Free Energy of BiPolytrope with <math>~(n_c, n_e) = (5, 1)</math>

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

Preliminaries

Mass Profile

The core has <math>~n_c = 5 \Rightarrow \gamma_c = 1+1/n_c = 6/5</math>. Referring to the general relation as established in our accompanying overview, and using <math>~\rho_0</math> to represent the central density, we can write,

<math>(\mathrm{For}~0 \leq x \leq q)</math> <math>~M_r </math>

<math> M_\mathrm{tot} \biggl( \frac{\nu}{q^3} \biggr) \biggl( \frac{\rho_0} {{\bar\rho}_\mathrm{core}}\biggr)_\mathrm{eq} \int_0^{x} 3 \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core} x^2 dx \, . </math>

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

<math>~\biggl[ \frac{\rho(\xi)}{\rho_0} \biggr]_\mathrm{core}</math>

<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-5/2} \, ,</math>

where the dimensionless radial coordinate is,

<math>~\biggl[ \frac{G \rho_0^{4/5}}{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} r \, .</math>

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 <math>~(\gamma_c = 6/5)</math> into all normalization parameters, we have,

<math>~R_\mathrm{norm} = \biggl[ \biggl(\frac{G}{K_c} \biggr) M_\mathrm{tot}^{2-\gamma} \biggr]^{1/(4-3\gamma)}</math>	<math>~\Rightarrow</math>	<math>~R_\mathrm{norm} = \biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \, ,</math>
<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl[ \frac{K_c^3}{G^3 M_\mathrm{tot}^2} \biggr]^{1/(4-3\gamma)}</math>	<math>~\Rightarrow</math>	<math>~\rho_\mathrm{norm} = \frac{3}{4\pi} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)^{5/2} \, .</math>

Hence, we can rewrite,

<math>~\xi</math>	<math>~=</math>	<math>~\biggl( \frac{r}{R_\mathrm{norm}} \biggr) \biggl( \frac{\rho_0}{\rho_\mathrm{norm}} \biggr)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} R_\mathrm{norm} \rho_\mathrm{norm}^{2/5}</math>
	<math>~=</math>	<math>~r^* (\rho_0^*)^{2/5} \biggl[ \frac{G }{K_c} \biggr]^{1/2} \biggl( \frac{2\pi}{3} \biggr)^{1/2} \biggl( \frac{G^5 M_\mathrm{tot}^4}{K_c^5} \biggr)^{1/2} \biggl( \frac{3}{4\pi} \biggr)^{2/5} \biggl( \frac{K_c^{3}}{G^3 M_\mathrm{tot}^2} \biggr)</math>
	<math>~=</math>	<math> ~r^* (\rho_0^)^{2/5} \biggl[ \biggl( \frac{2\pi}{3} \biggr)^{5} \biggl( \frac{3}{4\pi} \biggr)^{4} \biggr]^{1/10} = r^ (\rho_0^*)^{2/5} \biggl[ \frac{\pi}{2^3 \cdot 3}\biggr]^{1/10} \, . </math>

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

<math>~r^*</math>	<math>~\rightarrow~</math>	<math> ~ x \chi_\mathrm{eq} \, ; </math>
<math>~\rho_0^*</math>	<math>~\rightarrow~</math>	<math> \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \biggl( \frac{{\bar\rho}_\mathrm{core}}{\rho_\mathrm{norm}} \biggr) = \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \frac{\nu M_\mathrm{tot}/(q^3 R_\mathrm{edge}^3)_\mathrm{eq}}{M_\mathrm{tot}/R_\mathrm{norm}^3} = \frac{\nu}{q^3} \biggl[ \frac{\rho_0}{\bar\rho} \biggr]_\mathrm{core} \chi_\mathrm{eq}^{-3} \, . </math>

