Editing SSC/Stability/n1PolytropeLAWE/Pt4

__FORCETOC__
=Radial Oscillations of n = 1 Polytropic Spheres (Pt 4)=
<table border="1" align="center" width="100%" colspan="8">
<tr>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE|Part I: &nbsp; Search for Analytic Solutions]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE/Pt2|Part II:&nbsp; New Ideas]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue" width="25%"><br />[[SSC/Stability/n1PolytropeLAWE/Pt3|Part III:&nbsp; What About Bipolytropes?]]<br />&nbsp;</td>
  <td align="center" rowspan="1" bgcolor="lightblue"><br />[[SSC/Stability/n1PolytropeLAWE/Pt4|Part IV:&nbsp; Most General Structural Solution]]<br />&nbsp;</td>
</tr>
</table>

==Preamble Regarding Chatterji==
As far as we have been able to ascertain, the first technical examination of radial oscillation modes in <math>n=1</math> polytropes was performed &#8212; using numerical techniques &#8212; in 1951 by L. D. Chatterji; at the time, he was in the Mathematics Department of Allahabad University.  His two papers on this topic were published in, what  is now referred to as, the ''Proceedings of the Indian National Science Academy'' (PINSA).  The citations that immediately follow this opening paragraph provide inks to both of these papers by Chatterji, but the links may be insecure.  <!-- (Citations/links to articles that provide analyses of models having other polytropic indexes are provided at the [[#See_Also|bottom of this chapter]].) --> Apparently [https://www.springer.com/journal/43538 Springer] is archiving recent PINSA volumes, but their holdings do not date back as early as 1951.

* {{ Chatterji51full }}, ''Radial Oscillations of a Gaseous Star of Polytropic Index I'' 
* {{ Chatterji52full }}, ''Anharmonic Pulsations of a Polytropic Model of Index Unity''

A detailed review of {{ Chatterji51hereafter }} is provided in [[SSC/Structure/BiPolytropes/Analytic51Renormalize/Pt2#Setup|an accompanying discussion]].

==Equilibrium Structure==

When <math>n=1</math>, the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution|relevant Lane-Emden equation]] is,
<div align="center">
<math>\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H</math> ,
</div>

and we find that the [[SSC/Structure/Polytropes/Analytic#n_=_1_Polytrope|solution]] is, quite generally,

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

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
A \biggl[ \frac{\sin\xi}{\xi}\biggr] - B \biggl[\frac{\cos\xi}{\xi}\biggr]
=
\frac{1}{\xi}\biggl\{ 
A\sin\xi - B \cos\xi
\biggr\}
</math>
  </td>
  <td align="right">
</table>
in which case,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\frac{d \theta}{d\xi}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{d}{d\xi}~\biggl\{
A \biggl[ \frac{\sin\xi}{\xi}\biggr] - B \biggl[\frac{\cos\xi}{\xi}\biggr]
\biggr\}
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
A \biggl[ -\frac{\sin\xi}{\xi^2} + \frac{\cos\xi}{\xi}\biggr] 
+ 
B \biggl[\frac{\cos\xi}{\xi^2}+\frac{\sin\xi}{\xi}\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{\xi^2}\biggl[
-A\sin\xi + B\cos\xi
\biggr]
+
\frac{1}{\xi}\biggl[
A\cos\xi + B\sin\xi
\biggr] 
\, ,
</math>
  </td>
</tr>
</table>

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

<tr>
  <td align="right">
<math>Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\xi}{\theta}\cdot
\biggl\{
\frac{1}{\xi^2}\biggl[
-A\sin\xi + B\cos\xi
\biggr]
+
\frac{1}{\xi}\biggl[
A\cos\xi + B\sin\xi
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{1}{\xi\theta}\cdot
\biggl\{
\biggl[
A\sin\xi - B\cos\xi
\biggr]
+
\xi\biggl[
- A\cos\xi - B\sin\xi
\biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
1
-
\xi\biggl[
\frac{A\cos\xi + B\sin\xi}{A\sin\xi - B\cos\xi}
\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>
<span id="Beech88">If we set</span> <math>\beta \equiv \tan^{-1}(B/A)</math>, we can rewrite the expression for <math>\theta</math> as,

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

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{A}{\xi}\biggl\{ 
\sin\xi - \frac{B}{A} \cos\xi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{A}{\xi}\biggl\{ 
\sin\xi - \tan\beta \cos\xi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{A}{\xi\cos\beta}\biggl\{ 
\sin\xi \cos\beta - \cos\xi \sin\beta 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{A}{\cos\beta} \cdot \frac{\sin(\xi-\beta)}{\xi} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="center" colspan="3">
[[Appendix/References#Beech88|Beech88]], &sect;3, p. 221, Eq. (6)<br />
[[Appendix/References#EFC98|EFC98]], &sect;2, p. 831, Eq. (2)<br />
  </td>
</tr>
</table>

and the expression for <math>Q</math> as,

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

<tr>
  <td align="right">
<math>Q(\xi) </math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
1
-
\xi\biggl[
\frac{\cos\xi \cos\beta + \sin\xi\sin\beta}{\sin\xi\cos\beta - \cos\xi\sin\beta }
\biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
1
-
\xi\biggl[
\frac{\cos(\xi - \beta)}{\sin(\xi -\beta) }
\biggr] 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[
1
-
\xi\cot(\xi-\beta) 
\biggr] \, . 
</math>
  </td>
</tr>
</table>

