The masses of the neutron and mesotron

Abstract

1. The development of relativistic wave mechanics in the preceding paper makes it possible to calculate the mass of the neutron. References, unless otherwise stated, are to the sections of that paper. We have only to express the fact that a proton and electron by emitting a neutrino yield a neutron. We consider the proton and electron initially without interaction, since the interaction energy would in any case have to be recalculated after the emission of the neutrino. They are equivalent to an external and an internal particle of masses M = m 1 + m 2 , μ = m 1 m 2 / ( m 1 + m 2 ). These, like the electron and proton, are specified by complete momentum vectors having the full degree of degeneracy 10. I take it that the emission of a neutrino is a way of saying that one of the particles loses its spin. Accordingly its complete momentum vector, which (leaving aside the six dormant electrical components) consists of an ordinary momentum 4-vector and a spin 6-vector (§ 17), is reduced to a momentum 4-vector, and its degree of degeneracy is changed from 10 to 4. The particle which undergoes the modification is clearly the internal particle, for the neutron in its external relations resembles a hydrogen atom; indeed a spinless external momentum vector would represent a particle capable of existing only in one Lorentz frame and therefore immobile, since the spin 6-vector is necessary for a Lorentz transformation as Dirac’s pioneer investigation demonstrated.