Editing SSC/Structure/BiPolytropes/FreeEnergy51 (section)

=Free Energy of BiPolytrope with (n<sub>c</sub>, n<sub>e</sub>) = (5, 1)=
{| class="PGEclass" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
|- 
! style="height: 125px; width: 125px; background-color:white;" |<font size="-1">[[H_BookTiledMenu#MoreModels|<b>Free Energy<br />of<br />Bipolytropes</b>]]<br />(n<sub>c</sub>, n<sub>e</sub>) = (5, 1)</font>
|}
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 = 5</math>, and whose envelope has a polytropic index, <math>n_e = 1</math>.  The details presented here build upon an [[SSC/BipolytropeGeneralizationVersion2|overview of the free energy of bipolytropes that has been presented elsewhere]].
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />
&nbsp;<br />

==Preliminaries==

==Mass Profile==

===The Core===
The core has <math>~n_c = 5 \Rightarrow \gamma_c = 1+1/n_c = 6/5</math>.  Referring to the general relation [[SSC/BipolytropeGeneralizationVersion2#Partitioning_the_Mass|as established in our accompanying overview]], and using <math>~\rho_0</math> to represent the central density, we can write,
<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) \biggl( \frac{\rho_0} {{\bar\rho}_\mathrm{core}}\biggr)_\mathrm{eq} \int_0^{x}  3
\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}  
x^2 dx \, . 
</math>
  </td>
</tr>

</table>
</div>

Drawing on the derivation of [[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|detailed force-balance models of <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]], the density profile [[SSC/Structure/BiPolytropes/Analytic51#Step_4:__Throughout_the_core_.28.29|throughout the core]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(\xi)}{\rho_0} \biggr]_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,</math>
  </td>
</tr>
</table>
</div>
where the dimensionless radial coordinate is,
<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{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} r \, .</math>
  </td>
</tr>
</table>
</div>
Switching to the [[SSCpt1/Virial#Normalizations|normalizations that have been adopted in the broad context of our discussion of configurations in virial equilibrium]] and inserting the adiabatic index of the core <math>~(\gamma_c = 6/5)</math> into all normalization parameters, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~R_\mathrm{norm} = \biggl[ \biggl(\frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma} \biggr]^{1/(4-3\gamma)}</math>
  </td>
  <td align="center">
<math>~\Rightarrow</math>
  </td>
  <td align="left">
<math>~R_\mathrm{norm} = \biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{1/(4-3\gamma)}</math>
  </td>
  <td align="center">
<math>~\Rightarrow</math>
  </td>
  <td align="left">
<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)^{5/2} \, .</math>
  </td>
</tr>
</table>
</div>
Hence, we can rewrite,

<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{r}{R_\mathrm{norm}} \biggr) \biggl( \frac{\rho_0}{\rho_\mathrm{norm}} \biggr)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} 
\biggl( \frac{2\pi}{3} \biggr)^{1/2}  R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r^* (\rho_0^*)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} 
\biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~r^* (\rho_0^*)^{2/5} \biggl[ \biggl( \frac{2\pi}{3} \biggr)^{5} \biggl( \frac{3}{4\pi} \biggr)^{4} \biggr]^{1/10}
= r^* (\rho_0^*)^{2/5} \biggl[ \frac{\pi}{2^3 \cdot 3}\biggr]^{1/10} \, .
</math>
  </td>
</tr>
</table>
</div>
Now, following the same approach as was used in our [[SSCpt1/Virial#Separate_Time_.26_Space|introductory discussion]] and appreciating that our aim here is to redefine the coordinate, <math>~\xi</math>, in terms of normalized parameters evaluated ''in the equilibrium configuration,'' we will set,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
~ x \chi_\mathrm{eq} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_0^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \biggl( \frac{{\bar\rho}_\mathrm{core}}{\rho_\mathrm{norm}} \biggr)
= \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \frac{\nu M_\mathrm{tot}/(q^3 R_\mathrm{edge}^3)_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3} 
= \frac{\nu}{q^3} \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core}  \chi_\mathrm{eq}^{-3} \, .
</math>
  </td>
</tr>

