Why is it impossible to have a constant in the action via classical mechanics?

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Question
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So I don't  understand why there seems to be a general consensus that it is impossible to have a constant in the action via classical mechanics? Is there some no-go theorem I'm missing?

My Attempt
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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 while $\langle \tau \rangle$ should depend on the short range potential I see no reason it couldn't possibly have the term (after using the equation of state):

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

where $C$ is a constant?

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