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Considering a box of dimensions $V=L \times L \times L$ in the $xyz$ space which has $N \gg1$ particles of a certain gas, the velocity of a random particle is given by the Maxwell-Boltzmann distribution:

$$\rho_v(\vec{v})=\frac{1}{\sqrt{(2\pi \beta)^3}}e^{-\frac{v^2}{2\beta}},$$ where $\beta = k_BT/m$

With this data, I would like to calculate the probability distribution function with which a particle chosen at random collides against the wall located in the plane $z = L$.

I have considered two scenarios:

  1. A particle that has just collided with the wall at $z=L$ is moving with a velocity $v_z<0$ moving away from the wall. Hence, the probability that this particle has of colliding again is $0$, so its distribution is also $0$.

  2. A particle moving with positive $v_z$. Since the particles are constrained to move in that box, it will necessarily collide against the wall, and assuming a completely elastic collision, the velocity distribution in this case would be $2\rho_v(\vec{v})$ because after the collision the particle emerges with the same velocity it had before.

Therefore, the probability density for the velocity at which a randomly chosen particle collides with the wall contained in the plane would be

$$\rho_{collision}(\vec{v})=2\rho_v(\vec{v})\theta(v_z),$$

where $\theta(v_z)$ is the Heaviside theta function.

Would this approach be correct?

Thanks in advance!

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  • $\begingroup$ If you wait long enough the particle will hit the opposite wall, bounce off and travel back towards the wall you are interested in, so I don't think the quantity you have calculated is particularly physically meaningful. A more meaningful quantity would be the probability of a particle colliding with the wall in the next time $T$, say $\rho_{collision}(T)$ or the rate of collisions $\frac{\rho_{collision}(T)}{T}$. What you actually want will depend on what you want to do with the result $\endgroup$
    – By Symmetry
    Feb 13 at 12:09
  • $\begingroup$ Thanks, I'll bear that in mind to make this example more complete. However, back to the question, do you think the origin of the factor 2 is due to the fact that the particle hits the opposite wall and then travels back with vz>0 to the wall z=L? Would it be the right proof? $\endgroup$
    – user9867
    Feb 13 at 18:25
  • $\begingroup$ No what I mean is that the probability of a particle eventually hitting a given wall if you wait long enough is $1$ provided $v_z\ne 0$, and $0$ if $v_z = 0$. The issue is I don't think the question you seem to be trying to answer doesn't seem to be the question you actually want to know the answer to. I am trying to work out the question you want to be asking $\endgroup$
    – By Symmetry
    Feb 14 at 12:32

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