Free-Fall Collapse

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

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,

or, recognizing that <math>~v_r = dr/dt</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,

and the form of the Euler equation, as just derived, serves to describe the particle's governing equation of motion, namely,

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,

<math>~2\dot{r} \frac{d\dot{r}}{dt}</math>	<math>~=</math>	<math>~- \frac{2GM}{r^2} \cdot \frac{dr}{dt} </math>
<math>~\Rightarrow ~~~ d(\dot{r}^2)</math>	<math>~=</math>	<math>~2GM \cdot d(r^{-1}) \, ,</math>

which integrates once to give,

where, as an integration constant, <math>~k</math> is independent of time.

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,

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,

Taking the square root of both sides of our derived "kinetic energy" equation, we can write,

<math>~\frac{dr}{dt}</math>	<math>~=</math>	<math>~\pm \biggl[ \frac{2GM}{r} - k \biggr]^{1/2} </math>
<math>~\Rightarrow~~~ dt </math>	<math>~=</math>	<math>~ \pm \biggl[ \frac{2GM}{r} - k \biggr]^{-1/2} dr </math>

This can be integrated in closed form to give an analytic prescription for <math>~t(r)</math>. We'll consider three separate, physically interesting scenarios, all of which involve infall, so we will adopt the velocity root having only the negative sign.

Falling from rest at a finite distance

In this case, we set <math>~v_i = 0</math> in the definition of <math>~k</math>, so the relevant expression to be integrated is,

<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{-1/2} \biggl[ \biggl( \frac{r_i}{r} \biggr) - 1 \biggr]^{-1/2} dr \, .</math>

Customarily, this equation is integrated by first making the substitution,

<math>~\cos^2\zeta \equiv \frac{r}{r_i} \, ,</math>

which also means,

The relevant integral is, therefore,

<math>~+ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \int_0^\zeta \cos^2\zeta d\zeta \, ,</math>

where the limits of integration have been set to ensure that <math>~r/r_i = 1</math> at time <math>~t=0</math>. After integration, we have,

<math>~ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \biggl[ \frac{\zeta}{2} + \sin(2\zeta) \biggr] \, .</math>

The physically relevant portion of this formally periodic solution is the interval in time from when <math>~r/r_i = 1 ~ (\zeta = 0)</math> to when <math>~r/r_i \rightarrow 0</math> for the first time <math>~(\zeta = \pi/2)</math>. The particle's free-fall comes to an end at the time associated with <math>~\zeta = \pi/2</math>, that is, at the so-called "free-fall time,"

<math>~\tau_\mathrm{ff} </math>

<math>~\equiv</math>

<math>~ \biggl(\frac{2r_i^3}{GM} \biggr)^{1/2} \biggl[ \frac{\zeta}{2} + \sin(2\zeta) \biggr]_{\zeta=\pi/2} = \biggl(\frac{\pi^2 r_i^3}{8GM} \biggr)^{1/2} \, .</math>

The solution to this simplified, but dynamically relevant, problem is particularly interesting because it provides an analytic prescription for the function <math>~t(r)</math>. The inverted relation, <math>~r(t)</math>, is also known analytically, but only via the pair of parametric relations,

<math>~ \frac{r}{r_i} </math>	<math>~=</math>	<math>~ \cos^2\zeta </math>
<math>~ \frac{t}{\tau_\mathrm{ff}} </math>	<math>~=</math>	<math>~ \frac{4}{\pi} \biggl[ \frac{\zeta}{2} + \sin(2\zeta) \biggr] </math>

Falling from a finite distance with an initially nonzero velocity

<math>~ \pm k^{-1/2} \biggl[ \frac{(2GM/k)}{r} - 1 \biggr]^{-1/2} dr </math>

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