Broken Time Reversal Invariance in the Tsallis, Kaniadakis Cases and Correlations

Newton’s second law is time reversal invariant, but it is well-known that in Newtonian mechanics that one may have time reversal breaking processes such as heat production due to friction. These seem to involve symmetry breaking processes. For example, in the Maxwell-Boltzmann two body scattering scenario, there is a symmetry involved with elastic scattering and the uniform distribution indicates that any energy conserving two body collision has the same probability. This is quantitatively expressed through p(ei)p(ej) = p(ek)p(el) if ei+ej=ek+el. If p(ei)p(ej)=p(ek)p(el) for ei*ej =ek*el ((1)) for e/T>>1, (which holds in the Tsallis and Kaniadakis cases), then one does not have time reversal invariance for elastic scattering. In other words, the symmetry is gone. There is, however, a correlation in that ((1)) exists, even though it does not hold for before and after situations of a 2-body collision. In (1), we tried to show that this correlation is enough to ensure an equilibrium scenario which means that p(ei) does not change in time. As a result, ei*ej which breaks time reversal symmetry for an elastic collision, can restore it for an entire system because correlations ((1)) exist. In a scenario like heat loss due to friction, no such correlation exists anywhere making it seemingly impossible for an equilibrium to result. The basic purpose of this note is to suggest that the approach of finding solutions to p(ei)p(ej)=p(ek)p(el), using the simplest functions in the energy variable of the constraint, i.e. ei from Sum over i ei p(ei)=Eave, introduces a correlation which may be enough to ensure that p(ei,t)=p(ei) (which we loosely take as definition of equilibrium). One does not need to have product probabilities conserved at the elastic collision level (as in the Maxwell-Boltzmann case). It seems that the Tsallis and Kaniadakis cases show that the balance may be at a larger scale due to correlations quantitatively described by ei solutions to p(ei)p(ej)=p(ek)p(el).