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!

Summary

Setup

We begin with the traditional,

Adiabatic Wave (or Radial Pulsation) Equation

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

This linear, adiabatic wave equation (LAWE) can straightforwardly be rewritten in the form we will refer to as the,

Δ-Highlighted LAWE

	$0=\frac{d^{2}x}{d{r}_0^2}+\frac{1}{r_0}[4+\frac{d\ln P_0}{d\ln r_0}]\frac{dx}{d{r}_0}+\left[\right(3-\frac{4}{{\gamma }_g}\left)\frac{d\ln P_0}{d\ln r_0}\right]\frac{x}{r_0^2}-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]\frac{x}{r_0^2}$
where:    $\mathrm{\Delta }\equiv \frac{M_r}{4\pi r_0^3{\rho }_0},$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c}$

Multiplying this Δ-Highlighted LAWE through by $a_n^2=(r_0/\xi {)}^2$ and recognizing that, for polytropic configurations,

$\mathrm{\Delta }$

$=$

$-\frac{1}{{\xi }^2{\theta }^{n-1}}\frac{d\ln \theta }{d\ln \xi },$

$\frac{d\ln P_0}{d\ln r_0}$

$=$

$(n+1)\frac{d\ln \theta }{d\ln \xi },$

$\Rightarrow$

$\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$

$=$

$-(n+1){\xi }^2{\theta }^{n-1},$

we immediately obtain what we have frequently referred to 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,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$

Neutral-Mode Eigenfunction

In the preceding subsections of this chapter, we have demonstrated that if

${\gamma }_g=\frac{n+1}{n}$

$\Rightarrow$

$\alpha =\frac{3-n}{n+1},$

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

$x_{\mathrm{n}\mathrm{e}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}$

$=$

$\frac{3(n-3)}{2n}[\frac{n-1}{n-3}-\mathrm{\Delta }],$

to within an arbitrarily chosen leading scaling coefficient. More completely, if we let "LAWE" stand for the RHS of our Δ-Highlighted LAWE, then setting $x=x_{\mathrm{n}\mathrm{e}\mathrm{u}\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{l}}$ results in the expression,

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$(n+1)Q\cdot \frac{1}{{\theta }^n}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left[\frac{1}{\mathrm{\Delta }}\right(\frac{n-1}{n-3})-1]$
	$=$	$-(n+1)\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left[\frac{{\xi }^2}{\theta }\right(\frac{n-1}{n-3})-\frac{1}{{\theta }^n}\cdot \frac{d\ln \theta }{d\ln \xi }],$

which goes to zero if ${\sigma }_c^2=0$.

Parabolic Density Distribution

Here, we build upon our separate discussion of equilibrium configurations with a parabolic density distribution.

Equilibrium Structure

In an article titled, "Radial Oscillations of a Stellar Model," C. Prasad (1949, MNRAS, 109, 103) investigated the properties of an equilibrium configuration with a prescribed density distribution given by the expression,

${\rho }_0={\rho }_c[1-(\frac{r_0}{R}{)}^2],$

where, ${\rho }_c$ is the central density and, $R$ is the radius of the star. Both the mass distribution and the pressure distribution can be obtained analytically from this specified density distribution. Specifically,

$M_r$	$=$	$\underset{0}{\overset{r_0}{\int }}4\pi r_0^2{\rho }_{0}d{r}_0$
	$=$	$\frac{4\pi {\rho }_c{r}_0^3}{3}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2],$

in which case we can write,

$g_0\equiv \frac{G{M}_r}{r_0^2}$

$=$

$\frac{4\pi G{\rho }_c{r}_0}{3}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2],$

and,

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

$=$

$\frac{1}{3}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-(\frac{r_0}{R}{)}^2{]}^{-1}.$

Hence, proceeding via what we have labeled as "Technique 1", and enforcing the surface boundary condition, $P(R)=0$, Prasad (1949) determines that,

