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?


1 Answer 1


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.

  • $\begingroup$ -For massive particles L,R refer to chiralities, not helicities. $\endgroup$
    – SRS
    Jan 10, 2017 at 5:48
  • $\begingroup$ How did you get eqn. 1 and eqn. 2 for C-violation? I mean why can I not directly claim that they $\Gamma(p\to e^+\gamma)\neq\Gamma(\bar p\to e^-\gamma)$? $\endgroup$
    – SRS
    Jan 10, 2017 at 5:55
  • 3
    $\begingroup$ @SRS : typically at baryogenesis era particles are massless, so there is no problem with calling the helicity left or right. $\endgroup$
    – Name YYY
    Jan 10, 2017 at 9:47
  • $\begingroup$ @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. $\endgroup$
    – Name YYY
    Jan 10, 2017 at 9:49

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