SSC/StabilityConjecture/Bipolytrope51

Examine B-KB74 Conjecture in the Context of $(n_c,n_e)=(5,1)$ Bipolytropes

B-KB74 Conjecture RE: Bipolytrope (nc, ne) = (5, 1)

In §6 of their paper, G. S. Bisnovatyi-Kogan & S. I. Blinnikov (1974; hereafter, B-KB74) have suggested that "… a static configuration close to an extremum of the [mass-radius equilibrium] curve may be considered as a perturbed state of a model of the same mass situated on the other side of the extremum. The difference of the two models approximately represents the eigenfunction of the neutral mode." In an accompanying discussion we have demonstrated that this "B-KB74 conjecture" applies exactly in the context of an analysis of the stability of pressure-truncated, n = 5 polytropes. We know that it applies exactly in this case because, along the n = 5 mass-radius sequence, the eigenfunction of the fundamental mode of radial oscillation is known analytically.

Here we turn to the B-KB74 conjecture to assist us in examining the stability of models that lie along the sequence of bipolytropes with $(n_c,n_e)=(5,1)$. As Eggleton, Faulkner, and Cannon (1998, MNRAS, 298, 831) discovered — and we have independently detailed — the internal structure of these bipolytropes can be defined analytically. But, as far as we have been able to determine, nothing is known about the eigenvectors describing their natural modes of radial oscillation. Guided by the B-KB74 conjecture, we hope to be able to determine the eigenfunction of the fundamental mode of radial oscillation for the model that sits at the maximum-mass "turning point" along each sequence; our expectation is that each of these models is marginally [dynamically] unstable.

Properties of Equilibrium Models

Figure 1

Bipolytropic (5, 1) Equilibrium Sequences

Drawing from our accompanying detailed discussion, Figure 1 shows how the fractional core mass, $\nu \equiv M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ varies with the fractional core radius, $q\equiv r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/r_{\mathrm{t}\mathrm{o}\mathrm{t}}$, for seven equilibrium model sequences of bipolytropes having $(n_c,n_e)=(5,1)$. Along each sequence, the value of the radial location of the interface, ${\xi }_i$, varies while the mean-molecular-weight ratio at the interface, $({\mu }_e/{\mu }_c{)}_i\le 1$, is held fixed at the value that labels the sequence. A green circular marker has been placed at the maximum-mass "turning point" of each sequence for which $({\mu }_e/{\mu }_c{)}_i\le \frac{1}{3}$; no such point exists along sequences having $\frac{1}{3}<({\mu }_e/{\mu }_c{)}_i\le 1$.

Original Manipulations

As has been shown in our accompanying discussion, the value of ${\xi }_i$ at which the maximum-mass turning point resides along each sequence is given by a root of the analytic expression,

$(\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i)(1+{\ell }_i^2)[3+(1-m_3{)}^2(2-{\ell }_i^2){\ell }_i^2]$

$=$

$m_3{\ell }_i[(1-m_3){\ell }_i^4-(m_3^2-m_3+2){\ell }_i^2-3],$

where,

${\ell }_i$	$\equiv$	$\frac{{\xi }_i}{\sqrt{3}},$
$m_3$	$\equiv$	$3\left(\frac{{\mu }_e}{{\mu }_c}\right),$
${\mathrm{\Lambda }}_i$	$\equiv$	$\frac{1}{m_3{\ell }_i}[1+(1-m_3){\ell }_i^2].$

Table 1 details some example roots.

file = Dropbox/WorkFolder/Wiki edits/Bipolytrope/qAndNuMax.xlsx --- worksheet = Table Table 1 Properties of Models at the Maximum-Mass Turning Point
$\frac{{\mu }_e}{{\mu }_c}$	${\xi }_i$	${\mathrm{\Lambda }}_i$	LHS	RHS	$q$	$\nu$
$\frac{1}{3}$	$\mathrm{\infty }$	$$	$$	$$	$0.0$	$0.63661977$
$0.325$	$14.9254$	$0.339976$	$-50.490003$	$-99.589549$	$0.0555$	$0.453$
$0.320$	$11.95$	$$	$$	$$	$0.0643$	$0.4061$
$0.316943$	$10.74425$	$$	$$	$$	$0.06865$	$0.38238$
$0.310$	$9.01$	$0.59825$	$-19.112$	$+224,10$	$0.0755$	$0.3372$
$0.305$	$8.17$	$$	$$	$$	$0.0791$	$0.31$
$0.300$	$7.57$	$$	$$	$$	$0.0814$	$0.286$
$0.295$	$7.09$	$$	$$	$$	$0.0831$	$0.265$
$\frac{1}{4}$	$4.93827$	$$	$$	$$	$0.0848$	$0.1394$

New Manipulations

Following through the numbered steps that we have used to construct a bipolytrope with $(n_c,n_e)=(5,1)$, and adopting the substitute notation,

${\ell }_i\equiv \frac{{\xi }_i}{\sqrt{3}};$       and       $m_3\equiv 3\left(\frac{{\mu }_e}{{\mu }_c}\right),$

we seek expressions for $\nu (m_3,{\ell }_i)$ and $q(m_3,{\ell }_i)$. [Example #1 numerical evaluation is for ${\mu }_e/{\mu }_c=0.25$ and ${\xi }_i=0.5$, which implies that $m_3=0.75$ and ${\ell }_i=(12{)}^{-1/2}$.]

