Authors: D.J. Pons
Why is the neutron stable in the nucleus? Why is the free neutron unstable outside the atom? This paper applies the cordus conjecture to address these questions. The proposed explanation is that in the nucleus the discrete field structures (cordus HED) of the proton and neutron fulfil each other, thereby providing a joint stability. When the neutron is removed from the nucleus, its stability becomes compromised. By comparison the single proton on its own does not need the neutron, so it remains stable. The free neutron is able to maintain a dynamic stability by moving its field structures around. It can do this indefinitely. However it is in a compromised state, and vulnerable to perturbation by external fields. Two initiators are anticipated for decay. One is randomly occurring field fluctuations from the external fabric, and these are proposed for the conventional decay route. The second is impact by another particule. In both cases it is the external fields that cause the decay, by constraining the neutron so that it cannot dynamically adjust. Hence it is trapped in a state that leads to decay at its next frequency cycle. The second path could involve any particule with sufficient energy to disturb the neutron. Also, the impact of a neutrino is specifically identified as a potential initiator of decay. The implications if this is correct, are that the neutron has two separate decay paths, which are mixed together in what we perceive as the beta minus process. The first is determined by the local density of the (spacetime) fabric, and the second by the number of energetic particules and neutrinos encountered. The significance of the two decay paths is that neutron decay rates are predicted to be variable rather than constant. A general set of assumptions are extracted for stability and decay of particules in general.
Comments: 17 Pages.
[v1] 2011-12-01 15:24:55
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