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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?

Riemann
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1 Answers1

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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.

Andrew Steane
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  • 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 $\frac{v}{\Gamma(v)}=\frac{v}{0.75}=\frac43v$. – Riemann Oct 24 '22 at 14:02
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    @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. – Andrew Steane Oct 24 '22 at 14:16
  • @Riemann ... I just edited the answer so now it is not quite a complete answer, but I guess will be of some help. – Andrew Steane Oct 24 '22 at 14:21
  • I was also thinking about the integral formula for $\Gamma(v)$. How can we justify this, given that both the particle with speed $v$ and the other particle speed $v'$ are moving, and they collide if and only if there is some moment in time that they are on approximately the same location? – Riemann Oct 25 '22 at 08:04
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    @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. – Andrew Steane Oct 25 '22 at 08:48
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    @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. – Andrew Steane Nov 22 '22 at 11:10
  • I agree, good point! We have $\lambda(v)=v\tau(v)=v/\Gamma(v)$ by the exponential distribution. However, does this imply that $\langle \lambda(v)\rangle\neq\langle v\rangle /\langle \Gamma(v)\rangle$? – Riemann Nov 22 '22 at 12:09
  • @Riemann Correct. $\langle \lambda(v) \rangle = \langle v/\Gamma(v)\rangle \ne \langle v \rangle / \langle \Gamma(v) \rangle $. – Andrew Steane Nov 22 '22 at 14:01
  • Do you have a proof that the two values are unequal? – Riemann Nov 23 '22 at 10:46
  • @Riemann I did the integrals numerically. – Andrew Steane Nov 23 '22 at 18:26
  • That is a good idea! What was the ratio between the two results? – Riemann Nov 24 '22 at 18:59