I was attempting to work out the magnetic moments of the electron and proton, I will state that the relativity field of any logarithmic strain energy is:
Em/Es = (1 - v^2/c^2)^1/2
Where Em is the matter energy, and Es is the energy before logarithmic strain. Therefore the field of the electron and proton is very near to perfect c as a gravity-like strong force field, this c velocity gives the charge a magnetic moment, then the electron magnetic moment is typical:
ue = 1/2 * e * c * re/(2π) * (1 + (Ke^2)/(hc)) = 9.284 781E-24 A*m^2
Where ue is the magnetic moment of the electron e is the elementary charge c is the speed of light, re is the wavelength of the electron. The magnetic moment of the proton is then:
up = 1/2 * e * c * rp/(4π) * ln^2(g^2) * (1 - 1/(4πexp(2)))^2 = 1.410 536E-26 A*m^2
Where up is the magnetic moment of the proton, rp is the wavelength of the proton, ln is the natural logarithm, g is the electromagnetic coupling constant:
g = ((e^2)/(εђc))^1/2
Where ε is the permittivity of frees pace, ђ is the reduced Planck constant.
The proton uses a natural logarithm of the electromagnetic coupling constant similar to what was described for the electric field of the proton. It also uses a strong force at the wavelength drop in energy.
The magnetic moment of the neutron uses an extra negative electron, the electron has half the electromagnetic coupling constant squared. I will leave out the strong force factor for simplicity, but it's similar to the proton:
un = 1/2 * e * c * rn/(4π) * ln^2(g^2) + -1/2 * e * c * rn/(4π) * ln^2(1/2 * g^2) = 9.565E-27 A*m^2
Where un is the magnetic moment of the neutron, rn is the wavelength of the neutron.
The momentum of the particle might relate to the inductive reactance of the electric field as it is strained into matter energy and bound within the wavelength.
I have updated the PDF file in the OP.
Edited by devin553344, 06 July 2020 - 08:58 PM.