SSC/VariationalPrinciple/Pt2

Ledoux's Variational Principle

Part I:  Ledoux & Pekeris (1941)

Part II:  Exploratory Ideas

Related, Exploratory Ideas

Logarithmic Derivatives

Returning to our above discussion of the Ledoux & Walraven approach, we appreciate that the differential relation governing the Variational Principle is,

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

where,

$\alpha \equiv (3-\frac{4}{{\mathrm{\Gamma }}_1}).$

Pressure-Truncated Polytropes

Let's start with the integral expression derived in our discussion of the Ledoux & Walraven approach; insert the variable, $x$, in place of $\xi$; and adopt the boundary conditions,

$r=0$   at the center,

along with

$P=P_e$,   and  $\frac{d\ln x}{d\ln r}=-3$   at the surface (r = R).

That is, let's start with,

$\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$

$=$

$\underset{0}{\overset{R}{\int }}{r}^4{\mathrm{\Gamma }}_{1}P(\frac{dx}{dr}{)}^{2}dr-\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)r^3{x}^2(\frac{dP}{dr})dr+3{\mathrm{\Gamma }}_1{P}_e{R}^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2.$

Via Generalized Normalization

Next, we'll divide through by the normalization energy, as defined in an accompanying discussion,

$E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3=\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}},$

thereby making the integral relation dimensionless:

$0$	$=$	$-\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr+\left[\frac{1}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\right]\underset{0}{\overset{R}{\int }}{r}^4{\mathrm{\Gamma }}_{1}P(\frac{dx}{dr}{)}^{2}dr-[\frac{1}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}]\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)r^3{x}^2(\frac{dP}{dr})dr+[\frac{P_e{R}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}]3{\mathrm{\Gamma }}_1{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
	$=$	$-\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}^5{\rho }_c^2}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]\underset{0}{\overset{R}{\int }}{x}^2\left(\frac{{\sigma }^2}{G{\rho }_c}\right)\left(\frac{\rho }{{\rho }_c}\right)(\frac{r}{R}{)}^4\frac{dr}{R}+[\frac{P_c{R}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\left]\underset{0}{\overset{R}{\int }}\right(\frac{r}{R}{)}^4{\mathrm{\Gamma }}_1\left(\frac{P}{P_c}\right)[\frac{dx}{d(r/R)}{]}^2\frac{dr}{R}$
		$-\left[\frac{P_c{R}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\right]\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)\left(\frac{r}{R}{)}^3{x}^2\right[\frac{d(P/P_c)}{d(r/R)}]\frac{dr}{R}+[\frac{P_e{R}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}]3{\mathrm{\Gamma }}_1{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
	$=$	$-\left[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right[\left(\frac{3}{4\pi }\right)\frac{{\rho }_c}{\overline{\rho }}{]}^2{\chi }^{-1}\underset{0}{\overset{R}{\int }}{x}^2\left(\frac{{\sigma }^2}{G{\rho }_c}\right)\left(\frac{\rho }{{\rho }_c}\right)(\frac{r}{R}{)}^4\frac{dr}{R}+[\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}]3{\mathrm{\Gamma }}_1{\chi }^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
		$+\left[\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]{\chi }^3\underset{0}{\overset{R}{\int }}\left\{\right(\frac{r}{R}{)}^4{\mathrm{\Gamma }}_1\left(\frac{P}{P_c}\right)[\frac{dx}{d(r/R)}{]}^2-(3{\mathrm{\Gamma }}_1-4)(\frac{r}{R}{)}^3{x}^2\left[\frac{d(P/P_c)}{d(r/R)}\right]\}\frac{dr}{R},$

where,

$\chi \equiv \frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}.$

Note that we will ultimately insert the relation,

$\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}\right(\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\left){]}^{{\mathrm{\Gamma }}_1}\right(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-3{\mathrm{\Gamma }}_1}.$

But, for the time being, dividing through by $[P_c/P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}]{\chi }^3$ gives,

$0$	$=$	$-\left[\frac{P_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{-1}\right[\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\left[\right(\frac{3}{4\pi }\left)\frac{{\rho }_c}{\overline{\rho }}{]}^2{\chi }^{-4}\underset{0}{\overset{R}{\int }}{x}^2\right(\frac{{\sigma }^2}{G{\rho }_c}\left)\right(\frac{\rho }{{\rho }_c}\left)\right(\frac{r}{R}{)}^4\frac{dr}{R}$
		$+\left[\frac{P_e}{P_c}\right]3{\mathrm{\Gamma }}_1{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\underset{0}{\overset{R}{\int }}\left\{\right(\frac{r}{R}{)}^4{\mathrm{\Gamma }}_1\left(\frac{P}{P_c}\right)[\frac{dx}{d(r/R)}{]}^2-(3{\mathrm{\Gamma }}_1-4)(\frac{r}{R}{)}^3{x}^2\left[\frac{d(P/P_c)}{d(r/R)}\right]\}\frac{dr}{R},$

Now let's focus on the second line of this integral energy relation, evaluating it for pressure-truncated polytropic configurations, in which case, ${\mathrm{\Gamma }}_1\to (n+1)/n$,

$\frac{r}{R}\to \frac{\xi }{\tilde{\xi }}$

and

$\frac{P}{P_c}\to {\theta }^{n+1}.$

We have,

Second line of relation	$=$	$\left[\frac{P_e}{P_c}\right]3{\mathrm{\Gamma }}_1{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\underset{0}{\overset{R}{\int }}\left\{\right(\frac{r}{R}{)}^4{\mathrm{\Gamma }}_1\left(\frac{P}{P_c}\right)[\frac{dx}{d(r/R)}{]}^2-(3{\mathrm{\Gamma }}_1-4)(\frac{r}{R}{)}^3{x}^2\left[\frac{d(P/P_c)}{d(r/R)}\right]\}\frac{dr}{R}$
	$=$	$\left[\frac{P_e}{P_c}\right]\left[\frac{3(n+1)}{n}\right]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\right(\frac{\xi }{\tilde{\xi }}{)}^4\left(\frac{n+1}{n}\right){\theta }^{n+1}[\frac{dx}{d\xi }{]}^2{\tilde{\xi }}^2-(\frac{3-n}{n}\left)\right(\frac{\xi }{\tilde{\xi }}{)}^3{x}^2\left[\frac{d{\theta }^{n+1}}{d\xi }\right]\tilde{\xi }\}\frac{d\xi }{\tilde{\xi }}$
	$=$	$\left[\frac{P_e}{P_c}\right]\left[\frac{3(n+1)}{n}\right]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\frac{1}{n{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\{(n+1){\xi }^4{\theta }^{n+1}[\frac{dx}{d\xi }{]}^2-(3-n){\xi }^3{x}^2\left[\frac{d{\theta }^{n+1}}{d\xi }\right]\}d\xi$
	$=$	$\left[\frac{P_e}{P_c}\right]\left[\frac{3(n+1)}{n}\right]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\frac{1}{n{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\{(n+1){\xi }^4{\theta }^{n+1}[\frac{dx}{d\xi }{]}^2-(n+1)(3-n){\xi }^3{x}^2{\theta }^n{\theta }^{\text{'}}\}d\xi$
	$=$	$\left[\frac{P_e}{P_c}\right]\left[\frac{3(n+1)}{n}\right]x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\frac{(n+1)}{n{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{3}{2n}{)}^2\frac{\xi }{{\theta }^n}\right\{\xi \theta \left[\right(\frac{2n}{3})\xi {\theta }^n\cdot \frac{dx}{d\xi }{]}^2-(3-n)[\left(\frac{2n}{3}\right)\xi {\theta }^{n}x{]}^2{\theta }^{\text{'}}\}d\xi .$

Now, let's examine how these terms combine if we guess the analytically defined eigenfunction that applies to marginally unstable, pressure-truncated polytropic configurations, namely,

