Editing SSC/Structure/BiPolytropes/FreeEnergy00

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=Free Energy of BiPolytrope with <math>~(n_c, n_e) = (0, 0)</math>=

Here we present a specific example of the equilibrium structure of a bipolytrope as determined from a free-energy analysis.  The example is a bipolytrope whose core has a polytropic index, <math>~n_c = 0</math>, and whose envelope has a polytropic index, <math>~n_e = 0</math>.  The details presented here build upon an [[SSC/BipolytropeGeneralizationVersion2|overview of the free energy of bipolytropes that has been presented elsewhere]].

==Preliminaries==

==Mass Profile==

In this case, <math>~\rho_\mathrm{core}(x) = \rho_c = </math> constant &#8212; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{core} = 1</math> &#8212; and <math>~\rho_\mathrm{env}(x) = \rho_e = </math> constant &#8212; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{env} = 1</math> &#8212; but in general <math>~\rho_e \ne \rho_c</math>.  Performing the separate integrals to obtain expressions for <math>~M_r(r)</math> inside the core and the envelope, [[SSC/BipolytropeGeneralizationVersion2#Partitioning_the_Mass|as established in our accompanying overview]], we obtain:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>(\mathrm{For}~0 \leq x \leq q)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~M_r </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl( \frac{\nu}{q^3} \biggr) \int_0^{x}  3x^2 dx = \nu M_\mathrm{tot} \biggl( \frac{x}{q} \biggr)^3 \, ; 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>(\mathrm{For}~q \leq x \leq 1)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
<math>~M_r </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl\{\nu + \biggl( \frac{1-\nu}{1-q^3} \biggr) \int_{q}^{x}  3
x^2 dx \biggr\} 
= M_\mathrm{core}  + (1-\nu) M_\mathrm{tot}\biggl( \frac{x^3 - q^3}{1-q^3} \biggr)\, .
</math>
  </td>
</tr>
</table>
</div>
When <math>~x = q</math>, both expressions give,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{core} = \nu M_\mathrm{tot} \, ,</math>
  </td>
</tr>
</table>
</div>
as they should.  We deduce, as well, that the mass contained in the envelope is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~M_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~M_\mathrm{tot} - M_\mathrm{core} = (1-\nu) M_\mathrm{tot} \, ,</math>
  </td>
</tr>
</table>
</div>
and that the volumes occupied by the core and envelope are, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~V_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~q^3 R_\mathrm{edge}^3 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~V_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(1- q^3) R_\mathrm{edge}^3 \, .</math>
  </td>
</tr>

</table>
</div>

Hence, the ratio of envelope density to core density is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\rho_e}{\rho_c} = \frac{\bar\rho_\mathrm{env}}{\bar\rho_\mathrm{core}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{M_\mathrm{env}/V_\mathrm{env}}{M_\mathrm{core}/V_\mathrm{core}} = \frac{q^3(1-\nu)}{\nu(1-q^3)} \, .
</math>
  </td>
</tr>
</table>
</div>

These relations should be compared to &#8212; and ultimately must match &#8212; the prescriptions for <math>~M_r</math> that have been presented elsewhere in connection with [[SSC/Structure/BiPolytropes/Analytic00#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|detailed force-balance models of <math>~(n_c, n_e) = (0, 0)</math> bipolytropes]] and in our introductory discussion of [[SSC/VirialStability#Expressions_for_Mass|the virial stability of bipolytropes]].



==Gravitational Potential Energy==
Here we follow the steps that have been [[SSC/BipolytropeGeneralizationVersion2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in an accompanying overview]] to determine the separate contributions to the gravitational potential energy.  Let's do the core first.  In this case, <math>~\rho_\mathrm{core}(x) = \rho_c = </math> constant &#8212; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{core} = 1</math>.  As has been demonstrated above, the corresponding <math>~M_r</math> function is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}}\biggr]_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{\nu}{q^3} \biggr) x^3 \, .
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)
\int_0^{q} 3\biggl( \frac{\nu}{q^3} \biggr)x^4  dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl( \frac{3}{5} q^5 \biggr) 
</math>
  </td>
</tr>

