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Homework Help: The Hamiltonian

  1. Mar 8, 2006 #1
    This is a question, not a homework problem, as i am currently studying for my test on classical mechanics
    suppose [tex] H = \sum_{i} \dot{q_{i}}(p,q,t) p_{i} - L(p,q,t) [/tex]
    also i can prove that
    [tex] dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i}) - \frac{\partial L}{\partial t} dt [/tex]

    suppose [tex] \frac{\partial L}{\partial t} = \frac{\partial H}{\partial t} = 0 [/tex] then dH/dt = 0 i.e. H = E

    now if i sub in to the equation above i get
    [tex] dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i}) [/tex]

    how would i transform the above dH into dH/dt formally?

    Do i simply integrate by parts to get H and then differentiate wrt t??
    Last edited: Mar 8, 2006
  2. jcsd
  3. Mar 9, 2006 #2


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    Well, just use that

    [tex] dH=\frac{dH}{dt} dt [/tex]

    which is an equality between 2 differential 1-forms which is valid iff H=H(t) is the curve it's being evaluated on.

  4. Mar 9, 2006 #3
    oh alright

    what about going from this point
    [tex] dH = \sum_{i} (\dot{q_{i}}dp_{i} - \dot{p_{i}} dq_{i}) - \frac{\partial L}{\partial t} dt [/tex]

    to go on and say that
    [tex] \frac{dH}{dt} = \frac{\partial L}{\partial t} [/tex]
    what happens to the summed terms??

    q dot doesnt depend on p and p dot doesnt depend on q so
    what we get is
    [tex] \dot{q} p - \dot{p} q [/tex]
    is that zero??
  5. Mar 9, 2006 #4
    do i simply use the idea which you suggested, dexter, on the dp and dq terms? THat was we get dots over all and they cancel out
  6. Mar 9, 2006 #5
    You are "dividing" by dt in the whole expression, so you get terms of the form [tex]\dot{q_{i}} \dot{p_{i}} - \dot{p_{i}} \dot{q_{i}}[/tex] in the summation for all i. So what DOES happen to the summed terms?

  7. Mar 9, 2006 #6
    Simply because dH/dt = 0 it doesn't mean that H = E. E.g. See "Classical Mechanics 3rd Ed.," Goldstein, Safko and Poole page 334-346.

  8. Mar 9, 2006 #7
    let me correct that by saying H is constant with respect to time, but not necessarily the energy. I wil check Goldstein's book tomorrow

    since we 'dividing' by dt we get this right
    [tex] \frac{dH}{dt} dt = \sum_{i} (\dot{q_{i}}\frac{dp_{i}}{dt} dt - \dot{p_{i}} \frac{dq_{i}}{dt} dt) - \frac{\partial L}{\partial t} dt [/tex]

    and once we integrate we go [tex] \dot{p} \dot{q} - \dot{q} \dot{p} [/tex]
    so tahts why the summed terms are zero
  9. Mar 9, 2006 #8

    Physics Monkey

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    Hi stunner,

    Indeed, if L contains no explicit time dependence then neither does H, and this means H is conserved since [tex] dH/dt = \partial H/\partial t [/tex] by virtue of the equations of motion. The question of rather H is equal to the energy has little to do with the question of time dependence and depends rather on the structure of the kinetic energy term.
  10. Mar 9, 2006 #9
    isnt hte kinetic energy term for all purpose always p^2/2m?? Perhaps i have not come across situations where it is otherwise. When would it be different?
  11. Mar 9, 2006 #10

    Physics Monkey

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    Very often the kinetic energy is of the form [tex] \frac{1}{2} m v^2 [/tex], but your choice of generalized coordinates can change the formal structure of the Lagrangian. For example, consider a bead constrained to move on a rod that rotates in the plane at some frequency [tex] \omega [/tex]. The kinetic energy is just given by [tex] T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \dot{\theta}^2 [/tex]. You can solve the constraint immediately by setting [tex] \dot{\theta} = \omega [/tex] and then leave [tex] \theta [/tex] out of your generalized coordinates. The Lagrangian then looks like [tex] L = T = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \omega^2 [/tex]. If you construct the Hamiltonian you will find that it is [tex] H = \frac{p^2_r}{2 m} - \frac{1}{2} m r^2 \omega^2 [/tex] where [tex] p_r = m \dot{r} [/tex] which is clearly not the total kinetic energy. Nevertheless, the Hamiltonian is conserved and the total energy is not. The key here was how you treat your constraints and generalized coordinates. If you had left [tex] \theta [/tex] as a generalized coordinate and used the Lagrange multiplier scheme to deal with the constraint, then your Hamiltonian would be the total energy. However, the Hamiltonian wouldn't be conserved because of the time dependent constraint (an important caveat to the general result I quoted above).
    Last edited: Mar 9, 2006
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