User:Tohline/SSC/Structure/BiPolytropes/Analytic0 0

BiPolytrope with <math>n_c = 0</math> and <math>n_e=0</math>

Here we construct a bipolytrope in which both the core and the envelope have uniform densities, that is, the structure of both the core and the envelope will be modeled using an <math>n = 0</math> polytropic index. It should be possible for the entire structure to be described by closed-form, analytic expressions. Generally, we will follow the general solution steps for constructing a bipolytrope that we have outlined elsewhere. [On 1 February 2014, J. E. Tohline wrote: This particular system became of interest to me during discussions with Kundan Kadam about the relative stability of bipolytropes.]

Step 4: Throughout the core (<math>0 \le \chi \le \chi_i</math>)

Specify: <math>~P_0</math> and <math>\rho_0 ~\Rightarrow</math>
<math>~\rho</math>	<math>~=</math>	<math>~\rho_0</math>
<math>~P</math>	<math>~=</math>	<math>P_0 - \frac{2}{3} \pi G \rho_0^2 r^2</math>	<math>~=</math>	<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi^2 \biggr)</math>
<math>~r</math>	<math>~=</math>	<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math>	<math>~=</math>	<math>\biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{1/2} \chi</math>
<math>~M_r</math>	<math>~=</math>	<math>\frac{4\pi}{3} \rho_0 r^3</math>	<math>~=</math>	<math>\frac{4\pi}{3} \rho_0 \biggl[ \frac{P_0}{G \rho_0^2} \biggr]^{3/2} \chi^3 = \frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi^3</math>

Step 5: Interface Conditions

Specify: <math>~\chi_i</math> and <math>~\rho_e/\rho_0</math>, and demand …
<math>~P_{ei}</math>	<math>~=</math>	<math>~P_{ci}</math>	<math>~=</math>	<math>P_0 \biggl( 1 - \frac{2\pi}{3}\chi_i^2 \biggr)</math>

Step 6: Envelope Solution (<math>~\chi > \chi_i</math>)

<math>~\rho</math>	<math>~=</math>	<math>~\rho_e</math>
<math>~P</math>	<math>~=</math>	<math>P_{ei} + \biggl(\frac{2}{3} \pi G \rho_e\biggr) \biggl[ 2(\rho_0 - \rho_e) r_i^3\biggl( \frac{1}{r} - \frac{1}{r_i}\biggr) - \rho_e(r^2 - r_i^2) \biggr]</math>
	<math>~=</math>	<math>P_{ei} + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) P_0 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \chi_i^3\biggl( \frac{1}{\chi} - \frac{1}{\chi_i}\biggr) - \frac{\rho_e}{\rho_0} (\chi^2 - \chi_i^2) \biggr]</math>
<math>~\frac{P}{P_0}</math>	<math>~=</math>	<math>1 - \frac{2\pi}{3}\chi_i^2 + \frac{2\pi}{3} \biggl(\frac{\rho_e}{\rho_0}\biggr) \chi_i^2 \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( \frac{1}{\xi} - 1\biggr) - \frac{\rho_e}{\rho_0} (\xi^2 - 1) \biggr]</math>
<math>~M_r</math>	<math>~=</math>	<math>\frac{4\pi}{3} \biggl[ \rho_0 r_i^3 + \rho_e(r^3 - r_i^3) \biggr]</math>
	<math>~=</math>	<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \biggl[\chi_i^3 +\frac{\rho_e}{\rho_0} \biggl( \chi^3 - \chi_i^3 \biggr) \biggr]</math>
	<math>~=</math>	<math>\frac{4\pi}{3} \biggl[ \frac{P_0^3}{G^3 \rho_0^4} \biggr]^{1/2} \chi_i^3\biggl[1 +\frac{\rho_e}{\rho_0} \biggl( \xi^3 - 1\biggr) \biggr]</math>

Step 7: Surface Boundary Condition

At the surface (that is, at <math>r = R</math> and <math>M_r = M_\mathrm{tot}</math>), <math>P/P_0 = 0</math> and <math>\xi = \xi_s = R/r_i = 1/q</math>. Also, we can write,

<math> \chi_i = q\biggl[ \frac{G\rho_0^2 R^2}{P_0} \biggr]^{1/2} \, ; </math>

and, from earlier derivations,

<math> R^3 = \frac{3M_\mathrm{tot}}{4\pi \bar\rho} = \frac{3M_\mathrm{tot}}{4\pi \rho_0} \biggl( \frac{\nu}{q^3} \biggr) \, ; </math>

<math> \frac{\rho_e}{\rho_0} = \frac{q^3}{\nu} \biggl( \frac{1-\nu}{1-q^3}\biggr) \, . </math>

