SSC/FreeEnergy/PolytropesEmbedded/Pt1

Free Energy of Embedded Polytropes

Part I:   Synopsis	Part II:  Truncated Polytropes	Part III:  Free-Energy of Bipolytropes
		IIIA:  Focus on (5, 1) Bipolytropes	IIIB:  Focus on (0, 0) Bipolytropes	IIIC:  Overview

Free-Energy Synopsis

All of the self-gravitating configurations considered below have an associated Gibbs-like free-energy that can be expressed analytically as a power-law function of the dimensionless configuration radius, $x$. Specifically,

${\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}$

$=$

$-a{x}^{-1}+b{x}^{-3/n}+c{x}^{-3/j}+{\mathfrak{𝔊}}_0.$

Equilibrium Radii and Critical Radii

The first and second (partial) derivatives with respect to $x$ are, respectively,

$\frac{\partial {\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}}{\partial x}$	$=$	$a{x}^{-2}-\left(\frac{3b}{n}\right)x^{-3/n-1}-\left(\frac{3c}{j}\right)x^{-3/j-1}$
	$=$	$\frac{1}{x^2}[a-(\frac{3b}{n})x^{(n-3)/n}-(\frac{3c}{j}\left)x^{(j-3)/j}\right],$
$\frac{{\partial }^2{\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}}{\partial x^2}$	$=$	$-2a{x}^{-3}+\left(\frac{3b}{n}\right)\left(\frac{n+3}{n}\right)x^{-3/n-2}+\left(\frac{3c}{j}\right)\left(\frac{j+3}{j}\right)x^{-3/j-2}$
	$=$	$\frac{1}{x^3}\left\{\right(\frac{3b}{n}\left)\right(\frac{n+3}{n})x^{(n-3)/n}+(\frac{3c}{j}\left)\right(\frac{j+3}{j})x^{(j-3)/j}-2a\}.$

Equilibrium configurations are identified by setting the first derivative to zero. This gives,

$0$	$=$	$a-\left(\frac{3b}{n}\right)x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\left(\frac{3c}{j}\right)x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}$
$\Rightarrow x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$	$=$	$\left(\frac{n}{3b}\right)[a-(\frac{3c}{j}\left)x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}\right].$
$\Rightarrow \frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}+\frac{1}{j}\cdot x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}$	$=$	$0.$

We conclude, as well, that at this equilibrium radius, the second (partial) derivative assumes the value,

$[\frac{{\partial }^2{\mathfrak{𝔊}}_{\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}}^{*}}{\partial x^2}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{x_{\mathrm{e}\mathrm{q}}^3}\left\{\right(\frac{3b}{n}\left)\right(\frac{n+3}{n})x^{(n-3)/n}+(\frac{3c}{j}\left)\right(\frac{j+3}{j})x^{(j-3)/j}-2a{\}}_{\mathrm{e}\mathrm{q}}$
	$=$	$\frac{1}{x_{\mathrm{e}\mathrm{q}}^3}\left\{\right(\frac{n+3}{n}\left)\right[a-\left(\frac{3c}{j}\right)x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}]+(\frac{3c}{j}\left)\right(\frac{j+3}{j})x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}-2a\}$
	$=$	$\frac{1}{x_{\mathrm{e}\mathrm{q}}^3}\left\{\right(\frac{3c}{j}\left)\right[\left(\frac{j+3}{j}\right)-\left(\frac{n+3}{n}\right)]x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}+(\frac{3-n}{n}\left)a\right\}.$

Hence, equilibrium configurations for which the second (as well as first) derivative of the free energy is zero are found at "critical" radii given by the expression,

$0$	$=$	$\left(\frac{3c}{j}\right)\left[\right(\frac{j+3}{j})-(\frac{n+3}{n}\left)\right][x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}+(\frac{3-n}{n})a$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}^{(j-3)/j}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left[\frac{j^{2}a(n-3)}{3c}\right][n(j+3)-j(n+3){]}^{-1}$
	$=$	$\frac{a}{3^{2}c}\left[\frac{j^2(n-3)}{n-j}\right].$

Examples

Pressure-Truncated Polytropes

For pressure-truncated polytropes of index $n$, we set, $j=-1$, in which case,

$\frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}-x_{\mathrm{e}\mathrm{q}}^4$	$=$	$0;$
$\left[\right(\frac{3}{4\pi })\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{]}^{(n+1)/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{3}{20\pi }(\frac{n+1}{n}\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-x_{\mathrm{e}\mathrm{q}}^4$	$=$	$0;$
$x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$	$=$	$\left(\frac{n}{3b}\right)[a+3c{x}_{\mathrm{e}\mathrm{q}}^4];$
	and
$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$[\frac{a(n-3)}{3^{2}c(n+1)}{]}^{1/4}.$

Case M

More specifically, the expression that describes the "Case M" free-energy surface is,

${\mathfrak{𝔊}}_{K,M}^{*}\equiv \frac{{\mathfrak{𝔊}}_{K,M}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$ $=$ $-3{𝒜}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}+n{ℬ}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-3/n}+\left(\frac{4\pi }{3}\right)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}(\frac{R}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^3.$

Hence, we have,

$a$	$\equiv$	$3{𝒜}=\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2},$
$b$	$\equiv$	$n{ℬ}=n(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}},$
$c$	$\equiv$	$\frac{4\pi }{3}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right),$

where the structural form factors for pressure-truncated polytropes are precisely defined here. Therefore, the statement of virial equilibrium is,

$0$	$=$	$\frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}-x_{\mathrm{e}\mathrm{q}}^4$
$\Rightarrow \left(\frac{3}{4\pi }\right)c{x}_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left(\frac{3}{4\pi }\right)[\frac{b}{n}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3}]$
$\Rightarrow \left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)x_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left(\frac{3}{4\pi }\right)\left[\right(\frac{3}{4\pi }{)}^{1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{1}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}]$
	$=$	$(\frac{3}{4\pi }{)}^{(n+1)/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}.$