Focusing, first, on the core, we find,

${\theta }_i$	$=$	$(1+{\ell }_i^2{)}^{-1/2}=0.960768923$,
$(\frac{d\theta }{d\xi }{)}_i$	$=$	$-\frac{{\ell }_i}{\sqrt{3}}(1+{\ell }_i^2{)}^{-3/2}=-0.147810603$,

$[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}{r}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $=$ $(\frac{3^2}{2\pi }{)}^{1/2}{\ell }_i=0.345494149$, $[\frac{G^3{\rho }_0^{2/5}}{K_c^3}{]}^{1/2}{M}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $=$ $3^2(\frac{2}{\pi }{)}^{1/2}{\ell }_i^3(1+{\ell }_i^2{)}^{-3/2}=0.153203096$,

Then moving across the interface, through the envelope, and ultimately to the surface of the configuration, we find,

${\eta }_i$	$=$	$m_3{\ell }_i{\theta }_i^2=\frac{m_3{\ell }_i}{(1+{\ell }_i^2)}=0.199852016$,
$(\frac{d\varphi }{d\eta }{)}_i$	$=$	$3^{1/2}{\theta }_i^{-3}(\frac{d\theta }{d\xi }{)}_i=-0.288675135$,
${\mathrm{\Lambda }}_i$	$=$	$\frac{1}{{\eta }_i}+(\frac{d\varphi }{d\eta }{)}_i=\frac{1}{m_3{\theta }_i^2{\ell }_i}-{\ell }_i=\frac{1}{m_3{\ell }_i}[(1+{\ell }_i^2)-m_3{\ell }_i^2]=\frac{1}{m_3{\ell }_i}[1+(1-m_3){\ell }_i^2]=4.715027199,$
${\eta }_s$	$=$	$(\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i)+{\eta }_i=(\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i)+\frac{m_3{\ell }_i}{(1+{\ell }_i^2)}=3.132453649,$
$B$	$=$	${\eta }_s-\pi =-0.009139005,$
$A$	$=$	${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2}=\frac{m_3{\ell }_i}{(1+{\ell }_i^2)}\{1+\frac{1}{m_3^2{\ell }_i^2}[1+(1-m_3){\ell }_i^2{]}^2{\}}^{1/2}=0.963267676,$
$(\frac{d\varphi }{d\eta }{)}_s$	$=$	$\frac{A}{{\eta }_s^2}[{\eta }_s\cos (\pi )-\sin (\pi )]=-\frac{A}{{\eta }_s}=-0.307512188,$

$[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}R$ $=$ $(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}{\eta }_s=5.415228878$, $\Rightarrow q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$ $=$ $0.063800470$ $[\frac{G^3{\rho }_0^{2/5}}{K_c^3}{]}^{1/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}[-\frac{{\eta }_s^2}{{\theta }_i}\cdot (\frac{d\varphi }{d\eta }{)}_s]=40.0934$, $\Rightarrow \nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$ $=$ $0.003821153$

Now, putting all these steps together, we can generate the pair of desired model-parameter expressions:

$q(m_3,{\ell }_i)$	$=$	$\frac{m_3{\ell }_i{\theta }_i^2}{{\eta }_s}=\frac{m_3{\ell }_i}{(1+{\ell }_i^2)}\left\{\right(\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i)+\frac{m_3{\ell }_i}{(1+{\ell }_i^2)}{\}}^{-1}=\{[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i]\frac{(1+{\ell }_i^2)}{m_3{\ell }_i}+1{\}}^{-1}=0.063800470.$
$\nu (m_3,{\ell }_i)$	$=$	$\sqrt{3}\left[\frac{{\xi }_i^3{\theta }_i^4}{A{\eta }_s}\right]\frac{m_3^2}{3^2}=\frac{m_3{\ell }_i{\theta }_i^2}{{\eta }_s}\left[\frac{{\xi }_i^2{\theta }_i^2}{A}\right]\frac{m_3}{3}=q{\ell }_i\left[\frac{m_3{\ell }_i}{(1+{\ell }_i^2)A}\right]=q{\ell }_i\{1+\frac{1}{m_3^2{\ell }_i^2}[1+(1-m_3){\ell }_i^2{]}^2{\}}^{-1/2}$
	$=$	$q{m}_3{\ell }_i^2\{m_3^2{\ell }_i^2+[1+(1-m_3){\ell }_i^2{]}^2{\}}^{-1/2}=(0.059892291)q=0.00382116.$

See Also