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

Notice that the key components of this last term may be rewritten as,

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

So, for our trial eigenfunction, we have …

LAWE

$=$

$[2\mathrm{\Delta }-1]\frac{d\ln {\rho }_0}{d\ln r_0}-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+2\mathrm{\Delta }(\frac{d\ln {\rho }_0}{d\ln r_0}{)}^2+\frac{d\ln P_0}{d\ln r_0}\cdot \{[1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}]+\frac{(4-3{\gamma }_g)}{{\gamma }_g}[\frac{a}{b}-\mathrm{\Delta }]+\frac{{\sigma }_c^2}{6{\gamma }_g}\left(\frac{{\rho }_c}{{\rho }_0}\right)[\frac{a}{b\mathrm{\Delta }}-1]\}.$

Consider Polytropic Structures

Referring back to, for example, a separate review of polytropic structures, we recognize that,

$\mathrm{\Delta }=\frac{1}{4\pi G}\left(\frac{g_0}{r_0{\rho }_0}\right)$	$=$	$\frac{1}{{\xi }^3}\left[\right(-{\xi }^2\frac{d\theta }{d\xi }\left)\right]{\theta }^{-n}=\frac{1}{\xi }(-\frac{d\theta }{d\xi }){\theta }^{-n}=-\frac{{\theta }^{\text{'}}}{\xi {\theta }^n},$
$\frac{d\ln {\rho }_0}{d\ln r_0}$	$=$	$n,$
$\frac{d\ln P_0}{d\ln r_0}$	$=$	$(n+1).$

Also,

$\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}$	$=$	$\left(\frac{{\xi }^2}{{\rho }_c{\theta }^n}\right)\frac{d}{d\xi }\left[n{\rho }_c{\theta }^{n-1}{\theta }^{\text{'}}\right]$
	$=$	$\left(\frac{n{\xi }^2}{{\theta }^n}\right)[(n-1){\theta }^{n-2}({\theta }^{\text{'}}{)}^2+{\theta }^{n-1}{\theta }^{{\text{'}}^{\prime }}]$
	$=$	$\left(\frac{n{\xi }^2}{{\theta }^2}\right)[(n-1)({\theta }^{\text{'}}{)}^2+\theta \cdot {\theta }^{{\text{'}}^{\prime }}]$
	$=$	$\left(\frac{n{\xi }^2}{{\theta }^2}\right)[(n-1)({\theta }^{\text{'}}{)}^2-({\theta }^{n+1}+\frac{2\theta {\theta }^{\text{'}}}{\xi })]$
	$=$	$\left(\frac{n{\xi }^2}{{\theta }^2}\right)[(n-1)(\xi {\theta }^n\mathrm{\Delta }{)}^2+{\theta }^{n+1}]$
	$=$	$n(n-1)({\xi }^{n+1}{\theta }^{n-1}\mathrm{\Delta }{)}^2+n{\xi }^2{\theta }^{n-1}.$

Hence,

LAWE	$=$	$n(2\mathrm{\Delta }-1)-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+2n^2\mathrm{\Delta }+(n+1)[1-3\mathrm{\Delta }-n\mathrm{\Delta }]+(n+1)\left\{\frac{(4-3{\gamma }_g)}{{\gamma }_g}\right[\frac{a}{b}-\mathrm{\Delta }]+\frac{{\sigma }_c^2}{6{\gamma }_g}(\frac{{\rho }_c}{{\rho }_0}\left)\right[\frac{a}{b\mathrm{\Delta }}-1\left]\right\}$
	$=$	$2n\mathrm{\Delta }-n+2n^2\mathrm{\Delta }+n-3n\mathrm{\Delta }-n^2\mathrm{\Delta }+1-3\mathrm{\Delta }-n\mathrm{\Delta }-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+(n+1)\left\{\frac{(4-3{\gamma }_g)}{{\gamma }_g}\right[\frac{a}{b}-\mathrm{\Delta }]+\frac{{\sigma }_c^2}{6{\gamma }_g}(\frac{{\rho }_c}{{\rho }_0}\left)\right[\frac{a}{b\mathrm{\Delta }}-1\left]\right\}$
	$=$	$1+n^2\mathrm{\Delta }-(2n+3)\mathrm{\Delta }-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+(n+1)[\frac{4}{{\gamma }_g}-3][\frac{a}{b}-\mathrm{\Delta }]+(n+1)\left\{\frac{{\sigma }_c^2}{6{\gamma }_g}\right(\frac{{\rho }_c}{{\rho }_0}\left)\right[\frac{a}{b\mathrm{\Delta }}-1\left]\right\}.$

