Is the principle of least action actually a principle of least action or just one of stationary action? I think I read in Landau/Lifschitz that there are some examples where the action of an actual physical example is not minimal but only stationary. Does anybody of know such a physical example?

  • 5
    $\begingroup$ It is a principle of stationary action. Perhaps more excitingly, it can also be thought of as a principle of XTREME ACTION! $\endgroup$
    – wsc
    Jun 24, 2013 at 22:53
  • $\begingroup$ thank you for your affirmation. still, i would be highly interested in seeing an example that actually shows that it is not in general a principle of least action $\endgroup$
    – Xin Wang
    Jun 24, 2013 at 22:59
  • $\begingroup$ In General Relativity I think that it's a maximum. $\endgroup$
    – jinawee
    Jun 24, 2013 at 22:59
  • 2
    $\begingroup$ Classically, one may always change the overall sign of the action without affecting the Euler-Lagrange eqs., thereby turning a local minimum into a local maximum. The Lagrangian $L(q)=\cos(q)$ has both local minimums and local maximums. $\endgroup$
    – Qmechanic
    Jun 24, 2013 at 23:07
  • 1
    $\begingroup$ @wsc That's not really true. Sometimes the classical trajectory corresponds to a saddle point of the action, so the action is stationary but not XTREMEized. $\endgroup$
    – tparker
    Feb 4, 2019 at 3:05

1 Answer 1


By convention we try to set things up so that it's usually a minimum, but we can't make any definition of the action that would make that hold in all cases.

In optics, consider a situation in two dimensions where you have an ellipsoidal cavity with reflecting walls. If you release a ray of light from the center, along the major axis, it gets reflected back to the center by following a path of maximum time. If you start a ray from the center, along the minor axis, it comes back after following a path of minimum time. You can choose the action to be the time or minus the time, but no matter what, one of these rays will be a minimum of the action and the other will be a maximum.

In special relativity, you can take the action for a particle to be the proper time $s=\int ds$ (with $ds$ positive), or you can take it to be $-s$ or other possibilities such as $-mcs$ (Landau and Lifschitz's choice). It doesn't matter which sign you choose, because the physical predictions are the same either way.

In both of these examples (optics and SR), you can make a choice of sign such that the action is minimized for infinitesimally short trajectories in free space. However, the example of the ellipsoidal cavity shows that you cannot in general make it a minimum for all paths of finite length.

In relativity, the metric only defines $ds$ in absolute value, $ds^2=g_{ab}dx^a dx^b$. Also, we'd like to be able to talk about timelike, lightlike, and spacelike geodesics. If we choose the action for timelike geodesics to be real, then it has to be imaginary for spacelike ones -- or we could define it as $\int\sqrt{|ds^2|}$, but the absolute value could be a nuisance because it isn't a smooth function.

For L&L's discussion of this, see Mechanics (3rd ed.), section 2; and Classical theory of fields (2nd ed.), sections 8, 53, and 87.

  • $\begingroup$ More to the point, a single classical trajectory is sometimes a saddle point of the action, so it isn't a local extremum at all. $\endgroup$
    – tparker
    Feb 4, 2019 at 3:07

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