Okay, in the last video clip, we mentioned that we wanted to actually go back and do a quantitative calculation for the lag time of a set of moving clocks. In other words, when an observer like Alice, is looking a set of Bob's clocks that are moving past her. Or the other round if Bob is observing Alice's clocks then the front clock rags in a sequence of clocks as it's moving person. So we did the qualitative argument earlier in the course just to get the idea and the principle to remember that the front clock always lags. And now, we want to find out exactly how much it lags by, and so let's set up the situation here. We've got Bob's ship, is what we're going to be focusing on. And he has two clocks, one at the front and one of the rear of the ship. And a special apparatus that can shoot light beams out in both directions. And the length of the ship as far as Bob is concerned, is L sub B. And again, as far as Bob is concerned, he's at rest, he's not going any place. Because he's in his frame or reference, he doesn't see anything thing moving around him. He sees the ship just there. He can't tell whether he's moving or not. He assumes he is at At rest. So length of ship is L sub B, his front and rear clocks are synchronized, he's done that in advance, according to him. And this little device that shoots out light beams will shoot them out simultaneously, one going toward the front, one going toward the back. Since as far as Bob is concerned nothing is going on. He's not moving anything. He sees the front word going light beam traveling at C. It hits the front clock and maybe flash photograph's goes on takes a picture of that clock. At the same time, the rear word going light beam is traveling back and hits the rear clock. This is exactly positioned in the middle of ship and so clearly to Bob the two light beams are going to trigger the flash photographs of the back and front of his ship. At the same time, everything is synchronized so there really is nothing too interesting going on as far as he is concerned. Now though, let's look at Alice's perspective observing Bob's ship. And we will assume that Bob is moving left to right past Alice as we've done before. So, she is seeing his ship moving with velocity V in the positive X direction. She sees his Rear Clock and Front Clock. As before, she sees the light beams get shot out. We'll get back to those in a minute. Note that the length of the ship to her is L sub A, which is necessarily going to be the length of the ship to him because remember we have length contraction. So, we're going to use the length contraction A little bit later on, but for now we'll just say Bob measures his ship at L sub B, some value. Alice's perspective is, well it could be, although we know it isn't, her length of the ship is L sub A. It could be the same as L sub B, but as we know from, previously, a single length contraction occurs, and so we'll take that into account. But we'll do that later. For now we'll just say, L sub A is a length she sees for the ship. And the rear clock,front clock velocity v. Also we have to remember that Alice is doing all her measurements on her clocks. She has her own measuring stick with all the clocks. Here's Alice's. Alice is sitting in her ship. Perhaps as far as she's concerned she's the one who's at rest. It's Bob who's moving past her. And so, she has all her lattice of clocks. They're all synchronized in her frame of reference. So that's what she's going to do the measurements on. She's going to take flash photographs when the two light beams hit Bob's front and back clocks. Then she can do the calculations on her clocks and see what she would see on Bob's clocks at that point. Again, as far as Bob is concerned, there's nothing really too interesting As far as he's concerned, clearly the light beams hit the clocks simultaneously. The clocks are synchronized. Therefore, they'll stop at exactly the same time, or the flash photograph will occur at exactly the same time on each of them. But, let's see what Alice's perspective is. And really the key part here as we did qualitative of the first time is that the speed of light is constant to all observers okay. And if it wasn't that way then as the velocity was going that way the light beam here would take. Counter that velocity and so it slowed it down and speeded up for the other one. But in actual fact, Alice sees the speed of light as c, it does not change. You don't add the velocity v or subtract the velocity v from these light beams, they're just still c. And so, what happens here is as the front of the ship moves off in that direction. By velocity v, it takes this light beam longer to get to the front clock, then the backward going light beam to get to the rear clock because the rear clock is getting closer to that light beam as it's traveling toward it. And the front clock is traveling away from the frontward going light beam. So qualitatively we can see that there is a difference there. What we'd like to do is figure out quantitatively. What that might be. So we're going to erase Bob's picture here, because again that's fairly standard, and we still have Alice's version over here. And, do a little calculation here to see if we can figure out what's the actual difference that Alice sees in terms of this lag time. We know the lag time occurs, now we want From previous arguments we made in previous examples. So now let's see if we can actually calculate it. So from Alice's perspective let's look at the rearward going light beam first. And let's imagine that the ship is traveling along here a little bit, and she sees then a little bit later in time, the leftward, the rearward going light beam hit the clock. And let's say it just