7.4: Poisson’s Theorem
- Page ID
- 29571
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If \(f\) and \(g\) are two constants of the motion (i.e., they both have zero Poisson brackets with the Hamiltonian), then the Poisson bracket \(\begin{equation}
[f, g]
\end{equation}\) is also a constant of the motion. Of course, it could be trivial, like \(\begin{equation}
[p, q]=1
\end{equation}\) or it could be a function of the original variables. But sometimes it’s a new constant of motion. If f,g are time-independent, the proof follows immediately from Jacobi’s identity. A proof for time dependent functions is given in Landau—it’s not difficult.