Editing SSC/Structure/BiPolytropes/Analytic00 (section)

=Thermal Energy Content=
Knowing the pressure distribution throughout this bipolytropic configuration allows us to obtain an analytic expression for the configuration's total thermal content.  Specifically, the differential contribution to the total thermal energy that is made by each spherical shell is,
<div align="center">
<math>dS = \frac{3}{2} \biggl( \frac{P}{\rho} \biggr) dm 
= \frac{3}{2} \biggl( \frac{P}{\rho} \biggr) 4\pi \rho r^2 dr
= 6\pi R^3 P(x) x^2 dx \, ,</math>
</div>
where,
<div align="center">
<math>x \equiv \frac{r}{R} \, .</math>
</div>
(We are switching to a new normalization of the radial coordinate &#8212; using <math>~x</math> in preference to <math>~\chi</math> &#8212; because, at the interface between the core and the envelope, <math>~x=q</math>, while <math>~x=1 </math> at the surface of the bipolytropic configuration.)  Hence, the thermal content of the core and of the envelope will be given by performing the following integrals, respectively:
<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>6\pi R^3 \int_0^q  P_\mathrm{core}(x) x^2 dx \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 6\pi R^3 \int_q^1 P_\mathrm{env}(x) x^2 dx \, .</math>
  </td>
</tr>

</table>
</div>
Drawing from the expressions obtained in step #4 and step #6, above, the relevant functions <math>~P(x)</math> are,
<div align="center">
<table border="0">
<tr>
 <td align="right">
<math>~P_\mathrm{core}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_0 - \frac{2}{3} \pi G \rho_0^2 r^2 = P_i + \frac{2}{3} \pi G \rho_0^2(r_i^2 -  r^2)</math>
  </td>
</tr>

<tr>
 <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_i + \frac{2}{3} \pi G \rho_0^2 R^2 (q^2 -  x^2) = P_i + \frac{3}{2^3 \pi} 
\biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) (q^2 -  x^2) \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P_\mathrm{env}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_{i} - \biggl(\frac{2}{3} \pi G \rho_e\biggr) \biggl[ 2(\rho_0 - \rho_e) r_i^3\biggl( \frac{1}{r_i} - \frac{1}{r} \biggr) +
\rho_e(r^2 - r_i^2) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_{i} - \biggl(\frac{2}{3} \pi G \rho_0^2 R^2\biggr) \frac{\rho_e}{\rho_0}
\biggl[ 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{q} - \frac{1}{x} \biggr) +
\frac{\rho_e}{\rho_0} (x^2 - q^2) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_{i} - \frac{3}{2^3 \pi} 
\biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) \frac{\rho_e}{\rho_0}
\biggl[ 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{q} - \frac{1}{x} \biggr) +
\frac{\rho_e}{\rho_0} (x^2 - q^2) \biggr] \, .</math>
  </td>
</tr>
</table>
</div>


<div align="center">
<table border="1" cellpadding="8" width="85%">
<tr><td align="left">
Reminder:  For a given choice of the parameter set <math>~(M_\mathrm{tot}, R, \nu, q)</math>, 
&#8212; and remembering that choosing the parameter pair <math>~(\nu,q)</math> sets the density ratio via the expression, <math>~\rho_e/\rho_0 = [q^3(1-\nu)]/[\nu(1-q^3)]</math> &#8212; the value of the pressure at the interface, <math>~P_i</math>, is determined by setting boundary conditions at the surface of the configuration, namely, by setting <math>~x=1</math> and <math>~P_\mathrm{env} = 0</math> in the second expression.  Then, the central pressure <math>~(P_\mathrm{core} = P_0)</math> is determined by setting <math>~x=0</math> and inserting the determined value for <math>~P_i</math> in the first expression.
</td></tr>
</table>
</div>

Hence,
<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>6\pi R^3 \int_0^q  \biggl\{ P_i + \Pi (q^2 -  x^2) \biggr\} x^2 dx </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>6\pi R^3 \biggl\{ \biggl[ \biggl(P_i + q^2 \Pi \biggr)\frac{x^3}{3} \biggr]_0^q 
- \biggl[ \Pi \frac{x^5}{5} \biggr]_0^q \biggr\}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>\biggl( \frac{4\pi}{5} \biggr) R^3 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr)  \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~S_\mathrm{env}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 6\pi R^3 \int_q^1 \biggl\{ P_i - \Pi \frac{\rho_e}{\rho_0} \biggl( a_e + \frac{b_e}{x} + c_e x^2\biggr) \biggr\} x^2 dx </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 6\pi R^3 \biggl\{ \biggl[ P_i 
- a_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)\biggr] \frac{x^3}{3} \biggr|_q^1 
- \biggl[ b_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \frac{x^2}{2} \biggr]_q^1
- \biggl[ c_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \frac{x^5}{5} \biggr]_q^1 \biggr\}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> \frac{\pi R^3}{5} \biggl\{ 10 \biggl[ P_i 
- a_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr] (1-q^3) 
- 15 \biggl[ b_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) (1-q^2) \biggr]
- 6\biggl[ c_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) (1-q^5) \biggr] \biggr\}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> \frac{\pi R^3}{5} \biggl\{ 10 P_i  (1-q^3) - \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 10 a_e (1-q^3) 
+ 15 b_e (1-q^2) +  6 c_e  (1-q^5) \biggr] \biggr\}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
\frac{\pi R^3}{5} \biggl\{ 10 P_i  (1-q^3)  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>  
+\Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 
30 \biggl( \frac{\rho_e}{\rho_0} - \frac{2}{3} \biggr) q^2 (1-q^3)
+ 30 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3 (1-q^2) -  6 \frac{\rho_e}{\rho_0} (1-q^5) \biggr] \biggr\}  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
\frac{\pi R^3}{5} \biggl\{ 10 P_i  (1-q^3)  
+ 10 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) q^2 \biggl[  3 q (1-q^2) - 2 (1-q^3) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>  
  + 6 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ 5 q^2 (1-q^3)- 5  q^3 (1-q^2) -  (1-q^5) \biggr] \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
\frac{\pi R^3}{5} \biggl\{ 10 P_i  (1-q^3)  
+ 10 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2q^2 + 3q^3 - q^5  \biggr]
  + 6 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 +  q^5   \biggr] \biggr\} \, ,
</math>
  </td>
</tr>
</table>
</div>

<span id="PiDefinition">where,</span>
<div align="center">
<math>
\Pi \equiv \frac{3}{2^3 \pi} 
\biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) \, ,
</math>
</div>
and, in the expression for <math>~S_\mathrm{env}</math>, we temporarily used the shorthand notation,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>~a_e</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>3 \biggl(\frac{2}{3} - \frac{\rho_e}{\rho_0} \biggr) q^2 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~b_e</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>- 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3 \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~c_e</math>
  </td>
  <td align="center">
<math>~\equiv~</math>
  </td>
  <td align="left">
<math>\frac{\rho_e}{\rho_0}  \, .</math>
  </td>
</tr>

</table>
</div>

The total thermal energy (per unit volume) may therefore be written as,

<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>\biggl( \frac{5}{\pi R^3} \biggr) S_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
4 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr)  +
10 P_i  (1-q^3) - \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 10 a_e (1-q^3) 
+ 15 b_e (1-q^2) +  6 c_e  (1-q^5) \biggr]   
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
10P_i  + 2 \Pi \biggl[ 2q^5 - 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 )
  - 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 -  q^5 )  \biggr] \, ,
</math>
  </td>
</tr>

</table>
</div>
or, because <math>~P_0 = P_i + \Pi q^2</math>, we can reference the central pressure instead of the pressure at the interface and write,

<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>\biggl( \frac{5}{\pi R^3} \biggr) S_\mathrm{tot}</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math> 
10P_0 - 2\Pi \biggl[5q^2 -2q^5 + 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 )
  + 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 -  q^5 )  \biggr] \, .
</math>
  </td>
</tr>

</table>
</div>