SSC/Structure/BiPolytropes/FreeEnergy00

Free Energy of BiPolytrope with $(n_c,n_e)=(0,0)$

Here we present a specific example of the equilibrium structure of a bipolytrope as determined from a free-energy analysis. The example is a bipolytrope whose core has a polytropic index, $n_c=0$, and whose envelope has a polytropic index, $n_e=0$. The details presented here build upon an overview of the free energy of bipolytropes that has been presented elsewhere.

Mass Profile

In this case, ${\rho }_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(x)={\rho }_c=$ constant — hence, also, $[\rho (x)/\overline{\rho }{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=1$ — and ${\rho }_{\mathrm{e}\mathrm{n}\mathrm{v}}(x)={\rho }_e=$ constant — hence, also, $[\rho (x)/\overline{\rho }{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}=1$ — but in general ${\rho }_e\ne {\rho }_c$. Performing the separate integrals to obtain expressions for $M_r(r)$ inside the core and the envelope, as established in our accompanying overview, we obtain:

$(\mathrm{F}\mathrm{o}\mathrm{r}0\le x\le q)$       $M_r$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{\nu }{q^3}\right)\underset{0}{\overset{x}{\int }}3x^{2}dx=\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}(\frac{x}{q}{)}^3;$
$(\mathrm{F}\mathrm{o}\mathrm{r}q\le x\le 1)$       $M_r$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{\nu +(\frac{1-\nu }{1-q^3}\left)\underset{q}{\overset{x}{\int }}3x^{2}dx\right\}=M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+(1-\nu )M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{x^3-q^3}{1-q^3}\right).$

When $x=q$, both expressions give,

$M_r$

$=$

$M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}},$

as they should. We deduce, as well, that the mass contained in the envelope is,

$M_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}-M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=(1-\nu )M_{\mathrm{t}\mathrm{o}\mathrm{t}},$

and that the volumes occupied by the core and envelope are, respectively,

$V_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$q^3{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3,$
$V_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$(1-q^3)R_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3.$

Hence, the ratio of envelope density to core density is,

$\frac{{\rho }_e}{{\rho }_c}=\frac{{\overline{\rho }}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{{\overline{\rho }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}$

$=$

$\frac{M_{\mathrm{e}\mathrm{n}\mathrm{v}}/V_{\mathrm{e}\mathrm{n}\mathrm{v}}}{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}/V_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}=\frac{q^3(1-\nu )}{\nu (1-q^3)}.$

These relations should be compared to — and ultimately must match — the prescriptions for $M_r$ that have been presented elsewhere in connection with detailed force-balance models of $(n_c,n_e)=(0,0)$ bipolytropes and in our introductory discussion of the virial stability of bipolytropes.

Gravitational Potential Energy

Here we follow the steps that have been outlined in an accompanying overview to determine the separate contributions to the gravitational potential energy. Let's do the core first. In this case, ${\rho }_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(x)={\rho }_c=$ constant — hence, also, $[\rho (x)/\overline{\rho }{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=1$. As has been demonstrated above, the corresponding $M_r$ function is,

$[\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\left(\frac{\nu }{q^3}\right)x^3.$

Hence,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{\nu }{q^3}\right)\underset{0}{\overset{q}{\int }}3\left(\frac{\nu }{q^3}\right)x^{4}dx$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{\nu }{q^3}{)}^2\right(\frac{3}{5}{q}^5)$

Now, let's do the envelope. In this case, ${\rho }_{\mathrm{e}\mathrm{n}\mathrm{v}}(x)={\rho }_e=$ constant; hence, also, $[\rho (x)/\overline{\rho }{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}=1$. As shown elsewhere, the corresponding $M_r$ function is,

$[\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\nu +\left(\frac{1-\nu }{1-q^3}\right)(x^3-q^3).$