</table>
</div>
Then we can set,
<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>~(3a_\xi)^{1/2} x \, ,</math>
  </td>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-5/2} \, ,</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~(3a_\xi)^{1/2}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\chi_\mathrm{eq} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core}  \chi_\mathrm{eq}^{-3} \biggr]^{2/5} 
\biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} 
=\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} 
\biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~a_\xi</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\frac{1}{3} \biggl\{ \chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} 
\biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} \biggr\}^2
= \chi_\mathrm{eq}^{-2/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{4/5} 
\biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr)^{1/5} \, .
</math>
  </td>
</tr>
</table>
</div>
We therefore have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r\biggr|_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \int_0^{x}  3
\biggl( 1 + a_\xi x^2 \biggr)^{-5/2}  x^2 dx  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} 
\biggl[ x^3\biggl( 1 + a_\xi x^2 \biggr)^{-3/2}  \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>
Note that, when <math>~x \rightarrow q</math>, <math>~M_r|_\mathrm{core} \rightarrow M_\mathrm{core} = \nu M_\mathrm{tot}</math>. Hence, this last expression gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nu M_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} 
\biggl[ q^3\biggl( 1 + a_\xi q^2 \biggr)^{-3/2}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~\biggl[\biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( 1 + a_\xi q^2 \biggr)^{3/2}  \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, finally,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r\biggr|_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\nu M_\mathrm{tot} \biggl( \frac{x^3}{q^3} \biggr) 
\biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} \, ;
</math>
  </td>
</tr>

</table>
</div>
and, after the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been determined from the free-energy analysis, the coefficient, <math>~a_\xi</math>, can be determined via the relation,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\chi_\mathrm{eq}^{2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr)
\biggl( \frac{\nu}{q^3}  \biggr)^{4} \biggl( 1 + a_\xi q^2 \biggr)^{6}  a_\xi^{-5} \, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" cellpadding="5" align="center" width="90%">
<tr><td align="left">

MORE USEFUL:

Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~a_\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\ell_i}{q} \biggr)^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_\mathrm{Mcore}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\bar\rho}{\rho_c} \biggr)_\mathrm{core}
=
(1 + \ell_i^2)^{-3/2} \, .
</math>
  </td>
</tr>
</table>
</div>

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

===The Envelope===
The envelope has <math>~n_e = 1 \Rightarrow \gamma_e = 1+1/n_e = 2</math>.  Again, referring to the general relation [[SSC/BipolytropeGeneralizationVersion2#Partitioning_the_Mass|as established in our accompanying overview]], and continuing to use <math>~\rho_0</math> to represent the central density, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<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_{x_i}^{x}  3
\biggl[ \frac{\rho(x)}{\bar\rho} \biggr]_\mathrm{env}  
x^2 dx \biggr\} \, .
</math>
  </td>
</tr>

</table>
</div>

Drawing on the derivation of [[SSC/Structure/BiPolytropes/Analytic51#BiPolytrope_with_nc_.3D_5_and_ne_.3D_1|detailed force-balance models of <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]], the density profile [[SSC/Structure/BiPolytropes/Analytic51#Step_8:__Throughout_the_envelope_.28.2|throughout the envelope]] is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(\eta)}{\rho_0} \biggr]_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
where definitions of the constants <math>~A</math> and <math>~B</math> are given in an [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying table of parameter values]], and the dimensionless radial coordinate is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} r \, .</math>
  </td>
</tr>
</table>
</div>
Using the same radial and mass-density normalizations as defined, above, for the core, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~r^* (\rho_0^*)^{2/5}\biggl[ \frac{G}{K_c} \biggr]^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} 
R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~r^* (\rho_0^*)^{2/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5}  \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2} \, .
</math>
  </td>
</tr>
</table>
</div>
Next, we set,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~r^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
~ x \chi_\mathrm{eq} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\rho_0^*</math>
  </td>
  <td align="center">
<math>~\rightarrow~</math>
  </td>
  <td align="left">
<math>
\biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \biggl( \frac{{\bar\rho}_\mathrm{env}}{\rho_\mathrm{norm}} \biggr)
= \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{env} \frac{(1-\nu) M_\mathrm{tot}/[(1-q^3) R_\mathrm{edge}^3]_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3} 
= \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env}  \chi_\mathrm{eq}^{-3} \, .
</math>
  </td>
</tr>

