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Hamiltons equations of motion in terms of poisson bracket

  1. Mar 14, 2014 #1
    In Hamiltonian formulation there is an expression
    df / dt = { f , H } + ∂f / ∂t
    where f is function of q, p and t.
    While expressing Hamiltons equations of motion in terms of Poisson Bracket,
    i.e if the function f = q of p then its partial time derivative ∂f / ∂t becomes zero..
    Please explain why?
  2. jcsd
  3. Mar 14, 2014 #2


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    In the Hamilton formalism, by definition, the phase-space variables [itex](q,p)[/itex] are not explicitly time dependent. Of course, solving the Hamilton equations of motion leads to the trajectory of the system in phase space as function of time, but that's not what's meant in the phase-space formulation before the equations of motion are solved.
  4. Mar 14, 2014 #3
    p, q, and t are independent variables so [itex]\frac{\partial p}{\partial p}=1[/itex], [itex]\frac{\partial p}{\partial q}=0[/itex], [itex]\frac{\partial p}{\partial t}=0[/itex], [itex]\frac{\partial q}{\partial p}=0[/itex], [itex]\frac{\partial q}{\partial q}=1[/itex], [itex]\frac{\partial q}{\partial t}=0[/itex], [itex]\frac{\partial t}{\partial p}=0[/itex], [itex]\frac{\partial t}{\partial q}=0[/itex], [itex]\frac{\partial t}{\partial t}=1[/itex], by definition.
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