Appendix/Ramblings/LedouxVariationalPrinciple

Ledoux's Variational Principle (Supporting Derivations)

An 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.

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Material that appears after this point in our presentation is under development and therefore
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Our Initial Explorations

Review by Ledoux and Walraven (1958)

Here we are especially interested in understanding the origin of equation (59.10) of P. Ledoux & Th. Walraven (1958), which appears in §59 (pp. 464 - 466) of their Handbuch der Physik article.

From our accompanying summary of the set of nonlinear governing relations, we highlight the

Euler + Poisson Equations
$\frac{d^{2}r}{d{t}^2}=-4\pi r^2\frac{dP}{dm}-\frac{Gm}{r^2}$

Repeating a result from our separate derivation, linearization of the two terms on the righthand side of this equation gives,

$r^2\frac{dP}{dm}$	$\to$	$r_0^2[1+x{e}^{i\omega t}{]}^2\{\frac{d{P}_0}{dm}[1+p{e}^{i\omega t}]+P_0{e}^{i\omega t}\frac{dp}{dm}\}\approx r_0^2\frac{d{P}_0}{dm}[1+(2x+p)e^{i\omega t}]+P_0{r}_0^2{e}^{i\omega t}\frac{dp}{dm}$
$\frac{Gm}{r^2}$	$\to$	$\frac{Gm}{r_0^2}[1+x{e}^{i\omega t}{]}^{-2}\approx \frac{Gm}{r_0^2}[1-2x{e}^{i\omega t}].$

Adopting the terminology of Ledoux & Walraven (1958), the "variation" of each of these terms is obtained by subtracting off the leading order pieces — which presumably cancel in equilibrium. In particular, drawing a parallel with their equation (59.1), we can write,

$\delta (-\frac{Gm}{r^2})$

$\approx$

$\frac{Gm}{r_0^2}\left[2x{e}^{i\omega t}\right].$

And, drawing a parallel with their equation (59.2), we have,

$\delta (-\frac{1}{\rho }\cdot \frac{dP}{dr})=\delta (-4\pi r^2\frac{dP}{dm})$	$\approx$	$-4\pi r_0^2\frac{d{P}_0}{dm}[(2x+p)e^{i\omega t}]-4\pi P_0{r}_0^2{e}^{i\omega t}\frac{dp}{dm}$
	$=$	$\left\{\frac{Gm}{r_0^2}\right[(2x)]-4\pi r_0^2\frac{d{P}_0}{dm}[(p)]-4\pi P_0{r}_0^2\frac{dp}{dm}\}{e}^{i\omega t}$
	$=$	$\left\{\right(\frac{2Gm}{r_0^2})x-\frac{1}{{\rho }_0}\cdot \frac{d}{d{r}_0}[P_{0}p\left]\right\}{e}^{i\omega t}.$

Now, if we combined the linearized continuity equation and the linearized (adiabatic form of the) first law of thermodynamics, as derived elsewhere, we can write,

$p={\gamma }_{\mathrm{g}}d$	$=$	$-{\gamma }_{\mathrm{g}}[3x+r_0\frac{dx}{d{r}_0}]$
	$=$	$-\frac{{\gamma }_{\mathrm{g}}}{r_0^2}\frac{d}{d{r}_0}\left(r_0^{3}x\right).$

Hence,

$\delta (-\frac{1}{\rho }\cdot \frac{dP}{dr})$

$\approx$

$\left\{\right(\frac{2Gm}{r_0^2})x+\frac{1}{{\rho }_0}\cdot \frac{d}{d{r}_0}[\frac{{\gamma }_{\mathrm{g}}{P}_0}{r_0^2}\frac{d}{d{r}_0}\left(r_0^{3}x\right)\left]\right\}{e}^{i\omega t}.$

So, given that a mapping from our notation to that used by Ledoux & Walraven (1958) requires $x{e}^{i\omega t}\to \zeta /r_0$, I understand the origins of their equations (59.1) and (59.2). But I do not yet understand how … "Accordingly, the acting forces per unit volume can be considered as deriving from a potential density"

${\rho }_0{𝒱}$

$=$

$-\left(\frac{2Gm}{r_0^3}\right){\rho }_0{\zeta }^2+\frac{{\gamma }_{\mathrm{g}}{P}_0}{2}[\frac{1}{r_0^2}\frac{\partial (r_0^2\zeta )}{\partial r_0}{]}^2.$

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.,

${ℒ}={\rho }_0[{𝒦}-{𝒱}],$

becomes (see their equation 59.5),

${ℒ}$

$=$

$\frac{{\rho }_0}{2}{\dot{\zeta }}^2+\left(\frac{2Gm}{r_0^3}\right){\rho }_0{\zeta }^2-\frac{{\gamma }_{\mathrm{g}}{P}_0}{2}[\frac{1}{r_0^2}\frac{\partial (r_0^2\zeta )}{\partial r_0}{]}^2.$

