Bob's Relativistic Velocity Addition: 0.9c for Alice & Charlie

[DiracPool]

My name is Bob. I'm floating in deep space and I have these two powerful cannons that shoot my friends Alice and Charlie in opposite directions to me, one to the left and one to the right. I shoot each out at 0.9c. As I look to the left, I see that Alice is flying away from me at 0.9c. Then I turn to the right and see Charlie is also flying away from me at 0.9c. So I'm feeling pretty secure because I'm thinking, well, since I can only look at one at a time, nobody here is traveling faster than c. But then I think, well, Alice and Charlie are probably looking at each other here, and they must see themselves receding from each other at 1.8c. But then I'm thinking that there's some relativistic velocity addition thing going on between Alice and Charlie that I'm not privy too whereby they see themselves as traveling at less than c relative to each other.

Do I have this correct? Can I shoot out my friends out at 0.9 c relative to each other in opposite directions and see it this way without them seeing it this way?

 

[Orodruin]

Can I shoot out my friends out at 0.9 c relative to each other in opposite directions and see it this way without them seeing it this way?

Yes.

 

[mathman]

https://en.wikipedia.org/wiki/Velocity-addition_formula

For your case s=\frac{u+v}{1+\frac{uv}{c^2}}, where s is the observed speed by either one, and u and v are the speeds relative to you.

 

[Meir Achuz]

Friends who shoot friends aren't good friends.
Anyway, just use that equation, and they'll never be superluminal.

 

[harrylin]

It doesn't matter if you can see them both at the same time or not; it's easy to modify your scenario so that you see them fly past each other. In fact it has little to do with what you or they literally see.
Apart of that, yes you use the velocity transformation formula to calculate how each would measure the speed of the other if they set up a standard reference system in which they themselves are considered to be in rest.