In this video clip, we're going to take one more look at this whole idea of leading clocks, lagging a trailing clock. So this is take two, you may remember last time we did this derivation in terms of just looking at Bob. If he has a machine that gives off two photons, now Bob is on a space ship as far as he's concerned, he's at rest. As far as Alice's concerned down here, which we'll get to in a minute, she's going at velocity v from left to right. So on Bob's space ship, there's a little device that emits two photons or light waves. And one goes toward his front clock, and one goes toward the back clock. And we'll say to Alice's perspective, as Bob moves this way, this is the leading clock and this would be the trailing clock, the front clock or the rear clock, however you want to specify them. And for both perspective, remember both of these light ways are going to hit his clocks at the same time assuming that the device is right and exact center of his spaceship and the spaceship has length L as far as he's concerned. And therefore, when the light beams hit each clock, they trigger a flash photo take the picture and therefore you see these blocks is having the same time. Alice however remember as Bob goes by here, the key fact is that she certainly sees the light means go off in each difference. But because the speed of light is constant in frames of reference, it means that she sees this light being traveling at the speed of light C and she sees this one traveling at the speed of light C. But Bob's leading clock is moving away from that light beam and his trailing clock is moving toward that light beam. And therefore it takes longer for the light beam going toward the leader, leading clock to reach that and trigger the flash photo versus the one going backwards to the rear clock and triggering that flash photo. So on Bob's clocks, when the flash photos are triggered both of them show, both these photos, one here and one here, show his clocks having identical time. But in Alice's clocks down here again, remember that the lattice of clocks so that we always have to imagine in any frame. Bob also would have a whole lattice of clocks, but we only need two for this case. So imagine Alice's lattice of clocks all synchronized in her frame, just as Bob's clocks are synchronized in his frame. And we can see just qualitatively, we argue this awhile back that again, as far as Alice is concerned, even though Bob's 2 o'clock, the photo indicates flashes hit at the same time. As far as she's concerned, they aren't synchronized because she sees this flash occur after that one. And so Alice's perspective, she sees this flash occurs, records the time there and she can compare it to her clock here. And take this same flash photo of both clocks at that instant in space and time. And then a little while after that, she would see the light beam hit Bob's front clock, the leading clock. And she'd record a time for that as well so that this clock, to Alice's perspective is ahead of this clock. Or this clock lags this clock because Alice, as far as she's concerned, her clocks are synchronized. It's Bob who has a problem with the synchronization of his clocks. And so we capture that idea in the phrase Leading Clocks Lag, that this clock on the leading edge of Bob's spaceship lags behind this clock as far as Alice's observations are concerned. And so last time we sort of worked our way through derivation of that, because we're only doing that quantitatively. We wanted to actually get a result that we might be able to use in certain cases. And so we found that result was to Alice now, that the difference in Bob's two clocks was in at general at distance D, which here was the length L. But in general is going to be at distance D between the two clocks we're talking about, the relative velocity between the two frames Alice and Bob here and then the speed of light squared. So that was the result we came away with after a little bit of calculation. By this point in time, especially after the last few video clips, you might be thinking that these are getting a little long here in terms of the calculations we have to work through. But first of all, remember that it's just a little basic algebra. Once we sort of get this situation set up and analyze it, then we just work our way through the end and if it helps there's a nice quotation from Einstein. The story goes he was at a dinner honoring him as a national academy scientist dinner. And sort of halfway through listening to the speeches honoring him, he turned to his companion and said, I now have a new theory of eternity, obviously referring to the long winded speeches that had been going on. So maybe you feel like that a little bit as well, going through some of our analysis here, that it seems to take an eternal time to get to the end. But once we get to the end and again the analysis, we're not doing anything fancy, too fancy in terms of the math involved. But just some basic assumptions, principles that Einstein started off with. And then we're following them through in certain instances to see what the conclusions are. And one of the startling conclusions or the weird conclusions, we get is that when a spaceship like this is traveling by, it's got two clocks, one on either end or just any moving clocks. They could have a clock way out there and a clock way over here. In terms of Bob's lattice of clocks, Alice would see it in terms of her synchronized clocks as a time difference Dv over c squared. Now, note here that this is going to be a very small number in practice. And it's why we don't see it in common everyday life, because we've got a C squared on the bottom, a huge number and only a v on top here and the D. And normally, the distances involved are much smaller, even when you multiply distance times the velocity, much smaller than c squared here. So in general this is a very small number, but when you get up to speed near the speed of light and large distances, then this can have some interesting effects that can be observed. And we'll talk about that a little bit later on in our course. What we're going to do now though, in terms of Take 2, is use our Lorentz transformation to get this result, much more quickly than we did before. As an example of the power of the Lorentz transformation, remember where we got it from we're using variance of the interval and time dilation came in along the way and so on and so forth. But once we get to the end of it, based on those basic principles then we have this tool we can apply in situations like this. So we don't always have to go back sort of first principle in basic situations to get the answer that we might want. So let's see if we can apply the Lorentz transformation to this situation, and get our Dv over C squared result or Lv, if you're using the length L for the spaceship here. So let's just modify this slightly. So we're not going to use the light beams anymore here or the photons, if you want to call them that. So we're just going to look at Bob's ship, and let's actually label the clocks here, so we know what we're talking about. So this will be clock B1, and this will be clock B2, and I didn't quite line them up nicely here. But we'll say this is