# Lesson Video: Brownian Motion Physics

In this video, we will learn how to describe the Brownian motion of particles and how this explains the diffusion of gases.

14:35

### Video Transcript

In this video, we’re talking about Brownian motion. This is the name given to the motion of individual particles in a fluid, either a gas or a liquid. Brownian motion happens at such a small scale that it’s not something we would notice with our unaided eye. But viewed through a microscope or other magnifying device, this kind of motion is detectable.

One way to start understanding Brownian motion is to think of a game of billiards. Here, we’re looking top-down at a billiards table, and there are four billiard balls on the table. Now, if we set one of these balls in motion toward another one, we know that when they reach one another, they’ll collide, more or less elastically. As a result of the collision, the path that these two billiard balls follow will be altered.

We also know that at any one turn, a given billiard ball may interact with more than one other ball. For example, maybe this blue billiard ball has enough energy that it would ricochet off one side of the table, then bounce off another, and then run into the pink ball. Now, thinking in terms of collisions between billiard balls is a helpful way to understand interactions between particles in fluids, specifically in gases.

That said, there are important differences between the two. First, as we think about particles in a gas, we can say that those particles are all essentially the same. To emphasize that fact, we can color all of these particles the same. Another difference is that, unlike billiard balls, where typically one is set in motion at a time, which may set some others in motion, the particles in the gas are all in motion and all at the same time. And then, lastly, there are lots and lots of particles in a standard gas.

Despite all these differences though, the basic idea still holds that when gas particles collide, it’s a little bit like billiard balls running into one another. We could pick any one of these gas particles we’re seeing. And if we tracked its motion over time, we would see that it experiences many collisions with other particles. This is because there are so many particles in the space and they’re all in motion.

To get a sense for how this works, let’s track the motion of one of these particles of gas. Let’s pick this particle here. And even though this particle is no different from the other particles in the gas, let’s give it its own unique color so we can keep track of it more easily. What we’re looking at now is a zoomed-in view of molecules in a gas that are frozen in time. We can see where these particles would move. That’s indicated by the magnitude and direction of their velocity vectors. But for these particles to go through this motion, we’ll need to let time advance a bit.

Let’s say that we do that. Let’s say that we let time move forward so that each one of these particles moves to the end of its velocity vector. In that case then, the particle that we’re tracking would move like this. And then the rest of the particles in the gas advance forward as well. Now, unlike billiard balls on a billiard table, these gas molecules do not experience appreciable friction as they move throughout the gas. This means that unless a gas molecule runs into either a wall of the container it’s in or another molecule, it will keep moving in the same direction it already has and at the same speed.

Since we haven’t had any collisions yet among the particles we’re seeing, this means we can simply advance the velocity vectors for each one of the particles of our gas. And then we can let time advance again so that, once more, each particle moves to the end of its own velocity vector. Before we do that though, notice that during this next time step, the particle we’re tracking is going to collide with another particle in the gas. This will be the first case where the path of the particle we’re tracking will be disrupted.

Very shortly after this, the particle we’re tracking will experience a second collision, this time with this incoming particle. Once again, we let time advance, and we follow the path of the particle we’re tracking as it bounces off other gas molecules. And we also update the velocity vector for each particle. Advancing another time step and moving our velocity vectors forward, we continue to let time advance.

As we follow along with all this motion, we can see what’s happening. The particle we’re tracking is bouncing around in a seemingly random fashion. And note that the same thing is happening for all the other particles in this gas. As they collide and ricochet off one another, they follow this unpredictable motion path.

Now, here’s what’s interesting about this scenario. For each one of these particles individually, their motion is predictable. By that, we mean if we were to consider just one of these particles, say this one, in isolation, then based on its velocity vector, we could predict the line along which it will continue to move.

But as we can see, something interesting happens when we add in many more similar particles. Even if the motion of each particle all by itself is predictable, when we add lots of particles together, like we have in this gas. Then the interactions between all those particles as they collide make the motion of any one particle so difficult to predict that we say it’s apparently random. And that’s what we’ve seen here with the particle we’ve tracked, which has gone one way, then another, then another, then another, and then another. It’s this apparently random motion of particles arising from the interactions among particles whose motion individually is predictable that we call Brownian motion.

There are a few things we can point out about this motion. First, notice the distance of each leg of the journey that our tracked particle travels. The average length of each one of those distances is comparable to the average distance between particles in our gas. That makes sense because if we had very few particles, we would expect the path of the particle we’re tracking to change very infrequently. On the other hand, many, many particles densely packed in a small space would lead to lots of collisions and therefore many direction changes.

One other thing to notice about the motion of our tracked particle is that its start point, which was right here, and its end point are not separated by a very great distance. This is what we would expect if the particle really does move in a random way. That’s because over time, a randomly moving object will tend to return to its original position. And we can model these particles that way when they’re evenly surrounded by many other particles.

It’s the same thing with the motion of our gas particle. We don’t necessarily know which direction it will go next. But we can say that over the course of very many collisions with other particles in the gas, it will tend back to its original position. We could say then that for Brownian motion over a long amount of time, the distance a particle travels will be very large, while its displacement, the shortest distance between its starting and ending positions, will tend to be fairly small.

Now, this motion of particles and fluids has an important effect on the diffusion of gases. Imagine we have a container that’s divided into two separate chambers. Say further that we fill the first chamber with a certain kind of gas. But we leave the second chamber empty. It’s just a vacuum. Now, say that we remove the barrier between these two chambers. What will happen?

