# Sakharov condition of departure from equilibrium

I have a doubt with the third Sakharov condition which demands a departure from equilibrium. But why is that necessary when both baryon violating interactions and well as CP violation is present in a theory? In general, $CP$ invariance implies \begin{equation}\mathcal{M}(ab\rightarrow cd)=\mathcal{M}(\bar{a}\bar{b}\rightarrow\bar{c}\bar{d})\end{equation} Therefore, the violation of baryon number in process $ab\rightarrow cd$ will be compensated by the equal amount of violation in the conjugated process $\bar{a}\bar{b}\rightarrow\bar{c}\bar{d}$ occurring at the same rate. Therefore, we require a $CP$ violation to prevent this counterbalance. Now it looks like that after a finite amount of time there will be a net baryon excess. So shouldn't this be the end of the story?

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 initial conditions and have both $CP$ and $B$ violations -- and still your asymmetry will be washed away by simple thermodynamics.

In order to achieve a baryon asymmetry, we need $B$ violating processes. This is obvious.

As you state correctly, we also need $C$ and $CP$ violation, such that

$\Gamma(a\rightarrow b) \neq \Gamma(\bar{a}\rightarrow \bar{b}) \ .$

Now for the third condition. It is, like Kostya sais, in fact an additional, or "orthogonal" condition.

Lets consider a random process $X \rightarrow Y$ and its back reaction $Y \rightarrow X$. If the system is in equilibrium, the rates of the two processes are the same

$\Gamma(X \rightarrow Y) = \Gamma(Y \rightarrow X) \ .$

So what ever process you consider, be it $B$, $C$ or $CP$ violating, or nothing of that sort, the equal and opposite reaction will occur at the same rate in thermal equilibrium.