1
$\begingroup$

This question already has an answer here:

It's a somewhat theoretical question. In special relativity, The energy of a photon is given by $E = pc$. But, my argument is that, since photons have no mass, how can they have a momentum $p$? The energy $E$ turns out to be 0 always. So, why does this equation hold?

$\endgroup$

marked as duplicate by John Rennie, Brandon Enright, DavePhD, Neuneck, Kyle Kanos Jun 2 '14 at 17:03

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.

  • 3
    $\begingroup$ The energy and momentum of a photon depend only on its frequency (ν) or inversely, its wavelength (λ): $E=\hbar\omega=h\nu=\frac{hc}{\lambda}$ $\endgroup$ – Weasel Jun 2 '14 at 14:55
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/2229/2451 and links therein. $\endgroup$ – Qmechanic Jun 2 '14 at 15:15
3
$\begingroup$

Momentum in this case is: $p = h / \lambda$ for a massless particle. The momentum is related to the De Broglie wavelength of the particle with this formula. If you plug it in the equation you have stated you will get back the Energy equation of a massless particle:

$E = hc/\lambda = hf$

$\endgroup$
3
$\begingroup$

The relativistically correct relation between momentum $p$ and velocity $v$ is $$c^2 p = E v$$ This holds for non-relativistic massive particles ($E = m c^2$ and hence $p=m v$) as well as for massless particles like photons ($v = c$ and therefore $p=E/c$).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.