User:Tohline/SSC/VirialEquilibrium/UniformDensity

Uniform-Density Sphere

Review

In an introductory discussion of the virial equilibrium structure of spherically symmetric configurations — see especially the section titled, Energy Extrema — we deduced that a system's equilibrium radius, <math>~R_\mathrm{eq}</math>, measured relative to a reference length scale, <math>~R_0</math>, i.e., the dimensionless equilibrium radius,

<math>~\chi_\mathrm{eq} \equiv \frac{R_\mathrm{eq}}{R_0} \, ,</math>

is given by the root(s) of the following equation:

<math> 2C \chi^{-2} + ~ (1-\delta_{1\gamma_g})~3(\gamma_g-1) B\chi^{3 -3\gamma_g} +~ \delta_{1\gamma_g} B_I ~-~A\chi^{-1} -~ 3D\chi^3 = 0 \, , </math>

where the definitions of the various coefficients are,

<math>~A</math>	<math>~\equiv</math>	<math>\frac{3}{5} \frac{GM_\mathrm{tot} ^2}{R_0} \cdot \mathfrak{f}_W \, ,</math>
<math>~B</math>	<math>~\equiv</math>	<math> \frac{K M_\mathrm{tot} }{(\gamma_g-1)} \biggl( \frac{3M_\mathrm{tot} }{4\pi R_0^3} \biggr)^{\gamma_g - 1} \cdot \mathfrak{f}_A = \frac{\bar{c_s}^2 M_\mathrm{tot} }{(\gamma_g - 1)} \cdot \mathfrak{f}_A \, , </math>
<math>~B_I</math>	<math>~\equiv</math>	<math> 3c_s^2 M_\mathrm{tot} \cdot \mathfrak{f}_M \, , </math>
<math>~C</math>	<math>~\equiv</math>	<math> \frac{5J^2}{4M_\mathrm{tot} R_0^2} \cdot \mathfrak{f}_T \, , </math>
<math>~D</math>	<math>~\equiv</math>	<math> \frac{4}{3} \pi R_0^3 P_e \, . </math>

Once the pressure exerted by the external medium (<math>~P_e</math>), and the configuration's mass (<math>~M_\mathrm{tot}</math>), angular momentum (<math>~J</math>), and specific entropy (via <math>~K</math>) — or, in the isothermal case, sound speed (<math>~c_s</math>) — have been specified, the values of all of the coefficients are known and <math>~\chi_\mathrm{eq}</math> can be determined.

Adiabatic Evolution of an Isolated Sphere

Here we seek to determine the equilibrium radius of a non-rotating configuration (<math>~J = 0</math>) that undergoes adiabatic compression/expansion (<math>\delta_{1\gamma_g} =~0</math>) and that is not confined by an external medium (<math>P_e = 0~</math>).

Solution

In this case, the statement of virial equilibrium is simplified considerably. Specifically, <math>~\chi_\mathrm{eq}</math> is given by the root(s) of the equation,

<math>~A\chi_\mathrm{eq}^{-1} </math>	<math>~=</math>	<math>~3(\gamma_g-1) B\chi_\mathrm{eq}^{3 -3\gamma_g} </math>
<math>\Rightarrow ~~~~~\chi_\mathrm{eq}^{3\gamma_g-4} </math>	<math>~=</math>	<math>~\frac{3(\gamma_g-1) B}{A} </math>
	<math>~=</math>	<math> \biggl[ 3K M_\mathrm{tot} \biggl( \frac{3M_\mathrm{tot} }{4\pi R_0^3} \biggr)^{\gamma_g - 1} \cdot \mathfrak{f}_A \biggr] \biggl[ \frac{3}{5} \frac{GM_\mathrm{tot} ^2}{R_0} \cdot \mathfrak{f}_W \biggr] ^{-1} </math>
	<math>~=</math>	<math> \biggl( \frac{1}{R_0} \biggr)^{3\gamma_g-4} \biggl[ 5\biggl( \frac{3}{4\pi} \biggr)^{\gamma_g-1} \biggr(\frac{K}{G}\biggr) M^{(\gamma_g-2)} \cdot \frac{\mathfrak{f}_A}{\mathfrak{f}_W} \biggr] \, . </math>

In other words,

<math> R_\mathrm{eq} = \biggl[ 5\biggl( \frac{3}{4\pi} \biggr)^{\gamma_g-1} \biggr(\frac{K}{G}\biggr) M^{(\gamma_g-2)} \cdot \frac{\mathfrak{f}_A}{\mathfrak{f}_W} \biggr]^{1/(3\gamma_g-4)} \, . </math>

Discussion

Comparison with Detailed Force-Balance Solution

This derived solution will look more familiar if, instead of <math>~K</math>, we express the solution in terms of the central pressure,

