Tag Info

New answers tagged

0

I always consider that condition as kinda "orthogonal" to the first two. Thermal equilibrium means Boltzmann distribution. And we also have $CPT$ requiring $m_a=m_\bar{a}$. And the two straightforwardly lead you back to the $a$ v.s. $\bar{a}$ symmetry. That consideration even leads you to the conclusion that you could even start from $B$-asymmetric ...



Top 50 recent answers are included