Physics: Understanding Relative Motion

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Suppose I tell you that an object has a velocity of two meters per second. It's a simple statement. And by now I think we completely understand what it means. Velocity of two meters per second, I'd have to tell you what direction, say to the right, it's telling me the velocity, it's , that change in position with time. Seems nice and simple, but let's just think of a specific example. Imagine that there's a physicist sitting in a room doing physics experiments. Here's the physicist, and the ball is rolling away from her at two meters per second. So she's absolutely convinced that dr dt is two meters per second to the right. Now it turns out that this room is in a train. And the train is moving steadily to the right on a smooth straight track. And it's moving at ten meters per second.
So I'm sitting off to the side, and I'm a competitive physicist, so I'm going to make my own measurements. And what do I see? We're both focusing our attention on the ball. What I see is a ball, which is moving forward at two meters per second with respect to the train, and train, which is moving forward with ten meters per second with respect to me. So I see a relative motion. What I see when I just focus my attention on the ball, I'm not paying attention to the distraction of the train or anything, I see two meters per second to the right plus ten meters per second to the right. I see a motion of 12 meters per second. And I'm absolutely convinced that the real velocity of this ball is 12 meters per second. That is what the velocity of the ball is.
Well, wait a minute. My physics competitor is saying it's two meters per second. And she's absolutely convinced. We're both using a formula. Does this make any sense? Are we starting to realize that there's something wrong with the definition of velocity? How can a physical quantity have two different values? Well, there's nothing wrong with the equation. And we've already seen a situation when a perfectly physical quantity can have different values.
If you just place an object down right here, what's the value of its position? Well it seems like a perfectly reasonable physics question. If this is my coordinate system, the answer is negative one. It's at position negative one. Now I'm not going to move the object. It's exactly at the same position. But if I change my coordinate system, which after all it's a completely arbitrary choice. Now I would say it's at position zero. The number associated with the position is not the physics. It's really where it is that's the physics. It's the reference frame, or the coordinate system choice that determines the value of the number. And that's true for velocities as well as for positions. It's a little bit surprising. We haven't thought about this before, because even though we've allowed for the possibility of relocating our coordinate system at some different place, we never thought about allowing our coordinate system to move along. That's what was going on in this situation. This physicist's reference frame was cruising along with a constant ten meters per second.
So the story here is the story of relative motion. We want to understand motion of objects in two different frames and connect them. After all, if one physicist is doing lots of work uncovering laws and principles of physics, and she happens to be sitting in a train car, all that work shouldn't be lost. I should be able to use all of that work and just convert the answers to my reference frame, and get to use all that good physics. So it'd be nice if we could have some simple formulas that connect the numbers in one frame with the numbers in another frame.
Example, here's a car. It's sitting still. Later we're going to let it move. And here's yet another physicist, Iguanadon physicist, who sits in a train car which is moving northeast at some constant velocity. But right now, at some instant in time, we're taking a snapshot. I'm here in the corner of the page. And I have a coordinate system labeled x y, and here's my origin. I'm going to measure everything with respect to this coordinate system. Iguanadon is in a different reference frame. Iguanadon can choose any x and y-axis that it wants. And it's perfectly free to pick this coordinate system. And we'll just give it the name x prime y prime to indicate that it's a different coordinate system. Of course it could be tipped, Iguanadon's choice.
At this moment in time, what's the position of the car? As far as I'm concerned, the position of the car is this. Let me give it a good name. Let me call it r of the car in my reference frame, . I could put a little comma here. It's a slightly complicated notation, but it should be clear what I mean. It's the position of the car in my reference frame. There's a different measurement being made by Iguanadon, and that measurement is this one. That's r of the car, let's call it the train reference frame, position in the train. And there's one more vector, which I can draw, which is the position of the center of origin of this other reference frame as far as I'm concerned. It's the position r of the train with respect to me, or in my reference frame.
So there's these three vectors. Look at them. They add up. This one plus that one has to equal to that one. And that's really the formula that we were after. I wanted to connect these quantities in different reference frames. If you look carefully, this plus this equals rcm. Let me write it like so, rc in my frame is--you know when you add two vectors, A plus B is the same as B plus A. It's kind of convenient. It doesn't really matter to just write it in this order. It's . And this is the equation. It's a general equation. The position of some object in my reference frame will always be the position of that object in another reference frame plus the position of that reference frame in my frame. It's a little bit complicated, but just look at it . If you just stare at it, it makes sense. It's always correct. This is the equation we were after relating positions. But really I wasn't after positions. At the start of this story, I was talking about velocities.
