Editing Appendix/Ramblings/LedouxVariationalPrinciple

__FORCETOC__ 
<!-- __NOTOC__ will force TOC off -->

=Ledoux's Variational Principle (Supporting Derivations)=

An [[User:Tohline/SSC/VariationalPrinciple#Ledoux.27s_Variational_Principle|accompanying chapter]] presents a summary discussion of Ledoux's variational principle, emphasizing how it relates to stability analyses that are built upon free-energy arguments.  The detailed derivations presented below provide the foundation upon which that summary discussion has been built.

{{ SGFworkInProgress }}

=Our Initial Explorations=
==Review by Ledoux and Walraven (1958)==
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 

<div align="center">
<span id="PGE:Euler"><font color="#770000">'''Euler + Poisson Equations'''</font></span><br />
<math>\frac{d^2 r}{dt^2} = - 4\pi r^2 \frac{dP}{dm} - \frac{Gm}{r^2} </math><br />
</div>

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>
  <td align="right">
<math>
r^2 \frac{dP}{dm}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<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}
</math>
  </td>
</tr>
<tr>
  <td align="right">
<math>
\frac{Gm}{r^2}  
</math>
  </td>
  <td align="center">
<math>
\rightarrow
</math>
  </td>
  <td align="left">
<math>
\frac{Gm}{ r_0^2} \biggl[1 + x~ e^{i\omega t}  \biggr]^{-2} \approx \frac{Gm}{ r_0^2} 
\biggl[1 -2 x~ e^{i\omega t}  \biggr] \, .
</math>
  </td>
</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>
  <td align="right">
<math>
~\delta \biggl( - \frac{Gm}{r^2}  \biggr) 
</math>
  </td>
  <td align="center">
<math>
~\approx
</math>
  </td>
  <td align="left">
<math>
\frac{Gm}{ r_0^2} 
\biggl[2 x~ e^{i\omega t}  \biggr] \, .
</math>
  </td>
</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>
  <td align="right">
<math>
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr) = \delta \biggl( - 4\pi r^2 \frac{dP}{dm}  \biggr)
</math>
  </td>
  <td align="center">
<math>
~\approx
</math>
  </td>
  <td align="left">
<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}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>
~=
</math>
  </td>
  <td align="left">
<math>~\biggl\{ 
\frac{Gm}{r_0^2}\biggl[(2x)\biggr] 
-4\pi r_0^2 \frac{dP_0}{dm} \biggl[(p) \biggr] 
- 4\pi P_0 r_0^2 \frac{dp}{dm} \biggr\} e^{i\omega t} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&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 align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \gamma_\mathrm{g} \biggl[ 
3x + r_0 \frac{dx}{dr_0} 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- \frac{\gamma_\mathrm{g}}{r_0^2} \frac{d}{dr_0} \biggl( r_0^3 x \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

