SSC/Stability/BiPolytropes/RedGiantToPN/Pt4

Main Sequence to Red Giant to Planetary Nebula

Part I:  Background & Objective

Part II:

Part III:

Part IV:

Succinct

Generic

Adiabatic Wave (or Radial Pulsation) Equation

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-(\frac{g_0{\rho }_0}{P_0}\left)\right]\frac{dx}{d{r}_0}+\left(\frac{{\rho }_0}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]x=0$

may also be written as …

$0$

$=$

$\frac{d^{2}x}{dr{*}^2}+\{4-(\frac{{\rho }^{*}}{P^{*}}\left)\frac{M_r^{*}}{(r^{*})}\right\}\frac{1}{r^{*}}\frac{dx}{dr*}+\left(\frac{{\rho }^{*}}{P^{*}}\right)\{\frac{2\pi {\sigma }_c^2}{3{\gamma }_{\mathrm{g}}}-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\frac{M_r^{*}}{(r^{*}{)}^3}\right\}x.$

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 Polytropes

In a separate discussion, we have shown that configurations with a polytropic equation of state exhibit a characteristic length scale that is given by the expression,

$a_n$

$\equiv$

$[\frac{(n+1)K}{4\pi G}\cdot {\rho }_c^{(1-n)/n}{]}^{1/2};$

and, once the dimensionless polytropic temperature, $\theta (\xi )$, is known, the radial dependence of key physical variables is given by the expressions,

				if, as in a separate discussion, $n=5$ and $\theta =(1+{\xi }^2/3{)}^{-1/2}$ …
$r_0$	$=$	$a_n\xi ,$		$r_0$	$=$	$[\frac{K_5}{G}\cdot {\rho }_c^{-4/5}{]}^{1/2}(\frac{3}{2\pi }{)}^{1/2}\xi ,$
${\rho }_0$	$=$	${\rho }_c{\theta }^n,$		${\rho }_0$	$=$	${\rho }_c{\theta }^5,$
$P_0$	$=$	$K{\rho }_0^{(n+1)/n}=K{\rho }_c^{(n+1)/n}{\theta }^{n+1},$		$P_0$	$=$	$K_5{\rho }_c^{6/5}{\theta }^6,$
$M(r_0)$	$=$	$4\pi {\rho }_c{a}_n^3(-{\xi }^2\frac{d\theta }{d\xi })={\rho }_c^{(3-n)/(2n)}\left[\frac{(n+1{)}^3{K}^3}{4\pi G^3}{]}^{1/2}\right(-{\xi }^2\frac{d\theta }{d\xi }),$		$M(r_0)$	$=$	$[\frac{K_5^3}{G^3}\cdot {\rho }_c^{-2/5}{]}^{1/2}(\frac{2\cdot 3^3}{\pi }{)}^{1/2}(-{\xi }^2\frac{d\theta }{d\xi }),$
$g_0$	$=$	$\frac{GM(r_0)}{r_0^2}=\frac{G}{r_0^2}\left[4\pi a_n^3{\rho }_c\right(-{\xi }^2\frac{d\theta }{d\xi }\left)\right],$		$g_0$	$=$	$\left\{\right[\frac{K_5}{G}\cdot {\rho }_c^{-4/5}{]}^{1/2}\left(\frac{3}{2\pi }{)}^{1/2}\xi {\}}^{-2}\right[\frac{6K_5}{4\pi G}\cdot {\rho }_c^{-4/5}{]}^{3/2}\left[4\pi G{\rho }_c\right](-{\xi }^2\frac{d\theta }{d\xi })]$
					$=$	$\left(\frac{2^3{\pi }^2}{3}\right)\left[\frac{6}{4\pi }{]}^{3/2}\right[\frac{G}{K_5}\cdot {\rho }_c^{4/5}\left]\right[\frac{K_5}{G}\cdot {\rho }_c^{-4/5}{]}^{3/2}\left[G{\rho }_c\right](-\frac{d\theta }{d\xi })]$
					$=$	$(2^3\cdot 3\pi {)}^{1/2}[K_{5}G\cdot {\rho }_c^{6/5}{]}^{1/2}(-\frac{d\theta }{d\xi })].$

Combining variable expressions from the above right-hand column, we find that for $n=5$ polytropes,

