70
$\begingroup$

I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$. Okay, fine. You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations. Great. But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action. After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all: it's the "Principle of Stationary Action". Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional. Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.

But the problems don't stop there. Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$. As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.

From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way. Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.

Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas?

$\endgroup$
  • 2
    $\begingroup$ possible duplicate of Hamilton's Principle $\endgroup$ – Mark Eichenlaub Oct 19 '11 at 4:16
  • 7
    $\begingroup$ Lagrange's equation was originally discovered without the Principle of Least action, and can be derived directly from the Newtonian formulation of mechanics. Goldstein does it that way (and has a discussion of the history of stationary principles in classical physics). $\endgroup$ – dmckee Oct 19 '11 at 15:36
  • 2
    $\begingroup$ While I can see how this could be considered a duplicate of the other question, it's well written and its getting a good response so I don't feel particularly compelled to close it at this point. $\endgroup$ – David Z Oct 19 '11 at 22:12
  • 2
    $\begingroup$ A fresh article trying to answer such questions: motls.blogspot.com/2011/10/… $\endgroup$ – Luboš Motl Oct 22 '11 at 7:30
  • $\begingroup$ “Is it least or stationary action?” Has you said, is just required to be stationary. It is there in Wikipedia. However, after check for a few examples in nature, you realize that it tends to select the minimum effort or easy way. Ask yourself: “If any one proposed me two legal and moral ways to get rich just differing for one being harder than other. Would I choose the hard way?” Plants could grow oblique maximizing the stress over their base. Instead, they are just prone to grow vertically minimizing that. $\endgroup$ – J. Manuel Jul 6 '17 at 9:24

10 Answers 10

46
$\begingroup$

Could someone please convince me that there is something natural about the choice of the Lagrangian formulation...

If I ask a high school physics student, "I am swinging a ball on a string around my head in a circle. The string is cut. Which way does the ball go?", they will probably tell me that the ball goes straight out - along the direction the string was pointing when it was cut. This is not right; the ball actually goes along a tangent to the circle, not a radius. But the beginning student will probably think this is not natural. How do they lose this instinct? Probably not by one super-awesome explanation. Instead, it's by analyzing more problems, seeing the principles applied in new situations, learning to apply those principles themselves, and gradually, over the course of months or years, building what an undergraduate student considers to be common intuition.

So my guess is no, no one can convince you that the Lagrangian formulation is natural. You will be convinced of that as you continue to study more physics, and if you expect to be convinced of it all at once, you are going to be disappointed. It is enough for now that you understand what you've been taught, and it's good that you're thinking about it. But I doubt anyone can quickly change your mind. You'll have to change it for yourself over time.

That being said, I think the most intuitive way to approach action principles is through the principle of least (i.e. stationary) time in optics. Try Feynman's QED, which gives a good reason to believe that the principle of stationary time is quite natural.

You can go further mathematically by learning the path integral formulation of nonrelativistic quantum mechanics and seeing how it leads to high probability for paths of stationary action.

More importantly, just use Lagrangian mechanics as much as possible, and not just finding equations of motion for twenty different systems. Use it to do interesting things. Learn how to see the relationship between symmetries and conservation laws in the Lagrangian approach. Learn about relativity. Learn how to derive electromagnetism from an action principle - first by studying the Lagrangian for a particle in an electromagnetic field, then by studying the electromagnetic field itself as described by a Lagrange density. Try to explain it to someone - their questions will sharpen your understanding. Check out Leonard Susskind's lectures on YouTube (series 1 and 3 especially). They are the most intuitive source I know for this material.

Read some of the many questions here in the Lagrangian or Noether tags. See if you can figure out their answers, then read the answers people have provided to compare.

If you thought that the Lagrangian approach was wrong, then you might want someone to convince you otherwise. But if you just don't feel comfortable with it yet, you'd be robbing yourself of a great pleasure by not taking the time to learn its intricacies.

Finally, your question is very similar to this one, so check out the answers there as well.

