SSC/BipolytropeGeneralization

Bipolytrope Generalization (Pt 1)

Part I:  Bipolytrope Generalization

Part II:  Derivations

Part III:  Examples

Part IV:  Best of the Best

On 26 August 2014, Tohline finished rewriting the chapter titled "Bipolytrope Generalization" in a very concise manner (go here for this Version2 chapter) then set this chapter aside to provide a collection of older attempts at the derivations. While much of what follows is technically correct, it is overly detailed and cumbersome. Because it likely also contains some misguided steps, we label it in entirety as Work in Progress.

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Material that appears after this point in our presentation is under development and therefore
may contain incorrect mathematical equations and/or physical misinterpretations.
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Old Stuff

$\mathfrak{𝔊}$	$=$	$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+{\mathfrak{𝔖}}_A{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+{\mathfrak{𝔖}}_A{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$
	$=$	$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}+\left[\frac{2}{3({\gamma }_c-1)}\right]S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+\left[\frac{2}{3({\gamma }_e-1)}\right]S_{\mathrm{e}\mathrm{n}\mathrm{v}}.$

In addition to the gravitational potential energy, which is naturally written as,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$-\frac{3}{5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right)\cdot {\mathfrak{𝔣}}_{WM},$

it seems reasonable to write the separate thermal energy contributions as,

$S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\frac{3}{2}\left[M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right(\frac{P_{ic}}{{\rho }_{ic}}\left)\right]s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\frac{3}{2}\left[\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}{P}_{ic}\right(\frac{{\rho }_{ic}}{\overline{\rho }}{)}^{-1}\left(\frac{4\pi R^3}{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)]s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=2\pi R^3{P}_{ic}[q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}],$
$S_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\frac{3}{2}\left[M_{\mathrm{e}\mathrm{n}\mathrm{v}}\right(\frac{P_{ie}}{{\rho }_{ie}}\left)\right]s_{\mathrm{e}\mathrm{n}\mathrm{v}}=\frac{3}{2}[(1-\nu )M_{\mathrm{t}\mathrm{o}\mathrm{t}}{P}_{ie}(\frac{{\rho }_{ie}}{\overline{\rho }}{)}^{-1}\left(\frac{4\pi R^3}{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}\right)]s_{\mathrm{e}\mathrm{n}\mathrm{v}}=2\pi R^3{P}_{ie}[(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}],$

where the subscript "$i$" means "at the interface," and ${\mathfrak{𝔣}}_{WM},$ $s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}},$ and $s_{\mathrm{e}\mathrm{n}\mathrm{v}}$ are dimensionless functions of order unity (all three functions to be determined) akin to the structural form factors used in our examination of isolated polytropes.

While exploring how the free-energy function varies across parameter space, we choose to hold $M_{\mathrm{t}\mathrm{o}\mathrm{t}}$ and $K_c$ fixed. By dimensional analysis, it is therefore reasonable to normalize all energies, length scales, densities and pressures by, respectively,

$E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$[\frac{G^{3({\gamma }_c-1)}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{5{\gamma }_c-6}}{K_c}{]}^{1/(3{\gamma }_c-4)},$
$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$\left[\right(\frac{K_c}{G})M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c-2}{]}^{1/(3{\gamma }_c-4)},$
${\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$\frac{3}{4\pi }[\frac{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{K_c^3}{]}^{1/(3{\gamma }_c-4)},$
$P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$	$\equiv$	$[\frac{G^{3{\gamma }_c}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{2{\gamma }_c}}{K_c^4}{]}^{1/(3{\gamma }_c-4)}.$

As is detailed below — first, here, and via an independent derivation, here — quite generally the expression for the normalized free energy is,

${\mathfrak{𝔊}}^{*}\equiv \frac{\mathfrak{𝔊}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$-\frac{3}{5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\left(\frac{1}{R}\right)\cdot {\mathfrak{𝔣}}_{WM}+\left[\frac{4\pi q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{3({\gamma }_c-1)}\right]\left[\frac{R^3{P}_{ic}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]+\left[\frac{4\pi (1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3({\gamma }_e-1)}\right]\left[\frac{R^3{P}_{ie}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]$
	$=$	$-\frac{3\cdot {\mathfrak{𝔣}}_{WM}}{5}{\chi }^{-1}+\left[\frac{4\pi q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{3({\gamma }_c-1)}\right]\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\left[\right(\frac{P_{ic}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left){\chi }^3\right]$
		$+\left[\frac{4\pi (1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3({\gamma }_e-1)}\right]\left[\frac{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right]\left[\right(\frac{P_{ie}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left){\chi }^3\right]$
	$=$	$-\frac{3\cdot {\mathfrak{𝔣}}_{WM}}{5}{\chi }^{-1}+\left[\frac{4\pi q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{3({\gamma }_c-1)}\right][\frac{P_{ic}{\chi }^{3{\gamma }_c}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}{\chi }^{3-3{\gamma }_c}+[\frac{4\pi (1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3({\gamma }_e-1)}\left]\right[\frac{P_{ie}{\chi }^{3{\gamma }_e}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}{\chi }^{3-3{\gamma }_e}$

