SSC/FreeEnergy/PolytropesEmbedded/Pt3A

Background

Index to original, very long chapter

Free-Energy of Bipolytropes

In this case, the Gibbs-like free energy is given by the sum of four separate energies,

$\mathfrak{𝔊}$

$=$

$[W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+{\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+[W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+{\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}.$

In addition to specifying (generally) separate polytropic indexes for the core, $n_c$, and envelope, $n_e$, and an envelope-to-core mean molecular weight ratio, ${\mu }_e/{\mu }_c$, we will assume that the system is fully defined via specification of the following five physical parameters:

• Total mass, $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$;
• Total radius, $R$;
• Interface radius, $R_i$, and associated dimensionless interface marker, $q\equiv R_i/R$;
• 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,

$\mathfrak{𝔊}$

$=$

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

Order of Magnitude Derivation

Let's begin by providing very rough, approximate expressions for each of these four terms, assuming that $n_c=5$ and $n_e=1$.

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\approx$	$-{\mathfrak{𝔞}}_c\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}{M}_c}{(R_i/2)}\right]=-2{\mathfrak{𝔞}}_c\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right(\frac{\nu }{q}\left)\right];$
$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\approx$	$-{\mathfrak{𝔞}}_e\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}{M}_e}{(R_i+R)/2}\right]=-2{\mathfrak{𝔞}}_e\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right(\frac{1-\nu }{1+q}\left)\right];$
${\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=U_{\mathrm{i}\mathrm{n}\mathrm{t}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\approx$	${\mathfrak{𝔟}}_c\cdot n_c{K}_c{M}_c({\overline{\rho }}_c{)}^{1/n_c}=5{\mathfrak{𝔟}}_c\cdot K_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu [\frac{3M_c}{4\pi R_i^3}{]}^{1/5}$
	$=$	${\mathfrak{𝔟}}_c(\frac{3\cdot 5^5}{2^2\pi }{)}^{1/5}{K}_c(M_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu {)}^{6/5}(Rq{)}^{-3/5};$
${\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}=U_{\mathrm{i}\mathrm{n}\mathrm{t}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\approx$	${\mathfrak{𝔟}}_e\cdot n_e{K}_e{M}_{\mathrm{e}\mathrm{n}\mathrm{v}}({\overline{\rho }}_e{)}^{1/n_e}={\mathfrak{𝔟}}_e\cdot K_e{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}(1-\nu )[\frac{3M_{\mathrm{e}\mathrm{n}\mathrm{v}}}{4\pi (R^3-R_i^3)}]$
	$=$	${\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }\right)K_e[M_{\mathrm{t}\mathrm{o}\mathrm{t}}(1-\nu ){]}^2[R^3(1-q^3){]}^{-1}.$

In writing this last expression, it has been necessary to (temporarily) introduce a sixth physical parameter, namely, the polytropic constant that characterizes the envelope material, $K_e$. But this constant can be expressed in terms of $K_c$ via a relation that ensures continuity of pressure across the interface while taking into account the drop in mean molecular weight across the interface, that is,

$K_e({\overline{\rho }}_e{)}^{(n_e+1)/n_e}$	$\approx$	$K_c({\overline{\rho }}_c{)}^{(n_c+1)/n_c}$
$\Rightarrow K_e\left[\right(\frac{{\mu }_e}{{\mu }_c}){\overline{\rho }}_c{]}^2$	$\approx$	$K_c({\overline{\rho }}_c{)}^{6/5}$
$\Rightarrow \frac{K_e}{K_c}(\frac{{\mu }_e}{{\mu }_c}{)}^2$	$\approx$	$[\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu }{4\pi (Rq{)}^3}{]}^{-4/5}.$

Hence, the fourth energy term may be rewritten in the form,

${\mathfrak{𝔖}}_{\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{m}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}=U_{\mathrm{i}\mathrm{n}\mathrm{t}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\approx$	${\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }\right)\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{K}_c\right[\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu }{4\pi (Rq{)}^3}{]}^{-4/5}[M_{\mathrm{t}\mathrm{o}\mathrm{t}}(1-\nu ){]}^2[R^3(1-q^3){]}^{-1}$
	$=$	${\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }{)}^{1/5}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{K}_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6/5}{R}^{-3/5}[\frac{q^3}{\nu }{]}^{4/5}\frac{(1-\nu {)}^2}{(1-q^3)}.$