Then we can set,

in which case,

<math>~\biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}</math>

<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-5/2} \, ,</math>

where the coefficient,

<math>~(3a_\xi)^{1/2}</math>	<math>~\equiv</math>	<math>~ \chi_\mathrm{eq} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \chi_\mathrm{eq}^{-3} \biggr]^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} =\chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} </math>
<math>\Rightarrow~~~~a_\xi</math>	<math>~\equiv</math>	<math>~ \frac{1}{3} \biggl\{ \chi_\mathrm{eq}^{-1/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{2/5} \biggl( \frac{\pi}{2^3 \cdot 3}\biggr)^{1/10} \biggr\}^2 = \chi_\mathrm{eq}^{-2/5} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}^{4/5} \biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr)^{1/5} \, . </math>

We therefore have,

<math>(\mathrm{For}~0 \leq x \leq q)</math> <math>~M_r </math>	<math>~=</math>	<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \int_0^{x} 3 \biggl( 1 + a_\xi x^2 \biggr)^{-5/2} x^2 dx </math>
	<math>~=</math>	<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \biggl[ x^3\biggl( 1 + a_\xi x^2 \biggr)^{-3/2} \biggr] \, . </math>

Note that, when <math>~x \rightarrow q</math>, <math>~M_r \rightarrow M_\mathrm{core} = \nu M_\mathrm{tot}</math>. Hence, this last expression gives,

<math>~\nu M_\mathrm{tot}</math>	<math>~=</math>	<math> M_\mathrm{tot} \biggl[ \frac{\nu}{q^3} \biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq} \biggl[ q^3\biggl( 1 + a_\xi q^2 \biggr)^{-3/2} \biggr] </math>
<math>\Rightarrow~~~~\biggl[\biggl( \frac{\rho_0} Template:\bar\rho\biggr)_\mathrm{core} \biggr]_\mathrm{eq}</math>	<math>~=</math>	<math> \biggl( 1 + a_\xi q^2 \biggr)^{3/2} \, . </math>

Hence, finally,

<math>(\mathrm{For}~0 \leq x \leq q)</math> <math>~M_r </math>

<math> \nu M_\mathrm{tot} \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} \, ; </math>

and the coefficient, <math>~a_\xi</math>, will be determined only after the equilibrium radius, <math>~\chi_\mathrm{eq}</math>, has been determined, via the relation,

<math>~\chi_\mathrm{eq}^{2} </math>

<math>~\biggl( \frac{\pi}{2^3 \cdot 3^6}\biggr) \biggl( \frac{\nu}{q^3} \biggr)^{4} \biggl( 1 + a_\xi q^2 \biggr)^{6} a_\xi^{-5} \, . </math>

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, <math>~\rho_\mathrm{core}(x) = \rho_c = </math> constant — hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{core} = 1</math>. As has been demonstrated above, the corresponding <math>~M_r</math> function is,

<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}}\biggr]_\mathrm{core} </math>

<math> \biggl( \frac{\nu}{q^3} \biggr) x^3 \, . </math>

Hence,

<math>~W_\mathrm{grav}\biggr\|_\mathrm{core}</math>	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr) \int_0^{q} 3\biggl( \frac{\nu}{q^3} \biggr)x^4 dx </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl( \frac{3}{5} q^5 \biggr) </math>

Now, let's do the envelope. In this case, <math>~\rho_\mathrm{env}(x) = \rho_e = </math> constant; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{env} = 1</math>. As shown elsewhere, the corresponding <math>~M_r</math> function is,

<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env} </math>

<math> \nu + \biggl(\frac{1-\nu}{1-q^3} \biggr) (x^3 - q^3) \, . </math>

Hence,

<math>~W_\mathrm{grav}\biggr\|_\mathrm{env}</math>	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl\{ \int_{q}^{1} \biggl[ \nu -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr]3x dx + \int_{q}^{1} \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] 3x^4 dx \biggr\} </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr) \biggl\{ \frac{3}{2} \biggl[ \nu -q^3 \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^2) + \frac{3}{5} \biggl[ \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5) \biggr\} </math>
	<math>~=</math>	<math> - \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{1}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2} \biggl[ 1 - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (q-q^3) + \biggl[ \frac{q}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] (1-q^5) \biggr\} </math>
	<math>~=</math>	<math> - \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2} \biggl[ 1 - \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{q^3}{\nu} \biggl(\frac{1-\nu}{1-q^3} \biggr) \biggr] \biggl( \frac{1}{q^5}-1\biggr) \biggr\} \, . </math>

Realizing from the above mass segregation derivation that,

<math>~\frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr) = \frac{\rho_e}{\rho_c} \, ,</math>

this last expression can be rewritten as,

<math>~W_\mathrm{grav}\biggr\|_\mathrm{env}</math>	<math>~=</math>	<math> - \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\rho_e}{\rho_c} \biggr] \biggl\{ \frac{5}{2} \biggl[ 1 - \frac{\rho_e}{\rho_c} \biggr] \biggl(\frac{1}{q^2}-1 \biggr) + \biggl[ \frac{\rho_e}{\rho_c} \biggr] \biggl( \frac{1}{q^5}-1\biggr) \biggr\} </math>
	<math>~=</math>	<math> - \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl\{ \frac{5}{2}\biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \biggr\} \, . </math>

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

<math>~W_\mathrm{grav} = W_\mathrm{grav}\biggr|_\mathrm{core} + W_\mathrm{grav}\biggr|_\mathrm{env}</math>

<math> - \frac{3}{5} E_\mathrm{norm} \cdot \chi^{-1} \biggl(\frac{\nu^2}{q} \biggr) f(\nu,q) \, , </math>

where,

<math>~\equiv</math>

<math> 1+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \, . </math>

(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 <math>~(n_c, n_e) = (0, 0) </math> bipolytropes, in this case the pressure throughout the core is defined by the dimensionless function,

<math>~\biggl( \frac{2\pi}{3} \biggr) \xi^2 \, ,</math>

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

<math>\frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[ \frac{\rho_e}{\rho_0} (\xi^2 - \xi_i^2) - 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \xi_i^3\biggl( \frac{1}{\xi} - \frac{1}{\xi_i}\biggr) \biggr] \, , </math>

where, for both functions,

<math>~\xi</math>	<math>~\equiv</math>	<math>~\biggl[ \biggl( \frac{G\rho_0^2}{P_0} \biggr)^{1/2} R_\mathrm{edge} \biggr]_\mathrm{eq} x</math>
	<math>~=</math>	<math>~\biggl[ \frac{G R_\mathrm{edge}^2}{P_0} \biggl( \frac{3 \nu M_\mathrm{tot}}{4\pi q^3 R_\mathrm{edge}^3} \biggr)^2 \biggr]^{1/2}_\mathrm{eq} x</math>
	<math>~=</math>	<math>~\biggl[ \biggl( \frac{3^2}{2^4 \pi^2} \biggr) \frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2 \biggr]^{1/2}_\mathrm{eq} x</math>

So, defining the coefficient,

<math>~\equiv</math>

<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, ,</math>

such that,

<math>~\biggl( \frac{3}{2\pi} \cdot b_\xi \biggr)^{1/2} x \, ,</math>

and remembering that, at the interface, <math>~x \rightarrow x_i = q</math>, so <math>~\xi_i = (3b_\xi/2\pi)^{1/2} q</math>, the two dimensionless pressure functions become,

and,

<math>b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) - 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} - \frac{1}{q}\biggr) \biggr] \, . </math>

