The expression for the constant action term ( S_c ) provided is not correct, primarily due to dimensional inconsistencies and issues with how collision contributions to the action are modeled. Here's a breakdown:

1. Dimensional Analysis

The action ( S ) has units of energy × time (J·s). Let’s analyze the units of ( S_c ):

• ( N ): Dimensionless (count of particles).
• ( R ): Length...

Short answer: No. There is no universal, state‑independent “constant collision term” that should be added to the action of a classical gas. Collisions are already fully accounted for by the interparticle potential in the standard Lagrangian L = T − U, and any additive constant to L has no physical effect on the dynamics or thermodynamics.

Why a constant “collision action” does not arise

• Proper...

Evaluating the Constant Action Term for Gas Collisions

The question asks about the validity of a derived constant action term for a gas system that accounts for molecular collisions. Let me analyze this carefully.

Analysis of the Approach

The approach you've taken is interesting - attempting to derive a contribution to the action from finite-time collisions between gas molecules. Let me ex...

Short answer
No. In classical mechanics there is no extra “collision-term’’ that must be added to the action of a gas.
If the molecules interact through a pair potential

  V(r) = 0   for r > 2R
  V(r) = ∞  for r ≤ 2R (hard spheres)

then along every admissible trajectory one always has r>2R, so V≡0 and the action of N particles is simpl...

<think> We are given a question about the contribution to the action from collisions in a gas. The user has derived an expression for the collision action term and wants to know if it is correct.

The action for a system is defined as: [ S = \int (T - U) dt ] where ( T ) is the total kinetic energy and ( U ) is the total potential energy.

The user's derived expression for the collision act...

This is a very interesting and thought-provoking question. It delves into the fundamental connection between microscopic dynamics (described by the action) and macroscopic thermodynamics.

Let's break down your idea and analyze the expression for .

The short answer is: **No, the expression for as a contribution to the classical action is not correct in the standard fo...

Looking at your derivation, I can see several issues with the proposed constant action term from collisions:

Conceptual Issues

1. Action principles don't include collisions directly: In classical mechanics, the action integral is defined for smooth trajectories. Collisions represent discontinuities where the standard variational principle breaks down. The action formulation assumes differ...

A few remarks up front:

1. In classical mechanics any constant piece in the Lagrangian L→L+C
   does not change the equations of motion (it just adds an overall C · t to S, which drops out of δS=0). So whatever constant you pull out of the collision‐term will have no effect on the trajectories of your gas molecules.

2. If you do go ahead and compute
   S_c ≡ ∑₍collis...

In classical mechanics the dynamical information of a gas that is made up of (N) point particles of masses (m_i) is contained in the Lagrangian

[ L({\mathbf r_i,\dot{\mathbf r}i});=;T-U ;=;\sum{i=1}^{N}\frac{m_i}{2}\dot{\mathbf r}i^{2} ;-;\sum{i<j}V(r_{ij}), \qquad r_{ij}=|\mathbf r_i-\mathbf r_j|. ]

The action is the time–integral of the Lagrangia...