= Solution
In <canonical coordinates>, the <Poisson bracket> is
$$
\boxed{\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}
-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right).}
$$
In particular, $\{q_j,g\}=\partial_{p_j}g$, $\{p_j,g\}=-\partial_{q_j}g$, $\{f,q_j\}=-\partial_{p_j}f$ and $\{f,p_j\}=\partial_{q_j}f$. Hence $\{q_i,p_j\}=\delta_{ij}$ and the coordinate–coordinate and <momentum>–<momentum> brackets vanish.
The <chain rule> along a <Hamiltonian> trajectory gives
$$
\frac{df}{dt}=\partial_tf+\sum_i(\partial_{q_i}f\,\dot q_i+\partial_{p_i}f\,\dot p_i)
=\boxed{\partial_tf+\{f,H\}.}
$$
The <Poisson bracket of time-dependent conserved quantities> follows by defining $D_t=\partial_t+\{\,\cdot\,,H\}$. The ordinary <product rule> and <Jacobi identity> imply
$$
D_t\{f,g\}=\{D_tf,g\}+\{f,D_tg\}.
$$
To see the <Hamiltonian> part explicitly, Jacobi rearranges to $\{\{f,g\},H\}=\{\{f,H\},g\}+\{f,\{g,H\}\}$. Therefore $D_tf=D_tg=0$ implies \b[$D_t\{f,g\}=0$], even when the conserved functions depend explicitly on time.
Back to article page
