Editing SSC/Structure/BiPolytropes/Analytic00 (section)

==Expression for Free Energy==
Rewrite the expressions for <math>~S_\mathrm{core}</math> and <math>~S_\mathrm{env}</math>, recognizing that, although we won't know its value until the equilibrium radius of the configuration has been determined, during a radial perturbation the dimensionless ratio,
<div align="center">
<math>~\Lambda \equiv \frac{2 q^2\Pi}{5P_i} = \frac{3}{2^2\cdot 5 \pi} 
\biggl( \frac{GM_\mathrm{tot}^2}{R^4 P_i} \biggr) \biggl( \frac{\nu^2}{q^4} \biggr) = 
\biggl( \frac{2^2\pi}{3\cdot 5} \biggr) \frac{G\rho_0^2 r_i^2}{P_i} \, ,
</math> 
</div>
will remain unchanged.  Hence, we may write,

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

<tr>
  <td align="right">
<math>~S_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>2\pi q^3 R^3 P_{ic} \biggl (1 + \Lambda \biggr)  \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
\pi R^3 P_{ie} \biggl\{ 2 (1-q^3)  
+ 5 \Lambda \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2 + 3q - q^3  \biggr]
  + \frac{3 \Lambda}{q^2} \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 +  q^5   \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
The free energy is, therefore,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~\mathfrak{G}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~U_\mathrm{tot} + W</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl[ \frac{2}{3(\gamma_c -1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e -1)} \biggr] S_\mathrm{env} + W</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>B_\mathrm{core}R^3 P_{ic} + B_\mathrm{env} R^3 P_{ie}  - A_\mathrm{grav} R^{-1}  \, ,
</math>
  </td>
</tr>

</table>
</div>

where, for a given choice of the three parameters <math>~(M_\mathrm{tot}, \nu, q)</math>, the constant coefficients in this expression are,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~A_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
\frac{2^2 \pi}{5} \biggl[ \frac{3GM_\mathrm{tot}^2}{2^3\pi} \biggl(\frac{\nu^2}{q^6} \biggr) \biggr] \biggl\{
2q^5 + 5q^3\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2)
+ \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1-q^5) 
- 5q^3(1-q^2) \biggr] \biggr\} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~B_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
 \biggl[ \frac{4\pi q^3}{3(\gamma_c -1)} \biggr]\biggl(1 + \Lambda\biggr)  \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~B_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
 \biggl[ \frac{2\pi}{3(\gamma_e -1)} \biggr] \biggl\{ 2 (1-q^3)  
+ 5 \Lambda \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2 + 3q - q^3  \biggr]
  + \frac{3 \Lambda}{q^2} \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 +  q^5   \biggr] \biggr\}  \, .
</math>
  </td>
</tr>
</table>
</div>

Notice that, in the expression for <math>~\mathfrak{G}</math>, we have been careful to maintain the separate identities of the interface pressure, depending on whether it is set by the core <math>~(P_{ic})</math> or by the envelope <math>~(P_{ie})</math>, because they scale differently &#8212; along two separate adiabats &#8212; with density and, hence, with radius.  Specifically,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~P_{ic}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>K_c \rho_0^{\gamma_c} = K_c \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggr]^{\gamma_c} R^{-3\gamma_c} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_{ie}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>K_e \rho_e^{\gamma_e} = K_e \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} 
\biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr]^{\gamma_e} R^{-3\gamma_e} \, .</math>
  </td>
</tr>
</table>
</div>
With these pressure scalings in mind, the expression for the free energy becomes,

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

<tr>
  <td align="right">
<math>~\mathfrak{G}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
C_\mathrm{core} \biggl[ \frac{2}{3(\gamma_c - 1)} \biggr] R^{3-3\gamma_c} + 
C_\mathrm{env} \biggl[ \frac{2}{3(\gamma_e - 1)} \biggr] R^{3-3\gamma_e} 
-A_\mathrm{grav} R^{-1}\, ,
</math>
  </td>
</tr>

</table>
</div>

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

<tr>
  <td align="right">
<math>~C_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
2\pi q^3 K_c \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggr]^{\gamma_c} \biggl(1 + \Lambda\biggr) \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~C_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>
\pi K_e \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} 
\biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr]^{\gamma_e} 
\biggl\{ 2 (1-q^3)  
+ 5 \Lambda \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2 + 3q - q^3  \biggr]
  + \frac{3 \Lambda}{q^2} \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 +  q^5   \biggr] \biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>