Question
Take a ratio of relativistic rest energy, E=γmc2E = \gamma mc^2, to relativistic momentum, p=γmup = \gamma mu, and show that in the limit that mass approaches zero, you find Ep=c\dfrac{E}{p} = c.
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OpenStax College Physics, Chapter 29, Problem 46 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. In this question we take the expression for the total energy of a particle, which is this Lorentz factor which is 1 over the square root of 1 minus the particle's velocity squared over the speed of light squared, take that Lorentz factor times the mass, which is the rest mass times speed of light squared take that expression and divide it by the relativistic momentum γm times the speed of the particle and show that that ratio is the speed of light when the mass of the particle approaches zero or is zero. So we are going to divide the left sides so total energy divided by momentum equals the right sides divided the γ's cancel, the masses cancel and we have speed of light squared divided by the velocity of the particle. Now as the mass goes to zero, the velocity of the particle goes to the speed of light and you know, I can't really find a very good explanation for why exactly in the textbook so we are kind of taking that a bit on a leap of faith and I have a better expression down here for how to show that the total energy divided by momentum is the speed of light. In any case, if one makes this assumption that as mass goes to zero, the velocity goes to speed of light then u can be substituted with c and we have c squared over c, which equals c. But instead consider this other way of deriving the E over p equals c expression. When you take the total relativistic energy of a particle squared, it's going to equal the momentum times the speed of light all squared plus the rest energy mass times speed of light squared squared and then if m equals zero, this term is gone and then you have E squared equals pc squared and when you take the square root of both sides, you have total energy then is momentum times speed of light and you can divide both sides by p to get this same expression E over p equals c and this one is persuasive because we are starting with this equation, which is given to us.