SSC/FreeEnergy/PolytropesEmbedded/Pt3B

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

• Core mass, $M_c$, and associated dimensionless mass fraction, $\nu \equiv M_c/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$;
• Polytropic constant in the core, $K_c$.

In general, the warped free-energy surface drapes across a five-dimensional parameter "plane" such that,

$\mathfrak{𝔊}$

$=$

$\mathfrak{𝔊}(R,K_c,M_{\mathrm{t}\mathrm{o}\mathrm{t}},q,\nu ).$

Focus on Zero-Zero Free-Energy Expression

Here, we will draw heavily from the following accompanying chapters:

• Analytic Detailed Force Balance Models
• Free-Energy Analysis

From Detailed Force-Balance Models

Equilibrium Radius

First View

In an accompanying chapter we find,

$\frac{P_0{R}_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$

$=$

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

where,

$f$	$\equiv$	$1+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[\frac{1}{q^5}-1+\frac{5}{2}(1-\frac{1}{q^2})],$
$\mathfrak{𝔉}$	$\equiv$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[(-2q^2+3q^3-q^5)+\frac{3}{5}(\frac{{\rho }_e}{{\rho }_c})(-1+5q^2-5q^3+q^5)],$
$\frac{{\rho }_e}{{\rho }_c}$	$=$	$\frac{q^3(1-\nu )}{\nu (1-q^3)}.$

Here, we prefer to normalize the equilibrium radius to $R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$. So, let's replace the central pressure with its expression in terms of $K_c$. Specifically,

$P_0$	$=$	$K_c{\rho }_c^{{\gamma }_c}=K_c[\frac{3M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{4\pi R_i^3}{]}^{{\gamma }_c}=K_c[\frac{3\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi q^3{R}_{\mathrm{e}\mathrm{q}}^3}{]}^{(n_c+1)/n_c}\Rightarrow \frac{P_0}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left[\frac{3}{4\pi }\right(\frac{\nu }{q^3})\frac{1}{{\chi }_{\mathrm{e}\mathrm{q}}^3}{]}^{(n_c+1)/n_c}$
$\Rightarrow K_c[\frac{3\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi q^3{R}_{\mathrm{e}\mathrm{q}}^3}{]}^{(n_c+1)/n_c}\frac{R_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$	$=$	$\left(\frac{3}{2^3\pi }\right)\left(\frac{\nu }{q^3}{)}^2\right[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})]$
$\Rightarrow R_{\mathrm{e}\mathrm{q}}^{(n_c-3)/n_c}$	$=$	$\left(\frac{G}{K_c}\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{(n_c-1)/n_c}\left[\frac{3\nu }{4\pi q^3}{]}^{-(n_c+1)/n_c}\right(\frac{3}{2^3\pi }\left)\right(\frac{\nu }{q^3}{)}^2[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})]$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^{(n_c-3)/n_c}\equiv [\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}^{(n_c-3)/n_c}$	$=$	$\frac{1}{2}\left(\frac{4\pi }{3}{)}^{1/n_c}\right(\frac{\nu }{q^3}{)}^{(n_c-1)/n_c}[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})].$

Or, in terms of ${\gamma }_c$,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$

$=$

$\frac{1}{2}\left(\frac{3}{4\pi }{)}^{1-{\gamma }_c}\right(\frac{\nu }{q^3}{)}^{2-{\gamma }_c}[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})].$

Second View

Alternatively, from our derivation and discussion of analytic detailed force-balance models,

$\left[\frac{R^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right]P_0$

$=$

$\left(\frac{3}{2^3\pi }\right)\frac{{\nu }^2{g}^2}{q^4},$

where,

$[g(\nu ,q){]}^2$

$\equiv$

$1+\left(\frac{{\rho }_e}{{\rho }_0}\right)\left[2\right(1-\frac{{\rho }_e}{{\rho }_0}\left)\right(1-q)+\frac{{\rho }_e}{{\rho }_0}(\frac{1}{q^2}-1\left)\right].$

In order to show that this expression is the same as the other one, above, we need to show that,

