# Cosmology Lecture 08

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### Thermal Equilibrium vs. Decoupling (“Freeze-Out”)

The rule of thumb here is to compare the interaction rate ($\Gamma$ ) of the particle we are interested in to the expansion rate of the universe. We’ll examine two extremes: $\Gamma \gg H\Rightarrow TE$ , and $\Gamma \ll H\Rightarrow decoupled$ .

• "Ex:" Weak Interactions:

The cross-section for weak interaction goes as $\sigma \sim G_{F}^{2}T^{2}$ , where $G_{F}^{2}$ is the Fermi constant, ${G_{F}m_{o}^{2} \over (\hbar c)^{2}}\sim 10^{-5}$ . Thus, for $v\sim c$ (recall that $n\propto T^{3}$ in the relativistic limit):

$\Gamma _{weak}\sim n\sigma v\sim G_{F}^{2}T^{5}\,\!$ Let’s compare this to the Hubble expansion rate of early universe, when $\Omega \approx 1$ , so curvature is negligible. Therefore:

$H={\sqrt {{8\pi \over 3}G\rho }}\sim {\sqrt {Gg_{*}T^{4}}}\,\!$ where $G$ is Newton’s constant and $g_{*}$ is the effective degeneracy. In order for $\Gamma _{weak}\ll H$ , we need:

$G_{F}^{2}T^{5}\ll {\sqrt {g_{*}G}}T^{2}\,\!$ $T^{3}\ll {{\sqrt {g_{*}\hbar c}} \over G_{F}^{2}m_{planck}}\,\!$ We know $m_{planck}\sim {\sqrt {\hbar c \over G}}\sim 10^{19}GeV$ , so the temperature requirement for decoupling of the weak interaction is:

$\Gamma _{weak}<1MeV\,\!$ In general we can say that particles decouple after the rest mass stops being much more that $kT$ . We can compute the threshold temperature for particles based on their rest mass:

${\begin{matrix}Particles&mc^{2}&T_{thresh}={mc^{2} \over K}\\\tau &1.78GeV&2.1\cdot 10^{13}K\\n,p&.94GeV&1.1\cdot 10^{13}\\\pi &140MeV&1.6\cdot 10^{12}\\\mu &106MeV&1.2\cdot 10^{12}\\e&.511MeV&6\cdot 10^{9}\\\end{matrix}}\,\!$ ### Relating Temperature and Time in the Radiation-Dominated Era

Recall that the energy density of relativistic bosons is given by:

$u=\rho c^{2}={\pi ^{2} \over 30}g_{*}{(kT)^{4} \over (\hbar c)^{3}}\,\!$ We also have shown that in the radiation-dominated era, $a\propto t^{\frac {1}{2}}$ , so:

$H={{\dot {a}} \over a}={1 \over 2t}\,\!$ Now since $H^{2}={8\pi \over 3}G\rho$ :

$\left({1 \over 2t}\right)^{2}={8\pi \over 3}G{\pi ^{2} \over 30}{g_{*}(kT)^{4} \over \hbar ^{3}c^{5}}\,\!$ ${kT=\left({45\hbar ^{3}c^{5} \over 16\pi ^{3}g_{*}G}\right)^{1 \over 4}t^{-{\frac {1}{2}}}={0.86MeV \over {\sqrt {t(sec)}}}\left({10{3 \over 4} \over g_{*}}\right)^{1 \over 4}}\,\!$ ${T\approx {10^{10}Kelvin \over {\sqrt {t(sec)}}}\left({10.75 \over g_{*}}\right)^{1 \over 4}}\,\!$ A useful relation is that $1sec\sim 10^{10}Kelvin\sim 1MeV$ .

### The First 30 Minutes (in 6 frames)

• "Frame 1:" $T=10^{11}Kelvin$ , $t=0.01sec$ , $kT=8.6MeV$ , $z\sim 1+z={1 \over a}={T \over T_{0}}\sim 3\cdot 10^{10}$ . To put things in perspective, $m_{e} . The major players at this point in time are photons ($\gamma$ ), electrons and positrons ($e^{-},e^{+}$ ), neutrinos ($\nu _{i}{\bar {\nu }}_{i}$ ), and protons and neutrons ($p,n$ ). For $p,n$ , we’re assuming baryon asymmetry has occurred, and we should note that they only show up in small numbers. $\gamma ,e,\nu$ particles are all ultra-relativistic. Let’s figure out our $g_{*}$ :
$g_{*}\ composed\ of\ {\begin{cases}\gamma &g_{\gamma }=2\\\nu {\bar {\nu }}&g_{\nu }=3\cdot 2\cdot {7 \over 8}={21 \over 4}\\e^{+}e^{-}&g_{ee}2\cdot 2\cdot {7 \over 8}={7 \over 2}\\\end{cases}}\,\!$ $g_{*}=10{3 \over 4}\,\!$ Note that $g_{\nu }$ was computed as [3 species $\cdot$ 2 particle/antiparticle pairs with 1 spin state each $\cdot$ fermion factor]. We ignored $p,n$ because they were not relativistic. Their reactions are:

$n+\nu _{e}\to p+e^{-}\,\!$ $n+e^{+}\to p+{\bar {\nu }}_{e}\,\!$ $[n\to p+e^{-}+{\bar {\nu }}_{e}]\,\!$ The last reaction is negligible because it has timescale $\sim 15$ minutes. Remember that we derived the neutron-to-proton ratio:

${n_{n} \over n_{p}}=e^{-\Delta E \over kT}\,\!$ where $\Delta E\approx 1.293MeV$ . Thus:

${n_{n} \over n_{p}}\approx 0.86\,\!$ The neutron-to-baryon ratio is:

${n_{n} \over n_{B}}={n_{n} \over n_{p}+n_{n}}={0.86 \over 0.86+1}=0.46\,\!$ • "Frame 2:" $T=10^{10.5}Kelvin$ ,$t=0.1sec$ , $kT=2.72MeV$ , $\rho$ drops by a factor of 100. Our major player are the same, so:
${\begin{matrix}{n_{n} \over n_{p}}\approx 0.62,&{n_{n} \over n_{B}}\approx 0.38\end{matrix}}\,\!$ • "Frame 3:" $T=10^{10}Kelvin$ , $t=1sec$ , $kT=0.86MeV$ . Now the weak interaction rate falls below the Hubble time ($<1MeV$ ). Therefore, neutrinos are decoupling from the thermal bath. Before this decoupling occurred, recall that $\gamma ,e^{+},e^{-},\nu ,{\bar {\nu }}$ were all in thermal equilibrium, $g_{*}=10{3 \over 4}$ . After decoupling, $\gamma ,e^{+},e^{-}$ are in thermal equilibrium, so $g_{*}=2+{7 \over 2}={11 \over 2}$ . For the neutrinos, their Fermi-Dirac distribution is “frozen” in place:
$dn_{\nu }=g_{\nu }{d^{3}p \over h^{3}}{1 \over e^{pc \over kT}+1}\,\!$ • "Frame 4:" $T=10^{9.5}Kelvin$ , $t=14s$ , $kT=.272MeV$ . Because $T<0.511MeV$ , it’s hard to make new $e^{+},e^{-}$ . That is, $e^{+}e^{-}\to 2\gamma$ is a favored interaction over the reverse, so $e^{+}e^{-}$ start to disappear. As they annihilate, entropy is transferred to photons, so the entropy density $S=const\cdot g_{*}\cdot T^{3}$ is conserved (see Kolb and Turner, 66).
${\begin{matrix}{n_{n} \over n_{p}}=0.2,&g_{*}=2\end{matrix}}\,\!$ 