Hot Topic, Less Mumble, Jumble...

From the post "Hot Topic Mumble Jumble",

$T^2=T^2_c+T^2_g$

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,

$\Delta T=\Delta T_c+\Delta T_g$

$\Delta T=\Delta T_c$    $\because \Delta T_g=0$

$\Delta T=\Delta E=2qv^2_{tc}$   --- (*)

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.