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

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</div>
</div>


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>
</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">
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<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>
</div>
  <td align="center" colspan="3"><font color="maroon"><b>Foundational Variational Relation</b></font></td>
 
</tr>
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,
 
<table align="center" border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\sigma^2 \rho r^4 \xi^2</math>
r^2 \frac{dP}{dm} 
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
\rightarrow
</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}
-\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>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
<math>
<math>~\frac{\sigma^2 \rho \psi^2}{r^2}</math>
\frac{Gm}{r^2}
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
\rightarrow
</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t} \biggr]^{-2} \approx \frac{Gm}{ r_0^2}  
- \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]  
\biggl[1 -2 x~ e^{i\omega t} \biggr] \, .
- (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>


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,
<div align="center">
<table align="center" border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\delta \biggl( - \frac{Gm}{r^2}  \biggr)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~\approx
</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~
\frac{Gm}{ r_0^2}  
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr-3 \Gamma_1 P \psi ~\biggr]  
\biggl[2 x~ e^{i\omega t}  \biggr] \, .
- (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


And, drawing a parallel with their equation (59.2), we have,
<div align="center">
<table align="center" border="0" cellpadding="5">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr) = \delta \biggl( - 4\pi r^2 \frac{dP}{dm}  \biggr)
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~\approx
</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}
(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>
   </td>
   </td>
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   </td>
   </td>
   <td align="center">
   <td align="center">
<math>
<math>~=</math>
~=
</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\biggl\{  
<math>~
\frac{Gm}{r_0^2}\biggl[(2x)\biggr]
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
-4\pi r_0^2 \frac{dP_0}{dm} \biggl[(p) \biggr]  
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
- 4\pi P_0 r_0^2 \frac{dp}{dm} \biggr\} e^{i\omega t}  
- \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>
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   <td align="right">
   <td align="right">
&nbsp;
&nbsp;
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>~\biggl\{
\biggl( \frac{2 Gm}{r_0^2}\biggr) x
-\frac{1}{\rho_0} \cdot \frac{d}{dr_0}  \biggl[ P_0 p \biggr]
\biggr\} e^{i\omega t} \, .
</math>
  </td>
</tr>
</table>
</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,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~p = \gamma_\mathrm{g} d</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
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   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~- \gamma_\mathrm{g} \biggl[  
<math>~
3x + r_0 \frac{dx}{dr_0}  
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr)
\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>
   </td>
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   </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>~
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>
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</div>
</div>


Hence,
<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">
<div align="center">
<table align="center" border="0" cellpadding="5">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="left">
<math>
<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>; &nbsp; &nbsp; &nbsp; &nbsp;
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr)  
</math>
   </td>
   </td>
   <td align="center">
   <td align="left">
<math>
<math>~\frac{d\theta_5}{d\xi} = - \frac{\xi}{3}\biggl(\frac{3+\xi^2}{3}\biggr)^{-3 / 2}</math>; &nbsp; &nbsp; &nbsp; &nbsp;
~\approx
  </td>
</math>
</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>
   <td align="left">
   <td align="left">
<math>~\biggl\{  
<math>~
\biggl( \frac{2 Gm}{r_0^2}\biggr) x
\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) \, .
+\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]
\biggr\} e^{i\omega t} \, .
</math>
</math>
   </td>
   </td>
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</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>
----
 
