8
$\begingroup$

The mathematics of symplectic (as well as Hamiltonian) vector fields is something that has been quite clear to me for some time, but recently I have been thinking much more about what certain mathematical ideas are meant to capture from physics (as I do not have a physics background), and there is one notion that I just can't see: if some physical system admits a symplectic vector field, then what does this mean in terms of the physics of a system? What is the physical relevance of a vector field whose flow preserves the symplectic form?

EDIT: I've come to decide that there is not really any physical meaning for a symplectic vector field, other than, perhaps, a vector field that locally models the dynamics of the system, but not globally (as $X$ symplectic implies that $\iota_X\omega$ is closed, and so the Poincaré lemma says it is locally exact; hence $X$ is locally hamiltonian, i.e. its integral curves satisfy the equations of motion).

$\endgroup$
5
$\begingroup$

I will try to give a short (and surely partial) motivation. A bit of notation. Let $(M,\omega)$ be a symplectic manifold, with symplectic form $\omega$ and let $H:M\rightarrow \mathbb R$ be a smooth function (the hamiltonian). The hamiltonian vector field $X_H$ associated to $H$ is defined through

$$i_{X_H}\omega=dH.$$

An example: $(M,\omega)=(S^2, d\theta\wedge dh)$, with hamiltonian $H(\theta,h):=h$ (the height function). Then $X_H=\frac{\partial}{\partial\theta}$.

A symplectic vector field is a vector field $X$ on $M$ which preserves the symplectic form $\omega$, i.e. $\mathcal L_X\omega=0\Leftrightarrow i_X\omega$ is closed, by Cartan's magic formula.

In summary, $X$ is hamiltonian iff $i_X\omega$ is exact, while $X$ is symplectic iff $i_X\omega$ is closed. In general, not all symplectic fields are hamiltonian: the obstruction is given by non trivial $H^1(M)$. So I would see the presence of non trivial diff. geometry of the phase space as a motivation for dealing with symplectic vector fields. For example, given the 2-torus $(M, \omega) = (T^2, d\theta_1\wedge d \theta_2)$ the vector field $X = \frac{\partial}{\partial\theta_1}$ is symplectic but not hamiltonian.

$\endgroup$
5
  • $\begingroup$ First, thanks for the response! Yes, that's all perfectly clear to me, but I don't see what it means physically. Perhaps to highlight what I'm after, I have some idea of what a hamiltonian vector field is: the hamiltonian (the $H$ you mentioned) represents the energy of the system, and the integral curves are path in the phase space of the system, i.e. they describe evolution of the system in time, and $X_H$ being a hamiltonian vector field means that these integral curves satisfy Hamilton's equations, i.e. the equations of motion for the given hamiltonian. Further, one can show that … $\endgroup$
    – user101616
    Nov 10 '13 at 14:58
  • $\begingroup$ … $X_h$ is constant along its integral curves, which corresponds to conservation of energy. My question is: what is the corresponding physical content of a symplectic vector field? $X$ being hamiltonian means its integral curves satisfy the equations of motion, but what physical aspect of a system does $X$ being symplectic capture? $\endgroup$
    – user101616
    Nov 10 '13 at 14:59
  • $\begingroup$ You are right: but as there are in general obstructions, like in the 2-torus case, then focusing only on hamiltonian vector fields one cannot encode all dynamics of the system. In the 2-torus case the vector field $\frac{\partial}{\partial_\theta}$ is hamiltonian with flux given by translation by height $h$, while $\frac{\partial}{\partial_h}$ is symplectic not hamiltonian with flux given by rotation by $\theta$: I see then the study of sympl. vectors fields a way to consider all possible dynamics-and orbits- of the system. $\endgroup$
    – Avitus
    Nov 10 '13 at 15:17
  • $\begingroup$ Right, so sticking with your example of $X=\tfrac{\partial}{\partial h}$ on the 2-torus, we know that $X$ is symplectic and not hamiltonian. The fact that $X$ is not hamiltonian means that its integral curves do not satisfy the equations of motion; hence they do not represent possible evolutions of the system in time, right? So if the integral curves are not describing the system's evolution in time, then what aspect of the dynamics do they describe? (Sorry if this question already has a clear answer from what you've said thus far: like I said, I'm far from being a physicist! ;) ) $\endgroup$
    – user101616
    Nov 10 '13 at 15:35
  • $\begingroup$ Oops, I meant to edit my original post -- feel free to reject the edit I made to yours. :D $\endgroup$
    – user101616
    Nov 11 '13 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy