User:Tohline/SSC/UniformDensity

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Whitworth's (1981) Isothermal Free-Energy Surface
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Uniform-Density Sphere (structure)

Solution Technique 1

Adopting solution technique #1, we need to solve the integro-differential equation,

LSU Key.png

<math>~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}</math>

appreciating that,

<math> M_r \equiv \int_0^r 4\pi r^2 \rho dr </math> .

For a uniform-density configuration, <math>~\rho</math> = <math>\rho_c</math> = constant, so the density can be pulled outside the mass integral and the integral can be completed immediately to give,

<math> M_r = \frac{4\pi}{3}\rho_c r^3 </math> .

Hence, the differential equation describing hydrostatic balance becomes,

<math> \frac{dP}{dr} = - \frac{4\pi G}{3} \rho_c^2 r </math> .

Integrating this from the center of the configuration — where <math>r=0</math> and <math>P = P_c</math> — out to an arbitrary radius <math>r</math> that is still inside the configuration, we obtain,

<math> \int_{P_c}^P dP = - \frac{4\pi G}{3} \rho_c^2 \int_0^r r dr </math>
<math>\Rightarrow ~~~~ P = P_c - \frac{2\pi G}{3} \rho_c^2 r^2 </math>

We expect the pressure to drop to zero at the surface of our spherical configuration — that is, at <math>r=R</math> — so the central pressure must be,

<math>P_c = \frac{2\pi G}{3} \rho_c^2 R^2 = \frac{3G}{8\pi}\biggl( \frac{M^2}{R^4} \biggr)</math> ,

where <math>M</math> is the total mass of the configuration. Finally, then, we have,

<math>P(r) = P_c\biggl[1 - \biggl(\frac{r}{R}\biggr)^2 \biggr] </math> .

Solution Technique 3

Adopting solution technique #3, we need to solve the algebraic expression,

<math>H + \Phi = C_\mathrm{B}</math> .

in conjunction with the Poisson equation,

<math>\frac{1}{r^2} \frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr) = 4\pi G \rho </math> .

Appreciating that, as shown above, for a uniform density (<math>~\rho</math> = <math>\rho_c</math> = constant) configuration,

<math> M_r = \int_0^r 4\pi r^2 \rho dr = \frac{4\pi}{3}\rho_c r^3 </math> ,

we can integrate the Poisson equation once to give,

<math> \frac{d\Phi}{dr} = \frac{4\pi G}{3} \rho_c r </math> ,

everywhere inside the configuration. Integrating this expression from any point inside the configuration to the surface, we find that,

<math> \int_{\Phi(r)}^{\Phi_\mathrm{surf}} d\Phi = \frac{4\pi G}{3} \rho_c \int_r^R r dr </math>
<math>\Rightarrow ~~~~~ \Phi_\mathrm{surf} - \Phi(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr] </math>

Turning to the above algebraic condition, we will adopt the convention that <math>~H</math> is set to zero at the surface of a barotropic configuration, in which case the constant, <math>C_\mathrm{B}</math>, must be,

<math>C_\mathrm{B} = (H + \Phi)_\mathrm{surf} = \Phi_\mathrm{surf}</math> .

Therefore, everywhere inside the configuration <math>~H</math> must be given by the expression,

<math>H(r) = \Phi_\mathrm{surf} - \Phi(r)</math> .

Matching this with our solution of the Poisson equation, we conclude that, throughout the configuration,

<math> H(r) = \frac{2\pi G}{3} \rho_c R^2 \biggl[ 1- \biggl(\frac{r}{R} \biggr)^2 \biggr]</math> .

Comparing this result with the result we obtained using solution technique #1, it is clear that throughout a uniform-density, self-gravitating sphere,

<math>\frac{P}{H} = \rho</math> .


Whitworth's (1981) Isothermal Free-Energy Surface

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