Hamiltonian, Lagrangian and Newton formalism of mechanics

If my thinking is wrong please let me know. I have little knowledge on beyond-college physics.

For research purposes, I read a few introductions to these three formalisms of classical mechanics [1,2,part of 5](Hamiltonian, Lagrangian and Newton formalism). It seems to me that Newton formalisms is incapable of describing a quantum system(The uncertainty principle is not well addressed) while both Hamiltonian and Lagrangian are capable of describing a quantum system.So the Hamiltonian and the Lagrangian is simply a more general framework including Newton.

As Ben and Tobias pointed out in their answers, these three formalisms are equivalent their relationship are not simply inclusion but complementary. There are situations one of three systems that are particularly suitable to use.

For my purposes, I see there often seems to be one-to-one correspondence in-between Lagrangian construction and Hamiltonian construction for dynamic systems(OR in the simpler case the derived differential equations w.r.t. a chosen coordinate frame) and (sympletic) geometry when the concern is the dynamics on the manifold, say [3,4].

A curious question in my mind is that if these two physical formalisms are equivalent, then why only Hamiltonian is studied in most cases(say geometric analysis and sympletic geometry)?

(1)What is its(The Hamiltonian formalism) superiority over Lagrangian from mathematical perspective? Does it lead to a richer structure or more natural structure?(by structure I mean manifold structure over which the system is defined)

(2)Moreover, is there any example that is easily formalized in Hamiltonian formalism but too complicated/unnatural to formalize in Lagrangian?

Reference

[4]Koon, Wang Sang, and Jerrold E. Marsden. "The Hamiltonian and Lagrangian approaches to the dynamics of nonholonomic systems." Reports on Mathematical Physics 40.1 (1997): 21-62.

[5]Meyer, Kenneth, Glen Hall, and Dan Offin. Introduction to Hamiltonian dynamical systems and the N-body problem. Vol. 90. Springer Science & Business Media, 2008.

Motivation of OP

And as for my motivation, I primarily want to figure out why [Mumford&Michor] proposed Hamiltonian approach in (which looks not quite natural to me at first since they are just laying down a framework for dynamics on $Cur(\mathbb{R}^2)$.)

[Mumford&Michor]Michor, Peter W., and David Mumford. "An overview of the Riemannian metrics on spaces of curves using the Hamiltonian approach." Applied and Computational Harmonic Analysis 23.1 (2007): 74-113.

From these nice answers below, I feel that the most convincing reason why Hamiltonian approach is preferred in [Mumford&Michor] is that (1)the space of curves $Cur$ involves $\mathrm{Diff}(C)$ and Hamiltonian formalism is a convenient formalism to incorporate these transformations. (2)And the infinitesimal generators of $\mathrm{Diff}(C)$ can be used to describe the velocity field along the curves in $Cur$. (3)Combined with Tobias' answer, [Mumford&Michor] also said symmetry is a reason for Hamiltonian. Now I understand the sentence better.

...The Hamiltonian approach also provides a mechanism for converting symmetries of the underlying Riemannian manifold into conserved quantities, the momenta.[Mumford&Michor]

(Unless the authors disagree :)

Thanks again for everyone's input, I learnt a lot from you and willing to learn more!

• Clarification: I am aware of the condition when Lagragian can be formalized in Hamiltonian(mathoverflow.net/questions/258163/…), and I consider the case only the one-to-one correspondence exists. – Henry.L Mar 13 '17 at 20:37
• Henry.L, I've retagged. I think the tags classical mechanics and hamiltonian mechanics are more suitable. But if you disagree, feel free to rollback. – Konstantinos Kanakoglou Mar 16 '17 at 4:08

Each of the different formalism of classical mechanics has its advantages and disadvantages. However, in the end all three frameworks tend to be equivalent, and thus the following list is very subjective and there may be exceptions to all points.

