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
It’s mostly because p=mv is a classical equation, and doesn’t work in relativistic regimes. You should be starting with the equation of E=hc/(lambda), where you see that the energy of a photon is inversely proportional to its wavelength lambda, and then can combine it with Einstein’s energy equation to show that h^^2^^ c^^2^^ / (lambda)^^2^^ = (pc)^^2^^ -> p = h/(lambda) which is decidedly not mv.
> because p=mv and mass is still zero, p would also be zero
p=mv is from Newtonian mechanics, which is a good approximation to Relativistic physics, but only when the objects involved are massive and not moving at a significant fraction of lightspeed. E²=(pc)² (which further simplifies to E=pc) is not an indication that photons have no momentum, but an indication that the momentum of a massless photon is directly proportional to the photon’s energy.
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.
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*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.
You’re conflating relativity with Newtonian mechanics, Newtonian Mechanics is a good approximation of how things work but it’s just an approximation. P=mv is from Newtonian mechanics and thus isn’t applicable here.
The correct eqation is E² = m₀²c⁴+p²c², and since m₀ = 0, E=pc.
The energy of a photon is E = hc/λ where h is Planck’s constant, c is the speed of light, and λ is the wavelength.
Therefore pc = hc/λ which becomes p=h/λ (Momentum = Planck’s Constant / wavelength)
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