Noting that, $\dot{\zeta }=i\omega r_{0}x{e}^{i\omega t}$, this in turn gives,

$L\equiv \underset{0}{\overset{R}{\int }}4\pi r_0^2{ℒ}d{r}_0$	$=$	$4\pi \underset{0}{\overset{R}{\int }}\{\frac{{\rho }_0}{2}{\dot{\zeta }}^2+(\frac{2Gm}{r_0^3}){\rho }_0{\zeta }^2-\frac{{\gamma }_{\mathrm{g}}{P}_0}{2}[\frac{1}{r_0^2}\frac{\partial (r_0^2\zeta )}{\partial r_0}{]}^2\}{r}_0^{2}d{r}_0$
	$=$	$4\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{\frac{{\rho }_0}{2}(i\omega r_{0}x{)}^2+\left(\frac{2Gm}{r_0^3}\right){\rho }_0(r_{0}x{)}^2-\frac{{\gamma }_{\mathrm{g}}{P}_0}{2}[\frac{1}{r_0^2}\frac{\partial (r_0^{3}x)}{\partial r_0}{]}^2\}{r}_0^{2}d{r}_0$
	$=$	$2\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{-{\rho }_0{\omega }^2{r}_0^2{x}^2-4r_0{x}^2\frac{d{P}_0}{d{r}_0}-\frac{{\gamma }_{\mathrm{g}}{P}_0}{r_0^4}[3r_0^{2}x+r_0^3\frac{\partial x}{\partial r_0}{]}^2\}{r}_0^{2}d{r}_0$
	$=$	$2\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{-{\rho }_0{\omega }^2{r}_0^4{x}^2-{\gamma }_{\mathrm{g}}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^2-4r_0^3{x}^2\frac{d{P}_0}{d{r}_0}-3{\gamma }_{\mathrm{g}}{P}_0[3r_0^2{x}^2+2r_0^{3}x\frac{\partial x}{\partial r_0}]\}d{r}_0$
	$=$	$2\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{-{\rho }_0{\omega }^2{r}_0^4{x}^2-{\gamma }_{\mathrm{g}}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^2-4r_0^3{x}^2\frac{d{P}_0}{d{r}_0}+r_0^3{x}^2\frac{d}{d{r}_0}\left(3{\gamma }_{\mathrm{g}}{P}_0\right)-\frac{d}{d{r}_0}\left[3{\gamma }_{\mathrm{g}}{r}_0^3{x}^2{P}_0\right]\}d{r}_0$
	$=$	$2\pi e^{2i\omega t}\{-\underset{0}{\overset{R}{\int }}{\rho }_0{\omega }^2{r}_0^4{x}^{2}d{r}_0-\underset{0}{\overset{R}{\int }}{\gamma }_{\mathrm{g}}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0+\underset{0}{\overset{R}{\int }}{r}_0^3{x}^2\frac{d}{d{r}_0}[(3{\gamma }_{\mathrm{g}}-4)P_0]d{r}_0-\left[3{\gamma }_{\mathrm{g}}{r}_0^3{x}^2{P}_0{]}_0^R\right\}.$

The group of terms inside the curly braces, here, matches the group of terms inside the curly braces of Ledoux & Walraven's equation (59.8) if we acknowledge that:

1. Our ${\omega }^2$ has the same meaning as, but the opposite sign of, their ${\sigma }^2$.
2. Our last term goes to zero because, $r_0=0$ at the center, while $P_0=0$ at the surface.

LP41 Again

After setting the last term to zero, this last expression can be rewritten as,

$2e^{-2i\omega t}L$	$=$	$-{\omega }^2\underset{0}{\overset{R}{\int }}4\pi {\rho }_0{r}_0^4{x}^{2}d{r}_0-\underset{0}{\overset{R}{\int }}4\pi {\gamma }_{\mathrm{g}}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0-(3{\gamma }_{\mathrm{g}}-4)\underset{0}{\overset{R}{\int }}4\pi {\rho }_0{r}_0^3{x}^2(-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0})d{r}_0$
	$=$	$-{\omega }^2\underset{0}{\overset{R}{\int }}{x}^2{r}_0^{2}dm-{\gamma }_{\mathrm{g}}\underset{0}{\overset{R}{\int }}\left[r_0\right(\frac{\partial x}{\partial r_0}){]}^2{P}_{0}dV-(3{\gamma }_{\mathrm{g}}-4)\underset{0}{\overset{R}{\int }}{r}_0{x}^2(\frac{Gm}{r_0^2})dm$
$-\left[\frac{2e^{-2i\omega t}}{\underset{0}{\overset{R}{\int }}{x}^2{r}_0^{2}dm}\right]L$	$=$	${\omega }^2+\left\{\frac{{\gamma }_{\mathrm{g}}\underset{0}{\overset{R}{\int }}\left[r_0\right(\frac{\partial x}{\partial r_0}){]}^2{P}_{0}dV+(3{\gamma }_{\mathrm{g}}-4)\underset{0}{\overset{R}{\int }}{x}^2(\frac{Gm}{r_0})dm}{\underset{0}{\overset{R}{\int }}{x}^2{r}_0^{2}dm}\right\}.$

As is explained in detail in §59 (pp. 464 - 465) of Ledoux & Walraven (1958), and summarized in §1 of Ledoux & Pekeris (1941), the function inside the curly braces of this last expression will be minimized if the radially dependent displacement function, $x$, is set equal to the eigenfunction of the fundamental mode of radial oscillation, $x_0$; and, after evaluation, the minimum value of this expression will be equal to (the negative of) the square of the fundamental-mode oscillation frequency, ${\omega }^2$. This explicit mathematical statement is contained within equation (8) of Ledoux & Pekeris and within equation (59.10) of Ledoux & Walraven.

