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

This question already has an answer here:

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 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

## 1 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.

• 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
• I have found that form in Griffiths, so why cant i use that for photon? – zero_field Jun 16 '14 at 12:49
• @zero_field: You have just discovered why yourself in the question you asked. Why is is the fact that it is found in Griffiths relevant? Which Griffiths, anyway? (If it's Introduction to Electrodynamics, then he discusses just this issue for photons in section 10.2.3 in the 1981 ed.--sorry, I don't have the newer one on-hand right now.) – Stan Liou Jun 16 '14 at 12:55
• Yes it is Introduction to Electrodynamics, page no 509 and the topic is, relativistic energy and momentum. – zero_field Jun 16 '14 at 13:07