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Spherically Symmetric Configurations Synopsis

New Table Construction

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

<math>~dV = 4\pi r^2 dr</math> and    <math>~dM_r = \rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr</math> <math>~W_\mathrm{grav}</math> <math>~=</math> <math>~- \int_0^R \biggl(\frac{GM_r}{r}\biggr) dM_r ~~ \propto ~~ R^{-1}</math> <math>~U_\mathrm{int}</math> <math>~=</math> <math>~\frac{1}{(\gamma -1)} \int_0^R 4\pi r^2 P dr ~~ \propto ~~ R^{3-3\gamma}</math>
Equilibrium Structure
Detailed Force Balance	Free-Energy Analysis
Given a barotropic equation of state, <math>~P(\rho)</math>, solve the equation of Hydrostatic Balance <math>~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}</math> for the radial density distribution, <math>~\rho(r)</math>.	The Free-Energy is, <math>~\mathfrak{G}</math> <math>~=</math> <math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV</math>   <math>~=</math> <math>~-a R^{-1} + bR^{3-3\gamma}+ cR^3 \, .</math> Therefore, also, <math>~\frac{d\mathfrak{G}}{dR}</math> <math>~=</math> <math>~aR^{-2} +(3-3\gamma)bR^{2-3\gamma} + 3cR^2</math>   <math>~=</math> <math>~\frac{1}{R}\biggl[ -W_\mathrm{grav} - 3(\gamma-1)U_\mathrm{int} + 3P_eV\biggr]</math> 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, <math>~0</math> <math>~=</math> <math>~W_\mathrm{grav} + 3(\gamma-1)U_\mathrm{int} - 3P_eV\, .</math>
Virial Equilibrium
Multiply the hydrostatic-balance equation through by <math>~rdV</math> and integrate over the volume: <math>~0</math> <math>~=</math> <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>   <math>~=</math> <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>   <math>~=</math> <math>~-\int_0^R\biggl[ \frac{d}{dr}\biggl( 4\pi r^3P \biggr) - 12\pi r^2 P\biggr] dr + W_\mathrm{grav}</math>   <math>~=</math> <math>~\int_0^R 3\biggl[ 4\pi r^2 P \biggr]dr - \int_0^R \biggl[ d(3PV)\biggr] + W_\mathrm{grav}</math>   <math>~=</math> <math>~3(\gamma-1)U_\mathrm{int} + W_\mathrm{grav} - \biggl[ 3PV \biggr]_0^R \, .</math>
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, LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation <math>~0</math> <math>~=</math> <math>~ \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 </math> to find one or more radially dependent, radial-displacement eigenvectors, <math>~x \equiv \delta r/r</math>, along with (the square of) the corresponding oscillation eigenfrequency, <math>~\omega^2</math>.	The second derivative of the free-energy function is, <math>~\frac{d^2 \mathfrak{G}}{dR^2}</math> <math>~=</math> <math>~ -2aR^{-3} + (3-3\gamma)(2-3\gamma)b R^{1-3\gamma} + 6cR </math>   <math>~=</math> <math>~\frac{1}{R^2}\biggl[ 2W_\mathrm{grav} - 3(\gamma-1)(2-3\gamma)U_\mathrm{int} + 6P_e V \biggr] \, . </math> 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, <math>~3(\gamma-1)U_\mathrm{int}</math> <math>~=</math> <math>~3P_e V - W_\mathrm{grav} </math> <math>~\Rightarrow~~~ R^2 \biggl[\frac{d^2\mathfrak{G}}{dR^2}\biggr]_\mathrm{equil}</math> <math>~=</math> <math>~2W_\mathrm{grav} - (2-3\gamma)\biggl[3P_e V - W_\mathrm{grav} \biggr] + 6P_e V </math>   <math>~=</math> <math>~(4-3\gamma)W_\mathrm{grav} + 3^2\gamma P_e V \, . </math>
Variational Principle
Multiply the LAWE through by <math>~4\pi x dr</math>, and integrate over the volume of the configuration gives the, Governing Variational Relation <math>~0</math> <math>~=</math> <math>~ \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 </math>     <math>~ - 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 \, . </math>   <math>~=</math> <math>~ \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 </math>     <math>~ + \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 \, . </math> Now, by setting <math>~(d\ln x/d\ln r)_{r=R} = -3</math>, we can ensure that the pressure fluctuation is zero and, hence, <math>~P = P_e</math> at the surface, in which case this relation becomes, <math>~\omega^2</math> <math>~=</math> <math>~ \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} </math>
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, <math>~x</math> = constant, and the governing variational relation gives, <math>~\omega^2 \int_0^R r^2 dM_r</math> <math>~\approx</math> <math>~ (4- 3\gamma) W_\mathrm{grav}+ 3^2 \gamma P_eV \, . </math>

