StabilityVariationalPrinciple

Free-Energy Stability Analysis

Most General Case

Consider a free-energy function of the form,

${𝒢}$

$=$

$-a{\chi }^{-1}+b{\chi }^{-3/n}+c{\chi }^{-3/j}+{{𝒢}}_0,$

where, $a,b,c,$ and ${{𝒢}}_0$ are constants, and the dimensionless configuration radius,

$\chi \equiv \frac{R}{R_0},$

is defined in terms of a characteristic length, $R_0$, which is likely to be different for each type of problem.

Virial Equilibrium

The first variation (first derivative) of this function with respect to the configuration's radius is,

$\frac{d{𝒢}}{d\chi }$

$=$

$a{\chi }^{-2}-\left(\frac{3b}{n}\right){\chi }^{-3/n-1}-\left(\frac{3c}{j}\right){\chi }^{-3/j-1}.$

According to the virial theorem, the radius of an equilibrium configuration is obtained by setting $d{𝒢}/d\chi =0$ and identifying the roots of the resulting equation. For example, identifying roots of the polynomial expression,

$0$

$=$

$\frac{a}{3c}-\left(\frac{b}{nc}\right){\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\left(\frac{1}{j}\right){\chi }_{\mathrm{e}\mathrm{q}}^{(j-3)/j}.$

Stability

Let's rewrite the first variation of the free-energy function in terms of three coefficients $(e,f,g)$ which, in general, we will permit to have different values from the original three $(a,b,c)$,

${{𝒢}}^{\text{'}}$

$=$

$e{\chi }^{-2}-\left(\frac{3f}{n}\right){\chi }^{-3/n-1}-\left(\frac{3g}{j}\right){\chi }^{-3/j-1}.$

The first variation (first derivative) of this function with respect to the configuration's radius — which, in effect, represents the second variation of the free-energy function — gives,

$\frac{d{{𝒢}}^{\text{'}}}{d\chi }$

$=$

$-2e{\chi }^{-3}+(\frac{3}{n}+1)\left(\frac{3f}{n}\right){\chi }^{-3/n-2}+(\frac{3}{j}+1)\left(\frac{3g}{j}\right){\chi }^{-3/j-2}.$

If we evaluate this function by setting $\chi ={\chi }_{\mathrm{e}\mathrm{q}}$, the sign of the resulting expression should indicate stability (positive) or dynamical instability (negative); and the marginally unstable configuration is identified by the value of ${\chi }_{\mathrm{e}\mathrm{q}}$ for which $d{{𝒢}}^{\text{'}}/d\chi =0$.

Pressure-Truncated Configurations

Expectations

For pressure-truncated polytropes, we set $j=-1$ and let $n$ represent the chosen polytropic index. In this situation, then, we have,

Free-energy expression:		${𝒢}$	$=$	$-a{\chi }^{-1}+b{\chi }^{-3/n}+c{\chi }^3+{{𝒢}}_0;$
Virial equilibrium:		$0$	$=$	$\frac{a}{3c}-\left(\frac{b}{nc}\right){\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}+{\chi }_{\mathrm{e}\mathrm{q}}^4;$
Stability indicator:		$\frac{d{{𝒢}}^{\text{'}}}{d\chi }$	$=$	$-2e{\chi }^{-3}+(\frac{3}{n}+1)\left(\frac{3f}{n}\right){\chi }^{-3/n-2}+6g\chi .$

Hence, the (critical) equilibrium radius of the marginally unstable configuration is given by the expression,

