# What is the physical meaning of diffusion coefficient?

In Fick's first law, the diffusion coefficient is velocity, but I do not understand the two-dimensional concept of this velocity. Imagine that solutes are diffusing from one side of a tube to another (this would be the same as persons running from one side of a street) to unify the concentration across the tube.

Here we have a one-dimensional flow in x direction. The diffusion coefficient should define the velocity of solutes or persons across the tube or street direction. How the two-dimensional velocity does this? I wish to understand the concept to imagine the actual meaning of the diffusion coefficient.

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Do you mean that the dimenisionality of $D$ is $\text{length}^2/\text{time}$? That's not a velocity, it's in any case a surface velocity (rate of increase of area). Probably the most useful interpretation is just the straightforward one: it's the constant of proportionality between the gradient of concentration and the flux of substance. Look here for a dimensional analysis. – Jellby Jul 23 '12 at 11:22

Diffusion is a stochastic process where a single particle can move in each direction with the same probability. Another description of the diffusion coefficient is the following equation:

$$D = x^2/(2t)$$ where $t$ is the time and $x^2$ is the mean squared displacement of the particles at this time.

The mean squared displacement, $x^2$, can be interpreted as the statistical variance of the particle positions, so the diffusion coefficient can be interpreted as the rate at which the variance changes.

This idea was first proposed by Einstein (1905. Ann. Phys., 17, 549--560. http://www.zbp.univie.ac.at/dokumente/einstein2.pdf)

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Hi Helena. Welcome to Physics.SE. This site uses an unique TeX markup style called MathJax. Please have a look here for an intro or our FAQ for more info. For now, I'll help revising your post. – Waffle's Crazy Peanut Feb 4 '13 at 7:55
Note that this is only true for the displacement in $x$! If you have a three dimensional problem the equation is $D=L^2/(6t)$ where $L^2$ is the mean square displacement MSD. – dani Dec 3 '14 at 20:17

Let stick to your example of the people running around in the street.

this would be the same as persons running from one side of a street

If all people are running in the same direction, then this would be convection, and not diffusion.

Suppose you have a large room, with 100 guys running around in random directions. If you give them a random initial position somewhere in the room, not a lot will change. Except that each man will be at a different location the next time you look.

Now suppose you let all the guys starting at the left side of the room. You're standing with your back to the wall, exactly in the middle. Initially, you will, every once in a while, see a guy passing your line of sight from left to right. No man will cross from right to left, because there is no one at the right hand side. At some point, there will be that many guys on the right hand side, that you will also see man running from right to left, until, at some point, you will not see any difference between left and right.

This is basically the physical concept behind diffusion. If you replace the man by molecules, you get a gas or liquid. These molecules also have some random velocity and direction.

The diffusion coefficient in this context, is measure for how fast the left and right side 'mix', and thus also determines how long you have to wait before you don't note any difference. A larger diffusion coefficient does mean faster mixing.

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The diffusion coefficient $D$ is a constant relating the spreading $\left\langle x^{2}\right\rangle$ and the time $t$ it spread out. This relation can be clear seen in the diffusion of a single point source as follow.

Lets consider the homogeneous diffusion equation: $$u_t = D u_{xx} \tag{1}$$

The solution is given by: $$u(x,t) = \int K(x,t,x')u_o(x')dx' \tag{2}$$ where $u_o(x')$ is the initial condition and the $K(x,t,x')$ is the diffusion kernel (or Greens function): $$K(x,t,x')=\frac{1}{\sqrt{4\pi D t}}\exp\left[-\frac{(x-x')^2)}{4 D t}\right] \tag{3}$$

For a single point source $u_o(x')=\delta(x-x')$, we have the solution: $$u(x,t) = K(x,t,0) =\frac{1}{\sqrt{4\pi D t}}\exp\left[-\frac{x^2}{4 D t}\right] \tag{4}$$

The solution is shown in Fig. 1. Its second moment (same as variance since $\left\langle x\right\rangle=0$) is given by $$\left\langle x^{2}\right\rangle=\int x^2 u(x,t)dx=2Dt \tag{5}$$

Therefore, it clearly implies that the grow of the width square $\left\langle x^{2}\right\rangle$ of the Gaussian is linear proportional to the time $t$, with rate given by $2D$.

Fig. 1

The $x_{rms}=\sqrt{\left\langle x^{2}\right\rangle}$ defines a length scale of the spreading. If we have another length scale $\ell$, we would expect that when $x_{rms}\gg\ell$ or $Dt \gg \ell^2/2$, then the system behaviour is the same as a single Gaussian with the only one length scale $x_{rms}$.

Lets consider two point sources located at $\pm \ell/2$ so the solution is $$u(x,t) = \frac{1}{2}(K(x,t,-\ell/2) + K(x,t,\ell/2)) \tag{6}$$

The result is shown in the Fig. 2. When the time increase, those two peaks spread out and eventually merge into one when time is large enough. After a long time, the solution can be described approximately by: $$u(x,t) \approx K(x,t,0) \tag{7}$$

Fig. 2: Plot of Eq (6) when the time is $Dt=0.01, 0.1, 1$ of $\ell^2/2$ from top to bottom.

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