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

Hence,

Second line of relation	$=$	${\tilde{\theta }}^{n+1}\left[\frac{3(n+1)}{n}\right]\left\{\frac{3(n-1)}{2n}\right[1+\left(\frac{n-3}{n-1}\right)\frac{{\tilde{\theta }}^{\text{'}}}{\tilde{\xi }{\tilde{\theta }}^n}]{\}}^2$
		$+\frac{3^2(n+1)(3-n)}{2^2{n}^3{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{\xi }{{\theta }^n}\{\xi \theta (3-n)[{\theta }^n+\frac{3{\theta }^{\text{'}}}{\xi }+\frac{n({\theta }^{\text{'}}{)}^2}{\theta }{]}^2-[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2{\theta }^{\text{'}}\}d\xi$
	$=$	$\frac{1}{{\tilde{\xi }}^2{\tilde{\theta }}^{n+1}}\left[\frac{3^3(n+1)}{2^2{n}^3}\right][(n-1)\tilde{\xi }{\tilde{\theta }}^{n+1}+(n-3)\tilde{\theta }{\tilde{\theta }}^{\text{'}}{]}^2$
		$+\frac{3^2(n+1)(3-n)}{2^2{n}^3{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{1}{{\theta }^{n+1}}\{(3-n)[\xi {\theta }^{n+1}+3\theta {\theta }^{\text{'}}+n\xi ({\theta }^{\text{'}}{)}^2{]}^2-[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2\xi \theta {\theta }^{\text{'}}\}d\xi$
	$=$	$\frac{1}{{\tilde{\xi }}^2{\tilde{\theta }}^{n+1}}\left[\frac{3^3(n+1)}{2^2{n}^3}\right][(n-1)\tilde{\xi }{\tilde{\theta }}^{n+1}+(n-3)\tilde{\theta }{\tilde{\theta }}^{\text{'}}{]}^2$
		$+\frac{3^2(n+1)(3-n{)}^2}{2^2{n}^3{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{1}{{\theta }^{n+1}}\left\{\right[\xi {\theta }^{n+1}+3\theta {\theta }^{\text{'}}+n\xi ({\theta }^{\text{'}}{)}^2{]}^2+\frac{1}{(n-3)}[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2\xi \theta {\theta }^{\text{'}}\}d\xi$

Note that, in this derivation, we have inserted the expressions:

$[\xi {\theta }^{n+1}+3\theta {\theta }^{\text{'}}+n\xi ({\theta }^{\text{'}}{)}^2\left]\right[\xi {\theta }^{n+1}+3\theta {\theta }^{\text{'}}+n\xi ({\theta }^{\text{'}}{)}^2]={\xi }^2{\theta }^{2(n+1)}+6\xi {\theta }^{n+2}{\theta }^{\text{'}}+2n{\xi }^2{\theta }^{n+1}({\theta }^{\text{'}}{)}^2+6n\xi \theta ({\theta }^{\text{'}}{)}^3+n^2{\xi }^2({\theta }^{\text{'}}{)}^4$

$\frac{1}{(n-3)}[(n-1)\xi {\theta }^n+(n-3){\theta }^{\text{'}}{]}^2\xi \theta ({\theta }^{\text{'}})=[\frac{(n-1{)}^2}{(n-3)}]{\xi }^3{\theta }^{2n+1}({\theta }^{\text{'}})+2(n-1){\xi }^2{\theta }^{n+1}({\theta }^{\text{'}}{)}^2+(n-3)\xi \theta ({\theta }^{\text{'}}{)}^3$

