Usually, a typical example of the use of the action principle that I've read a lot is the derivation of Newton's equation (generalized to coordinate $q(t)$). However, in the classical mechanics interpretation, isn't this like tripping oneself up? Because we've already identified $T$ and $U$ as the kinetic energy and the potential energy, respectively which are derived, in fact, from Newton's equations.

  • 3
    $\begingroup$ First question(v1) is a possible duplicate of e.g. physics.stackexchange.com/q/4102/2451 Second question(v1) is a possible duplicate of e.g. physics.stackexchange.com/q/15899/2451 $\endgroup$ – Qmechanic Apr 15 '12 at 23:43
  • 1
    $\begingroup$ Hi Nesp - we generally prefer to have one actual question per post, unless they're very closely related (which these are not). Normally I would suggest splitting it into three posts, but since your first two are already addressed by other posts on the site, I'm going to just edit out the first two questions. Feel free to add some of the explanation text back in if you think it helps clarify your question, and especially to fix the title so it accurately describes what you want to ask. $\endgroup$ – David Z Apr 15 '12 at 23:52
  • 1
    $\begingroup$ As for the other two questions you had, if you read the posts Qmechanic linked to and they don't explain what you're trying to understand, you can definitely re-ask those questions as new posts. But make sure to say why the links Qmechanic gave you don't give you the answers you're looking for. (Your original posting is accessible in the edit history.) $\endgroup$ – David Z Apr 15 '12 at 23:57
  • $\begingroup$ Wow! Though the question still makes sense, and I can see the gardener's wisdom, you really pruned the question back hard! $\endgroup$ – Peter Morgan Apr 16 '12 at 0:15
  • 1
    $\begingroup$ Dear @Nesp, a circle as a topological concept means that one gets from point A to B along one curve and then from B to A along another curve. For example, if you tried to say that some equation A is inevitable because of B and B is inevitable because of A, it would be circular. But here - and in most of science - there is nothing circular here. A implies B and/or B implies A but one still needs the agreement with observations to show that either A or B is true. On the other hand, such an agreement of theory with facts is a near-proof of both A and B, too. $\endgroup$ – Luboš Motl Apr 16 '12 at 6:04

Answering the third question, in any mature branch of mathematics or physics, there are always equivalent or near-equivalent formulations of the same structure (it's pretty much a definition of maturity, that some of the relations of a new topic to older topics have been worked out). The Lagrangian can be regarded as a type of functional anti-derivative of a set of equations of motion (taking functional derivatives generates the equations of motion), which in turn can be regarded as a presentation of the detailed implications of Newton's law in the context of a particular model.

There may be reasons for taking some mathematical formulations to be more fundamental than others, typically because the map between one and another is not 1-1 or not onto, or, in modern times, because some formalisms present the symmetries of the dynamics more transparently. In the case of Newtonian mechanics, there are numerous different ways to specify a dynamics (perhaps as many as a dozen altogether, developed over three centuries, including, for example, the Hamilton-Jacobi formalism, although I couldn't name them all without looking them up).

Ultimately, if the empirical contents of different models are equivalent, a more empirical Physicist will say that the models are equivalent, so that the choice between using one or another rests in questions such as which is more tractable. However, realism about the role of symmetry groups in nature has led to many physicists making claims about the preeminence of Lagrangian approaches that rely to some extent on issues such as mathematical aesthetics. You will have to decide for yourself where on this spectrum you wish to place yourself, insofar as the decision is not based on experiment.

If you have a strong mathematical background, two references that come to mind that might broaden your perspective are Olver, "Applications of Lie groups to differential equations", or, slightly more accessibly, Marsden & Ratiu, "Introduction to Mechanics and Symmetry".

  • $\begingroup$ Thank you very much for your reply. If I understood correctly, I think you describe my question much like the classical link between statistical mechanics and classical thermodynamics. Before "knowing" each other, they tried to describe the same systems, although they can be derived from different principles (maximum entropy and the usual laws of thermodynamics, respectively). From there, some quantities of interest (e.g. temperature) seem to have mixed interpretations (temperature -> function of r.m.s velocity of particles)... $\endgroup$ – Néstor Apr 16 '12 at 2:15
  • $\begingroup$ ...however, I think I'm still confused. I argue that, once you derive the lagrangian formulation an you try to derive Newton's second law's equation, you usually do it by claiming that $T=mv^2/2$ and $U=U(q)$. From there, you naturally get $md^2q/dt^2=-U$, and it is usually claimed "Tada! There you go, Newton's equation!". However, I simply say that this is circular reasoning because in order to write $T=mv^2/2$ you actually use $\int md^2q/dt^2 dq$ (that's why, in the start, you know the functional form of $T$!). $\endgroup$ – Néstor Apr 16 '12 at 2:19
  • $\begingroup$ (Now that I think about it, maybe it isn't circular reasoning. You start from the lagrangian formulation and you say "well, let's try what we get if we put $T=mv^2/2$ and $U=U(q)$". You do the math and get Newton's equation back: magic? No! As you said: an equivalent (or nearly equivalent) formulation). $\endgroup$ – Néstor Apr 16 '12 at 2:23
  • $\begingroup$ Paraphrasing Luboš's comment, all the ways of formulating Newtonian mechanics are ultimately linked to experimental results. Insofar as there's an "ultimately", it's not a circle. There is a higher-level proviso, that deciding what experiments to do is partly driven by theoretical ideas of what experiments might be useful to try, but I take that not to be much at issue here, and it's not so much a circle as a dense network of interrelated choices. $\endgroup$ – Peter Morgan Apr 16 '12 at 12:20
  • 1
    $\begingroup$ @Néstor: It's not true that $T=mv^2/2$ has to come from $\int F dx$. That's just the order in which you've encountered the topics. $T=mv^2/2$ could come from experiment, and we could derive $F=ma$ from it. $\endgroup$ – user4552 Jun 5 '13 at 22:08

Potential energy is associated with a property of conservative forces that can be expressed as the gradient of a potential function, rather than coming from a definition using Newton's laws. Kinetic energy on the other hand can be derived from the definition of work done and, together with potential energy, expresses the conservation of energy which is why kinetic energy is defined the way it is. You can't derive the conservation of energy from just Newton's laws.

D'Alembert's principle comes partly from Newton's second law with the additional postulate that forces of constraint don't do work, which is true for some classical mechanics problems. You can't use Newton's laws to prove that the force of reaction of a table on a book resting on it is normal to the surface. D'Alembert's principle together with monogenic forces is used to derive the Euler-Lagrange equations and the principle of least action.

We can therefore conclude that the principle of least action is at least compatible with Newton's first and second laws for monogenic forces which includes gravity and electromagnetism.

  • $\begingroup$ MacVirgo, thanks for your reply. However, I really don't get your first pharagraph: you say that "Potential energy is associated with a property of conservative forces that can be expressed as (..)". This uses Newton's law: the concept of force is introduced/defined in his 2nd law. Your second argument is "Kinetic energy on the other hand can be derived from the definition of work done (...)". Again, the definition of work done is $W=\int \vec{F}\cdot d\vec{\ell}$, i.e., we use Newton's 2nd law again. $\endgroup$ – Néstor Apr 16 '12 at 2:05
  • 1
    $\begingroup$ @Nesp force is independent of the acceleration it causes and can be measured by its effect on the length of a spring. $\endgroup$ – John McVirgooo Apr 16 '12 at 12:43

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.