5
$\begingroup$

Consider the following question in classical mechanics

Are Newton's Second Law, Hamilton's Principle and Lagrange Equations equivalent for particles and system of particles?

  • If Yes, where can I find a complete proof?
    Are there certain conditions for this equivalence?

  • If No, which one is the most general one?

I couldn't find the answer of my question in the books since there are lots of sentences and no clear conclusion! Or at least I couldn't get it from the books! Maybe the reason is that physical books are not written axiomatically (like mathematics books). The book which I had my focus on was Classical Mechanics of Herbert Goldstein.

\begin{align*} \text{Newton's Second Law},\qquad\qquad &\mathbf{F}_j=m_j\mathbf{a}_j,\qquad j=1,\dots,N \\[0.9em] \text{Lagrange's Equations},\qquad\qquad &\frac{d}{dt}\frac{\partial T}{\partial\dot q_j}-\frac{\partial T}{\partial q_j}=Q_j,\qquad j=1,\dots,M \\ \text{Hamilton's Principle},\qquad\qquad &\delta\int_{t_1}^{t_2}L(q_1,\dots,q_M,\dot q_1,\dots,\dot q_M,t)dt=0 \end{align*}

where $N$ is the number of particles and $M$ is the number of generalized coordinates $q_j$. Interested readers may also read this post.

$\endgroup$

closed as too broad by AccidentalFourierTransform, Jon Custer, Gert, Cosmas Zachos, user36790 Nov 15 '16 at 3:27

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Related: physics.stackexchange.com/q/78138/2451 $\endgroup$ – Qmechanic Nov 14 '16 at 21:07
  • $\begingroup$ @AccidentalFourierTransform: It would be much nicer if you made some comments or edits on how to narrow the question instead of closing it! :) I don't know the answer of the question and so I don't know how to narrow it down since I am not a physicits! :( $\endgroup$ – H. R. Nov 15 '16 at 8:07
  • $\begingroup$ Without even touching on whether Hamilton's principle and the Euler-Lagrange equations are equivalent, it's easy to answer your question with "No" because Newton's II law is obviously not equivalent to either. Newton II only covers the change of momentum over time being equal to the force applied. Nothing about this law give you the true equations of motion for exotic systems, it doesn't include energy, it doesn't give a mechanism for application to other physics problems. Sure you can get Newton II from Hamilton, but saying they're equivalent is like saying sums are equivalent to integrals. $\endgroup$ – Jim Nov 15 '16 at 13:40
  • $\begingroup$ @Jim: See this link $\endgroup$ – H. R. Nov 15 '16 at 20:04
  • 1
    $\begingroup$ I don't understand how this question got closed. It is definitly not broad. Since the question is closed I cannot answer.. but the answer is simple: No. And in overall physics, there isn't a "general" either. There are systems only Hamilton can do. There are systems only newton can do. And so on. If you restrict only to classical mechanics (not relativity, not EM), then it is possible to prove that $P(H)\subset P(L)\subset P(N)$, where $P(N)$ is the set of all problems newton can solve, and so on. $\endgroup$ – Physicist137 Dec 2 '16 at 14:37
2
$\begingroup$

The earlier formulations of this question was quite broad. This answer is constructed as as a broad response within classical$^{\dagger}$ theories with some hopefully helpful navigation points:

  1. On one hand, the stationary action principle (= Hamilton's principle) and the Euler-Lagrange equations make sense far beyond the realm of Newtonian mechanics, e.g. in field theory or relativistic point mechanics.

  2. On the other hand, there are dissipative systems in Newtonian mechanics that have no action formulation, see e.g. this Phys.SE post.

  3. One may show that broad classes of Newtonian systems satisfy D'Alemberts principle, such as, e.g. rigid bodies, see this Phys.SE post.

  4. For the validity of D'Alembert's principle, see this & this Phys.SE post.

  5. One may show that D'Alembert's principle leads to Lagrange equations, cf. e.g. this Phys.SE post.

  6. Note that Lagrange equations are more general than Euler-Lagrange equations, cf. e.g. this Phys.SE post.

Within Newtonian mechanics, a comparison of various formulations is also discussed in this Phys.SE post.

