How does $\frac{\langle v\rangle}{\langle v_r\rangle}=\frac{1}{\sqrt2}$ imply the formula for the mean free path? In this question, it was asked how the formula $$l=\frac{1}{\sqrt 2n\sigma }$$ can be rigorously derived for a Maxwell-Boltzmann gas.
Here $l$ is the mean free path length in a gas, $n$ is the gas density (assumed to be homogeneous) and $\sigma$ is the cross section of colliding particles.
The answers to the question prove that $$\frac{\langle v\rangle}{\langle v_r\rangle}=\frac{1}{\sqrt2},$$ and conclude that the formula holds.
Here $v$ is the speed of a particle, and $v_r$ is the relative speed between two particles. For $\langle v_r\rangle$, we take the average over all pairs of particles.
The answers seem to use that $l=\frac{\langle v\rangle}{n\sigma \langle v_r\rangle}$. Can this be proven?
 A: It can't be rigorously derived because it is not true!
Using $v_r$ one can deduce the mean collision rate (after averaging over speed as well as over collisions times)
$$
\langle \Gamma(v) \rangle = \sqrt{2} n \sigma \bar{v}
$$
where $\bar{v}$ is the mean speed (equal to $(8 k_{\rm B}T/\pi m)^{1/2}$ for Maxwell-Boltzmann distribution).
The mean free path, after averaging over both path lengths and speeds, is
$$
\langle \lambda(v) \rangle = \int_0^\infty \lambda(v) f(v) dv
$$
where $f(v)$ is the speed distribution function and $\lambda(v)$ is the mean free path for molecules of speed $v$. The calculation of $\lambda(v)$ can be done via $\lambda(v) = v \tau(v)$ where $\tau(v)$ is mean collision time for molecules of speed $v$. In a previous version of this answer I had convinced myself that $\tau(v) = 1/\Gamma(v)$ but now I am not so sure! For mean collision rate at speed $v$ one has
$$
\Gamma(v) = n \sigma \iiint d^3{\bf v}' |{\bf v}' - {\bf v}| f_3({\bf v}')
$$
where $f_3$ is the velocity distribution function.
