Editing SSC/VirialStability (section)

=Virial Analysis=

==Isothermal Core Equilibrium Condition==
For a given set of fixed coefficients in the free energy expression, equilibria are identified by setting <math>\partial\mathfrak{G}/\partial\chi = 0</math>.  Generally,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\frac{\partial\mathfrak{G}}{\partial\chi}</math> 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>
A\chi^{-2} - C\chi^{-1} - \biggl(\frac{3}{n_e}\biggr)D\chi^{-(1+3/n_e)} \, .
</math>
  </td>
</tr>
</table>
So, for the case of <math>n_e = 3/2</math>, the equilibrium radius is given by the condition,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>A\chi_E^{-2}</math> 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>
 C\chi_E^{-1} + 2D\chi_E^{-3}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ C\chi_E^{2} - A\chi_E + 2D
</math> 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>0 \,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \chi_E
</math> 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>
\frac{1}{2C} \biggl[ A \pm \biggl( A^2 - 8DC \biggr)^{1/2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \chi_E
</math> 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>
\frac{A}{2C} \biggl[ 1 \pm \biggl( 1 - \frac{8DC}{A^2} \biggr)^{1/2}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp; 
  </td>
 <td align="center">
<math>= \,</math>
  </td>
 <td align="left">
<math>
\chi_0 [ 1 \pm \sqrt{1-\Pi} ] \, ,
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi_0 \,</math> 
  </td>
 <td align="center">
<math>\equiv \,</math>
  </td>
 <td align="left">
<math>
\frac{A}{2C}  \, ,
</math>
  </td>
  <td align="center">
&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;
  </td>

  <td align="right">
<math>\Pi \,</math> 
  </td>
 <td align="center">
<math>\equiv \,</math>
  </td>
 <td align="left">
<math>
\frac{8DC}{A^2}  \, .
</math>
  </td>
</tr>
</table>

==Combined Equilibrium Constraints==
But, from the condition on the temperature at the interface we also need,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\chi_E \,</math> 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c} \biggr)^{1/2} \biggl[ \frac{2\nu}{(1-\nu)} \biggl( \frac{D}{C} \biggr)\biggr]^{1/2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp; 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\chi_0 \biggl(\frac{\mu_e}{\mu_c} \biggr)^{1/2} \biggl[ \biggl(\frac{\nu}{1-\nu}\biggr) \Pi \biggr]^{1/2} \, .
</math>
  </td>
</tr>
</table>

Hence, the two conditions combined imply,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>1 \pm \sqrt{1-\Pi}</math> 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\lambda^{1/2} \Pi^{1/2} \, ,
</math>
  </td>
</tr>
</table>
where,
<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>
\biggl(\frac{\mu_e}{\mu_c} \biggr) \biggl(\frac{\nu}{1-\nu}\biggr) \, .
</math>
  </td>
</tr>
</table>
This, in turn, implies,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>1-\Pi \, </math> 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
(\lambda^{1/2} \Pi^{1/2} -1 )^2\, 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp; 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\lambda\Pi -2(\lambda\Pi)^{1/2} + 1 \, 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~2(\lambda\Pi)^{1/2} \,</math> 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\Pi(1+\lambda) \, 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~\Pi  \,</math> 
  </td>
 <td align="center">
<math>=\,</math>
  </td>
 <td align="left">
<math>
\frac{4\lambda}{(1+\lambda)^2} \, .
</math>
  </td>
</tr>
</table>

But notice, as well, that in order to have real roots of the equilibrium condition, we need <math>\Pi \le 1</math>.  This means,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>4\lambda \,</math> 
  </td>
 <td align="center">
<math>\le \,</math>
  </td>
 <td align="left">
<math>
(1+\lambda)^2 \, 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \lambda^2 - 2\lambda + 1  \,</math> 
  </td>
 <td align="center">
<math>\ge \,</math>
  </td>
 <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ (\lambda - 1)^2  \,</math> 
  </td>
 <td align="center">
<math>\ge \,</math>
  </td>
 <td align="left">
<math>
0 \, 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \lambda  \,</math> 
  </td>
 <td align="center">
<math>\ge \,</math>
  </td>
 <td align="left">
<math>
1 \,  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \frac{\mu_e}{\mu_c}  \,</math> 
  </td>
 <td align="center">
<math>\ge \,</math>
  </td>
 <td align="left">
<math>
\frac{1}{\nu} - 1 \,  
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \nu  \,</math> 
  </td>
 <td align="center">
<math>\ge \,</math>
  </td>
 <td align="left">
<math>
\biggl(\frac{\mu_e}{\mu_c} + 1\biggr)^{-1} \,  .
</math>
  </td>
</tr>
</table>

