# Pair production in complete vacuum

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?

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It is due to the fact that the photon has 0 mass whereas the electron positron pair will have a positive invariant mass. A third particle is needed with which the photon can interact so as not to come to a paradox. –  anna v Jan 12 '13 at 14:54
Yes, but could someone than show me how can we mathematically derive WHY pair production in vacuum is posible? –  71GA Jan 13 '13 at 10:40
But it is not possible to have pari production in vacuum. The photon needs to interact with another particle . –  anna v Jan 13 '13 at 12:00
You have already proven it yourself , it is called proof by "reductio ad absurdum" . When you reach conflicting solutions as you have, you have proven that the hypothesis, in this case that a real gamma can go into e+e-, is wrong. So you have proven it cannot. –  anna v Jan 13 '13 at 18:08
mathworld.wolfram.com/ReductioadAbsurdum.html :"A method of proof which proceeds by stating a proposition and then showing that it results in a contradiction, thus demonstrating the proposition to be false" –  anna v Jan 13 '13 at 18:15