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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    $\begingroup$ 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. $\endgroup$
    – Jellby
    Jul 23, 2012 at 11:22

4 Answers 4


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)

  • $\begingroup$ 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. $\endgroup$ Feb 4, 2013 at 7:55
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    $\begingroup$ 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. $\endgroup$
    – dani
    Dec 3, 2014 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.


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 $2D$.

enter image description here 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: $$ u(x,t) \approx K(x,t,0) \tag{7}$$

enter image description here

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


My mama taught me when you find something you’ve been looking for, you put it back in the first place you looked. Hence, Google-catalyzed question necromancy.

Square centimeters per second is the compressed unit for the diffusion coefficient, convenient for transport and storage. Thrown in a vacuum and allowed to expand, the full unit for the diffusion coefficient is:

grams flux per second for each change in concentration per centimeter.

Which I don’t like, let’s talk about the inverse of D.

Change in concentration per centimeter for each gram flux per second.

Water vapor’s D in 25 C air is 0.282 cm2/s. We’re watching one square centimeter that we’re absolutely fascinated with. We’re also studying the inverse of D, so:

If the concentration here is 1 gram per milliliter and 0.282 grams of water vapor are passing through our square centimeter, then the concentration one centimeter over there is 0 grams per mL and one centimeter in the opposite direction it’s 2 grams per mL. Or rather, the instantaneous concentration on top of our square centimeter is such. Then calculus roughs up some statistical distributions, less unprofessional chemists claw their eyes out, and I don’t care.

The time unit is the temporal distance to our second reference point for measuring flux. Measuring grams per hour is a very different D from measuring grams per second.

The first length unit is the linear distance to our second reference point for measuring concentration. If the concentration here is x, then D tells us one distance unit away the concentration is y. If the distance unit is meters instead of centimeters, naturally the concentration at y will be less and D as measured to that distance unit will be different.

Take these out and the rest of the unit for the non-inverted D is g/ua per g/mL, where ua is the incredibly rare microare, as in hectare. Without the per-second to make sense of it, it’s hard to imagine what grams per microare are. So let’s be crazy and go American:

pounds per square inch for each pound per cubic inch.

PSI is an obvious unit for pressure, and as long as we’re being crazy, g/mL is a unit for density Yes, I realize that a pound of weight is not a pound of mass, especially for gases outside STP, but I don’t think it hurts at all to think of diffusion as what happens after you stop holding back the pressure a fluid of a given density exerts on a given area.

Now back to sanity. We have g/cm2 per g/cm3. Dimensional analysis cancels it out to the second length unit. What is it? A scaling factor, because cubes go up with the cube and squares go up with the square, which is an ever growing ratio.

If the concentration is 1g/cc, and I convert it to 1000000g/m3, then the polite thing to do is to convert the flux from 1g/cm2 to 10000g/m2, which changes the ratio of density to pressure 100-fold, and subsequently the value of D in these units.

If a substance with a given ratio of diffusion density to diffusion pressure is let go, what does it look like one second later, one centimeter away, given that my scaling ratio is 1 cc per cm2?


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