I want to prove that pair production (electron-positron) cannot happen in complete vacuum. This is why I obeyed conservation of energy and got equation:
$$h \nu = m_e c^2 \Bigl[ \gamma(v_1) + \gamma(v_{2})\Bigl]$$
I did the same for conservation of momentum and got an equation which is different:
$$h \nu = m_e c \Bigl[ \gamma(v_1) \underbrace{v_{1} \cos \alpha}_{\neq c} + \gamma(v_{2}) \underbrace{v_{2} \cos \beta}_{\neq c} \Bigl]$$
I noticed that parts $v_1 \cos \alpha$ and $v_2 \cos \beta$ will never equal $c$, so I cannot get same equation as above.
QUESTION: Can I state now that pair production cannot happen? What here is the reason I can state this? I mean is it that energy of a photon should be the same in both cases?
Best Answer
The reply by Emilio Pisanty to your other question also pertains here.
But to prove the impossibility of pair production of a photon in vacuum it is not necessary to go into the mathematics of the Lorenz transformations further than you have already done. When one uses valid algebra and from two paths reaches a different answer one has already proven that the hypothesis, in this case possibility of decay in vacuum, is disproved. It is called proof by "reductio ad absurdum" , proof by reduction to contradiction, and is a widely used logic and mathematical tool, since euclidean geometry was established by Euclid.
This means that you have yourself proven that a photon, which has 0 rest mass, cannot decay in vacuum, by finding this contradicting result.
The story is different when other interactions are introduced and energy momentum balance can be sustained. It is also different in field theoretical formulations , Feynman diagrams, where the exchanged particles are off mass shell and called "virtual".