occurs, this has moved on a little bit here, let's say it occurs, let's do it on this one right here. So, she sees at that point in time She'll see a value on her clock T sub A and we'll do rear there we'll rewrite this in a second here. But the idea is at this instance in time she sees the light beams go off, ship is traveling this way and then say, okay. This light beam triggered the flash photograph at this point and on her clock, her set of synchronized clocks they all read the same thing at this instant in time for her. T sub A rear is the value that she sees on her clock there. OKay. Well Can we do a little calculation of that distance over time calculation the answer is yes. we know for example that the distance so let's look at the rearward light beam The light beam heading toward the back distance. Covered, is going to equal what? Well, we know that the length of the ship is LA to Alice, so it's going to cover. Half the distance here minus the distance the ship itself has traveled. So it's going to be LA over 2. The length divided by 2. But the back of the ship has been moving forward here by a value v, times whatever time it took to get there. So we get something like this. The distance covered is going to be half the length of the ship minus the velocity times T sub A rear, okay? That sound familiar to you, just sort of ponder that a little bit, it's clearly half the distance to the ship but then half the length of the ship. But the rear part is moving forward, well how far does it move froward in the given time until this is triggered and it hits the clock there? Well it's simply the velocity of the ship times that amount of time until it hits the backward or the rear clock there. Okay, so that's the distance covered and the next question is, how much time does it take to cover that distance? Well it's a light beam, it's going at speed c. So the elapse time for the rear where light beam is simply the distance covered, the distance it has to travel, maybe distance traveled would be a little more precise here, but distance covered divided by the speed of light, okay? But what is the lapsed time? The lapsed time is (TA)rear. So here is what we've got. We've got (TA). Again, this is all from Alice's perspective. She sees the light, the rearward going light beam go off, she sees them both go off simultaneously. Sees this one go this way, the ship is moving that way a little bit. It hits the rear clock at this point on her clocks, T sub A rear, whatever that happens to be. So that time is the distance covered from here to wherever it ends up, divided by c, the speed of light. Well, what's the distance covered? Distance covered is simply this. So here's where we can get an equation that says this. We say, (TA), the value on Alice's clock is this distance covered here so it's of L sub A, I'll write this a little bit bigger here. LA over 2- v times T sub A rear all over c. Now let's look at what we have. We have LA, Alice can get that value by measuring Bob's ship from her perspective, velocity we know, we've got speed of light (TA) rear appears here, appears here. That's really what we're after. I want to find what is the lapsed time on Alice's clock here? And if actually you, this is fairly simple algebra, we won't do it out here, we'll leave it as a little exercise, practice exercise for you to do. If you do that, here's what you find. You find, so you can, You can check your answer here. We can find, Alice's clock when, so this clock here, when the rearward going photon hits the rear clock on Bob's ship is LA over 2. 1 over C + v. So we go from there to there, a couple lines of algebra. Okay, so we say, great. We've got that part, and then of course, as we've being arguing qualitatively, she has to wait a little bit longer for the frontward going light beam to hit Bob's front clock. And, so, maybe that occurs a little bit later on, in terms of her lattice of all synchronized clocks here, and so maybe, we'll go out maybe even to our far one here, that occurs at, we'll call that T sub A front. And again, at that instant in time, all of Alice's clocks say (TA) front. They're all synchronized at zero, one, two, three, boom. Rearward one hits, record that, wait a little bit more here for the front going photon to catch up to the front clock. And boom, at that instant in time, do the flash. It'll read (TA) front, where (TA) front here is going to be a later time than (TA) rear. Okay, we can see that qualitatively, we want to figure out quantitatively. So, let's now do the similar type of thing but for the frontward going beam. We'll leave this one right here because we'll use it in a minute. That was our first result. Okay, so for the frontward. For the frontward light beam, It's really the same idea here, the distance covered, distance travelled by it, will be equal to what? Well, it's again, it has to go half the length of the ship LA over 2, L sub A over 2. Plus a little bit more. Because the front is traveling away from it. And that little bit more is going to be the velocity times however much time. It elapses between when the light beam was shot, and it actually gets there. So it's going to be LA over 2, that's half the ship, plus the extra part. Last time, this was a minus sign because the rear was getting closer to the light beam, the rearward going light beam. So, it's plus v times Alice's clock reading when it hits the front. So that's the distance covered. The elapsed time just like before, Is going to equal the distance covered, Divided by the speed of light, because we're talking about a light beam, that's how fast it's going. So it's just basic distance formula here, distance covered divided by speed gives me my elapsed