</table>
</div>
Hence, we can write,
<div align="center">
<math>~\eta = b_\eta x \, ,</math>
</div>
where,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
~\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{1-\nu}{1-q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{env}  \biggr]^{2/5} 
\biggl( \frac{3}{4\pi} \biggr)^{2/5}  \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2}  \, .
</math>
  </td>
</tr>
</table>
</div>
In which case,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
so,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
M_\mathrm{tot} \biggl\{\nu + \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} \int_{q}^{x}  3
A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr]  
x^2 dx \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} + 
M_\mathrm{tot}  \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr) 
\frac{3A\theta_i^5 }{b_\eta} 
\int_{q}^{x}  \sin(b_\eta x - B) x dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} + 
M_\mathrm{tot}  \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr) 
\frac{3A\theta_i^5 }{b_\eta} 
\biggl[ - \frac{\sin(B-b_\eta x) + b_\eta x \cos(B - b_\eta x)}{b_\eta^2} \biggr]_q^x 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} + 
M_\mathrm{tot}  \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}  \biggl( \frac{\mu_e}{\mu_c} \biggr) 
\frac{3A\theta_i^5 }{b_\eta^3} 
\biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, ,
</math>
  </td>
</tr>

</table>
</div>
where, <math>~C_1</math> is a constant obtained by evaluating the integral at the interface <math>~(x = x_i = q)</math>, specifically,
<div align="center">
<math>
C_1 \equiv \sin(B-b_\eta q)+ b_\eta q \cos(B - b_\eta q) \, .
</math>
</div>

Now, this expression can be significantly simplified by drawing on earlier results of this section as well as on attributes of the corresponding detailed force-balanced model.  First, independent of the specific density profiles that define the structure of a bipolytrope, the ratio of the ''mean'' densities of the two structural regions is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\bar\rho_e}{\bar\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 the bracketed pre-factor of the second term of the expression for <math>~M_r|_\mathrm{env}</math> may be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{\nu}{q^3}  \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \, .</math>
  </td>
</tr>
</table>
</div>
But, from the [[SSC/Structure/BiPolytropes/FreeEnergy51#The_Core|above derivation of the mass profile in the core]], we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\biggl[\biggl( \frac{\rho_0} {{\bar\rho}}\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( 1 + a_\xi q^2 \biggr)^{3/2} = \biggl( 1 + \frac{1}{3} \xi_i^2 \biggr)^{3/2} = \theta_i^{-3} \, ,
</math>
  </td>
</tr>
</table>
</div>
where the final step comes from knowledge of the expression for <math>~\theta_i</math> drawn from the detailed force-balanced model (see, for example, the associated [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|''Parameter Values'' table]]).  Hence, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} + 
M_\mathrm{tot}  \biggl( \frac{\mu_e}{\mu_c} \biggr) 
\frac{3\nu A\theta_i^2 }{(b_\eta q)^3} 
\biggl[ C_1-\sin(B-b_\eta x) -xb_\eta \cos(B - b_\eta x) \biggr] \, ,
</math>
  </td>
</tr>

</table>
</div>
and note that the expression for the coefficient, <math>~b_\eta</math>, becomes simpler as well, specifically,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~\chi_\mathrm{eq}^{-1/5} \biggl( \frac{3}{4\pi} \biggr)^{2/5}  \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 (2\pi)^{1/2}  \biggl( \frac{\nu}{q^3 \theta_i^3} \biggr)^{2/5} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~\chi_\mathrm{eq} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~b_\eta^{-5} \biggl( \frac{3^4  \pi}{2^3} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6}  \, .
</math>
  </td>
</tr>
</table>
</div>

Next &#8212; and, again, drawing from knowledge of the [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|internal structure of the detailed force-balanced model]], in particular, realizing that,
<div align="center">
<math>b_\eta q = \eta_i = 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^2 \xi_i \, ,</math>
</div>
&#8212; note that the constant, <math>~C_1</math>, can be rewritten as,

