Cosmic neutrino background temperature

Question

In my cosmology lecture course, we derived the temperature of the cosmic neutrino background as

\begin{equation} T_{\nu} = \left(\frac{4}{11}\right) ^ {1/3} T_{\gamma} \,. \end{equation} Since the temperature of the cosmic microwave background today is $T_{\gamma}=2.73K$, the temperature of the cosmic neutrino background today is $T_{\nu} = 1.95K$ according to the above equation. However, my lecturer mentioned that we believe that neutrinos have recently become non-relativistic. But doesn't the above equation assume that both temperatures are inversely proportional to the scale factor, which is true only for ultra-relativistic particles? So once neutrinos become non-relativistic, $T_{\nu}(a)$ changes and the above equation is not right and we cannot say that $T_{\nu}=1.95K$ today. What am I missing?

Answer 1

As the universe expands, the cosmic background neutrinos maintain a fixed relativistic Fermi-Dirac distribution as they cool because the neutrino number is approximately conserved. There is much lower occupation of high energy states and the distribution does not depend on the neutrino mass or how relativistic they are $$F(p,T) = \frac{1}{\exp(pc/kT_\nu) + 1}\ .$$ Both neutrino momentum and the neutrino "temperature" decrease with the scale factor as $(1+z)^{-1}$. i.e $T_\nu(a)$ doesn't change.

The calculation of the current neutrino temperature only assumes that the neutrino phase-space distribution retains the form above. The entropy of a gas governed by this distribution is $\propto g T^3$ and entropy conservation then leads to the $(4/11)^{1/3}$ factor.

Where the distinction between relativistic and non-relativistic neutrinos does come in is if you were to try and estimate an rms speed because it does not directly correspond to temperature as in a Maxwell-Boltzmann gas. Whilst the energy of relativistic particles falls as $a^{-1}$, the energy of non-relativistic particles falls as $a^{-2}$.