$6g{\chi }_{\mathrm{e}\mathrm{q}}^4$	$=$	$2e-(\frac{3}{n}+1)\left(\frac{3f}{n}\right){\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$
	$=$	$2e-\left[\frac{3f(n+3)}{n^2}\right]\left(\frac{nc}{b}\right)[\frac{a}{3c}+{\chi }_{\mathrm{e}\mathrm{q}}^4]$
$\Rightarrow 6g{\chi }_{\mathrm{e}\mathrm{q}}^4+\left[\frac{3f(n+3)}{n^2}\right]\left(\frac{nc}{b}\right){\chi }_{\mathrm{e}\mathrm{q}}^4$	$=$	$2e-\left[\frac{3f(n+3)}{n^2}\right]\left(\frac{nc}{b}\right)\left[\frac{a}{3c}\right]$
$\Rightarrow [6g+\frac{3cf(n+3)}{nb}]{\chi }_{\mathrm{e}\mathrm{q}}^4$	$=$	$2e-\left[\frac{af(n+3)}{nb}\right]$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^4{|}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left[\frac{2nbe-af(n+3)}{6nbg+3cf(n+3)}\right].$

Notice that, if $(e,f,g)\to (a,b,c)$, this gives,

${\chi }_{\mathrm{e}\mathrm{q}}^4{|}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left[\frac{2nba-ab(n+3)}{6nbc+3cb(n+3)}\right]$
	$=$	$\frac{a}{3^{2}c}\left[\frac{n-3}{n+1}\right].$

Energies and Structural Form Factors

Old Approach

As has been developed in, for example, our accompanying review, we adopt the following normalizations:

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$=$	$\left[\right(\frac{G}{K}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{n-1}{]}^{1/(n-3)},$
$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$=$	$[\frac{K^{4n}}{G^{3(n+1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2(n+1)}}{]}^{1/(n-3)},$
${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\equiv \frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}$	$=$	$\frac{3}{4\pi }[\frac{K^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}^{n/(n-3)},$
$E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$=$	$[K^n{G}^{-3}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{n-5}{]}^{1/(n-3)}.$

Then, from separate summaries — both here and here — we can write,

$\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left)\underset{0}{\overset{x}{\int }}3x^2\right[\frac{\rho (x)}{{\rho }_c}]dx,$
$\frac{P_{e}V}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{4\pi }{3}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }^3,$
$\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-{\chi }^{-1}\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left)\underset{0}{\overset{1}{\int }}3x\right[\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left]\right[\frac{\rho (x)}{{\rho }_c}]dx$
	$=$	$-\frac{3}{5}{\chi }^{-1}\left(\frac{{\rho }_c}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{q}}^2\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\underset{0}{\overset{1}{\int }}5x\left\{\underset{0}{\overset{x}{\int }}3x^2\right[\frac{\rho (x)}{{\rho }_c}\left]dx\right\}\left[\frac{\rho (x)}{{\rho }_c}\right]dx$
	$=$	$-\frac{3}{5}{\chi }^{-1}(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^2\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2},$
$\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{4\pi }{3({\gamma }_g-1)}\cdot {\chi }^{3-3\gamma }\left\{\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_c}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{q}}^{\gamma }\left(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{\gamma }\underset{0}{\overset{1}{\int }}3x^2\right[\frac{P(x)}{P_c}\left]dx\right\}$
	$=$	$\frac{4\pi n}{3}\cdot {\chi }^{-3/n}\left[\frac{3}{4\pi }\right(\frac{M_{\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}{]}_{\mathrm{e}\mathrm{q}}^{(n+1)/n}\cdot {\widetilde{\mathfrak{𝔣}}}_A,$

where the structural form factors are defined as follows:

${\mathfrak{𝔣}}_M$	$\equiv$	$\underset{0}{\overset{1}{\int }}3\left[\frac{\rho (x)}{{\rho }_c}\right]x^{2}dx=(\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{e}\mathrm{q}},$
${\mathfrak{𝔣}}_W$	$\equiv$	$3\cdot 5\underset{0}{\overset{1}{\int }}\left\{\underset{0}{\overset{x}{\int }}\right[\frac{\rho (x)}{{\rho }_c}\left]x^{2}dx\right\}\left[\frac{\rho (x)}{{\rho }_c}\right]xdx,$
${\mathfrak{𝔣}}_A$	$\equiv$	$\underset{0}{\overset{1}{\int }}3\left[\frac{P(x)}{P_c}\right]x^{2}dx.$