$P_0$

$=$

$\frac{4\pi G{\rho }_c^2{R}^2}{15}[1-(\frac{r_0}{R}{)}^2{]}^2[1-\frac{1}{2}(\frac{r_0}{R}{)}^2],$

where, it can readily be deduced, as well, that the central pressure is,

$P_c=\frac{4\pi }{15}G{\rho }_c^2{R}^2.$

Specific Entropy Distribution For purposes of later discussion, we find from a separate examination of specific entropy distributions, $s_0(r_0)$, that, $\frac{s_0}{\mathrm{\Re }/\overline{\mu }}$ $=$ $\frac{1}{({\gamma }_g-1)}\ln (\frac{{\tau }_0}{{\rho }_0}{)}^{{\gamma }_g}=\frac{1}{({\gamma }_g-1)}\ln [\frac{P_0}{({\gamma }_g-1){\rho }_0^{{\gamma }_g}}]$ $\Rightarrow \left[\frac{{\gamma }_g-1}{\mathrm{\Re }/\overline{\mu }}\right]s_0$ $=$ $\ln \left[\frac{P_c}{({\gamma }_g-1){\rho }_c^{{\gamma }_g}}\right]+\ln \left\{\right[1-\left(\frac{r_0}{R}{)}^2{]}^2\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]\}+\ln \{[1-(\frac{r_0}{R}{)}^2{]}^{-{\gamma }_g}\}$   $=$ $\ln \left[\frac{P_c}{({\gamma }_g-1){\rho }_c^{{\gamma }_g}}\right]+\ln \left\{\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]\}+(2-{\gamma }_g)\ln \{[1-(\frac{r_0}{R}{)}^2\left]\right\}.$ Notice that, independent of the value of ${\gamma }_g$, the specific entropy varies with $r_0$ throughout the structure. According to the Schwarzschild criterion, spherically symmetric equilibrium configurations will be stable against convection if the specific entropy increases outward, and unstable toward convection if the specific entropy decreases outward. Let's examine the slope, $d{s}_0/d{r}_0$, throughout configurations that have a parabolic density distribution. $\left[\frac{{\gamma }_g-1}{\mathrm{\Re }/\overline{\mu }}\right]\frac{d{s}_0}{d{r}_0}$ $=$ $-\frac{r_0}{R^2}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}-\frac{2(2-{\gamma }_g)r_0}{R^2}[1-(\frac{r_0}{R}{)}^2{]}^{-1}$ $\Rightarrow \frac{R^2}{r_0}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2\right]\left[\frac{{\gamma }_g-1}{\mathrm{\Re }/\overline{\mu }}\right]\frac{d{s}_0}{d{r}_0}$ $=$ $-[1-(\frac{r_0}{R}{)}^2]-2(2-{\gamma }_g)[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]$   $=$ $(2{\gamma }_g-5)+(3-{\gamma }_g)\left[\right(\frac{r_0}{R}{)}^2]$ The slope is zero when, $(\frac{r_0}{R}{)}^2$ $=$ $\frac{5-2{\gamma }_g}{3-{\gamma }_g}.$ Moving from the center of the configuration to its surface, $0<(r_0/R{)}^2<1$, the slope will go to zero — hence, the slope of the entropy will change sign

Some Relevant Structural Derivatives

We note for later use that,

$\frac{1}{P_c}\cdot \frac{d{P}_0}{d{r}_0}$	$=$	$[1-(\frac{r_0}{R}{)}^2{]}^2\frac{d}{d{r}_0}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2]+[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]\frac{d}{d{r}_0}[1-(\frac{r_0}{R}{)}^2{]}^2$
	$=$	$-\frac{1}{2R^2}[1-(\frac{r_0}{R}{)}^2{]}^2\frac{d}{d{r}_0}\left[r_0^2\right]+2[1-(\frac{r_0}{R}{)}^2\left]\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]\frac{d}{d{r}_0}[1-(\frac{r_0}{R}{)}^2]$
	$=$	$\{-\frac{1}{2R^2}[1-(\frac{r_0}{R}{)}^2{]}^2-\frac{2}{R^2}[1-\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2\left]\right\}\frac{d}{d{r}_0}\left[r_0^2\right]$
	$=$	$-\frac{r_0}{R^2}\left\{\right[1-(\frac{r_0}{R}{)}^2{]}^2+4[1-\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2\left]\right\}$
	$=$	$-\frac{r_0}{R^2}\left\{\right[1-\left(\frac{r_0}{R}{)}^2\right]+4[1-\frac{1}{2}(\frac{r_0}{R}{)}^2\left]\right\}[1-(\frac{r_0}{R}{)}^2]$
	$=$	$-\frac{5r_0}{R^2}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2\right].$