Now, as we have discussed separately — see, also, p. 64, Equation (12) of [C67] — the gravitational potential energy of the unperturbed configuration is given by the integral,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$-\underset{0}{\overset{M}{\int }}\left(\frac{Gm}{r_0}\right)dm;$

for adiabatic systems, the internal energy is,

$U_{\mathrm{i}\mathrm{n}\mathrm{t}}=\frac{1}{({\gamma }_g-1)}\underset{0}{\overset{R}{\int }}{P}_{0}dV;$

and — see the text at the top of p. 126 of Ledoux & Pekeris (1941) — the moment of inertia of the configuration about its center is,

$I=\underset{0}{\overset{M}{\int }}{r}_0^{2}dm.$

Hence, the function to be minimized may be written as,

$\left\{\frac{{\gamma }_{\mathrm{g}}({\gamma }_{\mathrm{g}}-1)\underset{0}{\overset{R}{\int }}\left[r_0\right(\frac{\partial x}{\partial r_0}){]}^{2}d{U}_{\mathrm{i}\mathrm{n}\mathrm{t}}-(3{\gamma }_{\mathrm{g}}-4)\underset{0}{\overset{R}{\int }}{x}^{2}d{W}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{\underset{0}{\overset{R}{\int }}{x}^{2}dI}\right\}.$

This expression appears in equation (9) of Ledoux & Pekeris (1941).

Chandrasekhar (1964)

In a paper titled, A General Variational Principle Governing the Radial and the Non-Radial Oscillations of Gaseous Masses, 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 & Pekeris. Returning to the second line of our effort to simplify the above definition of the Lagrangian, and making the substitution, $\psi \equiv r_0^{3}x$, we have,

$2L$	$=$	$4\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\left\{{\rho }_0\right(\frac{i\omega \psi }{r_0^2}{)}^2+\left(\frac{4Gm}{r_0^3}\right){\rho }_0(\frac{\psi }{r_0^2}{)}^2-{\gamma }_{\mathrm{g}}{P}_0[\frac{1}{r_0^2}\frac{\partial (\psi )}{\partial r_0}{]}^2\}{r}_0^{2}d{r}_0$
	$=$	$4\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{-{\omega }^2{\rho }_0(\frac{\psi }{r_0}{)}^2-4\left(\frac{d{P}_0}{d{r}_0}\right)\left(\frac{{\psi }^2}{r_0^3}\right)-{\gamma }_{\mathrm{g}}{P}_0\left[\frac{1}{r_0}\frac{\partial (\psi )}{\partial r_0}{]}^2\right\}d{r}_0$
	$=$	$-4\pi e^{2i\omega t}\underset{0}{\overset{R}{\int }}\{{\omega }^2{\rho }_0{\psi }^2+4(\frac{d{P}_0}{d{r}_0}\left)\right(\frac{{\psi }^2}{r_0})+{\gamma }_{\mathrm{g}}{P}_0[\frac{\partial \psi }{\partial r_0}{]}^2\}\frac{d{r}_0}{r_0^2}.$

This integral expression matches the integral expression that appears in equation (49) of Chandrasekhar (1964), if we accept that our squared frequency, ${\omega }^2$, has the opposite sign to Chandrasekhar's ${\sigma }^2$. 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 appears above in our review of the paper by Ledoux & Pekeris (1941), namely,

$0$

$=$

$\frac{d}{dr}\left[r^4{\mathrm{\Gamma }}_{1}P\frac{d\xi }{dr}\right]+[{\sigma }^2\rho r^4+(3{\mathrm{\Gamma }}_1-4)r^3\frac{dP}{dr}]\xi .$

We will develop the Lagrangian expression by following the guidance provided at the top of p. 666 of S. Chandrasekhar (1964, ApJ, 139, 664). First, we multiply the LAWE through by the fractional displacement, $\xi$; second, we make the substitution, $\xi \to \psi /r^3$, in order to shift to Chandrasekhar's variable notation; then we multiply through by $dr$ and integrate from the center $(r=0)$ to the surface $(r=R)$ of the configuration.

Multiplying through by the fractional displacement gives,

${\sigma }^2\rho r^4{\xi }^2$

$=$

$-\xi \cdot \frac{d}{dr}\left[r^4{\mathrm{\Gamma }}_{1}P\frac{d\xi }{dr}\right]-(3{\mathrm{\Gamma }}_1-4)r^3{\xi }^2\left(\frac{dP}{dr}\right).$