<table border="5" bordercolor="purple" align="center" cellpadding="10" width="80%">
<tr>
  <td align="left">

<div align="center">SUMMARY of EQUILIBRIUM STRUCTURE <br />and switching notation from <math>\theta(\xi)</math> to <math>\phi(\eta)</math>
</div>

When <math>n=1</math>, the [[SSC/Structure/Polytropes/Analytic#Primary_E-Type_Solution|relevant Lane-Emden equation]] is,
<div align="center">
<math>\frac{1}{\eta^2} \frac{d}{d\eta}\biggl( \eta^2 \frac{d\phi}{d\eta} \biggr) = - \phi</math> .
</div>
[[SSC/Structure/Polytropes/Analytic#NIST|Its solution]], quite generally, is

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

<tr>
  <td align="right">
<math>\phi</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
A \biggl[ \frac{\sin\eta}{\eta}\biggr] - B \biggl[\frac{\cos\eta}{\eta}\biggr] \, ,
</math>
  </td>
</tr>
</table>
where <math>A</math> and <math>B</math> are scalar constants, in which case,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>Q(\eta) \equiv - \frac{d \ln \phi}{d\ln \eta}</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl\{
1
-
\eta\biggl[
\frac{A\cos\eta + B\sin\eta}{A\sin\eta - B\cos\eta}
\biggr] 
\biggr\} \, .
</math>
  </td>
</tr>
</table>

----

Alternatively, drawing from Eq. (6) of [[Appendix/References#Beech88|Beech88]], this solution can be written in the form,

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

<tr>
  <td align="right">
<math>\phi</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\alpha_\mathrm{Beech} \cdot \frac{\sin(\eta-\beta_\mathrm{Beech})}{\eta} \, ,
</math>
  </td>
</tr>
</table>

in which case,

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

<tr>
  <td align="right">
<math>Q(\eta) </math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[
1
-
\eta\cot(\eta-\beta_\mathrm{Beech}) 
\biggr] \, , 
</math>
  </td>
</tr>
</table>
where, in terms of the coefficients <math>A</math> and <math>B</math>,

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

<tr>
  <td align="right">
<math>\beta_\mathrm{Beech} </math>
  </td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\tan^{-1}(B/A)
</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>\alpha_\mathrm{Beech} </math>
  </td>
  <td align="center"><math>\equiv</math></td>
  <td align="left">
<math>
\frac{A}{\cos\beta_\mathrm{Beech}}
=
A [ 1 + (B/A)^2 ]^{1 /2}
\, .
</math>
  </td>
</tr>
</table>

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

==Establish Relevant (n=1) LAWE==
From [[SSC/Stability/n1PolytropeLAWE#WorkInProgress|a related discussion]] &#8212; or [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|a broader overview of Instability Onset]] &#8212; we find the

<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>

Furthermore &#8212; see, for example, [[SSC/Stability/InstabilityOnsetOverview#Pressure_and_Density_Displacement_Functions|here]],
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="1"><font color="maroon"><b>Exact Solution to the Polytropic LAWE</b></font></td>
</tr>
<tr>
  <td align="left">
<math>x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] \, ,</math>
  </td>
</tr>
</table>
in which case for <math>n=1</math>,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="left">
<math>
x_P = -3 \biggl( \frac{1}{\xi \theta}\biggr) \frac{d\theta}{d\xi}
=
-3 \biggl( \frac{1}{\xi^2}\biggr) \frac{d\ln\theta}{d\ln\xi}
=
\frac{3}{\xi^2} Q \, .
</math>
  </td>
</tr>
</table>


===Isolated Sphere===
For an isolated n = 1 <math>(\gamma_g = 2, \alpha = 1)</math> polytrope, we know that,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\frac{\sin\xi}{\xi}</math>
  </td>
  <td align="right">
&nbsp; &nbsp; <math>\Rightarrow ~~~ Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>\biggl[1 - \xi\cot\xi\biggr] \, .</math>
  </td>
</tr>
</table>
Hence, the relevant LAWE is,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\biggl[4 - 2(1 - \xi\cot\xi)\biggr] \frac{1}{\xi} \cdot \frac{dx}{d\xi}
+
2 \biggl[\biggl( \frac{\sigma_c^2}{12}\biggr)\frac{\xi^3}{\sin\xi} - (1 - \xi\cot\xi)  \biggr]\frac{x}{\xi^2}
</math>
  </td>
</tr>
</table>