$\endgroup$
  • $\begingroup$ Beautiful response. It reminds me of the Rambam's response to the request to explain the Torah while standing on one foot. "Do not do to others what is hateful to you. The rest is commentary. Now go study." $\endgroup$ – AdamRedwine Oct 19 '11 at 13:28
  • 4
    $\begingroup$ A philosophical answer but a good one. ;-) $\endgroup$ – Luboš Motl Oct 22 '11 at 7:32
  • 9
    $\begingroup$ "In mathematics you don't understand things. You just get used to them." John von Neumann $\endgroup$ – Tom Sep 18 '13 at 9:35
  • 2
    $\begingroup$ I don't see why one needs to analyze a lot of problems to see that the ball we go tangentially as the string is cut off. A rational person will immediately get it in one go because there is a straight-forward rigorous proof to the claim that the ball will go tangentially. Developing an intuition for things based on your experience and not based on rigorous proofs is adopting a religion and not doing actual mathematical science. $\endgroup$ – Feynmans Out for Grumpy Cat Oct 7 '17 at 17:22
31
$\begingroup$

The intuition for the Lagrangian principle comes specific applications of Newton's laws, especially reversible systems with constraints, like nonspherical particles rolling along complicated surfaces. Newton's formulation of Newton's laws was not the end of the story, because there was more structure in the solutions of these types of problems than that which Newton made obvious.

One thing left unsaid by Newton is conservation of energy. Elastic processes are more fundamental than inelastic ones. But energy conservation is only part of the story. Suppose you have a bunch of masses connected by springs, and one of them is attached to a double-pendulum. You could theoretically have energy conservation in such a system by having all the energy leak out of the masses on the springs and go into the double pendulum. Perhaps every frictionless motion of the springs eventually settles all the energy into a single mode.

Your intuition is probably rebelling, telling you "that's infinitely unlikely! How could the pendulum move around and not set the springs vibrating!" But there is nothing in Newton's laws by themselves, even with the principle of conservation of energy, that prevents this sort of concentration of energy. But the solutions do not exhibit such phenomena, and there must be a reason why.

This intuition tells you that a perfect frictionless mechanical system is more than energy-conserving, it must conserve some notion of "motion-volume", so that if you alter the initial state by a certain amount, the final state should alter the same way. It can't concentrate all motion into one mode. This principle is the principle of conservation of phase-space volume, or the conservation of information. If all the motion got concentrated into one mode, the information about where everything was would have to get absurdly compressed into a tiny region of the phase space, the space of all possible motions.

The conservation of information is just about as fundamental as Newton's laws of motion--- it is revealing new facts about nature which are essential for the description of statistical and quantum systems. But it is nowhere to be found in Newton's formulation, because it does not follow from Newton's laws alone, even with the principle of conservation of energy added.

So you need to understand what type of law will give a law of conservation of information. There are two paths to go down, and both lead to the same structure, but from two different points of view, local in time and global in time.

One path is Hamiltonian: you consider formulating the law of motion as a set of symplectic equations for the position and momentum. This formulation clearly separates between reversible and irreversible dynamics, because it only works for reversible. It also explains the fundamental mathematical structure behind reversible classical mechanics, the symplectic geometry. The volume of symplectic geometry gives the precise law of information conservation, and further, the geometrical structure of systems with multiperiodic solutions, the integrable systems, is made clear.

But this point of view is centered on a time-slicing--- it describes things going from one instant of time to another. This is not playing very nice with relativity. So you also want to think about the solution globally, and consider the space of all solutions as the phase space. The initial position and velocities are good coordinates, and intuitive ones, because they determine the future. But if you want a global picture, you want coordinates which are symmetric between the final and initial state, since the dynamics are reversible. An explicit revesible description should treat the initial time and final time symmetrically. So you can use the initial positions and final positions, which also, generically, away from certain bad choices, determine the motion.

For these types of coordinates on phase space, you give the dynamical law as a condition on the trajectory between the intial and final positions. The condition should not be stated as a differential equation, because such a description is unnatural for boundary conditions of this sort. But when you have an action principle, you determine the trajectory by extremizing the action between the end points, you automatically have a notion of phase space volume, which is intuitive--- the phase space volume is defined by the change in the action of extremal trajectories with respect to changes in the initial velocities. This volume is the same as for the changes of the extremal trajectories with respect to changes in the final velocities. This is a straightforward consequence of the equivalence of Lagrangian and Hamiltonian formulation.

The full justification for both principles comes only with quantum mechanics. There you learn that the least action principle is a geometric optics Fermat principle for matter waves, and it is saying that the trajectories are perpendicular to constant-phase lines. But historically, the Lagrangian formulation was recognized to be more fundamental a century before Hamilton conjectured that classical mechanics was a wave mechanics, and this was many decades before Schrodinger. Still, with our modern point of view, it does not hurt to learn the quantum version of these formulations first, and it certainly provides a more solid motivation than the heuristic considerations I gave above.