Then,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 336: Line 316:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_0 \mathcal{V}</math>
<math>~\mathrm{RHS}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 343: Line 323:
   <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\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>
   </td>
   </td>
</tr>
</tr>
</table>
</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.,
<div align="center">
<math>~\mathcal{L} = \rho_0 [\mathcal{K} - \mathcal{V}] \, ,</math>
</div>
becomes (see their equation 59.5),
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\mathcal{L}</math>
<math>~\Rightarrow ~~~ \frac{\mathrm{RHS}}{P_c a_5^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 365: Line 338:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\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 \, .
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>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<span id="LDefinition">Noting that,</span> <math>~\dot\zeta = i\omega r_0 x e^{i\omega t}</math>, this in turn gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~L \equiv \int_0^R 4\pi r_0^2 \mathcal{L} dr_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 384: Line 354:
   </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
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>
   </td>
   </td>
Line 395: Line 366:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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
- \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>
   </td>
   </td>
Line 412: Line 383:
   </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
- 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>
   </td>
   </td>
Line 420: Line 393:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\Rightarrow ~~~ \frac{5 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 P_c a_5^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 426: Line 399:
   </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>~
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2   
- 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2  
- 4 r_0^3 x^2 \frac{dP_0}{dr_0}
+ \xi^2 (3+\xi^2) ( 3^2 - \xi^2 )^2
- 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
</math>
</math>
   </td>
   </td>
Line 439: Line 411:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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>~
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2  
- 3(3+\xi^2)^4 \biggl\{  
- 4 r_0^3 x^2 \frac{dP_0}{dr_0}
\xi^\biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr]  \frac{dx}{d\xi}
+r_0^3 x^2 \frac{d}{dr_0}\biggl(3\gamma_\mathrm{g}P_0\biggr)
+ 3\xi^2 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^3} \biggr]  
-\frac{d}{dr_0}\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]
+ \xi^3 x \biggl[ \frac{ -2\xi }{(3+\xi^2)^3} \biggr
\biggr\}dr_0
-2\cdot 3 \xi^4 x \biggl[ \frac{( 3^2 - \xi^2)}{(3+\xi^2)^4} \biggr]
\biggr\}  
</math>
</math>
   </td>
   </td>
Line 460: Line 433:
   </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
\xi^2 (3+\xi^2)  ( 3^2 - \xi^2 )^2
- \int_0^\gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
- 2^2 \cdot 3^2 \cdot 5~\xi^4 x^2  
+ \int_0^R r_0^3 x^2 \frac{d}{dr_0}\biggl[ (3\gamma_\mathrm{g} - 4)P_0\biggr]dr_0
- 3\xi^3 ( 3^2 - \xi^2) (3+\xi^2) \biggl[ - \frac{2\xi}{ 3\cdot 5} \biggr]
-\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]_0^{R}
\biggr\} \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</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:
# 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.
==LP41 Again==
After setting the last term to zero, this last expression can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2 e^{-2i\omega t} L </math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp;
   </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
- 3 \xi^2\biggl\{
- \int_0^R 4\pi \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
3 ( 3^2 - \xi^2) (3+\xi^2
- (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
-2 \xi^2 (3+\xi^2)
-2\cdot 3 \xi^2 ( 3^2 - \xi^2) 
\biggr\} x
</math>
</math>
   </td>
   </td>
Line 500: Line 461:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<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>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
- \omega^2 \int_0^R x^2 r_0^2 dm
5 (3+\xi^2)  ( 3^2 - \xi^2 )^2
- \gamma_\mathrm{g} \int_0^R  \biggl[ r_0 \biggl( \frac{\partial x}{\partial r_0}\biggr) \biggr]^2 P_0 dV
- 2^2~\xi^2 (15-\xi^2)^2
- (3\gamma_\mathrm{g} - 4) \int_0^R  r_0 x^2 \biggl( \frac{Gm}{r_0^2} \biggr)dm
+ 2 \xi^2  ( 3^2 - \xi^2) (3+\xi^2)  
</math>
</math>
   </td>
   </td>
Line 516: Line 477:
<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">
<math>~=</math>
&nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\omega^2
+ \biggl[
+ \biggl\{ \frac{\gamma_\mathrm{g} \int_0^R \bigl[ r_0 \bigl( \frac{\partial x}{\partial r_0}\bigr) \bigr]^2 P_0  dV
2 \xi^2 (3+\xi^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}
+ 2\cdot 3 \xi^2 ( 3^2 - \xi^2
\biggr\} \, .
- 3 ( 3^2 - \xi^2) (3+\xi^2
\biggr] (15-\xi^2)
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</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.
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>
<tr>
   <td align="right">
   <td align="right">
<math>~W_\mathrm{grav}</math>
&nbsp;
  </td>
</td>
   <td align="center">
   <td align="center">
<math>~=</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm \, ;</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>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


for adiabatic systems, the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Reservoir|internal energy]] is,
<tr>
<div align="center">
   <td align="right">
<math>
U_\mathrm{int}
=  \frac{1}{({\gamma_g}-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>
I =  \int_0^M r_0^2 dm
\, .</math>
</div>
Hence, the function to be minimized may be written as,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
   <td align="right">
&nbsp;
&nbsp;
   </td>
   </td>
Line 578: Line 517:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~  
<math>~
\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[ 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)
- (3\gamma_\mathrm{g} - 4) \int_0^R  x^2 dW_\mathrm{grav}}{\int_0^R x^2 dI}
\biggr] (15-\xi^2)
\biggr\} \, .
</math>
</math>
   </td>
   </td>
Line 587: Line 525:
</table>
</table>
</div>
</div>
This expression appears in equation (9) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)].


