SSC/Stability/n1PolytropeLAWE/Pt4

Radial Oscillations of n = 1 Polytropic Spheres (Pt 4)

Part I:   Search for Analytic Solutions

Part II:  New Ideas

Part III:  What About Bipolytropes?

Part IV:  Most General Structural Solution

Preamble Regarding Chatterji

As far as we have been able to ascertain, the first technical examination of radial oscillation modes in $n=1$ polytropes was performed — using numerical techniques — in 1951 by L. D. Chatterji; at the time, he was in the Mathematics Department of Allahabad University. His two papers on this topic were published in, what is now referred to as, the Proceedings of the Indian National Science Academy (PINSA). The citations that immediately follow this opening paragraph provide inks to both of these papers by Chatterji, but the links may be insecure. Apparently Springer is archiving recent PINSA volumes, but their holdings do not date back as early as 1951.

• 📚 L. D. Chatterji (1951, PINSA, Vol. 17, No. 6, pp. 467 - 470), Radial Oscillations of a Gaseous Star of Polytropic Index I
• 📚 L. D. Chatterji (1952, PINSA, Vol. 18, No. 3, pp. 187 - 191), Anharmonic Pulsations of a Polytropic Model of Index Unity

A detailed review of Chatterji51 is provided in an accompanying discussion.

Equilibrium Structure

When $n=1$, the relevant Lane-Emden equation is,

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

and we find that the solution is, quite generally,

$\theta$

=

$A\left[\frac{\sin \xi }{\xi }\right]-B\left[\frac{\cos \xi }{\xi }\right]=\frac{1}{\xi }\{A\sin \xi -B\cos \xi \}$

in which case,

$\frac{d\theta }{d\xi }$	$=$	$\frac{d}{d\xi }\left\{A\right[\frac{\sin \xi }{\xi }]-B[\frac{\cos \xi }{\xi }\left]\right\}$
	$=$	$A[-\frac{\sin \xi }{{\xi }^2}+\frac{\cos \xi }{\xi }]+B[\frac{\cos \xi }{{\xi }^2}+\frac{\sin \xi }{\xi }]$
	$=$	$\frac{1}{{\xi }^2}[-A\sin \xi +B\cos \xi ]+\frac{1}{\xi }[A\cos \xi +B\sin \xi ],$

and,

$Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi }$	$=$	$-\frac{\xi }{\theta }\cdot \left\{\frac{1}{{\xi }^2}\right[-A\sin \xi +B\cos \xi ]+\frac{1}{\xi }[A\cos \xi +B\sin \xi \left]\right\}$
	$=$	$\frac{1}{\xi \theta }\cdot \left\{\right[A\sin \xi -B\cos \xi ]+\xi [-A\cos \xi -B\sin \xi \left]\right\}$
	$=$	$\{1-\xi [\frac{A\cos \xi +B\sin \xi }{A\sin \xi -B\cos \xi }\left]\right\}.$

If we set $\beta \equiv {\tan }^{-1}(B/A)$, we can rewrite the expression for $\theta$ as,

$\theta$	=	$\frac{A}{\xi }\{\sin \xi -\frac{B}{A}\cos \xi \}$
	=	$\frac{A}{\xi }\{\sin \xi -\tan \beta \cos \xi \}$
	=	$\frac{A}{\xi \cos \beta }\{\sin \xi \cos \beta -\cos \xi \sin \beta \}$
	=	$\frac{A}{\cos \beta }\cdot \frac{\sin (\xi -\beta )}{\xi },$
Beech88, §3, p. 221, Eq. (6) EFC98, §2, p. 831, Eq. (2)

and the expression for $Q$ as,

$Q(\xi )$	$=$	$\{1-\xi [\frac{\cos \xi \cos \beta +\sin \xi \sin \beta }{\sin \xi \cos \beta -\cos \xi \sin \beta }\left]\right\}$
	$=$	$\{1-\xi [\frac{\cos (\xi -\beta )}{\sin (\xi -\beta )}\left]\right\}$
	$=$	$[1-\xi \cot (\xi -\beta )].$

