In 1D, elastic particles do not converge to 1D Maxwell-Boltzmann

Question

If we consider elastic particles in a 2D plane, their speed distribution will converge to a 2D Maxwell-Boltzmann distribution. For example, if the initial speed distribution is distributed uniformly, and the particles' initial direction is random, the speed distribution will converge to 2D Maxwell-Boltzmann over time.

However, if we consider the 1D case, where elastic particles are constrained to move along a line, it seems to me that the speed distribution will not converge to a 1D Maxwell-Boltzmann; if the particles start with a random uniform speed distribution, it will stay uniform over time. My reasoning here is that every elastic collision (for identical particles) simply swaps the speeds by conservation of momentum.

Clearly, I cannot use my 1D case above to derive a 1D Maxwell-Boltzmann distribution. But for 2D, I can. What exactly is the fundamental difference here?

Closely related: How does a gas of particles with uniform speed reach the Maxwell-Boltzmann distribution?, Necessary & sufficient conditions for an isolated system to approach equilibrium. — Quillo, Jun 20 '23 at 12:12
Might add the curious case in a 2D system where the container is perfectly round and perfectly elastic. The gas particles will conserve their rotational momentum, since there is no mechanism to transfer rotational momentum to the container. I expect that the rotational component of their speed will not conform to Maxwell and Boltzmann. And the density will be higher at the edges for speed distributions with a high rotational and low thermal component. — Menno van Lavieren, Jun 21 '23 at 16:11
@MennovanLavieren this is also true for rectangular containers... but only if the particles do not collide - the residual interaction responsible for establishing the thermodynamic equilibrium. — Roger V., Jun 21 '23 at 17:04

score 6 · Answer 1 · answered Jun 20 '23 at 13:30

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The 2D case corresponds to a chaotic system, i.e., a system with a mechanism allowing it to lose the memory of the initial conditions for almost every initial state (mixing dynamics).

The 1D case has not such an effective mechanism to equilibrate.

answered Jun 20 '23 at 13:30

GiorgioP-DoomsdayClockIsAt-90

34,706

Is there a formal proof that the 1D case is not chaotic and/or not mixing? Or is just that "all speeds equal" is a very peculiar initial condition? It's possible to find peculiar initial conditions that do not equilibrate also in the 2D case. – Quillo Jun 21 '23 at 01:40
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@Quillo an ideal 1D elastic collision between two particles swaps their velocities, such a process does not "mix up" the velocities. However, if the particles interact via a potential function that isn't infinitely rigid you can get three body interactions which allow thermalization. – Kevin Kostlan Jun 21 '23 at 03:34
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Does chaotic dynamics guarantee that the distribution converges to Boltzmann? – Roger V. Jun 21 '23 at 07:22
@KevinKostlan my question is slightly different: this "equal speed" initial condition is a peculiar (unlikely) state of the system. It is possible to find peculiar conditions that do not thermaloze also for the 2D. Therefore, I see no difference between the 1D and the 2D, unless there is a proof that the 1D is not "mixing" (for "almost every" initial condition), while the 2D is. Then, there is the important point of RogerVadim (and the relationship between chaos, mixing and convergence to Boltzmann). – Quillo Jun 21 '23 at 07:42
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@Quillo: The 1D case with idealized elastic collisions will almost always not thermalize, while the 2D case almost always will. – Kevin Kostlan Jun 21 '23 at 07:58
@Quillo I agree with Kevin here. In 1D, I could pick any initial state, and the speed distribution will not change over time, due to perfectly elastic collisions. In 2D, almost all initial states will eventually result in a Maxwell-Boltzmann speed distribution. A trivial exception is when all particles start in a grid, with a vertical velocity. Another is having a grid of synchronised oscillating crosses. But yes, in my original question, I am not doing anything special with the initial state - it can be almost anything. – natanijelvasic Jun 21 '23 at 13:02
@Quillo (Side note: In my original example, the particles don't all start with with same speed - they start with a random speed that is uniformly distributed. However, even if they did all start at the same speed, but random direction, all the points above remain valid. I will edit my original post.) – natanijelvasic Jun 21 '23 at 13:15

score 5 · Answer 2 · answered Jun 20 '23 at 11:44

In 2D case you have random direction - that is projections of speeds on x and y axes are not the same from the very beginning. This does not guarantee that the distribution would evolve to Boltzmann, but

The initial velocity projections are already distributed (rather than uniform)
Collisions result in exchange of velocity projections between molecules.

This is manifestly not the case in 1D. Note that in 1D the molecules, undergoing elastic collisions, cannot bypass each other (the phenomenon which in quantum case leads to emergence of rather new phenomena, such as Luttinger liquid instead of Fermi liquid.) Thus, their velocities merely fluctuate between $v_0$ and $-v_0$.

Related (non-interacting 1D case):

In 1D, elastic particles do not converge to 1D Maxwell-Boltzmann

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