Checking for detailed force-balance, we note that, $-\frac{1}{{\rho }_0}\cdot \frac{d{P}_0}{d{r}_0}$ $=$ $\frac{5r_0}{R^2}\left[\frac{4\pi G{\rho }_c^2{R}^2}{15}\right][1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2\right]\cdot \frac{1}{{\rho }_c}[1-(\frac{r_0}{R}{)}^2{]}^{-1}$   $=$ $\frac{4\pi G{\rho }_c{r}_0}{3}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2],$ which is exactly the expression that we have just derived for $g_0=G{M}_r/r_0^2$.

Hence,

$\frac{d\ln P_0}{d\ln r_0}=\frac{r_0}{P_0/P_c}[\frac{1}{P_c}\cdot \frac{d{P}_0}{d{r}_0}]$	$=$	$-\frac{5r_0^2}{R^2}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2\right]\left\{\right[1-\left(\frac{r_0}{R}{)}^2{]}^2\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]{\}}^{-1}$
	$=$	$-5\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-1}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1};$

and, given that,

${\mathrm{\Delta }}^{-1}$

$=$

$3[1-\frac{3}{5}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2],$

we can write,

$\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}$	$=$	$-15\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-1}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}[1-\frac{3}{5}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2]$
	$=$	$-15\left(\frac{r_0}{R}{)}^2\right[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}.$

Also,

$3\cdot \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$\frac{d}{d{r}_0}\left\{\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-1}\}$
	$=$	$[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\frac{d}{d{r}_0}\right[1-(\frac{r_0}{R}{)}^2{]}^{-1}+[1-\left(\frac{r_0}{R}{)}^2{]}^{-1}\frac{d}{d{r}_0}\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right]$
	$=$	$[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-(\frac{r_0}{R}{)}^2{]}^{-2}\frac{2r_0}{R^2}+[1-\left(\frac{r_0}{R}{)}^2{]}^{-1}\right[-\frac{6}{5}\cdot \frac{r_0}{R^2}]$
	$=$	$\frac{r_0}{R^2}\left\{2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-2}-\frac{6}{5}[1-(\frac{r_0}{R}{)}^2{]}^{-1}\}$
	$=$	$\frac{r_0}{5R^2}[1-(\frac{r_0}{R}{)}^2{]}^{-2}\left\{\right[10-6\left(\frac{r_0}{R}{)}^2\right]-[6-6(\frac{r_0}{R}{)}^2\left]\right\}$
	$=$	$\frac{4r_0}{5R^2}[1-(\frac{r_0}{R}{)}^2{]}^{-2}$
$\Rightarrow \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$\frac{4r_0}{15R^2}[1-(\frac{r_0}{R}{)}^2{]}^{-2};$

and,

$\frac{15R^2}{4}\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$[1-(\frac{r_0}{R}{)}^2{]}^{-2}-2r_0[1-(\frac{r_0}{R}{)}^2{]}^{-3}[-\frac{2r_0}{R^2}]$
	$=$	$[1-(\frac{r_0}{R}{)}^2{]}^{-3}\left\{\right[1-\left(\frac{r_0}{R}{)}^2\right]+\frac{4r_0^2}{R^2}\}$
	$=$	$[1-(\frac{r_0}{R}{)}^2{]}^{-3}[1+3(\frac{r_0}{R}{)}^2].$

