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The action

$$S=\int L \;\mathrm{d}t$$

is an important physical quantity. But can it be understood more intuitively? The Hamiltonian corresponds to the energy, whereas the action has dimension of energy × time, the same as angular momentum.

I've heard the action being described as a measure of change, although I don't know how this description can be justified.

Qmechanic
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Emerson
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    This is a strange stream of consciousness, as Stephen Hawking would say, http://motls.blogspot.com/2005/10/briefer-history-of-time.html - the statement that the Hamiltonian corresponds to energy is a vacuous tautology because the Hamiltonian is a technical synonym for energy. In the same way, you may say that the action intuitively corresponds to Wirkung because it's the same thing, too, and blame the energy for having "unnatural" units of action per unit time. Well, energy is intuitive because it's conserved, and the action is intuitive because it's minimized - what's the difference? – Luboš Motl May 10 '11 at 19:27
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    Of course, the only difference is that non-physicists don't learn to use the action at all. If the laymen's experience with a concept measures whether something is "intuitive", then the action simply is less intuitive and there is no reason to pretend otherwise. However, physicists learn that it's in some sense more fundamental than the energy. Well, the Hamiltonian is the key formula defining time evolution in the Hamiltonian picture while the action is the key formula to define the evolution in the nicer, covariant, "spacetime" picture, which is why HEP physicists use it all the time. – Luboš Motl May 10 '11 at 19:30
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    Otherwise, the main raison d'etre for the action is the principle of least action, http://en.wikipedia.org/wiki/Principle_of_least_action , which is what everyone should learn if he wants to know anything about the action itself. It makes no sense to learn about a quantity without learning about the defining "application" that makes it important in physics. Energy is defined so that it's conserved whenever the laws of Nature are time-translational symmetric; and action is defined as whatever is minimized by the history that the system ultimately takes to obey the same laws. – Luboš Motl May 10 '11 at 19:33
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    @Luboš I think your three comments could have made an answer, even though part of it challenges the amorphous notion of "intuition", I think this an important lesson that should be among the posted answers. – Approximist May 10 '11 at 23:09
  • OK, I will strip the "dialog" form and promote it to an answer, @Approximist. – Luboš Motl May 11 '11 at 09:13
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    Some comments on the principle of least action as it relates to Feynman - quite an interesting and very short read! – Cicero Jul 02 '15 at 05:07
  • There is the Hamilton's principle and the assosiated function, the Hamilton principal function. It's just a new mathematical representation for the motion of a dynamical system. – Hulkster Jan 30 '22 at 03:46

5 Answers5

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The action $S$ is not a well-known object for the laymen; however, when one seriously works as a physicist, it becomes as important and natural as the energy $H$. So the action is probably unintuitive for the inexperienced users - and there's no reason to hide it - but it is important for professional physicists, especially in particle and theoretical physics.

The OP's statement that the Hamiltonian corresponds to energy is a vacuous tautology because the Hamiltonian is a technical synonym for energy. In the same way, one might say that the action intuitively corresponds to Wirkung (a German name) because it's the same thing, too. Because it now has two names, it becomes more natural :-) and the OP could also blame the energy for having "unnatural" units of action per unit time. In other words, the question assumes that energy (and its unit) is more fundamental and intuitive than the action (and its unit) - so it shouldn't be surprising that using his assumptions, the OP may also "deduce" the conclusion that the energy is more fundamental and intuitive than the action. ;-)

But is the assumption = conclusion right? Well, energy is intuitive because it's conserved, and the action is intuitive because it's minimized - so there is no qualitative difference in their importance.

Of course, the only difference is that non-physicists don't learn to use the action at all. The energy may be imagined as "potatoes" which every can do; the action is an abstract score on the history that is only useful once we start to derive differential equations out of it - which almost no layman can imagine. If the laymen's experience with a concept measures whether something is "intuitive", then the action simply is less intuitive and there is no reason to pretend otherwise. However, physicists learn that it's in some sense more fundamental than the energy.

Well, the Hamiltonian is the key formula defining time evolution in the Hamiltonian picture while the action is the key formula to define the evolution in the nicer, covariant, "spacetime" picture, which is why HEP physicists use it all the time.

What the action is in general

Otherwise, the main raison d'etre for the action is the principle of least action, which is what everyone should learn if they want to know anything about the action itself.

Historically, this principle - and the concept of action - generalized various rules for the light rays that minimize time to get somewhere, and so on. It makes no sense to learn about a quantity without learning about the defining "application" that makes it important in physics. Energy is defined so that it's conserved whenever the laws of Nature are time-translational symmetric; and action is defined as whatever is minimized by the history that the system ultimately takes to obey the same laws.

The energy is a property of a system at a fixed moment of time - and because it's usually conserved, it has the same values at all moments. On the other hand, the action is not associated with the state of a physical object; it is associated with a history.

There is one point I need to re-emphasize. For particular systems, there may exist particular "defining" formulae for the Hamiltonian or the action, such as $E=mv^2/2$ or $S = \int dt(mv^2/2-kx^2/2)$. However, they're not the most universal and valid definitions of the concepts. These formulae don't explain why they were chosen in this particular way, what they're good for, and how to generalize them in other systems. And one shouldn't be surprised that one may derive the right equations of motion out of these formulae for $H$ or $S$.

