According to Einstein E²=(mc²)²+(pc)². In the case of photons m would be equal to zero since photons are massless, so the equation simplifies to E²=(pc)². p represents momentum but because p=mv and mass is still zero, p would also be zero. Therefore E should also be equal to zero, but this doesn’t make sense since light must have energy (eg. solar energy). I hope I’ve explained my thoughts clearly and thanks to anyone who can help me understand.
In: Physics
You’ve kind of got your equations the wrong way around. The energy–momentum relation you have used there is (or can be) derived from a definition of 4-momentum that works for massless particles. In fact, we can use that relationship to show how we would define momentum for a massless particle:
> E^2 = m^2 c^4 + p^2 c^2
> p^2 = (E/c)^2 – (mc)^2
So for m = 0, this becomes:
> p = E/c
And we have a sensible concept of momentum that works for massless particles.
This turns out to be really neat in Special Relativity, as we get 4-momentum (the 4-dimensional extension of the normal 3-momentum):
> **P** = (E/c, **p**)
where **p** is the normal 3-momentum (**p** = m**v**). So conservation of 4-momentum gives us both conservation of 3-momentum *and* conservation of energy. This definition of 4-momentum is where we get the energy-momentum relation (although again, it is worth noting that we get to this definition of 4-momentum by extending 3-momentum, and use it to derive the energy-momentum relation – not the other way around).
So for our photon, we can use the photon energy equation:
> E = hc / λ
to get an expression for its momentum:
> p = E/c = hc / λc = h/λ
and this lines up with our observations.
——–
*tl;dr*: photons have energy. So if the energy-momentum relationship holds, massless particles must have momentum. And we can use that relationship to define their momentum.
Latest Answers