I'm still wondering about the physical interpretation of the action integral of some mechanical system (classical theory here, to simplify things): \begin{equation}\tag{1} A = \int_{t_1}^{t_2} L(q, \, \dot{q}) \, dt, \end{equation} where $L = K - U$ is the system's lagrangian. I'm not interested in its variation and the principle of extremal action. I'm interested in the number given by (1), wathever the motion (extremal or not). I was always amazed by some analogies with the statistical definition of entropy, defined on phase space ($k_B = 1$ here): \begin{equation}\tag{2} S = -\int_{\Omega} \rho \, \ln{\rho} \, d\Omega, \end{equation} especially when the probability density is Boltzmanian : $\rho \propto e^{-\, \beta H}$ (for equilibrium macroscopic states). The usual standard interpretation of entropy is this:

Entropy $S$ is a measure of the lack of the statistical knowledge that you need to define the microscopic state of a system at a given time.

I'm wondering about a similar (inverted) interpretation for the action (1) (this is my own interpretation):

Action $A$ is a measure of the mechanical information that you already have on the state of a system and its evolution during some time interval.

Is it possible to make that statement more precise and rigorous? Could we define (or give a sense to) mechanical information as the action integral, a bit like entropy as a measure of "ignorance" (average of $\sigma = -\, \ln{\rho}$)?

I've checked the similar questions on Stack Exchange, like this one : What is the significance of action?, but all of the answers are systematically refering to the extremal action principle (or stationary action principle), and they aren't answering the question about the action itself.

I don't think that the "action" is just an abstract tool (with a fancy mysterious name) to find the classical trajectories in phase space. And I don't think that Quantum Mechanics is giving any answer on the interpretation (action as a "phase variable", which is just pushing the question under the carpet).

  • $\begingroup$ I think it is on you to convince others that $A$ is "a measure of the mechanical information". You cannot make up some interpretation and ask others to justify it for you, especially when that interpretation is not mainstream at all, and most likely wrong. What arguments do you have – beyond a presumed similarity to some other formula (which I sincerely don't see: they are just two integrals) – that suggest that $A$ is a measure of mechanical information? Why should $S$ have an interpretation at all? $\endgroup$ – AccidentalFourierTransform Dec 9 '18 at 18:05
  • $\begingroup$ @AccidentalFourierTransform, entropy $S$ does have a statistical interpretation (I guess you meant $A$ instead, just not to confuse with entropy). In QFT, there are many authors which are making similarities with action (using some weird Wick rotation to define euclidian path integrals) and statistical mechanics. $\endgroup$ – Cham Dec 9 '18 at 18:16
  • $\begingroup$ Related: physics.stackexchange.com/q/41138/2451 and links therein. $\endgroup$ – Qmechanic Dec 9 '18 at 18:18
  • $\begingroup$ Thanks Qmechanic. However, I don't think that your answer there really answers the OP's question. I agree with user2781942's viewpoint : Action should have a physical interpretation. It's not just an abstract tool to allow the stationary action principle to work. $\endgroup$ – Cham Dec 9 '18 at 18:28
  • 2
    $\begingroup$ @AccidentalFourierTransform, I don't agree that there are no similarities. Both integrals aren't just "ordinary" integrals. Both are scalar functionals : one is defined on the macroscopic states ($S$) while the other one is defined on microscopic states ($A$). Both are additive for independant sub-systems. And there's this weird euclidian paths stuff (that I don't understand very well!) which turns some action into entropy! And in classical relativistic theory, a single particle's action is equivalent to the proper time elapsed (so it does have an interpretation there). $\endgroup$ – Cham Dec 9 '18 at 18:43

In this answer will argue the following assertion:

The Action, while an indispensible part of the least action formulation, is purely mathematical apparatus, and has no physical interpretation. Other aspects of the least action formulation do have physical interpretation, but not the action.

The following is an abbreviated/adapted version of an answer that I wrote to a question titled 'What is the physical content of the principle of least action?'

To illustrate an elementary case: An object is thrown upward, acceleration is downward.

To make the numbers come out simple these input values:
- Total duration: 2 seconds (from t=-1 to t=1)
- Gravitational acceleration 2 $m/s^2$
- Mass of the object: 1 unit of mass

The height as a function of time is then:

$$ h(t) = -t^2 + 1 \qquad (1)$$

This trajectory is varied in the simplest way that varies the trajectory at all. The following expression introduces a single variational parameter $p_v$
So now the height is a function of two variables, time and $p_v$

$$ h(t,p_v) = (1+p_v)(-t^2+1) \qquad (2) $$

The image with the object reaching 1 unit of height is the true trajectory, the two smaller images represent variations.

Parabolic trajectory, true height Parabolic trajectory, too much height Parabolic trajectory, too little height

In the following three images the parabola on the left represents the trajectory of the object. On the right the red graph represents the kinetic energy as a function of time, the green line is a representation of minus the potential energy. (Zero height is taken as zero potential energy.)

True trajectory, graphs are parallel, Action minimal

Trajectory too high, higher action

Trajectory too low, higher action

Again, the green line represents the potential energy, but with a minus sign.

For the trajectory to obey the work-energy theorem the rate of change of kinetic energy must be a match for the rate of change of potential energy - everywhere along the trajectory. For the graphs that means that the green line must be parallel to the red line everywhere along the trajectory.

The velocity with respect to time is the derivative of (2): $-2(1+p_v)t \quad (3)$

Kinetic energy: $2(1+p_v)^2t^2 \quad (4) $

Potential energy: $2(1+p_v)(-t^2+1) \quad (5) $

The kinetic energy varies in proportion with the square of the variational parameter, whereas the potential energy varies in linear proportion to the variational parameter.

The task is: identify the trajectory with the property that at every point in time the rate of change of kinetic energy matches the rate of change of potential energy.

With the above the physics part of the task is completed, what remains doesn't involve physics, it is mathematical apparatus only.

The mathematical task: when you sweep through the variation, identify the point where the red line and the green line are parallel to each other - everywhere along the trajectory. The Action integral is the mathematical tool that allows us to identify that point. When the derivative of the Action wrt the variational parameter is stationary the red line and the green line are parallel to each other - everywhere along the trajectory. (Of course, a rigorous proof of this must show that it generalizes to all possible cases, but that is beyond the scope of this answer.)

Returning to the assertion at the start:
The Action, while an indispensible part of the least action formulation, is purely mathematical apparatus, and has no physical interpretation.

The physical content of the least action formulation is the work-energy theorem. The purpose of sweeping through the variation is to identify the trajectory that satisfies the work-energy theorem.


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