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Deriving the expression for the diffusion constant

  1. Sep 6, 2005 #1
    I have a question of the form, "how did they go from this step to that step?" I am following a paper from R. Kubo, called "The fluctuation-dissipation theorem". I suspect variable substitution and a simple assumption are used, but I don't know how. Here is the basic expression from which everything else follows.

    [tex]D=lim_{t\rightarrow\infty}\frac{1}{2t}<\{x(t)-x(0)\}^2>[/tex]​

    Since we have

    [tex]x(t)-x(0)=\int{v(t')dt'}[/tex]​

    where the limits of the integration are 0 to t, we substitute the second equation into the first equation to get

    [tex]D=lim_{t\rightarrow\infty}\frac{1}{2t}\int{\int{dt_{1}dt_{2}<v(t_{1})v(t_{2})}}>[/tex]​

    again, the limits of both integrals are from 0 to t. (Here comes the step I don't understand)

    [tex]D=lim_{t\rightarrow\infty}\frac{1}{t}\int{dt_{1}}\int{dt'<v(t_{1})v(t_{1}+t')}>[/tex]​

    where the limits of the outer integral are from 0 to t, and the limits of the inner integral are from 0 to [tex](t-t_{1})[/tex]

    It looks like the substitution [tex]t'=t_{2}-t_{1}[/tex] was made. But what happened to the upper limit of integration?
     
    Last edited: Sep 6, 2005
  2. jcsd
  3. Sep 10, 2005 #2
    I think it is not that difficult. let's call [tex]t_1=x[/tex] and [tex]t_2=y[/tex] and:

    [tex] <v(x) v(y)> = f(x,y) [/tex]​
    The integral

    [tex]I = \int \int dx dy f(x,y) [/tex]​
    is a 'volume integral' it should be called

    [tex] \int_{R(t)} f(x,y) [/tex]​
    Now notice that [tex] f(x,y) = f(y,x) [/tex], and since we want to integrate this over the box (bound by x=0, y=0, x=t and y=t) it is equal to integrate this twice over the triangle bound by x = 0, y=x, y=t. Then by Fubini's theorem

    [tex]I = 2 \int_{T(t)} f(x,y) = 2 \int_0^t dx \int_{x}^{t} dy f(x,y)[/tex]​
    Now substitute [tex] y=x+t' [/tex] then you should get the result.
     
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