# Is there anything natural about the principle of "stationary action"?

In Taylor's classical mechanics, he derived Lagrange equations and showed that they are equivalent to Newton's second law. Then, it was obvious that Lagrange equations are similar to the Euler-Lagrange equation, meaning that solving them makes "something" stationary. As a result, he introduced the principle of stationary action. Most YouTube videos/blogs talk about this principle, under the name of the principle of least action, being a natural thing that the nature tends to do, which I don't understand. Is there anything natural about this principle, or is it just a mathematical tool/model that happens to give the same results as Newton's second law?

• What is your definition of being natural? Possible duplicates: physics.stackexchange.com/q/15899/2451 , physics.stackexchange.com/q/9/2451 and links therein. May 30, 2020 at 12:41
• As the Newton second law shows, the "next time variables" x(t+dt) and v(t+dt) are determined with the "now variables" x(t) and v(t): v(t+dt) =F(t,x(t),v(t))dt/m, etc. There is no other "special" way to make a trajectory in the space. May 30, 2020 at 12:50
• You might learn in your Quantum Mechanics course (e.g. Feynman v III) about Dirac's 1933 derivation of the principle in the emergence of classical mechanics out of quantum mechanics, the true underlying theory of nature. In some sense, Newton's laws are a consequence of that principle, as it emerges from quantum mechanics. Just hold your breath... May 30, 2020 at 14:15
• @Qmechanic I take it that Karim Mohie is asking whether stationary action is purely a mathematical tool or that it has a physical interpretation. Phrased differently, does stationary action exist only in the realm of mathematical tools, or does it exist in Nature too? For comparison: in electric engineering, using complex numbers in circuit analysis. Instead of manipulating sines and cosines the equations are expressed in terms of exponentials. To that end in the equations imaginary current and imaginary voltage are introduced. Mathematical tool; no physical counterpart. Jun 1, 2020 at 5:06

I want to get some things out of the way, but feel free to jump to the section 'What the action is' as that section is the business end of this answer.

Hamilton's stationary action can be explained in terms of Newton's second law. Then again: in itself that does not necessarily say much: in any logical system there is great freedom to interchange axiom and theorem. (This interderivability is also discussed by SE contributor knzou in an answer about derivability in physics, answer posted Jan. 16, 2020)

Classical mechanics has various ways of expressing the physics taking place, and as we know they are interderivable. They are mathematically equivalent, so on that basis alone we don't have the means to judge which way of expressing the physics is the more fundamental.

The reason the work-energy theorem is referred to as a theorem (rather than principle) is that Newton's second law and the work-energy theorem are mathematically equivalent (as in: it's not another principle).

The second thing to get out of the way:
It is sometimes suggested that the concept of stationary action is accounted for in a deeper way in terms of quantum mechanics. However, as we know: when quantum mechanics was constructed the only way to make progress was to try only ideas that give formula's such that at macroscopic scale quantum mechanics reproduces classical mechanics. In order to make progress any trial formula that was found to lead to violation of the work-energy theorem was dropped. Ultimately, to suggest that quantum mechanics explains the work-energy theorem is circular reasoning.

What the action is

In the image below the curves represent the case of an object being thrown upwards against a potential that gives a 2 $$m/s^2$$ acceleration: it rises and descends again.

Black curve: trajectory of the object
red curve: kinetic energy
green curve: minus potential energy

When you sweep out a range of trial trajectories the curves of the energies change accordingly. (Further down is an animated GIF, 7 frames, 3 seconds per frame, that displays the process of sweeping out variation.)

The image displays the point in variation space where the trial trajectory hits the true trajectory: then the red curve (kinetic energy) and the green curve (minus potential energy) are parallel to each other.

So: what you need is a method to know where the slopes of the red/green curves are not equal, and where they are.

To that end we use the following property of the value of an integral: when the slope of a curve changes the value of that curve's integral changes in proportion. (The necessary and sufficient mathematical condition for that property is that the start point and the end point of the curve are fixed.)

In the image below the curves on the left are the same as the curves of the first image. The curves on the right are the corresponding functionals.

In the example presented here the potential increases linear with height. The green line on the right shows how the integral of the minus potential energy responds to variation of the trial trajectory. In this case the green functional is a straight line because in this case the potential energy increases linear with height. The red functional represents how the integral of the kinetic energy responds to variation of the trial trajectory. Since the kinetic energy is a quadratic function of the velocity the red functional is a quadratic function of the variational parameter.

There is only one point where the respective values of the two integrals change at the same rate (changing in opposite direction). That point is the point where the trial trajectory hits the true trajectory.

We see that 'stationary action' is purely a mathematical tool. The physical property that it corresponds to is the work-energy theorem: at every moment in time the rate of change of kinetic energy must be equal to the rate of change of minus potential energy.

Another way of seeing that 'stationary action' is purely a mathematical tool: the extremum that you find arises from the following two mathematical inputs that have no physical counterpart: the constraint that the start point and the end point are fixed, and the operation of integration.

This particular example was for the simplest case: a potential that increases linear with height. The reasoning generalizes to all cases.

An animated GIF, showing how the curves representing the energies respond to variation of the trial trajectory. The slider changes the variational parameter.

(The images illustrating this answer are my own work, they're from my own website.)

• Wait. I'm a beginner at this, but doesn't Cleonis's answer help provide a physical explanation? If the differentiation of the integral of the Lagranian with respect to epsilon is zero, then aren't we saying, in some sense, that the rate of change of the kinetic energy must equal the rate of change of the potential energy. To me, that's a logical physical interpretation. Jun 24, 2022 at 15:50
• @StevenDenenberg See also: new discussion (Oct. 2021) of Hamilton's stationary action, with new diagrams. The following is presented: derivation of Hamilton's stationary action from $F=ma$. First stage: derivation of the Work-Energy theorem from $F=ma$. Next: demonstration that in cases where work-energy theorem holds good Hamilton's stationary action will hold good also. The animated gifs are composited from screenshots of interactive diagrams that are available on my personal website. (Link to my website: on my stackexchange profile page.) Jun 25, 2022 at 4:22