Hence,
<div align="center">
<table align="center" border="0" cellpadding="5">
<tr>
  <td align="right">
<math>
~\delta \biggl( - \frac{1}{\rho} \cdot \frac{dP}{dr} \biggr) 
</math>
  </td>
  <td align="center">
<math>
~\approx
</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[ \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>
  </td>
</tr>
</table>
</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>
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\rho_0 \mathcal{V}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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 \, .
</math>
  </td>
</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>
  <td align="right">
<math>~\mathcal{L}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\rho_0}{2} {\dot\zeta}^2
+ \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 \, .
</math>
  </td>
</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>
  <td align="right">
<math>~L \equiv \int_0^R 4\pi r_0^2 \mathcal{L} dr_0</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 4\pi \int_0^R \biggl\{ \frac{\rho_0}{2} {\dot\zeta}^2
+ \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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 4\pi e^{2i\omega t} \int_0^R \biggl\{ \frac{\rho_0}{2} (i \omega r_0 x)^2
+ \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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^2 x^2
- 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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^4 x^2
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2  
- 4 r_0^3 x^2 \frac{dP_0}{dr_0} 
- 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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi e^{2i\omega t} \int_0^R \biggl\{ -\rho_0 \omega^2 r_0^4 x^2
- \gamma_\mathrm{g} P_0 r_0^4\biggl( \frac{\partial x}{\partial r_0}\biggr)^2  
- 4 r_0^3 x^2 \frac{dP_0}{dr_0} 
+r_0^3 x^2 \frac{d}{dr_0}\biggl(3\gamma_\mathrm{g}P_0\biggr)
-\frac{d}{dr_0}\biggl[3 \gamma_\mathrm{g} r_0^3 x^2 P_0\biggr]
\biggr\}dr_0
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi e^{2i\omega t} \biggl\{ 
- \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>
</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>
  <td align="right">
<math>~2 e^{-2i\omega t} L </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \omega^2 \int_0^R 4\pi\rho_0 r_0^4 x^2 dr_0
- \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\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
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \omega^2 \int_0^R x^2 r_0^2 dm
- \gamma_\mathrm{g} \int_0^R  \biggl[ r_0 \biggl( \frac{\partial x}{\partial r_0}\biggr) \biggr]^2 P_0  dV
- (3\gamma_\mathrm{g} - 4) \int_0^R  r_0 x^2 \biggl( \frac{Gm}{r_0^2} \biggr)dm
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~- \biggl[ \frac{2 e^{-2i\omega t}}{\int_0^R x^2 r_0^2 dm } \biggr] L </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
\omega^2 
+ \biggl\{ \frac{\gamma_\mathrm{g} \int_0^R  \bigl[ r_0 \bigl( \frac{\partial x}{\partial r_0}\bigr) \bigr]^2 P_0  dV
+ (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}
\biggr\} \, .
</math>
  </td>
</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.

<span id="EnergiesDefined">Now,</span> 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>
  <td align="right">
<math>~W_\mathrm{grav}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \int_0^{M} \biggl( \frac{Gm}{r_0} \biggr) dm  \, ;</math>
  </td>
</tr>
</table>
</div>

for adiabatic systems, the [[User:Tohline/SphericallySymmetricConfigurations/Virial#Reservoir|internal energy]] is,
<div align="center">
<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;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<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}
- (3\gamma_\mathrm{g} - 4) \int_0^R  x^2 dW_\mathrm{grav}}{\int_0^R x^2 dI}
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>
This expression appears in equation (9) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)].

==Chandrasekhar (1964)==
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.  

===The Lagrangian Expression using Chandrasekhar's Notation===
First, let's show that the Lagrangian expression derived by Chandrasekhar is, indeed, equivalent to the one presented by Ledoux &amp; Pekeris.  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">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2L </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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
+ \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>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~- 4\pi e^{2i\omega t} \int_0^R \biggl\{ \omega^2 \rho_0 \psi^2 
+ 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>
</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.

===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.

=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>
  <td align="right">
<math>~L </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2\pi e^{2i\omega t} \biggl\{ 
- \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>
</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>
  <td align="right">
<math>~\frac{L_{\{\}} }{P_c R_\mathrm{eq}^3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<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>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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
- \biggl[\frac{3 \gamma_\mathrm{g}P_e}{P_c} \biggr]x_\mathrm{surf}^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}
- \gamma_\mathrm{g} \theta^{n+1} \xi^4\biggl( \frac{\partial x}{\partial \xi}\biggr)^2  \biggr]d\xi
</math>
  </td>
</tr>
</table>
</div>

where, we have set the pressure at the (truncated) surface to the value, <math>~P_0|_\mathrm{surface} = P_e</math>.

==Chandra's Expression==
===Normalization===

Alternatively, starting from Chandrasekhar's expression,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~2L_{\{\}} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\int_0^R \biggl\{ \omega^2 \rho_0 (r_0^3 x)^2 
+ 4\biggl( \frac{dP_0}{dr_0}\biggr) \frac{(r_0^3 x)^2}{r_0} 
+ \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>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{2L_{\{\}} }{P_c R_\mathrm{eq}^3}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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 
+ 4 r_0^3 x^2 \cdot \frac{d}{dr_0}\biggl(\frac{P_0}{P_c}\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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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 
+ 4 \tilde\xi ~x^2 \biggl(\frac{\xi}{ {\tilde\xi}} \biggr)^3 \frac{d}{d\xi}\biggl(\frac{P_0}{P_c}\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>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <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 
+ \biggl(\frac{4 }{{\tilde\xi}^2}\biggr) ~x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \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}}
</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{\omega^2 \rho_c R_\mathrm{eq}^2}{{\tilde\xi}^2 P_c} \biggl( \theta^n \xi^4 \biggr) x^2 
+ 4x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
+ \gamma_\mathrm{g} \biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \biggl[ \frac{\partial (\xi^3 x)}{\partial \xi} \biggr]^2 \biggr\} d\xi \, .
</math>
  </td>
</tr>
</table>
</div>

