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Constructing a hamiltonian for a harmonic oscillator

  1. Aug 9, 2011 #1
    Hello:

    I am trying to understand how to build a hamiltonian for a general system and figure it is best to start with a simple system (e.g. a harmonic oscillator) first before moving on to a more abstract understanding. My end goal is to understand them enough so that I can move to symplectic transforms and then on to symplectic integration methods, but I plan on taking this one step at a time. From what I know and understand so far:

    [itex]T = \frac{1}{2} m v^{2}[/itex]

    [itex]V = \frac{1}{2} k q^{2}[/itex]

    [itex]L(q,\dot{q},t) = T + V[/itex]

    [itex]H(p,q,t) = p \dot{q} - L(q,\dot{q},t) [/itex]

    [itex]\dot{q} =\frac{\partial H}{\partial p}[/itex]

    [itex]\dot{p} = - \frac{\partial H}{\partial q}[/itex]


    I have been replacing [itex] v [/itex] with [itex] \dot{q} [/itex], but I don't believe I am getting the right answer. So my first questions are:

    1a. Are the terms [itex] p [/itex] and [itex] \dot{q} [/itex] the same thing, and if not why?
    1b. Are the [itex] \dot{q}, \dot{p} [/itex] and other dotted terms I see in many texts referring to the time derivative of that term? If so, why is [itex] \dot{p} [/itex] not referred to as [itex] \ddot{q} [/itex]? ​

    While I was working though the problem I tried above, I noticed that given [itex] q = a \sin(2\pi f t) [/itex], [itex] H [/itex] could be expressed as just a function of just [itex] t [/itex].

    2. So can the hamiltonian be function of only [itex] t [/itex]?​

    Thank for your help in advance,
    Dan
     
  2. jcsd
  3. Aug 9, 2011 #2
    No, they are not. They happen to be easily related in this particular case since they are proportional to each other. But generalized velocity and generalized momentum are not generally proportional to each other (although it might be true for most if not all example you encounter in class). A simple case to the contrary is a relativistic particle.

    When you really dwell into Hamiltonian mechanics, you realized that [itex]p[/itex] and [itex]q[/itex] (not [itex]\dot{q}[/itex]) are sort of interchangeable via what's called Canonical Transformation. In fact, you can already see that the two Hamiltonian equations are invariant under if you interchange [itex]p[/itex] and [itex]q[/itex] and make some clever sign change. Mathematically, [itex]\{p,q\}[/itex] are points of an geometric object called the Cotangent Bundle. But that's probably just formalism mumbo jumbo for you at this point.

    Because they are different thing. Again, this only works when [itex]p \propto \dot{q}[/itex], which is not necessarily true.

    This is a specific trajectory of the motion, of a particular initial condition. Hamiltonian mechanics is more interested in the general structure of the system itself. So in fact, you are throwing away information by making such a substitution: instead of studying all trajectories govern by this Hamiltonian, you are studying a particular trajectory that solves the Hamiltonian.
     
    Last edited: Aug 9, 2011
  4. Aug 10, 2011 #3
    Thank you for replying. I made a mistake in my first question and I should probably rephrase as well. I meant to ask, can [itex] m \dot{q} [/itex] be replaced by [itex] p [/itex]? I understand that if the mass was not constant then this would not be possible, but in a non-relativistic calculation, such as this harmonic approximation, are these terms (i.e. [itex] q, \dot{q}, p [/itex], etc.) treated the same way as in Newtonian mechanics?

    So if I understand you right, Hamiltonian mechanics is not generally used for finding trajectories of particles, just the quantities of a general system such as energy, average velocity, average position, etc.?
     
  5. Aug 10, 2011 #4
    Also, the quote above should be:

    [itex]L(q,\dot{q},t) = T - V[/itex]
     
  6. Aug 10, 2011 #5
    I suppose one can interpret generalized coordinate, its time derivative, and its conjugate momentum in Newtonian term if you want (and it is often good to do so). For you SHO problem, they are the same anyway. But in principle, there can be system with effective Hamiltonian in which this is not the case. For example, for a charged particle moving in a plane with a uniform magnetic field perpendicular to the plane. The conjugate momentum [itex]p[/itex] is not the same as the mechanical momentum [itex]m\dot{q}[/itex]

    There is no less power to solve for trajectory in Hamiltonian form than in the Newtonian form, when it can be done in closed form. And no, it is not correct to say that Hamiltonian mechanics can only solve for "average" something. Hamiltonian mechanics visualizes motion as flow in the phase space. So you end up with a picture of motion of an incompressible fluid in [itex](q,p)[/itex] space and you can trace any trajectory corresponding to any initial condition from that.
     
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