# Deriving the expression for the diffusion constant

1. Sep 6, 2005

### csk08

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.

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

Since we have

$$x(t)-x(0)=\int{v(t')dt'}$$​

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

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

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

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

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

It looks like the substitution $$t'=t_{2}-t_{1}$$ was made. But what happened to the upper limit of integration?

Last edited: Sep 6, 2005
2. Sep 10, 2005

### andyL

I think it is not that difficult. let's call $$t_1=x$$ and $$t_2=y$$ and:

$$<v(x) v(y)> = f(x,y)$$​
The integral

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

$$\int_{R(t)} f(x,y)$$​
Now notice that $$f(x,y) = f(y,x)$$, 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

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