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Need help seeing that phase-space area is action

  1. May 17, 2013 #1
    I see in descriptions of the "old quantum theory", action formulated as

    [itex]\oint_{H(p,q)=E} p \dot{x} dt[/itex]

    I'm struggling to see how that is equivalent to

    [itex]\int \mathcal{L} dt[/itex], which I thought was the usual definition of action.

    I know [itex]\mathcal{H}[/itex] is constructed so that [itex]p\dot{x} = \mathcal{H} + \mathcal{L}[/itex], but using that gives me an extra

    [itex]\oint_{H(p,q)=E} \mathcal{H} dt = E \times T[/itex],

    where T is the period of motion. Where am I going wrong?
  2. jcsd
  3. May 18, 2013 #2


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    Glidos, You're right, there are (at least!) two different quantities referred to as the action.

    In Lagrangian mechanics, the action integral is

    I = ∫t1t2 L dt

    This is what enters Hamilton's Principle, δI = 0, where δ means that t1 and t2 are held fixed.

    By contrast, in Hamiltonian mechanics, action is defined as

    A = ∫t1t2 ∑ pq· dt

    This is what enters the Principle of Least Action: in a system for which H is conserved, ΔA = 0 where Δ means that q is held fixed at the end points, but t1 and t2 are allowed to vary.

    As you point out, they differ by a term H (t2 - t1)
  4. May 21, 2013 #3
    Great, thanks Bill_K. Helps a lot to know they are actually different and that they are used with different parameters varying. I'd naively imagined that a set of phase states of constant H would form a one-dimensional loop and hence define exactly the motion of a system. My misunderstanding probably stems from only ever seeing phase space diagrams of one dimensional problems. It's very interesting that the principle of least action again picks out the natural motion, although the definition of action is slightly different.

    I'm still struggling to get a good intuition for action, but your comments help and I'll battle on,
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