SUMMARY of EQUILIBRIUM STRUCTURE and switching notation from $\theta (\xi )$ to $\varphi (\eta )$ When $n=1$, the relevant Lane-Emden equation is, $\frac{1}{{\eta }^2}\frac{d}{d\eta }\left({\eta }^2\frac{d\varphi }{d\eta }\right)=-\varphi$ . Its solution, quite generally, is $\varphi$ = $A\left[\frac{\sin \eta }{\eta }\right]-B\left[\frac{\cos \eta }{\eta }\right],$ where $A$ and $B$ are scalar constants, in which case, $Q(\eta )\equiv -\frac{d\ln \varphi }{d\ln \eta }$ $=$ $\{1-\eta [\frac{A\cos \eta +B\sin \eta }{A\sin \eta -B\cos \eta }\left]\right\}.$ Alternatively, drawing from Eq. (6) of Beech88, this solution can be written in the form, $\varphi$ = ${\alpha }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}}\cdot \frac{\sin (\eta -{\beta }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}})}{\eta },$ in which case, $Q(\eta )$ $=$ $[1-\eta \cot (\eta -{\beta }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}})],$ where, in terms of the coefficients $A$ and $B$, ${\beta }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}}$ $\equiv$ ${\tan }^{-1}(B/A)$       and       ${\alpha }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}}$ $\equiv$ $\frac{A}{\cos {\beta }_{\mathrm{B}\mathrm{e}\mathrm{e}\mathrm{c}\mathrm{h}}}=A[1+(B/A{)}^2{]}^{1/2}.$

Establish Relevant (n=1) LAWE

From a related discussion — or a broader overview of Instability Onset — we find the

Polytropic LAWE (linear adiabatic wave equation)

	$0=\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$
where:    $Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi },$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c},$     and,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$

Furthermore — see, for example, here,

Exact Solution to the Polytropic LAWE

$x_P\equiv \frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right],$

in which case for $n=1$,

$x_P=-3\left(\frac{1}{\xi \theta }\right)\frac{d\theta }{d\xi }=-3\left(\frac{1}{{\xi }^2}\right)\frac{d\ln \theta }{d\ln \xi }=\frac{3}{{\xi }^2}Q.$

Isolated Sphere

For an isolated n = 1 $({\gamma }_g=2,\alpha =1)$ polytrope, we know that,

$\theta$

=

$\frac{\sin \xi }{\xi }$

$\Rightarrow Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi }$

=

$[1-\xi \cot \xi ].$

Hence, the relevant LAWE is,

$0$

=

$\frac{d^{2}x}{d{\xi }^2}+[4-2(1-\xi \cot \xi )]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+2\left[\right(\frac{{\sigma }_c^2}{12})\frac{{\xi }^3}{\sin \xi }-(1-\xi \cot \xi )]\frac{x}{{\xi }^2}$

LAWE for n = 1 Polytrope

$0$

=

$\frac{d^{2}x}{d{\xi }^2}+\frac{2}{\xi }[1+\xi \cot \xi ]\frac{dx}{d\xi }+\frac{1}{2}\left[\right(\frac{{\sigma }_c^2}{3})\frac{\xi }{\sin \xi }-\frac{4}{{\xi }^2}(1-\xi \cot \xi \left)\right]x$

Surface boundary condition:

$-\frac{d\ln x}{d\ln \xi }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$\left(\frac{3-n}{n+1}\right)+\frac{n{\sigma }_c^2}{6(n+1)}[\frac{\xi }{{\theta }^{\prime }}{]}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$
$\Rightarrow -\frac{d\ln x}{d\ln \xi }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$1+\frac{{\sigma }_c^2}{12}[\frac{{\xi }^3}{(\xi \cos \xi -\sin \xi )}{]}_{\xi =\pi }=1-\frac{{\pi }^2{\sigma }_c^2}{12}$

Spherical Shell

In the context of a spherically symmetric n = 1 $({\gamma }_g=2,\alpha =1)$ shell (envelope) outside of a spherically symmetric bipolytropic core, we should adopt the more general Lane-Emden structural solution,

$\theta$

=

$A\left[\frac{\sin \xi }{\xi }\right]-B\left[\frac{\cos \xi }{\xi }\right]$

$\Rightarrow Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi }$

=

$[1-\xi \cot (\xi -\beta )]=\frac{{\xi }^2{x}_P}{3}.$

Reminder: the expression for $x_P$ is, $x_P=\frac{3}{{\xi }^2}[1-\xi \cot (\xi -\beta )]$. Playing around a bit, we find that, $\frac{{\xi }^2}{3}\cdot x_P$ $=$ $1-\xi \cdot \frac{\cos (\xi -\beta )}{\sin (\xi -\beta )}$   $=$ $1-\xi \left[\frac{\cos (\xi -\beta )}{\sin (\xi -\beta )}\right]\cdot \frac{(\xi -\beta )}{(\xi -\beta )}$   $=$ $1-\xi \left[\frac{\cos (\xi -\beta )}{(\xi -\beta )}\right]\left[\frac{(\xi -\beta )}{\sin (\xi -\beta )}\right]$

