Big Bang Nucleosynthesis

For its first three minutes the entire universe was a nuclear reactor. Big Bang Nucleosynthesis is the theory of what that reactor made — the earliest epoch we can still test directly, because the light nuclei it forged still float in the most pristine gas we can find. The setting is the radiation-dominated era, where the Friedmann equation ties the expansion rate to temperature as $H\propto T^2$ and the whole cosmos cools on one clock, the Hubble time. Around $T\sim1$ MeV ( $t\sim1$ s) two things happen almost together: the weak interactions that shuttle neutrons into protons and back can no longer keep pace with the expansion and drop out, freezing the neutron-to-proton ratio, and the neutrinos decouple to stream freely forever. Over the next few minutes the survivors are welded into deuterium, then helium, then a trace of lithium. Everything downstream is one competition — reaction rate $\Gamma$ against expansion rate $H$ . While $\Gamma\gg H$ a species stays in equilibrium; when $\Gamma$ falls below $H$ its abundance freezes. The remarkable payoff is that the yields depend on almost nothing: one number, $\eta\approx6\times10^{-10}$ , plus the neutron lifetime and the expansion rate. With that single $\eta$ the theory reproduces abundances spanning nine orders of magnitude — helium at a quarter of all mass, deuterium near $10^{-5}$ , lithium near $10^{-10}$ — and the value it needs agrees with the independent baryon density read off the cosmic microwave background. That concordance is a load-bearing pillar of the hot Big Bang.

Key takeaways

Big Bang Nucleosynthesis (BBN) forged the light nuclei $^2$ H, $^3$ He, $^4$ He and $^7$ Li in the first $\sim3$ minutes, as the universe cooled from $\sim10^{10}$ K to $\sim10^9$ K.
The physics is a race: a reaction tracks equilibrium while $\Gamma>H$ and freezes when $\Gamma<H$ . Weak rates $\propto T^5$ fall faster than $H\propto T^2$ , so $n/p$ freezes near $T\simeq0.8$ MeV at $\approx1/6$ , drifting to $\approx1/7$ as neutrons decay.
Nearly every surviving neutron is locked into $^4$ He, so the helium mass fraction is essentially a neutron counter, $Y_p\approx2(n/p)/(1+n/p)\approx0.247$ .
Synthesis is delayed to $T\simeq0.07$ MeV by the deuterium bottleneck — with a billion photons per baryon, deuterium is photo-dissociated long after it is bound.
Yields depend almost entirely on one number, $\eta\equiv n_b/n_\gamma\approx6\times10^{-10}$ ( $\Omega_b h^2\approx0.022$ ), in striking agreement with the CMB; the expansion rate makes $Y_p$ a precision counter of species $N_{\rm eff}\approx3$ .
Key idea: BBN is the race between weak/nuclear reaction rates and the Hubble expansion — $\Gamma=H$ sets neutron freeze-out near $T\simeq0.8$ MeV, the deuterium bottleneck delays synthesis to $T\simeq0.07$ MeV, and nearly all surviving neutrons are locked into $^4$ He, giving $Y_p\approx0.25$ as a near-parameter-free prediction.

The core insight

BBN is governed by the race between weak/nuclear reaction rates and the Hubble expansion: $\Gamma=H$ sets neutron freeze-out near $T\simeq0.8$ MeV, the deuterium bottleneck delays synthesis to $T\simeq0.07$ MeV, and nearly all surviving neutrons are then locked into $^4$ He, giving $Y_p\approx0.25$ as a near-parameter-free prediction.

