# How do photons have energy if they are massless particles?

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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 > 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. 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. 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) 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 halfway there, the energy of a photon is indeed p*c. However, the equation for momentum is not m*v for objects moving at (or near) the speed of that, that’s just an approximation that we use for low speed objects. For a photon, its momentum is planck’s constant divided by its wavelength p = mv is the nonrelativistic limit (v/c ≈ 0). Light obviously has to be treated relativistically, as light travels at… well… the speed of light.