As a result, the governing LAWE becomes,

$0$	$=$	$\frac{d^2{x}_P}{d{\xi }^2}+[4-2Q]\frac{1}{\xi }\cdot \frac{d{x}_P}{d\xi }+2[{\xcancel{\left(\frac{{\sigma }_c^2}{12}\right)}}^0\frac{{\xi }^2}{\theta }-Q]\frac{x_P}{{\xi }^2}$
	$=$	$\frac{d^2{x}_P}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d{x}_P}{d\xi }-\left[2Q\right]\frac{1}{\xi }\cdot \frac{d{x}_P}{d\xi }-\left[2Q\right]\frac{x_P}{{\xi }^2}$
	$=$	$\frac{d^2{x}_P}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d{x}_P}{d\xi }-2Q\{\frac{1}{\xi }\cdot \frac{d{x}_P}{d\xi }+\frac{x_P}{{\xi }^2}\}$
	$=$	$\frac{d^2{x}_P}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d{x}_P}{d\xi }-\frac{2{\xi }^2{x}_P}{3}\{\frac{1}{\xi }\cdot \frac{d{x}_P}{d\xi }+\frac{x_P}{{\xi }^2}\}$
	$=$	$\frac{d^2{x}_P}{d{\xi }^2}+[\frac{4}{\xi }-\frac{2\xi x_P}{3}]\cdot \frac{d{x}_P}{d\xi }-\frac{2x_P^2}{3}.$

Let's plug in the expression for $x_P$, namely, $x_P=3[1-\xi \cot (\xi -\beta )]/{\xi }^2$. We have, first of all,

$x_p^2$	$=$	$\frac{3^2}{{\xi }^4}[1-\xi \cot (\xi -\beta ){]}^2$
	$=$	$\frac{3^2}{{\xi }^4}[1-2\xi \cot (\xi -\beta )+{\xi }^2{\cot }^{}(\xi -\beta )];$
$\frac{d{x}_p}{d\xi }$	$=$	$\frac{d}{d\xi }\left\{\frac{3}{{\xi }^2}\right[1-\xi \cot (\xi -\beta )\left]\right\}$
	$=$	$[1-\xi \cot (\xi -\beta )]\frac{d}{d\xi }\left\{\frac{3}{{\xi }^2}\right\}-\frac{3}{\xi }\frac{d}{d\xi }[\cot (\xi -\beta )]-\frac{3}{{\xi }^2}[\cot (\xi -\beta )]$
	$=$	$-\frac{6}{{\xi }^3}[1-\xi \cot (\xi -\beta )]+\frac{3}{\xi }[1+{\cot }^{}(\xi -\beta )]-\frac{3}{{\xi }^2}[\cot (\xi -\beta )]$
	$=$	$-\frac{6}{{\xi }^3}+\frac{3}{{\xi }^2}[\cot (\xi -\beta )]+\frac{3}{\xi }+\frac{3}{\xi }[{\cot }^{}(\xi -\beta )].$

Note for later use that, $\frac{x_P^2}{r^2}(\frac{d\ln x_P}{d\ln r}{)}^2=(\frac{d{x}_P}{dr}{)}^2$ $=$ $\{-\frac{6}{{\xi }^3}+\frac{3}{{\xi }^2}[\cot (\xi -\beta )]+\frac{3}{\xi }+\frac{3}{\xi }[{\cot }^{}(\xi -\beta )]{\}}^2$

Recognize that we have used the trigonometric relations,

$\frac{d}{d\xi }[\cot (u)]$

$=$

$-\frac{1}{{\sin }^2(u)}\frac{du}{d\xi }=-[1+{\cot }^2(u)]\frac{du}{d\xi }.$

And,

$\frac{d^2{x}_p}{d{\xi }^2}$	$=$	$\frac{d}{d\xi }\{-\frac{6}{{\xi }^3}+\frac{3}{\xi }+\frac{3}{{\xi }^2}[\cot (\xi -\beta )]+\frac{3}{\xi }[{\cot }^{}(\xi -\beta )\left]\right\}$
	$=$	$\frac{18}{{\xi }^4}-\frac{3}{{\xi }^2}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{3}{{\xi }^2}[1+{\cot }^{}(\xi -\beta )]-\frac{3}{{\xi }^2}[{\cot }^{}(\xi -\beta )]-\frac{6}{\xi }[\cot (\xi -\beta )][1+{\cot }^{}(\xi -\beta )]$
	$=$	$\frac{18}{{\xi }^4}-\frac{6}{{\xi }^2}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }[\cot (\xi -\beta )]-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta ).$