==Chandrasekhar (1964)==
Coefficients of various powers of <math>~\xi</math>:
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 532:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2L </math>
<math>~\xi^0:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </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>~3^5\cdot 5 -3^5\cdot 5 = 0</math>
+ \biggl( \frac{4Gm}{r_0^3}\biggr) \rho_0 \biggl( \frac{\psi}{r_0^2} \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>
   </td>
</tr>
</tr>
Line 611: Line 544:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\xi^2:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </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>~-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>
- 4\biggl( \frac{dP_0}{dr_0}\biggr) \biggl( \frac{\psi^2}{r_0^3} \biggr)
- \gamma_\mathrm{g} P_0 \biggl[ \frac{1}{r_0} \frac{\partial(\psi)}{\partial r_0} \biggr]^2 \biggr\}dr_0
</math>
   </td>
   </td>
</tr>
</tr>
Line 629: Line 559:
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </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>~= 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>
+ 4\biggl( \frac{dP_0}{dr_0}\biggr) \biggl( \frac{\psi^2}{r_0} \biggr)
+ \gamma_\mathrm{g} P_0 \biggl[ \frac{\partial\psi}{\partial r_0} \biggr]^2 \biggr\} \frac{dr_0}{r_0^2} .
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</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=
 
==Ledoux's Expression==
Returning to the last line of our [[#LDefinition|above definition of the Lagrangian]], that is,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~L </math>
<math>~\xi^4:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~ 2\pi e^{2i\omega t} \biggl\{
<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>
- \int_0^R \rho_0 \omega^2 r_0^4 x^2 dr_0
- \int_0^R  \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
+ \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}
\biggr\} \, ,
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</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> and making the substitution, <math>~r_0/R_\mathrm{eq} \rightarrow \xi/\tilde\xi</math>, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\frac{L_{\{\}} }{P_c R_\mathrm{eq}^3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<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>
- \int_0^R \frac{\rho_0 \omega^2}{P_c R_\mathrm{eq}^3} r_0^4 x^2 dr_0
- \int_0^R  \gamma_\mathrm{g} \frac{P_0}{P_c R_\mathrm{eq}^3} r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2 dr_0
+ \int_0^R r_0^3 x^2 \frac{d}{dr_0}\biggl[ (3\gamma_\mathrm{g} - 4)\frac{P_0}{P_c R_\mathrm{eq}^3}\biggr]dr_0
- 3 \gamma_\mathrm{g} x_\mathrm{surf}^2 \frac{P_e}{P_c}
</math>
   </td>
   </td>
</tr>
</tr>
Line 693: Line 592:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\xi^6:</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=</math>
&nbsp; &nbsp;
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~5 - 2^2 -2 -2 +2\cdot 3 -3 = 0</math>
- \int_0^{\tilde\xi} \omega^2 \biggl[\frac{\rho_c  R_\mathrm{eq}^2}{P_c } \biggr] \biggl( \frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^4 x^2 \frac{d\xi}{\tilde\xi}
- \int_0^{\tilde\xi}  \gamma_\mathrm{g} \biggl(\frac{P_0}{P_c }\biggr) \biggl(\frac{\xi}{\tilde\xi}\biggr)^4\biggl[ \frac{\partial x}{\partial (\xi/\tilde\xi)}\biggr]^2 \frac{d\xi}{\tilde\xi}
+ \int_0^{\tilde\xi} \biggl(\frac{\xi}{\tilde\xi}\biggr) x^2 \frac{d}{d(\xi/\tilde\xi)}\biggl[ (3\gamma_\mathrm{g} - 4)\frac{P_0}{P_c }\biggr] \frac{d\xi}{\tilde\xi}
- 3 \gamma_\mathrm{g} x_\mathrm{surf}^2 \frac{P_e}{P_c}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
</td></tr>
  <td align="right">
</table>
&nbsp;
 
  </td>
<span id="ChandraEq49">Multiplying through by <math>~dr</math>, and integrating over the volume gives,</span>
  <td align="center">
 
