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I'm having trouble reconciling some conceptual issues of brownian motion.

Let's say we have a box with two compartments separated by a membrane. Solute is at a high concentration on one side, and at a low concentration on the other. We know that the solute will move down its concentration gradient until the concentration is equal.

Here is where I get confused. I have also been told that due to Brownian motion, the net displacement of any given particle is zero. If this is the case, the net displacement of a bunch of particles would be zero. This would not give us the diffusion that is expected as discussed in the preceding paragraph.

So is the net displacement of a given particle actually not zero? Or only zero in certain circumstances? Or have I made some other mistake?

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  • $\begingroup$ Average displacement may be zero, but average distance from starting point isn't! $\endgroup$ – Alan Sep 26 '15 at 10:19
  • $\begingroup$ Displacement is the final distance from starting point? Total distance traveled is nonzero, but I'm not sure that answers the question. $\endgroup$ – David Sep 26 '15 at 15:06
  • $\begingroup$ It doesn't answer the question, which is why it's just a comment. Anyway, the difference between distance and displacement is as follows: Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position. (I found that off a site and quoted it) Anyway, Tom-Tom is right: the boundary condition causes the net displacement of the particles to shift to the center, but displacement is NOT distance. $\endgroup$ – Alan Sep 28 '15 at 21:04
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In the situation of two boxes separated by a membrane, there is a very strong difference with the mathematical Brownian motion :

$$\text{The box is finite.}$$

This has a very strong consequcne : the mean displacement cannot be zero anymore.

To understand why, consider for instance a particle starting very close to the border of the box. It can not go beyond the border and therefore, the displacement can only be directed toward the opposite direction. After a long time, the position's probability distribution of any particle becomes uniform in the box, whatever the starting position was.

When there are now a huge number of particles, after an equilibration time, they will fill all the available volume with a uniform density, whatever was their initial distribution.

EDIT: uniform distribution in a finite domain

To see that the proability density of a Brownian motion tends to the uniform distribution, a simple way is to consider a segment $[0,L]$. The differential equation governing the pdf of the particle's position $p(x,t)$ is the heat equation, also called diffusion equation $$\frac{\partial p}{\partial t}(x,t)=D\frac{\partial^2p}{\partial x^2}(x,t).$$ We write $p(x,t)$ as a Fourier series in space $$p(x,t)=\sum_{k\in\mathbb Z}a_k(t)\mathrm e^{\mathrm i2\pi kx/L}.$$ A term $a_k$ corresponds to oscillations of $p$ that have a space periodicity $L/|k|$, so the larger $|k|$ the more $p$ oscillates rapidly with respect to position.

The diffusion equation splits into a series of ordinary differential equation, one for each $k$ $$\frac{\mathrm da_k}{\mathrm dt}(t)=-\frac{4\pi^2Dk^2}{L^2}a_k(t).$$

The solutions are $a_k(t)=a_k(0)\exp\left(-k^2\,4\pi^2Dt/L^2\right)$. We observe that only $a_0(t)$ is not decreasing, such that $p(x,t)\to a_0(0)$ as time $t\to\infty$. This proves that the distribution becomes uniform. We also learn that the larger is $|k|$, the faster the oscillations go to zero.

Spectral analysis of higher level shows that this statement remains true for any domain in any finite dimension, because the Laplacian $(-\Delta)$ has only positive eigenvalues and the uniform function $x\mapsto 1$ is always an eigenfunction, with eigenvalue $0$.

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  • $\begingroup$ Hi, thanks for this answer. I have come this statement (the distribution will be uniform even though there are boundaries) very frequently but do you happen to know of any notes or references where it's actually explained? Additionally, do we know of similar results when the box is finite and discrete? Many thanks in advance. $\endgroup$ – user929304 Jun 13 '17 at 10:31
  • $\begingroup$ When the position in a domain are discrete, we speak of a random walk rather than Brownian motion (which is a continuous process). If that's you mean, then the statement remains valid. You can probably prove it based on the edit I just made to address the reason why the distribution tend to be uniform for large times. $\endgroup$ – Tom-Tom Jun 13 '17 at 19:12
  • $\begingroup$ Thanks a lot for getting back to me on this, and for the edit, I still need time to think about this. Very interesting! So this is expected to be valid for all systems that behave diffusively, right? Do you happen to know of similar calculations being done for the discrete case in a piece of literature? $\endgroup$ – user929304 Jun 14 '17 at 8:32

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