Next, making the stated variable substitution gives,

$\frac{{\sigma }^2\rho {\psi }^2}{r^2}$	$=$	$-\left(\frac{\psi }{r^3}\right)\frac{d}{dr}\left[r^4{\mathrm{\Gamma }}_{1}P\frac{d}{dr}\right(\frac{\psi }{r^3}\left)\right]-(3{\mathrm{\Gamma }}_1-4)\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)$
	$=$	$-\left(\frac{\psi }{r^3}\right)\frac{d}{dr}[r{\mathrm{\Gamma }}_{1}P\frac{d\psi }{dr}-3{\mathrm{\Gamma }}_{1}P\psi ]-(3{\mathrm{\Gamma }}_1-4)\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)$
	$=$	$(4-3{\mathrm{\Gamma }}_1)\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)-\left(\frac{\psi }{r^3}\right)\frac{d}{dr}\left[r{\mathrm{\Gamma }}_{1}P\frac{d\psi }{dr}\right]+3{\mathrm{\Gamma }}_1\left(\frac{{\psi }^2}{r^3}\right)\frac{dP}{dr}+3{\mathrm{\Gamma }}_{1}P\left(\frac{\psi }{r^3}\right)\frac{d\psi }{dr}$
	$=$	$4\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)+3{\mathrm{\Gamma }}_{1}P\left(\frac{\psi }{r^3}\right)\frac{d\psi }{dr}-\left\{\frac{d}{dr}\right[r{\mathrm{\Gamma }}_{1}P\left(\frac{\psi }{r^3}\right)\frac{d\psi }{dr}]-r{\mathrm{\Gamma }}_{1}P\frac{d\psi }{dr}\cdot \frac{d}{dr}(\frac{\psi }{r^3}\left)\right\}$
	$=$	$4\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)+3{\mathrm{\Gamma }}_{1}P\left(\frac{\psi }{r^3}\right)\frac{d\psi }{dr}+\frac{{\mathrm{\Gamma }}_{1}P}{r^2}[\frac{d\psi }{dr}{]}^2-[\frac{3{\mathrm{\Gamma }}_{1}P\psi }{r^3}]\frac{d\psi }{dr}-\frac{d}{dr}[\frac{{\mathrm{\Gamma }}_{1}P\psi }{r^2}\left(\frac{d\psi }{dr}\right)]$
	$=$	$4\left(\frac{{\psi }^2}{r^3}\right)\left(\frac{dP}{dr}\right)+\frac{{\mathrm{\Gamma }}_{1}P}{r^2}[\frac{d\psi }{dr}{]}^2-\frac{d}{dr}[\frac{{\mathrm{\Gamma }}_{1}P\psi }{r^2}\left(\frac{d\psi }{dr}\right)].$

Finally, integrating over the volume gives,

$\underset{0}{\overset{R}{\int }}({\sigma }^2\rho {\psi }^2)\frac{dr}{r^2}$

$=$

$\underset{0}{\overset{R}{\int }}\left[{\mathrm{\Gamma }}_{1}P\right(\frac{d\psi }{dr}{)}^2+\frac{4{\psi }^2}{r}\left(\frac{dP}{dr}\right)]\frac{dr}{r^2}-[\frac{{\mathrm{\Gamma }}_{1}P\psi }{r^2}\left(\frac{d\psi }{dr}\right){]}_0^R,$

which is identical to equation (49) of Chandrasekhar (1964), if the last term — the difference of the central and surface boundary conditions — is set to zero.

Examples

Ledoux's Expression

Returning to the last line of our above definition of the Lagrangian, that is,

$L$

$=$

$2\pi e^{2i\omega t}\{-\underset{0}{\overset{R}{\int }}{\rho }_0{\omega }^2{r}_0^4{x}^{2}d{r}_0-\underset{0}{\overset{R}{\int }}{\gamma }_{\mathrm{g}}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0+\underset{0}{\overset{R}{\int }}{r}_0^3{x}^2\frac{d}{d{r}_0}[(3{\gamma }_{\mathrm{g}}-4)P_0]d{r}_0-\left[3{\gamma }_{\mathrm{g}}{r}_0^3{x}^2{P}_0{]}_0^R\right\},$

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 $P_c{R}_{\mathrm{e}\mathrm{q}}^3$ and making the substitution, $r_0/R_{\mathrm{e}\mathrm{q}}\to \xi /\tilde{\xi }$, gives,

$\frac{L_{\{\}}}{P_c{R}_{\mathrm{e}\mathrm{q}}^3}$	$=$	$-\underset{0}{\overset{R}{\int }}\frac{{\rho }_0{\omega }^2}{P_c{R}_{\mathrm{e}\mathrm{q}}^3}{r}_0^4{x}^{2}d{r}_0-\underset{0}{\overset{R}{\int }}{\gamma }_{\mathrm{g}}\frac{P_0}{P_c{R}_{\mathrm{e}\mathrm{q}}^3}{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0+\underset{0}{\overset{R}{\int }}{r}_0^3{x}^2\frac{d}{d{r}_0}[(3{\gamma }_{\mathrm{g}}-4)\frac{P_0}{P_c{R}_{\mathrm{e}\mathrm{q}}^3}]d{r}_0-3{\gamma }_{\mathrm{g}}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2\frac{P_e}{P_c}$
	$=$	$-\underset{0}{\overset{\tilde{\xi }}{\int }}{\omega }^2\left[\frac{{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{P_c}\right]\left(\frac{{\rho }_0}{{\rho }_c}\right)(\frac{\xi }{\tilde{\xi }}{)}^4{x}^2\frac{d\xi }{\tilde{\xi }}-\underset{0}{\overset{\tilde{\xi }}{\int }}{\gamma }_{\mathrm{g}}(\frac{P_0}{P_c}\left)\right(\frac{\xi }{\tilde{\xi }}{)}^4[\frac{\partial x}{\partial (\xi /\tilde{\xi })}{]}^2\frac{d\xi }{\tilde{\xi }}+\underset{0}{\overset{\tilde{\xi }}{\int }}(\frac{\xi }{\tilde{\xi }}\left)x^2\frac{d}{d(\xi /\tilde{\xi })}\right[(3{\gamma }_{\mathrm{g}}-4)\frac{P_0}{P_c}]\frac{d\xi }{\tilde{\xi }}-3{\gamma }_{\mathrm{g}}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2\frac{P_e}{P_c}$
	$=$	$-{\omega }^2\left[\frac{{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{P_c{\tilde{\xi }}^5}\right]\underset{0}{\overset{\tilde{\xi }}{\int }}{\theta }^n{\xi }^4{x}^{2}d\xi -\frac{{\gamma }_{\mathrm{g}}}{{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}{\theta }^{n+1}{\xi }^4[\frac{\partial x}{\partial \xi }{]}^{2}d\xi +\frac{(3{\gamma }_{\mathrm{g}}-4)}{\tilde{\xi }}\underset{0}{\overset{\tilde{\xi }}{\int }}\xi x^2\frac{d}{d\xi }[{\theta }^{n+1}]d\xi -[\frac{3{\gamma }_{\mathrm{g}}{P}_e}{P_c}]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2$
	$=$	$-{\omega }^2\left[\frac{{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{P_c{\tilde{\xi }}^5}\right]\underset{0}{\overset{\tilde{\xi }}{\int }}{\theta }^n{\xi }^4{x}^{2}d\xi -\left[\frac{3{\gamma }_{\mathrm{g}}{P}_e}{P_c}\right]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2+\frac{1}{{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}[(3{\gamma }_{\mathrm{g}}-4){\tilde{\xi }}^2\xi x^2\frac{d{\theta }^{n+1}}{d\xi }-{\gamma }_{\mathrm{g}}{\theta }^{n+1}{\xi }^4(\frac{\partial x}{\partial \xi }{)}^2]d\xi$

