SSC/Structure/BiPolytropes/FreeEnergy51

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

Free Energy of Bipolytropes (nc, ne) = (5, 1)

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, $n_c=5$, and whose envelope has a polytropic index, $n_e=1$. The details presented here build upon an overview of the free energy of bipolytropes that has been presented elsewhere.  
 
 
 
 

Mass Profile

The Core

The core has $n_c=5\Rightarrow {\gamma }_c=1+1/n_c=6/5$. Referring to the general relation as established in our accompanying overview, and using ${\rho }_0$ to represent the central density, we can write,

$(\mathrm{F}\mathrm{o}\mathrm{r}0\le x\le q)$       $M_r$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{\nu }{q^3}\right)\left(\frac{{\rho }_0}{{\overline{\rho }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{)}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{x}{\int }}3\right[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{x}^{2}dx.$

Drawing on the derivation of detailed force-balance models of $(n_c,n_e)=(5,1)$ bipolytropes, the density profile throughout the core is,

$[\frac{\rho (\xi )}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$(1+\frac{1}{3}{\xi }^2{)}^{-5/2},$

where the dimensionless radial coordinate is,

$\xi$

$=$

$\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{2\pi }{3}{)}^{1/2}r.$

Switching to the normalizations that have been adopted in the broad context of our discussion of configurations in virial equilibrium and inserting the adiabatic index of the core $({\gamma }_c=6/5)$ into all normalization parameters, we have,

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

Hence, we can rewrite,

$\xi$	$=$	$\left(\frac{r}{R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}\right)\left(\frac{{\rho }_0}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}^{2/5}\right[\frac{G}{K_c}{]}^{1/2}(\frac{2\pi }{3}{)}^{1/2}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{2/5}$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{G}{K_c}{]}^{1/2}\left(\frac{2\pi }{3}{)}^{1/2}\right(\frac{G^5{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^4}{K_c^5}{)}^{1/2}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{K_c^3}{G^3{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2})$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\left(\frac{2\pi }{3}{)}^5\right(\frac{3}{4\pi }{)}^4{]}^{1/10}=r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{\pi }{2^3\cdot 3}{]}^{1/10}.$

Now, following the same approach as was used in our introductory discussion and appreciating that our aim here is to redefine the coordinate, $\xi$, in terms of normalized parameters evaluated in the equilibrium configuration, we will set,

$r^{*}$	$\to$	$x{\chi }_{\mathrm{e}\mathrm{q}};$
${\rho }_0^{*}$	$\to$	$\left[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right(\frac{{\overline{\rho }}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})=[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\frac{\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}/(q^3{R}_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3{)}_{\mathrm{e}\mathrm{q}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{\nu }{q^3}[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}.$

Then we can set,

$\xi$

$=$

$(3a_{\xi }{)}^{1/2}x,$

in which case,

$[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$(1+a_{\xi }{x}^2{)}^{-5/2},$

where the coefficient,

$(3a_{\xi }{)}^{1/2}$	$\equiv$	${\chi }_{\mathrm{e}\mathrm{q}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}{]}^{2/5}(\frac{\pi }{2^3\cdot 3}{)}^{1/10}={\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}[\frac{\nu }{q^3}\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{2/5}\right(\frac{\pi }{2^3\cdot 3}{)}^{1/10}$
$\Rightarrow a_{\xi }$	$\equiv$	$\frac{1}{3}\left\{{\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\right[\frac{\nu }{q^3}\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{2/5}\right(\frac{\pi }{2^3\cdot 3}{)}^{1/10}{\}}^2={\chi }_{\mathrm{e}\mathrm{q}}^{-2/5}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}^{4/5}(\frac{\pi }{2^3\cdot 3^6}{)}^{1/5}.$

We therefore have,

$M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{x}{\int }}3(1+a_{\xi }{x}^2{)}^{-5/2}{x}^{2}dx$
	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\left[x^3\right(1+a_{\xi }{x}^2{)}^{-3/2}].$

Note that, when $x\to q$, $M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\to M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}$. Hence, this last expression gives,

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\left[q^3\right(1+a_{\xi }{q}^2{)}^{-3/2}]$
$\Rightarrow \left[\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}$	$=$	$(1+a_{\xi }{q}^2{)}^{3/2}.$

