c - 3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} </math>

 

<math>~=</math>

<math> \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} \, . </math>

Hence,

<math>~\Lambda</math>

<math>~\equiv~</math>

<math>

\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{-\gamma_c} 

\biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \, .</math>

Given that <math>~\rho_{ic}/\bar\rho = \nu/q^3</math> for the <math>~(n_c, n_e) = (0, 0)</math> bipolytrope, we can finally write,

<math>~\frac{R^3 P_i}{E_\mathrm{norm}}</math>

<math>~=</math>

<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c} \, ,</math>

and,

<math>~\Lambda</math>

<math>~\equiv~</math>

<math> \frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} = \frac{1}{5} \frac{\nu}{q} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} \, .</math>

Hence the renormalized gravitational potential energy becomes,

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

<math>~=</math>

<math> - \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f \, ;</math>

and the two, renormalized contributions to the thermal energy become,

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

<math>~=~</math>

<math> \frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c} \, ,</math>

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

<math>~=~</math>

<math> \frac{4\pi}{3(\gamma_e-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(3\gamma_c-3\gamma_e)/(3\gamma_c-4)} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_e} </math>

 

 

<math> \times \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, ,</math>

Finally, then, we can state that,

<math>~\mathfrak{f}_{WM}</math>

<math>~\equiv</math>

<math>~\frac{\nu^2}{q} \cdot f \, ,</math>

<math>~s_\mathrm{core}</math>

<math>~\equiv</math>

<math> 1 + \Lambda \, , </math>

<math>~(1-q^3) s_\mathrm{env}</math>

<math>~\equiv</math>

<math> (1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, .</math>

Virial Equilibrium and Stability Evaluation

With these expressions in hand, we can deduce the equilibrium radius and relativity stability of <math>~(n_c, n_e) = (0, 0)</math> bipolytropes using the generalized expressions provided above. For example, from the statement of virial equilibrium <math>~(2S_\mathrm{tot} = - W )</math> we obtain,

<math>~q^3 (1 + \Lambda) + (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) </math>

<math>~=~</math>

<math>~q^3 \Lambda \biggl[ 1 + \frac{5}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \frac{1}{2q^5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) \biggr] </math>

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

<math>~=~</math>

<math>\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3) + \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5) - \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

 

<math>~=~</math>

<math>\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q-q^3 + 2 -3q +q^3) + \frac{1}{2q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (2 - 5q^3 + 3q^5 +3 - 15q^2+15q^3 -3q^5) </math>

 

<math>~=~</math>

<math>\frac{5}{2}\biggl[ 2\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q) + \frac{1}{q^2} \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 (1 - 3q^2 + 2q^3 ) \biggr] </math>

 

<math>~=~</math>

<math>\frac{5}{2}(g^2-1) </math>

<math>\Rightarrow ~~~~ \biggl[ \frac{P_i}{GM_\mathrm{tot}^2} \biggr] R_\mathrm{eq}^4</math>

<math>~=~</math>

<math>\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \, . </math>

Or, given the above renormalization, this expression can be written as,

<math> \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{4-3\gamma_c } \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} </math>

<math>~=~</math>

<math>\biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) </math>

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

<math>~=~</math>

<math> \biggl\{ \biggl( \frac{3}{2^3 \pi } \biggr) \frac{\nu^2}{q^4} (g^2-1) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c} \biggr\}^{1/(4-3\gamma_c)} \, . </math>


And the condition for dynamical stability is,

<math>-\frac{W}{2}\biggl( \gamma_e - \frac{4}{3}\biggr) - (\gamma_e-\gamma_c) S_\mathrm{core} </math>

  <math>~>~</math> 

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

<math>\Rightarrow ~~~~ 2\pi q^3 \Lambda \biggl[ \biggl( \gamma_e - \frac{4}{3}\biggr) f - (\gamma_e-\gamma_c) \biggl( 1 + \frac{1}{\Lambda}\biggr) \biggr] </math>

  <math>~>~</math> 

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

<math>~\biggl( \gamma_e - \frac{4}{3} \biggr)f - (\gamma_e - \gamma_c) \biggl[1 + \frac{5}{2}(g^2-1) \biggr]</math>

  <math>~>~</math> 

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

(5, 1) Bipolytropes

In another accompanying discussion we have derived analytic expressions describing the equilibrium structure of bipolytropes with <math>~(n_c, n_e) = (5, 1)</math>. Can we perform a similar stability analysis of these configurations? Work in progress!