<math>P_c = K\rho_0^{\gamma_g} \, ,</math>

where, for this uniform-density sphere, <math>~\rho_0 = 3M_\mathrm{tot}/(4\pi R_\mathrm{eq}^3)</math>. Hence,

<math>~K</math>

<math>~=</math>

<math>P_c \biggl( \frac{4\pi R_\mathrm{eq}^3}{3M_\mathrm{tot}} \biggr)^{\gamma_g} \, ,</math>

and the solution takes the form,

<math>R_\mathrm{eq}^{3\gamma_g - 4}</math>	<math>~=</math>	<math>5\biggl( \frac{4\pi}{3} \biggr) \biggr(\frac{P_c R_\mathrm{eq}^{3\gamma_g}}{GM^2_\mathrm{tot}}\biggr) \cdot \frac{\mathfrak{f}_A}{\mathfrak{f}_W} </math>
<math>\Rightarrow ~~~~ R_\mathrm{eq}^{4} </math>	<math>~=</math>	<math>\biggl( \frac{3}{20\pi} \biggr) \biggr(\frac{GM^2_\mathrm{tot}}{P_c}\biggr) \cdot \frac{\mathfrak{f}_W}{\mathfrak{f}_A} \, .</math>

Or, solving for the central pressure,

<math>~P_c</math>

<math>~=</math>

<math>\biggl( \frac{3}{20\pi} \cdot \frac{\mathfrak{f}_W}{\mathfrak{f}_A} \biggr) \frac{GM^2_\mathrm{tot}}{R_\mathrm{eq}^{4} } \, .</math>

This should be compared with our detailed force-balance solution of the interior structure of an isolated, nonrotating, uniform-density sphere, which gives the precise expression,

<math>~P_c</math>

<math>~=</math>

<math>\biggl( \frac{3}{8\pi} \biggr) \frac{GM^2_\mathrm{tot}}{R_\mathrm{eq}^{4} } \, .</math>

The expression for <math>~P_c</math> derived from our identification of an extremum in the free energy is identical to the expression derived from the more precise, detailed force-balance analysis, except that the leading numerical coefficients differ by a factor of <math>~(5\mathfrak{f}_A/2\mathfrak{f}_W)</math>.

From a free-energy analysis alone, the best we can do is assume that both structural form factors, <math>~\mathfrak{f}_W</math> and <math>\mathfrak{f}_A</math>, are of order unity. But knowing the detailed force-balance solution allows us to evaluate both form factors. From our introductory discussion of the free energy function, their respective definitions are,

<math>~\mathfrak{f}_W</math>	<math>~\equiv</math>	<math>~ 3\cdot 5 \int_0^1 \biggl\{ \int_0^x \biggl[ \frac{\rho(x)}{\rho_c}\biggr] x^2 dx \biggr\} \biggl[ \frac{\rho(x)}{\rho_c}\biggr] x dx\, ,</math>
<math>~\mathfrak{f}_A</math>	<math>~\equiv</math>	<math>~ 3\int_0^1 \biggl[ \frac{P(x)}{P_c}\biggr] x^2 dx \, .</math>

Now, because the configuration under discussion has a uniform density, we should set <math>~\rho(x)/\rho_c = 1</math> in the definition of <math>~\mathfrak{f}_W</math> which, after evaluation of the nested integrals, gives <math>~\mathfrak{f}_W = 1</math>. But, instead of being uniform throughout the configuration, in the detailed force-balance model, the pressure drops from the center to the surface according to the relation,

<math>\frac{P(x)}{P_c} = 1 - x^2 \, .</math>

Integrating over this function, in accordance with the definition of <math>~\mathfrak{f}_A</math>, gives,

<math>~\mathfrak{f}_A</math>

<math>~\equiv</math>

<math>~ 3\int_0^1 (1-x^2) x^2 dx = 3 \biggl[ \frac{x^3}{3} - \frac{x^5}{5} \biggr]_0^1 = \frac{2}{5} \, .</math>

Hence, the ratio,

<math> \frac{5\mathfrak{f}_A}{2\mathfrak{f}_W} = 1 \, , </math>

which brings into perfect agreement the two separate determinations of the equilibrium expressions for <math>~R_\mathrm{eq}</math> and <math>~P_c</math> in terms of one another and the total mass.

This demonstrates that the free-energy approach to determining the equilibrium radius of a spherical configuration is only handicapped by its inability to precisely nail down values of the structural form factors. But this is not a severe limitation as the (dimensionless) form factors are generally of order unity. In contrast, the free-energy analysis brings with it a capability to readily evaluate the global stability of equilibrium configurations.

© 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