# 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

## 2 Answers

$3.36 = 2 + 3[2 × (7/8)(4/11)^{4/3}]$.

This is explained in this paper : bottom of page $15$, top of page $16$ + beginning of the discussion chapter $5$, page $14$

• Could you please make your answer self-contained? It's a link-only answer right now, and those aren't allowed here. – Danu Sep 9 '14 at 14:34
• Thanks for the answer. I edited it to explain the modification to the formula further. – Virgo Sep 9 '14 at 18:22

A quick and dirty answer: The effective degrees of freedom are compared to those of the photon (a boson). So at high temperature, you are right, that photons and neutrinos add up to 7.25. Normally when some heavy particles disappear, they annihilate equally into the remaining particles. So the relative difference between photons and neutrinos remain the same. However, this is not the case after the neutrinos decouple. Then, when the electrons and positrons annihilate they turn just into photons, but no new neutrinos are created. So while it seems like the neutrinos are getting fewer, it is the photons which increase in number. They increase from 2 to 5.5 (where the extra 3.5 degrees of freedom are from the electrons and positrons), such that the relative amount is 11/4, or reversely, that the number of neutrinos is 4/11 lower.