where, we have set the pressure at the (truncated) surface to the value, $P_0{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}=P_e$.

Chandra's Expression

Normalization

Alternatively, starting from Chandrasekhar's expression,

$2L_{\{\}}$	$=$	$\underset{0}{\overset{R}{\int }}\{{\omega }^2{\rho }_0(r_0^{3}x{)}^2+4\left(\frac{d{P}_0}{d{r}_0}\right)\frac{(r_0^{3}x{)}^2}{r_0}+{\gamma }_{\mathrm{g}}{P}_0\left[\frac{\partial (r_0^{3}x)}{\partial r_0}{]}^2\right\}\frac{d{r}_0}{r_0^2}$
$\Rightarrow \frac{2L_{\{\}}}{P_c{R}_{\mathrm{e}\mathrm{q}}^3}$	$=$	$\underset{0}{\overset{R}{\int }}\left\{\frac{{\omega }^2{\rho }_c}{P_c}\right(\frac{{\rho }_0}{{\rho }_c})r_0^4{x}^2+4r_0^3{x}^2\cdot \frac{d}{d{r}_0}(\frac{P_0}{P_c})+\frac{{\gamma }_{\mathrm{g}}{P}_0}{r_0^2{P}_c}[\frac{\partial (r_0^{3}x)}{\partial r_0}{]}^2\}\frac{d{r}_0}{R_{\mathrm{e}\mathrm{q}}^3}$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\frac{{\omega }^2{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{P_c}\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{\xi }{\tilde{\xi }}{)}^4{x}^2+4\tilde{\xi }{x}^2\left(\frac{\xi }{\tilde{\xi }}{)}^3\frac{d}{d\xi }\right(\frac{P_0}{P_c})+\frac{{\gamma }_{\mathrm{g}}{P}_0}{P_c}(\frac{\xi }{\tilde{\xi }}{)}^{-2}\left[\frac{\partial }{\partial (\xi /\tilde{\xi })}\right(\frac{{\xi }^{3}x}{{\tilde{\xi }}^3}\left){]}^2\right\}\frac{d\xi }{\tilde{\xi }}$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\frac{{\omega }^2{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{{\tilde{\xi }}^4{P}_c}\right({\theta }^n{\xi }^4)x^2+(\frac{4}{{\tilde{\xi }}^2}\left)x^2{\xi }^3\right[\frac{d{\theta }^{n+1}}{d\xi }]+\frac{{\gamma }_{\mathrm{g}}}{{\tilde{\xi }}^2}(\frac{{\theta }^{n+1}}{{\xi }^2}\left)\right[\frac{\partial ({\xi }^{3}x)}{\partial \xi }{]}^2\}\frac{d\xi }{\tilde{\xi }}$
	$=$	$\frac{1}{{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\frac{{\omega }^2{\rho }_c{R}_{\mathrm{e}\mathrm{q}}^2}{{\tilde{\xi }}^2{P}_c}\right({\theta }^n{\xi }^4)x^2+4x^2{\xi }^3[\frac{d{\theta }^{n+1}}{d\xi }]+{\gamma }_{\mathrm{g}}(\frac{{\theta }^{n+1}}{{\xi }^2}\left)\right[\frac{\partial ({\xi }^{3}x)}{\partial \xi }{]}^2\}d\xi .$

Known Analytic Eigenfunction

Now, let's plug in the known eigenfunction for the marginally unstable configuration, namely,

Exact Solution to the $(3\le n<\mathrm{\infty })$ Polytropic LAWE
${\sigma }_c^2=0$	and	$x_P\equiv \frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right].$

Given that, from the Lane-Emden equation,

$\frac{{\theta }^{{\text{'}}^{\prime }}}{{\theta }^n}=-1-\frac{2{\theta }^{\text{'}}}{\xi {\theta }^n},$

we recognize that,

$\frac{d({\xi }^3{x}_P)}{d\xi }$	$=$	$\frac{3(n-1)}{2n}\frac{d{\xi }^3}{d\xi }+\frac{3(n-3)}{2n}\frac{d}{d\xi }\left(\frac{{\xi }^2{\theta }^{\text{'}}}{{\theta }^n}\right)$
	$=$	$\frac{3^2(n-1){\xi }^2}{2n}+\frac{3(n-3){\xi }^2}{2n}[\frac{{\theta }^{{\text{'}}^{\prime }}}{{\theta }^n}+\frac{2{\theta }^{\text{'}}}{\xi {\theta }^n}-\frac{n({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}]$
	$=$	$\frac{3{\xi }^2}{2n}\{3(n-1)-(n-3)[1+\frac{n({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}\left]\right\}$
	$=$	$\frac{3{\xi }^2}{2}\{2+(3-n)[\frac{({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}\left]\right\}.$