Hence,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{1-\nu }{1-q^3}\right)\left\{\underset{q}{\overset{1}{\int }}\right[\nu -q^3\left(\frac{1-\nu }{1-q^3}\right)]3xdx+\underset{q}{\overset{1}{\int }}[\left(\frac{1-\nu }{1-q^3}\right)\left]3x^{4}dx\right\}$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{1-\nu }{1-q^3}\right)\left\{\frac{3}{2}\right[\nu -q^3\left(\frac{1-\nu }{1-q^3}\right)](1-q^2)+\frac{3}{5}[\left(\frac{1-\nu }{1-q^3}\right)](1-q^5)\}$
	$=$	$-\frac{3}{5}\left(\frac{{\nu }^2}{q}\right)E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\frac{1}{\nu }\right(\frac{1-\nu }{1-q^3}\left)\right]\left\{\frac{5}{2}\right[1-\frac{q^3}{\nu }\left(\frac{1-\nu }{1-q^3}\right)](q-q^3)+[\frac{q}{\nu }\left(\frac{1-\nu }{1-q^3}\right)](1-q^5)\}$
	$=$	$-\frac{3}{5}\left(\frac{{\nu }^2}{q}\right)E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\frac{q^3}{\nu }\right(\frac{1-\nu }{1-q^3}\left)\right]\left\{\frac{5}{2}\right[1-\frac{q^3}{\nu }\left(\frac{1-\nu }{1-q^3}\right)\left]\right(\frac{1}{q^2}-1)+[\frac{q^3}{\nu }\left(\frac{1-\nu }{1-q^3}\right)\left]\right(\frac{1}{q^5}-1\left)\right\}.$

Realizing from the above mass segregation derivation that,

$\frac{q^3}{\nu }\left(\frac{1-\nu }{1-q^3}\right)=\frac{{\rho }_e}{{\rho }_c},$

this last expression can be rewritten as,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$-\frac{3}{5}\left(\frac{{\nu }^2}{q}\right)E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\frac{{\rho }_e}{{\rho }_c}\right]\left\{\frac{5}{2}\right[1-\frac{{\rho }_e}{{\rho }_c}\left]\right(\frac{1}{q^2}-1)+[\frac{{\rho }_e}{{\rho }_c}\left]\right(\frac{1}{q^5}-1\left)\right\}$
	$=$	$-\frac{3}{5}\left(\frac{{\nu }^2}{q}\right)E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{{\rho }_e}{{\rho }_c}\right)\left\{\frac{5}{2}\right(\frac{1}{q^2}-1)+(\frac{{\rho }_e}{{\rho }_c}\left)\right[\frac{1}{q^5}-1+\frac{5}{2}(1-\frac{1}{q^2})\left]\right\}.$

So, when put together to obtain the total gravitational potential energy, we have,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}=W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$-\frac{3}{5}{E}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left(\frac{{\nu }^2}{q}\right)f(\nu ,q),$

where,

$f(\nu ,q)$

$\equiv$

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

(This result agrees with Tohline's earlier derivations in other sections of this H_Book, which may now be erased to avoid repetition.)

Thermodynamic Energy Reservoir

According to our derivation of the properties of detailed force-balance $(n_c,n_e)=(0,0)$ bipolytropes, in this case the pressure throughout the core is defined by the dimensionless function,

$p_c(x)$

$=$

$\left(\frac{2\pi }{3}\right){\xi }^2,$

and the pressure throughout the envelope is defined by the dimensionless function,

$p_e(x)$

$=$

$\frac{2\pi }{3}\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}[\frac{{\rho }_e}{{\rho }_0}({\xi }^2-{\xi }_i^2)-2(1-\frac{{\rho }_e}{{\rho }_0}\left){\xi }_i^3\right(\frac{1}{\xi }-\frac{1}{{\xi }_i}\left)\right],$

where, for both functions,

$\xi$	$\equiv$	$\left[\right(\frac{G{\rho }_0^2}{P_0}{)}^{1/2}{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}x$
	$=$	$\left[\frac{G{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^2}{P_0}\right(\frac{3\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi q^3{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3}{)}^2{]}_{\mathrm{e}\mathrm{q}}^{1/2}x$
	$=$	$\left[\right(\frac{3^2}{2^4{\pi }^2}\left)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}\right(\frac{\nu }{q^3}{)}^2{]}_{\mathrm{e}\mathrm{q}}^{1/2}x$

So, defining the coefficient,

$b_{\xi }$

$\equiv$

$\left(\frac{3}{2^3\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}(\frac{\nu }{q^3}{)}^2,$

such that,

$\xi$

$=$

$(\frac{3}{2\pi }\cdot b_{\xi }{)}^{1/2}x,$

and remembering that, at the interface, $x\to x_i=q$, so ${\xi }_i=(3b_{\xi }/2\pi {)}^{1/2}q$, the two dimensionless pressure functions become,

