SSC/Stability/BiPolytropes/RedGiantToPN

Main Sequence to Red Giant to Planetary Nebula

Following the Lead of Yabushita75

Here in the context of $(n_c,n_e)=(5,1)$ bipolytropes, we want to construct an interface-pressure versus volume plot; and mass-versus-central density plots like the ones displayed for truncated isothermal spheres in Figure 1 of an accompanying discussion, and as displayed for a $(n_c,n_e)=(\mathrm{\infty },3/2)$ bipolytrope in Figure 1 (p. 445) of 📚 S. Yabushita (1975, MNRAS, Vol. 172, pp. 441 - 453).

In our accompanying chapter that presents example models of $(n_c,n_e)=(5,1)$ bipolytropes, we have adopted the following normalizations:

${\rho }^{*}$	$\equiv$	$\frac{\rho }{{\rho }_0}$	;	$r^{*}$	$\equiv$	$\frac{r}{[K_c^{1/2}/(G^{1/2}{\rho }_0^{2/5})]}$
$P^{*}$	$\equiv$	$\frac{P}{K_c{\rho }_0^{6/5}}$	;	$M_r^{*}$	$\equiv$	$\frac{M_r}{[K_c^{3/2}/(G^{3/2}{\rho }_0^{1/5})]}$

Also, from the relevant interface conditions, we find,

$\left(\frac{K_e}{K_c}\right)$

$=$

${\rho }_0^{-4/5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}.$

Inverting this last expression gives,

${\rho }_0^{4/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}\right(\frac{K_e}{K_c}{)}^{-1}$
$\Rightarrow {\rho }_0^{1/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1/2}{\theta }_i^{-1}\right(\frac{K_e}{K_c}{)}^{-1/4}.$

Hence, we can rewrite the "normalized" expressions as follows:

$r$	$=$	$r^{*}\left[K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}\right]$
	$=$	$r^{*}\left\{K_c^{1/2}{G}^{-1/2}\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1/2}{\theta }_i^{-1}\right(\frac{K_e}{K_c}{)}^{-1/4}{]}^{-2}\}$
	$=$	$r^{*}\left\{K_c^{1/2}{G}^{-1/2}\right[\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2\left(\frac{K_e}{K_c}{)}^{1/2}\right]\}$
	$=$	$r^{*}\left[K_e^{1/2}{G}^{-1/2}\right]\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2.$

Fixed Interface Pressure

Start with the model relation,

$P_i$	$=$	$\left[K_c{\rho }_0^{6/5}\right]P_i^{*}$
	$=$	$\left[K_c{\rho }_0^{6/5}\right](1+\frac{1}{3}{\xi }_i^2{)}^{-3}$

Now, given that,

${\rho }_0^{1/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1/2}{\theta }_i^{-1}\right(\frac{K_e}{K_c}{)}^{-1/4}.$
$\Rightarrow {\rho }_0^{6/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\theta }_i^{-6}\right(\frac{K_e}{K_c}{)}^{-3/2}.$

Fixed Total Mass

Also, from the relevant interface conditions, we find,

$\left(\frac{K_e}{K_c}\right)$

$=$

${\rho }_0^{-4/5}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}.$

Inverting this last expression gives,

${\rho }_0^{4/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}\right(\frac{K_e}{K_c}{)}^{-1}$
$\Rightarrow {\rho }_0^{1/5}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1/2}{\theta }_i^{-1}\right(\frac{K_e}{K_c}{)}^{-1/4}.$

Hence, for a given specification of the interface location, ${\xi }_i$ — test values shown (in parentheses) assuming ${\mu }_e/{\mu }_c=1.0$ and ${\xi }_i=0.5$ — the desired expression for the central density is,

${\rho }_0$

$=$

$\left[K_e^{-5}{K}_c^5{]}^{1/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-5/2}{\theta }_i^{-5};$

and, drawing the expression for the normalized total mass from our accompanying table of parameter values, namely,

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{*}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}$

we find,

$M_r$	$=$	$M_r^{*}\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]$
	$=$	$M_r^{*}\left[K_c^{3/2}{G}^{-3/2}\right]\left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1/2}{\theta }_i^{-1}(\frac{K_e}{K_c}{)}^{-1/4}{\}}^{-1}$
	$=$	$M_r^{*}\left[K_c^{3/2}{G}^{-3/2}\right(\frac{K_e}{K_c}{)}^{1/4}\left]\right(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}{\theta }_i$
$\Rightarrow M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	$\left[K_c^{3/2}{G}^{-3/2}\right(\frac{K_e}{K_c}{)}^{1/4}\left]\right(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}{\theta }_i\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}$
	$=$	$\left[K_e{K}_c^{-5}{G}^{-6}{]}^{1/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}(\frac{2}{\pi }{)}^{1/2}A{\eta }_s,$

where — again, from our accompanying table of parameter values —

${\theta }_i$	$=$	$(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2};$		(0.96077)
${\eta }_i$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i;$		(0.79941)
${\mathrm{\Lambda }}_i$	$=$	$\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}};$		(0.96225)
$A$	$=$	${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2};$		(1.10940)
${\eta }_s$	$=$	${\eta }_i+\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i);$		(3.13637)
$\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{[K_e{K}_c^{-5}{G}^{-6}{]}^{1/4}}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}\right(\frac{2}{\pi }{)}^{1/2}A{\eta }_s;$		(2.77623)
$\frac{{\rho }_0}{[K_e^{-5}{K}_c^5{]}^{1/4}}$	$=$	$(\frac{{\mu }_e}{{\mu }_c}{)}^{-5/2}{\theta }_i^{-5}.$		(1.22153)

Building on Earlier Eigenfunction Details

In the heading of Figure 6 from our accompanying presentation of the properties of marginally unstable oscillation modes in $(n_c,n_e)=(5,1)$ bipolytropes, we point to the (Excel spreadsheet) "Data File" that contains most of the relevant model details. See specifically,

file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/LinearPerturbation/FaulknerBipolytrope1.xlsx --- worksheet = Mode0Ensemble Figure 6: Eigenfunctions Associated with the Fundamental-Mode of Radial Oscillation in Marginally Unstable Models having Various ${\mu }_e/{\mu }_c$

Related Discussions

• Instability Onset Overview
• Analytic $(n_c,n_e)=(5,1)$
  • Part 1
  • Part 2