SSC/Stability/BiPolytropes/RedGiantToPN

Main Sequence to Red Giant to Planetary Nebula

Part I:  Background & Objective

Part II:

Part III:

Part IV:

Preface

In terms of mass $(m)$, length $(\ell )$, and time $(t)$, the units of various physical constants and variables are: Mass-density $\sim$ $m{\ell }^{-3}$ Pressure (energy-density) $\sim$ $m{\ell }^{-1}{t}^{-2}$ Newtonian gravitational constant, $G$ $\sim$ $m^{-1}{\ell }^3{t}^{-2}$ The core's polytropic constant, $K_c$ $\sim$ $[m^{-1}{\ell }^{13}{t}^{-10}{]}^{1/5}$ The envelope's polytropic constant, $K_e$ $\sim$ $m^{-1}{\ell }^5{t}^{-2}$ As a result, for example (see details below), if we hold the central-density $({\rho }_0)$ — as well as $G$ and $K_c$ — constant along an equilibrium sequence, mass will scale as … Mass $\sim$ $\left[K_c^{3/2}{G}^{-3/2}{\rho }_0^{-1/5}\right]$   $\sim$ $\left\{\right[m^{-1}{\ell }^{13}{t}^{-10}{]}^{1/5}{\}}^{3/2}\left[m^{-1}{\ell }^3{t}^{-2}{]}^{-3/2}\right[m{\ell }^{-3}{]}^{-1/5}$   $\sim$ $\left[m^{-3/10+3/2-1/5}\right]\left[{\ell }^{39/10-9/2+3/5}\right]\left[t^{-3+3}\right]$   $\sim$ $\left[m^{+1}\right]\left[{\ell }^0\right]\left[t^0\right].$ If instead (see details below) we hold $K_e$ — as well as $G$ and $K_c$ — constant along an equilibrium sequence, mass will scale as … Mass $\sim$ $[K_c^5{K}_e{G}^{-6}{]}^{1/4}$   $\sim$ $\left\{\right[m^{-1}{\ell }^{13}{t}^{-10}\left]\right[m^{-1}{\ell }^5{t}^{-2}\left]\right[m^{-1}{\ell }^3{t}^{-2}{]}^{-6}{\}}^{1/4}$   $\sim$ $\left\{\right[m^{-1-1+6}\left]\right[{\ell }^{13+5-18}\left]\right[t^{-10-2+12}]{\}}^{1/4}$   $\sim$ $\left[m^{+1}\right]\left[{\ell }^0\right]\left[t^0\right].$

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 Interface Pressure

Equilibrium Sequence Expressions

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:

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

Sequence Plots

A plot of $M_{\mathrm{t}\mathrm{o}\mathrm{t}}[K_c^5{K}_e{G}^{-6}{]}^{-1/4}$ versus $R[K_e{G}^{-1}{]}^{-1/2}$ at fixed interface pressure will be generated via the relations,

Ordinate: $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$		Abscissa: $R$
$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}\right(\frac{2}{\pi }{)}^{1/2}A{\eta }_s$	vs	$\frac{{\eta }_s}{\sqrt{2\pi }}.$

Alternatively, a plot of $M_{\mathrm{t}\mathrm{o}\mathrm{t}}[K_c^5{K}_e{G}^{-6}{]}^{-1/4}$ versus ${\log }_{10}({\rho }_0)[\frac{K_e}{K_c}{]}^{5/4}$ at fixed interface pressure will be generated via the relations,

Ordinate: $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$		Abscissa: ${\log }_{10}({\rho }_0)$
$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3/2}\right(\frac{2}{\pi }{)}^{1/2}A{\eta }_s$	vs	${\log }_{10}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-5/2}{\theta }_i^{-5}]$

---

file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/TwoFirstOrderODEs/Bipolytrope51New.xlsx --- worksheet = SequenceMuRatio100 Example data values drawn from worksheet "SequenceMuRatio100" … $\mathrm{\Delta }\xi =(9.01499598-0.05)/99=0.0905551$ $n_{\mathrm{g}\mathrm{r}\mathrm{i}\mathrm{d}}$ ${\xi }_i=0.05+(n_{\mathrm{g}\mathrm{r}\mathrm{i}\mathrm{d}}-1)\cdot \mathrm{\Delta }\xi$ ${\theta }_i$ $A$ ${\eta }_s$ $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ $\log 10{\rho }_0$ $R$ 1 0.05 0.9995836 1.00124818 3.141592582 2.510 0.0009044 1.253 2 0.140555 0.9967235 1.0097655 3.141580334 2.531 0.0071264 1.253 18 1.5894375 0.7367887 1.539943947 2.821678456 3.467 0.663285301 1.126 19 1.6799927 0.7178117 1.566601145 2.775921455 3.470 0.719947375 1.107 20 1.7705478 0.6992927 1.591530391 2.728957898 3.465 0.7767049 1.089 100 9.0149598 0.1886798 1.973119305 0.841461698 1.325 3.62137 0.336

