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Given a barotropic equation of state, <math>~P(\rho)</math>, solve the equation of
Hydrostatic Balance
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<math>~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}</math>
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for the radial density distribution, <math>~\rho(r)</math>.
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Multiply the hydrostatic-balance equation through by <math>~rdV</math> and integrate over the volume:
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<math>~0</math>
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<math>~=</math>
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<math>~-\int_0^R r\biggl(\frac{dP}{dr}\biggr)dV - \int_0^R r\biggl(\frac{GM_r \rho}{r^2}\biggr)dV</math>
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<math>~=</math>
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<math>~-\int_0^R 4\pi r^3 \biggl(\frac{dP}{dr}\biggr) dr - \int_0^R \biggl(\frac{GM_r}{r}\biggr)dM_r</math>
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<math>~=</math>
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<math>~-\int_0^R 4\pi r^3 dP + W_\mathrm{grav}</math>
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<math>~=</math>
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<math>~\int_0^R 4\pi \biggl[ 3r^2 P dr - d(r^3P)\biggr] + W_\mathrm{grav}</math>
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<math>~=</math>
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<math>~\int_0^R 3\biggl[ 4\pi r^2 P dr \biggr] - \int_0^R \biggl[ d(3PV)\biggr] + W_\mathrm{grav}</math>
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<math>~=</math>
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<math>~3(\gamma-1)U_\mathrm{int} + W_\mathrm{grav} - \biggl[ 3PV \biggr]_0^R \, .</math>
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The Free-Energy is,
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<math>~\mathfrak{G}</math>
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<math>~=</math>
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<math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV</math>
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<math>~=</math>
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<math>~-a R^{-1} + bR^{3-3\gamma}+ cR^3 \, .</math>
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Therefore, also,
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<math>~\frac{d\mathfrak{G}}{dR}</math>
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<math>~=</math>
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<math>~aR^{-2} +(3-3\gamma)bR^{2-3\gamma} + 3cR^2</math>
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<math>~=</math>
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<math>~\frac{1}{R}\biggl[ -W_\mathrm{grav} - 3(\gamma-1)U_\mathrm{int} + 3P_eV\biggr]</math>
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Equilibrium configurations exist at extrema of the free-energy function, that is, they are identified by setting <math>~d\mathfrak{G}/dR = 0</math>. Hence, equilibria are defined by the condition,
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<math>~0</math>
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<math>~=</math>
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<math>~W_\mathrm{grav} + 3(\gamma-1)U_\mathrm{int} - 3P_eV\, .</math>
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