# Relativity: moving space shuttles [closed]

Two space shuttles $$A$$ and $$B$$ are moving in opposite directions. A person on Earth measures $$v_A = 0,75 c$$ and $$v_B = -0,85 c$$. What is the velocity of $$B$$ relative to $$A$$ (= $$v_{BA}$$)?

So my first thought was to say $$v_{BA} = v_{BW} + v_{WA} = -1.6c$$, but since we're working with velocity close to $$c$$ there needs to be a twist somewhere. Assuming $$A$$ and $$B$$ are moving along the $$x$$-axis, we can use $$V' = \frac{V-v}{1-\frac{vV}{c^2}}.$$ Now I'm stuck and I don't know which values to assign to the different parameters in this formula. I believe that I just don't really understand this formula. Could somebody help me out?

## closed as off-topic by John Rennie, Aaron Stevens, GiorgioP, Kyle Kanos, M. EnnsApr 1 at 15:04

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I replace "W" with "H", for Houston. They're space shuttles, after all. The velocity addition formula is for stacking up reference frames, not starting in the middle:

$$v_{AB} = \frac{v_{AH}+v_{HB}}{1+\frac{v_{AH}v_{HB}}{c^2}}$$

where $$v_{xy}$$ refers to $$y$$'s velocity in frame $$x$$.

You are given:

$$v_{HB} = -0.85c$$

and

$$v_{HA} = 0.75c$$

The key is that any 2 observes see each others' velocity as equal in magnitude, and opposite in direction, so that:

$$v_{AH} = - v_{HA}$$.

Of course, velocities do not add linearly. For future reference, what does is something called rapidity:

$$\omega = \tanh^{-1}{\frac v c}$$

so that:

$$\omega_{AB} = \omega_{AH} + \omega_{HB}$$

Ignoring relativitiy for a minute, $$v_{BA}=v_{BW}+v_{WA}$$ is the composition ["addition"] of velocities in Galilean physics.

Since $$v_{QP}=-v_{PQ}$$, then $$v_{BA}=v_{BW}+(-v_{AW})=v_{BW}-v_{AW}$$ is the formula for relative velocity in Galilean physics.
The picture that might be helpful to think of is
"relative angle as the difference in angles-from-a-reference-angle".

So, using the labeling you started with,
the special-relativistic equation for relative-velocity is $$v_{BA}=\frac{v_{BW}-v_{AW}}{1-v_{BW}v_{AW}/c^2}$$