Difference between revisions of "User:Tohline/SSC/VariationalPrinciple"

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We will draw heavily from the paper published by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], and from pp. 458-474 of the review by [http://adsabs.harvard.edu/abs/1958HDP....51..353L P. Ledoux &amp; Th. Walraven (1958)] in explaining how the ''variational principle'' can be used to identify the eigenvector of the fundamental mode of radial oscillation in spherically symmetric configurations.
We will draw heavily from the papers published by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)] and by [http://adsabs.harvard.edu/abs/1964ApJ...139..664C S. Chandrasekhar (1964)], as well as from pp. 458-474 of the review by [http://adsabs.harvard.edu/abs/1958HDP....51..353L P. Ledoux &amp; Th. Walraven (1958)] in explaining how the ''variational principle'' can be used to identify the eigenvector of the fundamental mode of radial oscillation in spherically symmetric configurations.  In an associated "Ramblings" appendix, we provide [[User:Tohline/Appendix/Ramblings/LedouxVariationalPrinciple#Ledoux.27s_Variational_Principle_.28Supporting_Derivations.29|various derivations that support]] this chapter's relatively abbreviated presentation.


==Ledoux and Pekeris (1941)==
==Ledoux and Pekeris (1941)==
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</div>


we can write,
<span id="RewrittenLAWE">we can write,</span>
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 107: Line 107:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
(checked for n = 5) ==>
   </td>
   </td>
   <td align="center">
   <td align="center">
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Assuming that <math>~\Gamma_1</math> is uniform throughout the configuration, this last expression is the same as equation (3) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], while the next-to-last expression is identical to equation (58.1) of [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)].
Assuming that <math>~\Gamma_1</math> is uniform throughout the configuration, this last expression is the same as equation (3) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], while the next-to-last expression is identical to equation (58.1) of [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)].


==Review by Ledoux and Walraven (1958)==
==Stability Based on Variational Principle==
Here we are especially interested in understanding the origin of equation (59.10) of [http://adsabs.harvard.edu/abs/1958HDP....51..353L P. Ledoux &amp; Th. Walraven (1958)], which appears in &sect;59 (pp. 464 - 466) of their ''Handbuch der Physik'' article.


From our [[User:Tohline/SSC/Perturbations#Summary_Set_of_Nonlinear_Governing_Relations|accompanying summary of the set of nonlinear governing relations]], we highlight the  
Here we derive the Lagrangian directly from the governing LAWE.  We begin with the next-to-last derived form of the LAWE that [[#RewrittenLAWE|appears above]] in our review of the paper by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)] and, following the guidance provided at the top of p. 666 of [http://adsabs.harvard.edu/abs/1964ApJ...139..664C S. Chandrasekhar (1964, ApJ, 139, 664)], we multiply the LAWE through by the fractional displacement, <math>~\xi</math>.  This gives, what we will henceforth refer to as, the,


<div align="center">
<div align="center" id="FoundationalVariationalRelation">
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<table border="0" cellpadding="5" align="center">
<math>\frac{d^2 r}{dt^2} = - 4\pi r^2 \frac{dP}{dm} - \frac{Gm}{r^2} </math><br />
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Foundational Variational Relation</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~\sigma^2 \rho r^4 \xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr]
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
</div>


Repeating a result from our [[User:Tohline/SSC/Perturbations#Euler_.2B_Poisson_Equations|separate derivation]], linearization of the two terms on the righthand side of this equation gives,
===Chandrasekhar's Approach===
 
Next, in an effort to adopt the notation used by [http://adsabs.harvard.edu/abs/1964ApJ...139..664C Chandrasekhar (1964)], we make the substitution, <math>~\xi \rightarrow \psi/r^3</math>, and regroup terms to obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">


<table align="center" border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\frac{\sigma^2 \rho \psi^2}{r^2}</math>
r^2 \frac{dP}{dm}
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d}{dr} \biggl( \frac{\psi}{r^3} \biggr) \biggr]
- (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
\rightarrow
  </td>
  <td align="left">
<math>~
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr}  -3 \Gamma_1 P \psi ~\biggr]
- (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
r_0^2 \biggl[1 + x~ e^{i\omega t} \biggr]^2 \biggl\{\frac{dP_0}{dm} \biggl[1 + p~ e^{i\omega t}  \biggr] + P_0~e^{i\omega t} \frac{dp}{dm}  \biggr\} \approx r_0^2 \frac{dP_0}{dm} \biggl[1 + (2x+p)~ e^{i\omega t} \biggr] + P_0 r_0^2~e^{i\omega t} \frac{dp}{dm}
(4-3\Gamma_1 ) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr}  \biggr]  
+ 3 \Gamma_1 \biggl( \frac{\psi^2}{r^3}\biggr) \frac{dP}{dr}
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
   <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
- \biggl\{
\frac{d}{dr}\biggl[ r \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}\biggr] -r\Gamma_1 P ~\frac{d\psi}{dr} \cdot \frac{d}{dr}\biggl( \frac{\psi}{r^3}\biggr)
\biggr\}
</math>
</math>
   </td>
   </td>
</tr>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
\frac{Gm}{r^2}
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
+ \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2
- \biggl[\frac{3\Gamma_1 P\psi}{r^3}\biggr]\frac{d\psi}{dr}
- \frac{d}{dr}\biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr]
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
\rightarrow
  </td>
  <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
+ \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2
- \frac{d}{dr}\biggl[ \frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr] \, .
</math>
</math>
   </td>
   </td>
</tr>
</table>
</div>
<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">
Let's check to see whether the terms in the RHS of this last expression sum to zero when we plug in the appropriate functions for the marginally unstable, n = 5 configuration.  In particular (replacing <math>~\xi</math> with <math>~x</math>, and setting <math>~r = a_5\xi</math>), we start with knowing,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
   <td align="left">
   <td align="left">
<math>
<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>; &nbsp; &nbsp; &nbsp; &nbsp;
\frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t} \biggr]^{-2} \approx \frac{Gm}{ r_0^2}  
  </td>
\biggl[1 -2 x~ e^{i\omega t} \biggr] \, .
  <td align="left">
<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
</tr>
 
<tr>
  <td align="left">
<math>~x = \biggl(\frac{3\cdot 5 - \xi^2}{3\cdot 5} \biggr)</math>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{3\cdot 5}</math>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
</tr>
 
<tr>
  <td align="left">
<math>~\psi = a_5^3 \xi^3 x</math>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{d\psi}{d\xi} = a_5^3 \biggl[ 3\xi^2 x + \xi^3 \biggl(\frac{dx}{d\xi}\biggr)\biggr] = \frac{a_5^3 \xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>


Adopting the terminology of [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)], the "variation" of each of these terms is obtained by subtracting off the leading order pieces &#8212; which presumably cancel in equilibrium.  In particular, drawing a parallel with their equation (59.1), we can write,
----
 
Then,
<div align="center">
<div align="center">
<table align="center" border="0" cellpadding="5">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathrm{RHS}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[ \frac{a_5^6 \xi^6 x^2}{a_5^3 \xi^3} \biggr] \frac{P_c}{a_5} \cdot \frac{d\theta^6}{d\xi}
+ P_c \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{a_5^4 \xi^2} \biggl\{ \frac{d\psi}{d\xi} \biggr\}^2
- \frac{P_c}{a_5} \cdot \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 a_5^3 \xi^3 x}{a_5^3 \xi^2} \biggl( \frac{d\psi}{d\xi} \biggr) \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\mathrm{RHS}}{P_c a_5^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[ \xi^3 x^2 \biggr]  \frac{d\theta^6}{d\xi}
+ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{ \xi^2} \biggl\{ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \biggr\}^2
- \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 \xi^3 x}{ \xi^2} \biggl[ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr)\biggr] \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2^3\cdot 3 \xi^3 x^2 \biggl(\frac{3+\xi^2}{3}\biggr)^{-5 / 2} \biggl[- \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggr]
+ \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{\xi}{3}\biggr)^2  \biggl(\frac{3+\xi^2}{3}\biggr)^{-3} ( 3^2 - \xi^2 )^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr)\biggl(\frac{3+\xi^2}{3}\biggr)^{-3} \frac{\xi^3 x}{ 3} ( 3^2 - \xi^2) \biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2^3 \cdot 3^4 ~\xi^4 x^2 \biggl(\frac{1}{3+\xi^2}\biggr)^{4}
+ \biggl(\frac{2\cdot 3^2}{5}\biggr) \xi^2  \biggl[ \frac{( 3^2 - \xi^2 )^2}{(3+\xi^2)^3} \biggr]
- \biggl(\frac{2\cdot 3^3}{5}\biggr)\frac{d}{d\xi}\biggl\{ \xi^3 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr]  \biggr\}
</math>
  </td>
</tr>
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\Rightarrow ~~~ \frac{5 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 P_c a_5^2}</math>
~\delta \biggl( - \frac{Gm}{r^2}  \biggr)  
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2
+ \xi^2 (3+\xi^2)  ( 3^2 - \xi^2 )^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 3(3+\xi^2)^4 \biggl\{
\xi^3  \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr]  \frac{dx}{d\xi}
+ 3\xi^2 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr] 
+ \xi^3 x \biggl[ \frac{ -2\xi }{(3+\xi^2)^3} \biggr] 
-2\cdot 3 \xi^4 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^4} \biggr]  
\biggr\}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^2 (3+\xi^2)  ( 3^2 - \xi^2 )^2
- 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2
- 3\xi^3  ( 3^2 - \xi^2) (3+\xi^2) \biggl[ - \frac{2\xi}{ 3\cdot 5} \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 3 \xi^2\biggl\{
3 ( 3^2 - \xi^2) (3+\xi^2) 
-2 \xi^2  (3+\xi^2)
-2\cdot 3 \xi^2 ( 3^2 - \xi^2) 
\biggr\} x
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{5^2 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 ~\xi^2 P_c a_5^2}</math>
</td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5 (3+\xi^2)  ( 3^2 - \xi^2 )^2
- 2^2~\xi^2 (15-\xi^2)^2
+ 2 \xi^2  ( 3^2 - \xi^2) (3+\xi^2)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[
2 \xi^2  (3+\xi^2)
+ 2\cdot 3 \xi^2 ( 3^2 - \xi^2) 
- 3 ( 3^2 - \xi^2) (3+\xi^2) 
\biggr] (15-\xi^2)
</math>
</math>
   </td>
   </td>
</tr>
<tr>
  <td align="right">
&nbsp;
</td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~\approx
  </td>
  <td align="left">
<math>~
(3\cdot 5 + 5\xi^2)  ( 3^4 - 2\cdot 3^2\xi^2 + \xi^4)
- 2^2~\xi^2 (3^2\cdot 5^2 - 2\cdot 3\cdot 5 \xi^2 + \xi^4)
+ 2 \xi^2  ( 3^3 + 2\cdot 3\xi^2 -\xi^4)
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{Gm}{ r_0^2}
+ \biggl[ 2\cdot 3 \xi^2 + 2\xi^4 + 2\cdot 3^3 \xi^2 - 2\cdot 3 \xi^4 - 3(3^3 + 2\cdot 3\xi^2 -\xi^4)
\biggl[2 x~ e^{i\omega t}  \biggr] \, .
\biggr] (15-\xi^2)
</math>
</math>
   </td>
   </td>
Line 212: Line 526:
</div>
</div>


And, drawing a parallel with their equation (59.2), we have,
Coefficients of various powers of <math>~\xi</math>:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\xi^0:</math>
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~3^5\cdot 5 -3^5\cdot 5 = 0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\xi^2:</math>
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~-2\cdot 3^3\cdot 5 + 3^4\cdot 5 -2^2 \cdot 3^2 \cdot 5^2 +2\cdot 3^3 + 2\cdot 3^2\cdot 5 + 2\cdot 3^4\cdot 5 - 2\cdot 3^3\cdot 5 + 3^4</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~= 3^2\cdot 5[-2\cdot 3 + 3^2  + 2 + 2\cdot 3^2 - 2\cdot 3] + 3^2[ 2\cdot 3 + 3^2 - 2^2 \cdot 5^2  ] = 3^2\cdot 5[17 ] - 3^2[5\cdot 17  ] = 0</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\xi^4:</math>
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~3\cdot 5 -2\cdot 3^2\cdot 5 +2^3\cdot 3\cdot 5 + 2^2\cdot 3 + 2\cdot 3 \cdot 5 - 2\cdot 3^2\cdot 5 - 2\cdot 3 -2\cdot 3^3 + 2\cdot 3^2 + 3^2\cdot 5</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~= 3\cdot 5[1 -2\cdot 3 +2^3 + 2 - 2\cdot 3 + 3] + 2\cdot 3[2 - 1 - 3^2 + 3] = 2\cdot 3\cdot 5 - 2\cdot 3\cdot 5 = 0 </math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\xi^6:</math>
  </td>
  <td align="center">
&nbsp; &nbsp;
  </td>
  <td align="left">
<math>~5 - 2^2 -2 -2 +2\cdot 3 -3 = 0</math>
  </td>
</tr>
</table>
</div>
 
</td></tr>
</table>
 
<span id="ChandraEq49">Multiplying through by <math>~dr</math>, and integrating over the volume gives,</span>
 
<div align="center">
<div align="center">
<table align="center" border="0" cellpadding="5">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\int_0^R (\sigma^2 \rho \psi^2)\frac{dr}{r^2}</math>
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr) = \delta \biggl( - 4\pi r^2 \frac{dP}{dm} \biggr)
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R \biggl[ \Gamma_1 P \biggl(\frac{d\psi}{dr} \biggr)^2
+ \frac{4\psi^2}{r} \biggl( \frac{dP}{dr} \biggr) \biggr]\frac{dr}{r^2}
- \biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr]_0^R \, ,
</math>
</math>
  </td>
</tr>
</table>
</div>
which is identical to equation (49) of [http://adsabs.harvard.edu/abs/1964ApJ...139..664C Chandrasekhar (1964)], if the last term &#8212; the difference of the central and surface boundary conditions &#8212; is set to zero.
Note that if we shift from the variable, <math>~\psi</math>, back to the fractional displacement function, <math>~\xi</math>, the last term in this expression may be written as,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~\approx
  </td>
  <td align="left">
<math>~
\Gamma_1 P r \xi \frac{d}{dr} \biggl[r^3 \xi\biggr]
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
-4\pi r_0^2 \frac{dP_0}{dm} \biggl[(2x+p)~ e^{i\omega t}  \biggr] - 4\pi P_0 r_0^2~e^{i\omega t} \frac{dp}{dm}
\Gamma_1 P r \xi \biggl[3r^2 \xi + r^3 \frac{d\xi}{dr}\biggr]
</math>
</math>
   </td>
   </td>
Line 238: Line 668:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~=
  </td>
  <td align="left">
<math>~
\Gamma_1 P r^3 \xi^2 \biggl[3 + \frac{d\ln\xi}{d\ln r}\biggr] \, .
</math>
</math>
  </td>
</tr>
</table>
</div>
So, as is pointed out by [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)] in connection with their equation (57.31), setting this expression to zero at the surface of the configuration is equivalent to setting the variation of the pressure to zero at the surface. Quite generally, this can be accomplished by demanding that,
<div align="center" id="SufaceBC">
<math>~\frac{d\ln\xi}{d\ln r}\biggr|_\mathrm{surface} = -3 \, .</math>
</div>
(An [[User:Tohline/SSC/Perturbations#Boundary_Conditions|accompanying chapter]] provides a broader discussion of this and other astrophysically reasonable boundary conditions that are associated with solutions to the LAWE.)
===Ledoux &amp; Walraven Approach===
Returning to the above [[#FoundationalVariationalRelation|''Foundational Variational Relation'']], we can also write,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2 \rho r^4 \xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~
\frac{Gm}{r_0^2}\biggl[(2x)\biggr]
-\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr]  
-4\pi r_0^2 \frac{dP_0}{dm} \biggl[(p) \biggr]  
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)
- 4\pi P_0 r_0^2 \frac{dp}{dm} \biggr\} e^{i\omega t}
</math>
</math>
   </td>
   </td>
Line 256: Line 710:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~=
  </td>
  <td align="left">
<math>~
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)
- \frac{d}{dr}\biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr]
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \int_0^R\sigma^2 \rho r^4 \xi^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~
\biggl( \frac{2 Gm}{r_0^2}\biggr) x
\int_0^R r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 dr
-\frac{1}{\rho_0} \cdot \frac{d}{dr_0} \biggl[ P_0 p \biggr]
- \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) dr
\biggr\} e^{i\omega t} \, .
- \biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr]_0^R
</math>
</math>
   </td>
   </td>
Line 271: Line 739:
</div>
</div>


Now, if we combined the linearized continuity equation and the linearized (adiabatic form of the) first law of thermodynamics, as [[User:Tohline/SSC/Perturbations#Summary_Set_of_Linearized_Equations|derived elsewhere]], we can write,
If the last term (boundary conditions) is set to zero, then we may also write,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr
- \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \bigl( \frac{dP}{dr} \bigr) dr}{\int_0^R \rho r^4 \xi^2 dr} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that, if the radial profile of the pressure and the density is known throughout a spherically symmetric, equilibrium configuration, and if, furthermore, the eigenfunction, <math>~\xi(r)</math>, of a radial oscillation mode is specified precisely, then this expression will give the (square of the) ''eigenfrequency'' of that oscillation mode.


By using formal ''variational principle'' techniques to derive this same expression, [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)] are able to offer a broader interpretation, which is encapsulated by their equation (59.10), viz.,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~p = \gamma_\mathrm{g} d</math>
<math>~\sigma_0^2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 283: Line 771:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \gamma_\mathrm{g} \biggl[
<math>~\mathrm{min}~
3x + r_0 \frac{dx}{dr_0}  
\frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr
\biggr]
- \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \bigl( \frac{dP}{dr} \bigr) dr}{\int_0^R \rho r^4 \xi^2 dr} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
This means that, if the exact radial eigenfunction, <math>~\xi(r)</math>, is not known, various approximate eigenfunctions can be tried.  The trial eigenfunction that ''minimizes'' the righthand-side of this expression will give the (square of the) eigenfrequency of the ''fundamental'' mode of oscillation (subscript zero).  Furthermore, via an evaluation of this righthand-side expression, any reasonable trial eigenfunction &#8212; for example, <math>~\xi</math> = constant &#8212; can provide an ''upper limit'' to  <math>~\sigma_0^2</math>.
===Ledoux &amp; Pekeris Approach===
Here we follow the lead of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)].  Returning to the integral expression just derived in our discussion of the ''Ledoux &amp; Walraven approach'', and multiplying through by <math>~4\pi</math>, we have,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\int_0^R 4\pi \sigma^2 \rho r^4 \xi^2 dr</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 298: Line 795:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \frac{\gamma_\mathrm{g}}{r_0^2} \frac{d}{dr_0} \biggl( r_0^3 x \biggr) \, .
<math>~
\int_0^R 4\pi r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 dr
- \int_0^R (3\Gamma_1 - 4) 4\pi r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) dr
- \biggr[4\pi r^3 \Gamma_1 P\xi^2 \biggl(\frac{d\ln \xi}{d\ln r}\biggr) \biggr]_0^R \, .
</math>
</math>
   </td>
   </td>
Line 304: Line 804:
</table>
</table>
</div>
</div>
If we acknowledge that:
* at the center of the configuration, <math>~r^3 = 0</math>;
* [[#SurfaceBC|as above]], the boundary condition at the surface is <math>~P = P_e</math> while <math>~(d\ln \xi/d\ln r) = -3</math>;
* the differential mass element is, <math>~dm = 4\pi r^2 \rho dr</math> and the corresponding differential volume element is, <math>~dV = 4\pi r^2 dr</math>; and
* a statement of detailed force balance is, <math>~dP/dr = - Gm\rho/r^2</math>,
this integral relation becomes,


Hence,
<div align="center">
<div align="center">
<table align="center" border="0" cellpadding="5">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~ \sigma^2 \int_0^R r^2 \xi^2 dm</math>
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr)  
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Gamma_1 \int_0^R \biggl[ r \biggl(\frac{d\xi}{dr}\biggr)\biggr]^2 P dV
+ (3\Gamma_1 - 4) \int_0^R  \xi^2 \biggl( \frac{Gm}{r} \biggr) dm
- \biggr[\Gamma_1 \xi_\mathrm{surface}^2 (3P_e V) \biggl(-3\biggr) \biggr] \, .
</math>
</math>
  </td>
</tr>
</table>
</div>
Now, as we have [[User:Tohline/SphericallySymmetricConfigurations/Virial#Wgrav|discussed separately]] &#8212; see, also, p. 64, Equation (12) of [<b>[[User:Tohline/Appendix/References#C67|<font color="red">C67</font>]]</b>] &#8212; the gravitational potential energy of the unperturbed configuration is given by the integral,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~W_\mathrm{grav}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm  \, ;</math>
  </td>
</tr>
</table>
</div>
for adiabatic systems, the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Reservoir|internal energy]] is,
<div align="center">
<math>
U_\mathrm{int}
=  \frac{1}{(\Gamma_1-1)} \int_0^R  P_0 dV
\, ;</math>
</div>
and &#8212; see the text at the top of p. 126 of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)] &#8212; the moment of inertia of the configuration about its center is,
<div align="center">
<math>
<math>
~\approx
I =  \int_0^M r_0^2 dm
</math>
\, .</math>
</div>
(Note that, defined in this way, <math>~I</math> is the same as [[User:Tohline/VE#Standard_Presentation_.5Bthe_Virial_of_Clausius_.281870.29.5D|what we have referred to elsewhere]] as the ''scalar moment of inertia'', which is obtained by taking the trace of the [[User:Tohline/VE#MOItensor|moment of inertia tensor]], <math>~I_{ij}</math>.)
<span id="GoverningIntegral">After inserting these expressions, we have what will henceforth be referred to as the,</span>
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Variational Principle's Governing Integral Relation</b></font></td>
</tr>
<tr>
  <td align="right">
<math>~ \sigma^2 \int_0^R \xi^2 dI</math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{
<math>~
\biggl( \frac{2 Gm}{r_0^2}\biggr) x
\Gamma_1 (\Gamma_1 - 1) \int_0^R \xi^2 \biggl[ \frac{d\ln\xi}{d\ln r}\biggr]^2 dU_\mathrm{int}
+\frac{1}{\rho_0} \cdot \frac{d}{dr_0}  \biggl[ \frac{\gamma_\mathrm{g} P_0}{r_0^2} \frac{d}{dr_0} \biggl( r_0^3 x \biggr) \biggr]
- (3\Gamma_1 - 4) \int_0^R  \xi^2 dW_\mathrm{grav}
\biggr\} e^{i\omega t} \, .
+ 3^2 \Gamma_1  P_e V \xi_\mathrm{surface}^2 \, .
</math>
</math>
   </td>
   </td>
Line 330: Line 889:
</div>
</div>


So, given that a mapping from our notation to that used by Ledoux &amp; Walraven (1958) requires <math>~xe^{i\omega t} \rightarrow \zeta/r_0</math>, I understand the origins of their equations (59.1) and (59.2).  But I do not yet understand how &hellip; <font color="darkgreen">"Accordingly, the acting forces per unit volume can be considered as deriving from a potential density"</font>
==Free-Energy Analysis==
 
<span id="Homologous">If we assume</span> the simplest approximation for the fundamental-mode eigenfunction, namely, <math>~\xi = \xi_0</math> = constant &#8212; that is, homologous expansion/contraction &#8212; then this last integral expression gives,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 336: Line 898:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_0 \mathcal{V}</math>
<math>~ \sigma^2 I</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 343: Line 905:
   <td align="left">
   <td align="left">
<math>~
<math>~
- \biggl( \frac{2Gm}{r_0^3}\biggr) \rho_0 \zeta^2 + \frac{\gamma_\mathrm{g} P_0}{2} \biggl[ \frac{1}{r_0^2} \frac{\partial(r_0^2 \zeta)}{\partial r_0} \biggr]^2 \, .
(4 - 3\Gamma_1) W_\mathrm{grav}
+ 3^2 \Gamma_1  P_e V  \, .
</math>
</math>
   </td>
   </td>
Line 349: Line 912:
</table>
</table>
</div>
</div>
It is clear that, once I understand the origin of this expression for the potential density, I will understand how the "Lagrangian density" as defined by their equation (47.8), viz.,
 
Contrast this result with the following free-energy analysis:
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\mathfrak{G}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV \, ,</math>
  </td>
</tr>
</table>
</div>
where, in terms of the configuration's (generally non-equilibrium) dimensionless radius, <math>~\chi \equiv R/R_0</math>,
<div align="center">
<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>~-a\chi^{-1}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~U_\mathrm{int}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~b\chi^{3-3\Gamma_1}</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~V</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{4\pi}{3} \chi^3 \, .</math>
  </td>
</tr>
</table>
</div>
Then,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial \mathfrak{G}}{\partial \chi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~+a \chi^{-2} + 3(1-\Gamma_1) b \chi^{2-3\Gamma_1} + 4\pi P_e \chi^{2} </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\chi^{-1} \biggl[- W_\mathrm{grav} + 3(1-\Gamma_1) U_\mathrm{int} + 3 P_e V \biggr] \, ,</math>
  </td>
</tr>
</table>
</div>
and,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2a \chi^{-3} + 3(1-\Gamma_1)(2-3\Gamma_1) b \chi^{1-3\Gamma_1} + 8\pi P_e \chi </math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\chi^{-2} \biggl[ 2W_\mathrm{grav} + 3(1-\Gamma_1)(2-3\Gamma_1) U_\mathrm{int}+ 6 P_e V \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
The equilibrium condition occurs when <math>~\partial \mathfrak{G}/\partial \chi = 0</math>, that is, when,
 
<div align="center">
<div align="center">
<math>~\mathcal{L} = \rho_0 [\mathcal{K} - \mathcal{V}] \, ,</math>
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~3(1-\Gamma_1) U_\mathrm{int}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~W_\mathrm{grav} - 3 P_e V \, ,</math>
  </td>
</tr>
</table>
</div>
</div>
becomes (see their equation 59.5),
in which case,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 359: Line 1,055:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{L}</math>
<math>~\chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 365: Line 1,061:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\rho_0}{2} {\dot\zeta}^2
<math>~2W_\mathrm{grav} + (2-3\Gamma_1) (W_\mathrm{grav} - 3P_eV) + 6 P_e V </math>
+ \biggl( \frac{2Gm}{r_0^3}\biggr) \rho_0 \zeta^2 - \frac{\gamma_\mathrm{g} P_0}{2} \biggl[ \frac{1}{r_0^2} \frac{\partial(r_0^2 \zeta)}{\partial r_0} \biggr]^2 \, .
  </td>
</math>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(4-3\Gamma_1)W_\mathrm{grav} + 3^2 \Gamma_1 P_e V \, .</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
<span id="LDefinition">Noting that,</span> <math>~\dot\zeta = i\omega r_0 x e^{i\omega t}</math>, this in turn gives,
 
Fantastic!  The righthand-side of this "free-energy-based" expression exactly matches the righthand-side of the [[#Homologous|above expression]] that has been derived from the variational principle, assuming homologous expansion/contraction (''i.e.,'' <math>~\xi</math> = constant).  In this case, we can make the direct association,
<div align="center">
<math>~\sigma^2 I = \chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2} \, .</math>
</div>
This also make sense in that the equilibrium configuration should be stable if <math>~\tfrac{\partial^2 \mathfrak{G}}{\partial \chi^2} > 0</math> &#8212; in which case, <math>~\sigma^2</math> is positive; whereas the equilibrium configuration should be ''unstable'' if <math>~\tfrac{\partial^2 \mathfrak{G}}{\partial \chi^2} < 0</math> &#8212; in which case, <math>~\sigma^2</math> is negative.
 
=Related, Exploratory Ideas=
<!-- OMIT
We can rewrite the [[#GoverningIntegral|Variational Principle's Governing Integral Relation]] as,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 378: Line 1,094:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~L \equiv \int_0^R 4\pi r_0^2 \mathcal{L} dr_0</math>
<math>~ 3^2 \Gamma_1  P_e V \xi_\mathrm{surface}^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 384: Line 1,100:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 4\pi \int_0^R \biggl\{ \frac{\rho_0}{2} {\dot\zeta}^2
<math>~
+ \biggl( \frac{2Gm}{r_0^3}\biggr) \rho_0 \zeta^2 - \frac{\gamma_\mathrm{g} P_0}{2} \biggl[ \frac{1}{r_0^2} \frac{\partial(r_0^2 \zeta)}{\partial r_0} \biggr]^2 \biggr\}r_0^2 dr_0
\int_0^R \xi^2 \biggl\{ \sigma^2 dI
+ (3\Gamma_1 - 4) dW_\mathrm{grav}
- \Gamma_1 (\Gamma_1 - 1) \biggl[ \frac{d\ln\xi}{d\ln r}\biggr]^2 dU_\mathrm{int}
\biggr\}
</math>
</math>
   </td>
   </td>
Line 398: Line 1,117:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 4\pi e^{2i\omega t} \int_0^R \biggl\{ \frac{\rho_0}{2} (i \omega r_0 x)^2
<math>~
+ \biggl( \frac{2Gm}{r_0^3}\biggr) \rho_0 (r_0 x)^2 - \frac{\gamma_\mathrm{g} P_0}{2} \biggl[ \frac{1}{r_0^2} \frac{\partial(r_0^3 x)}{\partial r_0} \biggr]^2 \biggr\}r_0^2 dr_0
\int_0^R \xi^2 \biggl\{ \sigma^2  dI
+ (3\Gamma_1 - 4) dW_\mathrm{grav}
- \Gamma_1 (\Gamma_1 - 1) \biggl[ \frac{d\ln\xi}{d\ln r}\biggr]^2 dU_\mathrm{int}
\biggr\}
</math>
  </td>
</tr>
</table>
</div>
END OMIT -->
==Logarithmic Derivatives==
 
Returning to our above discussion of the [[#Ledoux_.26_Walraven_Approach|Ledoux &amp; Walraven approach]], we appreciate that the ''differential'' relation governing the Variational Principle is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2 \rho r^4 \xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)
- \frac{d}{dr}\biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr]
</math>
</math>
   </td>
   </td>
Line 406: Line 1,152:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{d}{dr}\biggr[r^3 \Gamma_1 P\xi^2 \biggl(\frac{d\ln\xi}{d\ln r}\biggr) \biggr]
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 412: Line 1,159:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^2 x^2
<math>~
- 4 r_0 x^2 \frac{dP_0}{dr_0} - \frac{\gamma_\mathrm{g} P_0}{r_0^4} \biggl[ 3r_0^2 x + r_0^3 \frac{\partial x}{\partial r_0} \biggr]^2 \biggr\}r_0^2 dr_0
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)
- \sigma^2 \rho r^4 \xi^2
</math>
</math>
   </td>
   </td>
Line 426: Line 1,175:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^4 x^2
<math>~\xi^2 \biggl\{
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2   
r^2 \Gamma_1 P \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2   
- 4 r_0^3 x^2 \frac{dP_0}{dr_0}  
- (3\Gamma_1 - 4) r^3 \biggl( \frac{dP}{dr} \biggr)
- 3\gamma_\mathrm{g} P_0 \biggl[ 3r_0^2 x^2 + 2r_0^3 x \frac{\partial x}{\partial r_0} \biggr] \biggr\}dr_0
- \sigma^2 \rho r^4
\biggr\}
</math>
</math>
   </td>
   </td>
Line 442: Line 1,192:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^4 x^2
<math>~(r \xi)^2 P \biggl\{
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2   
\Gamma_1 \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2   
- 4 r_0^3 x^2 \frac{dP_0}{dr_0}
- (3\Gamma_1 - 4) \biggl( \frac{d\ln P}{d\ln r} \biggr)  
+r_0^3 x^2 \frac{d}{dr_0}\biggl(3\gamma_\mathrm{g}P_0\biggr)
- \frac{\sigma^2 \rho r^2}{P}  
-\frac{d}{dr_0}\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]
\biggr\}
\biggr\}dr_0
</math>
</math>
   </td>
   </td>
Line 460: Line 1,209:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \biggl\{  
<math>~\Gamma_1 (r \xi)^2 P \biggl\{
- \int_0^R \rho_0 \omega^2 r_0^4 x^2 dr_0
\biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2   
- \int_0^R  \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2  dr_0
-  \alpha \biggl( \frac{d\ln P}{d\ln r} \biggr)
+ \int_0^R r_0^3 x^2 \frac{d}{dr_0}\biggl[ (3\gamma_\mathrm{g} - 4)P_0\biggr]dr_0
- \frac{\sigma^2 \rho r^2}{\Gamma_1 P}  
-\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]_0^{R}
\biggr\} \, ,
\biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where,
<div align="center">
<math>~\alpha \equiv \biggl(3 - \frac{4}{\Gamma_1}\biggr) \, .</math>
</div>
</div>


The group of terms inside the curly braces, here, matches the group of terms inside the curly braces of Ledoux &amp; Walraven's equation (59.8) if we acknowledge that:
==Pressure-Truncated Polytropes==
# Our <math>~\omega^2</math> has the same meaning as, but the opposite sign of, their <math>~\sigma^2</math>.
 
