Mean free path

Remarks

Derivation: the particle travels a distance greater than $\ell$ without collision exactly when no obstacle centre lies within distance $d$ of the first $\ell$ units of its path. The set of these points, minus the ball of radius $d$ around the origin, has volume $\pi d^2 \ell = \sigma \ell$. For a Poisson point process, conditioned on the ball being empty, this has probability $e^{-n \sigma \ell}$. So the free path has the exponential distribution with rate $n\sigma$, and its mean is $1/(n\sigma)$.

Example: for hard-sphere molecules of diameter $d$, two molecules collide when their centres are at distance $d$, so $\sigma = \pi d^2$ and $\lambda = 1/(\pi n d^2)$ for a particle moving among fixed molecules.

Non-example: if the obstacles are not fixed but move with the Maxwell distribution of a gas in equilibrium, the formula $1/(n\sigma)$ is not exact. The relative speeds change the collision rate, and the standard result is $\lambda = 1/(\sqrt 2 \, n \sigma)$.

In the Boltzmann–Grad limit of $N$ hard spheres of diameter $\varepsilon$, the condition $N \varepsilon^2 = 1$ keeps the mean free path of order $1$.