This gives, specifically for specifically for pressure-truncated polytropic configurations,

${\widetilde{\mathfrak{𝔣}}}_M$	$=$	$(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}),$
${\widetilde{\mathfrak{𝔣}}}_W$	$=$	$\frac{3\cdot 5}{(5-n){\tilde{\xi }}^2}[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }],$
${\widetilde{\mathfrak{𝔣}}}_A$	$=$	$\frac{1}{(5-n)}\{6{\tilde{\theta }}^{n+1}+(n+1)[3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }]\}.$

New Approach

In order to accommodate the structural integrals required by the Ledoux variational principle, let's re-derive some of these key energy and form-factor expressions. Basically, we will be repeating some earlier derivations.

Mass

Defining $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ as the total mass of the isolated configuration, while $M\le M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ is the truncated configuration's mass; defining $R$ as the truncated configuration's (not necessarily equilibrium) radius; and being careful to define the mean density of the truncated configuration such that,

$\overline{\rho }\equiv \frac{3M}{4\pi R^3},$

we have,

$M_r(r)$	$=$	$\underset{0}{\overset{r}{\int }}4\pi r^2\rho dr$
$\Rightarrow \frac{M_r(r)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\frac{3}{4\pi }\underset{0}{\overset{r}{\int }}4\pi \left(\frac{r}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^2\right(\frac{\rho }{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})\frac{dr}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$
	$=$	$\left(\frac{{\rho }_c}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3\underset{0}{\overset{r}{\int }}3\right(\frac{r}{R}{)}^2\left(\frac{\rho }{{\rho }_c}\right)\frac{dr}{R}$
	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}\right)\left[\frac{\overline{\rho }}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3\underset{0}{\overset{\xi }{\int }}3\right(\frac{\xi }{\tilde{\xi }}{)}^2\left(\frac{\rho }{{\rho }_c}\right)\frac{d\xi }{\tilde{\xi }}$
	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}\right)\left[\frac{M/R^3}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3\underset{0}{\overset{\xi }{\int }}3\right(\frac{\xi }{\tilde{\xi }}{)}^2\left(\frac{\rho }{{\rho }_c}\right)\frac{d\xi }{\tilde{\xi }}$
	$=$	$\left(\frac{{\rho }_c}{\overline{\rho }}\right)\left(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right){\tilde{\xi }}^{-3}\underset{0}{\overset{\xi }{\int }}3{\xi }^2{\theta }^{n}d\xi .$

Acknowledging that $M_r\to M$ when the upper integration limit goes to $\tilde{\xi }$, we see that the "mass" form-factor is,

${\widetilde{\mathfrak{𝔣}}}_M$

$\equiv$

${\tilde{\xi }}^{-3}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\xi }^2{\theta }^{n}d\xi =\left(\frac{\overline{\rho }}{{\rho }_c}\right).$

Now, from the,

Polytropic Lane-Emden Equation

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$

we realize that,

$\frac{d}{d\xi }\left({\xi }^2{\theta }^{\text{'}}\right)$

$=$

$-{\xi }^2{\theta }^n.$

So these last two expressions may also be written as,

$\frac{M_r(r)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$\left(\frac{{\rho }_c}{\overline{\rho }}\right)\left(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right){\tilde{\xi }}^{-3}[-3{\xi }^2{\theta }^{\text{'}}];$

and,

${\widetilde{\mathfrak{𝔣}}}_M$

$\equiv$

$[-\frac{3{\theta }^{\text{'}}}{\xi }{]}_{\tilde{\xi }}.$

Modified Internal Energy

Now we want to develop the appropriately scaled integral definition of a "variational" internal energy having the form,

