# 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?

• Could you add definitions of $v, v_1$ and $v_r$ to your question? Oct 24, 2022 at 12:35
• Oct 24, 2022 at 23:07

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.
• Thank you for your answer! I think that $\lambda(v)$ is actually not equal to $\frac{v}{\Gamma(v)}$; for example, suppose that we have probability $0.5$ for a collision after $1$ second and probability $0.5$ for a collision after $2$ seconds. Then $\lambda(v)=1.5v$ while $\frac{v}{\Gamma(v)}=\frac{v}{0.75}=\frac43v$. Oct 24, 2022 at 14:02
• @Riemann I see what you mean. I now think one has $\lambda(v) = v \tau(v)$ where $\tau(v)$ is mean collision time for molecules of speed $v$, but I will have to think a bit more about rates. Oct 24, 2022 at 14:16
• @Riemann Adopt the rest-frame of particle at velocity $\bf v$. The other particles move at velocities ${\bf v}' - {\bf v}$. Those whose collision cylinders include the particle of interest will hit it. Oct 25, 2022 at 8:48
• @Riemann I had a further thought. In general, for some arbitrary distribution, the mean rate $\Gamma$ is not equal to inverse of mean time $\tau$. But for the exponential distribution they are equal. So I now think my original answer was correct. Interested to know if you agree. Nov 22, 2022 at 11:10