Putting all the terms together gives,

$\mathfrak{𝔊}$	$\approx$	$-2{\mathfrak{𝔞}}_c\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right(\frac{\nu }{q}\left)\right]-2{\mathfrak{𝔞}}_e\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right(\frac{1-\nu }{1+q}\left)\right]+{\mathfrak{𝔟}}_c(\frac{3\cdot 5^5}{2^2\pi }{)}^{1/5}{K}_c(M_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu {)}^{6/5}(Rq{)}^{-3/5}$
		$+{\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }{)}^{1/5}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{K}_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6/5}{R}^{-3/5}[\frac{q^3}{\nu }{]}^{4/5}\frac{(1-\nu {)}^2}{(1-q^3)}$
	$=$	$-2{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right]+{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}{K}_c[\frac{(\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}{)}^2}{qR}{]}^{3/5}$
$\Rightarrow \frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-2{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right](\frac{G^3}{K_c^5}{)}^{1/2}+{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}(\frac{{\nu }^2}{q}{)}^{3/5}{K}_c\left[\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}{]}^{3/5}\right(\frac{G^3}{K_c^5}{)}^{1/2}$
	$=$	$-2{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R}\right]+{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}\left(\frac{{\nu }^2}{q}{)}^{3/5}\right[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R}{]}^{3/5},$

where,

${{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}$	$\equiv$	$\left[{\mathfrak{𝔞}}_c\right(\frac{\nu }{q})+{\mathfrak{𝔞}}_e(\frac{1-\nu }{1+q}\left)\right],$
${{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}$	$\equiv$	$\left(\frac{3}{2^2\pi }{)}^{1/5}\right[5{\mathfrak{𝔟}}_c+{\mathfrak{𝔟}}_e\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\frac{q^3(1-\nu {)}^2}{{\nu }^2(1-q^3)}\right].$

Equilibrium Radius

Order of Magnitude Estimate

This means that,

$\frac{\partial \mathfrak{𝔊}}{\partial R}$

$=$

$+2{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}\left[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^2}\right]-\frac{3}{5}{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}{K}_c[\frac{{\nu }^2}{q}{]}^{3/5}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6/5}{R}^{-8/5}.$

Hence, because equilibrium radii are identified by setting $\partial \mathfrak{𝔊}/\partial R=0$, we have,

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

$=$

$\left(\frac{2\cdot 5}{3}{)}^{5/2}\right[\frac{{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}}{{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}}{]}^{5/2}(\frac{q}{{\nu }^2}{)}^{3/2}.$

Reconcile With Known Analytic Expression

From our earlier derivations, it appears as though,

${\chi }_{\mathrm{e}\mathrm{q}}\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\left(\frac{3^8}{2^5\pi }{)}^{-1/2}\right(\frac{3}{2^4}\left)\right(\frac{q}{{\ell }_i}{)}^5\left(\frac{\nu }{q^3}{)}^2\right(1+{\ell }_i^2{)}^3$
	$=$	$\left(\frac{2\cdot 5}{3}{)}^{5/2}\right(\frac{q}{{\nu }^2}{)}^{3/2}\left[\right(\frac{\pi }{2^8\cdot 3\cdot 5^5}{)}^{1/2}\left(\frac{{\nu }^2}{q}{)}^{5/2}\frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^5}\right].$

