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, $R_{\mathrm{e}\mathrm{q}}$, measured relative to a reference length scale, $R_0$, i.e., the dimensionless equilibrium radius,

${\chi }_{\mathrm{e}\mathrm{q}}\equiv \frac{R_{\mathrm{e}\mathrm{q}}}{R_0},$

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

$2C{\chi }^{-2}+(1-{\delta }_{1{\gamma }_g})3({\gamma }_g-1)B{\chi }^{3-3{\gamma }_g}+{\delta }_{1{\gamma }_g}{B}_I-A{\chi }^{-1}-3D{\chi }^3=0,$

where the definitions of the various coefficients are,

$A$	$\equiv$	$\frac{3}{5}\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_0}\cdot {\mathfrak{𝔣}}_W,$
$B$	$\equiv$	$\frac{K{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{({\gamma }_g-1)}(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_0^3}{)}^{{\gamma }_g-1}\cdot {\mathfrak{𝔣}}_A=\frac{{\overline{c_s}}^2{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}{({\gamma }_g-1)}\cdot {\mathfrak{𝔣}}_A,$
$B_I$	$\equiv$	$3c_s^2{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\cdot {\mathfrak{𝔣}}_M,$
$C$	$\equiv$	$\frac{5J^2}{4M_{\mathrm{t}\mathrm{o}\mathrm{t}}{R}_0^2}\cdot {\mathfrak{𝔣}}_T,$
$D$	$\equiv$	$\frac{4}{3}\pi R_0^3{P}_e.$

Once the pressure exerted by the external medium ($P_e$), and the configuration's mass ($M_{\mathrm{t}\mathrm{o}\mathrm{t}}$), angular momentum ($J$), and specific entropy (via $K$) — or, in the isothermal case, sound speed ($c_s$) — have been specified, the values of all of the coefficients are known and ${\chi }_{\mathrm{e}\mathrm{q}}$ can be determined.

Adiabatic Evolution of an Isolated Sphere

Here we seek to determine the equilibrium radius of a non-rotating configuration ($J=0$) that undergoes adiabatic compression/expansion (${\delta }_{1{\gamma }_g}=0$) and that is not confined by an external medium ($P_e=0$).

Solution

In this case, the statement of virial equilibrium is simplified considerably. Specifically, ${\chi }_{\mathrm{e}\mathrm{q}}$ is given by the root(s) of the equation,

$A{\chi }_{\mathrm{e}\mathrm{q}}^{-1}$	$=$	$3({\gamma }_g-1)B{\chi }_{\mathrm{e}\mathrm{q}}^{3-3{\gamma }_g}$
$\Rightarrow {\chi }_{\mathrm{e}\mathrm{q}}^{3{\gamma }_g-4}$	$=$	$\frac{3({\gamma }_g-1)B}{A}$
	$=$	$\left[3K{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\right(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_0^3}{)}^{{\gamma }_g-1}\cdot {\mathfrak{𝔣}}_A\left]\right[\frac{3}{5}\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_0}\cdot {\mathfrak{𝔣}}_W{]}^{-1}$
	$=$	$\left(\frac{1}{R_0}{)}^{3{\gamma }_g-4}\right[5\left(\frac{3}{4\pi }{)}^{{\gamma }_g-1}\right(\frac{K}{G})M^{({\gamma }_g-2)}\cdot \frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_W}].$

In other words,

$R_{\mathrm{e}\mathrm{q}}=\left[5\right(\frac{3}{4\pi }{)}^{{\gamma }_g-1}\left(\frac{K}{G}\right)M^{({\gamma }_g-2)}\cdot \frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_W}{]}^{1/(3{\gamma }_g-4)}.$

Comparison with Detailed Force-Balance Model

This derived solution will look more familiar if, instead of $K$, we express the solution in terms of the central pressure,

$P_c=K{\rho }_0^{{\gamma }_g},$

where, for this uniform-density sphere, ${\rho }_0=3M_{\mathrm{t}\mathrm{o}\mathrm{t}}/(4\pi R_{\mathrm{e}\mathrm{q}}^3)$. Hence,