<math>~=</math>
<div align="center">
  </td>
<table border="0" cellpadding="5" align="center">
  <td align="left">
<math>~
- \omega^2 \biggl[\frac{\rho_c  R_\mathrm{eq}^2}{P_c ~{\tilde\xi}^5} \biggr]  \int_0^{\tilde\xi} \theta^n \xi^4 x^2 d\xi
- \frac{\gamma_\mathrm{g}}{ {\tilde\xi}^3} \int_0^{\tilde\xi}  \theta^{n+1} \xi^4\biggl[ \frac{\partial x}{\partial \xi}\biggr]^2  d\xi
+ \frac{(3\gamma_\mathrm{g} - 4)}{\tilde\xi}\int_0^{\tilde\xi} \xi x^2 \frac{d}{d\xi}\biggl[ \theta^{n+1}\biggr] d\xi
- \biggl[\frac{3 \gamma_\mathrm{g}P_e}{P_c} \biggr]x_\mathrm{surf}^2
</math>
  </td>
</tr>


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\int_0^R (\sigma^2 \rho \psi^2)\frac{dr}{r^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 734: Line 621:
   <td align="left">
   <td align="left">
<math>~
<math>~
- \omega^2 \biggl[\frac{\rho_c  R_\mathrm{eq}^2}{P_c ~{\tilde\xi}^5} \biggr]  \int_0^{\tilde\xi} \theta^n \xi^4 x^2 d\xi
\int_0^R \biggl[ \Gamma_1 P \biggl(\frac{d\psi}{dr} \biggr)^2
- \biggl[\frac{3 \gamma_\mathrm{g}P_e}{P_c} \biggr]x_\mathrm{surf}^2  
+ \frac{4\psi^2}{r} \biggl( \frac{dP}{dr} \biggr) \biggr]\frac{dr}{r^2}
+ \frac{1}{ {\tilde\xi}^3} \int_0^{\tilde\xi} \biggl[ (3\gamma_\mathrm{g} - 4) {\tilde\xi}^2 \xi x^2 \frac{d\theta^{n+1}}{d\xi}
- \biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr]_0^R \, ,
- \gamma_\mathrm{g} \theta^{n+1} \xi^4\biggl( \frac{\partial x}{\partial \xi}\biggr)^2  \biggr]d\xi
</math>
</math>
   </td>
   </td>
Line 743: Line 629:
</table>
</table>
</div>
</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.


where, we have set the pressure at the (truncated) surface to the value, <math>~P_0|_\mathrm{surface} = P_e</math>.
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,
 
==Chandra's Expression==
Alternatively, starting from Chandrasekhar's expression,
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 753: Line 637:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~2L_{\{\}} </math>
<math>~\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr)</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 759: Line 643:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\int_0^R \biggl\{ \omega^2 \rho_0 (r_0^3 x)^2
<math>~
+ 4\biggl( \frac{dP_0}{dr_0}\biggr) \frac{(r_0^3 x)^2}{r_0}  
\Gamma_1 P r \xi \frac{d}{dr} \biggl[r^3 \xi\biggr]
+ \gamma_\mathrm{g} P_0 \biggl[ \frac{\partial (r_0^3 x)}{\partial r_0} \biggr]^2 \biggr\} \frac{dr_0}{r_0^2}
</math>
</math>
   </td>
   </td>
Line 768: Line 651:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ \frac{2L_{\{\}} }{P_c R_\mathrm{eq}^3}</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 774: Line 657:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\int_0^R \biggl\{ \frac{\omega^2 \rho_c}{P_c} \biggl( \frac{\rho_0}{\rho_c}\biggr) r_0^4 x^2
<math>~
+ 4 r_0^3 x^2 \cdot \frac{d}{dr_0}\biggl(\frac{P_0}{P_c}\biggr)
\Gamma_1 P r \xi \biggl[3r^2 \xi + r^3 \frac{d\xi}{dr}\biggr]
+ \frac{\gamma_\mathrm{g} P_0}{r_0^2 P_c} \biggl[ \frac{\partial (r_0^3 x)}{\partial r_0} \biggr]^2 \biggr\} \frac{dr_0 }{R_\mathrm{eq}^3}
</math>
</math>
   </td>
   </td>
Line 789: Line 671:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\int_0^{\tilde\xi} \biggl\{ \frac{\omega^2 \rho_c R_\mathrm{eq}^2}{P_c} \biggl( \frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{\xi}{ {\tilde\xi}} \biggr)^4 x^2
<math>~
+ 4 \tilde\xi ~x^2 \biggl(\frac{\xi}{ {\tilde\xi}} \biggr)^3 \frac{d}{d\xi}\biggl(\frac{P_0}{P_c}\biggr)
\Gamma_1 P r^3 \xi^2 \biggl[3 + \frac{d\ln\xi}{d\ln r}\biggr] \, .
+ \frac{\gamma_\mathrm{g} P_0}{ P_c} \biggl(\frac{\xi}{ {\tilde\xi}} \biggr)^{-2}\biggl[ \frac{\partial }{\partial (\xi/\tilde\xi)}\biggl(\frac{\xi^3 x}{ {\tilde\xi}^3} \biggr) \biggr]^2 \biggr\} \frac{d\xi }{ {\tilde\xi}}
</math>
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\sigma^2 \rho r^4 \xi^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 804: Line 698:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\int_0^{\tilde\xi} \biggl\{ \frac{\omega^2 \rho_c R_\mathrm{eq}^2}{{\tilde\xi}^4 P_c} \biggl( \theta^n \xi^4 \biggr) x^2
<math>~
+ \biggl(\frac{4 }{{\tilde\xi}^2}\biggr) ~x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
-\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr]  
+ \frac{\gamma_\mathrm{g} }{ {\tilde\xi}^2} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \biggl[ \frac{\partial (\xi^3 x)}{\partial \xi} \biggr]^2 \biggr\} \frac{d\xi }{ {\tilde\xi}}
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)  
</math>
</math>
   </td>
   </td>
Line 819: Line 713:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{1}{ {\tilde\xi}^3}\int_0^{\tilde\xi} \biggl\{ \frac{\omega^2 \rho_c R_\mathrm{eq}^2}{{\tilde\xi}^2 P_c} \biggl( \theta^n \xi^4 \biggr) x^2
<math>~
+ 4x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2
+ \gamma_\mathrm{g} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \biggl[ \frac{\partial (\xi^3 x)}{\partial \xi} \biggr]^2 \biggr\} d\xi \, .
- (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>
</tr>
</tr>
</table>
</div>
Now, let's plug in the [[User:Tohline/SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|known eigenfunction for the marginally unstable configuration]], namely,