And we conclude that,

$3c[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\frac{(n-3)}{5(n+1)}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
$\Rightarrow \left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
$\Rightarrow (\frac{3}{4\pi }{)}^{(n+1)/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\cdot [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}-\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$	$=$	$\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
$\Rightarrow (\frac{3}{4\pi }{)}^{(n+1)/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\cdot [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$	$=$	$\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}+\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
	$=$	$\frac{1}{20\pi }\left(\frac{4n}{n+1}\right)\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$	$=$	$\frac{1}{20\pi }\left(\frac{4\pi }{3}{)}^{(n+1)/n}\right(\frac{4n}{n+1})\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_A{\widetilde{\mathfrak{𝔣}}}_M^{(n-1)/n}}$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left[\frac{4n}{15(n+1)}\right(\frac{4\pi }{3}{)}^{1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_A{\widetilde{\mathfrak{𝔣}}}_M^{(n-1)/n}}{]}^{n/(n-3)}.$

ASIDE:  Let's see what this requires for the case of $n=5$, where everything is specifiable analytically. We have gathered together: Form factors from here. Hoerdt's equilibrium expressions from here. Conversion from Horedt's units to ours as specified here. ${\widetilde{\mathfrak{𝔣}}}_M$ $=$ $(1+{\ell }^2{)}^{-3/2}$ ${\widetilde{\mathfrak{𝔣}}}_W$ $=$ $\frac{5}{2^4}\cdot {\ell }^{-5}\left[\ell \right({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}+{\tan }^{-1}(\ell )]$ ${\widetilde{\mathfrak{𝔣}}}_A$ $=$ $\frac{3}{2^3}{\ell }^{-3}[{\tan }^{-1}(\ell )+\ell ({\ell }^2-1)(1+{\ell }^2{)}^{-2}]$ $\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}[\frac{4\pi }{(n+1{)}^n}{]}^{1/(n-3)}$ $=$ $\left\{3\right[\frac{({\xi }_e^2/3{)}^5}{(1+{\xi }_e^2/3{)}^6}\left]{\}}^{-1/2}\right[\frac{4\pi }{(n+1{)}^n}{]}^{1/(n-3)}$   $=$ $\left[\frac{(1+{\ell }^2{)}^3}{{\ell }^5}\right][\frac{\pi }{2^3\cdot 3^6}{]}^{1/2}$ $\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\frac{P_e}{P_{\mathrm{H}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{t}}}[\frac{(n+1{)}^3}{4\pi }{]}^{(n+1)/(n-3)}$ $=$ $3^3\left[\frac{({\xi }_e^2/3{)}^3}{(1+{\xi }_e^2/3{)}^4}{]}^3\right[\frac{(n+1{)}^3}{4\pi }{]}^{(n+1)/(n-3)}$   $=$ $\left[\frac{{\ell }^{18}}{(1+{\ell }^2{)}^{12}}\right][\frac{2\cdot 3^4}{\pi }{]}^3$ So, the radius of the critical equilibrium state should be, $[\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$ $=$ $\frac{(n-3)}{3\cdot 5(n+1)}\left(\frac{3}{2^2\pi }\right)(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$   $=$ $\frac{1}{2^2\cdot 3\cdot 5\pi }\left\{\frac{(1+{\ell }^2{)}^{12}}{{\ell }^{18}}\right[\frac{\pi }{2\cdot 3^4}{]}^3\}(1+{\ell }^2{)}^3\cdot \{\frac{5}{2^4}\cdot {\ell }^{-5}[\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)(1+{\ell }^2{)}^{-3}+{\tan }^{-1}(\ell )]\}$   $=$ $\frac{{\pi }^2}{2^9\cdot 3^{13}}\left\{\frac{(1+{\ell }^2{)}^{12}}{{\ell }^{23}}\right\}\cdot \left\{\right[\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)+(1+{\ell }^2{)}^3{\tan }^{-1}(\ell )]\};$ whereas, each equilibrium configuration has, $[\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^4$ $=$ $\frac{{\pi }^2}{2^6\cdot 3^{12}}\left[\frac{(1+{\ell }^2{)}^{12}}{{\ell }^{20}}\right].$ So the equilibrium state that marks the critical configuration must have a value of $\ell$ that satisfies the relation, $\frac{{\pi }^2}{2^6\cdot 3^{12}}\left[\frac{(1+{\ell }^2{)}^{12}}{{\ell }^{20}}\right]$ $=$ $\frac{{\pi }^2}{2^9\cdot 3^{13}}\left\{\frac{(1+{\ell }^2{)}^{12}}{{\ell }^{23}}\right\}\cdot \left\{\right[\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)+(1+{\ell }^2{)}^3{\tan }^{-1}(\ell )]\}$ $\Rightarrow 2^3\cdot 3{\ell }^3$ $=$ $\ell ({\ell }^4-\frac{8}{3}{\ell }^2-1)+(1+{\ell }^2{)}^3{\tan }^{-1}(\ell )$ $\Rightarrow \left[\frac{(1+{\ell }^2{)}^3}{\ell }\right]{\tan }^{-1}(\ell )$ $=$ $1+\frac{80}{3}\cdot {\ell }^2-{\ell }^4.$ The solution is: ${\ell }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}\approx 2.223175.$

In addition, we know from our dissection of Hoerdt's work on detailed force-balance models that, in the equilibrium state,

$\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^4$	$=$	$\left[\frac{{\tilde{\theta }}^{n+1}}{(4\pi )(n+1)(-{\tilde{\theta }}^{\prime }{)}^2}\right]$
$\Rightarrow 3c{x}_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left[\frac{{\tilde{\theta }}^{n+1}}{(n+1)(-{\tilde{\theta }}^{\prime }{)}^2}\right].$

This means that, for any chosen polytropic index, the critical equilibrium state is the equilibrium configuration for which (needs to be checked),

$2(9-2n){\tilde{\theta }}^{n+1}$

$=$

$3(n-3)[(-{\tilde{\theta }}^{\text{'}}{)}^2-\frac{\tilde{\theta }(-{\tilde{\theta }}^{\text{'}})}{\tilde{\xi }}].$

We note, as well, that by combining the Horedt expression for $x_{\mathrm{e}\mathrm{q}}$ with our virial equilibrium expression, we find (needs to be checked),

$x_{\mathrm{e}\mathrm{q}}^{n-3}$

$=$

$\frac{4\pi }{3}[\frac{3}{(n+1){\tilde{\xi }}^2}+\frac{{\widetilde{\mathfrak{𝔣}}}_W-{\widetilde{\mathfrak{𝔣}}}_M}{5{\widetilde{\mathfrak{𝔣}}}_A}{]}^n{\widetilde{\mathfrak{𝔣}}}_M^{1-n}.$

