Do gravitational many-body systems fall apart eventually?

Imagine an $$N$$-body problem with lots of particles of identical mass (billions of them).

I saw several simulations on the Internet, where the particles first form small clumps, then bigger clumps, then finally one huge globular cluster like clump around the center of mass.

Is this clump a stable feature? If we keep running the simulation forever, will this cluster stabilize and remain there forever, or will the particles gradually leak away so that the cluster dissolves?

Clarifications:

The particles attract each other gravitationally (so if the distance is $$r$$ between two particles, the attractive force between them is proportional $$1/r^2$$).

Initial conditions are unconstrained, and indeed if the particles are too fast then it can't be bound gravitationally. So probably it's a function of kinetic energy and gravitational potential energy in system, but what function?

If you want me to mention a particular situation, then consider a globular cluster, for example.

• The answer to this question surely depends on the properties of the system. As written, this question is not really answerable as it is too vague. Could you edit the post and focus in a particular type of particle with a particular set of properties? – DanielSank Aug 15 '16 at 16:16
• Obviously the answer is no if the interactions are repulsive. This question cannot be answered without some detail, an $n$-body system could be literally anything (with $n$ bodies). – Mark Mitchison Aug 15 '16 at 16:16
• @MarkMitchison That's a pretty big "if". Given that the particles form small clumps initially, it's also not a well founded "if" – Jim Aug 15 '16 at 16:21
• Hi Calmarius, this has kicked off a discussion in the chat room if you're interested. The consensus seems to be that it can't be proved either way. – John Rennie Aug 15 '16 at 16:56
• @JohnRennie answer below, since you seem to have a decent grasp of the relevant concepts I'd appreciate a once-over if you have a minute. – Kyle Oman Aug 15 '16 at 18:26

This answer deals with a particular class of systems, where these assumptions hold:

• Gravity is the only important force between bodies (this would also work for any other purely attractive inverse square law).
• Collisions (or encounters) in the sense of processes modelled by the Fokker-Planck equation (rather than the collisionless Boltzmann equation) occur. For e.g. a stellar system, this means that stars may gravitationally scatter off of each other.
• Inelastic collisions (direct hits between stars, also strong tidal interactions) occur rarely enough to be unimportant.

These assumptions are valid for stellar system like a globular cluster, or most galaxies. I'll touch on this validity a bit more below.

One more concept to introduce before I actually get to an answer. In stellar dynamics, "relaxation" refers to the process by which the orbits of individual stars are perturbed, which eventually tends to drive the system to a configuration with a dense stellar core and a diffuse "halo" of stars surrounding it. The characteristic timescale for this to occur is $t_{\rm rh}$:

$$t_{\rm rh} = \frac{0.17 N}{\ln(\lambda N)}\sqrt{\frac{r_{\rm h}^3}{GM}}$$

$N$ is the number of particles in the system, $\lambda$ is determined empirically and is typically taken to be $0.1$, $r_{\rm h}$ is the radius enclosing half the mass of the system, $M$ is the total mass of the system and $G$ is of course the gravitational constant.

Now the actual answer. As stars orbit in a stellar system, they interact and exchange energy. A particular star may happen to gradually gain energy through multiple interactions until it has positive energy (i.e. kinetic energy that exceeds the potential binding it to the system), it can then escape the system and go off "to infinity". The remaining particles are then less bound to the system, and gradually more and more particles can escape via the same process. Eventually only two particles (stars) remain, and are in a Keplerian orbit about each other.

A crude way to estimate the timescale for evaporation is to assume that stars with speeds exceeding the escape speed are removed on the timescale $t_{\rm rh}$. Assuming a Maxwellian speed distribution, the fraction of stars with speeds exceeding the escape speed is $\gamma=7.38\times10^{-3}$. The loss rate is then:

$$\frac{{\rm d}N}{{\rm d}t}=-\frac{\gamma N}{t_{\rm rh}}$$

The natural definition for the evaporation timescale is therefore $t_{\rm evap}=t_{\rm rh}/\gamma$, so the system evaporates in about $140$ relaxation timescales. A more detailed calculation gives a value closer to $t_{\rm evap}\approx 300 t_{\rm rh}$.

Regarding the rarity of inelastic encounters, the timescale for collisions in a typical globular cluster is $t_{\rm coll}\approx4\times10^{3}t_{\rm rh}$ (even longer in a galaxy), so generally these collisions can safely be neglected in the context of evaporation, but in some cases it can be as short as $t_{\rm coll}\approx20t_{\rm rh}$.

My reference for all of the above is Binney & Tremaine's Galactic Dynamics text (2 ed.). Chapter 7 goes through the evaporation of stellar systems, and many related processes, probably in more detail than you ever wanted to see. If you're unsatisfied with the above I suggest reading that, as I don't really want to produce a rehash of an entire book chapter here. If anything above requires clarification, though, please let me know!

• So do star clusters actually show Maxwellian velocity distributions? If true, please consider adding that as an answer here. – Emilio Pisanty Aug 15 '16 at 18:54
• @EmilioPisanty not sure I completely understand what you're looking for there. The cores of globular clusters are indeed approximately Maxwellian (approximate because the high-energy tail is different), but if you observe it instantaneously I think it still appears to be in TE. Does this fit what you're looking for? – Kyle Oman Aug 15 '16 at 19:07
• Yeah, that fits the bill on that one. – Emilio Pisanty Aug 15 '16 at 19:20
• The characteristic timescale formula doesn't seem to work with a small number of particles. If we plug in $N=10$, the time scale goes off to infinity, and for even fewer particles we get negative results. – John Dvorak Aug 15 '16 at 22:16
• @JanDvorak True, but $\lambda$ is determined empirically, like I said. The value of $0.1$ is for a typical globular cluster, but values of $0.2$ are better for other stellar systems. I would speculate that as $N\rightarrow 0$, $\lambda$ takes on a value closer to $1.0$. – Kyle Oman Aug 16 '16 at 0:02