If, ${\gamma }_g=(n+1)/n$, we can further simplify and obtain,

LAWE	$=$	$1+n^2\mathrm{\Delta }-(2n+3)\mathrm{\Delta }-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+[n-3][\frac{a}{b}-\mathrm{\Delta }]+(n+1)\left\{\frac{{\sigma }_c^2}{6{\gamma }_g}\right(\frac{{\rho }_c}{{\rho }_0}\left)\right[\frac{a}{b\mathrm{\Delta }}-1\left]\right\}$
	$=$	$1+(n-3)\frac{a}{b}+n^2\mathrm{\Delta }+(3-n)\mathrm{\Delta }-(2n+3)\mathrm{\Delta }-\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}+(n+1)\left\{\frac{{\sigma }_c^2}{6{\gamma }_g}\right(\frac{{\rho }_c}{{\rho }_0}\left)\right[\frac{a}{b\mathrm{\Delta }}-1\left]\right\}$

Try Again

General Form of Wave Equation

 

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

Employing the substitutions,

${\sigma }_c^2$	$\equiv$	$\frac{3{\omega }^2}{2\pi G{\rho }_c},$
$\alpha$	$\equiv$	$3-\frac{4}{{\gamma }_g}=\frac{3-n}{n+1},$
$g_0$	$\equiv$	$\frac{G{M}_r}{r_0^2}=-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0}\Rightarrow \frac{g_0{\rho }_0{r}_0}{P_0}=-\frac{d\ln P_0}{d\ln r_0},$
$\mathrm{\Delta }$	$\equiv$	$\frac{M_r}{4\pi r_0^3{\rho }_0}=\frac{1}{4\pi G}\left(\frac{g_0}{r_0{\rho }_0}\right)=-[\frac{P_0}{4\pi G{r}_0^2{\rho }_0^2}\cdot \frac{d\ln P_0}{d\ln r_0}],$

we have,

LAWE	$=$	$\frac{d^{2}x}{d{r}_0^2}+\frac{1}{r_0}[4-\frac{g_0{\rho }_0{r}_0}{P_0}]\frac{dx}{d{r}_0}+\left[\right(\frac{4}{{\gamma }_g}-3\left)\frac{g_0{\rho }_0{r}_0}{P_0}\right]\frac{x}{r_0^2}+\left(\frac{{\rho }_0}{P_0}\right)\left[4\pi G{\rho }_c\right(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]x$
	$=$	$\frac{d^{2}x}{d{r}_0^2}+\frac{1}{r_0}[4+\frac{d\ln P_0}{d\ln r_0}]\frac{dx}{d{r}_0}+[\alpha \cdot \frac{d\ln P_0}{d\ln r_0}]\frac{x}{r_0^2}-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]\frac{x}{r_0^2}.$

In the context of polytropic configurations (see more below), we appreciate that, $\frac{{\rho }_0}{{\rho }_c}$ $=$ ${\theta }^n,$ $\frac{d\ln P_0}{d\ln r_0}$ $=$ $(n+1)\frac{d\ln \theta }{d\ln \xi }=-(n+1)Q,$      and, $\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$ $=$ $-(n+1){\xi }^2{\theta }^{n-1}.$ Inserting these into the LAWE expression and multiplying through by the square of the polytropic length scale, $a_n^2$, we obtain, LAWE $=$ $\frac{d^{2}x}{d{\xi }^2}+\frac{1}{\xi }[4-(n+1)Q]\frac{dx}{d\xi }-[\alpha (n+1)Q]\frac{x}{{\xi }^2}+[(n+1)\frac{{\xi }^2}{\theta }(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]\frac{x}{{\xi }^2}.$ This is identical to what has been referred to in a separate discussion, as the Polytropic LAWE $0=\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$ 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,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$