<div align="center">
<font color="maroon"><b>LAWE for n = 1 Polytrope</b></font><br />
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\frac{d^2 x}{d\xi^2} 
+ 
\frac{2}{\xi} \biggl[1 +  \xi\cot\xi\biggr] \frac{dx}{d\xi}
+
\frac{1}{2} \biggl[\biggl( \frac{\sigma_c^2}{3}\biggr)\frac{\xi}{\sin\xi} - \frac{4}{\xi^2} \biggl(1 - \xi\cot\xi \biggr)  \biggr]x
</math>
  </td>
</tr>
</table>
</div>

[[SSC/Stability/Polytropes#Boundary_Conditions|Surface boundary condition]]:
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>- \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl( \frac{3-n}{n+1}\biggr) + \frac{n\sigma_c^2}{6(n+1)}
\biggl[ \frac{\xi}{\theta'}\biggr]_\mathrm{surf}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ - \frac{d\ln x}{d\ln \xi} \biggr|_\mathrm{surf}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
1 + \frac{\sigma_c^2}{12}
\biggl[ \frac{\xi^3}{(\xi \cos\xi - \sin\xi)}\biggr]_{\xi=\pi}
=
1 - \frac{\pi^2 \sigma_c^2}{12}
</math>
  </td>
</tr>
</table>
</td></tr></table>

===Spherical Shell===

In the context of a spherically symmetric n = 1 <math>(\gamma_g = 2, \alpha = 1)</math> shell (''envelope'') outside of a spherically symmetric bipolytropic ''core'', we should adopt the more general Lane-Emden structural solution,
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
<math>\theta</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>A \biggl[\frac{\sin\xi}{\xi}\biggr] - B \biggl[\frac{\cos\xi}{\xi}\biggr]</math>
  </td>
  <td align="right">
&nbsp; &nbsp; <math>\Rightarrow ~~~ Q(\xi) \equiv - \frac{d \ln \theta}{d\ln \xi}</math>
  </td>
  <td align="center">=</td>
  <td align="left">
<math>
\biggl[1 - \xi\cot(\xi-\beta)  \biggr]
=
\frac{\xi^2 x_P}{3} \, .
</math>
  </td>
</tr>
</table>

<table border="5" bordercolor="purple" align="center" cellpadding="10" width="80%">
<tr>
  <td align="left">
Reminder: the expression for <math>x_P</math> is, 
<div align="center"><math>x_P = \frac{3}{\xi^2}\biggl[1 - \xi\cot(\xi-\beta)\biggr]</math>.</div>

Playing around a bit, we find that,

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

<tr>
  <td align="right"><math>\frac{\xi^2}{3} \cdot x_P</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi \cdot \frac{\cos(\xi-\beta)}{\sin(\xi-\beta)}
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi 
\biggl[\frac{\cos(\xi-\beta)}{\sin(\xi-\beta)}\biggr] \cdot \frac{(\xi-\beta)}{(\xi-\beta)} 
</math></td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
1 - \xi 
\biggl[\frac{\cos(\xi-\beta)}{(\xi-\beta)}\biggr] 
\biggl[\frac{(\xi-\beta)}{\sin(\xi-\beta)}\biggr] 
</math></td>
</tr>
</table>

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

As a result, the governing LAWE becomes,
<!--
<div align="center">
<font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br />
{{ Math/EQ_RadialPulsation02 }}
</div>
-->

<table border="0" align="center" cellpadding="5">
<tr>
  <td align="right"><math>0</math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} + \biggl[4 - 2Q\biggr]\frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+ 2\biggl[\cancelto{0}{\biggl(\frac{\sigma_c^2}{12}\biggr)}\frac{\xi^2}{\theta} - Q\biggr]\frac{x_P}{\xi^2}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- \biggl[2Q\biggr]\frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
- \biggl[2Q\biggr]\frac{x_P}{\xi^2}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- 2Q \biggl\{  \frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+\frac{x_P}{\xi^2}\biggr\}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \frac{4}{\xi} \cdot \frac{dx_P}{d\xi} 
- \frac{2\xi^2 x_P}{3} \biggl\{  \frac{1}{\xi} \cdot \frac{dx_P}{d\xi} 
+\frac{x_P}{\xi^2}\biggr\}
</math></td>
</tr>
<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d^2x_P}{d\xi^2} 
+ \biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \frac{dx_P}{d\xi} 
- \frac{2 x_P^2}{3} 
\, .
</math></td>
</tr>
</table>

Let's plug in the expression for <math>x_P</math>, namely, <math>x_P = 3[1 - \xi\cot(\xi-\beta)]/\xi^2</math>.  We have, first of all,

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

<tr>
  <td align="right"><math>
x_p^2
  </math></td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{3^2}{\xi^4}\biggl[1 - \xi\cot(\xi-\beta) \biggr]^2
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{3^2}{\xi^4}\biggl[1 - 2\xi\cot(\xi-\beta) + \xi^2\cot^2 (\xi-\beta)\biggr] \, ;
</math></td>
</tr>