Where this lives today

BBN is a working precision tool. Its sharpest input is primordial deuterium, measured in the light of distant quasars passing through pristine, near-metal-free gas at high redshift; the best systems pin $\mathrm{D/H}=(2.53\pm0.03)\times10^{-5}$ , which through $\mathrm{D/H}\propto\eta^{-1.6}$ fixes $\Omega_b h^2$ to about one percent. The stunning result is that this agrees with the independent baryon density from the Planck CMB acoustic peaks — two probes separated by $\sim380{,}000$ years and entirely different physics, landing on the same $\eta\approx6\times10^{-10}$ . Because the helium yield tracks the expansion rate through $g_*$ , combining $Y_p$ and D/H bounds the neutrino species count to $N_{\rm eff}=2.9\pm0.3$ , closing the door on many light "dark radiation" particles. The one stubborn scar is the lithium problem: standard BBN with the CMB $\eta$ predicts $^7\mathrm{Li/H}\approx5\times10^{-10}$ , but ancient "Spite plateau" stars show only $\approx1.6\times10^{-10}$ , a factor-of-three deficit no one has convincingly closed. And because $Y_p$ depends on $\tau_n$ , primordial helium even cross-checks the persistent "beam" versus "bottle" discrepancy in the neutron lifetime.

The physical picture

Picture a furnace cooling on a fixed clock — the Hubble time $t\sim1/H$ . Any reaction fast compared with the cooling keeps its reactants in equilibrium; the moment it falls slower than the expansion, its product abundance "freezes." BBN is a cascade of these freeze-outs: first the weak interactions lose the race and fix the neutron-to-proton ratio; then, after a delay, the nuclear network lights and fixes the light-element yields. The first figure tracks the single most important quantity — the $n/p$ ratio — from equilibrium through freeze-out to its resting place inside helium.

Figure

The neutron-to-proton ratio through the first three minutes: it follows the equilibrium curve n/p = exp(−Δmc²/kT) down to weak freeze-out at T ≈ 0.8 MeV (n/p ≈ 1/6, t ≈ 1 s), drifts to ≈ 1/7 as free neutrons β-decay, then locks into ⁴He when the deuterium bottleneck breaks at T ≈ 0.07 MeV, giving Yₚ ≈ 0.25

The deuterium bottleneck is the crucial, counterintuitive twist. Helium-4 is so tightly bound (28.3 MeV total) that thermodynamics "wants" it already at $T\sim0.3$ MeV. But every road to $^4$ He runs through fragile deuterium, bound by only $Q=2.22$ MeV — and with a billion photons per baryon, even the sparse high-energy photons in the blackbody tail photo-dissociate deuterium as fast as it forms. The dam holds until $T$ falls to $\sim0.07$ MeV, then breaks and synthesis runs to completion within minutes. Helium then acts as a near-perfect neutron counter: almost every neutron alive at $T\sim0.07$ MeV is bound into $^4$ He, so $Y_p\approx2n/(n+p)$ depends mainly on the frozen $n/p$ and barely on the reaction rates. Deuterium and $^3$ He are the unburned leftovers — the more baryons, the more completely they burn. The second figure is the master diagram of the field: the abundances against the one free parameter $\eta$ .

Figure

The Schramm plot — standard BBN abundances versus the baryon-to-photon ratio η. Deuterium falls steeply (D/H ∝ η^−1.6), ⁴He rises gently to Yₚ ≈ 0.247, and ⁷Li traces a valley. All four meet observation at one value, η × 10¹⁰ ≈ 6.1 (Ω_b h² ≈ 0.022, yellow band) — the baryon density the CMB finds independently — while only ⁷Li sits a factor ≈ 3 high

The core equations

BBN runs on a handful of relations in two blocks: the expanding thermal bath (the clock) and the microphysics riding on it (the reactions). Begin with the clock. In the radiation era the Friedmann equation, with curvature and $\Lambda$ negligible at MeV temperatures, ties the expansion rate to a thermal energy density set by the species count.

H^2=\frac{8\pi G}{3}\,\rho_r,\qquad \rho_r=\frac{\pi^2}{30}\,g_*\,T^4 \qquad (1)

where: $H$ — Hubble parameter $\dot a/a$ ; $G$ — Newton's constant; $\rho_r$ — energy density of relativistic species; $g_*$ — effective degrees of freedom, $g_*=\sum_{\rm bosons}g_i+\tfrac{7}{8}\sum_{\rm fermions}g_i$ ; $T$ — photon temperature. Reading: $H\propto\sqrt{g_*}\,T^2$ ; at $T\sim1$ MeV the $\gamma,e^\pm,\nu,\bar\nu$ bath gives $g_*=2+\tfrac{7}{8}(4+6)=10.75$ , fixing $H$ by temperature alone. Integrating gives the cosmic clock $t(T)$ timestamping every milestone. In radiation domination $\rho\propto a^{-4}$ forces $a\propto t^{1/2}$ , so $H=1/(2t)$ , inverting to a time-temperature law.

