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Hamiltonian Question

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    The simple form H=T+U is true only if your generalized coordinates are "natural"(relation between generalized and underlying Cartesian coordinates is independent of time). If the generalized coordinates are not natural, you must use the definition H=Ʃpq'-L. To illustrate this point, consider the following: Two children are playing catch inside a railroad car that is moving with varying speed V along a straight horizontal track. For generalized coordinates you can use the position(x,y,z) of the ball relative to a point fixed in the car, but in setting up the Hamiltionian you must use coordinates in an inertial frame-a frame fixed to the ground. Find the Hamiltonian for the ball and show that it is not equal to T+U (neither as measured in the car, nor as measured in the ground-based frame).


    2. Relevant equations
    L=T-U
    H=Ʃpq'-L
    Px=[itex]\frac{dL}{dx'}[/itex]
    Py=[itex]\frac{dL}{dy'}[/itex]
    Pz=[itex]\frac{dL}{dz'}[/itex]

    ' is time derivative in this case

    3. The attempt at a solution
    T=.5m(x'2+y'2+z'2)
    U=mgz; z is the vertical axis
    L=.5m(x'2+y'2+z'2)-mgz

    Px=mx'
    Py=my'
    Pz=mz'
    Pv=mv

    H=Ʃpq'-L
    H=Pxx'+Pyy'+Pzz'+Pvx'-.5m(x'2+y'2+z'2)+mgz
    H=Px(Px/m)+Py(Py/m)+Pz(Pz/m)+mv(Px/m)-.5m(x'2+y'2+z'2)+mgz

    If I continue this way, I end up with
    H=(Px2+Py2+Pz2)/(2m)+Pxv+mgz
    but the correct answer is
    H=(Px2+Py2+Pz2)/(2m)-Pxv+mgz

    Does anyone see where I went wrong, or is my answer correct, just with different sign convention?
     
  2. jcsd
  3. Nov 12, 2013 #2

    vela

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    Shouldn't your expression for T depend on V?
     
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