SSC/Structure/BiPolytropes/FreeEnergy51/Pt3

Free Energy of BiPolytrope with (n_c, n_e) = (5, 1)

Part I:  Mass Profile

Part II:  Gravitational Potential Energy

Part III:  Thermal Energy Reservoir

Thermodynamic Energy Reservoir

The Core

From our introductory discussion of the free energy of bipolytropes, the energy contained in the core's thermodynamic reservoir may be written as,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\frac{2}{3({\gamma }_c-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_c}\right[\frac{2\pi P_{ic}{\chi }^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}\left[q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right],$

where,

$q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$\equiv$

$\underset{0}{\overset{q}{\int }}3\left[\frac{1-p_c(x)}{1-p_c(q)}\right]x^{2}dx,$

defines the relevant integral over the core's pressure distribution. According to our derivation of the properties of detailed force-balance $(n_c,n_e)=(5,1)$ bipolytropes — see also the relevant derivations in our accompanying overview — in this case the pressure throughout the core is defined by the dimensionless function,

$P^{*}\equiv \frac{P_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(\xi )}{P_0}$	$=$	$(1+\frac{1}{3}{\xi }^2{)}^{-3},$
$\Rightarrow 1-p_c(x)=\frac{P_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(x)}{P_0}$	$=$	$(1+a_{\xi }{x}^2{)}^{-3},$

where, $a_{\xi }$ is defined above in connection with our derivation of the mass profile. The desired integral over this pressure distribution therefore gives,

$q^3{s}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$3(1+a_{\xi }{q}^2{)}^3\underset{0}{\overset{q}{\int }}\frac{x^{2}dx}{(1+a_{\xi }{x}^2{)}^3}$
	$=$	$3(1+a_{\xi }{q}^2{)}^3\{\frac{{\tan }^{-1}[a_{\xi }^{1/2}q]}{2^3{a}_{\xi }^{3/2}}+\frac{q}{2^3{a}_{\xi }(a_{\xi }{q}^2+1)}-\frac{q}{2^2{a}_{\xi }(a_{\xi }{q}^2+1{)}^2}\}$
	$=$	$\frac{3}{2^3{a}_{\xi }^{3/2}}(1+a_{\xi }{q}^2{)}^3\{{\tan }^{-1}[a_{\xi }^{1/2}q]+\frac{a_{\xi }^{1/2}q}{(a_{\xi }{q}^2+1)}-\frac{2a_{\xi }^{1/2}q}{(a_{\xi }{q}^2+1{)}^2}\}$
	$=$	$\frac{3}{2^3{a}_{\xi }^{3/2}}(1+a_{\xi }{q}^2{)}^3[{\tan }^{-1}[a_{\xi }^{1/2}q]-a_{\xi }^{1/2}q\frac{(1-a_{\xi }{q}^2)}{(1+a_{\xi }{q}^2{)}^2}].$

Next, let's examine the factor in square brackets with an "eq" subscript. From our derivation of the properties of detailed force-balance $(n_c,n_e)=(5,1)$ bipolytropes, we know that,

$P_{ic}=K_c{\rho }_0^{6/5}(1+\frac{1}{3}{\xi }_i^2{)}^{-3},$

and,

${\chi }_{\mathrm{e}\mathrm{q}}=(\frac{R_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}=\frac{1}{q}(\frac{r_i}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{q}}=\frac{1}{q}\left[\frac{K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right](\frac{3}{2\pi }{)}^{1/2}{\xi }_i.$

Hence, the relevant factor may be rewritten as,

$\frac{2\pi P_{ic}{\chi }_{\mathrm{e}\mathrm{q}}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$	$=$	$2\pi \left[\frac{K_c{\rho }_0^{6/5}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right](1+\frac{1}{3}{\xi }_i^2{)}^{-3}\{\frac{1}{q}\left[\frac{K_c^{1/2}{G}^{-1/2}{\rho }_0^{-2/5}}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right](\frac{3}{2\pi }{)}^{1/2}{\xi }_i{\}}^3$
	$=$	$\left(\frac{3^3}{2\pi }{)}^{1/2}\right[\frac{K_c^{5/2}{G}^{-3/2}}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}\left]\right(1+\frac{1}{3}{\xi }_i^2{)}^{-3}(\frac{{\xi }_i}{q}{)}^3$
	$=$	$\left(\frac{3^6}{2\pi }{)}^{1/2}\right(1+a_{\xi }{q}^2{)}^{-3}{a}_{\xi }^{3/2},$

where, the last expression has been obtained by employing the substitution, defined above, ${\xi }_i=(3a_{\xi }{)}^{1/2}q$. Finally, then, we have,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\frac{2}{3({\gamma }_c-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_c}\right\{\left(\frac{3^8}{2^7\pi }{)}^{1/2}\right[{\tan }^{-1}[a_{\xi }^{1/2}q]-a_{\xi }^{1/2}q\frac{(1-a_{\xi }{q}^2)}{(1+a_{\xi }{q}^2{)}^2}\left]\right\}.$

As it should, the term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the core, $S_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}$, that has been derived elsewhere in conjunction with our discussion of the detailed force-balanced structure of this bipolytrope.

