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Homework Help: Tricky differentiation

  1. May 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\frac{dU_{ave}}{d\beta}[/tex]

    where

    [tex]U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right)[/tex]


    2. Relevant equations



    3. The attempt at a solution

    My answer is supposed to be [tex]-(U_{ave}^{2})+(U_{ave})^{2}[/tex]

    However I keep getting zero. I can only assume that the summation sign has something to do with the final form of the answer, but I have no idea how to deal with it. Can anyone offer a suggestion? Thanks
     
  2. jcsd
  3. May 26, 2008 #2

    Mute

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    Homework Helper

    I suspect you've written the problem incorrectly here. As you've written it,

    [tex]U_{ave}=\sum_{k}\left(\frac{U_{k}exp(-U_{k}\beta)}{exp(-U_{k}\beta)}\right) = \sum_k U_k [/tex]

    What it should be is

    [tex]U_{ave} = \frac{\sum_k U_k e^{-U_k \beta}}{\sum_k e^{-U_k \beta}}[/tex]

    This is a standard integral in statistical mechanics, and there's a neat trick that always goes along with solving it. Supposing U_k is indepent of beta, write the denominator as

    [tex]-\sum_k \frac{\partial}{\partial \beta} e^{-U_k \beta}[/tex]

    Pull the derivative outside the sum, and you note that what you have looks like (1/y) dy/dx = d(ln y)/dx. Then, take the second derivative of that result:

    [tex]-\frac{\partial}{\partial \beta} \frac{\partial \ln Z}{\partial \beta} = \mbox{answer}[/tex]

    where [itex]Z = \sum_k \exp(-U_k \beta)[/itex].
     
    Last edited: May 26, 2008
  4. May 26, 2008 #3
    Hmm, I've transcribed the problem straight from the question sheet. I too had that thought.

    I'm not sure I follow you with your methodology though. If I try to solve the equation using the standard quotient rule, it gives me zero - is there a reason for that?
     
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