==Free Energy Expression==
To within an additive constant, the free energy may now be written as,
<div align="center">
<math>
\mathfrak{G} = W + U  = - A \chi^{-1} + B_e \chi^{-3/n_e} + (1-\delta_{\infty n_c}) B_c \chi^{-3/n_c} - \delta_{\infty n_c} \biggl[ B_I \ln\chi  - \frac{1}{3}B_I \ln \biggl( \frac{\rho_c|_0}{\rho_0} \biggr) \biggr] \, ,
</math>
</div>
where, <math>\chi \equiv R/R_0</math> and,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>
A \,
</math>  
  </td>
  <td align="center">
<math>=\,</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3GM^2_\mathrm{tot}}{5R_0} \biggr) \nu^2 \xi_s \biggl\{ 1 
+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) (\xi_s^2 - 1)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ (\xi_s^5 - 1) - \frac{5}{2}(\xi_s^2-1) \biggr] \biggr\} 
= \biggl( \frac{3GM^2_\mathrm{tot}}{5R_0} \biggr) \frac{\nu^2}{q}  f(\nu, q) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B_e \,</math>  
</td>
  <td align="center">
<math>=\,</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot}(1-\nu) n_eK_e (\rho_e|_0)^{1/n_e} = M_\mathrm{tot} (1-\nu) K_e\rho_0^{1/n_e} n_e \biggl( \frac{\rho_e|_0}{\rho_0} \biggr)^{1/n_e} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B_c \,</math>  
</td>
  <td align="center">
<math>= \,</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \nu n_c K_c (\rho_c|_0)^{1/n_c} = M_\mathrm{tot} K_e\rho_0^{1/n_e} \nu n_c \biggl[\biggl(\frac{K_c}{K_e}\biggr) \rho_0^{1/n_c - 1/n_e} \biggr] \biggl( \frac{\rho_c|_0}{\rho_0}\biggr)^{1/n_c} 
= M_\mathrm{tot} K_e\rho_0^{1/n_e} \nu n_c \biggl[ \biggl(\frac{\rho_c}{\rho_0}\biggr)^{1/n_e - 1/n_c} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{1+1/n_e} \biggr] \biggl( \frac{\rho_c|_0}{\rho_0}\biggr)^{1/n_c}  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>B_I\,</math>  
</td>
  <td align="center">
<math>=\,</math>
  </td>
  <td align="left">
<math>
3 M_\mathrm{tot} c_s^2 \nu = M_\mathrm{tot} K_e\rho_0^{1/n_e} \biggl[ \frac{c_s^2}{K_e\rho_0^{1/n_e}  } \biggr] 3\nu 
= M_\mathrm{tot} K_e\rho_0^{1/n_e} \biggl[ \biggl(\frac{\rho_c}{\rho_0}\biggr)^{1/n_e} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{1+1/n_e}\biggr] 3\nu \, .
</math>
  </td>
</tr>
</table>

==Derivatives of Free Energy==
<div align="center">
<math>
\frac{\partial\mathfrak{G}}{\partial \chi} = A \chi^{-2} -\frac{3}{n_e} B_e \chi^{-(1+3/n_e)} - (1-\delta_{\infty n_c}) \frac{3}{n_c} B_c \chi^{-(1+3/n_c)} 
- \delta_{\infty n_c} B_I \chi^{-1}  \, ;
</math>

