# Does a constant in the action always have unobservable consequences in classical mechanics?

## Background

So in classical mechanics, my understanding is that for the action by using a the principle of least action one can get the equations of motion. Adding a constant to the action does not change anything.

However, the above argument relies on the notion of continuity and thus fails in collisions. For example, if my physical system is a ball seen at $$2$$ points under no other forces then the principle of least action predicts that the path taken must be a straight line (whereas it may have rebounded of a wall as well)

(where $$t$$ is time and $$x$$ is the position)

## Question

Now, I understand why is it that by adding a constant in the action density always have non-observable consequences if one assumes continuity. My question is how does one prove this when there are discontinuities as well?

## My attempt

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?

• Well, this is pushing the action concept; possibly the action concept is not applicable. I submit this requires examination of the action concept itself. In classical mechanics Hamilton's stationary action can be derived from F=ma The derivation proceeds in two stages: 1. Derivation of the Work-Energy theorem from $F=ma$ 2. Demonstration that in cases where the Work-Energy theorem holds good Hamilton's stationary action will hold good also. I expect this will give you the means to assess the case that you are examining. Jul 27, 2022 at 5:35
• No, the three so called adiabatic invariants of particle motion in a dipole magnetic field correspond to constant action yet they have observational signatures. See physics.stackexchange.com/a/670591/59023 Jul 27, 2022 at 11:34

There are three so called adiabatic invariants of single-particle motion under the influence of a quasi-static magnetic field. These are discussed in detail at: https://physics.stackexchange.com/a/670591/59023

These can be summarized as follows: \begin{align} \left( \pi \rho_{cs}^{2} B \right) & \equiv \text{magnetic flux conservation} \tag{0a} \\ \frac{ p_{\perp}^{2} }{ B } & \equiv \text{transverse momentum} \tag{0b} \\ \gamma \mu & \equiv \text{magnetic moment} \tag{0c} \\ \int_{a}^{b} \ ds \ p_{\parallel} & \equiv \text{parallel momentum} \tag{0d} \end{align} where $$e$$ is the fundamental charge [C], $$\gamma$$ is the Lorentz factor [N/A], $$B$$ is the magnitude of the magnetic field [T], $$m_{s}$$ is the mass of species $$s$$ [kg], $$v_{\perp}$$ is the speed orthogonal to the magnetic field vector [km/s], $$p_{j}$$ is the canonical momentum, and $$\mu = \tfrac{ e \ \Omega_{cs} \ \rho_{cs}^{2} }{ 2 \ c }$$ is the particle magnetic moment. Note that the gyroradius, $$\rho_{cs}$$, and cyclotron frequency, $$\Omega_{cs}$$, of species $$s$$ given by: \begin{align} \rho_{cs} & = \frac{ \gamma \ m_{s} \ v_{\perp} }{ e \ B } \tag{1a} \\ \Omega_{cs} & = \frac{ e \ B }{ \gamma \ m_{s} } \tag{1b} \end{align}

Does a constant in the action always have unobservable consequences in classical mechanics?

No, these adiabatic invariants of motion correspond to clearly observable signatures in particle data measured by spacecraft orbiting Earth. Sure, violation of these makes things easier to discern but when they hold the observations are also telling because the particle velocity distribution functions (VDFs) are effectively boring.

For instance, expressions ($$0$$b) and ($$0$$c) above can correspond to particles experiencing something called the ''mirror force.'' The name derives from, I think, one of Enrico Fermi's original papers back in the 1940s or 1950s on particles reflecting off of merging ''magnetic clouds'' in space (which led to what we now call Fermi acceleration or diffusive shock acceleration, e.g., see https://physics.stackexchange.com/a/252885/59023). The idea is that if 0b and 0c hold and if the magnetic field varies slowly enough (i.e., slower than the gyroperiod of the particle), then the total kinetic energy should be conserved. Then one can show that as a particle moves toward a region of higher magnetic field, it will move in velocity space to higher pitch-angles -- the angle between the particle velocity/momentum vector and the quasi-static magnetic field.

If we start with an isotropic (in velocity space) particle VDF (e.g., see What is the correct relativistic distribution function? for examples of a particle VDF), $$f\left( \mathbf{x}, \mathbf{v}, t \right)$$, as that VDF is transported to regions of higher magnetic field, it should actually remain isotropic if it is a continuous function and there are no losses. If it is discreet and/or there are losses, then one will see something called a ''loss-cone'' -- particle VDF where one side has a v-shaped depression in phase space density due to a loss of particles in the velocity direction parallel or anti-parallel to the magnetic field. In Earth's magnetosphere, we observe loss-cone VDFs a lot near the auroral regions at high latitudes and low altitudes. This is due to particles being lost to precipitation in Earth's atmosphere, i.e., they scatter off of charged or neutral particles and are not reflected by the mirror force.