# Relativistic momentum and photons

1. Nov 24, 2008

### FrankJ777

I have a question about photons and relativistic momentum. According to my Physics text (Serway & Beichner), the energy of a partle with zero mass, such as a photon, can be related by E$$^{}2$$=p$$^{}2$$c$$^{}2$$+(mc$$^{}2$$)$$^{}2$$. m=0 so the expression becomes E=pc. Since relativistic momentum is p=mu/$$\sqrt{}1(-u2^{}$$/c$$^{}2$$) and m=0 for a photon, how is there any momentum and thus energy?

2. Nov 24, 2008

### JesseM

That definition of momentum only works for things with nonzero rest mass, as you've noticed it becomes undefined (0/0) when you set m=0 and u=c. We can turn to quantum physics for the momentum of a photon--in QM we have E=hf where h is planck's constant and f is the frequency, so you can substitute this value of E into E=pc to find the photon's relativistic momentum.

3. Nov 24, 2008

### Staff: Mentor

Also note that in classical electrodynamics, an electromagnetic wave carries momentum in proportion to its energy, via E = pc, same as you get from the relativistic mass-energy-momentum relationship with m = 0.

4. Nov 24, 2008

### Naty1

I agree with the posts above. A major trick in physics, that usually comes after studying different pehenomena in different situations, is "which formula(s) applies to the situation I'm working on now"....

5. Nov 25, 2008

### bernhard.rothenstein

I would start with the transformation equations for the momentum and the energy of a particle moving with speed u' relative to I' presented as
p=gp'(1+V/u') (1)
E=gE'(1+Vu'/cc) (2)
In the case of a photon (u'=c) (1) and (2) become
p=gp'(1+V/c)
E=gE'(1+V/c)