---
title: Big Bang Nucleosynthesis
draft: false
module: The Hot Big Bang
moduleNumber: 12
lessonNumber: 2
order: 1202
summary: >
  In the first three minutes the weak interactions freeze out the
  neutron-to-proton ratio, and once deuterium survives photodissociation a fast
  reaction network converts nearly all free neutrons into helium-4. The primordial
  abundances of deuterium, helium-3, helium-4, and lithium-7 depend on a single
  free parameter, the baryon-to-photon ratio, so measuring them fixes the baryon
  density. The predictions match observation across nine decades of abundance,
  with a persistent discrepancy in lithium-7.
topics: [The Hot Big Bang]
sources:
  - book: Ryden
    ref: "Ch. 9 — Nucleosynthesis and the Early Universe"
  - book: Carroll & Ostlie
    ref: "Ch. 30 — The Early Universe; §30.2"
  - book: PDG
    ref: "Review of Particle Physics — Big-Bang Nucleosynthesis"
---

Between one second and roughly twenty minutes after the Big Bang, the universe
passed through the only epoch in its history when free nucleons could assemble
into nuclei in bulk. The outcome is fixed by nuclear and weak physics measured in
the laboratory, and it depends on essentially one cosmological number, the ratio
of baryons to photons. Big Bang nucleosynthesis (BBN) therefore does two things
at once: it predicts the primordial abundances of the light elements, and it
weighs the baryonic content of the universe. This lesson derives the
neutron-to-proton freeze-out, the deuterium bottleneck that delays nuclear
assembly, the resulting abundances as functions of the baryon-to-photon ratio,
and the concordance with observation together with its one sore point.

## The neutron-to-proton ratio

At temperatures above $k_B T \sim 1\ \text{MeV}$ the weak interactions
interconvert neutrons and protons rapidly,

$$
n + \nu_e \leftrightarrow p + e^-, \qquad
n + e^+ \leftrightarrow p + \bar\nu_e, \qquad
n \leftrightarrow p + e^- + \bar\nu_e ,
$$

and keep the two in chemical equilibrium. In equilibrium the ratio of number
densities follows the Boltzmann factor set by the neutron-proton mass difference
$Q = (m_n - m_p)c^2 = 1.293\ \text{MeV}$,

$$
\frac{n_n}{n_p} = \exp\!\left(-\frac{Q}{k_B T}\right).
$$

At high temperature the exponent is small and neutrons and protons are nearly
equal in number. As the universe cools the ratio falls, favoring the lighter
proton. If equilibrium held indefinitely the neutrons would disappear entirely;
they survive because the weak interactions freeze out.

The weak rate per nucleon scales as $\Gamma_{\text{weak}} \propto G_F^2 (k_B T)^5$,
while the expansion rate is $H \propto \sqrt{g_\ast}\,T^2$, so $\Gamma/H \propto
T^3$ and the interactions decouple at a freeze-out temperature $k_B T_{\text{fr}}
\approx 0.8\ \text{MeV}$. At that moment the equilibrium ratio is frozen in:

$$
\left.\frac{n_n}{n_p}\right|_{\text{fr}}
= \exp\!\left(-\frac{Q}{k_B T_{\text{fr}}}\right)
\approx \exp\!\left(-\frac{1.293}{0.8}\right) \approx \frac{1}{5}.
$$

$$
% caption: The equilibrium neutron fraction tracks the Boltzmann factor down to
% freeze-out near 0.8 MeV, where it locks at about one neutron per five protons,
% then decays slowly by beta decay until nucleosynthesis begins.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.6,0) node[right, black!70] {cooling to the right};
\draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {neutron to proton ratio};
% equilibrium track (falling)
\draw[acc, very thick] (0.4,4.0) .. controls (2.0,3.0) and (3.0,2.3) .. (4.0,1.9);
\node[acc, anchor=south] at (1.6,3.5) {equilibrium};
% freeze-out point and plateau (slow beta decay: gentle decline)
\fill[black!70] (4.0,1.9) circle (2pt);
\draw[black, densely dotted] (4.0,1.9) -- (4.0,0) node[below, black!70] {freeze-out};
\draw[acc, very thick, dashed] (4.0,1.9) -- (7.6,1.35);
\node[acc, anchor=south] at (5.9,1.5) {frozen, slow decay};
% onset of nucleosynthesis
\draw[black, densely dotted] (7.6,1.35) -- (7.6,0) node[below, black!70] {onset};
\node[black!70, anchor=west] at (7.7,1.9) {ratio near 1 to 7};
\end{tikzpicture}
$$

Freeze-out is not the whole story. Free neutrons are unstable, with a mean
lifetime $\tau_n = 880\ \text{s}$, so between freeze-out and the start of nuclear
assembly the ratio continues to fall by free decay,

$$
\frac{n_n}{n_p}(t) = \frac{(n_n/n_p)_{\text{fr}}\,e^{-t/\tau_n}}
{1 + (n_n/n_p)_{\text{fr}}\,(1 - e^{-t/\tau_n})} .
$$

By the time nucleosynthesis begins at $t \approx 200\ \text{s}$ the ratio has
dropped from $1/5$ to about $1/7$. This residual neutron fraction is what sets
the helium yield.

