User:Tohline/SSC/UniformDensity
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Uniform-Density Sphere (structure)
Here we're going to derive the interior structural properties of a uniform-density sphere using all three solution strategies. While deriving essentially the same solution three different ways might seem like a bit of overkill, this approach proves to be instructive because (a) it forces us to examine the structural behavior of a number of different physical parameters, and (b) it illustrates how to work through the different solution strategies for one model whose structure can in fact be derived analytically using any of the techniques. As we shall see when studying other self-gravitating configurations, the three strategies are not always equally fruitful.
Solution Technique 1
Adopting solution technique #1, we need to solve the integro-differential equation,
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<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> .
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© 2014 - 2021 by Joel E. Tohline |