research32 min studyLesson 599 in Physics

The Big Bang light-element network

Connect weak freezeout, neutron survival, and the deuterium bottleneck while preserving baryon accounting during neutron decay.

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What you’ll understand

  • Separate weak neutron–proton freezeout from the later onset of efficient light-nucleus assembly.
  • Estimate helium mass fraction from a stipulated neutron inventory.
  • Explain how expansion, photon abundance, and nuclear rates limit a simplified early-universe network.

Early-universe nucleosynthesis is a race between reactions and cosmic expansion. Weak processes first influence the neutron-to-proton ratio; surviving neutrons are later assembled into light nuclei after deuterium can persist. These stages occur under changing temperature and density, so a static equilibrium calculation is insufficient. A useful hand estimate of helium production comes from neutron counting, but that estimate must include the protons created when free neutrons decay and must state which lighter residual products are neglected.

Weak conversion and expansion set the initial inventory

At high temperature, neutron–proton weak interconversion can be rapid enough to track an equilibrium ratio related to their mass-energy difference. The simplest expression assumes small relevant chemical potentials and thermal particle distributions. As expansion and cooling reduce weak rates, the ratio departs gradually from equilibrium. The often used instantaneous freezeout estimate is a simplifying approximation to this evolving kinetic competition; it does not mean every weak event suddenly ceases at one exact temperature.

After weak interconversion becomes ineffective, free-neutron decay continues while the universe cools toward efficient nuclear assembly. Each neutron that decays becomes a proton while emitting leptons, conserving total baryon number. Thus both the numerator and denominator of the neutron-to-proton ratio change. Multiplying the ratio alone by a survival factor neglects the accompanying growth of the proton population. This can be acceptable in a stated approximation when the neutron fraction is small, but exact finite-inventory bookkeeping is straightforward and safer for the exercise.

(np)eq≃e−Δmc2/(kBT),Nn(t)=Nn0e−t/τn,Np(t)=Np0+Nn0(1−e−t/τn)\left(\frac np\right)_{\rm eq}\simeq e^{-\Delta m c^2/(k_BT)},\qquad N_n(t)=N_{n0}e^{-t/\tau_n},\qquad N_p(t)=N_{p0}+N_{n0}\left(1-e^{-t/\tau_n}\right)
The equilibrium ratio assumes thermal weak equilibrium with negligible relevant chemical potentials. The decay equations apply only to a later toy interval with free-neutron decay as the sole process changing the baryon species; tau_n is the mean neutron lifetime.↔ Scroll wide equations horizontally.

Why deuterium waits below its binding-energy scale

A neutron and a proton can form deuterium by radiative capture, but abundant photons can break it apart. The baryon-to-photon ratio is small, so even an exponentially rare energetic photon population can remain numerous compared with the nuclei being assembled. The deuterium bottleneck therefore persists until temperature is well below the deuterium binding-energy scale. A logarithmic tail estimate makes this qualitative point but does not supply the exact reaction-rate crossing or the full abundance history.

Once deuterium survives sufficiently, captures and rearrangements build helium three, tritium, and helium four. Most surviving neutrons can become locked into tightly bound helium four under the standard light-network conditions. Residual deuterium and helium three depend on incomplete burning as expansion dilutes the plasma. The absence of stable mass-five and mass-eight stepping stones, together with the short declining-density interval, suppresses sustained assembly of much heavier nuclei; the dense, long-lived helium-burning route of a star is unavailable here.

η−1e−BD/(kBT)∼1⟹kBTD∼BDln⁡(η−1)\eta^{-1}e^{-B_D/(k_BT)}\sim1\quad\Longrightarrow\quad k_BT_D\sim\frac{B_D}{\ln(\eta^{-1})}
Eta is the dimensionless baryon-to-photon ratio and BD is the deuterium binding energy. This tail-counting scale neglects kinetic prefactors, detailed photon integration, and nuclear depletion, so it is an onset estimate rather than an exact bottleneck temperature.↔ Scroll wide equations horizontally.

An abundance estimate is not a full precision prediction

If all surviving neutrons enter helium four and other bound products are neglected, each alpha requires two neutrons and two protons. Dividing the baryons in helium by the conserved total gives a simple approximate helium mass fraction. The approximation uses mass numbers rather than exact nuclear masses and assumes enough protons remain. It explains why neutron survival strongly affects helium, while it cannot predict residual deuterium or lithium from one scalar ratio alone.

