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"For classical (non-quantum) systems, the action is an extremum that can never be a maximum; that leaves us with a minimum or a saddle point, and both are possible."

The above statement is an excerpt from the "Introduction" (preface) of the book "THE PRINCIPLE OF LEAST ACTION - History and Physics" by ALBERTO ROJO & ANTHONY BLOCH.

I want to know whether the "For classical (non-quantum) systems, the action is an extremum that can never be a maximum" aspect of that statement is true because it looks pretty definitive. (definitive in the sense certain or assertive)

Note: Now I know there are a lot of related questions that look like this, but not any of them looks for a direct and definitive answer for this direct and definitive question, most are descriptive questions for descriptive scenarios and most answer's given are describing particular scenarios, incomplete ones or ones that asserting irrelevance of such question's for actual path determination as we only seek stationary action not whether that is minimum, maximum or inflexion point. (This is intended as to why this should not be labelled as a duplicate, not as a judgement on other questions or their answers as they serve their intended purpose. It is however important to differentiate between the scope of this question and other similar questions. I hope the Phys.SE community will respect the original poster's judgement on the relevance and uniqueness of their own questions unless there is overwhelming evidence to say otherwise.). I have already browsed similar questions as indicated by the system and have not found any definitive question or definitive answer. This definitive question clearly expects a definitive answer, so I hope it will remain a question, not a duplicate.

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  • $\begingroup$ Well, the EL equations are not changed if we scale the action with a non-zero constant. In particular, if we use a non-standard sign convention for the action, then it is easy to give counterexamples. $\endgroup$
    – Qmechanic
    Commented May 17, 2023 at 13:39
  • $\begingroup$ Thanks for the info. But if that is so aren't the new sign convention no longer permit the least action? I mean we can't have all three types of extremals using the same sign convention in classical mechanics. I think rather than worrying about sign convention, authors are trying to point out that we can't have all three types under the same framework (or convention) of classical mechanics. $\endgroup$ Commented May 17, 2023 at 14:00

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In this answer I will offer my thoughts as to why there is ambiguity.

The concept of stationary action is not a theory that stands on its own. Rather, stationary action is a recurring motif, seen in many different branches of physics.

In each branch of physics there is a bespoke Lagrangian. Those various Lagrangians do not necessarily have much in common.

There is no rule (or set of rules) to decide what counts as a Lagrangian and what doesn't. It's a judgement call.


In terms of general relativity: when a test mass is released to free motion the path it follows is referred to as 'geodesic of spacetime'. A geodesic of spacetime has the property that for that particular path (in the variation space of paths) the amount of proper time that elapses is the largest.

(Here 'free motion' means that the path of the test particle is affected by gravitational interaction only.)

So:
In the case of general relativity, if you choose to qualify elapsed proper time as an 'action', then you have a case where the extremum of that action is always a maximum.

It could be, I don't know, that Alberto Rojo and Anthony Bloch will say that the amount of proper time elapsed does not qualify as an action.

(Arguably, if Fermat time is accepted as an instance of action, then the GR proper time for motion along a geodesic should also be accepted as an action.)



I want to turn back to stationary action as a recurring motif.

'Stationary action' is very generic.

Another name for 'Principle of stationary action' with effectively the same meaning would be: 'Principle of differential equations'.

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  • $\begingroup$ The author's statement is about classical mechanics. $\endgroup$ Commented May 18, 2023 at 3:27
  • $\begingroup$ @DukeWilliam Rojo and Bloch place the transition away from classical mechanics at the introduction of quantum mechanics. That is: Rojo and Bloch classify all non-quantum as classical mechanics. It is quite common for physicists to take 'classical mechanics' as a wide category, putting relativistic mechanics in that category. There is no rule for such classification; it's a judgement call. $\endgroup$
    – Cleonis
    Commented May 18, 2023 at 5:01
  • $\begingroup$ There are no instances in GR where proper time is minimised. Only saddle points or maximum. Doesn't that imply some deeper truth, that we cannot have all three types when you consider action in the same field? $\endgroup$ Commented May 18, 2023 at 7:08
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  1. Well, the EL equations are not changed if we scale the action with a non-zero constant. In particular, if we use a non-standard sign convention for the action, then it is easy to give counterexamples.

  2. It should be pointed out that an extremum is by definition a maximum or a minimum. However many authors are misusing the word extremum to mean a stationary or critical point, i.e. they are including saddle points. Be aware that this misuse of terminology is quite common in the literature.

  3. Let us mention that the action $$ S~=~ -E_0 \Delta\tau\tag{1}$$ for a relativistic massive point particle is minus the rest energy $E_0=m_0c^2$ times the change $\Delta \tau=\tau_f-\tau_i$ in proper time. The minus sign is the standard convention and it is included to ensure the standard non-relativistic limit $$\begin{align} L~=~&-\frac{E_0}{\gamma}~=~-E_0\sqrt{1-\left(\frac{v}{c}\right)^2}\cr ~\approx~& \frac{1}{2}m_0 v^2 -E_0 \qquad\text{for}\qquad v\ll c.\end{align}\tag{2}$$ Interesting it means that the action is negative, and hence bounded from above! Nevertheless, this example still obeys the statement by Rojo & Bloch: The stationary path is never a maximum.

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  • $\begingroup$ About negative value of the action: In the context of non-relativistic classical mechanics: when sweeping out variation the value of the action ($E_k - E_p$) can dip down into negative range. That is: for the non-relativistic action ($E_k - E_p$) the variation space stretches across both positive and negative values of the action. I propose to summarize as follows: the value range of non-relativistic action ($E_k - E_p$) sweeps both negative and positive values, whereas the relativistic action $-E_0\Delta\tau$ is confined to negative values. $\endgroup$
    – Cleonis
    Commented May 18, 2023 at 14:01

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