SSC/Stability/NeutralMode

LAWE

Most General Form

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

Adiabatic Wave (or Radial Pulsation) Equation

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

where the gravitational acceleration,

$g_0$

$\equiv$

$\frac{G{M}_r}{r_0^2}=-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0}\Rightarrow \frac{g_0{\rho }_0{r}_0}{P_0}=-\frac{d\ln P_0}{d\ln r_0}.$

The solution to this equation gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. The boundary condition conventionally used in connection with the adiabatic wave equation is,

$r_0\frac{d\ln x}{d{r}_0}$

$=$

$\frac{1}{{\gamma }_g}(4-3{\gamma }_g+\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}})$        at         $r_0=R.$

Polytropic Configurations

Part 1

If the initial, unperturbed equilibrium configuration is a polytropic sphere whose internal structure is defined by the function, $\theta (\xi )$, that provides a solution to the,

Lane-Emden Equation

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$

then,

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

where,

$a_n$

$=$

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

Hence, after multiplying through by $a_n^2$, the above adiabatic wave equation can be rewritten in the form,

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

$=$

$0.$

In addition, given that,

$\frac{g_0}{a_n}$

$=$

$4\pi G{\rho }_c(-\frac{d\theta }{d\xi }),$

and,

$\frac{a_n^2{\rho }_0}{P_0}$

$=$

$\frac{(n+1)}{(4\pi G{\rho }_c)\theta }=\frac{a_n^2{\rho }_c}{P_c}\cdot \frac{{\theta }_c}{\theta },$

we can write,

$0$	$=$	$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }+\left[{\omega }^2\right(\frac{a_n^2{\rho }_c}{{\gamma }_g{P}_c})\frac{{\theta }_c}{\theta }-(3-\frac{4}{{\gamma }_g})\cdot \frac{(n+1)V(x)}{{\xi }^2}]x$
	$=$	$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }+(n+1)\left[{\omega }^2\right(\frac{a_n^2{\rho }_c}{{\gamma }_g{P}_c})\frac{{\xi }^2{\theta }_c}{(n+1)\theta }-(3-\frac{4}{{\gamma }_g})\cdot V(x)]\frac{x}{{\xi }^2}$

where we have adopted the function notation,

$V(\xi )$

$\equiv$

$-\frac{\xi }{\theta }\frac{d\theta }{d\xi }.$

Part 2

Drawing from an accompanying discussion, we have the following:

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

In order to reconcile with the "Part 1" expression, we note first that $V(\xi )\leftrightarrow Q(\xi )$. We note as well that since, $\left(\frac{a_n^2{\rho }_c}{P_c}\right){\theta }_c$ $=$ $\frac{(n+1)}{4\pi G{\rho }_c},$ we have, ${\omega }^2\left(\frac{a_n^2{\rho }_c}{{\gamma }_g{P}_c}\right)\frac{{\xi }^2{\theta }_c}{(n+1)\theta }$ $\leftrightarrow$ $\frac{{\omega }^2}{{\gamma }_g}\left[\frac{(n+1)}{4\pi G{\rho }_c}\right]\frac{{\xi }^2}{(n+1)\theta }=\frac{1}{6{\gamma }_g}\left[\frac{3{\omega }^2}{2\pi G{\rho }_c}\right]\frac{{\xi }^2}{\theta }=\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\frac{{\xi }^2}{\theta }.$

All physically reasonable solutions are subject to the inner boundary condition,

$\frac{dx}{d\xi }=0$         at         $\xi =0,$

but the relevant outer boundary condition depends on whether the underlying equilibrium configuration is isolated (surface pressure is zero), or whether it is a "pressure-truncated" configuration. As is the case with the pressure-truncated isothermal spheres, discussed above, if the polytropic configuration is truncated by the pressure, $P_e$, of a hot, tenuous external medium, then the solution to the LAWE is subject to the outer boundary condition,

$-\frac{d\ln x}{d\ln \xi }=3$         at         $\xi =\tilde{\xi }.$

But, for isolated polytropes, the sought-after solution is subject to the more conventional boundary condition,

$-\frac{d\ln x}{d\ln \xi }=\left(\frac{3-n}{n+1}\right)+\frac{n{\sigma }_c^2}{6(n+1)}\left[\frac{\xi }{{\theta }^{\text{'}}}\right]$         at         $\xi ={\xi }_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}.$

Radial Pulsation Neutral Mode

Background

The integro-differential version of the statement of hydrostatic balance is

$\frac{dP}{dr}=-\frac{G{M}_r\rho }{r^2}$

From our separate discussion, we have found that,

Exact Solution to the $(3\le n<\mathrm{\infty })$ Polytropic LAWE
${\sigma }_c^2=0$	and	$x_P\equiv \frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right].$

