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