In the Sakharov conditions for matter-antimatter asymmetry the requirement for the baryon number violation is often interchanged with lepton number violation. In the context of the Sakharov conditions, how is it that baryon number violation can be equivalent to lepton number violation?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ The reason is that one often assumes Baryogenesis via Leptogenesis . Have a look at the answeres here physics.stackexchange.com/questions/158943/… $\endgroup$– jakCommented May 17, 2016 at 11:48
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
In the Standard Model, the baryon and lepton number are accidental global symmetries. However, they are conserved only at the classical level: quantum corrections do not respect them, i.e., they are anomalous.
The interesting thing is that they are violated by exactly the same amount. In terms of currents we write:
$$\partial_\mu J_B^\mu=\partial_\mu J_L^\mu \neq 0$$
From here you see that the current $B-L$ is conserved (anomaly-free) but not $B+L$.
Therefore any violation of baryon number leads in principle to a violation of lepton number by the same amount (to keep $B-L$ conserved), and vice versa. This is mediated by the so-called sphalerons. However, sphaleron transition are extremely suppressed at energies below the electroweak scale so baryogenesis via leptogenesis scenarios must go to higher scales.
