22
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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, 2014 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, 2021 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, 2021 at 17:59

4 Answers 4

15
$\begingroup$

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:

  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.

$\endgroup$
2
  • 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, 2014 at 16:45
  • $\begingroup$ See also this related Phys.SE post. $\endgroup$
    – Qmechanic
    Jun 8, 2016 at 15:53
9
$\begingroup$

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.
$\endgroup$
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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, 2014 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, 2014 at 15:46
0
$\begingroup$

I'm aware this question was submitted in 2014, so it's quite likely that in the intervening years you have found the answer to it.

But just in case I'm submitting this answer anyway.

The background of the answer I'm submitting here is the exposition of Hamilton's stationary action that I submitted on physics.SE in october 2021.

In this answer I will discuss in which cases the true trajectory corresponds to a minimum of Hamilton's action, and in which cases the true trajectory corresponds to a maximum of Hamilton's action. I will also discuss which case is the critical case that is at the cusp of the flip from minimum to maximum


I will discuss the simplified case of motion in one spatial dimension; generalization to 3 spatial dimensions is straightforward.

I will in sequence discuss the following three cases:

  • the motion is subject to a uniform force, hence the potential increases linear with displacement
  • the motion is subject to a force that increases linear with displacement, hence the potential increases with the square of the displacement
  • the motion is subject to a force that increases quadratic with displacement, hence the potential increases with the cube of displacement

The variation of the trial trajectory is variation of the position coordinate.

As variation of the trial trajectory is applied the evaluation compares the response of the kinetic energy to the response of the potential energy.

The rate of change of the trial trajectory propagates to the velocity along the trial trajectory. As we know: the expresssion for kinetic energy proportional to the square of the velocity. Therefore the response of the kinetic energy to variaton is in all cases a quadratic function.


Potential increases linear with displacement

When the potential energy increases linear with displacement we have that the response of the potential energy to variation of the trial trajectory is linear.

Because of that: when the potential energy increases linear with displacement the true trajectory corresponds to a minimum of Hamilton's action.


Potential increases quadratic with displacement

As we know: in the idealized case of a force that increases in exact proportion to the displacement (perfect Hooke's law) the solution to the equation of motion is harmonic oscillation.

As we know: idealized harmonic oscillation has the following property: the period of oscillation is independent of the amplitude. Phrased differently: given a particular quadratic potential every amplitude of oscillation has the same period of oscillation.

Hamilton's stationary action reproduces the above amplitude property.

Evaluate a quadratic potential over a time interval that is equal to half a period of a full oscillation. So: if the period of oscillation is $2\pi$ seconds, then evaluate from $t=0$ to $t=\pi$ In that case, with the time interval equal to half a period, setting the starting position coordinate to zero means the ending position coordinate will be zero.

The evaluation then comes out as follows: for every amplitude of oscillation Hamilton's action evaluates to zero.

Hamilton's action evaluates to zero because in the case of harmonic oscillation both the kinetic energy and the potential energy respond quadratically to variation of the trial trajectory; the responses are equal.

Summerizing:
In the case of Hooke's law, evaluated for a time interval that is half a period of oscillation (or any integer multiple of that) we have that Hamilton's action evaluates to zero for every amplitude of oscillation.


It is a fundamental property of idealized harmonic oscillation that period of oscillation is independent of the amplitude. Hamilton's action complies with that: for every amplitude of oscillation Hamilton's action evaluates to zero

Therefore:
In order to calculate an actual amplitude of oscillation for some specific case an additional initial condition must be supplied. With just the boundary conditions the problem is underdetermined.


Potential increases in proportion to the cube of displacement

Let me make a comparison. In the physics of damping there is a natural subdivision in underdamping, critical damping, and overdamping. The cases of linear potential, quadratic potential, and potential-proportional-to-cube-of-displacement fall in an analogous subdivision.

Critical case:
Both the kinetic energy and the potential energy respond quadratically to variation of the trial trajectory

Under-critical:
Whenever the response of the potential energy to variation of the trial trajectory is of lower order than quadratic the true trajectory corresponds to a minimum of Hamilton's action.

Over-critical:
Whenever the response of the potential energy to variation of the trial trajectory is of higher order than quadratic the true trajectory corresponds to a maximum of Hamilton's action.

For emphasis let me repeat that:
There are classes of cases where the true trajectory corresponds to a maximum of Hamilton's action.


Note for instance that the Euler-Lagrange equation is agnostic as to whether the trajectory that it identifies corresponds to a minimum or a maximum of Hamilton's action. Whether Hamilton's action is a minimum or a maximum is not relevant. The only relevant property is that you are identifying the point where Hamilton's action is stationary.



There is a twist, however, in the over-critical case.
If the potential energy as a function of position is proportional to the cube (or higher) the evaluation of Hamilton's action must be performed over a sufficiently long time interval.

To see why, draw in the same diagram a quadratic function $f(x)=x^2$ and a cubic function $g(x)=x^3$. Given enough distance along the horizontal axis the cubic function will always outrun the quadratic function, but the cubic function is at the start slower.

So there is a scaling issue. Below some scale the quadratic function is steeper than the cubic function; for the cubic function to win the day the evaluation must extend over a sufficiently long scale.

Because of that scaling issue: when narrowed down to a sufficiently narrow specific interval of time Hamilton's action will be a minimum, even with a higher-than-quadratic potential.

My assessment is that Landau noticed this behavior in the mathematics, and he subsequently decided to assert it as a property, be it without supplying an explanation. I asssume that if Landau would have known the explanation he would have given the explanation.



'Stationary action' versus 'least action'

As we know, in a large majority of practical cases the response of potential energy to variation of the trial trajectory is of lower order than quadratic. The two biggies, gravity and the Coulomb force are inverse square force laws.

The critical case, harmonic oscillation, is ubiquitous too, of course. (Then again, it is the idealized case that is critical, and I'm not sure whether in classical mechanics there are instances where an oscillation is physically the idealized harmonic oscillation.)

Cases where the potential increases with the cube of displacement (or higher order) are rare, but they do exist.

The flip from minimum to maximum as you move from under-critical to over-critical demonstrates that fundamentally the criterium is to identify the point of stationary action.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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