Editing SSC/BipolytropeGeneralization

=Bipolytrope Generalization (Pt 1)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/BipolytropeGeneralization|Part I:&nbsp; Bipolytrope Generalization]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/BipolytropeGeneralization/Pt2|Part II:&nbsp; Derivations]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue" width="25%"><br />[[SSC/BipolytropeGeneralization/Pt3|Part III:&nbsp; Examples]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue"><br />[[SSC/BipolytropeGeneralization/Pt4|Part IV:&nbsp; Best of the Best]]
&nbsp;
  </td>
</tr>
</table>


On <font color="red">26 August 2014</font>, Tohline finished rewriting the chapter titled "Bipolytrope Generalization" in a very concise manner ([[SSC/BipolytropeGeneralizationVersion2#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.

{{ SGFworkInProgress }}

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
In addition to the gravitational potential energy, which is naturally written as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \cdot \mathfrak{f}_{WM} \, ,</math>
  </td>
</tr>
</table>
</div>
it seems reasonable to write the separate thermal energy contributions as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~S_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
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 [[SSCpt1/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, densities and pressures by, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~E_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{G^{3(\gamma_c-1)} M_\mathrm{tot}^{5\gamma_c-6}}{K_c}  \biggr]^{1/(3\gamma_c -4)} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~R_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} \biggr]^{1/(3\gamma_c -4)} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <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>
  </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>
</tr>
</table>
</div>

As is detailed below &#8212; [[#Detailed_Derivations|first, here]], and via [[#Another_Derivation_of_Free_Energy|an independent derivation, here]] &#8212; quite generally the expression for the normalized free energy is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{G}^* \equiv \frac{\mathfrak{G}}{E_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

</table>
</div>
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:
<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>
where,
<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{1}{5} \cdot \mathfrak{f}_{WM} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>

As is [[#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.''

==Simplest Bipolytrope==

===Familiar Setup===
As has been shown in [[#.280.2C_0.29_Bipolytropes|an accompanying presentation]], for an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope, 
<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_\mathrm{eq} \, ,
</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_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>

and where (see, for example, [[SSC/Structure/BiPolytropes/Analytic00#Expression_for_Free_Energy|in the context of its original definition]], or another,  [[SSC/Structure/BiPolytropes/Analytic00#LambdaDeff|separate derivation]]),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Lambda_\mathrm{eq}  
</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>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>

and where (see the [[SSC/VirialStability#Expressions_for_Mass|associated discussion of relevant mass integrals]]),
<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>

===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:

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

<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
where (see an [[SSC/Structure/BiPolytropes/Analytic00#Gravitational_Potential_Energy|associated discussion]] or the [[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>
  <td align="right">
<math>~\lambda_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{GM_\mathrm{tot}^2}{R^4 P_i} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathfrak{F} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[  (-2q^2 + 3q^3 - q^5) +
\frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</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 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 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
</tr>
</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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
</div>
and the virial is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2S_\mathrm{tot} + W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot (f - 1- \mathfrak{F} ) 
</math>
  </td>
</tr>

<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
f - 1- \mathfrak{F} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </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>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^2 \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} (g^2-1)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~ \frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</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>
  <td align="right">
<math>~P_i</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</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>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
</div>

Hence, the virial equilibrium condition gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \frac{R_\mathrm{eq}^4 P_0}{GM_\mathrm{tot}^2}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^{2} q^2 g^2  \, .
</math>
  </td>
</tr>
</table>
</div>

This result precisely matches [[SSC/Structure/BiPolytropes/Analytic00#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>
  <td align="right">
<math>~R_\mathrm{norm}^4 P_\mathrm{norm}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~GM_\mathrm{tot}^2 \, ,</math>
  </td>
</tr>
</table>
</div>
the expression alternatively can be rewritten as,
<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 align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</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>
  <td align="right">
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \chi_\mathrm{eq}^{4-3\gamma_c} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</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>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>

Note that, 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>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{3(\gamma_c - \gamma_e)} \biggl( \frac{\mathcal{C}}{\mathcal{B}} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</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>
  <td align="right">
<math>GM_\mathrm{tot}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~R_\mathrm{norm}^4 P_\mathrm{norm} = E_\mathrm{norm} R_\mathrm{norm}  \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>~\frac{\Pi}{P_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
<math>~\frac{K_c \rho_c^{\gamma_c} }{P_\mathrm{norm}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</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>
  <td align="right">
<math>~\mathcal{B}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>


And the free-energy coefficient,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{C}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
</div>

