Solving the Fick's second law with common boundary conditions and applying to the first law will result in a common expression of one-dimensional diffusion as

$$J_{(x=0)} = D\frac{\partial c(x,t)}{\partial t} = \frac{C^\infty\sqrt{D}}{\sqrt{\pi t}}$$

This equation says that the flux decreases by time as the concentration gradient weakens.

However, how much is the maximum flux? When $t \to 0$ (at very short times) then $J \to \infty$. It doesn't make sense.

  • $\begingroup$ It's infinite for an instant of time. The integral over time (to get the amount of mass that actually enters between time 0 and time t is finite, even at short times), and is proportional to $\sqrt{t}$ $\endgroup$ – Chet Miller Aug 24 '17 at 20:31
  • $\begingroup$ @ChesterMiller I understand the mathematical expression but I do not understand the physical meaning. $\endgroup$ – Kama Aug 24 '17 at 20:33
  • 1
    $\begingroup$ You have an infinite concentration gradient at exactly time t = 0, so the flux is infinite at that very instant. But the limit of the actual amount of mass that enters over a vanishingly short time is finite. $\endgroup$ – Chet Miller Aug 24 '17 at 20:52
  • $\begingroup$ @ChesterMiller that's exactly my question: how to calculate the maximum finite amount of mass passing at $x=0$? $\endgroup$ – Kama Aug 24 '17 at 21:51
  • $\begingroup$ You just take the integral of Jdt from zero to t. $\endgroup$ – Chet Miller Aug 24 '17 at 22:54

In general, Fick's first law states that the flux $\vec j$ is given by,

$$\vec j = -D \nabla \phi$$

where $\phi$ is the concentration, showing a concentration gradient sets up a flow, and as the gradient increases, the flux increases as well, in magnitude.

To find the maximum flux, let us consider an explicit example, known as the thin-film solution which consists of $N$ 'source' atoms placed at $x=0$ as an initial condition, from which one can then derive the solution,

$$\phi(x,t) = \frac{N}{\sqrt{4\pi Dt}}e^{-x^2/4Dt}.$$

Visually, this consists of the film spreading out in both directions, with the pillar at $x=0$, $t=0$ slowly diminishing. We can compute the flux field for this solution:

$$ j = \frac{Nx}{4t\sqrt{\pi D t}}e^{-x^2/4Dt}.$$

This obviously changes sign, since there is flow in the $+x$ and $-x$ direction. The maximum flow is indeed infinite at $x=0$ and $t=0$, because there is an infinite gradient in concentration. Everything is localised to one point, and the rest of space is empty.

However, the flow itself is not infinite. If we want to find the amount of the substance flowing per unit length (or area in higher dimensions), we can compute,

$$\int_0^t dt' \, j(x,t') = \frac{N}{2}\left(1-\mathrm{erf}\frac{x}{2\sqrt{Dt}} \right)$$

and in the limit $t\to 0$, the result is finite.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.