$\endgroup$
  • 1
    $\begingroup$ What do you mean by conservation of information? $\endgroup$ – Self-Made Man Nov 18 '13 at 14:06
  • $\begingroup$ @Self-MadeMan: When you don't know the initial conditions, you place a probability distribution $\rho$ on these, then you evolve $\rho$ by evolving the initial conditions according to Newton's laws. Then the information missing in the encoded ignorance of the probability distribution $\rho$, which up to an infinite log-divergent constant (depending on the phase space discretization), $\int \rho\log\rho dx dp$ over phase space, is constant. This is the 19th century law of conservation of entropy in classical reversible mechanics, basically uncovered by Boltzmann/Lorschmidt, Liouville's theorem. $\endgroup$ – Ron Maimon Apr 17 '15 at 16:57
  • 1
    $\begingroup$ @RonMaimon : Can you ellaborate on the reasoning for lagranges formalism, or do you know some texts on this particular subject? Why is a trajectory (which is a solution) an element of the phase space? $\endgroup$ – Quantumwhisp Dec 16 '17 at 23:47
13
$\begingroup$

OP wrote:

As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.

Too often, as a student, one is only shown how to derive Newton's 2nd law from Euler-Lagrange equations by postulating some particular Lagrangian $L$. If one believes that Newton's laws are more natural (within the context of non-relativistic classical mechanics), then perhaps it would be more satisfying to see a derivation in the other direction, i.e. to see Lagrange equations derived from Newton's laws. This is e.g. done in the first chapter of Herbert Goldstein, Classical Mechanics, cf. e.g. this Phys.SE post. (An important element in this derivation is to show that a large class of constraint forces do no virtual work, leading to D'Alembert's principle.)

Throwing a wrench into the works, let me finally mention that there exist equations of motion that have no action principle, cf. e.g. this Phys.SE post.

$\endgroup$
  • $\begingroup$ (+1) Is there a nice post on this site or any article on the equivalency of Lagrange equations, Hamilton's principle and Newton's second law!? $\endgroup$ – H. R. Nov 14 '16 at 19:05
  • $\begingroup$ Hence satisfing Newton equations is equivallent to be minima of the action integral? $\endgroup$ – user1 Jul 27 '18 at 10:28
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jul 27 '18 at 10:32
6
$\begingroup$

The main point of Lagrangian formulation of classical mechanics was to get rid of the constraint relations completely so that one does not have to bother about them while calculating anything (see this answer of mine. Remember all the valuable symmetries of the physical situation is automatically inbuilt in this formulation of mechanics.

Now after having such a powerful technique at our disposal, it is natural to ask the simple question. Can the classical theory of electromagnetism i.e. Maxwell's equations, be expressed as Euler-Lagrange equations by suitably defining a Lagrangian of the electromagnetic field, so that we may readily get all those beautiful results of the structure of this formulation (for example avoid annoying field constraints)? The answer is 'yes', provided we suitably define a Lagrangian.

In fact, it is observed that any general classical field equations can be expressed by Euler Lagrange equation of motion and in each case you need to define a Lagrangian (you get a corresponding "Action") and get everything for free.

This Lagrangian approach is so powerful that even quantum field theories exploit them fully and almost all modern theories of physics exploit them in some way.

However, it is not at all clear whether this approach is completely general. Can't it be the case for a future theory that the equations of the theory can't be expressed in Lagrangian formulation? I asked this question here. See some good answers.

$\endgroup$
3
$\begingroup$

What does "principle" mean? It has many definitions but in the context of this discussion it has the power that "axiom" has in mathematics: a very basic assumption, which, if changed, the whole construct theory shifts or is destroyed.

Physics has adopted this from the geometric observation that : the shortest distance between two points is a straight line which logically led to "minimum time taken" and the search for the shortest distance when unknown.

that it is so natural that we would not even dare abandon these ideas?

If somebody brilliant enough can come out with another principle for the system of mathematical formulation of classical mechanics and subsequently quantum field theory that does not follow least action but incorporates perfectly the large data base / existing equations etc there is no problem. It might even be adopted generally if it could predict new spectacular results. Otherwise, from economy of mental effort( another principle :) ) the system developed under the principle of least action will still prevail.

