Talk:Hamiltonian mechanics

Approach

Hmm, wouldn't an introduction to Hamiltonian mechanics WITHOUT starting from Lagrangians and starting from symplectic spaces and Poisson brackets be more natural? Phys 19:09, 5 Sep 2003 (UTC)

Er, NOT to most physicists chemists and engineers... :-) Linuxlad 19:01, 17 Mar 2005 (UTC)

Somewhere in the article it is written: "One thing which is not too obvious in this coordinate dependent formulation is that different generalized coordinates are really nothing more than different coordinatizations of the same symplectic manifold."

As an engineer I have no idea what a symplectic manifold is. It makes it rather difficult when one is trying to refresh one's ideas about Hamiltonians that one should now have to go and try and understand what "Symplectic manifolds are". Please - When writing articles proceed from the simple to the complex, from the concrete to the abstract and from the particular to the general and not vice versa Phillip (talk) 06:03, 21 April 2011 (UTC)..

Page move

The move of page title isn't a good idea. We generally prefer general titles (Hamiltonian mechanics), to more special ones, such as particular equations.

Charles Matthews 10:36, 16 Jun 2004 (UTC)

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For what it's worth, I agree with Charles. Also, can someone please explain the current mess of Talk pages involved? Especially Talk:ȧ£æžåŠ›å... (Mozilla won't let me type the whole of it).

Taral 17:42, 17 Jun 2004 (UTC)

What I want to know is why the Hamiltonian view of mechanics is more useful than the classical version? What can engineers and scientists do with it that they cannot do w/o it?

Good question. Hamiltonians are great for solving problems that involve transfer of energy, and momentum, between rigid bodies, and in some circumstances allow problems that would be hard to solve using just Newton to be solved in very few lines of equations. Examples include things like compound pendulums and so on. However, very few engineers use Hamiltonians in their day to day lives, to a large extent they have been replaced by kinematic programs like MSC ADAMS, which have the advantage of being able to easily handle flexible bodies, non ideal contact with damping, and friction. I did not study them at uni, and must confess that this page is absolutely no help at all to me, and I work in the area of dynamics. I'm sure it makes sense to physicists. Greglocock 04:48, 3 January 2007 (UTC)

Norm

Liouville's equation here shows total d/dt equal to the Poisson bracket. A physicist would expect to write partial d/dt here, because the essence of Liouville is that total d/dt, meaning the convective derivative taken with the particle, is zero. See main Liouville's theorem (Hamiltonian)

Linuxlad 12:28, 9 Nov 2004 (UTC)

Looking at this issue again, Goldstein's Classical Mechanics

(Later Note - this is Goldstein 1964 edition ie 1950 2nd reprint)

at the ready, there appear to be two differences from what I'd expect:-

In the absence of any further constraints on f, I'd expect (cf Goldstein eqn 8-58) that:-

a) the convective/total time derivative of f equalled the Poisson bracket of f & H PLUS the partial time derivative of f.

b) In the _particular_ case of phase space density (or probability) it is possible to show that the convective derivative is zero, so that the partial derivative equals minus the Poisson bracket (Goldstein eqn 8-84) - but note that this result does NOT follow trivially from the result for general f as implied.

(So I reckon that's 2/3 violations of my naive physicist's expectations) - I hereby give notice that I may edit accordingly :-)

Linuxlad 10:18, 10 Nov 2004 (UTC)

This page certainly needs some work. For example it doesn't give (and neither does the page linked to) the classical expression of the Poisson bracket. As far as I can see, though, the definitions are the standard ones, such as are given in Abraham and Marsden, Foundations of Mechanics, though.

Charles Matthews 11:08, 10 Nov 2004 (UTC)

Just to be clear, I'm looking at the first part of the section entitled Mathematical formalism. and specifically the two equations for f and for ρ

So

Equation 1 For general f :-

${\frac {df}{dt}}={\frac {\partial f}{\partial t}}+\sum _{i=1}^{N}\left[{\frac {\partial f}{\partial q_{i}}}{\dot {q}}_{i}+{\frac {\partial f}{\partial p_{i}}}{\dot {p}}_{i}\right]$

and on substituting Hamilton's equations for terms 2 & 3 we get the Poisson Bracket of f & H plus the partial time derivative.

So, (from standard maths methods viewpoint), first equation has partial df/dt missing.

2nd equation (for ρ) - Liouville's theorem is TOTAL d by dt of phase space density is zero, which does NOT follow directly from above, and is not what's given there anyway (you need to change total time derivative to partial)!

