Editing Appendix/Ramblings/LedouxVariationalPrinciple (section)

=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 [[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 [[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 [[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 [[SSCpt1/Virial#Wgrav|discussed separately]] &#8212; see, also, p. 64, Equation (12) of [<b>[[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 [[SSCpt1/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.