# Degrees of freedom in the early Universe

I am reading Dodelson's textbook on cosmology. On page 66 we find equation 3.26: $$\rho = \frac{\pi^2}{30}T^4\biggl[\sum_{i=\text{bosons}}g_i+\frac{7}{8}\sum_{i=\text{fermions}}g_i\biggr]\equiv g_\star \frac{\pi^2}{30}T^4 .\tag{3.26}$$ On the next page, there is the following statement:

By the time decays become important, electrons and positrons have annihlated, so $g_\star$ in Eq. (3.26) is 3.36

I don't understand where 3.36 comes from. If you take the above equation, the bosons are the photons and so they have $g=2$ and the fermions are the 3 generations of neutrinos and their anti-particles and so they have $g=6$ (these numbers are also given in the text). This then gives $g_\star=2+{7\over 8}\times 6=7.25$. Any help in understanding why $g_\star=3.36$ rather than 7.25 would be greatly appreciated.

• Could you please reproduce the relevant sections of the book rather than giving us a (too small) picture, which also contains a lot of redundant information? – Danu Sep 9 '14 at 8:05
• I will try to soon. In the mean time, you can right click on the image and choose open in a new window. That will show a zoomed in version. – Virgo Sep 9 '14 at 8:09
• Ok. I'm bored so I'll make the edit for you ;) – Danu Sep 9 '14 at 8:11
• I don't have the book anywhere nearby but this might be because the neutrinos have a different temperature. The effective degrees of freedom are then $g_{\nu\, eff} = g_{\nu} (T_\nu/T_\gamma)^4$. The number $3.36$ would correspond to $T_\nu/T_\gamma = 0.71$ which is reasonable. – Void Sep 9 '14 at 11:11

$3.36 = 2 + 3[2 × (7/8)(4/11)^{4/3}]$.
This is explained in this paper : bottom of page $15$, top of page $16$ + beginning of the discussion chapter $5$, page $14$