<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="center" colspan="3"><font color="maroon"><b>Exact Solution to the <math>~(3 \le n < \infty)</math> Polytropic LAWE</b></font></td>
</tr>
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\sigma_c^2 = 0</math>
<math>~\Rightarrow ~~~ \int_0^R\sigma^2 \rho r^4 \xi^2 dr</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp; &nbsp; &nbsp; and &nbsp; &nbsp; &nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~x_P \equiv \frac{3(n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\xi \theta^{n}}\biggr) \frac{d\theta}{d\xi}\biggr] \, .</math>
<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>
   </td>
</tr>
</tr>
Line 848: Line 739:
</div>
</div>


Given that, from the Lane-Emden equation,
If the last term (boundary conditions) is set to zero, then we may also write,
<div align="center">
<math>~\frac{\theta^{''}}{\theta^n} = - 1 -\frac{2\theta^'}{\xi \theta^n} \, ,</math>
</div>
 
we recognize that,
 
<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>~\frac{d(\xi^3 x_P)}{d\xi}</math>
<math>~\sigma^2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 867: Line 751:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{3(n-1)}{2n} \frac{d\xi^3}{d\xi}
\frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr
+\frac{3(n-3)}{2n} \frac{d}{d\xi} \biggl( \frac{\xi^2 \theta^'}{\theta^{n}}\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 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">
&nbsp;
<math>~\sigma_0^2 </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 881: Line 771:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\mathrm{min}~
\frac{3^2(n-1)\xi^2}{2n}
\frac{\int_0^R r^4 \Gamma_1 P \bigl(\frac{d\xi}{dr}\bigr)^2 dr
+\frac{3(n-3)\xi^2}{2n} \biggl[ \frac{\theta^{''}}{\theta^{n}} + \frac{2\theta^'}{\xi \theta^{n}} - \frac{n (\theta^')^2}{\theta^{n+1}}\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 897: Line 796:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{3\xi^2}{2n} \biggl\{ 3(n-1)
\int_0^R 4\pi r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2 dr
-(n-3)\biggl[ 1  + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr] \biggr\}
- \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>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~ \sigma^2 \int_0^R r^2 \xi^2 dm</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 912: Line 823:
   <td align="left">
   <td align="left">
<math>~
<math>~
\frac{3\xi^2}{2} \biggl\{ 2 + (3-n)\biggl[ \frac{(\theta^')^2}{\theta^{n+1}}\biggr]\biggr\} \, .
\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>
   </td>
Line 919: Line 832:
</div>
</div>


