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Jun 18, 2010 at 11:07 vote accept Jan Weidner
Jun 18, 2010 at 11:07
Jun 18, 2010 at 9:36 comment added Aasmund Ervik Ah. Yes, I stand corrected. I didn't want to get technical with this point (as you have in your comment), as I assumed the questioner was familiar with the symplectic structure, from the way he phrased the question. But I see my attempt at hand-waving failed, so thank you for your correction.
Jun 18, 2010 at 0:25 comment added José Figueroa-O'Farrill I realised that I ran out of room before defining the symplectic structure $\omega$ on $T^*Q$: $\omega = -d\theta$ (in my conventions).
Jun 17, 2010 at 23:20 comment added José Figueroa-O'Farrill "suppose that your coordinates q_i lie in a Riemannian manifold, Q. Then Q together with the two-form w, which is the exterior derivative of the so-called canonical one-form on the phase space, usually form a symplectic manifold." This is not quite right. Surely it is $T^*Q$ which admits a canonical symplectic structure. To define we start by defining the canonical 1-form $\theta$ on $T^*Q$. A point $\alpha \in T^*Q$ can be thought of as 1-form (also denoted) $\alpha$ on $Q$ at the point $\pi(\alpha)$, with $\pi:T^*Q \to Q$ the canonical projection. Finally, $\theta_\alpha = \pi^*\alpha$.
Jun 17, 2010 at 17:38 history answered Aasmund Ervik CC BY-SA 2.5