Case P

First Pass

Alternatively, let's examine the "Case P" free-energy surface. Drawing on Stahler's presentation, we adopt the following radius and mass normalizations:

$M_{\mathrm{S}\mathrm{W}\mathrm{S}}=(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]},$

$R_{\mathrm{S}\mathrm{W}\mathrm{S}}=(\frac{n+1}{n}{)}^{1/2}{G}^{-1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]}.$

In terms of these new normalizations, we have,

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\equiv \left[\right(\frac{G}{K}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{(n-1)}{]}^{1/(n-3)}$	$=$	$\left(\frac{G}{K}{)}^{n/(n-3)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{(n-1)/(n-3)}{R}_{\mathrm{S}\mathrm{W}\mathrm{S}}\right(\frac{n+1}{n}{)}^{-1/2}{G}^{1/2}{K}_n^{-n/(n+1)}{P}_{\mathrm{e}}^{-(1-n)/[2(n+1)]}$
		$+M_{\mathrm{S}\mathrm{W}\mathrm{S}}^{-(n-1)/(n-3)}\left[\right(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{]}^{(n-1)/(n-3)}$
	$=$	$R_{\mathrm{S}\mathrm{W}\mathrm{S}}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{[3(n-1)-(n-3)]/[2(n-3)]}{G}^{[2n+(n-3)-3(n-1)]/[2(n-3)]}$
		$+K_n^{n[2(n-1)-(n+1)-(n-3)]/[(n+1)(n-3)]}{P}_{\mathrm{e}}^{-(n-1)(3-n)/[2(n+1)(n-3)]}{P}_{\mathrm{e}}^{(n-1)(3-n)/[2(n+1)(n-3)]}$
	$=$	$R_{\mathrm{S}\mathrm{W}\mathrm{S}}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{n/(n-3)}.$

and,

$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\equiv [\frac{K^{4n}}{G^{3(n+1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2(n+1)}}{]}^{1/(n-3)}$	$=$	$\left[\frac{K^{4n}}{G^{3(n+1)}}{]}^{1/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-2(n+1)/(n-3)}\left\{\right(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{\}}^{-2(n+1)/(n-3)}$
	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-2(n+1)/(n-3)}\right(\frac{n+1}{n}{)}^{-3(n+1)/(n-3)}{K}^{4n/(n-3)}{G}^{-3(n+1)/(n-3)}$
		$\times G^{3(n+1)/(n-3)}{K}_n^{-4n/(n-3)}\{P_{\mathrm{e}}^{-(n-3)/[2(n+1)]}{\}}^{-2(n+1)/(n-3)}$
	$=$	$P_e\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-2(n+1)/(n-3)}\right(\frac{n+1}{n}{)}^{-3(n+1)/(n-3)}.$

Rewriting the expression for the free energy gives,

${\mathfrak{𝔊}}_{K,M}^{*}\equiv \frac{{\mathfrak{𝔊}}_{K,M}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-3{𝒜}\left(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}\right(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}})+n{ℬ}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3/n}+(\frac{4\pi }{3}\left)\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3}$
	$=$	$-3{𝒜}\left(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}\right[\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{n/(n-3)}]$
		$+n{ℬ}\left(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}\right[\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{n/(n-3)}{]}^{3/n}$
		$+\left(\frac{4\pi }{3}\right)\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{2(n+1)/(n-3)}\right(\frac{n+1}{n}{)}^{3(n+1)/(n-3)}\left(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3\right[\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{n/(n-3)}{]}^{-3}$
	$=$	$-3{𝒜}\left(\frac{n+1}{n}{)}^{n/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}+n{ℬ}(\frac{n+1}{n}{)}^{3/(n-3)}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]}\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}$
		$+\left(\frac{4\pi }{3}\right)\left(\frac{n+1}{n}{)}^{3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3.$

Therefore, in this case, we have,

$a$	$=$	$\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}{)}^{n/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)},$
$b$	$=$	$n\left(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{n+1}{n}{)}^{3/(n-3)}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]},$
$c$	$=$	$\frac{4\pi }{3}\left(\frac{n+1}{n}{)}^{3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)},$

where the structural form factors for pressure-truncated polytropes are precisely defined here. The statement of virial equilibrium is, therefore,

$x_{\mathrm{e}\mathrm{q}}^4+\alpha$

$=$

$\beta x_{\mathrm{e}\mathrm{q}}^{(n-3)/n},$

where,

$\alpha \equiv \frac{a}{3c}$	$=$	$\frac{1}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}{)}^{n/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\left\{\frac{3}{4\pi }\right(\frac{n+1}{n}{)}^{-3/(n-3)}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-5)/(n-3)}\right\}$
	$=$	$\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2$
	$=$	$\left(\frac{4\pi }{3\cdot 5}\right){\widetilde{\mathfrak{𝔣}}}_W\left(\frac{n+1}{n}\right){\mathfrak{𝔪}}^2,$
$\beta \equiv \frac{b}{nc}$	$=$	$\left(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{n+1}{n}{)}^{3/(n-3)}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]}\right\{\frac{3}{4\pi }\left(\frac{n+1}{n}{)}^{-3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-5)/(n-3)}\}$
	$=$	${\widetilde{\mathfrak{𝔣}}}_A{\mathfrak{𝔪}}^{(n+1)/n},$
$\mathfrak{𝔪}$	$\equiv$	$\left(\frac{3}{4\pi }\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right).$

From a previous derivation, we have,

$0$	$=$	$\frac{b}{nc}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{a}{3c}-x_{\mathrm{e}\mathrm{q}}^4$
	$=$	$\frac{3}{4\pi }\left(\frac{n+1}{n}{)}^{-3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-5)/(n-3)}\left\{\right(\frac{3}{4\pi }{)}^{1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\left(\frac{n+1}{n}{)}^{3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]}\}\cdot x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$
		$-\frac{3}{4\pi }\left(\frac{n+1}{n}{)}^{-3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-5)/(n-3)}\{\frac{1}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}(\frac{n+1}{n}{)}^{n/(n-3)}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}\right\}-x_{\mathrm{e}\mathrm{q}}^4$
$0$	$=$	$\left(\frac{3}{4\pi }{)}^{(n+1)/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2-x_{\mathrm{e}\mathrm{q}}^4$
	$=$	${\widetilde{\mathfrak{𝔣}}}_A\left[\right(\frac{3}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}){]}^{(n+1)/n}{x}_{\mathrm{e}\mathrm{q}}^{(n-3)/n}-\frac{1}{5}(\frac{4\pi }{3})\cdot {\widetilde{\mathfrak{𝔣}}}_W(\frac{n+1}{n}\left)\right[\left(\frac{3}{4\pi }\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right){]}^2-x_{\mathrm{e}\mathrm{q}}^4$

which, thankfully, matches! We conclude as well that the transition from stable to dynamically unstable configurations occurs at,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$

$=$

$\left[\frac{(n-3)}{3(n+1)}\right]\alpha .$

When combined with the statement of virial equilibrium at this critical point, we find,

$\left\{\right[\frac{(n-3)}{3(n+1)}]+1\}\frac{\alpha }{\beta }$	$=$	$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$
	$=$	$\left\{\right[\frac{(n-3)}{3(n+1)}]\alpha {\}}^{(n-3)/(4n)}$
$\Rightarrow \left[\frac{4n}{3(n+1)}{]}^{4n}\right(\frac{\alpha }{\beta }{)}^{4n}$	$=$	$[\frac{(n-3)}{3(n+1)}{]}^{(n-3)}{\alpha }^{(n-3)}$
$\Rightarrow \left[\frac{3n}{(n-3)}\right(\frac{n+1}{n}\left){]}^{(3-n)}\right[\frac{3}{4}\left(\frac{n+1}{n}\right){]}^{4n}$	$=$	${\alpha }^{3(n+1)}{\beta }^{-4n}$
	$=$	$\left\{\right(\frac{4\pi }{3\cdot 5}\left){\widetilde{\mathfrak{𝔣}}}_W\right(\frac{n+1}{n}\left){\mathfrak{𝔪}}^2{\}}^{3(n+1)}\right\{{\widetilde{\mathfrak{𝔣}}}_A{\mathfrak{𝔪}}^{(n+1)/n}{\}}^{-4n}$
	$=$	${\widetilde{\mathfrak{𝔣}}}_A^{-4n}\left[\right(\frac{4\pi }{3\cdot 5}\left){\widetilde{\mathfrak{𝔣}}}_W\right(\frac{n+1}{n}){]}^{3(n+1)}{\mathfrak{𝔪}}^{2(n+1)}$
$\Rightarrow {\mathfrak{𝔪}}^{2(n+1)}$	$=$	$\left[\right(\frac{4\pi }{3\cdot 5}\left){\widetilde{\mathfrak{𝔣}}}_W\right(\frac{n+1}{n}\left){]}^{-3(n+1)}\right[\frac{3n}{(n-3)}\left(\frac{n+1}{n}\right){]}^{(3-n)}\left[\frac{3{\widetilde{\mathfrak{𝔣}}}_A}{4}\right(\frac{n+1}{n}){]}^{4n}$
	$=$	$\left[\right(\frac{3\cdot 5}{4\pi }\left)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{3(n+1)}\right[\frac{3n}{(n-3)}{]}^{(3-n)}[\frac{3{\widetilde{\mathfrak{𝔣}}}_A}{4}{]}^{4n}$
	$=$	$[\frac{3^2\cdot 5n}{4\pi (n-3)}\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{(3-n)}[\left(\frac{3^2\cdot 5}{2^4\pi }\right)\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{4n}.$

This also means that the critical radius is,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\left[\frac{(n-3)}{3(n+1)}\right]\left(\frac{4\pi }{3\cdot 5}\right){\widetilde{\mathfrak{𝔣}}}_W\left(\frac{n+1}{n}\right){\mathfrak{𝔪}}^2$
	$=$	$[\frac{3^2\cdot 5n}{4\pi (n-3)}\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{-1}{\mathfrak{𝔪}}^2$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{4(n+1)}$	$=$	$[\frac{3^2\cdot 5n}{4\pi (n-3)}\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{-(n+1)}[\frac{3^2\cdot 5n}{4\pi (n-3)}\cdot \frac{1}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{(3-n)}\left[\right(\frac{3^2\cdot 5}{2^4\pi })\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{4n}$
	$=$	$[\frac{4\pi (n-3)}{3^2\cdot 5n}\cdot {\widetilde{\mathfrak{𝔣}}}_W{]}^{2(n-1)}[\left(\frac{3^2\cdot 5}{2^4\pi }\right)\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_W}{]}^{4n}$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{2(n+1)}$	$=$	$\left[\frac{n}{(n-3)}\right(\frac{3^2\cdot 5}{4\pi {\widetilde{\mathfrak{𝔣}}}_W}\left){]}^{(1-n)}\right[\left(\frac{3^2\cdot 5}{4\pi {\widetilde{\mathfrak{𝔣}}}_W}\right)\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{]}^{2n}$
	$=$	$\left(\frac{n}{n-3}{)}^{(1-n)}\right(\frac{3^2\cdot 5}{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{)}^{(n+1)}(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{2n}.$

The following parallel derivation was done independently. [Note that a factor of 2n/(n-1) appears to correct a mistake made during the original derivation.] Beginning with the virial expression,