## The deuterium bottleneck

Building helium requires deuterium as the first step,
$p + n \to d + \gamma$, and every heavier nucleus is assembled from deuterium.
The binding energy of deuterium is only $B_D = 2.22\ \text{MeV}$, small on the
scale of the photon bath. Even well below $k_B T = B_D$, the blackbody spectrum
has a vast number of photons in its high-energy tail, because photons outnumber
baryons by the enormous factor $\eta^{-1} \sim 10^9$. Deuterium is
photodissociated, $d + \gamma \to p + n$, as fast as it forms until the
temperature drops far enough that even the tail no longer contains a
dissociating photon per deuteron.

The condition for deuterium to survive is that the number of photons above
$B_D$ per baryon falls below unity. With a photon-to-baryon ratio $\eta$, the
relevant estimate sets

$$
\eta^{-1} \exp\!\left(-\frac{B_D}{k_B T}\right) \sim 1
\quad\Longrightarrow\quad
k_B T_{\text{nuc}} \approx \frac{B_D}{\ln(\eta^{-1})} \approx 0.07\ \text{MeV},
$$

corresponding to $T_{\text{nuc}} \approx 0.8 \times 10^9\ \text{K}$ and
$t \approx 200\ \text{s}$. The delay from $k_B T = B_D$ down to
$k_B T_{\text{nuc}}$ — the **deuterium bottleneck** — is entirely due to the large
photon-to-baryon ratio, and it costs about 20% of the frozen neutrons to decay in
the meantime.

$$
% caption: Deuterium forms readily but is destroyed by the high-energy photon
% tail; only when the temperature has dropped well below the binding energy do
% deuterons survive, and the reaction network then proceeds.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.6,0) node[right, black!70] {cooling to the right};
\draw[->, black] (0,0) -- (0,4.2) node[above, black!70] {deuterium abundance};
% suppressed then rising
\draw[acc, very thick] (0.4,0.35) -- (5.2,0.4)
  .. controls (5.9,0.5) and (6.1,3.2) .. (6.6,3.5) -- (9.0,3.5);
\node[black!70, anchor=south] at (2.6,0.45) {destroyed by photons};
\node[acc, anchor=south east] at (8.6,3.5) {deuterium survives};
% bottleneck marker
\draw[black, densely dotted] (5.9,0) node[below, black!70] {bottleneck ends} -- (5.9,3.0);
\end{tikzpicture}
$$

## The reaction network and the helium yield

Once deuterium survives, a fast chain of two-body reactions burns it into
helium-4, the most tightly bound of the light nuclei. The dominant paths are

$$
d + d \to {}^3\text{He} + n, \qquad d + d \to t + p,
$$
$$
{}^3\text{He} + d \to {}^4\text{He} + p, \qquad t + d \to {}^4\text{He} + n ,
$$

with $t$ the triton (${}^3\text{H}$). Because ${}^4\text{He}$ has by far the
largest binding energy per nucleon among the light species, and because there is
no stable nucleus of mass number 5 or 8 to bridge to heavier elements, the chain
piles essentially all available neutrons into ${}^4\text{He} $ and then stalls.

The helium mass fraction follows from a simple neutron count. Each helium-4
nucleus contains two neutrons and two protons; nearly every neutron present at
$T_{\text{nuc}}$ ends up in a helium nucleus, paired with an equal number of
protons. With a neutron-to-proton ratio $r = n_n/n_p \approx 1/7$ at onset, the
mass fraction in helium is

$$
Y_p = \frac{4\,n_{\text{He}}}{n_{\text{baryon}}}
= \frac{2 n_n}{n_n + n_p}
= \frac{2r}{1 + r}
\approx \frac{2(1/7)}{1 + 1/7} = \frac{2}{8} = 0.25 .
$$

Roughly a quarter of the baryonic mass emerges as helium-4, with the rest almost
entirely hydrogen. This value is insensitive to cosmological
parameters, because it depends only on the frozen $n/p$ ratio through $Q$, the
neutron lifetime, and $g_\ast$ — all laboratory quantities.

