Reciprocity XXV, #2 (Autumn, 1996), p. 25.
Considering how close Larson's calculated values are to the observed values for subatomic particles, it seems incongruous that both the muon and electron neutrinos should have such enormous error. In checking into the mass measurement procedure, I found that the observed values for both neutrinos should be correct, and concluded that there may be conceptual problems in Larson's interpretation of mass for these two particles.
The logic Larson uses to determine mass is, "The massless neutron [muon neutrino], the M ½-½-0 combination, has no effective rotation in the third dimension, but no rotation from the natural standpoint is rotation at unit speed from the standpoint of a fixed reference system. This rotational combination therefore has an initial unit of electric rotation, with a potential mass of 0.00057850, in addition to the mass of the two-dimensional basic rotation,…".1
As I understood the convention, a displacement of zero means a scalar value of unity—uniform motion, the natural datum. If "no rotation from the natural standpoint" is "rotation at unit speed" with potential mass, then every location not occupied by matter should exhibit a mass of "e", that of the electron or positron. This is not observed, and I submit that no rotation in any dimension is exactly that, no rotation, and no potential mass. Thus, since the muon neutrino has no rotation in the 3rd dimension, it contributes no mass to the particle.
Secondly, when Larson adapts the ½-½ convention over the 1-0 convention for the description of the massless neutron, he states, "If the addition to the rotational base is a magnetic unit rather than an electric unit,…" and "…half units do not exist, but a unit of two-dimensional rotation obviously occupies both dimensions."2
This makes the massless neutron, or muon neutrino, the two-dimensional version of a positron, having a single, two-dimensional temporal rotation instead of a single, one-dimensional temporal rotation, not necessarily occupying both dimensions, but distributed over both dimensions, and resulting in the appropriate ½-½-0 notation.
Since 12 = 1, the applicable mass is "e", not "p+m." And because this mass is distributed over two dimensions, the potential mass for the muon neutrino is e/2.
The new calculated mass is therefore e/2 times the conversion factor of natural units to unified atomic mass units (nu->u):3
e / 2 * (nu->u)= 0.00057870 / 2 * 0.999706441403 = 0.00028926691 u
Or, approximately 0.26945 MeV. Comparing to the observed value of "less than 0.27 MeV (CL = 90%)," is as close to perfect as can be expected, given the uncertainty of the observed value.
The electron neutrino, `½-½-(1), is the muon neutrino with an additional 1-D spatial (electric) rotation. This gives the particle no net motion, and hence no potential mass. Larson indicates, "But since the electric mass is independent of the basic rotation, and has its own initial unit, the neutrino has the same potential mass as the uncharged electron or positron, 0.00057870."1
I disagree with this statement for the neutrino. It may be true for the "p+m" mass conditions, but here we have "e-e", akin to a stable positron-electron combination due to the additional rotation in time on the positron component, and hence is massless.
But, the electron neutrino does have an observed mass of 5.1 eV. The measurement process deals primarily with charged particles, and I believe this observed mass is the mass due to the interaction of a charge on the neutrino with the charge on the atoms of the detector.
The charged neutrino has a mass of "c", the normal electron charge. The charge of atoms in the detector have a mass of "C", the mass of normal charge. Their interaction will be "C+c" (where "c" is positive, because we are on the same side of the unit boundary).4
Because charge is an effect of a "third region"5, the charge needs to be brought across the unit boundary to measure the mass effect. This is a relation similar to "equivalent space", and results in the effect being the square of the value, "(C+c)2".
The observed electron neutrino mass, due to charge interaction, is:
(C+c)2 * (nu->u) = (0.00004494+0.00002996)? * 0.999706 = 0.00000000560 u
Or, approximately 5.21 eV. The observed value is 5.1 eV, again, extremely close to the calculated value.
The corrected tables from Subatomic Mass Recalculated are:
(Note: (C+c) appears as (C-c) because c is negative.)