Calculating neutral velocity | Special relativity | Physics | Khan Academy

All right, we can now do the math to solve for v. So let me just simplify the right-hand side of this equation.

v minus negative e? Well, that's just going to be two v. One minus negative of v squared over c squared? Well, that's just one plus positive v squared over c squared. And let's see what can we do next.

Well, we can multiply—we're going to want to solve for v—we can multiply both sides of this equation by 1 plus v squared over c squared. So let's do that; 1 plus v squared over c squared on the right-hand side, those cancel.

On the left-hand side, we can distribute the 0.8, and we will get 0.8—or we can distribute the 0.8 c, I should say—0.8 c plus 0.8 c v squared over c squared. Well, the c in the numerator is going to cancel with one of these c's in the denominator, so it's going to be plus 0.8 v squared over c.

Then it's going to be equal to 2v. Remember, we want to solve for v, so we're setting up essentially a quadratic in v. So let's now find ourselves—give ourselves some real estate—so let's go right over there.

Let's subtract 2v from both sides. Actually, I'm going to write it in order of degree, so on the left-hand side, I have 0.8—I'll write that over c—v squared, and then minus 2v. I just subtracted 2v from both sides, plus c plus 0.8 c is equal to 0.

I guess we want to simplify a little bit. We can multiply both sides by c—I don't know if that'll, well, let me just go with that. If we multiply both sides by c, this will become 0.8 v squared minus 2 c v plus 0.8 c squared is equal to zero.

We could keep trying to algebraically manipulate this, but we could just go straight to the quadratic formula here to solve for v. I'll do that in a different color just for kicks.

v is going to be equal to negative b. So this right over here is b. Let me write this down. This is our b, this is our a, and this is our c. We're talking about the coefficient—or we're talking about the different variables—using the quadratic formula.

So v is going to be negative b, which is the negative of negative 2c. So it's going to be 2c plus or minus the square root of b squared. So negative 2c squared is positive 4c squared minus 4 times a—so minus 4 times 0.8 times c. So times 0.8 c squared, all of that over 2a.

So 0.8 times 2 is 1.6. And let's see, this is going to be equal to 2c plus or minus the square root. Now, let's see, I can factor out a 4c squared; 4c squared times 1 minus 0.8—0.8 is 0.64. All I did is I factored out a 4 c squared from both terms, of course, all of that over 1.6.

Scroll down a little bit, and this is going to be equal to—and this is all algebra at this point—2c plus or minus. If I take the 4c squared out of the radical, it's going to be 2c, so 2c plus or minus 2c times the square root of 1 minus 0.64. Well, that's 0.36.

Things are getting nicely simple now—all over 1.6. The square root of 0.36—the principal root of that is 0.6, so that's 0.6. Move down a little bit, so this is going to be equal to 2c.

I can see the end, plus or minus 2 times 0.6 is 1.2 c, all of that over 1.6. So we got two possible values for v, but we can tell if we add—if we do the plus up here, you're going to end up with 3.2 c divided by 1.6, which is going to be a speed of velocity greater than the speed of light.

So we can rule out the positive version of it. So we know the answer is going to have to be the negative version of it. So, 2c minus 1.2 c over 1.6, which is equal to 0.8 c over 1.6.

Well, 0.8 is half of 1.6, so this is 0.5 c. That was really, really, really cool because what's—what do we know now? We know now that if A—in A's frame of reference, A feels like it is stationary.

We have A's friend moving with a relative velocity of eight tenths of the speed of light. There could be a third party C that defines a frame of reference moving away from A with a velocity of half the speed of light—half the speed of light!

From A's frame of reference, it looks like C's velocity is closer to B's than it is to A's. But we're dealing in this relativistic—this Einsteinian world. Because if we then go into C's frame of reference, it actually looks like A and B are both leaving C with a speed of 0.5 c.

So here you can view the velocity as positive 0.5 c, and here you could say the velocity is negative 0.5—negative 0.5 c. This was really cool! We're able to find a frame of reference that you could think of as right in between.

What's going to be really cool about this is that we can actually— you know what? What some people don't like about Minkowski space-time diagrams is that it looks asymmetric. Even though if B is moving in a stream of reference with a velocity of positive eight tenths of the speed of light, well, then if you were in B's frame of reference, A is moving to the left with a speed of 0.8 times c.

But now we can define, I guess you could say, a neutral frame of reference where they're both moving in different directions with the same speed, which actually makes the interpreting the space-time diagram a little bit easier.

We will do that in future videos, but this is a fun problem in and of itself.