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,

$\{{\sum }^2{\}}_{x_P}$	$\equiv$	$\left\{4x^2{\xi }^3\right[\frac{d{\theta }^{n+1}}{d\xi }]+{\gamma }_{\mathrm{g}}(\frac{{\theta }^{n+1}}{{\xi }^2}\left)\right[\frac{d({\xi }^{3}x)}{d\xi }{]}^2{\}}_{x_P}$
	$=$	$4(n+1)\left\{\frac{3(n-1)}{2n}\right[1+\left(\frac{n-3}{n-1}\right)\left(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}\right)]{\}}^2{\xi }^3{\theta }^n{\theta }^{\text{'}}$
		$+\frac{(n+1)}{n}\left(\frac{{\theta }^{n+1}}{{\xi }^2}\right)\frac{3^2{\xi }^4}{2^2}\{2+(3-n)[\frac{({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}]{\}}^2$
	$=$	$4(n+1)\frac{3^2(n-1{)}^2}{2^2{n}^2}[1+(\frac{n-3}{n-1}\left)\right(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}){]}^2{\xi }^3{\theta }^n{\theta }^{\text{'}}$
		$+\frac{(n+1)}{n}\left(\frac{{\theta }^{n+1}}{{\xi }^2}\right)\frac{3^2{\xi }^4}{2^2}[2+\frac{(3-n)({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}{]}^2$
	$=$	$\frac{3^2(n+1){\xi }^2}{2^2{n}^2}\{2^2(n-1{)}^2[1+(\frac{n-3}{n-1}\left)\right(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}){]}^2\xi {\theta }^n{\theta }^{\text{'}}+n{\theta }^{n+1}[2+\frac{(3-n)({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}{]}^2\}$
	$=$	$\frac{3^2(n+1){\xi }^2}{2^2{n}^2}\left\{2^2\right[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}+\frac{n}{{\theta }^{n+1}}[2{\theta }^{n+1}+(3-n)({\theta }^{\text{'}}{)}^2{]}^2\}$
	$=$	$\frac{3^2(n+1)\xi }{2^2{n}^2{\theta }^{n+1}}\left\{n\xi \right[2{\theta }^{n+1}+(3-n)({\theta }^{\text{'}}{)}^2{]}^2+2^2\theta {\theta }^{\text{'}}[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2\}$
	$=$	$\frac{3^2(n+1)\xi }{2^2{n}^2{\theta }^{n+1}}\left\{n\xi \right[2^2{\theta }^{2(n+1)}+4(3-n){\theta }^{n+1}({\theta }^{\text{'}}{)}^2+(3-n{)}^2({\theta }^{\text{'}}{)}^4]$
		$+2^2\theta {\theta }^{\text{'}}[(n-1{)}^2{\xi }^2{\theta }^{2n}+2(n-1)(n-3)\xi {\theta }^n{\theta }^{\text{'}}+(n-3{)}^2({\theta }^{\text{'}}{)}^2\left]\right\}$
	$=$	$\frac{3^2(n+1)\xi }{2^2{n}^2{\theta }^{n+1}}\left\{n\xi \right[2^2{\theta }^{2(n+1)}+(3-n{)}^2({\theta }^{\text{'}}{)}^4]+2^2\theta {\theta }^{\text{'}}[(n-1{)}^2{\xi }^2{\theta }^{2n}+(n-3{)}^2({\theta }^{\text{'}}{)}^2]$
		$+[2^3(n-1)(n-3)-4n(n-3)]\xi {\theta }^{n+1}({\theta }^{\text{'}}{)}^2\}$
	$=$	$\frac{3^2(n+1)\xi }{2^2{n}^2{\theta }^{n+1}}\left\{n\xi \right[2^2{\theta }^{2(n+1)}+(3-n{)}^2({\theta }^{\text{'}}{)}^4]+2^2\theta {\theta }^{\text{'}}[(n-1{)}^2{\xi }^2{\theta }^{2n}+(n-3{)}^2({\theta }^{\text{'}}{)}^2]+2^2(n^2-5n+6)\xi {\theta }^{n+1}({\theta }^{\text{'}}{)}^2\}$

Application to n = 5 Polytropic Configuration

Let's try plugging in expressions for n = 5 configurations, for which the Lane-Emden function is known analytically. Specifically,

${\theta }_{n=5}=(1+\frac{{\xi }^2}{3}{)}^{-1/2}$       and       ${\theta }_{n=5}^{\text{'}}=-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}.$

