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Finding Probability(Stat mech)

  1. Jul 5, 2013 #1
    I am pre studying for Statistical Mechanics class in the fall and need help with this problem. I’ve already spent some time with it.

    Let the displacement x of an oscillator as a function of time t be given by X=Acos(wt+ϕ). Assume that the phase angle ϕ is equally likely to assume any value in the range 0 < ϕ < 2pi. The probability w(ϕ)d ϕ that ϕ lies in the range between ϕ and ϕ +d ϕ is then simply w(ϕ) dϕ=(2pi)^-1d ϕ. For any fixed time t, find the probability P(x)dx that x lies between x and x+dx by summing w(ϕ) over all angles ϕ for which x lies in this range. Express P(x) in termas of A and x.


    Relevant equations[/b]
    X=Acos(wt+ϕ).
    w(ϕ)d ϕ=(2pi)^-1d ϕ

    3. The attempt at a solution
    The only thing I can come up with is integrating
    ∫P(x)dx = ∫(2pi)^-1d ϕ and inegrating over x and x+dx
    Or ƩP((x)dx* w(ϕ) dϕ)/p(x)
     
  2. jcsd
  3. Jul 5, 2013 #2

    TSny

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    Hello catsonmars. Welcome to PF!

    You have a good start with ##w(\phi) = \frac{1}{2\pi}##.

    Since you are looking for the probability that ##x## lies in an infinitesimal range from ##x## to ##x+dx##, you will not need to integrate. The probability is just ##\small P(x)dx##. This is given by the probability ##w(\phi)|d\phi|## that ##\phi## lies in the range ##\phi## to ##\phi + d\phi##, where ##\phi## is the value of the phase angle that corresponds to ##x## and ##\phi + d\phi## corresponds to ##x+dx##. [Caution: think about whether or not there is more than one value of ##\phi## that corresponds to the same ##x##. If so, you will need to make an adjustment for that.]

    So, the probability that ##x## lies between ##x## and ##x+dx## could be expressed as ##\small P(x)dx## or as ##w(\phi)|d\phi|## (if there is only one value of ##\phi## that corresponds to a value of ##x##). That is, ##\small P(x)dx = ## ##w(\phi)|d\phi|## [I leave it to you to think about what to do if there is more than one value of ##\phi## corresponding to the same value of ##x##. Perhaps this has something to do with the word "summing" in the statement of the problem.]

    Since you already know how to express ##w(\phi)##, all you need to do is find an expression for ##d\phi## in terms of ##x## and ##dx##. Hint: ##d\phi = \frac{d\phi}{dx}dx##.
     
    Last edited: Jul 6, 2013
  4. Jul 12, 2013 #3
    There should be more x values than ϕ's because the range of ϕ is much smaller than x. Second I'm still not sure how I should right the summation. I have ϕ=(Ʃw(ϕ)dϕ)/(P(x)(dx)) but that still seems wrong. I've also thought about relating ϕ+dϕ to x+dx somehow but I'm can't think of what would make them equal so I can get ϕ in terms of x. Also I've looked at the answer key and I have no idea how the amplitue "A" would fit into the equation.
     
  5. Jul 12, 2013 #4

    TSny

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    Basically, you need to solve ##\small P(x)|dx| = w(\phi)|d\phi|## for ##\small P(x)##. That is,

    ##P(x) = w(\phi)\frac{d\phi}{dx}##

    Use ##x = Acos(\omega t + \phi)## to find ##\frac{d\phi}{dx}## as a function of ##A## and ##x##.

    There's the additional task of dealing with the fact that there might be two different values of ##\phi## corresponding to the same value of ##x##.
     
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