time. If I have to go 60 miles and I'm going 60 miles an hour, 60 miles divided by 60 miles an hour gives me 1 hour for my elapsed time and so on and so forth. And the elapsed time though is simply (TA) front. That's the elapsed time between when the light beam is shot out of here and when it actually reaches the clock and this is all according Alice's measurements on her clock. So I get just very similar to before, (TA) front equals the distance covered, which is LA over 2 + v (T sub A) front on Alice's clocks. All divided by c. And, if you do that, so notice we have (TA) front on this side, we have (TA) front on this side, we have v which is a given value, c we know, LA is a given value, the length, according to Alice of Bob's ship. And again, you manipulate a little bit this is too much algebra. Get the TAs is on one side and you will find out that we'll write it up here, because we'll want to see both of them here in a minute. You'll find out that. T A front equals very similar to what T A rear is, but not quite the same. L A over 2 times 1 over C- V. There's a C minus V here instead of a C plus V in the first case. And note also just to check this, I've got LA over 2 in both cases 1 over c- v compared to 1 over c + v. This is going to be a smaller number than this, right, as I'm taking c and subtracting some other velocity from it. So, this will be a small number. One over that will be bigger, than if I do, if I add the velocity, and that makes sense. If it, if we get something else, the opposite of that, it wouldn't make sense, because we know, T A front, is later in time, than T A rear. This is saying that the value of T A front is bigger, Then the value of TA rear. So that checks at least doesn't mean we couldn't have made some other mistake there perhaps. But at least we did that check, and it's consistent. So we're almost there. We say, again we're thinking about Alice here. So what is the time difference that Alice reads on her clocks between when the light beams hit the front Bob's front and rear clocks? Well, that time difference which often we use a delta symbol of course, so delta T. For Alice, is going to be simply T a front. The time when, the front light beam hits, minus, the time when the rear light beam hits. And so what is that? Well, we've got our two equations. So we've got L A, over 2. Time one over c minus v. That's from there. Minus our equation down there. La over two. One over c plus v. And if you again little algebra required here it's interesting what happens is because you just say look at this. And say well there's no much difference clearly I've got an lA over two we can pull out in each case, and then we just have the one over c minus v minus the minus the one over c plus v. Doesn't look like that would be too difficult. Or give us some interesting results or anything like that, but here's what happens. I'll leave a little exercise for you to take this and get to the next line here we're going to write down. You can check it against the answer here. Here's what you find. You get LA V all over c squared times 1 over 1 minus v squared over c squared. So when you actually takes about one or two lines there to get to here. And you look at that a little bit and hopefully, this term here reminds us of something. In fact, that is simply gamma squared, because again what should we, we should just write it over here. You should have this in your brains by now, but we'll write it one more time here 1 over the square root of 1- v squared over c squared. In fact, that's why we put it in this form, with the 1- v squared over c squared, in the denominator there. So that's gamma, gamma squared, then, is simply that square root squared, which just gives us 1- v squared over c squared in the denominator. So we could write this as La times v over c squared times gamma squared. Hey, now you sort of gotten some place. So that's our result for Delta T. Let's get rid of this for a minute. Get rid of that one because we have a little more to do here. We say okay that means that as far as Alice is concerned, she has the length of Bob's ship. Remember, this is not the length of Alice's ship, we're using L-As to stand for the length of Bob's ship as Alice observes it moving by. And C squared, of course, v is the velocity difference or just the velocity of Bob's ship, really the velocity difference between Alice and Bob, and the gamma squared there. Well, remember we said, we'd really like to put this in Bob's terms here a little bit, because that's often the easiest way to deal with it. We'll be given Bob's information Bob will say hey, I've got my clocks here, they're nicely synchronized, I've got the length of my ship, and we want to know how does Alice perceive that? Well, Alice will receive that again, the front clock is going to lag the rear clock because quantitatively again, we'll get to the answer here in a second. Quantitatively, we know the photon hits the rear clock first, and we see the value on that and then hits the front clock a little bit later on. And the actual value with Alice's numbers in here would be LA and then C squared V and gamma squared are all just the same for either Bob or Alice. But remember we said the length of Bob's ship was l b, l sub b to him. To Alice it's l sub a, length contraction applies here. As Bob's ship goes by Alice, Alice will see the length of Bob's ship to be smaller than Bob measures it to be. In fact l a Will be, now remember you have to think about this because it's easy to get it reversed. We say okay, we know it's length contraction so Alice is going to see a contracted length for Bob's ship. We know that gamma is a number bigger than one. Right? So, L A is going to be bigger than L