Directly to n = 5 Polytropic Configurations

Exact Demonstration of Variational Principle

$\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$	$=$	$\underset{0}{\overset{R}{\int }}{r}^4{\mathrm{\Gamma }}_{1}P(\frac{dx}{dr}{)}^{2}dr-\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)r^3{x}^2(\frac{dP}{dr})dr+3{\mathrm{\Gamma }}_1{P}_e{R}^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{1}{R^3{P}_c}\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$	$=$	$\underset{0}{\overset{R}{\int }}\left(\frac{r}{R}{)}^4\right(\frac{n+1}{n}\left)\right(\frac{P}{P_c}\left)\right[\frac{dx}{d(r/R)}{]}^2\frac{dr}{R}-\underset{0}{\overset{R}{\int }}\left[3\right(\frac{n+1}{n})-4]\left(\frac{r}{R}{)}^3{x}^2\right[\frac{d(P/P_c)}{d(r/R)}]\frac{dr}{R}+3(\frac{n+1}{n}\left)\right(\frac{P_e}{P_c})x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{6}{5}\left(\frac{\xi }{\tilde{\xi }}{)}^4{\theta }^6\right[\frac{dx}{d(\xi /\tilde{\xi })}{]}^2\frac{d\xi }{\tilde{\xi }}-\underset{0}{\overset{\tilde{\xi }}{\int }}(-\frac{2}{5})\left(\frac{\xi }{\tilde{\xi }}{)}^3{x}^2\right[\frac{d{\theta }^6}{d(\xi /\tilde{\xi })}]\frac{d\xi }{\tilde{\xi }}+(\frac{18}{5}){\tilde{\theta }}^6{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
	$=$	$\frac{1}{{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{6}{5}\right){\xi }^4{\theta }^6[\frac{dx}{d\xi }{]}^{2}d\xi +\frac{1}{{\tilde{\xi }}^3}\underset{0}{\overset{\tilde{\xi }}{\int }}(\frac{2}{5}\left){\xi }^3{x}^2\right[\frac{d{\theta }^6}{d\xi }]d\xi +(\frac{18}{5}){\tilde{\theta }}^6{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{5{\tilde{\xi }}^3}{2R^3{P}_c}\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}3{\xi }^4{\theta }^6[-\frac{2\xi }{15}{]}^{2}d\xi +\underset{0}{\overset{\tilde{\xi }}{\int }}6{\xi }^3[\frac{15-{\xi }^2}{15}{]}^2{\theta }^5\left[\frac{d\theta }{d\xi }\right]d\xi +9{\tilde{\xi }}^3{\tilde{\theta }}^6[\frac{15-{\tilde{\xi }}^2}{15}{]}^2$
	$=$	$\left(\frac{2^2}{3\cdot 5^2}\right)\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^6(\frac{3}{3+{\xi }^2}{)}^{3}d\xi +(\frac{2}{3\cdot 5^2}\left)\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^3\right[15-{\xi }^2{]}^2\left(\frac{3}{3+{\xi }^2}{)}^4\right[-\frac{\xi }{3}]d\xi +(\frac{1}{5^2}\left){\tilde{\xi }}^3\right(\frac{3}{3+{\tilde{\xi }}^2}{)}^3[15-{\tilde{\xi }}^2{]}^2$
	$=$	$\left(\frac{2^2\cdot 3^2}{5^2}\right)\underset{0}{\overset{\tilde{\xi }}{\int }}\left[\frac{{\xi }^6}{(3+{\xi }^2{)}^3}\right]d\xi -\left(\frac{2\cdot 3^2}{5^2}\right)\underset{0}{\overset{\tilde{\xi }}{\int }}\left[\frac{{\xi }^4(15-{\xi }^2{)}^2}{(3+{\xi }^2{)}^4}\right]d\xi +\left(\frac{3^3}{5^2}\right)\left[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}\right]$
$\Rightarrow \frac{5^3{\tilde{\xi }}^3}{2\cdot 3^2{R}^3{P}_c}\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left[\frac{4{\xi }^6(3+{\xi }^2)-2{\xi }^4(15-{\xi }^2{)}^2}{(3+{\xi }^2{)}^4}\right]d\xi +3\left[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}\right]$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\frac{2{\xi }^4[6{\xi }^2+2{\xi }^4-15^2+30{\xi }^2-{\xi }^4]}{(3+{\xi }^2{)}^4}\right\}d\xi +3\left[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}\right]$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left\{\frac{2{\xi }^4[{\xi }^4+36{\xi }^2-15^2]}{(3+{\xi }^2{)}^4}\right\}d\xi +3\left[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}\right]$
	$=$	$[\frac{2{\xi }^5({\xi }^2-15)}{({\xi }^2+3{)}^3}{]}_0^{\tilde{\xi }}+3[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}]$
	$=$	$\left[\frac{2{\tilde{\xi }}^5({\tilde{\xi }}^2-15)}{({\tilde{\xi }}^2+3{)}^3}\right]+3\left[\frac{{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{(3+{\tilde{\xi }}^2{)}^3}\right]$
	$=$	$\frac{2{\tilde{\xi }}^5({\tilde{\xi }}^2-15)+3{\tilde{\xi }}^3(15-{\tilde{\xi }}^2{)}^2}{({\tilde{\xi }}^2+3{)}^3}=\frac{5{\tilde{\xi }}^7-120{\tilde{\xi }}^5+3^3\cdot 5^2{\tilde{\xi }}^3}{({\tilde{\xi }}^2+3{)}^3},$

which equals zero if $\tilde{\xi }=3$. Hooray!!

