User:Tohline/SSC/FreeFall
Free-Fall Collapse
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In our broad study of the "dynamics of self-gravitating fluids," we are interested in examining how, in a wide variety of physical systems, unbalanced forces can lead to the development of fluid motions and structural changes that are of nonlinear amplitude. Here, we discuss the free-fall collapse of a spherically symmetric, uniform-density configuration. In the scheme of things, this is a simple example, but it proves to be powerfully illustrative.
Assembling the Key Relations
We begin with the set of time-dependent governing equations for spherically symmetric systems, namely,
Equation of Continuity
<math>\frac{d\rho}{dt} + \rho \biggl[\frac{1}{r^2}\frac{d(r^2 v_r)}{dr} \biggr] = 0 </math>
Euler Equation
<math>\frac{dv_r}{dt} = - \frac{1}{\rho}\frac{dP}{dr} - \frac{d\Phi}{dr} </math>
Adiabatic Form of the
First Law of Thermodynamics
<math>~\frac{d\epsilon}{dt} + P \frac{d}{dt} \biggl(\frac{1}{\rho}\biggr) = 0</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>
By definition, an element of fluid is in "free fall" if its motion in a gravitational field is unimpeded by pressure gradients. The most straightforward way to illustrate how such a system evolves is to set <math>~P = 0</math> in all of the governing equations. In doing this, the continuity equation and the Poisson equation are unchanged; the equation formulated by the first law of thermodynamics becomes irrelevant; and the Euler equation becomes,
<math>~\frac{dv_r}{dt} = - \frac{d\Phi}{dr} \, ,</math>
or, recognizing that <math>~v_r = dr/dt</math>,
<math>~\frac{d^2r}{dt^2} = - \frac{d\Phi}{dr} \, .</math>
Models of Increasing Complexity
Single Particle in a Point-Mass Potential
Suppose we examine the free-fall of a single (massless) particle, located a distance <math>~|\vec{r}|</math> from an immovable point-like object of mass, <math>~M</math>. The particle will feel a distance-dependent acceleration,
<math>~\frac{d\Phi}{dr} = \frac{GM}{r^2} \, ,</math>
and the form of the Euler equation, as just derived, serves to describe the particle's governing equation of motion, namely,
<math>~\ddot{r} = - \frac{GM}{r^2} \, ,</math>
where we have used dots to denote differentiation with respect to time. If we multiply this equation through by <math>~2\dot{r} = 2dr/dt</math>, we have,
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<math>~2\dot{r} \frac{d\dot{r}}{dt}</math> |
<math>~=</math> |
<math>~- \frac{2GM}{r^2} \cdot \frac{dr}{dt} </math> |
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<math>~\Rightarrow ~~~ d(\dot{r}^2)</math> |
<math>~=</math> |
<math>~2GM \cdot d(r^{-1}) \, ,</math> |
which integrates once to give,
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<math>~\dot{r}^2</math> |
<math>~=</math> |
<math>~\frac{2GM}{r} - k \, , </math> |
where, as an integration constant, <math>~k</math> is independent of time.
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ASIDE: Within the context of this particular physical problem, the constant, <math>~k</math>, should be used to specify the initial velocity, <math>~v_i</math>, of the particle that begins its collapse from the radial position, <math>~r_i</math>. Specifically, <math>~k = \frac{GM}{r_i} - v_i^2 \, .</math> Without this explicit specification, it should nevertheless be clear that, in order to ensure that <math>~\dot{r}^2</math> is positive — and, hence, <math>~\dot{r}</math> is real — the constant must be restricted to values, <math>~k \leq \frac{GM}{r_i} \, .</math> |
Taking the square root of both sides of our derived "kinetic energy" equation, we can write,
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<math>~\frac{dr}{dt}</math> |
<math>~=</math> |
<math>~\pm \biggl[ \frac{2GM}{r} - k \biggr]^{1/2} </math> |
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<math>~\Rightarrow~~~ dt </math> |
<math>~=</math> |
<math>~ \pm \biggl[ \frac{2GM}{r} - k \biggr]^{-1/2} dr \, .</math> |
Customarily, this equation is usually integrated by first making the substitution,
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© 2014 - 2021 by Joel E. Tohline |