Old Table Construction

Spherically Symmetric Configurations that undergo Adiabatic Compression/Expansion — adiabatic index, <math>~\gamma</math>
<math>~dV = 4\pi r^2 dr</math> and    <math>~dM_r = \rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr</math> <math>~W_\mathrm{grav}</math> <math>~=</math> <math>~- \int_0^R \biggl(\frac{GM_r}{r}\biggr) dM_r ~~ \propto ~~ R^{-1}</math> <math>~U_\mathrm{int}</math> <math>~=</math> <math>~\frac{1}{(\gamma -1)} \int_0^R 4\pi r^2 P dr ~~ \propto ~~ R^{3-3\gamma}</math>
Equilibrium Structure
Detailed Force Balance	Free-Energy Analysis
Given a barotropic equation of state, <math>~P(\rho)</math>, solve the equation of Hydrostatic Balance <math>~\frac{dP}{dr} = - \frac{GM_r \rho}{r^2}</math> for the radial density distribution, <math>~\rho(r)</math>.	The Free-Energy is, <math>~\mathfrak{G}</math> <math>~=</math> <math>~W_\mathrm{grav} + U_\mathrm{int} + P_eV</math>   <math>~=</math> <math>~-a R^{-1} + bR^{3-3\gamma}+ cR^3 \, .</math> Therefore, also, <math>~\frac{d\mathfrak{G}}{dR}</math> <math>~=</math> <math>~aR^{-2} +(3-3\gamma)bR^{2-3\gamma} + 3cR^2</math>   <math>~=</math> <math>~\frac{1}{R}\biggl[ -W_\mathrm{grav} - 3(\gamma-1)U_\mathrm{int} + 3P_eV\biggr]</math> 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, <math>~0</math> <math>~=</math> <math>~W_\mathrm{grav} + 3(\gamma-1)U_\mathrm{int} - 3P_eV\, .</math>
Virial Equilibrium
Multiply the hydrostatic-balance equation through by <math>~rdV</math> and integrate over the volume: <math>~0</math> <math>~=</math> <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>   <math>~=</math> <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>   <math>~=</math> <math>~-\int_0^R\biggl[ \frac{d}{dr}\biggl( 4\pi r^3P \biggr) - 12\pi r^2 P\biggr] dr + W_\mathrm{grav}</math>   <math>~=</math> <math>~\int_0^R 3\biggl[ 4\pi r^2 P \biggr]dr - \int_0^R \biggl[ d(3PV)\biggr] + W_\mathrm{grav}</math>   <math>~=</math> <math>~3(\gamma-1)U_\mathrm{int} + W_\mathrm{grav} - \biggl[ 3PV \biggr]_0^R \, .</math>
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, LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation <math>~0</math> <math>~=</math> <math>~ \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 </math> to find one or more radially dependent, radial-displacement eigenvectors, <math>~x \equiv \delta r/r</math>, along with (the square of) the corresponding oscillation eigenfrequency, <math>~\omega^2</math>.	The second derivative of the free-energy function is, <math>~\frac{d^2 \mathfrak{G}}{dR^2}</math> <math>~=</math> <math>~ -2aR^{-3} + (3-3\gamma)(2-3\gamma)b R^{1-3\gamma} + 6cR </math>   <math>~=</math> <math>~\frac{1}{R^2}\biggl[ 2W_\mathrm{grav} - 3(\gamma-1)(2-3\gamma)U_\mathrm{int} + 6P_e V \biggr] \, . </math> 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, <math>~3(\gamma-1)U_\mathrm{int}</math> <math>~=</math> <math>~3P_e V - W_\mathrm{grav} </math> <math>~\Rightarrow~~~ R^2 \biggl[\frac{d^2\mathfrak{G}}{dR^2}\biggr]_\mathrm{equil}</math> <math>~=</math> <math>~2W_\mathrm{grav} - (2-3\gamma)\biggl[3P_e V - W_\mathrm{grav} \biggr] + 6P_e V </math>   <math>~=</math> <math>~(4-3\gamma)W_\mathrm{grav} + 3^2\gamma P_e V \, . </math>
Variational Principle
Multiply the LAWE through by <math>~4\pi x dr</math>, and integrate over the volume of the configuration gives the, Governing Variational Relation <math>~0</math> <math>~=</math> <math>~ \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 </math>     <math>~ - 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 \, . </math>   <math>~=</math> <math>~ \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 </math>     <math>~ + \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 \, . </math> Now, by setting <math>~(d\ln x/d\ln r)_{r=R} = -3</math>, we can ensure that the pressure fluctuation is zero and, hence, <math>~P = P_e</math> at the surface, in which case this relation becomes, <math>~\omega^2</math> <math>~=</math> <math>~ \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} </math>
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, <math>~x</math> = constant, and the governing variational relation gives, <math>~\omega^2 \int_0^R r^2 dM_r</math> <math>~\approx</math> <math>~ (4- 3\gamma) W_\mathrm{grav}+ 3^2 \gamma P_eV \, . </math>

See Also

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation