Why deuterium waits for the universe to cool

This chapter follows When neutrons and protons stop turning into each other, which supplies the neutron supply and the relation $t\propto T^{-2}$. You need exponentials, logarithms and number densities. By the end you should be able to compute when deuterium can survive for a given baryon density, explain why helium depends only weakly on that density, and explain why deuterium depends on it strongly.

Photons per baryon

Write $n_b$ for the number density of baryons (protons and neutrons, free or bound) and $n_\gamma$ for that of photons. Their ratio is the baryon-to-photon ratio

Both densities fall as the cube of the scale factor once electrons and positrons have annihilated, so $\eta$ stays fixed through the minutes that matter here. The cosmic microwave background gives $\eta_{10}=6.12\pm0.04$ when BBN is used to fix the helium fraction in the CMB fit, and $6.104\pm0.058$ without that step (2025 Review of Particle Physics, Eq. 24.8 and the text after it). This chapter uses $6.12$: about $1.63\times10^9$ photons per baryon.

Blackbody photons have number density

Worked example: how dense is the universe at the bottleneck?

At $k_BT=0.07\,\mathrm{MeV}$, with $\hbar c=1.973\times10^{-11}\,\mathrm{MeV\,cm}$, the photon density is $n_\gamma=1.09\times10^{28}\,\mathrm{cm^{-3}}$. Multiplying by $\eta$ gives $n_b=6.7\times10^{18}\,\mathrm{cm^{-3}}$, or a baryon mass density of about $1.1\times10^{-5}\,\mathrm{g\,cm^{-3}}$. That is roughly a hundredth of the density of air at sea level, at a temperature of 800 million kelvin. Nuclear reactions in the first minutes run in a thin, very hot gas.

Balance between capture and break-up

Deuterons form in $p+n\to d+\gamma$ and are destroyed by the reverse reaction $\gamma+d\to p+n$. While both are fast, the three species share one temperature and their chemical potentials obey $\mu_d=\mu_n+\mu_p$. The non-relativistic number density of each species then gives the Saha relation

Here $g_d=3$ and $g_n=g_p=2$ count spin states, and $B_d=(m_n+m_p-m_d)c^2$ is the deuteron binding energy. From the 2020 atomic mass evaluation (AME2020 mass table) the mass excesses of the neutron, hydrogen-1 and hydrogen-2 give $B_d=2.2246\,\mathrm{MeV}$.

Abundances per baryon, $Y_i=n_i/n_b$, are more useful than densities because they do not change with the expansion. Use $m_d\simeq2m_N$ with $m_Nc^2=939\,\mathrm{MeV}$, $m_n\simeq m_p\simeq m_N$, and $n_b=\eta n_\gamma$:

Two factors fight. The small numbers $\eta\approx6\times10^{-10}$ and $(k_BT/m_Nc^2)^{3/2}\approx6\times10^{-7}$ express how rare baryons are among photons. The Boltzmann factor $\exp(B_d/k_BT)$ grows as the universe cools. Deuterium becomes common only when the exponential wins.

Worked example: when does the ratio reach one?

Set the right-hand side to one for $\eta_{10}=6.12$ and solve by bisection. The result is $k_BT=0.0665\,\mathrm{MeV}$, or $T_9=0.772$. Check it: $(0.0665/939)^{3/2}=5.96\times10^{-7}$ and $\exp(2.2246/0.0665)=3.37\times10^{14}$, so

At this point $B_d/k_BT=33.5$. The deuteron survives only once the thermal energy is about thirty times below its binding energy. Changing $\eta$ hardly moves this point, because $\eta$ sits outside the exponential:

A factor of 7.5 in baryon density shifts the temperature by 6 percent. Since $t\propto T^{-2}$, the time shifts by about 12 percent.