# Our last term goes to zero because, <math>~r_0 = 0</math> at the center, while <math>~P_0 = 0</math> at the surface.
Let's start with the integral expression derived in our discussion of the [[#Ledoux_.26_Walraven_Approach|Ledoux &amp; Walraven approach]]; insert the variable, <math>~x</math>, in place of <math>~\xi</math>; and adopt the boundary conditions,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~r = 0</math> &nbsp; at the center,
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; along with &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~P = P_e~</math>, &nbsp; and &nbsp;<math>\frac{d\ln x}{d\ln r} = -3</math> &nbsp; at the surface (r = R).
  </td>
</tr>
</table>
</div>
That is, let's start with,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr
- \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr
+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2 \, .
</math>
  </td>
</tr>
</table>
</div>


==LP41 Again==
===Via Generalized Normalization===
After setting the last term to zero, this last expression can be rewritten as,
Next, we'll divide through by the [[User:Tohline/StabilityVariationalPrincipal#Energies_and_Structural_Form_Factors|normalization energy, as defined in an accompanying discussion]],  
<div align="center">
<math>~E_\mathrm{norm} = P_\mathrm{norm}R_\mathrm{norm}^3 = \frac{GM_\mathrm{tot}^2}{R_\mathrm{norm}} \, ,</math>
</div>
thereby making the integral relation dimensionless:


<div align="center">
<div align="center">
Line 484: Line 1,278:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2 e^{-2i\omega t} L </math>
<math>~
0
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 490: Line 1,286:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
- \omega^2 \int_0^R 4\pi\rho_0 r_0^4 x^2 dr_0
- \biggl[\frac{R_\mathrm{norm}}{GM_\mathrm{tot}^2} \biggr] \int_0^R \sigma^2 \rho r^4 x^2 dr
- \int_0^R  4\pi \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
+\biggl[\frac{1}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr
- (3\gamma_\mathrm{g} - 4)\int_0^R 4\pi\rho_0 r_0^3 x^2 \biggl( -\frac{1}{\rho_0}\frac{dP_0}{dr_0} \biggr)dr_0
- \biggl[\frac{1}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr
+ \biggl[\frac{P_e R^3 }{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] 3\Gamma_1 x_\mathrm{surface}^2
</math>
</math>
   </td>
   </td>
Line 506: Line 1,303:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
- \omega^2 \int_0^R x^2 r_0^2 dm
- \biggl[\frac{R_\mathrm{norm} R^5 \rho_c^2}{M_\mathrm{tot}^2} \biggr] \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R}
- \gamma_\mathrm{g} \int_0^R \biggl[ r_0 \biggl( \frac{\partial x}{\partial r_0}\biggr) \biggr]^2 P_0  dV
+ \biggl[\frac{P_c R^3}{P_\mathrm{norm}R_\mathrm{norm}^3 } \biggr] \int_0^R \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R}
- (3\gamma_\mathrm{g} - 4) \int_0^R  r_0 x^2 \biggl( \frac{Gm}{r_0^2} \biggr)dm
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \biggl[\frac{P_c R^3}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}
+ \biggl[\frac{P_e R^3 }{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] 3\Gamma_1 x_\mathrm{surface}^2
</math>
</math>
   </td>
   </td>
Line 516: Line 1,327:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~- \biggl[ \frac{2 e^{-2i\omega t}}{\int_0^R x^2 r_0^2 dm } \biggr] L </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 522: Line 1,333:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\omega^2  
- \biggl[ \frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_c}{\bar\rho} \biggr]^2 \chi^{-1} \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R}
+ \biggl\{ \frac{\gamma_\mathrm{g} \int_0^R \bigl[ r_0 \bigl( \frac{\partial x}{\partial r_0}\bigr) \bigr]^2 P_0  dV
+ \biggl[\frac{P_e }{P_\mathrm{norm}} \biggr] 3\Gamma_1 \chi^3 x_\mathrm{surface}^2  
+ (3\gamma_\mathrm{g} - 4) \int_0^R x^2 \bigl( \frac{Gm}{r_0} \bigr)dm}{\int_0^R x^2 r_0^2 dm}
</math>
\biggr\} \, .
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{P_c}{P_\mathrm{norm} } \biggr] \chi^3 \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2
-  (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R} \, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
where,
<div align="center">
<math>~\chi \equiv \frac{R}{R_\mathrm{norm}} \, .</math>
</div>
</div>


As is explained in detail in &sect;59 (pp. 464 - 465) of [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &amp; Walraven (1958)], and summarized in &sect;1 of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], the function inside the curly braces of this last expression will be minimized if the radially dependent displacement function, <math>~x</math>, is set equal to the eigenfunction of the fundamental mode of radial oscillation, <math>~x_0</math>; and, after evaluation, the minimum value of this expression will be equal to (the negative of) the square of the fundamental-mode oscillation frequency, <math>~\omega^2</math>.  This explicit mathematical statement is contained within equation (8) of Ledoux &amp; Pekeris and within equation (59.10) of Ledoux &amp; Walraven.
Note that we will ultimately insert the relation,
<div align="center">
<math>~\frac{P_c}{P_\mathrm{norm}} = \biggl[\biggl( \frac{3}{4\pi}\biggr) \frac{\rho_c}{\bar\rho} \biggl( \frac{M}{M_\mathrm{tot}}\biggr)\biggr]^{\Gamma_1} \biggl( \frac{R}{R_\mathrm{norm}}\biggr)^{-3\Gamma_1} \, .</math>
</div>
But, for the time being, dividing through by <math>~[P_c/P_\mathrm{norm}]\chi^3</math> gives,


Now, as we have [[User:Tohline/SphericallySymmetricConfigurations/Virial#Wgrav|discussed separately]] &#8212; see, also, p. 64, Equation (12) of [<b>[[User:Tohline/Appendix/References#C67|<font color="red">C67</font>]]</b>] &#8212; the gravitational potential energy of the unperturbed configuration is given by the integral,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}</math>
<math>~0</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 546: Line 1,378:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm \, ;</math>
<math>~
- \biggl[\frac{P_c}{P_\mathrm{norm} } \biggr]^{-1} \biggl[ \frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_c}{\bar\rho} \biggr]^2 \chi^{-4} \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[\frac{P_e }{P_c} \biggr] 3\Gamma_1  x_\mathrm{surface}^2
+ \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2
- (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
Line 552: Line 1,402:
</div>
</div>


for adiabatic systems, the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Reservoir|internal energy]] is,
Now let's focus on the second line of this integral energy relation, evaluating it for pressure-truncated polytropic configurations, in which case, <math>~\Gamma_1 \rightarrow (n+1)/n</math>,
<div align="center">
<div align="center">
<math>
<table border="0" cellpadding="5" align="center">
U_\mathrm{int}  
 
= \frac{1}{({\gamma_g}-1)} \int_0^R  P_0 dV
<tr>
\, ;</math>
  <td align="right">
<math>~\frac{r}{R} \rightarrow \frac{\xi}{\tilde\xi}</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{P}{P_c} \rightarrow \theta^{n+1} \, .
</math>
  </td>
</tr>
</table>
</div>
</div>
and &#8212; see the text at the top of p. 126 of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)] &#8212; the moment of inertia of the configuration about its center is,
We have,
<div align="center">
<div align="center">
<math>
<table border="0" cellpadding="5" align="center">
I =  \int_0^M r_0^2 dm
 
  \, .</math>
<tr>
  <td align="right">
Second line of relation
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{P_e }{P_c} \biggr] 3\Gamma_1  x_\mathrm{surface}^2
+ \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2
-  (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
   <td align="left">
<math>~
\biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2
+ \int_0^{\tilde\xi} \biggl\{ \biggl( \frac{\xi}{\tilde\xi}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2 {\tilde\xi}^2
- \biggl(\frac{3-n}{n}\biggr) \biggl( \frac{\xi}{\tilde\xi} \biggr)^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr] \tilde\xi \biggr\}\frac{d\xi}{\tilde\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2
+ \frac{1}{n {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl\{ (n+1) \xi^4 \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2
-  (3-n) \xi^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]  \biggr\}d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2
+ \frac{1}{n {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl\{ (n+1) \xi^4 \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2
- (n+1) (3-n) \xi^3 x^2 \theta^n \theta^'  \biggr\}d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2
+ \frac{(n+1)}{n {\tilde\xi}^3}\int_0^{\tilde\xi}
\biggl(\frac{3}{2n}\biggr)^2\frac{\xi}{\theta^n} \biggl\{ \xi \theta \biggl[ \biggl( \frac{2n}{3}\biggr)\xi \theta^n \cdot \frac{dx}{d\xi}\biggr]^2
-  (3-n) \biggl[ \biggl( \frac{2n}{3}\biggr) \xi \theta^n x\biggr]^2 \theta^'  \biggr\}d\xi \, .
</math>
  </td>
</tr>
</table>
</div>
</div>
Hence, the function to be minimized may be written as,


Now, let's examine how these terms combine if we ''guess'' the [[User:Tohline/SSC/Stability/InstabilityOnsetOverview#Marginally_Unstable_Pressure-Truncated_Gas_Clouds|analytically defined eigenfunction that applies to marginally unstable, pressure-truncated polytropic configurations]], namely,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^'}{\xi \theta^{n} } \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{2n}{3}\biggr) \xi \theta^n x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[(n-1)\xi \theta^n  + (n-3) \theta^' \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[\frac{3(n-3)}{2n}\biggr] \biggl\{
\frac{\theta^{''}}{\xi \theta^{n}}
- \frac{\theta^'}{\xi^2 \theta^{n}}
- \frac{n(\theta^')^2}{\xi \theta^{(n+1)}}
\biggr\}
</math>
  </td>
</tr>


<tr>
<tr>
Line 575: Line 1,558:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
  </td>
  <td align="left">
<math>~- \biggl[\frac{3(n-3)}{2n}\biggr] \frac{1 }{\xi \theta^{n}} \biggl[
\theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta}
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{2n}{3}\biggr) \xi \theta^n\frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~(3-n)
\biggl\{ \frac{\gamma_\mathrm{g} (\gamma_\mathrm{g}-1) \int_0^R  \bigl[ r_0 \bigl( \frac{\partial x}{\partial r_0}\bigr) \bigr]^2 dU_\mathrm{int}
\biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr]
- (3\gamma_\mathrm{g} - 4) \int_0^R  x^2 dW_\mathrm{grav}}{\int_0^R x^2 dI}
\, .
\biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 587: Line 1,584:
</table>
</table>
</div>
</div>
This expression appears in equation (9) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)].
Hence,


==Chandrasekhar (1964)==
In a paper titled, ''A General Variational Principle Governing the Radial and the Non-Radial Oscillations of Gaseous Masses'', [http://adsabs.harvard.edu/abs/1964ApJ...139..664C S. Chandrasekhar (1964, ApJ, 139, 664)] independently derived the Ledoux-Pekeris Lagrangian.  Returning to the second line of our effort to simplify the [[#LDefinition|above definition of the Lagrangian]], and making the substitution, <math>~\psi \equiv r_0^3 x</math>, we have,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 596: Line 1,591:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2L </math>
Second line of relation
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 602: Line 1,597:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 4\pi e^{2i\omega t} \int_0^R \biggl\{ \rho_0 \biggl( \frac{i \omega \psi}{r_0^2} \biggr)^2
<math>~
+ \biggl( \frac{4Gm}{r_0^3}\biggr) \rho_0 \biggl( \frac{\psi}{r_0^2} \biggr)^2
{\tilde\theta}^{n+1} \biggl[ \frac{3( n+1)}{n} \biggr] \biggl\{ \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{ {\tilde\theta}^'}{\tilde\xi {\tilde\theta}^{n} } \biggr]  \biggr\}^2
- \gamma_\mathrm{g} P_0 \biggl[ \frac{1}{r_0^2} \frac{\partial(\psi)}{\partial r_0} \biggr]^2 \biggr\}r_0^2 dr_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3^2 (n+1)(3-n)}{2^2n^3 {\tilde\xi}^3}\int_0^{\tilde\xi}
\frac{\xi}{\theta^n} \biggl\{
\xi \theta (3-n)\biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr]^2
-  \biggl[(n-1)\xi \theta^n  + (n-3) \theta^' \biggr]^2 \theta^' 
\biggr\}d\xi
</math>
</math>
   </td>
   </td>
Line 617: Line 1,629:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 4\pi e^{2i\omega t} \int_0^R \biggl\{ -\omega^2 \rho_0 \biggl( \frac{\psi}{r_0} \biggr)^2
<math>~
- 4\biggl( \frac{dP_0}{dr_0}\biggr) \biggl( \frac{\psi^2}{r_0^3} \biggr)
\frac{1}{{\tilde\xi}^2 {\tilde\theta}^{n+1}} \biggl[ \frac{3^3( n+1)}{2^2n^3} \biggr] \biggl[(n-1) \tilde\xi {\tilde\theta}^{n+1} + (n-3) \tilde\theta {\tilde\theta}^' \biggr]^2
- \gamma_\mathrm{g} P_0 \biggl[ \frac{1}{r_0} \frac{\partial(\psi)}{\partial r_0} \biggr]^2 \biggr\}dr_0
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3^2 (n+1)(3-n)}{2^2n^3 {\tilde\xi}^3} \int_0^{\tilde\xi}
\frac{1}{\theta^{n+1}} \biggl\{
(3-n)\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]^2
\biggl[(n-1)\xi \theta^n  + (n-3) \theta^' \biggr]^2 \xi \theta \theta^' 
\biggr\}d\xi
</math>
</math>
   </td>
   </td>
Line 632: Line 1,661:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- 4\pi e^{2i\omega t} \int_0^R \biggl\{ \omega^2 \rho_0 \psi^2
<math>~
+ 4\biggl( \frac{dP_0}{dr_0}\biggr) \biggl( \frac{\psi^2}{r_0} \biggr)
\frac{1}{{\tilde\xi}^2 {\tilde\theta}^{n+1}} \biggl[ \frac{3^3( n+1)}{2^2n^3} \biggr] \biggl[(n-1) \tilde\xi {\tilde\theta}^{n+1} + (n-3) \tilde\theta {\tilde\theta}^' \biggr]^2
+ \gamma_\mathrm{g} P_0 \biggl[ \frac{\partial\psi}{\partial r_0} \biggr]^2 \biggr\} \frac{dr_0}{r_0^2} .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \frac{3^2 (n+1)(3-n)^2}{2^2n^3 {\tilde\xi}^3} \int_0^{\tilde\xi}
\frac{1}{\theta^{n+1}} \biggl\{  
\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]^2
+ \frac{1}{(n-3)}  \biggl[(n-1)\xi \theta^n  + (n-3) \theta^' \biggr]^2 \xi \theta \theta^' 
\biggr\}d\xi
</math>
</math>
   </td>
   </td>
Line 640: Line 1,686:
</table>
</table>
</div>
</div>
This integral expression matches the integral expression that appears in equation (49) of [http://adsabs.harvard.edu/abs/1964ApJ...139..664C Chandrasekhar (1964)], if we accept that our squared frequency, <math>~\omega^2</math>, has the opposite sign to Chandrasekhar's <math>~\sigma^2</math>.  Chandrasekhar acknowledged that, for radial modes of oscillation, his result was the same as that derived earlier by Ledoux and his collaborators.


