Editing SSC/VariationalPrinciple/Pt2

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


=Related, Exploratory Ideas=
<!-- OMIT
We can rewrite the [[#GoverningIntegral|Variational Principle's Governing Integral Relation]] as,

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

<tr>
  <td align="right">
<math>~ 3^2 \Gamma_1  P_e V \xi_\mathrm{surface}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>
</table>
</div>
END OMIT -->
==Logarithmic Derivatives==

Returning to our above discussion of the [[#Ledoux_.26_Walraven_Approach|Ledoux &amp; Walraven approach]], we appreciate that the ''differential'' relation governing the Variational Principle is,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma^2 \rho r^4 \xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2  
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) 
- \frac{d}{dr}\biggr[r^4 \Gamma_1 P\xi \biggl(\frac{d\xi}{dr}\biggr) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
r^4 \Gamma_1 P \biggl(\frac{d\xi}{dr}\biggr)^2  
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) 
- \sigma^2 \rho r^4 \xi^2
</math>
  </td>
</tr>

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

<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 [[StabilityVariationalPrinciple#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 [[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===

{| class="PGEclass" style="float:left; margin-right: 20px; border-style: solid; border-width: 3px border-color: black"
|- 
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<font size="-1">[[H_BookTiledMenu#MoreStabilityAnalyses|<b>''Exact''<br />Demonstration<br />of<br />Variational<br />Principle</b>]]</font>
|}

<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 [[StabilityVariationalPrinciple#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 [[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=

* [[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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