$\endgroup$
2
$\begingroup$

The Lagrangian is just a (special, functional kind of) anti-derivative of an equation of motion. In simple cases, the equation of motion of classical particles is a function $F(x,\dot x,\ddot x,...)$ of the positions, velocities, accelerations, etc. of the particles, for which the equation $F(x,\dot x,\ddot x,...)=0$ determines the trajectory. When a function $F(...)$ admits a Lagrangian anti-derivative, the maxima and minima of the Lagrangian anti-derivative determine where to find the zeros of the function $F(x,\dot x,\ddot x,...)$, the derivative of the Lagrangian.

There is not necessarily anything fundamental or natural about a Lagrangian. The Lagrangian has many formal properties that often make it extremely useful, and quite often make it rather simple or beautiful. We like beautiful things. Appreciating beauty is a tricky thing, to some extent a matter of experience, to some extent a matter of just seeing it. Mark and Anna are both pretty good guides.

$\endgroup$
1
$\begingroup$

Here is, at long last, my own attempt to thwack at this problem.

To understand it, we first need to, as with many things, take a bit of a step back. We should not so much at first be interested in the form of the Lagrangian inasmuch as its goal, which is this:

The Lagrangian lets us describe motion as an optimization process.

Now why might we want to do that? The answer is simple: Many things in the world involve some form of optimization process, across many, many domains. A familiar one from physics you quite likely have encountered directly in your life is that of a soap film. It takes the form it does because it seeks the minimum energy, or most even-handed distribution of forces. If they were off-balance, such would pull it into a different shape until that balance was had. The shapes of planets are another example: this is why they are all (near)spheres: that shape, again, optimizes the potential energy. In human civilization, we also try to seek the optimum in many things: e.g. we tend to want to find the least expensive solution to any problem - for however much better or for worse that that is. This, likely, is also rooted ultimately in biological optimization of our psychology or better psycho-cultural blend, through the process of evolution, optimizing for reproductive success.

These examples should convince you that looking at things in terms of seeking optima is a very reasonable thing to want to do, and thus we might ask ourselves as to whether or not motion could or should be treatable as such in some sense: is there some sense in which we can say that objects move on the paths they do because they are, in this sense, "optimal"? It may be that there isn't, or that it doesn't tell us anything illuminating, but on the other hand, it may be that there is.

And the Lagrangian is the answer to that, although it's not a 100% satisfactory one because the path that is taken need not strictly speaking be the truly least action path. We integrate it, it gives us a kind of "cost", so to speak, which is then (partially) optimized and that gives us the "right" path of motion that an object "really" takes. Of course, that still leaves open the question of why it happens to take the rather strange form

$$L(\mathbf{q}, \dot{\mathbf{q}}, t) := K(\dot{\mathbf{q}}) - U(\mathbf{q}, t)$$

so that the action is

$$S[\gamma] := \int_{t_i}^{t_f} L(\gamma(t), \dot{\gamma}(t), t)\ dt$$

over a parametric potential path of motion $\gamma$, beyond just "well, it reproduces the motions we see". While it would - and I'll get back to this at the end - not be a surprise if that, given how that this is a general framework, for rather complex and obscure phenomena, there might not be a straightforward interpretation, one's first exposure to this concept is still in the context of basic Newtonian mechanics, and thus there should be at least some way to make it intuitive to give a base on which to build those more complex concepts and applications.

And this is the best answer I could come up with. Since we are looking at motion as an optimization of path taken in terms of a "cost", we will seek ideally its minimization, given that generally speaking by intuition we tend to think in terms of savings, not in terms of spending, when it comes to "improvement" of doing something. To that end, let us consider something else that, hopefully, many people should be familiar with on at least some level: namely the money cost required to transport a parcel of goods from one point to another on the Earth's surface. In particular, we can imagine hauling the goods on a land vehicle like truck or train, and when we do so, we generally find that at least four factors influence the cost, as such:

  1. The mass of the freight: heavier loads are more expensive to transport and typically transport is often billed on a per-mass basis,
  2. The distance over which it is hauled: the farther we need to move it, the more expensive (e.g. more fuel, more risk, more pay to the driver, etc.),
  3. The time to haul: if we require it be delivered fast, expect to pay a premium (better vehicles, need to choose right shipping routes, driver has to stay up longer maybe...),
  4. The conditions of hauling: if the haul requires negotiating inclement weather, challenging terrain, or other such factors, expect, again, added cost at least somewhere along the line.

