Question

Prove that for any relative velocity $v$ between two observers, a beam of light sent from one to the other will approach at speed $c$ (provided that $v$ is less than $c$, of course).

Final Answer

Please see the solution video.

### Solution video

# OpenStax College Physics, Chapter 28, Problem 32 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. We are going to show that when two observers are approaching each other with a relative velocity of

*v*that when light is emitted from one observer with a speed of*c*of course, we are going to show using this formula that the other observer will also report a velocity for this light as*c*. So suppose this first observer is labeled with the velocity*v*that means the velocity for the light that they report is labeled*u prime*and we are told*u prime*is*c*so the question is what velocity does the second observer report for this light? So the velocity that the second observer reports is going to be*v*plus*u prime*over 1 plus*v*times*u prime*over*c squared*. So*u prime*is*c*so we plug in*c*, wherever we see*u prime*here and this*vc*over*c squared*becomes just*v*over*c*and then we can multiply top and bottom by*c*and on the top, let's just leave it outside of some brackets and on the bottom, the*c*gets distributed into this binomial and so we have*c*plus*c*times*v*over*c*, which is just*v*and this numerator and denominator are the same so this fraction is 1 and this ends up being*c*and so we have shown that this second observer will report the same thing as the first observer when they are talking about the speed of light.