Alice's clock, A1 and we're going to assume it's lined up with Bob's clock there, and then we also have Alice's clock, A2 over here. So the question we want to ask is, but before we get to the question, let's just say this. Let's also set remember Lorentz transformation, the basic equations. We have two equations. One is t sub A for Alice is going to equal gamma. This is the time equation now. Gamma times t sub B plus v over C squared x sub B. In other words, given the space time coordinates, the x coordinate and the time coordinate of something happening in Bob's frame. According to Bob's measurements, this time and distance measurements. Then I can find out the time coordinate in Alice's frame given this equation here. Where gamma our usual factor 1 over the square root of 1 minus V squared over C squared and V is relative velocity between 2 frames. So that is one, if we want to go the inverse transformation say, we're thinking about, if we were given Alice's coordinates and I want to find something, a time for Bob on one of his clocks, it'd be gamma T sub A minus v over C squared XA. Remember as we said before, usually it's best to memorize one form if you use it a while you'll just get it down, and then remember the inverse form is just the negative factor. So usually we try to remember the form where okay we got Alice stationary, Bob moving with velocity of v left to right. And then this equation here applies, and then we can just remember reverse everything sort of and get this form. And of course, there are comparable equations for xA and xB. We're not going to use them for this analysis, because we're just dealing with the clocks. So we know what those two equations are, let's remember that these equations apply when we have a reference point. In other words, at some point where we say okay, this is time A was 0 to Alice was zero, time for Bob was 0. X position was 0, the x location and the x location well for both of them, Alice and Bob. And so let's just assume for our analysis here that clock A1 is at TA1 equals 0. When we take a flash photograph of it, it's reading 0. And we'll also assume in that flash photograph, time for B1 equals 0. And we can set things up that way of course it's just one flash photograph there. We can have our clock set how we want them. Remember that as far as Alice is concerned all her clocks are synchronized. And all of Bob's clocks are synchronized as far as he is concerned. So we'll do that, and we'll also assume of course that this is maybe our 0 point. That XA equals 0 here and XB equals 0 right at that instant as Bob passes by Alice's and the 2 clocks B1 and A1 line up. So we've set that up, so clearly this is the leading clock here. B2 is the trailing clock and the question is, we take this flash photograph. Let's also at that instant in time, take a flash photograph back here. So because this is from Alice's perspective, because all of her clocks are synchronized, she'll see this, tA2 = 0. Those flash photographs both show her clocks reading 0. That's why we know it's from Alice's perspective because if this reads 0 and this clock reads 0, we know they're synchronized, we know at the same instant in time as far as Alice is concerned on in terms of reading from her clocks. So we also know for the first flash photograph, tB1 = 0 here, the key question then is what does tB2 equals what? That's what we like to find out and ideally, we'd like to see that if this is tB1 equal 0, and we just did the other derivation that leading clocks lag by an amount Dv over c squared. We're hoping to get that Dv over C squared factor for this trailing clock, or Lv over c squared, since we're talking about a length L between them or distance D if we want to call it the distance. Well let's see what we can do with this where after a time in Bob's reference frame, so let's use the tB equation here. So just start off from this one, and we can say okay, we've got tB2 equals what? This is what we're after, given this situation here, flash photograph, flash photograph at the same instant as far as Alice is concerned. This reads 0. This reads 0. This reads 0. What does this clock here read? The Lorentz transformation will give it to us. So we've got tB2 equals gamma tA2 minus v over C squared XA2. So if we know what tA2 is the time on Alice's second clock, and we know the position of Alice's second clock according to Alice then we're good to go here. Well, let's take a look at this. So clock tA2, here we are so right here is 0. We've set it up that way, and so this is just going to equal gamma times 0. We set our situation up that way, there are other ways we could set this up and essentially we get the same answer. So we know it's 0 minus v over c squared Now what is x A2 here? What is the position, the x location as far as Alice is concerned for clock A2? Well the only distance we have involved here is the length of Bob's spaceship according to Bob, and remember to Alice, that length Iis contracted, it's shorter. And so what we have for her length, really this position here. If this remember, is positioned right here, is X A1 = 0, we just set our 0 point there, then this is in the negative direction. And in fact, it's going to be minus L Over gamma. Thus the distance between those two clocks as far as Alice is concerned. So let's see what we have. We couldn't, we plugged in our numbers, remember we're using the length contraction factor there for that distance. because again, we're given length L in terms of Bob. And so this just becomes gamma and inside here I've got 0 minus, and I've got a negative v over c squared times a negative L over gamma. The negative signs cancel, so I get a positive, so I get vL over c squared gamma and then we can see the gammas cancel. Switch the order of the L and v we get Lv over c squared. In other words, the same result we got previously although it took a lot longer previously to get it. So it shows the power of something like the lens transformation. Once we have that tool in our toolbox, then we can apply it to situations like this. So we don't always have to go back to sort of the first principles and get the answer we want. So that's Take 2 on Leading Clocks Lag. So remember, first of all that Leading Clocks Lag, so when you have a situation where set of clocks is moving past an observer, then that observer for any two clocks in the steps that are moving pass the distance between them in the clocks frame and in this case Bob's frame. This would be L here, but it could be it doesn't have to be the length of a space ship, it could just be two distant clocks. And according to Bob's clocks, times velocity over c squared. The leading clock will lag the second clock, the trailing clock in whatever situation you're looking at there. So that's another derivation of that. Hopefully, that gives you a little more insight into not only just the Lv over c squared factor, a reminder of how this all works. But hence, a little bit of the power of the Lorentz transformation and how we can use it in situations like this. And the key thing really is the user character equation and think how things are set up here and get the right values in there and then hopefully of course, get the right answer.