Equilibrium Sequences of $(n_c,n_e)=(5,1)$ BiPolytropes Having ${\mu }_e/{\mu }_c=1.0$
Mass vs. Radius (Fixed Interface Pressure)	Mass vs. Central Density (Fixed Interface Pressure)

file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/TwoFirstOrderODEs/Bipolytrope51New.xlsx --- worksheet = MuRatio100Fund Data values drawn from worksheet "MuRatio100Fund" … Properties of the Marginally Unstable Model ${\xi }_i$ ${\theta }_i$ $A$ ${\eta }_s$ $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ $\log 10{\rho }_0$ $R$ 1.6639103 0.7211498 1.561995126 2.784147185 3.4698598 0.709872477 1.1107140

Fixed Total Mass

Equilibrium Sequence Expressions

Again, drawing from previous Examples in which ${\rho }_0$ — as well as $K_c$ and $G$ — is held fixed, equilibrium models obey the relations,

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

Let's invert the first expression in order to construct equilibrium sequences in which the total mass — rather than ${\rho }_0$ — is held fixed. We find that,

${\rho }_0^{1/5}$	$=$	$\left[K_c^{3/2}{G}^{-3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}$
$\Rightarrow R$	$=$	$\left[K_c^{1/2}{G}^{-1/2}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{{\eta }_s}{\sqrt{2\pi }{\theta }_i^2}\right\{\left[K_c^{3/2}{G}^{-3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}{\}}^{-2}$
	$=$	$\left[K_c^{1/2}{G}^{-1/2}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{{\eta }_s}{\sqrt{2\pi }{\theta }_i^2}\right[K_c^{3/2}{G}^{-3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{]}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{\pi }{2})\frac{{\theta }_i^2}{A^2{\eta }_s^2}$
	$=$	$\left[K_c^{-5/2}{G}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^3\right(\frac{\pi }{2^3}{)}^{1/2}\frac{1}{A^2{\eta }_s}.$

And,

$P_i$	$=$	$K_c{\theta }_i^6\left\{\right[K_c^{3/2}{G}^{-3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}\left]\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}{\}}^6$
	$=$	$K_c\left\{\right[K_c^{3/2}{G}^{-3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{]}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(\frac{2}{\pi }{)}^3{A}^6{\eta }_s^6\}$
	$=$	$\left[K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(\frac{2}{\pi }{)}^3{A}^6{\eta }_s^6.$

Note as well that,

$P_i\left(\frac{4\pi }{3}{R}^3\right)$	$=$	$\frac{4\pi }{3}\left[K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(\frac{2}{\pi }{)}^3{A}^6{\eta }_s^6\left\{\right[K_c^{-5/2}{G}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left]\right(\frac{{\mu }_e}{{\mu }_c}{)}^3(\frac{\pi }{2^3}{)}^{1/2}\frac{1}{A^2{\eta }_s}{\}}^3$
	$=$	$\frac{4\pi }{3}\left(\frac{2}{\pi }{)}^3\right(\frac{\pi }{2^3}{)}^{3/2}\left[K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\eta }_s^3\right\{\left[K_c^{-15/2}{G}^{15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right]\left(\frac{{\mu }_e}{{\mu }_c}{)}^9\right\}$
	$=$	$\frac{4\pi }{3}\left(\frac{2}{\pi }{)}^3\right(\frac{\pi }{2^3}{)}^{3/2}\left[K_c^{5/2}{G}^{-3/2}\right](\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_s^3$

Sequence Plots

A plot of $P_i\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right]$ versus $R^3[K_c^{5/2}{G}^{-5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}{]}^3$ at fixed interface pressure will be generated via the relations,

Ordinate		Abscissa
$\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(\frac{2}{\pi }{)}^3{A}^6{\eta }_s^6$	vs	$\left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^3(\frac{\pi }{2^3}{)}^{1/2}\frac{1}{A^2{\eta }_s}{\}}^3$

Pressure vs Volume

nu vs q

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