Explore BrainMass

Explore BrainMass

    Addition of velocities in special relativity

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    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.

    © BrainMass Inc. brainmass.com December 24, 2021, 6:51 pm ad1c9bdddf

    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)] ...

    Solution Summary

    A detailed solution is given.