$\frac{g_0{\rho }_0{r}_0}{P_0}$	$=$	$(2^3\cdot 3\pi {)}^{1/2}[K_{5}G\cdot {\rho }_c^{6/5}{]}^{1/2}(-\frac{d\theta }{d\xi })]\cdot {\rho }_c{\theta }^5\cdot [\frac{K_5}{G}\cdot {\rho }_c^{-4/5}{]}^{1/2}\left(\frac{3}{2\pi }{)}^{1/2}\xi \right[K_5{\rho }_c^{6/5}{\theta }^6{]}^{-1}$
	$=$	$6(-\frac{\xi }{\theta }\frac{d\theta }{d\xi }).$

More generally, combining variable expressions from the above left-hand column, we find, $\frac{g_0{\rho }_0{r}_0}{P_0}$ $=$ $\frac{G}{a_n^2{\xi }^2}\left[4\pi a_n^3{\rho }_c\right(-{\xi }^2\frac{d\theta }{d\xi }\left)\right]\cdot {\rho }_c{\theta }^n\cdot a_n\xi \cdot [K{\rho }_c^{(n+1)/n}{\theta }^{n+1}{]}^{-1}$   $=$ $\frac{4\pi G}{K}\left[{\rho }_c^{1-1/n}\right](-\xi \frac{d\theta }{d\xi })\cdot {\theta }^{-1}\cdot a_n^2$   $=$ $(n+1)(-\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi });$ $\frac{{\rho }_0{r}_0^2}{P_0}$ $=$ ${\rho }_c{\theta }^n\cdot (a_n\xi {)}^2\cdot [K{\rho }_c^{(n+1)/n}{\theta }^{n+1}{]}^{-1}$   $=$ $K^{-1}{\rho }_c^{-1/n}\cdot a_n^2\cdot \frac{{\xi }^2}{\theta }$   $=$ $\left[\frac{(n+1)}{4\pi G{\rho }_c}\right]\cdot \frac{{\xi }^2}{\theta }.$

As a result, for polytropes we can write,

$0$	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-(\frac{g_0{\rho }_0{r}_0}{P_0}\left)\right]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}+\left(\frac{{\rho }_0{r}_0^2}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]\frac{x}{r_0^2}$
	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-(\frac{g_0{\rho }_0{r}_0}{P_0}\left)\right]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}+\left[\frac{{\omega }^2}{{\gamma }_{\mathrm{g}}}\right(\frac{{\rho }_0{r}_0^2}{P_0})-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right(\frac{g_0{\rho }_0{r}_0}{P_0}\left)\right]\frac{x}{r_0^2}$
	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-(n+1)Q]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}+(n+1)\left[\frac{{\omega }^2}{{\gamma }_{\mathrm{g}}}\right[\frac{1}{4\pi G{\rho }_c}]\cdot \frac{{\xi }^2}{\theta }-(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)Q\right]\frac{x}{r_0^2}.$

Finally, multiplying through by $a_n^2$ — which everywhere converts $r_0$ to $\xi$ — gives, what we will refer to as the,

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,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$

BiPolytrope

Let's stick with the dimensional $(r_0)$ version and set ${\omega }^2=0$, in which case the Polytropic LAWE is,

$0$

$=$

$\frac{d^{2}x}{d{r}_0^2}+[4-(n+1)Q]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-[(n+1)\alpha Q]\frac{x}{r_0^2}.$

Core (n = 5)

For the $n=5$ core, we know that $\theta =(1+{\xi }^2/3{)}^{-1/2}$. Hence,

$\frac{d\theta }{d\xi }$	$=$	$-\frac{\xi }{3}\cdot (1+\frac{{\xi }^2}{3}{)}^{-3/2}$
$\Rightarrow Q_5\equiv -\frac{d\ln \theta }{d\ln \xi }=-\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi }$	$=$	$[\frac{\xi }{3}\cdot (1+\frac{{\xi }^2}{3}{)}^{-3/2}]\cdot \xi (1+\frac{{\xi }^2}{3}{)}^{1/2}$
	$=$	$[\frac{{\xi }^2}{3}\cdot (1+\frac{{\xi }^2}{3}{)}^{-1}]$
	$=$	$\left(\frac{{\xi }^2}{3+{\xi }^2}\right).$

