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Q & A: collision rates

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Most recent answer: 02/19/2008
Am a retired biologist with a question that should be easy, I would think, for a physicist to answer. Total momentum (or energy) is distributed over three particles in a box of fixed volume in a balanced way, say (5, 5, 5), and in an unbalanced way,say (9, 4, 2). Which has the higher collision frequency? The question may seem trivial, but it is not.
- Peter Calabria
Acapulco, Mexico
This is actually a very hard question, at least for me. First, I guess you must mean that what's divided up in different ways is the kinetic energy. The net momentum is a vector, not a number, and it changes on every elastic collison with a wall, so it can't be what you mean.
Assuming that there are many elastic collisions, after a while there will be a whole range of possible distributions of energies. Regardless of the starting values, on the average the distribution of values will follow a Maxwell-Boltzmann form, so in the long run the collision rates will be the same.
In the short run, it sounds like a harder problem. It would help to define a little better what's meant by the rate. Is it supposed to be the typical probability per time in some very short time of having a single collision? If so, I guess I could figure that out and answr your question, but then the rte will change after the first collison as the energy distribution changes.

Mike W.

p.s. I tried doing a calculation for a very simple case: 3 equal-mass particles, either with all the energy in one particle or with the energy evenly distributed.  At least first time through the calculation, I got that the even distribution case wins, with 2/31/2 times as high a net initial collision rate.

(published on 02/19/2008)

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