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So , everywhere I have looked for pair production it is stated that it cannnot happen in vacuum . Most proofs I have seen for that , state that the conservation of energy and momentum cannot be true at the same time without a body with which the photon will interact in the first place . In most of these proofs the momentum of the electron and the positron is considered the same value p . Why that ?

My explanation why pair production cannot happen in vacuum is based in the existence of a COM frame . We consider that a COM exists for every isolated system , so one will exist for the pair production in vacuum . But if that's true the spatial momentum of the photon will be zero , which is impossible . So if we consider that the photon first interacts with another object (for example with a nucleus that initially has zero velocity ) then we can find a COM . Is this explanation ok ?

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  • $\begingroup$ Your explanation is OK, it is a different way of using the conservations laws. The same momentum assumes the com system. $\endgroup$
    – anna v
    Sep 17, 2020 at 17:27

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In the center of mass frame, the momentum of the created electron positron pair (the magnitude of the combined momentum four vector) is 0.

This cannot be satisfied with a single photon.

There has to be either two photons (where the magnitude of the combined momentum four vector is 0), or as you say in your case, a nucleus and a single photon.

Given two photons of sufficient energy to yield at least the rest mass of an electron-positron pair, one finds that QED predicts a non-zero amplitude for the process γγ→e+e− to happen. That is all the theory tells us. No "fluctuation", no "virtual particles", nothing. Just a cold, hard, quantitative prediction of how likely such an event is.

How does gamma-gamma pair production really work?

So in the case as you say, when there is a nucleus nearby (or anything in the environment where the photon can steal momentum from), the nucleus is able to receive a recoil from the photon, and thus satisfy momentum conservation laws. In reality, the photon ceases to exist as photon, its momentum is transferred to the nucleus, and its energy is transformed to create an electron positron pair.

The flaw goes away if the photon can steal momentum from its environment. The inside of an atom has a strong electric field, which is made of (in quantum electrodynamics) "virtual" photons. Your real photon can exchange momentum with the atom by scattering from one of these virtual photons; that's where the pair creation happens.

Why is a nearby nucleus required for Pair Creation?

So your explanation is correct, and a nucleus is one example of how this process can satisfy momentum conservation laws.

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Whatever is the reference frame, one needs the energy of at least $2mc^2$ to create a pair. Moreover, the momentum conservation requires that the momenta of the electron and the positron sum up to zero.

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Any two massive particles always have a center of momentum frame, where the total momentum is zero. So for simplicity, you may as well analyze the pair production in that frame, which implies the two particles have four-momentum $(\sqrt{m_e^2+p^2},\vec{p})$ and $(\sqrt{m_e^2+p^2},-\vec{p})$. Then by conservation of four-momentum, the original photon must have four-momentum $(2\sqrt{m_e^2+p^2},0)$ and thus mass $2\sqrt{m_e^2+p^2}$. Since photons are massless, this is a contradiction.

Your explanation is equivalent. You need to analyze the four-momenta to determine if the energy is actually equivalent for pair production, however. The energy required is always greater than $2m_ec^2$ because some of the energy goes into recoil of the other object.

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