Why does dust gather in corners?

Consider the life of a single dust particle that randomly gets kicked up into the air, at which point it floats around for a while before being deposited on the floor again. A simple Markov model of this process might be as follows:

1. the probability of the particle getting kicked up in 1 second increases with the airspeed at it's location, and

2. once it's in the air, where it falls is (uniformly) random.

Then one would expect the particle to spend more time on the ground where the airspeed is low, and less where the airspeed is high, therefore accumulating in low-speed locations.

Simplified 2-state Markov model example

To illustrate the point, imagine the following simplified example:

• There are only two discrete locations the particle can be in: 'corner' and 'not corner', and time is divided into discrete units $t_1, t_2, \dots$

• If the particle is in the corner, then the probability of it getting kicked up and redistributed at any given time is 20%, whereas if the particle is in not corner, then the probability of it getting kicked up and redistributed is 60%

• once the particle gets kicked up, the probability of it getting distributed in one location vs. the other is 50/50.

Then, accounting for all the probabilities, the transition probability from $\textbf{corner} \rightarrow \textbf{not corner}$ is 10%, whereas the transition probability from $\textbf{not corner} \rightarrow \textbf{corner}$ is 30%. This is illustrated in the following image:

Now suppose each state (corner, not corner) starts with equal probability. What will happen to the probability after a series of timesteps?

In general, to get the vector of probabilities at the next timestep $t_{k+1}$, you multiply the vector of probabilities at the current timestep $t_k$ by the transition probability matrix.

$$\begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_{k+1} = \begin{bmatrix}0.9 & 0.3 \\ 0.1 & 0.7\end{bmatrix} \begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_{k}$$

Starting at time $t_0$ and repeatedly multiplying by the transition matrix, we get the following formula for the probabilities at the $N$'th timestep, \begin{align} \begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_{N} &= \overbrace{\begin{bmatrix}0.9 & 0.3 \\ 0.1 & 0.7\end{bmatrix} \dots \begin{bmatrix}0.9 & 0.3 \\ 0.1 & 0.7\end{bmatrix}}^{N \text{ times}}\begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_{0}\\ &= \begin{bmatrix}0.9 & 0.3 \\ 0.1 & 0.7\end{bmatrix}^N \begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_{0} \end{align}

From linear algebra, one can see that for enough timesteps $N$, the probability distribution will converge to a steady state distribution given by the eigenvector of the transition probability matrix associated with the largest eigenvalue, which in this case happens to be the following: $$\begin{bmatrix}P(\textbf{corner}) \\ P(\textbf{not corner})\end{bmatrix}_\text{steady state} = \begin{bmatrix}0.75 \\ 0.25\end{bmatrix}.$$

So then we would expect a single dust particle to spend 75% of its time in the corner, and 25% of the time not in the corner, for this simplified example. Or, supposing there are a lot of dust particles, after enough time we would expect 75% of them to be in the corner and 25% of them to not be in the corner.

Why do particles get kicked up more where the airspeed is higher?

When a particle of dust is lying on the floor, static electricity forces form between the dust and the floor, creating an energy barrier that the particle must overcome in order to be kicked up into the air.

Since the energy density of flowing air grows with the airspeed (actually, airspeed squared), the energy that the air is likely to impart on the particle will also grow with the airspeed. Therefore, the greater the airspeed at the location of a particle of dust, the more likely it is to be kicked up in a given amount of time.

Why is airspeed slower in corners?

If the probability of a particle getting kicked up is proportional the airspeed at its location, the question now becomes "why does the air move slower in corners?"

Empirically, (at least from walking around my apartment) it seems that the airflow is indeed slower in the corners, so this is a good sanity-check.

To really do this justice, you'd have to consider the full Navier-stokes equations, but those are hard, so lets consider a simplified model. Suppose the flow is an incompressible, slow, pressure-driven flow, with a fixed inflow at some air vents. In this case, the pressure field should solve the Laplace equation with no-flow boundary conditions at the walls, fixed flow boundary conditions the air vents, and fixed pressure conditions at the cracks under the door to match the outside pressure, etc. I.e.,

$$\begin{cases} -\Delta p = 0, & \text{inside the room}, \\ \nabla p \cdot n = 0, & \text{on the walls}, \\ \nabla p \cdot n = g_0, & \text{along air vents}, \\ p = p_0, & \text{along cracks under the door}. \end{cases}$$

We have some intuition about the solution to Laplace's equation as some sort of smooth rubber surface trying to be "as flat as possible". In particular, the solution changes smoothly from point to point - if you only move a little bit, the solution and its first derivatives will only change a little bit.

Now consider a corner between one wall along $x=0$, and another wall along $y=0$. If we are near enough to the corner, we expect to be within the realm of influence of both walls, so we expect both derivatives $dp/dx$ and $dp/dy$ to be small, since they are zero at the $y=0$ and $x=0$ walls respectively. Therefore the pressure gradient is small, so the velocity (which is proportional to the pressure gradient) is small in the corner, as expected.

Summary

A simple particle model predicts dust gets kicked up more often where the air is moving quickly, and less often where the air is moving more slowly. A simple Markov model predicts that dust will accumulate in places where it doesn't get kicked up as often. A simple pressure-driven flow model predicts flow to be slow near corners. Putting these pieces together, one is left with the conclusion that dust must accumulate in corners.

Air flow in a room will change direction from time to time. (Just the simple act of walking through the room in random directions will shift the air flow.)

The dust particle will be carried in the direction of the air current each time.

Eventually the dust particle will hit a wall and can obviously be carried no further in that direction.

New air currents, in slightly different directions, will then carry the dust particle in the direction of other walls. Because the dust particle is next to the wall, it will be difficult to create an air current that will move the dust particle in a direction 180 degrees in the opposite direction; therefore, the air current will have a tendency to move in the direction of one of the two perpendicular walls (assuming a square room for example's sake).

The interplay of air currents will continue to push the dust particle back and forth between the two perpendicular walls into it eventually arrives in the corner. This will happen because the closer the dust particle gets to the corner (as when it is pushed directly against a wall), the surface area available to capture air currents from the general direction of the corner DECREASES, while the sufrace area available to capture air currents from the perpendicular corner INCREASES. To say another way, the ability to capture an air current that carries the dust particle 180 degrees away from the corner exponentially decreases the closer it gets closer to the corner, while the ability to capture an air current that carries the dust particle 180 degrees towards the corner exponentially increases as it gets closer to the corner.

EVERYTHING in life is a matter of mathematics. There is not an organic or inorganic process anywhere in the known universe that can not be illustrated by mathematics (of course, there are many processes we do not yet know HOW to quantify mathematically, but that is a limitation on our part; not within the inherent reach of mathematics to explain it).

Okay, that makes for an interesting Saturday morning ... cheers.