Hence,

$\frac{d^2{x}_P}{d{\xi }^2}+[\frac{4}{\xi }-\frac{2\xi x_P}{3}]\cdot \frac{d{x}_P}{d\xi }-\frac{2x_P^2}{3}$	$=$	$\{\frac{18}{{\xi }^4}-\frac{6}{{\xi }^2}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }[\cot (\xi -\beta )]-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta )\}$
		$+[\frac{4}{\xi }-\frac{2\xi x_P}{3}]\cdot \{-\frac{6}{{\xi }^3}+\frac{3}{{\xi }^2}[\cot (\xi -\beta )]+\frac{3}{\xi }+\frac{3}{\xi }[{\cot }^{}(\xi -\beta )\left]\right\}$
		$-\left\{\frac{6}{{\xi }^4}\right[1-2\xi \cot (\xi -\beta )+{\xi }^2{\cot }^{}(\xi -\beta )\left]\right\}$
	$=$	$\frac{18}{{\xi }^4}-\frac{6}{{\xi }^2}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{\xi }[\cot (\xi -\beta )]-\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta )$
		$+\frac{4}{\xi }\{-\frac{6}{{\xi }^3}+\frac{3}{{\xi }^2}[\cot (\xi -\beta )]+\frac{3}{\xi }+\frac{3}{\xi }[{\cot }^{}(\xi -\beta )\left]\right\}$
		$+\left[\frac{2\xi x_P}{3}\right]\cdot \{\frac{6}{{\xi }^3}-\frac{3}{{\xi }^2}[\cot (\xi -\beta )]-\frac{3}{\xi }-\frac{3}{\xi }[{\cot }^{}(\xi -\beta )\left]\right\}$
		$+\frac{6}{{\xi }^4}[-1+2\xi \cot (\xi -\beta )-{\xi }^2{\cot }^{}(\xi -\beta )]$
	$=$	$\frac{18}{{\xi }^4}-\frac{6}{{\xi }^2}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{\xi }[\cot (\xi -\beta )]-\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta )$
		$-\frac{24}{{\xi }^4}+\frac{12}{{\xi }^3}[\cot (\xi -\beta )]+\frac{12}{{\xi }^2}+\frac{12}{{\xi }^2}[{\cot }^{}(\xi -\beta )]$
		$+\frac{2}{\xi }[1-\xi \cot (\xi -\beta )]\cdot \{\frac{6}{{\xi }^3}-\frac{3}{{\xi }^2}[\cot (\xi -\beta )]-\frac{3}{\xi }-\frac{3}{\xi }[{\cot }^{}(\xi -\beta )\left]\right\}$
		$+\frac{6}{{\xi }^4}[-1+2\xi \cot (\xi -\beta )-{\xi }^2{\cot }^{}(\xi -\beta )]$
	$=$	$-\frac{6}{{\xi }^4}+\frac{6}{{\xi }^2}+\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{\xi }[\cot (\xi -\beta )]+\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta )$
		$-\frac{12}{{\xi }^3}\cdot \cot (\xi -\beta )+\frac{6}{\xi }\cdot \cot (\xi -\beta )+\frac{6}{{\xi }^2}[{\cot }^{}(\xi -\beta )]+\frac{6}{\xi }[{\cot }^{}(\xi -\beta )]$
		$+\frac{12}{{\xi }^4}-\frac{6}{{\xi }^3}[\cot (\xi -\beta )]-\frac{6}{{\xi }^2}-\frac{6}{{\xi }^2}[{\cot }^{}(\xi -\beta )]$
		$-\frac{6}{{\xi }^4}+\frac{12}{{\xi }^3}\cdot \cot (\xi -\beta )-\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )$
	$=$	$-\frac{6}{\xi }[\cot (\xi -\beta )]+\frac{6}{{\xi }^2}\cdot {\cot }^{}(\xi -\beta )-\frac{6}{\xi }\cdot {\cot }^{}(\xi -\beta )$
		$+\frac{6}{\xi }\cdot \cot (\xi -\beta )+\frac{6}{{\xi }^2}[{\cot }^{}(\xi -\beta )]+\frac{6}{\xi }[{\cot }^{}(\xi -\beta )]$
		$-\frac{12}{{\xi }^2}[{\cot }^2(\xi -\beta )]$
	$=$	$0.$

---

Debugging LaTeX layout:

		$-\frac{12}{{\xi }^2}[{\cot }^2(\xi -\beta )]$
		$-\frac{12}{{\xi }^2}[{\cot }^3(\xi -\beta )]$
		$+[{\cot }^1(\xi -\beta )]+\frac{6}{{\xi }^2}[{\cot }^2(\xi -\beta {)}_m]+\frac{6}{\xi }[{\cot }^3(\xi -\beta )]$

Hydrostatic Balance and Virial Equilibrium

General Expression for Virial

Here we draw heavily from our accompanying "style sheet" synopsis of spherically symmetric configurations.

