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Addition of velocities in special relativity

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Suppose the speed of light were 100 mph. Trains on parallel tracks are approaching each
other at speeds of 92.64 mph and 87.89 mph. A person on one of the trains sees the two
trains approaching each other at what speed? Answer in units of mph.

After the trains pass a crazy person on the 92.64 mph train fires a rifle with muzzle velocity 92.64 mph back at the other train. With what speed does the person on that
train see the bullet approach? (A negative answer means the bullet is receding).
Answer in mph.

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Solution Preview

In c = 1 units the usual formulae are still valid. To convert the speeds to these units all you need to do is divide the speeds by the value of the speed of light in the original units. So, the (dimensionless) speeds in the c=1 units are just v1 = 0.9264 and v2= 0.8789. At the end of the calculation you just convert back to the original units by multiplying the speeds in c = 1 units by the value of c in the original units to obtain the value of the speeds in the original units.

You can "add up" the speeds using the formula:

v_rel = (v1 + v2)/(1+v1*v2/c^2)

in c = 1 units this is just:

v_rel = (v1 + v2)/(1+v1*v2)

Here positive v_rel means the trains are approaching each other. See below for the derivation.

You can just substitute c = 1 in the equation, or you can divide v_rel by c to convert that to c = 1 units:

v_rel/c = (v1/c + v2/c)/(1+v1*v2/c^2) =

(v1/c + v2/c)/[1+(v1/c)(v2/c)] ...

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