I have now edited acccordingly

Bob

Linuxlad 11:30, 10 Nov 2004 (UTC)

Hamiltonian systems

There is a redirection from "hamiltonian system" to this page, which is certainly better than nothing, but not really satisfying in my opinion of view.

There should be Hamiltonian systems, integral of motion, integrable systems, Lax pairs, ... Amateurs, please try to create some stubs... MFH 17:48, 17 Mar 2005 (UTC)

I utterly agree. A Hamiltonian system is never defined anywhere. I vaguely remember what it is from my lectures ten years ago but that's not enough. I try a stub please correct me! --131.220.68.177 16:34, 19 August 2005 (UTC)

Classical Relativistic Electromagnetic Hamiltonian

This article seems to have sprouted a section:

Classically (non-quantum mechanical), for a particle of charge q and mass m in an electromagnetic field with vector potential A and scalar potential φ, the relativistic Hamiltonian is

$H={\sqrt {(\mathbf {p} -q\mathbf {A} /c)^{2}c^{2}+m^{2}c^{4}}}+q\phi$

where the term qφ is potential energy, the radical term is total energy of the particle, in terms of the energy-momentum relation, with momentum replaced by what may be construed as a "gauge covariant momentum": it is called kinetic momentum.

However, I don't beleive this has anything to do with hamiltonian mechanics. Well, its an example of a hamiltonian, but its one of a thousand examples, and doesn't seem to be a particularly good one. If the goal is to add examples, something simpler should be given: e.g. ball on an inclined plane or something like that. Perhaps this belongs in its own article? linas 00:16, 26 October 2005 (UTC)

Don't fully agree with the above. Formulating the classical Hamiltonian for a particle in an e/m field, is a standard way (eg 3rd year undergraduate level, I recollect) of working to the important quantum mechanical case. Linuxlad 11:14, 11 November 2005 (UTC)

We have distinct articles for things like the simple harmonic oscillator and the hydrogen atom. I strongly urge you to start an independent article on the --classical relativistic electromagnetic Hamiltonian or maybe relativistic charged particle. Yes, I am well aware that this Hamiltonian is treated in many books, including the awe-inspiring development in Landau and Lifschitz. However, Hamiltonian mechanics has many, many facets to it that stretch far beyond what the third-year undergraduate is aware of. A distinct article for this will allow it to be expanded as it deserves (after all, L&L found 100+ pages of things to say about it). linas 22:36, 18 November 2005 (UTC)

I don't think you've made your case here! The example neatly extends the idea from the pure mechanical variables into e/m using probably the most important example of a single-particle system. But perhaps you're right - this is really a page for people whose searching for deeper mathematical insights should not be sullied with examples of the real world :-) Linuxlad

I frankly don't understand why we are arguing about this. There are thousands of Hamiltonians, and while relativistic charged particles are interesting and important for many reasons (and deserve their own article (hint hint)), I can't imagine why they are "important" for hamiltonian mechanics. Surely a ball running down an inclined plane is a "more important" example (as it illustrates contact geometry, sub-Riemannian geometry and lagrange multipliers all at the same time? Maybe a pendulum, as that can illustrate the transition to chaos, the KAM torus and the phase-locked loop at the same time? The lorenz attractor? An example from fluid dynamics? Why charged particles?

As to the quality of this article; at this current time, it is rather incomplete at this time. However, that's a different issue. linas 01:13, 19 November 2005 (UTC)

(Have you ever read the Landau and Lifshitz treatment of the classical relativistic charged particle?) linas 01:19, 19 November 2005 (UTC)

Proposed mergefrom Hamilton's equations

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section.

The result was merge sbandrews (t) 18:51, 23 June 2007 (UTC)

This article is supposed to be the category leader for Hamiltonian mechanics. But it offers no orientation for a general reader coming in who has never heard of either Hamiltonians or Lagrangians.

Such a reader might have been directed here by Equipartition theorem for instance (currently the object of scientific peer review at Wikipedia:Scientific_peer_review/Equipartition_theorem)

My belief is that this would be best fixed by merging in most of the content from Hamilton's equations at the top of the article, to give an overview, and an idea of where we're trying to get to, and why; before we get into the algebra fest.

The two articles appear to cover the same subject, namely what are Hamilton's equations, and where do they come from, and what new view of the world do they give one.