Now, given that,

$a_5^2$

$=$

$\frac{3}{2\pi }\left[K_5{G}^{-1}{\rho }_c^{-4/5}\right],$

we can everywhere make the substitution,

${\xi }^2$

$\to$

$(\frac{r_0}{a_5}{)}^2=\frac{2\pi }{3}[K_5^{-1}G{\rho }_c^{4/5}]r_0^2.$

Note, also, that throughout the core, the relevant LAWE is,

$0$

$=$

$\frac{d^{2}x}{d{r}_0^2}+[4-6Q_5]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-\left[6\alpha Q_5\right]\frac{x}{r_0^2}.$

Next, try the solution,

$x$	$=$	$[1-\frac{r_0^2}{15a_5^2}],$
$\Rightarrow \frac{dx}{d{r}_0}$	$=$	$-\frac{2r_0}{15a_5^2},$
$\Rightarrow \frac{d^{2}x}{d{r}_0^2}$	$=$	$-\frac{2}{15a_5^2},$

in which case,

$\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-6Q_5]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-\left[6\alpha Q_5\right]\frac{x}{r_0^2}$
	$=$	$-\frac{2}{15a_5^2}+[4-6Q_5][-\frac{2}{15a_5^2}]-\left[6\alpha Q_5\right]\frac{1}{r_0^2}\cdot [1-\frac{r_0^2}{15a_5^2}]$
	$=$	$-\frac{2}{15a_5^2}+[4-6Q_5][-\frac{2}{15a_5^2}]-\left[6\alpha Q_5\right]\frac{1}{r_0^2}\cdot \left[\frac{15a_5^2-r_0^2}{15a_5^2}\right]$
	$=$	$-\frac{2}{15a_5^2}+[4-6Q_5][-\frac{2}{15a_5^2}]-\left[6\alpha Q_5\right]\frac{1}{15a_5^2}\cdot \left[\frac{15-{\xi }^2}{{\xi }^2}\right]$
$\Rightarrow 15a_5^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$-10+12Q_5-\left[6\alpha Q_5\right]\cdot \left[\frac{15-{\xi }^2}{{\xi }^2}\right]$
	$=$	$-10+12\left(\frac{{\xi }^2}{3+{\xi }^2}\right)-\left[6\alpha \right(\frac{{\xi }^2}{3+{\xi }^2}\left)\right]\cdot \left[\frac{15-{\xi }^2}{{\xi }^2}\right]$
	$=$	$-10(3+{\xi }^2)+12{\xi }^2-\left[6\alpha \right(\frac{{\xi }^2}{3+{\xi }^2}\left)\right]\cdot \left[\frac{15-{\xi }^2}{{\xi }^2}\right]$
$\Rightarrow 15a_5^2(3+{\xi }^2)\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$-30+2{\xi }^2-6\alpha (15-{\xi }^2).$

Setting $\alpha =-1/3$ gives the desired result, namely,

$\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$

$=$

$0.$

Envelope (n = 1)

From the variable expressions in the right-hand column of Step 8 of the construction chapter,

$\frac{g_0{\rho }_0{r}_0}{P_0}=\frac{G{M}_r}{r_0^2}\cdot \frac{{\rho }_0{r}_0}{P_0}$	$=$	$G\left\{\right[\frac{K_5^3}{G^3{\rho }_c^{2/5}}{]}^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\right(\frac{2}{\pi }{)}^{1/2}(-{\eta }^2\frac{d\varphi }{d\eta })\}\cdot \{\left[\frac{K_5}{G{\rho }_c^{4/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta {\}}^{-1}\cdot \{{\rho }_c\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi \}\cdot \{K_5{\rho }_c^{6/5}{\theta }_i^6{\varphi }^2{\}}^{-1}$
	$=$	$\left\{\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta }\left)\right\}\cdot \left\{\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}{\eta }^{-1}\}\cdot \{\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi \}\cdot \{{\theta }_i^{-6}{\varphi }^{-2}\}$
	$=$	$2(-\frac{\eta }{\varphi }\cdot \frac{d\varphi }{d\eta }).$

