Editing SR/Piston (section)

=Hydrostatic Balance=

Drawing principally from [[SSCpt2/SolutionStrategies|an accompanying discussion]], we understand that hydrostatic balance throughout a self-gravitating sphere is given by the key relation,

<div align="center">
{{Math/EQ_SShydrostaticBalance01}}
</div>
where,
<div align="center">
<math>~M_r = \int_0^r 4\pi r^2 \rho dr</math> .
</div>


Now, according to [[SSC/Structure/BiPolytropes/Analytic51/Pt2#Examples|our analysis of the bipolytrope having <math>(n_c, n_e) = (5,1),</math>]] we have throughout the core, <math>\theta = (1 + \xi^2/3)^{-1 / 2}</math> and,

<table border="0" align="center" cellpadding="8">

<tr>
  <td  align="right"><math>r</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{K_c^{1 / 2}}{G^{1 / 2}\rho_c^{2 / 5}}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} ~ \xi
\, ;</math>
  </td>
</tr>

<tr>
  <td  align="right"><math>\rho</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\rho_c \theta^5 = \rho_c \biggl[1 + \frac{\xi^2}{3}\biggr]^{-5/2}
\, ;</math>
  </td>
</tr>

<tr>
  <td  align="right"><math>M_r</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{K_c^{3 / 2}}{G^{3 / 2}\rho_c^{1 / 5}}\biggl(\frac{6}{\pi}\biggr)^{1 / 2} ~ (\xi \theta)^3
=
\frac{K_c^{3 / 2}}{G^{3 / 2}\rho_c^{1 / 5}}\biggl(\frac{6}{\pi}\biggr)^{1 / 2} ~ \xi^3\biggl[1 + \frac{\xi^2}{3}\biggr]^{-3/2}
\, .</math>
  </td>
</tr>
</table>

Therefore the RHS of the hydrostatic-balance expression is,

<table border="0" align="center" cellpadding="8">

<tr>
  <td  align="right">RHS</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
G \biggl\{ \frac{K_c^{3 / 2}}{G^{3 / 2}\rho_c^{1 / 5}}\biggl(\frac{6}{\pi}\biggr)^{1 / 2} ~ \xi^3\biggl[1 + \frac{\xi^2}{3}\biggr]^{-3/2} \biggr\} 
\biggl\{ \rho_c \biggl[1 + \frac{\xi^2}{3}\biggr]^{-5/2} \biggr\} 
\biggl\{ \frac{K_c^{1 / 2}}{G^{1 / 2}\rho_c^{2 / 5}}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} ~ \xi \biggr\}^{-2}
</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
G \biggl\{ \frac{K_c^{3 / 2}}{G^{3 / 2}\rho_c^{1 / 5}}\biggl(\frac{6}{\pi}\biggr)^{1 / 2} ~ \xi^3\biggl[1 + \frac{\xi^2}{3}\biggr]^{-3/2} \biggr\} 
\biggl\{ \rho_c \biggl[1 + \frac{\xi^2}{3}\biggr]^{-5/2} \biggr\} 
\biggl\{ \frac{G\rho_c^{4 / 5}}{K_c}\biggl(\frac{2\pi}{3}\biggr) ~ \xi^{-2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
G^{1 / 2} K_c^{1 / 2} \rho_c^{8/5} \biggl(\frac{2\cdot 3}{\pi} \cdot \frac{2^2 \pi^2}{3^2}\biggr)^{1 / 2}
\biggl\{  ~ \xi^3\biggl[1 + \frac{\xi^2}{3}\biggr]^{-3/2} \biggr\} 
\biggl\{  \biggl[1 + \frac{\xi^2}{3}\biggr]^{-5/2} \biggr\} 
\biggl\{ \xi^{-2} \biggr\}
</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
G^{1 / 2} K_c^{1 / 2} \rho_c^{8/5} \biggl(\frac{2^3\pi}{3}\biggr)^{1 / 2}
\biggl[1 + \frac{\xi^2}{3}\biggr]^{-4} ~\xi
</math>
  </td>
</tr>
</table>

Integrating the hydrostatic-balance expression from the center, <math>\xi=0</math>, to a location, <math>\xi_i</math>, gives,

<table border="0" align="center" cellpadding="8">

<tr>
  <td  align="right"><math>\int_{P_c}^{P_i} dP</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>-~\int_0^{r_i} \mathrm{RHS} \cdot dr</math>
  </td>
</tr>

<tr>
  <td  align="right"><math>\Rightarrow ~~~ P_i - P_c</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>-~\frac{K_c^{1 / 2}}{G^{1 / 2}\rho_c^{2 / 5}}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} ~ \int_0^{\xi_i} \mathrm{RHS} \cdot d\xi</math>
  </td>
</tr>

<tr>
  <td  align="right"><math>\Rightarrow ~~~ P_i </math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>P_c - \frac{K_c^{1 / 2}}{G^{1 / 2}\rho_c^{2 / 5}}\biggl(\frac{3}{2\pi}\biggr)^{1 / 2} ~ \int_0^{\xi_i} 
G^{1 / 2} K_c^{1 / 2} \rho_c^{8/5} \biggl(\frac{2^3\pi}{3}\biggr)^{1 / 2}
\biggl[1 + \frac{\xi^2}{3}\biggr]^{-4} ~\xi
 \cdot d\xi</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>K_c\rho_c^{6/5} - 2K_c\rho_c^{6/5}~ \int_0^{\xi_i} 
\biggl[1 + \frac{\xi^2}{3}\biggr]^{-4} ~\xi
 \cdot d\xi</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>K_c\rho_c^{6/5} \biggl\{1 - 2\cdot 3^4~ \int_0^{\xi_i} 
(3+\xi^2)^{-4} \xi\cdot d\xi\biggr\}</math> 
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>K_c\rho_c^{6/5} \biggl\{1 + 2\cdot 3^4~ \biggl[ 
\frac{1}{6(3 + \xi^2)^3}
\biggr]_0^{\xi_i} \biggl\}</math>
  </td>
</tr>
<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>K_c\rho_c^{6/5} \biggl\{1 +  \biggl[ 
\frac{2\cdot 3^4~}{6(3 + \xi_i^2)^3} - \frac{2\cdot 3^4~}{6(3 )^3}
\biggr] \biggl\}</math> 
  </td>
</tr>

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

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>K_c\rho_c^{6/5}  \biggl(1 + \frac{\xi_i^2}{3} \biggr)^{-3}</math> 
  </td>
</tr>

</table>

<!--
This result should be compared with,

<table border="0" align="center" cellpadding="8">
<tr>
  <td  align="right"><math>P</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
K_c \rho_c^{6/5}\biggl(1 + \frac{\xi^2}{3} \biggr)^{-3}
\, ;</math>
  </td>
</tr>
</table>

March 13, 2026: &nbsp;<font color="red">This difference needs to be resolved!!!</font>
-->