===Known Analytic Eigenfunction===
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>
  <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_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>
  </td>
</tr>
</table>
</div>

Given that, from the Lane-Emden equation,
<div align="center">
<math>~\frac{\theta^{''}}{\theta^n} = - 1 -\frac{2\theta^'}{\xi \theta^n} \, ,</math>
</div>

we recognize that,

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3\xi^2}{2} \biggl\{ 2 + (3-n)\biggl[ \frac{(\theta^')^2}{\theta^{n+1}}\biggr]\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

===General Evaluation===
Therefore, returning to Chandrasekhar's expression for the Lagrangian and evaluating the sum of the last two terms inside the curly braces gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl\{\sum^2\biggr\}_{x_P}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl\{4x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
+ \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>
</tr>

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

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

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

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

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

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

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

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

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

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{3^2(n+1) \xi }{2^2n^2 \theta^{n+1}}\biggl\{ 
n\xi  \biggl[ 2^2\theta^{2(n+1)} +  (3-n)^2 (\theta^')^4\biggr] 
+2^2 \theta\theta^'\biggl[ (n-1)^2\xi^2 \theta^{2n} + (n-3)^2 (\theta^')^2 \biggr] 
+ 2^2 (n^2- 5n +6) \xi \theta^{n+1} (\theta^')^2
\biggr\}
</math>
  </td>
</tr>
</table>
</div>

===Application to n = 5 Polytropic Configuration===

Let's try plugging in expressions for n = 5 configurations, for which the [[User:Tohline/SSC/Structure/PolytropesEmbedded#Tabular_Summary_.28n.3D5.29|Lane-Emden function is known analytically]].  Specifically,
<div align="center">
<math>~\theta_{n=5} = \biggl(1 + \frac{\xi^2}{3}\biggr)^{-1 / 2}</math> &nbsp;  &nbsp;  &nbsp; and  &nbsp;  &nbsp;  &nbsp; <math>~\theta_{n=5}^' = - \frac{\xi}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2} \, .</math>
</div>

====First Attempt====
We have,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl\{\sum^2\biggr\}_{x_P}^{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3^3 \xi }{5^2 \theta^{6}}\biggl\{ 
5\xi  \biggl[ \theta^{12} +  (\theta^')^4\biggr] 
+ 2^4 \theta\theta^'\biggl[ \xi^2 \theta^{10} + (\theta^')^2 \biggr] 
+ 2\cdot 3 \xi \theta^{6} (\theta^')^2
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3^3 \xi }{5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{3}\biggl\{ 
5\xi  \biggl[ \biggl(1 + \frac{\xi^2}{3}\biggr)^{-6} +  \biggl[ - \frac{\xi}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2}\biggr]^4\biggr] 
</math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3^3 \xi }{5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{3}\biggl\{ 
5\xi  \biggl[ \biggl(1 + \frac{\xi^2}{3}\biggr)^{-6} +  \frac{\xi^4}{3^4}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-6} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
- \frac{2^4\xi}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-2} \biggl[ \xi^2 \biggl(1 + \frac{\xi^2}{3}\biggr)^{-5} + \frac{\xi^2}{3^2}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 } \biggr] 
+ \frac{2\xi^3}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-6 }
\biggr\}
</math>
  </td>
</tr>

