SSC/Stability/BiPolytropes/RedGiantToPN/Pt2

Main Sequence to Red Giant to Planetary Nebula (Part 2)

Part I:  Background & Objective

Part II:

Part III:

Part IV:

Foundation

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

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.

Introducing the dimensionless frequency-squared, ${\sigma }_c^2\equiv 3{\omega }^2/(2\pi G{\rho }_c)$, we can rewrite this LAWE as, $0$ $=$ $\frac{d^{2}x}{d{r}_0^2}+[4-(\frac{g_0}{r_0}\cdot \frac{{\rho }_0{r}_0^2}{P_0}\left)\right]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}+\left(\frac{{\rho }_0{r}_0^2}{P_0}\right)[\frac{2\pi G{\rho }_c{\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\frac{g_0}{r_0}\right]\frac{x}{r_0^2},$ where, as a reminder, $g_0\equiv GM(r_0)/r_0^2$. Now, for our $(n_c,n_e)=(5,1)$ bipolytrope, we have found it useful to adopt the following four dimensionless variables: ${\rho }^{*}$ $\equiv$ $\frac{{\rho }_0}{{\rho }_c}$ ;     $r^{*}$ $\equiv$ $\frac{r_0}{[K_c^{1/2}/(G^{1/2}{\rho }_c^{2/5})]}$ $P^{*}$ $\equiv$ $\frac{P_0}{K_c{\rho }_c^{6/5}}$ ;     $M_r^{*}$ $\equiv$ $\frac{M_r}{[K_c^{3/2}/(G^{3/2}{\rho }_c^{1/5})]}$ This means that, $\frac{g_0}{r_0}=\frac{GM(r_0)}{r_0^3}$ $=$ $G{M}_r^{*}\left[K_c^{3/2}{G}^{-3/2}{\rho }_c^{-1/5}\right]r_{*}^{-3}\left[K_c^{-3/2}{G}^{3/2}{\rho }_c^{6/5}\right]=\left[G{\rho }_c\right]M_r^{*}{r}_{*}^{-3};$ $\frac{{\rho }_0{r}_0^2}{P_0}$ $=$ ${\rho }^{*}{\rho }_c(r^{*}{)}^2[K_c{G}^{-1}{\rho }_c^{-4/5}](P^{*}{)}^{-1}\left[K_c^{-1}{\rho }_c^{-6/5}\right]=\left[G^{-1}{\rho }_c^{-1}\right]{\rho }^{*}(r^{*}{)}^2(P^{*}{)}^{-1};$ $\frac{g_0}{r_0}\cdot \frac{{\rho }_0{r}_0^2}{P_0}$ $=$ $\left[G{\rho }_c\right]M_r^{*}{r}_{*}^{-3}\cdot \left[G^{-1}{\rho }_c^{-1}\right]{\rho }^{*}(r^{*}{)}^2(P^{*}{)}^{-1}=\frac{M_r^{*}{\rho }^{*}}{P^{*}{r}^{*}}.$ Making these substitutions, the LAWE can be rewritten as, $0$ $=$ $\frac{d^{2}x}{d{r}_0^2}+[4-\frac{M_r^{*}{\rho }^{*}}{P^{*}{r}^{*}}]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}+\frac{1}{G{\rho }_c}\left[\frac{{\rho }^{*}(r^{*}{)}^2}{P^{*}}\right][\frac{2\pi G{\rho }_c{\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\frac{G{\rho }_c{M}_r^{*}}{(r^{*}{)}^3}\right]\frac{x}{r_0^2};$ then, multiplying through by $[K{G}^{-1}{\rho }_c^{-4/5}]$ allows us to everywhere switch from $(r_0{)}^2$ to $(r^{*}{)}^2$, namely, $0$ $=$ $\frac{d^{2}x}{d(r^{*}{)}^2}+[4-\frac{M_r^{*}{\rho }^{*}}{P^{*}{r}^{*}}]\frac{1}{r^{*}}\cdot \frac{dx}{d(r^{*})}+\left[\frac{{\rho }^{*}(r^{*}{)}^2}{P^{*}}\right][\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\frac{M_r^{*}}{(r^{*}{)}^3}\right]\frac{x}{(r^{*}{)}^2}.$

