A Roller Coaster Ride through Relativity/Appendix D

The Addition of Speeds

Consider a spaceship A of length l_A travelling past you at a speed v_A passing a second spaceship B of length l_B travelling in the opposite direction at a speed v_B (both speeds measured with reference to you). The question we need to answer is what speed v_X does B appear to be going past the astronauts in A?

We need to consider two vital events; first contact, when the nose cones of the two spaceships meet and second contact when the two tails part.

Two rockets passing each other

What is more, let us suppose that these two events occur at the same place from your point of view. This means that the spaceships have to be just the right length so that they both pass you in the same time. Note that from your point of view, both ships are contracted by the factors γ_A and γ_B respectively.

Now what does the commander of ship A see? He sees you travelling past at a speed v_A and also ship B travelling past at some greater speed v_X.

Note that to the occupants of ship A, B is length contracted by a factor γ_X.

Now commander A can calculate the time between first and second contact in two ways. First he sees you travelling a distance l_A at a speed v_A. Secondly, he sees ship B (whose length is contracted to l_B/γ_X) travel a distance equal to l_A + l_B/γ_X at a speed v_X. It follows that:

${l_{A} \over v_{A}}={l_{A}+l_{B}/\gamma _{X} \over v_{X}}$
$l_{A}v_{X}=l_{A}v_{A}+l_{B}v_{Z}/\gamma _{X}$
$l_{A}(v_{X}-v_{A})=l_{A}v_{B}/\gamma _{X}$

An identical argument from commander B's point of view leads to:

$l_{B}(v_{X}-v_{B})=l_{B}v_{A}/\gamma _{X}$

(It should be pointed out that because of the invariance of velocity, if commander A sees ship B moving at a speed v_X then commander B will see ship A moving at exactly the same speed and contracted by the same γ factor.)

Now multiplying the two equations together eliminates the lengths of the two ships and leaves us with a relation between the three velocities. The rest is just algebra.

$(v_{X}-v_{A})(v_{X}-v_{B})=v_{A}v_{B}/\gamma _{X}^{2}$

Now

$\gamma _{X}={1 \over {\sqrt {1-v^{2}/c^{2}}}}$

so:

$1/\gamma _{X}^{2}=1-v_{X}^{2}/c^{2}=(c^{2}-v_{X}^{2})/c^{2}$

hence:

$(v_{X}-v_{A})(v_{X}-v_{B})=v_{A}v_{B}(c^{2}-v_{X}^{2})/c^{2}$

from which we obtain by straightforward algebra the result we desire:

$v_{x}={v_{A}+v_{B} \over 1+v_{A}v_{B}/c^{2}}$

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