$\left(\frac{3}{2^3\pi }\right)\left(\frac{\nu }{q^3}{)}^2\right[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})]$	$=$	$\left(\frac{3}{2^3\pi }\right)\frac{{\nu }^2{g}^2}{q^4}$
$\Rightarrow f-1-\mathfrak{𝔉}$	$=$	$\frac{5}{2q^3}[g^2-1]$
	$=$	$\frac{5}{2q^3}\left(\frac{{\rho }_e}{{\rho }_0}\right)\left[2\right(1-\frac{{\rho }_e}{{\rho }_0}\left)\right(1-q)+\frac{{\rho }_e}{{\rho }_0}(\frac{1}{q^2}-1\left)\right]$
	$=$	$\frac{5}{2q^5}\left(\frac{{\rho }_e}{{\rho }_0}\right)\{2(q^2-q^3)+\frac{{\rho }_e}{{\rho }_0}[1-3q^2+2q^3\left]\right\}.$

Let's see …

$f-1-\mathfrak{𝔉}$	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[\frac{1}{q^5}-1+\frac{5}{2}(1-\frac{1}{q^2})]-\frac{5}{2}(\frac{{\rho }_e}{{\rho }_c}\left)\frac{1}{q^5}\right[(-2q^2+3q^3-q^5)+\frac{3}{5}\left(\frac{{\rho }_e}{{\rho }_c}\right)(-1+5q^2-5q^3+q^5)]$
	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)-\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[(-2q^2+3q^3-q^5)]$
		$-\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}\left[\frac{3}{5}\right(\frac{{\rho }_e}{{\rho }_c})(-1+5q^2-5q^3+q^5)]+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[\frac{1}{q^5}-1+\frac{5}{2}(1-\frac{1}{q^2})]$
	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}\{(q^3-q^5)+(2q^2-3q^3+q^5)\}$
		$+\frac{1}{2}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\frac{1}{q^5}\right[3(1-5q^2+5q^3-q^5)]+\frac{1}{2}(\frac{{\rho }_e}{{\rho }_c}{)}^2\frac{1}{q^5}[2-2q^5+5(q^5-q^3\left)\right]$
	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[(q^3-q^5)+(2q^2-3q^3+q^5)]+\frac{1}{2}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\frac{1}{q^5}\right[3(1-5q^2+5q^3-q^5)+2-2q^5+5(q^5-q^3)]$
	$=$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[2q^2-2q^3]+\frac{5}{2q^5}\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[1-3q^2+2q^3].$

Q.E.D.

Hence, the equilibrium radius can also be written as,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$

$=$

$\frac{1}{2}\left(\frac{3}{4\pi }{)}^{1-{\gamma }_c}\right(\frac{\nu }{q^3}{)}^{2-{\gamma }_c}{q}^2{g}^2;$

or, in terms of the polytropic index,

${\chi }_{\mathrm{e}\mathrm{q}}^{n_c-3}$

$=$

$\frac{1}{2^{n_c}}\left(\frac{4\pi }{3}\right)(\frac{\nu }{q^3}{)}^{n_c-1}(qg{)}^{2n_c}.$

Gravitational Potential Energy

Also from our accompanying discussion, we have,

$\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-{\mathrm{X}}^{-1}\left(\frac{3}{5}\right)\left(\frac{\nu }{q^3}{)}^2{q}^5\right[\frac{1}{2^{n_c}}\left(\frac{4\pi }{3}\right)(\frac{\nu }{q^3}{)}^{n_c-1}(qg{)}^{2n_c}{]}^{-1/(n_c-3)}f(\nu ,q)$
	$=$	$-{\mathrm{X}}^{-1}\left(\frac{6}{5}\right)q^{5}f\left[2^{n_c-(n_c-3)}\right(\frac{3}{4\pi }\left)\right(\frac{\nu }{q^3}{)}^{(1-n_c)+2(n_c-3)}{b}_{\xi }^{n_c}{]}^{1/(n_c-3)}$
	$=$	$-{\mathrm{X}}^{-1}\left(\frac{6}{5}\right)q^{5}f\left[\right(\frac{6}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{b}_{\xi }^{n_c}{]}^{1/(n_c-3)}.$

Internal Energy Components

First View

Before writing out the expressions for the internal energy of the core and of the envelope, we note from our separate detailed derivation that, in either case,

