On to Part 2 of the Lorentz Transformation. Let's remind ourselves what we just did in Part 1. Again, Alice and Bob, Alice stationary in this case, Bob moving with velocity v. And each of them have their lattice of the clocks. Synchronized. Now of course, as we know from relativity, simultaneity, etc., that they would not be synchronized completely. But at time t equals zero, we can setup such that at that instant in time as Bob passes Alice then their clocks agree at that instant and locations will say that the origin's the zero point as well. And we showed for an event of some t b, some time, t b measured by Bob on his lattice of clocks, and x sub b equals zero at his origin point. In other words, with him in his cockpit, in his ship. We imagine the flash of light, right there. So, that means. That's x b equals zero, when ever time that flash of light may all have occurred along here. If it occurred in this cockpit that's where his origin is remember. That's his last of clock's remember the prop's, props we use that showed that last clock's and as far as you concerned his at rest with this lot's of this clock's and that's the origin. That's the x b equals zero point of the flash and we showed given that special case then, and we know what Bob's time measurement is we can find t's of a for Alice, what her time would show on her clocks and x of a the location of Bob, in terms of her measuring system. Now, what we'd like to do, instead of having a flash right there, as we talked about, we'd like to have a flash occur any place, in terms of Bob's system, that he measures. And he says some. Hey, I saw a flash of light, an event, something occurred at x b equals something or other and time something or other by my lattice of clocks. In other words, some x b, t b.. And we'd like to find out, okay, for Alice what would be the x location of that and what would be the time showing on her a lattice of clocks. So in other words we need some formula. We have given x b and t b. We want to say okay, x equals something or other involving x b and t b and t a equals something or other involving x b and t b, sort of like what we have here. TA equals something involving t b, x a equals something involving t b, x v does not occur in these formulas because, remember, the special case we did had x v equals zero. Now, we need the general case where x v could be nonzero. And so, we have to figure out well, some formula couldn't there be infinite number of formulas that might apply here? We'd like to put some restrictions on this in terms of what we're looking for, and just to give you a couple exams of restrictions, so we're not just looking for an infinite number of formulas, we have to sort through and figure this out. First of all, let's just say we had a form like this. What if we said a very simple thing. What if x a equals x b squared plus t b or something like that. Well, let's even throw a gamma in there as well. We got some gammas around here. So, what if the formula had that form? That given x b and t b, multiply all this stuff together, add it together, and that would give us a value for x a. Note this can't be the right formula, because the units have to match up. x a is measured in meters or whatever length unit. We're using x b squared here, if we're using meters. That means this is in meters squared. You can't have meters over here and meters squared over here. And then gamma is dimensionless. It has no units. t is in time, it's seconds. And so, I'm adding meters squared plus seconds. You can't do that. It doesn't make any sense. And so, you've got sort of meters squared and seconds mixed up over there. You have got meters over here. That's not going to work. You can't have a formula like that. So that puts some restrictions on it. Another words, whatever the formula is, the units for this one over here have to be units of length. And the units of this formula whatever it is, have to end up being units of time there, we can't have something like that. Another thing, I just actually erased our formulas here, but when x b equals zero. Then I have to get my two equations from part one. So, these more general equations have to reduce down to part one equations, so that would be another restriction on it. But that still leaves a lot of possibilities. So we're going to do, it's not really a derivation here or a rigorous argument but we're going to try to motivate a certain form of these formulas that going to have here to make our life little easier. And here's a basic idea. If I'm writing x a equals whatever over here, and t a equals something over here involving x being t b. I can't have any x b squared, or t b squared or square roots of stuff like that over. I just have to have x b's and t b's in either case, why might that be a restriction? If it is a restriction, it's good. Because it will really reduce down the possibilities in terms of our formula. As we'll see. Let's just figure out, why not? There may be a t b squared in there or an x b squared or a square root or something like that. Let's consider our clocks here with t a and t b and let's just say let's let t a equals t b squared. Okay. Now I know, as I was just talking about the units aren't going to work there. So we'll assume, we have some other stuff there where we can get the units to work out, in fact, let's do this exactly here. So, well actually I don't, we could do something like this. So that's seconds squared on the top, I need to get seconds here, so I need the denominator, I'd have to put something that involved seconds there. And so, I certainly have a velocity. Now what, if I have sort of, if i just play around with this. If I have v involved there, velocity is meters per second. So, one over the velocity seconds per meter. And then, if I multiply that just by let me just say, it could be, okay, so x b divided by a velocity. So that's meters say, divided by meters per seconds, gives me units of seconds down here. I've got seconds squared on the top, so the whole thing turns into seconds. Okay. So, don't worry about it, if you don't quite get the point there. The point being is, I could make a formula over here such that I've got seconds here and seconds over there. But to make our lives a little easier, we're just going to assume that. We're going to leave aside the dimensions and units for a minute and just say we can do something like this. t a equals t b squared. Just to show that we can't have any t b squareds over here. Why not? Well, let's think about this a minute. Let's say that Bob. Okay? So we have t b. Equals zero, one, two, three. Okay, just in times t b. That means, by the formula we're trying out here t a would be t b squared, so it's zero squared is zero, one squared is one, two squared is four, three squared is nine. By that formula t b equals that, and plug in the values, we get the t a values there. It's a little strange here, and in fact, the point we're going to make is that the distances between ticks say, on our clocks, have to be uniform. Note the distances here. I've got one, one, one. Here, I've got one, three, five. Why do they have to be uniform? Well, let's do this Given here t sub b equal zero, one, two, three I can just as easily on Bob's clock we can also move clocks forward or back again right we can set them, we can calibrate them just like day light savings time or sometimes you do it depending what time zone living in what country you're living, as well whether you use day light saving time or not but very easily, you can re calibrate your clocks. Set the origin, set the zero point to whatever you want, and you're measuring elapsed times. It shouldn't effect anything. It shouldn't effect your answers, in terms of when you set your clocks, in terms of the elapsed time between one event and another event. So for Bob's clock so I could just, is easy reset it and move it forward a second so instead of zero, one, two, three, I have one, two, three, four. Okay, would it make any difference to Bob in terms of his motions but look what happens, when you then uses formula and put it into Alice's Clocks in her times, then it becomes well this is one squared four, nine and 16. So whereas, with Bob the distances between times between each event still the same. one second. one second. one second. So I got one second, one second, one second. No change between. If I had events at 0, 1, 2, 3, versus 1, 2, 3, 4. Just moving the origin for our timescale. But for Alice, it makes a huge difference that Bob just moved those over because. Here, I had one second between them, three seconds, and five seconds between those events. Now I have 3 seconds, instead of one second. Five seconds, instead of three seconds and seven seconds instead of five seconds. So if I have things like t b squared in there and square roots and things like that it messes up the time keeping system, we expect that this transformation as I have said, you may have time dilation as we've seen so, things get stretched out or compressed but the distances between cannot change. Okay, so if Bob has one second between events, it's fine to Alice to maybe have and and a half seconds between those events, but it's got to be one and a half, one and a half, one and a half to match up with Bob's one, one, one, or whatever the stretching or contracting factor happens to be. You can't have one and then three and then five and so on and so forth. So I realize, that's a little bit of a hand-waving argument, it's not a rigorous argument, but that's what we're going to use here, to say that, you can't have any t b squared, x b squares, cubes, square roots, anything like that. So, what this means is, assuming our little hand waving argument will accept that, For our equations, is something like this, and actually, let's right t a first. Okay, so means that t a is going to take this form. It's going to be a so called linear form, so for those of you who've had some math, either way back when, or more recently, not really too advanced math, but certainly a little bit beyond algebra perhaps, although you get this in algebra as well. All that being said, what am I talking about here? This is what it's going to look like t a is going to have the form sum capital g, we'll talk about what that is in a second. times x b plus h t b and x a the formula for x a in given x b and t b is going to have the form m times x b plus n times t b. And as I mentioned, this is so called a linear form because the s b's and t b's in the equations here are just x b and t b. You don't have and x b squared, t b squareds, square roots or anything like that. What are the g, h, m and n? Those are just coefficients, but they are just values that don't involve x b and t b. They may involve v and c. And they will, in fact, as we'll see in a few minutes, but not x b and t b. And so, it's a linear form because, technically, these are forms of lines and not parabolas and things like that and square root curves, etc., etc. So. And that comes from our little hand waving argument that the distances between times, and also works for distance as well, have to be uniform when you're transforming from one frame of reference to another frame of reference. It can be stretched or contracted but it can't change from one to the next as you go along with it, as we saw. Okay. So. Accepting that this is our form here. So these are, this is the form of our Lorentz transformation equations that we're going to go with. Now, our final thing we have to do, and it takes a bit of algebra here. But we need to figure out what g, h, m, and n are. And we can immediately, start off with going back to the case if x b equals zero. So we know if x b equals zero from part one, we know t a equals gamma t b. Our basic time dilation equation for that special case. And we know that x a equals gamma v t b those are the two equations that we came up with. And as we said before, these equations have to match this for this special case. So if x b equals zero, so if we take this and plug it into our equations here, that means that goes to 0 for this special case. This term goes to zero for this special case. And we know that x a has to, no let's do ta first. t a has to equal gamma t b. So t a equals, this equals Gamma t b and x a for this special case where x b equals zero has to equal gamma v t b. This equal gamma v t b. And lo and behold, we can immediately see what h and n are. We have already seen it. If t a equals gamma t b, and also equals h t b,, clearly h equals gamma, right? h t b equals gamma t b, so this implies h equals gamma. And for this one, x a equals gamma t b, which equals n t b, and our general form, linear form of. Of our transformation equations and therefore we have, this implies, n equals gamma v. Therefore, here's what we've got. So, the way we got this is we applied our special case x b equals zero to those general equations and therefore, we're halfway there because now we can write our general equation as this we can write okay t a equals we still don't know what g is but g x b plus h is gamma plus gamma t b and x a equals m Times x b plus n is gamma v t b. So those are two Lorentz transformation equations. As I said, we're halfway there now. We still need to find out what g and m are. But we found out, sort of half the form of equations. We had the argument that they need to be linear like this, they need to, in the case of x b equals zero, reproduce our part one equations, and so we've got that in there, and now in the next video clip what we'll do, is tackle finding g and m and we're going to go back and use our invariant interval equation to do that and through the algebra, and get our final form then of the Lorentz Transformation equations.