Say the world is governed by the Principle of Least Action (or Hamiltonian mechanics) and let's not worry about quantum mechanics too much.

Independently of any Lagrangian or Hamiltonian, does that tell us anything about the world? If yes, what?

To put it differently, is it possible to falsify the Principle of Least Action? What kind of experimental results would do so?

  • $\begingroup$ Long before there were the notions of Langrangian or Hamiltonians, it was known that the path of light through a medium acted in such a way as to minimize the time it took to get from point A to point B. I believe Hamilton used such studies in optics to motivate his own development of his principles. $\endgroup$
    – JQK
    Commented Jan 12, 2019 at 17:15
  • $\begingroup$ Given a particle with initial and final positions and velocities, there are many path connecting them. Least action is a constraint that picks out the path where $F = ma$ at each point along the path. See damtp.cam.ac.uk/user/tong/dynamics.html for a quick proof. $\endgroup$
    – mmesser314
    Commented Jan 12, 2019 at 19:52
  • $\begingroup$ It is always worth considering that The Euler-Lagrange equations can be obtained without the the least action assumption by applying the principle of virtual work. $\endgroup$ Commented Jan 12, 2019 at 23:53
  • $\begingroup$ @dmckee: In Wikipedia I read: "Suppose the force $F(r(t)+\epsilon h(t))$ is the same as $F(r(t))$." I do not understand this supposition. The gradient force term is also proportional to epsilon and makes a contribution. $\endgroup$ Commented Jan 13, 2019 at 10:35
  • $\begingroup$ @VladimirKalitvianski I have yet to find an explanation of the principle of virtual work that I like; I have not pursed the one in wikipedia, but I find them all baroque. $\endgroup$ Commented Jan 13, 2019 at 18:51

3 Answers 3


OP's question seems to be essentially a version of the inverse problem for Lagrangian mechanics, i.e. given a set of EOM$^1$ $$E_i(t)~\approx~ 0,\tag{1}$$ does there exist (or not) an action $S[q]$ such that the EOM (1) are the Euler-Lagrange (EL) equations $$\frac{\delta S}{\delta q^i(t)}~\approx~ 0,\tag{2}$$ possibly after rearrangements? This is in general an open problem. See however Douglas' theorem and the Helmholtz conditions mentioned on the Wikipedia page.

Physically, in the affirmative case, there is a functional Maxwell relation $$\frac{\delta E_i(t)}{\delta q^j(t^{\prime})}~=~\frac{\delta E_j(t^{\prime})}{\delta q^i(t)}. \tag{3}$$ See also e.g. this related Phys.SE post.


$^1$ The $\approx$ symbol means an on-shell equality, i.e. equality modulo EOM.

  • $\begingroup$ Interesting. So in general the problem is open, but how far have we progressed? There are still instances for which we know that the equations of motion cannot arise from an action, right? For example, motion with friction. $\endgroup$
    – user140255
    Commented Jan 14, 2019 at 4:38
  • $\begingroup$ Even friction is subtle, cf. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Commented Jan 14, 2019 at 9:46

Physically we have the Newton equations of the second order and two initial conditions. This helps us predict the future positions and velocities.

Unfortunately, the least action principle uses initial and final positions as if they were "known" physically. Thus the physical meaning is quite small and it is actually reduced to the statement that we deal with second order differential equations. No unique evolution is predicted by the equations themselves since they admit two-parametric liberty: the initial position and velocity. The "future variables" (i.e., taken at $t+dt$) are actually determined by the present: $p(t+dt)=p(t) + F(t)dt$, $x(t+dt)=x(t)+v(t)dt$. By the way, these relationships encode the so called "causality".

Falsifying happens when we understand and see the approximative character of these laws. There are some inequalities that make our equations "work". Outside these inequalities the equations do not work. For example, an approximation of point-like particles for finite size bodies, etc., etc.

  • $\begingroup$ Thank you, you bring elements of an answer. To your last paragraph : what I mean by falsifying could be to observe a trajectory from something as close as possible to an ideal point-like particle, that could still not be obtained from any Lagrangian. $\endgroup$
    – user140255
    Commented Jan 14, 2019 at 4:50
  • $\begingroup$ @Undead: The notion of a point-like particle itself stems from an inclusive picture of premanently developping complicated process depending on the observer too. $\endgroup$ Commented Jan 14, 2019 at 5:37

It would be big news the principle of least action was experimentally falsified. This, though, is unlikely to happen.

Rather we should expect something similar to the principle of least action to always hold.

Recall, that classical mechanics is deterministic and hence specifying that initial conditions we find a determined trajectory through which the system moves.

This can be alternatively phrased as saying there is a space of possible trajectories that a system can move in and specifying the initial condition means we pick out a specific trajectory.

Now, calculus allows us to pick out specific points on a curve where the gradient is at a minimum, these are the stationary points.

Thus we ought to expect that we can find a functional on the space of trajectories whose stationary point is the desired trajectory. This functional is the Lagrangian.

The physical content is that the Lagrangian is built simply from the difference of the kinetic and potential energy of the system and thus takes a very simple form.

It's probably worth pointing out that the principle is of archaic origin: Hero of Alexandria pointed out in 100AD that light reflecting from a mirror follows the path of least distance, this was rephrased by Damianus in 200AD as the path of least time and then much much later, Fermat generalised this in 1657 to any motion of light follows a path of least time.

  • $\begingroup$ So it means that the principle of least action is at least built from the assumption that classical mechanics is deterministic, which really is a physical proposition. If I was to find an intrinsically non deterministic trajectory, the principle would be "falsified". Maybe it wasn't 100% clear, but I'm asking about other assumptions hidden or built in the principle of leat action $\endgroup$
    – user140255
    Commented Jan 20, 2019 at 17:13
  • $\begingroup$ @undead: There are already non-deterministic paths in physics, ie in QM; but this doesn’t falsify what I wrote. You’re right though that what you wrote wasn’t particularly clear. At least you have that much right. $\endgroup$ Commented Jan 22, 2019 at 14:08
  • $\begingroup$ No need to get rude man. Take this as a definition of falsify : en.wikipedia.org/wiki/Falsifiability. The observation of non-deterministic trajectories in a quantum setup certainly falsifies the principle of least action (without entering in the Bohmian interpretation). Of course that doesn't mean the principle of least action is completely useless or not approximately true. Now I'm just asking about something else than non-determinism that would violate the principle. Qmechanic gave a reasonable enough answer, so I'll contempt myself with that I think. $\endgroup$
    – user140255
    Commented Jan 22, 2019 at 16:43
  • $\begingroup$ @undead: That’s just silly. Check out Feynmans path integral to see a use of least action methods in physics. $\endgroup$ Commented Jan 23, 2019 at 7:05
  • $\begingroup$ @undead: There are differing standards of rudeness, and as far as I am concerned a bunch of jargon thrown together is much worse than being rude. $\endgroup$ Commented Jan 23, 2019 at 7:08

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