t\simeq\frac{1}{2H}=\left(\frac{45}{16\pi^3 G\,g_*}\right)^{1/2}\frac{1}{T^2}\;\Rightarrow\;t\,[\mathrm{s}]\approx\frac{2.4}{\sqrt{g_*}}\left(\frac{T}{\mathrm{MeV}}\right)^{-2} \qquad (2)

where: $t$ — cosmic time since the singularity; $g_*$ — degrees of freedom at temperature $T$ . Reading: with $g_*=10.75$ this gives $t\approx0.74$ s at $T=1$ MeV — the famous "first second," pinning the schedule of all that follows. Now the microphysics. While the weak reactions $n+\nu_e\leftrightarrow p+e^-$ and $n+e^+\leftrightarrow p+\bar\nu_e$ are fast, neutrons and protons stay in equilibrium, their ratio a Boltzmann factor in the mass gap.

\left.\frac{n}{p}\right|_{\rm eq}=\exp\!\left(-\frac{\Delta m\,c^2}{k_BT}\right),\qquad \Delta m\,c^2=(m_n-m_p)c^2=1.293\ \mathrm{MeV} \qquad (3)

where: $n,p$ — neutron and proton number densities; $\Delta m$ — neutron-proton mass difference; $k_B$ — Boltzmann constant. Reading: at $T=1$ MeV, $n/p=e^{-1.293}\approx0.27$ , falling only while equilibrium holds. Equilibrium holds only while the weak rate beats the expansion. That rate carries a steep temperature dependence, so its ratio to $H$ collapses as the universe cools — and where the ratio crosses unity, $n/p$ freezes.

\Gamma_{n\leftrightarrow p}\sim G_F^2\,T^5,\qquad \frac{\Gamma}{H}\sim\left(\frac{T}{0.8\ \mathrm{MeV}}\right)^3 \qquad (4)

where: $\Gamma_{n\leftrightarrow p}$ — weak interconversion rate; $G_F$ — Fermi coupling constant; $H\propto T^2$ — Hubble rate. Reading: since $\Gamma/H\propto T^3$ falls fast, $\Gamma=H$ gives a sharp freeze-out at $T_{\rm fr}\approx0.7$ – $0.8$ MeV, locking $n/p\approx1/6$ . Between freeze-out and synthesis the free neutrons $\beta$ -decay, so the ratio entering helium is below the frozen one. Almost all surviving neutrons then end up in $^4$ He — the headline prediction.

Y_p=\frac{2(n/p)_{\rm BBN}}{1+(n/p)_{\rm BBN}}\Bigg|_{(n/p)_{\rm BBN}\simeq1/7}\approx0.247,\qquad (n/p)_{\rm BBN}=(n/p)_{\rm fr}\,e^{-t_{\rm nuc}/\tau_n} \qquad (5)

where: $Y_p$ — $^4$ He mass fraction (helium mass over total baryon mass); $\tau_n\approx879$ s — free-neutron lifetime; $t_{\rm nuc}\approx250$ s — time at synthesis. Reading: $\beta$ -decay drifts $n/p$ from $\approx1/6$ to $\approx1/7$ ; two neutrons and two protons per $^4$ He then give $Y_p=2n/(n+p)\approx0.25$ . The one genuinely free parameter is the baryon-to-photon ratio, fixing how much fuel there is and how completely the leftovers burn. It is often rescaled $\eta_{10}=\eta\times10^{10}$ .

\eta\equiv\frac{n_b}{n_\gamma}=2.74\times10^{-8}\,\Omega_b h^2\approx6.1\times10^{-10} \qquad (6)

where: $\eta$ — baryon-to-photon number ratio; $n_b,n_\gamma$ — baryon and photon number densities; $\Omega_b h^2\approx0.022$ — physical baryon density. Reading: about one baryon per $1.6$ billion photons — that tiny number is what starves and delays deuterium. Finally the deuterium bottleneck. (Aside: when $e^+e^-$ pairs annihilate they heat the photons but not the decoupled neutrinos, leaving $T_\nu/T_\gamma=(4/11)^{1/3}\approx0.714$ , which must be folded into $\rho_r$ and $g_*$ .) Equilibrium deuterium is exponentially favored at low $T$ but suppressed by the small $\eta$ , so synthesis waits until the exponent overcomes the photon flood.

