Editing SSC/Structure/PowerLawDensity (section)

==Isothermal Equation of State==

Suppose the gas is isothermal so that the relevant equation of state is,
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<math>
P = c_s^2 \rho ,
</math>
</div>
where <math>c_s</math> is the sound speed.  To determine what power-law density distribution will satisfy hydrostatic equilibrium in this case, it is better to return to the original statement of [[SSCpt2/SolutionStrategies#Spherically_Symmetric_Configurations_.28Part_II.29|hydrostatic balance for spherically symmetric configurations]],
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<math>
\frac{1}{\rho} \frac{dP}{dr} = -\frac{d\Phi}{dr} .
</math>
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Plugging in the isothermal equation of state and assuming a radial density distribution of the form, 
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<math>
\rho(r) = \rho_0 \biggl( \frac{r}{r_0} \biggr)^{-\beta} ,
</math>
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we obtain,
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<math>
\frac{d\Phi}{dr} = \beta \biggl(\frac{c_s^2}{r}\biggr) .
</math>
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Therefore, the Poisson equation gives,
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<math>
\frac{1}{r^2}\frac{d}{dr}\biggl[r^2 \frac{d\Phi}{dr}\biggr] = \beta \biggl(\frac{c_s^2}{r^2}\biggr)= 4\pi G \rho_0 \biggl( \frac{r}{r_0} \biggr)^{-\beta} .
</math>
</div>
This relation can be satisfied only if,
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<math>
\beta = 2 ~~~~~\mathrm{and}~~~~~ \rho_0 = \frac{c_s^2}{2\pi G r_0^2} .
</math>
</div>
Hence, hydrostatic balance can be achieved for an isothermal gas with a power-law density distribution of the form, <math>\rho \propto r^{-2}</math>.  Because an isothermal <math>P(\rho)</math> equation of state is obtained by setting <math>n = \infty</math> in the more general polytropic equation of state, the result just derived is consistent with the above, more general analysis which showed that, for values of the polytropic index <math>n \gg 1</math>, the equilibrium power-law density distribution tends toward a <math>\rho \propto r^{-2}</math> distribution.