> **Worked example.** The helium yield is a sensitive counter of relativistic
> species at freeze-out. A fourth light neutrino would raise $g_\ast$ from $10.75$
> to $12.5$, increasing the expansion rate $H \propto \sqrt{g_\ast}$ by a factor
> $\sqrt{12.5/10.75} \approx 1.08$. Faster expansion means earlier freeze-out at a
> higher $T_{\text{fr}}$, so the frozen ratio rises from
> $n_n/n_p = e^{-Q/k_B T_{\text{fr}}}$: a $2\%$ increase in $T_{\text{fr}}$ raises
> the ratio from $1/7$ to about $1/6$, lifting $Y_p = 2r/(1+r)$ from $0.25$ to
> about $0.27$. A change of a few percent in $Y_p$ is measurable, which is how BBN
> bounds $N_{\text{eff}}$ independently of the CMB. Its main dependence
is on the expansion rate at freeze-out: a larger $g_\ast$ (for instance, more
neutrino species) speeds the expansion, raises $T_{\text{fr}}$, freezes in more
neutrons, and increases $Y_p$. This is the sensitivity that lets BBN constrain
$N_{\text{eff}}$.

$$
% caption: The light-element network. Free neutrons and protons form deuterium,
% which burns through helium-3 and tritium into helium-4; the absence of stable
% mass-5 and mass-8 nuclei halts the chain, leaving trace lithium-7.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, node distance=1.4cm]
\definecolor{acc}{HTML}{4A6FA5}
\tikzset{iso/.style={draw, minimum width=0.9cm, minimum height=0.6cm, inner sep=2pt}}
\node[iso] (nn) at (0,3) {n};
\node[iso] (pp) at (0,1.4) {p};
\node[iso] (dd) at (2.4,2.2) {D};
\node[iso] (he3) at (5.0,3) {He-3};
\node[iso] (tt) at (5.0,1.4) {T};
\node[iso, draw=acc, fill=acc!10] (he4) at (7.6,2.2) {He-4};
\node[iso] (li7) at (10.0,2.2) {Li-7};
\draw[->] (nn) -- (dd);
\draw[->] (pp) -- (dd);
\draw[->] (dd) -- (he3);
\draw[->] (dd) -- (tt);
\draw[->] (he3) -- (he4);
\draw[->] (tt) -- (he4);
\draw[->, black, dashed] (he4) -- (li7);
\node[black, anchor=north] at (8.8,1.9) {rare};
\node[black, anchor=south, align=center] at (7.6,3.0) {no stable mass 5 or 8:\\ chain halts};
\end{tikzpicture}
$$

## Abundances as a function of the baryon-to-photon ratio

The single cosmological input to BBN is the **baryon-to-photon ratio**,

$$
\eta = \frac{n_b}{n_\gamma} \approx 6.1 \times 10^{-10},
$$

fixed and conserved once the photon and baryon numbers are set. Because
$n_\gamma$ is known from the CMB temperature, $\eta$ is equivalent to the baryon
density parameter,

$$
\Omega_b h^2 = 3.65 \times 10^7\,\eta \approx 0.022 .
$$

Each light-element abundance responds to $\eta$ in a characteristic way, and this
is what makes BBN a measurement rather than merely a consistency check:

- **Helium-4** ($Y_p \approx 0.247$) rises only logarithmically with $\eta$: a
  higher baryon density makes deuterium survive slightly earlier, so fewer
  neutrons decay and $Y_p$ increases weakly. It is the least sensitive probe but
  the best-measured.
- **Deuterium** falls steeply with $\eta$, roughly $\text{D/H} \propto
  \eta^{-1.6}$: more baryons burn deuterium more completely into helium, leaving
  less residual D. This steep slope makes deuterium the sharpest baryometer.
- **Helium-3** falls gently with $\eta$, tracking deuterium but partly replenished
  and partly destroyed in stars, which complicates its use.
- **Lithium-7** is non-monotonic, with a minimum near the observed $\eta$: at low
  $\eta$ it forms directly, at high $\eta$ it forms through ${}^7\text{Be}$ that
  later captures an electron, and the two channels produce a valley.