Derivatives of Δ

Here we evaluate the first derivative of $\mathrm{\Delta }$ with respect to $r_0$,

$\frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}\left\{\frac{M_r}{4\pi r_0^3{\rho }_0}\right\}$
	$=$	$\frac{1}{4\pi r_0^3{\rho }_0}\cdot \frac{d{M}_r}{d{r}_0}-\frac{3M_r}{4\pi r_0^4{\rho }_0}-\frac{M_r}{4\pi r_0^3{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}$
	$=$	$\frac{1}{r_0}-\frac{3M_r}{4\pi r_0^4{\rho }_0}-\frac{M_r}{4\pi r_0^3{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}$
	$=$	$\frac{1}{r_0}\{1-\mathrm{\Delta }[3+\frac{d\ln {\rho }_0}{d\ln r_0}\left]\right\}$
$\Rightarrow r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$[1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}];$

and the second derivative of $\mathrm{\Delta }$ with respect to $r_0$,

$\frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$\frac{d}{d{r}_0}\left\{\frac{1}{r_0}\right\}-\frac{d}{d{r}_0}\left\{\frac{3M_r}{4\pi r_0^4{\rho }_0}\right\}-\frac{d}{d{r}_0}\{\frac{M_r}{4\pi r_0^3{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}\}$
	$=$	$-\frac{1}{r_0^2}-\frac{3}{4\pi }\{\frac{1}{r_0^4{\rho }_0}\cdot \frac{d{M}_r}{d{r}_0}-\frac{4M_r}{r_0^5{\rho }_0}-\frac{M_r}{r_0^4{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}\}$
		$-\frac{1}{4\pi }\{\frac{1}{r_0^3{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}\cdot \frac{d{M}_r}{d{r}_0}-\frac{3M_r}{r_0^4{\rho }_0^2}\cdot \frac{d{\rho }_0}{d{r}_0}-\frac{2M_r}{r_0^3{\rho }_0^3}\cdot [\frac{d{\rho }_0}{d{r}_0}{]}^2+\frac{M_r}{r_0^3{\rho }_0^2}\cdot \frac{d^2{\rho }_0}{d{r}_0^2}\}$
	$=$	$-\frac{1}{r_0^2}-\frac{3}{r_0^2}\{1-\mathrm{\Delta }[4+\frac{d\ln {\rho }_0}{d\ln r_0}\left]\right\}-\frac{1}{r_0^2}\left\{\right[1-3\mathrm{\Delta }]\frac{d\ln {\rho }_0}{d\ln r_0}-2\mathrm{\Delta }\cdot [\frac{d\ln {\rho }_0}{d\ln r_0}{]}^2+\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}\}$
$\Rightarrow -r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$1+3-3\mathrm{\Delta }[4+\frac{d\ln {\rho }_0}{d\ln r_0}]+[1-3\mathrm{\Delta }]\frac{d\ln {\rho }_0}{d\ln r_0}-2\mathrm{\Delta }\cdot [\frac{d\ln {\rho }_0}{d\ln r_0}{]}^2+\mathrm{\Delta }(\frac{r_0^2}{{\rho }_0})\frac{d^2{\rho }_0}{d{r}_0^2}$
	$=$	$4-12\mathrm{\Delta }+[1-6\mathrm{\Delta }]\frac{d\ln {\rho }_0}{d\ln r_0}-2\mathrm{\Delta }\cdot [\frac{d\ln {\rho }_0}{d\ln r_0}{]}^2+\mathrm{\Delta }(\frac{r_0^2}{{\rho }_0})\frac{d^2{\rho }_0}{d{r}_0^2}.$