where we have introduced the parameter, $\nu \equiv M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/M_{\mathrm{t}\mathrm{o}\mathrm{t}}$. After defining the normalized (and dimensionless) configurarion radius, $\chi \equiv R/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$, we can write the normalized free energy of a bipolytrope in the following compact form:

${\mathfrak{𝔊}}^{*}$

$=$

$-3{𝒜}{\chi }^{-1}-\frac{{ℬ}}{(1-{\gamma }_c)}{\chi }^{3-3{\gamma }_c}-\frac{{𝒞}}{(1-{\gamma }_e)}{\chi }^{3-3{\gamma }_e},$

where,

${𝒜}$	$\equiv$	$\frac{1}{5}\cdot {\mathfrak{𝔣}}_{WM},$
${ℬ}$	$\equiv$	$\left(\frac{4\pi }{3}\right)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{P_{ic}{\chi }^{3{\gamma }_c}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}},$
${𝒞}$	$\equiv$	$\left(\frac{4\pi }{3}\right)(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{P_{ie}{\chi }^{3{\gamma }_e}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}.$

As is further detailed below, the second expression for the coefficient, ${𝒞}$, ensures that the pressure at the "surface" of the core matches the pressure at the "base" of the envelope; but it should only be employed after an equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, has been identified by locating an extremum in the free energy.

Simplest Bipolytrope

Familiar Setup

As has been shown in an accompanying presentation, for an $(n_c,n_e)=(0,0)$ bipolytrope,

${\mathfrak{𝔣}}_{WM}$	$\equiv$	$\frac{{\nu }^2}{q}\cdot f,$
$s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$\equiv$	$1+{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}},$
$(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$\equiv$	$(1-q^3)+\mathrm{\Lambda }\left[\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_c})(-2+3q-q^3)+\frac{3}{2q^2}(\frac{{\rho }_e}{{\rho }_c}{)}^2(-1+5q^2-5q^3+q^5)],$

and where (see, for example, in the context of its original definition, or another, separate derivation),

${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{1}{5}\left(\frac{\nu }{q}\right)\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{1-{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\frac{2}{5(g^2-1)}=\left\{\frac{5}{2}\right(\frac{{\rho }_e}{{\rho }_0}\left)\right[2(1-\frac{{\rho }_e}{{\rho }_0})(1-q)+\frac{{\rho }_e}{{\rho }_0}(\frac{1}{q^2}-1)]{\}}^{-1},$

and where (see the associated discussion of relevant mass integrals),

$\frac{{\rho }_c}{\overline{\rho }}=\frac{\nu }{q^3};\frac{{\rho }_e}{\overline{\rho }}=\frac{1-\nu }{1-q^3};\frac{{\rho }_e}{{\rho }_c}=\frac{q^3(1-\nu )}{\nu (1-q^3)}\Rightarrow \frac{q^3}{\nu }=\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-q^3)+q^3.$

Cleaner Virial Presentation

In an effort to show the similarity in structure among the several energy terms, we have also found it useful to write their expressions in the following forms:

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$	$=$	$-\frac{3}{5}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R}\right)\frac{{\nu }^2}{q}\cdot f=-4\pi P_i{R}^3\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}\cdot {\lambda }_{i}f,$
$S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$2\pi P_{ic}{R}^3[q^3+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q}\cdot {\lambda }_{ic}],$
$S_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$2\pi P_{ie}{R}^3[(1-q^3)+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q}\cdot {\lambda }_{ie}\mathfrak{𝔉}],$

where (see an associated discussion or the original derivation),

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

and where,

${\lambda }_i$	$\equiv$	$\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4{P}_i},$
$\mathfrak{𝔉}$	$\equiv$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[(-2q^2+3q^3-q^5)+\frac{3}{5}(\frac{{\rho }_e}{{\rho }_c})(-1+5q^2-5q^3+q^5)]$
	$=$	$\frac{1}{{\lambda }_{ie}}\left(\frac{2^2\cdot 5\pi }{3}\right)\frac{q(1-q^3)}{{\nu }^2}(s_{\mathrm{e}\mathrm{n}\mathrm{v}}-1).$