Newton:

• Easily includes dissipative systems and is the only formalism that can handle non-potential forces.
• Taken literally only applies to systems without constraints (but can be extended to include some constraints using d'Alembert's principle; nonetheless, it is usual not the right framework to handle constraints).
• Doesn't say much about conserved quantities for symmetries.

Lagrangian:

• Highlights the variational principles underlying classical mechanics by its direct connection with the principle of least action
• as such, it allows for direct extensions to relativistic systems and field theories.
• Noether's theorem provides a direct link between symmetries and conserved quantities.

Hamiltonian:

• There are systems described by symplectic manifolds which are not cotangent bundles and thus have no Lagrangian equivalent (for example, internal degrees of freedom like spin)
• The framework of "momentum maps" is often superior to Noether's theorem and symmetry reduction is better understood in the symplectic case.
• Breaks the covariant nature of the Lagrangian formalism and thus has problems with relativistic theories.
• The Hamiltonian is always conserved along the dynamical evolution. This has advantages in e.g. bifurcation theory but makes it (nearly) impossible to discuss dissipative systems.
• The variational nature is hidden (or on a more positive note encapsulated in the symplectic form).
• Symplectic manifolds are interesting in its own right, even without a Hamiltonian (this is the advantage of having the additional structure given by the symplectic form).

Note that the equivalence of the Lagrangian and Hamiltonian view depends on the Legendre transformation. Sometimes, especially when constraints are present, the Legendre transformation is not an isomorphism and thus both formalism might be not equivalent!

• Doesn't say much about conserved quantities for symmetries....Noether's theorem provides a direct link between symmetries and conserved quantities....The framework of "momentum maps" is often superior to Noether's theorem and symmetry reduction is better understood in the symplectic case. Could you add a ref in support of this point? And which Noether theorem are you referring to? Thanks! – Henry.L Mar 14 '17 at 10:35
• en.wikipedia.org/wiki/Noether%27s_theorem. For your other question, just search for "Lagrangian reduction" and "symplectic reduction". – Tobias Diez Mar 14 '17 at 20:34
• @Tobias Diez: In the last line of your answer you say that "Sometimes, especially when constraints are present, the Legendre transformation is not an isomorphism and thus both formalism might be not equivalent". I guess you refer to non-holonomic constraints ? – Konstantinos Kanakoglou Mar 15 '17 at 2:34
• In the end it depends on the second derivative of the Lagrangian/Hamiltonian whether the Legenedre transformation is a (local) diffeomorphism. Thus even without constraints both formalism can be not equivalent, see Section 9.4 & 9.6 in Abraham Marsden "Foundations of mechanics" for an example in the context of the Kepler problem. – Tobias Diez Mar 15 '17 at 9:34
• Well, in that level of generality I agree. But as far as I understand, if we keep ourselves confined to systems with holonomic constraints only and whose Lagrangian function is "standard" or "regular" (i.e.the Hessian of the Lagrangian w.r.t. to the generalized velocities is non-degenerate, see: mathoverflow.net/q/258163/85967) then Legendre's transformation is an isomorphism and consequently the Lagrangian and Hamiltonian formalisms are equivalent. (also: thanks for citing the example from Marsden's book; unfortunately I do not have access to the book right now). – Konstantinos Kanakoglou Mar 15 '17 at 17:37

The three formalisms of classical mechanics, i.e. the Newtonian, the Lagrangian (analytical mechanics) and the Hamiltonian (canonical formalism) are generally not equivalent to each other -at least not in some strict sense of the word "equivalent" which could be considered valid for the totality of physical systems. But in fact they are, for lots of systems of interest. In order to be more specific, let me try to sketch their logical interconnections:

$\blacktriangleright$ If we start from Newtonian mechanics and the assumptions:

1. there are holonomic constraints only, i.e. constraints of the form $f(x,y,z,t)=0$, which are independent of the velocity.
2. arbitrary virtual displacements are assumed to be in directions that are orthogonal to the constraint forces, i.e. the constraint forces do no work (this is equivalent to saying that the constraint forces are normal to the hypersurface determined by the constraints at a given instant $t=t_0$).

