says said:
Do particles not have energy-equivalence when they are in motion? Once the electron and positron collide couldn't we think of them at rest?
You can go to the reference frame where the electron-positron system is at rest. They (individually) will have some momentum ##\vec{p}_-## for electron and ##\vec{p}_+## for positron. The Center-of-Mass frame is given by ##\vec{p}_- = - \vec{p}_+## (or the particles have equal and opposite momenta,so that the total momentum is zero). In that frame you have that the initial 4-momentum is:
(p_+ + p_-)^\mu = \begin{pmatrix} 2E_e \\ \vec{0} \end{pmatrix}
since the energies of positron/electron are equal (they have the same mass and the same momentum magnitude).
What happens for the photon in that frame? Well the conservation of energy (in that frame) would imply that the photon has energy ##2E_e## (so far so good)... what about the momentum? The momentum of the photon will have to be zero.
That is impossible for a real photon, since the real photon should have equal momentum magnitude and energy -so that E^2 - |\vec{p}|^2= m_\gamma^2 =0.
That's why 1 photon alone cannot conserve energy/momentum.
If you have 2 photons you can assign their momenta/energy so that they cancel out ...that's why each photon would have a four momentum p_\mu^{(\gamma1)}= \begin{pmatrix}E_e \\ \vec{p}_\gamma \end{pmatrix} and p_\mu^{(\gamma2)}= \begin{pmatrix}E_e \\ -\vec{p}_\gamma \end{pmatrix} and so p^{\mu~(\gamma1)} + p^{\mu~(\gamma2)}= p_+^\mu + p_-^\mu. So in the center of mass frame of the electron/positron the photons go in opposite directions.
In a similar manner you can have more than 2 photons, however at each photon emission you get one factor of \alpha_{em} (fine-structure constant for electromagnetism), which suppresses the interaction by approximately a factor of 100.