This implies that,

$\frac{{{𝒜}}_{\mathrm{b}\mathrm{i}\mathrm{P}}}{{{ℬ}}_{\mathrm{b}\mathrm{i}\mathrm{P}}}$	$\approx$	$\left[\right(\frac{\pi }{2^8\cdot 3\cdot 5^5}{)}^{1/2}(\frac{{\nu }^2}{q}{)}^{5/2}\frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^5}{]}^{2/5}$
	$=$	$\left(\frac{{\nu }^2}{q}\right)(\frac{\pi }{2^8\cdot 3\cdot 5^5}{)}^{1/5}\frac{(1+{\ell }_i^2{)}^{6/5}}{{\ell }_i^2}$
$\Rightarrow \left[{\mathfrak{𝔞}}_c\right(\frac{\nu }{q})+{\mathfrak{𝔞}}_e(\frac{1-\nu }{1+q}\left)\right]$	$\approx$	$\frac{1}{2^2\cdot 5}\left(\frac{{\nu }^2}{q}\right)\frac{(1+{\ell }_i^2{)}^{6/5}}{{\ell }_i^2}[5{\mathfrak{𝔟}}_c+{\mathfrak{𝔟}}_e(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\frac{q^3(1-\nu {)}^2}{{\nu }^2(1-q^3)}]$
$\Rightarrow [{\mathfrak{𝔞}}_c+{\mathfrak{𝔞}}_e\cdot \frac{q(1-\nu )}{\nu (1+q)}]$	$\approx$	$\frac{\nu }{2^2\cdot 5}\frac{(1+{\ell }_i^2{)}^{6/5}}{{\ell }_i^2}[5{\mathfrak{𝔟}}_c+{\mathfrak{𝔟}}_e(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\frac{q^3(1-\nu {)}^2}{{\nu }^2(1-q^3)}]$

Focus on Five-One Free-Energy Expression

Approximate Expressions

Let's plug this equilibrium radius back into each term of the free-energy expression.

$\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\approx$	$-2{\mathfrak{𝔞}}_c\left(\frac{G^3}{K_c^5}{)}^{1/2}\right[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\left(\frac{\nu }{q}\right)]$
	$=$	$-2{\mathfrak{𝔞}}_c\left(\frac{\nu }{q}\right)\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{e}\mathrm{q}}}\right];$
$\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\approx$	$-2{\mathfrak{𝔞}}_e\left(\frac{G^3}{K_c^5}{)}^{1/2}\right[\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}}\left(\frac{1-\nu }{1+q}\right)]$
	$=$	$-2{\mathfrak{𝔞}}_e\left(\frac{1-\nu }{1+q}\right)\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{e}\mathrm{q}}}\right];$
$\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left[\frac{3({\gamma }_c-1)}{2}\right]\frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\approx$	$\left[\frac{3}{2\cdot 5}\right]{\mathfrak{𝔟}}_c\left(\frac{3\cdot 5^5}{2^2\pi }{)}^{1/5}\right(\frac{G^3}{K_c^5}{)}^{1/2}{K}_c(M_{\mathrm{t}\mathrm{o}\mathrm{t}}\nu {)}^{6/5}(R_{\mathrm{e}\mathrm{q}}q{)}^{-3/5}$
	$=$	$\left[\frac{3}{2\cdot 5}\right]{\mathfrak{𝔟}}_c\left(\frac{3\cdot 5^5}{2^2\pi }{)}^{1/5}\right(\frac{{\nu }^2}{q}{)}^{3/5}(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{e}\mathrm{q}}}{)}^{3/5};$
$\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}=\left[\frac{3({\gamma }_e-1)}{2}\right]\frac{U_{\mathrm{i}\mathrm{n}\mathrm{t}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\approx$	$\left[\frac{3}{2}\right]{\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }{)}^{1/5}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\left(\frac{G^3}{K_c^5}{)}^{1/2}{K}_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{6/5}{R}_{\mathrm{e}\mathrm{q}}^{-3/5}\right[\frac{q^3}{\nu }{]}^{4/5}\frac{(1-\nu {)}^2}{(1-q^3)}$
	$=$	$\left[\frac{3}{2}\right]{\mathfrak{𝔟}}_e\left(\frac{3}{2^2\pi }{)}^{1/5}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\left[\frac{q^3}{\nu }{]}^{4/5}\frac{(1-\nu {)}^2}{(1-q^3)}\right(\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{R_{\mathrm{e}\mathrm{q}}}{)}^{3/5}.$

From Detailed Force-Balance Models

In the following derivations, we will use the expression,

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

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}{)}^3\right(\frac{\pi }{2^3}{)}^{1/2}\frac{1}{A^2{\eta }_s}=(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}.$

