# Most probable free path length

The distribution of scattering time is

$nv\sigma e^{-nv\sigma t}$

where $\sigma$ is the collision-cross section, $n$ is the number density of molecules and $v$ is the average relative velocity. This distribution is strictly decreasing with $t$ so that it has maximum at $t=0$. Therefore the most probable free path should be $0$. Is this correct? Or have I misunderstood something?

• Don't let the zero bother you. The most probable value of a speckled laser reflection (en.wikipedia.org/wiki/Speckle_pattern) is also zero (for the same reason)--but that doesn't mean you can't see it.
– JEB
Commented Apr 29, 2018 at 21:35

For that distribution, the most probable single value is zero.

But the average value is $n\nu\sigma$, which is non-zero.

The underlying physical cause is that there's an equal probability of scattering in every cm travelled. If the particle doesn't scatter in the first cm, it'll get a chance in the second, then the third, etc. But if it does scatter in the 1st, it's no longer available to have it's first scatter in the 2nd or 3rd; it's already scattered. So the most probably place to scatter is the place with the most flux of unscattered particles, right at the start.

It’s easier to think about distance than about time. The distance between collision events is indeed exponentially distributed, $PDF\propto \exp (-n\sigma x)$, but the corresponding travel time depends on relative speed, which obeys the Maxwell-Boltzmann distribution. (It’s even possible that the scattering cross-section depends on relative velocity.) The mode of $\exp (-n\sigma x)$ is indeed at x=0, but the mean is more relevant than the mode for most purposes, e.g., transport phenomena.