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,

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

and 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} + \frac{1}{4}\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} + \frac{1}{4}\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{2}{\pi} \biggl[ \zeta + \frac{1}{2} \sin(2\zeta) \biggr] </math>

We note, as well, that the radially directed velocity is,

<math>~v_r = \frac{dr}{dt} </math>	<math>~=</math>	<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{1/2} \biggl[ \frac{1}{\cos^2\zeta} - 1 \biggr]^{1/2} </math>
	<math>~=</math>	<math>~ - \biggl(\frac{2GM}{r_i} \biggr)^{1/2} \tan\zeta \, , </math>

which formally becomes infinite in magnitude when <math>~\zeta \rightarrow \pi/2</math>, that is, when <math>~t \rightarrow \tau_\mathrm{ff}</math>.

Falling from rest at infinity …

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

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

Upon integration, this gives,

where, <math>~C_0</math> is an integration constant. In this case, it is useful to simply let <math>~t=0</math> mark the time at which <math>~r = 0</math> — hence, also, <math>~C_0 = 0</math> — so at all earlier times (<math>~t</math> intrinsically negative) we have,

<math>~- t </math>	<math>~=</math>	<math>~ \biggl( \frac{2r^3}{9GM} \biggr)^{1/2} </math>
<math>~\Rightarrow ~~~ r </math>	<math>~=</math>	<math>~ \biggl( \frac{9}{2} \cdot GMt^2 \biggr)^{1/3} \, .</math>

Falling from a finite distance with an initially nonzero velocity …

Here, we examine the case in which <math>~0 < r_i < \infty</math> and <math>~0 < v_i^2 < GM/r_i</math>, in which case, the constant <math>~k</math> is a nonzero, positive number. The relevant expression to be integrated is,

<math>~ - k^{-1/2}\biggl[ \frac{a}{r} - 1 \biggr]^{-1/2} dr \, ,</math>

where,

<math>~ a \equiv \frac{2GM}{k} \, .</math>

Using Wolfram Mathematica's online integrator, we find,

<math>~- \int \biggl[ \frac{a}{r} - 1 \biggr]^{-1/2} dr</math>

<math>~ r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(2r-a)(ar^{-1} - 1)^{1/2}}{2(r-a)} \biggr] \, .</math>

Hence, we find,

<math>~k^{1/2}(t + C_0)</math>	<math>~=</math>	<math>~ r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(2r-a)(ar^{-1} - 1)^{1/2}}{2(r-a)} \biggr] </math>
	<math>~=</math>	<math>~ r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(ar^{-1}-2)(ar^{-1} - 1)^{1/2}}{2(ar^{-1}-1)} \biggr] </math>
	<math>~=</math>	<math>~ r ( ar^{-1} -1 )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(ar^{-1}-2)}{2(ar^{-1}-1)^{1/2}} \biggr] </math>
	<math>~=</math>	<math>~ r k^{-1/2}( akr^{-1} -k )^{1/2} + \frac{a}{2} \tan^{-1} \biggl[ \frac{(akr^{-1}-2k)}{2k^{1/2}(akr^{-1}-k)^{1/2}} \biggr] \, .</math>

Let's determine the constant, <math>~C_0</math>. When <math>~t = 0</math>, we can write,

<math>~ \frac{2GM}{r_i} - \biggl[\frac{2GM}{r_i} - v_i^2 \biggr] = v_i^2 \, .</math>

Hence,

<math>~C_0</math>	<math>~=</math>	<math>~ r_i k^{-1}v_i + \biggl(\frac{a}{2k^{1/2}} \biggr) \tan^{-1} \biggl[ \frac{(v_i^2-k)}{2k^{1/2}v_i} \biggr] </math>
	<math>~=</math>	<math>~ GMk^{-3/2} \biggl\{ 2[\eta(1-\eta)]^{1/2} + \tan^{-1} \biggl[ \biggl( \eta - \frac{1}{2} \biggr) [ \eta(1-\eta)]^{-1/2} \biggr] \biggr\} \, , </math>

where, in this last expression,

<math>\eta \equiv \frac{v_i^2 r_i}{2GM} \, .</math>

(This last expression needs to be checked for errors, as it has been rather hastily derived.)

