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Hamilton's Equations of Motion

  1. Aug 23, 2007 #1
    Here's a quick one:

    If the generalized force F is not zero,
    does the equation
    dH/dq = -pdot

    dH/dq = F- pdot
  2. jcsd
  3. Aug 24, 2007 #2


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    What have generalized forces have to do with the hamiltonian formulation of dynamics ??
  4. Aug 24, 2007 #3
    Everything. Hamiltonian dynamics concerns itself with positions and generalized momenta. A generalized force is the derivative of a generalized momentum.
  5. Aug 24, 2007 #4
    Hamilton's equations are always:

    [tex]\dot{p} = -\frac{\partial H}{\partial q} = f[/tex]
    [tex]\dot{q} & = & \frac{\partial H}{\partial p} = v [/tex]
    Last edited: Aug 24, 2007
  6. Aug 27, 2007 #5
    Oh because dL/dq always equals d/dt[ dL/dqdot ]!
    no matter what the force is, the force is inclusive in dL/dq
    L = T - V
    dL/dq = dT/dq - dV/dq ,
    where the force, F = -dV/dq
    taking the partial derivative of L w/respect to position q takes care of our generalized force F, so we don't need to write it explicitly in the Hamiltonian equations. it's already covered by the Lagrangian in dL/dq, which we can express as the time derivative of the momentum, pdot.

    the idea is that the force is in the Lagrangian!

    Then when u derive the Hamiltonian equations from Lagrange's
    u have dH/dq = - dL/dq
    and using the above observation about dL/dq, we have
    dL/dq = d/dt[ dL/dqdot ]

    going to inside the brackets we have
    dL/dqdot = d/dqdot [ 1/2 m qdot^2] = m qdot = p
    which is always - always true.
    and that means its time derivative is
    d/dt[ dL/dqdot = p ] = pdot

    so the conclusion is the same
    the equation is always

    pdot = - dH/dq

    oh my god
    dT/dq = 0
    it's so obvious
    dL/dq = F
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