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Uniform-Density Sphere

Whitworth's (1981) Isothermal Free-Energy Surface
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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.

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

\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.


See Also


Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2019 by Joel E. Tohline
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