Editing SSC/FreeEnergy/PolytropesEmbedded/Pt1 (section)

===Zero-Zero Bipolytropes===

====General Form====
In this case, we retain full generality making the substitutions, <math>~n \rightarrow n_c</math> and <math>~j \rightarrow n_e</math>, to obtain,

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

<tr>
  <td align="right">
<math>~x^{(n_c-3)/n_c }_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{n_c}{ 3b} \biggl[a  -\biggl(\frac{3 c}{n_e}\biggr) x^{(n_e-3)/n_e}_\mathrm{eq} \biggr] \, ;</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{n_e^2(n_c-3)}{3[ n_c (n_e+3) - n_e(n_c+3)  ]}\biggr\} \frac{a}{c} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{n_e^2(n_c-3)}{3^2(n_c - n_e)}\biggr] \frac{a}{c} \, .
</math>
  </td>
</tr>
</table>
</div>


And here, [[#BiPolytrope00|the expression that describes the free-energy surface]] is,

<div align="center" id="FreeEnergy00">
<table border="1" cellpadding="5" align="center">
<tr><td align="center">

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

<tr>
  <td align="right">
<math>~\mathfrak{G}^*_{00} \equiv 5 \biggl(\frac{q}{\nu^2}\biggr) \chi_\mathrm{eq} 
\biggl[\frac{\mathfrak{G}_{00}}{E_\mathrm{norm}} \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{2q^3} \biggl[
n_c A_2\Chi^{-3/n_c} + n_e B_2\Chi^{-3/n_e} - 3C_2\Chi^{-1} \biggr] \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>
</div>
Hence, we have, 

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

<tr>
  <td align="right">
<math>~a \equiv 3\chi_\mathrm{eq} \biggl(\frac{5}{2q^3} \biggr) C_2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
3f \chi_\mathrm{eq}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b \equiv n_c \chi_\mathrm{eq}^{3/n_c} \biggl(\frac{5}{2q^3} \biggr) A_2 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
n_c \chi_\mathrm{eq}^{3/n_c}  \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]   \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c \equiv n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) B_2  </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ n_e \chi_\mathrm{eq}^{3/n_e} \biggl(\frac{5}{2q^3} \biggr) 
\biggl[\frac{2}{5} q^3  f - A_2\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
n_e \chi_\mathrm{eq}^{3/n_e}  \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\} \, , 
</math>
  </td>
</tr>
</table>
</div>

where the definitions of <math>~f</math> and <math>~\mathfrak{F}</math> are [[#BiPolytrope00|given below]].  We immediately deduce that the ''critical'' equilibrium state is identified by,

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

<tr>
  <td align="right">
<math>~[x_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{\frac{fn_e(n_c-3)}{3(n_c - n_e)}\biggr\} [\chi_\mathrm{eq}^{(n_e-3)/n_e}]_\mathrm{crit}  \biggl\{ f - \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \biggr\}^{-1} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 1 - \biggl[ \frac{n_e(n_c-3)}{3(n_c-n_e)} \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{n_c(3-n_e)}{3(n_c-n_e)} \, .</math>
  </td>
</tr>
</table>
</div>


From our [[#Equilibrium_Radius_2|associated detailed-force-balance derivation]], we know that the associated equilibrium radius is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{2-n_c} \nu^{n_c-1} q^{3-n_c} 
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{n_c} \biggr\}^{1/(n_c-3)}
\, .
</math>
  </td>
</tr>
</table>
</div>

<!--  Coefficient mistake, I think!
We have deduced that the system is unstable if,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{n_e}{3}\biggl[ \frac{3-n_e}{n_c-n_e} \biggr] </math>
  </td>
  <td align="center">
<math>~< </math>
  </td>
  <td align="left">
<math>~
\frac{A_2}{C_2}
= \frac{1}{f} \biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

-->

====Compare with Five-One====
It is worthwhile to set <math>~n_c = 5</math> and <math>~n_e = 1</math> in this expression and compare the result to the [[#FiveOneRadius|comparable expression shown above for the "Five-One" Bipolytrope]].  Here we have,

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

<tr>
  <td align="right">
<math>~\biggl[\chi_\mathrm{eq}\biggr]_{51}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl(\frac{\pi}{3}\biggr) 2^{-3} \nu^{4} q^{-2} 
\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5} \biggr\}^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\frac{1}{\sqrt{3}} \biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]^{5/2} \, ;
</math>
  </td>
</tr>
</table>
</div>
whereas, rewriting the [[#FiveOneRadius|above relation]] gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}\biggr|_{51}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot \frac{1}{\sqrt{3}} \biggl[\frac{(1+\ell_i^2)^{6/5}}{3\ell_i^2}\biggr]^{5/2} \, .</math>
  </td>
</tr>
</table>
</div>

And, here, we should conclude that the ''critical'' equilibrium configuration is associated with,

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

<tr>
  <td align="right">
<math>~\frac{1}{f}\biggl[1 + q^3 (f - 1-\mathfrak{F} ) \biggr]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{6} \, .</math>
  </td>
</tr>
<!--
<tr>
  <td align="right">
<math>~\Rightarrow~~~q^3 (f - 1-\mathfrak{F} )</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{5}{6} \cdot f - 1</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~\biggl[1 + \frac{2}{5} q^3(f-1-\mathfrak{F})\biggr]_\mathrm{crit} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 1 + \frac{2}{5}\biggl(\frac{5}{6} \cdot f - 1\biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{f}{3}  + \frac{3}{5}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ [\chi_\mathrm{eq}]_\mathrm{crit}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\pi}{2^3}\biggr)^{1/2} \frac{\nu^2}{q} \cdot
\frac{1}{\sqrt{3}} \biggl[\frac{f}{3}  + \frac{3}{5}\biggr]^{5/2} \, .
</math>
  </td>
</tr>
-->
</table>
</div>