Edit: The previous title didn't really ask the same thing as the question (sorry about that), so I've changed it. To clarify, I understand that the action isn't always a minimum. My questions are in the points 1. and 2. below.

I understand that "principle of least action" is somewhat of a misnomer, since we find that to determine the path taken by a system, we need only impose the condition that the action be stationary, i.e., that $\delta S$ should vanish to first order for small variations of the path, and this leads to the Euler-Lagrange equations.

In The Classical Theory of Fields, Landau & Lifshitz discuss the relativistic action for a free particle:

So for a free particle the action must have the form

$$S = -\alpha \int_a^b ds$$

(...) It is easy to see that $\alpha$ must be a positive quantity for all particles. In fact, as we saw [earlier], $\int_a^b ds$ has its maximum value along a straight world line; by integrating along a curved world line we can make the integral arbitrarily small. Thus the integral $\int_a^b ds$ with the positive sign cannot have a minimum; with the opposite sign it clearly has a minimum, along the straight world line.

There is also a footnote, addressed a couple of paragraphs earlier, but which is relevant:

Strictly speaking, the principle of least action asserts that the integral $S$ must be a minimum only for infinitesimal lengths of the path of integration. For paths of arbitrary length we can say only that $S$ must be an extremum, not necessarily a minimum.

I have two questions regarding this:

  1. How is the condition "the action must be a minimum for infinitesimal displacements" formulated? I've never heard of that outside of Landau & Lifshitz's books, and in Mechanics they mention it as well but doesn't go into detail. Is this discussed a bit more somewhere?

  2. If for the whole path the action only needs to be stationary, how can we make the argument for the negative sign? If the action had to be a minimum then it would make sense, but surely the fact that $\delta S$ = 0 isn't affected by an overall sign?

  • 1
    $\begingroup$ It is also possible to provide examples where the true minimum distance path doesn't solve the Euler-Lagrange problem. Take, for example, $\mathbb{R}^{2}$ with the unit circle and its interior removed, and try to extremize the path from $(-2,0)$ to $(2,0)$ $\endgroup$ Jun 30 '14 at 16:24
  • $\begingroup$ For posterity I'm leaving a link here to a numerical demonstration of the top two answers below (by Qmechanic and auxsvr). I solve for the motion of a simple harmonic oscillator in two ways: by integrating the Euler Lagrange ODE (typical) and by discretizing the action functional itself so it can be numerically minimized (think gradient descent, though I used SQP). $\endgroup$
    – jnez71
    Aug 6 '21 at 17:57
  • $\begingroup$ ^ These produce identical results when the simulation duration is less than the "characteristic time" that Qmechanic explained. Beyond that, the minimization approach yields nonsense, indicating that the physical solution always has stationary action but not always minimal action, even for very simple / fundamental systems. The (conjugate) point at which the minimization breaks is exactly the one theoretically calculated in auxsvr's answer. $\endgroup$
    – jnez71
    Aug 6 '21 at 17:59

Perhaps a simple example is in order. Consider a simple harmonic oscillator (SHO)

$$ S~=~\int_{t_i}^{t_f} \! dt~L, \qquad L~=~\frac{m}{2}\dot{x}^2 - \frac{k}{2}x^2, \tag{1}$$

with characteristic frequency

$$ \frac{2\pi}{T}~=~\omega~=~\sqrt{\frac{k}{m}}, \tag{2}$$

and Dirichlet boundary conditions

$$ x(t_i)~=~x_i \quad\text{and}\quad x(t_f)~=~x_f. \tag{3}$$

It can be shown that the classical path is only a minimum for the action (1) if the time period

$$ \Delta t~:=~t_f-t_i~\leq~ \frac{T}{2}\tag{4}$$

is smaller than a characteristic time scale $\frac{T}{2}$ of the problem. (If $\Delta t=\frac{T}{2}$ there is a zeromode.) For $\Delta t>\frac{T}{2}$ the classical path is no longer a minimum for the action (1), but only a saddle point. If we consider bigger and bigger $\Delta t$, a new negative mode/direction develops/appears each time $\Delta t$ crosses a multiple of $\frac{T}{2}$.

It is such examples that Ref. 1. has in mind when saying that the principle of least action is actually a principle of stationary action. The above phenomenon is quite general, and related to conjugated points/turning points and Morse theory. In semiclassical expansion of quantum mechanics, this behaviour affects the metaplectic correction/Maslov index. See e.g. Ref. 2 for further details.