The Envelope

Similarly, the energy contained in the envelope's thermodynamic reservoir may be written as,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\frac{2}{3({\gamma }_e-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_e}\right(\frac{P_{ie}}{P_{ic}}\left)\right[\frac{2\pi P_{ic}{\chi }^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}[(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}],$

where,

$(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$\equiv$

$\underset{q}{\overset{1}{\int }}3[1-p_e(x)]x^{2}dx,$

defines the relevant integral over the envelope's pressure distribution. According to our derivation of the properties of detailed force-balance $(n_c,n_e)=(5,1)$ bipolytropes — see also the relevant derivations in our accompanying overview — the pressure throughout the envelope is defined by the dimensionless function,

$P^{*}\equiv \frac{P_{\mathrm{e}\mathrm{n}\mathrm{v}}(\eta )}{P_0}$	$=$	${\theta }_i^6{\varphi }^2(\eta )={\theta }_i^6(\frac{A}{\eta }{)}^2{\sin }^2(\eta -B),$
$\Rightarrow 1-p_e(x)\equiv \frac{P_{\mathrm{e}\mathrm{n}\mathrm{v}}(x)}{P_{ie}}$	$=$	$\left(\frac{P_{ic}}{P_{ie}}\right)\left(\frac{P_0}{P_{ic}}\right)\frac{P_{\mathrm{e}\mathrm{n}\mathrm{v}}(x)}{P_0}$
	$=$	$\left(\frac{P_{ic}}{P_{ie}}\right)\left({\theta }_i^{-6}\right){\theta }_i^6(\frac{A}{b_{\eta }x}{)}^2{\sin }^2(b_{\eta }x-B),$
	$=$	$\left(\frac{P_{ic}}{P_{ie}}\right)(\frac{A}{b_{\eta }}{)}^2\frac{{\sin }^2(b_{\eta }x-B)}{x^2},$

where, $b_{\eta }$ has been defined above in connection with our derivation of the envelope's mass profile. The desired integral over this pressure distribution therefore gives,

$(1-q^3)s_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$3\left(\frac{P_{ic}}{P_{ie}}\right)(\frac{A}{b_{\eta }}{)}^2\underset{q}{\overset{1}{\int }}{\sin }^2(b_{\eta }x-B)dx$
	$=$	$\frac{3}{4}\left(\frac{P_{ic}}{P_{ie}}\right)\left(\frac{A^2}{b_{\eta }^3}\right)[2b_{\eta }x-\sin [2(b_{\eta }x-B)]{]}_q^1,$

where, as before, we have dropped the integration constant because it cancels upon insertion of the specified integration limits. Therefore, we have,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\frac{2}{3({\gamma }_e-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_e}\right[\frac{2\pi P_{ic}{\chi }^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{e}\mathrm{q}}\frac{3}{4}\left(\frac{A^2}{b_{\eta }^3}\right)[2b_{\eta }x-\sin [2(b_{\eta }x-B)]{]}_q^1.$

Now, drawing from our above derivation steps and discussion, we know that,

$b_{\eta }$

$=$

$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{{\theta }_i^2{\xi }_i}{q},$

and

$\frac{2\pi P_{ic}{\chi }_{\mathrm{e}\mathrm{q}}^3}{P_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}$

$=$

$\left(\frac{3^6}{2\pi }{)}^{1/2}\right(1+a_{\xi }{q}^2{)}^{-3}{a}_{\xi }^{3/2}=\left(\frac{3^3}{2\pi }{)}^{1/2}\right(\frac{{\theta }_i^2{\xi }_i}{q}{)}^3.$

Finally, then, we can write,

$(\frac{{\mathfrak{𝔖}}_A}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\frac{2}{3({\gamma }_e-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_e}\right\{\frac{3}{4}{A}^2\left[\right(\frac{3^3}{2\pi }{)}^{1/2}\left(\frac{{\theta }_i^2{\xi }_i}{q}{)}^3\right]\left[3^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}\left)\frac{{\theta }_i^2{\xi }_i}{q}{]}^{-3}\right[2b_{\eta }x-\sin [2(b_{\eta }x-B)]{]}_q^1\}$
	$=$	$\frac{2}{3({\gamma }_e-1)}\left(\frac{\chi }{{\chi }_{\mathrm{e}\mathrm{q}}}{)}^{3-3{\gamma }_e}\right\{\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}{A}^2\right(\frac{3^2}{2^5\pi }{)}^{1/2}[2b_{\eta }x-\sin [2(b_{\eta }x-B)]{]}_q^1\}.$

The term inside the curly brackets precisely matches the analytic expression for the dimensionless thermal energy of the envelope, $S_{\mathrm{e}\mathrm{n}\mathrm{v}}^{*}$, that has been derived elsewhere in conjunction with our discussion of the detailed force-balanced structure of this bipolytrope.

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