$[\frac{P_i{\chi }^{3\gamma }}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}{\chi }^{3-3\gamma }$	$=$	$\left[\right(\frac{P_i}{P_0}\left)\right(\frac{P_0}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left){\chi }^3{]}_{\mathrm{e}\mathrm{q}}\right[\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{]}^{3-3\gamma }$
	$=$	$\left\{\right(\frac{P_i}{P_0}\left)\right[\frac{3}{4\pi }\left(\frac{\nu }{q^3}\right){]}^{{\gamma }_c}{\chi }^{3-3{\gamma }_c}{\}}_{\mathrm{e}\mathrm{q}}{\mathrm{X}}^{3-3\gamma },$

where, in equilibrium,

$(\frac{P_i}{P_0}{)}_{\mathrm{e}\mathrm{q}}$	$=$	$1-b_{\xi }{q}^2$
$b_{\xi }{q}^2$	$=$	$\{\frac{2}{5}{q}^{3}f+[1-\frac{2}{5}{q}^3(1+\mathfrak{𝔉})]{\}}^{-1}$
	$=$	$[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}){]}^{-1}$

So, copying from our accompanying detailed derivation, we have,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\frac{4\pi /3}{({\gamma }_c-1)}\left\{\right(\frac{P_i}{P_0}\left)\right[\frac{3}{4\pi }\left(\frac{\nu }{q^3}\right){]}^{{\gamma }_c}{\chi }^{3-3{\gamma }_c}{\}}_{\mathrm{e}\mathrm{q}}{\mathrm{X}}^{3-3{\gamma }_c}\left\{\right(\frac{P_0}{P_{ic}}\left)\right[q^3-\left(\frac{3b_{\xi }}{5}\right)q^5\left]\right\}$
	$=$	$\frac{1}{({\gamma }_c-1)}\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]{\mathrm{X}}^{3-3{\gamma }_c}{q}^3[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right],$
$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\frac{4\pi /3}{({\gamma }_e-1)}\left\{\right(\frac{P_i}{P_0}\left)\right[\frac{3}{4\pi }\left(\frac{\nu }{q^3}\right){]}^{{\gamma }_c}{\chi }^{3-3{\gamma }_c}{\}}_{\mathrm{e}\mathrm{q}}{\mathrm{X}}^{3-3{\gamma }_e}\{(1-q^3)+b_{\xi }(\frac{P_0}{P_{ie}}\left)\right[\frac{2}{5}{q}^5\mathfrak{𝔉}\left]\right\}$
	$=$	$\frac{1}{({\gamma }_e-1)}\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]{\mathrm{X}}^{3-3{\gamma }_e}\left(\frac{P_i}{P_0}\right)\{(1-q^3)+b_{\xi }(\frac{P_0}{P_{ie}}\left)\right[\frac{2}{5}{q}^5\mathfrak{𝔉}\left]\right\}$
	$=$	$\frac{1}{({\gamma }_e-1)}\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]{\mathrm{X}}^{3-3{\gamma }_e}\{(1-b_{\xi }{q}^2)(1-q^3)+b_{\xi }[\frac{2}{5}{q}^5\mathfrak{𝔉}\left]\right\}$
	$=$	$\frac{1}{({\gamma }_e-1)}\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]{\mathrm{X}}^{3-3{\gamma }_e}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}.$

Furthermore,

$\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]$	$=$	$\left(\frac{3}{4\pi }{)}^{{\gamma }_c-1}\right(\frac{\nu }{q^3}{)}^{{\gamma }_c}\{{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}{\}}^{(3-3{\gamma }_c)/(4-3{\gamma }_c)}$
	$=$	$\left(\frac{3}{4\pi }{)}^{{\gamma }_c-1}\right(\frac{\nu }{q^3}{)}^{{\gamma }_c}\left\{\frac{1}{2}\right(\frac{3}{4\pi }{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{2-{\gamma }_c}\right[q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})]{\}}^{(3-3{\gamma }_c)/(4-3{\gamma }_c)}$
	$=$	$\left(\frac{3}{4\pi }{)}^{({\gamma }_c-1)/(4-3{\gamma }_c)}\right(\frac{\nu }{q^3}{)}^{(6-5{\gamma }_c)(4-3{\gamma }_c)}\left\{\frac{q^2}{2}\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})]{\}}^{(3-3{\gamma }_c)/(4-3{\gamma }_c)}$
	$=$	$\left(\frac{3}{4\pi }{)}^{1/(n_c-3)}\right(\frac{\nu }{q^3}{)}^{(n_c-5)(n_c-3)}\left\{\frac{q^2}{2}\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})]{\}}^{-3/(n_c-3)}$
	$=$	$\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^3{]}^{1/(n_c-3)}.$