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

<tr>
  <td align="right">
<math>~C_1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b_\eta q \cos(b_\eta q - B) - \sin(b_\eta q- B) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\eta_i \cos(\eta_i  - B) - \sin(\eta_i - B) </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\eta_i^2}{A}  \biggl( \frac{d\phi}{d\eta} \biggr)_i  = - \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \xi_i^3</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \frac{1}{A} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2 3^{1/2} \theta_i^4 \biggl[ 3^{-1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i^{-2} b_\eta q\biggr]^3 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
~- \frac{1}{3A\theta_i^2} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} (b_\eta q)^3 \, ,
</math>
  </td>
</tr>
</table>
</div>

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

<tr>
  <td align="right">
<math>~M_r \biggr|_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\nu M_\mathrm{tot} - 
\nu M_\mathrm{tot}  \biggl\{ 1 - \frac{1}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math> 
\frac{\nu M_\mathrm{tot}}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>

<table border="1" cellpadding="5" align="center" width="90%">
<tr><td align="left">

MORE USEFUL:

Letting, <math>\ell \equiv \xi/\sqrt{3}</math>,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~b_\eta = \eta_s</math>
  </td>
  <td align="center">
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
and
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
 </td>
  <td align="left">
<math>~b_\eta q = \eta_i = 3\biggl( \frac{\mu_e}{\mu_c} \biggr) \ell_i (1 + \ell_i^2)^{-1}  \, ,</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde\mathfrak{f}_\mathrm{Menv} \equiv \biggl( \frac{\bar\rho}{\rho_c}\biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{q^3(1-\nu)}{\nu(1-q^3)} \cdot \tilde\mathfrak{f}_\mathrm{Mcore}
</math>
  </td>
</tr>
</table>
</div>

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

==Gravitational Potential Energy==

===The Core===

Borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the core that has been [[SSC/BipolytropeGeneralizationVersion2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in our accompanying overview]] may be written as,
<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} \biggl(\frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}
\int_0^{q} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{core}  \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}  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} \biggl( 1 + a_\xi q^2 \biggr)^{3/2}  \biggr]_\mathrm{eq}
\int_0^{q} 3x \biggl\{ 
\nu \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} 
\biggr\} 
\biggl( 1 + a_\xi x^2 \biggr)^{-5/2}  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[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\int_0^{q} x^4 \biggl( 1 + a_\xi x^2 \biggr)^{-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[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl\{
\frac{a_\xi^{1/2} q(3a_\xi^2 q^4 - 8a_\xi q^2 - 3) + 3(a_\xi q^2 +1)^3 \tan^{-1}(a_\xi^{1/2} q)}{48 a_\xi^{5/2}(a_\xi q^2 + 1)^3}
\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[\biggl(\frac{3}{2^4}\biggr) a_\xi^{-5/2}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl[
a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q)
\biggr] \, .
</math>
  </td>
</tr>

</table>
</div>

<table border="1" cellpadding="5" align="center" width="90%">
<tr><td align="left">

MORE USEFUL:

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

<tr>
  <td align="right">
<math>~- \chi \biggl[ \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr]_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
\biggl[\biggl(\frac{3}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl[
\ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i
\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>\frac{3}{5}
\biggl[\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + \ell_i^2 \biggr)^{3}  \biggr]_\mathrm{eq}
\biggl(\frac{5}{2^4}\biggr) \biggl( \frac{q}{\ell_i}\biggr)^{5} \biggl[
\ell_i (\ell_i^4 - \frac{8}{3}\ell_i^2 - 1) (\ell_i^2 +1)^{-3} + \tan^{-1}\ell_i
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

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

But, also from our above discussion of the mass profile, we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~a_\xi^{-5/2} \biggl( \frac{\nu}{q^3} \biggr)^2 (1 + a_\xi q^2)^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\chi_\mathrm{eq} \biggl( \frac{2^3 \cdot 3^6}{\pi} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2}  
\biggl[
a_\xi^{1/2} q(a_\xi^2 q^4 - \frac{8}{3}a_\xi q^2 - 1) (a_\xi q^2 +1)^{-3} + \tan^{-1}(a_\xi^{1/2} q)
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
After making the substitution, <math>~(a_\xi^{1/2} q) \rightarrow x_i</math>, this expression agrees with a result for the dimensionless energy, <math>~W^*_\mathrm{core}</math>, [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|derived by Tohline in the context of detailed force-balanced bipolytropes]].