First Attempt

We have,

$\{{\sum }^2{\}}_{x_P}^{n=5}$	$=$	$\frac{2\cdot 3^3\xi }{5^2{\theta }^6}\left\{5\xi \right[{\theta }^{12}+({\theta }^{\text{'}}{)}^4]+2^4\theta {\theta }^{\text{'}}[{\xi }^2{\theta }^{10}+({\theta }^{\text{'}}{)}^2]+2\cdot 3\xi {\theta }^6({\theta }^{\text{'}}{)}^2\}$
	$=$	$\frac{2\cdot 3^3\xi }{5^2}(1+\frac{{\xi }^2}{3}{)}^3\{5\xi \left[\right(1+\frac{{\xi }^2}{3}{)}^{-6}+[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{]}^4]$
		$+2^4(1+\frac{{\xi }^2}{3}{)}^{-1/2}[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}]\left[{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-5}+[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{]}^2]$
		$+2\cdot 3\xi (1+\frac{{\xi }^2}{3}{)}^{-3}[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{]}^2\}$
	$=$	$\frac{2\cdot 3^3\xi }{5^2}(1+\frac{{\xi }^2}{3}{)}^3\{5\xi \left[\right(1+\frac{{\xi }^2}{3}{)}^{-6}+\frac{{\xi }^4}{3^4}(1+\frac{{\xi }^2}{3}{)}^{-6}]$
		$-\frac{2^4\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-2}[{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-5}+\frac{{\xi }^2}{3^2}(1+\frac{{\xi }^2}{3}{)}^{-3}]+\frac{2{\xi }^3}{3}(1+\frac{{\xi }^2}{3}{)}^{-6}\}$
	$=$	$\frac{2\cdot 3^3{\xi }^2}{5^2}(1+\frac{{\xi }^2}{3}{)}^{-4}\{5(1+\frac{{\xi }^2}{3})[1+\frac{{\xi }^4}{3^4}]+\frac{2{\xi }^2}{3}(1+\frac{{\xi }^2}{3})-\frac{2^4{\xi }^2}{3}[1+\frac{1}{3^2}(1+\frac{{\xi }^2}{3}{)}^2\left]\right\}$
	$=$	$\frac{2\cdot 3^3{\xi }^2}{5^2}(1+\frac{{\xi }^2}{3}{)}^{-4}\{\frac{5}{3^5}[3^5+3^4{\xi }^2+3{\xi }^4+{\xi }^6]+\frac{2}{3^2}(3{\xi }^2+{\xi }^4)-\frac{2^4{\xi }^2}{3^5}[3^4+3^2+2\cdot 3{\xi }^2+{\xi }^4]\}$
	$=$	$\frac{2{\xi }^2}{3^2\cdot 5^2}(1+\frac{{\xi }^2}{3}{)}^{-4}\{5[3^5+3^4{\xi }^2+3{\xi }^4+{\xi }^6]+2\cdot 3^3(3{\xi }^2+{\xi }^4)-2^4[2\cdot 3^2\cdot 5{\xi }^2+2\cdot 3{\xi }^4+{\xi }^6]\}$
	$=$	$\frac{2{\xi }^2}{3^2\cdot 5^2}(1+\frac{{\xi }^2}{3}{)}^{-4}\{3^5\cdot 5+3^2{\xi }^2[3^2\cdot 5+2\cdot 3^2-2^5\cdot 5]+3{\xi }^4[5+2\cdot 3^2-2^5]+{\xi }^6[5-2^4]\}$
	$=$	$\frac{2{\xi }^2}{3^2\cdot 5^2}(1+\frac{{\xi }^2}{3}{)}^{-4}[3^5\cdot 5-3^2\cdot 97{\xi }^2-3^3{\xi }^4-11{\xi }^6].$

Second Attempt

First, we evaluate,

$\{x_P{\}}^{n=5}$	$=$	$\frac{2\cdot 3}{5}[1-\frac{1}{2\xi }(1+\frac{{\xi }^2}{3}{)}^{5/2}\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}]$
	$=$	$\frac{1}{5}[6-(1+\frac{{\xi }^2}{3}\left)\right]$
	$=$	$1-\frac{{\xi }^2}{3\cdot 5};$

and,

$\{\frac{d({\xi }^3{x}_P)}{d\xi }{\}}^{n=5}$	$=$	$3{\xi }^2\{1-[\frac{({\theta }^{\text{'}}{)}^2}{{\theta }^6}\left]\right\}$
	$=$	$3{\xi }^2\{1-[(1+\frac{{\xi }^2}{3}{)}^{-1/2}{]}^{-6}[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}{]}^2\}$
	$=$	$3{\xi }^2(1-\frac{{\xi }^2}{3^2}).$

Then, returning to Chandrasekhar's expression for the Lagrangian and evaluating the sum of the last two terms inside the curly braces gives,

