I'm aware that with Lagrangian mechanics, the path of the system is one that makes the action stationary. I've also read that its possible to find a choice of Lagrangian such that minimization is sufficient to find the path it takes for a short period of time, cf. e.g. this Phys.SE post. And I know of examples where, given the typical Lagrangian of T-V, the true path makes the action stationary.

Is there a proof that, for some system described in Newtonian physics, that there is no possible action such that minimization can't find the true path? In other word, a proof that one does indeed need to look for paths that make the action stationary, not just paths that minimize it.

I can find questions like this one where the answers show how to go from a stationary action to Newtons laws, but I'm looking at the opposite direction: from Newton's laws to a minimization problem. Apparently that doesn't always work, but I haven't found a proof of that.

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/20188/2451 , physics.stackexchange.com/q/20298/2451 , physics.stackexchange.com/q/78138/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 5 at 7:33
  • 1
    $\begingroup$ I don't think this is a duplicate. The post is not asking whether extremisation of the action always works, it is asking if it is always possible to find a form for the Lagrangian such that minimisation of the action always works. $\endgroup$ Apr 5 at 7:37
  • 1
    $\begingroup$ Cort Ammon and @John Rennie: If that's the question, then it should stressed in the title, e.g. Is there a proof that a least action principle (as opposed to a stationary action principle) cannot model some physical systems? $\endgroup$
    – Qmechanic
    Apr 5 at 7:43
  • $\begingroup$ @Qmechanic Like that? $\endgroup$ Apr 5 at 7:49
  • 1
    $\begingroup$ @John Rennie. Technically speaking no: The word extremizing is only minimizing or maximizing, which excludes saddle points. $\endgroup$
    – Qmechanic
    Apr 5 at 8:00

3 Answers 3


On the question whether it is possible to go from Newton's laws to Hamilton's stationary action:

Yes, that is possible.

(As pointed out by physics stackexchange contributor Kevin Zhou: in physics you can often run derivations in both directions. Newton-Hamilton is an instance of that.)

That said: the Newton-to-Hamilton derivation arrives at the concept of stationary action; no involvement of minimization.

In the following section I discuss why the forward derivation arrives at 'stationary action' rather than at 'minimized action'.

(If you just want the link to the Newton-to-Hamilton path, scroll down.)

The point is:
The following criterion is sufficient:

The true trajectory corresponds to the point in variation space such that the derivative of Hamilton's action (wrt variation) is zero.

To take minimization as a factor is not a necessity.

Looking for a 'minimization' or an 'extremization' is superfluous; it does not contribute to the descriptive power of the theory; to cover all cases that are in scope the derivative-is-zero criterion is already sufficient.

In this situation:
Relinquishing the unnecessary assumption of involvement of minimization is progress.

The path from Newton's laws to Hamilton's stationary action proceeds in two stages:

  1. Derivation of the work-energy theorem from $F=ma$
  2. Demonstration that under the conditions where the work-energy theorem holds good Hamilton's stationary action will hold good also.

A maximum of Hamilton's action

In the case of Hamilton's stationary action: there are classes of cases such that the true trajectory corresponds to a maximum of Hamilton's action.

We have: any assertion that the true trajectory can never correspond to a maximum of Hamilton's action is refuted.

(There are mathematical tricks to squeeze out the desired result, but if you go down that road you are twisting facts to suit theories, instead of the other way round.)

Link to the derivation

The derivation is in an answer I posted in october 2021
from Newton's laws to Hamilton's stationary action

Additional remarks:

There are interesting parallels between Fermat's stationary time and Hamilton's stationary action.

Fermat's stationary time:
The path of light is such that the derivative of the total transit time (wrt to variation) is zero.

In the case of refraction: when a refractive interface is sufficiently curved then there is true path that is the path of largest time.
A, Tan, A. Ranasinghe and V. M. Edwards, 2014
On Principle of Fermat in refraction of light

Reflection exhibits that too: for a sufficiently convex reflective surface there is a true path that is the path of largest time.