$\frac{U_{\mathrm{{\rm Y}}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$\equiv$	$\frac{1}{({\gamma }_{\mathrm{g}}-1)}\underset{0}{\overset{R}{\int }}4\pi {\mathrm{{\rm Y}}}_U(r)\left(\frac{r}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^2\right(\frac{P}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left)\right(\frac{dr}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})$
	$=$	$\frac{1}{({\gamma }_{\mathrm{g}}-1)}\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^{{\gamma }_{\mathrm{g}}}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_{\mathrm{g}}}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3-3{\gamma }_{\mathrm{g}}}\underset{0}{\overset{R}{\int }}4\pi {\mathrm{{\rm Y}}}_U(r)(\frac{r}{R}{)}^2\left(\frac{P}{P_c}\right)\left(\frac{dr}{R}\right)$
	$=$	$\frac{4\pi n}{3}\left[\right(\frac{3}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}){]}^{(n+1)/n}{\chi }^{-3/n}{\tilde{\xi }}^{-3}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\mathrm{{\rm Y}}}_U(\xi ){\xi }^2{\theta }^{n+1}d\xi .$

Hence, the coefficient, $f$, in the free-energy expression is,

$f={\chi }^{3/n}\left[\frac{U_{\mathrm{{\rm Y}}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$\frac{4\pi n}{3}\left[\right(\frac{3}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{(n+1)/n}\right\{{\tilde{\xi }}^{-3}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\mathrm{{\rm Y}}}_U(\xi ){\xi }^2{\theta }^{n+1}d\xi \};$

or, if ${\mathrm{{\rm Y}}}_U(\xi )=1$, then,

$f\to b={\chi }^{3/n}\left[\frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$\frac{4\pi n}{3}\left[\right(\frac{3}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}){]}^{(n+1)/n}{\widetilde{\mathfrak{𝔣}}}_A;$

where,

${\widetilde{\mathfrak{𝔣}}}_A$

$=$

$\left\{{\tilde{\xi }}^{-3}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\xi }^2{\theta }^{n+1}d\xi \right\}.$

When ${\mathrm{{\rm Y}}}_U(\xi )=1$, then according to Viala & Horedt (1974), this integral over polytropic functions becomes,

$\underset{0}{\overset{\tilde{\xi }}{\int }}3{\xi }^2{\theta }^{n+1}d\xi$	$=$	$\frac{(n+1)}{(5-n)}[\frac{6}{(n+1)}\cdot {\tilde{\xi }}^3{\tilde{\theta }}^{n+1}+3{\tilde{\xi }}^3({\tilde{\theta }}^{\text{'}}{)}^2-3(-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}})\tilde{\theta }]$
$\Rightarrow {\widetilde{\mathfrak{𝔣}}}_A\equiv {\tilde{\xi }}^{-3}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\xi }^2{\theta }^{n+1}d\xi$	$=$	$\frac{(n+1)}{(5-n)}[\frac{6{\tilde{\theta }}^{n+1}}{(n+1)}+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }],$

which matches the expression for ${\widetilde{\mathfrak{𝔣}}}_A$ derived earlier.

Modified Gravitational Potential Energy

Similarly, we have,

$\frac{W_{\mathrm{{\rm Y}}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\underset{0}{\overset{R}{\int }}{\mathrm{{\rm Y}}}_W(r)\left(\frac{G{M}_r}{r}\right)4\pi r^2\rho dr$
	$=$	$-\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\rho }_c{R}^2}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\underset{0}{\overset{R}{\int }}4\pi {\mathrm{{\rm Y}}}_W(r)\left(\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)\left(\frac{\rho }{{\rho }_c}\right)\frac{rdr}{R^2}$
	$=$	$-\frac{{\rho }_c}{\overline{\rho }}\left(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right){\chi }^{-1}\underset{0}{\overset{R}{\int }}3{\mathrm{{\rm Y}}}_W(r)\left[\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right]\left(\frac{\rho }{{\rho }_c}\right)\frac{rdr}{R^2}$
	$=$	$-\left[\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}){]}^2{\chi }^{-1}{\tilde{\xi }}^{-5}\underset{0}{\overset{\tilde{\xi }}{\int }}3{\mathrm{{\rm Y}}}_W(\xi )[-3{\xi }^2{\theta }^{\text{'}}]{\theta }^n\xi d\xi$
	$=$	$-\frac{3}{5}\left[\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}){]}^2{\chi }^{-1}{\tilde{\xi }}^{-5}\underset{0}{\overset{\tilde{\xi }}{\int }}5{\mathrm{{\rm Y}}}_W(\xi )[-3{\xi }^2{\theta }^{\text{'}}]{\theta }^n\xi d\xi .$

