Hi, I have one question,(adsbygoogle = window.adsbygoogle || []).push({});

The rel. dofs are:

[itex]g_* (T) = \sum_{i=bosons} g_i \Big( \frac{T_i}{T} \Big)^4 + \frac{7}{8} \sum_{j=fermions} g_j \Big( \frac{T_j}{T} \Big)^4 [/itex]

Before the neutrinos decoupling, the relativistic degrees of freedom are:

[itex]g_*=g_\gamma + \frac{7}{8} (g_e + g_\nu)= 10.75 [/itex]

This is clear to me...

Now what are the relativistic degrees of freedom before and after the electron annihilation?

I think that before we have thermal equilibrium between photons and electrons, and thus their temperatures are the same... The neutrinos are still relativistic but their temperature is not the same as the rest of the photons, so:

[itex]g_{*}(T)= g_\gamma + \frac{7}{8} g_e + \frac{7}{8} g_\nu \Big( \frac{T_\nu}{T} \Big)^4 = 5.5 + 5.25 \Big( \frac{T_\nu}{T} \Big)^4[/itex]

However I haven't seen people using this expression... They instead take the case: neutrino decoupling- after electron decoupling and find the result:

[itex]g_{*}(T)= g_\gamma + \frac{7}{8} g_\nu \Big( \frac{T_\nu}{T} \Big)^4 = 2 + 5.25 \Big( \frac{T_\nu}{T} \Big)^4 [/itex]

(Check Equations 14 to 15 : http://www.helsinki.fi/~hkurkisu/cosmology/Cosmo6.pdf ... he may have the entropy but nothing changes between the definitions except for the powers of the temperature)

Any idea?

I think my problem lies in when the neutrinos decouple... they say that they decouple at [itex]T \sim 1 ~MeV[/itex] - isn't that the same decoupling temperature for electrons since the annihilation channel is then suppressed by the photons' energy?

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# Relativistic Degrees of freedom g(T)

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