Neutral Mode

Again, adopting the trial eigenfunction,

$x_t$

$=$

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

from the,

Δ-Highlighted LAWE

	$0=\frac{d^{2}x}{d{r}_0^2}+\frac{1}{r_0}[4+\frac{d\ln P_0}{d\ln r_0}]\frac{dx}{d{r}_0}+\left[\right(3-\frac{4}{{\gamma }_g}\left)\frac{d\ln P_0}{d\ln r_0}\right]\frac{x}{r_0^2}-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right]\frac{x}{r_0^2}$
where:    $\mathrm{\Delta }\equiv \frac{M_r}{4\pi r_0^3{\rho }_0},$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c}$

we can write,

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

$=$

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

First Attempt

$\Rightarrow \frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}-r_0[4+\frac{d\ln P_0}{d\ln r_0}]\frac{d\mathrm{\Delta }}{d{r}_0}+\alpha \left[\frac{d\ln P_0}{d\ln r_0}\right](\frac{a}{b}-\mathrm{\Delta })-\frac{1}{\mathrm{\Delta }}[\frac{d\ln P_0}{d\ln r_0}\cdot \frac{{\rho }_c}{{\rho }_0}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right](\frac{a}{b}-\mathrm{\Delta })$
	$=$	$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$
		$-r_0[4+\frac{d\ln P_0}{d\ln r_0}]\frac{d\mathrm{\Delta }}{d{r}_0}$
		$+\alpha \left[\frac{d\ln P_0}{d\ln r_0}\right](\frac{a}{b}-\mathrm{\Delta })$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$-\frac{4}{15}\left(\frac{r_0}{R}{)}^2\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-3}\right[1+3\left(\frac{r_0}{R}{)}^2\right]$
		$-\frac{16}{15}\left(\frac{r_0}{R}{)}^2\right[1-(\frac{r_0}{R}{)}^2{]}^{-2}-\frac{4}{15}(\frac{r_0}{R}{)}^2\{-5(\frac{r_0}{R}{)}^2[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-1}\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2{]}^{-1}\right\}[1-(\frac{r_0}{R}{)}^2{]}^{-2}$
		$+\alpha \{-5(\frac{r_0}{R}{)}^2[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-1}\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2{]}^{-1}\right\}\{\frac{a}{b}-\mathrm{\Delta }\}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$-\frac{4}{3}\left(\frac{r_0}{R}{)}^2\right[1-\frac{1}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-3}$
		$+\frac{4}{3}\left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2{]}^{-3}$
		$+\frac{5\alpha }{3}\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-2}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}\{1-\frac{3a}{b}-\frac{3}{5}(\frac{r_0}{R}{)}^2\}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$-\frac{4}{3}\left(\frac{r_0}{R}{)}^2\right[1-\frac{1}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-3}+\frac{5\alpha }{3}\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-2}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}(1-\frac{3a}{b})$
		$+\frac{4}{3}\left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2{]}^{-3}-\alpha \left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-(\frac{r_0}{R}{)}^2{]}^{-2}[1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$

