Talk:Hamiltonian vector field

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This article was automatically assessed because at least one WikiProject had rated the article as start, and the rating on other projects was brought up to start class. BetacommandBot 09:53, 10 November 2007 (UTC)[reply]

Poisson bracet

I changed the eq. $X_{\{f,g\}}=[X_{f},X_{g}]$ to $X_{\{f,g\}}=-[X_{f},X_{g}]$ .

If we define

\{f,g\}=\omega (X_{f},X_{g})=df(X_{g})={\mathcal {L}}_{X_{g}}f\ ,

then

X_{\{f,g\}}=-[X_{f},X_{g}]

since

{\begin{aligned}&\iota _{[X_{f},X_{g}]}\omega \\=&({\mathcal {L}}_{X_{f}}\circ \iota _{X_{g}}-\iota _{X_{g}}\circ {\mathcal {L}}_{X_{f}})\omega \\=&{\mathcal {L}}_{X_{f}}\circ \iota _{X_{g}}\omega \\=&(\iota _{X_{f}}\circ d+d\circ \iota _{X_{f}})dg\\=&d(\iota _{X_{f}})dg)\\=&-d\{f,g\}\\\end{aligned}}

--刻意 06:52, 10 November 2008 (UTC)[reply]

The sign convention used in this article for Hamiltonian vector fields is somewhat unusual - at least for mathematical physicists. For the commonly used convention $X_{\{f,g\}}=[X_{f},X_{g}]$ is indeed true. 93.222.25.88 (talk) 13:12, 2 May 2013 (UTC

I think the sign convention should be consistent with the convention used in the Poisson Bracket page. Any thoughts? Pratyush Sarkar (talk) 17:03, 3 February 2015 (UTC)[reply]

Yes, it should be. Based on a quick skim of the above, plus a quick skim of the article, it appears that the article is not even consistent with itself, right now. I'm too busy to double-check. If you are reading this, then please check and correct as necessary, and respond here as to the status. 67.198.37.16 (talk) 04:59, 3 November 2020 (UTC)[reply]

Suggestions

I'm not an expert in this area, but I know some mathematics. This web page isn't very helpful in defining a Hamiltonian vector fields. Could somebody please first explain the conditions for a hamiltonian vector field before introducing the generalizations to general manifolds? —Preceding unsigned comment added by 64.178.98.24 (talk) 02:59, 19 April 2010 (UTC)[reply]

I agree, this article is more confusing than helpful. For example

X_{H}

is inconsistent acreoss the article — Preceding unsigned comment added by 193.206.68.168 (talk) 08:59, 23 April 2012 (UTC)[reply]

The condition for a vector field to be Hamiltonian is that the contraction of the symplectic structure with it is an exact 1-form. It is locally Hamiltonian if that 1-form is closed. If you want to understand the reason for this condition I would advise to study some Hamiltonian mechanics from a physical perspective. You should get out that in local Darboux coordinates the integral curve equations for Hamiltonian vector fields are Hamilton's equations of motion. 93.222.25.88 (talk) 13:57, 2 May 2013 (UTC)[reply]

Flows as symplectomorphisms

I do not see why flows of Hamiltonian vector fields should be considered canonical transformations. Even though they are always symplectic maps, they may not be injective and thus cannot qualify as diffeomorphisms. 93.222.25.88 (talk) 13:46, 2 May 2013 (UTC)[reply]

Nevermind. The injectivity follows from the fact that the maximal integral curves do not intersect each other unless they coincide and the derivative of the flow has full rank since it is symplectic. Under the condition that the Hamiltonian flow is globally defined for the given time, it is indeed a symplectic automorphism. 87.146.143.246 (talk) 09:13, 13 May 2013 (UTC)[reply]

Geodesic flows on Riemannian manifolds are "famously" Hamiltonian flows (because cotangent bundles are always symplectic manifolds.) The geodesically complete manifolds (such as symmetric spaces) are called geodesic manifolds. 67.198.37.16 (talk) 05:06, 3 November 2020 (UTC)[reply]