In shorthand, we can rewrite this equation in the form,

$0$

$=$

$x^{″}+\frac{{\mathscr{H}}}{r^{*}}{x}^{\prime }+{𝒦}x,$

where,

$x^{\prime }$

$=$

$\frac{dx}{d{r}^{*}}$

and

$x^{″}$

$=$

$\frac{d^{2}x}{d(r^{*}{)}^2};$

and,

${𝒦}\equiv \left(\frac{{\rho }^{*}}{P^{*}}\right)\left[\right(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}})\frac{2\pi }{3}-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\frac{M_r^{*}}{(r^{*}{)}^3}\right];$

and,

${\mathscr{H}}$

$\equiv$

$\{4-(\frac{{\rho }^{*}}{P^{*}}\left)\frac{M_r^{*}}{(r^{*})}\right\}.$

Specific Case of (n_c, n_e) = (5,1)

Drawing from our "Table 2" profiles, let's evaluate ${\mathscr{H}}$ and ${𝒦}$ for the two separate regions of bipolytrope model.

The n_c = 5 Core

${\displaystyle }$

$r^{*}=(\frac{3}{2\pi }{)}^{1/2}\xi$

$\frac{{\rho }^{*}}{P^{*}}$	$=$	$(1+\frac{{\xi }^2}{3}{)}^{-5/2}(1+\frac{{\xi }^2}{3}{)}^{6/2}=(1+\frac{{\xi }^2}{3}{)}^{1/2}$
$\frac{M^{*}}{r^{*}}$	$=$	$\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-3/2}](\frac{2\pi }{3}{)}^{1/2}{\xi }^{-1}=2{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
$\Rightarrow {\mathscr{H}}$	$=$	$4-2{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-3/2}(1+\frac{{\xi }^2}{3}{)}^{1/2}=4-2{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-1}.$

Also,

$\Rightarrow {𝒦}$

$=$

$(1+\frac{{\xi }^2}{3}{)}^{1/2}\{\left(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}}\right)\frac{2\pi }{3}-(3-\frac{4}{{\gamma }_{\mathrm{g}}})2{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-3/2}(\frac{2\pi }{3}\left){\xi }^{-2}\right\}=\frac{2\pi }{3}(1+\frac{{\xi }^2}{3}{)}^{1/2}\{\left(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}}\right)-2(3-\frac{4}{{\gamma }_{\mathrm{g}}})(1+\frac{{\xi }^2}{3}{)}^{-3/2}\}$

