Editing SSC/Structure/BiiPolytropes/FreeEnergy51

__FORCETOC__
=Free Energy of BiPolytrope with (n<sub>c</sub>, n<sub>e</sub>) = (5, 1)=
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  <td align="center" bgcolor="lightblue" width="33%"><br />[[SSC/Structure/BiPolytropes/FreeEnergy51|Part I:&nbsp; Mass Profile]]
&nbsp;
  </td>
  <td align="center" bgcolor="lightblue"3 width="33%"><br />[[SSC/Structure/BiPolytropes/FreeEnergy51/Pt2|Part II:&nbsp; Gravitational Potential Energy]]
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  <td align="center" bgcolor="lightblue"><br />[[SSC/Structure/BiPolytropes/FreeEnergy51/Pt3|Part III:&nbsp; Thermal Energy Reservoir]]
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</tr>
</table>


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|- 
! 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]].
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==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>

=See Also=

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