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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).

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    $\begingroup$ 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, 2018 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, 2018 at 18:16
  • $\begingroup$ Related: physics.stackexchange.com/q/41138/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Dec 9, 2018 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, 2018 at 18:28
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    $\begingroup$ @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, 2018 at 18:43

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