Hence, the LAWE becomes,

$0$

$=$

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

Multiplying through by $3{\xi }^2/(2\pi )$ gives,

$0$

$=$

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

Let's compare this with the equivalent expression presented separately, namely, Polytropic LAWE (linear adiabatic wave equation) $0=\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$ where:    $Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi },$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c},$     and,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$ The primary E-type solution for n = 5 polytropes states that, ${\theta }_{n=5}$ $=$ $[1+\frac{{\xi }^2}{3}{]}^{-1/2}.$ $\Rightarrow Q=-\frac{\xi }{\theta }\frac{d\theta }{d\xi }$ $=$ $-\xi [1+\frac{{\xi }^2}{3}{]}^{1/2}\frac{d}{d\xi }[1+\frac{{\xi }^2}{3}{]}^{-1/2}$   $=$ $+\xi [1+\frac{{\xi }^2}{3}{]}^{1/2}\{\frac{\xi }{3}[1+\frac{{\xi }^2}{3}{]}^{-3/2}\}$   $=$ $\frac{{\xi }^2}{3}[1+\frac{{\xi }^2}{3}{]}^{-1}.$ Hence, the LAWE may be written as, $0$ $=$ $\frac{d^{2}x}{d{\xi }^2}+[4-6Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+6\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{{\xi }^2}{\theta }-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)Q\right]\frac{x}{{\xi }^2}$   $=$ $\frac{d^{2}x}{d{\xi }^2}+\{4-2{\xi }^2[1+\frac{{\xi }^2}{3}{]}^{-1}\}\frac{1}{\xi }\cdot \frac{dx}{d\xi }+\{\left(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}}\right){\xi }^2[1+\frac{{\xi }^2}{3}{]}^{1/2}-2{\xi }^2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\frac{{\xi }^2}{3}{]}^{-1}\}\frac{x}{{\xi }^2}$   $=$ $\frac{d^{2}x}{d{\xi }^2}+\{4-2{\xi }^2[1+\frac{{\xi }^2}{3}{]}^{-1}\}\frac{1}{\xi }\cdot \frac{dx}{d\xi }+\{\left(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}}\right)[1+\frac{{\xi }^2}{3}{]}^{1/2}-2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\frac{{\xi }^2}{3}{]}^{-1}\}x$ Versus above, $0$ $=$ ${\xi }^2\frac{d^{2}x}{d{\xi }^2}+[4\xi -2{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-1}]\frac{dx}{d\xi }+{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{1/2}\left\{\right(\frac{{\sigma }_c^2}{{\gamma }_{\mathrm{g}}})-2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right(1+\frac{{\xi }^2}{3}{)}^{-3/2}\}x.$

If we set ${\gamma }_{\mathrm{g}}=6/5$ and we set ${\sigma }_c^2=0$, this becomes,

$0$

$=$

${\xi }^2\frac{d^{2}x}{d{\xi }^2}+[4\xi -2{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-1}]\frac{dx}{d\xi }+\frac{2}{3}{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-1}x$

Next, try the solution, $x=(1-{\xi }^2/15)\Rightarrow dx/d\xi =-2\xi /15$ and $d^{2}x/d{x}^2=-2/15$:

LAWE	$=$	$-\frac{2}{15}{\xi }^2-[4\xi -2{\xi }^3(1+\frac{{\xi }^2}{3}{)}^{-1}]\frac{2\xi }{15}+\frac{2}{3}{\xi }^2(1+\frac{{\xi }^2}{3}{)}^{-1}[1-{\xi }^2/15]$
$\Rightarrow 15(1+\frac{{\xi }^2}{3})$  LAWE	$=$	$-2{\xi }^2(1+\frac{{\xi }^2}{3})-\left[60\xi \right(1+\frac{{\xi }^2}{3})-30{\xi }^3]\frac{2\xi }{15}+10{\xi }^2[1-{\xi }^2/15]$
	$=$	$-2{\xi }^2-\frac{2{\xi }^4}{3}-[60{\xi }^2-10{\xi }^4]\frac{2}{15}+\frac{10}{15}[15{\xi }^2-{\xi }^4]$
	$=$	$-2{\xi }^2-\frac{2{\xi }^4}{3}-[4{\xi }^2-\frac{2}{3}{\xi }^4]2+[10{\xi }^2-\frac{2}{3}{\xi }^4]$
	$=$	$-2{\xi }^2-\frac{2{\xi }^4}{3}-8{\xi }^2+\frac{4}{3}{\xi }^4+[10{\xi }^2-\frac{2}{3}{\xi }^4]$
	$=$	$0.$

