# Spherically Symmetric Configurations Synopsis

Spherically Symmetric Configurations that undergo Adiabatic Compression/Expansion — adiabatic index, $~\gamma$

 $~dV = 4\pi r^2 dr$ and $~dM_r = \rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr$ $~W_\mathrm{grav}$ $~=$ $~- \int_0^R \biggl(\frac{GM_r}{r}\biggr) dM_r ~~ \propto ~~ R^{-1}$ $~U_\mathrm{int}$ $~=$ $~\frac{1}{(\gamma -1)} \int_0^R 4\pi r^2 P dr ~~ \propto ~~ R^{3-3\gamma}$

Equilibrium Structure

Detailed Force Balance

Free-Energy Analysis

Given a barotropic equation of state, $~P(\rho)$, solve the equation of

Hydrostatic Balance

 $~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}$

for the radial density distribution, $~\rho(r)$.

The Free-Energy is,

 $~\mathfrak{G}$ $~=$ $~W_\mathrm{grav} + U_\mathrm{int} + P_eV$ $~=$ $~-a R^{-1} + bR^{3-3\gamma}+ cR^3 \, .$

Therefore, also,

 $~\frac{d\mathfrak{G}}{dR}$ $~=$ $~aR^{-2} +(3-3\gamma)bR^{2-3\gamma} + 3cR^2$ $~=$ $~\frac{1}{R}\biggl[ -W_\mathrm{grav} - 3(\gamma-1)U_\mathrm{int} + 3P_eV\biggr]$

Equilibrium configurations exist at extrema of the free-energy function, that is, they are identified by setting $~d\mathfrak{G}/dR = 0$. Hence, equilibria are defined by the condition,

 $~0$ $~=$ $~W_\mathrm{grav} + 3(\gamma-1)U_\mathrm{int} - 3P_eV\, .$
Virial Equilibrium

Multiply the hydrostatic-balance equation through by $~rdV$ and integrate over the volume:

 $~0$ $~=$ $~-\int_0^R r\biggl(\frac{dP}{dr}\biggr)dV - \int_0^R r\biggl(\frac{GM_r \rho}{r^2}\biggr)dV$ $~=$ $~-\int_0^R 4\pi r^3 \biggl(\frac{dP}{dr}\biggr) dr - \int_0^R \biggl(\frac{GM_r}{r}\biggr)dM_r$ $~=$ $~-\int_0^R\biggl[ \frac{d}{dr}\biggl( 4\pi r^3P \biggr) - 12\pi r^2 P\biggr] dr + W_\mathrm{grav}$ $~=$ $~\int_0^R 3\biggl[ 4\pi r^2 P \biggr]dr - \int_0^R \biggl[ d(3PV)\biggr] + W_\mathrm{grav}$ $~=$ $~3(\gamma-1)U_\mathrm{int} + W_\mathrm{grav} - \biggl[ 3PV \biggr]_0^R \, .$

Stability Analysis

Perturbation Theory

Free-Energy Analysis

Given the radial profile of the density and pressure in the equilibrium configuration, solve the eigenvalue problem defined by the,

 $~0$ $~=$ $~ \frac{d}{dr}\biggl[ r^4 \gamma P ~\frac{dx}{dr} \biggr] +\biggl[ \omega^2 \rho r^4 + (3\gamma - 4) r^3 \frac{dP}{dr} \biggr] x$

to find one or more radially dependent, radial-displacement eigenvectors, $~x \equiv \delta r/r$, along with (the square of) the corresponding oscillation eigenfrequency, $~\omega^2$.

The second derivative of the free-energy function is,

 $~\frac{d^2 \mathfrak{G}}{dR^2}$ $~=$ $~ -2aR^{-3} + (3-3\gamma)(2-3\gamma)b R^{1-3\gamma} + 6cR$ $~=$ $~\frac{1}{R^2}\biggl[ 2W_\mathrm{grav} - 3(\gamma-1)(2-3\gamma)U_\mathrm{int} + 6P_e V \biggr] \, .$

Evaluating this second derivative for an equilibrium configuration — that is by calling upon the (virial) equilibrium condition to set the value of the internal energy — we have,

 $~3(\gamma-1)U_\mathrm{int}$ $~=$ $~3P_e V - W_\mathrm{grav}$ $~\Rightarrow~~~ R^2 \biggl[\frac{d^2\mathfrak{G}}{dR^2}\biggr]_\mathrm{equil}$ $~=$ $~2W_\mathrm{grav} - (2-3\gamma)\biggl[3P_e V - W_\mathrm{grav} \biggr] + 6P_e V$ $~=$ $~(4-3\gamma)W_\mathrm{grav} + 3^2\gamma P_e V \, .$

Variational Principle

Multiply the LAWE through by $~4\pi x dr$, and integrate over the volume of the configuration gives the,

Governing Variational Relation

 $~0$ $~=$ $~ \int_0^R 4\pi r^4 \gamma P \biggl(\frac{dx}{dr}\biggr)^2 dr - \int_0^R 4\pi (3\gamma - 4) r^3 x^2 \biggl( \frac{dP}{dr} \biggr) dr$ $~ - 4\pi \biggr[r^4 \gamma Px \biggl(\frac{dx}{dr}\biggr) \biggr]_0^R - \int_0^R 4\pi \omega^2 \rho r^4 x^2 dr \, .$ $~=$ $~ \int_0^R x^2 \biggl(\frac{d\ln x}{d\ln r}\biggr)^2 \gamma 4\pi r^2P dr - \int_0^R (3\gamma - 4)x^2 \biggl( - \frac{GM_r}{r} \biggr) 4\pi \rho r^2 dr$ $~ + \biggr[\gamma 4\pi r^3 Px^2 \biggl(-\frac{d\ln x}{d\ln r}\biggr) \biggr]_0^R - \int_0^R 4\pi \omega^2 \rho r^4 x^2 dr \, .$

Now, by setting $~(d\ln x/d\ln r)_{r=R} = -3$, we can ensure that the pressure fluctuation is zero and, hence, $~P = P_e$ at the surface, in which case this relation becomes,

 $~\omega^2$ $~=$ $~ \frac{\gamma (\gamma -1) \int_0^R x^2 \bigl(\frac{d\ln x}{d\ln r}\bigr)^2 dU_\mathrm{int} - \int_0^R (3\gamma - 4)x^2 dW_\mathrm{grav} + 3^2 \gamma x^2 P_eV}{ \int_0^R x^2 r^2 dM_r}$

Approximation:   Homologous Expansion/Contraction

If we guess that radial oscillations about the equilibrium state involve purely homologous expansion/contraction, then the radial-displacement eigenfunction is, $~x$ = constant, and the governing variational relation gives,

 $~\omega^2 \int_0^R r^2 dM_r$ $~\approx$ $~ (4- 3\gamma) W_\mathrm{grav}+ 3^2 \gamma P_eV \, .$