Keep in mind, as well — as derived in an accompanying discussion — that,

$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

$(m_3^2{\ell }_i^3)(1+{\ell }_i^2{)}^{-1/2}[1+(1-m_3{)}^2{\ell }_i^2{]}^{-1/2}[m_3{\ell }_i+(1+{\ell }_i^2)(\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i){]}^{-1},$

where,

$m_3\equiv 3\left(\frac{{\mu }_e}{{\mu }_c}\right).$

From the accompanying Table 1 parameter values, we also can write,

$q$	$=$	$\frac{{\eta }_i}{{\eta }_s}={\eta }_i\{\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}[\frac{1}{{\eta }_i}-{\ell }_i]{\}}^{-1}$
	$=$	${\eta }_i\{{\eta }_i+{\cot }^{-1}[{\ell }_i-\frac{1}{{\eta }_i}]{\}}^{-1},$

where,

${\eta }_i$

$=$

$m_3\left[\frac{{\ell }_i}{(1+{\ell }_i^2)}\right].$

Let's also define the following shorthand notation:

${\mathfrak{𝔏}}_i$	$\equiv$	$\frac{({\ell }_i^4-1)}{{\ell }_i^2}+\frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^3}\cdot {\tan }^{-1}{\ell }_i;$
${\mathfrak{𝔎}}_i$	$\equiv$	$\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}.$

Gravitational Potential Energy of the Core

Pulling from our detailed derivations,

$[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i)].$
$\Rightarrow -{\chi }_{\mathrm{e}\mathrm{q}}[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i)](\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}$
	$=$	$\left(\frac{3}{2^4}\right)\frac{{\nu }^2}{q}\cdot \frac{1}{{\ell }_i^5}\left[{\ell }_i\right({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)+(1+{\ell }_i^2{)}^3{\tan }^{-1}({\ell }_i)]$

Out of equilibrium, then, we should expect,

$\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-{\chi }^{-1}\left(\frac{3}{2^4}\right)\frac{{\nu }^2}{q}\cdot \frac{1}{{\ell }_i^5}\left[{\ell }_i\right({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)+(1+{\ell }_i^2{)}^3{\tan }^{-1}({\ell }_i)]$
	$=$	$-{\chi }^{-1}\left(\frac{3}{2^4}\right)\frac{{\nu }^2}{q}\cdot \frac{1}{{\ell }_i^2}[{\mathfrak{𝔏}}_i-\frac{8}{3}],$

which, in comparison with our above approximate expression, implies,

${\mathfrak{𝔞}}_c$

$=$

$\left(\frac{3}{2^5}\right)\frac{\nu }{{\ell }_i^5}\left[{\ell }_i\right({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)+(1+{\ell }_i^2{)}^3{\tan }^{-1}({\ell }_i)].$

Thermal Energy of the Core

Again, pulling from our detailed derivations,

$[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{2}\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i({\ell }_i^4-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i)]$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^3[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}^5$	$=$	$\frac{1}{2^5}\left(\frac{3^8}{2^5\pi }{)}^{5/2}\right[{\ell }_i({\ell }_i^4-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i){]}^5[(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}{]}^3$
	$=$	$\frac{1}{\pi }\left(\frac{3}{2^2}{)}^{11}\right(\frac{{\nu }^2}{q}{)}^3[{\ell }_i({\ell }_i^4-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i){]}^5\left[\frac{(1+{\ell }_i^2{)}^9}{{\ell }_i^{15}}\right].$

Out of equilibrium, we should then expect,

$\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$\left(\frac{3}{2^2\pi }{)}^{1/5}\right[{\chi }^{-1}\left(\frac{{\nu }^2}{q}\right)\frac{1}{(1+{\ell }_i^2{)}^2}{]}^{3/5}(\frac{3}{2^2}{)}^2{\mathfrak{𝔏}}_i.$