<tr>
  <td align="right">
<math>
\frac{dx_p}{d\xi}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d}{d\xi}\biggl\{
\frac{3}{\xi^2}\biggl[1 - \xi\cot(\xi-\beta)\biggr]
\biggr\}
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl[1 - \xi\cot(\xi-\beta)\biggr]
\frac{d}{d\xi}\biggl\{
\frac{3}{\xi^2}
\biggr\}
-
\frac{3}{\xi}
\frac{d}{d\xi}
\biggl[\cot(\xi-\beta)\biggr]
-
\frac{3}{\xi^2}
\biggl[\cot(\xi-\beta)\biggr]
</math></td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr] \, .
</math></td>
</tr>
</table>


<table border="5" bordercolor="purple" align="center" cellpadding="10" width="80%">
<tr>
  <td align="left">
Note for later use that,

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

<tr>
  <td align="right">
<math>
\frac{x_P^2}{r^2} \biggl(\frac{d\ln x_P}{d \ln r}\biggr)^2 = \biggl(\frac{dx_P}{dr} \biggr)^2
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr] \biggr\}^2

</math>
  </td>
</tr>
</table>
  </td>
</tr>
</table>

Recognize that we have used the trigonometric relations, 

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

<tr>
  <td align="right">
<math>\frac{d}{d\xi}\biggl[\cot(u)\biggr]</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- ~\frac{1}{\sin^2(u)} ~\frac{du}{d\xi}
=
- ~ \biggl[1 + \cot^2(u)\biggr] \frac{du}{d\xi}
\, .
</math></td>
</tr>
</table>
And,

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

<tr>
  <td align="right">
<math>
\frac{d^2x_p}{d\xi^2}
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{d}{d\xi}\biggl\{
- \frac{6}{\xi^3}
+
\frac{3}{\xi} 
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+ 
\frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\}
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{3}{\xi^2} 
- \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{3}{\xi^2}\biggl[1 + \cot^2(\xi-\beta)\biggr]
- \frac{3}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
- \frac{6}{\xi}\biggl[\cot(\xi-\beta)\biggr]\biggl[1 + \cot^2(\xi-\beta) \biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] - \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
\, .
</math></td>
</tr>
</table>
Hence,

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

<tr>
  <td align="right">
<math>
\frac{d^2x_P}{d\xi^2} 
+ \biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \frac{dx_P}{d\xi} 
- \frac{2 x_P^2}{3} 
</math>
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\biggl\{
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] - \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
\biggr\}
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\biggl[ \frac{4}{\xi}- \frac{2\xi x_P}{3}  \biggr] \cdot \biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
-
\biggl\{
\frac{6}{\xi^4}\biggl[1 - 2\xi\cot(\xi-\beta) + \xi^2\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\frac{4}{\xi} \biggl\{
- \frac{6}{\xi^3}
+ \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{3}{\xi} + \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+
\biggl[ \frac{2\xi x_P}{3}  \biggr] \cdot \biggl\{
 \frac{6}{\xi^3}
- \frac{3}{\xi^2}\biggl[\cot(\xi-\beta)\biggr]
-
\frac{3}{\xi} - \frac{3}{\xi}\biggl[\cot^2(\xi-\beta)\biggr]
\biggr\} 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi^4}\biggl[-1 + 2\xi\cot(\xi-\beta) - \xi^2\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
\frac{18}{\xi^4} - \frac{6}{\xi^2}  - \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
- \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
-
\frac{24}{\xi^4}
+ \frac{12}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
+
\frac{12}{\xi^2} + \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
</math></td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi^4}\biggl[-1 + 2\xi\cot(\xi-\beta) - \xi^2\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
-\frac{6}{\xi^4} + \frac{6}{\xi^2}  + \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
+ \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^3}\cdot \cot(\xi-\beta) 
+ \frac{6}{\xi} \cdot\cot(\xi-\beta) 
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{12}{\xi^4}
- \frac{6}{\xi^3}\biggl[\cot(\xi-\beta)\biggr]
-
\frac{6}{\xi^2} - \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{6}{\xi^4} + \frac{12}{\xi^3}\cdot \cot(\xi-\beta) 
- \frac{6}{\xi^2}\cdot \cot^2(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
- \frac{6}{\xi} \biggl[\cot(\xi-\beta) \biggr] 
+ \frac{6}{\xi^2} \cdot \cot^2(\xi-\beta)
- \frac{6}{\xi}\cdot \cot^3(\xi-\beta)
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
+ \frac{6}{\xi} \cdot\cot(\xi-\beta) 
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center"><math>=</math></td>
  <td align="left"><math>
0 \, .</math></td>
</tr>
</table>

<hr>

<b>Debugging LaTeX layout:</b>
<table border="0" align="center" cellpadding="5">

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^2(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left"><math>
- \frac{12}{\xi^2}\biggl[\cot^3(\xi-\beta)\biggr]
</math></td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">&nbsp;</td>
  <td align="left">
<math>
+ \biggl[\cot^1(\xi-\beta)\biggr]
+ \frac{6}{\xi^2}\biggl[\cot^2(\xi-\beta)_{m}\biggr] 
+ \frac{6}{\xi}\biggl[\cot^3(\xi-\beta)\biggr] 
</math>
  </td>
</tr>
</table>

=Hydrostatic Balance and Virial Equilibrium=

==General Expression for Virial==
Here we draw heavily from our accompanying [[SSC/SynopsisStyleSheet|"style sheet" synopsis of spherically symmetric configurations]].

