SSC/Stability/GammaVariation

How Does Stability Change with γ_g?

Isolated Uniform-Density Configuration

Our Setup

From our separate discussion, the relevant LAWE is,

$\frac{1}{(1-{\chi }_0^2)}\{(1-{\chi }_0^2)\frac{d^{2}x}{d{\chi }_0^2}+\frac{4}{{\chi }_0}[1-\frac{3}{2}{\chi }_0^2]\frac{dx}{d{\chi }_0}+\mathfrak{𝔉}x\}$

$=$

$0,$

where, ${\chi }_0\equiv r_0/R$, $\alpha \equiv (3-4/{\gamma }_{\mathrm{g}})$, and

$\mathfrak{𝔉}$

$\equiv$

$[\frac{3{\omega }^2}{2\pi {\gamma }_{g}G\overline{\rho }}-2(3-\frac{4}{{\gamma }_{\mathrm{g}}}\left)\right]\Rightarrow \frac{{\gamma }_{\mathrm{g}}\mathfrak{𝔉}}{2}=[\frac{3{\omega }^2}{4\pi G\overline{\rho }}+4-3{\gamma }_{\mathrm{g}}]$

Also, the two relevant boundary conditions are,

$\frac{dx}{d{\chi }_0}=0$        at         ${\chi }_0=0;$

and,

$\frac{d\ln x}{d{\chi }_0}$

$=$

$\frac{1}{{\gamma }_g}(4-3{\gamma }_g+\frac{3{\omega }^2}{4\pi G\overline{\rho }})$        at         ${\chi }_0=1.$

Alternatively, this last expression may be written as,

$\frac{d\ln x}{d{\chi }_0}{|}_{{\chi }_0=1}$

$=$

$\frac{\mathfrak{𝔉}}{2}.$

The Sterne37 Solution

From the general solution derived by 📚 T. E. Sterne (1937, MNRAS, Vol. 97, pp. 582 - 593), we have …

Sterne (1937)

Sterne (1937)

The first few solutions are displayed in the following boxed-in image that has been extracted directly from §2 (p. 587) of 📚 Sterne (1937); to the right of his table, we have added a column that expressly records the value of the square of the normalized eigenfrequency that corresponds to each of the solutions presented by Sterne37.

Based on exact eigenvector expressions extracted from §2 (p. 587) of … T. E. Sterne (1937) Models of Radial Oscillation Monthly Notices of the Royal Astronomical Society, Vol. 97, pp. 582 - 593	$\frac{{\omega }^2}{4\pi G\overline{\rho }}$
$j=0;$      $\mathfrak{𝔉}=0;$       $x=1$	$\gamma -4/3$
$j=1;$      $\mathfrak{𝔉}=14;$      $x=1-(7/5){\chi }_0^2$	$2(5\gamma -2)/3$
$j=2;$      $\mathfrak{𝔉}=36;$      $x=1-(18/5){\chi }_0^2+(99/35){\chi }_0^4$	$7\gamma -4/3$
$j=3;$      $\mathfrak{𝔉}=66;$      $x=1-(33/5){\chi }_0^2+(429/35){\chi }_0^4-(143/21){\chi }_0^6$	$12\gamma -4/3$

Cross-Check

Check j = 0:    The eigenvector is $x=1$, that is, homologous contraction/expansion, in which case both the first and the second derivative of $x$ are zero. Hence, this eigenvector is a solution to the LAWE only if $\mathfrak{𝔉}=0$. What about the two boundary conditions? Well, the central boundary condition is immediately satisfied; for the outer boundary condition, we see that the logarithmic derivative of $x$ is supposed to be zero, which it is because it equals $\mathfrak{𝔉}/2$. Finally, since $\mathfrak{𝔉}=0$, we see that the oscillation frequency is given by the expression,

$\frac{{\omega }^2}{4\pi G\overline{\rho }}={\gamma }_{\mathrm{g}}-4/3.$

Check j = 1:    The eigenvector is $x=1-\frac{7}{5}{\chi }_0^2$, hence, $dx/d{\chi }_0=-\frac{14}{5}{\chi }_0$, and, $d^{2}x/d{\chi }_0^2=-\frac{14}{5}.$ This means that,