This also means that the three key terms used as shorthand notation in the above expressions for the three energy terms have the following definitions: ${\mathfrak{𝔣}}_{WM}$ $\equiv$ $\frac{{\nu }^2}{q}\cdot f,$ $s_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $\equiv$ $1+\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q^4}\cdot {\lambda }_{ic},$ $s_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $\equiv$ $1+\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q(1-q^3)}\cdot {\lambda }_{ie}\mathfrak{𝔉},$

Hence, if all the interface pressures are equal — that is, if $P_i=P_{ic}=P_{ie}$ and, hence also, ${\lambda }_i={\lambda }_{ic}={\lambda }_{ie}$ — then the total thermal energy is,

$S_{\mathrm{t}\mathrm{o}\mathrm{t}}=S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+S_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$2\pi P_i{R}^3[1+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q}\cdot {\lambda }_i(1+\mathfrak{𝔉})];$

and the virial is,

$2S_{\mathrm{t}\mathrm{o}\mathrm{t}}+W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$

$=$

$4\pi P_i{R}^3[1+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q}\cdot {\lambda }_i(1+\mathfrak{𝔉}-f)].$

The virial should sum to zero in equilibrium, which means,

$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}\cdot (f-1-\mathfrak{𝔉})$
$\Rightarrow \left[\right(\frac{2^2\cdot 5\pi }{3}\left)\frac{q}{{\nu }^2}\right]\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$	$=$	$f-1-\mathfrak{𝔉}$
$\Rightarrow \left(\frac{{\rho }_e}{{\rho }_c}{)}^{-1}\right[\left(\frac{2^3\pi }{3}\right)\frac{q^6}{{\nu }^2}]\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$	$=$	$[(q^3-q^5)+(\frac{{\rho }_e}{{\rho }_c}\left)\right(\frac{2}{5}-q^3+\frac{3}{5}{q}^5\left)\right]-[(-2q^2+3q^3-q^5)+(\frac{{\rho }_e}{{\rho }_c})(-\frac{3}{5}+3q^2-3q^3+\frac{3}{5}{q}^5)]$
	$=$	$2q^2(1-q)+\left(\frac{{\rho }_e}{{\rho }_c}\right)(1-3q^2+2q^3)$
	$=$	$q^2(\frac{{\rho }_e}{{\rho }_c}{)}^{-1}(g^2-1)$
$\Rightarrow \frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$	$=$	$\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2(g^2-1).$

Shift to Central Pressure Normalization

Let's rework the definition of ${\lambda }_i$ in two ways: (1) Normalize $R_{\mathrm{e}\mathrm{q}}$ to $R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$ and normalize the pressure to $P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$; (2) shift the referenced pressure from the pressure at the interface $(P_i)$ to the central pressure $(P_0)$, because it is $P_0$ that is directly related to $K_c$ and ${\rho }_c$; specifically, $P_0=K_c{\rho }_c^{{\gamma }_c}$. Appreciating that, in equilibrium,

$P_i$

$=$

$P_0-q^2{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}=K_c{\rho }_c^{{\gamma }_c}-\frac{3}{2^3\pi }\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\right)\left(\frac{{\nu }^2}{q^6}\right)q^2,$

the left-hand-side of the last expression, above, can be rewritten as,

$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$	$\equiv$	$\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$
	$=$	$\frac{R_{\mathrm{e}\mathrm{q}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}[P_0-\frac{3}{2^3\pi }(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\left)\right(\frac{{\nu }^2}{q^6}\left)q^2\right]$
	$=$	$\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_0}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}-\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2.$

Hence, the virial equilibrium condition gives,

$\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_0}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}-\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2$	$=$	$\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2(g^2-1)$
$\Rightarrow \frac{R_{\mathrm{e}\mathrm{q}}^4{P}_0}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$	$=$	$\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2{g}^2.$

This result precisely matches the result obtained via the detailed force-balanced conditions imposed through hydrostatic equilibrium.

