Editing SSC/Structure/BiPolytropes/Analytic00 (section)

=BiPolytrope with (n<sub>c</sub>, n<sub>e</sub>) = (0, 0)=

Here we construct a [[SSC/Structure/BiPolytropes#BiPolytropes|bipolytrope]] in which both the core and the envelope have uniform densities, that is, the structure of both the core and the envelope will be modeled using an <math>n = 0</math> polytropic index.  It should be possible for the entire structure to be described by closed-form, analytic expressions.  Generally, we will follow the [[SSC/Structure/BiPolytropes#Solution_Steps|general solution steps for constructing a bipolytrope]] that we have outlined elsewhere.  [On '''<font color="red">1 February 2014</font>''', J. E. Tohline wrote:  This particular system became of interest to me during discussions with Kundan Kadam about the relative stability of bipolytropes.]   

==Step 4:  Throughout the core (0 &le; &chi; &le; &chi;<sub>i</sub>)==
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="center" colspan="3">
Specify:  <math>~P_0</math> and <math>\rho_0 ~\Rightarrow</math>
  </td>
  <td colspan="2">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
<math>~\rho</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>~\rho_0</math>
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
&nbsp;
  </td>
</tr>

<tr>
  <td align="right">
<math>~P</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</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi^2 \biggr)</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~r</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
 </td>
  <td align="left">
<math>\frac{4\pi}{3} \rho_0 r^3</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\frac{4\pi}{3} \rho_0 \biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{3/2} \chi^3
= \frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi^3</math>
  </td>
</tr>
</table>

</div>

==Step 5:  Interface Conditions==
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="center" colspan="3">
Specify:  <math>~\chi_i</math> and <math>~\rho_e/\rho_0</math>, and demand &hellip;
  </td>
  <td colspan="2">
&nbsp;
  </td>
</tr>
<tr>
  <td align="right">
<math>~P_{ei}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>~P_{ci}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi_i^2 \biggr)</math>
  </td>
</tr>
</table>

</div>

==Step 6:  Envelope Solution (&chi; &ge; &chi;<sub>i</sub>)==
<div align="center">
<table border="0" cellpadding="3">
<tr>
  <td align="right">
<math>~\rho</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>~\rho_e</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~P</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>P_{ei} + \biggl(\frac{2}{3} \pi G \rho_e\biggr) \biggl[ 2(\rho_0 - \rho_e) r_i^3\biggl( \frac{1}{r} - \frac{1}{r_i}\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_{ei} + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) P_0 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \chi_i^3\biggl( \frac{1}{\chi} - 
\frac{1}{\chi_i}\biggr) - \frac{\rho_e}{\rho_0} (\chi^2 - \chi_i^2) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{P}{P_0}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>1 - \frac{2\pi}{3}\chi_i^2  + 
\frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \chi_i^2 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( \frac{1}{\xi} - 
1\biggr) - \frac{\rho_e}{\rho_0} (\xi^2 - 1) \biggr]</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~M_r</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
 </td>
  <td align="left">
<math>\frac{4\pi}{3} \biggl[ \rho_0 r_i^3 + \rho_e(r^3 - r_i^3) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
 </td>
  <td align="left">
<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \biggl[\chi_i^3 +\frac{\rho_e}{\rho_0} 
\biggl( \chi^3 - \chi_i^3 \biggr) \biggr]</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
 </td>
  <td align="left">
<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi_i^3\biggl[1 +\frac{\rho_e}{\rho_0} 
\biggl( \xi^3 - 1\biggr) \biggr]</math>
  </td>
</tr>
</table>

</div>


==Step 7:  Surface Boundary Condition==
At the surface (that is, at <math>r = R</math> and <math>M_r = M_\mathrm{tot}</math>), <math>P/P_0 = 0</math> and <math>\xi = \xi_s = R/r_i = 1/q</math>.  Also, we can write,
<div align="center">
<math>
\chi_i = q\biggl[  \frac{G\rho_0^2 R^2}{P_0} \biggr]^{1/2} \, ;
</math>
</div>
and, from earlier derivations,
<div align="center">
<math>
R^3 = \frac{3M_\mathrm{tot}}{4\pi \bar\rho} = \frac{3M_\mathrm{tot}}{4\pi \rho_0} \biggl( \frac{\nu}{q^3} \biggr) \, ;
</math>

<math>
\frac{\rho_e}{\rho_0} = \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3}\biggr) \, .
</math>
</div>
Therefore, setting the pressure to zero at the surface means,
<div align="center">
<table cellpadding="5">

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

<tr>
  <td align="right">
<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) q^{-2} \biggl[  \frac{P_0}{G\rho_0^2 R^2} \biggr]</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>1  + 
\biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + 
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) \biggl( \frac{4\pi}{3} \biggr)^{2/3} \nu^{-2/3} 
\biggl[  \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr]^{1/3}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>1  + 
\biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + 
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~~~\biggl\{ \biggl( \frac{6}{\pi} \biggr) \frac{1}{\nu^{2} }
\biggl[  \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr] \biggr\}^{1/3}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>1  + 
\biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + 
\frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]
</math>
  </td>
</tr>
</table>
</div>
<span id="gdefinition">It therefore seems prudent to define a function,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
<math>
\Rightarrow ~~~~ q^2 g^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
q^2  + \biggl(\frac{\rho_e}{\rho_0}\biggr)  \biggl[ 
2q^2(1-q) + \biggl(\frac{\rho_e}{\rho_0}\biggr) (1-3q^2 + 2q^3)
\biggr]  \, ,
</math>
  </td>
</tr>
</table>
</div>

<span id="MassRadius">in which case the expressions for the equilibrium radius and equilibrium total mass are, respectively,</span>
<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>
\biggl[ \frac{G\rho_0^2}{P_0} \biggr]^{1/2} R</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \frac{1}{q g} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\biggl[ \frac{G^3\rho_0^4}{P_0^3} \biggr]^{1/2} M_\mathrm{tot}</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\biggl( \frac{6}{\pi} \biggr)^{1/2} \frac{1}{\nu g^3} \, .</math>
  </td>
</tr>
</table>
</div>
Note that this means that,
<div align="center">
<math>
\chi_i^2 = \biggl( \frac{3}{2\pi}\biggr) \frac{1}{g^2}  \, .
</math>
</div>
<span id="CentralPressure">We can also combine these two expressions and eliminate direct reference to the central density,</span> <math>\rho_0</math>, obtaining,
<div align="center">
<table border="0">
<tr>
  <td align="right">
<math>
\biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr] P_0</math>
  </td>
  <td align="center">
&nbsp; <math>~=</math>&nbsp; 
  </td>
  <td align="left">
<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} </math>
  </td>
</tr>

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