In comparison with our above approximate expression, we therefore have,

$\left[\right(\frac{3}{2\cdot 5}\left){\mathfrak{𝔟}}_c\right(\frac{3\cdot 5^5}{2^2\pi }{)}^{1/5}(\frac{{\nu }^2}{q}{)}^{3/5}{]}^5$	$=$	$\frac{1}{\pi }\left(\frac{3}{2^2}{)}^{11}\right(\frac{{\nu }^2}{q}{)}^3[{\ell }_i({\ell }_i^4-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i){]}^5\left[\frac{(1+{\ell }_i^2{)}^9}{{\ell }_i^{15}}\right]$
$\Rightarrow {\mathfrak{𝔟}}_c$	$=$	$\frac{3}{2^3{\ell }_i^3(1+{\ell }_i^2{)}^{6/5}}[{\ell }_i({\ell }_i^4-1)+(1+{\ell }_i^2{)}^3{\tan }^{-1}({\ell }_i)].$

Gravitational Potential Energy of the Envelope

Again, pulling from our detailed derivations and appreciating, in particular, that (see, for example, our notes on equilibrium conditions),

$A$	$=$	$\frac{{\eta }_i}{\sin ({\eta }_i-B)},$
$({\eta }_s-B)$	$=$	$\pi ,$
${\eta }_i-B$	$=$	$\frac{\pi }{2}-{\tan }^{-1}({\mathrm{\Lambda }}_i),$

$\Rightarrow \sin ({\eta }_i-B)=(1+{\mathrm{\Lambda }}_i^2{)}^{-1/2}$

and

$\sin [2({\eta }_i-B)]=2{\mathrm{\Lambda }}_i(1+{\mathrm{\Lambda }}_i^2{)}^{-1},$

we have,

$[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\left(\frac{1}{2^3\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{A}^2\left\{\right[6({\eta }_s-B)-3\sin [2({\eta }_s-B)]-4{\eta }_s{\sin }^{}({\eta }_s-B)+4B]$
		$-[6({\eta }_i-B)-3\sin [2({\eta }_i-B)]-4{\eta }_i{\sin }^{}({\eta }_i-B)+4B]\}$
	$=$	$-\left(\frac{1}{2^3\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}\left[\frac{{\eta }_i}{\sin ({\eta }_i-B)}{]}^2\right\{6\pi -[6({\eta }_i-B)-3\sin [2({\eta }_i-B)]-4{\eta }_i{\sin }^{}({\eta }_i-B)]\}$
	$=$	$-\left(\frac{1}{2^3\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2(1+{\mathrm{\Lambda }}_i^2)\{6\pi -6[\frac{\pi }{2}-{\tan }^{-1}({\mathrm{\Lambda }}_i)]+6[\frac{{\mathrm{\Lambda }}_i}{(1+{\mathrm{\Lambda }}_i^2)}]+4{\eta }_i[\frac{1}{(1+{\mathrm{\Lambda }}_i^2)}\left]\right\}$
	$=$	$-\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i+\frac{2}{3}\cdot {\eta }_i\}.$

So, in equilibrium we can write,

$-{\chi }_{\mathrm{e}\mathrm{q}}[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i+\frac{2}{3}\cdot {\eta }_i\}(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}$
	$=$	$\frac{3}{2^2}\left(\frac{{\eta }_i}{m_3}{)}^3\right\{\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}+\frac{2}{3}\}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^5}$
	$=$	$\frac{3}{2^2}\left(\frac{{\nu }^2}{q}\right)\frac{1}{{\ell }_i^2}\left\{\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}\right[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}+\frac{2}{3}\}.$

And out of equilibrium,

$\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$-{\chi }^{-1}\cdot \frac{3}{2^2}\left(\frac{{\nu }^2}{q}\right)\frac{1}{{\ell }_i^2}[{\mathfrak{𝔎}}_i+\frac{2}{3}].$

This, in turn, implies that both in and out of equilibrium,

${\mathfrak{𝔞}}_e$

$=$

$\frac{3}{2^3}\left[\frac{{\nu }^2(1+q)}{q(1-\nu )}\right]\frac{1}{{\ell }_i^2}\left\{\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}\right[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}+\frac{2}{3}\}.$