$\beta x_{\mathrm{e}\mathrm{q}}^{(n-3)/n}$

$=$

$\alpha +x_{\mathrm{e}\mathrm{q}}^4$

$\left(\frac{3}{4\pi }{)}^{(n+1)/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$	$=$	$\frac{3}{20\pi }\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2+\frac{(n-3)}{20\pi n}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$
	$=$	$\frac{(n-1)}{10\pi n}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\right[\frac{2n}{(n-1)}]$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$	$=$	$\frac{2(n-1)}{15n}(\frac{4\pi }{3}{)}^{1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_A{\widetilde{\mathfrak{𝔣}}}_M^{(n-1)/n}}\cdot (\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/n}\left[\frac{2n}{(n-1)}\right]$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)}$	$=$	$\left[\frac{2(n-1)}{15n}{]}^n\right(\frac{4\pi }{3}\left)\frac{{\widetilde{\mathfrak{𝔣}}}_W^n}{{\widetilde{\mathfrak{𝔣}}}_A^n{\widetilde{\mathfrak{𝔣}}}_M^{(n-1)}}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)}[\frac{2n}{(n-1)}{]}^n$
	$=$	$\left[\frac{2(n-1)}{15n}{]}^n\right(\frac{4\pi }{3}\left)\frac{{\widetilde{\mathfrak{𝔣}}}_W^n}{{\widetilde{\mathfrak{𝔣}}}_A^n{\widetilde{\mathfrak{𝔣}}}_M^{(n-1)}}\right\{\left[\frac{20\pi n}{(n-3)}{]}^{(n-1)/2}\right(\frac{{\widetilde{\mathfrak{𝔣}}}_M^2}{{\widetilde{\mathfrak{𝔣}}}_W}{)}^{(n-1)/2}[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{2(n-1)}\}[\frac{2n}{(n-1)}{]}^n$
	$=$	$\left[\frac{2(n-1)}{15n}{]}^n\right(\frac{4\pi }{3}\left)\frac{{\widetilde{\mathfrak{𝔣}}}_W^{(n+1)/2}}{{\widetilde{\mathfrak{𝔣}}}_A^n}\right\{[\frac{20\pi n}{(n-3)}{]}^{(n-1)/2}[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{2(n-1)}\left\}\right[\frac{2n}{(n-1)}{]}^n$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n+1)}$	$=$	$\left(\frac{3}{4\pi }\right)\left[\frac{15n}{2(n-1)}{]}^n\right[\frac{(n-3)}{20\pi n}{]}^{(n-1)/2}\frac{{\widetilde{\mathfrak{𝔣}}}_A^n}{{\widetilde{\mathfrak{𝔣}}}_W^{(n+1)/2}}[\frac{(n-1)}{2n}{]}^n$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n+1)}$	$=$	$\left(\frac{3}{4\pi }\right)\left[\frac{15}{2^2}{]}^n\right[\frac{(n-3)}{20\pi n}{]}^{(n-1)/2}\frac{{\widetilde{\mathfrak{𝔣}}}_A^n}{{\widetilde{\mathfrak{𝔣}}}_W^{(n+1)/2}}$

Also from Stahler's work we know that the equilibrium mass and radius are,

$\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left(\frac{n^3}{4\pi }{)}^{1/2}\right[{\tilde{\theta }}_n^{(n-3)/2}{\tilde{\xi }}^2(-{\tilde{\theta }}^{\text{'}})],$
$\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}$	$=$	$\left(\frac{n}{4\pi }{)}^{1/2}\right[\tilde{\xi }{\tilde{\theta }}_n^{(n-1)/2}].$

Additional details in support of an associated PowerPoint presentation can be found here.

Reconcile

$[\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$

$=$

$\left[\frac{(n-3)}{20\pi (n+1)}\right]\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$

$(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$

$=$

$\frac{1}{20\pi }\left(\frac{n-3}{n+1}\right)(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}\frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}$

Taking the ratio, the RHS is,

$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})$	$=$	$P_e{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left[\frac{G^{3(n+1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2(n+1)}}{K^{4n}}{]}^{1/(n-3)}\right[\left(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{]}^{-2}\right(\frac{n+1}{n})$
	$=$	$\left(\frac{n+1}{n}{)}^{-2}{P}_e{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\right[G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2{]}^{(n+1)/(n-3)}{K}_n^{-4n/(n-3)}\left[G^3{K}_n^{-4n/(n+1)}{P}_{\mathrm{e}}^{(n-3)/(n+1)}\right]$
	$=$	$\left(\frac{n+1}{n}{)}^{-2}\right[G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2{]}^{[(n-3)+(n+1)]/(n-3)}[K_n^{[(n+1)+(n-3)]/[(n+1)(n-3)]}{]}^{-4n}{P}_{\mathrm{e}}^{[(n+1)+(n-3)]/(n+1)}$
	$=$	$(\frac{n+1}{n}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{4(n-1)/(n-3)}{G}^{[6(n-1)]/(n-3)}{K}_n^{-8(n-1)/[(n+1)(n-3)]}{P}_{\mathrm{e}}^{2(n-1)/(n+1)};$

while the LHS is,

$(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$	$=$	$\left[\right(\frac{G}{K}{)}^n{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{(n-1)}{]}^{4/(n-3)}\left\{\right(\frac{n+1}{n}{)}^{1/2}{G}^{-1/2}{K}_n^{n/(n+1)}{P}_{\mathrm{e}}^{(1-n)/[2(n+1)]}{\}}^{-4}$
	$=$	$(\frac{n+1}{n}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{4(n-1)/(n-3)}{G}^{[6(n-1)]/(n-3)}{K}^{-8n(n-1)/[(n-3)(n+1)]}{P}_{\mathrm{e}}^{2(n-1)/(n+1)}.$

Q.E.D.

Now, given that,

$M_{\mathrm{S}\mathrm{W}\mathrm{S}}^{-4(n-1)/(n-3)}$	$=$	$\left[\right(\frac{n+1}{n}{)}^{3/2}{G}^{-3/2}{K}_n^{2n/(n+1)}{P}_{\mathrm{e}}^{(3-n)/[2(n+1)]}{]}^{-4(n-1)/(n-3)}$
	$=$	$(\frac{n+1}{n}{)}^{-6(n-1)/(n-3)}{G}^{6(n-1)/(n-3)}{K}_n^{-8n(n-1)/[(n+1)(n-3)]}{P}_{\mathrm{e}}^{2(n-1)/(n+1)}$

we have,

$(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^4$	$=$	$\left(\frac{n+1}{n}{)}^{-2}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{4(n-1)/(n-3)}(\frac{n+1}{n}{)}^{6(n-1)/(n-3)}$
	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{4(n-1)/(n-3)}\right(\frac{n+1}{n}{)}^{4n/(n-3)}$
$\Rightarrow (\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{n-3}$	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{n-1}\right(\frac{n+1}{n}{)}^n$

By inspection, in the specific case of $n=5$ (see above), this critical configuration appears to coincide with one of the "turning points" identified by Kimura. Specifically, it appears to coincide with the "extremal in r₁" along an M₁ sequence, which satisfies the condition,

$[\frac{n-3}{n-1}{]}_{n=5}$	$=$	$[\frac{\tilde{\xi }{\tilde{\theta }}^n}{(-{\tilde{\theta }}^{\text{'}})}{]}_{n=5}$
$\Rightarrow \frac{1}{2}$	$=$	$3^{1/2}\ell [(1+{\ell }^2{)}^{-1/2}{]}^5[\frac{\ell }{3^{1/2}}(1+{\ell }^2{)}^{-3/2}{]}^{-1}$
	$=$	$3(1+{\ell }^2{)}^{-1}$
$\Rightarrow \ell$	$=$	$5^{1/2}.$

Hence, according to Kimura, the turning point associated with the configuration with the largest equilibrium radius, corresponds to the equilibrium configuration having,

$\ell {|}_{{\mathrm{R}}_{\mathrm{m}\mathrm{a}\mathrm{x}}}=\sqrt{5}\approx 2.2360680.$

This is, indeed, very close to — but decidedly different from — the value of ${\ell }_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$ determined, above!

