User:Tohline/SSC/FreeFall

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

<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,

<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,

<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.

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{2GM}{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{2GM}{r_i} \, .</math>

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>~dt </math>

<math>~=</math>

<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,

<math>~dr = - 2r_i \sin\zeta \cos\zeta d\zeta \, .</math>

The relevant integral is, therefore,

<math>~\int_0^t dt </math>

<math>~=</math>

<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>~ t </math>

<math>~=</math>

<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>

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>~dt </math>

<math>~=</math>

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

Upon integration, this gives,

<math>~t + C_0 </math>

<math>~=</math>

<math>~ - \frac{2}{3}(2GM)^{-1/2} r^{3/2} \, ,</math>

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>~ dt</math>

<math>~=</math>

<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>~=</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>~[akr^{-1} - k]_{t=0}</math>

<math>~=</math>

<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.)

© 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | 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