=Examples=
Note that, in this derivation, we have inserted the expressions:
Returning to the last line of our [[#LDefinition|above definition of the Lagrangian]], that is,
<div align="center">
<math>~
\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr] =
\xi^2 \theta^{2(n+1)} + 6\xi \theta^{n+2}\theta^' + 2n\xi^2 \theta^{n+1} (\theta^')^2 + 6n\xi\theta (\theta^')^3 + n^2 \xi^2 (\theta^')^4
</math>
</div>
<div align="center">
<math>~
\frac{1}{(n-3)} \biggl[(n-1)\xi \theta^n  + (n-3) \theta^' \biggr]^2 \xi\theta (\theta^')=
\biggl[ \frac{(n-1)^2}{(n-3)}\biggr] \xi^3 \theta^{2n+1}(\theta^') + 2(n-1)\xi^2 \theta^{n+1} (\theta^' )^2 + (n-3) \xi\theta (\theta^')^3
</math>
</div>
 
===Directly to n = 5 Polytropic Configurations===
 


<div align="center">
<div align="center">
Line 650: Line 1,709:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~L </math>
<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr
- \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr
+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R \biggl(\frac{r}{R}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{P}{P_c}\biggr) \biggl[\frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R}
-  \int_0^R \biggl[3\biggl(\frac{n+1}{n}\biggr)  - 4\biggr] \biggl(\frac{r}{R}\biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}
+3\biggl(\frac{n+1}{n}\biggr) \biggl( \frac{P_e}{P_c}\biggr)  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \frac{6}{5} \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 \theta^6 \biggl[\frac{dx}{d(\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi}
-  \int_0^{\tilde\xi} \biggl( - \frac{2}{5}\biggr)  \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 x^2 \biggl[ \frac{d\theta^{6}}{d(\xi/\tilde\xi)} \biggr] \frac{d\xi}{\tilde\xi}
+\biggl(\frac{18}{5}\biggr) {\tilde\theta}^6  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{ {\tilde\xi}^3} \int_0^{\tilde\xi} \biggl( \frac{6}{5}\biggr) \xi^4 \theta^6 \biggl[\frac{dx}{d\xi}\biggr]^2 d\xi
+  \frac{1}{ {\tilde\xi}^3} \int_0^{\tilde\xi} \biggl(\frac{2}{5}\biggr)  \xi^3 x^2 \biggl[ \frac{d\theta^{6}}{d\xi} \biggr] d\xi
+\biggl(\frac{18}{5}\biggr) {\tilde\theta}^6  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{5 {\tilde\xi}^3 }{2R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} 3\xi^4 \theta^6 \biggl[  - \frac{2\xi}{15} \biggr]^2 d\xi
+  \int_0^{\tilde\xi} 6\xi^3 \biggl[\frac{15-\xi^2}{15}\biggr]^2 \theta^5\biggl[ \frac{d\theta}{d\xi} \biggr] d\xi
+9 {\tilde\xi}^3 {\tilde\theta}^6  \biggl[\frac{15- {\tilde\xi}^2}{15}\biggr]^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{ 2^2}{3\cdot 5^2 }  \biggr) \int_0^{\tilde\xi} \xi^6 \biggl( \frac{3}{3+\xi^2}\biggr)^3  d\xi
+  \biggl(\frac{ 2}{3\cdot 5^2 }  \biggr) \int_0^{\tilde\xi} \xi^3 \biggl[15-\xi^2\biggr]^2 \biggl( \frac{3}{3+\xi^2}\biggr)^{4} \biggl[- \frac{\xi}{3}\biggr] d\xi
+ \biggl( \frac{1}{5^2} \biggr) {\tilde\xi}^3 \biggl( \frac{3}{3+ {\tilde\xi}^2}\biggr)^3  \biggl[15- {\tilde\xi}^2\biggr]^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{ 2^2\cdot 3^2}{5^2 }  \biggr) \int_0^{\tilde\xi} \biggl[ \frac{\xi^6 }{(3+\xi^2)^3}\biggr]  d\xi
~~- ~~ \biggl(\frac{ 2\cdot 3^2}{5^2 }  \biggr) \int_0^{\tilde\xi} \biggl[ \frac{\xi^4 (15-\xi^2)^2}{(3+\xi^2)^4}\biggr] d\xi
~~ + ~~ \biggl( \frac{3^3}{5^2} \biggr)  \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{5^3 {\tilde\xi}^3 }{2\cdot 3^2R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \biggl[ \frac{4\xi^6(3+\xi^2)-2\xi^4 (15-\xi^2)^2}{(3+\xi^2)^4}\biggr] d\xi
~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \biggl\{ \frac{2\xi^4 [6\xi^2 + 2\xi^4 -15^2 + 30\xi^2 - \xi^4] }{(3+\xi^2)^4}\biggr\} d\xi
~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \biggl\{ \frac{2\xi^4 [\xi^4 + 36\xi^2  -15^2 ] }{(3+\xi^2)^4}\biggr\} d\xi
~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 656: Line 1,872:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \biggl\{  
<math>~
- \int_0^R \rho_0 \omega^2 r_0^4 x^2 dr_0
\biggl[ \frac{2\xi^5(\xi^2-15)}{(\xi^2+3)^3} \biggr]_0^{\tilde\xi}
- \int_0^\gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
+ \int_0^R r_0^3 x^2 \frac{d}{dr_0}\biggl[ (3\gamma_\mathrm{g} - 4)P_0\biggr]dr_0
</math>
-\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]_0^{R}
  </td>
\biggr\} \, ,
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{2{\tilde\xi}^5({\tilde\xi}^2-15)}{({\tilde\xi}^2+3)^3} \biggr]
~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] 
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2{\tilde\xi}^5({\tilde\xi}^2-15) + 3{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{({\tilde\xi}^2+3)^3}
=
\frac{5{\tilde\xi}^7 - 120{\tilde\xi}^5 + 3^3\cdot 5^2{\tilde\xi}^3 }{({\tilde\xi}^2+3)^3} \, ,
</math>
</math>
   </td>
   </td>
Line 667: Line 1,911:
</table>
</table>
</div>
</div>
let's attempt to evaluate the terms inside the curly braces for the case of pressure-truncated polytropic configurations because, as has been discussed separately, we have an analytic expression for the eigenvector of the fundamental-mode of radial oscillation.  Dividing through by <math>~P_c R_\mathrm{eq}^3</math> gives,
which equals zero if <math>~\tilde\xi = 3</math>.  <font size="+1" color="red"><b>Hooray!!</b></font>
 
===For All Polytropic Indexes===


====Generalized Governing Integral Relation====
Given that the derivation just completed works for the special case of n = 5, let's generalize it to all polytropic indexes
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 674: Line 1,922:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{L_{\{\}} }{P_c R_\mathrm{eq}^3}</math>
<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr
- \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr
+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{R^5 \rho_c}{R^3 P_c}\int_0^R \sigma^2 \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{r}{R}\biggr)^4 x^2 \frac{dr}{R}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R \biggl(\frac{r}{R}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{P}{P_c}\biggr) \biggl[\frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R}
-  \int_0^R \biggl[3\biggl(\frac{n+1}{n}\biggr)  - 4\biggr] \biggl(\frac{r}{R}\biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}
+3\biggl(\frac{n+1}{n}\biggr) \biggl( \frac{P_e}{P_c}\biggr)  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{R^2 \rho_c}{P_c} \int_0^{\tilde\xi} \sigma^2 \theta^n \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 x^2 \frac{d\xi}{\tilde\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \biggl(\frac{\xi}{{\tilde\xi}}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \theta^{n+1}  \biggl[\frac{dx}{d(\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi}
~+  \int_0^{\tilde\xi} \biggl(\frac{n-3}{n}\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d(\xi/\tilde\xi)} \biggr] \frac{d\xi}{\tilde\xi}
~+~3\biggl(\frac{n+1}{n}\biggr) {\tilde\theta}^{n+1}  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{n R^2\rho_c}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \sigma^2 \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^4 \theta^{n+1}  \biggl[\frac{dx}{d\xi}\biggr]^2 d\xi
~+  \int_0^{\tilde\xi} (n-3) \xi^3 \theta^n x^2 \biggl[ \frac{d\theta}{d\xi} \biggr] d\xi
~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 d\xi
~+  \int_0^{\tilde\xi} (n-3) \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] d\xi
~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 681: Line 2,009:
   <td align="left">
   <td align="left">
<math>~
<math>~
- \int_0^R \rho_0 \omega^2 r_0^4 x^2 dr_0
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2  
- \int_0^\gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2  dr_0
+ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3) \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi
+ \int_0^R r_0^3 x^2 \frac{d}{dr_0}\biggl[ (3\gamma_\mathrm{g} - 4)P_0\biggr]dr_0
-\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]_0^{R}
</math>
</math>
   </td>
   </td>
Line 691: Line 2,017:
</div>
</div>


where, we have set the pressure at the (truncated) surface to the value, <math>~P_0|_\mathrm{surface} = P_e</math>.
For additional clarification, let's rewrite the leading coefficient on the lefthand-side of this expression.
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G R_\mathrm{norm}^2}{P_\mathrm{norm}} \biggr]
\biggl(\frac{R}{R_\mathrm{norm}^2}\biggr) \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2
\biggl[ \frac{3M}{4\pi R^3}\biggr]^2
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr)
\biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G M_\mathrm{tot}^2}{P_\mathrm{norm}R_\mathrm{norm}^4} \biggr]
\biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2
\biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr)
\biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] 
\biggl(\frac{P_\mathrm{norm}}{P_e} \biggr)
\biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2
\biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2
\biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
</table>
</div>


We will adopt the normalizations used in the [[User:Tohline/SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler presentation]].  Specifically,
Now, from an [[User:Tohline/StabilityVariationalPrincipal#Test_Virial_Equilibrium_Condition|accompanying discussion]], we know that, in equilibrium,
<div align="center">
<div align="center">
<table border="0" cellpadding="3">
<table border="0" cellpadding="3">
Line 700: Line 2,093:
   <td align="right">
   <td align="right">
<math>
<math>
~M
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
</math>
   </td>
   </td>
Line 707: Line 2,100:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
\biggl[(n+1)^{-n} ( 4\pi )\biggr]^{1/(n-3)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)/(n-3)}
\tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)}
\, ,
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
<math>
M_\mathrm{SWS} \biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl\{ \theta_n^{(n-3)/2} \xi^2
~\frac{P_e}{P_\mathrm{norm}}  
\biggl| \frac{d\theta_n}{d\xi} \biggr| \biggr\}_{\tilde\xi}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)/(n-3)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)/(n-3)}
\tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)}  
\, ,
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Hence,
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>
~R_\mathrm{eq}
~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)}
\tilde\theta_n^{(n+1)(n-3)}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)} \biggr\}^{1/(n-3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~\times
\biggl\{\biggl[(n+1)^{-n} ( 4\pi )\biggr] \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)}
\tilde\xi^{(n-3)} ( -\tilde\xi^2 \tilde\theta' )^{(1-n)} \biggr\}^{4/(n-3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)}
\biggl\{ (n+1)^{3(n+1)} ( 4\pi )^{(-n-1)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2n-2}
( -\tilde\xi^2 \tilde\theta' )^{2n+2} (n+1)^{-4n} ( 4\pi )^4 \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(4n-4)}
( -\tilde\xi^2 \tilde\theta' )^{(4-4n)}
\biggr\}^{1/(n-3)}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)}
\biggl\{ (n+1)^{(3-n)} ( 4\pi )^{(3-n)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2(n-3)}
( -\tilde\xi^2 \tilde\theta' )^{2(3-n)}
\biggr\}^{1/(n-3)}
</math>
</math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 724: Line 2,205:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
R_\mathrm{SWS} \biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl\{ \xi \theta_n^{(n-1)/2} \biggr\}_{\tilde\xi}
(n+1)^{-1} ( 4\pi )^{(-1)}  \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2} \tilde\xi^{4} \tilde\theta_n^{(n+1)}( -\tilde\xi^2 \tilde\theta' )^{-2} \, .
</math>
  </td>
</tr>
</table>
</div>
This means that, in equilibrium,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
LHS
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ \frac{n}{(n+1)} \biggr] 
\biggl\{ (n+1) ( 4\pi )  \tilde\xi^{-4} \tilde\theta_n^{-(n+1)}( -\tilde\xi^2 \tilde\theta' )^{2} \biggr\}
\biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2
\biggl(\frac{3}{4\pi}\biggr)^2
\biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr]
\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi \, .</math>
  </td>
</tr>
</table>
</div>
 
In summary, then, we have,
 
<div align="center" id="PolytropeRelation">
<table border="1" align="center" cellpadding="8">
<tr><td align="center">
 
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2
+ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3)  \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi \, .
</math>
  </td>
</tr>
</table>
 
</td></tr>
</table>
</div>
 
Perhaps this looks better if the terms are rearranged to give,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
3 {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2\theta^{n+1} x^2 \biggl\{ \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \frac{\xi^2}{\theta}
- \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3)  \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr] \biggr\} d\xi \, .
</math>
</math>
   </td>
   </td>
Line 731: Line 2,296:
</table>
</table>
</div>
</div>
where,
 
====Plug in Known Marginally Unstable Solution====
 
As has been summarized in an [[User:Tohline/SSC/Stability/InstabilityOnsetOverview#Marginally_Unstable_Pressure-Truncated_Gas_Clouds|accompanying discussion]], we have found that, for marginally unstable pressure-truncated polytropic configurations, the eigenvector associated with the fundamental mode of radial oscillation is prescribed analytically by the following eigenfrequency-eigenfunction pair:
<div align="center">
<div align="center">
<math>M_\mathrm{SWS} =
<table border="0" cellpadding="5" align="center">
\biggl( \frac{n+1}{nG} \biggr)^{3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \, ,</math>
<tr>
  <td align="right">
<math>~\sigma_c^2 = 0</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~x = \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>
</div>
</div>
This means that,
<div align="center">
<div align="center">
<math>
<table border="0" cellpadding="5" align="center">
R_\mathrm{SWS} = \biggl( \frac{n+1}{nG} \biggr)^{1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, .
 
<tr>
  <td align="right">
<math>~\biggl[ \frac{2n}{3(n-1)} \biggr] \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \frac{d}{d\xi}\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \biggl[
\frac{\theta^{''}}{\xi \theta^{n}}
- \frac{\theta^'}{\xi^2 \theta^{n}}
- \frac{n (\theta^')^2}{\xi \theta^{n+1}}
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{n-3}{n-1}\biggr) \biggl[
- \frac{1}{\xi \theta^{n}} \biggl( \theta^n + \frac{2\theta^'}{\xi} \biggr)
- \frac{\theta^'}{\xi^2 \theta^{n}}
- \frac{n (\theta^')^2}{\xi \theta^{n+1}}
\biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3-n}{n-1}\biggr) \biggl[
\frac{1}{\xi }
+ \frac{3\theta^'}{\xi^2 \theta^{n}}
+ \frac{n (\theta^')^2}{\xi \theta^{n+1}}
\biggr] \, .
</math>
</math>
  </td>
</tr>
</table>
</div>
</div>
In addition, from our [[User:Tohline/SSC/Structure/Polytropes#Lane-Emden_Equation|introductory discussion of polytropes]], we have,
Hence, also,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 747: Line 2,394:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_0</math>
<math>~\frac{d\ln x}{d\ln \xi} = \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{3-n}{n-1}\biggr) \biggl[1 
+ \frac{3\theta^'}{\xi \theta^{n}}
+ \frac{n (\theta^')^2}{\theta^{n+1}}
\biggr] \biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~\equiv</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\rho_c \theta^n </math>
<math>~
\biggl(\frac{3-n}{n-1}\biggr) \biggl(\frac{n-3}{n-1}\biggr)^{-1} \biggl[1 
+ \frac{3\theta^'}{\xi \theta^{n}}
+ \frac{n (\theta^')^2}{\theta^{n+1}}
\biggr] \biggl[\biggl(\frac{n-1}{n-3}\biggr)  + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1}
</math>
   </td>
   </td>
</tr>
</tr>
Line 759: Line 2,428:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ P_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 765: Line 2,434:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~K_n \biggl[ \rho_c \theta^n \biggr]^{(n+1)/n} </math>
<math>~
- \biggl[
+ \frac{3\theta^'}{\xi \theta^{n}}
+ \frac{n (\theta^')^2}{\theta^{n+1}}
\biggr] \biggl[\biggl(\frac{n-1}{n-3}\biggr) + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1} \, .
</math>
   </td>
   </td>
</tr>
</tr>
Line 771: Line 2,445:
</div>
</div>


