Editing SSC/BipolytropeGeneralization/Pt4

=Bipolytrope Generalization (Pt 4)=
<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>

==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, [[SSC/Structure/BiPolytropes/Analytic00#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 [[SSC/Structure/BiPolytropes/Analytic00#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 align="center">
<math>~5.8964 \times 10^{-2}</math>
  </td>
  <td align="center">
<math>~0.032196</math>
  </td>
</tr>
</table>

=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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