$$
% caption: Predicted primordial abundances versus the baryon-to-photon ratio.
% Helium-4 rises weakly, deuterium and helium-3 fall steeply, and lithium-7 dips
% then rises; the vertical band marks the value fixed by the CMB.
\begin{tikzpicture}[scale=1.0, font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (9.6,0) node[right, black!70] {baryon to photon ratio};
\draw[->, black] (0,0) -- (0,5.0) node[above, black!70] {abundance (log)};
% consistency band
\fill[acc!10] (5.2,0) rectangle (6.0,5.0);
\node[black, anchor=south, rotate=90] at (5.6,2.5) {CMB value};
% helium-4: nearly flat, slight rise
\draw[black, very thick] (0.4,4.4) .. controls (4.0,4.5) and (7.0,4.65) .. (9.0,4.75);
\node[anchor=south east] at (3.2,4.5) {He-4};
% deuterium: steep fall
\draw[black, very thick, dashed] (0.6,4.2) .. controls (2.5,3.0) and (4.5,1.8) .. (9.0,0.8);
\node[black, anchor=west] at (6.4,1.2) {D};
% helium-3: gentle fall
\draw[black, very thick, densely dotted] (0.6,3.2) .. controls (3.5,2.6) and (6.0,2.2) .. (9.0,1.9);
\node[black, anchor=south] at (7.8,1.95) {He-3};
% lithium-7: valley
\draw[black, very thick] (0.6,3.0) .. controls (3.0,1.6) and (4.5,1.3) .. (5.6,1.5)
  .. controls (7.0,1.8) and (8.2,2.6) .. (9.0,3.2);
\node[anchor=north] at (4.4,1.25) {Li-7};
\end{tikzpicture}
$$

The predicted curves cross the observed abundances at a common value of $\eta$ —
the defining success of BBN. The deuterium abundance, measured in the light of
distant quasars absorbed by unprocessed intergalactic gas, gives
$\text{D/H} \approx 2.5 \times 10^{-5}$ and pins $\eta$ to a few percent. That
$\eta$ agrees with the completely independent value from the CMB acoustic peaks,
a concordance across physics separated by 380,000 years of cosmic history.

## Concordance and the lithium problem

The measured primordial abundances span nine orders of magnitude, from $Y_p
\approx 0.25$ down to $\text{Li/H} \sim 10^{-10}$, and BBN reproduces them from
one parameter fixed elsewhere.

| Species | BBN prediction | observation | agreement |
| --- | --- | --- | --- |
| ${}^4\text{He}$ ($Y_p$) | $0.247$ | $0.245 \pm 0.003$ | excellent |
| D/H | $2.5 \times 10^{-5}$ | $2.5 \times 10^{-5}$ | excellent |
| ${}^3\text{He}$/H | $1.0 \times 10^{-5}$ | $\sim 10^{-5}$ | consistent (stellar processing) |
| ${}^7\text{Li}$/H | $5 \times 10^{-10}$ | $1.6 \times 10^{-10}$ | factor-of-3 low |

The helium and deuterium agreement is the strongest quantitative evidence that
the universe was once hot and dense, and that its physics at $k_B T \sim
1\ \text{MeV}$ was the physics measured in the laboratory. The one discrepancy is
the **lithium problem**: the abundance of ${}^7\text{Li}$ measured in the
atmospheres of old, metal-poor halo stars — the Spite plateau — is a factor of
about three below the BBN prediction for the CMB value of $\eta$. Proposed
resolutions fall into three classes: depletion of lithium in the stellar
atmospheres over billions of years, systematic errors in the nuclear reaction
rates feeding ${}^7\text{Be}$, or new physics altering the expansion or particle
content during BBN. None is established, and the lithium problem remains open.[^pdg-bbn]

BBN also constrains parameters beyond $\eta$. Because the helium yield depends on
the expansion rate at freeze-out, the concordance limits the effective number of
relativistic neutrino species to $N_{\text{eff}} = 2.9 \pm 0.3$, consistent with
the three Standard Model neutrinos and independent of the CMB determination. Any
extra light species — a fourth neutrino, or other relativistic relics — would
speed the expansion, raise $Y_p$, and break the agreement.

The baryon density fixed here, $\Omega_b \approx 0.05$, is far below the total
matter density $\Omega_m \approx 0.31$ inferred from [galaxy rotation curves and
clusters](/astrophysics-cosmology/galaxies/galaxy-rotation-curves-and-dark-matter).
The difference is the first cosmological argument that most of the matter is
non-baryonic, a conclusion the CMB independently confirms in the next lessons on
[recombination](/astrophysics-cosmology/the-hot-big-bang/recombination-and-the-cosmic-microwave-background)
and the [acoustic peaks](/astrophysics-cosmology/the-hot-big-bang/cmb-anisotropies-and-cosmological-parameters).

[^pdg-bbn]: Particle Data Group, Review of Particle Physics, Big-Bang Nucleosynthesis — the neutron-proton freeze-out, the deuterium bottleneck, the abundance predictions versus $\eta$, the concordance value $\Omega_b h^2 \approx 0.022$, and the status of the lithium problem. https://pdg.lbl.gov