Streamlined

Let's copy the expression for the "Case P" free energy derived above, then factor out a common term:

$\frac{{\mathfrak{𝔊}}_{K,M}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-3{𝒜}\left(\frac{n+1}{n}{)}^{n/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n-1)/(n-3)}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}+n{ℬ}(\frac{n+1}{n}{)}^{3/(n-3)}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{3(n-1)/[n(n-3)]}\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}$
		$+\left(\frac{4\pi }{3}\right)\left(\frac{n+1}{n}{)}^{3/(n-3)}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3.$
	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)}\right(\frac{n+1}{n}{)}^{3/(n-3)}\{-3{𝒜}(\frac{n+1}{n}\left)\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}+n{ℬ}(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}+\frac{4\pi }{3}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3\}$

Defining a new normalization energy,

$E_{\mathrm{S}\mathrm{W}\mathrm{S}}$	$\equiv$	$E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(5-n)/(n-3)}\right(\frac{n+1}{n}{)}^{3/(n-3)}$
	$=$	$(\frac{n+1}{n}{)}^{3/2}{K}^{3n/(n+1)}{G}^{-3/2}{P}_e^{(5-n)/[2(n+1)]},$

we can write,

${\mathfrak{𝔊}}_{K,M}^{*}\equiv \frac{{\mathfrak{𝔊}}_{K,M}}{E_{\mathrm{S}\mathrm{W}\mathrm{S}}}$

$=$

$-3{𝒜}\left(\frac{n+1}{n}\right)\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-1}+n{ℬ}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}\right(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{-3/n}+\frac{4\pi }{3}(\frac{R}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^3,$

in which case the coefficients of the generic free-energy expression are,

$a$	$=$	$\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2}\left(\frac{n+1}{n}\right)(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^2=\frac{3}{5}\cdot (\frac{4\pi }{3}{)}^2\left(\frac{n+1}{n}\right){\widetilde{\mathfrak{𝔣}}}_W{\mathfrak{𝔪}}^2$
$b$	$=$	$n\left(\frac{3}{4\pi }{)}^{1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{(n+1)/n}=\left(\frac{4\pi n}{3}\right){\widetilde{\mathfrak{𝔣}}}_A{\mathfrak{𝔪}}^{(n+1)/n}$
$c$	$=$	$\frac{4\pi }{3},$

where, as above,

$\mathfrak{𝔪}$

$\equiv$

$\left(\frac{3}{4\pi }\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right).$

Now, if we define the pair of parameters,

$\alpha$	$\equiv$	$\frac{a}{3c}$
$\beta$	$\equiv$	$\frac{b}{nc},$

then the statement of virial equilibrium is,

$x_{\mathrm{e}\mathrm{q}}^4+\alpha$

$=$

$\beta x_{\mathrm{e}\mathrm{q}}^{(n-3)/n},$

and the boundary between dynamical stability and instability occurs at,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$

$=$

$\left[\frac{n-3}{3(n+1)}\right]\alpha .$

Combining these last two expressions means that the boundary between dynamical stability and instability is associated with the parameter condition,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)/n}$	$=$	$[\frac{n-3}{3(n+1)}+1]\frac{\alpha }{\beta }$
$\Rightarrow \left\{\right[\frac{n-3}{3(n+1)}]\alpha {\}}^{(n-3)/(4n)}$	$=$	$\left[\frac{4n}{3(n+1)}\right]\frac{\alpha }{\beta }$
$\Rightarrow \beta {\alpha }^{-3(n+1)/(4n)}$	$=$	$\left[\frac{4n}{3(n+1)}\right][\frac{n-3}{n}\cdot \frac{n}{3(n+1)}{]}^{(3-n)/(4n)}$
	$=$	$4\left[\frac{n}{3(n+1)}{]}^{3(n+1)/(4n)}\right[\frac{n-3}{n}{]}^{(3-n)/(4n)}$
$\Rightarrow (\frac{\beta }{4}{)}^{4n}{\alpha }^{-3(n+1)}$	$=$	$\left[\frac{n}{3(n+1)}{]}^{3(n+1)}\right[\frac{n-3}{n}{]}^{(3-n)}$
$\Rightarrow (\frac{\beta }{4}{)}^{4n}$	$=$	$\left[\frac{n\alpha }{3(n+1)}{]}^{3(n+1)}\right[\frac{n}{n-3}{]}^{n-3}.$

Case M

$a$	$\equiv$	$3{𝒜}=\frac{3}{5}\cdot \frac{{\widetilde{\mathfrak{𝔣}}}_W}{{\widetilde{\mathfrak{𝔣}}}_M^2},$
$b$	$\equiv$	$n{ℬ}=n(\frac{4\pi }{3}{)}^{-1/n}\frac{{\widetilde{\mathfrak{𝔣}}}_A}{{\widetilde{\mathfrak{𝔣}}}_M^{(n+1)/n}},$
$c$	$\equiv$	$\frac{4\pi }{3}\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right).$

Hence,

$\alpha$	$=$	$\left(\frac{4\pi }{15}\right){\widetilde{\mathfrak{𝔣}}}_W\left(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^2\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}$
$\beta$	$=$	${\widetilde{\mathfrak{𝔣}}}_A\left(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{(n+1)/n}\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-1}.$

So the dynamical stability conditions are:

$\left(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\left(\frac{n}{n-3}\right)[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$