Adopting our new variable normalizations and realizing, in particular, that,

$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}$

$=$

$G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2,$

the expression alternatively can be rewritten as,

$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$	$\equiv$	$\frac{R_{\mathrm{e}\mathrm{q}}^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}={\chi }_{\mathrm{e}\mathrm{q}}^4\left(\frac{P_i}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)$
	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^4\{\frac{K_c{\rho }_c^{{\gamma }_c}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}-\frac{2\pi }{3}[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^2{q}^2\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\}$
	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^4\left\{\frac{K_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right[\frac{{\rho }_c}{\overline{\rho }}\left(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\right){\chi }_{\mathrm{e}\mathrm{q}}^{-3}{]}^{{\gamma }_c}-\frac{2\pi }{3}\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^2{q}^2{\chi }_{\mathrm{e}\mathrm{q}}^{-4}\right\}$
	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\left[\right(\frac{3}{4\pi }\left)\frac{\nu }{q^3}{]}^{{\gamma }_c}\frac{K_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{3{\gamma }_c}})-\frac{2\pi }{3}[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^2{q}^2$
	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{{\gamma }_c}-\frac{2\pi }{3}[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^2{q}^2.$

Normalized in this manner, the virial equilibrium (as well as the hydrostatic balance) condition gives,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{{\gamma }_c}-\frac{2\pi }{3}[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^2{q}^2$	$=$	$\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2(g^2-1)$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$=$	$\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{2-{\gamma }_c}{q}^2{g}^2.$

Free-Energy Coefficients

Therefore, for an $(n_c,n_e)=(0,0)$ bipolytrope, the coefficients in the normalized free-energy function are,

${𝒜}$	$=$	$\frac{{\nu }^2}{5q}\cdot f=\frac{1}{5}\left(\frac{\nu }{q^3}{)}^2\right[q^5+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)q^3(1-q^2)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right(1-\frac{5}{2}{q}^3+\frac{3}{2}{q}^5\left)\right],$
${ℬ}$	$\equiv$	$\left(\frac{4\pi }{3}\right)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{P_{ic}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c}=(\frac{4\pi }{3}\left)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\left(\frac{4\pi }{3}\right)q^3[1+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q^4}\cdot {\lambda }_{ic}][\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}=(\frac{4\pi }{3}\left)q^3\right[\frac{1}{{\lambda }_{ic}}+\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q^4}]{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\left(\frac{4\pi }{3}\right)q^3\left\{{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\right[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c}-\frac{2\pi }{3}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^2{q}^2+(\frac{3}{2^2\cdot 5\pi }\left)\frac{{\nu }^2}{q^4}\right\}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\left(\frac{4\pi }{3}\right)q^3\left\{\right[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c}+{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}[\frac{3}{2^2\cdot 5\pi }-\frac{3}{2^3\pi }]\frac{{\nu }^2}{q^4}\}$
	$=$	$\left\{\nu \right[\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}-{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}\left(\frac{3^2}{2^3\cdot 5\pi }\right)\frac{{\nu }^2}{q^4}\left(\frac{4\pi }{3}\right)q^3\}=\nu [\left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}{]}^{{\gamma }_c-1}-{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}\left(\frac{3}{10}\right)\frac{{\nu }^2}{q}$
${𝒞}$	$\equiv$	$\left(\frac{4\pi }{3}\right)(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{P_{ie}{\chi }^{3{\gamma }_e}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}.$

Note that, because $P_{ie}=P_{ic}$ in equilibrium, the ratio of coefficients,

$\frac{{𝒞}}{{ℬ}}$	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_e-{\gamma }_c)}\left\{\frac{(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}\right\}$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_c-{\gamma }_e)}\left(\frac{{𝒞}}{{ℬ}}\right)$	$=$	$\frac{20\pi q(1-q^3){\lambda }_i^{-1}+3{\nu }^2\mathfrak{𝔉}}{20\pi q^4{\lambda }_i^{-1}+3{\nu }^2}.$

The equilibrium condition is,

$\frac{{𝒜}}{{ℬ}+{{𝒞}}^{\text{'}}}={\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c},$

where,

${{𝒞}}^{\text{'}}\equiv {𝒞}{\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_c-{\gamma }_e)}.$