Continuing …

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$\left(\frac{r_0}{R}{)}^2\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-3}\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2{]}^{-1}\right\{\frac{5\alpha }{3}(1-\frac{3a}{b})[1-\frac{3}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\left(\frac{r_0}{R}{)}^2\right]-\frac{4}{3}[1-\frac{1}{5}(\frac{r_0}{R}{)}^2\left]\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2\right]\}$
		$+(\frac{4}{3}-\alpha )\left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2{]}^{-3}$
		$-\alpha \left(\frac{r_0}{R}{)}^6\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right][1-\frac{1}{2}(\frac{r_0}{R}{)}^2{]}^{-1}[1-(\frac{r_0}{R}{)}^2{]}^{-3}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$\left(\frac{r_0}{R}{)}^2\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2{]}^{-1}\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-3}\right\{\frac{5\alpha }{3}(1-\frac{3a}{b})[1-\frac{8}{5}(\frac{r_0}{R}{)}^2+\frac{3}{5}\left(\frac{r_0}{R}{)}^4\right]-\frac{4}{3}[1-\frac{7}{5}(\frac{r_0}{R}{)}^2+\frac{1}{10}\left(\frac{r_0}{R}{)}^4\right]$
		$+(\frac{4}{3}-\alpha )\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right]-\alpha \left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right]\}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$
	$=$	$\left(\frac{r_0}{R}{)}^2\right[1-\frac{1}{2}\left(\frac{r_0}{R}{)}^2{]}^{-1}\right[1-\left(\frac{r_0}{R}{)}^2{]}^{-3}\right\{\left[\frac{5\alpha }{3}\right(1-\frac{3a}{b})-\frac{4}{3}]$
		$+\left(\frac{r_0}{R}{)}^2\right[-\frac{8\alpha }{3}(1-\frac{3a}{b})+\frac{28}{15}]+(\frac{r_0}{R}{)}^4\left[\alpha \right(1-\frac{3a}{b})-\frac{2}{15}]$
		$+(\frac{4}{3}-\alpha )\left(\frac{r_0}{R}{)}^2\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right]-\alpha \left(\frac{r_0}{R}{)}^4\right[1-\frac{3}{5}\left(\frac{r_0}{R}{)}^2\right]\}$
		$+\frac{{\rho }_c}{{\rho }_0}\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\left\{\right[\frac{d\ln P_0}{d\ln r_0}]-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\cdot \frac{d\ln P_0}{d\ln r_0}\left]\right\}$

Second Attempt

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$

$=$

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

where,

$\frac{{\rho }_0}{{\rho }_c}$	$=$	$(1-x^2),$
$\mathrm{\Delta }$	$=$	$\frac{1}{3}(1-\frac{3}{5}{x}^2)(1-x^2{)}^{-1},$
$\frac{d\ln P_0}{d\ln r_0}$	$=$	$-5x^2(1-\frac{3}{5}{x}^2)(1-x^2{)}^{-1}(1-\frac{1}{2}{x}^2{)}^{-1}=-15x^2(1-\frac{1}{2}{x}^2{)}^{-1}\mathrm{\Delta },$
$r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}$	$=$	$\frac{4}{15}{x}^2(1-x^2{)}^{-2},$
$r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}$	$=$	$\frac{4}{15}{x}^2(1-x^2{)}^{-3}(1+3x^2).$

Hence,

$\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-r_0^2\cdot \frac{d^2\mathrm{\Delta }}{d{r}_0^2}-r_0\left[4\right]\frac{d\mathrm{\Delta }}{d{r}_0}$
		$+\left[\frac{d\ln P_0}{d\ln r_0}\right][-r_0\cdot \frac{d\mathrm{\Delta }}{d{r}_0}+\alpha (\frac{a}{b}-\mathrm{\Delta }\left)\right]$
		$+\left[\frac{d\ln P_0}{d\ln r_0}\right]\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\{1-\frac{a}{b}[\frac{1}{\mathrm{\Delta }}\left]\right\}\frac{{\rho }_c}{{\rho }_0}$
	$=$	$-\frac{4}{15}{x}^2(1-x^2{)}^{-3}(1+3x^2)-\frac{16}{15}{x}^2(1-x^2{)}^{-2}$
		$+\left[\frac{d\ln P_0}{d\ln r_0}\right][\frac{a\alpha }{b}-\frac{4}{15}{x}^2(1-x^2{)}^{-2}-\frac{\alpha }{3}(1-\frac{3}{5}{x}^2)(1-x^2{)}^{-1}]$
		$+\left[\frac{d\ln P_0}{d\ln r_0}\right]\left(\frac{{\sigma }_c^2}{6{\gamma }_g}\right)\{1-\frac{3a}{b}(1-\frac{3}{5}{x}^2{)}^{-1}(1-x^2)\left\}\right(1-x^2{)}^{-1}.$

