\(T^2=T^2_c+T^2_g\)
there is really no reason why \(T\) cannot be a linear sum, so instead,
\(T=T_c+T_g\)
and
\(T_c=\cfrac{1}{2}qv^2_{rc}\)
\(T_g=\cfrac{1}{2}mv^2_{rg}\)
then,
\(T=\cfrac{1}{2}(qv^2_{rc}+mv^2_{rg})\)
If the above is valid, a matter/antimatter collision along the charge time axis
\(qv_{tc} + q(-v_{tc}) = 0\)
that results in a release of energy,
\(\Delta E=2qv^2_{tc}\) from \(E=mc^2\)
The temperature component about the charge time axis changes by,
\(\Delta T_c=\Delta E=2qv^2_{tc}\)
and in total,
\(\Delta T=\Delta T_c+\Delta T_g\)
\(\Delta T=\Delta T_c\) \(\because \Delta T_g=0\)
where \(q\), and \(v_{tc}\) are constants.
The change in temperature is roughly a constant, given an overall constant collision rate.
All these are hypothetical. Temperature has actually been redefined as a energy term; no longer a potential term. The rotational quantities, \(v_{rc}\) and \(v_{rg}\) are not measurable immediately. It is not surprising that the expression (*) allows \(v_{tc}\), time speed along the charge time axis, to be estimated. By symmetry, since the axes are assigned arbitrarily, \(v_{tc}=v_{tg}\), time speed along the gravitation time axis is the same as time speed along the charge time axis. There no reason for the two time speeds to be different. \(v_{tc}\) can possibly be derived from the plasma temperature vs time curve, obtained at a constant collision/annihilation rate.
This rotational value, \(v_{rc}\), is not equal to \(v_{rg}\); the temperature associated with each time axis can be different. Eventually these two temperatures/rotations/spins might equalized, as already there is a mechanism by which the axes transfer energy (conservation of energy, \(v^2_t+v^2_c=c^2\)).
Please remember all these are hypothetical.
