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This is a very simple-minded question. Why is it difficult to understand the phenomena of chaos in Newtonian mechanics and one has to turn to Hamiltonian formulation? I haven't read much about chaos except listening about it here and there. In particular, David Tong's lecture mention

...it obscures certain features of dynamics so that concepts such as chaos theory took over 200 years to discover; and it's not at all clear what the relationship is between Newton's classical laws and quantum physics.

SRS
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  • In fact it is not even well defined: https://en.wikipedia.org/wiki/Chaos_theory BTW the Hamiltonian formulation is not incompatible with Newtonian mechanics. – Miguel Aug 13 '17 at 14:17
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    @user163104, the chaos meant here is also deterministic. The indetermination comes about "only" in practice , where arbitrary precision isn't achievable. – stafusa Aug 13 '17 at 18:04

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Newtonian and Hamiltonian formulations are fully equivalent, and Newtonian mechanics does not introduce any additional difficulty in understanding chaos than Hamiltonian mechanics. Hamiltonian formalism does introduce much better understanding of reality, but none of that is directly related to chaos.

(Deterministic) Chaos arises when solving difference or differential equations that have nonlinear components, and a key feature of chaos is sensitive dependence on initial conditions. It matters not what the equations describe or where they came from. It can be a Hamiltonian system (e.g. solar system) or even a weather simplification (e.g. Lorenz equations). If the systems are nonlinear, chaos is very likely to arise. And, to study chaos you turn to tools unrelated to the origin of your differential equations, like e.g. Lyapunov exponents. Side comment: Poincare surfaces of section are also unrelated to the origin of your equations and can be applied just as well to weather systems.

By the way, there has been some comments (see @OON) that "you cannot have a Hamiltonian system with dissipation". Every Hamiltonian system where the Hamiltonian depends on time is equivalent to having a positive or negative dissipation, as one learns in introductory University physics (see any of the available textbooks on mechanics). For the experts: Notice that this can only apply to systems which respect the symplectic structure, since by definition a Hamiltonian system can be only symplectic. Of course, you can write aaaaany kind of model that is Newtonian but not symplectic. Have you ever seen such a thing in experiments though...?

EDIT: About Tong's lecture: (I have not watched these lectures) I am guessing that Tong is referring to the concept of the phase-space, which is not given special attention in Newtonian formalism (since, as @stafusa said in his answer, momentum and position are not treated on equal footing). Although the phase-space is not directly related to the existence of chaos (because stable systems also have a phase-space!) it does allow one to immediately understand whether a Hamiltonian system will be chaotic: in the Hamiltonian case the phase-space will be mixed or fully chaotic (ergodic). Of course, this requires some techniques of visualizing the phase-space, which is very easy with a computer.

An important side-note in the discovery of chaos: Chaos was first confronted by Poincare, when trying to answer the question "Is the solar system stable?", posed by King Oscar II of Sweden+Norway. At one point of his research, Poincare assumed that "a small change in the initial conditions will only have a small change in the evolved orbit". To his devastation, he realized later that this was not true. The concept of phase-space may have been useful for him to understand that. However, chaos was not truly appreciated until Lorenz came around 70 years later, and more specifically, computers came, which could solve unsolvable ODEs. Being able to solve these equations numerically was, at least for me, the crucial point in chaos theory history, and this does not depend on the origin of your equations.

George Datseris
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  • No problem. Can you write a time-dependent Hamiltonian not for the textbook damped oscillator but for e.g. $\ddot{\vec{x}}=-\lambda f(|\dot{\vec{x}}|)\dot{\vec{x}}-\nabla U$? If you can for $f$ being function what about anisotropic drag? – OON Aug 13 '17 at 23:58
  • Can you include in your answer how you interpret D. Tong's comment? @GeorgeDatseris – SRS Aug 14 '17 at 00:10
  • @SRS I have added 2 more paragraphs that firstly address Tong's comment and secondly give a historic argument on why in my opinion chaos does not really have anything to do with the origin of your equations. – George Datseris Aug 15 '17 at 06:56
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It's not.

Both formulations are equivalent, so the equations one eventually solve with the computer in order to simulate the system are the same no matter which formulation is used $-$ but the Hamiltonian point of view, with its phase space characterization (see below) and an arguably weaker emphasis in individual solutions is indeed important.