Hence, the coefficient, $e$, in the free-energy expression is,

$e=-\chi \left[\frac{W_{\mathrm{{\rm Y}}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$\frac{3}{5}\left[\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^2\right\{{\tilde{\xi }}^{-5}\underset{0}{\overset{\tilde{\xi }}{\int }}5{\mathrm{{\rm Y}}}_W(\xi )[-3{\xi }^2{\theta }^{\text{'}}]{\theta }^n\xi d\xi \};$

or, if ${\mathrm{{\rm Y}}}_W(\xi )=1$, then,

$e\to a=-\chi \left[\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$\frac{3}{5}\left[\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}){]}^2{\widetilde{\mathfrak{𝔣}}}_W;$

where,

${\widetilde{\mathfrak{𝔣}}}_W$

$\equiv$

$\left\{{\tilde{\xi }}^{-5}\underset{0}{\overset{\tilde{\xi }}{\int }}5\right[-3{\xi }^2{\theta }^{\text{'}}\left]{\theta }^n\xi d\xi \right\}.$

Now, according to Viala & Horedt (1974), when ${\mathrm{{\rm Y}}}_W(\xi )=1$, this integral over polytropic functions becomes,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$-\frac{(4\pi {)}^2}{(5-n)}\cdot G{\rho }_c^2{a}_n^5[{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}+3{\tilde{\xi }}^3({\tilde{\theta }}^{\text{'}}{)}^2-3(-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}})\tilde{\theta }]$
$\Rightarrow \frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-\frac{1}{(5-n)}[{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}+3{\tilde{\xi }}^3({\tilde{\theta }}^{\text{'}}{)}^2-3(-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}})\tilde{\theta }\left]\right[(-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}}{)}_{{\xi }_1}^{(5-n)}\cdot \frac{(n+1{)}^n}{4\pi }{]}^{1/(n-3)}.$

As we have detailed elsewhere, from this, we have deduced that, for polytropic configurations,

${\widetilde{\mathfrak{𝔣}}}_W$	$=$	${\tilde{\xi }}^{-5}\underset{0}{\overset{\tilde{\xi }}{\int }}5[-3{\xi }^2{\theta }^{\text{'}}]{\theta }^n\xi d\xi$
	$=$	$\frac{3\cdot 5}{(5-n){\tilde{\xi }}^2}[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }].$

Test Virial Equilibrium Condition

If the correct normalized equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, is specified, our expectation regarding virial equilibrium is that,

$3nc{\chi }_{\mathrm{e}\mathrm{q}}^4-3b{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}+an$

$=$

$0.$

Let's see if this expression is valid when we plug in the expressions for the equilibrium parameter pair — $R_{\mathrm{e}\mathrm{q}}$ and $P_e$ — that has been given by Horedt (1970), namely,

${\chi }_{\mathrm{e}\mathrm{q}}=\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\cdot \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}$	$=$	$[(n+1{)}^{-n}(4\pi ){]}^{1/(n-3)}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{(n-1)/(n-3)}\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)},$
$\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\cdot \frac{P_{\mathrm{e}}}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}$	$=$	$[(n+1{)}^3(4\pi {)}^{-1}{]}^{(n+1)/(n-3)}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-2(n+1)/(n-3)}{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)},$

