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In his book on Classical Mechanics, Prof. Feynman asserts that it just does. But if this is really what happens (& if the Principle of Least Action is more fundamental than Newton's Laws), then don't we run into some severe problems regarding causality? In Newtonian Mechanics, a particle's position right now is a result of all the forces that acted on it in the past. It's entirely deterministic in the sense that given position & velocity right now, I can predict the future using Newton's laws. But the principle of least action seems to reframe the question by saying that if the particle ends up in some arbitrary position, then it would take a certain path (namely one minimises the action). But that means that the particle already knows where it'll be and it "naturally" takes the path that minimises the action.

Is there any deeper reason for why this is true? In fact principle of least action seems so arbitrary that it's hard to see why it manages to replicate Newton's Equations? If any of you have any insight into this, please share because I just cannot get my head around it.

Note - Please keep in mind, my question is regarding the principle itself, not the equations that result from that principle.

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Seems like you're essentially asking Why the Principle of Least Action? – ACuriousMind Mar 26 at 23:26
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Essentially a duplicate of physics.stackexchange.com/q/38348/2451 , physics.stackexchange.com/q/161586/2451 and links therein. – Qmechanic Mar 26 at 23:35

The particle doesn't 'know' anything. The principle of least action is used when we already know the endpoints of the path, and we want to find out how the particle got from the initial to the final position. We need to specify the final position in advance.

On the face of it, this makes it look like least action can't make predictions about the future. However, that's not a problem because given a least action principle, you can instantly derive a local differential equation, called the Euler-Lagrange equation, that holds at every point for every legal path. In almost all practical calculations, we work with Euler-Lagrange, not the full action.

Maybe Feynman neglected to mention this because he created the path integral formulation of quantum mechanics, in which the particle actually is smart, and does 'sniff out' nearby paths. (There are ways to extract local equations from the path integral formalism too, such as the Schwinger-Dyson equations. But from what I've read of Feynman, he really liked the global picture with a mysteriously smart particle.)

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The interesting thing is, though we frame the development in terms of known boundary conditions the result you get out is completely local. You don't need to know the ends of the path to use Lagrange's equation because it simply supplies you with a set of equations of motion just as Newton's Laws do. It helps to know that Lagrange's equation can be arrived at without Hamilton's principle. Indeed Hamilton enunciated his principle many years after Lagrange arrived at his equation. – dmckee Mar 27 at 2:01

Imagine you have an initial position.

Then there are lots of different possible initial velocities and those different velocities might have you end up in different places. So a different initial velocity might give you a different final position.

So instead of describing the different initial velocities you could instead describe the different final positions.

That's what the principle of extremal action does. Instead of fixing the initial velocity it fixes the final position. In the end you get an equation of motion, and you never even had to say what the final position was. And so as long as there was a final position (i.e. the particle doesn't cease to exist before $t=t_f$) then everything works out fine.

And not every potential plus initial position plus initial velocity gives a unique solution according to Newton's Laws. You could try to propose that as Newton's zeroeth law but firstly, that's historically misleading, and even so what do you restrict, certain initial positions, certain initial velocities, certain potentials?

But neither the principle of extremal action nor Newton's laws of motion require that there be a unique solution, they just predict the actual solution satisfies an equation.

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I like this answer because it shows that different questions can be asked about the same system and some formulation of the same laws are more suited in some cases. As a matter of fact, your answer recalls that the final position is known but not by the particle (presumably) but most likely by the person asking the question. And there is of course nothing wrong with that. – gatsu Mar 30 at 20:23

Do not take Feynman's metaphorical language at face value. There are neither classical "particles" nor classical "causality" in quantum theory, which presumably describes what "really" happens, both are artifacts of the classical description. And in classical description the only physically relevant fact is that classical trajectories have to obey classical laws, equivalently expressed in Newton's or least action form. The rest like "deterministic" or "particle knows" are just literary devices used to explain what those laws state. We can infer particle's possible future states from the current one from Newton's laws, but we can also solve Newton's equations in reversed time, and infer its possible past states instead. This does not "reframe" Newton's laws from causality to teleology or vice versa, nor are they "reframed" when we re-express them in the form of an extremal principle, see Disputes about possible teleological aspects.

And "the principle of least action" is a misnomer, the trajectory must be an extremal of least action, not necessarily minimize it (the difference is the same as between critical points and maxima/minima for functions). And the former unlike the latter is a local condition, i.e. the particle need not "know" its destination to extremize the action, which is why the trajectories can be computed locally from Newton's laws.

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I did not mean that computed trajectories have to be unique, only that extremal principle and Newton's laws constrain them equivalently. Here is a long post I wrote a while ago explaining that non-Lipschitz classical mechanics is indeterministic, and it was known already in 19th century. hsm.stackexchange.com/questions/2678/… – Conifold Mar 31 at 4:42

Imagine rainwater falling on a crinkly mountain. Each raindrop follows deterministic Newtonian mechanics, and follows the 'easiest' path downhill, with successive drops forming 'trajectories' and flowing into streams.

This is basically the Principle of Least Action. Each drop obeys local forces, but over a long interval of time - the time integral of Action is minimal, when compared with nearby alternative paths.

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Over a long enough time interval the action can actually be a saddle point, not a minimum. You might say the words "minimize the action" but really you just set the variation to zero, which means it could be a maximum or a saddle point. – Timaeus Mar 27 at 2:29

I'm fairly sure that Feynman's point has nothing to do with comparing Newton's Laws to Lagrange's equation or to the Hamiltonian equations of motion in the context of classical mechanics and everything to do with the ability of Lagrangian and Hamiltonian mechanics to be extended to cover field theories like those of Maxwell and of quantum mechanics without invoking new principles.

That is Lagrangian mechanics can describe in the same formalism and on the conceptual basis both Newtonian mechanics and Maxwell's Equations. The Lagrangians of fields use exactly the same formalism as those of discrete systems (though they have infinite dimensional phase spaces).

If the quote were simply about classical mechanics then any claim of priority for either version would have a hard time because you can arrive at Lagrange's equation starting with nothing more than Newton's Laws (that is without Hamilton's Principle, which is how Lagrange did it; see Goldstein for instance) or at Newton's Laws starting from Lagrange's equation. Thus, the implication of the quote has to be understood in a wider context.

Finally, the application of Lagrange's equation absolutely does not require any foreknowledge at all. The equation is completely local, just like Newton's Laws. The framing in terms of paths minimizing the action simply implies that the rule continues to be applicable as time passes.

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protected by Qmechanic Mar 27 at 0:44

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