$\{{\sum }^2{\}}_{x_P}^{n=5}$	$=$	$\left\{4x^2{\xi }^3\right[\frac{d{\theta }^{n+1}}{d\xi }]+\frac{n+1}{n}(\frac{{\theta }^{n+1}}{{\xi }^2}\left)\right[\frac{d({\xi }^{3}x)}{d\xi }{]}^2{\}}_{x_P}^{n=5}$
	$=$	$\left[{\theta }_{n=5}{]}^5\right\{2^3\cdot 3{\xi }^3\left[x_P{]}^2\right[\frac{d\theta }{d\xi }]+\frac{2\cdot 3}{5{\xi }^2}[\theta \left]\right[\frac{d({\xi }^3{x}_P)}{d\xi }{]}^2{\}}^{n=5}$
	$=$	$\left[\right(1+\frac{{\xi }^2}{3}{)}^{-1/2}{]}^5\left\{\frac{2\cdot 3}{5{\xi }^2}\right[(1+\frac{{\xi }^2}{3}{)}^{-1/2}]\left[3{\xi }^2\right(1-\frac{{\xi }^2}{3^2}){]}^2+2^3\cdot 3{\xi }^3[3{\xi }^2(1-\frac{{\xi }^2}{3^2}){]}^2[-\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}\left]\right\}$
	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-4}\{\frac{2\cdot 3^3{\xi }^2}{5}(1-\frac{{\xi }^2}{3^2}{)}^2(1+\frac{{\xi }^2}{3})-2^3\cdot 3^2{\xi }^8(1-\frac{{\xi }^2}{3^2}{)}^2\}$
	$=$	$\frac{1}{3\cdot 5}(1+\frac{{\xi }^2}{3}{)}^{-4}(1-\frac{{\xi }^2}{3^2}{)}^2\{2\cdot 3^3{\xi }^2(3+{\xi }^2)-2^3\cdot 3^3\cdot 5{\xi }^8\}$
	$=$	$\frac{2\cdot 3^2{\xi }^2}{5}(1+\frac{{\xi }^2}{3}{)}^{-4}(1-\frac{{\xi }^2}{3^2}{)}^2\{3+{\xi }^2-2^2\cdot 5{\xi }^6\}.$

Normalizations

Returning to the last line of our derivation of the Ledoux & Walraven Lagrangian, we can write,

$\left[\frac{2}{{\gamma }_{\mathrm{g}}}\right]e^{-2i\omega t}L+4\pi \underset{0}{\overset{R}{\int }}{\rho }_0\left(\frac{{\omega }^2}{{\gamma }_{\mathrm{g}}}\right)r_0^4{x}^{2}d{r}_0$	$=$	$4\pi \{-\underset{0}{\overset{R}{\int }}{P}_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0+\underset{0}{\overset{R}{\int }}{r}_0^3{x}^2\frac{d}{d{r}_0}\left[\right(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)P_0\right]d{r}_0-\left[3r_0^3{x}^2{P}_0{]}_0^R\right\}$
	$=$	$-\underset{0}{\overset{R}{\int }}\left[r_0\right(\frac{\partial x}{\partial r_0}){]}^{2}4\pi P_0{r}_0^{2}d{r}_0-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\underset{0}{\overset{R}{\int }}\right[x^2\left]\right(-\frac{r_0}{{\rho }_0}\frac{d{P}_0}{d{r}_0})4\pi r_0^2{\rho }_{0}d{r}_0-4\pi [3r_0^3{x}^2{P}_0{]}_0^R$
	$=$	$-\underset{0}{\overset{R}{\int }}\left[r_0\right(\frac{\partial x}{\partial r_0}){]}^{2}4\pi P_0{r}_0^{2}d{r}_0-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\underset{0}{\overset{R}{\int }}\right[x^2\left]\right(\frac{G{M}_r}{r_0})4\pi r_0^2{\rho }_{0}d{r}_0-[3x{]}^2\frac{4\pi R^3}{3}{P}_e.$

Alternate Approach:   Integrate Over LAWE

As we have demonstrated, above, if we assume that ${\mathrm{\Gamma }}_1$ is constant throughout the configuration, our version of the LAWE can be straightforwardly rearranged to give equation (58.1) of Ledoux & Pekeris (1941), that is,

$0$

$=$

$\frac{d}{dr}\left[r^4{\mathrm{\Gamma }}_{1}P\frac{dx}{dr}\right]+[{\sigma }^2\rho r^4+(3{\mathrm{\Gamma }}_1-4)r^3\frac{dP}{dr}]x,$

where, we are using $x$ in place of $\xi$ to represent the fractional Lagrangian displacement, $\delta r/r$. If we multiply this expression through by $4\pi dr$ and integrate over the entire volume of the configuration, we have,

$0$	$=$	$4\pi \underset{0}{\overset{R}{\int }}\frac{d}{dr}\left[r^4{\mathrm{\Gamma }}_{1}P\frac{dx}{dr}\right]dr+\underset{0}{\overset{R}{\int }}[{\sigma }^{2}x{r}^2+(3{\mathrm{\Gamma }}_1-4)x(\frac{r}{\rho }\frac{dP}{dr}\left)\right]4\pi r^2\rho dr$
	$=$	$\left[4\pi r^4{\mathrm{\Gamma }}_{1}P\right(\frac{dx}{dr}){]}_0^R+\underset{0}{\overset{R}{\int }}[{\sigma }^{2}x{r}^2+(3{\mathrm{\Gamma }}_1-4)x(-\frac{Gm}{r})]dm$
	$=$	$\left[4\pi r^4{\mathrm{\Gamma }}_{1}P\right(\frac{dx}{dr}){]}_0^R+{\sigma }^2\underset{0}{\overset{R}{\int }}xdI+\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)xd{W}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}},$
$\Rightarrow {\sigma }^2$	$=$	$-\{\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)xd{W}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+[4\pi r^4{\mathrm{\Gamma }}_{1}P\left(\frac{dx}{dr}\right){]}_0^R\left\}\right[\underset{0}{\overset{R}{\int }}xdI{]}^{-1},$

where the definitions of $d{W}_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$ and $dI$ are as provided, above. This last expression is the same as equation (59.17) of Ledoux & Pekeris (1941), except that:   these authors have retained a term allowing for radial variation of ${\mathrm{\Gamma }}_1$, whereas we have not; and we have retained a boundary term that can accommodate a nonzero surface pressure, whereas Ledoux & Pekeris have not.

See Also

• Summary discussion of Ledoux's Variational Principle