(\(T^+\), \(p^+\))
(\(p^+\))
(\(g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\), \(g^+\)) and (\(T^+\), \(p^+\), \(g^+\), \(T^+\))
(\(p^+\), \(g^+\)) and (\(p^+\), \(g^+\), \(T^+\))
(\(g^+\), \(T^+\), \(p^+\), \(g^+\)) and (\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))
the spins of \(p^{+}\) particles do not contribute to the mass of the nuclei, only the presence of \(g^{+}\) particle. In which case, hydrogen isotopes have zero mass, mass of one \(g^{+}\) and the mass of two \(g^{+}\). The relative abundance of these stable isotopes give rise to the decimals in hydrogen mass that can not be factored.
The notion of \(p^{+}\) having mass \(g^{+}\) is the result of having to add the tuple (\(g^+\), \(T^+\), \(p^+\)) to any nucleus ending with a \(p^{+}\) particle in the cyclic permutation set. The tuple must contain a \(g^{+}\) particle. Such an array of stable isotopes with different positions of \(p^{+}\) in the nucleus may also add to the decimal points in experimental isotope mass measurements, if the weak \(g\) field generated by the spinning \(p^{+}\) particles also contribute to mass. What about spins? Spins seem to be associated with \(g^{+}\) particles only.
Unstable nuclei are not any members of the cyclic permutation set. For example,
(\(3g^+\), \(T^+\), \(p^+\))
(\(T^+\), \(p^+\), \(3g^+\)) and (\(T^+\), \(p^+\), \(3g^+\), \(T^+\))
(\(p^+\), \(3g^+\)) and (\(p^+\), \(3g^+\), \(T^+\))
are all Hydrogen-3, \(^3H\) with spin \(\small{\cfrac{1}{2}}^{+}\); the group \(3g^{+}\) spins as one. What about \(^3H\) with \(2^{-}\) spin?
What is \(g^{+}\) and how can it transmute to a charge?
The time axes, \(t_g\) and \(t_c\) of a \(g^{+}\) particle swapped. The result is an electron that leaves the nucleus; \(\beta^-\) decay.
How does such a swap occurs? 'Til next time...