<math>
\frac{\partial^2\mathfrak{G}}{\partial \chi^2} = -2 A \chi^{-3} + \frac{3}{n_e} \biggl(1+\frac{3}{n_e}\biggr) B_e \chi^{-(2+3/n_e)} + (1-\delta_{\infty n_c}) \frac{3}{n_c} \biggl(1+\frac{3}{n_c}\biggr) B_c \chi^{-(2+3/n_c)} 
+ \delta_{\infty n_c} B_I \chi^{-2} \, .
</math>
</div>

==Equilibrium Condition==
We obtain the equilibrium radius, <math>\chi_E</math>, when <math>\partial\mathfrak{G}/\partial\chi = 0</math>.  Hence,
the relation governing the equilibrium radius is,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
A \chi_E^{-2} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(1-\delta_{\infty n_c}) \frac{3}{n_c} B_c \chi_E^{-1- 3/n_c} 
+\delta_{\infty n_c} B_I \chi_E^{-1} +\frac{3}{n_e} B_e \chi_E^{-1-3/n_e}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~~ \frac{n_e A}{3B_e}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
(1-\delta_{\infty n_c}) \frac{n_e B_c}{n_c B_e} \chi_E^{1- 3/n_c} 
+\delta_{\infty n_c} \frac{n_e B_I}{3B_e} \chi_E + \chi_E^{1-3/n_e}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~~ \chi_E^{1-3/n_e}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\alpha - (1-\delta_{\infty n_c}) \beta \chi_E^{1- 3/n_c} - \delta_{\infty n_c} \beta_I \chi_E \, ,
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\alpha \,</math> 
  </td>
  <td align="center">
<math>\equiv \,</math>
  </td>
  <td align="left">
<math>
\frac{n_e A}{3B_e} = \biggl[ \frac{GM_\mathrm{tot}}{5R_0 K_e (\rho_e|_0)^{1/n_e} } \biggr]  \frac{\nu^2 f(\nu, q)}{q} 
= \biggl[ \frac{GM_\mathrm{tot}}{5R_0 K_e \rho_0^{1/n_e} } \biggr]\biggl( \frac{\rho_e|_0}{\rho_0} \biggr)^{-1/n_e}  \frac{\nu^2 f(\nu, q)}{q} 
 \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\beta \,</math> 
  </td>
  <td align="center">
<math>\equiv \,</math>
  </td>
  <td align="left">
<math>\frac{n_e B_c}{n_c B_e} = \nu \biggl[ \biggl(\frac{\rho_c}{\rho_0}\biggr)^{1/n_e - 1/n_c}  \biggl( \frac{\rho_c|_0}{\rho_0}\biggr)^{1/n_c} 
\biggl( \frac{\rho_e|_0}{\rho_0} \biggr)^{-1/n_e} \biggr] \biggl( \frac{\mu_e}{\mu_c} \biggr)^{1+1/n_e} \, ; </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\beta_I \,</math> 
  </td>
  <td align="center">
<math>\equiv \,</math>
  </td>
  <td align="left">
<math>
\frac{n_e B_I}{3B_e} = \frac{c_s^2 \nu}{(1-\nu)K_e (\rho_e|_0)^{1/n_e}} = 
\nu \biggl[ \biggl(\frac{\rho_c}{\rho_0}\biggr)^{1/n_e} \biggl( \frac{\rho_e|_0}{\rho_0} \biggr)^{-1/n_e}
\biggr] \biggl( \frac{\mu_e}{\mu_c} \biggr)^{1+1/n_e}
\, .
</math>
  </td>
</tr>
</table>
</div>

===Isothermal Core===
In the case of an isothermal core (<math>\delta_{\infty n_c} = 1</math>),

<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
A \chi_E^{-2} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
B_I \chi_E^{-1} +\frac{3}{n_e} B_e \chi_E^{-1-3/n_e}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ B_I \chi_E^{3/n_e} -  A \chi_E^{3/n_e-1} +\frac{3}{n_e} B_e 
</math>
  </td>
  <td align="center"><math>=\,</math></td>
  <td align="left">
<math>
0 \, .
</math>
  </td>
</tr>

