$\beta^{+}$ Decay

If this is $\beta^{+}$ decay,

$p^+$+$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.