Best of the Best

One Derivation of Free Energy

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

<math>~=</math>

<math> - \frac{3\cdot \mathfrak{f}_{WM}}{5} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} + \frac{\nu s_\mathrm{core} }{(\gamma_c - 1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_c-1)} </math>

 

 

<math> ~+ \frac{(1-\nu) s_\mathrm{env} }{(\gamma_e - 1)} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl[ \frac{R}{R_\mathrm{norm}} \biggr]^{-3(\gamma_e-1)} \, . </math>

Another Derivation of Free Energy

Hence the renormalized gravitational potential energy becomes,

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

<math>~=</math>

<math> - \biggl( \frac{3}{5} \biggr) \frac{\nu^2}{q} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{-1} \cdot f \, ;</math>

and the two, renormalized contributions to the thermal energy become,

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

<math>~=~</math>

<math> \frac{4\pi q^3 (1 + \Lambda) }{3(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c} \biggl( \frac{R}{R_\mathrm{norm}} \biggr)^{3-3\gamma_c} </math>

 

<math>~=~</math>

<math> \frac{\nu (1 + \Lambda) }{(\gamma_c-1)} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{\gamma_c-1} \chi^{3-3\gamma_c} \, ,</math>

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

<math>~=~</math>

<math> \frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{R^3 P_{ie}}{ E_\mathrm{norm} } \biggr] \biggl[ (1-q^3) + \frac{5}{2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \Lambda \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

 

<math>~=~</math>

<math> \frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] R^3 K_e \rho_{ie}^{\gamma_e} </math>

 

<math>~=~</math>

<math> \frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] R^3 K_e \rho_\mathrm{norm}^{\gamma_e} \biggl( \frac{\rho_{ie}}{\bar\rho} \biggr)^{\gamma_e} \biggl( \frac{\bar\rho}{\rho_\mathrm{norm}} \biggr)^{\gamma_e} </math>

 

<math>~=~</math>

<math> \frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{\mathrm{BigTerm}}{E_\mathrm{norm}} \biggr] ( \rho_\mathrm{norm} R_\mathrm{norm}^3) K_e \rho_\mathrm{norm}^{\gamma_e-1} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e} </math>

 

<math>~=~</math>

<math> \frac{2 (2\pi) }{3(\gamma_e-1)} \biggl[ \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e} \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl( \frac{3M_\mathrm{tot}}{4\pi} \biggr) \frac{K_e}{E_\mathrm{norm}} \biggl( \frac{3}{4\pi} \biggr)^{\gamma_e-1} \biggl[ \frac{G^3 M_\mathrm{tot}^2 }{K_c^3}\biggr]^{(\gamma_e-1)/(3\gamma_c-4)} </math>

 

<math>~=~</math>

<math> \frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \frac{K_c M_\mathrm{tot}}{E_\mathrm{norm}} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)} </math>

 

<math>~=~</math>

<math> \frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)} \biggl[ \frac{K_c^{3\gamma_c-4} M_\mathrm{tot}^{3\gamma_c-4}}{ G^{3\gamma_c-3} M_\mathrm{tot}^{5\gamma_c-6} K_c^{-1}} \biggr]^{1/(3\gamma_c-4)} </math>

 

<math>~=~</math>

<math> \frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \chi^{3-3\gamma_e} \biggl[ \mathrm{BigTerm}\biggr] \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(1-\gamma_e)/(3\gamma_c-4)} \biggl[ \frac{K_c^{3}}{G^{3} M_\mathrm{tot}^{2}} \biggr]^{(\gamma_c-1)/(3\gamma_c-4)} </math>

 