Hence, finally,

$M_r{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{x^3}{q^3}\right)[\frac{1+a_{\xi }{x}^2}{1+a_{\xi }{q}^2}{]}^{-3/2};$

and, after the equilibrium radius, ${\chi }_{\mathrm{e}\mathrm{q}}$, has been determined from the free-energy analysis, the coefficient, $a_{\xi }$, can be determined via the relation,

${\chi }_{\mathrm{e}\mathrm{q}}^2$

$=$

$\left(\frac{\pi }{2^3\cdot 3^6}\right)\left(\frac{\nu }{q^3}{)}^4\right(1+a_{\xi }{q}^2{)}^6{a}_{\xi }^{-5}.$

MORE USEFUL: Letting, $\ell \equiv \xi /\sqrt{3}$, $a_{\xi }$ $=$ $(\frac{{\ell }_i}{q}{)}^2,$ ${\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $=$ $(\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}=(1+{\ell }_i^2{)}^{-3/2}.$

The Envelope

The envelope has $n_e=1\Rightarrow {\gamma }_e=1+1/n_e=2$. Again, referring to the general relation as established in our accompanying overview, and continuing to use ${\rho }_0$ to represent the central density, we can write,

$(\mathrm{F}\mathrm{o}\mathrm{r}q\le x\le 1)$       $M_r$

$=$

$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{\nu +(\frac{1-\nu }{1-q^3}\left)\underset{x_i}{\overset{x}{\int }}3\right[\frac{\rho (x)}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}{x}^{2}dx\}.$

Drawing on the derivation of detailed force-balance models of $(n_c,n_e)=(5,1)$ bipolytropes, the density profile throughout the envelope is,

$[\frac{\rho (\eta )}{{\rho }_0}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (\eta -B)}{\eta }\right],$

where definitions of the constants $A$ and $B$ are given in an accompanying table of parameter values, and the dimensionless radial coordinate is,

$\eta$

$=$

$\left[\frac{G{\rho }_0^{4/5}}{K_c}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}r.$

Using the same radial and mass-density normalizations as defined, above, for the core, we can write,

$\eta$	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}[\frac{G}{K_c}{]}^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2(2\pi {)}^{1/2}{R}_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^{2/5}$
	$=$	$r^{*}({\rho }_0^{*}{)}^{2/5}(\frac{3}{4\pi }{)}^{2/5}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2(2\pi {)}^{1/2}.$

Next, we set,

$r^{*}$	$\to$	$x{\chi }_{\mathrm{e}\mathrm{q}};$
${\rho }_0^{*}$	$\to$	$\left[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right(\frac{{\overline{\rho }}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{{\rho }_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}})=[\frac{{\rho }_0}{\overline{\rho }}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\frac{(1-\nu )M_{\mathrm{t}\mathrm{o}\mathrm{t}}/[(1-q^3)R_{\mathrm{e}\mathrm{d}\mathrm{g}\mathrm{e}}^3{]}_{\mathrm{e}\mathrm{q}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}/R_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}^3}=\frac{1-\nu }{1-q^3}(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{\chi }_{\mathrm{e}\mathrm{q}}^{-3}.$

Hence, we can write,

$\eta =b_{\eta }x,$

where,

$b_{\eta }$

$\equiv$

${\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\left[\frac{1-\nu }{1-q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}^{2/5}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}.$

In which case,

$[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (b_{\eta }x-B)}{b_{\eta }x}\right],$

so,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{\nu +[\left(\frac{1-\nu }{1-q^3}\right)\left(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\underset{q}{\overset{x}{\int }}3A\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^5\right[\frac{\sin (b_{\eta }x-B)}{b_{\eta }x}\left]x^{2}dx\right\}$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }}\underset{q}{\overset{x}{\int }}\sin (b_{\eta }x-B)xdx$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }}[-\frac{\sin (B-b_{\eta }x)+b_{\eta }x\cos (B-b_{\eta }x)}{b_{\eta }^2}{]}_q^x$
	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3A{\theta }_i^5}{b_{\eta }^3}[C_1-\sin (B-b_{\eta }x)-x{b}_{\eta }\cos (B-b_{\eta }x)],$

where, $C_1$ is a constant obtained by evaluating the integral at the interface $(x=x_i=q)$, specifically,

$C_1\equiv \sin (B-b_{\eta }q)+b_{\eta }q\cos (B-b_{\eta }q).$

Now, this expression can be significantly simplified by drawing on earlier results of this section as well as on attributes of the corresponding detailed force-balanced model. First, independent of the specific density profiles that define the structure of a bipolytrope, the ratio of the mean densities of the two structural regions is,

