Mathematical Physics

   

Neutronization Equations

Authors: Antonio Puccini

With the neutronization protons(Ps) and electrons(es) fuse together to form neutrons(Ns) and release electronic neutrinos(e): e + P  N + e(1). This is possible because the electron in the neutronization equation is equipped with a very high energy, just provided by a collapsing star, or by a neutron star. In those extreme conditions electrons become relativistic, since they acquire a 200MeV energy, so as to fill the conspicuous energy gap between N and P. However, Eq.(1) appears incomplete because it does not explain the ex abrupto appearance of the e. One may wonder: how was it produced? As it is known matter and antimatter particles are always produced as a couple. Where is the antiparticle of e, i.e. the ῡe, not represented in Eq.(1)? In our opinion, Eq.(1) implies some intermediate steps not represented. A phenomenon associated with neutronization is photoannihilation, characterized by the materialization of the electromagnetic radiation(γ), with consequent production of pairs, such as: γ  ῡe + e(2). If we enter Eq.(2) in Eq.(1), we have: e+ P  e + P + γ  e + P + ῡe + e  N + e(3), that is:e + P + ῡe + e ↔ N + e(4), i.e.:e + P + ῡe ↔ N(5). From Eq.(5) it emerges that to N corresponds a compound of 3 particles, i.e. a multiplet:[e, P, ῡe]. This is in agreement with Spin Statistics, as well as with Quantum Mechanics, since the relativistic electron has an energy>140MeV. Furthermore, let's try to read Eq.(5) in reverse: N  e + P + ῡe(6). It is surprising: Eq.(6) shows exactly the decay products of N, corresponding precisely to the famous Fermi equation describing the N decay, providing a counter-test to Eq.(5)

Comments: 14 Pages.

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[v1] 2018-07-30 02:19:14

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