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Gas in a box with Maxwell-Boltzmann distribution
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[QUOTE="TSny, post: 6854632, member: 229090"] Hello, [USER=730670]@rogdal[/USER]. I don't understand your argument. The particle does not emerge from the collision with the same velocity it had just before the collision. Since ##v_z## changes in the collision, the final velocity vector is not the same as the initial velocity vector. If ##v_z## just changes sign in the collision with no change in ##v_x## or ##v_y##, then the initial and final speeds would be the same. But I don't see why this would give the factor of 2 in the answer. I'm not completely sure of the interpretation of the statement of the problem: Two possible interpretations come to my mind: (1) Consider some arbitrary interval of time that is long enough for many particles to strike the wall. Imagine making a list of the incoming velocity ##\vec v## for each particle that strikes the wall during the time interval. What is the probability distribution ##\rho_{\rm collision}(\vec v)## associated with this list? (2) Pick an arbitrary particle of the gas. Wait until it collides with the wall. (It could be a long wait, but that's ok.) Write down the incoming velocity ##\vec v## for this collision. Repeat this many times. What is the probability distribution ##\rho_{\rm collision}(\vec v)## associated with this list? I favor the second interpretation. It seems to be more in line with the way I would literally interpret the wording of the problem statement. Also, I think the second interpretation will give the answer that's stated in the problem; whereas, I don't believe the first interpretation would. [/QUOTE]
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Gas in a box with Maxwell-Boltzmann distribution
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