In the previous video clip we looked at what was not suspect. Previously, we'd seen time was suspect. Length was suspect with time dilation and length contraction. But last time, we saw that the transverse dimensions, when you're moving along in a certain direction, that the transverse or perpendicular dimensions, width and height, are not affected by that motion. We get the same results for any observer looking at that, observing that. And as I also mentioned, that another name which really never caught on for the Theory of Relativity was the Theory of Invariance. And so, what we want to consider here, it is important that you have these invariant quantities. And we want to look at a quantity called the invariant interval because it's going to be useful for us later on when we look at the so-called Lorenz transformation, which is the relativistic version of the Galilean transformation we considered earlier. So, let's set up the situation here. We've got Kris and Bob and Alice. And we're not using C for Chris, but K, because we don't want to get it mixed up with c for the speed of light. That's the only reason there. So, we're going to consider the time and distance between two events as observed in different frames of reference. So here's the situation. Alice, this time, will be the stationary observer, at least from her perspective, of course. And Bob is in a spaceship traveling with velocity v. And Kris is in a spaceship traveling with velocity 3v, or it just doesn't really matter what that is, I just picked 3. It's a velocity that is faster than Bob is traveling, okay? So, that's the set up here. On Bob's ship, we're going to look at a light clock that he has. And from his perspective, of course, he's just stationary, he sees the light clock bouncing up and down. And the two events we're going to consider are when that light pulse starts going up, so maybe a little flash of light or something will indicate, hey, the light pulse is going up. Hits the top mirror, and then comes back down again and hits the bottom mirror, and that will be event 2. So event 1 is the light pulse going upwards on Bob's clock. Here's event 1. Event 2 is the light pulse coming back down, maybe just a little flash of light or something to indicate at that point in space and time, according to Bob, those were the two events. Now, let's consider what Alice would observe looking at those two events on Bob's clock. Let's assume the first event here, that upward flash or the flash as the light pulse went up, occurs right here. They're all together. And then, of course, in the instance after that, Bob is moving that way. Kris is moving even faster that way. So, Alice is now looking at the light clock on Bob's ship. What does she observe? Well, she observes our classic triangular diagram we've seen before. She's going to see Bob's light clock moving that direction. So here's snapshot one she could take when the light clock just, or the light pulse just takes off. So, there's event 1. She'll see the light pulse moving at velocity c here, let's put that in, of course, just to remind ourselves. That's velocity c there, same thing here, c, c. And we'll get to Kris in a minute. So, she sees the light pulse going up, hitting the upper mirror, takes another snapshot there again. Then the light clock is here, here, and then here. And then the third snapshot when the light pulse comes back down again, that would be event 2. So she sees, according to her frame of reference, her lattice of clocks and measuring system, she sees, of course, event 1 and event 2 at different places. Bob sees event 1 and event 2 occurring at the exact same place, right, at point zero on his clock because he's measuring the system from his frame of reference. And then, what does Kris see? Well, remember, Kris is going, say, 3 times as fast as Bob. So, the first event occurs here. And she makes note of that. Then, however, of course, when the second event occurs, so Bob goes along here. Maybe the second event occurs when he's right here, right? Light pulse goes up, comes back down again right here. Where is Kris at that point? She's actually about 3 times farther down the road there. And, therefore, she will see event in her frame of reference, she will see event 1 here, right? There's event 1. But event 2 will occur behind event 1, okay, because imagine, she's gone farther. She's looking back there to Bob and sees event 2, the light pulse come back down again. So her diagram, again, it's a triangle. But she will observe the light going in a different direction than Alice will. So again, this is c here, speed of light, speed of light. And so, she sees event 1 occur here. Watching Bob's light pulse, looking behind her, she'll see it go up, hit the top mirror, come back down again. And then, again she's way down here now, looking back at where event 2 occurs behind her. So that's what she will see. Let's put a couple of other things in here because we want to do a little analysis of what's going on. I put the height of the light clock. We'll call it h this time. As we talked about in the last video, and I just mentioned a minute ago, those transverse or perpendicular dimensions are not affected by motion in this direction. So, h here, the height of the light clock, will be the same for Alice and will also be the same for Kris. They will all see the height of Bob's light clock as h. That will not change, and that is a key point here in what we're going to derive in a minute. But we need to define another quantity, and that is, we want to define the distance between the two events. Clearly, for Bob, the distance between events is essentially zero. We drew them a little bit apart there. But we'll assume they're exactly. Light pulse goes up, comes back down, hits the same spot there. For Alice, however, we'll say this distance here between the two events in Alice's frame of reference, we'll call that x sub A, all right? Obviously, A for Alice there. And similarly for Kris, distance between the two events in Kris's frame of reference, we'll say it's like this, x sub K. And that's why we want Kris with a K and not a C, because it might get confusing with the speed of