So it seems to me a fine idea to merge them, and gather together the best content from each. Jheald 14:09, 3 April 2007 (UTC)

That sounds like a good idea, and I'd support the move. — Rebelguys2 ^talk 03:54, 11 April 2007 (UTC)

make sense, I'll have a go... sbandrews (t) 10:36, 10 June 2007 (UTC)

ok I'm done, comments or reverts welcome... sbandrews (t) 11:00, 10 June 2007 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

Derivation 1

Isn't the form of Lagrange's equations given there plain wrong? Why is the generalized force entering the equation at all? 131.215.123.218 (talk) 23:03, 10 June 2008 (UTC)

It seems that derivation is wrong. The lagrange equation is not

${\frac {\mathrm {d} }{\mathrm {d} t}}{\frac {\partial {\mathcal {L}}}{\partial {{\dot {q}}_{i}}}}-{\frac {\partial {\mathcal {L}}}{\partial q_{i}}}=F_{i}$

this but

${d \over dt}{\partial T \over \partial {\dot {q}}_{\sigma }}-{\partial T \over \partial q_{\sigma }}=F_{\sigma }\qquad \sigma =1,\;\cdots ,\;3N-k$,

where T is kinetic energy. see Lagrange's Equations -- from Eric Weisstein's World of Physics.

Also, the Hamilton's equation seems wrong. The right form is

${\frac {\partial {\mathcal {H}}}{\partial q_{j}}}=-{\dot {p}}_{j},\qquad {\frac {\partial {\mathcal {H}}}{\partial p_{j}}}={\dot {q}}_{j},\qquad$.

see Hamilton Equations -- from Wolfram MathWorld. --StarLight (talk) 10:52, 25 July 2008 (UTC)

Mnemotechnics

One of the subsections on this page is titled "Basic physical interpretation, mnemotechnics". Can anyone explain the relevance of mnemotechnics to this section? AstroMark (talk) 13:39, 15 August 2008 (UTC)

No? I shall remove it then. AstroMark (talk) 14:03, 27 August 2008 (UTC)

Example?

Would it be useful to provide an example or two, to demonstrate how Hamiltonian mechanics is used, and what insight they provide? Perhaps the examples from Lagrangian mechanics could be brought over and redone, highlighting the difference between the two. I would do it, but I don't really know the answers to my own questions. User:!jim^talk _contribs 22:45, 13 December 2008 (UTC)

DS1000's added material

I again removed an erroneous statement added to the section Hamiltonian mechanics#Charged particle in an electromagnetic field by DS1000 (talk · contribs). He said "The above remains valid< ref>http://academic.reed.edu/physics/courses/Physics411/html/411/page2/files/Lecture.10.pdf< /ref> at relativistic speeds with the observation that: $p_{i}=\gamma _{i}m_{i}{\dot {x}}_{i}$" (I broke the "ref" to keep this in-line). The erroneous part of this statement is that the "above remains valid" which is only true for ${\mathcal {H}}=\sum _{i}{\dot {x}}_{i}p_{i}-{\mathcal {L}}\,,$ but not for the remainder of the section which is non-relativistic. The reference given uses a non-conventional definition of the dot (time derivative) which makes it the derivative with respect to proper time rather than coordinate time. It is possible that he got the erroneous formulas from the part of the reference which contains (10.36) which is a different (from the usual relativistic Hamiltonian) which the author calls "functionally equivalent". But that is for his special purpose. Please do not re-add this material without discussing it here first. JRSpriggs (talk) 05:12, 18 December 2008 (UTC)

You aren't making any sense, first you admit that it is correct for relativistic domain and then you complain that it isn't applicable for the non-relativistic domain.Of course it isn't.What is with the stuff you are writing about not using the derivative wrt proper time? The derivative wrt proper time is gamma*derivative wrt coordinate time. Didn't you see the gamma? This is a repeat of the situation we had with your editing the relativistic force derivation . Is it that you want the relativistic Hamiltonian separated out? I did that, I hope that this is satisfactory for you. DS1000 (talk) 05:36, 18 December 2008 (UTC)

Your many editings are completely incorrect, please stop to use the wikipedia to defend your incorrect points in sci.physics.relativity < ref>http://groups.google.com/group/sci.physics.relativity/msg/e80d792ecfc7b42f< /ref>. First, the square Hamiltonian is not valid for relativistic velocities, second your momentum $p_{j}=\gamma _{j}m_{j}{\dot {x}}_{j}$ is completely wrong for the given Lagrangian, because lacks the magnetic contribution. You were explained this before in both USENET and BLOGGER and corrected. You were explained what is the correct momentum and relativistic Hamiltonian JuanR (talk) 14:19, 18 December 2008 (UTC).