\frac{Q}{k_BT}\sim\ln\!\left(\eta^{-1}\right)\sim26\;\Rightarrow\;T_{\rm nuc}\approx\frac{2.22\ \mathrm{MeV}}{26}\approx0.07\ \mathrm{MeV} \qquad (7)

where: $Q=2.22$ MeV — deuterium binding energy; $\eta\sim10^{-9}$ — baryon-to-photon ratio; $T_{\rm nuc}$ — temperature at which deuterium survives and synthesis begins. Reading: the logarithm of the photon-to-baryon ratio pushes synthesis a factor of $\sim30$ below the naive binding scale.

Derivation ANeutron freeze-out from the rate-versus-expansion race

The neutron-to-proton ratio that ultimately fixes the helium yield is decided by a single competition: the weak interactions that interconvert neutrons and protons against the Hubble expansion that cools everything. The goal is to locate the temperature at which that competition tips — the freeze-out temperature — by writing both rates as powers of temperature, forming their ratio, setting it to unity, and reading off the frozen $n/p$ from the equilibrium Boltzmann factor there.

Step 1. Write the weak rate. In the regime $k_BT\gg\Delta m c^2$ , the reactions $n+\nu_e\leftrightarrow p+e^-$ and $n+e^+\leftrightarrow p+\bar\nu_e$ scale as $\Gamma_{\rm wk}\sim G_F^2\,T^5$ — the Fermi coupling squared times five powers of temperature from the lepton phase space. Why: the steep $T^5$ makes the weak rate plunge through the expansion rate rather than crossing it gently, so freeze-out is sharp.

Step 2. Write the expansion rate. From (1), $H=\sqrt{(8\pi G/3)(\pi^2/30)\,g_*}\;T^2$ with $g_*=10.75$ for the $\gamma,e^\pm,3\nu$ bath, so $H\propto T^2$ . Why: the competitors carry different powers of temperature — $T^5$ against $T^2$ — and that mismatch guarantees a crossover as the universe cools.

Step 3. Form the ratio and balance it. Dividing, $\Gamma/H\propto T^3$ , falling by a factor of eight per halving of temperature. Setting $\Gamma/H=1$ isolates a narrow band, $T_{\rm fr}\approx0.7$ – $0.8$ MeV, below which the weak interactions can no longer keep up. Why: because the ratio scales as $T^3$ , the crossing point is insensitive to the prefactors and depends only weakly on $\tau_n$ and $g_*$ — why the prediction is robust.

Step 4. Read the frozen ratio. Evaluating (3) at $T_{\rm fr}$ , $n/p=e^{-1.293/0.8}\approx0.20$ , roughly $1/6$ . Why: this single number, carried through neutron decay, later determines the helium mass fraction almost by itself. *Check: $\Gamma/H\propto T^3$ ; setting $\Gamma=H$ gives $T_{\rm fr}\approx0.7$ – $0.8$ MeV; evaluating the Boltzmann factor there, $e^{-1.293/0.8}=0.199\approx1/6$ .*

Go deeper · Derivation B · 1 minThe primordial helium mass fraction as a neutron counter

Derivation BThe primordial helium mass fraction as a neutron counter

Having frozen the neutron-to-proton ratio, we now turn it into the observable helium mass fraction $Y_p$ . The goal is to follow the surviving neutrons through two stages — $\beta$ -decay during the bottleneck delay, then near-total capture into $^4$ He — and to show that because $^4$ He is a bin collecting every available neutron, the helium abundance is fixed by the neutron budget, nearly insensitive to the nuclear network or the baryon density.

Step 1. Start from freeze-out with $(n/p)_{\rm fr}\approx1/6$ . This is the neutron inventory to work with; nothing after this creates neutrons, it can only decay or bind them. Why: fixing the starting inventory separates the two fates of a neutron — decay or capture — into clean successive steps.