The criterion that always holds good is:
The path of light is such that the derivative of the total transit time (wrt to variation) is zero.

  • $\begingroup$ Your linked answer is very helpful. I haven't found a good derivation so the walk through is much appreciated. And the plot near the end of f(x)-g(x) provides a lot of intuitive clarity. For reference, the reason I ask about minimization is one of psychologial/philosophical value. Minimization is something that's done all the time, and is intuitively simple, even if it has more constraints than finding the stationary path. I'm curious what one would miss out on if one tried to do everything with minimization only. $\endgroup$
    – Cort Ammon
    Apr 8 at 15:15
  • $\begingroup$ @CortAmmon Well, on what you would miss out on: Specific instance: in the case of Hamilton's stationary action: there are classes of cases such that the true trajectory corresponds to a maximum of Hamilton's action. With obligatory minimization those cases cannot be handled. Incidentally, the Euler-Lagrange equation is agnostic as to whether the derivative-is-zero condition is occuring at a minimum or a maximum. In the discussion on my website I show for which classes of cases the true trajectory corresponds to a maximum. (A link to my website is available on my stackexchange profile page.) $\endgroup$
    – Cleonis
    Apr 8 at 23:00

If you divide the action integral by the fixed time between the two fixed points in space, the variation principle states, that all paths, invariant with respect to the time mean of the diffenence between kinetic energy and potential energy, are solutions of the Lagrangian equations of motion.

Consider the set of closed paths of Kepler ellipses with the same large axis and the same period passing through a given point.

This is the normal case, its point in the space of orbits with derivative zero wrt to small deviations and an undefined collection of both signs of the second derivative.

Evidently, following the path of a solution with lower velocity for the first half of the period and a higher velocity on the second, always increases the the action in a potential problem by the constant sum of partial times.

On the other hand, by taking a short cut in a closed loop always decreases the action.

Besides the case of a free straight path (and the high energy/short path paradigma) the 'minimum principle in point mechanics' is a philosophical idea induced by Leibniz ideas of the best of all worlds, realized by minimizing costs during transport.


Yes it is typically the case for field theories. A simple construction involves many harmonic oscillators with large frequencies. When there are an infinite number, the frequencies can be unbounded which means that the half periods cannot have a positive lower bound.

A simple example is to consider d'Alembert's equation: $$ \partial_t^2\phi-\partial_x^2\phi = 0 $$ To make things very concrete, you can consider a finite domain $x\in(0,1)$ with Dirichlet boundary conditions: $$ \phi(x=0,1)=0 $$ Solutions are stationary points of the action: $$ S = \frac12\int_0^T\int_0^1(\partial_t\phi^2-\partial_x\phi^2)dxdt $$

The modes are indexed on $n\in\mathbb N^*$ of the form: $$ \phi_n = \sin(k_nx) \\ k_n = n\pi $$ with associated frequencies $\omega_n = k_n$. The time at which the trajectory is a minimum is at $\inf_{n\in\mathbb N^*} \frac\pi{\omega_n} = 0$.

Equivalently, $S$ is quadratic and diagonalisable in the eigenbasis for $m,n\in\mathbb N^*$: $$ \phi_{mn} = \sin(\Omega_mt)\sin(k_nx) \\ \Omega_m = \frac{m\pi}T $$ and corresponding eigenvalue: $$ \lambda_{mn} = \frac{\omega_m^2-k_n^2}2 $$ and no matter how small $T$ is, the spectrum contains negative eigenvalues for $m<n$. Therefore, the stationary point is always a saddle point, with an unstable space of infinite dimension.

Note that for finite dimensional phase space, you have general theorems for the existence of a small $T$ for which the action is minimised thanks to the convexity of the kinetic part, which is why I looked at examples in field theories.

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.