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

Finite-Difference Expressions

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 $x_i$.

A:     Pick ${\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}$; this will give analytic expressions for ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}$, $B$, and for ${\eta }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$, as well as analytic expressions for $(r_0{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ and $(r_0{)}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$.

B:     Divide the radial coordinate grid into 99 spherical shells $\Rightarrow {\mathrm{\Delta }}_r=[(r_0{)}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}-(r_0{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}]/99.$ Then tabulate 100 values of $(r_0{)}_i,{\eta }_i,(Q_1{)}_i=[1-\eta \cot (\eta -B){]}_i$.

Generally speaking, after multiplying through by $r_0^2$, the finite-difference representation of the envelope's LAWE takes the form,

$0$	$=$	$r_0^2\left[\frac{x_{+}-2x_i+x_{-}}{{\mathrm{\Delta }}_r^2}\right]+[4-2Q_1]r_0\left[\frac{x_{+}-x_{-}}{2{\mathrm{\Delta }}_r}\right]-\left[2Q_1\right]x_i$
	$=$	$x_{+}\{\frac{r_0^2}{{\mathrm{\Delta }}_r^2}+(4-2Q_1)\frac{r_0}{2{\mathrm{\Delta }}_r}\}+x_i\{-\frac{2r_0^2}{{\mathrm{\Delta }}_r^2}-2Q_1\}+x_{-}\{\frac{r_0^2}{{\mathrm{\Delta }}_r^2}-(4-2Q_1)\frac{r_0}{2{\mathrm{\Delta }}_r}\}$

Multiplying through by $({\mathrm{\Delta }}_r^2/r_0^2)$ and solving for $x_{+}$ gives,

$0$	$=$	$x_{+}\{1+(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}\}-2x_i\{1+Q_1(\frac{{\mathrm{\Delta }}_r^2}{r_0^2}\left)\right\}+x_{-}\{1-(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}\}$
$\Rightarrow x_{+}\{1+(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}\}$	$=$	$2x_i\{1+Q_1(\frac{{\mathrm{\Delta }}_r^2}{r_0^2}\left)\right\}-x_{-}\{1-(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}\}$
$\Rightarrow x_{+}$	$=$	$\left\{2x_i\right[1+Q_1\left(\frac{{\mathrm{\Delta }}_r^2}{r_0^2}\right)]-x_{-}[1-(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}\left]\right\}[1+(4-2Q_1)\frac{{\mathrm{\Delta }}_r}{2r_0}{]}^{-1}.$

Now, at the interface, $x_i$, we don't know the value of $x_{-}$, but we know analytically the value of the first derivative, namely,

$[\frac{dx}{d{r}_0}{]}_{\mathrm{i}\mathrm{n}\mathrm{t}}$

$=$

$\frac{x_{\mathrm{i}\mathrm{n}\mathrm{t}}}{r_0}\cdot \{\frac{d\ln x}{d\ln r_0}{|}_{\mathrm{i}\mathrm{n}\mathrm{t}}{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