$=$

$\left[\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}\right](\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^2;$

and,

$\left(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{4(n+1)}(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-4n}$	$=$	$\left[\frac{n}{3(n+1)}{]}^{3(n+1)}\right[\frac{n}{n-3}{]}^{n-3}\left(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{15}{)}^{3(n+1)}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{6(n+1)}(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{-3(n+1)}$
$\Rightarrow (\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}$	$=$	$\left[\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{3(n+1)}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{2(n+1)}\left[\frac{n}{n-3}\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){]}^{n-3}$
$\Rightarrow \left[\frac{n}{n-3}\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}){]}^{n-3}$	$=$	$\left(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}\right[\left(\frac{n}{n+1}\right)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{-3(n+1)}(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{-2(n+1)}.$

Together, then,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{4(n-3)}$	$=$	$\left[\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{n-3}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{2(n-3)}\left[\right(\frac{P_e}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})\frac{n}{n-3}{]}^{-(n-3)}$
	$=$	$\left[\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{n-3}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{2(n-3)}\left(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{-4n}\right[\left(\frac{n}{n+1}\right)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{3(n+1)}(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{2(n+1)}$
	$=$	$\left[\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^{4n}\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{4(n-1)}(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{-4n}$
$\Rightarrow [x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{(n-3)}$	$=$	$\left[\frac{4}{{\widetilde{\mathfrak{𝔣}}}_A}\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^n\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{(n-1)}.$

Case P

$a$	$=$	$\frac{3}{5}\cdot \left(\frac{4\pi }{3}{)}^2\right(\frac{n+1}{n}){\widetilde{\mathfrak{𝔣}}}_W{\mathfrak{𝔪}}^2$
$b$	$=$	$\left(\frac{4\pi n}{3}\right){\widetilde{\mathfrak{𝔣}}}_A{\mathfrak{𝔪}}^{(n+1)/n}$
$c$	$=$	$\frac{4\pi }{3},$

where, as above,

$\mathfrak{𝔪}$

$\equiv$

$\left(\frac{3}{4\pi }\right)\frac{1}{{\widetilde{\mathfrak{𝔣}}}_M}\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{M_{\mathrm{S}\mathrm{W}\mathrm{S}}}\right).$

Hence,

$\alpha$	$=$	$\frac{1}{5}\cdot \left(\frac{4\pi }{3}\right)\left(\frac{n+1}{n}\right){\widetilde{\mathfrak{𝔣}}}_W{\mathfrak{𝔪}}^2$
$\beta$	$=$	${\widetilde{\mathfrak{𝔣}}}_A{\mathfrak{𝔪}}^{(n+1)/n}.$

So the dynamical stability conditions are:

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^4$	$=$	$\left[\frac{n}{3(n+1)}\right]\left[\frac{n-3}{n}\right]\frac{1}{5}\cdot \left(\frac{4\pi }{3}\right)\left(\frac{n+1}{n}\right){\widetilde{\mathfrak{𝔣}}}_W{\mathfrak{𝔪}}^2$
	$=$	$\left[\frac{n-3}{n}\right]\left(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}\right){\mathfrak{𝔪}}^2$

and,

$(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}{\mathfrak{𝔪}}^{4(n+1)}$	$=$	$\left[\right(\frac{4\pi }{3^2\cdot 5}\left){\widetilde{\mathfrak{𝔣}}}_W{\mathfrak{𝔪}}^2{]}^{3(n+1)}\right[\frac{n}{n-3}{]}^{n-3}$
$\Rightarrow {\mathfrak{𝔪}}^{2(n+1)}$	$=$	$\left[\right(\frac{4\pi }{3^2\cdot 5}\left){\widetilde{\mathfrak{𝔣}}}_W{]}^{-3(n+1)}\right[\frac{n}{n-3}{]}^{-(n-3)}(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}.$

Together, then,

$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{4(n+1)}$	$=$	$\left[\frac{n-3}{n}{]}^{(n+1)}\right(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{)}^{(n+1)}\left(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{)}^{-3(n+1)}\right[\frac{n-3}{n}{]}^{(n-3)}(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}$
	$=$	$\left[\frac{n-3}{n}{]}^{2(n-1)}\right(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{)}^{-2(n+1)}(\frac{{\widetilde{\mathfrak{𝔣}}}_A}{4}{)}^{4n}.$

Compare

Let's see if the two cases, in fact, provide the same answer.

$(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{S}\mathrm{W}\mathrm{S}}}{)}^{n-3}=[\frac{x_{\mathrm{P}}}{x_{\mathrm{M}}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{n-3}$	$=$	$\left\{\right[\frac{n-3}{n}\left]\right(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}\left){\mathfrak{𝔪}}^2{\}}^{(n-3)/4}\right\{\left[\frac{4}{{\widetilde{\mathfrak{𝔣}}}_A}\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^n\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{(n-1)}{\}}^{-1}$
	$=$	$\left\{\right[\frac{n-3}{n}\left]\right(\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}\left){\mathfrak{𝔪}}^2{\}}^{(n-3)/4}\right\{\left[\frac{4}{{\widetilde{\mathfrak{𝔣}}}_A}\right(\frac{n}{n+1}\left)\frac{4\pi {\widetilde{\mathfrak{𝔣}}}_W}{3^2\cdot 5}{]}^n\right(\frac{3}{4\pi {\widetilde{\mathfrak{𝔣}}}_M}{)}^{(n-1)}{\}}^{-1}$

Five-One Bipolytropes

For analytically prescribed, "five-one" bipolytropes, $n=5$ and $j=1$, in which case,

$x_{\mathrm{e}\mathrm{q}}^{2/5}$	$=$	$\left(\frac{5}{3b}\right)[a-3c{x}_{\mathrm{e}\mathrm{q}}^{-2}];$
	and
$[x_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$[\frac{18c}{a}{]}^{1/2}.$

More specifically, the expression that describes the free-energy surface is,

${\mathfrak{𝔊}}_{51}^{*}\equiv 2^4\left(\frac{q}{{\nu }^2}\right){\chi }_{\mathrm{e}\mathrm{q}}\left[\frac{{\mathfrak{𝔊}}_{51}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$ $=$ $\frac{1}{{\ell }_i^2}[{\mathrm{X}}^{-3/5}(5{\mathfrak{𝔏}}_i)+{\mathrm{X}}^{-3}(4{\mathfrak{𝔎}}_i)-{\mathrm{X}}^{-1}(3{\mathfrak{𝔏}}_i+12{\mathfrak{𝔎}}_i)].$

