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In some books I found they are writing photon momentum $mc$ rather than $E/c $. Which one is right? But if I put $v=c$ in relativistic mass equation, $$m=\frac{m_0}{\sqrt{1-v^2/c^2}},$$ it becomes infinity.

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    $\begingroup$ BTW, 0/0 isn't infinity, it's indeterminate. $\endgroup$
    – PM 2Ring
    Mar 8 '19 at 11:58
  • $\begingroup$ @PM2Ring It depends on if the OP is assuming $m_0=0$ for the photon or not I guess. It seems like they might not be $\endgroup$ Mar 8 '19 at 12:04
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The equation $p=E/c$ comes from the idea that $m=0$ for the photon, since in general $$E^2=p^2c^2+m^2c^4$$

Also, relativistic mass isn't really used anymore, so I wouldn't even get hung up on thinking about it. But saying $p=mc$ is just not correct.

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A few things:

  1. Whatever book you're using that writes the photon momentum in terms of a mass, throw those books away. Photons do not have any mass. I repeat, photons do not have mass.
  2. The concept of relativistic mass is pretty outdated and isn't really used anymore. Relativistic mass only applies for massive particles anyway, for which the energy of that particle can be written $E=Mc^2$, where $M$ is the relativistic mass. For massless particles, this isn't even a concept that makes sense.
  3. The relationship between the mass of an object, its momentum, and its energy in special relativity is $$E^2=m^2c^4+p^2c^2.$$ This immediately tells you that, for any object with no mass, the momentum is given by $|p|=E/c$.
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