# Do photons have any relativistic mass?

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.

• BTW, 0/0 isn't infinity, it's indeterminate. Mar 8, 2019 at 11:58
• @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 Mar 8, 2019 at 12:04
• @PM2Ring: That's misleading. "0/0" is a string with 3 symbols that is meaningless (i.e. not defined in any reasonable system), so you cannot write "0/0" without the quotes. Don't propagate the common confusion between form and value. A limit expression may have the form "0/0" but "0/0" does not have a value. Similarly, it is reasonable to define define "0^0" to have value 1, which is totally separate from the fact that some limit expressions have form "0^0" but have no value. Yes, that wikipedia article is misleading, so don't refer to it! Apr 4, 2022 at 15:19

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.
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 The concept of relativistic mass is pretty outdated and isn't really used anymore.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 $$|p|=E/c$$.