Step 2. Apply $\beta$ -decay. Between freeze-out ( $t\sim1$ s) and synthesis ( $t_{\rm nuc}\approx250$ s) free neutrons decay with lifetime $\tau_n\approx879$ s, multiplying the ratio by $e^{-t_{\rm nuc}/\tau_n}$ ; with capture biting before the full $250$ s, the ratio drifts from $\approx1/6$ to $\approx1/7$ . Why: this is the one place the neutron lifetime enters $Y_p$ , why primordial helium doubles as a cosmological measurement of $\tau_n$ .

Step 3. Bind the neutrons into helium. Once the bottleneck breaks, deuterium, tritium and $^3$ He funnel almost every surviving neutron into $^4$ He. With $n<p$ the neutrons are the limiting reagent, so the number of $^4$ He nuclei is $n/2$ . Why: helium being the tightest-bound light nucleus makes it the thermodynamic sink, so the flow runs there and stops, leaving only traces of the intermediates.

Step 4. Form the mass fraction. The helium mass is $4\times(n/2)=2n$ out of a total baryon mass $n+p$ , so

Y_p=\frac{2n}{n+p}=\frac{2(n/p)}{1+(n/p)}.

Substituting $n/p=1/7$ gives $Y_p=2/8=0.25$ ; detailed codes refine this to $\approx0.247$ . Why: the algebra shows $Y_p$ depends only on $n/p$ , not on any reaction cross-section — why the prediction is so clean and constrains new physics through $g_*$ . *Check: $Y_p=2(n/p)/(1+n/p)=0.25$ at $n/p=1/7$ ; the observed value is $Y_p=0.245\pm0.003$ .*

Go deeper · Derivation C · 1 minThe cosmic clock — time and temperature in the radiation era

Derivation CThe cosmic clock — time and temperature in the radiation era

Every BBN milestone, from neutron freeze-out to the breaking of the deuterium bottleneck, is quoted as both a temperature and a time. The goal is to build the function $t(T)$ converting between the two — taking the Friedmann equation in the radiation era, using the density scaling to fix how the scale factor grows, and integrating to get elapsed time as an explicit function of temperature with a usable coefficient.

Step 1. Friedmann with radiation. At MeV temperatures curvature and $\Lambda$ are negligible, leaving $H^2=(8\pi G/3)(\pi^2/30)\,g_*\,T^4$ , so $H\propto T^2$ . Why: dropping them is not convenience but a fact of the era — radiation density exceeds everything else by many orders of magnitude.

Step 2. Fix the growth of the scale factor. Radiation redshifts as $\rho\propto a^{-4}$ ; feeding that into $H^2\propto\rho$ gives $\dot a/a\propto a^{-2}$ , so $a\propto t^{1/2}$ and $H=1/(2t)$ . Why: the $a^{-4}$ scaling — one power beyond matter's $a^{-3}$ , from each photon redshifting — makes the expansion law $t^{1/2}$ rather than $t^{2/3}$ .

Step 3. Invert to get time. Combining $t=1/(2H)$ with $H(T)$ gives

t=\left(\frac{45}{16\pi^3 G\,g_*}\right)^{1/2}\frac{1}{T^2}\approx\frac{2.4}{\sqrt{g_*}}\left(\frac{k_BT}{\mathrm{MeV}}\right)^{-2}\ \mathrm{s}.

Why: with time going as $T^{-2}$ , halving the temperature quadruples the age — the universe spends most of the interval at its coolest, which is why synthesis, though it starts late, still finishes inside three minutes.

