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Why is diffusion coefficient = 1/2?

  1. Mar 21, 2012 #1
    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
    [tex]
    \frac{\partial }
    {{\partial t}}p\left( {x,t} \right) = D\frac{{\partial ^2 }}
    {{\partial x^2 }}p\left( {x,t} \right)
    [/tex]


    3. The attempt at a solution
     
  2. jcsd
  3. Mar 22, 2012 #2
    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
  4. Mar 22, 2012 #3
    Thank you, but can you expand on your answer a bit? Not sure I understand what you mean in your second sentence.
     
  5. Mar 22, 2012 #4
    You know how to solve diffusion equation with Fourier transforms, right? The result is
    [itex] p = \frac{1}{\sqrt{4\pi Dt}} \exp(-x^2/4Dt) [/itex]
    so if you choose D = 1/2, this becomes a normal distribution with mean 0 and variance t.
     
  6. Mar 22, 2012 #5
    Of course, now I get it! Thank you.
     
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