Hence, we have,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$n_c\left[\right(\frac{4\pi }{3}{)}^{1-{\gamma }_c}\left(\frac{\nu }{q^3}{)}^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_c}\right]{\mathrm{X}}^{-3/n_c}{q}^3[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]$
	$=$	$n_c\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^3{]}^{1/(n_c-3)}{\mathrm{X}}^{-3/n_c}{q}^3[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right],$
$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$n_e\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^3{]}^{1/(n_c-3)}{\mathrm{X}}^{-3/n_e}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}.$

Second View

In our accompanying discussion of energies associated with detailed force balance models, we used the notation,

$\mathrm{\Pi }$

$\equiv$

$\left(\frac{3}{2^3\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4}(\frac{\nu }{q^3}{)}^2=P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\chi }^{-4}(\frac{3}{2^3\pi }\left)\right(\frac{\nu }{q^3}{)}^2,$

which allows us to rewrite the above quoted relationship between the central pressure and the radius of the bipolytrope as,

$P_0=\mathrm{\Pi }(qg{)}^2.$

We also showed that, in equilibrium, the relationship between the central pressure and the interface pressure is,

$P_0=P_i+{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}{q}^2.$

This means that, in equilibrium, the ratio of the interface pressure to the central pressure is,

$(\frac{P_i}{P_0}{)}_{\mathrm{e}\mathrm{q}}$

$=$

$1-\frac{{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}{q}^2}{P_0}=1-\frac{1}{g^2},$

or given that (see above),

$\frac{5}{2q^3}[g^2-1]$	$=$	$f-1-\mathfrak{𝔉}$
$\Rightarrow g^2$	$=$	$1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}),$

we have,

$(\frac{P_i}{P_0}{)}_{\mathrm{e}\mathrm{q}}$

$=$

$1-\frac{{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}{q}^2}{P_0}=1-[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉}){]}^{-1}.$

This is exactly the pressure-ratio expression presented in our "first view" and unveils the notation association,

$b_{\xi }{q}^2$

$\leftrightarrow$

$\frac{1}{g^2}.$

From our separate derivation, we have, in equilibrium,

${\mathfrak{𝔊}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\left(\frac{2n_c}{3}\right)S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\left(\frac{2n_c}{3}\right)\left(\frac{4\pi }{5}\right)R_{\mathrm{e}\mathrm{q}}^3{q}^5(\frac{5P_i}{2q^2}+\mathrm{\Pi }{)}_{\mathrm{e}\mathrm{q}}$
	$=$	$\left(\frac{q^5{n}_c}{5}\right)R_{\mathrm{e}\mathrm{q}}^3\left(\frac{2^3\pi }{3}\right){\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}\left[\frac{5}{2q^2}\right(\frac{P_i}{\mathrm{\Pi }}{)}_{\mathrm{e}\mathrm{q}}+1]$
	$=$	$\left(\frac{n_c}{5}\right)\left[R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\right]{\chi }_{\mathrm{e}\mathrm{q}}^{-1}\left(\frac{{\nu }^2}{q}\right)\left[\frac{5}{2q^2}\right(\frac{P_i}{P_0}{)}_{\mathrm{e}\mathrm{q}}(\frac{P_0}{\mathrm{\Pi }}{)}_{\mathrm{e}\mathrm{q}}+1]$
$\Rightarrow [\frac{{\mathfrak{𝔊}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{n_c}{5}\right)\left(\frac{{\nu }^2}{q}\right)\left[\frac{5}{2q^2}\right(1-\frac{1}{g^2}\left)\right(q^2{g}^2)+1]{\chi }_{\mathrm{e}\mathrm{q}}^{-1}$
	$=$	$\left(\frac{n_c}{2}\right)\left(\frac{{\nu }^2}{q}\right)[g^2-\frac{3}{5}]\left\{\frac{1}{2^{n_c}}\right(\frac{4\pi }{3}\left)\right(\frac{\nu }{q^3}{)}^{n_c-1}(qg{)}^{2n_c}{\}}^{-1/(n_c-3)}$
	$=$	$n_c[1-(\frac{3}{5}\left)\frac{1}{g^2}\right]\left(\frac{1}{2}\right)\left(\frac{{\nu }^2}{q}\right)g^2\left\{2^{n_c}\right(\frac{3}{4\pi }\left)\right(\frac{\nu }{q^3}{)}^{1-n_c}(qg{)}^{-2n_c}{\}}^{1/(n_c-3)}$
	$=$	$n_c[1-(\frac{3}{5}\left)\frac{1}{g^2}\right]\{2^{n_c}\cdot 2^{(3-n_c)}(\frac{3}{4\pi }\left)\right(\frac{\nu }{q^3}{)}^{1-n_c}(\frac{\nu }{q^3}{)}^{2(n_c-3)}{q}^{5(n_c-3)}{q}^{-2n_c}{g}^{-2n_c}{g}^{2(n_c-3)}{\}}^{1/(n_c-3)}$
	$=$	$n_c[1-(\frac{3}{5}\left)\frac{1}{g^2}\right]\left\{\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{q}^{3n_c-15}{g}^{-6}{\}}^{1/(n_c-3)}.$

