SSC/Stability/BiPolytropes/RedGiantToPN

Main Sequence to Red Giant to Planetary Nebula

Preface

In terms of mass $(m)$, length $(\ell )$, and time $(t)$, the units of various physical variables are: Pressure $\sim$ $m{\ell }^{-1}{t}^{-2}$

Original Model Construction

Fixed Central Density

From Examples, we find,

$M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]$	$=$	$\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right](\frac{6}{\pi }{)}^{1/2}({\xi }_i{\theta }_i{)}^3;$
$M_{\mathrm{t}\mathrm{o}\mathrm{t}}=M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{*}\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]$	$=$	$\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i};$
$r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}\left[K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}\right]$	$=$	$\left[K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}\right](\frac{3}{2\pi }{)}^{1/2}{\xi }_i;$
$R=r_s^{*}\left[K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}\right]$	$=$	$\left[K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}\right](\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{{\eta }_s}{\sqrt{2\pi }{\theta }_i^2};$

where, rewriting the relevant expressions in terms of the parameters,

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

we find,

${\theta }_i$	$=$	$(1+{\ell }_i^2{)}^{-1/2}$
${\eta }_i$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i=m_3\left(\frac{{\ell }_i}{1+{\ell }_i^2}\right)$
${\mathrm{\Lambda }}_i$	$=$	$\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}}=\left[\frac{1+{\ell }_i^2}{m_3{\ell }_i}\right]-{\ell }_i=\frac{1}{m_3{\ell }_i}[1+(1-m_3){\ell }_i^2]$
$A^2$	$=$	${\eta }_i^2(1+{\mathrm{\Lambda }}_i^2)$
	$=$	$m_3^2(\frac{{\ell }_i}{1+{\ell }_i^2}{)}^2\cdot \frac{(1+{\ell }_i^2)}{m_3^2{\ell }_i^2}[1+(1-m_3{)}^2{\ell }_i^2]$
	$=$	$\left[\frac{1+(1-m_3{)}^2{\ell }_i^2}{1+{\ell }_i^2}\right]$
${\eta }_s$	$=$	$\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i)$

Fixed K_c and K_e

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,

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

Hence, keeping $K_c$ and $K_e$ constant, we have,

$M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\left[K_c^{3/2}{G}^{-3/2}\right(\frac{K_e}{K_c}{)}^{1/4}\left]\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}{\theta }_i\right](\frac{6}{\pi }{)}^{1/2}({\xi }_i{\theta }_i{)}^3$
	$=$	$\left[K_c^5{K}_e{G}^{-6}{]}^{1/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}(\frac{6}{\pi }{)}^{1/2}{\xi }_i^3{\theta }_i^4;$
$M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	$\left[K_c^{3/2}{G}^{-3/2}\right(\frac{K_e}{K_c}{)}^{1/4}\left]\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}{\theta }_i\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}$
	$=$	$\left[K_c^5{K}_e{G}^{-6}{]}^{1/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}(\frac{2}{\pi }{)}^{1/2}A{\eta }_s;$
$r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\left[K_e{G}^{-1}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}\left)\right(\frac{3}{2\pi }{)}^{1/2}{\xi }_i{\theta }_i^2;$
$R$	$=$	$[K_e{G}^{-1}{]}^{1/2}\frac{{\eta }_s}{\sqrt{2\pi }};$
${\rho }_0$	$=$	$\left[\frac{K_e}{K_c}{]}^{-5/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-5/2}{\theta }_i^{-5};$
$P_i$	$=$	$\left[K_c\right]\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\theta }_i^{-6}\left(\frac{K_e}{K_c}{)}^{-3/2}\right]{\theta }_i^6$
	$=$	$\left[K_c^5{K}_e^{-3}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}.$

This last expression shows that if $K_c$ and $K_e$ are both held fixed, then the interface pressure, $P_i$, will be constant along the sequence of equilibrium models.

Note also:

$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left\{\right[K_c^5{K}_e{G}^{-6}{]}^{1/4}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{1/2}\right(\frac{6}{\pi }{)}^{1/2}{\xi }_i^3{\theta }_i^4\left\}\right\{\left[K_c^5{K}_e{G}^{-6}{]}^{1/4}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}(\frac{2}{\pi }{)}^{1/2}A{\eta }_s{\}}^{-1}$
	$=$	$(\frac{{\mu }_e}{{\mu }_c}{)}^2\frac{\sqrt{3}{\xi }_i^3{\theta }_i^4}{A{\eta }_s};$
$q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$	$=$	$\left\{\right[K_e{G}^{-1}{]}^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right)\left(\frac{3}{2\pi }{)}^{1/2}{\xi }_i{\theta }_i^2\right\}\left\{\right[K_e{G}^{-1}{]}^{1/2}\frac{{\eta }_s}{\sqrt{2\pi }}{\}}^{-1}$
	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{\sqrt{3}{\xi }_i{\theta }_i^2}{{\eta }_s}.$

Fixed Total Mass

In order to build a sequence along which $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ is held fixed, we must set

$C_{\mathrm{M}}\equiv A{\eta }_s$

$=$

$\left[\frac{1+(1-m_3{)}^2{\ell }_i^2}{1+{\ell }_i^2}{]}^{1/2}\right[\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i)]=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{.}$

This means,

${\tan }^{-1}({\mathrm{\Lambda }}_i)$	$=$	$[\frac{C_M^2(1+{\ell }_i^2)}{1+(1-m_3{)}^2{\ell }_i^2}{]}^{1/2}-\frac{\pi }{2}-m_3(\frac{{\ell }_i}{1+{\ell }_i^2})$
$\Rightarrow {\mathrm{\Lambda }}_i$	$=$	$\tan \left\{\right[\frac{C_M^2(1+{\ell }_i^2)}{1+(1-m_3{)}^2{\ell }_i^2}{]}^{1/2}-\frac{\pi }{2}-m_3\left(\frac{{\ell }_i}{1+{\ell }_i^2}\right)\}$
$\Rightarrow \frac{1}{m_3{\ell }_i}[1+(1-m_3){\ell }_i^2]$	$=$	$\tan \left\{\right[\frac{C_M^2(1+{\ell }_i^2)}{1+(1-m_3{)}^2{\ell }_i^2}{]}^{1/2}-\frac{\pi }{2}-m_3\left(\frac{{\ell }_i}{1+{\ell }_i^2}\right)\}$

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