# Why is diffusion coefficient = 1/2?

1. Mar 21, 2012

### Old Guy

1. The problem statement, all variables and given/known data
I'm working with a 0ne-dimensional random walk and looking at Fick's second law, below. All I've read seems to take D = 1/2 as a given for ordinary diffusion, but where does this come from? Is there a way to derive it?

2. Relevant equations
$$\frac{\partial } {{\partial t}}p\left( {x,t} \right) = D\frac{{\partial ^2 }} {{\partial x^2 }}p\left( {x,t} \right)$$

3. The attempt at a solution

2. Mar 22, 2012

### clamtrox

Essentially D defines your units, so you are allowed to choose any numerical value. I guess people choose 1/2 because then the "usual" solution is a Gaussian with variance t (instead of √(2Dt)).

Last edited: Mar 22, 2012
3. Mar 22, 2012

### Old Guy

Thank you, but can you expand on your answer a bit? Not sure I understand what you mean in your second sentence.

4. Mar 22, 2012

### clamtrox

You know how to solve diffusion equation with Fourier transforms, right? The result is
$p = \frac{1}{\sqrt{4\pi Dt}} \exp(-x^2/4Dt)$
so if you choose D = 1/2, this becomes a normal distribution with mean 0 and variance t.

5. Mar 22, 2012

### Old Guy

Of course, now I get it! Thank you.