$K$

$=$

$P_c(\frac{4\pi R_{\mathrm{e}\mathrm{q}}^3}{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g},$

and the solution takes the form,

$R_{\mathrm{e}\mathrm{q}}^{3{\gamma }_g-4}$	$=$	$5\left(\frac{4\pi }{3}\right)\left(\frac{P_c{R}_{\mathrm{e}\mathrm{q}}^{3{\gamma }_g}}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}\right)\cdot \frac{{\mathfrak{𝔣}}_A}{{\mathfrak{𝔣}}_W}$
$\Rightarrow R_{\mathrm{e}\mathrm{q}}^4$	$=$	$\left(\frac{3}{20\pi }\right)\left(\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_c}\right)\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A}.$

Or, solving for the central pressure,

$P_c$

$=$

$(\frac{3}{20\pi }\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A})\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}.$

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,

$P_c$

$=$

$\left(\frac{3}{8\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}.$

The expression for $P_c$ 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 $(5{\mathfrak{𝔣}}_A/2{\mathfrak{𝔣}}_W)$.

From a free-energy analysis alone, the best we can do is assume that both structural form factors, ${\mathfrak{𝔣}}_W$ and ${\mathfrak{𝔣}}_A$, 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,

${\mathfrak{𝔣}}_W$	$\equiv$	$3\cdot 5\underset{0}{\overset{1}{\int }}\left\{\underset{0}{\overset{x}{\int }}\right[\frac{\rho (x)}{{\rho }_c}\left]x^{2}dx\right\}\left[\frac{\rho (x)}{{\rho }_c}\right]xdx,$
${\mathfrak{𝔣}}_A$	$\equiv$	$3\underset{0}{\overset{1}{\int }}\left[\frac{P(x)}{P_c}\right]x^{2}dx.$

Now, because the configuration under discussion has a uniform density, we should set $\rho (x)/{\rho }_c=1$ in the definition of ${\mathfrak{𝔣}}_W$ which, after evaluation of the nested integrals, gives ${\mathfrak{𝔣}}_W=1$. 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,

$\frac{P(x)}{P_c}=1-x^2.$

Integrating over this function, in accordance with the definition of ${\mathfrak{𝔣}}_A$, gives,

${\mathfrak{𝔣}}_A$

$\equiv$

$3\underset{0}{\overset{1}{\int }}(1-x^2)x^{2}dx=3[\frac{x^3}{3}-\frac{x^5}{5}{]}_0^1=\frac{2}{5}.$

Hence, the ratio,

$\frac{5{\mathfrak{𝔣}}_A}{2{\mathfrak{𝔣}}_W}=1,$

which brings into perfect agreement the two separate determinations of the equilibrium expressions for $R_{\mathrm{e}\mathrm{q}}$ and $P_c$ 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.

Adiabatic Evolution of Pressure-truncated Sphere

Here we seek to determine the equilibrium radius of a non-rotating configuration ($J=0$) that undergoes adiabatic compression/expansion (${\delta }_{1{\gamma }_g}=0$) and that is embedded in a hot, tenuous external medium whose confining pressure, $P_e$, truncates the configuration.

Solution

In this case, virial equilibrium implies that ${\chi }_{\mathrm{e}\mathrm{q}}$ is given by the root(s) of the equation,

$3({\gamma }_g-1)B{\chi }^{3-3{\gamma }_g}-A{\chi }^{-1}-3D{\chi }^3=0.$

Hence,

$D$	$=$	$({\gamma }_g-1)B{\chi }_{\mathrm{e}\mathrm{q}}^{-3{\gamma }_g}-\frac{A}{3}{\chi }_{\mathrm{e}\mathrm{q}}^{-4}$
$\Rightarrow \frac{4\pi }{3}{R}_0^3{P}_e$	$=$	$\left[K{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\right(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_0^3}{)}^{{\gamma }_g-1}\cdot {\mathfrak{𝔣}}_A]{\chi }_{\mathrm{e}\mathrm{q}}^{-3{\gamma }_g}-[\frac{1}{5}\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_0}\cdot {\mathfrak{𝔣}}_W]{\chi }_{\mathrm{e}\mathrm{q}}^{-4}$
$\Rightarrow P_e$	$=$	$\left[K\right(\frac{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{4\pi R_{\mathrm{e}\mathrm{q}}^3}{)}^{{\gamma }_g}\cdot {\mathfrak{𝔣}}_A]-[\left(\frac{3}{20\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\cdot {\mathfrak{𝔣}}_W]$
	$=$	$P_c\cdot {\mathfrak{𝔣}}_A-\left(\frac{3}{20\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}\cdot {\mathfrak{𝔣}}_W,$

where, in the last step as was recognized above, we have set,

$K$

$=$

$P_c(\frac{4\pi R_{\mathrm{e}\mathrm{q}}^3}{3M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{{\gamma }_g}.$