<div align="center" id="DerivationTable">
<table border="1" align="center" cellpadding="10">
<tr>
  <td align="center">
OLD DERIVATION
<div align="center">
<math>P_{ic} = K_c \rho_c^{\gamma_c}</math>
</div>
  </td>
  <td align="center">
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>
</tr>


<tr>
  <td align="center" colspan="2">
&hellip; therefore &hellip;
  </td>
</tr>


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

<tr>
  <td align="right">
<math>~\mathcal{B}_\mathrm{OLD}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{B}_\mathrm{NEW}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

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



<tr>
  <td align="center" colspan="2">
&hellip; and, enforcing in equilibrium <math>~P_{ie} = P_{ic}</math> &hellip;
  </td>
</tr>


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

<tr>
  <td align="right">
<math>~\mathcal{C}_\mathrm{OLD}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{C}_\mathrm{NEW}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

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



<tr>
  <td align="center" colspan="2">
&hellip; and, also &hellip;
  </td>
</tr>


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

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c}  
</math>
  </td>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} -  \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{1}{\Lambda_\mathrm{eq}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{5q}{\nu} \biggr) \sigma^{\gamma_c - 1} \chi_\mathrm{eq}^{4 - 3\gamma_c}  - \frac{5}{2}
</math>
  </td>
</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>
  <td align="right">
<math>~\mathcal{A}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{C} \chi_\mathrm{eq}^{4-3\gamma_e} \, .</math>
  </td>
</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>
</tr>
</table>
</div>
When put together, these two relations imply,
<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>~\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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>

But the definition of <math>~\mathcal{B}</math> gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\mathcal{B} \chi_\mathrm{eq}^{4-3\gamma_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{4\pi }{3} \biggr) q^3 s_\mathrm{core}   
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq} \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, extrema occur when,
<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>~\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>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>  
\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq}   \biggl[ q^3 s_\mathrm{core}  +  (1-q^3) s_\mathrm{env} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~\biggl[ \frac{1}{\lambda_{ic}} \biggr]_\mathrm{eq}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>  
\biggl( \frac{3}{2^2 \cdot 5\pi } \biggr) \frac{\nu^2}{q} \cdot (f - 1 - \mathfrak{F})
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>  
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2 - 1 ) \, .
</math>
  </td>
</tr>
</table>
</div>

In what follows, keep in mind that,
<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>
\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>
</tr>

<tr>
  <td align="right">
<math>~K_c \rho_c^{\gamma_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Pi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>



<div align="center">
<table border="1" align="center" cellpadding="10">
<tr>
  <td align="center">
OLD DERIVATION
<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 align="center">
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>
</tr>


<tr>
  <td align="center" colspan="2">
&hellip; hence, as derived in the above table &hellip;
  </td>
</tr>


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

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} 
</math>
  </td>
</tr>
</table>
  </td>
  <td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\sigma^{\gamma_c} \chi_\mathrm{eq}^{4 - 3\gamma_c} -  \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2 
</math>
  </td>
</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>
  <td align="right">
<math>~\chi_\mathrm{eq}^{4 - 3\gamma_c}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 (g^2 - 1 )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4}  P_i }{GM_\mathrm{tot}^{2} } </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 (g^2-1)
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\sigma^{\gamma_c}\chi_\mathrm{eq}^{4 - 3\gamma_c} -  \biggl( \frac{2\pi}{3} \biggr) q^2 \sigma^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^2 q^2 (g^2 - 1 )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \chi_\mathrm{eq}^{4 - 3\gamma_c} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ R_\mathrm{eq}^{4-3\gamma_c} \biggl( \frac{K_c}{G} \biggr) M_\mathrm{tot}^{\gamma_c-2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2-\gamma_c} q^2 g^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{ R_\mathrm{eq}^{4}  P_0 }{GM_\mathrm{tot}^{2} } </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{2\pi}{3 } \biggr) \sigma^{2} q^2 g^2
</math>
  </td>
</tr>

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

<tr>
  <td align="center" colspan="2">
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>
</tr>


</table>
</div>

==Examples==

<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>
  <td align="center">
<math>~\nu</math>
  </td>
  <td align="center">
<math>~q</math>
  </td>
  <td align="center">
<math>~ \frac{\rho_e}{\rho_c} </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>
  <td align="center">
MIN/MAX
  </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>~0.5</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>
  <td align="center">
MIN
  </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>~0.5</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>
  <td align="center">
MIN
  </td>
</tr>

