# Fick's law should contain a factor 2, but it doesn't. Why? Consider the above system. We will drive the Fick's law from it.

Let $$\sigma(x,y)$$ be the concentration inside the box centered at $$(x,y)$$. Then, (using some physical argument which I will skip in here). Let $$j(x+dx/2,y)$$ be the flux across the section located at $$(x+dx/2,y)$$.

$$j_x (x+dx/2, y) = D \frac{\sigma(x,y) - \sigma(x+dx,y)}{dx}$$

$$j_x (x-dx/2, y) = D \frac{\sigma(x-dx,y) - \sigma(x,y)}{dx}$$

Then the net flux along the x-axis

$$j_x (x-dx/2, y) + j_x (x+dx/2, y) = -D \frac{\sigma(x-dx,y) - \sigma(x+dx,y)}{dx}$$

In the limit $$dx \to 0$$,

$$2 j(x,y) = -D \partial_x \sigma$$

However, original Fick's law do not contain that factor 2. Why? What am I doing wrong?

## 1 Answer

The right hand side of the finite difference

$$j_x (x-dx/2, y) + j_x (x+dx/2, y) = -D \frac{\sigma(x-dx,y) - \sigma(x+dx,y)}{dx}$$

contains the difference between the concentration at points that are distant $$2 dx$$, but you divide by $$dx$$ only. Therefore, in the limit of $$dx$$ that goes to zero, you obtain $$2 \partial_x \sigma$$. This factor of 2 appears now on both sides of your equation and you can simplify it.