Let's now do the math to calculate the equal speed at which both A and B could be traveling away from a "neutral" observer.
- [Voiceover] Alright, 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 v. 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 want to solve for v. We can multiply both sides of this equation by one plus v squared over c squared. So let's do that. One plus v squared over c squared. On the right hand side, those cancel. And 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. And then, it is going to be equal to two v. And remember, we want to solve for v. So we're setting up essentially a quadratic in v. So let's now find ourselves some real estate, so let's go right over there. Let's subtract two v from both sides. And actually, I'm going to write it in order of degree. So on the left hand side, I have 0.8 over c v squared, and then minus two v, I subtracted two v from both sides, Plus 0.8 c, plus 0.8 c is equal to zero. I guess if we want to simplify a little bit, we can multiply both sides by c. If we multiply both sides by c, this will become 0.8 v squared minus two c v plus 0.8c squared equals zero. And 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. This is our a. And this is our c. We're talking about the different variables using the quadratic formula. So v is going to be negative b. So the negative of negative two c. So it's going to be two c, plus or minus the square root of b squared, so negative two c squared is positive four c squared, minus four, times a, so minus four times 0.8 times c, so times 0.8 c squared. All of that over two a, so 0.8 times two is 1.6. Now let's see this is going to be equal to two c plus or minus the square root -- now I can factor out a four c squared. Four c squared times one minus, 0.8 times 0.8 is 0.64. All I did was factor out a four 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 -- this is all algebra at this point -- two c plus or minus -- if I take the four c squared out of the radical, its going to be two c. So two c plus or minus two c times the square root of one 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 principle root of that 0.6. So that's 0.6. And move down a little bit. So this is going to be equal to two c -- and I can see the end -- plus or minus 1.2 c, all of that over 1.6. So we got two possible values for v, but we can tell if we add this up here, you're going to end up with 3.2 c divided by 1.6, which is going to be a speed, a velocity, greater than the speed of light. So we can rule out the positive version of it. So we know the answer's going have to be negative version of it. So two c 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 do we know now? We know now that A in A's frame of reference 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 the velocity of half the speed of light. Half the speed of light. And 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 Einstein-ian 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 c. This was really cool. We were able to find a frame of reference that you could think of it as right in between. Now what's going to be really cool about this, is that we can actually -- you know what some people don't like about Minkowski space-time diagrams is that it looks asymmetric. Even though B is moving in A's frame of reference with a velocity of positive eight-tenths of the speed of light, 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 a neutral frame of reference, where they're both moving in different directions with the same speed. Which actually makes interpreting the space-time diagram a little bit easier. We will do that in future videos. But this was a fun problem in it of itself.