\(p^+\)+\(P_{g^{+}}\)\(\rightarrow\)\(T^{-}+e^{+}+g^{+}\)
where the collision of photon, \(P_{g^{+}}\) with \(p^{+}\) reverses \(t_c\) and \(t_T\) on \(p^{+}\) changing it into a \(T^{-}\) particle that is detected as the electron neutrino. The photon is completely slowed in space to give an \(g^{+}\) particle and, to provide more energy to the collision, a \(e^{+}\) particle is produced also. On the \(e^{+}\) particle, velocity along \(t_g\) is completely reversed.
The problem with \(\beta^{+}\) decay is that it normally occurs with the emission of two \(g^{+}\) particles also. For example,
(\(T^+\), \(p^+\), \(g^+\), \(T^+\))\(\rightarrow\)(\(2T^{+})+2g^{+}+e^{+}+T^{-}\)
Only when the \(p^{+}\) particle involved are at the innermost end of the nucleus set, eg.
(\(p^+\), \(g^+\), \(T^+\))\(\rightarrow\)(\(2g^{+},\,\,T^{+})+e^{+}+T^{-}\)
does \(\beta^{+}\) decay not emit the two \(g^{+}\) particle and it would seem that the \(p^{+}\) particle has converted into a \(g^{+}\) particle.
What? Radioactive decays do not involved photons? Maybe.
