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Particle in a potential

  1. Mar 7, 2006 #1
    Particle of mass m moves in two dimensionas un der the influence of a potential V(x,y) = kxy where K > 0 (not a cetnral potential)

    a) Determine the momenta conjugate to the coordiante x,y of this system and Hamiltonian punction of this system (in terms of the coordiante and conjugate momenta). Explian whether the Hamilton is a constant of motion


    [tex] H = \frac{1}{2m} (p_{x}^2 + p_{y}^2} + kxy [/tex]

    wouldnt hte momenta simply be [tex] p_{q} = m\dot{q}[/tex] where q is a coordinate?
    well the Hamitlon doesnt dependa on timte epxlicitly so it is constnat of motion

    so far so good?

    thank you for ANY help
     
  2. jcsd
  3. Mar 7, 2006 #2
    q is a generalized coordinate and so will not always have units of distance. So p is not automatically m(dq/dt). In general:
    [tex]\dot p_j=-\frac{\partial H}{\partial q_j}[/tex].

    Yes, if H has no explicit time dependence, then H is a constant of the motion.

    -Dan
     
  4. Mar 7, 2006 #3

    so do i use the lagrangian which is L = T-V to get the Hamiltonian

    doing that yielded
    [tex] H = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} m \dot{y}^2 + kxy [/tex]
    if i did it for x and x dot
    the signs of the first tow terms were switched when i did it for y
     
  5. Mar 7, 2006 #4
    Are you looking for the mechanical linear momentum or the canonical momentum? If the later then obtain the Lagrangian for this system from the Hamiltonian. Express all varibles in terms of generalized velocities. From this you can take the partial derivative of the Lagrangian with respect to the generalized velocity conjugate to the appropriate coordinate.

    That should give you enough to start. It may be a bit difficult for you to do this transformation (Which I believe is called an inverse Legendre transformation), Good luck.

    Pete
     
  6. Mar 7, 2006 #5
    i am loking for hte canonical momentum (i think) but i dont believe we have touched on inverse legendre tranformations

    however
    [tex] L = \frac{1}{2} m\dot{r}^2 - kxy [/tex]
    where r(t) = (x(t),y(t))

    right?
    so do i do
    [tex] H = \dot{r} \frac{\partial L}{\partial \dot{r}} - L [/tex]??

    but what about hte x and y?
     
  7. Mar 8, 2006 #6
    That is correct. The hint to that is that they asked for the momentum conjugate to x and y. That means that canonical momentum.
    Its a tricky buisness in general but in this case its a snap. You have the kinetic energy and the potential energy. You must have the Lagrangian expressed in terms of vx and vy. The Lagrangian is

    [tex]L = \frac{1}{2m}\dot{x}^2 + \dot{y}^2 - kxy[/tex]

    Now use the definition of conjugate momenta

    [tex]p_x = \frac{\partial L}{\partial \dot{x}}[/tex]

    [tex]p_y = \frac{\partial L}{\partial \dot{y}}[/tex]

    Good luck and let me know if you have more quetions. It'd be good for you to look up the term "Legendre transformation" in a classical mechanics text such as Goldstein's.

    Pete
     
    Last edited: Mar 8, 2006
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