A similar phenomenon takes place in geometrical optics, where it is straightforward to construct examples of light paths that do not minimize the time, cf. Fermat's principle of least time.


  1. Landau and Lifshitz, Vol.2, The Classical Theory of Fields, p. 24.

  2. W. Dittrich and M. Reuter, Classical and Quantum Dynamics, 1992, Chapter 3.

  • 1
    $\begingroup$ Concerning OP's second question (v2): In quantum mechanics, the overall sign of the action $S$ in the path integral is tied to unitarity, i.e. that the Hamiltonian should be bounded from below. For non-standard sign conventions, see e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Jul 2 '14 at 16:45
  • $\begingroup$ See also this related Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 8 '16 at 15:53

Your questions are answered in Calculus of variations, Gelfand, 2000, section 36.2. First we need a theorem:

The functional $S[x] = \int_a^b L(t,x,\dot{x}) dt$, $x(a) = A$, $x(b) = B$ must satisfy the following conditions in order to have a weak minimum for $x = x(t)$:

  1. The curve $x(t)$ satisfies the Euler-Lagrange equation, namely it is an extremal,
  2. $\partial_{\dot{x}}\partial_{\dot{x}} L |_{x(t)} >0$,
  3. The interval $[a,b]$ contains no points conjugate to $a$.

The definition of conjugate points is in p.114.

  1. An example that illustrates this is the harmonic oscillator, $$m\ddot{x} + kx = 0, \quad x(0)=0, \dot{x}(0) = 1$$ with solution $x(t) = \frac{1}{\omega}\sin(\omega t)$, $\omega \equiv \sqrt{\frac{k}{m}}$ and action $$S[x] = \frac{1}{2}\int_a^b m\dot{x}^2 - kx^2 dt.$$ Points $(t=\pi/\omega,x=0)$ and $(t=0,x=0)$ are conjugate, because every extremal starting from $x(0)=0$ intersects the aforementioned solution at $(\pi/\omega,0)$. The conditions of the previous theorem for a minimum are satisfied for $0\leq a \leq t < \pi/\omega$ and not for enlarged intervals.
  2. On page 161 Gelfand shows that for a vibrating string with fixed ends there is no time interval without a pair of conjugate points, therefore we cannot guarantee that the solution of the wave equation minimizes the action at all. Then, he states that this is the reason that we replace the principle of least action with the principle of stationary action for mechanical systems.

I think this might be some dope Landau sign convention because in principle we can set:

$$\int_a^b ds = \int_a^b \frac{ds}{dp}dp = \int_a^b \pm \sqrt{\pm\eta_{\mu \nu} \frac{d x^\mu}{dp}\frac{d x^\nu}{dp}}dp$$

Since we are meddling with the sign inside the square root based on whether investigating space-like/time-like curves $x^\mu (p)$, we can as well put a minus in front to denote we are working with a different kind of "length" than in normal space. In that case we would truly get a minimum with the factor $-\alpha$.

This whole "minimum of action" thing is more of a historical relict from natural philosophy and is true only for special Lagrangians. For example in gravitational lensing (i.e. null-geodesics in relativity), multiple images are obtained by multiple extremal paths of which at least one is a maximum (in the case of multiple images, for a single image, it is a minimum).

However, to practically investigate the minimum, you can expand the Lagrangian as in the usual derivation of Euler-Lagrange equations but to second order in the variation of the trajectory $\delta x^\mu$. To the first order you obtain the Euler-Lagrange equations which you must generally solve. The solution is then substituted into the second order expansion which causes the first order to vanish and you must then investigate the sign of the resulting expression.

  • $\begingroup$ I believe that it's more than a historical relict because it fixed the sign of the mass on the non-relativistic free particle Lagrangian, thus fixing the signs of everything that comes after it. $\endgroup$
    – Hydro Guy
    Jun 30 '14 at 15:36
  • 2
    $\begingroup$ Any kind of sign, or in fact, any kind of constant be it positive, negative or complex, multiplying the Lagrangian, does not change the actual physics it predicts. Just write down the Euler-Lagrange equations - the constant can always be cancelled out. As I mentioned already for the lensing, there might even not strictly be just one type of extremum for a given Lagrangian, so trying to settle the convention on either a minimum or maximum is just pointless. $\endgroup$
    – Void
    Jun 30 '14 at 15:46

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