And rather very interestingly it turns out that, at least in simple cases, we can, surprisingly, interpret the "weird" Lagrangian in virtually almost the very same way:

The action is the "price" that Nature will pay to haul its goods.

To do that, we first have to narrow our attention a bit and then rewrite it in a form that a physicist, at least trained on the standard "model", might find peculiar, but which is mathematically 100% kosher. The narrowing of our attention is basically to the case of a single particle, and we use the usual coordinates for the motion. In this setup, the action principle takes the form, which you can verify, from the (opaque) "kinetic minus potential" business,

$$S[\gamma] = \int_{t_i}^{t_f} \left(\frac{1}{2} m[\dot{\gamma}(t)]^2\right) - U(\gamma(t))\ dt$$

which we can then rewrite as

$$S[\gamma] = \int_{t_i}^{t_f} \left(\frac{1}{2} m[\dot{\gamma}(t)]^2\right)\ dt + \int_{t_i}^{t_f} [-U(\gamma(t))]\ dt$$

Now, note the following trick: $\dot{\gamma} = \frac{d\gamma}{dt}$ is just the velocity, $\mathbf{v}(t)$, a vector, since it's the time derivative of the path in ordinary coordinates, by setup. Its square is thus the square of the speed: $[v(t)]^2$, where speed $v$ equals $\frac{ds}{dt}$, the rate at which arc length ($s$) is covered as time is elapsed. Using that allows us to transform the first integral to:

$$\int_{t_i}^{t_f} \left(\frac{1}{2} m[\dot{\gamma}(t)]^2\right)\ dt = \frac{1}{2} m \left(\int_{t_i}^{t_f} \frac{ds}{dt} \left[\frac{ds}{dt}\ dt\right]\right)$$

which by "unashamed mashing of differentials" (i.e. the chain rule and change of variable) becomes

$$\frac{1}{2} m \left(\int_{0}^{d_\mathrm{tot}} v(s)\ ds\right)$$

(Note if $s$ is a function of $t$, $ds$ becomes $s'\ dt$, and $v(s(t))$ is just $v(t)$, which is just what we had before.)

where $d_\mathrm{tot} = d_\mathrm{trav}$ is the total distance covered over the complete motion and we have switched to measuring the progress of the motion in terms of the distance covered so far. Even more suggestively, noting the usual definition of averages from calculus, we can thus rewrite the above kinetic term, and hence the whole action, via the average speed,

$$S[\gamma] = \frac{1}{2} mv_\mathrm{avg}d_\mathrm{trav} + \int_{t_i}^{t_f} [-U(\gamma(t), t)]\ dt$$

Moreover, we can do the same for the potential term on the right using the average potential encountered and the journeying time $t_\mathrm{journ} := t_f - t_i$:

$$S[\gamma] = \frac{1}{2} [mv_\mathrm{avg}d_\mathrm{trav}] + [-U_\mathrm{avg} t_\mathrm{journ}]$$

And we see that this expression, then, very, very well coheres with the intuitions we just discussed about transport cost: the first term is basically the cost of movement, that is, the cost inherent to moving a certain distance at a certain speed. The cost increases in proportion to mass transported, the speed of transport, and the distance: exactly as we might think (though in our human world the relation is seldom so simple as an exact proportionality like this - but such is the elegance of basic principles of the Universe). Moreover, the meaning of the potential term, and that all-vexing minus sign, comes into play: this term addresses the fourth factor (hence why I chose it, because I worked this out ahead of writing this post), which is the environment, or perhaps, "terrain cost". Remember that, unless our particle is in deep, intergalactic space, free from virtually all other influences, it is going to be subject to the actions of forces which will be competing to influence its motion. The potential term basically says "stay for as little time as possible at as shallow a depth as possible in any attractive wells", or in terms of cost, that you will be "billed" more for staying longer and deeper. The reason for the negative sign is just that: as a potential well goes deeper, its potential decreases. To make that deeper depth cost more, we must flip the sign on the potential, so it is negative. Perhaps not quite how we'd set up the cost, but it should be understandable and sensible in its own way.

Insofar as why (intuitively) a stationary point and not always a minimum? Well, not always can we get the cheapest we might want in every situation, but we can get something that's kinda cheap - at the very least, if something goes a bit wrong and we have to take a little detour, it shouldn't hurt the cost too much.

