User:Tohline/SphericallySymmetricConfigurations/SolutionStrategies

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Whitworth's (1981) Isothermal Free-Energy Surface
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This page is intended to accompany and extend our discussion of Spherically Symmetric Configurations.

SUMMARY: The pair of simplified governing differential equations that must be solved in concert with one another to determine the equilibrium structure of spherically symmetric, self-gravitating configurations is:

Hydrostatic Balance

<math>\frac{1}{\rho}\frac{dP}{dr} =- \frac{d\Phi}{dr} </math> ,

Poisson Equation

<math>\frac{1}{r^2} \biggl[\frac{d }{dr} \biggl( r^2 \frac{d \Phi}{dr} \biggr) \biggr] = 4\pi G \rho </math>


Solution Strategies

When attempting to solve the identified pair of simplified governing differential equations, it will be useful to note that, in a spherically symmetric configuration (where <math>~\rho</math> is not a function of <math>\theta</math> or <math>\varphi</math>), the differential mass <math>dm_r</math> that is enclosed within a spherical shell of thickness <math>dr</math> is,

<math>dm_r = \rho dr \oint dS = r^2 \rho dr \int_0^\pi \sin\theta d\theta \int_0^{2\pi} d\varphi = 4\pi r^2 \rho dr</math> ,

where we have pulled from the Wikipedia discussion of integration and differentiation in spherical coordinates to define the spherical surface element <math>dS</math>. Integrating from the center of the spherical configuration (<math>r=0</math>) out to some finite radius <math>r</math> that is still inside the configuration gives the mass enclosed within that radius, <math>M_r</math>; specifically,

<math>M_r \equiv \int_0^r dm_r = \int_0^r 4\pi r^2 \rho dr</math> .

We can also state that,

<math>\frac{dm_r}{dr} = 4\pi r^2 \rho </math> .

This differential relation is often identified as a statement of mass conservation that replaces the equation of continuity for spherically symmetric, static equilibrium structures.

Technique #1

Integrating the Poisson equation once, from the center of the configuration (<math>r=0</math>) out to some finite radius <math>r</math> that is still inside the configuration, gives,

<math> \int_0^r d\biggl( r^2 \frac{d \Phi}{dr} \biggr) = \int_0^r 4\pi G r^2 \rho dr </math>

<math> \Rightarrow ~~~~~ r^2 \frac{d \Phi}{dr} \biggr|_0^r = GM_r </math> .

Now, as long as <math>d\Phi/dr</math> increases less steeply than <math>r^{-2}</math> as we move toward the center of the configuration — indeed, we will find that it usually goes smoothly to zero at the center — the term on the left-hand-side of this last expression will go to zero at <math>r=0</math>. Hence, this first integration of the Poisson equation gives,

<math> \frac{d \Phi}{dr} = \frac{G M_r}{r^2} </math> .

Substituting this expression into the hydrostatic equilibrium equation gives,

<math>\frac{1}{\rho}\frac{dP}{dr} =- \frac{G M_r}{r^2} </math> ,

that is, a single governing integro-differential equation which depends only on the two unknown functions, <math>~P</math> and <math>~\rho</math> .


Whitworth's (1981) Isothermal Free-Energy Surface

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