For All Polytropic Indexes

Generalized Governing Integral Relation

Given that the derivation just completed works for the special case of n = 5, let's generalize it to all polytropic indexes

$\underset{0}{\overset{R}{\int }}{\sigma }^2\rho r^4{x}^{2}dr$	$=$	$\underset{0}{\overset{R}{\int }}{r}^4{\mathrm{\Gamma }}_{1}P(\frac{dx}{dr}{)}^{2}dr-\underset{0}{\overset{R}{\int }}(3{\mathrm{\Gamma }}_1-4)r^3{x}^2(\frac{dP}{dr})dr+3{\mathrm{\Gamma }}_1{P}_e{R}^3{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{R^5{\rho }_c}{R^3{P}_c}\underset{0}{\overset{R}{\int }}{\sigma }^2\left(\frac{\rho }{{\rho }_c}\right)(\frac{r}{R}{)}^4{x}^2\frac{dr}{R}$	$=$	$\underset{0}{\overset{R}{\int }}\left(\frac{r}{R}{)}^4\right(\frac{n+1}{n}\left)\right(\frac{P}{P_c}\left)\right[\frac{dx}{d(r/R)}{]}^2\frac{dr}{R}-\underset{0}{\overset{R}{\int }}\left[3\right(\frac{n+1}{n})-4]\left(\frac{r}{R}{)}^3{x}^2\right[\frac{d(P/P_c)}{d(r/R)}]\frac{dr}{R}+3(\frac{n+1}{n}\left)\right(\frac{P_e}{P_c})x_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{R^2{\rho }_c}{P_c}\underset{0}{\overset{\tilde{\xi }}{\int }}{\sigma }^2{\theta }^n(\frac{\xi }{\tilde{\xi }}{)}^4{x}^2\frac{d\xi }{\tilde{\xi }}$	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{\xi }{\tilde{\xi }}{)}^4\right(\frac{n+1}{n}\left){\theta }^{n+1}\right[\frac{dx}{d(\xi /\tilde{\xi })}{]}^2\frac{d\xi }{\tilde{\xi }}+\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{n-3}{n}\right)\left(\frac{\xi }{\tilde{\xi }}{)}^3{x}^2\right[\frac{d{\theta }^{n+1}}{d(\xi /\tilde{\xi })}]\frac{d\xi }{\tilde{\xi }}+3(\frac{n+1}{n}){\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{n{R}^2{\rho }_c}{(n+1){\tilde{\xi }}^2{P}_c}\underset{0}{\overset{\tilde{\xi }}{\int }}{\sigma }^2{\theta }^n{\xi }^4{x}^{2}d\xi$	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^4{\theta }^{n+1}[\frac{dx}{d\xi }{]}^{2}d\xi +\underset{0}{\overset{\tilde{\xi }}{\int }}(n-3){\xi }^3{\theta }^n{x}^2[\frac{d\theta }{d\xi }]d\xi +3{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
$\Rightarrow \frac{n{R}^{2}G{\rho }_c^2}{(n+1){\tilde{\xi }}^2{P}_c}\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{{\sigma }^2}{G{\rho }_c}\right){\theta }^n{\xi }^4{x}^{2}d\xi$	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}{x}^2[\frac{\xi }{x}\cdot \frac{dx}{d\xi }{]}^{2}d\xi +\underset{0}{\overset{\tilde{\xi }}{\int }}(n-3){\xi }^2{\theta }^{n+1}{x}^2[\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi }]d\xi +3{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$
	$=$	$3{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}{x}^2\left\{\right[\frac{\xi }{x}\cdot \frac{dx}{d\xi }{]}^2+(n-3)[\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi }]\}d\xi$

For additional clarification, let's rewrite the leading coefficient on the lefthand-side of this expression.