What are you rambling about? You aren't making any sense. DS1000 (talk) 16:42, 18 December 2008 (UTC)

Of your incorrect editing of relativistic Hamiltonian deleted by JRSpriggs (talk · contribs) at 21:16, 16 December 2008, you resubmitted and was deleted again by by JRSpriggs (talk · contribs) at 04:54, 18 December 2008. This is also archived in this section above where JRSpriggs (talk · contribs) wrote I again removed an erroneous statement added to the section Hamiltonian mechanics#Charged particle in an electromagnetic field by DS1000 (talk · contribs).

The more recent version does not contain the wrong Hamiltonians and momentum you wrote in many previous incorrect versions but now contains the correct momentum and square root Hamiltonian was given to you so early as the day 12 Dec. However, you more recent editing is not still acceptable.

First, you use notation is incompatible with the rest of the page. for example the relativistic Hamiltonian in Hamiltonian mechanics#CRelativistic_charged_particle_in_an_electromagnetic_field does not reduce to the square root Hamiltonian given in section Hamiltonian mechanics#Charged particle in an electromagnetic field because you are using a different definition for $p_{j}$. There is not reason to change notation in the middle of page. Momentum can be given as $p_{j}=m\gamma {\dot {x}}_{j}+eA_{j}$ and for low velocities reduces to the $p_{j}=m{\dot {x}}_{j}+eA_{j}$ given in previous section.

Second, your claimed equivalent and much nicer representation as function of the relativistic momentum makes no sense. You do not understand the Legendre transformation. The state variable in the Hamiltonian is your given $P$ not your $p$. Moreover your 'representation' $H=\gamma mc^{2}+e\Phi$ is not a function of $p$ as you claim because $\gamma$ is a function of your $P$. Again you are mixing apples and oranges.

Third, your claim that the Hamiltonian can be viewed as the sum between the total relativistic energy E and the potential energy is also incorrect. The total relativistic energy $E$ is obtained from the zero component of the four momentum in a covariant approach. I.e. $p^{\mu }=(E/c,{\vec {p}})$. Evidently $E$ is not $H-V$, except for a free particle $V=0$.

Please stop from submitting more incorrect editings of this page, your 47 incorrect versions given in the last 4 days are enough!! JuanR (talk) 12:38, 19 December 2008 (UTC)

I see, you are some kind of stalker-crackpot, you can't understand the math and you compensate by stalking. Find a different place to conduct your stalking, ok? —Preceding unsigned comment added by DS1000 (talk • contribs) 14:48, 19 December 2008 (UTC)

It is evident you cannot learn but fortunately JRSpriggs (talk · contribs) continues correcting your wrong editings. He has today corrected your incorrect claim that $E$ is the total energy < ref>http://en.wikipedia.org/w/index.php?title=Hamiltonian_mechanics&oldid=259112015< ref> as I warned in this talk before.

However, the notation is still inconsistent with that given in previous non-relativistic sections. Notation in all the page would be unified! Moreover, symbol $E$ usually denotes the zero component of four-momentum. The Hamiltonian is better splinted as $H=mc^{2}+T+V$, where $T$ is relativistic kinetic energy (see Goldstein or any other textbook for details).

The claim that exist an equivalent representation as function of the relativistic (kinetic) momentum is still not acceptable. The Legendre transformation changes state variables $(x,v)$ into your $(x,P)$. The kinetic momentum $p$ is not state variable in Hamiltonian mechanics (except when there is not magnetic potential, of course because then $P=p$), the final expression is not a Hamiltonian and the symbol $H$ would not be used for that function. One alternative notation could be $H=H(x,P)$ for the Hamiltonian and $h=h(x,p)$ for the last expression.

I apologize by not editing by myself the article, but I still need to practice more with this system. JuanR (talk) 13:57, 20 December 2008 (UTC)

The derivation is a matter of very simple math, it is not my fault that you are unable to follow it. Your sole reason for registering seems to be trolling DS1000 (talk) 18:33, 20 December 2008 (UTC)

I do not understand why you insist on either signed or unsigned ad hominem and insults when this talk is about the physics and math in the article. As reported above your last editing < ref>http://en.wikipedia.org/w/index.php?title=Hamiltonian_mechanics&oldid=259115585< /ref> still contains some incorrect claims reflecting your misunderstandings about the Legendre transformation, the state variables, and Hamiltonian mechanics in general. I have suggested several modifications to your last editing. Also the notation you choose would be revised because is inconsistent with the rest of the article. I apologize by not editing by myself the article, but I still need to practice more with this system. JuanR (talk) 13:39, 21 December 2008 (UTC)

Now that you have separated the material into its own section, the new version (but not the old one) appears to be correct. However, it would be good to say something about what the Lagrangian is in this case. See Lagrangian#Special relativistic test particle with electromagnetism.