Uniform-Density Sphere

Now, let's consider the (pressure-free) collapse, from rest, of a uniform-density sphere of total mass <math>~M_\mathrm{tot}</math> and radius, <math>~R(t)</math>. If we use a subscript "0" to label the radius of the sphere at time <math>~t=0</math>, then the initial mass-density throughout the sphere is,

<math>~\rho_0 = \frac{3M_\mathrm{tot}}{4\pi R_0^3} \, .</math>

If we not only assume that the total mass of this configuration remains constant but that all of the mass remains fully enclosed within the surface of radius, <math>~R(t)</math>, throughout the collapse (the validity of this second assumption will be critically assessed shortly), then at all points across the surface of the configuration, the acceleration will be given — analogous to the single-particle case, above — by,

<math>~\frac{d\Phi}{dR} = \frac{GM_\mathrm{tot}}{R^2} \, ,</math>

and the equation of motion for the surface is, as before,

<math>~\ddot{R} = - \frac{GM_\mathrm{tot}}{R^2} \, .</math>

As in the single-particle case, above, this 2^nd-order ODE can be integrated twice to give the following parametric relationship between the sphere's radius, and time:

<math>~ \frac{R}{R_0} </math>	<math>~=</math>	<math>~ \cos^2\zeta </math>
<math>~ \frac{t}{\tau_\mathrm{ff}} </math>	<math>~=</math>	<math>~ \frac{2}{\pi} \biggl[ \zeta + \frac{1}{2}\sin(2\zeta) \biggr] </math>
where, <math>\tau_\mathrm{ff} \equiv \biggl(\frac{\pi^2 R_0^3}{8GM_\mathrm{tot}} \biggr)^{1/2} = \biggl[ \frac{3\pi}{32G\rho_0} \biggr]^{1/2}</math>

It is important to notice, from this result, that the timescale for collapse, <math>~\tau_\mathrm{ff}</math>, depends only on the density of the configuration in its initial state. It is important to realize, as well, that the derived parametric solution that gives the ratio <math>~R/R_0</math> as a function of time applies for all positions within the sphere. In this more general way of interpreting the solution, <math>~R</math> represents any radial position, <math>~R_0</math> represents the value of that <math>~R</math> at time <math>~t=0</math>, and the relevant mass is the mass interior to that position, rather than the configuration's total mass. This works because, for spherically symmetric configurations, the acceleration only depends on the mass interior to each position. The ultimate result is that the free-fall collapse of an initially uniform-density sphere happens homologously. This happens because, independent of <math>~R</math>, the timescale for collapse only depends on <math>~\rho_0</math> and, by design, <math>~\rho_0</math> is independent of <math>~R</math>.

Because the pressure-free collapse of an initially uniform-density sphere proceeds in an homologous fashion, the sphere remains uniform in density and the mass interior to any radial shell remains constant. This fully justifies the assumption of constant mass that was made earlier in this derivation.

The expression for the time-dependent velocity that was obtained, above, in the context of a particle falling from rest at a finite distance can also be generalized here to the case of a collapsing uniform-density sphere. A radial shell initially at any position, <math>~R_i \le R_0</math>, within the sphere will enclose a mass, <math>M_i = 4\pi \rho_0 R_i^3/3</math>. Hence the radially directed velocity of that shell at any time, <math>~t</math> (specified via the parameter, <math>~\zeta</math>), will be,

<math>~v_r</math>	<math>~=</math>	<math>~ - \biggl(\frac{2GM_i}{R_i} \biggr)^{1/2} \tan\zeta = - R_i \biggl[ \biggl(\frac{8\pi G\rho_0}{3} \biggr)^{1/2} \tan\zeta \biggr] </math>
	<math>~=</math>	<math>~ - R \biggl[ \biggl(\frac{8\pi G\rho_0}{3} \biggr)^{1/2} \frac{\sin\zeta}{\cos^3\zeta} \biggr] \, .</math>

Because everything inside the square brackets of this last expression is independent of space, the expression tells us that, at any time during the collapse, the radially directed velocity is linearly proportional to the (initial) radial coordinate of the shell. From this function we can deduce, as well, that,

<math>~\nabla\cdot\vec{v} = \frac{1}{R^2} \frac{d}{dR} \biggl( R^2 v_r \biggr)</math>	<math>~=</math>	<math>~ - \biggl[ \biggl(\frac{8\pi G\rho_0}{3} \biggr)^{1/2} \frac{\sin\zeta}{\cos^3\zeta} \biggr] \frac{1}{R^2} \frac{d}{dR} \biggl( R^3 \biggr)</math>
	<math>~=</math>	<math>~ - 3\biggl[ \biggl(\frac{8\pi G\rho_0}{3} \biggr)^{1/2} \frac{\sin\zeta}{\cos^3\zeta} \biggr] = - \biggl(\frac{3\pi}{2\tau_\mathrm{ff}} \biggr) \frac{\sin\zeta}{\cos^3\zeta}\, .</math>

User:Tohline/SSC/FreeFall

Contents