LAWE	$=$	$-\frac{14}{5}(1-{\chi }_0^2)-[1-\frac{3}{2}{\chi }_0^2]\frac{56}{5}+\mathfrak{𝔉}[1-\frac{7}{5}{\chi }_0^2]$
	$=$	${\chi }_0^2[\frac{14}{5}+\frac{168}{10}-\frac{7}{5}\mathfrak{𝔉}]+[-\frac{14}{5}-\frac{56}{5}+\mathfrak{𝔉}]$
	$=$	$\frac{7}{5}{\chi }_0^2[14-\mathfrak{𝔉}]+[\mathfrak{𝔉}-14],$

which goes to zero if $\mathfrak{𝔉}=14$, in which case,

$\frac{{\omega }^2}{4\pi G\overline{\rho }}$

$=$

$\frac{1}{3}[\frac{{\gamma }_{\mathrm{g}}\mathfrak{𝔉}}{2}-4+3{\gamma }_{\mathrm{g}}]=\frac{2}{3}[5{\gamma }_{\mathrm{g}}-2].$

Is the surface boundary condition satisfied? Well …

$\frac{d\ln x}{d{\chi }_0}{|}_{{\chi }_0=1}=[\frac{1}{x}\cdot \frac{dx}{d{\chi }_0}{]}_{{\chi }_0=1}$

$=$

$\left[\right(1-\frac{7}{5}{\chi }_0^2{)}^{-1}(-\frac{14}{5}){\chi }_0{]}_{{\chi }_0=1}=\left[\right(-\frac{2}{5}{)}^{-1}(-\frac{14}{5})]=+7,$

which matches the desired logarithmic slope, $\mathfrak{𝔉}/2$.

Entropy Distribution

According to our discussions with P. Motl, to within an additive constant, the entropy distribution is given by the expression,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}$

$=$

$\frac{1}{({\gamma }_g-1)}\ln \left[\frac{P/P_c}{({\gamma }_g-1)(\rho /{\rho }_c{)}^{{\gamma }_g}}\right].$

Now, from the derived properties of a uniform-density sphere, we know that, $\rho /{\rho }_c=1$, and,

$\frac{P}{P_c}$

$=$

$(1-{\chi }_0^2).$

Hence, again to within an additive constant,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}$

$=$

$\frac{1}{({\gamma }_g-1)}\{\ln [\frac{P}{P_c}\left]\right\}=\ln [(1-{\chi }_0^2{)}^{1/({\gamma }_{\mathrm{g}}-1)}].$

Notice that, if ${\gamma }_{\mathrm{g}}<1$, the entropy is an increasing function of the fractional radius, ${\chi }_0$, and is therefore stable against convection according to the Schwarzschild criterion.

Comments on Uniform-Density Configurations

According to Sterne's stability analysis, the square of the oscillation frequency, ${\omega }^2/(4\pi G{\rho }_c)$, of the fundamental mode is negative for all values of ${\gamma }_{\mathrm{g}}<\frac{4}{3}$. All models with ${\gamma }_{\mathrm{g}}<\frac{4}{3}$ are therefore dynamically unstable toward collapse with a radial-displacement eigenfunction given by that of the fundamental mode. We appreciate as well that all models with ${\gamma }_{\mathrm{g}}<\frac{2}{5}$ are (also) dynamically unstable toward collapse with a radial-displacement eigenfunction given by the 1^st overtone mode.

At the same time, an examination of each model's entropy distribution indicates that models with ${\gamma }_{\mathrm{g}}>1$ are unstable toward convection throughout their entire volume. Hence, we identify the following model regimes:

$\mathrm{\infty }\ge {\gamma }_{\mathrm{g}}>\frac{4}{3}$	Dynamically stable against collapse, but unstable toward convection throughout.
$\frac{4}{3}>{\gamma }_{\mathrm{g}}>1$	Unstable toward convection throughout and, simultaneously dynamically unstable toward collapse with the eigenfunction provided by the fundamental mode. (All other radial overtone modes are dynamically stable against collapse.)
$1>{\gamma }_{\mathrm{g}}>\frac{2}{5}$	Stable against convection, but dynamically unstable toward collapse with the eigenfunction provided by the fundamental mode. (All other radial overtone modes are dynamically stable against collapse.)
$\frac{2}{5}>{\gamma }_{\mathrm{g}}$	Stable against convection, but dynamically unstable simultaneously toward collapse due to the fundamental and 1st overtone modes.