$\frac{{\overline{\rho }}_e}{{\overline{\rho }}_c}$

$=$

$\frac{q^3(1-\nu )}{\nu (1-q^3)}.$

Hence the bracketed pre-factor of the second term of the expression for $M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ may be rewritten as,

$\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}$

$=$

$\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}.$

But, from the above derivation of the mass profile in the core, we know that,

$\left[\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}$

$=$

$(1+a_{\xi }{q}^2{)}^{3/2}=(1+\frac{1}{3}{\xi }_i^2{)}^{3/2}={\theta }_i^{-3},$

where the final step comes from knowledge of the expression for ${\theta }_i$ drawn from the detailed force-balanced model (see, for example, the associated Parameter Values table). Hence, we can write,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}+M_{\mathrm{t}\mathrm{o}\mathrm{t}}\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{3\nu A{\theta }_i^2}{(b_{\eta }q{)}^3}[C_1-\sin (B-b_{\eta }x)-x{b}_{\eta }\cos (B-b_{\eta }x)],$

and note that the expression for the coefficient, $b_{\eta }$, becomes simpler as well, specifically,

$b_{\eta }$	$=$	${\chi }_{\mathrm{e}\mathrm{q}}^{-1/5}\left(\frac{3}{4\pi }{)}^{2/5}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2(2\pi {)}^{1/2}(\frac{\nu }{q^3{\theta }_i^3}{)}^{2/5}$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}$	$=$	$b_{\eta }^{-5}\left(\frac{3^4\pi }{2^3}{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^5\frac{{\nu }^2{\theta }_i^4}{q^6}.$

Next — and, again, drawing from knowledge of the internal structure of the detailed force-balanced model, in particular, realizing that,

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

— note that the constant, $C_1$, can be rewritten as,

$C_1$	$=$	$b_{\eta }q\cos (b_{\eta }q-B)-\sin (b_{\eta }q-B)$
	$=$	${\eta }_i\cos ({\eta }_i-B)-\sin ({\eta }_i-B)$
	$=$	$\frac{{\eta }_i^2}{A}(\frac{d\varphi }{d\eta }{)}_i=-\frac{1}{A}(\frac{{\mu }_e}{{\mu }_c}{)}^2{3}^{1/2}{\theta }_i^4{\xi }_i^3$
	$=$	$-\frac{1}{A}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2{3}^{1/2}{\theta }_i^4\right[3^{-1/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}{b}_{\eta }q{]}^3$
	$=$	$-\frac{1}{3A{\theta }_i^2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}(b_{\eta }q{)}^3,$

which means,

$M_r{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}-\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}\{1-\frac{1}{C_1}[\sin (B-b_{\eta }x)+x{b}_{\eta }\cos (B-b_{\eta }x)\left]\right\}$
	$=$	$\frac{\nu M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{C_1}[\sin (B-b_{\eta }x)+x{b}_{\eta }\cos (B-b_{\eta }x)].$

MORE USEFUL: Letting, $\ell \equiv \xi /\sqrt{3}$, $b_{\eta }={\eta }_s$        and        $b_{\eta }q={\eta }_i=3\left(\frac{{\mu }_e}{{\mu }_c}\right){\ell }_i(1+{\ell }_i^2{)}^{-1},$ ${\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{e}\mathrm{n}\mathrm{v}}\equiv (\frac{\overline{\rho }}{{\rho }_c}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $=$ $\frac{q^3(1-\nu )}{\nu (1-q^3)}\cdot {\widetilde{\mathfrak{𝔣}}}_{\mathrm{M}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

Gravitational Potential Energy

The Core

Borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the core that has been outlined in our accompanying overview may be written as,

$W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\frac{\nu }{q^3}\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}{]}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{q}{\int }}3x\left[\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}dx$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\frac{\nu }{q^3}\right(1+a_{\xi }{q}^2{)}^{3/2}{]}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{q}{\int }}3x\left\{\nu \right(\frac{x^3}{q^3}\left)\right[\frac{1+a_{\xi }{x}^2}{1+a_{\xi }{q}^2}{]}^{-3/2}\left\}\right(1+a_{\xi }{x}^2{)}^{-5/2}dx$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[3\right(\frac{\nu }{q^3}{)}^2(1+a_{\xi }{q}^2{)}^3{]}_{\mathrm{e}\mathrm{q}}\underset{0}{\overset{q}{\int }}{x}^4(1+a_{\xi }{x}^2{)}^{-4}dx$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[3\right(\frac{\nu }{q^3}{)}^2(1+a_{\xi }{q}^2{)}^3{]}_{\mathrm{e}\mathrm{q}}\{\frac{a_{\xi }^{1/2}q(3a_{\xi }^2{q}^4-8a_{\xi }{q}^2-3)+3(a_{\xi }{q}^2+1{)}^3{\tan }^{-1}(a_{\xi }^{1/2}q)}{48a_{\xi }^{5/2}(a_{\xi }{q}^2+1{)}^3}\}$
	$=$	$-E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}\cdot {\chi }^{-1}\left[\right(\frac{3}{2^4}\left)a_{\xi }^{-5/2}\right(\frac{\nu }{q^3}{)}^2(1+a_{\xi }{q}^2{)}^3{]}_{\mathrm{e}\mathrm{q}}[a_{\xi }^{1/2}q(a_{\xi }^2{q}^4-\frac{8}{3}{a}_{\xi }{q}^2-1)(a_{\xi }{q}^2+1{)}^{-3}+{\tan }^{-1}(a_{\xi }^{1/2}q)].$

MORE USEFUL: $-\chi [\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ $=$ $\left[\right(\frac{3}{2^4}\left)\right(\frac{q}{{\ell }_i}{)}^5\left(\frac{\nu }{q^3}{)}^2\right(1+{\ell }_i^2{)}^3{]}_{\mathrm{e}\mathrm{q}}[{\ell }_i({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)({\ell }_i^2+1{)}^{-3}+{\tan }^{-1}{\ell }_i]$   $=$ $\frac{3}{5}\left[\right(\frac{\nu }{q^3}{)}^2(1+{\ell }_i^2{)}^3{]}_{\mathrm{e}\mathrm{q}}(\frac{5}{2^4}\left)\right(\frac{q}{{\ell }_i}{)}^5[{\ell }_i({\ell }_i^4-\frac{8}{3}{\ell }_i^2-1)({\ell }_i^2+1{)}^{-3}+{\tan }^{-1}{\ell }_i].$

But, also from our above discussion of the mass profile, we can write,

$a_{\xi }^{-5/2}(\frac{\nu }{q^3}{)}^2(1+a_{\xi }{q}^2{)}^3$

$=$

${\chi }_{\mathrm{e}\mathrm{q}}(\frac{2^3\cdot 3^6}{\pi }{)}^{1/2}.$

Hence,

$(\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$-\frac{{\chi }_{\mathrm{e}\mathrm{q}}}{\chi }\left(\frac{3^8}{2^5\pi }{)}^{1/2}\right[a_{\xi }^{1/2}q(a_{\xi }^2{q}^4-\frac{8}{3}{a}_{\xi }{q}^2-1)(a_{\xi }{q}^2+1{)}^{-3}+{\tan }^{-1}(a_{\xi }^{1/2}q)].$

After making the substitution, $(a_{\xi }^{1/2}q)\to x_i$, this expression agrees with a result for the dimensionless energy, $W_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}$, derived by Tohline in the context of detailed force-balanced bipolytropes.

The Envelope

Again, borrowing from our derivation, above, of the mass distribution in this type of bipolytrope, the expression for the gravitational potential energy in the envelope that has been outlined in our accompanying overview may be written as,

