# Cosmology Lecture 08

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

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

• "Ex:" Weak Interactions:

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

${\displaystyle \Gamma _{weak}\sim n\sigma v\sim G_{F}^{2}T^{5}\,\!}$

Let’s compare this to the Hubble expansion rate of early universe, when ${\displaystyle \Omega \approx 1}$, so curvature is negligible. Therefore:

${\displaystyle H={\sqrt {{8\pi \over 3}G\rho }}\sim {\sqrt {Gg_{*}T^{4}}}\,\!}$

where ${\displaystyle G}$ is Newton’s constant and ${\displaystyle g_{*}}$ is the effective degeneracy. In order for ${\displaystyle \Gamma _{weak}\ll H}$, we need:

${\displaystyle G_{F}^{2}T^{5}\ll {\sqrt {g_{*}G}}T^{2}\,\!}$
${\displaystyle T^{3}\ll {{\sqrt {g_{*}\hbar c}} \over G_{F}^{2}m_{planck}}\,\!}$

We know ${\displaystyle m_{planck}\sim {\sqrt {\hbar c \over G}}\sim 10^{19}GeV}$, so the temperature requirement for decoupling of the weak interaction is:

${\displaystyle \Gamma _{weak}<1MeV\,\!}$

In general we can say that particles decouple after the rest mass stops being much more that ${\displaystyle kT}$. We can compute the threshold temperature for particles based on their rest mass:

${\displaystyle {\begin{matrix}Particles&mc^{2}&T_{thresh}={mc^{2} \over K}\\\tau &1.78GeV&2.1\cdot 10^{13}K\\n,p&.94GeV&1.1\cdot 10^{13}\\\pi &140MeV&1.6\cdot 10^{12}\\\mu &106MeV&1.2\cdot 10^{12}\\e&.511MeV&6\cdot 10^{9}\\\end{matrix}}\,\!}$

### Relating Temperature and Time in the Radiation-Dominated Era

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

${\displaystyle u=\rho c^{2}={\pi ^{2} \over 30}g_{*}{(kT)^{4} \over (\hbar c)^{3}}\,\!}$

We also have shown that in the radiation-dominated era, ${\displaystyle a\propto t^{\frac {1}{2}}}$, so:

${\displaystyle H={{\dot {a}} \over a}={1 \over 2t}\,\!}$

Now since ${\displaystyle H^{2}={8\pi \over 3}G\rho }$:

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

A useful relation is that ${\displaystyle 1sec\sim 10^{10}Kelvin\sim 1MeV}$.

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

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

Note that ${\displaystyle g_{\nu }}$ was computed as [3 species ${\displaystyle \cdot }$ 2 particle/antiparticle pairs with 1 spin state each ${\displaystyle \cdot }$ fermion factor]. We ignored ${\displaystyle p,n}$ because they were not relativistic. Their reactions are:

${\displaystyle n+\nu _{e}\to p+e^{-}\,\!}$
${\displaystyle n+e^{+}\to p+{\bar {\nu }}_{e}\,\!}$
${\displaystyle [n\to p+e^{-}+{\bar {\nu }}_{e}]\,\!}$

The last reaction is negligible because it has timescale ${\displaystyle \sim 15}$ minutes. Remember that we derived the neutron-to-proton ratio:

${\displaystyle {n_{n} \over n_{p}}=e^{-\Delta E \over kT}\,\!}$

where ${\displaystyle \Delta E\approx 1.293MeV}$. Thus:

${\displaystyle {n_{n} \over n_{p}}\approx 0.86\,\!}$

The neutron-to-baryon ratio is:

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