Lane-Emden in Terms of Various Physical Quantities

In a separate discussion we derived the,

Lane-Emden Equation

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

which governs the hydrostatic structure of spherically symmetric polytropes. In this differential equation,

${\mathrm{\Theta }}_H$

$=$

$\frac{H}{H_c}=(\frac{\rho }{{\rho }_c}{)}^{1/n}=(\frac{P}{P_c}{)}^{1/(n+1)}$

so the Lane-Emden equation readily can be rewritten in terms of the dimensionless density or the dimensionless pressure.

For n = 1, in Terms of Pressure

In terms of the dimensionless pressure, $p\equiv P/P_c$, the Lane-Emden equation becomes,

$-p^{n/(n+1)}$	$=$	$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left[{\xi }^2\frac{d{p}^{1/(n+1)}}{d\xi }\right]$
	$=$	$\left[\frac{1}{(n+1)}\right]\frac{1}{{\xi }^2}\frac{d}{d\xi }\left[{\xi }^2{p}^{(-n)/(n+1)}\frac{dp}{d\xi }\right]$
	$=$	$\left[\frac{1}{(n+1)}\right]\frac{1}{{\xi }^2}\{2\xi p^{(-n)/(n+1)}\frac{dp}{d\xi }+{\xi }^2[\frac{(-n)}{(n+1)}\left]p^{(-2n-1)/(n+1)}\right(\frac{dp}{d\xi }{)}^2+{\xi }^2{p}^{(-n)/(n+1)}\frac{d^{2}p}{d{\xi }^2}\}$
	$=$	$\left[\frac{1}{(n+1{)}^2}\right]p^{(-n)/(n+1)}\{\frac{2(n+1)}{\xi }\cdot \frac{dp}{d\xi }-n{p}^{-1}(\frac{dp}{d\xi }{)}^2+(n+1)\frac{d^{2}p}{d{\xi }^2}\};$
$\Rightarrow -(n+1{)}^2{p}^{2n/(n+1)}$	$=$	$\{\frac{2(n+1)}{\xi }\cdot \frac{dp}{d\xi }-n{p}^{-1}(\frac{dp}{d\xi }{)}^2+(n+1)\frac{d^{2}p}{d{\xi }^2}\}.$

Let's not set n = 1 yet. Instead, let's first insert the functional behavior of $p(\xi )$ that we know is the proper function for an isolated n = 1 polytrope, namely,

$p$	$=$	$(\frac{\sin \xi }{\xi }{)}^2;$
$\Rightarrow \frac{dp}{d\xi }$	$=$	$\frac{2\sin \xi \cos \xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^3};$
$\Rightarrow \frac{d^{2}p}{d{\xi }^2}$	$=$	$\frac{2{\cos }^2\xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^2}-\frac{8\sin \xi \cos \xi }{{\xi }^3}+\frac{6{\sin }^2\xi }{{\xi }^4};$
and,      $[\frac{dp}{d\xi }{]}^2$	$=$	$[\frac{2\sin \xi \cos \xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^3}{]}^2=\frac{4{\sin }^2\xi {\cos }^2\xi }{{\xi }^4}-\frac{8{\sin }^3\xi \cos \xi }{{\xi }^5}+\frac{4{\sin }^4\xi }{{\xi }^6}.$

Inside the curly braces on the RHS of the Lane-Emden equation, we therefore have,

