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Gas in a box with Maxwell-Boltzmann distribution
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[QUOTE="rogdal, post: 6854798, member: 730670"] Hello [USER=229090]@TSny[/USER]. Many thanks for your answer. Yes, obviously, after thinking it a bit, the argument I used in my 2nd scenario for justifying the origin of the 2 extra factor makes no sense. Let me try to think about the probability distributions from your interpretations: (1) In this case, we would obtain exactly the Maxwell-Boltzmann distribution ##\rho_{collision}(\vec{v}) = \rho_v(\vec{v})##. However, since we are only considering positive ##v_z## from, let's say, 0 to infinity, in this particular case, ##\rho_v(\vec{v})## wouldn't be normalized. Integrating, instead of 1 one would obtain 0.5, which is not correct. But if we normalize it by adding a factor of 2, we would get exactly ##\rho_{collision}(\vec{v}) = 2\rho_v(\vec{v})\theta(v_z)##. Does this make sense? (2) Instead the Maxwell-Boltzmann distribution for a group of particles, this would be the Maxwell-Boltzmann distribution just for a single particle, right? This distribution would be given by: $$\rho_{collision}(\vec{v}) = \left(\frac{m}{2kT}\right)^{3/2} 4\pi v^2 e^{-\frac{mv^2}{2kT}}$$ The factor of 4πv^2 would arise from the fact that the velocity distribution is for a single particle rather than for a group of particles.The velocity would be proportional to the surface area of a sphere of radius v around the particle. How would this expression be linked to the final answer that I'm told to prove? Thanks! [/QUOTE]
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Gas in a box with Maxwell-Boltzmann distribution
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