Lorentz transformation derivation part 1 | Special relativity | Physics | Khan Academy

So, in all of our videos on special relativity so far, we've had this little thought experiment where I'm floating in space and, right at time equals zero, a friend passes by in her spaceship. She's traveling in the positive x direction; velocity is equal to v. We draw space-time diagrams for both of us. First, I draw my space-time diagram in white, and then I overlay her space-time diagram. The angle that is formed between the time axes and the position axes, right over there, that's going to be dictated by how fast v is, how fast she is actually traveling. We give the space-time diagram from her frame of reference. We see with the little primes right over there.

Now, one thing that you might have been thinking about throughout this entire series was, well, if I perceive her traveling with a velocity of v in the positive x direction, if we took her point of view—and that's what I have right over here—if she views herself as just floating in space, what she will see me as right at time equals zero? Actually, I'd say right at t prime is equal to zero. We're saying that t prime and x prime, t prime and t equals zero are coinciding.

Right at that moment, she will see me fly by at negative v, going in the pod in the negative x direction. Once again, there's no absolute frame of reference; these frames of reference are all absolute. These frames of reference are all relative. And so, you could imagine what it would look like. If we drew her space-time diagram where her ct prime axis and x-prime axes, that they are perpendicular to each other, and then based on that, my space-time diagram would be at an angle.

It's at an angle like this. You can kind of see the positive ct axis is in the second quadrant here because I'm traveling with the velocity of negative v. But these angles are going to be the same. This is going to be alpha, and that is going to be—let me write this—is going to be alpha, and this right over here is going to be alpha.

Now, what I want to do in this video is use this symmetry, use these two ideas, to give us a derivation of the Lorentz transformation or the Lorentz transformations. The way we might start—and this is actually a reasonable way that the Lorentz transformations were stumbled upon—is to say, all right, we could start with the Galilean transformation, where we could say, all right, the Galilean transformation would be x prime is equal to—it's going to be equal to x minus vt, v times t.

Now, we already know that if you just use the Galilean transformation, then the speed of light would not be absolute; it would not be the same in every frame of reference. So we had to let go of the constraints that time and space are absolute, and so there's going to be some type of scaling factor involved. We can call that scaling factor gamma, so we could say, all right, let's just postulate that there's going to—if we assume the speed of light is absolute, it's going to be some scaling factor gamma times x minus vt.

Well, you can make the same argument the other way around. If you view it from her frame of reference and you're trying to translate into my coordinates, you could say, well, x—instead of just using the Galilean transformation—x is going to be equal to x prime. Now, instead of a v, we have a negative v, right? So if you subtract a negative v—in fact, let me just write it that way—x minus negative v times t prime. That would be the Galilean transformation but whatever scaling factor we used here.

There's a symmetry here; I shouldn't have to use a different scaling factor if I assume a different kind of frame of reference. So if we assume the absoluteness of the speed of light, we're going to have some other scaling factor just like that. Or we could rewrite this as—let's do that same color—we could rewrite it as x is equal to this scaling factor. I'm really having trouble changing colors today.

It's going to be equal to that scaling factor times x prime subtract a negative plus vt prime. If you ignore the scaling factor right over here, this is the Galilean transformation from the primed frame of reference to the non-primed frame of reference. So, an interesting thing is, well, what is this scaling factor? How do we figure out what that scaling factor is going to be?

And so we can do a little bit of interesting algebra here. What we could do is—actually, let me just write what I just wrote; let me write it right below here. So we could say that x, again changing colors is difficult, we could write that x is equal to our scaling factor gamma times x prime plus vt prime.

Now, what I'm going to do, in order to have myself an equation that involves all of the interesting variables, I'm going to multiply both sides of this equation by—I guess I could—one way to think about it, I'm going to multiply both sides of this top equation by x. If I multiply the left-hand side by x, I'm going to have x times x prime. And then, the right-hand side of the equation, I can multiply by x—but x is the same thing; I'm saying it's the same thing as gamma times all of this business.

So I'm just going to multiply the left-hand sides of the equation, and I'm going to multiply the right-hand sides of the equation. So if I multiply the right-hand sides of the equation, I am going to get gamma squared times—I'm going to have a big expression here. And so, just really applying the distributive property twice, x times x prime, x times x prime, and then x times positive vt—so, whoops, that prime doesn't look like a prime.

x times x prime plus x times positive vt, plus x times actually positive vt prime, I should say. Got to be careful here! x times positive vt prime, and then I have negative vt times x prime. So it's going to be negative vt times x prime times x prime, and then finally I'll have negative vt times positive vt prime. So that's going to be—I could write that as negative—let's write that as vt squared, v—sorry, negative v squared, and actually, let me delete this parentheses. I don't want to force myself to squeeze for no reason.

So I'm going to have negative v times v, so that's negative v squared times t times t prime times t prime. And now let me place my parentheses. So how can I use all of this craziness here to actually solve for gamma? Here, we're going to go back to one of the fundamental postulates, one of the assumptions of special relativity, and that's the speed of light is absolute. You're going to measure it to be the same in any frame of reference.

To think about that, let's imagine an event that is connected with the origin with a light beam. So let's say, right at time and t prime is equal to zero, I were to shoot my flashlight and let's say it hits something at some point, right? So there we look at some event right over there, and they're connected by a light beam by a photon.

So let me connect them. So let me connect them, and so if you say, once again, this could be me turning on my flashlight and the photon at some future, at some forward distance and some forward time—it could just be at some position or maybe it hits something; it triggers some type of reaction. Who knows what it does? But we're going to talk about this event right over there. That event's in my frame of reference; its coordinates are going to be x and ct.

Since we know the speed of light is absolute, and the way that we've set up these diagrams, any path of light is always going to be at a 45-degree or a negative 45-degree angle. We know that x is going to be equal to ct. x is going to be equal to ct for this particular case. I could draw it on this—I could draw it on this diagram as well, if I'd like, just to show that I can. So let me draw that. So it would look like this; it would look like this.

And we would once again have x equaling ct. How would you read that? Well, to get the x coordinate, you go parallel to the ct axis. So that would be the x coordinate on this diagram, and then the ct coordinate you go parallel to the x axis, just like that. But once again, x is going to be equal to ct, and similarly, because the speed of light is going to be absolute in any frame of reference, if we look at x prime, x prime is going to be the same—is going to need to be equal to ct prime, ct prime.

If we look at over here, x prime is going to be equal to ct prime—once again, because light is going to be at a 45-degree angle, so x prime is equal to ct prime. They're connected by light events. So if you take your change in x divided by change of time, it's going to be the speed of light.

So what we can do is use this information for this particular event. If gamma is going to be true for all transformations, it definitely should be true for this particular event. I can use this; I can use this information to substitute back in and then solve for gamma, and that's exactly what I'm going to do in the next video. Although I encourage you to try it on your own before you watch the next video.