Applying Einstein velocity addition | Special relativity | Physics | Khan Academy

Now let's apply the formula we came up in the last video, sometimes known as the Einstein velocity addition formula, and we'll see that it's a pretty neat thing.

So let's say, this is once again me floating in space. My frame of reference is just the S frame of reference. Let's say my friend over here, she's flying away from me. Her velocity, let's give it some numbers: let's say she's flying away from me with a velocity of, let’s say, 0.7 C—so 7/10 of the speed of light or 70% of the speed of light.

And let's say this third character right over here, let's say they’re flying towards me. So let me change the direction of the velocity vector. Let me erase that. See, that doesn't look like a pure black, but I think it looks fairly erased, so there you go. Let me give the velocity vector in the other direction.

So the velocity vector this character is flying towards me at, let's say, so U is going to be equal to, since they are flying towards me, X is decreasing/increasing with time. It’s let's say it's negative 0.5 C—so flying towards me at half the speed of light. And once again, both of these velocities are given in my frame of reference.

Now if we were in a Newtonian world, or if we use the Galilean transformations, and this is related to something on your highway, if I'm on a highway and if I'm going in one direction at 70 miles an hour and someone is coming towards me at 50 miles an hour, it'll look to me like they are coming towards me at 120 miles per hour. We would essentially add these two together.

So if we were in a Newtonian world, this person is flying at 0.7 C and this person is flying this way at 0.5 C. You would, in a Newtonian world, say okay, from this point of view it will look like this blue character is approaching at 0.7 C plus 0.5 C, or at 1.2 times the speed of light, which we know violates the laws of the universe.

And that's where this formula is really handy. And actually, let me even make the spaceship point in this direction just so we don't get confused. So the spaceship is going in that direction, just like that. Once again, the magnitude of the velocity coming towards me is half the speed of light. I put the negative there to show it's coming towards me—that it's going, it's that X is decreasing over time.

Well, lucky for us, in the last video we just came up with this Einstein velocity addition formula. So let's apply it from this friend's frame of reference, the S Prime frame of reference. The velocity of this character, which is change in X Prime over change in T, is going to be this.

So what do we have here? So U is the blue velocity in my frame of reference, so that is negative 0.5 C. V is my friend's frame of reference, the frame of reference that we're trying to get the velocity in, so that's 0.7 C.

Let me do it in that color, so that is 0.7 C. And then on top of UV over C^2. So once again, U is negative 0.5 C; I know I wrote it really small. And then V is 0.7 C.

So what is this going to be equal to? Well, in the numerator, negative 0.5 C minus 0.7 C is going to be negative. Our numerator right over here is going to be negative 1.2 C. This is the velocity that you would expect if we were dealing in a Newtonian world. This person says, "Hey, this person looks like they're coming towards me at 1.2 times the speed of light!" That would—that's what our Galilean transformations would give us.

But luckily we have all this business at the bottom that keeps us from violating the absoluteness of the velocity of light, this notion that nothing can travel faster than the speed of light, because in our denominator we're going to get 1 minus, let’s see, negative 0.5 C * 0.7 C.

Let’s see, 0.5 * 0.7 is going to be 0.35, so 0.35 C^2, and it'll be positive. I have negative times a positive, so I could put the negative there, but if I'm subtracting a negative, that's just going to be positive.

So I'm just going to divide by C^2. And so that cancels out. Notice we're going to be dividing negative 1.2 C by something that's slightly larger than 1.2.

So this is going to be negative 1.2 C over 1.35. And lucky for us, this is going to be less than C. Or its absolute value is going to be less than C.

So it's going to be let's get a calculator out, so if we have, I'll just… so if we have 1.2 divided by 1.35, this is equal to approximately 0.89. So this is going to be approximately negative 0.89 C, which is cool.

It kind of goes with our—let me write it over here—so this delta X Prime over delta T Prime is approximately negative 0.89 C. So it makes sense that from our friend's frame of reference, this ship will look like it's approaching her faster than it looks like it's approaching me.

In my frame of reference, it looks like it's coming to me at half the speed of light; in her frame of reference, it looks like it's coming at 0.89 times the speed of light. The negative is just specifying the direction, but we didn't violate the fundamental postulate of special relativity that nothing can go faster than the speed of light and the speed of light is absolute.

So this is pretty cool! I've often even thought about making a video game somehow where you leverage this, where things are flying in different relative velocities, but special relativity applies.