Therefore, evaluating the sum of the last two terms inside the curly braces 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>~\biggl\{4x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
<math>~W_\mathrm{grav}</math>
+ \gamma_\mathrm{g} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \biggl[ \frac{d (\xi^3 x)}{d \xi} \biggr]^2\biggr\}_{x_P} </math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 932: Line 843:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm \, ;</math>
4(n+1) \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 \xi^3 \theta^n \theta^'
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<tr>
for adiabatic systems, the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Reservoir|internal energy]] is,
  <td align="right">
<div align="center">
&nbsp;
<math>
  </td>
U_\mathrm{int}
  <td align="center">
=  \frac{1}{(\Gamma_1-1)} \int_0^R  P_0 dV
&nbsp;
\, ;</math>
  </td>
</div>
  <td align="left">
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,
<math>~
<div align="center">
+ \frac{(n+1)}{n} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \frac{3^2\xi^4}{2^2} \biggl\{ 2 + (3-n)\biggl[ \frac{(\theta^')^2}{\theta^{n+1}}\biggr]\biggr\}^2
<math>
</math>
I =  \int_0^M r_0^2 dm
   </td>
\, .</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>
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~ \sigma^2 \int_0^R \xi^2 dI</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 961: Line 880:
   <td align="left">
   <td align="left">
<math>~
<math>~
4(n+1) \frac{3^2(n-1)^2}{2^2n^2}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 \xi^3 \theta^n \theta^'
\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^\xi^2 dW_\mathrm{grav}
+ 3^2 \Gamma_1  P_e V \xi_\mathrm{surface}^2 \, .
</math>
</math>
   </td>
   </td>
</tr>
</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 &#8212; that is, homologous expansion/contraction &#8212; then this last integral expression gives,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~ \sigma^2 I</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~
+ \frac{(n+1)}{n} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \frac{3^2\xi^4}{2^2} \biggl[ 2 + \frac{(3-n)(\theta^')^2}{\theta^{n+1}}\biggr]^2
(4 - 3\Gamma_1) W_\mathrm{grav}
+ 3^2 \Gamma_1  P_e V  \, .
</math>
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
Contrast this result with the following free-energy analysis:
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\mathfrak{G}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 988: Line 925:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3^2(n+1) \xi^2 }{2^2n^2}\biggl\{
<math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV \, ,</math>
2^2(n-1)^2 \biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2 \xi \theta^n \theta^'
+ n \theta^{n+1} \biggl[ 2 + \frac{(3-n)(\theta^')^2}{\theta^{n+1}}\biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~W_\mathrm{grav}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,003: Line 942:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3^2(n+1) \xi^2 }{2^2n^2}\biggl\{
<math>~-a\chi^{-1}</math>
2^2 \biggl[(n-1)\xi \theta^{n} + (n-3)\theta^' \biggr]^2 \frac{ \theta^'}{\xi \theta^{n}}
+ \frac{n}{\theta^{n+1} } \biggl[ 2\theta^{n+1} + (3-n)(\theta^')^2\biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,012: Line 948:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~U_\mathrm{int}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,018: Line 954:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{3^2(n+1) \xi }{2^2n^2 \theta^{n+1}}\biggl\{
<math>~b\chi^{3-3\Gamma_1}</math>
2^2 \biggl[(n-1)\xi \theta^{n} + (n-3)\theta^' \biggr]^2 \theta\theta^'
+ n\xi  \biggl[ 2\theta^{n+1} + (3-n)(\theta^')^2\biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>


<!-- OLD INCORRECT DERIVATION
<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">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,033: Line 977:
<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\partial \mathfrak{G}}{\partial \chi}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,039: Line 983:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~+a \chi^{-2} + 3(1-\Gamma_1) b \chi^{2-3\Gamma_1} + 4\pi P_e \chi^{2} </math>
4(n+1) \biggl[\frac{3(n-3)}{2n} \biggr]^2
\biggl\{ \biggl[\frac{3(n-1)}{2n}\cdot \frac{2n}{3(n-3)}\biggr]^2\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2  \xi^3 \theta^n \theta^'
</math>
   </td>
   </td>
</tr>
</tr>
 
