Editing Appendix/Ramblings/LedouxVariationalPrinciple (section)

===Chandrasekhar's Independent Derivation===
Now, let's follow Chandrasekhar's lead and derive the Lagrangian directly from the governing LAWE. We begin with a version 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)], namely,

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

<tr>
  <td align="right">
<math>~0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] 
+\biggl[ \sigma^2 \rho r^4 + (3\Gamma_1 - 4) r^3 \frac{dP}{dr} \biggr] \xi \, .
</math>
  </td>
</tr>
</table>
</div>
We will develop the Lagrangian expression by 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)].  First, we multiply the LAWE through by the fractional displacement, <math>~\xi</math>; second, we make the substitution, <math>~\xi \rightarrow \psi/r^3</math>, in order to shift to Chandrasekhar's variable notation; then we multiply through by <math>~dr</math> and integrate from the center <math>~(r = 0)</math> to the surface <math>~(r = R)</math> of the configuration.

Multiplying through by the fractional displacement gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\sigma^2 \rho r^4 \xi^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\xi \cdot \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{d\xi}{dr} \biggr] 
- (3\Gamma_1 - 4) r^3 \xi^2 \biggl( \frac{dP}{dr} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>
Next, making the stated variable substitution gives,
<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>
Finally, integrating over the volume gives,

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

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