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3^3 \xi^2 }{5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}\biggl\{ 
\frac{5}{3^5}\biggl[3^5  + 3^4\xi^2 + 3\xi^4 + \xi^6   \biggr]  
+ \frac{2}{3^2}  \biggl(3\xi^2 + \xi^4\biggr)
- \frac{2^4\xi^2}{3^5}  \biggl[ 3^4 + 3^2 + 2\cdot 3\xi^2 + \xi^4 \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 \xi^2 }{3^2\cdot 5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}\biggl\{ 
5 \biggl[3^5  + 3^4\xi^2 + 3\xi^4 + \xi^6   \biggr]  
+ 2\cdot 3^3  \biggl(3\xi^2 + \xi^4\biggr)
- 2^4  \biggl[ 2\cdot 3^2\cdot 5 \xi^2 + 2\cdot 3\xi^4 + \xi^6 \biggr] 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 \xi^2 }{3^2\cdot 5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}\biggl\{ 
3^5\cdot 5 + 3^2\xi^2 [3^2\cdot 5 + 2\cdot 3^2 - 2^5 \cdot 5 ] + 3\xi^4 [ 5  + 2\cdot 3^2 - 2^5]  + \xi^6 [5  -2^4]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2 \xi^2 }{3^2\cdot 5^2 } \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4}\biggl[ 
3^5\cdot 5 - 3^2 \cdot 97\xi^2 - 3^3 \xi^4   - 11 \xi^6 
\biggr] \, .
</math>
  </td>
</tr>
</table>
</div>

====Second Attempt====

First, we evaluate,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\biggl\{ x_P \biggr\}^{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3}{5}\biggl[1 - \frac{1}{2\xi} \biggl(1 + \frac{\xi^2}{3}\biggr)^{5 / 2} \frac{\xi}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2} \biggr]  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{5}\biggl[6 -  \biggl(1 + \frac{\xi^2}{3}\biggr) \biggr]  </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~1 -   \frac{\xi^2}{3\cdot 5} \, ;  </math>
  </td>
</tr>
</table>
</div>

and,

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

<tr>
  <td align="right">
<math>~\biggl\{ \frac{d(\xi^3 x_P)}{d\xi} \biggr\}^{n=5}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\xi^2 \biggl\{ 1 -\biggl[ \frac{(\theta^')^2}{\theta^{6}}\biggr]\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\xi^2 \biggl\{ 1 - \biggl[ \biggl(1 + \frac{\xi^2}{3}\biggr)^{-1 / 2} \biggr]^{-6} \biggl[ - \frac{\xi}{3}  \biggl(1 + \frac{\xi^2}{3}\biggr)^{-3 / 2} \biggr]^2 \biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
3\xi^2 \biggl( 1 - \frac{\xi^2}{3^2} \biggr) \, .
</math>
  </td>
</tr>
</table>
</div>

Then, returning to Chandrasekhar's expression for the Lagrangian and evaluating the sum of the last two terms inside the curly braces gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl\{\sum^2\biggr\}^{n=5}_{x_P}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl\{4x^2 \xi^3 \biggl[ \frac{d\theta^{n+1}}{d\xi} \biggr]
+ \frac{n+1}{n}\biggl( \frac{\theta^{n+1}}{\xi^2}\biggr) \biggl[ \frac{d (\xi^3 x)}{d \xi} \biggr]^2\biggr\}^{n=5}_{x_P} </math>
  </td>
</tr>

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

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

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl(1 + \frac{\xi^2}{3}\biggr)^{-4} \biggl\{ 
\frac{2\cdot 3^3 \xi^2}{5} \biggl( 1 - \frac{\xi^2}{3^2} \biggr)^2  \biggl(1 + \frac{\xi^2}{3}\biggr) 
- 2^3\cdot 3^2 \xi^8 \biggl( 1 - \frac{\xi^2}{3^2} \biggr)^2 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{3\cdot 5} \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4} \biggl( 1 - \frac{\xi^2}{3^2} \biggr)^2 \biggl\{ 
2\cdot 3^3 \xi^2  (3 + \xi^2 ) 
- 2^3\cdot 3^3\cdot 5 \xi^8  
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2\cdot 3^2 \xi^2}{5} \biggl(1 + \frac{\xi^2}{3}\biggr)^{-4} \biggl( 1 - \frac{\xi^2}{3^2} \biggr)^2 \biggl\{ 
3 + \xi^2 - 2^2\cdot 5 \xi^6  
\biggr\} \, .
</math>
  </td>
</tr>
</table>
</div>

