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 LM 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:
Vexp=12μv2rel
where
Vexp is the potential experienced,
μ is the reduced mass and
vrel is the relative velocity. The new action density when collisions are included is given by:
S(p)→S(p)+Sc
where p is the momentum, S(p) is the action when only kinetic energies are considered and Sc is the action contributed by the potential energy. Now, if I assume a short ranged interaction:
Sc=∫Lcdt≈Vexpτ
where
τ is the collision duration.
Now for a gas, the
number of collisions per
4 volume is given by:
dNc=12ρ2A|⟨vrel⟩|dtdxdydz
where the
ρ is the density,
A is the area of the molecule and
dt,
dx,
dy,
dz are infinitesimals.
Hence, the action density
˜Sc for the entire gas is given by:
˜Sc≈14ρ2A|⟨vrel⟩|μ⟨v2rel⟩⟨τ⟩
Now while
⟨τ⟩ should depend on the short range potential I see no reason it couldn't possibly have the term (after using the equation of state):
⟨τ⟩=Cρ2|⟨vrel⟩|⟨v2rel⟩+…
where
C is a constant?