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

Notice that, $\frac{g_0{\rho }_0{r}_0}{P_0}$ $=$ $\frac{G}{a_n^2{\xi }^2}\left[4\pi a_n^3{\rho }_c\right(-{\xi }^2\frac{d\theta }{d\xi }\left)\right]\cdot {\rho }_c{\theta }^n\cdot a_n\xi \cdot [K{\rho }_c^{(n+1)/n}{\theta }^{n+1}{]}^{-1}$   $=$ $\frac{4\pi G}{K}\left[{\rho }_c^{1-1/n}\right](-\xi \frac{d\theta }{d\xi })\cdot {\theta }^{-1}\cdot a_n^2$   $=$ $(n+1)(-\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi });$ $\frac{{\rho }_0{r}_0^2}{P_0}$ $=$ ${\rho }_c{\theta }^n\cdot (a_n\xi {)}^2\cdot [K{\rho }_c^{(n+1)/n}{\theta }^{n+1}{]}^{-1}$   $=$ $K^{-1}{\rho }_c^{-1/n}\cdot a_n^2\cdot \frac{{\xi }^2}{\theta }$   $=$ $\left[\frac{(n+1)}{4\pi G{\rho }_c}\right]\cdot \frac{{\xi }^2}{\theta }.$

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,

$0$

$=$

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

Envelope (n = 1)

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

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

Related Discussions

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  • Part 2