The desired integrals over these pressure distributions therefore give,

<math>~\int_0^q \biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr] x^2 dx</math>	<math>~=</math>	<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \int_0^q (1-b_\xi x^2)x^2 dx</math>
	<math>~=</math>	<math>~\biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ \frac{1}{3}\cdot q^3 - \biggl( \frac{b_\xi}{5} \biggr) q^5 \biggr] </math>
	<math>~=</math>	<math>~\frac{q^3}{3} \biggl[ \frac{1}{1-b_\xi q^2} \biggr] \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr] = \frac{q^3}{3} \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \biggl( \frac{3b_\xi}{5} \biggr) q^2 \biggr] \, ;</math>
<math>~\int_q^1 \biggl[1 - p_e(x) \biggr] x^2 dx</math>	<math>~=</math>	<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \int_q^1 \biggl[ \frac{\rho_e}{\rho_0} (x^2 - q^2) - 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{x} - \frac{1}{q}\biggr) \biggr] x^2 dx</math>
	<math>~=</math>	<math>~\frac{1}{3}(1-q^3) - b_\xi\biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{x^5}{5} - \frac{q^2 x^3}{3} \biggr) - 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{x^2}{2} - \frac{x^3}{3q}\biggr) \biggr]_q^1</math>
	<math>~=</math>	<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{ \biggl[ \frac{\rho_e}{\rho_0} \biggl( \frac{3}{5} - q^2 \biggr) - \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^2\biggl( 3q - 2\biggr) \biggr] </math>
		<math>~ - \biggl[ \frac{\rho_e}{\rho_0} \biggl( -\frac{2}{5} \biggr)q^5 - \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^5 \biggr] \biggr\} </math>
	<math>~=</math>	<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl\{ \biggl[ q^2(2-3q) +q^5\biggr] + \frac{\rho_e}{\rho_0}\biggl[ \biggl( \frac{3}{5} - q^2 \biggr) + q^2\biggl( 3q - 2\biggr) +\frac{2q^5}{5} -q^5\biggr] \biggl\} </math>
	<math>~=</math>	<math>~\frac{1}{3}(1-q^3) - \frac{b_\xi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \frac{P_0}{P_{ie} } \biggl[ (2q^2 - 3q^3 +q^5) + \frac{3}{5} \cdot \frac{\rho_e}{\rho_0} ( 1 - 5q^2 + 5q^3 - q^5 ) \biggr] </math>
	<math>~=</math>	<math>~\frac{1}{3}\biggl\{ (1-q^3) + b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} \, , </math>

where,

<math>~\mathfrak{F} </math>

<math>~\equiv</math>

<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, . </math>

Finally, then, we have,

<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{core}</math>	<math>~=</math>	<math> \frac{4\pi/3 }{({\gamma_c}-1)} \biggl[ \frac{P_{ic} \chi^{3\gamma_c}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_c} \biggl\{ \biggl( \frac{P_0}{P_{ic}} \biggr) \biggl[ q^3 - \biggl( \frac{3b_\xi}{5} \biggr) q^5 \biggr] \biggr\} </math>
<math>~\biggl( \frac{\mathfrak{S}_A}{E_\mathrm{norm}} \biggr)_\mathrm{env}</math>	<math>~=</math>	<math> \frac{4\pi/3 }{({\gamma_e}-1)} \biggl[ \frac{P_{ie} \chi^{3\gamma_e}}{P_\mathrm{norm}} \biggr]_\mathrm{eq} \chi^{3-3\gamma_e} \biggl\{ (1-q^3) + b_\xi \biggl(\frac{P_0}{P_{ie} } \biggr) \biggl[\frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} \, . </math>

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,

<math>~\mathcal{A}</math>	<math>~=</math>	<math>~\mathcal{B}_\mathrm{core} \chi_\mathrm{eq}^{4-3\gamma_c} + \mathcal{B}_\mathrm{env} \chi_\mathrm{eq}^{4-3\gamma_e} </math>
	<math>~=</math>	<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} [ q^3 s_\mathrm{core} + (1-q^3) s_\mathrm{env} ] \, . </math>

For <math>~(n_c, n_e) = (0, 0) </math> bipolytropes, the relevant coefficient functions are,

<math>~\mathcal{A}</math>	<math>~=</math>	<math>~\frac{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f \, ,</math>
<math>~q^3 s_\mathrm{core}</math>	<math>~=</math>	<math>~ q^3 \biggl(\frac{P_0}{P_{ic}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] \, , </math>
<math>~(1-q^3) s_\mathrm{env}</math>	<math>~=</math>	<math>~ (1-q^3) + \biggl(\frac{P_0}{P_{ie}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \, , </math>

where,

<math>~f</math>	<math>~\equiv</math>	<math> 1+ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl(\frac{1}{q^2}-1 \biggr) +\biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ \frac{1}{q^5}-1 + \frac{5}{2} \biggl( 1-\frac{1}{q^2}\biggr)\biggr] \, , </math>
<math>~\mathfrak{F} </math>	<math>~\equiv</math>	<math>~ \frac{5}{2} \biggl( \frac{\rho_e}{\rho_c} \biggr) \frac{1}{q^5} \biggl[ (-2q^2 + 3q^3 - q^5) + \frac{3}{5} \biggl( \frac{\rho_e}{\rho_c}\biggr) (-1 +5q^2 - 5q^3 + q^5) \biggr] \, , </math>
<math>~\frac{P_{ic}}{P_0}</math>	<math>~=</math>	<math>~1- p_c(q) = 1 - b_\xi q^2 \, ,</math>
<math>~b_\xi</math>	<math>~\equiv</math>	<math>~\biggl( \frac{3}{2^3 \pi} \biggr) \frac{G M_\mathrm{tot}^2 }{P_0 R_\mathrm{edge}^4} \biggl( \frac{\nu}{q^3}\biggr)^2\, .</math>

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

<math>~\frac{1}{5} \biggl(\frac{\nu^2}{q}\biggr) f</math>	<math>~=</math>	<math>~\frac{4\pi}{3} \biggl[ \frac{P_i R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ q^3 \biggl(\frac{P_0}{P_{i}} \biggr) \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] + (1-q^3) + \biggl(\frac{P_0}{P_{i}} \biggr) \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} </math>
	<math>~=</math>	<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ q^3 \biggl[ 1 - \frac{3}{5}q^2 b_\xi\biggr] + (1-q^3)( 1- b_\xi q^2) + \frac{2}{5} q^5 \mathfrak{F} b_\xi \biggr\} </math>
	<math>~=</math>	<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl\{ 1 - b_\xi \biggl[ \frac{3}{5}q^5 + q^2(1-q^3) - \frac{2}{5} q^5 \mathfrak{F} \biggr] \biggr\} </math>
	<math>~=</math>	<math>~\frac{4\pi}{3} \biggl[ \frac{P_0 R_\mathrm{edge}^4}{GM_\mathrm{tot}^2} \biggr]_\mathrm{eq} \biggl[ \frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] b_\xi </math>
	<math>~=</math>	<math>~\frac{1}{2} \biggl[ \frac{1}{b_\xi} - q^2 + \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] \biggl( \frac{\nu}{q^3}\biggr)^2 </math>
<math>\Rightarrow~~~~\frac{1}{b_\xi}</math>	<math>~=</math>	<math>~ \frac{2}{5}q^5 f + \biggl[q^2 - \frac{2}{5} q^5( 1+\mathfrak{F} ) \biggr] </math>
<math>\Rightarrow~~~~\biggl( \frac{2^3 \pi}{3} \biggr) \frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } \biggl( \frac{q^3}{\nu}\biggr)^2</math>	<math>~=</math>	<math>~ q^2 + \frac{2}{5} q^5( f - 1-\mathfrak{F} ) </math>
<math>\Rightarrow ~~~~ \frac{P_0 R_\mathrm{edge}^4}{G M_\mathrm{tot}^2 } </math>	<math>~=</math>	<math>~ \biggl( \frac{3}{2^3 \pi} \biggr) \biggl( \frac{\nu}{q^3}\biggr)^2 \biggl\{ q^2 + \biggl( \frac{\rho_e}{\rho_c} \biggr) \biggl[ 2q^2(1-q) + \biggl( \frac{\rho_e}{\rho_c} \biggr) (1-3q^2 + 2q^3) \biggr] \biggr\} \, .</math>

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

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Free-energy determination of equilibirum if BiPolytrope with <math>~n_c = 0</math> and <math>~n_e=0</math>.
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User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5 1

Contents

Free Energy of BiPolytrope with <math>~(n_c, n_e) = (5, 1)</math>

Preliminaries

Mass Profile

Gravitational Potential Energy

Thermodynamic Energy Reservoir

Virial Theorem

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