</table>
</div>

Now, let's do the envelope.  In this case, <math>~\rho_\mathrm{env}(x) = \rho_e = </math> constant; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{env} = 1</math>.  As [[SSC/BipolytropeGeneralization#ExampleMass|shown elsewhere]], the corresponding <math>~M_r</math> function is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\nu  + \biggl(\frac{1-\nu}{1-q^3} \biggr) (x^3 - q^3) \, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr)
\biggl\{ \int_{q}^{1}  
\biggl[ \nu  -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr]3x dx + \int_{q}^{1}   \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] 3x^4 dx
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr)
\biggl\{ \frac{3}{2}
\biggl[ \nu  -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^2) + \frac{3}{5}  \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} 
\biggl[ \frac{1}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
\biggl[ 1  - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (q-q^3) + \biggl[ \frac{q}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5)
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} 
\biggl[ \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
\biggl[ 1  - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) 
\biggr] \biggl( \frac{1}{q^5}-1\biggr)  \biggr\} \, .
</math>
  </td>
</tr>

</table>
</div>
Realizing from the above mass segregation derivation that,
<div align="center">
<math>~\frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr) = \frac{\rho_e}{\rho_c} \, ,</math>
</div>
this last expression can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} 
\biggl[ \frac{\rho_e}{\rho_c} \biggr] \biggl\{ \frac{5}{2}
\biggl[ 1  - \frac{\rho_e}{\rho_c}  \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{\rho_e}{\rho_c}  
\biggr] \biggl( \frac{1}{q^5}-1\biggr)  \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} 
\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl\{ \frac{5}{2}\biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c}  
\biggr) \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]  \biggr\} \, .
</math>
  </td>
</tr>

</table>
</div>
So, when put together to obtain the total gravitational potential energy, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~W_\mathrm{grav} = W_\mathrm{grav}\biggr|_\mathrm{core} + W_\mathrm{grav}\biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{3}{5} E_\mathrm{norm} \cdot \chi^{-1} \biggl(\frac{\nu^2}{q} \biggr) f(\nu,q) \, ,
</math>
  </td>
</tr>

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

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

</table>
</div>
(This result agrees with Tohline's earlier derivations in other sections of this H_Book, which may now be erased to avoid repetition.)

==Thermodynamic Energy Reservoir==

According to our [[SSC/Structure/BiPolytropes/Analytic00#Step_4:__Throughout_the_core_.28.29|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (0, 0) </math> bipolytropes]], in this case the pressure throughout the core is defined by the dimensionless function,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~p_c(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{2\pi}{3} \biggr) \xi^2 \, ,</math>
  </td>
</tr>
</table>
</div>
and the pressure throughout the envelope is defined by the dimensionless function,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~p_e(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } 
\biggl[ \frac{\rho_e}{\rho_0} (\xi^2 - \xi_i^2) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \xi_i^3\biggl( \frac{1}{\xi} - 
\frac{1}{\xi_i}\biggr) 
\biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>

where, for both functions,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\xi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{G\rho_0^2}{P_0} \biggr)^{1/2} R_\mathrm{edge} \biggr]_\mathrm{eq} x</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 
\frac{G R_\mathrm{edge}^2}{P_0} \biggl( \frac{3 \nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{edge}^3} \biggr)^2
\biggr]^{1/2}_\mathrm{eq} x</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \biggl( \frac{3^2}{2^4 \pi^2} \biggr) 
\frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2
\biggr]^{1/2}_\mathrm{eq} x</math>
  </td>
</tr>

</table>
</div>
So, defining the coefficient,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~b_\xi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) 
\frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, ,</math>
  </td>
</tr>
</table>
</div>
such that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\xi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{2\pi} \cdot b_\xi \biggr)^{1/2} x \, ,</math>
  </td>
</tr>
</table>
</div>
and remembering that, at the interface, <math>~x \rightarrow x_i = q</math>, so <math>~\xi_i = (3b_\xi/2\pi)^{1/2} q</math>, the two dimensionless pressure functions become,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~p_c(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b_\xi x^2 \, ,</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~p_e(x)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } 
\biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} - 
\frac{1}{q}\biggr) 
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
The desired integrals over these pressure distributions therefore give,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\int_0^q  \biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr]  x^2 dx</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \int_0^q (1-b_\xi x^2)x^2 dx</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ \frac{1}{3}\cdot q^3 - \biggl( \frac{b_\xi}{5} \biggr) q^5 \biggr] </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{q^3}{3} \biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr] 
= \frac{q^3}{3} \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr]
\, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\int_q^1  \biggl[1 - p_e(x) \biggr]  x^2 dx</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \int_q^1  
\biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} - 
\frac{1}{q}\biggr) 
\biggr]   x^2 dx</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} }
\biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{x^5}{5} - \frac{q^2 x^3}{3} \biggr) -
2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{x^2}{2} - 
\frac{x^3}{3q}\biggr) 
\biggr]_q^1</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{
\biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{3}{5} - q^2  \biggr) -
\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^2\biggl( 3q - 2\biggr) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[ \frac{\rho_e}{\rho_0} \biggl( -\frac{2}{5} \biggr)q^5 -
\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^5 
\biggr] \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{
\biggl[ q^2(2-3q) +q^5\biggr] 
+ \frac{\rho_e}{\rho_0}\biggl[ \biggl( \frac{3}{5} - q^2  \biggr) + q^2\biggl( 3q - 2\biggr) +\frac{2q^5}{5} -q^5\biggr] \biggl\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[
(2q^2 - 3q^3 +q^5) 
+ \frac{3}{5} \cdot \frac{\rho_e}{\rho_0} ( 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}{3}\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr] \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<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>
</table>
</div>