If the constraints are scleronomic then the actual displacements $d\mathbf{s}$ are virtual displacements $\delta\mathbf{s}$.
($\Sigma$ is the hypersurface determined by the constraints $f(x,y,z)=0$, $\ R \$ the constraint force, $F$
is the resultant non-constraint force and $\mathbf{s}$ the position vector).
then Newton's $2^{nd}$ law is equivalent to the D'Alembert's principle which in turn implies the Euler-Lagrange's equations of motion i.e. the analytical mechanics formalism.
(Note that, systems sattisfying the second of the above assumptions are sometimes called "mechanical" or "pure mechanical" systems in the literature.
On the other hand, one does not have to go far in order to find systems violating one or both of the above assumptions: rolling without slipping is a common system with non-holonomic constraints and generally systems with resistance forces -various friction forces for example- violate the second of the above assumptions).

The converse implication, that is starting from the Euler-Lagrange equations and deriving D'Alembert's principle and thus Newton's $2^{nd}$ law is relatively straightforward.

$\blacktriangleright$ If we start from a Lagrangian and its Euler-Lagrange equations of motion, under the assumptions:

1. there are holonomic constraints only

2. the Lagrangian function is "standard" or "regular" in the sense that $\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0$, i.e.the Hessian of the Lagrangian w.r.t. to the generalized velocities is non-degenerate (see: When does a Lagrangian dynamical system have an equivalent Hamiltonian description? for more details on this point)

then we can derive the Hamiltonian and the canonical formalism of Hamilton's equations through a Legendre transformation. Note that Legendre's transformation, transforms functions on a vector space to functions on the dual space. In this case, it transforms the Lagrangian function (on the tangent bundle of the configuration space manifold) to the Hamiltonian function (on the cotangent bundle of the configuration space manifold).

Summarizing the above discussion: $$\small{ \left\{ \begin{array}{c} \text{Newtonian} \\ \text{mechanics} \end{array} \right\} \underset{\begin{array}{c} \text{pure} \\ \text{mechanical} \\ \text{system} \end{array}}{\overset{\begin{array}{c} \text{holonomic} \\ \text{constraints} \end{array}}{\mathbf{\leftrightsquigarrow}}} \left\{ \begin{array}{c} \text{d'Alembert's} \\ \text{principle} \end{array} \right\} \leftrightsquigarrow \left\{ \begin{array}{c} \text{Lagrangian} \\ \text{mechanics} \end{array} \right\} \underset{\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0}{\overset{\begin{array}{c} \text{holonomic} \\ \text{constraints} \end{array}}{\mathbf{\leftrightsquigarrow}}} \left\{ \begin{array}{c} \text{Hamiltonian} \\ \text{mechanics} \end{array} \right\} }$$ Remarks:
(a). in the framework of Newtonian mechanics, forces may depend on positions and velocities but not on accelerations and
(b). throughout the preceding disussion, we consider all non-constraint forces to be derivable from generalized scalar potentials depending on coordinates and -at most- linearly on velocities: $V=U(q_i,t)+A_j(q_i,t)\dot{q}_j$. Such systems are more general than conservative systems and fall into the class of monogenic systems.

Now, regarding your first question: I don't think it would be suitable to speak about "superiority" or "richer structure". From a technical aspect, Euler-Lagrange's equations are a system of $n$, second order, ODE and the Lagrangian function involved "lives" on the tangent bundle while The Hamiltonian formulation comprises of a system of $2n$, first order, ODE, involving the Hamiltonian function which "lives" on the cotangent bundle (of the configuration space manifold).
If we consider the passage to quantum mechanics, then both formalisms are suitable to handle the elementary aspects of the quantisation problem (at both levels, first and second quantization as well): Hamilton's equations have almost the same typical form in quantum mechanics with their classical counterparts, although their interpretation is quite different in the quantum case -but this is an apparently different story. The road to quantisation through the hamiltonian formalism is generally refered to as canonical quantisation while the road through Lagrangian formalism is known as path integral quantization.