</table>
</div>

Or, alternatively, via a dimensionless treatment,

<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_E^{1-3/n_e}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\alpha - \beta_I \chi_E \, .
</math>
  </td>
</tr>

</table>
</div>
===Adiabatic Core===
In the case of an adiabatice core (<math>\delta_{\infty n_c} = 0</math>),

<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_E^{1-3/n_e}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\alpha - \beta \chi_E^{1- 3/n_c}  \, .
</math>
  </td>
</tr>

</table>
</div>

==Stability==
At this equilibrium radius, the second derivative of the free energy has the value,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_E^3 \biggl( \frac{n_e}{3B_e} \biggr) \frac{\partial^2\mathfrak{G}}{\partial \chi^2} \biggr|_E
</math>
  </td>
  <td align="right">
<math>
=
</math>
  </td>
  <td align="left">
<math>
-2 A\biggl( \frac{n_e}{3B_e} \biggr)  + (1-\delta_{\infty n_c}) \biggl( \frac{n_e}{3B_e} \biggr) \frac{3}{n_c} \biggl(1+\frac{3}{n_c}\biggr) B_c \chi_E^{1-3/n_c} 
+ \delta_{\infty n_c} \biggl( \frac{n_e}{3B_e} \biggr) B_I \chi_E + \frac{3}{n_e}\biggl( \frac{n_e}{3B_e} \biggr)  \biggl(1+\frac{3}{n_e}\biggr) B_e \chi_E^{1-3/n_e}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
<math>
=
</math>
  </td>
  <td align="left">
<math>
-2 \alpha + (1-\delta_{\infty n_c}) \beta \biggl(1+\frac{3}{n_c}\biggr) \chi_E^{1-3/n_c} 
+ \delta_{\infty n_c} \beta_I \chi_E + \biggl(1+\frac{3}{n_e}\biggr) \chi_E^{1-3/n_e} \, ,
</math>
  </td>
</tr>

</table>
</div>
which, when combined with the condition for equilibrium gives,
<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_E^3 \biggl( \frac{n_e}{3B_e} \biggr) \frac{\partial^2\mathfrak{G}}{\partial \chi^2} \biggr|_E
</math>
  </td>
  <td align="right">
<math>
=
</math>
  </td>
  <td align="left">
<math>
-2 \alpha + (1-\delta_{\infty n_c}) \beta \biggl(1+\frac{3}{n_c}\biggr) \chi_E^{1-3/n_c} 
+ \delta_{\infty n_c} \beta_I \chi_E + \biggl(1+\frac{3}{n_e}\biggr) \biggl[  \alpha - (1-\delta_{\infty n_c}) \beta \chi_E^{1- 3/n_c} - \delta_{\infty n_c} \beta_I \chi_E \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="right">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\alpha \biggl(\frac{3}{n_e}-1\biggr) + (1-\delta_{\infty n_c}) \beta \biggl(\frac{3}{n_c}-\frac{3}{n_e}\biggr) \chi_E^{1-3/n_c} 
-  \delta_{\infty n_c} \beta_I \biggl(\frac{3}{n_e}\biggr)\chi_E 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ \chi_E^3 \biggl( \frac{n_e^2}{3B_e} \biggr) \frac{\partial^2\mathfrak{G}}{\partial \chi^2} \biggr|_E
</math>
  </td>
  <td align="right">
<math>
=
</math>
  </td>
  <td align="left">
<math>
\alpha (3-n_e) - (1-\delta_{\infty n_c}) 3\beta \biggl(1 - \frac{n_e}{n_c} \biggr) \chi_E^{1-3/n_c} -  \delta_{\infty n_c} 3\beta_I \chi_E \, .
</math>
  </td>
</tr>

</table>
</div>

The equilibrium configuration is stable as long as this second derivative is positive.  

