# Cosmology Lecture 09

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

At this point, neutron decay becomes important (${\displaystyle {n_{n} \over n_{p}}\sim 0.16}$):

${\displaystyle p+n\to D+\overbrace {\gamma } ^{2.2MeV}\,\!}$

There isn’t much deuterium in the universe because ${\displaystyle n_{p},n_{n}}$ are small and ${\displaystyle n_{\gamma }}$ 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:

{\displaystyle {\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 ${\displaystyle 1.5\cdot 10^{7}K}$ in the core of the sun. The early universe, this temperature occurred about 9 days in. Back then, the dominant energy-releasing reaction was:

${\displaystyle p+n\to D+\gamma \,\!}$

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 (${\displaystyle \sim 100{g \over cm^{3}}}$) is much greater than the density of the early universe (${\displaystyle \sim 10^{-8}{g \over cm^{3}}}$). 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:" ${\displaystyle t=3.75min}$. Here T finally gets low enough for deuterium to persist. Nucleosynthesis begins. ${\displaystyle {n_{n} \over n_{p}}\sim 0.15}$.
• "Frame 6:" ${\displaystyle T=10^{8.5}K}$, ${\displaystyle t\sim 30min}$, ${\displaystyle kT=.027MeV}$. The major players are still ${\displaystyle \gamma }$, with a little more p, ${\displaystyle e^{-}}$. Free neutrons are now mostly in ${\displaystyle He}$.

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

### Frame 4 Revisited

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

${\displaystyle \overbrace {e^{+}+e^{-}} ^{g_{*}={11 \over 2}}\to \overbrace {2\gamma } ^{g_{*}=2}\,\!}$
${\displaystyle {\begin{matrix}{11 \over 2}T_{before}^{3}=2T_{after}^{3}&\Rightarrow &{T_{after} \over T_{before}}=\left({11 \over 4}\right)^{1 \over 3}\end{matrix}}\,\!}$

Since the neutrinos have already decoupled, they are unaffected by this heating. Therefore:

${\displaystyle {T_{\nu }=\left({4 \over 11}\right)^{1 \over 3}T_{\gamma }}\,\!}$

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:

{\displaystyle {\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 ${\displaystyle H}$, then ${\displaystyle He}$. The abundance of ${\displaystyle ^{4}He}$ can be calculated:

${\displaystyle Y\equiv {mass\ in\ ^{4}He \over mass\ of\ all\ baryons}={4m_{p}\cdot n_{^{4}He} \over m_{p}(n_{p}+n_{n})}\,\!}$

We can estimate that ${\displaystyle n_{^{4}He}\approx {n_{n} \over 2}}$, since all other n-containing elements are rare. Thus:

${\displaystyle {Y={2{n_{n} \over n_{p}} \over 1+{n_{n} \over n_{p}}}}\,\!}$

We can use an estimate of ${\displaystyle {n_{n} \over n_{p}}\sim {1 \over 7}}$.

### ${\displaystyle Y}$ dependencies

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

This affects how many neutrons have decayed. We can establish a fitting formula for Y:

${\displaystyle Y=0.23+0.025\log _{10}({\eta \over 10^{10}})+0.0075(g_{*}-10.75)+0.014(\tau -10.6min)\,\!}$

This shows, for example, that an increase in ${\displaystyle N_{\nu }}$ of 1 would change ${\displaystyle g_{*}}$:

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