I said that the dot was used unconventionally in the reference (the pdf file). I did not say that you used it that way in your addition; I saw the gamma. JRSpriggs (talk) 06:12, 18 December 2008 (UTC)

section on deformable bodies

First, there are no sources cited for this section, and second, shouldn't this be in the article on Hamilton's principle? —Preceding unsigned comment added by 72.10.133.106 (talk) 01:15, 12 January 2009 (UTC)

derivation by time

Hello. You have mistake in the article - the right equation is that the total derivative of H by time is minus the partial derivative of L by time. 23:31, 10 February 2009 (UTC) —Preceding unsigned comment added by ירון (talk • contribs)

Article about hamiltonian

Am I blind or isn't there any article about classical hamiltonian itself (a function)? Magdulewicz (talk) 13:49, 23 June 2010 (UTC)

I think that the function (in classical physics) is adequately covered in this article. Is it not? What else would you have us say about it which is not already in this article? JRSpriggs (talk) 06:39, 24 June 2010 (UTC)

I guess I'd talk about effective potentials, but that's just my two cents. Effective potentials using the hamiltonian are an interesting thing that you can do without talking about the dynamics of the situation.--24.7.197.142 (talk) 08:53, 3 February 2011 (UTC)

Symbols

Does anyone know the origin of the abbreviations T and V that are "traditionally" used for kinetic and potential operators? The closest guess I can come up with is translational T, but I haven't a clue for V. Laburke (talk) 21:07, 12 January 2011 (UTC)

According to Marsden & Ratiu (1999) p. 231 the notation T and V seems to go back (at least) to Lagrange. In a footnote they refer to further possible sources on historical background. -- Crowsnest (talk) 06:45, 14 January 2011 (UTC)

Merci, mais je reste avec la même question, pourquoi V? <<un champ Vectoriel?>>. Thank you for looking into this, but I didn't see what V stands for. Perhaps "champ Vectoriel"?Laburke (talk) 20:22, 13 March 2011 (UTC)

Frankly at this point in time, it does not matter for what word "V" is short, if indeed there is any such word. Given the immense multitude of concepts and the limited number of mathematical symbols available to represent them, re-use of symbols (overloading) and arbitrary choice of symbols (merely because they are different from some other symbols one has chosen to use) are bound to occur. JRSpriggs (talk) 02:22, 14 March 2011 (UTC)

Relating the time derivatives of the lagrangian and the hamiltonian.

This is really nitpicky. I'm fairly sure you can prove that the total derivative of the hamiltonian equals the negative of the partial derivative of the lagrangian. I don't mean to be weird about this because I look at the proof given in the article, and I'm convinced of the statement about the partial derivatives of the two equaling each other. Though, I'm pretty sure it is actually the way that I have written it here. This is a much deeper statement. I think this should be changed :). Anybody interested in finding a source for this?

${\frac {d{\mathcal {H}}}{dt}}=-{\partial {\mathcal {L}} \over \partial t}$ --24.7.197.142 (talk) 08:49, 3 February 2011 (UTC)

Well, I'm stupid. Somebody else mentioned this problem in this discussion page. I'm feeling stupid now for not reading through this discussion page more carefully, but I think I'll just leave this here.--24.7.197.142 (talk) 08:55, 3 February 2011 (UTC)

I did more research and the partial of the hamiltonian is equal to the total derivative of the hamiltonian in time. So, everybody is right :). Here's what it is:

${\frac {d{\mathcal {H}}}{dt}}=-{\partial {\mathcal {L}} \over \partial t}={\frac {\partial {\mathcal {H}}}{\partial t}}$--24.7.197.142 (talk) 09:00, 3 February 2011 (UTC)

See V.I. Arnold (1978) Mathematical methods of classical mechanics, p. 67, for the total time derivative of ${\mathcal {H}}({\boldsymbol {p}}(t),{\boldsymbol {q}}(t),t)$:

${\frac {d{\mathcal {H}}}{dt}}={\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}{\dot {\boldsymbol {p}}}+{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}{\dot {\boldsymbol {q}}}+{\frac {\partial {\mathcal {H}}}{\partial t}}={\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}\left(-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}\right)+{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}}+{\frac {\partial {\mathcal {H}}}{\partial t}}={\frac {\partial {\mathcal {H}}}{\partial t}}.$

-- Crowsnest (talk) 09:20, 3 February 2011 (UTC)