First, we pull the equation for
<div align="center">
<font color="maroon"><b>Hydrostatic Balance</b></font><br />
{{ Math/EQ_SShydrostaticBalance01 }}
</div>

from subsection <b><font color="maroon" size="+1">&#x2460;</font></b> of the synopsis; then, guided by subsection <b><font color="maroon" size="+1">&#x2461;</font></b>, we multiply both sides through by <math>r dV = 4\pi r^3 dr</math> and integrate over the volume.  This gives,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\int_0^R r\biggl(\frac{dP}{dr}\biggr)dV - \int_0^R r\biggl(\frac{GM_r \rho}{r^2}\biggr)dV</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>-\int_0^R 4\pi r^3 \biggl(\frac{dP}{dr}\biggr) dr - \int_0^R \biggl(\frac{GM_r}{r}\biggr)dM_r \, ,</math>
  </td>
</tr>
</table>
where we have used the relations,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>dV = 4\pi r^2 dr </math> 
  </td>
  <td align="center">
and,
  </td>
  <td align="left">
&nbsp;&nbsp;&nbsp;<math>dM_r = \rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr \, .</math>
  </td>
</tr>
</table>
Now, given that,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>\frac{d}{dr}\biggl(4\pi r^3 P\biggr)</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>4\pi r^3 \biggl( \frac{dP}{dr} \biggr) + 12\pi P r^2 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ 
-~4\pi r^3 \biggl( \frac{dP}{dr} \biggr)
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
12\pi r^2 P
-
\frac{d}{dr}\biggl(4\pi r^3 P\biggr)
\, ,</math>
  </td>
</tr>
</table>
we can rewrite the integral expression in the form,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\int_0^R\biggl[12\pi r^2 P - \frac{d}{dr}\biggl( 4\pi r^3P \biggr) \biggr] dr 
- \int_0^R \biggl(\frac{GM_r}{r}\biggr)dM_r
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>\int_0^R 3\biggl[ 4\pi r^2 P \biggr]dr 
- \int_0^R \biggl(\frac{GM_r}{r}\biggr)dM_r
- \int_0^R \biggl[ d(3PV)\biggr] 
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>3(\gamma-1)U_\mathrm{int}  + W_\mathrm{grav} - \biggl[ 3PV \biggr]_0^R \, ,</math>
  </td>
</tr>
</table>
where,

<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>- \int_0^R \biggl(\frac{GM_r}{r}\biggr) dM_r</math>
  </td>

<td align="center">&nbsp;&nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp;</td>
  <td align="right">
<math>U_\mathrm{int}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{1}{(\gamma -1)} \int_{0}^{R} 4\pi r^2 P dr
\, .</math>
  </td>
</tr>
</table>
[[VE#Relationship_to_the_System's_Internal_Energy|Note as well]] that <math>U_\mathrm{int} = 2S_\mathrm{therm}/[3(\gamma - 1)]</math>.

==Calculate Relevant Energy Expressions==

Adopting the energy normalization shown [[SSC/Structure/BiPolytropes/Analytic51/Pt4#Expression_for_Free_Energy|here01]] along with the other variable normalizations defined [[SSC/Structure/BiPolytropes/Analytic51/Pt2#Normalization|here02]], we have &hellip;

===Thermal Energy===
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>S^* \equiv \frac{S}{[K_c^5/G^3]^{1 / 2}} = \frac{3}{2}(\gamma-1) \cdot \frac{U_\mathrm{int}}{[K_c^5/G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2[K_c^5/G^3]^{1 / 2}} \int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 4\pi r^2 P dr
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2}\biggl\{ K_c^{- 5/ 2} G^{3 / 2} \biggr\} \int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 
\biggl\{ K_c^{3 / 2} G^{-3 / 2}\rho_0^{-6 / 5} \cdot K_c\rho_0^{6 / 5}\biggr\}
4\pi \biggl(r^*)^2 P^* dr^*
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2} \int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 
4\pi \biggl(r^*)^2 P^* dr^* \, .
</math>
  </td>
</tr>
</table>

Plugging in the [[SSC/Structure/BiPolytropes/Analytic51/Pt2#Profile|derived radial profiles]] for <math>r^*</math> and <math>P^*</math>, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>S^*</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} \int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2}~ \frac{3^{3 / 2}}{2^3}
\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
8\cdot \frac{\xi^2}{3} \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  \frac{d\xi}{3^{1 / 2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[
(2^2 \cdot 3^2 \pi^2) \biggl(\frac{3^3}{2^3 \pi^3}\biggr) \biggl(\frac{3^3}{2^6}\biggr)
\biggr]^{1 / 2}
\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
8\cdot \frac{\xi^2}{3} \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  \frac{d\xi}{3^{1 / 2}} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl(\frac{3^8}{2^7 \pi}\biggr)^{1 / 2}
\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
8\cdot \frac{\xi^2}{3} \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  \frac{d\xi}{3^{1 / 2}} \, .
</math>
  </td>
</tr>
</table>

After making the substitution, <math>x \equiv \xi/\sqrt{3}</math>, this expression matches the expression for <math>S^*_\mathrm{core}</math> obtained [[SSC/Structure/BiPolytropes/Analytic51/Pt4#Expression_for_Free_Energy|separately]].

