Primordial Helium Abundance and the Deuterium Bottleneck

Primordial helium is now a precision cross-check on the whole hot Big Bang. Observers measure $Y_p$ from recombination lines of hydrogen and helium in metal-poor extragalactic HII regions — gas in nearby dwarf galaxies that has processed almost nothing through stars — extrapolating the small stellar contribution to zero metallicity via the correlation of helium with oxygen and nitrogen. Modern analyses combining optical and near-infrared He I lines report $Y_p^{\mathrm{obs}} = 0.245 \pm 0.003$, in striking agreement with the standard-BBN prediction $0.247$ evaluated at the baryon density Planck reads independently off the CMB acoustic peaks — a consistency between the first three minutes and last scattering spanning a factor of a thousand in redshift. Because $Y_p$ responds to the expansion rate, it is a cosmic speedometer for the radiation era: the combined BBN-plus-CMB fit gives $N_{\mathrm{eff}} = 2.9 \pm 0.2$, consistent with three Standard Model neutrinos and constraining extra "dark radiation." The same measurement bounds early-universe variation of $\tau_n$, $G_F$, and the neutron–proton mass gap, and anchors work on the stubborn lithium problem, where $^7\mathrm{Li}$ — unlike helium and deuterium — refuses to match theory.

Picture deuterium as a fragile intermediate on a factory line. A deuteron binds a neutron to a proton with only $2.22\,\mathrm{MeV}$ — feeble by nuclear standards — and it is the mandatory first product of the assembly line, since there is no way to jump straight from free nucleons to $^3\mathrm{He}$ or $^4\mathrm{He}$ without passing through $D$. Around it seethes a photon gas outnumbering baryons by a billion to one. Even after the temperature has fallen well below $B_D$, that billion-fold head start means the thin exponential tail of photons above $2.22\,\mathrm{MeV}$ is still numerous enough to shatter every deuteron the moment it forms. Synthesis must wait until the photon bath cools enough that formation finally outpaces destruction — the delay that defines the bottleneck. The first figure follows the currency that the whole story trades in: the neutron-to-proton ratio. While weak reactions ($n + \nu_e \leftrightarrow p + e^-$ and friends) stay faster than the expansion, $n/p$ tracks its equilibrium value $\exp(-Q/kT)$ and falls as the universe cools. Near $T \approx 0.8\,\mathrm{MeV}$ those reactions freeze out, stranding $n/p \approx 1/6$. From then on the only thing that changes $n/p$ is free-neutron decay, which slowly erodes it toward $1/7$ during the long wait for the bottleneck to break.

The second figure is the quantitative heart of the delay. It plots the equilibrium deuterium fraction $n_D/n_p$ that the Saha equation predicts, and the point is dramatic: for any temperature comfortably above $0.07\,\mathrm{MeV}$ the fraction sits astronomically below unity — a deuteron is destroyed almost as soon as it appears — even though the exponential Boltzmann factor $e^{B_D/kT}$ alone would scream that binding is favourable. Only when $T$ has fallen to about a thirtieth of $B_D$ does the curve rocket up through one, and synthesis switches on almost like a step function. The wall is the bottleneck made visible; its position is set by the competition between the Boltzmann factor and the tiny prefactor $\eta$.

The whole calculation rests on a chain of four ideas: the equilibrium neutron fraction, the temperature where the weak reactions abandon it, the Saha condition that fixes when deuterium survives, and the counting rule that turns the surviving neutrons into helium. We take them in that order. While the weak interactions keep neutrons and protons interconverting faster than the universe expands, their ratio sits at the Boltzmann value set by the rest-mass gap between them. This is the starting reservoir from which everything else is drawn.

where: $n,p$ — number densities of neutrons and protons; $Q$ — neutron–proton mass difference, $1.293\,\mathrm{MeV}$; $T$ — photon temperature; $k_B$ — Boltzmann constant. Reading it: because neutrons are heavier, they are always the minority, and the deficit deepens as the universe cools. The equilibrium cannot be maintained forever. When the weak interconversion rate $\Gamma_{\mathrm{weak}} \sim G_F^2 T^5$ falls below the Hubble rate $H$, the ratio stops tracking equation (1) and freezes at whatever value it held at that instant.

where: $T_{\mathrm{fr}}$ — freeze-out temperature, where $\Gamma_{\mathrm{weak}} \simeq H$; $G_F$ — Fermi constant. Reading it: the frozen neutron fraction is the single most important number in the problem, because helium is built from neutrons. Deuterium formation and destruction, $p + n \leftrightarrow D + \gamma$, sit in their own equilibrium described by the Saha (nuclear statistical) equation. It balances the binding-energy bonus against the phase-space cost of assembling two particles into one.

where: $n_D,n_n,n_p$ — number densities of deuterium, neutrons, protons; $g_D = 3,\ g_n = g_p = 2$ — spin degeneracies; $B_D = 2.22\,\mathrm{MeV}$ — deuterium binding energy; $m_D \approx 2m_N$ — deuteron mass. Reading it: the factor $e^{B_D/k_B T}$ favours binding, but the prefactor buries an implicit dependence on the baryon density. Expressing the nucleon densities through the baryon-to-photon ratio makes the buried dependence explicit, and it is the whole story: the fraction of nucleons in deuterium is suppressed by $\eta$, so it reaches order unity only far below $B_D$.