Of course, as I said, this is all for a relatively simple case. And yes, as you get to more complicated cases and more difficult physical phenomena, it becomes less clear how the action relates to how we would account for cost, but that's not a surprise: we're not, now, dealing with simple point-to-point movement. The purpose of this exercise is to first get you the base intuition that action is a cost of movement - and this is something that might be familiar to anyone who has played certain role-playing games: very often this notion of "action cost" finds its way there. Naturally, to describe other, more complicated, phenomena, we have to define the action differently, describing them in terms of other costs than these. This is no different from working in terms of forces, where the intuition is presumably somewhat clearer. For example, in electromagnetism (though I might not have gotten this part quite right), we can similarly describe the action as having to blend and account in an appropriate way the costs of building up and/or tearing down an electromagnetic field, the rate of such construction and/or demolition, and the maintenance cost of holding a nonzero field.

And that, finally, answers your question of, essentially, how do you "think" in Lagrangian mechanics so as to get the right result and not be led astray: yes, it is a different framework, and unfortunately, like in many cases of dealing with novel frameworks, it seems the dominant approach to it is to essentially use it "by conversion" (similar problem as in teaching Americans the metric system, or teaching a foreign language by how to translate or relate words/phrases to your native language, or any one of a number of other instances of this common failing that are just not coming to the top of my head right now) instead of "on its own terms". To really use it proficiently, you essentially have to give the notion of "forces" a heave-ho and think afresh, from the get-go, in terms of actions instead: what actions happen here and what may impede or facilitate them (e.g. dipping into a potential well), and what are the right formulae by which to describe the costs those actions have. Thus, had we gone this route from the beginning, e.g. perhaps had Newton had the thinking of a economist or service provider seeking to improve productivity, he might have started by postulating that linear steady motion, i.e. no potential wells, would have a linear cost formula. (And he would then have had to develop the calculus of variations first, whereas we treat the instantaneous calculus as primary instead ... I wonder how that would have influenced the development of mathematics ... so many roads not taken, all throughout history, where could they go? What possibilities lurk therein? I wonder ...)

In conclusion, we see that, as Maupertuis said (who formulated a somewhat weaker action principle before Hamilton's Principle), indeed,

Nature is thrifty in all its actions.

but with our more sophisticated understanding, we should perhaps gently persuade him of:

Nature prefereth thrift in action, except when need shouldst compel that it spend more, in which case, next best shall do.

(One more thing: You might ask why there's a $\frac{1}{2}$ term, from this point of view, in the kinetic term [motional action cost]. One way to say it might be that Nature likes to make terrain twice as important as movement, but you should note that with a suitable choice of units, the term can be made to disappear: we could measure in half mass-units, or in double energy-units, noting though this may seem to break the coherence of how these units are usually defined. However, we could also, then, perhaps that that is "on us" in that we derive our energy units from force considerations as primary: remember that a "joule" is "one Newton of force for one metre of distance". Indeed, were we to take Lagrangian as primary, which may make more sense given its more fundamental role, we may seek that the energy unit is "the energy that 'gets stuff done' at a rate of one action unit per unit of time", and define mass however we need as a separate quantity. Indeed, with this choice made, we finally arrive at the "so hard" formulation, for Newtonian mechanics in the case of conservative forces:

$$ \mbox{"Action Cost"}\\ is\\ \mbox{Mass to Move}\ \ times\ \ \mbox{Distance of Transport}\ \ times\ \ \mbox{Speed Required}\\ plus\\ \mbox{Obstacle Negotiation Cost}\ \ times\ \ \mbox{Obstacle Negotiating Time} $$

Almost as easy as $\mathbf{F} = m\mathbf{a}$!)

$\endgroup$
0
$\begingroup$

Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics?

No, because it's not natural at all! That's why it took hundreds of years and two of the most brilliant minds in history (Lagrange and Hamilton) to come up with it!

But the fact remains that every regime of physics - Newtonian mechanics, fluid mechanics, electromagnetism, nonrelativistic quantum mechanics, particle physics, relativistic quantum field theory, condensed matter physics, general relativity - can be formulating as extremizing some action which is an integral of a local Lagrangian. (Systems of finitely many interacting classical point particles is arguably the one exception). So whether or not you like the idea, apparently Nature does, and you need to accept it if you want to understand the universe.

