Deriving Lorentz transformation part 2 | Special relativity | Physics | Khan Academy

We left off in the last video trying to solve for gamma. We set up this equation, and then we had the inside that, well, look, we could pick a particular event that is connected by a light signal. In that case, X would be equal to CT, but also X Prime would be equal to CT Prime. If gamma's going to hold for any transformations between events between x and x Prime and T and T prime, it should definitely hold for this particular event.

So maybe we could use this to substitute back in and solve for gamma. That's exactly what we're going to do right now. For all the X's, I'm going to substitute it with the CT's. So I'm going to substitute it with the CT. So X becomes CT, X becomes CT, and that's it.

All the X primes I'm going to substitute with a CT Prime. So X Prime becomes CT Prime, X Prime becomes CT Prime, and then I have an X Prime here, so it's going to be CT Prime. Let's simplify, and now I'm going to switch to a neutral color.

I'm now going to have C * C * T * T Prime, so that's going to be c^2. Actually, let me just keep using the t and t primes. I'll still do a little bit of color coding. T Prime Prime is equal to gamma squared times, so it's going to be c^2 times T * T Prime.

Then we have plus C * V * T * T Prime. Plus C * V, I'll just write it this way: c times T times V times T Prime. Then we have minus, minus, let's see, we're going to have a c here, so minus C minus c * T * T times V * T Prime times V * T Prime.

I wrote this V in blue just so it matches up with this, and we see something interesting is about to happen. Finally, we have minus V^2 minus V^2 * T * T Prime times T * T Prime. It doesn't look that much simpler, but we're about to simplify it a good bit.

We're going to get these two middle terms to cancel out. So plus ctvt Prime minus ctvt Prime, so those are going to cancel out. Then every other term has a T T Prime in it, so let's divide both sides of this equation by T T Prime.

If we divide the left-hand side by T T Prime, you're just going to be left with c^2. Then we're just going to divide everything by T T Prime, and our whole thing has simplified quite nicely. Our equation is now, I'll continue it over here: our equation now is c^2 is equal to gamma squared times c^2 minus v^2.

Now we can divide both sides by c^2 minus v^2, and we would get gamma squared. I'm going to swap the sides too, so gamma squared is equal to c^2 over c^2 minus v^2. I'll write it all in one color now: c^2 minus v^2.

Now, if we like, we can divide the numerator and the denominator by c^2, in which case this will be equal to 1 over 1 minus v^2 over c^2. Now we are in the home stretch! We can just take the square root of both sides, and we get, we deserve a little bit of a drum roll. Actually, 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: 1 minus v^2 over c^2. Hopefully, you found that as satisfying as I did, because all we did, we just thought about, well, the symmetry.

If x Prime is going to 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, we use that. 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, and that X Prime over T Prime is going to be equal to C for some event that's associated with a light beam.

We used that to substitute back in, and we were able to solve for gamma. So this looks pretty neat! Some of you all might be saying, well, what about, what about our... So we've been able to do the derivation for the x coordinates, but what about the Lorentz transformation for the T and T Prime coordinates?

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.