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Homework Help: The Hamiltonian vs. the energy function

  1. Nov 26, 2009 #1
    1. The problem statement, all variables and given/known data

    The mechanics of a system are described by the Lagrangian:
    [tex] L = \frac{1}{2}\dot{x}^2 + \dot{x}t [/tex]

    2. Relevant equations

    (a) Write the Energy (Jacobi function) for the system.
    (b) Show that [tex] \frac{dh}{dt} \neq \frac{\partial h}{\partial t} [/tex]
    (c) Write an expression for the Hamiltonian of the system.
    (d) Recall that [tex] \frac{dH}{dt} = \frac{\partial H}{\partial t} [/tex] allways.
    explain why[tex] \frac{dH}{dt} = \frac{\partial H}{\partial t}, \frac{dh}{dt} \neq \frac{\partial h}{\partial t} [/tex] , even though H and h are equal in value.

    3. The attempt at a solution

    a. [tex] \frac{\partial L}{\partial \dot{x}} = \dot{x} + t [/tex]
    and we get
    [tex] h = \frac{\partial L}{\partial \dot{x}}\dot{x} - L = \frac{1}{2}\dot{x}^2 [/tex]

    b. [tex] \frac{\partial h}{\partial t} = 0, \frac{dh}{dt} \dot{x}\ddot{x} [/tex]

    c. This is what I don't understand..
    They both have the same expression... what is the difference between the two sections..
  2. jcsd
  3. Nov 26, 2009 #2
    Ah, but they don't have the same expression. The Hamiltonian ([itex]H[/itex]) is written in terms of the [itex]q[/itex]'s and [itex]p[/itex]'s, while the pre-Hamiltonian ([itex]h[/itex]) is written in terms of [itex]q[/itex]'s and [itex]\dot{q}[/itex]'s. Because of that little change, there is a difference.

    In this case,




    where we used the Legendre transform (part a) to get the momentum (recall [itex]p_q=\partial L/\partial\dot{q}[/itex]).
  4. Nov 27, 2009 #3
    Thank you for clarifying this.
    Could you elaborate more about what you meant with the legendere transform?
  5. Nov 27, 2009 #4
    Sure. The Legendre transform is a mathematical operation that transforms one set of coordinates into another set. In the case of Hamiltonian mechanics, you are turning velocity coordinates ([itex]\dot{q}[/itex]) into momentum coordinates ([itex]p[/itex]):

    p_q=\frac{\partial L}{\partial\dot{q}}
  6. Nov 30, 2009 #5
    Thanks :smile:
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