Difference between revisions of "User:Tohline/SSC/Synopsis"

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<math>~dV = 4\pi r^2 dr</math>&nbsp;&nbsp;&nbsp; and &nbsp;&nbsp;&nbsp; <math>~dM_r</math>
<math>~dV = 4\pi r^2 dr</math>  
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<math>~=~</math>    
and    
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<math>\rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr</math>
&nbsp;&nbsp;&nbsp;<math>~dM_r = \rho dV ~~~\Rightarrow ~~~M_r = 4\pi \int_0^r \rho r^2 dr</math>
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Revision as of 04:34, 18 June 2017


Spherically Symmetric Configurations Synopsis

Whitworth's (1981) Isothermal Free-Energy Surface
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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>

Detailed Force Balance

Free-Energy Analysis

Given a barotropic equation of state, <math>~P(\rho)</math>, solve the equation of

Hydrostatic Balance

LSU Key.png

<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>

See Also

Whitworth's (1981) Isothermal Free-Energy Surface

© 2014 - 2021 by Joel E. Tohline
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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