Let us consider a 3D box of volume $V$, containing $N$ identical Brownian particles. The diffusion coefficient of the particles is noted $D$. Inside this box there is a square surface of area $L^2$.

To keep everything as simple as possible, the box dimensions are much larger than that of the surface, and the surface dimensions are much larger than that of the Brownian particles. Moreover, the Brownian particles do not interact with each other and do not interact with the surface either.

What is the rate $j$ at which particles cross the surface?

enter image description here

Attempt 1: I have tried to combined intuition and dimensional analysis. $j$ physical dimensions are number/time. If the surface area is twice larger, $j$ is twice larger as well, and $j \propto L^2$. If there are twice more particles, $j$ will be twice larger, and $j \propto N$. If the box volume is twice smaller, it is equivalent than the number of particles being twice larger, and $j \propto 1/V$. The last two considerations amount to $j$ being proportional to the particle concentration $N/V$. To match the correct physical dimensions of the quantity $j$, I also need a quantity that involves time. $D$ is the only time-dependent quantity in my problem, and is expressed as length^2/time. Therefore I must have $j \propto D$. In summary I have

$$ j \propto \frac{N}{V}L^2D $$

This is wrong since the dimensions are not matching: left hand side dimension is 1/time and right hand side is length/time.

I think the correct answer is $j \propto \frac{N}{V}L D$ but it does not make sense to me, since in my opinion $j$ must scale with $L^2$.

What am I missing? Maybe the mean free path is also relevant?

Attempt 2. I also tried to solve the problem from a simulation point of view and discretized both time and space. So now the $N$ particles are randomly distributed on a 3D lattice which sites are cubes of dimension $a$. The surface is aligned with the lattice axis for simplicity. Every time step $\Delta t$, particles jump on a randomly selected neighbour site. The concentration of particles is small enough so that the probability that two particles occupy the same site is negligible.

During one time step $\Delta t$, the only particles that have a chance to cross the surface are the one located on sites that are adjacent to the surface. The total number of such sites is $$ 2 \frac{L^2}{a^2} \ . $$ The probability that any given site is occupied by a particle is $$ \frac{N a^3}{V} \ . $$ For any given particle located on a site adjacent to the surface, the probability to cross the surface during $\Delta t$ is $1/6$ (since the particles randomly choose one direction out of the six available directions).

Therefore the average number of particles crossing the surface during $\Delta t$ is $$ \frac{N a L^2}{3 V} \ . $$ Since every time step is equivalent, the number of particles crossing the surface, per unit time $\Delta t$, is $$ \frac{N a L^2}{3 V \Delta t} \ . $$

Now we need to relate the diffusion coefficient $D$ to the lattice dimension $a$ and the time step $\Delta t$. This a discrete random walk problem and the solution is $$ D = { \frac{a^2}{2 \Delta t}} \ . $$

We therefore find the rate $j$ of crossing: $$ j=\frac{2}{3} \frac{N}{V} \frac{L^2}{a} D $$

In my opinion the lattice size $a$ can be interpreted as the mean free path.

Now I recover the scaling with $L^2$, but I am uncomfortable with this result, as $D$ is a function of $a$ and $\Delta t$.

  • $\begingroup$ Here is what I think: If there are only a finite number of particles in an infinite box they will forever diffuse away from each other. As time passes the average flux through a finite area (no matter how large) must go to zero. If instead you have a constant number concentration of particles $n$ then on average the flux of particles across any surface will be zero for all times (because diffusion occurs only when there concentration differences), but of of course there will be fluctuations. $\endgroup$ – Deep Dec 20 '17 at 5:21
  • $\begingroup$ Hi @Deep, thanks your time. I think you are mistaking the rate at which particles cross the surface, and the net flux. The net flux is indeed related to the concentration gradient, and is zero in my problem (in average). That is because in a given duration there are as many particles crossing in one direction as particles crossing in the opposite direction. What I am trying to find here is the total number of crossing events, independently of the crossing direction. This number is definitely not zero. $\endgroup$ – J-D Dec 20 '17 at 11:01
  • $\begingroup$ Ok now I understand your question. But if you have finite particles in an infinite box won't the rate of crossing go to zero as time increases, because the particles tend to move apart from each other? If I understand correctly your formula indeed seems to say so. Perhaps the question should be either in terms of number density, or the box walls must be made periodic. $\endgroup$ – Deep Dec 21 '17 at 4:48
  • $\begingroup$ My bad, you're right. By "box infinitely larger than..." I actually meant "box much larger than...". I have now corrected this mistake. $\endgroup$ – J-D Dec 21 '17 at 10:29
  • $\begingroup$ I think that if a particle crosses the surface during time $dt$, it will cross it infinitely many times during that same time $dt$ (almost certainly). Defining the rate $j$ is not so easy. Dimensionnal analysis may be tricky. There is something like "the number of times it crosses the surface is proportionnal to $\sqrt{dt}$". $\endgroup$ – Benoit Sep 12 at 13:07

There is no scale for the velocity of the Brownian particles in your consideration. Whereas the velocity of the Brownian particle is known to be distributed according to the Maxwell distribution. The system, a media and the Brownian particles, is at equilibrium. Because of this I think that coefficient of diffusion $D$ is irrelevant in the problem under consideration. In my opinion the right expression is $$ j \approx 2 L^2 \frac{N}{V} \sqrt{\frac{kT}{2\pi m}}. $$ Here $m$ is the mass of the Brownian particles, $T$ - temperature of media, $k$ - Boltzmann constant. The factor 2 takes into account two sides of the square surface.

Edited 1. Look at Kinetic theory of gases, and at Collisions with container in particular. I took the formula from there and multiplied it by $2L^2$. The particles of a gas and the Brownian particles both have velocities distributed according to the Maxwell distribution.

Edited 2. About Maxwell distribution for the velocity of the Brownian particles look at https://en.wikipedia.org/wiki/Brownian_motion#Physics, for example. First pages of http://jfi.uchicago.edu/~leop/Physics%20352/Chicago%20course%20lectures/Chicago%20Course%20Lectures%20/Part%205%20Momentum%20Hops.pdf can also be useful.

| cite | improve this answer | |
  • $\begingroup$ Can you please expand on how you get to this? $\endgroup$ – J-D Dec 20 '17 at 17:19
  • 1
    $\begingroup$ I tried to. Look at Edited. $\endgroup$ – Gec Dec 20 '17 at 17:30

For a pure mathematical Brownian motion (the way you define it), I think the answer is :

$$ j \propto \frac{N}{V}L^2\sqrt{\frac{D}{\Delta t}} $$

where ${\Delta t}$ is the minimal time allowed to count multiple passings for a single particle: all passings of the same particle happenning in a time interval smaller than this cutoff are counted just once. It's entirely arbitrary here, and a convention you need to choose.

The problem it that any particule going through the surface will do it infinitely many times in any arbitrary small time interval around this event. That's why you have to choose a "cut-off" time. I can write my draft derivation and maybe find the propotionality constant is you're interested in this perspective.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.