# Cosmology Lecture 09

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• "Frame 5:" $T=10^{9}K$ , $t=3min$ , $kT=0.086MeV$ . As was started in Frame 4, the major player is $\gamma$ , with only small contributions from (p,n,$e^{-}$ ). Therefore:
$g_{*}=2\,\!$ At this point, neutron decay becomes important (${n_{n} \over n_{p}}\sim 0.16$ ):

$p+n\to D+\overbrace {\gamma } ^{2.2MeV}\,\!$ There isn’t much deuterium in the universe because $n_{p},n_{n}$ are small and $n_{\gamma }$ is large, so deuterium gets broken up as soon as it’s made.

Digression: The primary channel for energy generation in the center of stars is the pp chain, whereby H is fused into He. The pp chain generates about 85% of the energy in the sun. This chain goes as follows:

{\begin{aligned}&(1)p+p\to D+e^{+}+\nu _{e}\\&(2)p+D\to ^{3}He+\gamma \\&(3)^{3}He+^{3}He\to ^{4}He+2p\\&(net)4p\to ^{4}He+2e^{+}+2\nu _{e}+energy\\\end{aligned}}\,\! This generates a temperature of about $1.5\cdot 10^{7}K$ in the core of the sun. The early universe, this temperature occurred about 9 days in. Back then, the dominant energy-releasing reaction was:

$p+n\to D+\gamma \,\!$ Why are the primary reactions different for the center of a star and the early universe? Well, in the center of a star, there aren’t so many free neutrons floating around, so the early universe reaction doesn’t work there. Also, the density of the core of a star ($\sim 100{g \over cm^{3}}$ ) is much greater than the density of the early universe ($\sim 10^{-8}{g \over cm^{3}}$ ). Notice that one of the reactions for the sun generates a neutrino, so these reactions involve weak interactions. We already discussed that the universe’s density was out of the realm of weak interactions after about 1 sec. Now back to our regularly scheduled program.

• "Frame 5.5:" $t=3.75min$ . Here T finally gets low enough for deuterium to persist. Nucleosynthesis begins. ${n_{n} \over n_{p}}\sim 0.15$ .
• "Frame 6:" $T=10^{8.5}K$ , $t\sim 30min$ , $kT=.027MeV$ . The major players are still $\gamma$ , with a little more p, $e^{-}$ . Free neutrons are now mostly in $He$ .

After Frame 6, the next important epoch is matter-radiation equality. After that, it’s the last scattering of $e^{-}$ off $\gamma$ . The CMB epoch is when recombination occurs.

### Frame 4 Revisited

This is the epoch of $e^{-},e^{+}$ annihilation. Energy conservation dictates that $S$ (the entropy density) be conserved. Thus, the entropy is transferred into $\gamma$ as $e^{-},e^{+}$ annihilate. Since $S=constant\cdot g_{*}T^{3}$ , we can calculate how much the photon gas was heated up in this epoch:

$\overbrace {e^{+}+e^{-}} ^{g_{*}={11 \over 2}}\to \overbrace {2\gamma } ^{g_{*}=2}\,\!$ ${\begin{matrix}{11 \over 2}T_{before}^{3}=2T_{after}^{3}&\Rightarrow &{T_{after} \over T_{before}}=\left({11 \over 4}\right)^{1 \over 3}\end{matrix}}\,\!$ Since the neutrinos have already decoupled, they are unaffected by this heating. Therefore:

${T_{\nu }=\left({4 \over 11}\right)^{1 \over 3}T_{\gamma }}\,\!$ This is true even today.

### Frame 5.5 Revisited

Recall that this was the epoch of nucleosynthesis. In Frame 5, there was a bottleneck where elements heavier that H could not be made because D kept being destroyed by radiation. In 5.5, that bottleneck has been lifted, and heavier elements are synthesized:

{\begin{aligned}D+D&\to ^{3}He+n\\D+D&\to ^{4}He+\gamma \\n+D&\to ^{3}H+\gamma \\D+^{3}H&\to ^{4}He+n\\^{4}He+^{3}H&\to ^{7}Li+\gamma \\^{4}He+^{3}He&\to ^{7}Be+\gamma \\^{7}Be+e^{-}&\to ^{7}Li+\nu _{e}\\\end{aligned}}\,\! This tree continues ad infinitum. The most abundant element in the universe is $H$ , then $He$ . The abundance of $^{4}He$ can be calculated:

$Y\equiv {mass\ in\ ^{4}He \over mass\ of\ all\ baryons}={4m_{p}\cdot n_{^{4}He} \over m_{p}(n_{p}+n_{n})}\,\!$ We can estimate that $n_{^{4}He}\approx {n_{n} \over 2}$ , since all other n-containing elements are rare. Thus:

${Y={2{n_{n} \over n_{p}} \over 1+{n_{n} \over n_{p}}}}\,\!$ We can use an estimate of ${n_{n} \over n_{p}}\sim {1 \over 7}$ .

### $Y$ dependencies

• $Y$ depends on the baryon-to-photon ratio=$\eta \propto \Omega _{b}h^{2}$ . As $\eta$ increases, $p+n\to D+\gamma$ is favored over photo-dissociation, so D is more stable and the bottleneck breaks earlier, so $n_{n} \over n_{p}$ is higher when $He$ forms. Thus $Y$ would be larger.
• $Y$ depends on the lifetime of neutrons: $\tau _{n}$ .
• $Y$ depends on the # of species of neutrinos, $N_{\nu }$ . This parameter affects the energy density of the universe. A higher $N_{\nu }$ causes the expansion of the universe to increase, since:
$H\propto {\sqrt {\rho }}\propto {\sqrt {g_{*}}}T^{2}\,\!$ This affects how many neutrons have decayed. We can establish a fitting formula for Y:

$Y=0.23+0.025\log _{10}({\eta \over 10^{10}})+0.0075(g_{*}-10.75)+0.014(\tau -10.6min)\,\!$ This shows, for example, that an increase in $N_{\nu }$ of 1 would change $g_{*}$ :

${\begin{matrix}\Delta g_{*}=2\cdot {7 \over 8}={7 \over 4}&\Rightarrow &\Delta Y\approx 0.013\end{matrix}}\,\!$ 