and,
Rather, let's try:
<div align="center">
<div align="center">
<math>
<table border="0" cellpadding="5" align="center">
r_0 \equiv a_\mathrm{n} \xi ,
<tr>
  <td align="right">
<math>~
\xi^2 x^2 \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3)  \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 \xi^2 \biggl( \frac{\xi}{x}\cdot \frac{dx}{d\xi}\biggr)^2 + (n-3) x^2 \xi^2 \biggl( \frac{\xi}{\theta} \cdot \frac{d\theta}{d\xi} \biggr)
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \biggl\{ \frac{dx}{d\xi}\biggr\}^2 + (n-3) \biggl[ \frac{\xi^3 \theta^'}{\theta} \biggr] x^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\xi^4 \biggl\{ \frac{3(n-1)}{2n}\biggl(\frac{3-n}{n-1}\biggr) \biggl[
\frac{1}{\xi }
+ \frac{3\theta^'}{\xi^2 \theta^{n}}
+ \frac{n (\theta^')^2}{\xi \theta^{n+1}}
\biggr]\biggr\}^2 + (n-3) \biggl[ \frac{\xi^3 \theta^'}{\theta} \biggr]
\biggl\{ \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr] \biggr\}^2
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\xi^2 (n-3) \biggl[ \frac{3}{2n} \biggr]^2\biggl\{
(n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2
+\xi \biggl( \frac{ \theta^'}{\theta} \biggr)
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2
\biggr\}
</math>
</math>
  </td>
</tr>
</table>
</div>
</div>
where,
Hence, after setting <math>~\sigma^2 = 0</math>, the [[#PolytropeRelation|above rearranged integral relation]] becomes,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
<math>~
a^2_\mathrm{n} = \frac{(n+1)K_n}{4\pi G} \cdot \rho_c^{(1-n)/n}  \, .
- \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 \biggr]
</math>
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^2 \theta^{n+1} \biggl\{
(n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2
+\xi \biggl( \frac{ \theta^'}{\theta} \biggr)
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2
\biggr\} d\xi
</math>
  </td>
</tr>
</table>
</div>
<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">
Let's check to see whether the terms in this last expression balance out when we plug in the functions that are appropriate for the marginally unstable, n = 5 configuration, namely,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="left">
<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>,
  </td>
  <td align="center">
&nbsp; &nbsp;&nbsp; and &nbsp;&nbsp;&nbsp;
  </td>
  <td align="left">
<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>.
  </td>
</tr>
</table>
</div>
</div>
Also, from our [[User:Tohline/SphericallySymmetricConfigurations/Virial#Structural_Form_Factors|definition of the structural form factors]] and our evaluation of them in the context of [[User:Tohline/SSC/Virial/PolytropesEmbedded/SecondEffortAgain#Structural_Form_Factors|pressure-truncated configurations]], we know that,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 789: Line 2,564:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_c</math>
RHS Term 1
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(n-3) \int_0^{\tilde\xi} \xi^2 \theta^{n+1}
\biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \int_0^{\tilde\xi} \xi^2 \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2} \biggr]^{6} \biggl\{
1 - \biggl[ \xi \biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggr]\frac{1}{\xi }\biggl(\frac{3+\xi^2}{3}\biggr)^{5 / 2}
+5  \biggl[ \frac{\xi^2}{3^2}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3} \biggr] \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{3} \biggr]
\biggr\}^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2 \int_0^{\tilde\xi} \frac{3^3 \xi^2}{(3+\xi^2)^3}  \biggl\{
1 -  \biggl(\frac{3+\xi^2}{3}\biggr)
+ \frac{5\xi^2}{3^2} 
\biggr\}^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3} \int_0^{\tilde\xi} \frac{\xi^6 ~d\xi}{(3+\xi^2)^3}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3}  \biggl[
\frac{27\xi}{8(3+\xi^2)} - \frac{9\xi}{4(3+\xi^2)^2} + \xi - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr)
\biggr]_0^{3} = \frac{2^3}{3}  \biggl[ \frac{3^5}{2^6} - \frac{3^{1 / 2}\cdot 5\pi}{2^3} \biggr] \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
RHS Term 2
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^{\tilde\xi} \xi^3 \theta^{n} \theta^'
\biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \int_0^{\tilde\xi} \xi^3 \biggl(\frac{3+\xi^2}{3}\biggr)^{-5 / 2} \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}
\biggl\{ 4 - 2\frac{1}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}\biggl(\frac{3+\xi^2}{3}\biggr)^{5/2}
\biggr\}^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{3}\int_0^{\tilde\xi} \biggl(\frac{3\xi}{3+\xi^2}\biggr)^{4}
\biggl\{ 4 - \frac{2}{3}\biggl(\frac{3+\xi^2}{3}\biggr)
\biggr\}^2 d\xi
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 795: Line 2,696:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\bar\rho}{ {\tilde{f}}_M}</math>
<math>~
- \frac{2^3}{3}\int_0^{\tilde\xi} \frac{1}{2} \biggl(\frac{\xi}{3+\xi^2}\biggr)^{4}  
\biggl\{ 15-\xi^2
\biggr\}^2 d\xi
</math>
   </td>
   </td>
</tr>
</tr>
Line 807: Line 2,712:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl[\frac{3M}{4\pi R_\mathrm{eq}^3}\biggr] \biggl[ \frac{3}{\xi} \biggl(- \frac{d\theta}{d\xi}\biggr) \biggr]_{\tilde\xi}^{-1}</math>
<math>~
-\frac{2^3}{3}  \biggl[  
\frac{123\xi}{8(3+\xi^2)} - \frac{243\xi}{4(3+\xi^2)^2} + \frac{162\xi}{2(3+\xi^2)^3} + \frac{\xi}{2} - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr)
\biggr]_0^{3}  
</math>
   </td>
   </td>
</tr>
</tr>
Line 819: Line 2,728:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{4\pi} \biggl[\frac{M/M_\mathrm{SWS}}{(R/R_\mathrm{SWS})^3}\biggr] \biggl[ \frac{M_\mathrm{SWS}}{ R_\mathrm{SWS}^3} \biggr]\biggl(- \frac{d\theta}{d\xi}\biggr)_{\tilde\xi}^{-1} \tilde\xi</math>
<math>~
\frac{2^3}{3} \cdot \frac{5}{2^5}  \biggl[ 2^2\cdot 3^{1 / 2} \pi - 3^3\biggr] = -\frac{2^3}{3} \biggl[ \frac{3^3\cdot 5}{2^5} - \frac{3^{1 / 2} \cdot 5\pi}{2^3} \bigg] \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>~\Rightarrow ~~~</math> RHS Total
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2^3}{3}  \biggl[ \frac{3^5}{2^6} - \frac{3^3\cdot 5}{2^5} \biggr]
= \frac{3^2}{2^3}  \biggl[ 3^2 - 2\cdot 5  \biggr] = - \frac{3^2}{2^3}  \, .
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
LHS
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 \biggr]
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{2^2 5^2}{2\cdot 3} \biggl[ 3^3 \biggl(\frac{3}{3+3^2}\biggr)^{3} \frac{2^2}{5^2} \biggr]
=
- \frac{2^3 }{3} \biggl[\biggl(\frac{3}{2^2}\biggr)^{3} \biggr]
=
- \frac{3^2 }{2^3} \, .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
Hence, the LHS = RHS. &nbsp; <font size="+1" color="red"><b>Hooray!</b></font>
</td></tr>
</table>


=See Also=
=See Also=
* [[User:Tohline/Appendix/Ramblings/LedouxVariationalPrinciple#Ledoux.27s_Variational_Principle_.28Supporting_Derivations.29|Derivations that support this chapter's discussion of the Ledoux Variational Principle]]
* [https://ui.adsabs.harvard.edu/abs/1967MNRAS.136..293L/abstract D. Lynden-Bell &amp; J. P. Ostriker (1967)], MNRAS, 136, 293:  ''On the stability of differentially rotating bodies''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%">&nbsp;</td><td align="left">
<font color="green">A variational principle of great power is derived.  It is naturally adapted for computers, and may be used to determine the stability of any fluid flow including those in differentially-rotating, self-gravitating stars and galaxies.  The method also provides a powerful theoretical tool for studying general properties of eigenfunctions, and the relationships between secular and ordinary stability.  In particular we prove the anti-sprial theorem indicating that no stable (or steady( mode can have a spiral structure</font>.
</td></tr></table>
* [https://ui.adsabs.harvard.edu/abs/1972ApJS...24..319S/abstract B. F. Schutz, Jr. (1972)], ApJSuppl., 24, 319:  ''Linear Pulsations and Stability of Differentially Rotating Stellar Models. I. Newtonian Analysis''
<table border="0" align="center" width="100%" cellpadding="1"><tr>
<td align="center" width="5%">&nbsp;</td><td align="left">
<font color="green">A systematic method is presented for deriving the Lagrangian governing the evolution of small perturbations of arbitrary flows of a self-gravitating perfect fluid. The method is applied to a differentially rotating stellar model; the result is a Lagrangian equivalent to that of [https://ui.adsabs.harvard.edu/abs/1967MNRAS.136..293L/abstract D. Lynden-Bell &amp; J. P. Ostriker (1967)]. A sufficient condition for stability of rotating stars, derived from this Lagrangian, is simplified greatly by using as trial functions not the three components of the Lagrangian displacement vector, but three scalar functions &hellip; This change of variables saves one from integrating twice over the star to find the effect of the perturbed gravitational field.
&hellip; we examine the special cases of (i) axially symmetric perturbations of a rotating star (as treated by [https://ui.adsabs.harvard.edu/abs/1968ApJ...152..267C/abstract S. Chandrasekhar &amp; N. R. Lebovitz 1968]); and (ii) perturbations of a nonrotating star (as treated by [https://ui.adsabs.harvard.edu/abs/1964ApJ...140.1517C/abstract Chandrasekhar and Lebovitz 1964)].  We find that the stability criteria for those cases can also be simplified &hellip;</font>
</td></tr></table>


{{LSU_HBook_footer}}
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Latest revision as of 00:32, 29 June 2019


Ledoux's Variational Principle

Whitworth's (1981) Isothermal Free-Energy Surface
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All of the discussion in this chapter will build upon our derivation elsewhere of the,

LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation

LSU Key.png

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 </math>

We will draw heavily from the papers published by Ledoux & Pekeris (1941) and by S. Chandrasekhar (1964), as well as from pp. 458-474 of the review by P. Ledoux & Th. Walraven (1958) in explaining how the variational principle can be used to identify the eigenvector of the fundamental mode of radial oscillation in spherically symmetric configurations. In an associated "Ramblings" appendix, we provide various derivations that support this chapter's relatively abbreviated presentation.

Ledoux and Pekeris (1941)

Historically, by the 1940s, the LAWE was a relatively familiar one to astrophysicists. For example, the opening paragraph of a 1941 paper by Ledoux & Pekeris (1941, ApJ, 94, 124), reads:

Paragraph extracted from P. Ledoux & C. L. Pekeris (1941)

"Radial Pulsations of Stars"

ApJ, vol. 94, pp. 124-135 © American Astronomical Society

Ledoux & Pekeris (1941, ApJ, 94, 124)

If we divide their equation (1) through by <math>~Xr = \Gamma_1 P r</math> and recognize that,

<math> \frac{dX}{dr} = \frac{dX}{dm}\frac{dm}{dr} = - \Gamma_1 g_0 \rho \, , </math>

we obtain,

<math> \frac{d^2\xi}{dr^2} + \biggl[ \frac{4}{r} - \frac{g_0 \rho}{P} \biggr] \frac{d\xi}{dr} +\frac{\rho}{\Gamma_1 P} \biggl[ \sigma^2 + (4 - 3\Gamma_1) \frac{g_0}{r} \biggr] \xi = 0 \, . </math>

Clearly, this 2nd-order, ordinary differential equation is the same as our derived LAWE, but with a more general definition of the adiabatic exponent that allows consideration of a situation where the total pressure is a sum of both gas and radiation pressure.

Multiplying this last equation through by <math>~\Gamma_1 P r^4</math>, and recognizing that,

<math>~(r^4 \Gamma_1 P)\frac{d^2\xi}{dr^2} </math>

<math>~=</math>

<math>~ \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] - \frac{d\xi}{dr} \cdot \frac{d}{dr} \biggl[ r^4 \Gamma_1 P\biggr] </math>

we can write,

<math>~0</math>

<math>~=</math>

<math>~ \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] - \frac{d\xi}{dr} \cdot \frac{d}{dr} \biggl[ r^4 \Gamma_1 P\biggr] + ( \Gamma_1 P r^4 ) \biggl[ \frac{4}{r} - \frac{g_0 \rho}{P} \biggr] \frac{d\xi}{dr} +\rho \biggl[ \sigma^2 r^4 + (4 - 3\Gamma_1) g_0 r^3\biggr] \xi </math>

 

<math>~=</math>

<math>~ \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] - \biggl[4r^3\Gamma_1 P + \Gamma_1 r^4 \frac{dP}{dr} \biggr] \frac{d\xi}{dr} + \biggl[ 4 r^3\Gamma_1 P + \Gamma_1 r^4 \frac{dP}{dr}\biggr] \frac{d\xi}{dr} +\biggl[ \sigma^2 \rho r^4 - (4 - 3\Gamma_1) r^3 \frac{dP}{dr} \biggr] \xi </math>

(checked for n = 5) ==>

<math>~=</math>

<math>~ \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] +\biggl[ \sigma^2 \rho r^4 + (3\Gamma_1 - 4) r^3 \frac{dP}{dr} \biggr] \xi </math>

 

<math>~=</math>

<math>~ \frac{d}{dr}\biggl[ \Gamma_1 P r^4 ~\frac{d\xi}{dr} \biggr] +\biggl[ \sigma^2 r^4 \rho + 4 Gm (r ) r \rho + 3\Gamma_1 r^3 \frac{dP}{dr} \biggr] \xi \, . </math>

Assuming that <math>~\Gamma_1</math> is uniform throughout the configuration, this last expression is the same as equation (3) of Ledoux & Pekeris (1941), while the next-to-last expression is identical to equation (58.1) of Ledoux & Walraven (1958).

Stability Based on Variational Principle

Here we derive the Lagrangian directly from the governing LAWE. We begin with the next-to-last derived form of the LAWE that appears above in our review of the paper by Ledoux & Pekeris (1941) and, following the guidance provided at the top of p. 666 of S. Chandrasekhar (1964, ApJ, 139, 664), we multiply the LAWE through by the fractional displacement, <math>~\xi</math>. This gives, what we will henceforth refer to as, the,

Foundational Variational Relation

<math>~\sigma^2 \rho r^4 \xi^2</math>

<math>~=</math>

<math>~ -\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] - (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) \, . </math>

Chandrasekhar's Approach

Next, in an effort to adopt the notation used by Chandrasekhar (1964), we make the substitution, <math>~\xi \rightarrow \psi/r^3</math>, and regroup terms to obtain,

<math>~\frac{\sigma^2 \rho \psi^2}{r^2}</math>

<math>~=</math>

<math>~ - \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d}{dr} \biggl( \frac{\psi}{r^3} \biggr) \biggr] - (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) </math>

 

<math>~=</math>

<math>~ - \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr} -3 \Gamma_1 P \psi ~\biggr] - (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) </math>

 

<math>~=</math>

<math>~ (4-3\Gamma_1 ) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) - \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr} \biggr] + 3 \Gamma_1 \biggl( \frac{\psi^2}{r^3}\biggr) \frac{dP}{dr} +3 \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr} </math>

 

<math>~=</math>

<math>~ 4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) +3 \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr} - \biggl\{ \frac{d}{dr}\biggl[ r \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}\biggr] -r\Gamma_1 P ~\frac{d\psi}{dr} \cdot \frac{d}{dr}\biggl( \frac{\psi}{r^3}\biggr) \biggr\} </math>

 

<math>~=</math>

<math>~ 4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) +3 \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr} + \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2 - \biggl[\frac{3\Gamma_1 P\psi}{r^3}\biggr]\frac{d\psi}{dr} - \frac{d}{dr}\biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr] </math>

 

<math>~=</math>

<math>~ 4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) + \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2 - \frac{d}{dr}\biggl[ \frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr] \, . </math>

Let's check to see whether the terms in the RHS of this last expression sum to zero when we plug in the appropriate functions for the marginally unstable, n = 5 configuration. In particular (replacing <math>~\xi</math> with <math>~x</math>, and setting <math>~r = a_5\xi</math>), we start with knowing,

<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>;        

<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>;        

<math>~x = \biggl(\frac{3\cdot 5 - \xi^2}{3\cdot 5} \biggr)</math>;        

<math>~\frac{dx}{d\xi} = -\frac{2\xi}{3\cdot 5}</math>;        

<math>~\psi = a_5^3 \xi^3 x</math>;        

<math>~ \frac{d\psi}{d\xi} = a_5^3 \biggl[ 3\xi^2 x + \xi^3 \biggl(\frac{dx}{d\xi}\biggr)\biggr] = \frac{a_5^3 \xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \, . </math>


Then,

<math>~\mathrm{RHS}</math>

<math>~=</math>

<math>~ 4\biggl[ \frac{a_5^6 \xi^6 x^2}{a_5^3 \xi^3} \biggr] \frac{P_c}{a_5} \cdot \frac{d\theta^6}{d\xi} + P_c \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{a_5^4 \xi^2} \biggl\{ \frac{d\psi}{d\xi} \biggr\}^2 - \frac{P_c}{a_5} \cdot \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 a_5^3 \xi^3 x}{a_5^3 \xi^2} \biggl( \frac{d\psi}{d\xi} \biggr) \biggr\} </math>

<math>~\Rightarrow ~~~ \frac{\mathrm{RHS}}{P_c a_5^2}</math>

<math>~=</math>

<math>~ 4\biggl[ \xi^3 x^2 \biggr] \frac{d\theta^6}{d\xi} + \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{ \xi^2} \biggl\{ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \biggr\}^2 - \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 \xi^3 x}{ \xi^2} \biggl[ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr)\biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ 2^3\cdot 3 \xi^3 x^2 \biggl(\frac{3+\xi^2}{3}\biggr)^{-5 / 2} \biggl[- \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggr] + \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{\xi}{3}\biggr)^2 \biggl(\frac{3+\xi^2}{3}\biggr)^{-3} ( 3^2 - \xi^2 )^2 </math>

 

 

<math>~ - \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr)\biggl(\frac{3+\xi^2}{3}\biggr)^{-3} \frac{\xi^3 x}{ 3} ( 3^2 - \xi^2) \biggr\} </math>

 

<math>~=</math>

<math>~ - 2^3 \cdot 3^4 ~\xi^4 x^2 \biggl(\frac{1}{3+\xi^2}\biggr)^{4} + \biggl(\frac{2\cdot 3^2}{5}\biggr) \xi^2 \biggl[ \frac{( 3^2 - \xi^2 )^2}{(3+\xi^2)^3} \biggr] - \biggl(\frac{2\cdot 3^3}{5}\biggr)\frac{d}{d\xi}\biggl\{ \xi^3 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr] \biggr\} </math>