A precision network must evolve nuclear and weak rates, expansion, thermodynamic distributions, and the baryon density consistently. Observations require additional astrophysical interpretation because later stellar processing can destroy or produce light isotopes. Discrepancies in an isotope such as lithium should therefore be framed as a comparison between specified calculations and inferred primordial abundances, with their uncertainties; the hand-counting model below cannot diagnose a unique explanation. Its purpose is to expose conservation and timing before those more detailed comparisons are attempted.

X4≃4NαNb=2NnNn+Np=2r1+r,r=NnNpX_4\simeq\frac{4N_\alpha}{N_b}=\frac{2N_n}{N_n+N_p}=\frac{2r}{1+r},\qquad r=\frac{N_n}{N_p}
This approximate helium mass fraction assumes all available neutrons enter alpha particles, enough protons are present, and other bound species and binding corrections in the mass normalization are negligible. X4 is not the number abundance of helium nuclei.↔ Scroll wide equations horizontally.

Neutrons lost to decay become protons

A toy comoving parcel starts a decay-only interval with 300 free neutrons and 1500 free protons. One third of the neutrons decay before assembly. Then put every surviving neutron into helium four, neglecting other nuclei. Find the final alpha count, leftover free protons, and approximate helium mass fraction.

  1. One hundred neutrons decay, leaving 200 neutrons and creating 100 protons. The proton count becomes 1600, while total baryon count stays 300+1500=1800300+1500=1800.

  2. Two hundred surviving neutrons form 100 alpha particles and consume 200 protons. The leftover free-proton count is therefore 1600−200=14001600-200=1400. This step checks that enough protons are available.

  3. The helium nuclei contain 4(100)=4004(100)=400 baryons. Dividing by the conserved 1800 gives X4=400/1800=2/9≃0.2222X_4=400/1800=2/9\simeq0.2222. Equivalently, the preassembly ratio is 200/1600=1/8200/1600=1/8, which gives the same result in the ratio formula.

The result

The parcel makes 100 alpha particles and retains 1400 free protons; the approximate helium mass fraction is two ninths.

Explore the model

Interactive lab

Interactive lab · markov

A one-way species conversion conserves total probability

Use state zero only as an analogy for an initially tagged neutron and state one for its converted proton. Reset and set p to 0.2 and q to zero. After two Step actions predict probabilities 0.64 and 0.36. This discrete chain conserves the sum while transferring population. It does not model weak freezeout, a measured neutron lifetime, expansion, nuclear assembly, or an early-universe reaction network.

A one-way species conversion conserves total probabilityExact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Current readings: P(state 1) at step 0: 0%; Stationary P(state 1): 60%; Chain behavior: Monotone convergence. All parameters are adjustable with the labeled controls below.Two-state Markov distribution after 0 steps: state zero probability 1, state one probability 0.Initial distribution (1, 0) · step 0p = 0.3q = 0.201100%0%1−p1−qP(0)P(1)Monotone convergence
P(state 0)P(state 1)Stationary distribution
P(state 1) at step 0
0%
Stationary P(state 1)
60%
Chain behavior
Monotone convergence
t = 0 s

Exact finite-step probabilities for a two-state chain starting in state 0, not sampled trajectories. Playback advances two transitions per second. When p = q = 0 every distribution is stationary; when p = q = 1 the chain alternates forever. A single zero transition creates an absorbing state. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA decayed neutron leaves the neutron inventory and enters the proton inventory.

Check your understanding

In a decay-only interval, which bookkeeping rule is correct for neutron-to-proton evolution?

Try it yourself

Use the illustrative tail estimate with BD=2.2 MeVB_D=2.2\,\mathrm{MeV} and first η=10−9\eta=10^{-9}, then η=10−8\eta=10^{-8}. Taking ln⁡10≃2.3\ln10\simeq2.3, estimate both thermal onset scales and explain the direction of the change.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The logarithms are approximately 20.7 and 18.4. A larger baryon-to-photon ratio means fewer photons per baryon.

Compare with the solution

The estimates are kBTD≃2.2/20.7=0.106 MeVk_BT_D\simeq2.2/20.7=0.106\,\mathrm{MeV} and 2.2/18.4=0.120 MeV2.2/18.4=0.120\,\mathrm{MeV}. Increasing eta lets deuterium survive at a higher approximate temperature because there are fewer destructive high-energy photons per baryon. These values illustrate the logarithmic dependence; an actual onset and final abundance require the time-dependent reaction and expansion calculation.

Take it with you

  • Weak freezeout and deuterium survival are separate stages of an expanding reaction network.
  • Neutron decay preserves baryon number by adding protons while removing neutrons.
  • The helium inventory estimate is useful but does not replace a precision light-element network.

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