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Homework Help: Time average vs. phase space average

  1. Jan 23, 2010 #1
    1. The problem statement, all variables and given/known data
    For a given total energy E0 compute and compare a time average and a phase space average of x2 for the harmonic oscillator. The one-dimensional Hamiltonian is

    [tex] H = \frac{p^2}{2m}+\frac{m\omega^2}{2}x^2 [/tex]

    Reminder: the time average is defined as

    [tex] \langle x^2\rangle =\frac{1}{t}\int_0^t x^2\tau\,d\tau [/tex]

    we will be mostly interested in the long time limit. The phase space average is

    [tex] \overline{x}^2=\frac{\int\delta (E_0-H)x^2\,dx\,dp}{\int\delta (E_0-H)\,dx\,dp} [/tex]


    2. Relevant equations



    3. The attempt at a solution First, for the time average, all I can think of is that for a harmonic oscillator [tex] x = a\cos (\sqrt{k/m}t+\phi ) [/tex]. I can then substitute this in the given integral for time average, which I can then evaluate. The problem is that I don't know what a and [tex] \phi [/tex] are given the information in the problem.

    Any hints/suggestions would be greatly appreciated.
     
  2. jcsd
  3. Jan 24, 2010 #2
    The amplitude a can be calclated since the energy E0 is given.
    The initial phase [tex] \phi [/tex] doesn't affect the average x2.

    By the way, the definition of the time average has wrong dimension. Is it a typing error?
     
  4. Jan 24, 2010 #3
    Thank you for the reply.

    Yes, I've now found out that [tex] E = (1/2)m\omega^2 A^2. [/tex]

    Well, I've copied it exactly the way it shows up on the problem set, so maybe the professor made a typo?
     
  5. Jan 24, 2010 #4
    The common time average definition is:

    [tex]
    <f(t)> = \frac{1}{t} \int_0^t f(\tau)\,d\tau.
    [/tex]

    May be the brackets around the [tex] \tau [/tex] are missed in your definition?
     
  6. Jan 24, 2010 #5
    Okay, I see what you're saying; I agree it should be

    [tex] \frac{1}{t}\int_0^t x^2(\tau)\,d\tau [/tex]

    Thanks, I got it now.
     
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