User:Tohline/SSC/Structure/BiPolytropes/FreeEnergy5 1

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.

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>~=</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>~=</math>

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

where the dimensionless radial coordinate is,

<math>~\xi</math>

<math>~=</math>

<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,

<math>~\xi</math>

<math>~=</math>

<math>~(3a_\xi)^{1/2} x \, ,</math>

in which case,

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

<math>~=</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>~=</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>~=</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>~=</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>~=</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) = (5, 1)</math> bipolytropes — see also the relevant derivations in our accompanying overview — in this case the pressure throughout the core is defined by the dimensionless function,

<math>~P^* \equiv \frac{P_\mathrm{core}(\xi)}{P_0} </math>	<math>~=</math>	<math>~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3} \, ,</math>
<math>\Rightarrow ~~~~ 1-p_c(x) = \frac{P_\mathrm{core}(x)}{P_0} </math>	<math>~=</math>	<math>~\biggl( 1 + a_\xi x^2 \biggr)^{-3} \, ,</math>

where, <math>~a_\xi</math> is defined above in connection with our derivation of the mass profile. The desired integral over this pressure distribution therefore gives,

<math>~q^3 s_\mathrm{core} </math>	<math>~\equiv</math>	<math>~ \int_0^q 3\biggl[\frac{1 - p_c(x)}{1-p_c(q)} \biggr] x^2 dx </math>
	<math>~=</math>	<math>~ 3\biggl( 1 + a_\xi q^2 \biggr)^{3} \int_0^q \frac{x^2 dx}{(1+a_\xi x^2)^3} </math>
	<math>~=</math>	<math>~ 3\biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{ \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} \biggr\} </math>
	<math>~=</math>	<math>~ \frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl\{ \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} \biggr\} </math>
	<math>~=</math>	<math>~ \frac{3}{2^3 a_\xi^{3/2}} \biggl( 1 + a_\xi q^2 \biggr)^{3} \biggl[ \tan^{-1}[a_\xi^{1/2}q] - a_\xi^{1/2}q ~\frac{(1 - a_\xi q^2)}{(1 + a_\xi q^2)^2} \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.

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation