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I find it hard to see the differences between Brownian motion and diffusion.

As I understand, both are represented by the diffusion equation – am I right? And if I'm not, how is Brownian motion described?

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  • $\begingroup$ en.wikipedia.org/wiki/Wiener_process $\endgroup$
    – tsufli
    May 26 '18 at 15:36
  • $\begingroup$ @tsufli , So when I want to get a brownian motion out of a random walk I just take as a variable the sum of n steps and devide it by sqrt(n)? How does it help me? $\endgroup$
    – Ilay
    May 26 '18 at 16:01
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Diffusion comes about as the result of Fick's law in continuous mediums.

Fick's law briefly: if you have a "stuff" (which can be any conserved quantity) and you allow two containers of that stuff together so that they can share it, then the flow of the stuff will be from the container with greater concentration to the container of less concentration, and the flow will be proportional to the concentration difference. Some examples: my stove works by creating a burner with a high concentration of thermal energy; the "concentration" of this energy is measured as a temperature, and so when I put a colder pot on the burner heat flows into the pot, proportional to the temperature difference. Another "stuff" is momentum in the $x$-direction, so when I am in a canoe on a lake and I put my paddle into a body of water, as I move it through in one direction, it starts to pull the water along with it in that direction, sharing momentum with the water; this is how I move forward by pushing the water backwards. To move faster, I need to put more speed into my paddle.

Now if you have a lot of particles collectively undergoing Brownian motion, they will naturally diffuse according to Fick's law: divide a fluid into conceptual boxes with a coordinate system that is flowing downstream with the fluid. Each box holds an amount of particles proportional to their concentration at that point, and assuming the directions are entirely randomized, if you look at two boxes side-by-side then the number crossing from A to B will be proportional to A's concentration, while the number crossing from B to A will be proportional to B's concentration, and so the net effect will be a flow from higher to lower concentration, proportional to that difference in concentration.

But also note that Fick's law does not directly answer why these things are randomly moving in the first place; diffusion just emerges as the consequence of the fact that they are moving. And diffusion makes sense for "stuffs" that are harder to conceptualize as particles, like momentum-in-the-x-direction, or energy.

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  • $\begingroup$ Thank you! So it sounds like as a result from a lot of brownian motions, I will get a diffusion. But now I am not sure about how do I describe only one brownian particle, and how do I derive the diffusion equation out of it. $\endgroup$
    – Ilay
    May 26 '18 at 16:14
  • $\begingroup$ @Ilay an interesting question! A single particle in Brownian motion has a motion over space and time which abruptly changes for no reason: it is nondifferentiable. You might say "I better give up, I can't do calculus on nondifferentiable things!" but you'll be surprised to know that you can, if you do it carefully. This is the subject of stochastic analysis, historically developed in two forms: Itô's calculus abandons the chain rule and its formulae are computer simulations; Stratonovich's calculus preserves the chain rule but it's harder to see how to use the resulting expressions. $\endgroup$
    – CR Drost
    May 26 '18 at 16:54
  • $\begingroup$ With all of that said you can skip these in-depth ideas if you instead just treat the one particle as a very dense cluster of particles, at which point you still have the diffusion equation again -- the diffused concentration then needs to be regarded as a probability density function for the resulting particle, as you've only got one and you're not sure which of these particles it was. $\endgroup$
    – CR Drost
    May 26 '18 at 16:56

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