When neutrons and protons stop turning into each other

Read How temperature and expansion speed set helium and deuterium first for the overall picture. The mathematics needed is exponentials, natural logarithms, ratios and rates per second. After this chapter you should be able to compute the equilibrium ratio at a given temperature, compare a reaction rate with the expansion rate, and turn a neutron-to-proton ratio at a known time into a helium fraction.

Temperature, energy and the cost of a neutron

Cosmologists quote the temperature of the early universe in two ways. $T_9$ is the temperature in units of $10^9\,\mathrm K$. The typical thermal energy is $k_BT$, and

At $T_9=10$ the thermal energy is $0.862\,\mathrm{MeV}$. The neutron is heavier than the proton by

When the weak reactions $n+e^+\leftrightarrow p+\bar\nu_e$, $n+\nu_e\leftrightarrow p+e^-$ and neutron decay run fast in both directions, neutrons and protons share one temperature. Both are spin one-half particles, so the ratio of their number densities $n$ and $p$ is a Boltzmann factor:

The mass prefactor is 1.002 and is dropped. The ratio uses the neutrino and electron chemical potentials set to zero, the standard assumption in this tutorial. The table shows how fast the equilibrium ratio falls.

If equilibrium held all the way down, there would be almost no neutrons left by the time nuclei could form, and almost no helium. Something must stop the ratio from following this table.

How fast the universe expands

The expansion rate $H$ is the fractional rate at which distances grow, in $\mathrm{s^{-1}}$. In the first minutes the energy density is dominated by particles moving close to the speed of light. Each contributes in proportion to $T^4$, weighted by its number of internal states, and fermions count with a factor $7/8$. The Friedmann equation then gives

Before electrons and positrons annihilate, $g_*=2+\tfrac78(4+2N_\nu)$: two photon states, four electron and positron states, and two states for each of $N_\nu$ neutrino species. For $N_\nu=3$ this is $10.75$, the same as the $5.5+\tfrac74N_\nu$ quoted in the 2025 Review of Particle Physics, Section 24.4.

Worked example: the clock at one MeV

In natural units the same equation reads $H=1.66\sqrt{g_*}\,(k_BT)^2/M_{\mathrm{Pl}}$, with the Planck mass $M_{\mathrm{Pl}}c^2=1.221\times10^{19}\,\mathrm{GeV}$. At $k_BT=1\,\mathrm{MeV}$,

The last factor converts an energy into a rate with $\hbar$. For radiation domination the scale factor grows as $a\propto t^{1/2}$, so $H=\dot a/a=1/(2t)$ and

The tutorial network reaches $k_BT=1\,\mathrm{MeV}$ at 0.74 seconds as well. Doubling the temperature makes $H$ four times larger, so a hotter universe is a younger one: $t\propto T^{-2}$ while $g_*$ stays fixed.

Conversion rate against expansion rate

Let $\lambda_{np}$ be the rate per neutron of all reactions that turn a neutron into a proton, and $\lambda_{pn}$ the reverse rate per proton. The ratio relaxes toward equilibrium at the total rate $\Gamma=\lambda_{np}+\lambda_{pn}$. Far above $Q$ and the electron mass, $\Gamma$ grows roughly as $T^5$, while $H$ grows as $T^2$. Their ratio therefore falls roughly as $T^3$ as the universe cools.

The tutorial computes both rates from the Born approximation, the leading-order weak-interaction calculation described in the PRIMAT paper by Pitrou, Coc, Uzan and Vangioni. The rates are normalised to a neutron lifetime $\tau_n=878.4\,\mathrm s$.

The last column measures $\lambda_{np}$ in units of the free decay rate $1/\tau_n$. At $T_9=30$ conversions are a hundred thousand times faster than decay. By $T_9=1$ the column has fallen to one: the only process still turning neutrons into protons is free decay. That is why $\Gamma/H$ rises again at the bottom of the table. The decay rate stays fixed while $H$ keeps falling.

$\Gamma/H$ crosses one at $T_9=7.71$, where $k_BT=0.664\,\mathrm{MeV}$ and $t=1.71\,\mathrm s$. Equilibrium would give $n/p=0.143$ there. The network instead has $n/p=0.269$. The ratio was already lagging at $T_9=10$, where $\Gamma/H\approx2$ and the network gives $0.313$ against an equilibrium $0.223$. Keeping up with a target that keeps moving takes many conversions per expansion time, not one. The crossing $\Gamma=H$ marks the middle of a gradual departure, not a switch.

After the crossing, conversions continue at a reduced pace. In the network the ratio falls to $0.198$ at 12.6 seconds ($T_9=3$), $0.186$ at 32 seconds ($T_9=2$) and $0.153$ at 168 seconds ($T_9=1$). The early part of this drift is weak conversion; the late part is mainly decay.