B. Alice is going to see it contracted. Actually, you have to be careful, I said it the other way around here. So, L A is actually going to be a smaller number than L B. In other words, L B over gamma, this is why is just useful to think, okay length contraction, what does that mean? It means Alice sees the length of to be smaller than Bob sees it to be. And we know gamma is always bigger than 1 and there for Lb divided by gamma will give me a smaller number for For LA. So we can plug that into our formula here, and remember this is the time difference. That Alice perceives, that Alice observes, between the two clocks. So we've got LA, L sub A becomes LB over gamma. So this becomes LB V over gamma, c squared times gamma squared, in fact I'm going to write it up here now, ok? So we've did all the arithmetic here, so we'll keep the basic formula we want. Let's bring it up a little bit so we can see it better. So we've found now that this equals, and that we can cancel one of the gammas. We've got a gamma denominator, a gamma up there, so we get l b v over c squared And a gamma involved here. Okay, so now we have it in terms of Bob's length. This is very important, the next step is that this equation is on Alice's clocks. This is what Alice sees on her clocks when those photos get taken. Really what we'd like to know is what are on Bob's clocks? In other words when Alice takes those photographs right here, boom and then a little bit later boom, she looks at her clocks and sees this time difference. Either in this form or a previous form or this form. What's the time difference on Bob's clocks? Well, we just use length contraction. Remember, the other thing is time dilation. Moving clocks run slow to an observer who is observing those moving clocks. So, from Alice's perspective, she's just found the time difference on her clocks, between when the two light beams hit the front and rear clocks. The time difference on Bobs clocks is going to be less, right? Because Bobs clocks are running slower than Alice's clock as Alice observes them. And that time dilation factor again is gamma and it's going to be one over gamma. All right. Because again, one way we can write this is, if we have delta t for Alice, a time length for Alice. Is going to be in terms of Bob's clocks, let's write like this so it'll be easier to say this way because that's how we've been saying it. For Bob's clocks, a time difference is going to be smaller than for Alice's clocks. Alice perceives Bob's clocks, or observes Bob's clocks. Running more slowly, and so it's going to be delta t for Alice, whatever time difference her is, divided by gamma. And what we have here, is a delta t, a time difference, length of time, elapsed time for Alice. So to get Bob's version on his clocks, we have to take this and divide it by gamma. And therefore, so, this implies, that the delta t, on Bob's clocks, is going to be this divided by gamma. And though we have a gamma here when we divide by gamma, we'll lose the gamma. And so we get, lbv over c squared. And this is actually what we're looking for after all that, okay. So again, let's review just what we've found here. Qualitatively, we made the argument qualitatively here, we did it before as well, that Alice will see. Bob's front clock lagging the rear clock. Okay, even though the Bob's perspective, they're perfectly synchronized. Okay. To Alice's perspective, they're not synchronized. The hits of the two photons, or light beams, occur at different times on her clocks. And she sees Bob's clocks as not synchronized. And the front clock lags the rear clock by how much? By this much. If I know what the length of Bob's ship is to Bob, and I know what v is, then the lag on, the difference that Alice sees on Bob's clocks is the length According to Bob, times v, divided by c squared so. And this is true then, here we're just using the length of Bob's ship. This is true for any distance where we have something in Bob's frame of reference and converting into Alice's perspective here. So Bob has whole line of synchronized clocks, his whole lattice of synchronized clocks moving along here you need two clocks on that huge measuring stick of synchronized clocks when Alice is looking at those clocks as they go by at speed g. She will say give me two of those clocks on that line I'll know the distance between them. So we actually write this in general. The distance times velocity over c squared, where d is the distance between the two clocks in Bob's frame of reference. So Bob tells me, hey, the distance is this. I'm traveling at speed v, I will be able therefore to figure out, Bob's clocks are not synchronized. And from my perspective the front clock lagging the rear clock by this amount here, this dv over c squared amount. So two things to remember, one is the front clock lags. We knew that before. Now we know exactly how much it lags by If Bob tells us the distance between the two clocks in his frame, it's going to be dv over c squared. Now we went through this to show that we want the result of course, but there's nothing to profound here in one sense, but something very profound in another sense. In other words, this all comes from the fact, really three key facts we use. One is, we use the fact that c is the same to everybody. That a light beam will always have the speed c, relative velocities will not affect the speed of light. And therefore we get the lag time effect. And then to get the actual, quantitative value, we use time dilation, and length contraction to end up with this. In the next video clip, we'll do a shorter version of this, and see how we can get it using it the Lorentz transformation. Just to verify we can also get it from that as well.