Let's rewrite the significant functional term in this expressions in terms of basic variables. That is,

$\left(\frac{1}{\xi {\theta }^n}\right)\frac{d\theta }{d\xi }$	$=$	$-\left(\frac{a_n{\rho }_c}{r_0{\rho }_0}\right)\frac{g_0}{4\pi G{\rho }_c{a}_n}$
	$=$	$-\frac{M(r_0)}{4\pi r_0^3{\rho }_0}.$

Trial Eigenfunction & Its Derivatives

Let's adopt the following trial solution:

$x_t$

$=$

$a-\frac{b{M}_r}{4\pi r_0^3{\rho }_0}=a-\frac{b{g}_0}{4\pi G{r}_0{\rho }_0}.$

Then we have,

$-\left(\frac{1}{b}\right)\frac{d{x}_t}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}\left[\frac{M_r}{4\pi r_0^3{\rho }_0}\right]$
	$=$	$\left[\frac{1}{4\pi r_0^3{\rho }_0}\right]\frac{d{M}_r}{d{r}_0}-\left[\frac{M_r}{4\pi r_0^3{\rho }_0^2}\right]\frac{d{\rho }_0}{d{r}_0}-\left[\frac{3M_r}{4\pi r_0^4{\rho }_0}\right]$
	$=$	$\frac{1}{r_0}-\left[\frac{3M_r}{4\pi r_0^4{\rho }_0}\right]-\left[\frac{M_r}{4\pi r_0^3{\rho }_0^2}\right]\frac{d{\rho }_0}{d{r}_0}$
$-\left(\frac{1}{b}\right)\frac{d^2{x}_t}{d{r}_0^2}$	$=$	$\frac{d}{d{r}_0}\{\frac{1}{r_0}-[\frac{3M_r}{4\pi r_0^4{\rho }_0}]-[\frac{M_r}{4\pi r_0^3{\rho }_0^2}\left]\frac{d{\rho }_0}{d{r}_0}\right\}$
	$=$	$-\frac{1}{r_0^2}-\left[\frac{3}{4\pi r_0^4{\rho }_0}\right]\frac{d{M}_r}{d{r}_0}+\left[\frac{3M_r}{4\pi r_0^4{\rho }_0^2}\right]\frac{d{\rho }_0}{d{r}_0}+4\left[\frac{3M_r}{4\pi r_0^5{\rho }_0}\right]$
		$-\left[\frac{M_r}{4\pi r_0^3{\rho }_0^2}\right]\frac{d^2{\rho }_0}{d{r}_0^2}-\left[\frac{1}{4\pi r_0^3{\rho }_0^2}\right]\frac{d{\rho }_0}{d{r}_0}\cdot \frac{d{M}_r}{d{r}_0}+\left[\frac{3M_r}{4\pi r_0^4{\rho }_0^2}\right]\frac{d{\rho }_0}{d{r}_0}+\left[\frac{2M_r}{4\pi r_0^3{\rho }_0^3}\right](\frac{d{\rho }_0}{d{r}_0}{)}^2$
	$=$	$-\frac{4}{r_0^2}+\frac{3M_r}{4\pi r_0^5{\rho }_0}[4+\frac{d\ln {\rho }_0}{d\ln r_0}]+\frac{1}{r_0^2}[\frac{3M_r}{4\pi r_0^3{\rho }_0}-1]\frac{d\ln {\rho }_0}{d\ln r_0}-\left[\frac{M_r}{4\pi r_0^3{\rho }_0^2}\right]\frac{d^2{\rho }_0}{d{r}_0^2}+\left[\frac{2M_r}{4\pi r_0^5{\rho }_0}\right](\frac{d\ln {\rho }_0}{d\ln r_0}{)}^2.$

Given that,

$\mathrm{\Delta }\equiv \frac{M_r}{4\pi r_0^3{\rho }_0}$

$=$

$\frac{1}{4\pi G}\left(\frac{g_0}{r_0{\rho }_0}\right)=-[\frac{P_0}{4\pi G{r}_0^2{\rho }_0^2}\cdot \frac{d\ln P_0}{d\ln r_0}],$

these expression can be rewritten as,

$-\left(\frac{r_0^2}{b}\right)\frac{d{x}_t}{d{r}_0}$

$=$

$r_0\{1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}\},$

and,

$-\left(\frac{r_0^2}{b}\right)\frac{d^2{x}_t}{d{r}_0^2}$

$=$

$-4+3\mathrm{\Delta }[4+\frac{d\ln {\rho }_0}{d\ln r_0}]+[3\mathrm{\Delta }-1]\frac{d\ln {\rho }_0}{d\ln r_0}-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+2\mathrm{\Delta }(\frac{d\ln {\rho }_0}{d\ln r_0}{)}^2.$

