# Time between two collisions in a 2D gas

I'm trying to code a simulation of a 2D gas in a box. The molecules are represented by circles of radius $R$ having elastic collisions with the walls only (for now).

I read that the average time step to the next collision (considering they happen at regular intervals) is :

$$\langle\Delta t\rangle = \frac{\tau}{N}$$

where $\tau$ is the average duration of a collision, and N the number of molecules.

This formula is given without further justification, and I don't really understand it. I can see why the time step would decrease with the number of particles but I don't understand how the duration of a collision landed here.

How do you justify this particular relation ?

EDIT (detailed development):

Initial state : positions and velocities randomly distributed, the norms of the velocities are identical

$$||\vec{v_i}|| = v_0 = \sqrt{\langle v^2\rangle}$$

The time step is variable and defined as :

$$\Delta t = t_c-t$$

where $t$ is the current time and $t_c$ the time of the next collision.

Considering the collisions happen at regular intervals, the mean value of the time step is estimated to be:

$$\langle\Delta t\rangle = \frac{\tau}{N}$$ (which is what I don't understand)

where $\tau$ is the average duration of a collision, and N the number of molecules.

Taking (for convenience) the Length of the box to be twice the width :

$$L = 2l$$

Hence, when the particles travel $2l$ horizontally (Ox), they hit a wall once and when they travel the same distance vertically (Oy) they hit a wall twice.

Since

$$\langle v_x^2\rangle = \langle v_y^2\rangle = \frac{\langle v_x^2\rangle + \langle v_y^2\rangle}{2} = \frac{\langle v^2\rangle}{2}$$

(Equipartition)

A particle undergoes on average 3 collisions during : $\tau_3 = \frac{2l}{\sqrt{\langle v_x^2\rangle}}$

(Neglecting the $R$ length restriction on each wall)

Finally:

$$\tau = \frac{\tau_3}{3} = \frac{2l}{3\sqrt{\langle v_x^2\rangle}} = \frac{2\sqrt{2}l}{3\sqrt{\langle v^2\rangle}}$$

and

$$\langle \Delta t\rangle = \frac{2\sqrt{2}l}{3N\sqrt{\langle v^2\rangle}}$$

So in the end the time step does depend on the RMS-velocity. But what I don't understand is the reasoning that leads to acknowledge that :

$$\langle\Delta t\rangle = \frac{\tau}{N}$$

• Can you cite the source for that equation & definition of tau? – Kyle Kanos Nov 21 '15 at 23:13
• It's a French book: "Thermodynamique - Fondements et applications" by José-Philippe Pérez. But I misread, it's actually not exactly the time between two collisions but rather the time step: $\Delta t= t_c - t$ with $t$ the current time and $t_c$ the time of the next collision. I still don't understand the relation though... – mwa1 Nov 21 '15 at 23:28
• the speed of particles wouldn't count ? I don't believe that. – Fabrice NEYRET Nov 22 '15 at 0:59
• It does implicitly, I'll detail the development. – mwa1 Nov 22 '15 at 1:14
• From what you wrote, I would say $\tau$ is the (average) time interval between two collisions between a 'particular particle' and the wall. Then $\Delta t$ is the time interval between two collisions of 'any particle' with the wall (i.e., the interval between two times the wall get hit by some particle). [This has nothing to do with the time interval between two collisions between the particles themselves, which must depend explicitly on R]. – chau Nov 23 '15 at 15:33