From the Tong's lecture introduction, it seems he means the fact that in Newton's mechanics, momentum isn't obviously as important as position, since it appears simply as a derived quantity ($\mathbf{p}\equiv m\,d\mathbf{x}/dt$) $-$ while, in Hamilton's, they gain equal footing and the mechanical system state lives in the phase space (with its position-momentum coordinates). The point being that the phase space is where the system's underlying structures actually live.

stafusa
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    No, they are not equivalent. You can easily introduce a force leading to the non-Hamiltonian dynamics. You can even easier write a Hamiltonian which will lead to equations not looking like $F=ma$. The Hamiltonian picture however provides you with lots and lots of powerful methods. – OON Aug 13 '17 at 19:57
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    You're mistaken. Lagrangian, Hamiltonian, and Newtonian all describe the same physics. You might use symplectic formalism outside mechanics and even physics, but OP's question is about mechanics and that's a physics forum. – stafusa Aug 13 '17 at 20:10
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    Dear self-constrained physicist. Can you describe for me give Largangian and Hamiltonian mechanics for dissipative and anholonomic systems (that definetely admit Newtonian description)? Or do you think that those are purely mathematical constructs never arising in nature? Do you think that any Lagrangian/Hamiltonian be it fundamental or effective corresponds to the Newtonian $F=ma$? Probably it's not me who is mistaken. – OON Aug 13 '17 at 21:07
  • This is a pointless argument. Of course you can write mathematical equations that don't correspond to anything physically observable. Of course different formulations of the same subset of physics don't have exactly the same limitations at their boundaries. So what? If you really believe that there "ought to be" one mathematical formulation that includes everything, keep dreaming about what you will do when you get a Nobel prize for discovering it! – alephzero Aug 13 '17 at 21:48
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    @OON, no need to get funny with your comments. Let's keep it civil. See, as far as I know, Lagrangian formalism for dissipative systems (and with any kind of force) is commonplace (see, e.g., its Wikipedia entry or https://physics.stackexchange.com/a/8462/75633 ), and the Hamiltonian is a Legendre transformation of the Lagrangian. So how there could be some nonequivalence? Also, the statement that the three are simply different mathematical formalisms of the same theory is found in many books on the subject. An online example is https://goo.gl/S2xKU2. – stafusa Aug 13 '17 at 22:23
  • @stafusa Well, most modern physicist don't work with such classical systems and I suspect some authors of the textbooks may even not know of such nonequivalencies (e.g. that was a major criticism of Landau-Lifschitz textbook by Fock) You can describe by certain time-dependent Lagrangians or Hamiltonians a large class of one-dimensional models and the models with dissipative forces linear in velocities. But as far as I know there's no variational or Hamiltonian description of the generic dissipative forces. Also you can't describe in such a way the systems with nonholonomic constraints. – OON Aug 13 '17 at 23:51
  • @stafusa Now there's a funny thing about nonholonomic constraints. You could try to introduce them into the variational principle simply as a constraint multiplied on the Lagrange multiplier. Job's done, right? Wrong. It gives wrong virtual displacement for the actual mechanical systems with nonholonomic constraints. As far as I know it's impossible to describe the proper ones with a variational principle. – OON Aug 13 '17 at 23:55
  • @OON, Thanks for the replies. I think my mechanics is too pedestrian to dispute your claims, especially if classical textbooks and Q.mechanic answer ( https://physics.stackexchange.com/a/8462/75633 as I understand it) are wrong on that. The "funny" part was referring to "Dear self-constrained physicist", which could be a provocation - but it's mild and was a reply to a comment of mine where I could have been more friendly. It doesn't matter. More importantly: do you have a classical mechanics reference that backs your claims? – stafusa Aug 14 '17 at 00:21
  • @stafusa Well, concerning the nonholonomic constraints I think that should be somewhere in A.M.Bloch et al. "Nonholonomic Mechanics and Control". I sadly can't look into it right now, so without any pages. The variant where the nonholonomic constraint is introduced through the Lagrange constraint is called "vakonomic". Of course this can actually appear e.g. in gauge theories. – OON Aug 18 '17 at 11:29