Editing SR/Piston

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=Piston Model=
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<font size="-1">[[H_BookTiledMenu#Context|<b>KW94<br />Piston Model</b>]]</font>
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Here we draw principally from the discussion of a simple piston model as presented in &sect;2.7 and &sect;6.6 of [<b>[[Appendix/References#KW94|<font color="red">KW94</font>]]</b>].  

An ''[[SR#Equation_of_State|ideal gas]]'' of mass <math>m^*</math> is held in a vertical container with a movable piston resting on top of &#8212; and confining &#8212; the gas; the mass of the piston is <math>M^*</math>.  A vertically directed gravitational acceleration, <math>g</math>, acts on the piston, in which case the weight of the piston is given by the expression,

<div align="center"><math>G^* = g M^*.</math></div>

<font color="darkgreen">"In the case of hydrostatic equilibrium, the gas pressure <math>P</math> adjusts in such a way that the weight per unit area is balanced by the pressure:"</font> 

<div align="center"><math>P = \frac{G^*}{A}.</math></div>

<font color="darkgreen">"If the forces do not compensate each other, the piston is accelerated in the vertical direction according to the equation of motion,"
<div align="center">
<math>M^*\frac{d^2 h}{dt^2} = -G^* + PA.</math>
</div>
</font>

=Bipolytropes=

If we consider only the structure and oscillations of the core, we should set the "external" pressure, <math>P_e</math>, equal to the pressure, <math>P_i</math>, at the core-envelope interface ''as viewed from the perspective of the envelope.  

==Extra Relations==
Keep in mind that, in hydrostatic balance,

<div align="center">
{{Math/EQ_SShydrostaticBalance01}}
</div>

Otherwise,
<div align="center">

<span id="PGE:Euler"><font color="#770000">'''Euler Equation'''</font></span><br />
('''Momentum Conservation''')

{{ Math/EQ_Euler01 }}
</div>
In equilibrium, the pressure at the core-envelope interface is, 
<div align="center">
<math>P_i = K_c \rho_c^{6/5}\biggl(1 + \frac{\xi_i^2}{3} \biggr)^{-3}</math>.
</div>


<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_n \rho^{1 + 1/n}</math>
  </td>
</tr>
</table>

==Use Surface Area==

According to the "piston model," it should be true that,
<table border="0" align="center" cellpadding="8">
<tr>
  <td  align="right"><math>P_e = \frac{G^*}{A}</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>\frac{g_i M^*}{4\pi r_i^2} \, ,</math>
  </td>
</tr>
</table>

where (the magnitude of) the acceleration at the interface is,
<table border="0" align="center" cellpadding="8">
<tr>
  <td  align="right"><math>g_i</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>\frac{GM_\mathrm{core}}{r_i^2} \, .</math>
  </td>
</tr>
</table>

This means that,
<table border="0" align="center" cellpadding="8">
<tr>
  <td  align="right"><math>M^*</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{4\pi r_i^2 P_e}{g_i} 
= 
\frac{4\pi r_i^4 P_e}{GM_\mathrm{core}}
\, .</math>
  </td>
</tr>
</table>

==Insert Interface Details==

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,

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

<tr>
  <td  align="right"><math>P_i</math></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>

<tr>
  <td  align="right"><math>r_i</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_i
\, ;</math>
  </td>
</tr>

<tr>
  <td  align="right"><math>M_\mathrm{core}</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
\, .</math>
  </td>
</tr>
</table>

Hence, we find that,
<table border="0" align="center" cellpadding="8">

<tr>
  <td  align="right"><math>M^*</math></td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{4\pi }{G} \biggl[ r_i^4 \biggr] 
\biggl[P_i\biggr]
\biggl[ M_\mathrm{core} \biggr]^{-1}
</math>
  </td>
</tr>

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

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{4\pi }{G} \biggl[ \frac{K_c^{ 2}}{G^{ 2}\rho_c^{8 / 5}}\biggl(\frac{3}{2\pi}\biggr)^{ 2} ~ \xi_i^4 \biggr] 
\biggl[ K_c \rho_c^{6/5}\biggl(1 + \frac{\xi_i^2}{3} \biggr)^{-3} \biggr]
\biggl[ \frac{G^{3 / 2}\rho_c^{1 / 5}}{K_c^{3 / 2}}\biggl(\frac{\pi}{6}\biggr)^{1 / 2} ~ (\xi \theta)^{-3} \biggr]
</math>
  </td>
</tr>

<tr>
  <td  align="right">&nbsp;</td>
  <td  align="center"><math>=</math></td>
  <td  align="left">
<math>
\frac{4\pi }{G} \biggl[ \frac{K_c^{ 2}}{G^{ 2}\rho_c^{8 / 5}}\biggr] 
\biggl[ K_c \rho_c^{6/5}\biggr]
\biggl[ \frac{G^{3 / 2}\rho_c^{1 / 5}}{K_c^{3 / 2}}\biggr]
\biggl(\frac{3}{2\pi}\biggr)^{ 2} ~ \xi_i^4 \biggl(1 + \frac{\xi_i^2}{3} \biggr)^{-3} \biggl(\frac{\pi}{6}\biggr)^{1 / 2} ~ (\xi \theta)^{-3} 
</math>
  </td>
</tr>

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

=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>
-->

=See Also=

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