cosmology - Number $g(T)$ of relativistic degrees of freedom as a function of temperature $T$

Let us consider the total number of relativistic degrees of freedom $g(T)$ for particle species in our universe:

$$g(T)=\left(\sum_Bg_B\right)+\frac{7}{8}\left(\sum_Fg_F\right)$$

Where the sums are over the degrees of freedom for bosons ($B$) and and fermions ($F$) which are relativistic when the universe has temperature $T$ (meaning $T$ > their mass energy). For example the photon contributes a $g_{ph}=2$ for the two polarization degrees of freedom it has.

Now, I heard of the following rough estimates for $g(T)$:

When $T_1\geq 1GeV$ we have $g(T_1)\approx 100$.

When $100MeV\geq T_2\geq 1MeV$ we have $g(T_2)\approx 10$.

When $0.1MeV\geq T_3$ we have $g(T_3)\approx 3$.

I am trying to reproduce these estimates by counting all relativistic particles at the specific $T$ values and summing up their degrees of freedom. However, there seem to be contradictions and unclarities here.

For example, the lowest of the three values $g(T_3)\approx3$ is supposedly due to the 2 polarizations of photons and 1 spin degree of freedom of the electron neutrino. However, shouldn't we also count the spin degree of freedom of the electron anti-neutrino? And what about the other two neutrino species? Why include electron neutrino but leave out the others?

Similarly, for $g(T_2)\approx 10$ I would expect to count 2 photon polarizations, 1 spin d.o.f. for neutrinos and anti-neutrinos (6 d.o.f.s in total), 2 spins for electrons and anti-electrons and muons and anti-muons (8 in total), again 2 spins for up, down and strange quark particle anti-particle pairs (12 in total). I am not sure if I missed any particle species here, but we already have $g(T_2)\approx30$ instead of $10$.

Could someone explain to me how to do this counting properly and why some species appear to be missing from consideration even though they should count as relativistic?

Answer

First, note that the equation you use is only valid when all relativistic particles are in thermal equilibrium. The more general equation, which allows for particles with different temperatures, is $$g(T) = \sum_B g_B\left(\frac{T_B}{T}\right)^4 + \frac{7}{8}\sum_F g_F\left(\frac{T_F}{T}\right)^4$$ where $T$ is the photon temperature and $T_B$, $T_F$ are the temperatures of each boson and fermion.

The degrees of freedom for all Standard Model particles are listed in the table below (source: http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf):

At temperatures $T\sim 200\;\text{GeV}$, all particles are present, relativistic, and in thermal equilibrium, so we find $$g(T) = 28 + \frac{7}{8}\cdot 90 = 106.75$$ When $T\sim 1\;\text{GeV}$, the temperature has dropped below the rest energy of the $t$, $b$, $c$, $\tau$, $W^+$, $W^-$, $Z^0$, and $H^0$ particles, therefore these are no longer relativistic (and will have annihilated) and we have to take them out of the equation. We are left with $$g(T) = 18 + \frac{7}{8}\cdot 50 = 61.75$$ When $T$ drops below $100\;\text{MeV}$, the remaining quarks and gluons are locked up in non-relativistic hadrons, and the muons have annihilated. All that's left are photons, electrons, positrons, neutrinos and anti-neutrinos, so that $$g(T) = 2 + \frac{7}{8}\cdot 10 = 10.75$$ So far, all relativistic particles were in thermal equilibrium. However, as the temperature drops to $1\;\text{MeV}$, the neutrinos decouple and move freely, which means their temperature will start to diverge from the photon temperature. At $T < 500\;\text{keV}$, the electrons and positrons are no longer relativistic, so only the photons and neutrinos remain, and $$g(T) = 2 + \frac{7}{8}\cdot 6\left(\frac{T_\nu}{T}\right)^4,$$ where $T_\nu$ is the neutrino temperature. Calculating this requires a bit of work.

The Second Law of Thermodynamics implies that the entropy density $s(T)$ is given by $$s(T) = \frac{\rho(T) + P(T)}{T},$$ where $\rho$ is the energy density and $P$ the pressure. Using the Fermi-Dirac and Bose-Einstein distributions, one finds that for relativistic particles $$\rho(T) = \begin{cases} \dfrac{g_B}{2}a_B\, T^4&\text{bosons}\\ \dfrac{7g_F}{16}a_B\, T^4&\text{fermions} \end{cases}$$ and $P = \rho/3$, so that $s(T) = 4\rho(T)/3T$. Let us now consider the entropy density of the photons and the electrons and positrons at high temperatures, when they are still relativistic: $$s(T_\text{high}) = \frac{2}{3}a_B\,T_\text{high}^3\left(2 + \frac{7}{8}\cdot 4\right) = \frac{4}{3}a_B\,T_\text{high}^3\left(\frac{11}{4}\right).$$ At low temperatures, the electrons and positrons become non-relativistic, most annihilate and the remaining particles have negligible contribution to the entropy, therefore $$s(T_\text{low}) = \frac{4}{3}a_B\,T_\text{low}^3\,.$$ However, thermal equilibrium implies that the entropy in a comoving volume remains constant: $$s(T)a^3 = \text{constant}.$$ Also, the temperature of the neutrinos drops off as $T_\nu \sim 1/a$ after they decouple. Combining these results, we find $$\left(\frac{T_\text{low}}{T_{\nu,\text{low}}}\right)^3 = \frac{11}{4} \left(\frac{T_\text{high}}{T_{\nu,\text{high}}}\right)^3.$$ At high temperatures, the neutrinos are still in thermal equilibrium with the photons, i.e. $T_{\nu,\text{high}}=T_\text{high}\,$, so finally we obtain $$T_\nu = \left(\frac{4}{11}\right)^{1/3}T$$ at low temperatures. Therefore, $$g(T) = 2 + \frac{7}{8}\cdot 6\left(\frac{4}{11}\right)^{4/3} = 3.36.$$ A more detailed treatment is given in the same link from where I took the table.