I've noticed when sweeping the floor that dust gathers particularly in the corners. I assume there is a fluid mechanics reason for this. Does anyone know what it is?

Edit: No, really, this is a mathematical question. Air blows around the room, which constitutes a vector field. Let's say it blows through a square room with doors near bottom-left and top-right. Air blows from bottom-left to top-right.

There are dust particles in the room, let's say uniformly distributed at first. But after application of this flow they are not uniformly distributed. They pile in the corners. Maybe that's because vortices form more in the corners, maybe some other reason.

It's not because of where I start sweeping.

It is a physics question, but obviously knowledge of Newtonian mechanics won't solve it. It comes down to fluid flow and vector fields, which is math.

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    $\begingroup$ It's harder for brooms to reach corners? I'm not convinced that this is a mathematical question. I suppose any air currents in a room will tend to blow dust to the boundaries of the room, but this is a basic observation. $\endgroup$ – Qiaochu Yuan Dec 11 '10 at 7:56
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    $\begingroup$ Re: your edited question: isn't this easily explained by the fact that a dust particle near the middle of the room will be blown from all sides (assuming the air currents are sufficiently random), but a dust particle near a corner will tend to only be blown toward the corner? I don't think there's anything particularly deep going on here. $\endgroup$ – Qiaochu Yuan Dec 11 '10 at 8:12
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    $\begingroup$ @Qiaochu: I think you should post that as an answer, even if it's a basic observation and not particularly deep. $\endgroup$ – ShreevatsaR Dec 11 '10 at 8:16
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    $\begingroup$ You'll probably need to model additional things like the deposition/erosion of dust on the boundaries, or maybe the inertia of dust particles with respect to the air flow. If you just have massless dust particles advected passively through an incompressible air flow, the distribution of dust will remain uniform. $\endgroup$ – Rahul Dec 11 '10 at 8:17
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    $\begingroup$ @Qiaochu: Why wouldn't a dust particle near a corner have equal probability of being blown towards or away from the corner? If what you said was true, there would tend to be a net flow of air away from the middle of the room, which isn't physically sound. I would agree that the air currents are likely to be weaker near the corner due to the presence of boundary conditions, but I don't know if you can conclude much more than that. $\endgroup$ – Rahul Dec 11 '10 at 8:20

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.

  • $\begingroup$ I especially like your the statement that EVERYTHING is a matter of mathematics. Upvote just for that. I would add that quantification is not necessary to be mathematical, for instance consider ordering relations and topological relations. $\endgroup$ – isomorphismes Dec 14 '10 at 2:54
  • $\begingroup$ @Guess Your answer explains why dust would be at the walls but I'm not totally convinced that it explains why dust goes to the corners. If a dust particle is taking a random walk back and forth between the corner to its "left" and the corner to its "right" then it has 100% chance of returning to the center. $\endgroup$ – isomorphismes Dec 19 '10 at 20:04
  • $\begingroup$ In particular why is surface area related to air that can blow from / towards a corner? $\endgroup$ – isomorphismes Dec 20 '10 at 1:00

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:

Transition probabilities for dust particle

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.


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.

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    $\begingroup$ Your first sentence makes for the a great first line in a really, really depressing novel :) $\endgroup$ – Mariano Suárez-Álvarez Sep 23 '11 at 17:05
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    $\begingroup$ Also, I believe similar logic explains why, after you buy a set of pens and pencils they evaporate from your desk, accumulating in weird places where you don't use them like your car. $\endgroup$ – Nick Alger Mar 10 '12 at 10:02

First, we must assume a more or less constant airflow in the room, for as everyone knows,in a room that is closed up for a long time with no airflow, dust accumulates evenly throughout the room. The simplest air flow type is random. Like water, or anything tbat flows, air flows through the path of least resistance. For air, the path of least resistance is a straight line. Resistance to flow is proportional to the angular change of direction produced by an obstacle; in this case a corner, which does not neccessarily imply a 90 degree corner. There are two types of corners: inside corners (less than 180 degrees and outside corners (greater than 180 degrees). Outside corners do not gather dust, simply because they do not force the air flow to change its direction. Inside corners, however present a physical barrier to straight-line air flow. The smaller the angle of deflection, the greater the resistance. The air simply avoids the highly flow resistant corner by curving around it. In doing so, it loses some of its energy, slowing its flowrate or speed. Of course, some of the air particles pass closer to the corner than others. The closer particles have to turn more sharply to avoid the black hole of the corner, thus slowing down more, thus losing more of their ability to hold a dust particle aloft, thus losing more dust particles before returning to the (assumed) random air flow of the room. In a simple rectangular or square room, this action results in dusty corners. This is the simplest and least technical answer I can think of. Such things as Navier-stokes equations, Laplace equations and pressure derivatives have their place, but are superflous, irrelevant overkill for answering such a simple question. "KISS" always applies. If you want a solution to the dusty corner problem (and seeking the cause of a problem is, after all, the first step in seeking a solution), simply place a contantly running vaccumn inlet in each corner with the (filtered) outlet in the center of the room, directed in all directions, and you will have constantly self-dusting corners. Then, (not so simply) eliminate all dust generators and dust attractors, and replace all objects that have inside corners (or modify them so that they have inside curves instead), and you will have a constantly self-dusting room. However, you should not complain if, upon enterimg this room, you friends declare, "This room really sucks!". I hope this clears the air (and the corners, of course).

  • $\begingroup$ I should have specified that I am talking about 90 degree corners. $\endgroup$ – isomorphismes Dec 19 '10 at 19:57
  • $\begingroup$ How do you know that $\mathrm{resistance } \propto \Delta \mathrm{\, direction}$? $\endgroup$ – isomorphismes Dec 19 '10 at 19:58
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    $\begingroup$ Is it really true that a sealed-up room accumulates a constant amount of dust? I don't have experience one way or the other. $\endgroup$ – isomorphismes Dec 20 '10 at 1:05

If there is a net airflow in any particular direction at floor-level for any reason (doors/windows/radiators etc.), then the dust will tend to accumulate in the corner that is furthest 'down-stream'.

  • $\begingroup$ I've also noticed, peculiarly, that dust accumulates as well near doorways that are not in corners. $\endgroup$ – isomorphismes Dec 18 '10 at 7:44

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