light there. So, x sub K, x sub A, we could also define x sub B, which would be zero. Distance between events, as far as Bob is concerned, is zero. So, that's the set up. We can clearly see they have different perspectives on those things. What we want to do now, is use a little bit of the Pythagorean Theorem, right? So, we'll remind ourselves of that, and see where that leads us here. So I'm going to erase this, we have that now. So, let's consider Alice's case here first, right? We've done similar analysis,when we did the light clock, first light clock analysis. Let's sort of redraw the triangle here for Alice, roughly. Looks something like this, more or less, right? And we said this distance here is x sub A, so I'll put it in here, x sub A. And we said that speed of light is c, okay? And we will assume that one tick of the clock is a round trip, up and back again, okay? So, we didn't put that in there but we'll say t sub A, we'll just define it here. t sub A = One tick of the clock from Alice's perspective. Well, what is that one tick? It's up here and then back down again. What is this distance? It's the speed of light times half a tick, right, because this is half a tick here, this is a full tick, and so we can say speed of light, velocity, times time is distance. So this distance here is cta divided by 2. That leg of our triangle. And by symmetry, this other way is the same thing. That's cta over 2. And then this distance here is XA. So now we've defined the three legs of this triangle. So this is the triangle we're looking at, right, I've just redrawn it up here. This is the distance XA, just by definition, we say the distance between the two events, that's XA and the time of travel for the light pulse up and back down again. Well, the time of travel is tA, up and down, so tA over 2 is up and tA over 2 is down. And the distance is the speed of light times that time of travel. So CtA/2 is the distance of that part of the leg, XA's the distance here, ctA over 2. So just as that part of the leg. Now, you may remember of course that the Pythagorean theorem applies to right triangles. And we don't have a right triangle here but we can draw, cut it in two here and that's a nice right angle there for us. There will be an XA there. So now we have another triangle we can actually work with here. So this is the triangle we're going to apply the Pythagorean theorem to. It's essentially half of that triangle. So, this distance here is half of the whole distance. By symmetry, we cut it down the middle here. So this Xa over 2, this leg here is ctA over 2, speed of light times half a tick by definition and what about this part here? Well, remember that's just the height. That is the height of our light clock, Bob's light clock. And so this is h and this is a right angle. So now we can apply the Pythagorean theorem. Let's do that. I don't think we need Bob and Chris's picture here any more here. We can sort of remember what's happening here with that, with Alice. Okay. So the Pythagorean theorem applied to this triangle. So we'll bring it over here. And we say, okay, this side squared plus that side squared equals that side squared. So what do we get? We get h squared = XA over 2 squared plus, oops sorry. That's not what I want. Hypotenuse of course. Hope you caught that. Try this again here. So get rid of the equals sign, let's do this. h squared + XA over 2 squared. h squared + XA over 2 squared = the hypotenuse squared. I got ahead of myself there so equals Hypotenuse (ctA / 2) squared. All right, let's just multiply that out a little bit. So, we'll do every step here, h squared + xA squared over 4 = c squared, tA squared over 4. Squaring each of those terms, and then let's multiple through by 4. I'd like to get rid of these 4s in the denominator and just clean it up a little bit. So multiply everything by 4, so you get 4, if I do it on both sides of course it's legitimate, 4h squared and then plus 4 times that just gives me the XA squared and I get rid of that one as well. In other words I can have this, 4h squared + XA squared = c squared, tA squared, okay? And I gotta do one more thing here. I want to get the xA and the tA on one side. So we're going to write this as 4h squared = c squared, tA squared- xA squared. So, that's my analysis. Just, we started with Alice here, we just said, okay, we'll define the distance between the two events as xA, we know the speed of light is c, so we know this distance it covers. If we define one clock tick, tA, the time from here to here, just as c tA / 2, half of the time. And then when we do the triangle and split it in two, we get xA over 2 here then we just apply the Pythagorean theorem, realizing that we defined the height to be h to our manipulation there, and we get this. You're saying, well, big deal. What's the deal here? So that's for Alice. Note, that we could do the exact same analysis for Kris here. Now, hers is going backwards, but we have a triangle for her. We'll just do it on the triangle here. We will assume that t sub k, here t sub a was one tick for Alice, so here for t sub K = one tick for Kris, okay, this was for Alice. Again, they're all observing Bob's clock, so this is one tick on Bob's clock from Alice's perspective for frame of reference. 