Mathematically, the resulting wave after the time axes swap is equivalent to the wave before the swap. The time axes variables are arbitrary, but we know that, either,
\(t_c=i.t_T\) ot \(t_T=i.t_c\)
that the time axes are orthogonal.
How much energy is expended when we multiply a time axis \(t\) by \(i\)? For the swap process as a whole, nothing, since mathematically the waves are equivalent, no net energy input is required.
Which brings us to the other colliding particle,
On collision, the photon \(P_{g^{+}}\) stops along the space dimension \(x\) but has speed \(v=c\) along \(t_c\). It then splits into two along the \(t_g\) axis. Part of it \(g^{+}\) travels along the positive \(t_g\), \(v=c\) and the other part travels along \(t_g\) negative; ie. \(v=-c\). The momentum of the photon is either split equally between the particle and the antiparticle or, only the anti-particle has \(v=c\) in space when \(g^{+}\) is captured (\(p^{+}\) is originally in orbit).
Why should one particle be sent back in time? This particle is otherwise an electron, the negative particle of \(p^{+}\).
One particle is sent back in time because \(P_{g^{+}}\) is stationary along \(t_c\) before the collision and \(p^{+}\) is at light speed along \(t_c\). After the collision, part of \(P_{g^{+}}\) is propelled forward, the other part is repelled backwards along \(t_c\). Then the time axes swapped, \(t_g\leftrightarrow t_c\). \(t_g\) now has two velocity vectors positive and negative that sum to zero, and \(t_c\) has a vector component \(v=c\).
On the \(p^{+}\) particle \(t_c\leftrightarrow t_T\) and on the photon, \(P_{g^{+}}\), \(t_c\leftrightarrow t_g\).
In the case of \(\beta^{-}\) decay, there is only one time axis swap,
and the photon stopped along the space dimension after the collision,
In this case both colliding particles have speed \(v=c\) along \(t_c\) when they collided. The photon \(P_{g^{-}}\) has \(v=0\) on \(t_T\), in the particle \(g^{+}\), \(t_T\) is the oscillatory component of the wave. In the previous case, the photon \(P_{g^{+}}\) is stationary along \(t_c\), \(v=0\) and the particle \(p^{+}\) is at light speed, \(v=c\) along \(t_c\). Momentum along \(t_c\) is split into two, a negative part and a positive part, they sum to zero.
In both cases, the oscillatory components remain intact. After the collision, except for the photon/particle that slowed, all non oscillatory time axes swapped. If one of the colliding, non oscillatory time axes has zero velocity, momentum along that time axes splits into a negative and positive part and an anti-particle is created.
In summary...
In both cases, the particles involved in the decays in the nucleus is first identified. And appropriate photon that provides for the resultant particles after the decay is made to collide with the nucleus particle. The nucleus particle is transmuted by swapping non oscillatory time axes. If between the photon and the nucleus particle, any of the non-oscillatory time axes has zero velocity, a anti-particle is produced. This anti-particle is identified after the time axis swap.
Is this a scheme for general radioactive decay? Maybe.
