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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.

Hosein Rahnama
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  • 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. – Jim Nov 15 '16 at 13:40
  • @Jim: See this link – Hosein Rahnama Nov 15 '16 at 20:04
  • @H.R. They supposedly showed Lagrangian mechanics are equivalent to Newtonian mechanics, which is not technically the same as equivalence to solely Newton's second law. Furthermore, they showed this by taking the derivative of both sides of two equations and showing that the LHS's are equivalent as are the RHS's. This is not what I'd call rigorous. Saying the derivatives are equivalent does not mean the original expressions are equivalent. The x derivative of $x^2+1$ equals that of $x^2+e^y$ but they are hardly equivalent expressions. – Jim Nov 16 '16 at 13:14
  • Hamilton's principle can be utilized to explore the physics of quantum field theories; Newton's second law cannot. Therefore, there must be a fundamental difference between them, which both makes one compatible with higher-level physics like QFT but not the other AND, by the very existence of a difference between them, makes them not equivalent. If they were truly equivalent, you could use Newton II in place of Hamilton's principle everywhere without there being any differences. Show me this is possible and I'll admit they are the same thing. – Jim Nov 16 '16 at 13:26
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    I think that this is a very justified specific question that merits an answer by the community. – freecharly Nov 16 '16 at 18:31
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    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. – Physicist137 Dec 02 '16 at 14:37
  • @H.R. I Would do if I could. But I have insufficient reputation to do such a thing. Well, the proof is a "consequence" of the construction. For instance, when you move from newton to lagrange $L = T-V$, you assume a potential function exists. What if it does not exist? What if $\nabla\times\mathbf F\neq 0$? Then newton is more general since it can solve a problem without need of potential. (This is restricting to classical mechanics only). In overall physics, there is no such thing as "more general" as I told before. =D. – Physicist137 Dec 02 '16 at 15:10
  • @Physicist137: mmmmmm, We can derive Lagrange equations for non-conservative systems from Newton! Can't we? – Hosein Rahnama Dec 02 '16 at 15:13
  • @H.R. As you can. But it won't have $L = T-V$ since a generalized force will appear. But then the Euler-Lagrange equation won't be the one with $L$, but the one with $T$ only and $Q$ (the generalized force). This case yes, is equivalent. But then, a lagrangian function $L$ might or might not exist. Ie, you cannot jump to hamilton by legendre transform if $L$ does not exist. And since a lagrangian might not exist, I wouldn't call it "Lagrangian System".. – Physicist137 Dec 02 '16 at 15:16

2 Answers2

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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.

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

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Qmechanic
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    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. – Physicist137 Dec 02 '16 at 14:17
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    @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. – freecharly Dec 02 '16 at 19:54
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    @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? – Physicist137 Dec 02 '16 at 19:59
  • 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. – Qmechanic Aug 08 '17 at 17:02
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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.

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freecharly
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  • (+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. – Hosein Rahnama Nov 15 '16 at 07:59
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    @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. http://users.physics.harvard.edu/~morii/phys151/lectures/Lecture03.pdf – freecharly Nov 15 '16 at 14:18
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    @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. – K7PEH Nov 16 '16 at 16:17
  • 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. – Physicist137 Dec 02 '16 at 14:47
  • @I pointed out in my answer that the equivalences might be restricted to certain conditions. I think that the Lagrange equations with non-conservative (generalized) force terms are fully equivalent to Newton's equations. See Goldstein, Classical Mechanics. Maybe you can point out an example where Newtons equations are more general. – freecharly Dec 02 '16 at 16:08
  • Oh.. They are equivalent (ie, the one with generalized force). But I don't think it is a Lagrangian system since there is no Lagrangian function. (Ie, $L = T-V$). I could be wrong... – Physicist137 Dec 02 '16 at 20:02