Relative velocity addition

Question

Suppose $A$ is an observer at rest. $B$ is moving relative to $A$ at a speed of $u$ to the right and C is moving relative to $A$ with speed $v$ to the right and relative to $B$ with speed $w$ to the right.

Then I know that $$w = \frac{v-u}{1-\frac{uv}{c^2}}$$ However, if now $B$ and $C$ were moving in opposite directions, I am struggling to work out what the new formula would be and to think about it intuitively.

I think it would be $$w = \frac{-v-u}{1+\frac{uv}{c^2}}$$ but I am not sure. Could someone explain this and give an intuitive explanation for why this is?

This is a very simple question, but what would be the Galilean velocity addition for $B$ and $C$ relative to $A$ and why?

Answer 1

Your equations are essentially correct... but possibly confusing.
When you deal with speeds [magnitudes], you should use absolute-values.

I think it might help to use [x-components of] velocities rather than "speeds [to the right, or left]" and to use a more descriptive notation.

Let $v_{BA}$ denote "B is moving relative to A at a speed of u to the right".
Let $v_{CA}$ denote "C is moving relative to A with speed v to the right".
Let $v_{CB}$ denote "[C is moving] relative to B with speed w to the right.".

Then the relative-velocity formula is $$v_{CB} = \frac{v_{CA}-v_{BA}}{1-\frac{v_{CA}v_{BA}}{c^2}},$$ holds, regardless of the whether folks are moving to the right or the left. (In particular, $v_{CB}$ can be negative.)

If you insist on using "speeds" [magnitudes], then signs get introduced if "moving to the left". So, if C is [my quotes] "moving relative to A with speed $V$ to the left", then $v_{CA}=-V$. So, the relative-velocity would be $$v_{CB} = \frac{(-V)-v_{BA}}{1-\frac{(-V)v_{BA}}{c^2}},$$ and the relative-speed would be its absolute-value.

In the Galilean limit (when relative-velocities have magnitudes much smaller than light-speed $c$), the denominator is practically "1". So, we obtain the Galilean relative-velocity formula: $$v_{CB} \stackrel{\rm Gal}{=} v_{CA}-v_{BA}.$$ For the Galilean the [addition] composition-of-velocities formula, solve for $v_{CA}$ to obtain $$v_{CA} \stackrel{\rm Gal}{=} v_{CB}+v_{BA}.$$