<math>~=~</math>

<math> \frac{(1-\nu)}{ (1-q^3) (\gamma_e-1)} \biggl[ \frac{3}{4\pi} \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl( \frac{K_e}{K_c} \biggr) \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2}\biggr]^{(\gamma_c-\gamma_e)/(3\gamma_c-4)} \biggl[ \mathrm{BigTerm}\biggr] \chi^{3-3\gamma_e} </math>

Finally, then, we can state that,

<math>~\mathfrak{f}_{WM}</math>

<math>~\equiv</math>

<math>~\frac{\nu^2}{q} \cdot f \, ,</math>

<math>~s_\mathrm{core}</math>

<math>~\equiv</math>

<math> 1 + \Lambda \, , </math>

<math>~(1-q^3) s_\mathrm{env}</math>

<math>~\equiv</math>

<math> (1-q^3) + \Lambda\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_0}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_0}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \, .</math>

Note,

<math>~\Lambda</math>

<math>~\equiv~</math>

<math>

\frac{3}{2^2 \cdot 5\pi} \frac{\nu^2}{q^4} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{-\gamma_c} 

\biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^{3\gamma_c - 4} = \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4} \, .</math>

We also want to ensure that envelope pressure matches the core pressure at the interface. This means,

<math>~K_e \rho_{ie}^{\gamma_e}</math>

<math>~=</math>

<math>~K_c \rho_{ic}^{\gamma_c}</math>

<math>\Rightarrow ~~~~\frac{K_e}{K_c} </math>

<math>~=</math>

<math>~\rho_{ic}^{\gamma_c} \rho_{ie}^{-\gamma_e} </math>

 

<math>~=</math>

<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \rho_\mathrm{norm}^{\gamma_c - \gamma_e}</math>

 

<math>~=</math>

<math>~\biggl[ \frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl\{ \frac{3}{4\pi} \biggl[ \frac{G^3 M_\mathrm{tot}^2}{K_c^3} \biggr]^{1/(3\gamma_c -4)} \biggr\}^{\gamma_c - \gamma_e}</math>

<math>\Rightarrow ~~~~\frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1}</math>

<math>~=</math>

<math>~\biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\rho_\mathrm{norm}} \biggr]^{\gamma_c} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ie}}{\rho_\mathrm{norm}} \biggr]^{-\gamma_e} \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_{ie}}{\bar\rho} \biggr]^{\gamma_e-1} </math>

 

<math>~=</math>

<math> \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{\rho_\mathrm{norm}}{ \bar\rho } \biggr)^{\gamma_e - \gamma_c} </math>

 

<math>~=</math>

<math> \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\rho_{ic}}{\bar\rho} \biggr]^{\gamma_c-1} \biggl( \frac{\rho_{ic}}{\rho_{ie}} \biggr) \biggl( \frac{ R}{R_\mathrm{norm}} \biggr)^{3(\gamma_e - \gamma_c)} </math>

Keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and

<math> \frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} \, . </math>

Summary

Understanding Free-Energy Behavior

Step 1: Pick values for the separate coefficients, <math>\mathcal{A}, \mathcal{B},</math> and <math>\mathcal{C},</math> of the three terms in the normalized free-energy expression,

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

<math>~=</math>

<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} </math>

then plot the function, <math>\mathfrak{G}^*(\chi)</math>, and identify the value(s) of <math>~\chi_\mathrm{eq}</math> at which the function has an extremum (or multiple extrema).

Step 2: Note that,

<math>~\mathcal{A}</math>

<math>~\equiv</math>

<math>~\frac{\nu^2}{5q} \biggl\{ 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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\}</math>

 

<math>~=</math>

<math>~\frac{1}{5} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl[ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \biggr]</math>

<math>~\mathcal{B}</math>

<math>~\equiv</math>

<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl[ 1+\Lambda_\mathrm{eq} \biggr]</math>

<math>~\mathcal{C}</math>

<math>~\equiv</math>

<math>~(1-\nu)\biggl( \frac{K_e}{K_c} \biggr)^* \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{(1-\nu)}{(1-q^3)} \biggr]^{\gamma_e-1} \biggl\{ 1 + \frac{\Lambda_\mathrm{eq}}{(1-q^3)}\biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\}</math>

 

