[tex]Q[/tex] is a conserved charge if [tex]\{Q, H\} = 0[/tex](adsbygoogle = window.adsbygoogle || []).push({});

Show that [tex]q+\epsilon\delta q[/tex] satisfies the equation of motion.

[tex]\delta q = \{q, Q\}[/tex]

I couldn't find the proof. Anybody knows?

My workings:

[tex]\delta q = \{q, Q\}[/tex]

[tex]\delta\dot{q} = \{\{q,Q\},H\} = - \{\{Q,H\},q\} - \{\{H,q\},Q\}[/tex]

[tex]\delta\dot{q} = \{\{q,Q\},H\} = \{\{q,H\},Q\}[/tex]

[tex]\delta\dot{q} = \{\{q,Q\},H\} = \{\dot{q},Q\}[/tex]

Let [tex]q' = q+\epsilon\delta q[/tex]

Show that [tex]\dot{q'} = \{q,H\}|_{q'}[/tex]

L.H.S. is [tex]\dot{q}+\epsilon\delta\dot{q}[/tex]

R.H.S is [tex]\{q,H\}+\epsilon\partial_q\{q,H\}\delta q[/tex]

Therefore we have to prove [tex]\{\dot{q},Q\} = \partial_q\{q,H\}\{q,Q\}[/tex]

................

No clue how to continue

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# Homework Help: Conserved charge generates symmetry transformation in Hamiltonian Mechanics

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