First, we pull the equation for

Hydrostatic Balance

$\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$

from subsection ① of the synopsis; then, guided by subsection ②, we multiply both sides through by $rdV=4\pi r^{3}dr$ and integrate over the volume. This gives,

$0$	$=$	$-\underset{0}{\overset{R}{\int }}r\left(\frac{dP}{dr}\right)dV-\underset{0}{\overset{R}{\int }}r\left(\frac{G{M}_r\rho }{r^2}\right)dV$
	$=$	$-\underset{0}{\overset{R}{\int }}4\pi r^3\left(\frac{dP}{dr}\right)dr-\underset{0}{\overset{R}{\int }}\left(\frac{G{M}_r}{r}\right)d{M}_r,$

where we have used the relations,

$dV=4\pi r^{2}dr$

and,

$d{M}_r=\rho dV\Rightarrow M_r=4\pi \underset{0}{\overset{r}{\int }}\rho r^{2}dr.$

Now, given that,

$\frac{d}{dr}\left(4\pi r^{3}P\right)$	$=$	$4\pi r^3\left(\frac{dP}{dr}\right)+12\pi P{r}^2$
$\Rightarrow -4\pi r^3\left(\frac{dP}{dr}\right)$	$=$	$12\pi r^{2}P-\frac{d}{dr}\left(4\pi r^{3}P\right),$

we can rewrite the integral expression in the form,

$0$	$=$	$\underset{0}{\overset{R}{\int }}[12\pi r^{2}P-\frac{d}{dr}(4\pi r^{3}P\left)\right]dr-\underset{0}{\overset{R}{\int }}\left(\frac{G{M}_r}{r}\right)d{M}_r$
	$=$	$\underset{0}{\overset{R}{\int }}3\left[4\pi r^{2}P\right]dr-\underset{0}{\overset{R}{\int }}\left(\frac{G{M}_r}{r}\right)d{M}_r-\underset{0}{\overset{R}{\int }}[d(3PV)]$
	$=$	$3(\gamma -1)U_{\mathrm{i}\mathrm{n}\mathrm{t}}+W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}-[3PV{]}_0^R,$

where,

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

$=$

$-\underset{0}{\overset{R}{\int }}\left(\frac{G{M}_r}{r}\right)d{M}_r$

and,

$U_{\mathrm{i}\mathrm{n}\mathrm{t}}$

$=$

$\frac{1}{(\gamma -1)}\underset{0}{\overset{R}{\int }}4\pi r^{2}Pdr.$

Note as well that $U_{\mathrm{i}\mathrm{n}\mathrm{t}}=2S_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}/[3(\gamma -1)]$.

Calculate Relevant Energy Expressions

Adopting the energy normalization shown here01 along with the other variable normalizations defined here02, we have …

Thermal Energy

$S^{*}\equiv \frac{S}{[K_c^5/G^3{]}^{1/2}}=\frac{3}{2}(\gamma -1)\cdot \frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$\frac{3}{2[K_c^5/G^3{]}^{1/2}}\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}4\pi r^{2}Pdr$
	$=$	$\frac{3}{2}\left\{K_c^{-5/2}{G}^{3/2}\right\}\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}\{K_c^{3/2}{G}^{-3/2}{\rho }_0^{-6/5}\cdot K_c{\rho }_0^{6/5}\}4\pi (r^{*}{)}^2{P}^{*}d{r}^{*}$
	$=$	$\frac{3}{2}\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}4\pi (r^{*}{)}^2{P}^{*}d{r}^{*}.$

Plugging in the derived radial profiles for $r^{*}$ and $P^{*}$, we have,

$S^{*}$	$=$	$6\pi \left(\frac{3}{2\pi }{)}^{3/2}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi$
	$=$	$6\pi (\frac{3}{2\pi }{)}^{3/2}\frac{3^{3/2}}{2^3}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}8\cdot \frac{{\xi }^2}{3}(1+\frac{{\xi }^2}{3}{)}^{-3}\frac{d\xi }{3^{1/2}}$
	$=$	$[(2^2\cdot 3^2{\pi }^2)(\frac{3^3}{2^3{\pi }^3}\left)\right(\frac{3^3}{2^6}){]}^{1/2}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}8\cdot \frac{{\xi }^2}{3}(1+\frac{{\xi }^2}{3}{)}^{-3}\frac{d\xi }{3^{1/2}}$
	$=$	$(\frac{3^8}{2^7\pi }{)}^{1/2}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}8\cdot \frac{{\xi }^2}{3}(1+\frac{{\xi }^2}{3}{)}^{-3}\frac{d\xi }{3^{1/2}}.$

After making the substitution, $x\equiv \xi /\sqrt{3}$, this expression matches the expression for $S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}$ obtained separately.

For later use, we note that, $d{S}^{*}$ $=$ $\frac{3}{2}\left[\frac{4\pi r^{2}Pdr}{[K_c^5/G^3{]}^{1/2}}\right]=6\pi \left(\frac{3}{2\pi }{)}^{3/2}{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi .$

Gravitational Potential Energy

$W^{*}=\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{[K_c^5/G^3]}$	$=$	$[K_c^{-5/2}{G}^{3/2}]\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}(-\frac{G{M}_r}{r})d{M}_r$
	$=$	$-[K_c^{-5/2}{G}^{3/2}]\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}4\pi G{M}_r\rho rdr$
	$=$	$-[K_c^{-5/2}{G}^{3/2}]\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}4\pi G\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]M_r^{*}\left[{\rho }^{*}{\rho }_0\right]\left[K_c{G}^{-1}{\rho }_0^{-4/5}\right]r^{*}d{r}^{*}$
	$=$	$-\underset{r_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{r_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}4\pi M_r^{*}{\rho }^{*}{r}^{*}d{r}^{*}.$

Plugging in the derived radial profiles for $r^{*}$, ${\rho }^{*}$ and $M_r^{*}$, we have,

$W^{*}$	$=$	$-4\pi \underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}](1+\frac{{\xi }^2}{3}{)}^{-5/2}(\frac{3}{2\pi })\xi d\xi$
	$=$	$-4\pi \left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right(\frac{3}{2\pi }\left)\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}\right[{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}](1+\frac{{\xi }^2}{3}{)}^{-5/2}\xi d\xi$
	$=$	$-\left(2^4{\pi }^2{)}^{1/2}\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left(\frac{3^2}{2^2{\pi }^2}{)}^{1/2}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}\right[{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi$
	$=$	$-[\frac{2^3\cdot 3^3}{\pi }{]}^{1/2}\underset{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{n}\mathrm{e}\mathrm{r}}}{\overset{{\xi }_{\mathrm{o}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{r}}}{\int }}{3}^{5/2}\cdot [{\chi }^4(1+{\chi }^2{)}^{-4}]d\chi .$

This expression matches the expression for $W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}$ obtained separately.

For later use, we note that, $d{W}^{*}$ $=$ $[K_c^{-5/2}{G}^{3/2}](-\frac{G{M}_r}{r})4\pi r^2\rho dr=-\left[\frac{2^3\cdot 3^3}{\pi }{]}^{1/2}\right[{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi .$

Stability Analysis

Here, as well, we draw heavily from our accompanying "style sheet" synopsis of spherically symmetric configurations.

This time, we pull the

LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation

$0$

$=$

$\frac{d}{dr}\left[r^4\gamma P\frac{dx}{dr}\right]+[{\omega }^2\rho r^4+(3\gamma -4)r^3\frac{dP}{dr}]x,$

from subsection ④ of the synopsis; then, guided by subsection ⑤, we multiply both sides through by $4\pi xdr$ to obtain,

$-[4\pi x^2({\omega }^2\rho r^4)]dr$

$=$

$4\pi x\left\{\frac{d}{dr}\right[r^4\gamma P\left(\frac{dx}{dr}\right)\left]\right\}dr+\left\{\right[(3\gamma -4)4\pi x^2{r}^3\left(\frac{dP}{dr}\right)\left]\right\}dr.$

Now, given that,

$\frac{d}{dr}\left[4\pi x\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]$	$=$	$4\pi x\frac{d}{dr}\left[\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]+\left[\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]\frac{d}{dr}\left[4\pi x\right]$
$\Rightarrow 4\pi x\frac{d}{dr}\left[\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]$	$=$	$\frac{d}{dr}\left[4\pi x\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]-\left[4\pi \gamma r^{4}P\right(\frac{dx}{dr}{)}^2],$

we can rewrite this last expression in the form,

$-[4\pi x^2({\omega }^2\rho r^4)]dr$	$=$	$\frac{d}{dr}\left[4\pi x\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]dr-\left[4\pi \gamma r^{4}P\right(\frac{dx}{dr}{)}^2]dr+\{[(3\gamma -4)4\pi x^2{r}^3(\frac{dP}{dr}\left)\right]\}dr$
$\Rightarrow [4\pi x^2({\omega }^2\rho r^4)]dr$	$=$	$-d\left[4\pi x\gamma r^{4}P\right(\frac{dx}{dr}\left)\right]+\left[4\pi \gamma r^{4}P\right(\frac{dx}{dr}{)}^2]dr-[(3\gamma -4)4\pi x^2{r}^3(-\frac{G{M}_r\rho }{r^2})]dr$

Note that, in order to obtain the last term on the RHS of this expression, we used the hydrostatic balance relation to replace the pressure gradient in terms of the gravitational potential. Finally, integrating over the volume of the configuration gives,

$\underset{0}{\overset{R}{\int }}[4\pi x^2({\omega }^2\rho r^4)]dr$

$=$

$\underset{0}{\overset{R}{\int }}\left[4\pi \gamma r^{4}P\right(\frac{dx}{dr}{)}^2]dr-\underset{0}{\overset{R}{\int }}[(3\gamma -4)x^2(-\frac{G{M}_r}{r})4\pi \rho r^2]dr-[4\pi x\gamma r^{4}P\left(\frac{dx}{dr}\right){]}_0^R,$

or,

$\underset{0}{\overset{R}{\int }}[4\pi x^2({\omega }^2\rho r^4)]dr$

$=$

$\underset{0}{\overset{R}{\int }}\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{\overbrace{\left[x^2\right(\frac{d\ln x}{d\ln r}{)}^{2}4\pi \gamma r^{2}P]dr}}-\underset{0}{\overset{R}{\int }}\underset{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{\underbrace{[(3\gamma -4)x^2(-\frac{G{M}_r}{r}\left)4\pi \rho r^2\right]dr}}+\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{3}}{\overbrace{\left[\gamma 4\pi x^2{r}^{2}P\right(-\frac{d\ln x}{d\ln r}){]}_0^R}}.$

TERM1

Given that (from above),

$d{S}^{*}$

$=$

$\frac{3}{2}\left[\frac{4\pi r^{2}Pdr}{[K_c^5/G^3{]}^{1/2}}\right]=6\pi \left(\frac{3}{2\pi }{)}^{3/2}{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi ,$

we can write,

$\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$\gamma x_P^2\left(\frac{d\ln x_P}{d\ln r}{)}^2\right[\frac{4\pi r^{2}Pdr}{[K_c^5/G^3{]}^{1/2}}]$
	$=$	$\gamma x_P^2\left(\frac{d\ln x_P}{d\ln r}{)}^2\right\{\left(\frac{2}{3}\right)6\pi \left(\frac{3}{2\pi }{)}^{3/2}{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi \},$

Furthermore, given that for a truncated $n=5$ configuration,

$x_p$	$=$	$1-\frac{{\xi }^2}{15}$
$\Rightarrow \frac{d{x}_p}{d\xi }$	$=$	$-\frac{2}{15}\xi$
$\Rightarrow \frac{d\ln x_p}{d\ln \xi }$	$=$	$-\frac{\xi }{x_P}\left[\frac{2}{15}\xi \right]=-\left[\frac{2}{15}{\xi }^2\right]\left[\frac{15}{15-{\xi }^2}\right]=-\left[\frac{2{\xi }^2}{15-{\xi }^2}\right]$
$\Rightarrow x_P^2(\frac{d\ln x_p}{d\ln \xi }{)}^2$	$=$	$\left[\frac{15-{\xi }^2}{15}{]}^2\right[\frac{2{\xi }^2}{15-{\xi }^2}{]}^2=[\frac{2{\xi }^2}{15}{]}^2,$

we have,

$\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$\gamma \left[\frac{2{\xi }^2}{15}{]}^2\right\{\left(\frac{2}{3}\right)6\pi \left(\frac{3}{2\pi }{)}^{3/2}{\xi }^2\right(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi \}$
	$=$	$\gamma \left[\frac{2^2}{3^2{5}^2}\right]\left(\frac{2}{3}\right)6\pi \left(\frac{3}{2\pi }{)}^{3/2}\right\{{\xi }^6(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi \}$
	$=$	$\gamma \left[\frac{2^4\pi }{3^2{5}^2}\right]\left[\frac{3^3}{2^3{\pi }^3}{]}^{1/2}\right\{{\xi }^6(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi \}$
	$=$	$\gamma \left[\frac{2^5}{3\cdot 5^4\pi }{]}^{1/2}\right\{{\xi }^6(1+\frac{{\xi }^2}{3}{)}^{-3}d\xi \}.$