Therefore, setting the pressure to zero at the surface means,

<math>~\frac{3}{2\pi}\chi_i^{-2}</math>	<math>~=</math>	<math>1 - \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( q - 1\biggr) - \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr]</math>
<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) q^{-2} \biggl[ \frac{P_0}{G\rho_0^2 R^2} \biggr]</math>	<math>~=</math>	<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>
<math>\Rightarrow ~~~~~\biggl( \frac{3}{2\pi} \biggr) \biggl( \frac{4\pi}{3} \biggr)^{2/3} \nu^{-2/3} \biggl[ \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr]^{1/3}</math>	<math>~=</math>	<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>
<math>\Rightarrow ~~~~~\biggl\{ \biggl( \frac{6}{\pi} \biggr) \frac{1}{\nu^{2} } \biggl[ \frac{P_0^3}{G^3\rho_0^4 M_\mathrm{tot}^2} \biggr] \biggr\}^{1/3}</math>	<math>~=</math>	<math>1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1- q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] </math>

It therefore seems prudent to define a function,

<math> g(\nu,q) \equiv \biggl\{ 1 + \biggl(\frac{\rho_e}{\rho_0}\biggr) \biggl[ 2 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) \biggl( 1-q \biggr) + \frac{\rho_e}{\rho_0} \biggl(\frac{1}{q^2} - 1\biggr) \biggr] \biggr\}^{1/2}\, , </math>

in which case the expressions for the equilibrium radius and equilibrium total mass are, respectively,

<math> \biggl[ \frac{G\rho_0^2}{P_0} \biggr]^{1/2} R</math>	<math>~=</math>	<math>\biggl( \frac{3}{2\pi} \biggr)^{1/2} \frac{1}{q g} \, ;</math>
<math> \biggl[ \frac{G^3\rho_0^4}{P_0^3} \biggr]^{1/2} M_\mathrm{tot}</math>	<math>~=</math>	<math>\biggl( \frac{6}{\pi} \biggr)^{1/2} \frac{1}{\nu g^3} \, .</math>

We can also combine these two expressions and eliminate direct reference to the central density, <math>\rho_0</math>, obtaining,

<math> \biggl[ \frac{R^4}{GM_\mathrm{tot}^2} \biggr] P_0</math>	<math>~=</math>	<math>\biggl( \frac{3}{2^3\pi} \biggr) \frac{\nu^2 g^2}{q^4} </math>
	<math>~=~</math>	<math> \biggl( \frac{3}{2^3\pi}\biggr) \frac{\nu^2}{q^6} \biggl[ q^2 + 2\biggl( \frac{\rho_e}{\rho_0} \biggr) q^2(1-q) + \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 3q^2 + 2q^3) \biggr] \, . </math>

Thermal Energy Content

Knowing the pressure distribution throughout this bipolytropic configuration allows us to obtain an analytic expression for the configuration's total thermal content. Specifically, the differential contribution to the total thermal energy that is made by each spherical shell is,

<math>dS = \frac{3}{2} \biggl( \frac{P}{\rho} \biggr) dm = \frac{3}{2} \biggl( \frac{P}{\rho} \biggr) 4\pi \rho r^2 dr = 6\pi R^3 P(x) x^2 dx \, ,</math>

where,

<math>x \equiv \frac{r}{R} \, .</math>

(We are switching to a new normalization of the radial coordinate — using <math>~x</math> in preference to <math>~\chi</math> — because, at the interface between the core and the envelope, <math>~x=q</math>, while <math>~x=1 </math> at the surface of the bipolytropic configuration.) Hence, the thermal content of the core and of the envelope will be given by performing the following integrals, respectively:

<math>~S_\mathrm{core}</math>	<math>~=~</math>	<math>6\pi R^3 \int_0^q P_\mathrm{core}(x) x^2 dx \, ;</math>
<math>~S_\mathrm{env}</math>	<math>~=~</math>	<math> 6\pi R^3 \int_q^1 P_\mathrm{env}(x) x^2 dx \, .</math>

Drawing from the expressions obtained in step #4 and step #6, above, the relevant functions <math>~P(x)</math> are,

