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?


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:

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

  • $\begingroup$ This doesn't look right: as $T$ increases processes generally speed up but the result suggests that the rate of collisions goes down? $\endgroup$
    – Themis
    Commented Sep 25, 2022 at 19:18
  • $\begingroup$ But the collision duration is usually negligible are the some processes whose modelling require finite duration? $\endgroup$ Commented Sep 25, 2022 at 19:22
  • $\begingroup$ 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)? $\endgroup$ Commented Nov 14, 2022 at 22:33


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