If Thermal Gravity is anti-gravity, and gravity is along the positive time axis, then thermal gravity is on the negative time axis.
And we experience anti-matter as heat, or temperature; just as a proton and an electron feel each other hot when they collide.
Temperature can still be spins on the time axis. We are experiencing the effects of anti-matter not the anti-matter itself.
Consider the change in rotational energy,
\(\Delta E_{r}=\cfrac{1}{2}mv^2_f-\cfrac{1}{2}mv^2\)
Lost in KE along the time axis equals change in rotational energy for one particle,
\(mc^2=\Delta E_{\omega}=\cfrac{1}{2}m(v^2_f-v^2)\)
\(c^2=\cfrac{1}{2}(v^2_f-v^2)\)
\(v^2_f=2c^2+v^2\), \(v^2_f=(\sqrt{2}c)^2+v^2\)
If we expand the Big Band to include the time axis as well, this is consistent with the fact that the first kinetic energy lost is the result of collision of particles with speed \(\sqrt{2}c\), from the post "If The Universe Is A Mochi". It is not surprising, for both cases started with \(E=m^2\). So, the result of the very first temperate increment is,
\(v^2_i=v^2_f=(\sqrt{2}c)^2+0\)
\(v_i=\sqrt{2}c\)
If temperature is defined as energy, and
\(m = 2m_p\)
where \(m_p\) is the particle when the universe reaches maximum entropy, and \(2m_p\) is after the first coalescence, after which we have masses at zero velocity, common masses we interact with day to day. The rotation KE for a mass \(2m_p\) about the time axis,
\(\cfrac{1}{2}.2m_p.v_i^2=2m_pc^2\)
This is consistent with an matter/anti-matter collision of 2 particles of mass \(m_p\) on the time axis. So, the first temperature increment is given by,
\(\Delta T_1=2m_pc^2=2.\cfrac{1}{2}.2m_pc^2\)
And each particle has a rotational kinetic energy of , temperature of,
\(T_i=KE_r=\cfrac{1}{2}.2m_pc^2\)
this is the smallest temperature increment possible.
What? Point mass has not rotational concerns? That is the point, point mass does rotate and it its temperature. Intuitively, when the distance from a rotational axis collapses, \(r\rightarrow0\) (from the \(I=r^2m\) fame), space collapses and that leaves the particle on the time axis. Rotational energy is still defined but rotational moment is at this moment, not.
Matter/anti-matter interaction, is a head on collision on the time axis, the energy released is the total loss in translational kinetic energy on the time axis,
\(E=mc^2\)
where \(m\) is the total mass \(m=m_m+m_{anti-m}=2m_m\),
\(E\) is the gain in rotational energy about the time axis of the two particles, now stationary on the time axis, and so, disappeared from our existence as we move forward in time. We experience this rotational KE as temperature. This suggests that the decay as a result of matter/antimatter interaction must be slow. The remaining mass of the interaction over time, carries the temperature with it and we experience that as plasma. Mass that interacted has time speed zero, falls back in time and disappear from our time. It is as if mass has been converted to energy. If the interaction is instantaneous then both mass and temperature would have disappeared instantaneously.
Anti-matter then, is mass travelling in the negative time direction. A proton is an anti-matter. We are in the same direction as an electron on the time axis. An charged anti matter can be captured by a very fast rotating opposite charge and be brought into our reality, just as electrons in orbits around a proton nucleus.
What's the energy need for containment? Consider the case of a hydrogen atom...