--

$^{\dagger}$ By the word classical we will mean $\hbar=0$.

$\endgroup$
  • 1
    $\begingroup$ Broad? Why broad? It is either equivalent, or it is not. Which the answer is obviously no. And to nicely answer it, a single counter example would suffice as formal proof. $\endgroup$ – Physicist137 Dec 2 '16 at 14:17
  • $\begingroup$ @Physicist137 - This is a pretty good answer to the equivalence conditions! Equivalences can depend on conditions. Thus your stipulation "It is equivalent, or not." is not correct. $\endgroup$ – freecharly Dec 2 '16 at 19:54
  • 1
    $\begingroup$ @freecharly Well.. When I've read the question, I think I got the impression OP was asking mathematically. Ie, is set of equation $A$ equivalent to set of equations $B$, in the sense that $A\Longleftrightarrow B$? From math logic, either this sentence is true, or false. So..... And OP's comments seems to confirm it. I mean.. what other interpretation that is? $\endgroup$ – Physicist137 Dec 2 '16 at 19:59
  • $\begingroup$ The equivalence is not true in general. It seems like a hopeless task to try and formulate precise sufficient & necessary conditions for equivalence. The above answer only points out one-way implication arrows, not bi-implications. $\endgroup$ – Qmechanic Aug 8 '17 at 17:02
1
$\begingroup$

The equivalence of Newtons 2nd law with Hamilton's principle and Lagrange's equations means that you can (mathematically) derive Hamilton's principle and Lagrange's equations from Newtons law, and conversely that you can derive Newtons law from Hamilton's principle and Lagrange's equations.

First, the variational Hamilton's principle of stationary action is equivalent to the Euler-Lagrange equations (Lagrange equations of second kind)Hamilton's Principle, i.e., each follows from the other. Second, from Newtons laws follow the Lagrange equations. On the other hand, it can be easily seen that Newtons law follows from the Lagrange equations for cartesian coordinates.See e.g.Equivalence Newton and Lagrange

Thus Newtons law, Hamilton's principle and Lagrange's equations are equivalent, because they can mutually can be derived from each other. However, these equivalences might be restricted to certain conditions, like, e.g., assumption of conservative forces derived from a potential while the validity of the Lagrange equations or Hamilton's principle might be more general.

$\endgroup$
  • $\begingroup$ (+1) Thanks for your clear answer, but I am interested in the details under which the equivalence is valid. Can you kindly add some details. $\endgroup$ – H. R. Nov 15 '16 at 7:59
  • 1
    $\begingroup$ @H.R. - I don't know a monograph where the most general conditions are given. For the validity of Lagrange's equations sometimes a monogenic potential is stated, i.e., that all forces are derived from a potential that depends on generalized coordinates, velocitities, and time, and that constraints are holonomic, constraint forces do no work and thus friction is excluded. See these Harvard lectures (starting with 3) for a derivation of equivalences starting from Newton's law. users.physics.harvard.edu/~morii/phys151/lectures/Lecture03.pdf $\endgroup$ – freecharly Nov 15 '16 at 14:18
  • 1
    $\begingroup$ @H.R. Not sure if it would totally answer your questions but you should consider spending time in the book "Mathematical Methods of Classical Mechanics" by V. I. Arnold. It delves more deeply into the mathematical arguments of classical mechanics than many other books on the subject. At least give it some time and then you may be able to more suitably ask the particular detailed question that helps you gain the understanding you want. $\endgroup$ – K7PEH Nov 16 '16 at 16:17
  • $\begingroup$ @K7PEH: Thanks for the advice. :) My question is put on hold for a ridiculous reason! :( $\endgroup$ – H. R. Nov 16 '16 at 17:02
  • $\begingroup$ I disagree. They are not equivalent. When you move from newton to lagrange, you make restrictions, thus newton remains more general. That is, there are systems only newton can solve. $\endgroup$ – Physicist137 Dec 2 '16 at 14:47

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