Hence, we have,

$a$	$\equiv$	$3{\chi }_{\mathrm{e}\mathrm{q}}({\mathfrak{𝔏}}_i+4{\mathfrak{𝔎}}_i),$
$b$	$\equiv$	$5{\mathfrak{𝔏}}_i{\chi }_{\mathrm{e}\mathrm{q}}^{3/5},$
$c$	$\equiv$	$4{\mathfrak{𝔎}}_i{\chi }_{\mathrm{e}\mathrm{q}}^3,$

and conclude that,

$[{\chi }_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$[\frac{18(4{\mathfrak{𝔎}}_i{\chi }_{\mathrm{e}\mathrm{q}}^3)}{3{\chi }_{\mathrm{e}\mathrm{q}}({\mathfrak{𝔏}}_i+4{\mathfrak{𝔎}}_i)}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}^{1/2}$
	$=$	$[{\chi }_{\mathrm{e}\mathrm{q}}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}[\frac{24{\mathfrak{𝔎}}_i}{({\mathfrak{𝔏}}_i+4{\mathfrak{𝔎}}_i)}{]}^{1/2}$
$\Rightarrow [\frac{24{\mathfrak{𝔎}}_i}{({\mathfrak{𝔏}}_i+4{\mathfrak{𝔎}}_i)}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$1$
$\Rightarrow [\frac{{\mathfrak{𝔏}}_i}{{\mathfrak{𝔎}}_i}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$20.$

Also, from our detailed force balance derivations, we know that,

${\chi }_{\mathrm{e}\mathrm{q}}\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}.$

Zero-Zero Bipolytropes

General Form

In this case, we retain full generality making the substitutions, $n\to n_c$ and $j\to n_e$, to obtain,

$x_{\mathrm{e}\mathrm{q}}^{(n_c-3)/n_c}$	$=$	$\frac{n_c}{3b}[a-(\frac{3c}{n_e}\left)x_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}\right];$
	and
$[x_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left\{\frac{n_e^2(n_c-3)}{3[n_c(n_e+3)-n_e(n_c+3)]}\right\}\frac{a}{c}$
	$=$	$\left[\frac{n_e^2(n_c-3)}{3^2(n_c-n_e)}\right]\frac{a}{c}.$

And here, the expression that describes the free-energy surface is,

${\mathfrak{𝔊}}_{00}^{*}\equiv 5\left(\frac{q}{{\nu }^2}\right){\chi }_{\mathrm{e}\mathrm{q}}\left[\frac{{\mathfrak{𝔊}}_{00}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$ $=$ $\frac{5}{2q^3}[n_c{A}_2{\mathrm{X}}^{-3/n_c}+n_e{B}_2{\mathrm{X}}^{-3/n_e}-3C_2{\mathrm{X}}^{-1}].$

Hence, we have,

$a\equiv 3{\chi }_{\mathrm{e}\mathrm{q}}\left(\frac{5}{2q^3}\right)C_2$	$=$	$3f{\chi }_{\mathrm{e}\mathrm{q}},$
$b\equiv n_c{\chi }_{\mathrm{e}\mathrm{q}}^{3/n_c}\left(\frac{5}{2q^3}\right)A_2$	$\equiv$	$n_c{\chi }_{\mathrm{e}\mathrm{q}}^{3/n_c}[1+q^3(f-1-\mathfrak{𝔉})],$
$c\equiv n_e{\chi }_{\mathrm{e}\mathrm{q}}^{3/n_e}\left(\frac{5}{2q^3}\right)B_2$	$\equiv$	$n_e{\chi }_{\mathrm{e}\mathrm{q}}^{3/n_e}\left(\frac{5}{2q^3}\right)[\frac{2}{5}{q}^{3}f-A_2]$
	$=$	$n_e{\chi }_{\mathrm{e}\mathrm{q}}^{3/n_e}\{f-[1+q^3(f-1-\mathfrak{𝔉})\left]\right\},$

where the definitions of $f$ and $\mathfrak{𝔉}$ are given below. We immediately deduce that the critical equilibrium state is identified by,

$[x_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}$	$=$	$\left\{\frac{f{n}_e(n_c-3)}{3(n_c-n_e)}\right\}[{\chi }_{\mathrm{e}\mathrm{q}}^{(n_e-3)/n_e}{]}_{\mathrm{c}\mathrm{r}\mathrm{i}\mathrm{t}}\{f-[1+q^3(f-1-\mathfrak{𝔉})]{\}}^{-1}$
$\Rightarrow \frac{1}{f}[1+q^3(f-1-\mathfrak{𝔉})]$	$=$	$1-\left[\frac{n_e(n_c-3)}{3(n_c-n_e)}\right]$
	$=$	$\frac{n_c(3-n_e)}{3(n_c-n_e)}.$

From our associated detailed-force-balance derivation, we know that the associated equilibrium radius is,

${\chi }_{\mathrm{e}\mathrm{q}}$

$=$

$\left\{\right(\frac{\pi }{3}\left)2^{2-n_c}{\nu }^{n_c-1}{q}^{3-n_c}\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}){]}^{n_c}{\}}^{1/(n_c-3)}.$

Compare with Five-One

It is worthwhile to set $n_c=5$ and $n_e=1$ in this expression and compare the result to the comparable expression shown above for the "Five-One" Bipolytrope. Here we have,

$[{\chi }_{\mathrm{e}\mathrm{q}}{]}_{51}$	$=$	$\left\{\right(\frac{\pi }{3}\left)2^{-3}{\nu }^4{q}^{-2}\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}){]}^5{\}}^{1/2}$
	$=$	$(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{1}{\sqrt{3}}[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}){]}^{5/2};$

whereas, rewriting the above relation gives,

${\chi }_{\mathrm{e}\mathrm{q}}{|}_{51}$

$=$

$(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{1}{\sqrt{3}}[\frac{(1+{\ell }_i^2{)}^{6/5}}{3{\ell }_i^2}{]}^{5/2}.$

And, here, we should conclude that the critical equilibrium configuration is associated with,

$\frac{1}{f}[1+q^3(f-1-\mathfrak{𝔉})]$

$=$

$\frac{5}{6}.$

See Also

• Index to a Variety of Free-Energy and/or Virial Analyses