# Heaviside step function related to Boltzmann distribution

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?

• 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 Commented Feb 13, 2023 at 12:09
• 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 Commented Feb 14, 2023 at 12:32