Thermal Energy of the Envelope

Again, pulling from our detailed derivations,

$[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{1}{2^5\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{A}^2\left\{\right[6({\eta }_s-B)-3\sin [2({\eta }_s-B)]]-[6({\eta }_i-B)-3\sin [2({\eta }_i-B)]\left]\right\}$
	$=$	$\left(\frac{1}{2^5\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}\left[\frac{{\eta }_i}{\sin ({\eta }_i-B)}{]}^2\right\{6\pi -6({\eta }_i-B)+3\sin [2({\eta }_i-B)]\}$
	$=$	$\left(\frac{1}{2^5\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2(1+{\mathrm{\Lambda }}_i^2)\left\{6\right[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+6[{\mathrm{\Lambda }}_i(1+{\mathrm{\Lambda }}_i^2{)}^{-1}]\}$
	$=$	$\frac{1}{2}\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i\}.$

So, in equilibrium we can write,

${\chi }_{\mathrm{e}\mathrm{q}}^3[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{2}\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i\}\left[\right(\frac{\pi }{2^3}{)}^{1/2}\frac{{\nu }^2}{q}\cdot \frac{(1+{\ell }_i^2{)}^3}{3^3{\ell }_i^5}{]}^3$
	$=$	$\left(\frac{{\nu }^2}{q}{)}^3\right(\frac{3^2{\pi }^2}{2^{12}}{)}^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^3\right\{\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}\left\}\right[\frac{(1+{\ell }_i^2{)}^9}{3^9{\ell }_i^{15}}]$
	$=$	$\left(\frac{{\nu }^2}{q}{)}^3\right(\frac{\pi }{2^6\cdot 3^5}\left)\right[\frac{(1+{\ell }_i^2{)}^6}{{\ell }_i^{12}}\left]\right\{\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}\}.$

And, out of equilibrium,

$[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$

$=$

${\chi }^{-3}\left(\frac{{\nu }^2}{q}{)}^3\right(\frac{\pi }{2^6\cdot 3^5}\left)\right[\frac{(1+{\ell }_i^2{)}^6}{{\ell }_i^{12}}]\mathfrak{𝔎}.$

Combined in Equilibrium

Notice that, in combination,

$[\frac{2S_{\mathrm{e}\mathrm{n}\mathrm{v}}+W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\frac{2}{3}\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^3$
	$=$	$-\frac{2}{3}\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}\left[3\right(\frac{{\mu }_e}{{\mu }_c}\left){\ell }_i\right(1+{\ell }_i^2{)}^{-1}{]}^3$
	$=$	$-\left(\frac{2\cdot 3^6}{\pi }{)}^{1/2}\right[\frac{{\ell }_i^3}{(1+{\ell }_i^2{)}^3}].$

Also, from above,

$[\frac{2S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i(-\frac{8}{3}{\ell }_i^2)(1+{\ell }_i^2{)}^{-3}]$
	$=$	$+\left(\frac{2\cdot 3^6}{\pi }{)}^{1/2}\right[\frac{{\ell }_i^3}{(1+{\ell }_i^2{)}^3}].$

So, in equilibrium, these terms from the core and envelope sum to zero, as they should.

Out of Equilibrium

And now, in combination out of equilibrium,

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

$=$

$\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-1}\right\{[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}\}+(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-3/5}\left(\frac{2n_c}{3}\right)[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-3}\left(\frac{2n_e}{3}\right)[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}.$

Hence, quite generally out of equilibrium,

$\frac{\partial }{\partial \chi }\left[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$-{\chi }^{-1}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-1}\right\{[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}\}-\frac{3}{5}{\chi }^{-1}(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-3/5}\left(\frac{10}{3}\right)[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}-3{\chi }^{-1}(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{-3}\left(\frac{2}{3}\right)[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}.$

Let's see what the value of this derivative is if the dimensionless radius, $\chi$, is set to the value that has been determined, via a detailed force-balanced analysis, to be the equilibrium radius, namely, $\chi ={\chi }_{\mathrm{e}\mathrm{q}}$. In this case, we have,

$\left\{\frac{\partial }{\partial \chi }\right[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}]{\}}_{\mathrm{\chi }\mathrm{\to }{\mathrm{\chi }}_{\mathrm{e}\mathrm{q}}}$

$=$

$-{\chi }_{\mathrm{e}\mathrm{q}}^{-1}\left\{\right[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+2[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}+2\left[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}\right\}.$

But, according to the virial theorem — and, as we have just demonstrated — the four terms inside the curly braces sum to zero. So this demonstrates that the derivative of our out-of-equilibrium free-energy expression does go to zero at the equilibrium radius, as it should!