<math>~\Rightarrow ~~~ \frac{5 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 P_c a_5^2}</math>

<math>~=</math>

<math>~ - 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2 + \xi^2 (3+\xi^2) ( 3^2 - \xi^2 )^2 </math>

 

 

<math>~ - 3(3+\xi^2)^4 \biggl\{ \xi^3 \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr] \frac{dx}{d\xi} + 3\xi^2 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr] + \xi^3 x \biggl[ \frac{ -2\xi }{(3+\xi^2)^3} \biggr] -2\cdot 3 \xi^4 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^4} \biggr] \biggr\} </math>

 

<math>~=</math>

<math>~ \xi^2 (3+\xi^2) ( 3^2 - \xi^2 )^2 - 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2 - 3\xi^3 ( 3^2 - \xi^2) (3+\xi^2) \biggl[ - \frac{2\xi}{ 3\cdot 5} \biggr] </math>

 

 

<math>~ - 3 \xi^2\biggl\{ 3 ( 3^2 - \xi^2) (3+\xi^2) -2 \xi^2 (3+\xi^2) -2\cdot 3 \xi^2 ( 3^2 - \xi^2) \biggr\} x </math>

<math>~\Rightarrow ~~~ \frac{5^2 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 ~\xi^2 P_c a_5^2}</math>

<math>~=</math>

<math>~ 5 (3+\xi^2) ( 3^2 - \xi^2 )^2 - 2^2~\xi^2 (15-\xi^2)^2 + 2 \xi^2 ( 3^2 - \xi^2) (3+\xi^2) </math>

 

 

<math>~ + \biggl[ 2 \xi^2 (3+\xi^2) + 2\cdot 3 \xi^2 ( 3^2 - \xi^2) - 3 ( 3^2 - \xi^2) (3+\xi^2) \biggr] (15-\xi^2) </math>

 

<math>~=</math>

<math>~ (3\cdot 5 + 5\xi^2) ( 3^4 - 2\cdot 3^2\xi^2 + \xi^4) - 2^2~\xi^2 (3^2\cdot 5^2 - 2\cdot 3\cdot 5 \xi^2 + \xi^4) + 2 \xi^2 ( 3^3 + 2\cdot 3\xi^2 -\xi^4) </math>

 

 

<math>~ + \biggl[ 2\cdot 3 \xi^2 + 2\xi^4 + 2\cdot 3^3 \xi^2 - 2\cdot 3 \xi^4 - 3(3^3 + 2\cdot 3\xi^2 -\xi^4) \biggr] (15-\xi^2) </math>

Coefficients of various powers of <math>~\xi</math>:

<math>~\xi^0:</math>

   

<math>~3^5\cdot 5 -3^5\cdot 5 = 0</math>

<math>~\xi^2:</math>

   

<math>~-2\cdot 3^3\cdot 5 + 3^4\cdot 5 -2^2 \cdot 3^2 \cdot 5^2 +2\cdot 3^3 + 2\cdot 3^2\cdot 5 + 2\cdot 3^4\cdot 5 - 2\cdot 3^3\cdot 5 + 3^4</math>

 

   

<math>~= 3^2\cdot 5[-2\cdot 3 + 3^2 + 2 + 2\cdot 3^2 - 2\cdot 3] + 3^2[ 2\cdot 3 + 3^2 - 2^2 \cdot 5^2 ] = 3^2\cdot 5[17 ] - 3^2[5\cdot 17 ] = 0</math>

<math>~\xi^4:</math>

   

<math>~3\cdot 5 -2\cdot 3^2\cdot 5 +2^3\cdot 3\cdot 5 + 2^2\cdot 3 + 2\cdot 3 \cdot 5 - 2\cdot 3^2\cdot 5 - 2\cdot 3 -2\cdot 3^3 + 2\cdot 3^2 + 3^2\cdot 5</math>

 

   

<math>~= 3\cdot 5[1 -2\cdot 3 +2^3 + 2 - 2\cdot 3 + 3] + 2\cdot 3[2 - 1 - 3^2 + 3] = 2\cdot 3\cdot 5 - 2\cdot 3\cdot 5 = 0 </math>

<math>~\xi^6:</math>

   

<math>~5 - 2^2 -2 -2 +2\cdot 3 -3 = 0</math>

Multiplying through by <math>~dr</math>, and integrating over the volume gives,

<math>~\int_0^R (\sigma^2 \rho \psi^2)\frac{dr}{r^2}</math>

<math>~=</math>

<math>~ \int_0^R \biggl[ \Gamma_1 P \biggl(\frac{d\psi}{dr} \biggr)^2 + \frac{4\psi^2}{r} \biggl( \frac{dP}{dr} \biggr) \biggr]\frac{dr}{r^2} - \biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr]_0^R \, , </math>

which is identical to equation (49) of Chandrasekhar (1964), if the last term — the difference of the central and surface boundary conditions — is set to zero.

Note that if we shift from the variable, <math>~\psi</math>, back to the fractional displacement function, <math>~\xi</math>, the last term in this expression may be written as,

<math>~\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr)</math>

<math>~=</math>

<math>~ \Gamma_1 P r \xi \frac{d}{dr} \biggl[r^3 \xi\biggr] </math>

 

<math>~=</math>

<math>~ \Gamma_1 P r \xi \biggl[3r^2 \xi + r^3 \frac{d\xi}{dr}\biggr] </math>

 

<math>~=</math>

<math>~ \Gamma_1 P r^3 \xi^2 \biggl[3 + \frac{d\ln\xi}{d\ln r}\biggr] \, . </math>

So, as is pointed out by Ledoux & Walraven (1958) in connection with their equation (57.31), setting this expression to zero at the surface of the configuration is equivalent to setting the variation of the pressure to zero at the surface. Quite generally, this can be accomplished by demanding that,

<math>~\frac{d\ln\xi}{d\ln r}\biggr|_\mathrm{surface} = -3 \, .</math>

(An accompanying chapter provides a broader discussion of this and other astrophysically reasonable boundary conditions that are associated with solutions to the LAWE.)

Ledoux & Walraven Approach

Returning to the above Foundational Variational Relation, we can also write,

<math>~\sigma^2 \rho r^4 \xi^2</math>

<math>~=</math>

<math>~ -\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] - (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) </math>

 

<math>~=</math>

<math>~ r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 - (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) - \frac{d}{dr}\biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr] </math>

<math>~\Rightarrow ~~~ \int_0^R\sigma^2 \rho r^4 \xi^2 dr</math>

<math>~=</math>

<math>~ \int_0^R r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) dr - \biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr]_0^R </math>

If the last term (boundary conditions) is set to zero, then we may also write,

<math>~\sigma^2 </math>

<math>~=</math>

<math>~ \frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \bigl( \frac{dP}{dr} \bigr) dr}{\int_0^R \rho r^4 \xi^2 dr} \, . </math>

This means that, if the radial profile of the pressure and the density is known throughout a spherically symmetric, equilibrium configuration, and if, furthermore, the eigenfunction, <math>~\xi(r)</math>, of a radial oscillation mode is specified precisely, then this expression will give the (square of the) eigenfrequency of that oscillation mode.

By using formal variational principle techniques to derive this same expression, Ledoux & Walraven (1958) are able to offer a broader interpretation, which is encapsulated by their equation (59.10), viz.,

<math>~\sigma_0^2 </math>

<math>~=</math>

<math>~\mathrm{min}~ \frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 \xi^2 \bigl( \frac{dP}{dr} \bigr) dr}{\int_0^R \rho r^4 \xi^2 dr} \, . </math>

This means that, if the exact radial eigenfunction, <math>~\xi(r)</math>, is not known, various approximate eigenfunctions can be tried. The trial eigenfunction that minimizes the righthand-side of this expression will give the (square of the) eigenfrequency of the fundamental mode of oscillation (subscript zero). Furthermore, via an evaluation of this righthand-side expression, any reasonable trial eigenfunction — for example, <math>~\xi</math> = constant — can provide an upper limit to <math>~\sigma_0^2</math>.

Ledoux & Pekeris Approach

Here we follow the lead of Ledoux & Pekeris (1941). Returning to the integral expression just derived in our discussion of the Ledoux & Walraven approach, and multiplying through by <math>~4\pi</math>, we have,

<math>~\int_0^R 4\pi \sigma^2 \rho r^4 \xi^2 dr</math>

<math>~=</math>

<math>~ \int_0^R 4\pi r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 dr - \int_0^R (3\Gamma_1 - 4) 4\pi r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) dr - \biggr[4\pi r^3 \Gamma_1 P\xi^2 \biggl(\frac{d\ln \xi}{d\ln r}\biggr) \biggr]_0^R \, . </math>

If we acknowledge that:

  • at the center of the configuration, <math>~r^3 = 0</math>;
  • as above, the boundary condition at the surface is <math>~P = P_e</math> while <math>~(d\ln \xi/d\ln r) = -3</math>;
  • the differential mass element is, <math>~dm = 4\pi r^2 \rho dr</math> and the corresponding differential volume element is, <math>~dV = 4\pi r^2 dr</math>; and
  • a statement of detailed force balance is, <math>~dP/dr = - Gm\rho/r^2</math>,

this integral relation becomes,

<math>~ \sigma^2 \int_0^R r^2 \xi^2 dm</math>

<math>~=</math>

<math>~ \Gamma_1 \int_0^R \biggl[ r \biggl(\frac{d\xi}{dr}\biggr)\biggr]^2 P dV + (3\Gamma_1 - 4) \int_0^R \xi^2 \biggl( \frac{Gm}{r} \biggr) dm - \biggr[\Gamma_1 \xi_\mathrm{surface}^2 (3P_e V) \biggl(-3\biggr) \biggr] \, . </math>

Now, as we have discussed separately — see, also, p. 64, Equation (12) of [C67] — the gravitational potential energy of the unperturbed configuration is given by the integral,

<math>~W_\mathrm{grav}</math>

<math>~=</math>

<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm \, ;</math>

for adiabatic systems, the internal energy is,

<math> U_\mathrm{int} = \frac{1}{(\Gamma_1-1)} \int_0^R P_0 dV

\, ;</math>

and — see the text at the top of p. 126 of Ledoux & Pekeris (1941) — the moment of inertia of the configuration about its center is,

<math> I = \int_0^M r_0^2 dm

\, .</math>

(Note that, defined in this way, <math>~I</math> is the same as what we have referred to elsewhere as the scalar moment of inertia, which is obtained by taking the trace of the moment of inertia tensor, <math>~I_{ij}</math>.) After inserting these expressions, we have what will henceforth be referred to as the,

Variational Principle's Governing Integral Relation

<math>~ \sigma^2 \int_0^R \xi^2 dI</math>

<math>~=</math>

<math>~ \Gamma_1 (\Gamma_1 - 1) \int_0^R \xi^2 \biggl[ \frac{d\ln\xi}{d\ln r}\biggr]^2 dU_\mathrm{int} - (3\Gamma_1 - 4) \int_0^R \xi^2 dW_\mathrm{grav} + 3^2 \Gamma_1 P_e V \xi_\mathrm{surface}^2 \, . </math>

Free-Energy Analysis

If we assume the simplest approximation for the fundamental-mode eigenfunction, namely, <math>~\xi = \xi_0</math> = constant — that is, homologous expansion/contraction — then this last integral expression gives,

<math>~ \sigma^2 I</math>

<math>~=</math>

<math>~ (4 - 3\Gamma_1) W_\mathrm{grav} + 3^2 \Gamma_1 P_e V \, . </math>

Contrast this result with the following free-energy analysis:

<math>~\mathfrak{G}</math>

<math>~=</math>

<math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV \, ,</math>

where, in terms of the configuration's (generally non-equilibrium) dimensionless radius, <math>~\chi \equiv R/R_0</math>,

<math>~W_\mathrm{grav}</math>

<math>~=</math>

<math>~-a\chi^{-1}</math>

<math>~U_\mathrm{int}</math>

<math>~=</math>

<math>~b\chi^{3-3\Gamma_1}</math>

<math>~V</math>

<math>~=</math>

<math>~\frac{4\pi}{3} \chi^3 \, .</math>

Then,

<math>~\frac{\partial \mathfrak{G}}{\partial \chi}</math>

<math>~=</math>

<math>~+a \chi^{-2} + 3(1-\Gamma_1) b \chi^{2-3\Gamma_1} + 4\pi P_e \chi^{2} </math>

 

<math>~=</math>

<math>~\chi^{-1} \biggl[- W_\mathrm{grav} + 3(1-\Gamma_1) U_\mathrm{int} + 3 P_e V \biggr] \, ,</math>

and,

<math>~\frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>

<math>~=</math>

<math>~-2a \chi^{-3} + 3(1-\Gamma_1)(2-3\Gamma_1) b \chi^{1-3\Gamma_1} + 8\pi P_e \chi </math>

 

<math>~=</math>

<math>~\chi^{-2} \biggl[ 2W_\mathrm{grav} + 3(1-\Gamma_1)(2-3\Gamma_1) U_\mathrm{int}+ 6 P_e V \biggr] \, .</math>

The equilibrium condition occurs when <math>~\partial \mathfrak{G}/\partial \chi = 0</math>, that is, when,

<math>~3(1-\Gamma_1) U_\mathrm{int}</math>

<math>~=</math>

<math>~W_\mathrm{grav} - 3 P_e V \, ,</math>

in which case,

<math>~\chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>

<math>~=</math>

<math>~2W_\mathrm{grav} + (2-3\Gamma_1) (W_\mathrm{grav} - 3P_eV) + 6 P_e V </math>

 

<math>~=</math>

<math>~(4-3\Gamma_1)W_\mathrm{grav} + 3^2 \Gamma_1 P_e V \, .</math>

Fantastic! The righthand-side of this "free-energy-based" expression exactly matches the righthand-side of the above expression that has been derived from the variational principle, assuming homologous expansion/contraction (i.e., <math>~\xi</math> = constant). In this case, we can make the direct association,

<math>~\sigma^2 I = \chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2} \, .</math>

This also make sense in that the equilibrium configuration should be stable if <math>~\tfrac{\partial^2 \mathfrak{G}}{\partial \chi^2} > 0</math> — in which case, <math>~\sigma^2</math> is positive; whereas the equilibrium configuration should be unstable if <math>~\tfrac{\partial^2 \mathfrak{G}}{\partial \chi^2} < 0</math> — in which case, <math>~\sigma^2</math> is negative.

Related, Exploratory Ideas

Logarithmic Derivatives

Returning to our above discussion of the Ledoux & Walraven approach, we appreciate that the differential relation governing the Variational Principle is,

<math>~\sigma^2 \rho r^4 \xi^2</math>

<math>~=</math>

<math>~ r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 - (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) - \frac{d}{dr}\biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr] </math>

<math>~\Rightarrow ~~~ \frac{d}{dr}\biggr[r^3 \Gamma_1 P\xi^2 \biggl(\frac{d\ln\xi}{d\ln r}\biggr) \biggr] </math>

<math>~=</math>

<math>~ r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 - (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) - \sigma^2 \rho r^4 \xi^2 </math>

 

<math>~=</math>

<math>~\xi^2 \biggl\{ r^2 \Gamma_1 P \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2 - (3\Gamma_1 - 4) r^3 \biggl( \frac{dP}{dr} \biggr) - \sigma^2 \rho r^4 \biggr\} </math>

 

<math>~=</math>

<math>~(r \xi)^2 P \biggl\{ \Gamma_1 \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2 - (3\Gamma_1 - 4) \biggl( \frac{d\ln P}{d\ln r} \biggr) - \frac{\sigma^2 \rho r^2}{P} \biggr\} </math>

 

<math>~=</math>

<math>~\Gamma_1 (r \xi)^2 P \biggl\{ \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^2 - \alpha \biggl( \frac{d\ln P}{d\ln r} \biggr) - \frac{\sigma^2 \rho r^2}{\Gamma_1 P} \biggr\} \, , </math>

where,

<math>~\alpha \equiv \biggl(3 - \frac{4}{\Gamma_1}\biggr) \, .</math>

Pressure-Truncated Polytropes

Let's start with the integral expression derived in our discussion of the Ledoux & Walraven approach; insert the variable, <math>~x</math>, in place of <math>~\xi</math>; and adopt the boundary conditions,

<math>~r = 0</math>   at the center,

        along with        

<math>~P = P_e~</math>,   and  <math>\frac{d\ln x}{d\ln r} = -3</math>   at the surface (r = R).

That is, let's start with,

<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr

+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2 \, .