From the ratio to helium

Write the neutron fraction as $X_n=n/(n+p)=(n/p)/(1+n/p)$. When deuterium finally survives, almost every free neutron ends up in helium-4. A helium-4 nucleus holds two neutrons and has four mass units, so the helium mass fraction is

Here $Y_p$ is four times the number of helium-4 nuclei per baryon, the convention used in the tutorial; it differs from the true mass fraction by the small nuclear binding energy.

Worked example: decay from one billion kelvin to the bottleneck

Start at $T_9=1$, where the network gives $n/p=0.1528$ at $t=168.1\,\mathrm s$, so $X_n=0.1326$. From here on only decay removes neutrons. The network's helium is half formed at $t=247.3\,\mathrm s$; take that as the moment the bottleneck opens. Then

and $Y_p\simeq2\times0.1211=0.242$. The full network gives $0.2424$. The two numbers agree because by $T_9=1$ the only remaining process is the one this estimate keeps.

The same estimate started at $T_9=3$ goes wrong. With $n/p=0.198$ at 12.6 seconds, pure decay to 247.3 seconds gives $X_n=0.1263$ and $Y_p\simeq0.253$, about 4 percent too high. Table 2.2 shows the reason: at $T_9=3$ the neutron-to-proton conversion is still $4.4$ times faster than decay. Electron and positron captures remove neutrons that a decay-only estimate keeps.

The companion files hold every value used here: teaching-values.json lists the clock, the conversion table and the freeze-out point, and the method guide Recompute a small network from a frozen rate grid and propagate rate errors shows how they were produced.

Lab task: change the neutron lifetime

1. Predict. The neutron lifetime enters twice: shorter lifetimes mean faster decay and, because every weak rate is normalised to $1/\tau_n$, stronger conversion. Decide whether a longer lifetime raises or lowers $Y_p$, and say which of the two effects you expect to matter more.
2. Change. Open the advanced controls, set the lifetime to $860\,\mathrm s$ and then to $900\,\mathrm s$, and record $Y_p$ and the neutron-to-proton ratio at $10^9\,\mathrm K$.
3. Compare. The saved runs give $Y_p=0.2386$ and $0.2468$, and $n/p$ at $T_9=1$ of $0.1504$ and $0.1556$. The Lab uses a coarser time step and can differ in the last digit.
4. Explain. A longer lifetime raises helium. The ratio at $T_9=1$ moves by about 3.5 percent and the final helium by about 3.4 percent, so most of the change is already in place at one billion kelvin. That points to the conversion strength during freeze-out rather than the last minute of decay.

Exercises

1. Compute the equilibrium ratio at $k_BT=1\,\mathrm{MeV}$. Explain why the ratio is not zero even though making a neutron costs energy.
2. Assume $\Gamma\propto T^5$ and $H\propto T^2$. Starting from $\Gamma/H=1.99$ at $T_9=10$, predict where $\Gamma/H=1$. Compare with Table 2.2 and explain the difference.
3. Start the decay-only estimate at $T_9=2$ instead of $T_9=3$. Is it closer to the network? Why?
4. Two experiments report neutron lifetimes that differ by ten seconds. Use the Lab task numbers to estimate the resulting difference in $Y_p$, and compare it with the observational uncertainty of 0.003 quoted in the 2025 review.
5. Transfer. Imagine a universe in which $Q$ is $1.0\,\mathrm{MeV}$ instead of $1.293\,\mathrm{MeV}$ and the freeze-out temperature stays at $0.66\,\mathrm{MeV}$. Predict the direction of the helium change, and name one thing that a real calculation would also have to change.

Where the simple picture stops

The tutorial switches neutrinos off from the plasma instantly, so it uses $N_{\mathrm{eff}}=3$ rather than the $3.044$ that follows from a gradual decoupling, and its weak rates stop at the Born approximation. Table V of the PRIMAT paper shows what the omitted pieces are worth. With the Born rates alone PRIMAT finds $Y_p=0.24262$; radiative corrections add $0.0032$ and finite nucleon mass with weak magnetism another $0.0013$; incomplete neutrino decoupling, thermal radiative and plasma terms together change the result by less than $0.0001$, for a final $0.24709$. The tutorial network reproduces the Born row to $0.02$ percent. Its $0.242$ is therefore expected to sit about $0.005$ below the $0.247$ of the CMB-plus-BBN prediction of Fields, Olive, Yeh and Young. None of this changes the mechanism of this chapter. The next chapter, Why deuterium waits for the universe to cool, explains why the neutrons had to wait until about 247 seconds.