===The Envelope===

Again, borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the envelope that has been [[SSC/BipolytropeGeneralizationVersion2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in our accompanying overview]] may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \chi^{-1} \biggl[ \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl( \frac{\rho_0}{\bar\rho}\biggr)_\mathrm{env} \biggr]_\mathrm{eq}
\int_{q}^{1} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env}  \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{env}  dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \chi^{-1} \biggl[ \frac{\nu}{q^3 \theta_i^3} \biggr]
\int_{q}^{1} 3x \biggl\{ \frac{\nu}{C_1} \biggl[ \sin(B-b_\eta x) + xb_\eta \cos(B - b_\eta x) \biggr]\biggr\}  
\biggl\{ A \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta_i^5 \biggl[ \frac{\sin(b_\eta x - B)}{b_\eta x} \biggr] \biggr\}  dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)  \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr] \frac{1}{C_1} 
\int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B)  ]\sin(b_\eta x - B)   dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)  \biggl[ \frac{3\nu^2 A \theta_i^2}{b_\eta q^3 } \biggr] 
\biggl[ - 3A\theta_i^2 \biggl( \frac{\mu_e}{\mu_c} \biggr) (b_\eta q)^{-3} \biggr]
\int_{q}^{1} [ xb_\eta \cos(b_\eta x-B) - \sin(b_\eta x- B)  ]\sin(b_\eta x - B)   dx 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \chi^{-1} \biggl( \frac{\mu_e}{\mu_c} \biggr)^2  \biggl[ \frac{3^2 A^2 }{b_\eta^4 } \biggr] \biggl( \frac{\nu^2 \theta_i^4}{q^6} \biggr)
\int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B)   dx \, .
</math>
  </td>
</tr>

</table>
</div>
But we also know, [[SSC/Structure/BiPolytropes/FreeEnergy51#The_Envelope|from above]], that,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\chi_\mathrm{eq} = b_\eta^{-5} \biggl( \frac{3^4  \pi}{2^3} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^5 \frac{\nu^2 \theta_i^4}{q^6}</math>
  </td>
  <td align="center">
&nbsp;&nbsp;&nbsp;&nbsp;<math>~~~~\Rightarrow ~~~~</math>&nbsp;&nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>
~\frac{\nu^2 \theta_i^4}{q^6} = \chi_\mathrm{eq} b_\eta^{5} \biggl( \frac{2^3}{3^4  \pi} \biggr)^{1/2}  \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-5}   \, .
</math>
  </td>
</tr>
</table>
</div>
So we have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3}  \biggl( \frac{2^3}{\pi} \biggr)^{1/2} A^2 b_\eta
\int_{q}^{1} [\sin(b_\eta x- B) - xb_\eta \cos(b_\eta x-B) ]\sin(b_\eta x - B)   dx \, .
</math>
  </td>
</tr>

</table>
</div>


The integral can be broken into two separate parts:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\int_{q}^{1} \sin^2(b_\eta x - B) dx
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{4b_\eta} \biggl\{2 b_\eta x - \sin[2(b_\eta x-B)] \biggr\}_q^1 \, ,
</math>
  </td>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
- \int_{q}^{1} xb_\eta \cos(b_\eta x-B)\sin(b_\eta x - B) dx
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)]  - \sin[2(b_\eta x-B)]  \biggr\}_q^1 \, .</math>
  </td>
</tr>
</table>
</div>
(Note:  We have dropped integration constants that might result from carrying out an indefinite integral because such constants would disappear upon application of our specified limits of integration.)   When added together, they give,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\int_q^1 \ldots ~dx</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{8b_\eta} \biggl\{2b_\eta x \cos[2(b_\eta x - B)] - 3\sin[2(b_\eta x-B)]  + 4 b_\eta x \biggr\}_q^1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{8b_\eta} \biggl\{2b_\eta x \biggl[ 1 - 2\sin^2(b_\eta x - B) \biggr] - 3\sin[2(b_\eta x-B)] +  4 b_\eta x \biggr\}_q^1
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{1}{8b_\eta} \biggl[6 b_\eta x  - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B)  \biggr]_q^1 \, .
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
 <td align="left">
<math>
- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3}  \biggl( \frac{1}{2^3\pi} \biggr)^{1/2} A^2 
\biggl[6 b_\eta x  - 3\sin[2(b_\eta x-B)] - 4b_\eta x \sin^2(b_\eta x - B)  \biggr]_q^1 \, .
</math>
  </td>
</tr>