$\{{\}}_{\mathrm{R}\mathrm{H}\mathrm{S}}$	$=$	$\frac{2(n+1)}{\xi }\cdot \frac{dp}{d\xi }+(n+1)\frac{d^{2}p}{d{\xi }^2}-\frac{n{\xi }^2}{{\sin }^2\xi }(\frac{dp}{d\xi }{)}^2$
	$=$	$(n+1)[\frac{4\sin \xi \cos \xi }{{\xi }^3}-\frac{4{\sin }^2\xi }{{\xi }^4}]+(n+1)[\frac{2{\cos }^2\xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^2}-\frac{8\sin \xi \cos \xi }{{\xi }^3}+\frac{6{\sin }^2\xi }{{\xi }^4}]-\frac{n{\xi }^2}{{\sin }^2\xi }[\frac{4{\sin }^2\xi {\cos }^2\xi }{{\xi }^4}-\frac{8{\sin }^3\xi \cos \xi }{{\xi }^5}+\frac{4{\sin }^4\xi }{{\xi }^6}]$
	$=$	$(n+1)[\frac{2{\cos }^2\xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^2}-\frac{4\sin \xi \cos \xi }{{\xi }^3}+\frac{2{\sin }^2\xi }{{\xi }^4}]-n[\frac{4{\cos }^2\xi }{{\xi }^2}-\frac{8\sin \xi \cos \xi }{{\xi }^3}+\frac{4{\sin }^2\xi }{{\xi }^4}]$
	$=$	$[\frac{2{\cos }^2\xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^2}-\frac{4\sin \xi \cos \xi }{{\xi }^3}+\frac{2{\sin }^2\xi }{{\xi }^4}]+n[-\frac{2{\cos }^2\xi }{{\xi }^2}-\frac{2{\sin }^2\xi }{{\xi }^2}+\frac{4\sin \xi \cos \xi }{{\xi }^3}-\frac{2{\sin }^2\xi }{{\xi }^4}].$

Now, if we set n = 1, this expression collapses substantially to give,

$\{{\}}_{\mathrm{R}\mathrm{H}\mathrm{S}}{|}_{n=1}$

$=$

$-\frac{4{\sin }^2\xi }{{\xi }^2}.$

Simultaneously, the LHS of the Lane-Emden expression becomes,

$-[(n+1{)}^2{p}^{2n/(n+1)}{]}_{n=1}$

$=$

$-[(n+1{)}^2(\frac{\sin \xi }{\xi }{)}^{4n/(n+1)}{]}_{n=1}=-\frac{4{\sin }^2\xi }{{\xi }^2}.$

So, the two sides of the expression prove to be identical.

What About in Terms of Entropy?

First Try

What about, in terms of the entropy? Well, from above, once the value of ${\gamma }_{\mathrm{g}}$ has been specified, to within an additive constant, the dimensionless entropy, $\mathrm{\Sigma }$, is given by the relation,

$\mathrm{\Sigma }\equiv \frac{s}{\mathrm{\Re }/\overline{\mu }}$	$=$	$\frac{1}{({\gamma }_g-1)}\ln \left[\right(\frac{P}{P_c}\left)\right(\frac{\rho }{{\rho }_c}{)}^{-{\gamma }_{\mathrm{g}}}]$
	$=$	$\frac{1}{({\gamma }_g-1)}\ln \left[{\mathrm{\Theta }}_H^{n+1}\right({\mathrm{\Theta }}_H^n{)}^{-{\gamma }_{\mathrm{g}}}]$
	$=$	$\ln [{\mathrm{\Theta }}_H^{n+1-n{\gamma }_{\mathrm{g}}}{]}^{1/({\gamma }_{\mathrm{g}}-1)}$
$\Rightarrow e^{\mathrm{\Sigma }}$	$=$	$\left[{\mathrm{\Theta }}_H^{(n+1-n{\gamma }_{\mathrm{g}})/({\gamma }_{\mathrm{g}}-1)}\right]$
$\Rightarrow {\mathrm{\Theta }}_{\mathrm{H}}$	$=$	$e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}.$

NOTE:   If we set ${\gamma }_{\mathrm{g}}=(1+1/n)$, then the exponent, $[\frac{n+1-n{\gamma }_{\mathrm{g}}}{{\gamma }_{\mathrm{g}}-1}{]}_{{\gamma }_{\mathrm{g}}=(1+1/n)}$ $=$ $n[n+1-(n+1)]=0,$ which means that, independent of the functional behavior of the dimensionless enthalpy, $e^{\mathrm{\Sigma }}$ $=$ ${\mathrm{\Theta }}_H^0=\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t},$ that is, the entropy is uniform throughout the equilibrium configuration.

