# Which 'annihilation' processes dump energy to the cosmological plasma?

I have been considering a generalisation of the cosmological process involved in computing the temperature of the cosmic neutrino background.

It is well-known that (simplistically) this temperature $T_{\nu}$ differs from the temperature of the CMB $T_{\gamma}$ by a factor of $(4/11)^{1/3}$ due to the fact that shortly after neutrino-decoupling at 0.8 MeV, electrons and positrons annihilate and dump energy to the photon plasma which is still at equilibrium, while leaving the temperature of the decoupled neutrinos constant, as they no longer interact with the plasma.

I was wondering what would happen if the neutrinos decoupled before any particles became non-relativistic ($T > 100$ GeV) such that they had to go through all of the 'annihilation' processes as follows: top quarks, Higgs + W + Z, bottom quarks, charm quarks, tau leptons, (QCD phase transition), pions, muons, and electrons. Specifically, I am wondering what happens for the annihilation processes which are not of the form

$$X^+ + X^- \rightarrow \gamma \gamma,$$

such as the annihilation of Higgs, W bosons and Z bosons, and what happens at the QCD phase transition (between tau annihilation and pion annihilation). These processes do change the number of relativistic degrees of freedom in entropy, so if we just used the conservation of entropy to compute the temperature we would treat them the same way as the other processes. However do these processes actually dump energy to the plasma? I can't see why they would.

In general, if the neutrinos had decoupled at a temperature $T_\text{dc}$, the temperature of the neutrinos would, at any later time, be given by: $$T_\nu = \left(\frac{g_{*S}(T)}{g_{*S}(T_\text{dc})}\right)^{1/3}T,$$ where the effective number of relativistic degrees of freedom, $g_{*S}$, is given by $$g_{*S} = \sum_{\text{bosons}}^{\text{relativistic}} g_i \left(\frac{T_i}{T}\right)^3 + \frac{7}{8} \sum_{\text{fermions}}^{\text{relativistic}} g_i \left(\frac{T_i}{T}\right)^3.$$
• Yes, the evolution during the QCD phase transition will be complicated, but as you say, if you want the temperature afterwards you just apply the formula. But I disagree that there are no annihilations in this process, virtually all the protons and antiprotons annihilate to leave a tiny ($10^{-10}$)remnant of protons (because of a tiny initial asymmetry between particles and antiparticles). – Ihle May 28 '16 at 14:59