1 t sub K one tick on Bob's clock, from Kris' perspective. Of course, one tick, again, is all the way up there, all the way down there. So, she will see slower ticks than Alice will. And Alice sees, they both see, slower ticks than Bob does, the time dilation effect there. But, we have the exact same triangle again. We can split this in two. This whole thing was Xk so the triangle we're going to be looking at here is, I'll do it in green just to show it there, it'll be that triangle. This is the height. That's the same. This becomes Xk over 2. That's a k there. And this distance here is the speed of light times half a tick, because half a tick up, half a tick down. So this becomes ct k over 2. Exact same little triangle as with Alice, just different values here. Okay? We have h is the same. That's going to be important here in a minute. We have ckt over 2, versus cta over 2, and xk over 2, versus xa over 2, here. We could do the exact same analysis, draw the little triangle, do the Pythagorean theorem. We'd get the exact same result, except instead of ta and xa, we have tk, t sub k, and x sub k. And so, let's see, where are we going to put this? Okay. So, here we can get rid of this now. You know what we were just doing there. Okay. We'll leave Alice's triangle up for a minute, there is Chris's triangle. So, Chris's result would also be 4 h squared. h is the same in every case here. 4h squared = ctk squared- xk squared. Okay? And what about Bob? Let's consider Bob here, I guess we'll get rid of Alice's triangle here. So Alice's triangle. Would be that, that's what we analyze there. Well, Bob doesn't have a triangle, right? [LAUGH] His distance x here is Zero, but note for him, oops, and I forgot, gotta get a c squared there, hope you noticed that. Easy to lose these things as you're doing these things, always good to go back and double check. Make sure you haven't missed anything or dropped anything like that. So, both of them, 4h squared, c squared, tK squared, minus xK squared, c squared tA squared with minus xA squared. For Bob, what would he have here? Well, he has the up trip and down trip, okay so it's h up and h down, right? So that's 2h. So we know that light travels, now for him we'll say t sub B equals one tick on his clock according to him. Well according to Bob. The time to go up, is therefore c, times half of t sub B, right? And, c times t sub B, times half of that. So, the total distance here, for Bob, is 2h, okay? H up and h down or we could also say it is c times, c times tB over 2, half a tick plus another c times tB, oops, C times tB over 2, see that? So, h up and h down, so 2h total for the distance. And then thinking about the speed of light, it is going velocity c up, velocity c down and half a tick is tB over 2 because one full tick is tB. So CtB over 2 plus ctB over 2, that just equals ctB. So I've got 2h=ctB, and we'll squeeze this in here. If we square each side, which we can do, everything is positive here, we get 4h squared equals c squared tB squared for Bob's. Let's highlight these. So this is Bob's here, this is Chris' result, and this is Alice's result. And note how I have written them, you may have hopefully caught on by now that's going on here, or have an idea. 4h squared, 4h squared, 4h squared, each is the same for all of them, therefore this equals that, equals that. And qualitatively you can just see that, you've got two events here, event one and event two, h is the same in all cases. H is an invariant quantity, fancier way to say that, it does not change in all these frames of reference. All of them see h as the same thing. And therefore, 4h squared is squared the same for all of them. And so what does that mean? It means, c squared t sub K squared minus x K squared, is the same as c squared, tA squared, minus xA squared. Even though, xA clearly is very different from xK and, t sub K, is very different from t sub A. Because, t sub K, is one tick for Chris, clearly it's longer, than Alice's tick, tA being one tick for Alice. Which is clearly longer than, one tick for Bob here, tB there. Yet this quantity, and it's known as the invariant interval. This quantity is invariant, from one frame of reference to another frame of reference. So again, we have two events, at different space and time locations, depending on the frame of reference. But if you take the location of the event, and the time between the events, okay, so here's what's important here. We've taken the distance between the events, in each of the frames. We've taken the time between events in each of the frames, so for Chris, she measures the distance between those two flashes and she measures time on it, according to her perspective here. Looking at Bob's clock, Alice does the same. And we'll get to Bob in a minute, he's slightly a special case here. But, if they actually take their measurements, t sub K, and x sub K, and t sub A and x sub A, even thought they're different values and you do this little operation, c squared, t squared, minus x squared, you get the same value every time. It is an invariant quantity, it's called the invariant interval. And as I mentioned previously, it will be a very useful quantity when we have to look at the Lorentz transformation, when we want to actually derive the so-called Lorentz transformation. What about Bob though here, let's just say a word about Bob, because you might say well this doesn't look like Alice and Chris. This is different, it doesn't have that second term there. But note that for Bob, the distance between the two events is zero. So really he has a -xB squared here. His general form, you could say is minus xB squared, just like we have a minus xK squared and a minus xA squared there. But for him xB would be a zero. The distance between the two events is zero, so that's why it's not appearing in this example, because it's zero. So in general, what we have here is measure the distance between two events and the time between two events, in any given frame of reference. Okay, so this is Chris' frame of reference. She is measuring the tick of Bob's clock, based on her own light clock, she could have with her. Alice is doing the same thing, based on her light clock. And they're getting different values for the t and x, okay? Clearly, we can see x is different here, xA versus xK. And we can also see tK is going to be different as well versus tA, because again, different times for the ticks here. And then you've got Bob of course being his tB, the shortest of all. And x is zero in his case, x would be equal to zero. Even though the time and distances between the two events are different for each of the frames of reference, this quantity, when you do that operation, will always give you the same answer. And it's really based ont the fact that h is in variant, the height, the transverse dimension here, will not be affected by motion in, say the x direction, as we've drawn here. Okay, so that's the invariant interval. And as I've said several times now, we'll see it again when we consider the Lorentz transformation. Which is coming up next week, I believe.