<tr>
  <td align="center">
<math>~0.473473</math>
  </td>
  <td align="center">
<math>~0.681838</math>
  </td>
  <td align="center">
<math>~0.516107</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<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 align="center">
<font color="red">MAX</font>
  </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>~0.5</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>
  <td align="center">
MIN
  </td>
</tr>

<tr>
  <td align="center">
<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 align="center">
<math>~0.032196</math>
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
<font color="red">MAX</font>
  </td>
</tr>
</table>




<table border="1" cellpadding="5" align="center">
<tr>
  <th align="center" colspan="13">
<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 align="center">
<math>~q</math>
  </td>
  <td align="center">
<math>~ \frac{\rho_e}{\rho_c} </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>~\frac{1}{\lambda_{i}}\biggr|_\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}_\mathrm{NEW}</math>
  </td>
  <td align="center">
<math>~\mathcal{C}_\mathrm{NEW}</math>
  </td>
  <td align="center">
<math>~G^*</math>
  </td>
  <td align="center">
<font color="red">Stability</font>
  </td>
  <td align="center">
MIN/MAX
  </td>
</tr>

<tr>
  <td align="center">
<math>~0.5</math>
  </td>
  <td align="center">
<math>~\biggl( \frac{1}{3} \biggr)^{1/3}</math>
  </td>
  <td align="center">
<math>~0.5</math>
  </td>
  <td align="center">
<math>~2.985115</math>
  </td>
  <td align="center">
<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 align="center">
MIN
  </td>
</tr>

<tr>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<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>
</tr>


<tr>
  <td align="center">
<math>~0.6</math>
  </td>
  <td align="center">
<math>~\biggl( \frac{3}{7} \biggr)^{1/3}</math>
  </td>
  <td align="center">
<math>~0.5</math>
  </td>
  <td align="center">
<math>~2.2507129</math>
  </td>
  <td align="center">
<math>~1.31282895</math>
  </td>
  <td align="center">
<math>~0.04160318</math>
  </td>
  <td align="center">
<math>~0.3411545</math>
  </td>
  <td align="center">
<math>~0.21493717</math>
  </td>
  <td align="center">
<math>~0.26165939</math>
  </td>
  <td align="center">
<math>~5.208750 \times 10^{-3}</math>
  </td>
  <td align="center">
<math>~+ 0.73532249</math>
  </td>
  <td align="center">
<math>~+0.0935217</math>
  </td>
  <td align="center">
MIN
  </td>
</tr>

<tr>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<math>~0.431745</math>
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
"
  </td>
  <td align="center">
<math>~+0.7367797</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
<font color="red">MAX</font>
  </td>
</tr>

<tr>
  <td align="center">
<math>~0.7</math>
  </td>
  <td align="center">
<math>~\biggl( \frac{7}{13} \biggr)^{1/3}</math>
  </td>
  <td align="center">
<math>~0.5</math>
  </td>
  <td align="center">
<math>~1.7707809</math>
  </td>
  <td align="center">
<math>~1.2209446</math>
  </td>
  <td align="center">
<math>~0.029500</math>
  </td>
  <td align="center">
<math>~0.3589388</math>
  </td>
  <td align="center">
<math>~0.21330744</math>
  </td>
  <td align="center">
<math>~0.28172532</math>
  </td>
  <td align="center">
<math>~3.389793 \times 10^{-3}</math>
  </td>
  <td align="center">
<math>~+ 0.8953395</math>
  </td>
  <td align="center">
<math>~- 0.0767108</math>
  </td>
  <td align="center">
<font color="red">MAX</font>
  </td>
</tr>

<tr>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&hellip;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </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, [[#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">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{1}{\nu}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{1}{q^3} - 1 \biggr) \, .</math>
  </td>
</tr>
</table>
</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">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}\biggr|_\mathrm{guess} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\mathcal{C}}{\mathcal{C}^'} \biggr)^{1/(3\gamma_e - 3\gamma_c)} \, .</math>
  </td>
</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>
  <td align="right">
<math>~\chi_\mathrm{eq}\biggr|_\mathrm{virial} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\mathcal{A}}{\mathcal{B} + \mathcal{C}^'}  \biggr]^{1/(4 - 3\gamma_c)} \, .</math>
  </td>
</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. 

=Related Discussions=
* [[SSC/BipolytropeGeneralizationVersion2|Newer, Version2 of ''Bipolytrope Generalization'' derivations]].
* [[SSC/Structure/BiPolytropes/Analytic00#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution]] with <math>n_c = 0</math> and <math>n_e=0</math>.
* [[SSC/Structure/BiPolytropes/Analytic51#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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