How much deuterium is enough

Helium does not wait for $Y_d/(Y_nY_p)=1$. Once a few deuterons per thousand baryons exist, the fast reactions $d+d$, $d+p$, $t+d$ and ${}^3\mathrm{He}+d$ pass them on to helium-4 almost as quickly as they form. The tutorial network shows this directly. Take the moment its helium is half formed and read off the deuterium abundance:

The network temperature and the equilibrium temperature for the same deuterium ratio agree within 1 percent at the two higher densities. Deuterium is tracking its equilibrium value while helium forms. At $\eta_{10}=2$ the network lags by about 2.5 percent: in thinner matter, neutron capture is slower and deuterium falls behind equilibrium.

The table also explains why helium depends on the baryon density at all. Denser matter opens the bottleneck earlier, partly because the equilibrium shifts and partly because a smaller deuterium ratio is enough to feed faster reactions. Fewer neutrons have decayed by then. The network gives $Y_p=0.2300$, $0.2424$ and $0.2504$ for the three densities: a change of only 9 percent in helium for a factor of 7.5 in density. These values come from the method guide's saved runs; see Recompute a small network from a frozen rate grid and propagate rate errors.

Why denser matter leaves less deuterium

After the bottleneck opens, deuterium is both made and destroyed. Neutron capture makes it; $d(p,\gamma){}^3\mathrm{He}$, $d(d,n){}^3\mathrm{He}$ and $d(d,p)t$ destroy it, and the products are burned on to helium-4. The rate at which one deuteron is destroyed is proportional to the density of its partners, so to $\eta$. Destruction stops when this rate falls below the expansion rate. In denser matter the burning goes on longer, and the deuterium that remains when it stops is smaller.

The network's final deuterium-to-hydrogen ratios are

From the lowest to the highest density, D/H falls by a factor of 32. Near our universe it scales as $\eta^{-1.6}$; the next chapter defines this exponent and compares it with published calculations. That steep dependence is why deuterium is the baryon density meter. Helium-3 falls more slowly, because it is both a product of deuterium burning and a fuel for helium-4.

Lab task: watch the bottleneck move

1. Predict. Before touching the controls, decide which preset will form its helium earlier, Sparse baryons keep deuterium ($\eta_{10}=2$) or Dense baryons burn deuterium ($\eta_{10}=15$), and which will end with more deuterium.
2. Change. Select each preset in turn. Open the advanced group and record the time when half the helium has formed, together with $Y_p$ and $\mathrm{D/H}$.
3. Compare. Compare your readings with Table 3.2 and Table 3.3. The Lab's time step is coarser than the saved runs, so expect small differences in the last digit.
4. Explain. Use the equilibrium relation to explain the change in timing, and the burning argument to explain the change in deuterium. Then explain why the two changes move helium only a little but move deuterium a lot.

Exercises

1. A student argues that at $k_BT=0.2\,\mathrm{MeV}$ the photons are already ten times less energetic than the binding energy, so deuterium should be stable. Evaluate $Y_d/(Y_nY_p)$ at that temperature for $\eta_{10}=6.12$ and answer the student.
2. Differentiate the logarithm of the Saha ratio with respect to $k_BT$ to estimate how much the temperature at which the ratio equals one shifts when $\eta$ grows by a factor of $2.45$. Compare with Table 3.1.
3. Use $t\propto T^{-2}$ and your answer to exercise 2 to estimate the bottleneck time for $\eta_{10}=15$, starting from 247 seconds. The network gives 219 seconds. Explain the gap.
4. Transfer. Suppose the deuteron binding energy were 10 percent larger while everything else, including the neutron supply at the end of the first chapter, stayed the same. Estimate the bottleneck temperature and time, and the resulting change in helium. Some proposed solutions to the lithium problem involve a varying deuteron binding energy; see Reading the BBN evidence on extra radiation and the lithium problem.

What the equilibrium picture leaves out

The Saha relation knows nothing about reaction speed. It says where deuterium would sit if capture and break-up were infinitely fast, and Table 3.2 shows that this is a good guide only while neutron capture is fast compared with the expansion. The final deuterium abundance is set by burning rates, which require measured cross sections. The tutorial uses the fits of Serpico and collaborators, and the next chapter, How baryon density and extra radiation move the light elements, shows how their uncertainties propagate into the predicted abundances.