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Part I: Ledoux & Pekeris (1941)
Part II: Exploratory Ideas

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Created page with "__FORCETOC__  =Ledoux's Variational Principle= <table border="1" align="center" width="100%" colspan="8"> <tr> <td align="center" bgcolor="lightblue" width="50%"><br />Part I: Ledoux & Pekeris (1941) </td> <td align="center" bgcolor="lightblue"><br />Part II: Exploratory Ideas </td> </tr> </table> {| class="PGEclass" style="float:lef..."

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__FORCETOC__

=Ledoux's Variational Principle=
<table border="1" align="center" width="100%" colspan="8">
<tr>
<td align="center" bgcolor="lightblue" width="50%"><br />[[SSC/VariationalPrinciple|Part I:  Ledoux & Pekeris (1941)]]
 
</td>
<td align="center" bgcolor="lightblue"><br />[[SSC/VariationalPrinciple/Pt2|Part II:  Exploratory Ideas]]
 
</td>
</tr>
</table>

{| class="PGEclass" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
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<font size="-1">[[H_BookTiledMenu#Stability_Analysis|<b>Variational<br />Principle</b>]]</font>
|}
All of the discussion in this chapter will build upon our [[SSC/Perturbations#2ndOrderODE|derivation elsewhere]] of the,
<div align="center" id="2ndOrderODE">
<font color="#770000">'''LAWE:   Linear Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br />

{{Math/EQ_RadialPulsation01}}
</div>

We will draw heavily from the papers published by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux & 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 & 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 [[Appendix/Ramblings/LedouxVariationalPrinciple#Ledoux.27s_Variational_Principle_.28Supporting_Derivations.29|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 [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux & Pekeris] (1941, ApJ, 94, 124), reads:
<div align="center">
<table border="1" cellpadding="5">
<tr><td align="center">
Paragraph extracted from [http://adsabs.harvard.edu/abs/1941ApJ....94..124L P. Ledoux & C. L. Pekeris (1941)]<p></p>
"''Radial Pulsations of Stars''"<p></p>
ApJ, vol. 94, pp. 124-135 © American Astronomical Society
</td></tr>
<tr>
<td>
[[File:LedouxPekeris1941.jpg|600px|center|Ledoux & Pekeris (1941, ApJ, 94, 124)]]
</td>
</tr>
</table>
</div>

If we divide their equation (1) through by <math>~Xr = \Gamma_1 P r</math> and recognize that,
<div align="center">
<math>
\frac{dX}{dr} = \frac{dX}{dm}\frac{dm}{dr} = - \Gamma_1 g_0 \rho \, ,
</math>
</div>

we obtain,

<div align="center">
<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>
</div>
Clearly, this 2<sup>nd</sup>-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,

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

<tr>
<td align="right">
<math>~(r^4 \Gamma_1 P)\frac{d^2\xi}{dr^2} </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>
</table>
</div>

<span id="RewrittenLAWE">we can write,</span>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
<td align="right">
<math>~0</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\frac{d}{dr}\biggl[ r^4 \Gamma_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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>

<tr>
<td align="right">
(checked for n = 5) ==>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>
</table>
</div>
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 & Pekeris (1941)], while the next-to-last expression is identical to equation (58.1) of [http://adsabs.harvard.edu/abs/1958HDP....51..353L 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 [[#RewrittenLAWE|appears above]] in our review of the paper by [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux & 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" id="FoundationalVariationalRelation">
<table border="0" cellpadding="5" align="center">
<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>

===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">

<tr>
<td align="right">
<math>~\frac{\sigma^2 \rho \psi^2}{r^2}</math>
</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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>

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

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

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</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>
</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">
<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>;        
</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>

<tr>
<td align="left">
<math>~x = \biggl(\frac{3\cdot 5 - \xi^2}{3\cdot 5} \biggr)</math>;        
</td>
<td align="left">
<math>~\frac{dx}{d\xi} = -\frac{2\xi}{3\cdot 5}</math>;        
</td>
</tr>

<tr>
<td align="left">
<math>~\psi = a_5^3 \xi^3 x</math>;        
</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>
</td>
</tr>
</table>
</div>

----

Then,
<div align="center">
<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">
 
</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">
 
</td>
<td align="center">
 
</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">
 
</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>
<td align="right">
<math>~\Rightarrow ~~~ \frac{5 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 P_c a_5^2}</math>
</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">
 
</td>
<td align="center">
 
</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">
 
</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">
 
</td>
<td align="center">
 
</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">
 
</td>
<td align="center">
 
</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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</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>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
 
</td>
<td align="left">
<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>
</td>
</tr>
</table>
</div>

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

<tr>
<td align="right">
<math>~\int_0^R (\sigma^2 \rho \psi^2)\frac{dr}{r^2}</math>
</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>
</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 — 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,
<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 align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Gamma_1 P r \xi \frac{d}{dr} \biggl[r^3 \xi\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~
\Gamma_1 P r \xi \biggl[3r^2 \xi + r^3 \frac{d\xi}{dr}\biggr]
</math>
</td>
</tr>

<tr>
<td align="right">
 
</td>
<td align="center">
<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>
</td>
</tr>
</table>
</div>
So, as is pointed out by [http://adsabs.harvard.edu/abs/1958HDP....51..353L 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,
<div align="center" id="SufaceBC">
<math>~\frac{d\ln\xi}{d\ln r}\biggr|_\mathrm{surface} = -3 \, .</math>
</div>

(An [[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 & 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 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>

<tr>
<td align="right">
 
</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>
</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 align="left">
<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>
</td>
</tr>
</table>
</div>

If the last term (boundary conditions) is set to zero, then we may also write,
<div 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 & 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>
<td align="right">
<math>~\sigma_0^2 </math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</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 — 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 [http://adsabs.harvard.edu/abs/1941ApJ....94..124L 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,

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

<tr>
<td align="right">
<math>~\int_0^R 4\pi \sigma^2 \rho r^4 \xi^2 dr</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<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>
</td>
</tr>
</table>
</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,

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

<tr>
<td align="right">
<math>~ \sigma^2 \int_0^R r^2 \xi^2 dm</math>
</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>
</td>
</tr>
</table>
</div>

Now, as we have [[SSCpt1/Virial#Wgrav|discussed separately]] — see, also, p. 64, Equation (12) of [<b>[[Appendix/References#C67|<font color="red">C67</font>]]</b>] — 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 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 [[SSCpt1/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 — see the text at the top of p. 126 of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux & Pekeris (1941)] — the moment of inertia of the configuration about its center is,
<div align="center">
<math>
I = \int_0^M r_0^2 dm
\, .</math>
</div>
(Note that, defined in this way, <math>~I</math> is the same as [[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 [[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 align="left">
<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>
</td>
</tr>
</table>
</div>

==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 — that is, homologous expansion/contraction — then this last integral expression gives,

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

<tr>
<td align="right">
<math>~ \sigma^2 I</math>
</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>
</tr>
</table>
</div>

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">
 
</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">
 
</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">
<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>
in which case,

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

<tr>
<td align="right">
<math>~\chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>
</td>
<td align="center">
<math>~=</math>
</td>
<td align="left">
<math>~2W_\mathrm{grav} + (2-3\Gamma_1) (W_\mathrm{grav} - 3P_eV) + 6 P_e V </math>
</td>
</tr>

<tr>
<td align="right">
 
</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>
</tr>
</table>
</div>

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> — 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.

=See Also=

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Part II: Exploratory Ideas

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