Regarding your second question, and since you are asking for references, this article appears to discuss examples of classical Hamiltonian systems possessing no Lagrangian formulation. On the other hand, the Hamiltonian and the Lagrangian formulation of geometrical optics -as has already been mentioned in another answer- is a well known classical system possessing no meaningful description at the level of Newton's laws.
From one point of view, one may say that the Hamiltonian formalism is "wider" in the sense that it includes several systems of ODE not being directly or obviously related to classical mechanics (an example is the lagrangian/hamiltonian description of Maxwell's laws of the electromagnetic field) or to physics at all. In practise, the hamiltonian is frequently considered to be some abstract function while the Lagrangian is more intimately related to the concepts of kinetic and potential energy of some system. I think that you might also find some interest in the article Classical Mechanics Is Lagrangian; It Is Not Hamiltonian by Erik Curiel.
However, from another point of view, Newtonian formulation is "wider", in the sense that if one is to consider, frictions, dissipative forces, non-holonomic constraints etc then -although there are various formal generalizations of Lagrange's and Hamilton's equations in order to deal with such systems- usually one turns to the fundamentals that is Newton's laws.
So, concluding, i agree that there are no simple proper subset relationships between these three formulations of classical mechanics.

Some further References:

1. Mathematical methods of classical Mechanics, by V.I. Arnold, and
2. A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, by E.T. Whittaker,

are invaluable -imo- sources, delving into the foundations, not trying to hide the pitfalls under the carpet, with an eye on details (either in the computations or the arguments).

1. This: https://physics.stackexchange.com/q/89035/130499 question in physics.stackexchange has a natural overlap with the OP and has lots of answers. You might be interested in taking a look there as well.

P.S.: One last thing: the above disussion and the diagram provided, aimed at emphasizing the logical interconnections of these three formalisms and the usual assumptions upon which one is being derived from or imply the other(s). However, it should be noted that both the Euler-Lagrange's and the Hamilton's equations, can be derived -each one of them seperately and independently of Newton's laws- in an "axiomatic" (or should I say "ad hoc" ?) manner through suitable variational principles: these are the Hamilton's principle (from which euler-lagrange's eqs are derived) and the modified Hamilton's principle (from which the canonical eqs of hamiltonian mechanics can be derived). They both determine the corresponding equations of motion and thus the evolution of the system, through the demand that the trajectories are such that the corresponding action-functional $S$ gets a stationary value (its $\delta$-variation becomes zero): $$\delta S=0$$ In the former case $S=\int_{t_1}^{t_2} Ldt$ and the principle is applied in the configuration space, while in the later case $S=\int_{t_1}^{t_2} (\dot{q}p-H)dt$ and the corresponding principle is applied in the phase space.

It is actually this possibility of independent foundation of these formalisms that inspired/enabled the extension of these formalism to non-mechanical systems such as classical fields.

• Thanks so much for your nice answer/response! I need some time to digest all these materials! My deep appreciation! – Henry.L Mar 15 '17 at 20:19
• Thanks for the revision and added ref. I think this is a really beneficial post to me and everyone involved! – Henry.L Mar 16 '17 at 2:57

It seems to me that Newton formalisms is incapable of describing a quantum system(The uncertainty principle is not well addressed) while both Hamiltonian and Lagrangian are capable of describing a quantum system. So the Hamiltonian and the Lagrangian is simply a more general framework including Newton.

This part is not correct. The Newton, Hamilton, and Lagrangian formulations are all classical, and they are all essentially equivalent. None of them is a generalization of any of the others. Given a description of any classical system in one of these systems, you can normally find its description in the other systems. (There are cases of systems that can be described in one of these formalisms but not another, e.g., optics can be described in the Hamiltonian formalism but not using Newton's laws. But I don't think any of these are simple proper-subset relationships.)