<math>~P_\mathrm{core}</math>	<math>~=</math>	<math>P_0 - \frac{2}{3} \pi G \rho_0^2 r^2 = P_i + \frac{2}{3} \pi G \rho_0^2(r_i^2 - r^2)</math>
	<math>~=</math>	<math>P_i + \frac{2}{3} \pi G \rho_0^2 R^2 (q^2 - x^2) = P_i + \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) (q^2 - x^2) \, ;</math>
<math>~P_\mathrm{env}</math>	<math>~=</math>	<math>P_{i} - \biggl(\frac{2}{3} \pi G \rho_e\biggr) \biggl[ 2(\rho_0 - \rho_e) r_i^3\biggl( \frac{1}{r_i} - \frac{1}{r} \biggr) + \rho_e(r^2 - r_i^2) \biggr]</math>
	<math>~=</math>	<math>P_{i} - \biggl(\frac{2}{3} \pi G \rho_0^2 R^2\biggr) \frac{\rho_e}{\rho_0} \biggl[ 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{q} - \frac{1}{x} \biggr) + \frac{\rho_e}{\rho_0} (x^2 - q^2) \biggr]</math>
	<math>~=</math>	<math>P_{i} - \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) \frac{\rho_e}{\rho_0} \biggl[ 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3\biggl( \frac{1}{q} - \frac{1}{x} \biggr) + \frac{\rho_e}{\rho_0} (x^2 - q^2) \biggr] \, .</math>

Reminder: For a given choice of the parameter set <math>~(M_\mathrm{tot}, R, \nu, q)</math>, — and remembering that choosing the parameter pair <math>~(\nu,q)</math> sets the density ratio via the expression, <math>~\rho_e/\rho_0 = [q^3(1-\nu)]/[\nu(1-q^3)]</math> — the value of the pressure at the interface, <math>~P_i</math>, is determined by setting boundary conditions at the surface of the configuration, namely, by setting <math>~x=1</math> and <math>~P_\mathrm{env} = 0</math> in the second expression. Then, the central pressure <math>~(P_\mathrm{core} = P_0)</math> is determined by setting <math>~x=0</math> and inserting the determined value for <math>~P_i</math> in the first expression.

Hence,

<math>~S_\mathrm{core}</math>	<math>~=~</math>	<math>6\pi R^3 \int_0^q \biggl\{ P_i + \Pi (q^2 - x^2) \biggr\} x^2 dx </math>
	<math>~=~</math>	<math>6\pi R^3 \biggl\{ \biggl[ \biggl(P_i + q^2 \Pi \biggr)\frac{x^3}{3} \biggr]_0^q - \biggl[ \Pi \frac{x^5}{5} \biggr]_0^q \biggr\} </math>
	<math>~=~</math>	<math>\biggl( \frac{4\pi}{5} \biggr) R^3 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr) \, ,</math>
<math>~S_\mathrm{env}</math>	<math>~=~</math>	<math> 6\pi R^3 \int_q^1 \biggl\{ P_i - \Pi \frac{\rho_e}{\rho_0} \biggl( a_e + \frac{b_e}{x} + c_e x^2\biggr) \biggr\} x^2 dx </math>
	<math>~=~</math>	<math> 6\pi R^3 \biggl\{ \biggl[ P_i - a_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)\biggr] \frac{x^3}{3} \biggr|_q^1 - \biggl[ b_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \frac{x^2}{2} \biggr]_q^1 - \biggl[ c_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \frac{x^5}{5} \biggr]_q^1 \biggr\} </math>
	<math>~=~</math>	<math> \frac{\pi R^3}{5} \biggl\{ 10 \biggl[ P_i - a_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr] (1-q^3) - 15 \biggl[ b_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) (1-q^2) \biggr] - 6\biggl[ c_e \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) (1-q^5) \biggr] \biggr\} </math>
	<math>~=~</math>	<math> \frac{\pi R^3}{5} \biggl\{ 10 P_i (1-q^3) - \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 10 a_e (1-q^3) + 15 b_e (1-q^2) + 6 c_e (1-q^5) \biggr] \biggr\} </math>
	<math>~=~</math>	<math> \frac{\pi R^3}{5} \biggl\{ 10 P_i (1-q^3) </math>
		<math> +\Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 30 \biggl( \frac{\rho_e}{\rho_0} - \frac{2}{3} \biggr) q^2 (1-q^3) + 30 \biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3 (1-q^2) - 6 \frac{\rho_e}{\rho_0} (1-q^5) \biggr] \biggr\} </math>
	<math>~=~</math>	<math> \frac{\pi R^3}{5} \biggl\{ 10 P_i (1-q^3) + 10 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) q^2 \biggl[ 3 q (1-q^2) - 2 (1-q^3) \biggr] </math>
		<math> + 6 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ 5 q^2 (1-q^3)- 5 q^3 (1-q^2) - (1-q^5) \biggr] \biggr\} </math>
	<math>~=~</math>	<math> \frac{\pi R^3}{5} \biggl\{ 10 P_i (1-q^3) + 10 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2q^2 + 3q^3 - q^5 \biggr] + 6 \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 + q^5 \biggr] \biggr\} \, , </math>

where,

<math> \Pi \equiv \frac{3}{2^3 \pi} \biggl( \frac{GM_\mathrm{tot}^2}{R^4} \biggr) \biggl( \frac{\nu^2}{q^6} \biggr) \, , </math>

and, in the expression for <math>~S_\mathrm{env}</math>, we temporarily used the shorthand notation,

