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Conserved charge generates symmetry transformation in Hamiltonian Mechanics

  1. Aug 15, 2006 #1
    [tex]Q[/tex] is a conserved charge if [tex]\{Q, H\} = 0[/tex]
    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
     
  2. jcsd
  3. Aug 15, 2006 #2
    Solved! I neglected the change in p
     
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