# How to get the accurate relativistic momentum form for photons? [duplicate]

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?

## marked as duplicate by John Rennie, Kyle Kanos, BMS, Qmechanic♦Jun 16 '14 at 14:00

• possible duplicate of If photons have no mass, how can they have momentum? – John Rennie Jun 16 '14 at 13:04
• You have to take the double limit $m \to 0$ and $v \to 1$ (I'm using $c=1$ units), don't simply set $m = 0$ and $v = 1$. You know how the the mass $m$ is bound to energy and momentum from the energy-momentum relation. So, in $p = mv/\sqrt{1 - v^2}$ replace $m$ by $\sqrt{E^2 - p^2}$ and let $v$ be $1$, in this way you will recover $p = E$, but starting from energy-momentum relation and setting $m = 0$ is much simpler. – giordano Jun 16 '14 at 13:48
• can you plz explain without setting c=1. from the equation $\vec{p}=\frac{m \vec{v}}{\sqrt{1-\frac{v^2}{c^2}}}$ to E =pc please? – zero_field Jun 16 '14 at 15:52

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.
• I understand how to get this form, but $p = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} mv$ is it not valid for photon momentum ? why not? – zero_field Jun 16 '14 at 12:22
• @zero_field: it's not valid because photons have zero mass and $0/0$ doesn't mean anything. However, having the constraint of some particular energy enables you to take the limit in a way that gets the correct result--with the numerator and denominator changing in a consistent way. – Stan Liou Jun 16 '14 at 12:33