Finally, switching from the $g$ notation to the $b_{\xi }$ notation gives,

$[\frac{{\mathfrak{𝔊}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$n_c[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]\left\{\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{q}^{3n_c-15}{b}_{\xi }^3{q}^6{\}}^{1/(n_c-3)}$
	$=$	$n_c{q}^3[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]\left\{\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{b}_{\xi }^3{\}}^{1/(n_c-3)},$

which, after setting $\mathrm{X}=1$, precisely matches the above, "first view" expression. Also from our previous derivation, we can write,

${\mathfrak{𝔊}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\left(\frac{2n_e}{3}\right)S_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$2\pi \left(\frac{2n_e}{3}\right)R_{\mathrm{e}\mathrm{q}}^3{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}\{(1-q^3)(\frac{P_i}{\mathrm{\Pi }}{)}_{\mathrm{e}\mathrm{q}}+\left(\frac{{\rho }_e}{{\rho }_0}\right)[(-2q^2+3q^3-q^5)+\frac{3}{5}(\frac{{\rho }_e}{{\rho }_0})(-1+5q^2-5q^3+q^5)]\}$
	$=$	$2\pi \left(\frac{2n_e}{3}\right)R_{\mathrm{e}\mathrm{q}}^3\left[P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\chi }^{-4}\right(\frac{3}{2^3\pi }\left)\right(\frac{\nu }{q^3}{)}^2{]}_{\mathrm{e}\mathrm{q}}\{(1-q^3)q^2(g^2-1)+(\frac{2}{5}\left)q^5\mathfrak{𝔉}\right\}$
	$=$	$\left[P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3\right]\frac{n_e}{2}\left(\frac{{\nu }^2}{q^4}\right)(1-q^3)\{(g^2-1)+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}{\chi }_{\mathrm{e}\mathrm{q}}^{-1}$
$\Rightarrow [\frac{{\mathfrak{𝔊}}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$n_e(1-q^3)\{(g^2-1)+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}\frac{q^2}{2}\left(\frac{\nu }{q^3}{)}^2\right[\frac{1}{2^{n_c}}\left(\frac{4\pi }{3}\right)(\frac{\nu }{q^3}{)}^{n_c-1}(qg{)}^{2n_c}{]}^{-1/(n_c-3)}$
	$=$	$n_e(1-q^3)\{(g^2-1)+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}\left[2^{[n_c-(n_c-3)]}\right(\frac{3}{4\pi }\left)\right(\frac{\nu }{q^3}{)}^{(1-n_c)+2(n_c-3)}{q}^{2(n_c-3)-2n_c}{g}^{-2n_c}{]}^{1/(n_c-3)}$
	$=$	$n_e(1-q^3)\{(g^2-1)+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{q}^{-6}{g}^{-2n_c}{]}^{1/(n_c-3)}.$