</table>
</div>

This expression matches in detail the expression for the gravitational potential energy of the envelope derived in the context of our [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|derivation of detailed force-balanced models of this bipolytrope]].

==Thermodynamic Energy Reservoir==

===The Core===

From our [[SSC/BipolytropeGeneralizationVersion2#Separate_Thermodynamic_Energy_Reservoirs|introductory discussion of the free energy of bipolytropes]], the energy contained in the core's thermodynamic reservoir may be written as,
<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{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c} \biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  
\biggl[ q^3 s_\mathrm{core} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>~q^3 s_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 
\int_0^q  3\biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr]  x^2 dx \, ,
</math>
  </td>
</tr>
</table>
</div>
defines the relevant integral over the core's pressure distribution.  According to our [[SSC/Structure/BiPolytropes/Analytic51#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]] &#8212; see also the relevant derivations [[SSC/BipolytropeGeneralizationVersion2#Bipolytrope_Generalization|in our accompanying overview]] &#8212;  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^* \equiv \frac{P_\mathrm{core}(\xi)}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ 1-p_c(x) = \frac{P_\mathrm{core}(x)}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-3} \, ,</math>
  </td>
</tr>
</table>
</div>
where, <math>~a_\xi</math> is defined above in connection with our [[SSC/Structure/BiPolytropes/FreeEnergy51#Mass_Profile|derivation of the mass profile]].
The desired integral over this pressure distribution therefore gives,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>~q^3 s_\mathrm{core} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
3\biggl( 1 + a_\xi q^2 \biggr)^{3} \int_0^q \frac{x^2 dx}{(1+a_\xi x^2)^3}
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl[ 
\tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2}
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

Next, let's examine the factor in square brackets with an "eq" subscript.   From our [[SSC/Structure/BiPolytropes/Analytic51#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]], we know that,
<div align="center">
<math>P_{ic} = K_c \rho_0^{6/5} \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} \, ,</math>
</div>
and,
<div align="center">
<math>
\chi_\mathrm{eq} = \biggl(\frac{R_\mathrm{edge}}{R_\mathrm{norm}} \biggr)_\mathrm{eq} 
= \frac{1}{q}\biggl(\frac{r_i}{R_\mathrm{norm}}\biggr)_\mathrm{eq}
= \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \, .
</math>
</div>

Hence, the relevant factor may be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
2\pi \biggl[ \frac{K_c \rho_0^{6/5}}{P_\mathrm{norm}}\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} 
\biggl\{ \frac{1}{q}\biggl[ \frac{K_c^{1/2} G^{-1/2} \rho_0^{-2/5}}{R_\mathrm{norm}}\biggr] \biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi_i \biggr\}^3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl[ \frac{K_c^{5/2} G^{-3/2}}{P_\mathrm{norm} R_\mathrm{norm}^3 }\biggr] \biggl(1 + \frac{1}{3}\xi_i^2 \biggr)^{-3} 
\biggl( \frac{\xi_i }{q}\biggr)^3
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, the last expression has been obtained by employing the substitution, defined above, <math>~\xi_i = (3a_\xi)^{1/2}q</math>.  Finally, then, we have,
<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{2}{3({\gamma_c}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_c} 
\biggl\{\biggl( \frac{3^8}{2^7\pi} \biggr)^{1/2} \biggl[ 
\tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2} \biggr]
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
As it should, the term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the core, <math>~S^*_\mathrm{core}</math>, that has been derived elsewhere in conjunction with our [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|discussion of the detailed force-balanced structure of this bipolytrope]].