Hence, after making the replacement $\chi \equiv \xi /\sqrt{3}\Rightarrow \xi =3^{1/2}\chi$, we find that,

$\underset{0}{\overset{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{\int }}\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$\gamma [\frac{2^5}{3\cdot 5^4\pi }{]}^{1/2}\cdot 3^{7/2}\underset{0}{\overset{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{\int }}\{{\chi }^6(1+{\chi }^2{)}^{-3}d\chi \}$
	$=$	$\gamma [\frac{2^5}{3\cdot 5^4\pi }{]}^{1/2}\cdot 3^{7/2}\cdot \frac{1}{8}\{\frac{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}(8{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^4+25{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2+15)}{(1+{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2{)}^2}-15{\tan }^{-1}({\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}})\}$

where, we have completed the integral with the assistance of the WolframAlpha online integrator:

Mathematica Integral

TERM2

Given that (from above),

$d{W}^{*}$

$=$

$[K_c^{-5/2}{G}^{3/2}](-\frac{G{M}_r}{r})4\pi r^2\rho dr=-\left[\frac{2^3\cdot 3^3}{\pi }{]}^{1/2}\right[{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi ,$

we can write,

$\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$[K_c^{-5/2}{G}^{3/2}][(3\gamma -4)x^2(-\frac{G{M}_r}{r}\left)4\pi \rho r^2\right]dr$
	$=$	$-(3\gamma -4)x_P^2\left\{\right[\frac{2^3\cdot 3^3}{\pi }{]}^{1/2}\left[{\xi }^4\right(1+\frac{{\xi }^2}{3}{)}^{-4}\left]d\xi \right\}.$

Furthermore, given that (as above) for a truncated $n=5$ configuration,

$x_p$

$=$

$\frac{15-{\xi }^2}{15},$

we have,

$\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$-(3\gamma -4)[\frac{2^3\cdot 3^3}{\pi }\cdot \frac{1}{3^4\cdot 5^4}{]}^{1/2}\{[(15-{\xi }^2{)}^2{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi \}$
	$=$	$-(3\gamma -4)\left[\frac{2^3}{3\cdot 5^4\pi }{]}^{1/2}\right\{[(15-{\xi }^2{)}^2{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi \}.$

Hence, after making the replacement $\chi \equiv \xi /\sqrt{3}\Rightarrow \xi =3^{1/2}\chi$, we find that,

$\underset{0}{\overset{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{\int }}\frac{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{[K_c^5/G^3{]}^{1/2}}$	$=$	$-(3\gamma -4)\left[\frac{2^3}{3\cdot 5^4\pi }{]}^{1/2}\underset{0}{\overset{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{\int }}\right\{[(15-{\xi }^2{)}^2{\xi }^4(1+\frac{{\xi }^2}{3}{)}^{-4}]d\xi \}$
	$=$	$-(3\gamma -4)[\frac{2^3}{3\cdot 5^4\pi }{]}^{1/2}\cdot 3^{9/2}\underset{0}{\overset{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{\int }}\{[(5-{\chi }^2{)}^2{\chi }^4(1+{\chi }^2{)}^{-4}]d\chi \}$
	$=$	$-(3\gamma -4)[\frac{2^3\cdot 3^8}{5^4\pi }{]}^{1/2}\cdot \frac{1}{4}\{\frac{{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}(4{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6+53{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^4+40{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2+15)}{(1+{\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2{)}^3}-15{\tan }^{-1}({\chi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}})\}$

where, we have completed the integral with the assistance of the WolframAlpha online integrator:

Mathematica Integral

See Also

• Equilibrium Model Parameter Profiles (Table of Parameters)
• Instability Onset Overview
• Index Less that 3
• Radial Oscillations of n=1 Polytropic Spheres
  • Search for Analytic Solutions
  • Most General Structural Solution
• Synopsis Style Sheet
• Free Energy of (n_c, n_e) = (5, 1) Bipolytrope
  • One Discussion
  • Another Discussion <== Very Useful Analytic Integrals