Summary51

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

$\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}+\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}+\left(\frac{2n_c}{3}\right)\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}+\left(\frac{2n_e}{3}\right)\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$
	$=$	$-{\chi }^{-1}\left(\frac{3}{2^4}\right)\frac{{\nu }^2}{q}\cdot \frac{1}{{\ell }_i^2}[{\mathfrak{𝔏}}_i-\frac{8}{3}]-{\chi }^{-1}\cdot \frac{3}{2^2}\left(\frac{{\nu }^2}{q}\right)\frac{1}{{\ell }_i^2}[{\mathfrak{𝔎}}_i+\frac{2}{3}]$
		$+\left(\frac{2\cdot 5}{3}\right)\left(\frac{3}{2^2\pi }{)}^{1/5}\right[{\chi }^{-1}\left(\frac{{\nu }^2}{q}\right)\frac{1}{(1+{\ell }_i^2{)}^2}{]}^{3/5}(\frac{3}{2^2}{)}^2{\mathfrak{𝔏}}_i+(\frac{2}{3}\left){\chi }^{-3}\right(\frac{{\nu }^2}{q}{)}^3\left(\frac{\pi }{2^6\cdot 3^5}\right)\left[\frac{(1+{\ell }_i^2{)}^6}{{\ell }_i^{12}}\right]\mathfrak{𝔎}$
	$=$	$-\left(\frac{3}{2^4}\right)[{\chi }^{-1}\frac{{\nu }^2}{q}\cdot \frac{1}{{\ell }_i^2}][{\mathfrak{𝔏}}_i+4{\mathfrak{𝔎}}_i]+\left(\frac{3}{2^2\pi }{)}^{1/5}\right(\frac{3\cdot 5}{2^3}\left)\right[{\chi }^{-1}\left(\frac{{\nu }^2}{q}\right)\frac{1}{(1+{\ell }_i^2{)}^2}{]}^{3/5}{\mathfrak{𝔏}}_i$
		$+\left(\frac{\pi }{2^5\cdot 3^6}\right)\left[{\chi }^{-1}\right(\frac{{\nu }^2}{q})\frac{(1+{\ell }_i^2{)}^2}{{\ell }_i^4}{]}^3\mathfrak{𝔎}.$

Or, in terms of the ratio,

$\mathrm{X}\equiv \frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}},$

and pulling from the above expressions,

$[\frac{W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i)]$
	$=$	$-\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[\frac{{\ell }_i}{(1+{\ell }_i^2)}{]}^3[{\mathfrak{𝔏}}_i-\frac{8}{3}]$
$[\frac{W_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$-\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i+\frac{2}{3}\cdot {\eta }_i\}$
	$=$	$-\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[\frac{{\ell }_i}{(1+{\ell }_i^2)}{]}^3[4{\mathfrak{𝔎}}_i+\frac{8}{3}]$
$[\frac{S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{2}\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[{\ell }_i({\ell }_i^4-1)(1+{\ell }_i^2{)}^{-3}+{\tan }^{-1}({\ell }_i)]$
	$=$	$\frac{1}{2}\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[\frac{{\ell }_i}{(1+{\ell }_i^2)}{]}^3{\mathfrak{𝔏}}_i$
$[\frac{S_{\mathrm{e}\mathrm{n}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{2}\left(\frac{3^2}{2\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{\eta }_i^2\{(1+{\mathrm{\Lambda }}_i^2)[\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i)]+{\mathrm{\Lambda }}_i\}$
	$=$	$\frac{1}{2}\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[\frac{{\ell }_i}{(1+{\ell }_i^2)}{]}^3(4{\mathfrak{𝔎}}_i),$

we have the streamlined,

$\left(\frac{2^5\pi }{3^6}{)}^{1/2}\right[\frac{(1+{\ell }_i^2)}{{\ell }_i}{]}^3\left[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$

