Editing SSC/BipolytropeGeneralization (section)

===Cleaner Virial Presentation===

In an effort to show the similarity in structure among the several energy terms, we have also found it useful to write their expressions in the following forms:

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

<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{3}{5} \biggl( \frac{GM_\mathrm{tot}^2}{R} \biggr) \frac{\nu^2}{q} \cdot f 
= - 4\pi P_i R^3 \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_i f \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi P_{ic} R^3 \biggl[ q^3 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ic} \biggr]   \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2\pi P_{ie} R^3 \biggl[ (1-q^3) + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{ie} \mathfrak{F} \biggr]   \, ,</math>
  </td>
</tr>
</table>
</div>
where (see an [[SSC/Structure/BiPolytropes/Analytic00#Gravitational_Potential_Energy|associated discussion]] or the [[SSC/VirialStability#Energy_Expressions|original derivation]]),
<div align="center">
<math>
f\biggl(q, \frac{\rho_e}{\rho_c}\biggr) = 1 + \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (q^3 - q^5 )
+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] \, ,
</math>
</div>

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

<tr>
  <td align="right">
<math>~\lambda_i</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{GM_\mathrm{tot}^2}{R^4 P_i} \, ,</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">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\lambda_{ie}} \biggl( \frac{2^2 \cdot 5\pi}{3} \biggr) \frac{q(1-q^3)}{\nu^2} (s_\mathrm{env} -1)
 \, .</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="1" cellpadding="10" width="80%">
<tr><td align="left">
This also means that the three key terms used as shorthand notation in the above expressions for the three energy terms have the following definitions:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathfrak{f}_{WM}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\frac{\nu^2}{q} \cdot f \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~s_\mathrm{core}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q^4} \cdot \lambda_{ic} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~s_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>
1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q(1-q^3)} \cdot \lambda_{ie} \mathfrak{F} 
\, ,</math>
  </td>
</tr>
</table>
</div>

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

Hence, if all the interface pressures are equal &#8212; that is, if <math>~P_i = P_{ic} = P_{ie}</math> and, hence also, <math>~\lambda_{i} = \lambda_{ic} = \lambda_{ie}</math> &#8212; then the total thermal energy is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~S_\mathrm{tot} = S_\mathrm{core} + S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
2\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F}) \biggr] \, ;
</math>
  </td>
</tr>
</table>
</div>
and the virial is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~2S_\mathrm{tot} + W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
4\pi P_{i} R^3 \biggl[ 1 + \biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot \lambda_{i} (1+\mathfrak{F} - f ) \biggr] 
\, .
</math>
  </td>
</tr>
</table>
</div>

The virial should sum to zero in equilibrium, which means,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl( \frac{3}{2^2\cdot 5\pi} \biggr) \frac{\nu^2}{q} \cdot (f - 1- \mathfrak{F} ) 
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~~ \biggl( \frac{\rho_e}{\rho_c} \biggr)^{-1} \biggl[ \biggl( \frac{2^3\pi}{3} \biggr) \frac{q^6}{\nu^2} \biggr] \frac{R_\mathrm{eq}^4 P_i}{GM_\mathrm{tot}^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ (q^3 - q^5 )+ \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl( \frac{2}{5} - q^3 + \frac{3}{5}q^5 \biggr) \biggr] -
\biggl[  (-2q^2 + 3q^3 - q^5) + \biggl( \frac{\rho_e}{\rho_c}\biggr) (-\frac{3}{5}  +3q^2 - 3q^3 + \frac{3}{5} q^5) \biggr]
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
<math>\Rightarrow~~~~ \frac{1}{\lambda_i} \biggr|_\mathrm{eq}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{2\pi}{3} \biggl[ \biggl( \frac{3}{4\pi} \biggr) \frac{\nu}{q^3} \biggr]^2 q^2 (g^2-1)   \, .
</math>
  </td>
</tr>
</table>
</div>