Instead, the energy is universally defined in such a way that it is conserved as a result of the time-translational symmetry; and the action is defined in such a way that the condition $\delta S = 0$ (stationarity of the action) is equivalent to the equations of motion. These are the general conditions that define the concepts in general and that make them important; particular formulae for the energy or action are just particular applications of the general rules.

In the text above, I was talking about classical i.e. non-quantum physics. In quantum physics, the action doesn't pick the only allowed history; instead, one calculates the probability amplitudes as sums over all histories weighted by $\exp(iS/\hbar)$ which may be easily seen to reduce to the classical predictions in the classical limit. A stationary action of a history means that the nearby histories have a similar phase and they constructively interfere with each other, making the classically allowed history more important than others.

hft
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Luboš Motl
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    This answer is really long and spends most of the time explaining why “action is as intuitive as energy, you just don't know it” and “energy is defined as something that is conserved” but doesn't introduce any useful intuition about what action is. I mean – I know how action is defined and that it's stationary, I came here for the intuition. And didn't get any. – m93a Jan 20 '19 at 14:33
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    Too bad to hear. There's probably no intuition in the sense that you're looking for, as some picture with teddy bears for children in the kindergarten. The action is something that allows the equations of motion to be derived from the elegant condition that the action is minimized. One should go over the examples why it works and through the derivations of the Euler-Lagrange equations from the action. That's what matters and it's all the intuition. Objects evolve so that something that may be calculated is minimized. That's it. There are no teddy bears hiding behind what I just wrote. – Luboš Motl Jan 20 '19 at 17:57
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    The first examples could include the motion along the straight path. Objects move on straight lines and what could be minimized is the length of the path - and the action is somewhat similar to it although not identical in general. Then one may derive the laws of reflection and Snell's law of refraction - paths of photons - from minimizing or maximizing other things. The fun is that such a quantity exists whenever the equations may also be derived from the energy/Hamiltonian. The action is a mathematically defined object and it's unreasonable to try to think about it non-mathematically. – Luboš Motl Jan 20 '19 at 18:00
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    You can say that of all physics. Nevertheless, it is generally the case that greater understanding can be achieved by appeal to examples, existing concepts, and, yes, intuition; and furthermore that most large advances in physics (quantum mechanics being the odd one out) started not with playing with the math, but with a purely conceptual insight. – linkhyrule5 Oct 23 '19 at 05:46
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    Any intuition that is any valuable is equivalent to a mathematical idea or argument. It is your mathematico-phobia that will prevent you from getting these things. The maximization or minimization of anything is a mathematical operation, and without that, the action is completely devoid of meaning in classical physics. Of course I can say roughly the same thing about all of physics - because it is true in all of physics. The laws of physics can only be defined in the language of mathematics. One may have at most different levels of precision and rigor of that mathematics but it's maths. – Luboš Motl Oct 24 '19 at 06:25
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    I also had the same question recently. What is the intuitive meaning of action and was even more motivated towards finding an interpretation from the observation that uncertainty principle can be stated as "minimum action required to make an observation is $\hbar/2$". This followed from dimensional analysis. Given that all the natural dynamics is hidden in the principle of least action and path integral formulation, I suppose there is more to action then we presently understand. – prateek Mar 13 '21 at 12:47
  • Yes, it is totally possible that there is even more to it - assuming that the formula that determines all the evolution of everything in time isn't enough LOL. The minimum action to make an observation is a fun concept. I intuitively agree that something like that should be right, some minimum change is needed... But the additive shift of the action is kind of unphysical so I am not so sure that even "S=0" is too special and well-defined in general. – Luboš Motl Mar 14 '21 at 15:22
  • Fitch, speaking of the definition of continuity, noted,

    ”Mathematical continuity, in the technical sense, is a precisely definable mathematical notion which refers to certain properties of numbers and number sequences. The continuity of the physi- cal world, on the other hand, is rather different from mathematical continuity, since it is a directly experienced attribute of nature and does not require, for being understood, any mathematical theory of properties of numbers.” - F. B. Fitch. Physical Continuity. Philosophy of Science, 3(4):486–493, October 1936.

    – ernie Dec 31 '23 at 23:14
  • Ok, this is precisely an example of a philosopher who doesn't have the slightest idea about physics, mathematics, or their relationship. First, nothing conceptual about the real world can be read from the experiments directly. Physics is a science where one formulates and tests hypotheses. Continuity and discreteness are competing hypotheses. Second, all hypotheses in physics must be formulated in the language of mathematics. And that is why when a mathematical notion like continuity is used, it has the same meaning as in mathematics. – Luboš Motl Jan 02 '24 at 02:40
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Here comes an attempt at an intuitive analogy:

You must have heard the classic "the shortest distance between two points is a straight line" , in euclidean geometry.