where we have taken into account the shift in normalization factors,

Switch from Hoerdt's (1970) Normalization
$[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-(n-1)/(n-3)}\frac{R_{\mathrm{H}\mathrm{o}\mathrm{e}\mathrm{r}\mathrm{d}\mathrm{t}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left[\frac{(\gamma -1)}{\gamma }\right(4\pi {)}^{\gamma -1}{]}^{1/(4-3\gamma )}=[(n+1{)}^{-1}(4\pi {)}^{1/n}{]}^{n/(n-3)}=[(n+1{)}^{-n}(4\pi ){]}^{1/(n-3)};$
$[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{2(n+1)/(n-3)}\frac{P_{\mathrm{H}\mathrm{o}\mathrm{e}\mathrm{r}\mathrm{d}\mathrm{t}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left\{\right[\frac{\gamma }{(\gamma -1)}{]}^3\left(\frac{1}{4\pi }\right){\}}^{\gamma /(4-3\gamma )}=[(n+1{)}^3(4\pi {)}^{-1}{]}^{(n+1)/(n-3)}.$

We therefore have:

First Term

$3n\left[\frac{4\pi }{3}\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left)\right]{\chi }_{\mathrm{e}\mathrm{q}}^4$	$=$	$4\pi n[(n+1{)}^3(4\pi {)}^{-1}{]}^{(n+1)/(n-3)}{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{2(n+1)/(n-3)}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-2(n+1)/(n-3)}$
		$\times \left\{\right[(n+1{)}^{-n}(4\pi ){]}^{1/(n-3)}\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{\}}^4[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{4(n-1)/(n-3)}$
$$	$=$	$4\pi n[(n+1{)}^{[3(n+1)-4n]}(4\pi {)}^{[4-(n+1)]}{]}^{1/(n-3)}{\tilde{\xi }}^4{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{[2(n+1)+4(1-n)]/(n-3)}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{[4(n-1)-2(n+1)]/(n-3)}$
$$	$=$	$\left[\frac{n}{(n+1)}\right]{\tilde{\xi }}^4{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2.$

Second Term

$3b{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$	$=$	$4\pi n\left[\right(\frac{3}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{(n+1)/n}\frac{(n+1)}{(5-n)}\right[\frac{6{\tilde{\theta }}^{n+1}}{(n+1)}+3({\tilde{\theta }}^{\text{'}}{)}^2-{\widetilde{\mathfrak{𝔣}}}_M\tilde{\theta }]$
		$\times \left\{\right[(n+1{)}^{-n}(4\pi ){]}^{1/(n-3)}\tilde{\xi }(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/(n-3)}{\}}^{(n-3)/n}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{(n-1)/n}$
	$=$	$\frac{4\pi n}{(5-n)}\left[\frac{1}{4\pi }\right(-\frac{\tilde{\xi }}{{\tilde{\theta }}^{\text{'}}}\left)\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{(n+1)/n}\right[6{\tilde{\theta }}^{n+1}+3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2-(n+1)(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right]$
		$\times (n+1{)}^{-1}(4\pi {)}^{1/n}{\tilde{\xi }}^{(n-3)/n}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{(1-n)/n}[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{(n-1)/n}$
	$=$	$\frac{n}{(5-n)(n+1)}\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right[6{\tilde{\theta }}^{n+1}+3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2-(n+1)(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right]$
		$\times {\tilde{\xi }}^{[(n-3)/n+3(n+1)/n]}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{[(1-n)/n-(n+1)/n]}$
	$=$	$\frac{n}{(5-n)(n+1)}\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right[6{\tilde{\theta }}^{n+1}+3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2-(n+1)(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right]{\tilde{\xi }}^4(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}.$

Third Term

$an$	$=$	$\frac{3}{5}\left[\right(-\frac{\tilde{\xi }}{3{\tilde{\theta }}^{\text{'}}}\left)\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^2\frac{3\cdot 5n}{(5-n){\tilde{\xi }}^2}\right[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right]$
	$=$	$\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\frac{n{\tilde{\xi }}^4}{(5-n)}\right[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right](-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}}{)}^{-2}.$

