Appendix/Ramblings/BiPolytrope51ContinueSearch

Continue Search for Marginally Unstable (5,1) Bipolytropes

This Ramblings Appendix chapter — see also, various trials — provides some detailed trial derivations in support of the accompanying, thorough discussion of this topic.

Key Differential Equation

In an accompanying discussion, we derived the so-called,

Linear Adiabatic Wave (or Radial Pulsation) Equation

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-(\frac{g_0{\rho }_0}{P_0}\left)\right]\frac{dx}{d{r}_0}+\left(\frac{{\rho }_0}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]x=0$

whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. After adopting an appropriate set of variable normalizations — as detailed here — this becomes,

$0$

$=$

$\frac{d^{2}x}{dr{*}^2}+\{4-(\frac{{\rho }^{*}}{P^{*}}\left)\frac{M_r^{*}}{(r^{*})}\right\}\frac{1}{r^{*}}\frac{dx}{dr*}+\left(\frac{{\rho }^{*}}{P^{*}}\right)\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-\frac{{\alpha }_{\mathrm{g}}{M}_r^{*}}{(r^{*}{)}^3}\}x,$

where, ${\alpha }_g\equiv (3-4/{\gamma }_g)$. Alternatively — see, for example, our introductory discussion — for polytropic configurations we may write,

$0$

$=$

$\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)(-\frac{d\ln \theta }{d\ln \xi }\left)\right]\frac{1}{\xi }\frac{dx}{d\xi }+\left\{\frac{(n+1)}{\theta }\right[\frac{{\sigma }_c^2}{6{\gamma }_g}]-(n+1)(-\frac{d\ln \theta }{d\ln \xi }\left)\frac{{\alpha }_g}{{\xi }^2}\right\}x.$

Applied to the Core

As we have already summarized in an accompanying discussion, throughout the core we have,

$r^{*}$

$=$

$(\frac{3}{2\pi }{)}^{1/2}\xi ;$

$\frac{{\rho }^{*}}{P^{*}}$

$=$

$(1+\frac{1}{3}{\xi }^2{)}^{1/2};$

$\frac{M_r^{*}}{r^{*}}$

$=$

$2{\xi }^2(1+\frac{1}{3}{\xi }^2{)}^{-3/2}.$

So the relevant core LAWE becomes,

$0$	$=$	$\left(\frac{2\pi }{3}\right)\frac{d^{2}x}{d{\xi }^2}+\left(\frac{2\pi }{3}\right)\{4-(1+\frac{1}{3}{\xi }^2{)}^{1/2}\left[2{\xi }^2\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]\right\}\frac{1}{\xi }\frac{dx}{d\xi }+(1+\frac{1}{3}{\xi }^2{)}^{1/2}\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-\left(\frac{2\pi }{3}\right)\frac{{\alpha }_{\mathrm{g}}}{{\xi }^2}\left[2{\xi }^2\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]\right\}x$
$\Rightarrow \left(\frac{3}{4\pi }\right)\cdot 0$	$=$	$\frac{1}{2}\cdot \frac{d^{2}x}{d{\xi }^2}+[2-{\xi }^2(1+\frac{1}{3}{\xi }^2{)}^{-1}]\frac{1}{\xi }\frac{dx}{d\xi }+(1+\frac{1}{3}{\xi }^2{)}^{1/2}[\frac{{\sigma }_c^2}{2{\gamma }_{\mathrm{g}}}-{\alpha }_{\mathrm{g}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]x.$

Now, following our separate discussion of an analytic solution to this LAWE, we try,

$x_P{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\equiv$	$1-\frac{{\xi }^2}{15}$
$\Rightarrow \frac{d{x}_P}{d\xi }{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\equiv$	$-\frac{2\xi }{15}$
$\Rightarrow \frac{d\ln x_P}{d\ln \xi }{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\equiv$	$-\frac{2{\xi }^2}{15}[\frac{(15-{\xi }^2)}{15}{]}^{-1}=-\frac{2{\xi }^2}{(15-{\xi }^2)}.$