The n_e = 1 Envelope

Let's compare this with the equivalent expression presented separately, namely, Polytropic LAWE (linear adiabatic wave equation) $0=\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$ where:    $Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi },$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c},$     and,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$ The equilibrium, off-center equilibrium solution for n = 1 polytropes states that, ${\varphi }_{n=1}$ $=$ $-\frac{A}{\eta }\cdot \sin (B-\eta );$ $\frac{d\varphi }{d\eta }$ $=$ $-A\frac{d}{d\eta }\{{\eta }^{-1}\cdot \sin (B-\eta )\}$   $=$ $A\{{\eta }^{-2}\cdot \sin (B-\eta )+{\eta }^{-1}\cdot \cos (B-\eta )\};$ $\Rightarrow Q=-\frac{\eta }{\varphi }\frac{d\varphi }{d\eta }$ $=$ $-\eta [-\frac{A}{\eta }\cdot \sin (B-\eta ){]}^{-1}A\{{\eta }^{-2}\cdot \sin (B-\eta )+{\eta }^{-1}\cdot \cos (B-\eta )\}$   $=$ $\eta \left[\frac{\eta }{\sin (B-\eta )}\right]\{{\eta }^{-2}\cdot \sin (B-\eta )+{\eta }^{-1}\cdot \cos (B-\eta )\}$   $=$ $[1+\eta \cdot \cot (B-\eta )]$ Hence, the LAWE may be written as, $0$ $=$ $\frac{d^{2}x}{d{\eta }^2}+\{4-2[1+\eta \cdot \cot (B-\eta )\left]\right\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+2\left\{\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{{\eta }^2}{\varphi }-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\eta \cdot \cot (B-\eta )\left]\right\}\frac{x}{{\eta }^2}$

First Try

$x$

$=$

${\eta }^{-m}\Rightarrow \frac{dx}{d\eta }=-m{\eta }^{-m-1}$       and     $\frac{d^{2}x}{d{\eta }^2}=-m(-m-1){\eta }^{-m-2},$

in which case,

LAWE	$=$	$-m(-m-1){\eta }^{-m-2}+\{4-2[1+\eta \cdot \cot (B-\eta )\left]\right\}\frac{1}{\eta }\cdot [-m{\eta }^{-m-1}]+2\left\{\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{{\eta }^2}{\varphi }-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\eta \cdot \cot (B-\eta )\left]\right\}{\eta }^{-m-2}$
$\Rightarrow {\eta }^{m+2}\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$m(m+1)-m\{4-2[1+\eta \cdot \cot (B-\eta )\left]\right\}+2\left\{\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{{\eta }^2}{\varphi }-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\eta \cdot \cot (B-\eta )\left]\right\}.$

Now set ${\sigma }_c^2=0$ and set ${\gamma }_{\mathrm{g}}=2$:

$\Rightarrow {\eta }^{m+2}\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$m(m+1)-m\{4-2[1+\eta \cdot \cot (B-\eta )\left]\right\}-2\left\{\right[1+\eta \cdot \cot (B-\eta )\left]\right\}$
	$=$	$m(m+1)-4m+2m[1+\eta \cdot \cot (B-\eta )]-2[1+\eta \cdot \cot (B-\eta )]$

We see that the complexity of the LAWE reduces substantially if we set $m=+1$; specifically, this choice gives,

$[{\eta }^{m+2}\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{]}_{m\to 1}$

$=$

$-2.$

Close, but no cigar!

Second Try

Next, let's set ${\sigma }_c^2=0$ but let's leave ${\gamma }_{\mathrm{g}}$ unspecified:

$\Rightarrow {\eta }^{m+2}\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$m(m+1)-m\{4-2[1+\eta \cdot \cot (B-\eta )\left]\right\}-2\left\{\right(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right[1+\eta \cdot \cot (B-\eta )\left]\right\}$
	$=$	$m(m+1)-4m+2m[1+\eta \cdot \cot (B-\eta )]-2(3-\frac{4}{{\gamma }_{\mathrm{g}}})[1+\eta \cdot \cot (B-\eta )]$
	$=$	$m(m-3)+\{2m-2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right\}[1+\eta \cdot \cot (B-\eta )].$

The first term goes to zero if we set $m=3$; then, in order for the second term to go to zero, we need …

$0$	$=$	$\{6-2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right\}$
$\Rightarrow {\gamma }_{\mathrm{g}}$	$=$	$\mathrm{\infty }.$

This means that the envelope is incompressible.

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