===The Envelope===

Similarly, the energy contained in the envelope's thermodynamic reservoir may be written as,
<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{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl( \frac{P_{ie}}{P_{ic}} \biggr)
\biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  
\biggl[ (1-q^3) s_\mathrm{env} \biggr] \, ,
</math>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>~(1-q^3) s_\mathrm{env} </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~ 
\int_q^1  3\biggl[ 1 - p_e(x) \biggr]  x^2 dx \, ,
</math>
  </td>
</tr>
</table>
</div>
defines the relevant integral over the envelope's pressure distribution.  According to our [[SSC/Structure/BiPolytropes/Analytic51#Profile|derivation of the properties of detailed force-balance <math>~(n_c, n_e) = (5, 1)</math> bipolytropes]] &#8212; see also the relevant derivations [[SSC/BipolytropeGeneralizationVersion2#Bipolytrope_Generalization|in our accompanying overview]] &#8212;  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^* \equiv \frac{P_\mathrm{env}(\eta)}{P_0} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\theta_i^6 \phi^2(\eta) = \theta_i^6 \biggl( \frac{A}{\eta} \biggr)^2 \sin^2(\eta-B) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ 1-p_e(x) \equiv \frac{P_\mathrm{env}(x)}{P_{ie}} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \frac{P_0}{P_{ic}}  \biggr) \frac{P_\mathrm{env}(x)}{P_0} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \theta_i^{-6} \biggr) \theta_i^6 \biggl( \frac{A}{b_\eta x} \biggr)^2 \sin^2(b_\eta x-B) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \frac{\sin^2(b_\eta x-B)}{x^2} \, ,
</math>
  </td>
</tr>
</table>
</div>
where, <math>~b_\eta</math> has been defined above in connection with our [[SSC/Structure/BiPolytropes/FreeEnergy51#The_Envelope|derivation of the envelope's mass profile]].  The desired integral over this pressure distribution therefore gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(1-q^3)s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 \biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \frac{A}{b_\eta} \biggr)^2 \int_q^1 \sin^2(b_\eta x-B) dx
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{4} \biggl( \frac{P_{ic}}{P_{ie}}  \biggr) \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1 \, ,
</math>
  </td>
</tr>
</table>
</div>
where, as before, we have dropped the integration constant because it cancels upon insertion of the specified integration limits.  Therefore, we have,
<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{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} 
\biggl[ \frac{2\pi P_{ic} \chi^3}{P_\mathrm{norm}} \biggr]_\mathrm{eq}  
\frac{3}{4} \biggl( \frac{A^2}{b_\eta^3} \biggr) \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1  \, .
</math>
  </td>
</tr>
</table>
</div>
Now, drawing from our above derivation steps and discussion, we know that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~b_\eta</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \, ,</math>
  </td>
</tr>
</table>
</div>
and
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>\frac{2\pi P_{ic} \chi_\mathrm{eq}^3}{P_\mathrm{norm}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3^6}{2\pi} \biggr)^{1/2} \biggl(1 + a_\xi q^2 \biggr)^{-3} a_\xi^{3/2} = \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \, .
</math>
  </td>
</tr>
</table>
</div>

Finally, then, we can write,

<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{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{
\frac{3}{4} A^2\biggl[ \biggl( \frac{3^3}{2\pi} \biggr)^{1/2} \biggl( \frac{\theta_i^2\xi_i }{q}\biggr)^3 \biggr] 
\biggl[ 3^{1/2} \biggl( \frac{\mu_e}{\mu_c} \biggr) \frac{\theta_i^2 \xi_i}{q} \biggr]^{-3} \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1  \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2}{3({\gamma_e}-1)} \biggl( \frac{\chi}{\chi_\mathrm{eq}} \biggr)^{3-3\gamma_e} \biggl\{
\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-3} A^2  \biggl( \frac{3^2}{2^5\pi} \biggr)^{1/2}  \biggl[ 2b_\eta x -\sin[2(b_\eta x - B)] \biggr]_q^1  \biggr\} \ .
</math>
  </td>
</tr>
</table>
</div>
The term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the envelope, <math>~S^*_\mathrm{env}</math>, that has been derived elsewhere in conjunction with our [[SSC/Structure/BiPolytropes/Analytic51#Expression_for_Free_Energy|discussion of the detailed force-balanced structure of this bipolytrope]].

==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.