where: $\eta \equiv n_b/n_\gamma \approx 6\times10^{-10}$ — baryon-to-photon ratio; $m_N$ — nucleon mass. Reading it: the tiny $\eta$ out front is exactly the billion-to-one photon glut, and it is what postpones synthesis. Setting $n_D/n_p \sim 1$ and taking logarithms turns the exponential into an algebraic condition for the temperature at which the bottleneck breaks. Because the balance is logarithmic, $T_{\mathrm{nuc}}$ is far below $B_D$ but only weakly sensitive to any of the inputs.

where: $T_{\mathrm{nuc}}$ — temperature at which deuterium accumulates and synthesis proceeds. Reading it: the denominator is dominated by $\ln(\eta^{-1}) \approx 21$, which is the mathematical fingerprint of the entropy per baryon. Once deuterium survives, the network races to $^4\mathrm{He}$ and stops there. Counting nucleons — two neutrons and two protons per helium nucleus, with the leftover protons remaining as hydrogen — gives the mass fraction directly in terms of the neutron ratio.

where: $Y_p$ — mass fraction of $^4\mathrm{He}$; $n_{\mathrm{He}},n_{\mathrm{H}}$ — number densities of helium and hydrogen nuclei; $n/p$ — the ratio evaluated at $T_{\mathrm{nuc}}$. Reading it: helium counts the neutrons, so anything that changes $n/p$ changes $Y_p$ almost one-for-one. The ratio fed into equation (6) is not the freeze-out value but the smaller one that survives the wait, because free neutrons $\beta$-decay throughout the bottleneck while bound neutrons in $D$ and $^4\mathrm{He}$ do not.

where: $\tau_n = 879.4\,\mathrm{s}$ — free-neutron mean lifetime; $t_{\mathrm{nuc}} \approx 200\text{–}300\,\mathrm{s}$ — cosmic time at bottleneck breaking. Reading it: the delay is a clock, and the longer it runs the fewer neutrons — hence less helium — remain.

How the helium yield and the bottleneck respond to the parameters that control them:

1 · The helium mass fraction from first principles (fully worked)

Take the freeze-out temperature $k_B T_{\mathrm{fr}} = 0.8\,\mathrm{MeV}$, the mass gap $Q = 1.293\,\mathrm{MeV}$, the neutron lifetime $\tau_n = 879.4\,\mathrm{s}$, and a bottleneck-breaking time $t_{\mathrm{nuc}} \approx 300\,\mathrm{s}$. Predict the primordial helium mass fraction $Y_p$.

Step 1. Freeze-out ratio from equation (2): $(n/p)_{\mathrm{fr}} = \exp(-Q/k_B T_{\mathrm{fr}}) = \exp(-1.293/0.8) = \exp(-1.616) = \mathbf{0.199}$. This is close to $1/5$; the value conventionally quoted as "about $1/6$" already anticipates that freeze-out is gradual and that a little decay has begun.

Step 2. Apply neutron decay over the delay (equation 7): the survival factor is $e^{-t_{\mathrm{nuc}}/\tau_n} = e^{-300/879.4} = e^{-0.341} = 0.711$, so $(n/p)_{\mathrm{nuc}} = 0.199 \times 0.711 = \mathbf{0.141} \approx 1/7$.

Step 3. Convert to helium with equation (6): $Y_p = \dfrac{2(n/p)_{\mathrm{nuc}}}{1 + (n/p)_{\mathrm{nuc}}} = \dfrac{2(0.141)}{1.141} = \dfrac{0.283}{1.141} = \mathbf{0.247}$.

Step 4. Sanity check against the clean fraction: with the rounded $n/p = 1/7$ exactly, $Y_p = (2/7)/(8/7) = 1/4 = 0.250$. The two agree to within the rounding, and both sit right on the network-code value $Y_p = 0.2470 \pm 0.0002$ and the observed $Y_p^{\mathrm{obs}} = 0.245 \pm 0.003$. A four-line hand estimate reproduces the measured primordial helium abundance to better than 1%.

2 · Where and when the bottleneck breaks (fully worked)

Using $\eta = 6\times10^{-10}$, $B_D = 2.22\,\mathrm{MeV}$, and $m_N c^2 = 939\,\mathrm{MeV}$, estimate the temperature $T_{\mathrm{nuc}}$ at which deuterium starts to accumulate and the corresponding cosmic time.

Step 1. Evaluate the two pieces of the denominator in equation (5). The entropy term is $\ln(\eta^{-1}) = \ln(1.67\times10^{9}) = \mathbf{21.2}$; the phase-space term is $\tfrac32\ln(m_N/B_D) = \tfrac32\ln(939/2.22) = \tfrac32\ln(423) = \tfrac32(6.05) = \mathbf{9.1}$.