<span id="InternalEnergies">Finally, then, we have,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi/3 }{({\gamma_c}-1)}  \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_c} 
\biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{4\pi/3 }{({\gamma_e}-1)}  \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_e} 
\biggl\{ (1-q^3) +  b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F}
\biggr]  \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

==Virial Theorem==
As has been shown in our [[SSC/BipolytropeGeneralizationVersion2#More_Utilitarian_Form|accompanying overview]], the condition for equilibrium based on a free-energy analysis &#8212; that is, the virial theorem &#8212; is,
<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}_\mathrm{core} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{B}_\mathrm{env} \chi_\mathrm{eq}^{4-3\gamma_e} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} [
q^3 s_\mathrm{core} + (1-q^3) s_\mathrm{env} ] \, .
</math>
  </td>
</tr>

</table>
</div>

For <math>~(n_c, n_e) = (0, 0) </math> bipolytropes, the relevant coefficient functions 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{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~q^3 s_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
q^3 \biggl(\frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~(1-q^3) s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(1-q^3) +  \biggl(\frac{P_0}{P_{ie}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
 <td align="left">
<math>
1+
\frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr)  \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c}  
\biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr]  \, ,
</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">
<math>~\frac{P_{ic}}{P_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1- p_c(q) = 1 - b_\xi q^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b_\xi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{3}{2^3 \pi} \biggr) 
\frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, .</math>
  </td>
</tr>

</table>
</div>

Plugging these expressions into the equilibrium condition shown above, and setting the interface pressures equal to one another, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{
q^3 \biggl(\frac{P_0}{P_{i}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr]  
+ (1-q^3) +  \biggl(\frac{P_0}{P_{i}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{
q^3 \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr]  
+ (1-q^3)( 1- b_\xi q^2)  +  \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{
1 - b_\xi \biggl[ \frac{3}{5}q^5  + q^2(1-q^3) - \frac{2}{5} q^5 \mathfrak{F}  \biggr]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq}  \biggl[
\frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] b_\xi
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}  \biggl[
\frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] \biggl( \frac{\nu}{q^3}\biggr)^2
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow~~~~\biggl( \frac{2^3 \pi}{3} \biggr) 
\frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } \biggl( \frac{q^3}{\nu}\biggr)^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} )
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~
\frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl\{ q^2 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[
2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-3q^2 + 2q^3) \biggr] \biggr\} \, .</math>
  </td>
</tr>
</table>
</div>

This exactly matches the equilibrium relation that was derived from our [[SSC/Structure/BiPolytropes/Analytic00#CentralPressure|detailed force-balance analysis of]] <math>~(n_c, n_e) = (0, 0)</math> bipolytropes.

=Related Discussions=
* [[SSC/Structure/BiPolytropes/FreeEnergy00|Free-energy determination of equilibrium configurations for BiPolytropes]] with <math>~n_c = 0</math> and <math>~n_e=0</math>.
* [[SSC/Structure/BiPolytropes/FreeEnergy51#Free_Energy_of_BiPolytrope_with|Free-energy determination of equilibrium configurations for BiPolytropes]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[SSC/Structure/BiPolytropes/Analytic00#BiPolytrope_with_nc_.3D_0_and_ne_.3D_0|Analytic solution of Detailed-Force-Balance BiPolytrope]] 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 of Detailed-Force-Balance BiPolytrope]] with <math>~n_c = 5</math> and <math>~n_e=1</math>.
* [[SSC/BipolytropeGeneralization|Old ''Bipolytrope Generalization'' derivations]].



=See Also=
<ul>
<li>[[SphericallySymmetricConfigurations/IndexFreeEnergy#Index_to_Free-Energy_Analyses|Index to a Variety of Free-Energy and/or Virial Analyses]]</li>
</ul>


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