</math>

Via Generalized Normalization

Next, we'll divide through by the normalization energy, as defined in an accompanying discussion,

<math>~E_\mathrm{norm} = P_\mathrm{norm}R_\mathrm{norm}^3 = \frac{GM_\mathrm{tot}^2}{R_\mathrm{norm}} \, ,</math>

thereby making the integral relation dimensionless:

<math>~ 0 </math>

<math>~=</math>

<math>~ - \biggl[\frac{R_\mathrm{norm}}{GM_\mathrm{tot}^2} \biggr] \int_0^R \sigma^2 \rho r^4 x^2 dr +\biggl[\frac{1}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr - \biggl[\frac{1}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr + \biggl[\frac{P_e R^3 }{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] 3\Gamma_1 x_\mathrm{surface}^2 </math>

 

<math>~=</math>

<math>~ - \biggl[\frac{R_\mathrm{norm} R^5 \rho_c^2}{M_\mathrm{tot}^2} \biggr] \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R} + \biggl[\frac{P_c R^3}{P_\mathrm{norm}R_\mathrm{norm}^3 } \biggr] \int_0^R \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R} </math>

 

 

<math>~ - \biggl[\frac{P_c R^3}{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] \int_0^R (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R} + \biggl[\frac{P_e R^3 }{P_\mathrm{norm}R_\mathrm{norm}^3} \biggr] 3\Gamma_1 x_\mathrm{surface}^2 </math>

 

<math>~=</math>

<math>~ - \biggl[ \frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_c}{\bar\rho} \biggr]^2 \chi^{-1} \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R} + \biggl[\frac{P_e }{P_\mathrm{norm}} \biggr] 3\Gamma_1 \chi^3 x_\mathrm{surface}^2 </math>

 

 

<math>~ + \biggl[\frac{P_c}{P_\mathrm{norm} } \biggr] \chi^3 \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2 - (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R} \, , </math>

where,

<math>~\chi \equiv \frac{R}{R_\mathrm{norm}} \, .</math>

Note that we will ultimately insert the relation,

<math>~\frac{P_c}{P_\mathrm{norm}} = \biggl[\biggl( \frac{3}{4\pi}\biggr) \frac{\rho_c}{\bar\rho} \biggl( \frac{M}{M_\mathrm{tot}}\biggr)\biggr]^{\Gamma_1} \biggl( \frac{R}{R_\mathrm{norm}}\biggr)^{-3\Gamma_1} \, .</math>

But, for the time being, dividing through by <math>~[P_c/P_\mathrm{norm}]\chi^3</math> gives,

<math>~0</math>

<math>~=</math>

<math>~ - \biggl[\frac{P_c}{P_\mathrm{norm} } \biggr]^{-1} \biggl[ \frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl[\biggl( \frac{3}{4\pi} \biggr) \frac{\rho_c}{\bar\rho} \biggr]^2 \chi^{-4} \int_0^R x^2 \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \biggl( \frac{\rho}{\rho_c} \biggr) \biggl(\frac{r}{R}\biggr)^4 \frac{dr}{R} </math>

 

 

<math>~ + \biggl[\frac{P_e }{P_c} \biggr] 3\Gamma_1 x_\mathrm{surface}^2 + \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2 - (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R} \, , </math>

Now let's focus on the second line of this integral energy relation, evaluating it for pressure-truncated polytropic configurations, in which case, <math>~\Gamma_1 \rightarrow (n+1)/n</math>,

<math>~\frac{r}{R} \rightarrow \frac{\xi}{\tilde\xi}</math>

        and        

<math>~ \frac{P}{P_c} \rightarrow \theta^{n+1} \, . </math>

We have,

Second line of relation

<math>~=</math>

<math>~ \biggl[\frac{P_e }{P_c} \biggr] 3\Gamma_1 x_\mathrm{surface}^2 + \int_0^R \biggl\{ \biggl( \frac{r}{R}\biggr)^4 \Gamma_1\biggl(\frac{ P }{P_c} \biggr) \biggl[ \frac{dx}{d(r/R)}\biggr]^2 - (3\Gamma_1 - 4) \biggl( \frac{r}{R} \biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \biggr\}\frac{dr}{R} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2 + \int_0^{\tilde\xi} \biggl\{ \biggl( \frac{\xi}{\tilde\xi}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2 {\tilde\xi}^2 - \biggl(\frac{3-n}{n}\biggr) \biggl( \frac{\xi}{\tilde\xi} \biggr)^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr] \tilde\xi \biggr\}\frac{d\xi}{\tilde\xi} </math>

 

<math>~=</math>

<math>~ \biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2 + \frac{1}{n {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl\{ (n+1) \xi^4 \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2 - (3-n) \xi^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr] \biggr\}d\xi </math>

 

<math>~=</math>

<math>~ \biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2 + \frac{1}{n {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl\{ (n+1) \xi^4 \theta^{n+1} \biggl[ \frac{dx}{d\xi}\biggr]^2 - (n+1) (3-n) \xi^3 x^2 \theta^n \theta^' \biggr\}d\xi </math>

 

<math>~=</math>

<math>~ \biggl[\frac{P_e }{P_c} \biggr] \biggl[ \frac{3( n+1)}{n} \biggr] x_\mathrm{surface}^2 + \frac{(n+1)}{n {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl(\frac{3}{2n}\biggr)^2\frac{\xi}{\theta^n} \biggl\{ \xi \theta \biggl[ \biggl( \frac{2n}{3}\biggr)\xi \theta^n \cdot \frac{dx}{d\xi}\biggr]^2 - (3-n) \biggl[ \biggl( \frac{2n}{3}\biggr) \xi \theta^n x\biggr]^2 \theta^' \biggr\}d\xi \, . </math>

Now, let's examine how these terms combine if we guess the analytically defined eigenfunction that applies to marginally unstable, pressure-truncated polytropic configurations, namely,

<math>~x</math>

<math>~=</math>

<math>~\frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{\theta^'}{\xi \theta^{n} } \biggr] </math>

<math>~\Rightarrow ~~~ \biggl( \frac{2n}{3}\biggr) \xi \theta^n x</math>

<math>~=</math>

<math>~\biggl[(n-1)\xi \theta^n + (n-3) \theta^' \biggr] </math>

<math>~\Rightarrow ~~~ \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~\biggl[\frac{3(n-3)}{2n}\biggr] \biggl\{ \frac{\theta^{}}{\xi \theta^{n}} - \frac{\theta^'}{\xi^2 \theta^{n}} - \frac{n(\theta^')^2}{\xi \theta^{(n+1)}} \biggr\} </math>

 

<math>~=</math>

<math>~- \biggl[\frac{3(n-3)}{2n}\biggr] \frac{1 }{\xi \theta^{n}} \biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr] </math>

<math>~\Rightarrow ~~~ \biggl( \frac{2n}{3}\biggr) \xi \theta^n\frac{dx}{d\xi}</math>

<math>~=</math>

<math>~(3-n) \biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr] \, . </math>

Hence,

Second line of relation

<math>~=</math>

<math>~ {\tilde\theta}^{n+1} \biggl[ \frac{3( n+1)}{n} \biggr] \biggl\{ \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \frac{ {\tilde\theta}^'}{\tilde\xi {\tilde\theta}^{n} } \biggr] \biggr\}^2 </math>

 

 

<math>~ + \frac{3^2 (n+1)(3-n)}{2^2n^3 {\tilde\xi}^3}\int_0^{\tilde\xi} \frac{\xi}{\theta^n} \biggl\{ \xi \theta (3-n)\biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr]^2 - \biggl[(n-1)\xi \theta^n + (n-3) \theta^' \biggr]^2 \theta^' \biggr\}d\xi </math>

 

<math>~=</math>

<math>~ \frac{1}{{\tilde\xi}^2 {\tilde\theta}^{n+1}} \biggl[ \frac{3^3( n+1)}{2^2n^3} \biggr] \biggl[(n-1) \tilde\xi {\tilde\theta}^{n+1} + (n-3) \tilde\theta {\tilde\theta}^' \biggr]^2 </math>

 

 

<math>~ + \frac{3^2 (n+1)(3-n)}{2^2n^3 {\tilde\xi}^3} \int_0^{\tilde\xi} \frac{1}{\theta^{n+1}} \biggl\{ (3-n)\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]^2 - \biggl[(n-1)\xi \theta^n + (n-3) \theta^' \biggr]^2 \xi \theta \theta^' \biggr\}d\xi </math>

 

<math>~=</math>

<math>~ \frac{1}{{\tilde\xi}^2 {\tilde\theta}^{n+1}} \biggl[ \frac{3^3( n+1)}{2^2n^3} \biggr] \biggl[(n-1) \tilde\xi {\tilde\theta}^{n+1} + (n-3) \tilde\theta {\tilde\theta}^' \biggr]^2 </math>

 

 

<math>~ + \frac{3^2 (n+1)(3-n)^2}{2^2n^3 {\tilde\xi}^3} \int_0^{\tilde\xi} \frac{1}{\theta^{n+1}} \biggl\{ \biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]^2 + \frac{1}{(n-3)} \biggl[(n-1)\xi \theta^n + (n-3) \theta^' \biggr]^2 \xi \theta \theta^' \biggr\}d\xi </math>

Note that, in this derivation, we have inserted the expressions:

<math>~ \biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr]\biggl[ \xi \theta^{n+1} + 3\theta \theta^' + n\xi (\theta^')^2 \biggr] = \xi^2 \theta^{2(n+1)} + 6\xi \theta^{n+2}\theta^' + 2n\xi^2 \theta^{n+1} (\theta^')^2 + 6n\xi\theta (\theta^')^3 + n^2 \xi^2 (\theta^')^4 </math>

<math>~ \frac{1}{(n-3)} \biggl[(n-1)\xi \theta^n + (n-3) \theta^' \biggr]^2 \xi\theta (\theta^')= \biggl[ \frac{(n-1)^2}{(n-3)}\biggr] \xi^3 \theta^{2n+1}(\theta^') + 2(n-1)\xi^2 \theta^{n+1} (\theta^' )^2 + (n-3) \xi\theta (\theta^')^3 </math>

Directly to n = 5 Polytropic Configurations

<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr

+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2 

</math>

<math>~\Rightarrow ~~~ \frac{1}{R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^R \biggl(\frac{r}{R}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{P}{P_c}\biggr) \biggl[\frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R} - \int_0^R \biggl[3\biggl(\frac{n+1}{n}\biggr) - 4\biggr] \biggl(\frac{r}{R}\biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}

+3\biggl(\frac{n+1}{n}\biggr) \biggl( \frac{P_e}{P_c}\biggr)  x_\mathrm{surface}^2 

</math>

 

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \frac{6}{5} \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 \theta^6 \biggl[\frac{dx}{d(\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi} - \int_0^{\tilde\xi} \biggl( - \frac{2}{5}\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 x^2 \biggl[ \frac{d\theta^{6}}{d(\xi/\tilde\xi)} \biggr] \frac{d\xi}{\tilde\xi}

+\biggl(\frac{18}{5}\biggr) {\tilde\theta}^6  x_\mathrm{surface}^2 

</math>

 

<math>~=</math>

<math>~ \frac{1}{ {\tilde\xi}^3} \int_0^{\tilde\xi} \biggl( \frac{6}{5}\biggr) \xi^4 \theta^6 \biggl[\frac{dx}{d\xi}\biggr]^2 d\xi + \frac{1}{ {\tilde\xi}^3} \int_0^{\tilde\xi} \biggl(\frac{2}{5}\biggr) \xi^3 x^2 \biggl[ \frac{d\theta^{6}}{d\xi} \biggr] d\xi

+\biggl(\frac{18}{5}\biggr) {\tilde\theta}^6  x_\mathrm{surface}^2 

</math>

<math>~\Rightarrow ~~~ \frac{5 {\tilde\xi}^3 }{2R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} 3\xi^4 \theta^6 \biggl[ - \frac{2\xi}{15} \biggr]^2 d\xi + \int_0^{\tilde\xi} 6\xi^3 \biggl[\frac{15-\xi^2}{15}\biggr]^2 \theta^5\biggl[ \frac{d\theta}{d\xi} \biggr] d\xi

+9 {\tilde\xi}^3 {\tilde\theta}^6  \biggl[\frac{15- {\tilde\xi}^2}{15}\biggr]^2 

</math>

 

<math>~=</math>

<math>~ \biggl(\frac{ 2^2}{3\cdot 5^2 } \biggr) \int_0^{\tilde\xi} \xi^6 \biggl( \frac{3}{3+\xi^2}\biggr)^3 d\xi + \biggl(\frac{ 2}{3\cdot 5^2 } \biggr) \int_0^{\tilde\xi} \xi^3 \biggl[15-\xi^2\biggr]^2 \biggl( \frac{3}{3+\xi^2}\biggr)^{4} \biggl[- \frac{\xi}{3}\biggr] d\xi

+ \biggl( \frac{1}{5^2} \biggr) {\tilde\xi}^3 \biggl( \frac{3}{3+ {\tilde\xi}^2}\biggr)^3  \biggl[15- {\tilde\xi}^2\biggr]^2 

</math>

 

<math>~=</math>

<math>~ \biggl(\frac{ 2^2\cdot 3^2}{5^2 } \biggr) \int_0^{\tilde\xi} \biggl[ \frac{\xi^6 }{(3+\xi^2)^3}\biggr] d\xi ~~- ~~ \biggl(\frac{ 2\cdot 3^2}{5^2 } \biggr) \int_0^{\tilde\xi} \biggl[ \frac{\xi^4 (15-\xi^2)^2}{(3+\xi^2)^4}\biggr] d\xi ~~ + ~~ \biggl( \frac{3^3}{5^2} \biggr) \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

<math>~\Rightarrow ~~~ \frac{5^3 {\tilde\xi}^3 }{2\cdot 3^2R^3 P_c}\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \biggl[ \frac{4\xi^6(3+\xi^2)-2\xi^4 (15-\xi^2)^2}{(3+\xi^2)^4}\biggr] d\xi ~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

 

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \biggl\{ \frac{2\xi^4 [6\xi^2 + 2\xi^4 -15^2 + 30\xi^2 - \xi^4] }{(3+\xi^2)^4}\biggr\} d\xi ~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

 

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \biggl\{ \frac{2\xi^4 [\xi^4 + 36\xi^2 -15^2 ] }{(3+\xi^2)^4}\biggr\} d\xi ~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{2\xi^5(\xi^2-15)}{(\xi^2+3)^3} \biggr]_0^{\tilde\xi} ~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

 

<math>~=</math>

<math>~ \biggl[ \frac{2{\tilde\xi}^5({\tilde\xi}^2-15)}{({\tilde\xi}^2+3)^3} \biggr] ~ + ~ 3 \biggl[ \frac{{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{(3+ {\tilde\xi}^2)^3}\biggr] </math>

 

<math>~=</math>

<math>~ \frac{2{\tilde\xi}^5({\tilde\xi}^2-15) + 3{\tilde\xi}^3(15- {\tilde\xi}^2)^2}{({\tilde\xi}^2+3)^3} = \frac{5{\tilde\xi}^7 - 120{\tilde\xi}^5 + 3^3\cdot 5^2{\tilde\xi}^3 }{({\tilde\xi}^2+3)^3} \, , </math>

which equals zero if <math>~\tilde\xi = 3</math>. Hooray!!

For All Polytropic Indexes

Generalized Governing Integral Relation

Given that the derivation just completed works for the special case of n = 5, let's generalize it to all polytropic indexes

<math>~\int_0^R \sigma^2 \rho r^4 x^2 dr</math>

<math>~=</math>

<math>~ \int_0^R r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr)^2 dr - \int_0^R (3\Gamma_1 - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr

+3\Gamma_1 P_e R^3 x_\mathrm{surface}^2 

</math>

<math>~\Rightarrow ~~~ \frac{R^5 \rho_c}{R^3 P_c}\int_0^R \sigma^2 \biggl( \frac{\rho}{\rho_c}\biggr) \biggl(\frac{r}{R}\biggr)^4 x^2 \frac{dr}{R}</math>

<math>~=</math>

<math>~ \int_0^R \biggl(\frac{r}{R}\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \biggl(\frac{P}{P_c}\biggr) \biggl[\frac{dx}{d(r/R)}\biggr]^2 \frac{dr}{R} - \int_0^R \biggl[3\biggl(\frac{n+1}{n}\biggr) - 4\biggr] \biggl(\frac{r}{R}\biggr)^3 x^2 \biggl[ \frac{d(P/P_c)}{d(r/R)} \biggr] \frac{dr}{R}

+3\biggl(\frac{n+1}{n}\biggr) \biggl( \frac{P_e}{P_c}\biggr)  x_\mathrm{surface}^2 

</math>

<math>~\Rightarrow ~~~ \frac{R^2 \rho_c}{P_c} \int_0^{\tilde\xi} \sigma^2 \theta^n \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 x^2 \frac{d\xi}{\tilde\xi}</math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \biggl(\frac{\xi}Template:\tilde\xi\biggr)^4 \biggl(\frac{n+1}{n}\biggr) \theta^{n+1} \biggl[\frac{dx}{d(\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi} ~+ \int_0^{\tilde\xi} \biggl(\frac{n-3}{n}\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^3 x^2 \biggl[ \frac{d\theta^{n+1}}{d(\xi/\tilde\xi)} \biggr] \frac{d\xi}{\tilde\xi} ~+~3\biggl(\frac{n+1}{n}\biggr) {\tilde\theta}^{n+1} x_\mathrm{surface}^2 </math>

<math>~\Rightarrow ~~~ \frac{n R^2\rho_c}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \sigma^2 \theta^n \xi^4 x^2 d\xi</math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \xi^4 \theta^{n+1} \biggl[\frac{dx}{d\xi}\biggr]^2 d\xi ~+ \int_0^{\tilde\xi} (n-3) \xi^3 \theta^n x^2 \biggl[ \frac{d\theta}{d\xi} \biggr] d\xi ~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 </math>

<math>~\Rightarrow ~~~ \frac{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 d\xi ~+ \int_0^{\tilde\xi} (n-3) \xi^2 \theta^{n+1} x^2 \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] d\xi ~+~3 {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 </math>

 

<math>~=</math>

<math>~ 3 {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 + \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3) \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi </math>

For additional clarification, let's rewrite the leading coefficient on the lefthand-side of this expression.