$\Rightarrow \frac{1}{x^2}(1-x^2)\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-\frac{4}{15}(1-x^2{)}^{-2}(1+3x^2)-\frac{16}{15}(1-x^2{)}^{-1}$
		$-5(1-\frac{3}{5}{x}^2)(1-\frac{1}{2}{x}^2{)}^{-1}[\frac{a\alpha }{b}-\frac{4}{15}{x}^2(1-x^2{)}^{-2}-\frac{\alpha }{3}(1-\frac{3}{5}{x}^2\left)\right(1-x^2{)}^{-1}]$
		$-5(1-\frac{3}{5}{x}^2)(1-\frac{1}{2}{x}^2{)}^{-1}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right\{1-\frac{3a}{b}(1-\frac{3}{5}{x}^2{)}^{-1}(1-x^2\left)\right\}(1-x^2{)}^{-1}$
$\Rightarrow \frac{1}{x^2}(1-x^2{)}^3\frac{1}{b}[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-\frac{4}{15}(1+3x^2)-\frac{16}{15}(1-x^2)$
		$-5(1-\frac{3}{5}{x}^2)(1-\frac{1}{2}{x}^2{)}^{-1}[\frac{a\alpha }{b}(1-x^2{)}^2-\frac{4}{15}{x}^2-\frac{\alpha }{3}(1-\frac{3}{5}{x}^2\left)\right(1-x^2\left)\right]$
		$-5(1-\frac{1}{2}{x}^2{)}^{-1}(\frac{{\sigma }_c^2}{6{\gamma }_g}\left)\right\{(1-\frac{3}{5}{x}^2)-\frac{3a}{b}(1-x^2)\left\}\right(1-x^2)$
$\Rightarrow \frac{15}{x^2}(1-x^2{)}^3(1-\frac{1}{2}{x}^2\left)\frac{1}{b}\right[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$-2(2-x^2)(5-x^2)$
		$-(5-3x^2)\left[\frac{15a\alpha }{b}\right(1-2x^2+x^4)-4x^2-\alpha (5-8x^2+3x^4\left)\right]$
		$-\left(\frac{5{\sigma }_c^2}{2{\gamma }_g}\right)\left\{\right(5-3x^2)-\frac{15a}{b}(1-x^2\left)\right\}(1-x^2)$
$\Rightarrow -\frac{15}{x^2}(1-x^2{)}^3(1-\frac{1}{2}{x}^2\left)\frac{1}{b}\right[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$20-14x^2+2x^4+(5-3x^2)\left[\right(\frac{15a\alpha }{b}-5\alpha )+x^2(\frac{30a\alpha }{b}-4+8\alpha )+\alpha x^4(\frac{15a}{b}-3\left)\right]$
		$+\left(\frac{5{\sigma }_c^2}{2{\gamma }_g}\right)\left[\right(5-\frac{15a}{b})+x^2(-3+\frac{15a}{b}\left)\right](1-x^2).$

Now, if we set $(15a/b)=3$, this last expression reduces to,

$-\frac{15}{x^2}(1-x^2{)}^3(1-\frac{1}{2}{x}^2\left)\frac{1}{b}\right[r_0^2\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}]$	$=$	$20-14x^2+2x^4-2(5-3x^2)[\alpha +x^2(2-7\alpha )]+\left(\frac{5{\sigma }_c^2}{{\gamma }_g}\right)(1-x^2)$
	$=$	$20-14x^2+2x^4-2[5\alpha +x^2(10-38\alpha )+x^4(21\alpha -6)]+\left(\frac{5{\sigma }_c^2}{{\gamma }_g}\right)(1-x^2)$
	$=$	$x^0[20-10\alpha +(\frac{5{\sigma }_c^2}{{\gamma }_g}\left)\right]+x^2[76\alpha -34-(\frac{5{\sigma }_c^2}{{\gamma }_g}\left)\right]+14x^4[1-3\alpha ]$

See Also