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I'm running some simulations of particle collisions in two dimensions with discretised time and space. In essence, particles only collide if they occupy the same location (cell) at the same time step. The particles are in a 2D box and collide with both each other and the walls, we also assume a closed system (so no gravity etc).

I want to use the mean free path $\lambda$ (the average distance travelled by a particle between collisions) to determine the best values for number of particles $N$, rms velocity $V_{rms}$, and box length $L$; as such I want to know how to calculate mean free path in two dimensions.

So far I have been able to find a formula on Wikipedia for three dimensions: $$ \lambda=\frac{k_BT}{\sqrt{2}\pi d^2p} $$ where d is the diameter of the particle and p is the pressure.
which I can easily turn into: $$ \lambda=\frac{mv_{rms}^2}{2\sqrt{2}\pi d^2p} $$ but again this is for three dimensions not the two dimensions that I need.

So the question is, can this be converted to two dimensions? or does a two dimensional form already exist?

Any help with this would be greatly appreciated.

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Sure, you can generalize the mean free path to a different number of dimensions. But first, let's understand the derivation in 3D.

  1. A particle will collide with any other particle that it comes within a distance $d$ of. So if it moves a length $\ell$, it will collide if there is another particle in a cylindrical volume $\pi d^2 \ell$. Call this the volume swept out by the particle.

  2. Imagine a region of volume $V$ filled with gas, and let $\bar{v}$ be the mean speed of the particles of that gas in the rest frame of the entire volume. Then, in the reference frame of one of the gas particles, the other particles have an average speed of $\sqrt{2}\bar{v}$. (See this derivation of the factor of $\sqrt{2}$, and note the presence of $\bar{v}$ instead of $v_\text{rms}$ - see this question for details on that)

  3. In the reference frame of that one particle, the total volume swept out by all the other particles in a time $\Delta t$ is $N\pi d^2 \sqrt{2}\bar{v}\Delta t$, where $N$ is the number of particles.

  4. The probability that the chosen at-rest particle experiences an interaction during $\Delta t$ is equal to the fraction of the total volume ($V$) swept out, namely

    $$P(\text{int.}) = \frac{N\pi d^2 \sqrt{2}\bar{v}\Delta t}{V}$$

  5. To calculate the mean free time $\tau$, I technically should find the probability distribution of interaction times and compute its mean. But to make the calculation simple, I'll take advantage of a handy coincidence (which I am not justifying here): $\tau$ happens to be equal to the time after which this probability would reach 1 if it increased at a fixed rate over time. So I can replace $\Delta t$ with $\tau$ and $P(\text{int.})$ with $1$, and I get

    $$\tau = \frac{V}{N\pi d^2 \sqrt{2}\bar{v}}$$

    The mean free path is then given by

    $$\lambda = \bar{v}\tau = \frac{V}{\sqrt{2}N\pi d^2}$$

If you assume the particles follow the ideal gas law, you can replace $\frac{V}{N} = \frac{k_B T}{p}$ and recover the formula from Wikipedia.

To modify this to 2D, we just need to change step 1 and follow the argument from there.

  1. Instead of a particle sweeping out a volume $\pi d^2\ell$, it sweeps out an area $2d\ell$.

  2. No change

  3. The total area swept out by all $N$ particles is then $Nd\sqrt{8}\bar{v}\Delta t$

  4. The probability is equal to the fraction of the total area,

    $$P(\text{int.}) = \frac{Nd \sqrt{8}\bar{v}\Delta t}{A}$$

  5. The mean free time is again the time after which this probability would reach 1,

    $$\tau = \frac{A}{Nd\sqrt{8}\bar{v}}$$

    and the mean free path is

    $$\lambda = \bar{v}\tau = \frac{A}{\sqrt{8}Nd}$$

Bear in mind that in 2D, the ideal gas law would be modified; it would have area instead of volume, and you would have to use a 2D version of pressure, which would be force per unit length, not per unit area. It may be easier to just work with $\lambda = \frac{A}{\sqrt{8}Nd}$ directly, since you can just set $A = L^2$ if you know the box side length.

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  • $\begingroup$ @MichaelLevy I'll make an edit, but I'm curious: why did you comment about $V$ and not any of the other variables I used without explicitly defining them? $\endgroup$
    – David Z
    Sep 19 at 6:24
  • $\begingroup$ Yeah, fair enough. I agree that being more clear about definitions is better than less clear. $\endgroup$
    – David Z
    Sep 21 at 20:29
  • $\begingroup$ Let me know if the latest version looks better. $\endgroup$
    – David Z
    Sep 21 at 20:41
  • $\begingroup$ Eh personally I don't like that particular change. $\endgroup$
    – David Z
    Sep 21 at 23:18

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