Generally, then, in terms of the dimensionless entropy, the Lane-Emden equation may be rewritten as,

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left[{\xi }^2\frac{d}{d\xi }\right(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}\left)\right]$

$=$

$-[e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}{]}^n.$

That is,

$-e^{n\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}$	$=$	$\left[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\right]\frac{1}{{\xi }^2}\frac{d}{d\xi }\left[{\xi }^2\right(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}\left)\frac{d\mathrm{\Sigma }}{d\xi }\right]$
	$=$	$\left[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\right]\frac{1}{{\xi }^2}\left\{2\xi \right(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})})\frac{d\mathrm{\Sigma }}{d\xi }+{\xi }^2(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})})\frac{d^2\mathrm{\Sigma }}{d{\xi }^2}+{\xi }^2[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\left]\right(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})}\left)\right(\frac{d\mathrm{\Sigma }}{d\xi }{)}^2\}$
	$=$	$\left[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\right]\{\frac{2}{\xi }\cdot \frac{d\mathrm{\Sigma }}{d\xi }+\frac{d^2\mathrm{\Sigma }}{d{\xi }^2}+[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\left]\right(\frac{d\mathrm{\Sigma }}{d\xi }{)}^2\left\}\right(e^{\mathrm{\Sigma }({\gamma }_{\mathrm{g}}-1)/(n+1-n{\gamma }_{\mathrm{g}})})$
$\Rightarrow -e^n$	$=$	$\left[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\right]\{\frac{2}{\xi }\cdot \frac{d\mathrm{\Sigma }}{d\xi }+\frac{d^2\mathrm{\Sigma }}{d{\xi }^2}+[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\left]\right(\frac{d\mathrm{\Sigma }}{d\xi }{)}^2\}.$

Setting,

$\mathrm{{\rm Y}}\equiv \left[\frac{{\gamma }_{\mathrm{g}}-1}{n+1-n{\gamma }_{\mathrm{g}}}\right]\mathrm{\Sigma },$

the statement of hydrostatic balance becomes,

$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+(\frac{d\mathrm{{\rm Y}}}{d\xi }{)}^2+\frac{2}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }$

$=$

$-e^n.$

What do I do with this???

In our above discussion of uniform-density configurations, we found that,

$\mathrm{\Sigma }=\frac{s}{\mathrm{\Re }/\overline{\mu }}$

$=$

$({\gamma }_{\mathrm{g}}-1{)}^{-1}\ln [(1-{\xi }^2/6)],$

where we have made the substitution, ${\chi }_0^2={\xi }^2/6$. For this situation, we can write,

$\mathrm{{\rm Y}}$

$=$

$m^{-1}\ln [1-\frac{{\xi }^2}{6}],$

where,

$m$

$\equiv$

$n+1-n{\gamma }_{\mathrm{g}}.$

In this case,

$\frac{d\mathrm{{\rm Y}}}{d\xi }$	$=$	$m^{-1}[1-\frac{{\xi }^2}{6}{]}^{-1}\cdot (-\frac{\xi }{3});$
$(\frac{d\mathrm{{\rm Y}}}{d\xi }{)}^2$	$=$	$\frac{{\xi }^2}{3^2{m}^2}[1-\frac{{\xi }^2}{6}{]}^{-2};$
$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}$	$=$	$\frac{\xi }{3m}[1-\frac{{\xi }^2}{6}{]}^{-2}\cdot (-\frac{\xi }{3})-\frac{1}{3m}[1-\frac{{\xi }^2}{6}{]}^{-1};$

that is to say,

$-e^n$	$=$	$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+(\frac{d\mathrm{{\rm Y}}}{d\xi }{)}^2+\frac{2}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }$
	$=$	$\left\{\frac{\xi }{3m}\right[1-\frac{{\xi }^2}{6}{]}^{-2}\cdot (-\frac{\xi }{3})-\frac{1}{3m}[1-\frac{{\xi }^2}{6}{]}^{-1}\}+\left\{\frac{{\xi }^2}{3^2{m}^2}\right[1-\frac{{\xi }^2}{6}{]}^{-2}\}+\frac{2}{\xi }\cdot \{m^{-1}[1-\frac{{\xi }^2}{6}{]}^{-1}\cdot (-\frac{\xi }{3}\left)\right\}$
	$=$	$[1-\frac{{\xi }^2}{6}{]}^{-2}\{\frac{\xi }{3m}(-\frac{\xi }{3})-\frac{1}{3m}[1-\frac{{\xi }^2}{6}]+\frac{{\xi }^2}{3^2{m}^2}+\frac{2}{\xi }\cdot m^{-1}[1-\frac{{\xi }^2}{6}]\cdot (-\frac{\xi }{3})\}$
	$=$	$[1-\frac{{\xi }^2}{6}{]}^{-2}\{-\frac{{\xi }^2}{3^{2}m}-\frac{1}{3m}+\frac{{\xi }^2}{2\cdot 3^{2}m}+\frac{{\xi }^2}{3^2{m}^2}-\frac{2}{3m}+\frac{{\xi }^2}{3^{2}m}\}$
	$=$	$[1-\frac{{\xi }^2}{6}{]}^{-2}\{-\frac{1}{m}+\frac{{\xi }^2}{2\cdot 3^{2}m}+\frac{{\xi }^2}{3^2{m}^2}\}$
	$=$	$[1-\frac{{\xi }^2}{6}{]}^{-2}\{-m+\frac{{\xi }^2}{2\cdot 3^2}[m+2]\}\frac{1}{m^2}.$