More General Derivation of Free-Energy Coefficients B and C

Keep in mind that, generally,

$G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2$	$=$	$R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^4{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}=E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}};$
$\frac{1}{{\lambda }_i}$	$\equiv$	$\frac{R^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}=\left(\frac{P_i}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right){\chi }^4$   … and, note that …    $\frac{1}{\mathrm{\Lambda }}=\left(\frac{3\cdot 5}{2^2\pi }\right)\frac{1}{{\lambda }_i}\cdot \frac{1}{q^2{\sigma }^2};$
$\frac{\mathrm{\Pi }}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{3}{2^3\pi }\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}^4}\right)\frac{{\nu }^2}{q^6}=\left(\frac{2\pi }{3}\right){\sigma }^2{\chi }^{-4};$
$\frac{K_c{\rho }_c^{{\gamma }_c}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$\frac{K_c}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left(\frac{{\rho }_c}{\overline{\rho }}{)}^{{\gamma }_c}\right[\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R^3}{]}^{{\gamma }_c}=\frac{K_c{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{3{\gamma }_c}{P}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\left[\right(\frac{3}{4\pi })\frac{\nu }{q^3}{]}^{{\gamma }_c}{\chi }^{-3{\gamma }_c}={\sigma }^{{\gamma }_c}{\chi }^{-3{\gamma }_c},$

where we have introduced the notation,

$\sigma \equiv \left(\frac{3}{4\pi }\right)\frac{\nu }{q^3}.$

So, the free-energy coefficient,

${ℬ}$	$=$	$\left(\frac{4\pi }{3}\right)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{P_{ic}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c}=(\frac{4\pi }{3}\left)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\left(\frac{4\pi }{3}\right)q^3[1+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q^4}\cdot {\lambda }_{ic}{]}_{\mathrm{e}\mathrm{q}}[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}=\left(\frac{4\pi }{3}\right)q^3[\frac{1}{{\lambda }_{ic}}+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q^4}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$
	$=$	$\left(\frac{4\pi }{3}\right)q^3\left[{\chi }_{\mathrm{e}\mathrm{q}}^4\right(\frac{P_{ic}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}+\left(\frac{4\pi }{3\cdot 5}\right)q^2{\sigma }^2]{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}.$

And the free-energy coefficient,

${𝒞}$	$\equiv$	$\left(\frac{4\pi }{3}\right)(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{P_{ie}{\chi }^{3{\gamma }_e}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}=(\frac{4\pi }{3})(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{1}{{\lambda }_{ie}}{]}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}$
	$=$	$\left(\frac{4\pi }{3}\right)(1-q^3)\{\frac{1}{{\lambda }_{ie}}+(\frac{3}{2^2\cdot 5\pi })\frac{{\nu }^2}{q(1-q^3)}\cdot \mathfrak{𝔉}{\}}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}$
	$=$	$\left(\frac{4\pi }{3}\right)\{(1-q^3){\chi }_{\mathrm{e}\mathrm{q}}^4(\frac{P_{ie}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}+\left(\frac{2\pi }{3}\right){\sigma }^2\left[\frac{2}{5}{q}^5\mathfrak{𝔉}\right]{\}}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}.$

OLD DERIVATION $P_{ic}=K_c{\rho }_c^{{\gamma }_c}$	NEW DERIVATION $P_{ic}=P_0-q^2\mathrm{\Pi }=K_c{\rho }_c^{{\gamma }_c}-q^2\mathrm{\Pi }$
… therefore …
${{ℬ}}_{\mathrm{O}\mathrm{L}\mathrm{D}}$ $=$ $\left(\frac{4\pi }{3}\right)q^3[{\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}+(\frac{4\pi }{3\cdot 5}\left)q^2{\sigma }^2\right]{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$   $=$ $\left(\frac{4\pi }{3}\right)q^3[{\sigma }^{{\gamma }_c}+(\frac{4\pi }{3\cdot 5}\left)q^2{\sigma }^2{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}\right]$	${{ℬ}}_{\mathrm{N}\mathrm{E}\mathrm{W}}$ $=$ $\left(\frac{4\pi }{3}\right)q^3[{\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-(\frac{2\pi }{3})q^2{\sigma }^2+(\frac{4\pi }{3\cdot 5}\left)q^2{\sigma }^2\right]{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}$   $=$ $\left(\frac{4\pi }{3}\right)q^3[{\sigma }^{{\gamma }_c}-(\frac{2\pi }{5}\left)q^2{\sigma }^2{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_c-4}\right]$
… and, enforcing in equilibrium $P_{ie}=P_{ic}$ …
${{𝒞}}_{\mathrm{O}\mathrm{L}\mathrm{D}}$ $=$ $\left(\frac{4\pi }{3}\right)\{(1-q^3)[{\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}]+(\frac{2\pi }{3}\left){\sigma }^2\right[\frac{2}{5}{q}^5\mathfrak{𝔉}]{\}}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}$   $=$ $\left(\frac{4\pi }{3}\right)[(1-q^3){\sigma }^{{\gamma }_c}+(\frac{2\pi }{3}\left){\sigma }^2\right(\frac{2}{5}{q}^5\mathfrak{𝔉}\left){\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}\right]$	${{𝒞}}_{\mathrm{N}\mathrm{E}\mathrm{W}}$ $=$ $\left(\frac{4\pi }{3}\right)\{(1-q^3)[{\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-\left(\frac{2\pi }{3}\right)q^2{\sigma }^2]+(\frac{2\pi }{3}\left){\sigma }^2\right[\frac{2}{5}{q}^5\mathfrak{𝔉}]{\}}_{\mathrm{e}\mathrm{q}}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}$   $=$ $\left(\frac{4\pi }{3}\right)\{(1-q^3){\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-3{\gamma }_c}+(\frac{2\pi }{3}\left){\sigma }^2\right[\left(\frac{2}{5}{q}^5\mathfrak{𝔉}\right)-q^2(1-q^3)\left]{\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_e-4}\right\}$
… and, also …
$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$ $=$ ${\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$ $\Rightarrow \frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}$ $=$ $\frac{1}{q^2}\left(\frac{3\cdot 5}{2^2\pi }\right){\sigma }^{{\gamma }_c-2}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$   $=$ $\left(\frac{5q}{\nu }\right){\sigma }^{{\gamma }_c-1}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$ $=$ ${\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-\left(\frac{2\pi }{3}\right)q^2{\sigma }^2$ $\Rightarrow \frac{1}{{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}}$ $=$ $\frac{1}{q^2{\sigma }^2}\left(\frac{3\cdot 5}{2^2\pi }\right)[{\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-(\frac{2\pi }{3}\left)q^2{\sigma }^2\right]$   $=$ $\left(\frac{5q}{\nu }\right){\sigma }^{{\gamma }_c-1}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-\frac{5}{2}$