Steps

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 ${\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}=1.668646016$ $\Rightarrow (r_0{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}[K_5^{-1}G{\rho }_c^{4/5}{]}^{1/2}=1.153014872.$ ${\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $[1+\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2}{3}{]}^{-1/2};$ (0.720165375) $(\frac{d\theta }{d\xi }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $-\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}}{3}[1+\frac{{\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2}{3}{]}^{-3/2};$ (- 0.207749350) ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}^2\cdot {\xi }_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (1.498957494) $(\frac{d\varphi }{d\eta }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $3^{1/2}{\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}^{-3}\cdot (\frac{d\theta }{d\xi }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (- 0.963393227) ${\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}$ = $\frac{1}{{\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}}+(\frac{d\varphi }{d\eta }{)}_{\mathrm{i}\mathrm{n}\mathrm{t}};$ (- 0.296262902) $A$ = ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}(1+{\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}^2{)}^{1/2};$ (1.563357124) $B$ = ${\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_{\mathrm{i}\mathrm{n}\mathrm{t}}).$ (- 0.359863580) ${\eta }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$ = $B+\pi .$ (2.781729074) 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, $x$ $=$ $[1-\frac{r_0^2}{15a_5^2}]=[1-\frac{{\xi }^2}{15}];$ (0.814374698) $\Rightarrow r_0\cdot \frac{dx}{d{r}_0}$ $=$ $-\frac{2r_0^2}{15a_5^2}=-\frac{2{\xi }^2}{15};$ (- 0.371250604) $\Rightarrow r_0^2\cdot \frac{d^{2}x}{d{r}_0^2}$ $=$ $-\frac{2r_0^2}{15a_5^2}=-\frac{2{\xi }^2}{15};$ (- 0.371250604) also …     $\{\frac{d\ln x}{d\ln \xi }{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\{\frac{d\ln x}{d\ln r_0}{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\frac{r_0}{x}\cdot \frac{dx}{d{r}_0}$ $=$ $\left[\frac{15}{15-{\xi }^2}\right]\cdot [-\frac{2{\xi }^2}{15}]=\left[\frac{2{\xi }^2}{{\xi }^2-15}\right].$ (-0.455871977)† STEP #4:    From the determination of the logarithmic slope of the displacement function at the edge of the core — i.e., at the core-envelope interface — determine the slope as viewed from the perspective of the envelope. $\{\frac{d\ln x}{d\ln r_0}{|}_{\mathrm{i}\mathrm{n}\mathrm{t}}{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\{\frac{d\ln x}{d\ln \eta }{|}_{\mathrm{i}\mathrm{n}\mathrm{t}}{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $=$ $3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}\{\frac{d\ln x}{d\ln \xi }{|}_{\mathrm{i}\mathrm{n}\mathrm{t}}{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}.$ (-1.473523186)† †This analytically determined value matches the previous determination that was obtained via numerical integration of the LAWE.

Throughout the envelope — that is, over the range, $({\eta }_{\mathrm{i}\mathrm{n}\mathrm{t}}\le \eta \le {\eta }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}})$ — the radial coordinate, $r_0$, is a linear function of $\eta$ and takes on values given by the expression,

$r_0[K_5^{-1}G{\rho }_c^{4/5}{]}^{1/2}$	$=$	$\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_{\mathrm{i}\mathrm{n}\mathrm{t}}^{-2}(2\pi {)}^{-1/2}]\cdot \eta$	(0.769211186 × η)
$\Rightarrow 1.153014872$	$\le r_0\le$	$2.139737121.$

From our earlier discussions,

Here we examine some of the properties of the fundamental-mode eigenfunctions that we have found are associated with marginally unstable, $(n_c,n_e)=(5,1)$ bipolytropes.

Figure 5

Consider the model on the ${\mu }_e/{\mu }_c=1$ sequence for which ${\sigma }_c^2=0$; key properties of this specific equilibrium model are enumerated in the first row of numbers provided in Table 2, above. Figure 5 shows how our numerically derived, fundamental-mode eigenfunction, $x=\delta r/r_0$, varies with the fractional radius over the entire range, $0\le r/R\le 1$. By prescription, the eigenfunction has a value of unity and a slope of zero at the center $(r/R=0)$. Integrating the LAWE outward from the center, through the model's core (blue curve segment), $x$ drops smoothly to the value $x_i=0.81437$ at the interface $({\xi }_i=1.6686460157\Rightarrow q=r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/R_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}=0.53885819)$. Our numerical integration of the LAWE showed that, at the interface, the logarithmic slope of the core (blue) segment of the eigenfunction is,

$\{\frac{d\ln x}{d\ln r}{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$-0.455872.$

Next, following the above discussion of matching conditions at the interface, we determined that, from the perspective of the envelope, the slope of the eigenfunction at the interface must therefore be,

$\{\frac{d\ln x}{d\ln r}{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\{\frac{d\ln x}{d\ln \eta }{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=-1.47352.$

Adopting this "env" slope along with the amplitude, $x_i=0.81437$, as the appropriate interface boundary conditions, we integrated the LAWE from the interface to the surface, obtaining the green-colored segment of the eigenfunction that is shown in Figure 5. The amplitude continued to steadily decrease, reaching a value of $x_s=0.38203$, at the model's surface $(r/R=1)$. At the surface, this envelope (green) segment of the eigenfunction exhibits a logarithmic slope that matches to eight significant digits the value that is expected from astrophysical arguments for this marginally unstable $({\sigma }_c^2=0)$ model, namely,

$\frac{d\ln x}{d\ln \eta }{|}_s=\left[\right(\frac{{\rho }_c}{\overline{\rho }})\frac{{\xcancel{{\sigma }_c^2}}^0}{2{\gamma }_e}-(3-\frac{4}{{\gamma }_e}\left)\right]=-1.$

Related Discussions

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