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

$~\chi_\mathrm{eq} \equiv \frac{R_\mathrm{eq}}{R_0} \, ,$

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,

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

Once the pressure exerted by the external medium ($~P_e$), and the configuration's mass ($~M_\mathrm{tot}$), 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 $~\chi_\mathrm{eq}$ 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, $~\chi_\mathrm{eq}$ is given by the root(s) of the equation,

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

In other words,

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

### 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, $~\rho_0 = 3M_\mathrm{tot}/(4\pi R_\mathrm{eq}^3)$. Hence,

 $~K$ $~=$ $P_c \biggl( \frac{4\pi R_\mathrm{eq}^3}{3M_\mathrm{tot}} \biggr)^{\gamma_g} \, ,$

and the solution takes the form,

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

Or, solving for the central pressure,

 $~P_c$ $~=$ $\biggl( \frac{3}{20\pi} \cdot \frac{\mathfrak{f}_W}{\mathfrak{f}_A} \biggr) \frac{GM^2_\mathrm{tot}}{R_\mathrm{eq}^{4} } \, .$

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,

 $~P_c$ $~=$ $\biggl( \frac{3}{8\pi} \biggr) \frac{GM^2_\mathrm{tot}}{R_\mathrm{eq}^{4} } \, .$

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,

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

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,

 $~\mathfrak{f}_A$ $~\equiv$ $~ 3\int_0^1 (1-x^2) x^2 dx = 3 \biggl[ \frac{x^3}{3} - \frac{x^5}{5} \biggr]_0^1 = \frac{2}{5} \, .$

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 $~R_\mathrm{eq}$ 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 $~\chi_\mathrm{eq}$ 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,

 $~D$ $~=$ $(\gamma_g-1) B\chi_\mathrm{eq}^{-3\gamma_g} - \frac{A}{3}\chi_\mathrm{eq}^{-4}$ $\Rightarrow ~~~~\frac{4\pi}{3} R_0^3 P_e$ $~=$ $\biggl[ K M_\mathrm{tot} \biggl( \frac{3M_\mathrm{tot} }{4\pi R_0^3} \biggr)^{\gamma_g - 1} \cdot \mathfrak{f}_A \biggr] \chi_\mathrm{eq}^{-3\gamma_g} - \biggl[ \frac{1}{5} \frac{GM_\mathrm{tot} ^2}{R_0} \cdot \mathfrak{f}_W \biggr] \chi_\mathrm{eq}^{-4}$ $\Rightarrow ~~~~ P_e$ $~=$ $\biggl[ K \biggl( \frac{3M_\mathrm{tot} }{4\pi R_\mathrm{eq}^3} \biggr)^{\gamma_g} \cdot \mathfrak{f}_A \biggr] - \biggl[ \biggl(\frac{3}{20\pi} \biggr) \frac{GM_\mathrm{tot} ^2}{R_\mathrm{eq}^4} \cdot \mathfrak{f}_W \biggr]$ $~=$ $P_c \cdot \mathfrak{f}_A - \biggl(\frac{3}{20\pi} \biggr) \frac{GM_\mathrm{tot} ^2}{R_\mathrm{eq}^4} \cdot \mathfrak{f}_W \, ,$

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

 $~K$ $~=$ $P_c \biggl( \frac{4\pi R_\mathrm{eq}^3}{3M_\mathrm{tot}} \biggr)^{\gamma_g} \, .$

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

 $~R_\mathrm{eq} $ $~=$ $\biggl[ \biggl(\frac{3}{20\pi} \biggr) \frac{GM_\mathrm{tot} ^2}{P_c} \cdot \frac{\mathfrak{f}_W}{\mathfrak{f}_A} \biggl( 1 - \frac{P_e}{P_c} \cdot \frac{1}{\mathfrak{f}_A} \biggr)^{-1} \biggr]^{1/4} \, .$

### 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.

$\mathfrak{W}_A = \frac{1}{({\gamma_g}-1)} \int_0^R 4\pi r^2 P dr \, ,$


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,

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

we have,

 $~\mathfrak{W}_A$ $~=$ $\frac{4\pi P_c R_0^3}{({\gamma_g}-1)} \int_0^{x_e} ( 1-x^2 ) x^2 dx$ $~=$ $\frac{4\pi P_c R_0^3}{({\gamma_g}-1)} \biggl[ \frac{x^3}{3}-\frac{x^5}{5} \biggr]_0^{x_e} = \frac{P_c }{({\gamma_g}-1)} \biggl( \frac{4\pi R_e^3}{3} \biggr) \biggl[ 1-\frac{3}{5}x_e^2 \biggr]$ $~=$ $\frac{M_\mathrm{tot} }{({\gamma_g}-1)} \biggl( \frac{P_c}{\rho_c} \biggr) \biggl[ 1-\frac{3}{5}\biggl(1 - \frac{P_e}{P_c}\biggr) \biggr] \, .$

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

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

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

 $~\biggl(\frac{3}{20\pi} \biggr) \frac{GM_\mathrm{tot} ^2}{R_\mathrm{eq}^4}$ $~=$ $~P_c \cdot \biggl[ \frac{2}{5} + \frac{3}{5}\frac{P_e}{P_c} \biggr] - P_e$ $~=$ $~\frac{2}{5} P_c \biggl( 1 - \frac{P_e}{P_c} \biggr)$ $\Rightarrow ~~~~ R_\mathrm{eq}$ $~=$ $\biggl[ \biggl( \frac{3}{2^3\pi} \biggr) \frac{G M^2}{P_c} \biggl( 1 - \frac{P_e}{P_c} \biggr)^{-1} \biggr]^{1/4} \, .$

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