# Why are both C and CP violation necessary for baryogenesis?

Sakharov condition states both C and CP violation is necessary for baryogenesis. Now consider, for example, a theory with B-number violating interaction and C-violation. Therefore, if $p\to e^+\gamma$ is an allowed B-violating process, C-violation would imply that the C-conjugated process $\bar{p}\to e^-\gamma$ would occur at a different rate. Shouldn't therefore, C-violation be sufficient for baryogenesis? Why do we also need CP-violation?

Assume You have only $$C$$-violation. Then it implies that the rate $$\Gamma$$ of hypothetical process $$p^{-} \to e^{-}\gamma$$ won't be equal to the rate of hypothetical process $$p^{+} \to e^{+}\gamma$$, but only for the given helicities $$L/R$$. Say, $$\tag 1 \Gamma\big(p_{L}^{+} \to e^{+}_{R}\gamma_{L}\big) \neq \Gamma(p_{L}^{-} \to e^{-}_{R}\gamma_{L}),$$ and $$\tag 2 \Gamma\big(p_{R}^{+} \to e^{+}_{L}\gamma_{R}\big) \neq \Gamma(p_{R}^{-} \to e^{-}_{L}\gamma_{R})$$ (note that the "photon" $$\gamma$$ has helicities $$\pm 1$$ while the "electron" $$e$$ and the "proton" $$p$$ have helicites $$\pm \frac{1}{2}$$). (Added) This is because the $$C$$-transformation changes the particle on corresponding antiparticle without changing the helicity.

But let's assume that such processes respect $$CP$$-symmetry, (added) under which the left/right particle is changed on right/left antipatrticle. Then there must be $$\tag 3 \Gamma (p^{-}_{L} \to e^{-}_{R}\gamma_{L}) = \Gamma (p^{+}_{R} \to e_{L}^{+}\gamma_{R})$$ Let's add separately the left and the right hand-sides of $$(1)$$ and $$(2)$$ and use $$(3)$$. We obtain $$\Gamma\big(p_{R}^{+} \to e^{+}_{L}\gamma_{R}\big) + \Gamma\big(p_{L}^{+} \to e^{+}_{R}\gamma_{L}\big) = \Gamma(p_{L}^{-} \to e^{-}_{R}\gamma_{L})+ \Gamma(p_{R}^{-} \to e^{-}_{L}\gamma_{R}),$$ and no total baryon-asymmetry will be generated.

Therefore we require the CP-asymmetry.

• -For massive particles L,R refer to chiralities, not helicities.
– SRS
Jan 10, 2017 at 5:48
• How did you get eqn. 1 and eqn. 2 for C-violation? I mean why can I not directly claim that they $\Gamma(p\to e^+\gamma)\neq\Gamma(\bar p\to e^-\gamma)$?
– SRS
Jan 10, 2017 at 5:55
• @SRS : typically at baryogenesis era particles are massless, so there is no problem with calling the helicity left or right. Jan 10, 2017 at 9:47
• @srs : as for Your second question, note that charge conjugation converts a particle in the corresponding antiparticle with the same chirality (helicity). Therefore, if only C-symmetry is violated then You can talk only about the violation of equalities $$\Gamma(p^{+}_{L/R}\to e^{+}_{R/L}+\gamma_{L/R}) = \Gamma (p^{-}_{L/R}\to e^{-}_{R/L}+\gamma_{L/R})$$ separately for each given helicity. Jan 10, 2017 at 9:49