<math>~\equiv</math>

<math>~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)}</math>

where (see, for example, in the context of its original definition),

<math>~\Lambda_\mathrm{eq} \equiv

\frac{3}{2^2\pi \cdot 5} \biggl( \frac{GM_\mathrm{tot}^2}{R_\mathrm{eq}^4 P_i} \biggr) \frac{\nu^2}{q^4} 

</math>

<math>~=~</math>

<math>

\frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} 

\chi_\mathrm{eq}^{3\gamma_c - 4} </math>

and, where,

<math>\biggl( \frac{K_e}{K_c} \biggr)^* \equiv \frac{K_e}{K_c} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{(\gamma_c - \gamma_e)/(3\gamma_c -4)} </math>

<math>~=</math>

<math> \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{1-\nu}{1-q^3} \biggr]^{-\gamma_e} \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c} \chi_\mathrm{eq}^{3(\gamma_e - \gamma_c)} \, . </math>

Also, keep in mind that, if the envelope and core both have uniform (but different) densities, then <math>~\rho_{ic} = \rho_c</math>, <math>~\rho_{ie} = \rho_e</math>, and

<math> \frac{\rho_c}{\bar\rho} = \frac{\nu}{q^3} \, ; ~~~~~ \frac{\rho_e}{\bar\rho} = \frac{1-\nu}{1-q^3} \, ; ~~~~~ \frac{\rho_e}{\rho_c} = \frac{q^3(1-\nu)}{\nu (1-q^3)} ~~\Rightarrow ~~~ \frac{q^3}{\nu} = \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \, . </math>

Step 3: An analytic evaluation tells us that the following should happen. Using the numerically derived value for <math>~\chi_\mathrm{eq}</math>, define,

<math>~\mathcal{C}^' \equiv \mathcal{C} \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \, .</math>

We should then discover that,

<math>\frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'} = \chi_\mathrm{eq}^{4-3\gamma_c} = \frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{1}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \, .</math>

Check It

<math>~\mathcal{B} + \mathcal{C}^'</math>

<math>~=</math>

<math>~ \nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ \biggl[ 1+\Lambda_\mathrm{eq} \biggr] + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] \biggr\} </math>

<math>\Rightarrow~~~~\mathcal{A} \biggl[ \Lambda_\mathrm{eq} \cdot 5\biggl( \frac{q}{\nu^2} \biggr) \biggr]</math>

<math>~=</math>

<math>~ 1+\Lambda_\mathrm{eq} + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

<math> \Rightarrow~~~~\Lambda_\mathrm{eq} \biggl\{ 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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\} </math>

<math>~=</math>

<math>~ 1+\Lambda_\mathrm{eq} + \frac{(1-q^3)}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

<math> \Rightarrow~~~~\Lambda_\mathrm{eq} \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\} </math>

<math>~=</math>

<math>~ \frac{1}{q^3} + \frac{\Lambda_\mathrm{eq}}{q^3} \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

<math>\Rightarrow~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>

<math>~=</math>

<math> q^3 \biggl\{ \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[ \biggl(\frac{1}{q^5} - 1 \biggr) - \frac{5}{2}\biggl(\frac{1}{q^2} - 1 \biggr) \biggr] \biggr\} - \biggl[ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) + \frac{3}{2q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

<math>\Rightarrow~~~~ \frac{2}{\Lambda_\mathrm{eq}} </math>

<math>~=</math>

<math> 5\biggl( \frac{\rho_e}{\rho_c} \biggr) ( q - q^3 ) + \frac{2}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (1 - q^5 ) - \frac{5}{2}( q^3 - q^5 ) \biggr] - 5 \biggl( \frac{\rho_e}{\rho_c}\biggr) (-2 + 3q - q^3) - \frac{3}{q^2} \biggl( \frac{\rho_e}{\rho_c}\biggr)^2 (-1 +5q^2 - 5q^3 + q^5) </math>

 

<math>~=</math>

<math> 5\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ ( q - q^3 ) - (-2 + 3q - q^3) \biggr] + \frac{1}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1 - q^5 ) - 5( q^3 - q^5 ) - 3(-1 +5q^2 - 5q^3 + q^5) \biggr] </math>