$(\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$-{\chi }^{-1}\left[\right(\frac{1-\nu }{1-q^3}\left)\right(\frac{{\rho }_0}{\overline{\rho }}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}{]}_{\mathrm{e}\mathrm{q}}\underset{q}{\overset{1}{\int }}3x\left[\frac{M_r(x)}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right[\frac{\rho (x)}{{\rho }_0}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}dx$
	$=$	$-{\chi }^{-1}\left[\frac{\nu }{q^3{\theta }_i^3}\right]\underset{q}{\overset{1}{\int }}3x\left\{\frac{\nu }{C_1}\right[\sin (B-b_{\eta }x)+x{b}_{\eta }\cos (B-b_{\eta }x)\left]\right\}\left\{A\right(\frac{{\mu }_e}{{\mu }_c}\left){\theta }_i^5\right[\frac{\sin (b_{\eta }x-B)}{b_{\eta }x}\left]\right\}dx$
	$=$	$-{\chi }^{-1}\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{3{\nu }^{2}A{\theta }_i^2}{b_{\eta }{q}^3}\right]\frac{1}{C_1}\underset{q}{\overset{1}{\int }}[x{b}_{\eta }\cos (b_{\eta }x-B)-\sin (b_{\eta }x-B)]\sin (b_{\eta }x-B)dx$
	$=$	$-{\chi }^{-1}\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{3{\nu }^{2}A{\theta }_i^2}{b_{\eta }{q}^3}\right][-3A{\theta }_i^2(\frac{{\mu }_e}{{\mu }_c})(b_{\eta }q{)}^{-3}]\underset{q}{\overset{1}{\int }}[x{b}_{\eta }\cos (b_{\eta }x-B)-\sin (b_{\eta }x-B)]\sin (b_{\eta }x-B)dx$
	$=$	$-{\chi }^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right[\frac{3^2{A}^2}{b_{\eta }^4}\left]\right(\frac{{\nu }^2{\theta }_i^4}{q^6})\underset{q}{\overset{1}{\int }}[\sin (b_{\eta }x-B)-x{b}_{\eta }\cos (b_{\eta }x-B)]\sin (b_{\eta }x-B)dx.$

But we also know, from above, that,

${\chi }_{\mathrm{e}\mathrm{q}}=b_{\eta }^{-5}\left(\frac{3^4\pi }{2^3}{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^5\frac{{\nu }^2{\theta }_i^4}{q^6}$

$\Rightarrow$

$\frac{{\nu }^2{\theta }_i^4}{q^6}={\chi }_{\mathrm{e}\mathrm{q}}{b}_{\eta }^5\left(\frac{2^3}{3^4\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-5}.$

So we have,

$(\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$-\frac{{\chi }_{\mathrm{e}\mathrm{q}}}{\chi }\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}\right(\frac{2^3}{\pi }{)}^{1/2}{A}^2{b}_{\eta }\underset{q}{\overset{1}{\int }}[\sin (b_{\eta }x-B)-x{b}_{\eta }\cos (b_{\eta }x-B)]\sin (b_{\eta }x-B)dx.$

The integral can be broken into two separate parts:

$\underset{q}{\overset{1}{\int }}{\sin }^2(b_{\eta }x-B)dx$

$=$

$\frac{1}{4b_{\eta }}\{2b_{\eta }x-\sin [2(b_{\eta }x-B)]{\}}_q^1,$

and,

$-\underset{q}{\overset{1}{\int }}x{b}_{\eta }\cos (b_{\eta }x-B)\sin (b_{\eta }x-B)dx$

$=$

$\frac{1}{8b_{\eta }}\{2b_{\eta }x\cos [2(b_{\eta }x-B)]-\sin [2(b_{\eta }x-B)]{\}}_q^1.$

(Note: We have dropped integration constants that might result from carrying out an indefinite integral because such constants would disappear upon application of our specified limits of integration.) When added together, they give,

$\underset{q}{\overset{1}{\int }}\dots dx$	$=$	$\frac{1}{8b_{\eta }}\{2b_{\eta }x\cos [2(b_{\eta }x-B)]-3\sin [2(b_{\eta }x-B)]+4b_{\eta }x{\}}_q^1$
	$=$	$\frac{1}{8b_{\eta }}\left\{2b_{\eta }x\right[1-2{\sin }^2(b_{\eta }x-B)]-3\sin [2(b_{\eta }x-B)]+4b_{\eta }x{\}}_q^1$
	$=$	$\frac{1}{8b_{\eta }}[6b_{\eta }x-3\sin [2(b_{\eta }x-B)]-4b_{\eta }x{\sin }^{}(b_{\eta }x-B){]}_q^1.$

Hence,

$(\frac{W_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}}{E_{\mathrm{n}\mathrm{o}\mathrm{r}\mathrm{m}}}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$-\frac{{\chi }_{\mathrm{e}\mathrm{q}}}{\chi }\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-3}\right(\frac{1}{2^3\pi }{)}^{1/2}{A}^2[6b_{\eta }x-3\sin [2(b_{\eta }x-B)]-4b_{\eta }x{\sin }^{}(b_{\eta }x-B){]}_q^1.$

This expression matches in detail the expression for the gravitational potential energy of the envelope derived in the context of our derivation of detailed force-balanced models of this bipolytrope.

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

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