Plugging this trial function into the relevant LAWE gives,

LAWE	$=$	$\frac{1}{2}(-\frac{2}{3\cdot 5})+(-\frac{2}{3\cdot 5})[2-{\xi }^2(1+\frac{1}{3}{\xi }^2{)}^{-1}]+(1+\frac{1}{3}{\xi }^2{)}^{1/2}[\frac{{\sigma }_c^2}{2{\gamma }_{\mathrm{g}}}-{\alpha }_{\mathrm{g}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]\right[1-\frac{{\xi }^2}{15}]$
	$=$	$-\frac{1}{3}+\left(\frac{2}{3\cdot 5}\right)\left[{\xi }^2\right(1+\frac{1}{3}{\xi }^2{)}^{-1}]+(1+\frac{1}{3}{\xi }^2{)}^{1/2}[\frac{{\sigma }_c^2}{2{\gamma }_{\mathrm{g}}}-{\alpha }_{\mathrm{g}}(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]\right[1-\frac{{\xi }^2}{15}]$

Now, if we set ${\sigma }_c^2=0$ and ${\gamma }_g={\gamma }_c=\frac{6}{5}\Rightarrow {\alpha }_g=-1/3$, we find that the terms on the RHS sum to zero. It therefore appears that we have identified a dimensionless displacement function that satisfies the core LAWE.

Applied to the Envelope

And as we have also summarized in the same accompanying discussion, throughout the envelope we have,

$r^{*}$

$=$

$(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ;$

$\frac{{\rho }^{*}}{P^{*}}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{-1}\varphi (\eta {)}^{-1};$

$\frac{M_r^{*}}{r^{*}}$

$=$

$2\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i\right(-\eta \frac{d\varphi }{d\eta }).$

So the relevant envelope LAWE becomes,

$\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^{2}x}{dr{*}^2}+\{4-(\frac{{\rho }^{*}}{P^{*}}\left)\frac{M_r^{*}}{r^{*}}\right\}\frac{1}{r^{*}}\frac{dx}{dr*}+\left(\frac{{\rho }^{*}}{P^{*}}\right)\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-\frac{{\alpha }_{\mathrm{g}}{M}_r^{*}}{(r^{*}{)}^3}\}x$
	$=$	$\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}{]}^{-2}\frac{d^{2}x}{d{\eta }^2}+\{4-\left[\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^{-1}{\varphi }^{-1}\right]\left[2\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i(-\eta \frac{d\varphi }{d\eta })\left]\right\}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}{]}^{-2}\frac{1}{\eta }\frac{dx}{d\eta }$
		$+\left[\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^{-1}{\varphi }^{-1}\right]\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-\frac{{\alpha }_{\mathrm{g}}}{{\eta }^2}[2\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i\right(-\eta \frac{d\varphi }{d\eta }\left)\right]\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}{]}^{-2}\}x$
$\Rightarrow \left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i^4(2\pi ){]}^{-1}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^{2}x}{d{\eta }^2}+\{4-[\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{-1}{\varphi }^{-1}\left]\right[2\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i\right(-\eta \frac{d\varphi }{d\eta }\left)\right]\}\frac{1}{\eta }\frac{dx}{d\eta }$
		$+\left[\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^{-1}{\varphi }^{-1}\right]\left\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}\right[(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i^4(2\pi ){]}^{-1}-\frac{{\alpha }_{\mathrm{g}}}{{\eta }^2}[2\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i\right(-\eta \frac{d\varphi }{d\eta }\left)\right]\}x$
	$=$	$\frac{d^{2}x}{d{\eta }^2}+\{4-[2(-\frac{d\ln \varphi }{d\ln \eta })\left]\right\}\frac{1}{\eta }\frac{dx}{d\eta }+\left\{\frac{{\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-5}{\varphi }^{-1}\right]-\frac{{\alpha }_{\mathrm{g}}}{{\eta }^2}\left[2\right(-\frac{d\ln \varphi }{d\ln \eta }\left)\right]\}x$
	$=$	$\frac{d^{2}x}{d{\eta }^2}+\{4-2Q_{\eta }\}\frac{1}{\eta }\frac{dx}{d\eta }+\left\{\frac{{\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-5}{\varphi }^{-1}\right]-(2Q_{\eta })\frac{{\alpha }_{\mathrm{g}}}{{\eta }^2}\}x$

where,

$\varphi (\eta )$

$=$

$\frac{A\sin (\eta -B)}{\eta }$

and

$Q_{\eta }$

$\equiv$

$-\frac{d\ln \varphi }{d\ln \eta }=[1-\eta \cot (\eta -B)].$

If we set ${\sigma }_c^2=0$ and ${\gamma }_g={\gamma }_e=2\Rightarrow {\alpha }_g=+1$, the envelope LAWE simplifies to the form,

