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If a proton is supposed to decay in a positron and gamma ray photons is possible to obtain the opposite process colliding enough energetic photons with a positron and create a proton ?

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  • $\begingroup$ Most (it used to be all, but kaons and D-mesons and stuff...) microscopic processes in physics exhibit time-reversal invariance, but as a practical matter anything involving proton decay has such vanishingly small cross-section that we would no more be able to go after the inverse processes experimentally than we can go after evidence of string theory by a naive search for the structure of the leptons. $\endgroup$ Commented Mar 3, 2014 at 15:18

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I'm guessing you're referring to the original model of proton decay that arose from the first grand unified theories:

Proton decay

(picture from here). The proton decays to a positron and a neutral pion, and the pion then decays to two gamma particles with a half life of $8.4 \times 10^{−17}$ seconds. So to make protons you would have to get two gamma ray photons to produce a pion, then within $8.4 \times 10^{−17}$ seconds that pion would have to react with a positron to create a leptoquark and that in turn would have to decay to the two up quarks. Even if leptoquarks exist, this would be a very, very improbable process.

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Proton Decay is currently a completely hypothetical process, as far as recall there isn't actually any experimental evidence to back up the prediction.

If however we assume that proton decay is possible into photons and a positron. Then yes it would be possible to do the reverse. The problem with this however is how to hit a positron with enough photons simultaneously, if one photon hits the electron before the others it will cause it to Compton scatter away and thus the other photons will miss.

This a purely hypothetical experiment however and it would be hard to prove it true or false. I suppose you can also assume that if the forward process is possible the reverse will be by time reversal.

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  • $\begingroup$ While it is purely hypothetical it is present in all the leading contenders for a next generation theory (and because it pops up, not because people are putting it there). That's one of the reasons for the persistent interest in looking for it: it seems to be necessary. $\endgroup$ Commented Mar 3, 2014 at 15:21

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