LHS

<math>~=</math>

<math>~\frac{n R^2 G \rho_c^2}{(n+1){\tilde\xi}^2 P_c}\int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

 

<math>~=</math>

<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G R_\mathrm{norm}^2}{P_\mathrm{norm}} \biggr] \biggl(\frac{R}{R_\mathrm{norm}^2}\biggr) \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2 \biggl[ \frac{3M}{4\pi R^3}\biggr]^2 \biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) \biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr] \int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

 

<math>~=</math>

<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl[ \frac{G M_\mathrm{tot}^2}{P_\mathrm{norm}R_\mathrm{norm}^4} \biggr] \biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( \frac{\rho_c}{ {\bar\rho}}\biggr)^2 \biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) \biggl(\frac{P_e}{P_c} \biggr) \biggl[ \frac{1}{{\tilde\xi}^2} \biggr] \int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

 

<math>~=</math>

<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl(\frac{P_\mathrm{norm}}{P_e} \biggr) \biggl(\frac{R_\mathrm{norm}}{R}\biggr)^4 \biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2 \biggl[ \biggl(\frac{3}{4\pi}\biggr)\frac{M}{M_\mathrm{tot}}\biggr]^2 \biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr] \int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

Now, from an accompanying discussion, we know that, in equilibrium,

<math> ~\frac{R_\mathrm{eq}}{R_\mathrm{norm}} </math>

<math>~=~</math>

<math>~ \biggl[(n+1)^{-n} ( 4\pi )\biggr]^{1/(n-3)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)/(n-3)} \tilde\xi ( -\tilde\xi^2 \tilde\theta' )^{(1-n)/(n-3)} \, , </math>

<math> ~\frac{P_e}{P_\mathrm{norm}} </math>

<math>~=~</math>

<math>~ \biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)/(n-3)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)/(n-3)} \tilde\theta_n^{n+1}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)/(n-3)} \, , </math>

Hence,

<math> ~\biggl( \frac{P_e}{P_\mathrm{norm}} \biggr) \biggl( \frac{R_\mathrm{eq}}{R_\mathrm{norm}} \biggr)^4 </math>

<math>~=~</math>

<math>~ \biggl\{ \biggl[(n+1)^{3} ( 4\pi )^{-1} \biggr]^{(n+1)}\biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2(n+1)} \tilde\theta_n^{(n+1)(n-3)}( -\tilde\xi^2 \tilde\theta' )^{2(n+1)} \biggr\}^{1/(n-3)} </math>

 

 

<math>~\times \biggl\{\biggl[(n+1)^{-n} ( 4\pi )\biggr] \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(n-1)} \tilde\xi^{(n-3)} ( -\tilde\xi^2 \tilde\theta' )^{(1-n)} \biggr\}^{4/(n-3)} </math>

 

<math>~=~</math>

<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)} \biggl\{ (n+1)^{3(n+1)} ( 4\pi )^{(-n-1)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{-2n-2} ( -\tilde\xi^2 \tilde\theta' )^{2n+2} (n+1)^{-4n} ( 4\pi )^4 \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{(4n-4)} ( -\tilde\xi^2 \tilde\theta' )^{(4-4n)} \biggr\}^{1/(n-3)} </math>

 

<math>~=~</math>

<math>~ \tilde\xi^{4} \tilde\theta_n^{(n+1)} \biggl\{ (n+1)^{(3-n)} ( 4\pi )^{(3-n)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2(n-3)} ( -\tilde\xi^2 \tilde\theta' )^{2(3-n)} \biggr\}^{1/(n-3)} </math>

 

<math>~=~</math>

<math>~ (n+1)^{-1} ( 4\pi )^{(-1)} \biggl[\frac{M}{M_\mathrm{tot}} \biggr]^{2} \tilde\xi^{4} \tilde\theta_n^{(n+1)}( -\tilde\xi^2 \tilde\theta' )^{-2} \, . </math>

This means that, in equilibrium,

LHS

<math>~=</math>

<math>~\biggl[ \frac{n}{(n+1)} \biggr] \biggl\{ (n+1) ( 4\pi ) \tilde\xi^{-4} \tilde\theta_n^{-(n+1)}( -\tilde\xi^2 \tilde\theta' )^{2} \biggr\} \biggl( - \frac{\tilde\xi}{3 {\tilde\theta}^'}\biggr)^2 \biggl(\frac{3}{4\pi}\biggr)^2 \biggl[ \frac{{\tilde\theta}^{n+1}}{{\tilde\xi}^2} \biggr] \int_0^{\tilde\xi} \biggl( \frac{\sigma^2}{G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

 

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi \, .</math>

In summary, then, we have,

<math>~\int_0^{\tilde\xi} \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \theta^n \xi^4 x^2 d\xi</math>

<math>~=</math>

<math>~ 3 {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 + \int_0^{\tilde\xi} \xi^2 \theta^{n+1} x^2 \biggl\{ \biggl[\frac{\xi}{x} \cdot \frac{dx}{d\xi}\biggr]^2 + (n-3) \biggl[\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr] \biggr\} d\xi \, . </math>

Perhaps this looks better if the terms are rearranged to give,

<math>~ 3 {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 </math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \xi^2\theta^{n+1} x^2 \biggl\{ \biggl( \frac{n \sigma^2}{4\pi G\rho_c}\biggr) \frac{\xi^2}{\theta} - \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3) \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr] \biggr\} d\xi \, . </math>

Plug in Known Marginally Unstable Solution

As has been summarized in an accompanying discussion, we have found that, for marginally unstable pressure-truncated polytropic configurations, the eigenvector associated with the fundamental mode of radial oscillation is prescribed analytically by the following eigenfrequency-eigenfunction pair:

<math>~\sigma_c^2 = 0</math>

      and      

<math>~x = \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>

This means that,

<math>~\biggl[ \frac{2n}{3(n-1)} \biggr] \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~ \biggl(\frac{n-3}{n-1}\biggr) \frac{d}{d\xi}\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) </math>

 

<math>~=</math>

<math>~ \biggl(\frac{n-3}{n-1}\biggr) \biggl[ \frac{\theta^{}}{\xi \theta^{n}} - \frac{\theta^'}{\xi^2 \theta^{n}} - \frac{n (\theta^')^2}{\xi \theta^{n+1}} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl(\frac{n-3}{n-1}\biggr) \biggl[ - \frac{1}{\xi \theta^{n}} \biggl( \theta^n + \frac{2\theta^'}{\xi} \biggr) - \frac{\theta^'}{\xi^2 \theta^{n}} - \frac{n (\theta^')^2}{\xi \theta^{n+1}} \biggr] </math>

 

<math>~=</math>

<math>~ \biggl(\frac{3-n}{n-1}\biggr) \biggl[ \frac{1}{\xi } + \frac{3\theta^'}{\xi^2 \theta^{n}} + \frac{n (\theta^')^2}{\xi \theta^{n+1}} \biggr] \, . </math>

Hence, also,

<math>~\frac{d\ln x}{d\ln \xi} = \frac{\xi}{x} \cdot \frac{dx}{d\xi}</math>

<math>~=</math>

<math>~ \biggl(\frac{3-n}{n-1}\biggr) \biggl[1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}} \biggr] \biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1} </math>

 

<math>~=</math>

<math>~ \biggl(\frac{3-n}{n-1}\biggr) \biggl(\frac{n-3}{n-1}\biggr)^{-1} \biggl[1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}} \biggr] \biggl[\biggl(\frac{n-1}{n-3}\biggr) + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1} </math>

 

<math>~=</math>

<math>~ - \biggl[1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}} \biggr] \biggl[\biggl(\frac{n-1}{n-3}\biggr) + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^{-1} \, . </math>

Rather, let's try:

<math>~ \xi^2 x^2 \biggl[ \biggl( \frac{d\ln x}{d\ln \xi}\biggr)^2 + (n-3) \biggl( \frac{d\ln\theta}{d\ln\xi} \biggr) \biggr] </math>

<math>~=</math>

<math>~ x^2 \xi^2 \biggl( \frac{\xi}{x}\cdot \frac{dx}{d\xi}\biggr)^2 + (n-3) x^2 \xi^2 \biggl( \frac{\xi}{\theta} \cdot \frac{d\theta}{d\xi} \biggr) </math>

 

<math>~=</math>

<math>~ \xi^4 \biggl\{ \frac{dx}{d\xi}\biggr\}^2 + (n-3) \biggl[ \frac{\xi^3 \theta^'}{\theta} \biggr] x^2 </math>

 

<math>~=</math>

<math>~ \xi^4 \biggl\{ \frac{3(n-1)}{2n}\biggl(\frac{3-n}{n-1}\biggr) \biggl[ \frac{1}{\xi } + \frac{3\theta^'}{\xi^2 \theta^{n}} + \frac{n (\theta^')^2}{\xi \theta^{n+1}} \biggr]\biggr\}^2 + (n-3) \biggl[ \frac{\xi^3 \theta^'}{\theta} \biggr] \biggl\{ \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr] \biggr\}^2 </math>

 

<math>~=</math>

<math>~\xi^2 (n-3) \biggl[ \frac{3}{2n} \biggr]^2\biggl\{ (n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 +\xi \biggl( \frac{ \theta^'}{\theta} \biggr) \biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 \biggr\} </math>

Hence, after setting <math>~\sigma^2 = 0</math>, the above rearranged integral relation becomes,

<math>~ - \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 \biggr] </math>

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \xi^2 \theta^{n+1} \biggl\{ (n-3) \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 +\xi \biggl( \frac{ \theta^'}{\theta} \biggr) \biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 \biggr\} d\xi </math>


Let's check to see whether the terms in this last expression balance out when we plug in the functions that are appropriate for the marginally unstable, n = 5 configuration, namely,

<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>,

     and    

<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>.

RHS Term 1

<math>~=</math>

<math>~ (n-3) \int_0^{\tilde\xi} \xi^2 \theta^{n+1} \biggl[ 1 + \frac{3\theta^'}{\xi \theta^{n}} + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 d\xi </math>

 

<math>~=</math>

<math>~ 2 \int_0^{\tilde\xi} \xi^2 \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2} \biggr]^{6} \biggl\{ 1 - \biggl[ \xi \biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggr]\frac{1}{\xi }\biggl(\frac{3+\xi^2}{3}\biggr)^{5 / 2} +5 \biggl[ \frac{\xi^2}{3^2}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3} \biggr] \biggl[ \biggl(\frac{3+\xi^2}{3}\biggr)^{3} \biggr] \biggr\}^2 d\xi </math>

 

<math>~=</math>

<math>~ 2 \int_0^{\tilde\xi} \frac{3^3 \xi^2}{(3+\xi^2)^3} \biggl\{ 1 - \biggl(\frac{3+\xi^2}{3}\biggr) + \frac{5\xi^2}{3^2} \biggr\}^2 d\xi </math>

 

<math>~=</math>

<math>~ \frac{2^3}{3} \int_0^{\tilde\xi} \frac{\xi^6 ~d\xi}{(3+\xi^2)^3} </math>

 

<math>~=</math>

<math>~ \frac{2^3}{3} \biggl[ \frac{27\xi}{8(3+\xi^2)} - \frac{9\xi}{4(3+\xi^2)^2} + \xi - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr) \biggr]_0^{3} = \frac{2^3}{3} \biggl[ \frac{3^5}{2^6} - \frac{3^{1 / 2}\cdot 5\pi}{2^3} \biggr] \, . </math>

RHS Term 2

<math>~=</math>

<math>~ \int_0^{\tilde\xi} \xi^3 \theta^{n} \theta^' \biggl[(n-1) + (n-3)\biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 d\xi </math>

 

<math>~=</math>

<math>~ - \int_0^{\tilde\xi} \xi^3 \biggl(\frac{3+\xi^2}{3}\biggr)^{-5 / 2} \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2} \biggl\{ 4 - 2\frac{1}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}\biggl(\frac{3+\xi^2}{3}\biggr)^{5/2} \biggr\}^2 d\xi </math>

 

<math>~=</math>

<math>~ - \frac{1}{3}\int_0^{\tilde\xi} \biggl(\frac{3\xi}{3+\xi^2}\biggr)^{4} \biggl\{ 4 - \frac{2}{3}\biggl(\frac{3+\xi^2}{3}\biggr) \biggr\}^2 d\xi </math>

 

<math>~=</math>

<math>~ - \frac{2^3}{3}\int_0^{\tilde\xi} \frac{1}{2} \biggl(\frac{\xi}{3+\xi^2}\biggr)^{4} \biggl\{ 15-\xi^2 \biggr\}^2 d\xi </math>

 

<math>~=</math>

<math>~ -\frac{2^3}{3} \biggl[ \frac{123\xi}{8(3+\xi^2)} - \frac{243\xi}{4(3+\xi^2)^2} + \frac{162\xi}{2(3+\xi^2)^3} + \frac{\xi}{2} - \biggl(\frac{3^{3/2}\cdot 5}{2^3} \biggr)\tan^{-1}\biggl(\frac{\xi}{3^{1 / 2}}\biggr) \biggr]_0^{3} </math>

 

<math>~=</math>

<math>~ \frac{2^3}{3} \cdot \frac{5}{2^5} \biggl[ 2^2\cdot 3^{1 / 2} \pi - 3^3\biggr] = -\frac{2^3}{3} \biggl[ \frac{3^3\cdot 5}{2^5} - \frac{3^{1 / 2} \cdot 5\pi}{2^3} \bigg] \, . </math>

<math>~\Rightarrow ~~~</math> RHS Total

<math>~=</math>

<math>~ \frac{2^3}{3} \biggl[ \frac{3^5}{2^6} - \frac{3^3\cdot 5}{2^5} \biggr] = \frac{3^2}{2^3} \biggl[ 3^2 - 2\cdot 5 \biggr] = - \frac{3^2}{2^3} \, . </math>

LHS

<math>~=</math>

<math>~ - \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1} x_\mathrm{surface}^2 \biggr] </math>

 

<math>~=</math>

<math>~ - \frac{2^2 5^2}{2\cdot 3} \biggl[ 3^3 \biggl(\frac{3}{3+3^2}\biggr)^{3} \frac{2^2}{5^2} \biggr] = - \frac{2^3 }{3} \biggl[\biggl(\frac{3}{2^2}\biggr)^{3} \biggr] = - \frac{3^2 }{2^3} \, . </math>

Hence, the LHS = RHS.   Hooray!

See Also

 

A variational principle of great power is derived. It is naturally adapted for computers, and may be used to determine the stability of any fluid flow including those in differentially-rotating, self-gravitating stars and galaxies. The method also provides a powerful theoretical tool for studying general properties of eigenfunctions, and the relationships between secular and ordinary stability. In particular we prove the anti-sprial theorem indicating that no stable (or steady( mode can have a spiral structure.

  • B. F. Schutz, Jr. (1972), ApJSuppl., 24, 319: Linear Pulsations and Stability of Differentially Rotating Stellar Models. I. Newtonian Analysis
 

A systematic method is presented for deriving the Lagrangian governing the evolution of small perturbations of arbitrary flows of a self-gravitating perfect fluid. The method is applied to a differentially rotating stellar model; the result is a Lagrangian equivalent to that of D. Lynden-Bell & J. P. Ostriker (1967). A sufficient condition for stability of rotating stars, derived from this Lagrangian, is simplified greatly by using as trial functions not the three components of the Lagrangian displacement vector, but three scalar functions … This change of variables saves one from integrating twice over the star to find the effect of the perturbed gravitational field.

… we examine the special cases of (i) axially symmetric perturbations of a rotating star (as treated by S. Chandrasekhar & N. R. Lebovitz 1968); and (ii) perturbations of a nonrotating star (as treated by Chandrasekhar and Lebovitz 1964). We find that the stability criteria for those cases can also be simplified …


Whitworth's (1981) Isothermal Free-Energy Surface

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Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation