Baryosynthesis and Antimatter Generation 145Substituting this퐻and the expression in Equation (6.109) into Equation (6.108), the
Universe is out of equilibrium when
퐴푀X≲√
푔∗(푇)
푇^2
푀P
at 푇=푀X, (6.111)where all the constants have been lumped into퐴. Solving for the temperature squared,
we find
푇^2 ≳퐴푀X푀P
√
푔∗(푇)
. (6.112)
At temperature푇X, the effective degrees of freedom푔∗are approximately 100. The
condition in Equation (6.112) then gives a lower limit to the X boson mass,
푀X≳퐴′
푀P
√
푔∗(푇)
=퐴′
1. 2 × 1019 GeV
√
10 675≃퐴′× 1018 GeV, (6.113)where퐴′includes all constants not cited explicitly.
Thus, if the mass푀Xis heavier than퐴′× 1018 GeV, the X bosons are stable at ener-
gies above푀X. Let us assume that this is the case. As the energy drops below푀X,
the X andX bosons start to decay, producing the net baryon number required. The
interactions must be such that the decays really take place out of equilibrium, that is,
the temperature of decoupling should be above푀X. Typically, bosons decouple from
annihilation at about푀X∕20, so it is not trivial to satisfy this requirement.
Let us now see how C and CP violation can be invoked to produce a net BB-
asymmetry in X andX decays. The effect of the discrete operator C calledcharge con-
jugationon a particle state is to reverse all flavours, lepton numbers and the baryon
number. The mirror reflection in three-space is calledparity transformation,andthe
correspondingparity operatoris denoted P. Obviously, every vector풗in a right-handed
coordinate system is transformed by P into its negative in a left-handed coordinate
system, CP transforms left-handed leptons into right-handed antileptons.
We can limit ourselves to the case when the only decay channels are Equa-
tions (6.105) and (6.106), and correspondingly for theX channels. For these chan-
nels we tabulate in Table A.7 the net baryon number change훥퐵and the푖th branching
fractions훤(푋→channel푖)∕훤(푋→all channels)in terms of two unknown parameters
푟and푟.
The baryon number produced in the decay of one pair of X,X vector bosons
weighted by the different branching fractions is then
훥퐵=푟훥퐵 1 +( 1 −푟)훥퐵 2 +푟훥퐵 3 +( 1 −푟)훥퐵 4 =푟−푟. (6.114)If C and CP symmetry are violated,푟and 푟are different, and we obtain the
desired result훥퐵≠0. Similar arguments can be made for the production of a net
lepton–anti-lepton asymmetry, but nothing is yet known about leptonic CP violation.
Suppose that the number density of X andX bosons is푁X. We now want to generate
a net baryon number density
푁B=훥퐵푁X≃훥퐵푁훾