advanced20 min studyLesson 857 in Physics

Nucleosynthesis timing and baryon density

Connect weak conversion, neutron survival, and the deuterium bottleneck to light-element abundances.

Before you begin

On this page

The question for today

What you’ll understand

Explain why weak freezeout precedes efficient nuclear assembly.
Track proton production when free neutrons decay.
Separate simple helium bookkeeping from a quantitative nuclear network.

Big-bang nucleosynthesis connects microscopic weak and nuclear reactions to an expanding radiation background. Its outcome depends on the neutron supply, the delay before deuterium survives, and the rates that process light nuclei after that delay. An approximate helium fraction can be computed by counting surviving neutrons, but precise deuterium and helium predictions require a coupled network, measured nuclear rates, and a specified cosmology. We use natural units in thermal formulas and explicitly restored seconds for a neutron-decay illustration. Illustrative abundances below are calculations under stated inputs, rather than new measurements or a claim that every observed element matches standard predictions.

Weak conversion fixes a moving neutron target

When electron and neutrino weak reactions are rapid, they interconvert neutrons and protons. With negligible relevant chemical potentials and nonrelativistic nucleons, the equilibrium neutron-to-proton ratio is approximately an exponential of the neutron-proton mass difference divided by temperature. Neutrons are heavier, so cooling favors protons. The weak conversion rate falls faster than the radiation-era expansion rate, causing the actual ratio to depart gradually from its equilibrium target near the MeV scale.

A sharp freezeout temperature is only an estimate: residual weak conversions, neutrino spectral changes, and neutron decay all affect the ratio afterward. Extra relativistic energy raises the expansion rate at fixed photon temperature and can shift the available time for conversion and decay. Lepton asymmetries can alter the weak equilibrium relation itself. These distinct effects explain why nucleosynthesis constrains more than a single nuclear binding energy or a single expansion parameter.

r_{\rm eq}=\frac{n_n}{n_p}\simeq e^{-Q/T},\qquad Q=m_n-m_p\simeq1.293\,\mathrm{MeV},\qquad \Gamma_{n\leftrightarrow p}\sim G_F^2T^5,\quad H\propto\sqrt{g_*}\,T^2

The equilibrium ratio neglects chemical-potential and small nucleon-mass prefactor corrections. The rate expressions indicate scaling, not a precision freezeout calculation;

g_*

is the radiation energy count.↔ Scroll wide equations horizontally.

Deuterium must survive before helium can assemble

The first essential bridge is neutron capture on a proton to form deuterium. Its binding energy is about $2.22\,\mathrm{MeV}$ , but efficient survival begins far below that temperature. Photons greatly outnumber baryons, so even an exponentially small high-energy tail can photodissociate newly made deuterium. The rough number of above-threshold photons per baryon scales as $\eta^{-1}e^{-B_D/T}$ , with $\eta$ the baryon-to-photon ratio and $B_D$ the deuteron binding energy.

Once the bottleneck opens near a tenth of an MeV in standard estimates, reactions can rapidly move most surviving neutrons into helium-4. The delay leaves free neutrons time to decay. Each neutron decay produces a proton, so reducing the neutron numerator while holding the proton denominator fixed gives the wrong later ratio. Baryon number remains conserved; it is redistributed among nucleons and then among nuclei. Higher nuclei are limited by falling density and gaps in the stable light-nucleus chain.

N_n=fN_{n,i},\qquad N_p=N_{p,i}+(1-f)N_{n,i},\qquad f=e^{-\Delta t/\tau_n},\qquad Y_p\simeq\frac{2N_n}{N_n+N_p}=\frac{2r}{1+r}

Here

N_n,N_p

are comoving nucleon counts before assembly, and

Y_p

is the approximate helium-4 mass fraction if nearly all neutrons enter helium. Binding-energy and nucleon-mass differences are neglected in this bookkeeping formula.↔ Scroll wide equations horizontally.

Different elements probe different parts of the network

Increasing baryon density raises nuclear encounter rates at a given temperature. More efficient processing generally burns residual deuterium into heavier light nuclei, making deuterium a sensitive baryometer in the standard network. The helium mass fraction depends strongly on the available neutrons and more weakly on baryon density. These trends should not be confused with exact power laws valid for arbitrary densities or nonstandard expansion histories; a full calculation follows all important forward and inverse reactions.

