Revision as of 18:55, 29 August 2014

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

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,

<math>~W_\mathrm{grav}\biggr\|_\mathrm{core}</math>	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl(\frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq} \int_0^{q} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{core} \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core} dx </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl( 1 + a_\xi q^2 \biggr)^{3/2} \biggr]_\mathrm{eq} \int_0^{q} 3x \biggl\{ \nu \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2} \biggr\} \biggl( 1 + a_\xi x^2 \biggr)^{-5/2} dx </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \int_0^{q} x^4 \biggl( 1 + a_\xi x^2 \biggr)^{-4} dx </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl\{ \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)}{48 a_\xi^{5/2}(a_\xi q^2 + 1)^3} \biggr\} </math>
	<math>~=</math>	<math> - E_\mathrm{norm} \cdot \chi^{-1} \biggl[\biggl(\frac{3}{2^4}\biggr) a_\xi^{-5/2}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggr]_\mathrm{eq} \biggl[ 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) \biggr] \, . </math>

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

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

<math>~\chi_\mathrm{eq} \biggl( \frac{2^3 \cdot 3^6}{\pi} \biggr)^{1/2} \, .</math>

Hence,

<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{core} </math>

<math> - \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2} \biggl[ 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) \biggr] \, . </math>

After making the substitution, <math>~(a_\xi^{1/2} q) \rightarrow x_i</math>, this expression agrees with a result for the dimensionless energy, <math>~W^*_\mathrm{core}</math>, derived by Tohline in the context of detailed force-balanced bipolytropes.

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.

Related Discussions

Free-energy determination of equilibirum if BiPolytrope with <math>~n_c = 0</math> and <math>~n_e=0</math>.
Free-energy determination of equilibirum if BiPolytrope with <math>~n_c = 5</math> and <math>~n_e=1</math>.
Analytic solution of Detailed-Force-Balance BiPolytrope with <math>~n_c = 0</math> and <math>~n_e=0</math>.
Analytic solution of Detailed-Force-Balance BiPolytrope with <math>~n_c = 5</math> and <math>~n_e=1</math>.
Old Bipolytrope Generalization derivations.

@@ Line 363: / Line 363: @@
 ==Gravitational Potential Energy==
-Here we follow the steps that have been [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Contributions_to_Gravitational_Potential_Energy|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 &#8212; hence, also, <math>~[\rho(x)/\bar\rho]_\mathrm{core} = 1</math>.  As has been demonstrated above, the corresponding <math>~M_r</math> function is,
-<div align="center">
-<table border="0" cellpadding="5" align="center">
-<tr>
+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 [[User:Tohline/SSC/BipolytropeGeneralization_Version2#Separate_Contributions_to_Gravitational_Potential_Energy|outlined in our accompanying overview]] may be written as,
-  <td align="right">
-<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}}\biggr]_\mathrm{core} </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>
-\biggl( \frac{\nu}{q^3} \biggr) x^3 \, .
-</math>
-  </td>
-</tr>
-</table>
-</div>
-Hence,
 <div align="center">
 <table border="0" cellpadding="5" align="center">
@@ Line 396: / Line 377: @@
   <td align="left">
 <math>
-- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)
+- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl(\frac{\rho_0}{\bar\rho} \biggr)_\mathrm{core} \biggr]_\mathrm{eq}
-\int_0^{q} 3\biggl( \frac{\nu}{q^3} \biggr)x^4  dx
+\int_0^{q} 3x \biggl[\frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{core}  \biggl[ \frac{\rho(x)}{\rho_0} \biggr]_\mathrm{core}  dx
 </math>
    </td>
@@ Line 411: / Line 392: @@
   <td align="left">
 <math>
-- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{\nu}{q^3} \biggr)^2 \biggl( \frac{3}{5} q^5 \biggr)
+- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ \frac{\nu}{q^3} \biggl( 1 + a_\xi q^2 \biggr)^{3/2}  \biggr]_\mathrm{eq}
-</math>
+\int_0^{q} 3x \biggl\{
-  </td>
+\nu \biggl( \frac{x^3}{q^3} \biggr) \biggl[ \frac{ 1 + a_\xi x^2 }{ 1 + a_\xi q^2 } \biggr]^{-3/2}
-</tr>
-</table>
-</div>
-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 [[User:Tohline/SSC/BipolytropeGeneralization#ExampleMass|shown elsewhere]], the corresponding <math>~M_r</math> function is,
-<div align="center">
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~\biggl[ \frac{M_r(x)}{M_\mathrm{tot}} \biggr]_\mathrm{env} </math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
-  <td align="left">
-<math>
-\nu  + \biggl(\frac{1-\nu}{1-q^3} \biggr) (x^3 - q^3) \, .
-</math>
-  </td>
-</tr>
-</table>
-</div>
-Hence,
-<div align="center">
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
- <td align="left">
-<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\}
+\biggl( 1 + a_\xi x^2 \biggr)^{-5/2}  dx
 </math>
    </td>
@@ Line 468: / Line 410: @@
   <td align="left">
 <math>
-- E_\mathrm{norm} \cdot \chi^{-1} \biggl( \frac{1-\nu}{1-q^3} \biggr)
+- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
-\biggl\{ \frac{3}{2}
+\int_0^{q} x^4 \biggl( 1 + a_\xi x^2 \biggr)^{-4}  dx
-\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>
    </td>
@@ Line 485: / Line 425: @@
   <td align="left">
 <math>
-- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}
+- E_\mathrm{norm} \cdot \chi^{-1} \biggl[ 3\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
-\biggl[ \frac{1}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
+\biggl\{
-\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)
+\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)}{48 a_\xi^{5/2}(a_\xi q^2 + 1)^3}
 \biggr\}
 </math>
@@ Line 502: / Line 442: @@
   <td align="left">
 <math>
-- \frac{3}{5} \biggl(\frac{\nu^2}{q} \biggr) E_\mathrm{norm} \cdot \chi^{-1}
+- E_\mathrm{norm} \cdot \chi^{-1} \biggl[\biggl(\frac{3}{2^4}\biggr) a_\xi^{-5/2}\biggl(\frac{\nu}{q^3} \biggr)^2 \biggl( 1 + a_\xi q^2 \biggr)^{3}  \biggr]_\mathrm{eq}
-\biggl[ \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr)\biggr] \biggl\{ \frac{5}{2}
+\biggl[
-\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)
+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)
-\biggr] \biggl( \frac{1}{q^5}-1\biggr)  \biggr\} \, .
+\biggr] \, .
 </math>
    </td>
@@ Line 512: / Line 452: @@
 </table>
 </div>
-Realizing from the above mass segregation derivation that,
+But, also from our above discussion of the mass profile, we can write,
-<div align="center">
-<math>~\frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3} \biggr) = \frac{\rho_e}{\rho_c} \, ,</math>
-</div>
-this last expression can be rewritten as,
 <div align="center">
 <table border="0" cellpadding="5" align="center">
 <tr>
    <td align="right">
-<math>~W_\mathrm{grav}\biggr|_\mathrm{env}</math>
+<math>~a_\xi^{-5/2} \biggl( \frac{\nu}{q^3} \biggr)^2 (1 + a_\xi q^2)^3</math>
    </td>
    <td align="center">
 <math>~=</math>
    </td>
- <td align="left">
+  <td align="left">
-<math>
+<math>~\chi_\mathrm{eq} \biggl( \frac{2^3 \cdot 3^6}{\pi} \biggr)^{1/2} \, .</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>
    </td>
 </tr>
-<tr>
-  <td align="right">
-&nbsp;
-  </td>
-  <td align="center">
-<math>~=</math>
-  </td>
- <td align="left">
-<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>
-  </td>
-</tr>
 </table>
 </div>
-So, when put together to obtain the total gravitational potential energy, we have,
+Hence,
 <div align="center">
 <table border="0" cellpadding="5" align="center">
@@ Line 561: / Line 474: @@
 <tr>
    <td align="right">
-<math>~W_\mathrm{grav} = W_\mathrm{grav}\biggr|_\mathrm{core} + W_\mathrm{grav}\biggr|_\mathrm{env}</math>
+<math>~\biggl( \frac{W_\mathrm{grav}}{E_\mathrm{norm}} \biggr)_\mathrm{core} </math>
    </td>
    <td align="center">
@@ Line 568: / Line 481: @@
   <td align="left">
 <math>
-- \frac{3}{5} E_\mathrm{norm} \cdot \chi^{-1} \biggl(\frac{\nu^2}{q} \biggr) f(\nu,q) \, ,
+- \frac{\chi_\mathrm{eq}}{\chi} \biggl( \frac{3^8}{2^5\pi} \biggr)^{1/2}
-</math>
+\biggl[
-  </td>
+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)
-</tr>
+\biggr] \, .
-</table>
-</div>
-where,
-<div align="center">
-<table border="0" cellpadding="5" align="center">
-<tr>
-  <td align="right">
-<math>~f(\nu,q)</math>
-  </td>
-  <td align="center">
-<math>~\equiv</math>
-  </td>
- <td align="left">
-<math>
-+
-\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>
    </td>
 </tr>
 </table>
 </div>
-(This result agrees with Tohline's earlier derivations in other sections of this H_Book, which may now be erased to avoid repetition.)
+After making the substitution, <math>~(a_\xi^{1/2} q) \rightarrow x_i</math>, this expression agrees with a result for the dimensionless energy, <math>~W^*_\mathrm{core}</math>, [[User:Tohline/SSC/Structure/BiPolytropes/Analytic5_1#Expression_for_Free_Energy|derived by Tohline in the context of detailed force-balanced bipolytropes]].
 ==Thermodynamic Energy Reservoir==

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