Plug Trial Eigenfunction Into LAWE

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

Plugging our trial radial displacement function, $x_t$, into the LAWE gives,

LAWE	$=$	$-\left(\frac{r_0^2}{b}\right)\frac{d^2{x}_t}{d{r}_0^2}-\left(\frac{r_0}{b}\right)[4+\frac{d\ln P_0}{d\ln r_0}]\frac{d{x}_t}{d{r}_0}-\left(\frac{r_0^2}{b}\right)\left(\frac{{\rho }_0}{{\gamma }_g{P}_0}\right)[(4-3{\gamma }_g)\frac{g_0}{r_0}+{\sigma }_c^2(\frac{2\pi G{\rho }_c}{3}\left)\right]x_t$
	$=$	$-\left(\frac{r_0^2}{b}\right)\frac{d^2{x}_t}{d{r}_0^2}-\left(\frac{r_0}{b}\right)[4+\frac{d\ln P_0}{d\ln r_0}]\frac{d{x}_t}{d{r}_0}+\left(\frac{1}{b}\right)\frac{1}{{\gamma }_g}\left(\frac{d\ln P_0}{d\ln r_0}\right)(4-3{\gamma }_g)x_t-\left(\frac{r_0^2}{b}\right)\left(\frac{{\rho }_0}{{\gamma }_g{P}_0}\right)\left[{\sigma }_c^2\right(\frac{2\pi G{\rho }_c}{3}\left)\right]x_t$
	$=$	$-4+3\mathrm{\Delta }[4+\frac{d\ln {\rho }_0}{d\ln r_0}]+[3\mathrm{\Delta }-1]\frac{d\ln {\rho }_0}{d\ln r_0}-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+2\mathrm{\Delta }(\frac{d\ln {\rho }_0}{d\ln r_0}{)}^2$
		$+[4+\frac{d\ln P_0}{d\ln r_0}]\{1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}\}+\left(\frac{1}{b}\right)\frac{1}{{\gamma }_g}\left(\frac{d\ln P_0}{d\ln r_0}\right)(4-3{\gamma }_g)(a-b\mathrm{\Delta })-\left(\frac{1}{b}\right)\left(\frac{{\rho }_0{r}_0^2}{{\gamma }_g{P}_0}\right)\left[{\sigma }_c^2\right(\frac{2\pi G{\rho }_c}{3}\left)\right](a-b\mathrm{\Delta }).$

Now, if we set ${\sigma }_c^2=0$ and $d\ln P_0/d\ln r_0={\gamma }_g(d\ln {\rho }_0/d\ln r_0)$, this expression becomes,

LAWE	$=$	$-4+3\mathrm{\Delta }[4+\frac{d\ln {\rho }_0}{d\ln r_0}]+[3\mathrm{\Delta }-1]\frac{d\ln {\rho }_0}{d\ln r_0}-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+2\mathrm{\Delta }(\frac{d\ln {\rho }_0}{d\ln r_0}{)}^2$
		$+[4+\frac{d\ln P_0}{d\ln r_0}]\{1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}\}+\left(\frac{1}{b}\right)\frac{1}{{\gamma }_g}\left(\frac{d\ln P_0}{d\ln r_0}\right)(4-3{\gamma }_g)(a-b\mathrm{\Delta })-\left(\frac{1}{b}\right)\left(\frac{{\rho }_0{r}_0^2}{{\gamma }_g{P}_0}\right)\left[{\sigma }_c^2\right(\frac{2\pi G{\rho }_c}{3}\left)\right](a-b\mathrm{\Delta }).$

Now, the key components of this last term can be rewritten as,

$\left(\frac{{\rho }_0{r}_0^2}{{\gamma }_g{P}_0}\right)\left[{\sigma }_c^2\right(\frac{2\pi G{\rho }_c}{3}\left)\right]$	$=$	$\left(\frac{4\pi G{\rho }_0^2{r}_0^2}{P_0}\right)\left[\frac{{\sigma }_c^2}{6{\gamma }_g}\right(\frac{{\rho }_c}{{\rho }_0}\left)\right]$
	$=$	$-\left(\frac{1}{\mathrm{\Delta }}\right)\frac{d\ln P_0}{d\ln r_0}\left[\frac{{\sigma }_c^2}{6{\gamma }_g}\right(\frac{{\rho }_c}{{\rho }_0}\left)\right]$

See Also