special relativity - How to get the accurate relativistic momentum form for photons?

I have studied from Griffiths, the relativistic form of momentum is $$p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} m_0v$$

Now when I evaluate the momentum for photon, I just insert $v=c$ and $m_0=0$ and I get $p= 0/0$. How does it make sense?

Can you tell me that where I am wrong?

Answer

You should consider a particle with some finite energy $E$ and use that constraint to take the $v\rightarrow c$ limit.

With Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$, the relativistic total energy is $E = \gamma mc^2$. Therefore, $p/E = v/c^2$. With the particular case of $v = c$, it follows that $E = pc$.

Although really, you should simply consider $E = pc$ for massless particles to be more fundamental. The general relation is $(mc^2)^2 = E^2 - (pc)^2$, which corresponds to the the norm-squared of the four-momentum vector in relativity.