The action integral is analogous to distance in the generalized coordinates of the system under study. Its minimization gives the shortest path in configuration space going from time t1 to time t2 and is called the principle of least action. Applying it leads directly to Lagrange's equations, from which the equations of motion of the system are derived.

Chapter 2 of H. Goldstein's "classical mechanics" (available on the web) is a good introduction to this and to variational principles.

anna v
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    If $\int L dt$ is analogous to distance (in generalized coordinates), does that mean $L$ is in some sense analogous to a speed (in generalized coordinates)? – Museful Jul 17 '22 at 15:08
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    @Museful the analogy should not be carried too far – anna v Jul 17 '22 at 18:01
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As a layman, I can't offer too much, but I can offer how I think about action.

$$S=\int L \;\mathrm{d}t$$

The key distinction in classical mechanics between the Hamiltonian and the Lagrangian is that Hamiltonian (H) is the sum of kinetic energy (T) and potential energy (V), where as the Lagrangian (L) is the difference:

$$H=T+V$$

$$L=T-V$$

In the case of simple harmonic motion (a mass, spring system for instance) we can use the following equations for the different types of energy:

$$T=mv^2/2$$

$$V=kx^2/2$$

I would refer you to most elementary texts, but the solution for position as a function of time for the simple harmonic system described is:

$$x=A \cos({t\sqrt{k/m} + \phi})$$

and for velocity:

$$v=-A(\sqrt{k/m}) \sin({t\sqrt{k/m} + \phi})$$

Setting $A = 2$, and $m=k=1$ we can plot the following graph over time:

Hamiltonian vs Lagrangian

In this graph, the highlighted area, the area under the curve plotted by the lagrangian, is the action. The area under the curve depends on the interval of interest (however, for integer multiples of the wave period, and for infinite time, the action would equal zero [see note]). In any case, this should give an intuitive notion for action, and for the concept of a functional...the area under the curve for some interval is a number determined by the integral of a function:

$$S=\int L \;\mathrm{d}t$$

When it comes to the least action principle (as discussed by others), we set:

$$\delta S = 0$$

Which is done by considering the effects of a infinitesimal change in position $\epsilon$ so that $x$ becomes $x+\epsilon$ (I will refer you to chapter 2 of Quantum Field Theory Demystified for a good discussion on this).

In any case, the objective is that the equation that one desires (newton's equation of motion, or what I like to think of as a force equation) is one that satisifies the $\delta S = 0$ constraint, which in this example, has the effect of keeping the value of the action (the area under the curve) constant.

I hope that helps, I will edit for more content if needed.

[Note: I would add that if one is clever, they can make the half cycle infinitely long, which would make the action non-zero in the case of a SHO]

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    This is a nice compliment to the others answers but I feel it necessary to point out that the book you cite (QFT Demystified) contains a huge number of errors and in many sections over-simplifies the theory. It should not be considered as a suitable introductory text to the subject, or even one that should be trusted. – qftme May 12 '11 at 10:58
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    Thanks for the nice comment. As far as the book, I would agree that its not a textbook, and I think the author makes a good show of saying that up front, however, for me it was one of the books that was what I consider a "connector" from the world of popular "noise" to the world of QFT, so it will always have a high place in my personal library despite its admitted flaws. – Unassuminglymeek May 17 '11 at 23:22
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But can it be understood more intuitively?

As other answers have pointed out, the action is a fundamental quantity, just as fundamental as energy. "An intuitive explanation" just means we can find something similar in our ordinary lives to compare it to.

Imagine asking for an intuitive explanation of energy. Energy could be defined in dynamics as just that thing which is conserved. But why do we accept this as intuitive? In daily life we encounter situations where quantities are conserved, or the exchange of one quantity/material into another type of quantity/material happens at a fixed conversion rate.

So how to define the action? It is just that thing which is minimized in a path over time. What type of life situation is like this? Well people in general like pleasure and do not like pain. So when trying to determine the course of action to take they will minimize: $$S=\int (\text{Pain}-\text{Pleasure})dt$$

Or, organisms like to conserve "potential energy" in the form of fat and avoid "kinetic energy" in the form of exercise (they are lazy). Sometimes it might be worth it to spend some kinetic energy to get more potential energy.

$$S=\int (\text{Movement}-\text{Stored Food})dt = \int (\text{Potential}-\text{Kinetic})dt$$

Jbag1212
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Action is a function of the upper limit of integration (if $L(t)$ is specified), so it grows from zero when $t_2$ goes farther and farther from $t_1$. In this sense, action is a "measure of change".

But normally we do not calculate the action values with solutions inserted in $L$. Action is a functional of possible trajectories before varying $q(t)$ and $\dot{q} (t)$. Before varying, $q(t)$ and $\dot{q} (t)$ are arbitrary, not specific solutions of equations.

An intuitive understanding of action might be the following: on real trajectories its growth rate might be minimal but it is not the case, unfortunately. Instead, we have the least action requirement: $\delta S = 0$ between two fixed points.

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    It should probably be stressed that $q$ and $\dot{q}$ are not independent variables in the action $S[q]$. – Qmechanic May 10 '11 at 22:06