# When is the principle of stationary action not the principle of least action?

I've only had a very brief introduction to Lagrangian mechanics. In a physics course I took last year, we briefly covered the principle of stationary action --- we looked at it, derived some equations of motion with it, and moved on.

While the lecturer often referred to it as the principle of least action, he always reminded us that it wasn't actually least action, but stationary action --- a minimum, maximum, or point of inflexion, rather than just a minimum. He never, however, gave an example of a system where we didn't seek the least action.

Why is it, then, the principle of stationary action, instead of least action? What is an example of a system which we would seek a maximum instead of a minimum?

• Stationary doesn't mean maximum instead of minimum — it means vanishing derivative. Example of such point is $x=0$ for $y=x^3$. Nov 2, 2014 at 10:37
• I'm aware of that. I guess I phrased this poorly then. If we're looking for a vanishing derivative, then when (in a physical situation, of course) does that happen if not at a minimum? Nov 2, 2014 at 10:39
• Possible duplicates: physics.stackexchange.com/q/907/2451 , physics.stackexchange.com/q/69077/2451 , physics.stackexchange.com/q/122486/2451 and links therein. Nov 2, 2014 at 10:40

If we solve equations of motion for a particle with mass $m=1$ in some potential, e.g. $U=x^4-4x^3+4.5x^2$, fixing two points like $x(0)=0$ and $x(1)=2.651$, we'll get infinite number of solutions, here're some of them:

They differ by initial velocity. Now each of them satisfies equations of motion, but only one makes the action minimal, here's how $S$ depends on $v_0$ for such paths in this example:

As all they satisfy equations of motion, they make the action stationary, and it'd be wrong to just throw them away because they don't minimize the action.

• This goes against most of my physics education. I was always taught that the solution of second order differential equations contain exactly 2 free parameters. Why is your solution not completely fixed once you have fixed $X(0)$ and $X(1)$? Apr 16, 2015 at 14:03
• @StevenMathey Because there's no uniqueness theorem for this type of problem. You can satisfy the boundary conditions by multiple couples of free parameters (e.g. different pairs of initial position and velocity). Cauchy problem, OTOH, does guarantee uniqueness of the solution provided some conditions are met. Apr 16, 2015 at 15:24
• Hm... That's a surprise. Did you have to fine tune the value of $x(1)$ to $2.651$ for this to happen or is this phenomena generic? What causes it? Does it have to do with your choice of potential? Is this feature visible in the Hamiltonian flow? Does it cause problems with causality? Apr 16, 2015 at 20:29
• @JakubSkórka See the discussion in the comments above. Boundary value problems don't in general have a unique solution (c.f. Sturm-Liouville problem). Lagrange equations only specify which paths will satisfy equations of motion, and there may be multiple such which also satisfy boundary conditions. Apr 14, 2017 at 6:36
• @StevenMathey For a counterexample to your claim "I was always taught that the solution of second order differential equations contain exactly 2 free parameters," just consider a quantum-mechanical particle in a box. The wavefunction satisfies the linear second-order ODE $-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi$ with fixed boundary values $\psi(0) = \psi(a) = 0$, but there are infinitely many solutions - the energy eigenstates. Feb 4, 2019 at 3:22

In optics, you can take the example of a concave mirror : the optical path chosen by the light to join two fixed points A and B is a maximum.

• for example in classical mechanics the stationary action is indeed minimum, whereas in GR the geodesic is found when the action is stationry but maximum (maximum time) Nov 2, 2014 at 11:43