<table border="1" align="center" width="80%" cellpadding="8">
<tr><td align="left">
For later use, we note that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>dS^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2} \biggl[ \frac{4\pi r^2 P dr}{[K_c^5/G^3]^{1 / 2}} \biggr] 
=
6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi \, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>

===Gravitational Potential Energy===

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

<tr>
  <td align="right">
<math>W^* = \frac{W_\mathrm{grav}}{[K_c^5/G^3]}</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>[K_c^{-5/2} G^{3/2}]\int_{r_\mathrm{inner}}^{r_\mathrm{outer}} \biggl(-\frac{GM_r}{r}\biggr) dM_r</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- [K_c^{-5/2} G^{3/2}]\int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 4\pi GM_r \rho r dr</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- [K_c^{-5/2} G^{3/2}]\int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 4\pi G 
\biggl[ K_c^{3 / 2} G^{-3 / 2}\rho_0^{-1 / 5} \biggr] M_r^*
\biggl[ \rho^* \rho_0 \biggr] 
\biggl[ K_c G^{-1} \rho_0^{-4 / 5} \biggr]r^* dr^*</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \int_{r_\mathrm{inner}}^{r_\mathrm{outer}} 4\pi M_r^*\rho^* r^* dr^* \, .</math>
  </td>
</tr>
</table>

Plugging in the [[SSC/Structure/BiPolytropes/Analytic51/Pt2#Profile|derived radial profiles]] for <math>r^*</math>, <math>\rho^*</math> and <math>M_r^*</math>, we have,

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

<tr>
  <td align="right">
<math>W^* </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- 4\pi \int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
\biggl(\frac{2\cdot 3}{\pi}\biggr)^{1 / 2}\biggl[\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  \biggr]
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2}
\biggl(\frac{3}{2\pi}\biggr)\xi d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- 4\pi \biggl(\frac{2\cdot 3}{\pi}\biggr)^{1 / 2} \biggl(\frac{3}{2\pi}\biggr)\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
\biggl[\xi^3 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}  \biggr]
\biggl(1 + \frac{\xi^2}{3}\biggr)^{-5 / 2}
\xi d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
- \biggl(2^4\pi^2 \biggr)^{1 / 2} \biggl(\frac{2\cdot 3}{\pi}\biggr)^{1 / 2} \biggl(\frac{3^2}{2^2\pi^2}\biggr)^{1 / 2}
\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>- \biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\int_{\xi_\mathrm{inner}}^{\xi_\mathrm{outer}} 
3^{5 / 2} \cdot \biggl[\chi^4 \biggl(1 + \chi^2\biggr)^{-4}  \biggr] d\chi \, . 
</math>
  </td>
</tr>
</table>
This expression matches the expression for <math>W^*_\mathrm{core}</math> obtained [[SSC/Structure/BiPolytropes/Analytic51/Pt4#Expression_for_Free_Energy|separately]].

<table border="1" align="center" width="80%" cellpadding="8">
<tr><td align="left">
For later use, we note that,

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

<tr>
  <td align="right"><math>dW^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
[K_c^{-5/2} G^{3/2}] \biggl(-\frac{GM_r}{r}\biggr) 4\pi r^2 \rho dr 
=
- \biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi 
\, .
</math>
  </td>
</tr>
</table>
</td></tr>
</table>

=Stability Analysis=
Here, as well, we draw heavily from our accompanying [[SSC/SynopsisStyleSheet#Stability|"style sheet" synopsis of spherically symmetric configurations]].

This time, we pull the
<div align="center">
<font color="#770000">'''LAWE: &nbsp; Linear Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>0</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dr}\biggl[ r^4 \gamma P ~\frac{dx}{dr} \biggr] 
+\biggl[ \omega^2 \rho r^4 + (3\gamma - 4) r^3 \frac{dP}{dr} \biggr] x 
\, ,
</math>
  </td>
</tr>
</table>
</div>

from subsection <b><font color="maroon" size="+1">&#x2463;</font></b> of the synopsis; then, guided by subsection <b><font color="maroon" size="+1">&#x2464;</font></b>, we multiply both sides through by <math>4\pi x dr</math> to obtain,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>- \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi x\biggl\{ \frac{d}{dr}\biggl[ r^4 \gamma P \biggl( \frac{dx}{dr}\biggr) \biggr] \biggr\} dr
+ \biggl\{ \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( \frac{dP}{dr}\biggr) \biggr] \biggr\} dr
\, .
</math>
  </td>
</tr>