Combined

Combining the three terms, the virial expression becomes,

$(5-n)(n+1)\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-2}\right[3nc{\chi }_{\mathrm{e}\mathrm{q}}^4+an-3b{\chi }_{\mathrm{e}\mathrm{q}}^{(n-3)/n}]$	$=$	$n(5-n){\tilde{\xi }}^4{\tilde{\theta }}_n^{n+1}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}+n(n+1){\tilde{\xi }}^4[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }}\left)\tilde{\theta }\right](-{\tilde{\xi }}^2{\tilde{\theta }}^{\text{'}}{)}^{-2}$
		$-n[6{\tilde{\theta }}^{n+1}+3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2-(n+1)(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }})\tilde{\theta }]{\tilde{\xi }}^4(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}$
	$=$	$n(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}{\tilde{\xi }}^4\{(5-n){\tilde{\theta }}_n^{n+1}+(n+1)[{\tilde{\theta }}^{n+1}+3({\tilde{\theta }}^{\text{'}}{)}^2-(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }})\tilde{\theta }]$
		$-[6{\tilde{\theta }}^{n+1}+3(n+1)({\tilde{\theta }}^{\text{'}}{)}^2-(n+1)(-\frac{3{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }})\tilde{\theta }\left]\right\}$
	$=$	$n(n+1)(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^{-2}{\tilde{\xi }}^4\{0\}.$

Q. E. D.

The Ledoux Variational Principle

Drawing from a separate presentation of Ledoux's variational principle, let's normalize his Lagrangian using the same normalizations that have been used, above. His expression is …

$L$

$=$

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

Our normalization produces,

$\frac{L_{\{\}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-\underset{0}{\overset{R}{\int }}\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]{\rho }_0{\omega }^2{r}_0^4{x}^{2}d{r}_0-{\gamma }_{\mathrm{g}}\underset{0}{\overset{R}{\int }}\left[\frac{1}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\right]P_0{r}_0^4(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0$
		$-(3{\gamma }_{\mathrm{g}}-4)\underset{0}{\overset{R}{\int }}\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]r_0^3{\rho }_0{x}^2(-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0})d{r}_0-[\frac{3{\gamma }_{\mathrm{g}}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}{r}_0^3{x}^2{P}_0{]}_0^R$
	$=$	$-\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\underset{0}{\overset{R}{\int }}{x}^2\right[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}^5}{G{\rho }_c}\left]\right[\frac{3{\rho }_c}{4\pi \overline{\rho }{R}^3}{]}^2\left(\frac{{\rho }_0}{{\rho }_c}\right){\omega }^2(\frac{r_0}{R}{)}^4\frac{d{r}_0}{R}-{\gamma }_{\mathrm{g}}\underset{0}{\overset{R}{\int }}[\frac{1}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\left]P_0{r}_0^4\right(\frac{\partial x}{\partial r_0}{)}^{2}d{r}_0$
		$-(3{\gamma }_{\mathrm{g}}-4)\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\underset{0}{\overset{R}{\int }}{x}^2\right[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}^2}{G{M}^2}\left]\right(\frac{r_0}{R}{)}^3{\rho }_0\left(\frac{G{M}_r{R}^2}{r_0^2}\right)\frac{d{r}_0}{R}-3{\gamma }_{\mathrm{g}}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2\left[\frac{R^3}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$
	$=$	$-\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^2\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}{x}^2\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{r_0}{R}{)}^4\frac{d{r}_0}{R}-{\gamma }_{\mathrm{g}}\left[\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3\underset{0}{\overset{R}{\int }}\right[\left(\frac{r_0}{R}\right)\frac{\partial x}{\partial (r_0/R)}{]}^2\left(\frac{P_0}{P_c}\right)(\frac{r_0}{R}{)}^2\frac{d{r}_0}{R}$
		$-(3{\gamma }_{\mathrm{g}}-4)\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}\right]\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}{x}^2\left(\frac{{\rho }_0}{{\rho }_c}\right)\left(\frac{r_0}{R}\right)\left(\frac{M_r}{M}\right)\frac{d{r}_0}{R}-3{\gamma }_{\mathrm{g}}\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2\right[\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}].$

