Recent content by Guido Diforti

in Einstein's own words: "[...]In every elementary process of absorption or emission a momentum of absolute magnitude ## \frac{h \nu}{c} ## is transferred to the molecule " The momentum is in the direction of the movement of the photon during the absorption and in the opposite direction...

Thank you, you've been very helpful! Sorry about the mistakes, I'm new to special relativity.

ok so one is at rest and has an energy associated: ϒmc2+mc2=2pc

Energy conservation: ϒmc2=2P c cos θ is it correct?

Momentum in the y direction is 0 before and after since the movement is in the x direction : γmvy=0=E/c sin θ + E/c sin(-θ)=0

Thanks for moving my post. Here's my attempt: With the kinetic energy K i was able to calculate the velocity v of the electron: K=ϒmc2-mc2 From the equation above : v= 0,9893c I'm having trouble with the conservation of the linear momentum P, which I guess I must do the conservation in X...

I need help with a problem and I would appreciate if you help me with it... An electron with kinetic energy K=1Mev collides with a positron at rest. In the collision, both particles mutually annihiliate, creating two gamma rays with equal energy, each of them traveling in equal angles θ in the...