</table>
Now, given that,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
4\pi x \frac{d}{dr}\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
+
\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr)\biggr]\frac{d}{dr}\biggl[4\pi x \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>
\Rightarrow ~~~ 
4\pi x \frac{d}{dr}\biggl[\gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 

</math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
-
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] \, ,
</math>
  </td>
</tr>
</table>
we can rewrite this last expression in the form,

<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>- \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\frac{d}{dr}\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr]dr 
-
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
+ \biggl\{ \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( \frac{dP}{dr}\biggr) \biggr] \biggr\} dr
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow ~~~ \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
-d\biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr] 
+
\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
- \biggl[ (3\gamma - 4) 4\pi x^2 r^3 \biggl( - \frac{GM_r \rho}{r^2}\biggr) \biggr] dr
</math>
  </td>
</tr>

</table>
Note that, in order to obtain the last term on the RHS of this expression, we used the hydrostatic balance relation to replace the pressure gradient in terms of the gravitational potential.  Finally, integrating over the volume of the configuration gives,

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

<tr>
  <td align="right">
<math>\int_0^R \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^R\biggl[4\pi \gamma r^4 P \biggl(\frac{dx}{dr}\biggr)^2\biggr] dr
- \int_0^R \biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr
- \biggl[4\pi x \gamma r^4 P \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R \, ,
</math>
  </td>
</tr>

</table>
or,

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

<tr>
  <td align="right">
<math>\int_0^R \biggl[ 4\pi x^2 (\omega^2 \rho r^4 ) \biggr]dr </math>
  </td>
  <td align="center">
<math>=</math>
  </td>
  <td align="left">
<math>
\int_0^R \overbrace{\biggl[x^2 \biggl(\frac{d\ln x}{d\ln r}\biggr)^2 4\pi \gamma r^2 P \biggr] dr}^{\mathrm{TERM1}} 
- \int_0^R \underbrace{\biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr}_{\mathrm{TERM2}}
+ \overbrace{\biggl[\gamma 4\pi x^2 r^2 P \biggl(-\frac{d\ln x}{d\ln r}\biggr) \biggr]_0^R}^{\mathrm{TERM3}} \, .
</math>
  </td>
</tr>

</table>

==TERM1==
Given that (from above),
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>dS^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{3}{2} \biggl[ \frac{4\pi r^2 P dr}{[K_c^5/G^3]^{1 / 2}} \biggr] 
=
6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi \, ,
</math>
  </td>
</tr>
</table>
we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma x_P^2 \biggl(\frac{d\ln x_P}{d\ln r} \biggr)^2 \biggl[ \frac{4\pi r^2 P dr}{[K_c^5/G^3]^{1 / 2}} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma x_P^2 \biggl(\frac{d\ln x_P}{d\ln r} \biggr)^2 \biggl\{
\biggl( \frac{2}{3} \biggr) 6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} \, ,
</math>
  </td>
</tr>
</table>
Furthermore, [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|given that]] for a truncated <math>n=5</math> configuration,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>x_p</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
1 - \frac{\xi^2}{15}
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{dx_p}{d\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{2}{15}~\xi 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ \frac{d\ln x_p}{d\ln\xi}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- \frac{\xi}{x_P}\biggl[\frac{2}{15}~\xi \biggr]
=
- \biggl[\frac{2}{15}~\xi^2 \biggr]\biggl[ \frac{15}{15-\xi^2}\biggr]
=
- \biggl[ \frac{2\xi^2}{15-\xi^2}\biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right"><math>\Rightarrow ~~~ x_P^2 \biggl(\frac{d\ln x_p}{d\ln\xi}\biggr)^2</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\biggl[\frac{15 - \xi^2}{15}\biggr]^2
\biggl[ \frac{2\xi^2}{15-\xi^2}\biggr]^2 
=
\biggl[ \frac{2\xi^2}{15} \biggr]^2
\, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right"><math>\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2\xi^2}{15} \biggr]^2 \biggl\{
\biggl( \frac{2}{3} \biggr) 6\pi \biggl( \frac{3}{2\pi} \biggr)^{3 / 2} ~ 
\xi^2 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\biggl\{
\xi^6 \biggl(1+\frac{\xi^2}{3}\biggr)^{-3}  d\xi\biggr\} \, .
</math>
  </td>
</tr>
</table>