 

<math>~=</math>

<math> 10\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 1-q \biggr] + \frac{5}{q^2}\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 1 - 3q^2 + 2q^3 \biggr] </math>

<math>\Rightarrow~~~~ \frac{1}{\Lambda_\mathrm{eq}} \biggl[ \frac{2q^2}{5} \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggr] </math>

<math>~=</math>

<math> 2q^2 (1-q )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) ( 1 - 3q^2 + 2q^3 ) </math>

Fortunately, this precisely matches our earlier derivation, which states that,

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

  <math>~=</math> 

<math>\frac{5}{2}(g^2-1) = \frac{5}{2}\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>

Playing With One Example

By setting,

<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

2.5

1.0

2.0

a plot of <math>~\mathfrak{G}^*</math> versus <math>~\chi</math> exhibits the following, two extrema:

extremum

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

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

 

<math>~\chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)}</math>

<math>~\mathcal{C}^'</math>

<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>

<math>~\frac{\mathcal{A}}{\mathcal{B} +\mathcal{C}^'}</math>

MIN

<math>1.1824</math>

<math>-0.611367</math>

 <math>~~~~\Rightarrow~~~~</math> 

<math>0.66891</math>

<math>1.3378</math>

<math>1.0693</math>

<math>1.0694</math>

MAX

<math>9.6722</math>

<math>+0.508104</math>

<math>~\Rightarrow</math>

<math>0.004313</math>

<math>0.008625</math>

<math>2.4786</math>

<math>2.4786</math>

The last two columns of this table confirm the internal consistency of the relationships presented in Step 3, above. But what does this mean in terms of the values of <math>~\nu</math>, <math>~q</math>, and the related ratio of densities at the interface, <math>~\rho_e/\rho_c</math>?

Let's assume that what we're trying to display and examine is the behavior of the free-energy surface for a fixed value of the ratio of densities at the interface. Once the value of <math>~\rho_e/\rho_c</math> has been specified, it is clear that the value of <math>~q</math> (and, hence, also <math>~\nu</math>) is set because <math>~\mathcal{A}</math> has also been specified. But our specification of <math>~\mathcal{B}</math> along with <math>~\rho_e/\rho_c</math> also forces a particular value of <math>~q</math>. It is unlikely that these two values of <math>~q</math> will be the same. In reality, once <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> have both been specified, they force a particular <math>~(\nu, q)</math> pair. How do we (easily) figure out what this pair is?

Let's begin by rewriting the expressions for <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math> in terms of just <math>~q</math> and the ratio, <math>~\rho_e/\rho_c</math>, keeping in mind that, for the case of a uniform-density core (of density, <math>~\rho_c</math>) and a uniform-density envelope (of density, <math>~\rho_e</math>),

<math>~\frac{\rho_e}{\rho_c}</math>

<math>~=</math>

<math>~\frac{q^3(1-\nu)}{\nu(1-q^3)} \, ,</math>

hence,

<math>~\nu</math>

<math>~=</math>

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

   and    

<math>~\frac{q^3}{\nu}</math>

<math>~=</math>

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

Putting the expression for <math>~\mathcal{A}</math> in the desired form is simple because <math>~\nu</math> only appears as a leading factor. Specifically, we have,

<math>~\mathcal{A}</math>

<math>~=</math>

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

 

<math>~=</math>

<math>~\frac{\pi }{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \biggl\{ q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3 - q^5 ) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[1-\frac{5}{2} q^3 + \frac{3}{2}q^5 \biggr] \biggr\} \, .</math>

The expression for <math>~\mathcal{B}</math> can be written in the form,

<math>~\mathcal{B}</math>

<math>~=</math>

<math>~\nu \biggl[ \biggl( \frac{3}{4\pi} \biggr)\frac{\nu}{q^3} \biggr]^{\gamma_c-1} \biggl\{ 1+\frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} \chi_\mathrm{eq}^{3\gamma_c - 4} \biggr\}</math>

 