<tr>
<tr>
   <td align="right">
   <td align="right">
Line 1,051: Line 992:
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~\chi^{-1} \biggl[- W_\mathrm{grav} + 3(1-\Gamma_1) U_\mathrm{int} + 3 P_e V \biggr] \, ,</math>
+ \frac{1}{4n} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \xi^4\biggl[ 1  + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
and,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,068: Line 1,013:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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>
4(n+1) \biggl[\frac{3(n-3)}{2n} \biggr]^2
\biggl\{\biggl[\frac{(n-1)}{(n-3)} + \biggl( \frac{\theta^'}{\xi \theta^{n}}\biggr) \biggr]^2   \xi^3 \theta^n \theta^'
+ \biggl( \frac{\xi^2 \theta^{n+1}}{4n}\biggr) \biggl[ 1  + \frac{n (\theta^')^2}{\theta^{n+1}}\biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,084: Line 1,025:
   </td>
   </td>
   <td align="left">
   <td align="left">
<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>
4(n+1) \biggl[\frac{3(n-3)}{2n} \biggr]^2
\biggl\{ \frac{\xi\theta^'}{(n-3) \theta^{n}} \biggl[(n-1)\xi \theta^{n} + (n-3)\theta^' \biggr]^2 
+ \biggl( \frac{\xi^2 }{4n\theta^{n+1}}\biggr) \biggl[ \theta^{n+1} + n (\theta^')^2 \biggr]^2 \biggr\}
</math>
   </td>
   </td>
</tr>
</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>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~3(1-\Gamma_1) U_\mathrm{int}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,100: Line 1,043:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~W_\mathrm{grav} - 3 P_e V \, ,</math>
\frac{4(n+1)}{4n(n-3)} \biggl[\frac{3(n-3)}{2n} \biggr]^2
\biggl\{ \frac{4n\xi\theta^'}{ \theta^{n}} \biggl[(n-1)^2\xi^2 \theta^{2n} + 2(n-1)(n-3) \xi \theta^{n}\theta^' + (n-3)^2 (\theta^')^2 \biggr] 
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
in which case,
<div align="center">
<table border="0" cellpadding="5" align="center">


<tr>
<tr>
   <td align="right">
   <td align="right">
&nbsp;
<math>~\chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2}</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
&nbsp;
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~
<math>~2W_\mathrm{grav} + (2-3\Gamma_1) (W_\mathrm{grav} - 3P_eV) + 6 P_e V </math>
+ \frac{(n-3)\xi^2 }{\theta^{n+1}} \biggl[ \theta^{2(n+1)} + 2n\theta^{n+1} (\theta^')^2  + n^2 (\theta^')^4 \biggr] \biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</div>
END OLD INCORRECT DERIVATION -->
==Normalizations==
We will adopt the normalizations used in the [[User:Tohline/SSC/Structure/PolytropesEmbedded#Stahler.27s_Presentation|Stahler presentation]].  Specifically,
<div align="center">
<table border="0" cellpadding="3">


<tr>
<tr>
   <td align="right">
   <td align="right">
<math>
&nbsp;
~M
</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
<math>~=~</math>
<math>~=</math>
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>
<math>~(4-3\Gamma_1)W_\mathrm{grav} + 3^2 \Gamma_1 P_e V \, .</math>
M_\mathrm{SWS} \biggl( \frac{n^3}{4\pi} \biggr)^{1/2} \biggl\{ \theta_n^{(n-3)/2} \xi^2
\biggl| \frac{d\theta_n}{d\xi} \biggr| \biggr\}_{\tilde\xi}
</math>
  </td>
</tr>
 
<tr>
  <td align="right">
<math>
~R_\mathrm{eq}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<math>
R_\mathrm{SWS} \biggl( \frac{n}{4\pi} \biggr)^{1/2} \biggl\{ \xi \theta_n^{(n-1)/2} \biggr\}_{\tilde\xi}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</div>
where,
 
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">
<div align="center">
<math>M_\mathrm{SWS} =  
<math>~\sigma^2 I = \chi^2 \cdot \frac{\partial^2 \mathfrak{G}}{\partial \chi^2} \, .</math>
\biggl( \frac{n+1}{nG} \biggr)^{3/2} K_n^{2n/(n+1)} P_\mathrm{e}^{(3-n)/[2(n+1)]} \, ,</math>
</div>
</div>
<div align="center">
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.
<math>
 
R_\mathrm{SWS} = \biggl( \frac{n+1}{nG} \biggr)^{1/2} K_n^{n/(n+1)} P_\mathrm{e}^{(1-n)/[2(n+1)]} \, .
=Related, Exploratory Ideas=
</math>
<!-- OMIT
</div>
We can rewrite the [[#GoverningIntegral|Variational Principle's Governing Integral Relation]] as,
In addition, from our [[User:Tohline/SSC/Structure/Polytropes#Lane-Emden_Equation|introductory discussion of polytropes]], we have,
 