<math>~a_e</math>	<math>~\equiv~</math>	<math>3 \biggl(\frac{2}{3} - \frac{\rho_e}{\rho_0} \biggr) q^2 \, ,</math>
<math>~b_e</math>	<math>~\equiv~</math>	<math>- 2\biggl(1 - \frac{\rho_e}{\rho_0} \biggr) q^3 \, ,</math>
<math>~c_e</math>	<math>~\equiv~</math>	<math>\frac{\rho_e}{\rho_0} \, .</math>

The total thermal energy (per unit volume) may therefore be written as,

<math>\biggl( \frac{5}{\pi R^3} \biggr) S_\mathrm{tot}</math>	<math>~=~</math>	<math> 4 q^5 \biggl (\frac{5P_i}{2q^2} + \Pi \biggr) + 10 P_i (1-q^3) - \Pi \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[ 10 a_e (1-q^3) + 15 b_e (1-q^2) + 6 c_e (1-q^5) \biggr] </math>
	<math>~=~</math>	<math> 10P_i + 2 \Pi \biggl[ 2q^5 - 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 ) - 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 - q^5 ) \biggr] \, , </math>

or, because <math>~P_0 = P_i + \Pi q^2</math>, we can reference the central pressure instead of the pressure at the interface and write,

<math>\biggl( \frac{5}{\pi R^3} \biggr) S_\mathrm{tot}</math>

<math>~=~</math>

<math> 10P_0 - 2\Pi \biggl[5q^2 -2q^5 + 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 ) + 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 - q^5 ) \biggr] \, . </math>

Gravitational Potential Energy

In a separate discussion, we derived the following expression for the total gravitational potential energy of an <math>~(n_c, n_e) = (0, 0)</math> bipolytrope:

<math>~W</math>

<math> ~= - \frac{3GM^2_\mathrm{tot}}{5R} \biggl( \frac{\nu^2}{q} \biggr) f(q, \rho_e/\rho_c) \, , </math>

where,

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

Adopting the parameters and normalizations used above in the context of the total thermal energy derivation, this can be rewritten as,

<math>~\biggl( \frac{5}{\pi R^3} \biggr) W</math>

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

Virial Equilibrium

In order for this system to be in virial equilibrium, we must have <math>(2S_\mathrm{tot} + W) = 0</math>, or,

<math>S_\mathrm{tot} = - \frac{W}{2} \, .</math>

The central pressure that is required in order to establish this virial equilibrium condition is,

<math>~10P_0</math>	<math>~=~</math>	<math> 2\Pi \biggl[5q^2 -2q^5 + 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 ) + 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 - q^5 ) \biggr] </math>
		<math> + 2\Pi \biggl\{ 2q^5 + 5q^3\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1-q^5) - 5q^3(1-q^2) \biggr] \biggr\} </math>
<math>\Rightarrow ~~~\frac{5P_0}{\Pi}</math>	<math>~=~</math>	<math> 5q^2 + 5\biggl( \frac{\rho_e}{\rho_0} \biggr) (2q^2 - 3q^3 + q^5 ) + 3 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 5q^2 + 5q^3 - q^5 ) </math>
		<math> + 5 \biggl( \frac{\rho_e}{\rho_c} \biggr) (q^3-q^5) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 ( 2 - 5q^3 + 3q^5 ) </math>
<math>\Rightarrow ~~~\frac{P_0}{\Pi}</math>	<math>~=~</math>	<math> q^2 + 2\biggl( \frac{\rho_e}{\rho_0} \biggr) q^2(1-q) + \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 3q^2 + 2q^3) </math>
<math>\Rightarrow ~~~\biggl[\frac{R^4}{GM_\mathrm{tot}^2}\biggr] P_0</math>	<math>~=~</math>	<math> \biggl( \frac{3}{2^3\pi}\biggr) \frac{\nu^2}{q^6} \biggl[ q^2 + 2\biggl( \frac{\rho_e}{\rho_0} \biggr) q^2(1-q) + \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 ( 1 - 3q^2 + 2q^3) \biggr] \, . </math>

Hooray! This precisely matches the expression for the central pressure, <math>~P_0</math>, that was obtained above at the end of step #7 of the detailed force balance derivation. It is very satisfying to see that the central pressure can be derived from demanding a balance of the thermal and gravitational potential energies via the virial equilibrium expression.

