Conserved charge generates symmetry transformation in Hamiltonian Mechanics

1. Aug 15, 2006

kakarukeys

$$Q$$ is a conserved charge if $$\{Q, H\} = 0$$
Show that $$q+\epsilon\delta q$$ satisfies the equation of motion.
$$\delta q = \{q, Q\}$$

I couldn't find the proof. Anybody knows?
My workings:
$$\delta q = \{q, Q\}$$
$$\delta\dot{q} = \{\{q,Q\},H\} = - \{\{Q,H\},q\} - \{\{H,q\},Q\}$$
$$\delta\dot{q} = \{\{q,Q\},H\} = \{\{q,H\},Q\}$$
$$\delta\dot{q} = \{\{q,Q\},H\} = \{\dot{q},Q\}$$
Let $$q' = q+\epsilon\delta q$$
Show that $$\dot{q'} = \{q,H\}|_{q'}$$
L.H.S. is $$\dot{q}+\epsilon\delta\dot{q}$$
R.H.S is $$\{q,H\}+\epsilon\partial_q\{q,H\}\delta q$$
Therefore we have to prove $$\{\dot{q},Q\} = \partial_q\{q,H\}\{q,Q\}$$
................
No clue how to continue

2. Aug 15, 2006

kakarukeys

Solved! I neglected the change in p

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