Imagine two equivalent (e.g. same frequency) photons colliding with each other head-on. The linear momentum of the system is $0$ because each photon's momentum has the same magnitude but is pointing in opposite directions.

If these two photons collide and form a single photon (and nothing else), then conservation of linear momentum would mean that this photon has $0$ momentum. But a photon is, definitionally, always moving at the speed of light, and its energy is purely from its momentum (in special relativity, $E^2=p^2{c}^2+m^2{c}^4$ with $m=0$). For it to have zero momentum, it would have to also have zero energy which would violate conservation of energy and correspond to their not being anything.

Even for a more glancing collision, we'd either lose energy from the cancelled out parallel portions of the momentum, or we'd need to increase momentum in the perpendicular direction to make up for it which would violate conservation of linear momentum in that direction.

Thus, to satisfy conservation of energy and conservation of linear momentum simultaneously, we either need multiple particles moving at the speed of light afterwards, or the photons need to annihilate and produce a massive particle which can have zero momentum and absorb the energy as its mass (or some mixture of massive and massless particles whose net momentum is zero).

posted almost 2 years ago

CC BY-SA 4.0

2y ago

Derek Elkins‭

331 reputation 0 8 33 5