Free Energy

Now, let's see if we can derive the virial equilibrium condition — and, hence the correct central pressure — from the free energy.

<math>~\mathfrak{G}</math>	<math>~=~</math>	<math>~U_\mathrm{tot} + W</math>
	<math>~=~</math>	<math>\biggl[ \frac{2}{3(\gamma_c -1)} \biggr] S_\mathrm{core} + \biggl[ \frac{2}{3(\gamma_e -1)} \biggr] S_\mathrm{env} + W</math>
	<math>~=~</math>	<math>(A_\mathrm{core} + A_\mathrm{env} -A_\mathrm{grav}) R^{-1} + B_\mathrm{core}R^3 P_{ic} + B_\mathrm{env} R^3 P_{ie} \, , </math>

where, for a given choice of the three parameters <math>~(M_\mathrm{tot}, \nu, q)</math>, the constant coefficients in this expression are,

<math>~A_\mathrm{grav}</math>	<math>~\equiv~</math>	<math> 2^2 \biggl[ \frac{3GM_\mathrm{tot}^2}{2^3\pi} \biggl(\frac{\nu^2}{q^6} \biggr) \biggr] \biggl\{ 2q^5 + 5q^3\biggl( \frac{\rho_e}{\rho_c} \biggr) (1-q^2) + \biggl( \frac{\rho_e}{\rho_c} \biggr)^2 \biggl[ 2(1-q^5) - 5q^3(1-q^2) \biggr] \biggr\} \, , </math>
<math>~A_\mathrm{core}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{3GM_\mathrm{tot}^2}{2^3\pi} \biggl(\frac{\nu^2}{q^6} \biggr) \biggr] \biggl[ \frac{2}{3(\gamma_c -1)} \biggr] \biggl( \frac{4\pi}{5} \biggr) q^5 \, , </math>
<math>~A_\mathrm{env}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{3GM_\mathrm{tot}^2}{2^3\pi} \biggl(\frac{\nu^2}{q^6} \biggr) \biggr] \biggl[ \frac{2}{3(\gamma_e -1)} \biggr] \biggl( \frac{\pi}{5} \biggr) \biggl\{ 10 \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggl[-2q^2 + 3q^3 - q^5 \biggr] + 6 \biggl( \frac{\rho_e}{\rho_0} \biggr)^2 \biggl[ -1 + 5q^2 -5q^3 + q^5 \biggr] \biggr\} \, , </math>
<math>~B_\mathrm{core}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{4\pi q^3}{3(\gamma_c -1)} \biggr] \, , </math>
<math>~B_\mathrm{env}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{4\pi (1-q^3)}{3(\gamma_e -1)} \biggr] \, . </math>

Notice that, in the expression for <math>~\mathfrak{G}</math>, we have been careful to maintain the separate identities of the interface pressure, depending on whether it is set by the core <math>~(P_{ic})</math> or by the envelope <math>~(P_{ie})</math>, because they scale differently — along two separate adiabats — with density and, hence, with radius. Specifically,

<math>~P_{ic}</math>	<math>~=~</math>	<math>K_c \rho_0^{\gamma_c} = K_c \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggr]^{\gamma_c} R^{-3\gamma_c} \, ,</math>
<math>~P_{ie}</math>	<math>~=~</math>	<math>K_e \rho_e^{\gamma_e} = K_e \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr]^{\gamma_e} R^{-3\gamma_e} \, .</math>

With these pressure scalings in mind, the expression for the free energy becomes,

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

<math>~=~</math>

<math>(A_\mathrm{core} + A_\mathrm{env} -A_\mathrm{grav}) R^{-1} + C_\mathrm{core}R^{3-3\gamma_c} + C_\mathrm{env} R^{3-3\gamma_e} \, , </math>

where,

<math>~C_\mathrm{core}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{4\pi q^3}{3(\gamma_c -1)} \biggr] K_c \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggr]^{\gamma_c} \, , </math>
<math>~C_\mathrm{env}</math>	<math>~\equiv~</math>	<math> \biggl[ \frac{4\pi (1-q^3)}{3(\gamma_e -1)} \biggr] K_e \biggl[ \frac{3M_\mathrm{tot} \nu}{4\pi q^3} \biggl( \frac{\rho_e}{\rho_0} \biggr) \biggr]^{\gamma_e} \, . </math>

Related Discussions

• Analytic solution with <math>~n_c = 5</math> and <math>~n_e=1</math>.

© 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