LHS	$=$	$\frac{n{R}^{2}G{\rho }_c^2}{(n+1){\tilde{\xi }}^2{P}_c}\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{{\sigma }^2}{G{\rho }_c}\right){\theta }^n{\xi }^4{x}^{2}d\xi$
	$=$	$\left[\frac{n}{(n+1)}\right]\left[\frac{G{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\left(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^2}\right)\left(\frac{{\rho }_c}{\overline{\rho }}{)}^2\right[\frac{3M}{4\pi R^3}{]}^2\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_e}\right)\left(\frac{P_e}{P_c}\right)\left[\frac{1}{{\tilde{\xi }}^2}\right]\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{{\sigma }^2}{G{\rho }_c}\right){\theta }^n{\xi }^4{x}^{2}d\xi$
	$=$	$\left[\frac{n}{(n+1)}\right]\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4}\right]\left(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R}{)}^4\right(\frac{{\rho }_c}{\overline{\rho }}{)}^2\left[\right(\frac{3}{4\pi }\left)\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_e}\left)\right(\frac{P_e}{P_c}\left)\right[\frac{1}{{\tilde{\xi }}^2}\left]\underset{0}{\overset{\tilde{\xi }}{\int }}\right(\frac{{\sigma }^2}{G{\rho }_c}){\theta }^n{\xi }^4{x}^{2}d\xi$
	$=$	$\left[\frac{n}{(n+1)}\right]\left(\frac{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{P_e}\right)\left(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R}{)}^4\right(-\frac{\tilde{\xi }}{3{\tilde{\theta }}^{\text{'}}}{)}^2\left[\right(\frac{3}{4\pi }\left)\frac{M}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^2\right[\frac{{\tilde{\theta }}^{n+1}}{{\tilde{\xi }}^2}\left]\underset{0}{\overset{\tilde{\xi }}{\int }}\right(\frac{{\sigma }^2}{G{\rho }_c}){\theta }^n{\xi }^4{x}^{2}d\xi$

Now, from an accompanying discussion, we know that, in equilibrium,

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

Hence,

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

This means that, in equilibrium,

LHS	$=$	$\left[\frac{n}{(n+1)}\right]\{(n+1)(4\pi ){\tilde{\xi }}^{-4}{\tilde{\theta }}_n^{-(n+1)}(-{\tilde{\xi }}^2{\tilde{\theta }}^{\prime }{)}^2\left\}\right(-\frac{\tilde{\xi }}{3{\tilde{\theta }}^{\text{'}}}{)}^2\left(\frac{3}{4\pi }{)}^2\right[\frac{{\tilde{\theta }}^{n+1}}{{\tilde{\xi }}^2}\left]\underset{0}{\overset{\tilde{\xi }}{\int }}\right(\frac{{\sigma }^2}{G{\rho }_c}){\theta }^n{\xi }^4{x}^{2}d\xi$
	$=$	$\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{n{\sigma }^2}{4\pi G{\rho }_c}\right){\theta }^n{\xi }^4{x}^{2}d\xi .$

In summary, then, we have,

$\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{n{\sigma }^2}{4\pi G{\rho }_c}\right){\theta }^n{\xi }^4{x}^{2}d\xi$ $=$ $3{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2+\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}{x}^2\left\{\right[\frac{\xi }{x}\cdot \frac{dx}{d\xi }{]}^2+(n-3)[\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi }]\}d\xi .$

Perhaps this looks better if the terms are rearranged to give,

$3{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2$

$=$

$\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}{x}^2\left\{\right(\frac{n{\sigma }^2}{4\pi G{\rho }_c})\frac{{\xi }^2}{\theta }-[(\frac{d\ln x}{d\ln \xi }{)}^2+(n-3)(\frac{d\ln \theta }{d\ln \xi }\left)\right]\}d\xi .$

Plug in Known Marginally Unstable Solution

As has been summarized in an accompanying discussion, we have found that, for marginally unstable pressure-truncated polytropic configurations, the eigenvector associated with the fundamental mode of radial oscillation is prescribed analytically by the following eigenfrequency-eigenfunction pair:

${\sigma }_c^2=0$

and

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

This means that,

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

Hence, also,

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

Rather, let's try:

${\xi }^2{x}^2\left[\right(\frac{d\ln x}{d\ln \xi }{)}^2+(n-3)\left(\frac{d\ln \theta }{d\ln \xi }\right)]$	$=$	$x^2{\xi }^2(\frac{\xi }{x}\cdot \frac{dx}{d\xi }{)}^2+(n-3)x^2{\xi }^2(\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi })$
	$=$	${\xi }^4\{\frac{dx}{d\xi }{\}}^2+(n-3)[\frac{{\xi }^3{\theta }^{\text{'}}}{\theta }]x^2$
	$=$	${\xi }^4\left\{\frac{3(n-1)}{2n}\right(\frac{3-n}{n-1}\left)\right[\frac{1}{\xi }+\frac{3{\theta }^{\text{'}}}{{\xi }^2{\theta }^n}+\frac{n({\theta }^{\text{'}}{)}^2}{\xi {\theta }^{n+1}}]{\}}^2+(n-3)[\frac{{\xi }^3{\theta }^{\text{'}}}{\theta }\left]\right\{\frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}\left)\right]{\}}^2$
	$=$	${\xi }^2(n-3)\left[\frac{3}{2n}{]}^2\right\{(n-3)[1+\frac{3{\theta }^{\text{'}}}{\xi {\theta }^n}+\frac{n({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}{]}^2+\xi (\frac{{\theta }^{\text{'}}}{\theta }\left)\right[(n-1)+(n-3)\left(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}\right){]}^2\}$

Hence, after setting ${\sigma }^2=0$, the above rearranged integral relation becomes,

$-\frac{2^2{n}^2}{3(n-3)}\left[{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2\right]$

$=$

$\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}\{(n-3)[1+\frac{3{\theta }^{\text{'}}}{\xi {\theta }^n}+\frac{n({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}{]}^2+\xi \left(\frac{{\theta }^{\text{'}}}{\theta }\right)[(n-1)+(n-3)(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}\left){]}^2\right\}d\xi$