For the $n=1$ envelope, we know from separate work that,

$\varphi$	$=$	$A\left[\frac{\sin (\eta -B)}{\eta }\right]$
$\Rightarrow \frac{d\varphi }{d\eta }$	$=$	$\frac{A}{{\eta }^2}[\eta \cos (\eta -B)-\sin (\eta -B)]$
$\Rightarrow Q_1\equiv -\frac{d\ln \varphi }{d\ln \eta }=-\frac{\eta }{\varphi }\cdot \frac{d\varphi }{d\eta }$	$=$	$-\frac{\eta }{A}\left[\frac{\eta }{\sin (\eta -B)}\right]\cdot \frac{A}{{\eta }^2}[\eta \cos (\eta -B)-\sin (\eta -B)]$
	$=$	$[1-\eta \cot (\eta -B)].$

$0$

$=$

$\frac{d^{2}x}{d{r}_0^2}+[4-(n+1)Q_1]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-[(n+1)\alpha Q_1]\frac{x}{r_0^2}.$

Numerical Integration Through Envelope

The discussion in this subsection is guided by our previous attempt at numerical integration.

Here, we focus on the LAWE that is relevant to the envelope, namely,

$0$	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-(n+1)Q_1]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-[(n+1)\alpha Q_1]\frac{x}{r_0^2},$
	$=$	$\frac{d^{2}x}{d{r}_0^2}+[4-2Q_1]\frac{1}{r_0}\cdot \frac{dx}{d{r}_0}-\left[2Q_1\right]\frac{x}{r_0^2},$

where we have plugged in the values, $(n,\alpha )=(1,1)$. Using the general finite-difference approach described separately, we make the substitutions,

$[\frac{dx}{d{r}_0}{]}_i$

$\approx$

$\frac{x_{+}-x_{-}}{2{\mathrm{\Delta }}_r};$

and,

$[\frac{d^{2}x}{d{r}_0^2}{]}_i$

$\approx$

$\frac{x_{+}-2x_i+x_{-}}{{\mathrm{\Delta }}_r^2};$

which will provide an approximate expression for $x_{+}\equiv x_{i+1}$, given the values of $x_{-}\equiv x_{i-1}$ and $x_i$.

STEP 1:  Specify the interface location from the perspective of the core; that is, specify ${\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}$, in which case, $(r_0{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}=a_5\cdot {\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $\left[K_5{G}^{-1}{\rho }_c^{-4/5}{]}^{1/2}\right(\frac{3}{2\pi }{)}^{1/2}{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}.$ STEP 2:   Adopting the normalization ${\varphi }_{\mathrm{i}\mathrm{n}\mathrm{t}}=1$, determine numerous additional equilibrium properties at the interface, such as … Example numerical values inside parentheses assume $({\mu }_e/{\mu }_c)=1$ and ${\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}=2$ ${\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $[1+\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2}{3}{]}^{-1/2};$ (0.654654) $(\frac{d\theta }{d\xi }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $-\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}}{3}[1+\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2}{3}{]}^{-3/2};$ (- 0.187044) ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2\cdot {\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (1.484615) $(\frac{d\varphi }{d\eta }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $3^{1/2}{\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}^{-3}\cdot (\frac{d\theta }{d\xi }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (- 0.090895) ${\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $\frac{1}{{\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}}+(\frac{d\varphi }{d\eta }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (0.582681) $A$ = ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}(1+{\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}^2{)}^{1/2};$ (1.718256) $B$ = ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}).$ (0.441406) STEP 3:   Throughout the core — that is, at all radial positions, $0\le r_0\le (r_0{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ — the displacement amplitude and its first and second derivatives, respectively, are given by the expressions, $x$ $=$ $[1-\frac{r_0^2}{15a_5^2}];$ $\Rightarrow \frac{dx}{d{r}_0}$ $=$ $-\frac{2r_0}{15a_5^2}\Rightarrow \frac{d\ln x}{d\ln r_0}=\left[\frac{15a_5^2}{15s_5^2-r_0^2}\right]\cdot [-\frac{2r_0^2}{15a_5^2}]=[-\frac{2r_0^2}{15s_5^2-r_0^2}]=[-\frac{2(r_0^2/a_5^2)}{1-(r_0^2/a_5^2)}],$ $\Rightarrow \frac{d^{2}x}{d{r}_0^2}$ $=$ $-\frac{2}{15a_5^2}.$

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