That is,

$-m+\frac{{\xi }^2}{2\cdot 3^2}[m+2]$

$=$

$-m^2{e}^n[1-\frac{{\xi }^2}{6}{]}^2.$

What do I do with this???

Second Try

From above — and, for simplicity, removing the subscript $(\mathrm{g})$ on ${\gamma }_{\mathrm{g}}$ — we have,

$\mathrm{\Sigma }\equiv \frac{s}{\mathrm{\Re }/\overline{\mu }}$	$=$	$\frac{1}{(\gamma -1)}\ln \left[\frac{P/P_c}{(\gamma -1)(\rho /{\rho }_c{)}^{{\gamma }_g}}\right]$
$\Rightarrow (\gamma -1)e^{(\gamma -1)\mathrm{\Sigma }}$	$=$	$(\frac{\rho }{{\rho }_c}{)}^{-\gamma }\frac{P}{P_c}={\mathrm{\Theta }}^{-n\gamma }\cdot {\mathrm{\Theta }}^{(n+1)}={\mathrm{\Theta }}^{(n+1-n\gamma )}$
$\Rightarrow \mathrm{\Theta }$	$=$	$[(\gamma -1)e^{(\gamma -1)\mathrm{\Sigma }}{]}^{1/(n+1-n\gamma )}=(\gamma -1{)}^{1/(n+1-n\gamma )}{e}^{(\gamma -1)\mathrm{\Sigma }/(n+1-n\gamma )}=A{e}^{a\mathrm{\Sigma }},$

where,

$A$

$\equiv$

$(\gamma -1{)}^{1/(n+1-n\gamma )}$

and,

$a$

$\equiv$

$\frac{(\gamma -1)}{(n+1-n\gamma )}.$

Hence, in terms of the configuration's entropy profile, the Lane-Emden equation becomes,

$-A^n{e}^{an\mathrm{\Sigma }}$

$=$

$\frac{1}{{\xi }^2}\cdot \frac{d}{d\xi }[{\xi }^2\cdot \frac{d(A{e}^{a\mathrm{\Sigma }})}{d\xi }]$

In Bipolytropes, How Does Stability Change with ξ_i

Taken from an accompanying discussion.

file = Dropbox/WorkFolder/Wiki edits/BiPolytrope/TwoFirstOrderODEs/Bipolytrope51New.xlsx --- worksheet = Fun031FirstOvertone Variation of Oscillation Frequency with ${\xi }_i$ for $(5,1)$ Bipolytropes Having ${\mu }_e/{\mu }_c=0.310$

${\xi }_i$ Fundamental (red) 1st Overtone (blue) ${\mathrm{\Omega }}^2$ ${\sigma }_c^2$ $\frac{{\rho }_c}{\overline{\rho }}$ ${\mathrm{\Omega }}^2\equiv \frac{{\sigma }_c^2}{2}\left(\frac{{\rho }_c}{\overline{\rho }}\right)$ 1.60 3.8944 0.498473 58.398587 14.555059 2.00 3.81053 0.236047 108.69129 12.828126 2.40 2.79491 0.0870005 199.16363 8.6636677 2.609509754 0.00000 0.048214 270.5922 6.5231608 3.00 - 13.287 0.0232907 468.15 5.4517612 3.50 - 44.63801 0.0117478 902.64028 5.3020065 4.00 - 98.215 0.0064276 1656.926 5.3250395 5.00 --- 0.0022154 4900.105 5.4278831 6.00 --- 0.0008785 12544.67 5.5100707 9.014959766 --- 9.61 × 10-5 116641.6 5.6036778 12.0 --- 1.86 × 10-5 6.01 × 10+5 5.5796084