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:
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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?
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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?
Best Answer
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.
References:
Landau and Lifshitz, Vol.2, The Classical Theory of Fields, p. 24.
W. Dittrich and M. Reuter, Classical and Quantum Dynamics, 1992, Chapter 3.