Deriving Lorentz transformation part 2

0:00- [Voiceover] We left off in the last video trying to solve for gamma. We set up this equation and then we had the insight that look, we could particular, we could pick a particular event that is connected by light signal, and in that case x would be equal to c t, but also x prime would be equal to c t prime.

0:20And if gamma's gonna hold for any transformations between events, between x and x prime and t and t prime, it should definitely hold for this particular event, and so maybe we can use this to substitute back in and solve for gamma, so that's exactly what we're gonna do right now.

0:35So for all the xes I'm gonna substitute it, with the c ts, so I'm gonna substitute with a c t, so x, substitute ct. x, I'm gonna substitute c t. x, substitute c t, and that's it. And then all the x primes I'm gonna substitute with a c t prime, with a c t prime.

0:56So x prime, c t prime. x prime, c t prime. And then I have an x prime here, so it's gonna be c t prime. So let's simplify, and now I'm gonna switch to a neutral color. So I'm gonna now have, I'm now gonna have c times t times t times t prime. So that's gonna be c squared, actually let me just keep using the t, t primes, t, t primes.

1:27OK, I'll still do a little bit of color coding. T prime is equal to gamma squared, gamma squared times, so it's gonna be c squared c times c is c, let me do that in the yellow color, so c squared, t times t prime, so t times t prime. and then we have plus c times v times t times t prime.

2:05plus c times v, I'll just write it this way. C times t, times v, times t prime. And then we have minus, minus, let's see, we're gonna have a c here. So minus c, minus c times t t times v, times t prime times v times t prime. I wrote this v in blue just so it matches up with this.

2:45And we see something interesting is about to happen. And then finally we have minus v squared. Minus v squared, times t times t prime. Times t times t prime. It doesn't look that much simpler, but we're about to simplify it a good bit. And so we're gonna get these two middle terms to cancel out.

3:08So plus c t v t prime, minus c t v t prime. So those are going to cancel out. And then every other term has a t, t prime in it. So let's divide both sides of this equation by t t prime. And so we're gonna get, if we divide the left-hand side by t t prime, we're just gonna be left with c squared, and then we're just gonna divide everything by t t prime, and there our whole thing has simplified quite nicely.

3:35Our equation is now, I'll continue it over here. Our equation now is, c squared is equal to gamma squared, is equal to gamma squared, times c squared, times c squared minus v squared. Minus v squared. Minus v squared. Close the parentheses. And now we can divide both sides by c squared minus v squared, and we would get gamma squared, and I'm gonna swap the sides, too.

4:13So gamma squared is equal to c squared, c squared over, c squared, I'll write it all in one color now. c squared minus v squared. Now if we like we can divide the numerator and the denominator by c squared, in which case this will be equal to one over c squared divided by c squared is one, and then, v squared divided by c squared, and we are in the home stretch now, we can just take the square root of both sides, and we get, we deserve a little bit of a drum roll.

4:45Actually let me continue it up here where I have some real estate. We get gamma is equal to the square root of this, well the square root of one is just one, over the square root of the denominator. square root of one minus v squared over c squared. So hopefully you found that as satisfying as I did, because all we did, we just thought about the symmetry of x prime is gonna be some scaling factor times the traditional Galilean transformation, and x is going to be some scaling factor times the traditional Galilean transformation from the prime coordinates.

5:20We use that and it's important that we use one of the fundamental assumptions of special relativity, that the speed of light is absolute in either frame of reference, that x divided by t is c, that x prime divided by t prime is going to be equal to c, for some event that's associated with a, a light beam.

5:39We use that to substitute back in, and we were able to solve for gamma. So this looks pretty neat. And so some of y'all might be saying, well what about, what about, so we've been able to do the derivation for the x coordinates, but what about the, the Lorentz transformation for the t and t prime coordinates?

5:59And I'll let you think about how we do that. And I'll give you a clue, it's just going to be a little bit more algebra, and we're going to do that in the next video.