Deriving the expression for the diffusion constant

In summary, the conversation discusses the process of deriving an expression from a paper on the fluctuation-dissipation theorem. The speaker is unsure of how a substitution and assumption were made to ultimately get the final expression. The expert explains that the substitution t'=t2-t1 was made and the upper limit of the integral was changed accordingly. The expert also uses Fubini's theorem to simplify the integral.
  • #1
csk08
1
0
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?
 
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  • #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.
 
  • #3


To derive the expression for the diffusion constant, we start with the basic expression:

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

This expression can be rewritten as:

D=lim_{t\rightarrow\infty}\frac{1}{2t}<\{\int{v(t')dt'}\}^2>

where v(t) represents the velocity of the particle at time t. Now, in order to simplify the expression, 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})}}>

The limits of both integrals are from 0 to t. Now, in order to further simplify this expression, we make a substitution t'=t_{2}-t_{1}. This substitution allows us to rewrite the inner integral as:

D=lim_{t\rightarrow\infty}\frac{1}{2t}\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}). This substitution allows us to simplify the expression and get rid of the second time variable t_{2}. The upper limit of the inner integral is now (t-t_{1}) instead of t, because as t_{1} increases, t_{2} decreases, and we want to keep the total time t constant. This is why the upper limit of the inner integral disappears.

So, in summary, the substitution t'=t_{2}-t_{1} was made to simplify the expression and get rid of the second time variable. The upper limit of the inner integral was changed to (t-t_{1}) to keep the total time t constant. I hope this helps clarify the steps used in this derivation.
 

1. What is diffusion and why is it important to understand?

Diffusion is the process by which particles move from an area of high concentration to an area of low concentration. It is important to understand because it plays a crucial role in many natural and industrial processes, such as gas exchange in the lungs and the spread of pollutants in the environment.

2. What is the expression for the diffusion constant?

The expression for the diffusion constant is D = (kT)/6πηr, where k is the Boltzmann constant, T is the temperature, η is the viscosity of the medium, and r is the radius of the diffusing particle.

3. How is the diffusion constant derived?

The expression for the diffusion constant is derived using the Stokes-Einstein equation, which relates the diffusion constant to the particle's radius, temperature, and viscosity of the medium. This equation is based on the theory of Brownian motion, which describes the random movement of particles in a fluid.

4. What are the assumptions made in deriving the expression for the diffusion constant?

The derivation of the expression for the diffusion constant assumes that the particles are small enough to be considered point-like, that the medium is homogeneous and isotropic, and that there are no external forces acting on the particles. It also assumes that the particles are in thermal equilibrium with the surrounding medium.

5. How is the diffusion constant related to the rate of diffusion?

The diffusion constant is directly proportional to the rate of diffusion. This means that as the diffusion constant increases, the rate of diffusion also increases. This relationship is described by Fick's first law of diffusion, which states that the rate of diffusion is equal to the diffusion constant multiplied by the concentration gradient.

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