Let's check to see whether the terms in this last expression balance out when we plug in the functions that are appropriate for the marginally unstable, n = 5 configuration, namely, ${\theta }_5=(\frac{3+{\xi }^2}{3}{)}^{-1/2}$,      and     $\frac{d{\theta }_5}{d\xi }=-\frac{\xi }{3}(\frac{3+{\xi }^2}{3}{)}^{-3/2}$. RHS Term 1 $=$ $(n-3)\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2{\theta }^{n+1}[1+\frac{3{\theta }^{\text{'}}}{\xi {\theta }^n}+\frac{n({\theta }^{\text{'}}{)}^2}{{\theta }^{n+1}}{]}^{2}d\xi$   $=$ $2\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^2\left[\right(\frac{3+{\xi }^2}{3}{)}^{-1/2}{]}^6\{1-[\xi \left(\frac{3+{\xi }^2}{3}{)}^{-3/2}\right]\frac{1}{\xi }(\frac{3+{\xi }^2}{3}{)}^{5/2}+5[\frac{{\xi }^2}{3^2}\left(\frac{3+{\xi }^2}{3}{)}^{-3}\right]\left[\right(\frac{3+{\xi }^2}{3}{)}^3]{\}}^{2}d\xi$   $=$ $2\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{3^3{\xi }^2}{(3+{\xi }^2{)}^3}\{1-(\frac{3+{\xi }^2}{3})+\frac{5{\xi }^2}{3^2}{\}}^{2}d\xi$   $=$ $\frac{2^3}{3}\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{{\xi }^{6}d\xi }{(3+{\xi }^2{)}^3}$   $=$ $\frac{2^3}{3}[\frac{27\xi }{8(3+{\xi }^2)}-\frac{9\xi }{4(3+{\xi }^2{)}^2}+\xi -(\frac{3^{3/2}\cdot 5}{2^3}\left){\tan }^{-1}\right(\frac{\xi }{3^{1/2}}){]}_0^3=\frac{2^3}{3}[\frac{3^5}{2^6}-\frac{3^{1/2}\cdot 5\pi }{2^3}].$ RHS Term 2 $=$ $\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^3{\theta }^n{\theta }^{\text{'}}[(n-1)+(n-3)(\frac{{\theta }^{\text{'}}}{\xi {\theta }^n}){]}^{2}d\xi$   $=$ $-\underset{0}{\overset{\tilde{\xi }}{\int }}{\xi }^3\left(\frac{3+{\xi }^2}{3}{)}^{-5/2}\frac{\xi }{3}\right(\frac{3+{\xi }^2}{3}{)}^{-3/2}\{4-2\frac{1}{3}(\frac{3+{\xi }^2}{3}{)}^{-3/2}(\frac{3+{\xi }^2}{3}{)}^{5/2}{\}}^{2}d\xi$   $=$ $-\frac{1}{3}\underset{0}{\overset{\tilde{\xi }}{\int }}\left(\frac{3\xi }{3+{\xi }^2}{)}^4\right\{4-\frac{2}{3}\left(\frac{3+{\xi }^2}{3}\right){\}}^{2}d\xi$   $=$ $-\frac{2^3}{3}\underset{0}{\overset{\tilde{\xi }}{\int }}\frac{1}{2}\left(\frac{\xi }{3+{\xi }^2}{)}^4\right\{15-{\xi }^2{\}}^{2}d\xi$   $=$ $-\frac{2^3}{3}[\frac{123\xi }{8(3+{\xi }^2)}-\frac{243\xi }{4(3+{\xi }^2{)}^2}+\frac{162\xi }{2(3+{\xi }^2{)}^3}+\frac{\xi }{2}-(\frac{3^{3/2}\cdot 5}{2^3}\left){\tan }^{-1}\right(\frac{\xi }{3^{1/2}}){]}_0^3$   $=$ $\frac{2^3}{3}\cdot \frac{5}{2^5}[2^2\cdot 3^{1/2}\pi -3^3]=-\frac{2^3}{3}[\frac{3^3\cdot 5}{2^5}-\frac{3^{1/2}\cdot 5\pi }{2^3}].$ $\Rightarrow$ RHS Total $=$ $\frac{2^3}{3}[\frac{3^5}{2^6}-\frac{3^3\cdot 5}{2^5}]=\frac{3^2}{2^3}[3^2-2\cdot 5]=-\frac{3^2}{2^3}.$ LHS $=$ $-\frac{2^2{n}^2}{3(n-3)}\left[{\tilde{\xi }}^3{\tilde{\theta }}^{n+1}{x}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}\mathrm{a}\mathrm{c}\mathrm{e}}^2\right]$   $=$ $-\frac{2^2{5}^2}{2\cdot 3}\left[3^3\right(\frac{3}{3+3^2}{)}^3\frac{2^2}{5^2}]=-\frac{2^3}{3}[\left(\frac{3}{2^2}{)}^3\right]=-\frac{3^2}{2^3}.$ Hence, the LHS = RHS.   Hooray!

See Also

• Derivations that support this chapter's discussion of the Ledoux Variational Principle
• D. Lynden-Bell & J. P. Ostriker (1967), MNRAS, 136, 293: On the stability of differentially rotating bodies

A variational principle of great power is derived. It is naturally adapted for computers, and may be used to determine the stability of any fluid flow including those in differentially-rotating, self-gravitating stars and galaxies. The method also provides a powerful theoretical tool for studying general properties of eigenfunctions, and the relationships between secular and ordinary stability. In particular we prove the anti-sprial theorem indicating that no stable (or steady( mode can have a spiral structure.

• B. F. Schutz, Jr. (1972), ApJSuppl., 24, 319: Linear Pulsations and Stability of Differentially Rotating Stellar Models. I. Newtonian Analysis

A systematic method is presented for deriving the Lagrangian governing the evolution of small perturbations of arbitrary flows of a self-gravitating perfect fluid. The method is applied to a differentially rotating stellar model; the result is a Lagrangian equivalent to that of D. Lynden-Bell & J. P. Ostriker (1967). A sufficient condition for stability of rotating stars, derived from this Lagrangian, is simplified greatly by using as trial functions not the three components of the Lagrangian displacement vector, but three scalar functions … This change of variables saves one from integrating twice over the star to find the effect of the perturbed gravitational field. … we examine the special cases of (i) axially symmetric perturbations of a rotating star (as treated by S. Chandrasekhar & N. R. Lebovitz 1968); and (ii) perturbations of a nonrotating star (as treated by Chandrasekhar and Lebovitz 1964). We find that the stability criteria for those cases can also be simplified …