And, finally, switching from the $g$ notation to the $b_{\xi }$ notation gives,

$[\frac{{\mathfrak{𝔊}}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$n_e(1-q^3)(b_{\xi }{q}^2{)}^{-1}\{1-[1-\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right]b_{\xi }{q}^2\left\}\right[\left(\frac{2\cdot 3}{\pi }\right)(\frac{\nu }{q^3}{)}^{n_c-5}{q}^{-6}(b_{\xi }{q}^2{)}^{n_c}{]}^{1/(n_c-3)}$
	$=$	$n_e(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{q}^{-6-2(n_c-3)+2n_c}{b}_{\xi }^{3-n_c+n_c}{]}^{1/(n_c-3)}$
	$=$	$n_e\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{b}_{\xi }^3{]}^{1/(n_c-3)}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\},$

which, after setting $\mathrm{X}=1$, precisely matches the above, "first view" expression.

Summary00

In summary, the desired out of equilibrium free-energy expression is,

$\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$A_0{\mathrm{X}}^{-3/n_c}+B_0{\mathrm{X}}^{-3/n_e}-C_0{\mathrm{X}}^{-1}$

where,

$A_0\equiv (\frac{{\mathfrak{𝔖}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{n_c}{b_{\xi }}\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^{n_c}{]}^{1/(n_c-3)}{q}^3[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right],$
$B_0\equiv (\frac{{\mathfrak{𝔖}}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{n_e}{b_{\xi }}\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^{n_c}{]}^{1/(n_c-3)}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\},$
$C_0\equiv (\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{6}{5}\right)q^{5}f\left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{n_c-5}{b}_{\xi }^{n_c}{]}^{1/(n_c-3)}.$

Or, in a more compact form,

${\mathfrak{𝔊}}^{*}\equiv \left[\right(\frac{2\cdot 3}{\pi }\left)\right(\frac{\nu }{q^3}{)}^{(n_c-5)}{b}_{\xi }^{n_c}{]}^{-1/(n_c-3)}\left[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$n_c{A}_1{\mathrm{X}}^{-3/n_c}+n_e{B}_1{\mathrm{X}}^{-3/n_e}-3C_1{\mathrm{X}}^{-1}$

where,

$A_1$	$\equiv$	$\frac{1}{b_{\xi }}(q^3)[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right],$
$B_1$	$\equiv$	$\frac{1}{b_{\xi }}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\},$
$C_1$	$\equiv$	$\left(\frac{2}{5}\right)q^{5}f.$

Let's examine the behavior of the first radial derivative.

$\frac{\partial {\mathfrak{𝔊}}^{*}}{\partial \mathrm{X}}$

$=$

$\frac{3}{\mathrm{X}}[-A_1{\mathrm{X}}^{-3/n_c}-B_1{\mathrm{X}}^{-3/n_e}+C_1{\mathrm{X}}^{-1}].$

Let's see whether the sum of terms inside the square brackets is zero at the derived equilibrium radius, that is, when $\mathrm{X}=1$ and, hence, when

$\chi ={\chi }_{\mathrm{e}\mathrm{q}}$	$=$	$\left[\frac{1}{2^{n_c}}\right(\frac{4\pi }{3}\left)\right(\frac{\nu }{q^3}{)}^{n_c-1}(qg{)}^{2n_c}{]}^{1/(n_c-3)}$
	$=$	$\left[\frac{1}{2^{n_c}}\right(\frac{4\pi }{3}\left)\right(\frac{\nu }{q^3}{)}^{n_c-1}{b}_{\xi }^{-n_c}{]}^{1/(n_c-3)}.$