Step 2. Sum them: the denominator is $21.2 + 9.1 = \mathbf{30.3}$, dominated by the photon-to-baryon logarithm. So $k_B T_{\mathrm{nuc}} = B_D/30.3 = 2.22/30.3 = \mathbf{0.073\,MeV}$.

Step 3. Confirm by back-substitution into equation (4): at $k_B T = 0.073\,\mathrm{MeV}$, $e^{B_D/k_B T} = e^{30.4} = 1.6\times10^{13}$ and $\eta(k_B T/m_N c^2)^{3/2} = 6\times10^{-10}(7.8\times10^{-5})^{3/2} = 4\times10^{-16}\cdot\ldots$, whose product is of order unity — the balance we imposed.

Step 4. Convert to time with the radiation-era relation $t \approx (2.4/\sqrt{g_*})(k_B T/\mathrm{MeV})^{-2}\,\mathrm{s}$ and $g_* \approx 3.36$: $t_{\mathrm{nuc}} = (2.4/1.83)(0.073)^{-2} = 1.31 \times 188 = \mathbf{246\,s} \approx 4\,\mathrm{min}$. Synthesis begins near kT ≈ 0.07 MeV at an age of roughly four minutes — a factor of thirty in temperature below the naive B_D estimate.

3 · The imprint of an extra neutrino species

A hypothetical fourth light neutrino would raise $N_{\mathrm{eff}}$ from $3.046$ to $4.046$. Estimate the shift in $Y_p$ and confirm the sign through the physics.

Step 1. Use the empirical sensitivity from the limiting-cases table: $\partial Y_p/\partial N_{\mathrm{eff}} \approx +0.013$ per species, so $\Delta Y_p \approx +0.013$, raising the prediction from $0.247$ to about $\mathbf{0.260}$.

Step 2. Trace the mechanism: an extra relativistic species increases $g_*$, hence $H \propto \sqrt{g_*}$. A faster expansion makes the weak reactions lose the race sooner, so freeze-out happens at a higher $T_{\mathrm{fr}}$ where equation (1) gives a larger $n/p$.

Step 3. More neutrons at freeze-out means more neutrons survive to synthesis, and since $Y_p$ counts neutrons, more helium. A shift of $+0.013$ is comparable to the observational error bar of $\pm 0.003$, which is exactly why $Y_p$ is a competitive constraint on $N_{\mathrm{eff}}$. One extra neutrino species raises Y_p by about 0.013 — several times the measurement uncertainty, making helium a sensitive counter of relativistic degrees of freedom.

Primordial helium is where nuclear physics, particle physics, and cosmology first fused into a single quantitative prediction. The number $Y_p \approx 0.25$ was argued into existence decades before the reaction network could be solved on a computer, and the man whose doctoral thesis started it all watched the field grow up around him.

Texts. Ryden, Introduction to Cosmology, 2nd ed., ch. 10 (Nucleosynthesis and the Early Universe) · Dodelson, Modern Cosmology, ch. 3 (The Origin of Species) · Carroll & Ostlie, An Introduction to Modern Astrophysics, 2nd ed., ch. 29–30 · Kolb & Turner, The Early Universe, ch. 4 · Weinberg, Cosmology, ch. 3 · Mo, van den Bosch & White, Galaxy Formation and Evolution, §3.4 · Particle Data Group Review, Big Bang Nucleosynthesis. Numerical values used throughout: $Q = 1.293\,\mathrm{MeV}$, $B_D = 2.224\,\mathrm{MeV}$, $\tau_n = 879.4\,\mathrm{s}$ (PDG), $\eta \approx 6.1\times10^{-10}$ ($\Omega_b h^2 \approx 0.022$), $T_{\mathrm{fr}} \approx 0.8\,\mathrm{MeV}$, $T_{\mathrm{nuc}} \approx 0.07\text{–}0.08\,\mathrm{MeV}$, $t_{\mathrm{nuc}} \approx 200\text{–}300\,\mathrm{s}$, $Y_p^{\mathrm{BBN}} = 0.2470$, $Y_p^{\mathrm{obs}} = 0.245 \pm 0.003$. Key checks: $(n/p)_{\mathrm{fr}} = e^{-Q/k_B T_{\mathrm{fr}}} = 0.199$; the Saha bottleneck denominator $\ln(\eta^{-1}) + \tfrac32\ln(m_N/B_D) = 30.3$, giving $T_{\mathrm{nuc}} = 0.073\,\mathrm{MeV}$; and $Y_p = 2(n/p)/(1+n/p)$. Note that the compact bottleneck formula uses a plus sign on the phase-space log term — the entropy and phase-space contributions add.

Prerequisites. Big Bang Nucleosynthesis · Equilibrium Thermodynamics of the Early Universe · Decoupling and Freeze-Out · Thermal History of the Universe. Leads to. The Lithium Problem · Effective Number of Relativistic Degrees of Freedom · Recombination and the Saha Equation · Neutrino Decoupling and the Cosmic Neutrino Background.

Ask the tutor: Ask me to walk through why the same tiny number $\eta$ that delays deuterium by a factor of thirty in temperature leaves the final helium abundance almost untouched.