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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?

natanijelvasic
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2 Answers2

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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.

GiorgioP-DoomsdayClockIsAt-89
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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):

Roger V.
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