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 $A=\cos \beta$ and $B=\sin \beta$ — in which case, $\beta ={\tan }^{-1}(B/A)$ — we can rewrite this last expression as,

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

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

Trial01

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

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,

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

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,

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

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Debugging LaTeX layout:

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

See Also