$(\frac{r^{*}}{\eta }{)}^2\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$

$=$

$\frac{d^{2}x}{d{\eta }^2}+\{4-2Q_{\eta }\}\frac{1}{\eta }\frac{dx}{d\eta }-\left\{\frac{2Q_{\eta }}{{\eta }^2}\right\}x.$

In yet another Ramblings Appendix derivation we have explored a trial dimensionless displacement for the envelope of the form,

$x_P{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=\frac{3c_0}{{\eta }^2}\cdot Q_{\eta }.$

In this case,

$\frac{1}{3c_0}\cdot \frac{d{x}_P}{d\eta }$	$=$	$\frac{1}{{\eta }^2}\frac{d{Q}_{\eta }}{d\eta }-\frac{2Q_{\eta }}{{\eta }^3}$
	$=$	$\frac{1}{{\eta }^2}[\eta -\cot (\eta -B)+\eta {\cot }^{}(\eta -B)]-\frac{2Q_{\eta }}{{\eta }^3}$
	$=$	$\frac{1}{{\eta }^3{\sin }^2(\eta -B)}[{\eta }^2+\eta \sin (\eta -B)\cos (\eta -B)-2{\sin }^{}(\eta -B)]$
	$=$	$\frac{1}{\eta {\sin }^2(\eta -B)}+\frac{\cot (\eta -B)}{{\eta }^2}-\frac{2}{{\eta }^3};$
$\frac{1}{3c_0}\cdot \frac{d^2{x}_P}{d{\eta }^2}$	$=$	$\frac{d}{d\eta }[\frac{1}{\eta {\sin }^2(\eta -B)}+\frac{\cot (\eta -B)}{{\eta }^2}-\frac{2}{{\eta }^3}]$
	$=$	$\frac{6}{{\eta }^4}-\frac{2\cot (\eta -B)}{{\eta }^3}-\frac{2}{{\eta }^2{\sin }^2(\eta -B)}-\frac{2\cos (\eta -B)}{\eta {\sin }^3(\eta -B)},$

and it can be shown that the simplified envelope LAWE is perfectly satisfied. Notice that, with this adopted segment of the eigenfunction for the envelope, we have,

$\frac{d\ln x_P}{d\ln \eta }{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\frac{{\eta }^3}{3c_0{Q}_{\eta }}\cdot \frac{d{x}_P}{d\eta }$	$=$	$\frac{{\eta }^3}{Q_{\eta }}\left\{\frac{1}{{\eta }^2}\right[\eta -\cot (\eta -B)+\eta {\cot }^{}(\eta -B)]-\frac{2Q_{\eta }}{{\eta }^3}\}$
	$=$	$\frac{1}{Q_{\eta }}[{\eta }^2-\eta \cot (\eta -B)+{\eta }^2{\cot }^{}(\eta -B)]-2$
	$=$	$\frac{[{\eta }^2-2+\eta \cot (\eta -B)+{\eta }^2{\cot }^{}(\eta -B)]}{[1-\eta \cot (\eta -B)]}$
	$=$	$\frac{[{\eta }^2-2{\sin }^2(\eta -B)+\eta \sin (\eta -B)\cos (\eta -B)]}{[{\sin }^2(\eta -B)-\eta \sin (\eta -B)\cos (\eta -B)]}.$

Interface Matching

According to our accompanying discussion of the interface matching condition — as we presently understand it — the proper eigenfunction will exhibit a discontinuity in the slope of the dimensionless displacement function such that,

$\frac{d\ln x_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\ln \eta }{|}_{\eta ={\eta }_i}$	$=$	$3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}(\frac{d\ln x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\ln \xi }{)}_{\xi ={\xi }_i}$
	$=$	$\frac{3}{5}\left[\right(\frac{d\ln x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\ln \xi }{)}_{\xi ={\xi }_i}-2].$

See Also

• K. De et al. (12 October 2018, Science, Vol. 362, No. 6411, pp. 201 - 206), A Hot and Fast Ultra-stripped Supernova that likely formed a Compact Neutron Star Binary.