Editing SSC/VariationalPrinciple (section)

===Chandrasekhar's Approach===

Next, in an effort to adopt the notation used by [http://adsabs.harvard.edu/abs/1964ApJ...139..664C Chandrasekhar (1964)], we make the substitution, <math>~\xi \rightarrow \psi/r^3</math>, and regroup terms to obtain,
<div align="center">
<table border="0" cellpadding="5" align="center">

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr}  -3 \Gamma_1 P \psi ~\biggr] 
- (3\Gamma_1 - 4) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(4-3\Gamma_1 ) \biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) 
- \biggl( \frac{\psi}{r^3}\biggr) \frac{d}{dr}\biggl[ r \Gamma_1 P ~\frac{d\psi}{dr}  \biggr] 
+ 3 \Gamma_1 \biggl( \frac{\psi^2}{r^3}\biggr) \frac{dP}{dr} 
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) 
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
- \biggl\{ 
\frac{d}{dr}\biggl[ r \Gamma_1 P \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}\biggr] -r\Gamma_1 P ~\frac{d\psi}{dr} \cdot \frac{d}{dr}\biggl( \frac{\psi}{r^3}\biggr)
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) 
+3 \Gamma_1 P  \biggl( \frac{\psi}{r^3}\biggr) \frac{d\psi}{dr}
+ \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2
- \biggl[\frac{3\Gamma_1 P\psi}{r^3}\biggr]\frac{d\psi}{dr} 
- \frac{d}{dr}\biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl( \frac{\psi^2}{r^3} \biggr) \biggl( \frac{dP}{dr} \biggr) 
+ \frac{\Gamma_1 P}{r^2} \biggl[\frac{d\psi}{dr} \biggr]^2
- \frac{d}{dr}\biggl[ \frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

<table border="1" align="center" width="80%" cellpadding="5">
<tr><td align="left">
Let's check to see whether the terms in the RHS of this last expression sum to zero when we plug in the appropriate functions for the marginally unstable, n = 5 configuration.  In particular (replacing <math>~\xi</math> with <math>~x</math>, and setting <math>~r = a_5\xi</math>), we start with knowing,
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<math>~\theta_5 = \biggl(\frac{3+\xi^2}{3}\biggr)^{-1 / 2}</math>; &nbsp; &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>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
</tr>

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

<tr>
  <td align="left">
<math>~\psi = a_5^3 \xi^3 x</math>; &nbsp; &nbsp; &nbsp; &nbsp;
  </td>
  <td align="left">
<math>~
\frac{d\psi}{d\xi} = a_5^3 \biggl[ 3\xi^2 x + \xi^3 \biggl(\frac{dx}{d\xi}\biggr)\biggr] = \frac{a_5^3 \xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

----

Then,
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<math>~\mathrm{RHS}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[ \frac{a_5^6 \xi^6 x^2}{a_5^3 \xi^3} \biggr] \frac{P_c}{a_5} \cdot \frac{d\theta^6}{d\xi} 
+ P_c \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{a_5^4 \xi^2} \biggl\{ \frac{d\psi}{d\xi} \biggr\}^2
- \frac{P_c}{a_5} \cdot \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 a_5^3 \xi^3 x}{a_5^3 \xi^2} \biggl( \frac{d\psi}{d\xi} \biggr) \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\mathrm{RHS}}{P_c a_5^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4\biggl[ \xi^3 x^2 \biggr]  \frac{d\theta^6}{d\xi} 
+ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6}{ \xi^2} \biggl\{ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr) \biggr\}^2
- \frac{d}{d\xi}\biggl\{ \biggl(\frac{n+1}{n}\biggr) \frac{\theta^6 \xi^3 x}{ \xi^2} \biggl[ \frac{\xi^2}{3}\biggl( 3^2 - \xi^2 \biggr)\biggr] \biggr\} 
</math>
  </td>
</tr>