Given that,

$\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{\gamma }\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-3\gamma },$

and,

$\frac{M_r}{M}=\left(\frac{{\rho }_c}{\overline{\rho }}\right)\underset{0}{\overset{R}{\int }}3\left(\frac{r_0}{R}{)}^2\right(\frac{{\rho }_0}{{\rho }_c})\frac{d{r}_0}{R},$

this expression for the Lagrangian becomes,

$\frac{L_{\{\}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^2\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}{x}^2\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{r_0}{R}{)}^4\frac{d{r}_0}{R}-{\gamma }_{\mathrm{g}}\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{\gamma }\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3-3\gamma }\underset{0}{\overset{R}{\int }}\left[\right(\frac{r_0}{R}\left)\frac{\partial x}{\partial (r_0/R)}{]}^2\right(\frac{P_0}{P_c}\left)\right(\frac{r_0}{R}{)}^2\frac{d{r}_0}{R}$
		$-(3{\gamma }_{\mathrm{g}}-4)\left(\frac{3}{4\pi }\right)\left[\frac{{\rho }_c}{\overline{\rho }}{]}^2\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}{x}^2\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{r_0}{R}\left)\right\{\underset{0}{\overset{R}{\int }}3\left(\frac{r_0}{R}{)}^2\right(\frac{{\rho }_0}{{\rho }_c}\left)\frac{d{r}_0}{R}\right\}\frac{d{r}_0}{R}-3{\gamma }_{\mathrm{g}}\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2\right[\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}].$

In an effort to help identify the various terms in this expression as well as the relationship between the entire expression and our unperturbed free energy expression, let's ignore all terms involving the radial eigenfunction, $x$, and its derivative. In this case, we have,

$\frac{L_{\{\}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{u}\mathrm{n}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{b}\mathrm{e}\mathrm{d}}$	$=$	$-\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^2\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left[\frac{{\omega }^2}{G{\rho }_c}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{r_0}{R}{)}^4\frac{d{r}_0}{R}-\frac{{\gamma }_{\mathrm{g}}({\gamma }_{\mathrm{g}}-1)}{4\pi }\left\{\frac{4\pi }{3({\gamma }_{\mathrm{g}}-1)}\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_c}{\overline{\rho }}\left(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right){]}^{\gamma }\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{3-3\gamma }\underset{0}{\overset{R}{\int }}3\right(\frac{P_0}{P_c}\left)\right(\frac{r_0}{R}{)}^2\frac{d{r}_0}{R}\}$
		$+(3{\gamma }_{\mathrm{g}}-4)\left(\frac{1}{4\pi }\right)\{-\frac{3}{5}[\frac{{\rho }_c}{\overline{\rho }}{]}^2\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}5\left(\frac{{\rho }_0}{{\rho }_c}\right)\left(\frac{r_0}{R}\right)\left[\underset{0}{\overset{R}{\int }}3\right(\frac{r_0}{R}{)}^2\left(\frac{{\rho }_0}{{\rho }_c}\right)\frac{d{r}_0}{R}\left]\frac{d{r}_0}{R}\right\}-\frac{3^2{\gamma }_{\mathrm{g}}}{4\pi }\left\{\frac{4\pi }{3}\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3\left[\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\}$
$\Rightarrow \frac{4\pi L_{\{\}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{u}\mathrm{n}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{b}\mathrm{e}\mathrm{d}}$	$=$	$-\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^2\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left[\frac{4\pi {\omega }^2}{G{\rho }_c}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\underset{0}{\overset{R}{\int }}\right(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{r_0}{R}{)}^4\frac{d{r}_0}{R}-{\gamma }_{\mathrm{g}}({\gamma }_{\mathrm{g}}-1)\left\{\frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right\}+(3{\gamma }_{\mathrm{g}}-4)\left\{\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right\}-3^2{\gamma }_{\mathrm{g}}\left\{\frac{P_{e}V}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right\}.$

See Also