<math>~=</math>

<math>~\nu \biggl[ \biggl( \frac{4\pi}{3} \biggr)\frac{q^3}{\nu} \biggr]^{1-\gamma_c} +\frac{\pi q^5}{5} \biggl( \frac{\nu^2}{q^6} \biggr) \chi_\mathrm{eq}^{3\gamma_c - 4} </math>

 

<math>~=</math>

<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \frac{q^3}{\nu} \biggr]^{-\gamma_c} +\frac{\pi q^5}{5} \biggl( \frac{q^3}{\nu} \biggr)^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} </math>

 

<math>~=</math>

<math>~q^3 \biggl( \frac{4\pi}{3} \biggr)^{1-\gamma_c} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-\gamma_c} +\frac{\pi q^5}{5} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^{-2} \chi_\mathrm{eq}^{3\gamma_c - 4} \, . </math>

Generally speaking, the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, which appears in the expression for <math>~\mathcal{B}</math>, is not known ahead of time. Indeed, as is illustrated in our simple example immediately above, the normal path is to pick values for the coefficients, <math>~\mathcal{A}</math>, <math>~\mathcal{B}</math>, and <math>~\mathcal{C}</math>, and determine the equilibrium radius by looking for extrema in the free-energy function. And because <math>~\chi_\mathrm{eq}</math> is not known ahead of time, it isn't clear how to (easily) figure out what pair of physical parameter values, <math>~(\nu, q)</math>, give self-consistent values for the coefficient pair, <math>~(\mathcal{A}, \mathcal{B})</math>.

Because we are using a uniform density core and uniform density envelope as our base model, however, we do know the analytic solution for <math>~\chi_\mathrm{eq}</math>. As stated above, it is,

<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>

<math>~=</math>

<math>~\frac{1}{\Lambda_\mathrm{eq}} \cdot \frac{\pi}{5} \biggl( \frac{\nu}{q} \biggr) \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{1-\gamma_c} </math>

 

<math>~=</math>

<math>~\frac{\pi q^2}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c}\biggl( \frac{\nu}{q^3} \biggr) \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_c} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \cdot \biggl[ \frac{q^3}{\nu} \biggr]^{\gamma_c-1} </math>

 

<math>~=</math>

<math>~\frac{\pi}{2} \biggl( \frac{3}{4\pi} \biggr)^{1-\gamma_c} \biggl(\frac{\rho_e}{\rho_c}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_c} \biggr) ( q^2-q^3 ) + \frac{\rho_e}{\rho_c} (1 - q^2) \biggr] \cdot \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3\biggr]^{\gamma_c-2} </math>

Combining this expression with the one for <math>~\mathcal{B}</math> gives us the desired result — although, strictly speaking, it is cheating! We can now methodically choose <math>~(\nu, q)</math> pairs and map them into the corresponding values of <math>~\mathcal{A}</math> and <math>~\mathcal{B}</math>. And, via an analogous "cheat," the choice of <math>~(\nu, q)</math> also gives us the self-consistent value of <math>~\mathcal{C}</math>. In this manner, we should be able to map out the free-energy surface for any desired set of physical parameters.

Second Example

Explain Logic

FreeEnergyExample.jpg

The figure presented here, on the right, shows a plot of the free energy, as a function of the dimensionless radius, <math>~\mathfrak{G}^*(\chi)</math>, where,

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

<math>~=</math>

<math>~ -~ 3\mathcal{A} \chi^{-1} - \frac{\mathcal{B}}{(1-\gamma_c)} ~\chi^{3-3\gamma_c} - \frac{\mathcal{C}}{(1-\gamma_e)} ~\chi^{3-3\gamma_e} \, ,</math>

and, where we have used the parameter values,

<math>~\gamma_c = 6/5; ~~~ \gamma_e = 2</math>

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

0.201707

0.0896

0.002484

Directly from this plot we deduce that this free-energy function exhibits a minimum at <math>~\chi_\mathrm{eq} = 0.1235</math> and that, at this equilibrium radius, the configuration has a free-energy value, <math>~\mathfrak{G}^*(\chi_\mathrm{eq} ) = -2.0097</math>. Via the steps described below, we demonstrate that this identified equilibrium radius is appropriate for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope (with the just-specified core and envelope adiabatic indexes) that has the following physical properties:

  • Fractional core mass, <math>~\nu = 0.1</math>;
  • Core-envelope interface located at <math>~r_i/R = q = 0.435</math>;
  • Density jump at the core-envelope interface, <math>~\rho_e/\rho_c = 0.8</math>.