Step 4. Evaluate the landmarks. With $g_*=10.75$ , $T=1$ MeV gives $t\approx0.74$ s. At synthesis, $T\approx0.07$ MeV, the $e^\pm$ pairs have annihilated so $g_*\approx3.36$ , giving $t\approx250$ – $300$ s. Why: dropping $g_*$ after annihilation is essential — keeping $10.75$ would misdate synthesis by nearly a factor of two. *Check: the prefactor $2.4/\sqrt{g_*}$ s gives $t(1\,\mathrm{MeV})\approx0.74$ s and, with $g_*\approx3.36$ after $e^+e^-$ annihilation, $t(0.07\,\mathrm{MeV})\approx250$ – $300$ s.*

Limiting cases

BBN yields respond to a few physical knobs, each moving the abundances in a characteristic direction:

Case	Result	Meaning
Faster expansion (larger $g_*$ / extra $N_{\rm eff}$ )	$H\propto\sqrt{g_*}\,T^2$	more species raise $H$ , freezing $n/p$ earlier and higher, so $Y_p$ rises; pins $N_{\rm eff}\approx3$
Larger baryon density $\eta$	$\mathrm{D/H}\propto\eta^{-1.6}$	more baryons burn deuterium more completely: D/H and $^3$ He fall steeply, $Y_p$ creeps up
Longer neutron lifetime $\tau_n$	$Y_p\simeq Y_p^0(1+0.2\,\Delta\tau_n/\tau_n)$	slower $\beta$ -decay leaves more neutrons, raising $Y_p$ ; PDG $\tau_n\approx879$ s
Zeroth-order helium	$Y_p\approx\dfrac{2(n/p)}{1+(n/p)}\Big\|_{n/p\simeq1/7}\approx0.25$	with all neutrons in $^4$ He at $n/p\simeq1/7$ , $Y_p$ is nearly independent of $\eta$

Common trapConfusing the freeze-out temperature with the synthesis temperature

The single most common error is to imagine helium forms at its binding scale of several MeV. It does not. There are two very different temperatures: the $n/p$ ratio freezes near $T\simeq0.8$ MeV, but nucleosynthesis waits until $T\simeq0.07$ MeV — ten times cooler and a few hundred seconds later. The gap is the deuterium bottleneck: with a billion photons per baryon, the blackbody tail photo-dissociates deuterium long after it is bound, and every path to helium runs through deuterium. A subtler trap is to set $T_\nu=T_\gamma$ after $e^+e^-$ annihilation reheats the photons — you must carry $T_\nu/T_\gamma=(4/11)^{1/3}$ into $\rho_r$ and $g_*$ , and remember that $g_*$ drops from $10.75$ to $\approx3.36$ across the window, or the clock will misdate every milestone.

Worked examples

1 · From freeze-out to the helium abundance (fully worked)

Take $\Delta m c^2=1.293$ MeV, freeze-out temperature $T_{\rm fr}=0.8$ MeV, neutron lifetime $\tau_n=879$ s, and synthesis time $t_{\rm nuc}=250$ s. Predict the primordial helium mass fraction $Y_p$ .

Step 1. Frozen ratio. Evaluate the equilibrium factor at freeze-out: $(n/p)_{\rm fr}=e^{-1.293/0.8}=\mathbf{0.199}$ , roughly $1/6$ . This is the neutron inventory entering the delay.

Step 2. Neutron decay. The survival factor is $e^{-250/879}=\mathbf{0.75}$ , so the after-decay ratio is $0.199\times0.75=0.150$ . Network codes, in which capture bites before the full $250$ s, give an effective $(n/p)_{\rm BBN}\approx1/7\approx\mathbf{0.143}$ .

Step 3. Lock into helium. With neutrons the limiting reagent, $Y_p=\dfrac{2(0.143)}{1.143}=\mathbf{0.250}$ . Full BBN codes refine this to $Y_p\approx0.247$ , agreeing with the observed $Y_p=0.245\pm0.003$ from metal-poor HII regions. The prediction is $Y_p\approx0.25$ , set almost entirely by the frozen $n/p$ ratio.

2 · Timing the reactor with the cosmic clock

Using $t\approx(2.4/\sqrt{g_*})(T/\mathrm{MeV})^{-2}$ s, find the age at (a) freeze-out, $T=0.7$ MeV with $g_*=10.75$ , and (b) synthesis, $T=0.07$ MeV where $e^+e^-$ have annihilated so $g_*\approx3.36$ .

Step 1. Freeze-out. The prefactor is $2.4/\sqrt{10.75}=0.732$ s, so $t=0.732\times(0.7)^{-2}=\mathbf{1.5\ s}$ — the famous "first second."

