Einstein velocity addition
- [Voiceover] So let's revisit a scenario that we have seen in several videos, especially the last video, where we tried to find this neutral frame of reference. Let's say we're in spaceship A. We are in an inertial frame of reference. And let's say right at time equal zero in our frame of reference, spaceship B is exactly where we are, but it's traveling in the positive x-direction at eight tenths the speed of light. We've already seen that we can overlay the spacetime diagrams for each of these frames of reference. If we do our axes, the x is one direction of spacetime that we associate with the x-direction of space and then the ct, this vertical axis, this is another direction of spacetime that we perceive as the passage of time. Well, you can then overlay B's frame of reference, and you kind of have these skewed axes. And you could look at a change in time. So for example, if you were to compare the event right when B passes us up to let's say some amount of time later, some amount of, so let's say this right over there is the change in ct. In our frame of reference, these two events are happening at the exact same place. They're just separated, in our frame of reference, we would perceive them as being separated by time. But if you want to say, well how much time seems to separate these events in B's frame of reference? Well then, you would wanna go parallel to the x prime axis. We're viewing B's frame of reference as a primed frame of reference and see where we intersect the ct prime, the ct prime axis. So this is our change in ct while this is our change in ct, this looks like our change in ct prime. And it looks longer but we have to remember that we haven't scaled these things. And actually the scales change depending on the relative velocities. But we can actually verify algebraically that our change in ct prime is going to be larger than our change in ct. And we just have to look at Lorentz transformations to realize that. So our change in ct prime is going to be equal to the Lorentz factor times our change in ct minus beta times our change in x. We've seen that multiple times before. Well our change in x, our change in x is zero. It looks stationary in our frame of reference so that term is zero. So our change in ct prime is going to be equal to the Lorentz factor times, I don't have to use this parentheses anymore, times our change in ct. And our Lorentz factor is going to be greater than one. I could actually calculate that, let's do it. So the Lorentz factor, the Lorentz factor here. So gamma is going to be equal to one over the square root of one minus, well B's relative velocity so 0.8c over the speed of light squared. Well what is this going to simplify to? The c's cancel out. 0.8 squared is 0.64. One minus that is 0.36. This is going to be, and then you take the square root of that, that's going to be one over 0.6 which is equal to one over six tenths which is the same thing as 10 over six which is equal to the same thing as five thirds which is equal to one and two thirds. So you can see our change in ct prime is going to be one and two thirds times the change in ct. Now you might wanna just say, well do these two look like one and two thirds? And it might look a little bit like that the way I draw it, but you can't just look at it purely on, you can't just take a ruler for this length and a ruler for this length because the scales are different and I haven't marked off the scales. Well this at least helps us visualize. But let's think about the other way around. Let's imagine the change in ct prime between right where the spaceships pass by and a little bit later. Now, I'll do this in a different color 'cause this is actually a different event in spacetime than the one that we were focusing on right now. And we're gonna be viewing it from B's frame of reference so that's our change in ct prime. But what is going to be our change in ct? Well, to think about it, we could go parallel to the x-axis, the x-axis right over there, so you'll parallel, parallel to the x-axis and you get right over there and so our change in ct looks like it's more. And once again, we can algebraically verify it by really doing the same thing. Our change in ct is going to be equal to gamma times our change in ct prime minus beta times delta x prime. And if you were to actually, if we actually did have a change in x prime here, the beta, the velocity now has a different direction so it would all be the negative but our change in x prime from B's point of view, these two events are happening in the same place so our change in x prime is zero. So you have change in ct is equal to gamma times change, once again I don't need my parentheses, times change in ct prime. And it's going to be the same gamma because remember, we're taking v over c and so whether it's either v or negative v, when you square it, it gets the same value so once again, gamma is going to be one and two thirds. So it's going to be one and two thirds. So it seems a little bit strange. You know, I have some passage of time in my frame of reference where it looks, you know, it's something that looks stationary, two events that look like they're happening in the same place but one after another. It looks like their change in time, it takes longer for those two events to happen in the ct prime in the moving frame of reference. But then, if we have some event that looks stationary in B's frame of reference and they're separated by a change in ct prime, it looks like the change in ct between those two events is even larger. So it looks like this somewhat bizarre phenomenon. And to help us reconcile these and to visualize a little bit better, actually even to be able to put ct and ct prime on the same scale, we can look for this neutral frame of reference which we did in actually the last video I made. I don't know if it's the last video you've seen. Or we could say look, if A and B are travelling with a relative velocity of 0.8 times of the speed of light relative to each other, B is travelling 0.8c in the positive x direction relative to a stationary A or A is travelling 0.8c in the negative x direction relative to a stationary B. You can find a frame of reference where A and B were in that frame of reference to a stationary observer in that frame of reference, A and B are both travelling outwards at half the speed of light. And we figured that out where we did those videos on a neutral frame of reference. And what's neat about that is if you make what looks like a Minkowski spacetime diagram in that neutral frame of reference then the ct and ct prime, A and B's frame of references, get equally skewed to the, you know, if you think about the time axis to the left and the right. And since they're equally skewed, the time dilation relative to this rest frame is the scaling is going to be the same. So we could put both of these on, both of these on the same scale. And you could see that this neutral frame of reference, this ct prime prime frame of reference, I drew them over here, that if you look at either of A or B's frame of reference, they're going to be in between A and B's frame of reference. But here it's the neutral one. It's the one where we're drawing the time or we're drawing the two axes being perpendicular to each other. And now if you look at these two events, so if you look at this first event, where you have our delta ct, so you look at this first event where you have a delta in ct between right when the spaceships pass and right over there. You see that if you were to look at that, if you were to look at the ct prime coordinate for that event, when you go parallel to the x prime axis, you get right over there. So this is, that is your change in ct prime. And likewise, if you have that other event that I drew in that blue-green color, right over here. And this is your change in ct prime. This is your change in ct prime, well then if you think of it from A's frame of reference, well we just follow the x-axis not the x prime. We go parallel to it. We end up right over there. Now what's really interesting about this, what's really interesting about this is from A's frame of reference, the yellow event happens before the blue event. But in B's frame of reference, the blue event happens before the yellow event. So really, really, really fascinating things going on but what I really like about this diagram is that A and B's frame of reference are going to have the same scale since they're both equally skewed to the left and the right if we're thinking about the time axis. And this type of diagram is called a Loedel diagram. Loedel, Loedel diagram which is really a variation of a Minkowski diagram but it lets us really appreciate the symmetry between these frames of reference.