Well, we know that the particles of our gas are all moving in random directions. Just like before, we can show these particles moving along. And as they do, see what’s happening. The particles of this gas are tending to spread out throughout both sides of the container.

Originally, in our first chamber, on the left-hand side of our container, we had a relatively high gas concentration. And then on the other side, the concentration started out low. When we removed the barrier though and allowed this gas to diffuse, overall, the gas moved from the area of high concentration to low concentration. That’s what it means for a gas to diffuse. It adds particles to areas of low concentration by taking them away from areas of high concentration.

We see then that Brownian motion in a gas has an averaging effect on gas density. When the density of a gas is not uniform, just as the density across our two chambers wasn’t uniform right after we removed the barrier. Then particles tend to move into the less dense spaces rather than to loop back on their starting positions via collisions with other particles.

Knowing all this about Brownian motion, let’s get some practice with these ideas now through an example.

If an object exhibits Brownian motion that has an average speed of 1.2 centimeters per second, which of the following is most likely to be the magnitude of its displacement after 60 seconds? (A) Zero centimeters, (B) 1.2 centimeters, (C) 34 centimeters, (D) 72 centimeters.

All right, so in this exercise, we’re starting with some object. And let’s say this is our object. And we’re told that it exhibits Brownian motion. This has a very specific meaning. It refers to the apparently random motion of particles in fluids. So, for example, if our object were a particle of a gas among many, many gas particles, then through collisions with other particles, this object would have its direction of travel changed over and over and over again. If we were to track the path followed by our particle as it goes through all these motions, that path would seem to be random. This is the Brownian motion that our problem statement is talking about.

We can see that over the course of all its motion, our object will have some average speed. And that average speed is given as 1.2 centimeters per second. This means that if we were to measure the entire distance traveled by our particle as it moves and then divide that distance by the time it takes our object to move that far, the result, the average speed of the object, would be 1.2 centimeters per second.

Our question then goes on to say, if this were to continue for 60 seconds, then which of the options we’re given is most likely to be the magnitude of our object’s displacement? Now, let’s recall that the displacement of an object is the straight-line distance between its starting and ending locations. So if our object started here, and then we’ll say that it ends here, then the displacement of our object is equal to this distance in that direction.

Recall that displacement is a vector having magnitude and direction. But then in our case, we’re just focusing on the absolute value of our displacement. We just want to know its magnitude. So if we let an object move apparently randomly at an average speed of 1.2 centimeters per second for 60 seconds, the question is, what will the magnitude of its displacement most likely be?

Now, often, when we think of object motion taking place, we tend to think of that motion possibly happening in three different dimensions. The object could move up and down. That’s one dimension. Or it could move right and left. That’s another. Or it could move what we could call into or out of the screen. A particle of gas, for example, in a three-dimensional container could move in all these ways as well as in any combination of these directions.

But let’s say the object we’re working with is constrained to move only to the left or to the right. This is only for the sake of example, by the way. And our analysis will work just as well for an object moving in three dimensions. So if we say that our object starts here, then the only way it can move from that point is either to the left or to the right. And let’s say further that the length of each one of these arrows we’ve drawn in represents the distance our object would move in one second. In other words, each of these distances is 1.2 centimeters, since our object moves at 1.2 centimeters every second.

Okay, so here’s where our object starts. And let’s say in the first second it moves to the right, so this way. Now, it was equally likely that the object move to the left because it’s exhibiting Brownian motion. And that means that when our object is here, after one second has elapsed, once again, it can only either move right or left. And the probability of it moving in either of those directions is the same.

Let’s say that, for this next interval of one second, the object again moves to the right. We can see that at this instant in time, its total displacement is 2.4 centimeters to the right. But only two of our 60 seconds has elapsed, so we’ll let time continue to unfold. When our object is at this point, it once again can move either to the left or to the right. And let’s say that now it moves to the left, 1.2 centimeters back towards where it started. So our object is now back here, and its total displacement has decreased.

Now, we could do this for 57 more seconds. But before we get that far along, we can start to see a general principle for the motion of this object. At any given point in time, it’s equally likely to be moving to the left as it is to the right. That has a very important implication. It means that if we were to total up all the distance to the right that our object travels over this time interval of 60 seconds. Say that that was equal to this distance. Then we would expect over the same time interval to have the object move the same distance in the opposite direction to the left.

The reason for this, as we said, is that the particle is equally likely to move in either direction. So over this relatively longer time interval, our particle would move some total distance to the right. And then it would likely move that same total distance back to the left. And therefore, it’s most likely that this object ends up in the same location where it began. We see that that agrees with option (A), which says that the magnitude of this object’s displacement after 60 seconds is most likely to be zero centimeters.

Now, note that these other answer options are certainly possibilities for where the object could be. For example, if over this 60-second interval the object always moved in the same direction away from its original location, then it would end up 72 centimeters from that spot. So it’s possible that its displacement is 72 centimeters. But as we saw, it’s more likely that each movement to the right is effectively canceled out by an equivalent movement to the left. And therefore, for this object exhibiting Brownian motion, the most likely magnitude of its displacement is zero centimeters.

Let’s summarize now what we’ve learned about Brownian motion. In this lesson, we saw that Brownian motion is the apparently random movement of particles in a liquid or gas. This motion is caused by collisions between moving particles. And lastly, we learned that Brownian motion in gases results in diffusion of the gas from areas of high concentration to areas of low concentration. This is a summary of Brownian motion.