Step 2. Synthesis. Now $g_*$ has dropped, so the prefactor is $2.4/\sqrt{3.36}=1.31$ s and $t=1.31\times(0.07)^{-2}=\mathbf{267\ s}$ . Keeping $g_*=10.75$ would have given $150$ s — misdating synthesis by nearly a factor of two.

Step 3. The window. Synthesis runs from $t\sim1$ s to $t\sim250$ – $300$ s — the celebrated "first three minutes." Freeze-out at $\approx1.5$ s, synthesis at $\approx270$ s: the reactor's whole life is under five minutes.

Physics of the portrait

BBN is where nuclear physics, particle physics and cosmology were first forced to speak one language — beginning with an audacious idea: that the chemical elements are fossils of a hot, dense beginning.

Origins · 1 minGeorge Gamow and how this was discovered

George Gamow

1904–1968 · hot Big Bang cosmology and primordial nucleosynthesis

George Gamow took the expanding universe of Friedmann and Lemaître and asked what it was made of. Trained under Friedmann himself, and already famous for explaining alpha decay by quantum tunneling in 1928, Gamow turned in the 1940s to the origin of the elements. His insight was that a hot, dense early universe would be a natural furnace: start with a sea of neutrons and build the periodic table by successive captures. In 1948 he and his student Ralph Alpher published the calculation, and Gamow — never able to resist a joke — added Hans Bethe's name so the author list read Alpher, Bethe, Gamow, the "αβγ" paper. Its flaw was real: with no stable nucleus at mass 5 or 8, neutron capture stalls at helium and cannot bridge to the heavy elements, which we now know are forged in stars. But for the light nuclei the furnace was exactly right — and Alpher and Robert Herman, working out the consequences, predicted that same year that the universe should still glow with relic radiation at a few kelvin, the cosmic microwave background found by accident seventeen years later. Hayashi corrected the neutron-proton budget in 1950; Peebles and, independently, Wagoner, Fowler and Hoyle turned it into a precision network calculation in the 1960s. What Gamow began as a bold guess is now our most stringent test of the first seconds.

History

1946

— Gamow proposes that the elements were cooked in a hot, dense early universe

1948

— Alpher, Bethe and Gamow publish the αβγ paper; Alpher and Herman predict relic radiation

1950

— Hayashi shows the weak interactions equilibrate the neutron-to-proton ratio

1966

— Peebles recomputes the primordial helium abundance for the hot Big Bang

1967

— Wagoner, Fowler and Hoyle build the full light-element reaction network

Going further

Texts. Misner, Thorne & Wheeler, Gravitation (1973), §28 · Wald, General Relativity (1984), Ch. 5 · Carroll, Spacetime and Geometry (2004), §8.4 · Hartle, Gravity (2003), Ch. 19 · Schutz, A First Course in General Relativity, 2nd ed. (2009), Ch. 12 · Kolb & Turner, The Early Universe (1990), Ch. 4 · Weinberg, Cosmology (2008), §3.2 · Particle Data Group, Review of Particle Physics — Big-Bang Nucleosynthesis. Key numerical anchors: $g_*=2+\tfrac{7}{8}(4+6)=10.75$ for $\gamma,e^\pm,3\nu$ , dropping to $\approx3.36$ after $e^+e^-$ annihilation; the clock gives $t\approx(2.4/\sqrt{g_*})$ s, so $t\approx0.74$ s at $T=1$ MeV; $(n/p)_{\rm eq}=e^{-\Delta m c^2/k_BT}$ with $\Delta m c^2=1.293$ MeV gives $\approx1/6$ at freeze-out; $Y_p=2(n/p)/(1+n/p)=0.25$ at $n/p=1/7$ , refined to $\approx0.247$ ; $T_\nu/T_\gamma=(4/11)^{1/3}=0.714$ ; $\eta=2.74\times10^{-8}\,\Omega_b h^2\approx6\times10^{-10}$ . Standard values: $Q_D=2.22$ MeV, $\tau_n\approx879$ s, $Y_p^{\rm obs}=0.245\pm0.003$ , $\mathrm{D/H}\approx2.5\times10^{-5}$ , $N_{\rm eff}\approx3$ .