$p_c(x)$

$=$

$b_{\xi }{x}^2,$

and,

$p_e(x)$

$=$

$b_{\xi }\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}[\frac{{\rho }_e}{{\rho }_0}(x^2-q^2)-2(1-\frac{{\rho }_e}{{\rho }_0}\left)q^3\right(\frac{1}{x}-\frac{1}{q}\left)\right].$

The desired integrals over these pressure distributions therefore give,

$\underset{0}{\overset{q}{\int }}\left[\frac{1-p_c(x)}{1-p_c(q)}\right]x^{2}dx$	$=$	$\left[\frac{1}{1-b_{\xi }{q}^2}\right]\underset{0}{\overset{q}{\int }}(1-b_{\xi }{x}^2)x^{2}dx$
	$=$	$\left[\frac{1}{1-b_{\xi }{q}^2}\right][\frac{1}{3}\cdot q^3-(\frac{b_{\xi }}{5}\left)q^5\right]$
	$=$	$\frac{q^3}{3}\left[\frac{1}{1-b_{\xi }{q}^2}\right][1-(\frac{3b_{\xi }}{5}\left)q^2\right]=\frac{q^3}{3}\left(\frac{P_0}{P_{ic}}\right)[1-(\frac{3b_{\xi }}{5}\left)q^2\right];$
$\underset{q}{\overset{1}{\int }}[1-p_e(x)]x^{2}dx$	$=$	$\frac{1}{3}(1-q^3)-b_{\xi }\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}\underset{q}{\overset{1}{\int }}[\frac{{\rho }_e}{{\rho }_0}(x^2-q^2)-2(1-\frac{{\rho }_e}{{\rho }_0}\left)q^3\right(\frac{1}{x}-\frac{1}{q}\left)\right]x^{2}dx$
	$=$	$\frac{1}{3}(1-q^3)-b_{\xi }\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}\left[\frac{{\rho }_e}{{\rho }_0}\right(\frac{x^5}{5}-\frac{q^2{x}^3}{3})-2(1-\frac{{\rho }_e}{{\rho }_0}\left)q^3\right(\frac{x^2}{2}-\frac{x^3}{3q}){]}_q^1$
	$=$	$\frac{1}{3}(1-q^3)-\frac{b_{\xi }}{3}\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}\left\{\right[\frac{{\rho }_e}{{\rho }_0}(\frac{3}{5}-q^2)-(1-\frac{{\rho }_e}{{\rho }_0})q^2(3q-2)]$
		$-\left[\frac{{\rho }_e}{{\rho }_0}\right(-\frac{2}{5})q^5-(1-\frac{{\rho }_e}{{\rho }_0}\left)q^5\right]\}$
	$=$	$\frac{1}{3}(1-q^3)-\frac{b_{\xi }}{3}\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}\left\{\right[q^2(2-3q)+q^5]+\frac{{\rho }_e}{{\rho }_0}[(\frac{3}{5}-q^2)+q^2(3q-2)+\frac{2q^5}{5}-q^5\left]\right\}$
	$=$	$\frac{1}{3}(1-q^3)-\frac{b_{\xi }}{3}\left(\frac{{\rho }_e}{{\rho }_0}\right)\frac{P_0}{P_{ie}}[(2q^2-3q^3+q^5)+\frac{3}{5}\cdot \frac{{\rho }_e}{{\rho }_0}(1-5q^2+5q^3-q^5)]$
	$=$	$\frac{1}{3}\{(1-q^3)+b_{\xi }(\frac{P_0}{P_{ie}}\left)\right[\frac{2}{5}{q}^5\mathfrak{𝔉}\left]\right\},$

where,

$\mathfrak{𝔉}$

$\equiv$

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

Finally, then, we have,

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

Virial Theorem

As has been shown in our accompanying overview, the condition for equilibrium based on a free-energy analysis — that is, the virial theorem — is,

${𝒜}$	$=$	${{ℬ}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_c}+{{ℬ}}_{\mathrm{e}\mathrm{n}\mathrm{v}}{\chi }_{\mathrm{e}\mathrm{q}}^{4-3{\gamma }_e}$
	$=$	$\frac{4\pi }{3}[\frac{P_i{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}_{\mathrm{e}\mathrm{q}}[q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}+(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}].$

