1. Apr 22, 2013

### CallumC

Ok so the equation E= hf about the energy of a photon, I'm having a problem understanding energies to do with photons.
Since, E=hf
∴ 1/2mv2=hf. But if m=0 how can a photon have energy?

2. Apr 22, 2013

### ZapperZ

Staff Emeritus
hf is NOT the "kinetic energy" of photons. So you cannot equate things that are different.

Zz.

Last edited by a moderator: May 6, 2017
3. Apr 22, 2013

### Staff: Mentor

(1/2)mv^2 doesn't work for energy in relativity. The general relationship between energy, mass and momentum in relativity is

$E^2 = (mc^2)^2 + (pc)^2$

where m is the invariant mass which is sometimes called "rest mass." For photons, m = 0 so E = pc, that is, they can have both energy and momentum even though they don't have ("rest") mass.

4. Apr 22, 2013

### CallumC

But if p=mv then are we not back at the same problem, sorry for my ignorance I just wish to understand it.

5. Apr 22, 2013

### ahaanomegas

The point is that $p=mv$ does not hold for a massless object. If it would, then $v=\infty$ and we have nonsense. I recommend Giancoli's 6th Edition of Physics for a deeper understanding of Photons.

6. Apr 22, 2013

### dextercioby

Callum, p=mv is valid in Newtonian physics only (in the form $\displaystyle{\vec{p}=m\vec{v}}$). You may have seen p=mv with m the so-called <relativistic mass> which is a highly useless and missleading notion, now abandoned even in introductory texts.