And we can go from position to velocity very easily. It's the formula that we had up before. Velocity is dr dt. Dr dt, just take the derivative with respect to time of this equation. And what you find is velocity of the car, with respect to me, is equal to the velocity of the car, with respect to the train, plus the velocity of the train, with respect to me. The equation looks almost the same. The order of all of the subscripts is the same. I've just replaced r with v, because I took one derivative of everything.
While I'm at it, let me take another derivative. Let's look at the acceleration. The acceleration of the car, with respect to me, is equal to the acceleration of the car, with respect to the train. What's the acceleration of the train, with respect to me? In this whole story, it turns out that it's really going to be nice if the train is just moving steadily, constant velocity, uniform motion so that it's not accelerating. If it's accelerating, the whole story gets complicated. So as long as the two reference frames, mine and the train's reference frame, are moving with uniform relative velocity. The acceleration of the object we're looking at, the car, is the same. Remember I've been telling you that acceleration is where the action is. We're going to be seeing this coming up soon. Acceleration is really where all of the deep and interesting physics is. And that's already a nice thing that the acceleration is the same no matter which reference frame you measure it in as long as the two reference frames have constant relative motion.
This equation is nice. It's pretty much what we were after. If the person in the train makes a physics measurement of the velocity of the car, and I want to know what it is, I merely need to know what's the velocity of the train with respect to me. And I'm all done.
Let's do an example where we have relative motion. Instead of a train, there doesn't have to be another observer. For instance, imagine the following situation. I've got a river. And the river water is flowing. Here I am. And I see the velocity of the river, with respect to me, is ten meters per second. And if I call this x and y, I would call this î. There's a boater. And the boater wants to get from a dock here to a dock straight across. In my reference frame, the boater wants to go in the j direction. So this is the goal of the boater. But the boater is in the water. The boater looks around, sees water. As far as the boater is concerned, she is living in this reference frame, in the reference frame of the river. So what the goal is, is that the velocity of the boat, with respect to me, with respect to the shore, should be equal to--let's suppose they want to get across also at ten meters per second in the j direction. That's the goal, but what does the boater really have to do? What's the motion in their reference frame? In the reference frame of the river, what's v of the boat, with respect to the river? Well look at our master equation. I've changed symbols on you. So we used to have v of a car, with respect to me. It was v of the car, with respect to train, plus v of the train, with respect to me. I've changed car to boat. And I've changed train to river. This is the term that I'm after, so I can solve for it. It's this minus that. It's v, in this case, of the boat, with respect to me, minus v of the river, with respect to me. Think about the symbols. Connect them up here, and you'll convince yourself this comes straight from that equation. And now I can work this out. I know vbm. It's ten meters per second j. I know vrm. It's ten meters per second î. If I subtract those two vectors, let me put the î first. It's being subtracted, minus ten meters per second î plus ten meters per second j. That's the answer that we were after, the velocity that the boater has to have, with respect to the river. Let's draw this vector. Negative ten in the x direction plus ten in the j direction. So negative ten in the x direction, plus ten in the j direction. The boater has to tip their boat upstream and paddle fast. In the Pythagorean theorem this would be something like 14 meters per second northwest. So she's paddling upstream. The river is going to the right. And the net resulting relative motion as far as I'm concerned is ten meters per second straight across.
This is a nice tool. You can use it to connect motions in different reference frames. It's convenient, and there's actually some very deep physics here. This principle of relative motion was invented or discovered by Galileo. And it's called Galilean relativity. Nothing especially deep about it, it all seems fairly reasonable. We pretty much just wrote it down. But what Galileo was telling us is that the laws of physics, the equations of physics, are the same in different reference frames. And you can really check that. You can look at all of our equations of motion that we've written down and write them in one reference frame. Write them in another reference frame. Use these connections and you'll see that the equations themselves, v final equals v initial plus AT. That equation will be the same. The numbers you plug in will be different in different reference frames, but the equation is the same. Physics is the same in different reference frames.
Albert Einstein thought about this idea, and he generalized it. The special theory of relativity was the first generalization, and finally the general theory of relativity. And at this point, we understand that all laws of physics are frame independent. It's a nice thing. It means that you can do physics in any reference frame you like. And we can always just use these simple equations to convert the results, the details, to any other reference frame that we might be interested in.
Kinematics
Relative Motion and Reference Frames
Understanding Relative Motion Page [1 of 3]