Hamiltonian and Lagrangian mechanics are not descriptions of a quantum-mechanical system. The classical description of a system can often be converted into such a description, using various heuristic recipes such as canonical quantization. Sometimes these methods of quantization fail (as with quantum gravity), and sometimes they are ambiguous or need creative tweaking.

The reason the Hamiltonian formalism is nice when quantizing a system is that the Hamiltonian is the infinitesimal generator of time evolution. The Hamiltonian approach makes manifest the unitary nature of quantum mechanics.

One reason that the Lagrangian formalism is nice is that the Lagrangian transforms relativistically as a scalar, so that by specifying a Lagrangian, we are guaranteed to get equations of motion that are coordinate-independent. This is different from the Hamiltonian approach, in which time has a special status, setting it apart from the spatial coordinates.

• Thanks so much for the input! (1) Is there any ref. that supplement your comment that "There are cases of systems that can be described in one of these formalisms ..." as answer to (2) in OP. (2)But Hamiltonian is also coordinate-independent according to Carlo's answer, right? (3)And I do not quite understand your comment "Sometimes these methods of quantization fail (as with quantum gravity), and sometimes they are ambiguous or need creative tweaking." Is there any ref. you can point me to? Thanks a lot for this input! – Henry.L Mar 14 '17 at 1:01
• It seems that there is a lot underlying physical material(say, quantum gravity...totally new to me) that I should know. Since my focus is not math-physics but random matrices, is there any ref that you think a math person can cope with? – Henry.L Mar 14 '17 at 1:02
• And due to your first paragraph, how should I understand p.5's comment of macs.hw.ac.uk/~simonm/mechanics.pdf? Again thanks so much for your kindness. – Henry.L Mar 14 '17 at 1:05
• You can understand they are equivalent. But just because you CAN describe a complicated system with many constraints with Newton's equations doesn't meet you should. You should use Lagrange or Hamiltonian then. – AHusain Mar 14 '17 at 1:19
• @Henry.L: But Hamiltonian is also coordinate-independent according to Carlo's answer, right? In relativity, we have four-dimensional spacetime with four coordinates, and it's OK to make any change of coordinates that is a diffeomorphism. In the Hamiltonian formalism, time is treated as a parameter, not a coordinate. – Ben Crowell Mar 14 '17 at 2:29

In response to (1), a key advantage of Hamilton's equations of motion is that they remain invariant under a large class of "canonical" transformations, $(x,p)\mapsto (Q(x,p),P(x,p))$ for some scalar function $S(x,P)$, in which both coordinates and momenta are transformed. For the Lagrange equations of motion only coordinate transformations, $x\mapsto Q(x)$, are allowed. This larger freedom exists because phase space has twice the number of dimensions of configuration space. One way to use this extra freedom is to find conservation laws, $\dot{Q}=0$, by searching for a canonical transformation with vanishing Poisson bracket $\{Q,H\}=0$.

• Could you specify more details about "canonical transformations" and refs.? So basically Hamiltonian systems are "more transform-invariant" than Lagrangian systems? Is there an example to illustrate this point? – Henry.L Mar 13 '17 at 21:58
• This might be a helpful link – Carlo Beenakker Mar 13 '17 at 22:02
• They are not any more coordinate free, it's just easier to list transformations that leave the system invariant, is what I believe Carlo is saying. – Ryan Budney Mar 13 '17 at 22:02
• @RyanBudney --- certainly, the two formalisms are fully equivalent, with the same conservation laws, so it's ultimately a matter of convenience of one formulation over another. – Carlo Beenakker Mar 13 '17 at 22:04
• Here is an i-python notebook setting up a double pendulum with friction, in Lagrangian formalism. github.com/delooper/math-phys.248-2017/blob/master/Week.9/… – Ryan Budney Mar 13 '17 at 22:11