===Isothermal Core===
Hence, for a bipolytrope with an isothermal core (<math>\delta_{\infty n_c} = 1</math>), the configuration is stable as long as,
<div align="center">
<math>
\chi_E <  \frac{\alpha (3-n_e)}{3\beta_I} \, .
</math>
</div>

===Adiabatic Core===
In the adiabatic case (<math>\delta_{\infty n_c} = 0</math>), the configuration is stable as long as,
<div align="center">
<math>
\chi_E^{1-3/n_c} <  \frac{\alpha n_c (3-n_e)}{3\beta (n_c-n_e)} \, .
</math>
</div>


==Examples==
===Isothermal Core with <math>n_e=3/2</math>===
Consider the case examined by [http://adsabs.harvard.edu/abs/1942ApJ....96..161S Sch&ouml;nberg &amp; Chandrasekhar (1942)], that is, the case of an isothermal core and an envelope with <math>n_e = 3/2</math>.  The equilibrium radius is given by the expression,

<div align="center">
<table border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
\chi_E^{-1}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\alpha - \beta_I \chi_E \,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~~ \beta_I \chi_E^{2} -\alpha \chi_E + 1
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
0 \,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~~ \chi_E
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{2\beta_I} \biggl[\alpha \pm \biggr(\alpha^2-4\beta_I \biggr)^{1/2}\biggr] \,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{\alpha}{2\beta_I} \biggl[1 \pm \biggr(1-\frac{4\beta_I}{\alpha^2} \biggr)^{1/2}\biggr] \, .
</math>
  </td>
</tr>

</table>
</div>
And the system is stable when,
<div align="center">
<math>
\chi_E < \chi_0 \equiv  \frac{\alpha }{2\beta_I} \, .
</math>
</div>

A couple of physical attributes are now clear:
*Physical configurations only exist for <math>(4\beta_I/\alpha^2) \le 1</math>.
*For each value of <math>(4\beta_I/\alpha^2) < 1 \,</math>, there are two equilibrium configurations, given by the <math>\pm</math> roots of the quadratic equation for <math>\chi_E</math>; the "negative" branch is stable but the "positive" branch is unstable.

Note that,
<div align="center">
<math>
\chi_0 \equiv \frac{\alpha }{2\beta_I}  = \biggl( \frac{GM_\mathrm{tot}}{10R_0 c_s^2} \biggr) \frac{\nu f(\nu,q)}{q} ~~~\Rightarrow ~~~ 
\biggl( \frac{10R_0 c_s^2}{GM_\mathrm{tot}} \biggr) = \frac{\nu f(\nu,q)}{q \chi_0}  \, ,
</math>
</div>
and,
<div align="center">
<math>
\frac{4\beta_I}{\alpha^2 }  = \biggl( \frac{10 R_0 c_s^2 }{GM_\mathrm{tot}} \biggr)^2 \biggl[ \frac{K_e (\rho_e|_0)^{1/n_e}}{c_s^2} \biggr] \frac{q^2(1-\nu)}{\nu^3 f^2(\nu,q)}  
= \biggl[ \frac{K_e (\rho_e|_0)^{1/n_e}}{c_s^2} \biggr] \frac{q^2(1-\nu)}{\nu^3 f^2(\nu,q)} \biggl[\frac{\nu f(\nu,q)}{q \chi_0} \biggr]^2 
= \biggl[ \frac{\mu_c}{\mu_e} \biggr] \biggl( \frac{1}{\nu} - 1 \biggr) \, .
</math>
</div>
But, this last expression must be less than or equal to unity, which implies,
<div align="center">
<math>
\frac{1}{\nu} \le 1 + \frac{\mu_e}{\mu_c} ~~~\Rightarrow ~~~ \nu \ge \biggl(1 + \frac{\mu_e}{\mu_c}  \biggr)^{-1}
</math>
This doesn't seem to have the correct behavior, for example, the smaller values of <math>\nu</math> should be the stable ones, so there must be a mistake in the derivation.
</div>

===Adiabatic Core with <math>n_c = 5</math> and <math>n_e=1</math>===
Consider the case with an analytical structure derived by  [http://adsabs.harvard.edu/abs/1998MNRAS.298..831E Eagleton, Faulkner, and Cannon] (1998, MNRAS, 298, 831), that is, the case of an adiabatic core having <math>n_c=5</math> and an envelope with <math>n_e = 1</math>.  The equilibrium radius is,