Compare theoretical abundances with observations that attempt to recover primordial composition rather than material substantially altered by stars. Nuclear-rate uncertainties and astrophysical abundance systematics are different sources of error. The Particle Data Group's 2024 nucleosynthesis review discusses the broadly successful helium and deuterium framework alongside the unresolved primordial lithium discrepancy. Agreement in selected light elements supports a hot thermal history, while the lithium tension prevents an honest summary from saying that all standard abundances are observationally settled.

Work it through

Decay changes both sides of the neutron-proton ratio

At a chosen weak-departure point, normalized comoving abundances are $N_{n,i}=1$ and $N_{p,i}=5$ . Before nuclear assembly, three quarters of the neutrons survive; ignore further interconversion. Then put every remaining neutron into helium-4. Find the new neutron-to-proton ratio, the approximate helium mass fraction, and the elapsed time if a stipulated lifetime is $\tau_n=880\,\mathrm{s}$ .

The surviving neutron abundance is $3/4$ , and the decayed abundance is $1/4$ . Add the decay-produced protons to obtain $N_p=5+1/4=21/4$ . The total nucleon abundance remains six in the chosen normalization.
Divide the new abundances to obtain $r=(3/4)/(21/4)=1/7$ . Each helium nucleus contains two neutrons and four nucleons, so the approximate mass fraction is $2(3/4)/6=1/4$ . This assumes enough protons and complete incorporation of the remaining neutrons.
Invert the survival law: $\Delta t=\tau_n\ln(4/3)\simeq253.2\,\mathrm{s}$ . This timespan and the initial abundances were stipulated; they do not independently derive a realistic freezeout temperature or nuclear reaction history.

The result

The pre-assembly ratio becomes $1/7$ , the idealized helium mass fraction is $0.25$ , and the decay interval is about $253\,\mathrm{s}$ . Keeping the initial proton abundance fixed would have violated the modeled decay bookkeeping.

The living notebook

Explore the model

Interactive lab

Interactive lab · probability

An independent-survival analogy

Reset and set both controls to success probability $p=0.75$ , trials $n=4$ : the exact binomial mean is $3$ and variance $0.75$ . Reset and set both controls to $p=0.5$ , $n=4$ : the mean is $2$ and variance $1$ . Interpret success only as independent survival in a toy set. Bars are exact probabilities, not resampled neutron measurements; the model has no proton conversion, weak equilibrium, or nuclear reaction network.

Exact probabilityHighlighted outcomeMean np

Mean · np: 5
Variance · np(1−p): 2.5
Total probability: 1

Success probability p0.5Trials n10

t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 survival probability controls expected remaining particles, but nuclear abundances also require conservation and subsequent reactions.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The relevant comparison includes the photon-to-baryon ratio and the energetic tail, not just a typical photon energy against one binding energy.

Your turn

Try it yourself

Use a stipulated illustrative fit $\mathrm{D/H}=2.5\times10^{-5}(\eta_{10}/6)^{-1.6}$ only for this exercise, where $\eta_{10}=10^{10}\eta$ . Compare the predicted ratio at $\eta_{10}=3$ with its value at six. Take $2^{1.6}\simeq3.03$ . 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

Halving the argument of a negative power increases the result; distinguish surviving deuterium from total nuclear processing.

Compare with the solution

At six the exercise gives $2.5\times10^{-5}$ . At three it gives $2.5\times10^{-5}\,2^{1.6}\simeq7.58\times10^{-5}$ , about three times larger. Lower baryon density makes subsequent deuterium destruction less effective, leaving more residual deuterium. This supplied local power law is a teaching approximation, not a new abundance observation or a replacement for a nuclear network with specified reaction rates.

Keep the idea

Take it with you

Weak conversion, neutron survival, and nuclear assembly occur over distinct intervals.
A neutron decay increases the proton count while preserving baryon number.
Light-element agreement and discrepancies must be evaluated with nuclear and astrophysical uncertainties.

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