Trial Eigenfunction

As above, let's adopt a trial eigenfunction of the form,

$x_t$

$=$

$a-\frac{b{M}_r}{4\pi r_0^3{\rho }_0}=a-b\mathrm{\Delta }.$

Then we have,

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{]}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}-[4+\frac{d\ln P_0}{d\ln r_0}]r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}+[\alpha \cdot \frac{d\ln P_0}{d\ln r_0}](\frac{a}{b}-\mathrm{\Delta })-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$\{4-12\mathrm{\Delta }+[1-6\mathrm{\Delta }]\frac{d\ln {\rho }_0}{d\ln r_0}-2\mathrm{\Delta }\cdot [\frac{d\ln {\rho }_0}{d\ln r_0}{]}^2+\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}\}-[4+\frac{d\ln P_0}{d\ln r_0}\left]\right[1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}]$
		$+[\alpha \cdot \frac{d\ln P_0}{d\ln r_0}](\frac{a}{b}-\mathrm{\Delta })-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta }).$

Assume Polytropic Relations

If we assume that the equilibrium models are polytropes, then we know that,

${\rho }_0\propto {\theta }^n$	$\Rightarrow$	$\frac{d\ln {\rho }_0}{d\ln r_0}=n\cdot \frac{d\ln \theta }{d\ln \xi };$
$P_0\propto {\theta }^{n+1}$	$\Rightarrow$	$\frac{d\ln P_0}{d\ln r_0}=(n+1)\cdot \frac{d\ln \theta }{d\ln \xi }.$

We also deduce that,

$\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}$	$=$	$\left(\frac{{\xi }^2}{{\theta }^n}\right)\frac{d}{d\xi }\left[\frac{d{\theta }^n}{d\xi }\right]$
	$=$	$\left(\frac{{\xi }^2}{{\theta }^n}\right)\frac{d}{d\xi }\left[n{\theta }^{n-1}{\theta }^{\text{'}}\right]$
	$=$	$\left(\frac{n{\xi }^2}{{\theta }^n}\right)[(n-1){\theta }^{n-2}({\theta }^{\text{'}}{)}^2+{\theta }^{n-1}{\theta }^{{\text{'}}^{\prime }}]$
	$=$	$\left[\frac{n(n-1){\xi }^2}{{\theta }^2}\right]({\theta }^{\text{'}}{)}^2-(\frac{n{\xi }^2}{\theta }\left)\right[{\theta }^n+\frac{2}{\xi }{\theta }^{\text{'}}]$
	$=$	$n(n-1)\cdot [\frac{d\ln \theta }{d\ln \xi }{]}^2-n{\xi }^2{\theta }^{n-1}-2n\cdot \frac{d\ln \theta }{d\ln \xi }$
	$=$	$n(n-1)\cdot {\mathrm{\Xi }}^2-n{\xi }^2{\theta }^{n-1}-2n\cdot \mathrm{\Xi },$

where we have introduced the shorthand notation,

$\mathrm{\Xi }$

$\equiv$

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

Drawing from our accompanying discussion, for example, we note as well that,

$\mathrm{\Delta }$	$=$	$-[\frac{P_0}{4\pi G{r}_0^2{\rho }_0^2}\cdot \frac{d\ln P_0}{d\ln r_0}]$
$\Rightarrow -\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$	$=$	$\frac{4\pi G{r}_0^2{\rho }_0^2}{P_0}$
	$=$	$4\pi G(a_n^2{\xi }^2)({\rho }_c{\theta }^n{)}^2[K^{-1}{\rho }_c^{-(n+1)/n}{\theta }^{-(n+1)}]$
	$=$	$4\pi G[\frac{(n+1)K}{4\pi G}\cdot {\rho }_c^{(1-n)/n}]({\rho }_c{)}^2[K^{-1}{\rho }_c^{-(n+1)/n}]{\xi }^2{\theta }^{(n-1)}$
	$=$	$(n+1){\xi }^2{\theta }^{(n-1)}$
$\Rightarrow \mathrm{\Delta }\cdot {\xi }^2{\theta }^{n-1}$	$=$	$-\frac{1}{(n+1)}\cdot \frac{d\ln P_0}{d\ln r_0}=-\mathrm{\Xi }.$