=Normalizations=
Returning to the last line of our derivation of the [[#LDefinition|Ledoux &amp; Walraven Lagrangian]], we can write,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\biggl[\frac{2}{\gamma_\mathrm{g}} \biggr]e^{-2i\omega t}  L + 4\pi \int_0^R \rho_0 \biggl(\frac{\omega^2}{\gamma_\mathrm{g}}\biggr) r_0^4 x^2 dr_0
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 4\pi \biggl\{ 
- \int_0^R  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[ \biggl(3 - \frac{4}{\gamma_\mathrm{g}} \biggr)P_0\biggr]dr_0
-\biggl[3 r_0^3 x^2 P_0\biggr]_0^{R}
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \int_0^R   \biggl[r_0\biggl( \frac{\partial x}{\partial r_0}\biggr) \biggr]^2 4\pi P_0 r_0^2 dr_0
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}} \biggr) \int_0^R \biggl[ x^2 \biggr] \biggl(- \frac{r_0}{\rho_0} ~\frac{dP_0}{dr_0}\biggr) 4\pi r_0^2 \rho_0 dr_0
- 4\pi \biggl[3  r_0^3 x^2 P_0\biggr]_0^{R}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 
- \int_0^R   \biggl[r_0\biggl( \frac{\partial x}{\partial r_0}\biggr) \biggr]^2 4\pi P_0 r_0^2 dr_0
- \biggl(3 - \frac{4}{\gamma_\mathrm{g}} \biggr) \int_0^R \biggl[ x^2 \biggr] \biggl(\frac{GM_r}{r_0}\biggr) 4\pi r_0^2 \rho_0 dr_0
- \biggl[ 3x \biggr]^2 \frac{4\pi  R^3}{3}  P_e \, .
</math>
  </td>
</tr>
</table>
</div>

=Alternate Approach: &nbsp; Integrate Over LAWE=

As we have demonstrated, [[#Ledoux_and_Pekeris_.281941.29|above]], if we assume that <math>~\Gamma_1</math> is constant throughout the configuration, our version of the LAWE can be straightforwardly rearranged to give equation (58.1) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], that is,
<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{dx}{dr} \biggr] 
+\biggl[ \sigma^2 \rho r^4 + (3\Gamma_1 - 4) r^3 \frac{dP}{dr} \biggr] x \, ,
</math>
  </td>
</tr>
</table>
</div>
where, we are using <math>~x</math> in place of <math>~\xi</math> to represent the fractional Lagrangian displacement, <math>~\delta r/r</math>.  If we multiply this expression through by <math>~4 \pi dr</math> and integrate over the entire volume of the configuration, we have,
<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>~
4\pi \int_0^R \frac{d}{dr}\biggl[ r^4 \Gamma_1 P ~\frac{dx}{dr} \biggr] dr
+ \int_0^R\biggl[ \sigma^2  x r^2 + (3\Gamma_1 - 4) x \biggl( \frac{r}{\rho} \frac{dP}{dr}\biggr) \biggr] 4\pi r^2 \rho dr 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[4\pi  r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R
+ \int_0^R\biggl[ \sigma^2  x r^2 + (3\Gamma_1 - 4) x \biggl( -\frac{Gm}{r}\biggr) \biggr] dm 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[4\pi  r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R
+ \sigma^2 \int_0^R  x ~dI
+ \int_0^R (3\Gamma_1 - 4) x ~dW_\mathrm{grav} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \sigma^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl\{ 
\int_0^R (3\Gamma_1 - 4) x ~dW_\mathrm{grav} + \biggl[4\pi  r^4 \Gamma_1 P \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R \biggr\} \biggl[\int_0^R  x ~dI 
\biggr]^{-1} \, ,
</math>
  </td>
</tr>
</table>
</div>
where the definitions of <math>~dW_\mathrm{grav}</math> and <math>~dI</math> are as [[#EnergiesDefined|provided, above]].  This last expression is the same as equation (59.17) of [http://adsabs.harvard.edu/abs/1941ApJ....94..124L Ledoux &amp; Pekeris (1941)], except that: &nbsp; these authors have retained a term allowing for radial variation of <math>~\Gamma_1</math>, whereas we have not; and we have retained a ''boundary'' term that can accommodate a nonzero surface pressure, whereas Ledoux &amp; Pekeris have not.

=See Also=

* [[SSC/VariationalPrinciple#Ledoux.27s_Variational_Principle|Summary discussion of Ledoux's Variational Principle]]


{{ SGFfooter }}