Hence, for any external pressure, $P_e<P_c$, the pressure-confined equilibrium radius is,

$R_{\mathrm{e}\mathrm{q}}$

$=$

$\left[\right(\frac{3}{20\pi })\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{P_c}\cdot \frac{{\mathfrak{𝔣}}_W}{{\mathfrak{𝔣}}_A}(1-\frac{P_e}{P_c}\cdot \frac{1}{{\mathfrak{𝔣}}_A}{)}^{-1}{]}^{1/4}.$

Comparison with Detailed Force-Balance Model

It is reasonable to ask how close this virial expression for the equilibrium radius is to the exact result. As before, from a free-energy analysis alone, the best we can do is assume that both structural form factors, ${\mathfrak{𝔣}}_W$ and ${\mathfrak{𝔣}}_A$, are of order unity. But we can do better than this. To begin with, because $\rho$ is uniform throughout the configuration, ${\mathfrak{𝔣}}_W=1$, even though the configuration is truncated by the imposed external pressure. We need to reassess how ${\mathfrak{𝔣}}_A$ is evaluated, however, because the pressure does not drop to zero at the surface of the configuration.

Going back to our original definition of the thermodynamic energy reservoir for spherically symmetric adiabatic systems,

${\mathfrak{𝔚}}_A=\frac{1}{({\gamma }_g-1)}\underset{0}{\overset{R}{\int }}4\pi r^{2}Pdr,$

we begin by normalizing the radial coordinate to $R_0$, the radius of the isolated (i.e., not truncated) sphere, because we know from the detailed force-balanced solution that, structurally, the pressure varies with $r$ inside the configuration as,

$\frac{P(x)}{P_c}=1-x^2,$

where, $x\equiv r/R_0$. Integrating only out to the edge of the truncated sphere, which we will identify as $R_e$ and, correspondingly,

$x_e\equiv \frac{R_e}{R_0}=(1-\frac{P_e}{P_c}{)}^{1/2},$

we have,

${\mathfrak{𝔚}}_A$	$=$	$\frac{4\pi P_c{R}_0^3}{({\gamma }_g-1)}\underset{0}{\overset{x_e}{\int }}(1-x^2)x^{2}dx$
	$=$	$\frac{4\pi P_c{R}_0^3}{({\gamma }_g-1)}[\frac{x^3}{3}-\frac{x^5}{5}{]}_0^{x_e}=\frac{P_c}{({\gamma }_g-1)}(\frac{4\pi R_e^3}{3}\left)\right[1-\frac{3}{5}{x}_e^2]$
	$=$	$\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{({\gamma }_g-1)}\left(\frac{P_c}{{\rho }_c}\right)[1-\frac{3}{5}(1-\frac{P_e}{P_c}\left)\right].$

Hence, in the case of a pressure-truncated, uniform-density sphere, we surmise that the relevant structural form factor is,

${\mathfrak{𝔣}}_A=1-\frac{3}{5}(1-\frac{P_e}{P_c})=\frac{2}{5}+\frac{3}{5}\frac{P_e}{P_c}.$

Plugging this expression for ${\mathfrak{𝔣}}_A$ along with ${\mathfrak{𝔣}}_W=1$ into the just-derived virial equilibrium solution gives,

$\left(\frac{3}{20\pi }\right)\frac{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2}{R_{\mathrm{e}\mathrm{q}}^4}$	$=$	$P_c\cdot [\frac{2}{5}+\frac{3}{5}\frac{P_e}{P_c}]-P_e$
	$=$	$\frac{2}{5}{P}_c(1-\frac{P_e}{P_c})$
$\Rightarrow R_{\mathrm{e}\mathrm{q}}$	$=$	$\left[\right(\frac{3}{2^3\pi }\left)\frac{G{M}^2}{P_c}\right(1-\frac{P_e}{P_c}{)}^{-1}{]}^{1/4}.$

This result exactly matches the solution for the equilibrium radius of a pressure-truncated, uniform-density sphere that has been derived elsewhere.

See Also

• Index to a Variety of Free-Energy and/or Virial Analyses