Hence,

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{]}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	$4-12\mathrm{\Delta }+[1-6\mathrm{\Delta }]n\mathrm{\Xi }-2\mathrm{\Delta }\cdot [n\mathrm{\Xi }{]}^2+\mathrm{\Delta }[n(n-1)\cdot {\mathrm{\Xi }}^2-n{\xi }^2{\theta }^{n-1}-2n\cdot \mathrm{\Xi }]-[4+(n+1)\mathrm{\Xi }\left]\right[1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot n\mathrm{\Xi }]$
		$+[\alpha \cdot (n+1)\mathrm{\Xi }](\frac{a}{b}-\mathrm{\Delta })-\frac{1}{\mathrm{\Delta }}[(n+1)\mathrm{\Xi }\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$n\mathrm{\Xi }-6\mathrm{\Delta }n\mathrm{\Xi }-2\mathrm{\Delta }\cdot [n\mathrm{\Xi }{]}^2+\mathrm{\Delta }[n(n-1)\cdot {\mathrm{\Xi }}^2-n{\xi }^2{\theta }^{n-1}+2n\cdot \mathrm{\Xi }]+2(n+1)\mathrm{\Xi }+n(n+1){\mathrm{\Xi }}^2\cdot \mathrm{\Delta }$
		$+[\alpha \cdot (n+1)\mathrm{\Xi }]\left(\frac{a}{b}\right)-[\alpha \cdot (n+1)\mathrm{\Xi }]\mathrm{\Delta }-\frac{1}{\mathrm{\Delta }}[(n+1)\mathrm{\Xi }\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$\mathrm{\Xi }[1+n+2(n+1)+\alpha \cdot (n+1)(\frac{a}{b}\left)\right]+\mathrm{\Delta }[-[4n+\alpha \cdot (n+1)]\mathrm{\Xi }+n(n-1)\cdot {\mathrm{\Xi }}^2+n(n+1){\mathrm{\Xi }}^2-2n^2{\mathrm{\Xi }}^2]$
		$-\frac{1}{\mathrm{\Delta }}[(n+1)\mathrm{\Xi }\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$\mathrm{\Xi }[3(n+1)+\alpha \cdot (n+1)(\frac{a}{b}\left)\right]-\mathrm{\Delta }[4n+\alpha \cdot (n+1)]\mathrm{\Xi }-\frac{1}{\mathrm{\Delta }}[(n+1)\mathrm{\Xi }\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$

Third Time

General Relations

Various general relations taken from above derivations:

LAWE

$=$

$\frac{d^{2}x}{d{r}_0^2}+\frac{1}{r_0}[4+\frac{d\ln P_0}{d\ln r_0}]\frac{dx}{d{r}_0}+[\alpha \cdot \frac{d\ln P_0}{d\ln r_0}]\frac{x}{r_0^2}-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]\frac{x}{r_0^2},$

where,

$\mathrm{\Delta }$	$\equiv$	$\frac{M_r}{4\pi r_0^3{\rho }_0}$
$\Rightarrow \frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$	$=$	$-\frac{4\pi G{r}_0^2{\rho }_0^2}{P_0};$
$r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$[1-3\mathrm{\Delta }-\mathrm{\Delta }\cdot \frac{d\ln {\rho }_0}{d\ln r_0}];$
$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$4-12\mathrm{\Delta }+[1-6\mathrm{\Delta }]\frac{d\ln {\rho }_0}{d\ln r_0}-2\mathrm{\Delta }\cdot [\frac{d\ln {\rho }_0}{d\ln r_0}{]}^2+\mathrm{\Delta }(\frac{r_0^2}{{\rho }_0})\frac{d^2{\rho }_0}{d{r}_0^2}.$

Polytropes

If polytropic relations are adopted:

$\mathrm{\Delta }$	$=$	$\frac{Q}{{\xi }^2{\theta }^{n-1}};$
$\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$	$=$	$-(n+1){\xi }^2{\theta }^{n-1};$
$r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$r_0^2\frac{d^{2}x}{d{r}_0^2}+[4-(n+1)Q]r_0\cdot \frac{dx}{d{r}_0}+[(n-3)Q]x+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]x;$
$r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$[1-3\mathrm{\Delta }+n\mathrm{\Delta }Q]=1+(nQ-3)\mathrm{\Delta };$
$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$4-12\mathrm{\Delta }-n[1-6\mathrm{\Delta }]Q-2n^2\mathrm{\Delta }{Q}^2+\mathrm{\Delta }\left(\frac{r_0^2}{{\rho }_0}\right)\frac{d^2{\rho }_0}{d{r}_0^2}$
	$=$	$4-12\mathrm{\Delta }-n[1-6\mathrm{\Delta }]Q-2n^2\mathrm{\Delta }{Q}^2+\mathrm{\Delta }[n(n-1)Q^2+2nQ-\frac{nQ}{\mathrm{\Delta }}]$
	$=$	$4-nQ-12\mathrm{\Delta }+6nQ\mathrm{\Delta }-2n^2\mathrm{\Delta }{Q}^2+\mathrm{\Delta }[n(n-1)Q^2+2nQ]-nQ$
	$=$	$4-2nQ+\mathrm{\Delta }[6nQ-2n^2{Q}^2+n(n-1)Q^2+2nQ-12].$
	$=$	$4-2nQ+\mathrm{\Delta }[8nQ-n^2{Q}^2-n{Q}^2-12].$

Eigenfunction Choice

Again, let's try the trial eigenfunction,

$x_t$

$=$

$a-b\mathrm{\Delta },$

in which case,

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-r_0^2\frac{d^2\mathrm{\Delta }}{d{r}_0^2}-[4-(n+1)Q]r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}+[(n-3)Q](\frac{a}{b}-\mathrm{\Delta })+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$4-2nQ+\mathrm{\Delta }[8nQ-n^2{Q}^2-n{Q}^2-12]-[4-(n+1)Q][1+(nQ-3)\mathrm{\Delta }]+[(n-3)Q](\frac{a}{b}-\mathrm{\Delta })$
		$+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$4-2nQ+\mathrm{\Delta }[8nQ-n^2{Q}^2-n{Q}^2-12]-4-(4nQ-12)\mathrm{\Delta }+(n+1)Q+(n+1)(n{Q}^2-3Q)\mathrm{\Delta }+(n-3)Q\left(\frac{a}{b}\right)-(n-3)Q\mathrm{\Delta }$
		$+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$[(n+1)-2n+(n-3)(\frac{a}{b}\left)\right]Q+\mathrm{\Delta }[8nQ-n^2{Q}^2-n{Q}^2-12-4nQ+12+(n+1)(n{Q}^2)-3Q(n+1)+(3-n)Q]$
		$+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$[(1-n)+(n-3)(\frac{a}{b}\left)\right]Q+\mathrm{\Delta }\left[0\right]+\frac{1}{\mathrm{\Delta }}[(n+1)Q\cdot \frac{1}{{\theta }^n}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta }).$

Hence, we are left with only the ${\sigma }_c^2$ term if we set,

$0$	$=$	$(1-n)+(n-3)\left(\frac{a}{b}\right)$
$\Rightarrow \left(\frac{a}{b}\right)$	$=$	$\frac{n-1}{n-3}.$

We conclude, therefore, that the radial displacement function (i.e., the eigenfunction) for the neutral $({\sigma }_c^2=0)$ mode of all polytropic configurations is,

$x_{\mathrm{n}\mathrm{e}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}$	$=$	$\frac{1}{b}[\frac{n-1}{n-3}-\mathrm{\Delta }]=\frac{1}{b}[\frac{n-1}{n-3}-\frac{Q}{{\xi }^2{\theta }^{n-1}}]$
	$=$	$\frac{1}{b}\left(\frac{n-1}{n-3}\right)[1+(\frac{n-3}{n-1})\frac{1}{{\xi }^2{\theta }^{n-1}}\cdot \frac{d\ln \theta }{d\ln \xi }]$
	$=$	$\frac{1}{b}\left(\frac{n-1}{n-3}\right)[1+(\frac{n-3}{n-1})\frac{1}{\xi {\theta }^n}\cdot \frac{d\theta }{d\xi }].$

This last expression exactly matches our earlier result found for polytropic configurations if we choose an overall amplitude coefficient of the form,

$b$

$=$

$\frac{2n}{3(n-3)}.$

Hooray!

See Also