Consider this,
\(F_T=\cfrac{T_aT_b}{4\pi\tau_o r^2}\)
but \(T\) is energy,
\(T=mc^2\)
so,
\(\cfrac{m_ac^2m_bc^2}{4\pi\tau_o r^2}\)
\(\cfrac{m_am_bc^4}{4\pi\tau_o r^2}\)
but,
\(F_G=-G\cfrac{m_am_b}{r^2}\)
maybe,
\(\cfrac{c^4}{4\pi \tau_o}=-G\)
the negative sign is to indicate that thermal gravity is repulsive,
\(\tau_o=\cfrac{c^4}{4\pi G}\), numerically
There is of course no reason to equate the two forces, the point is that \(\tau_o\) can have a consistent unit.
Then again, since if the universe started as a packet of energy, \(mc^2\) that condenses to particles \(2^n.m_p \quad n=0,1,2,3...\), energy equates to mass and its kinetic energy when splits and coalescence are considered as a whole,
\(F_T=F_G\), gravity equals thermal gravity,
why not?