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

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- 3 \xi^2\biggl\{ 
3 ( 3^2 - \xi^2) (3+\xi^2)   
-2 \xi^2  (3+\xi^2) 
-2\cdot 3 \xi^2 ( 3^2 - \xi^2)   
\biggr\} x
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{5^2 (3+\xi^2)^4 [\mathrm{RHS} ]}{ 2\cdot 3^2 ~\xi^2 P_c a_5^2}</math>
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
5 (3+\xi^2)  ( 3^2 - \xi^2 )^2
- 2^2~\xi^2 (15-\xi^2)^2 
+ 2 \xi^2  ( 3^2 - \xi^2) (3+\xi^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 
2 \xi^2  (3+\xi^2) 
+ 2\cdot 3 \xi^2 ( 3^2 - \xi^2)   
- 3 ( 3^2 - \xi^2) (3+\xi^2)   
\biggr] (15-\xi^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
 </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(3\cdot 5 + 5\xi^2)  ( 3^4 - 2\cdot 3^2\xi^2 + \xi^4)
- 2^2~\xi^2 (3^2\cdot 5^2 - 2\cdot 3\cdot 5 \xi^2 + \xi^4)
+ 2 \xi^2  ( 3^3 + 2\cdot 3\xi^2 -\xi^4)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \biggl[ 2\cdot 3 \xi^2 + 2\xi^4 + 2\cdot 3^3 \xi^2 - 2\cdot 3 \xi^4 - 3(3^3 + 2\cdot 3\xi^2 -\xi^4)
\biggr] (15-\xi^2)
</math>
  </td>
</tr>
</table>
</div>

Coefficients of various powers of <math>~\xi</math>:
<div align="center">
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<tr>
  <td align="right">
<math>~\xi^0:</math>
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~3^5\cdot 5 -3^5\cdot 5 = 0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\xi^2:</math>
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~-2\cdot 3^3\cdot 5 + 3^4\cdot 5 -2^2 \cdot 3^2 \cdot 5^2 +2\cdot 3^3 + 2\cdot 3^2\cdot 5 + 2\cdot 3^4\cdot 5 - 2\cdot 3^3\cdot 5 + 3^4</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~= 3^2\cdot 5[-2\cdot 3 + 3^2  + 2 + 2\cdot 3^2 - 2\cdot 3] + 3^2[ 2\cdot 3 + 3^2 - 2^2 \cdot 5^2  ] = 3^2\cdot 5[17 ] - 3^2[5\cdot 17  ] = 0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\xi^4:</math>
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~3\cdot 5 -2\cdot 3^2\cdot 5 +2^3\cdot 3\cdot 5 + 2^2\cdot 3 + 2\cdot 3 \cdot 5 - 2\cdot 3^2\cdot 5 - 2\cdot 3 -2\cdot 3^3 + 2\cdot 3^2 + 3^2\cdot 5</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~= 3\cdot 5[1 -2\cdot 3 +2^3 + 2 - 2\cdot 3 + 3] + 2\cdot 3[2 - 1 - 3^2 + 3] = 2\cdot 3\cdot 5 - 2\cdot 3\cdot 5 = 0 </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\xi^6:</math>
  </td>
  <td align="center">
&nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~5 - 2^2 -2 -2 +2\cdot 3 -3 = 0</math>
  </td>
</tr>
</table>
</div>

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

<span id="ChandraEq49">Multiplying through by <math>~dr</math>, and integrating over the volume gives,</span>

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  <td align="right">
<math>~\int_0^R (\sigma^2 \rho \psi^2)\frac{dr}{r^2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\int_0^R \biggl[ \Gamma_1 P \biggl(\frac{d\psi}{dr} \biggr)^2
+ \frac{4\psi^2}{r} \biggl( \frac{dP}{dr} \biggr) \biggr]\frac{dr}{r^2}
- \biggl[\frac{\Gamma_1 P \psi}{r^2} \biggl( \frac{d\psi}{dr} \biggr) \biggr]_0^R \, , 
</math>
  </td>
</tr>
</table>
</div>
which is identical to equation (49) of [http://adsabs.harvard.edu/abs/1964ApJ...139..664C Chandrasekhar (1964)], if the last term &#8212; the difference of the central and surface boundary conditions &#8212; is set to zero.

Note that if we shift from the variable, <math>~\psi</math>, back to the fractional displacement function, <math>~\xi</math>, the last term in this expression may be written as,
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<table border="0" cellpadding="5" align="center">

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\Gamma_1 P r^3 \xi^2 \biggl[3 + \frac{d\ln\xi}{d\ln r}\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
So, as is pointed out by [http://adsabs.harvard.edu/abs/1958HDP....51..353L Ledoux &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 [[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.)