Extrema

Extrema in the free energy occur when,

${𝒜}$

$=$

${ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}+{𝒞}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_e}.$

Also, as stated above, because $P_{ie}=P_{ic}$ in equilibrium, the ratio of coefficients,

$\frac{{𝒞}}{{ℬ}}$

$=$

${\chi }_{\mathrm{e}\mathrm{q}}^{3({\gamma }_e-{\gamma }_c)}\left[\frac{(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}\right].$

When put together, these two relations imply,

${𝒜}$	$=$	${ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}+{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}{ℬ}\left[\frac{(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}\right]$
	$=$	${ℬ}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}[1+\frac{(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}].$

But the definition of ${ℬ}$ gives,

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

$=$

$\left(\frac{4\pi }{3}\right)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}.$

Hence, extrema occur when,

${𝒜}$	$=$	$\left(\frac{4\pi }{3}\right)q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\left[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}\right[1+\frac{(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}}{q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}]$
$\Rightarrow \left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}\cdot f$	$=$	$\left[\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}\right[q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}]$
	$=$	$\frac{q^3}{[{\lambda }_i{]}_{\mathrm{e}\mathrm{q}}}+\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}+\frac{(1-q^3)}{[{\lambda }_i{]}_{\mathrm{e}\mathrm{q}}}+\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}\cdot \mathfrak{𝔉}$
$\Rightarrow [\frac{1}{{\lambda }_{ic}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$\left(\frac{3}{2^2\cdot 5\pi }\right)\frac{{\nu }^2}{q}\cdot (f-1-\mathfrak{𝔉})$
	$=$	$\left(\frac{2\pi }{3}\right){\sigma }^2{q}^2(g^2-1).$

In what follows, keep in mind that,

${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$=$	$(\frac{R_{\mathrm{e}\mathrm{q}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{4-3{\gamma }_c}=R_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}(\frac{K_c}{G})M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c-2};$
$K_c{\rho }_c^{{\gamma }_c}$	$=$	$K_c\left(\frac{{\rho }_c}{\overline{\rho }}{)}^{{\gamma }_c}\right[\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R^3}{]}^{{\gamma }_c}=K_c{\sigma }^{{\gamma }_c}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c}{R}^{-3{\gamma }_c};$
$\mathrm{\Pi }$	$=$	$\frac{3}{2^3\pi }\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4}\right)\frac{{\nu }^2}{q^6}=\frac{2\pi }{3}\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R^4}\right){\sigma }^2.$