On a less flippant note: if you're asking this question then you've probably only seen the action principle formulated in the context of Newtonian mechanics. In this case it's best for efficiently incorporating constraints - particles confined to move on certain surfaces and so on. In this context, I would agree that it might best to think of Newton's laws as being more fundamental than a choice of a particular Lagrangian, which usually describes an extremely specific system.

But as you go on to learn field theory and the concepts of coarse-graining, renormalization, and universality, you'll see that the low-energy properties of a huge array of systems consisting of enormous numbers of microscopic degrees of freedom with local interactions can be described by field theories specified by an action. In fact, pretty much any system can be so described, and these kind of systems come up in a huge number of different contexts, from condensed matter to quantum gravity. All you need to know about about the microscopic degrees of freedom is their symmetries and perhaps a few very basic facts like whether they're bosons or fermions.

In fact, the general consensus among many physicists these days is that we have pretty much no clue what goes on at the Planck scale, but we can give fairly precise and quantitative arguments for why it doesn't matter what happens there in order for us to be able to validly use quantum field theory (defined by an action!) to describe physics on lengths scales that range over many orders of magnitude.

P.S. You are completely correct that say that the "Principle of Least Action" is just wrong. The action is stationary at the configurations that satisfy the physical equations of motion, but it can be a maximum, minimum, or saddle point. A. Zee's book on GR contains a problem demonstrating that even for the simple harmonic oscillator, the action is often maximized rather than minimized along the equations of motion.

$\endgroup$
  • $\begingroup$ "So whether or not you like the idea, apparently Nature does, and you need to accept it if you want to understand the universe." --- But I would say that understanding why Nature does like the idea is part of understanding the universe. I can accept it and at the same time ask "Why?". $\endgroup$ – Jonathan Gleason Jul 7 '17 at 17:27
  • $\begingroup$ @JonathanGleason Certainly, but I think that's more of a philosophy than a physics question. At some point, you need to start from some purely empirical postulates - otherwise you have nothing to go on. That having been said, I think my comments on field theory and renormalization give reasonably good motivation. $\endgroup$ – tparker Jul 7 '17 at 17:36
0
$\begingroup$

Not an answer, but too long for a comment. I wanted to show that, terminology aside, in general a stationary action is neither minimised nor maximised, so in theory we should speak of the stationary action principle. Consider the simplest model of falling for which a unit mass has Lagrangian $\dot{z}^2-gz$ so the equation of motion is $\ddot{z}=-g$. Suppose the mass falls through a height $h$ for a time $\tau:=\sqrt{\frac{2h}{g}}$, viz. $z=h\left(1-\left(\frac{t}{\tau}\right)^n\right)$ for $t\in\left[0,\,\tau\right]$ with $n=2$. I deliberately replaced an exponent with a free parameter because, theoretically, a mass that fell through the same height in the same period could use any $n>0$ were it not for the equation of motion. The point is we can show that $n=2$ neither minimises nor maximises the action obtained over the period of falling. We have $\dot{z}=-\frac{nht^{n-1}}{\tau^n}$ so $$S=\int_0^\tau\left(\dot{z}^2-gz\right)dt=\int_0^\tau\left(\frac{n^2h^2t^{2n-2}}{\tau^{2n}}-gh+\frac{ght^n}{\tau^n}\right)dt=f\left(n\right)\frac{h^2}{\tau}$$ with $$f\left(n\right):=\frac{n^2}{2n-1}-\frac{g\tau^2}{h}\left(1-\frac{1}{n+1}\right)=\frac{n^2}{2n-1}-\frac{2n}{n+1}.$$Thus $f\left(2\right)=\frac{4}{3}-\frac{4}{3}=0$. But as you can see from the graph of $f\left( x\right)$ here, values of $n>0$ exist for which $f\left( n\right)<0$ - to be precise, the solution set is $\left(0,\,\frac{1}{2}\right)$. (The rightmost asymptote of $f$ play a role in this.) In fact, no finite $n$ not on an asymptote minimises or maximises $f$, but exactly one $n>0$ makes $f$ stationary, namely $n=2$.

$\endgroup$
-3
$\begingroup$

The functional integral in quantum mechanics includes a summation over all possible paths in the configuration space a quantum system might take . These paths are weighted by an exponential imaginary function whose phase is the action .Using the method of steepest descent , one can pass to the classical limit which shows that the Euler-lagrange equations should hold for the classical path .

$\endgroup$

protected by ACuriousMind Mar 20 '15 at 13:52

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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