$C_1-A_1-B_1$	$=$	$\left(\frac{2}{5}\right)q^{5}f-\frac{1}{b_{\xi }}(q^3)[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]-\frac{1}{b_{\xi }}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}$
	$=$	$\left(\frac{2}{5}\right)q^{5}f-\frac{1}{b_{\xi }}\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}+\frac{q^3}{b_{\xi }}\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}-\frac{q^3}{b_{\xi }}[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]$
	$=$	$\left(\frac{2}{5}\right)q^{5}f-\frac{1}{b_{\xi }}+[1-\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right]q^2+\frac{q^3}{b_{\xi }}-[1-\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right]q^5-\frac{q^3}{b_{\xi }}+\left(\frac{3}{5}\right)q^5$
	$=$	$q^2\left\{\right(\frac{2}{5})q^{3}f-\frac{1}{b_{\xi }{q}^2}+[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}](1-q^3)+(\frac{3}{5}\left)q^3\right\}$
	$=$	$q^2\left\{\right(\frac{2}{5})q^{3}f-[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})]+[(1-q^3)-\frac{2}{5}{q}^3\mathfrak{𝔉}]+(\frac{3}{5}\left)q^3\right\}$
	$=$	$q^2\left\{0\right\}.$

Q.E.D.

Even slightly better:

$\frac{1}{q^2}\left[\right(\frac{\pi }{2\cdot 3}\left)\right(\frac{\nu }{q^3}{)}^{(5-n_c)}{b}_{\xi }^{-n_c}{]}^{1/(n_c-3)}\left[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$n_c{A}_2{\mathrm{X}}^{-3/n_c}+n_e{B}_2{\mathrm{X}}^{-3/n_e}-3C_2{\mathrm{X}}^{-1},$

or, better yet,

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with Structural $(n_c,n_e)=(0,0)$
$2\left(\frac{q^2}{\nu }{)}^2{\chi }_{\mathrm{e}\mathrm{q}}\right[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}]$ $=$ $n_c{A}_2{\mathrm{X}}^{-3/n_c}+n_e{B}_2{\mathrm{X}}^{-3/n_e}-3C_2{\mathrm{X}}^{-1}$

where, keeping in mind that,

$\frac{1}{(b_{\xi }{q}^2)}$

$=$

$[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})],$

we have,

$A_2\equiv \frac{A_1}{q^2}$	$\equiv$	$\frac{q^3}{(b_{\xi }{q}^2)}[1-(\frac{3}{5}\left)b_{\xi }{q}^2\right]$
	$=$	$q^3\left\{\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})]-(\frac{3}{5}\left)\right\}$
	$=$	$\frac{2}{5}{q}^3[1+q^3(f-1-\mathfrak{𝔉})],$
$B_2\equiv \frac{B_1}{q^2}$	$\equiv$	$\frac{1}{(b_{\xi }{q}^2)}(1-q^3)\{1-[1-\frac{2}{5}\left(\frac{q^3}{1-q^3}\right)\mathfrak{𝔉}\left]b_{\xi }{q}^2\right\}$
	$=$	$(1-q^3)\{\frac{1}{(b_{\xi }{q}^2)}-1+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}$
	$=$	$(1-q^3)\left\{\right[1+\frac{2}{5}{q}^3(f-1-\mathfrak{𝔉})]-1+\frac{2}{5}(\frac{q^3}{1-q^3}\left)\mathfrak{𝔉}\right\}$
	$=$	$\frac{2}{5}{q}^3\{(1-q^3)(f-1-\mathfrak{𝔉})+\mathfrak{𝔉}\}$
	$=$	$\frac{2}{5}{q}^3\{f-[1+q^3(f-1-\mathfrak{𝔉})\left]\right\}$
	$=$	$\frac{2}{5}{q}^{3}f-A_2,$
$C_2\equiv \frac{C_1}{q^2}$	$\equiv$	$\frac{2}{5}{q}^{3}f.$

As before, the equilibrium system is dynamically unstable if ${\partial }^2\mathfrak{𝔊}/\partial {\mathrm{X}}^2<0$. We have deduced that the system is unstable if,

$\frac{n_e}{3}\left[\frac{3-n_e}{n_c-n_e}\right]$

$<$

$\frac{A_2}{C_2}=\frac{1}{f}[1+q^3(f-1-\mathfrak{𝔉})].$

See Also

In October 2023, this very long chapter was subdivided in order to more effectively accommodate edits. Here is a list of the resulting set of shorter chapters:

1. Free-Energy Synopsis
2. Free-Energy of Truncated Polytropes
3. Free-Energy of BiPolytropes
   • Focus on Five-One Free-Energy Expression
   • Focus on Zero-Zero Free-Energy Expression
   • Overview