Astroparticle Physics

10.4 Neutrino Decoupling, Positron Annihilation, and Neutron Decay 219and therefore they still contribute to the effective number ofdegrees of freedom.For T ≫ m e ≈ 0.5 MeV, the reaction e + e − ↔ γγproceeds at the same rate in both directions. As the temperaturedrops below the electron mass, however, the photonsno longer have enough energy to allow γγ → e + e − .Thepositrons as well as all but a small fraction of the initiallypresent electrons annihilate through e + e − → γγ.Theynolonger make a significant contribution to the total energydensity, and therefore they do not contribute towards g ∗ .The fact that the reaction e + e − → γγ produces photonsmeans that the photon temperature decreases less quicklythan it otherwise would. The neutrinos, however, are obliviousto this, and the neutrino temperature continues to scaleas T ν ∼ 1/R. Using thermodynamic arguments based onconservation of entropy, one can show that after positronannihilation the neutrino temperature is lower than that ofthe photons by a factor [17]annihilation and creationneutrino temperature( )T ν 4 1/3= ≈ 0.714 . (10.15)T γ 11In the following T is always assumed to mean the photontemperature.So, to obtain the total energy density, one needs to takeinto account that the neutrinos have a slightly lower temperature,and thus one has to use (9.19),photon temperatureg ∗ = g γ + 7 8 2N Tν4νT 4 = 2 + 7 ( ) 4 4/38 2N ν ≈ 3.36 ,11(10.16)where for the final value N ν = 3 was used.This new value of g ∗ alters the relation between expansionrate and temperature, and therefore also between timeand temperature. Using (10.2) with g ∗ = 3.36 now givestT 2 = 1.32 s MeV 2 . (10.17)In the next section it will be shown that the neutrons, presentat freeze-out, will be able to decay until deuterium productionbegins at around T = 0.085 MeV. From (10.17) thistakes place at a time t ≈ 180 s.To first approximation one can say that the neutronscan decay for around 3 minutes, after which time they arefreeze-out temperature