SSC/Structure/BiPolytropes/FreeEnergy51/Pt2

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

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.

See Also