<div align="center">
<div align="center">
<table border="0" cellpadding="5" align="center">
<table border="0" cellpadding="5" align="center">
Line 1,179: Line 1,094:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_0</math>
<math>~ 3^2 \Gamma_1  P_e V \xi_\mathrm{surface}^2</math>
   </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>~
\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>
   </td>
</tr>
</tr>
Line 1,191: Line 1,111:
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\Rightarrow ~~~ P_0</math>
&nbsp;
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,197: Line 1,117:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~K_n \biggl[ \rho_c \theta^n \biggr]^{(n+1)/n} </math>
<math>~
\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>
   </td>
</tr>
</tr>
</table>
</table>
</div>
</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,


and,
<div align="center">
<math>
r_0 \equiv a_\mathrm{n} \xi ,
</math>
</div>
where,
<div align="center">
<math>~
a^2_\mathrm{n} = \frac{(n+1)K_n}{4\pi G} \cdot \rho_c^{(1-n)/n}  \, .
</math>
</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">
<tr>
<tr>
   <td align="right">
   <td align="right">
<math>~\rho_c</math>
<math>~\sigma^2 \rho r^4 \xi^2</math>
   </td>
   </td>
   <td align="center">
   <td align="center">
Line 1,227: Line 1,142:
   </td>
   </td>
   <td align="left">
   <td align="left">
<math>~\frac{\bar\rho}{ {\tilde{f}}_M}</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>
   </td>
   </td>
</tr>
</tr>
Line 1,233: 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 1,239: Line 1,159:
   </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>~
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr)  
- \sigma^2 \rho r^4 \xi^2
</math>
   </td>
   </td>
</tr>
</tr>
Line 1,251: Line 1,175:
   </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>~\xi^2 \biggl\{
r^2 \Gamma_1 P \biggl(\frac{d\ln\xi}{d\ln r}\biggr)^
- (3\Gamma_1 - 4) r^3 \biggl( \frac{dP}{dr} \biggr)  
- \sigma^2 \rho r^4
\biggr\}
</math>
   </td>
   </td>
</tr>
</tr>
</table>
 
</div>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\alpha \equiv \biggl(3 - \frac{4}{\Gamma_1}\biggr) \, .</math>
</div>
 
==Pressure-Truncated Polytropes==
 
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>
 
===Via Generalized Normalization===
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">
<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>~
- \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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
</table>
</div>
where,
<div align="center">
<math>~\chi \equiv \frac{R}{R_\mathrm{norm}} \, .</math>
</div>
 
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,
 
<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>~
- \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>
</tr>
</table>
</div>
 
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">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <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>
We have,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<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>
 
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">
<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>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<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 align="left">
<math>~(3-n)
\biggl[ \theta^n + \frac{3\theta^'}{\xi} + \frac{n(\theta^')^2}{\theta} \biggr]
\, .
</math>
  </td>
</tr>
</table>
</div>
Hence,
 
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
Second line of relation
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
</table>
</div>
 
Note that, in this derivation, we have inserted the expressions:
<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">
<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>
 
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</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>
  </td>
</tr>
</table>
</div>
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">
<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>
 
<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 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>
</div>
 
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>
 
Now, from an [[User:Tohline/StabilityVariationalPrincipal#Test_Virial_Equilibrium_Condition|accompanying discussion]], we know that, in equilibrium,
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>
~\frac{R_\mathrm{eq}}{R_\mathrm{norm}}
</math>
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <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>
~\frac{P_e}{P_\mathrm{norm}}
</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>
  </td>
</tr>
</table>
</div>
 
Hence,
<div align="center">
<table border="0" cellpadding="3">
 
<tr>
  <td align="right">
<math>
~\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>
  </td>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=~</math>
  </td>
  <td align="left">
<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>
  </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>
  </td>
</tr>
</table>
</div>
 
====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">
<table border="0" cellpadding="5" align="center">
<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>
This means that,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<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>
  </td>
</tr>
</table>
</div>
Hence, also,
<div align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
  </td>
</tr>
</table>
</div>
 
Rather, let's try:
<div align="center">
<table border="0" cellpadding="5" align="center">
<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>
  </td>
</tr>
</table>
</div>
Hence, after setting <math>~\sigma^2 = 0</math>, the [[#PolytropeRelation|above rearranged integral relation]] becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~
- \frac{2^2 n^2}{3(n-3)} \biggl[ {\tilde\xi}^3 {\tilde\theta}^{n+1}  x_\mathrm{surface}^2 \biggr]
</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 align="center">
<table border="0" cellpadding="5" align="center">
 
<tr>
  <td align="right">
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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
 
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
</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>


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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

© 2014 - 2021 by Joel E. Tohline
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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