Poisson Bracket

When I first learned about the Poisson bracket in classical mechanics I was very excited. I had learned about the commutator in quantum first and this looked a lot alike. For any dynamical quantity f\in F(T^*M) that we’re interested in,

\frac{df}{dt}=\{f,H\}+\partial_t f

Where H is the Hamiltonian. But here’s where I was a bit confused. If we put in the coordinates (on T^*M) in for f then,


So I wondered, what happened to the \partial_t \xi (where \xi can stand for either position or momentum)? I was flummoxed because my experience has been to find these letters as functions of time,

q=q(t) and p=p(t)

where they seem to be explicit functions of time. But they are not. We’re really working on a cotangent bundle, which the Hamiltonian is a function on. Hamilton’s equations define a vector field on T^*M and what we’re trying to find are integral curves. The vector field:


Generates a flow \phi_t which acts on T^*M. So if our initial data is q_0, p_0 then there is a curve \gamma(t)=\phi_t (q_0,p_0) that satisfies Hamilton’s equations for our system and initial conditions. We can describe \gamma in local coordinates \gamma(t)=(q^1(t), q^2(t),...,p_1(t), p_2(t),...) where q^i(t) stands for the way that component of \gamma evolves with time.

Its a subtle point that I’m still digesting but it does resolve the above issue. Comments?


About because0fbeauty

Fascinated by the way mathematics and physics interact, captivated by visual and tactile mathematics and hoping to become a better expositor of these things is why I blog...occasionally...when I remember.
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