$=$

$+{\mathrm{X}}^{-3/5}(5{\mathfrak{𝔏}}_i)+{\mathrm{X}}^{-3}(4{\mathfrak{𝔎}}_i)-{\mathrm{X}}^{-1}(3{\mathfrak{𝔏}}_i+12{\mathfrak{𝔎}}_i)$

or, better yet,

Out-of-Equilibrium, Free-Energy Expression for BiPolytropes with $(n_c,n_e)=(5,1)$
$2^4\left(\frac{q{\ell }_i^2}{{\nu }^2}\right){\chi }_{\mathrm{e}\mathrm{q}}\left[\frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$ $=$ ${\mathrm{X}}^{-3/5}(5{\mathfrak{𝔏}}_i)+{\mathrm{X}}^{-3}(4{\mathfrak{𝔎}}_i)-{\mathrm{X}}^{-1}(3{\mathfrak{𝔏}}_i+12{\mathfrak{𝔎}}_i)$

where,

${\mathfrak{𝔏}}_i$	$\equiv$	$\frac{({\ell }_i^4-1)}{{\ell }_i^2}+\frac{(1+{\ell }_i^2{)}^3}{{\ell }_i^3}\cdot {\tan }^{-1}{\ell }_i,$
${\mathfrak{𝔎}}_i$	$\equiv$	$\frac{{\mathrm{\Lambda }}_i}{{\eta }_i}+\frac{(1+{\mathrm{\Lambda }}_i^2)}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i],$
${\mathrm{\Lambda }}_i$	$\equiv$	$\frac{1}{{\eta }_i}-{\ell }_i,$
${\eta }_i$	$=$	$3\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{{\ell }_i}{(1+{\ell }_i^2)}\right].$

From the accompanying Table 1 parameter values, we also can write,

$\frac{1}{q}$	$=$	$\frac{{\eta }_s}{{\eta }_i}=1+\frac{1}{{\eta }_i}[\frac{\pi }{2}+{\tan }^{-1}{\mathrm{\Lambda }}_i],$
$\nu$	$=$	$\frac{{\ell }_{i}q}{(1+{\mathrm{\Lambda }}_i^2{)}^{1/2}}.$

Radial Derivatives

$\frac{\partial {\mathfrak{𝔊}}^{*}}{\partial \mathrm{X}}$ $=$ $-{\mathrm{X}}^{-8/5}(3{\mathfrak{𝔏}}_i)-{\mathrm{X}}^{-4}(12{\mathfrak{𝔎}}_i)+{\mathrm{X}}^{-2}(3{\mathfrak{𝔏}}_i+12{\mathfrak{𝔎}}_i)$ $\frac{{\partial }^2{\mathfrak{𝔊}}^{*}}{\partial {\mathrm{X}}^2}$ $=$ $\frac{3}{5}[{\mathrm{X}}^{-13/5}(8{\mathfrak{𝔏}}_i)+{\mathrm{X}}^{-5}(80{\mathfrak{𝔎}}_i)-{\mathrm{X}}^{-1}(10{\mathfrak{𝔏}}_i+40{\mathfrak{𝔎}}_i)]$

Consistent with our generic discussion of the stability of bipolytropes and the specific discussion of the stability of bipolytropes having $(n_c,n_e)=(5,1)$, it can straightforwardly be shown that $\partial \mathfrak{𝔊}/\partial \chi =0$ is satisfied by setting $\mathrm{X}=1$; that is, the equilibrium condition is,

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

$=$

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

Furthermore, the equilibrium configuration is unstable whenever ${\partial }^2\mathfrak{𝔊}/\partial {\chi }^2<0$, that is, it is unstable whenever,

$\frac{{\mathfrak{𝔏}}_i}{{\mathfrak{𝔎}}_i}$

$>$

$20.$

Table 1 of an accompanying chapter — and the red-dashed curve in the figure adjacent to that table — identifies some key properties of the model that marks the transition from stable to unstable configurations along equilibrium sequences that have various values of the mean-molecular weight ratio, ${\mu }_e/{\mu }_c$.

See Also