OLD DERIVATION $P_i=K_c{\rho }_c^{{\gamma }_c}$ $\Rightarrow K_c=P_i{\sigma }^{-{\gamma }_c}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-{\gamma }_c}{R}^{+3{\gamma }_c}$	NEW DERIVATION $P_0=K_c{\rho }_c^{{\gamma }_c}$ $\Rightarrow K_c=P_0{\sigma }^{-{\gamma }_c}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-{\gamma }_c}{R}^{+3{\gamma }_c}$
… hence, as derived in the above table …
$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$ $=$ ${\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$	$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$ $=$ ${\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-\left(\frac{2\pi }{3}\right)q^2{\sigma }^2$
… which, when combined with the condition that identifies extrema, gives …
${\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^{2-{\gamma }_c}{q}^2(g^2-1)$ $\Rightarrow R_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\left(\frac{K_c}{G}\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c-2}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^{2-{\gamma }_c}{q}^2(g^2-1)$ $\Rightarrow \frac{R_{\mathrm{e}\mathrm{q}}^4{P}_i}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^2{q}^2(g^2-1)$	${\sigma }^{{\gamma }_c}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}-\left(\frac{2\pi }{3}\right)q^2{\sigma }^2$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^2{q}^2(g^2-1)$ $\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^{2-{\gamma }_c}{q}^2{g}^2$ $\Rightarrow R_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}\left(\frac{K_c}{G}\right)M_{\mathrm{t}\mathrm{o}\mathrm{t}}^{{\gamma }_c-2}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^{2-{\gamma }_c}{q}^2{g}^2$ $\Rightarrow \frac{R_{\mathrm{e}\mathrm{q}}^4{P}_0}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$ $=$ $\left(\frac{2\pi }{3}\right){\sigma }^2{q}^2{g}^2$
These are consistent results because they result in the detailed force-balance relation, $P_0-P_i=q^2{\mathrm{\Pi }}_{\mathrm{e}\mathrm{q}}.$

Examples

Identification of Local Extrema in Free Energy
$\nu$	$q$	$\frac{{\rho }_e}{{\rho }_c}$	$f(q,\frac{{\rho }_e}{{\rho }_c})$	$g^2(q,\frac{{\rho }_e}{{\rho }_c})$	${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}$	${\chi }_{\mathrm{e}\mathrm{q}}$	${𝒜}$	${ℬ}$	${𝒞}$	MIN/MAX
$0.2$	$9^{-1/3}=0.48075$	$0.5$	$12.5644$	$2.091312$	$0.366531$	$0.037453$	$0.2090801$	$0.2308269$	$2.06252\times 10^{-4}$	MIN
$0.4$	$4^{-1/3}=0.62996$	$0.5$	$4.21974$	$1.56498$	$0.707989$	$0.0220475$	$0.2143496$	$0.5635746$	$4.4626\times 10^{-5}$	MIN
$0.473473$	$0.681838$	$0.516107$			$0.462927$	$0.08255$	"	"	"	MAX
$0.5$	$3^{-1/3}=0.693361$	$0.5$	$2.985115$	$1.42334$	$0.9448663$	$0.0152116$	$0.2152641$	$0.791882$	$1.5464\times 10^{-5}$	MIN
$0.559839$	$0.729581$	$0.499188$			$0.75089$	$0.032196$	"	"	"	MAX

Free Energy Extrema when:     $\frac{{\rho }_e}{{\rho }_c}=\frac{1}{2}\Rightarrow q^3=\frac{\nu }{2-\nu }$
$\nu$	$q$	$\frac{{\rho }_e}{{\rho }_c}$	$f(q,\frac{{\rho }_e}{{\rho }_c})$	$g^2(q,\frac{{\rho }_e}{{\rho }_c})$	$\frac{1}{{\lambda }_i}{|}_{\mathrm{e}\mathrm{q}}$	${\chi }_{\mathrm{e}\mathrm{q}}$	${𝒜}$	${{ℬ}}_{\mathrm{N}\mathrm{E}\mathrm{W}}$	${{𝒞}}_{\mathrm{N}\mathrm{E}\mathrm{W}}$	$G^{*}$	Stability	MIN/MAX
$0.5$	$(\frac{1}{3}{)}^{1/3}$	$0.5$	$2.985115$	$1.423340$	$0.05466039$	$0.3152983$	$0.21526406$	$0.23552725$	$6.643899\times 10^{-3}$	$+0.5176146$	$+0.429245$	MIN
…	…	…				$0.6674$	"	"	"	$+0.55572115$		MAX
$0.6$	$(\frac{3}{7}{)}^{1/3}$	$0.5$	$2.2507129$	$1.31282895$	$0.04160318$	$0.3411545$	$0.21493717$	$0.26165939$	$5.208750\times 10^{-3}$	$+0.73532249$	$+0.0935217$	MIN
…	…	…				$0.431745$	"	"	"	$+0.7367797$		MAX
$0.7$	$(\frac{7}{13}{)}^{1/3}$	$0.5$	$1.7707809$	$1.2209446$	$0.029500$	$0.3589388$	$0.21330744$	$0.28172532$	$3.389793\times 10^{-3}$	$+0.8953395$	$-0.0767108$	MAX
…	…	…				$0.270615$	"	"	"	$+0.89227216$		MIN
System should be stable (with free energy minimum) if:     $\frac{({\gamma }_e-\frac{4}{3})}{({\gamma }_e-{\gamma }_c)}f-[1+\frac{5}{2}(g^2-1)]>0$