Step 1: Because the ratio, <math>~q^3/\nu</math>, is a linear function of the density ratio, <math>~\rho_e/\rho_c</math>, the full definition of the free-energy coefficient, <math>~\mathcal{A}</math>, can be restructured into a quadratic equation that gives the density ratio for any choice of the parameter pair, <math>~(q, \mathcal{A})</math>. Specifically,

<math>~5 \biggl( \frac{q^3}{\nu} \biggr)^2 \mathcal{A}</math>

<math>~=</math>

<math> q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) </math>

<math>~\Rightarrow ~~~~ 5\mathcal{A} \biggl[ \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^3) + q^3 \biggr]^2 </math>

<math>~=</math>

<math> q^5 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (1 - q^2 )q^3 + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl( 1 - \frac{5}{2}q^3 + \frac{3}{2}q^5\biggr) \, , </math>

and this can be written in the form,

<math> \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 a + \biggl( \frac{\rho_e}{\rho_c} \biggr) b + c</math>

<math>~=</math>

<math> 0 \, , </math>

where,

<math>~a</math>

<math>~\equiv</math>

<math>~5\mathcal{A} (1-q^3)^2 - 1 + \frac{5}{2}q^3 - \frac{3}{2}q^5 \, ,</math>

<math>~b</math>

<math>~\equiv</math>

<math>~10\mathcal{A} q^3(1-q^3) - \frac{5}{2}q^3 (1-q^2) \, ,</math>

<math>~c</math>

<math>~\equiv</math>

<math>~5\mathcal{A} q^6 - q^5 \, .</math>

Hence,

<math>~\frac{\rho_e}{\rho_c}</math>

<math>~=</math>

<math>~\frac{1}{2a} \biggl[\pm ( b^2 - 4ac)^{1/2} - b \biggr] \, .</math>

(For our physical problem it appears as though only the positive root is relevant.) For the purposes of this example, we set <math>~\mathcal{A} = 0.2017</math> and examined a range of values of <math>~q</math> to find a physically interesting value for the density ratio. We picked:

<math>~\mathcal{A}</math>

<math>~q</math>

 

<math>~a</math>

<math>~b</math>

<math>~c</math>

 

<math>~\frac{\rho_e}{\rho_c}</math>

 

<math>~\nu</math>

<math>~0.2017</math>

<math>~0.435</math>

<math>\Rightarrow</math>

<math>~0.03173</math>

<math>~-0.01448</math>

<math>~-0.008743</math>

<math>\Rightarrow</math>

<math>~0.80068</math>

<math>\Rightarrow</math>

<math>~0.10074</math>


Step 2: Next, we chose the parameter pair,

<math> ~\biggl(q, \frac{\rho_e}{\rho_c} \biggr) = (0.43500, 0.80000) </math>

and determined the following parameter values from the known analytic solution:

<math>~\nu</math>

<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

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

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

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

<math>~0.100816</math>

<math>~43.16365</math>

<math>~3.923017</math>

<math>~0.13684</math>

<math>~0.12349</math>

<math>~0.201707</math>

<math>~0.089625</math>

<math>~0.002484</math>

Construction Multiple Curves to Define a Free-Energy Surface

Okay. Now that we have the hang of this, let's construct a sequence of curves that represent physical evolution at a fixed interface-density ratio, <math>~\rho_e/\rho_c</math>, but for steadily increasing core-to-total mass ratio, <math>~\nu</math>. Specifically, we choose,

<math> ~\frac{\rho_e}{\rho_c} = \frac{1}{2} \, . </math>

From the known analytic solution, here are parameters defining several different equilibrium models:

Identification of Local Minimum in Free Energy

<math>~\nu</math>

<math>~q</math>

<math>~f\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

<math>~g^2\biggl(q, \frac{\rho_e}{\rho_c} \biggr)</math>

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

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

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

<math>~0.2</math>

<math>~9^{-1/3} = 0.48075</math>

<math>~12.5644</math>

<math>~2.091312</math>

<math>~0.366531</math>

<math>~0.037453</math>

<math>~0.2090801</math>

<math>~0.2308269</math>

<math>~2.06252 \times 10^{-4}</math>

<math>~0.4</math>

<math>~4^{-1/3} = 0.62996</math>

<math>~4.21974</math>

<math>~1.56498</math>

<math>~0.707989</math>

<math>~0.0220475</math>

<math>~0.2143496</math>

<math>~0.5635746</math>

<math>~4.4626 \times 10^{-5}</math>

<math>~0.5</math>

<math>~3^{-1/3} = 0.693361</math>

<math>~2.985115</math>

<math>~1.42334</math>

<math>~0.9448663</math>

<math>~0.0152116</math>

<math>~0.2152641</math>

<math>~0.791882</math>

<math>~1.5464 \times 10^{-5}</math>

Here we are examining the behavior of the free-energy function for bipolytropic models having <math>~(n_c, n_e) = (0, 0)</math>, <math>~(\gamma_c, \gamma_e) = (6/5, 2)</math>, and a density ratio at the core-envelope interface of <math>~\rho_e/\rho_c = 1/2</math>. The figure shown here, on the right, displays the three separate free-energy curves, <math>~\mathfrak{G}^*(\chi)</math> — where, <math>~\chi \equiv R/R_\mathrm{norm}</math> is the normalized configuration radius — that correspond to the three values of <math>~\nu \equiv M_\mathrm{core}/M_\mathrm{tot}</math> given in the first column of the above table. Along each curve, the local free-energy minimum corresponds to the the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, recorded in column 6 of the above table.

ThreeFreeEnergyCurves.png

Each of the free-energy curves shown above has been entirely defined by our specification of the three coefficients in the free-energy function, <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math>. In each case, the values of these three coefficients was judiciously chosen to produce a curve with a local minimum at the correct value of <math>~\chi_\mathrm{eq}</math> corresponding to an equilibrium configuration having the desired <math>~(\nu, \rho_e/\rho_c)</math> model parameters. Upon plotting these three curves, we noticed that two of the curves — curves for <math>~\nu = 0.4</math> and <math>~\nu = 0.5</math> — also display a local maximum. Presumably, these maxima also identify equilibrium configurations, albeit unstable ones. From a careful inspection of the plotted curves, we have identified the value of <math>~\chi_\mathrm{eq}</math> that corresponds to the two newly discovered (unstable) equilibrium models; these values are recorded in the table that immediately follows this paragraph. By construction, we also know what values of <math>~\mathcal{A}, \mathcal{B}</math>, and <math>~\mathcal{C}</math> are associated with these two identified equilibria; these values also have been recorded in the table. But it is not immediately obvious what the values are of the <math>~(\nu, \rho_e/\rho_c)</math> model parameters that correspond to these two equilibrium models.

Subsequently Identified Local Energy Maxima

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

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

<math>~\chi_\mathrm{eq}^{4-3\gamma_c}</math>

<math>~\therefore</math>

<math>~\mathcal{A}</math>

<math>~\mathcal{B}</math>

<math>~\mathcal{C}</math>

<math>~\mathcal{C}^' = \mathcal{A} \chi_\mathrm{eq}^{3\gamma_c-4} - \mathcal{B}</math>

<math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)}</math>

<math>~0.08255</math>

<math>~+ 4.87562</math>

<math>~0.368715</math>

<math>~0</math>

<math>~0.2143496</math>

<math>~0.5635746</math>

<math>~4.4626 \times 10^{-5}</math>

<math>~1.7768 \times 10^{-2}</math>

<math>~0.08254</math>

<math>~0.032196</math>

<math>~+11.5187</math>

<math>~0.25300</math>

<math>~0</math>

<math>~0.2152641</math>

<math>~0.791882</math>

<math>~1.5464 \times 10^{-5}</math>

<math>~5.8964 \times 10^{-2}</math>

<math>~0.032196</math>

Related Discussions

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