# Is the average collision duration $\propto T^{1/2}/ P^2$?

## Question

So I was reading an old question of mine and realized the average collision duration $$\langle \tau \rangle$$ must be proportional to:

$$\langle \tau \rangle \propto \frac{T^{1/2}}{P^2}$$

where $$P$$ is pressure and $$T$$ is temperature. Does my reasoning seem correct? Is this experimentally testable?

## Reasoning

Consider the Lagrangian $$\mathcal{L_M}$$ for a gas. Generally in the gas ideal model only the kinetic energies are considered but let us think of the potential energy of a collision and not assume the collision is an event in spacetime but has finite duration. The turning point can be thought as a consequence of regularisation.

The potential experienced by $$2$$ objects when they collide is given by: $$V_{exp} = \frac{1}{2} \mu v_{rel}^2$$

where $$V_{exp}$$ is the potential experienced, $$\mu$$ is the reduced mass and $$v_{rel}$$ is >the relative velocity. The new action density when collisions are included is given by:

$$S(p) \to S(p) + S_c$$

where $$p$$ is the momentum, $$S(p)$$ is the action when only kinetic energies are considered >and $$S_c$$ is the action contributed by the potential energy. Now, if I assume a short ranged interaction:

$$S_c = \int L_c dt \approx V_{exp} \tau$$ where $$\tau$$ is the collision duration. Now for a gas, the number of collisions per $$4$$ volume is given by:

$$d N_c = \frac{1}{2}\rho^2 A |\langle v_{rel} \rangle | dt dx dy dz$$ where the $$\rho$$ is the density, $$A$$ is the area of the molecule and $$dt$$, $$dx$$, $$dy$$, $$dz$$ are infinitesimals. Hence, the action density $$\tilde S_c$$ for the entire gas is given by:

$$\tilde S_c \approx \frac{1}{4} \rho^2 A |\langle v_{rel} \rangle | \mu\langle v_{rel}^2 \rangle \langle \tau \rangle$$

Now if $$\langle \tau \rangle$$ was short range potential of any other form than:

$$\langle \tau \rangle = \frac{C}{\rho^2 |\langle v_{rel} \rangle | \langle v^{2}_{rel} \rangle }$$

(with $$C$$ being a constant) Then the equation of state would give rise to an unwanted observable consequence which are not observed in a gas in thermal equilibrium. Note:

$$v_{rel} \propto \sqrt{T}$$

• This doesn't look right: as $T$ increases processes generally speed up but the result suggests that the rate of collisions goes down? Commented Sep 25, 2022 at 19:18
• But the collision duration is usually negligible are the some processes whose modelling require finite duration? Commented Sep 25, 2022 at 19:22
• Isn't the typical Coulomb collision rate expressed as $\nu \propto n \ T^{-3/2}$, where $n$ is the number density and $T$ is the temperature (e.g., see physics.stackexchange.com/a/268594/59023)? Commented Nov 14, 2022 at 22:33