Solution Strategy

For a given set of free-energy coefficients, ${𝒜},{ℬ},$ and ${𝒞}$, along with a choice of the two adiabatic exponents $({\gamma }_c,{\gamma }_e)$, here's how to determine all of the physical parameters that are detailed in the above example table.

• Step 1: Guess a value of $0<q<1$.

• Step 2: Given the pair of parameter values, $({𝒜},q)$, determine the interface-density ratio, ${\rho }_e/{\rho }_c$, by finding the appropriate root of the expression that defines the function, ${𝒜}(q,{\rho }_e/{\rho }_c)$. This can be straightforwardly accomplished because, as demonstrated below, the relevant expression can be written as a quadratic function of $({\rho }_e/{\rho }_c)$.

• Step 3: Given the pair of parameter values, $(q,{\rho }_e/{\rho }_c)$, determine the value of the core-to-total mass ratio, $\nu$, from the expression that was obtained from an integration over the mass, namely,

$\frac{1}{\nu }$

$=$

$1+\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^3}-1).$

• Step 4: Given the value of ${ℬ}$ along with the pair of parameter values, $(q,\nu )$, the above expression that defines ${ℬ}$ can be solved to give the relevant value of the dimensionless parameter, ${\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}}.$

• Step 5: The value of ${{𝒞}}^{\text{'}}$ — the coefficient that appears on the right-hand-side of the above expression that defines ${𝒞}$ — can be determined, given the values of parameter triplet, $(q,\nu ,{\mathrm{\Lambda }}_{\mathrm{e}\mathrm{q}})$.

• Step 6: Given the value of ${𝒞}$ and the just-determined value of the coefficient ${{𝒞}}^{\text{'}}$, the normalized equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}},$ that corresponds to the value of $q$ that was guessed in Step #1 can be determined from the above definition of ${𝒞}$, specifically,

${\chi }_{\mathrm{e}\mathrm{q}}{|}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}$

$=$

$(\frac{{𝒞}}{{{𝒞}}^{\text{'}}}{)}^{1/(3{\gamma }_e-3{\gamma }_c)}.$

• Step 7: But, independent of this guessed value of ${\chi }_{\mathrm{e}\mathrm{q}},$ the condition for virial equilibrium — which identifies extrema in the free-energy function — gives the following expression for the normalized equilibrium radius:

${\chi }_{\mathrm{e}\mathrm{q}}{|}_{\mathrm{v}\mathrm{i}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$

$=$

$[\frac{{𝒜}}{{ℬ}+{{𝒞}}^{\text{'}}}{]}^{1/(4-3{\gamma }_c)}.$

• Step 8: If ${\chi }_{\mathrm{e}\mathrm{q}}{|}_{\mathrm{g}\mathrm{u}\mathrm{e}\mathrm{s}\mathrm{s}}\ne {\chi }_{\mathrm{e}\mathrm{q}}{|}_{\mathrm{v}\mathrm{i}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$, return to Step #1 and guess a different value of $q$. Repeat Steps #1 through #7 until the two independently derived values of the normalized radius match, to a desired level of precision.

• Keep in mind: (A) A graphical representation of the free-energy function, $\mathfrak{𝔊}(\chi )$, can also be used to identify the "correct" value of ${\chi }_{\mathrm{e}\mathrm{q}}$ and, ultimately, the above-described iteration loop should converge on this value. (B) The free-energy function may exhibit more than one (or, actually, no) extrema, in which case more than one (or no) value of $q$ should lead to convergence of the above-described iteration loop.

Related Discussions

• Newer, Version2 of Bipolytrope Generalization derivations.
• Analytic solution with $n_c=0$ and $n_e=0$.
• Analytic solution with $n_c=5$ and $n_e=1$.