For $(n_c,n_e)=(0,0)$ bipolytropes, the relevant coefficient functions are,

${𝒜}$	$=$	$\frac{1}{5}\left(\frac{{\nu }^2}{q}\right)f,$
$q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$q^3\left(\frac{P_0}{P_{ic}}\right)[1-\frac{3}{5}{q}^2{b}_{\xi }],$
$(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$(1-q^3)+\left(\frac{P_0}{P_{ie}}\right)\frac{2}{5}{q}^5\mathfrak{𝔉}{b}_{\xi },$

where,

$f$	$\equiv$	$1+\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)(\frac{1}{q^2}-1)+\left(\frac{{\rho }_e}{{\rho }_c}{)}^2\right[\frac{1}{q^5}-1+\frac{5}{2}(1-\frac{1}{q^2})],$
$\mathfrak{𝔉}$	$\equiv$	$\frac{5}{2}\left(\frac{{\rho }_e}{{\rho }_c}\right)\frac{1}{q^5}[(-2q^2+3q^3-q^5)+\frac{3}{5}(\frac{{\rho }_e}{{\rho }_c})(-1+5q^2-5q^3+q^5)],$
$\frac{P_{ic}}{P_0}$	$=$	$1-p_c(q)=1-b_{\xi }{q}^2,$
$b_{\xi }$	$\equiv$	$\left(\frac{3}{2^3\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}(\frac{\nu }{q^3}{)}^2.$

Plugging these expressions into the equilibrium condition shown above, and setting the interface pressures equal to one another, gives,

$\frac{1}{5}\left(\frac{{\nu }^2}{q}\right)f$	$=$	$\frac{4\pi }{3}\left[\frac{P_i{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}_{\mathrm{e}\mathrm{q}}\right\{q^3\left(\frac{P_0}{P_i}\right)[1-\frac{3}{5}{q}^2{b}_{\xi }]+(1-q^3)+\left(\frac{P_0}{P_i}\right)\frac{2}{5}{q}^5\mathfrak{𝔉}{b}_{\xi }\}$
	$=$	$\frac{4\pi }{3}\left[\frac{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}_{\mathrm{e}\mathrm{q}}\right\{q^3[1-\frac{3}{5}{q}^2{b}_{\xi }]+(1-q^3)(1-b_{\xi }{q}^2)+\frac{2}{5}{q}^5\mathfrak{𝔉}{b}_{\xi }\}$
	$=$	$\frac{4\pi }{3}\left[\frac{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}_{\mathrm{e}\mathrm{q}}\right\{1-b_{\xi }[\frac{3}{5}{q}^5+q^2(1-q^3)-\frac{2}{5}{q}^5\mathfrak{𝔉}]\}$
	$=$	$\frac{4\pi }{3}\left[\frac{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{]}_{\mathrm{e}\mathrm{q}}\right[\frac{1}{b_{\xi }}-q^2+\frac{2}{5}{q}^5(1+\mathfrak{𝔉})]b_{\xi }$
	$=$	$\frac{1}{2}[\frac{1}{b_{\xi }}-q^2+\frac{2}{5}{q}^5(1+\mathfrak{𝔉})](\frac{\nu }{q^3}{)}^2$
$\Rightarrow \frac{1}{b_{\xi }}$	$=$	$\frac{2}{5}{q}^{5}f+[q^2-\frac{2}{5}{q}^5(1+\mathfrak{𝔉})]$
$\Rightarrow \left(\frac{2^3\pi }{3}\right)\frac{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}(\frac{q^3}{\nu }{)}^2$	$=$	$q^2+\frac{2}{5}{q}^5(f-1-\mathfrak{𝔉})$
$\Rightarrow \frac{P_0{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^4}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}$	$=$	$\left(\frac{3}{2^3\pi }\right)\left(\frac{\nu }{q^3}{)}^2\right\{q^2+\left(\frac{{\rho }_e}{{\rho }_c}\right)[2q^2(1-q)+(\frac{{\rho }_e}{{\rho }_c})(1-3q^2+2q^3)]\}.$

This exactly matches the equilibrium relation that was derived from our detailed force-balance analysis of $(n_c,n_e)=(0,0)$ bipolytropes.

Related Discussions

• Free-energy determination of equilibrium configurations for BiPolytropes with $n_c=0$ and $n_e=0$.
• Free-energy determination of equilibrium configurations for BiPolytropes with $n_c=5$ and $n_e=1$.
• Analytic solution of Detailed-Force-Balance BiPolytrope with $n_c=0$ and $n_e=0$.
• Analytic solution of Detailed-Force-Balance BiPolytrope with $n_c=5$ and $n_e=1$.
• Old Bipolytrope Generalization derivations.

See Also

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