Hence, after making the replacement <math>\chi \equiv \xi/\sqrt{3} ~\Rightarrow ~ \xi = 3^{1 / 2}\chi</math>, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\int_0^{\chi_\mathrm{surf}}\frac{\mathrm{TERM1}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\cdot 3^{7/2}\int_0^{\chi_\mathrm{surf}}\biggl\{
\chi^6 \biggl(1+\chi^2\biggr)^{-3} d\chi
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\gamma \biggl[ \frac{2^5 }{3 \cdot 5^4 \pi} \biggr]^{1 /2}
\cdot 3^{7/2}
\cdot \frac{1}{8}\biggl\{
\frac{\chi_\mathrm{surf}(8\chi^4_\mathrm{surf} + 25 \chi^2_\mathrm{surf} 
+ 15)}{(1+\chi^2_\mathrm{surf})^2} - 15\tan^{-1}(\chi_\mathrm{surf})
\biggr\} 
</math>
  </td>
</tr>
</table>
where, we have completed the integral with the assistance of the [https://wolframalpha.com WolframAlpha] online integrator:
[[File:Mathematica05.png|500px|center|Mathematica Integral]]

==TERM2==
Given that (from above),

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

<tr>
  <td align="right"><math>dW^* </math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
[K_c^{-5/2} G^{3/2}] \biggl(-\frac{GM_r}{r}\biggr) 4\pi r^2 \rho dr 
=
- \biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi 
\, ,
</math>
  </td>
</tr>
</table>
we can write,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
[K_c^{-5 / 2} G^{3 / 2}]\biggl[ (3\gamma - 4) x^2 \biggl( - \frac{GM_r }{r}\biggr)4\pi \rho r^2 \biggr] dr
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) x_P^2 \biggl\{
\biggl[ \frac{2^3 \cdot 3^3}{\pi} \biggr]^{1 / 2}
\biggl[\xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi\biggr\}
\, .
</math>
  </td>
</tr>
</table>
Furthermore, [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|given that]] (as above) for a truncated <math>n=5</math> configuration,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>x_p</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
\frac{15 - \xi^2}{15} \, ,
</math>
  </td>
</tr>
</table>
we have,

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

<tr>
  <td align="right"><math>\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 \cdot 3^3}{\pi} \cdot \frac{1}{3^4\cdot 5^4}\biggr]^{1 / 2}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}
\, .
</math>
  </td>
</tr>
</table>

Hence, after making the replacement <math>\chi \equiv \xi/\sqrt{3} ~\Rightarrow ~ \xi = 3^{1 / 2}\chi</math>, we find that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right"><math>\int_0^{\chi_\mathrm{surf}}\frac{\mathrm{TERM2}}{[K_c^5 / G^3]^{1 / 2}}</math></td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2}
\int_0^{\chi_\mathrm{surf}}
\biggl\{
\biggl[(15 - \xi^2)^2 \xi^4 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}  \biggr] d\xi
\biggr\}

</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 }{3\cdot 5^4\pi} \biggr]^{1 / 2} \cdot 3^{9 / 2}
\int_0^{\chi_\mathrm{surf}}
\biggl\{
\biggl[(5 - \chi^2)^2 \chi^4 \biggl(1 + \chi^2\biggr)^{-4}  \biggr] d\chi
\biggr\}

</math>
  </td>
</tr>

<tr>
  <td align="right">&nbsp;</td>
  <td align="center"><math>=</math></td>
  <td align="left">
<math>
- (3\gamma - 4) \biggl[ \frac{2^3 \cdot 3^8 }{5^4\pi} \biggr]^{1 / 2} 
\cdot \frac{1}{4}\biggl\{
\frac{\chi_\mathrm{surf}(4\chi^6_\mathrm{surf} + 53\chi^4_\mathrm{surf} + 40 \chi^2_\mathrm{surf} 
+ 15)}{(1+\chi^2_\mathrm{surf})^3} - 15\tan^{-1}(\chi_\mathrm{surf})
\biggr\} 
</math>
  </td>
</tr>
</table>
where, we have completed the integral with the assistance of the [https://wolframalpha.com WolframAlpha] online integrator:
[[File:Mathematica06.png|500px|center|Mathematica Integral]]

=See Also=

<ul>
  <li>[[SSC/Structure/BiPolytropes/Analytic51/Pt2#Parameter_Values|Equilibrium Model Parameter Profiles (Table of Parameters)]]</li>
  <li>[[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|Instability Onset Overview]]</li>
  <li>[[SSC/Stability/InstabilityOnsetOverview#Configurations_Having_an_Index_Less_Than_Three|Index Less that 3]]</li>
  <li>Radial Oscillations of n=1 Polytropic Spheres
    <ul>
      <li>[[SSC/Stability/n1PolytropeLAWE|Search for Analytic Solutions]]</li>
      <li>[[SSC/Stability/n1PolytropeLAWE/Pt4|Most General Structural Solution]]</li>
    </ul>
    </li>
  <li>[[SSC/SynopsisStyleSheet|Synopsis Style Sheet]]</li>
  <li>Free Energy of (n<sub>c</sub>, n<sub>e</sub>) = (5, 1) Bipolytrope
    <ul>
      <li>[[SSC/Structure/BiiPolytropes/FreeEnergy51|One Discussion]]</li>
      <li>[[SSC/Structure/BiPolytropes/Analytic51/Pt4|Another Discussion]] <font color="red"> <==  Very Useful Analytic Integrals</font></li>
    </ul>
  </li>
</ul>

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