Prerequisites. The Friedmann Equations · The Radiation-Dominated Era · The Cosmological Fluid and Equation of State · Cosmological Redshift · Critical Density and the Density Parameters. Leads to. The Matter-Dominated Era · The Cosmic Microwave Background.

Ask the tutor: Ask me to walk through how a fourth neutrino species would shift $Y_p$ , or to trace a single neutron from freeze-out into a helium nucleus.

Why does BBN produce almost no elements heavier than helium?

Because the ladder has two broken rungs. Building heavier nuclei must pass through mass numbers 5 and 8, but there is no stable nucleus at either: $^5$ He and $^5$ Li fall apart in $\sim10^{-21}$ s, $^8$ Be in $\sim10^{-16}$ s. In a star, densities and timescales let three helium nuclei fuse directly to carbon (triple-alpha), leaping the gap. In the early universe the density is far too low and the reactor cools out of business in minutes, so the flow piles up at $^4$ He with only a whisker of $^7$ Li across. Everything heavier had to wait for the first stars.

How can one number η fit abundances spanning nine orders of magnitude?

This is exactly what makes BBN so powerful a test. The four light nuclei respond to $\eta$ in different, calculable ways: $^4$ He barely moves (it is set by $n/p$ ), deuterium and $^3$ He fall steeply as more baryons burn them, and $^7$ Li traces a valley. If BBN were wrong there would be no single $\eta$ matching a helium fraction near $0.25$ , a deuterium abundance near $10^{-5}$ , and a lithium abundance near $10^{-10}$ at once. Instead one value, $\eta\approx6\times10^{-10}$ , threads them all (bar the lithium tension) — and the same value reappears in the CMB, making the result concordant, not merely consistent.

Is the freeze-out really sharp, or is "freeze-out temperature" a fiction?

It is an idealization, but a very good one. In reality the weak rates taper over a range of temperatures rather than switching off at a knife-edge. But the ratio $\Gamma/H\propto T^3$ drops so fast — a factor of eight per halving of $T$ — that the transition from equilibrium to frozen happens within a narrow band around $0.7$ – $0.8$ MeV. Treating it as a single temperature captures the physics to the accuracy of the toy model; the residual smoothness and the ongoing neutron decay are what full network codes integrate to sharpen $Y_p=0.25$ to $0.247$ .

Why does the helium abundance constrain the number of neutrino species?

Because helium is a stopwatch for the early expansion. Adding a light species — a fourth neutrino, or any "dark radiation" — raises $g_*$ , and since $H\propto\sqrt{g_*}\,T^2$ the universe expands faster at fixed temperature. Faster expansion means the weak $n/p$ conversions lose their race sooner, at a higher freeze-out temperature, freezing in more neutrons and hence more helium. Quantitatively, each extra thermalized neutrino species raises $Y_p$ by about $0.013$ . That is why BBN bounds $N_{\rm eff}$ to roughly $\pm0.3$ — a constraint on new physics at one second no accelerator can reach.

If deuterium is so fragile, why is any of it left over at all?

Because burning stops before it can finish. Once the bottleneck breaks at $T\simeq0.07$ MeV, deuterium is rapidly converted onward to $^3$ He, $^3$ H and then $^4$ He — but the universe keeps expanding and cooling. Below $T\sim0.03$ MeV the densities fall too low to keep the reactions going, and whatever deuterium is unprocessed is frozen in as a relic. The leftover thus measures how efficiently burning ran, set by the baryon density — the more baryons, the more completely deuterium is consumed, hence $\mathrm{D/H}\propto\eta^{-1.6}$ .

What exactly is the lithium problem, and is it a crisis?

Standard BBN, fed the baryon density the CMB and deuterium both prefer, predicts a primordial $^7$ Li abundance near $5\times10^{-10}$ . But lithium in the oldest, most metal-poor stars — the "Spite plateau," thought to preserve primordial gas — sits at only $\approx1.6\times10^{-10}$ , roughly three times too little. It is not a crisis for the hot Big Bang, since helium and deuterium agree beautifully; it is a localized tension. The leading suspects are astrophysical (lithium slowly sinking below the observable surface layers of these ancient stars) and nuclear (an underappreciated $^7$ Be-destruction rate), with new physics a minority hope.