Cosmology · Primordial nuclei and light-element tests

Neutrons, protons and the competition of rates

Separate weak interconversion from later free-neutron decay, and keep the daughter protons in a conserving comoving nucleon account.

Preparation: ratios, powers of ten and ordinary arithmetic. Physical quantities and notation are introduced as they are needed.

Back to the chapter

After reading, you can

Explain how interconversion, expansion, decay and nuclear capture affect neutron and proton populations.
Calculate a supplied rate comparison without treating it as a complete abundance solution.
Track both neutron loss and proton production in an isolated free-decay interval.

The neutron-to-proton balance is dynamical

Early light-element formation depends on how many neutrons and protons are available and which reactions can bind them into nuclei. Neutrons and protons can interconvert through weak reactions involving leptons, while free neutrons can also undergo beta decay.

Examples include n + ν_e ↔ p + e⁻ and n + e⁺ ↔ p + anti-ν_e. Free-neutron beta decay produces a proton, an electron and an electron antineutrino. A neutron disappearing from the free-neutron column does not mean its nucleon has vanished.

Under suitable equilibrium and small-chemical-potential approximations, an approximate neutron/proton ratio contains exp[−(m_n − m_p)c²/(k_B T)]. The heavier neutron is increasingly disfavoured as thermal energy falls. This simplified target ratio is not automatically the actual evolving ratio.

Actual populations depend on reaction rates, expansion, decay, nuclear capture and their initial conditions. A rate equality or one equilibrium expression cannot replace the coupled evolution needed for a nucleosynthesis prediction.

A temperature scaling gives a diagnostic

To isolate a timescale comparison, put θ = (k_B T)/(k_B T_star), where T_star is a supplied reference temperature with no physical numerical value assigned. Let H_star > 0 have units of inverse time.

Over 1/2 ≤ θ ≤ 2, choose Γ = H_star θ⁵ and H = H_star θ², with constant effective radiation degrees of freedom. These powers illustrate the approximate weak-interaction and radiation-expansion trends in a restricted regime; no absolute weak coefficient is calculated.

Their ratio is Γ/H = θ³: it is 8 at θ = 2, 1 at θ = 1 and 1/8 at θ = 1/2. Cooling makes this supplied interconversion rate less effective relative to expansion.

Equality at θ = 1 marks comparable timescales in this chosen account. It supplies neither an exact freeze-out surface nor a final neutron/proton ratio. A finite-rate calculation can depart gradually from the equilibrium target.

Keep a mechanism example separate from a joined thermal history

Now consider a separate isolated interval after interconversion and nuclear capture have been switched off by assumption. Start with comoving counts, or proportional expected counts, N_n = 24 and N_p = 96. This is not claimed as the endpoint of the preceding rate panel.

Let τ be a supplied mean free-neutron lifetime used as the time unit and set u = Δt/τ. Free decay gives N_n = 24e^(−u). Every decay adds a proton, so N_p = 96 + 24[1 − e^(−u)] = 120 − N_n.

The common nucleon count N_n + N_p stays 120. Comoving counts remove the shared expansion dilution; the lepton products are not represented by these two nucleon columns. Their charge and energy accounts have not been replaced by a nucleon-only energy conservation claim.

The ratio is N_n/N_p, with the changing proton denominator included. Multiplying the initial ratio by e^(−u) while holding its denominator fixed would miss the daughter protons.

A separate chosen free-decay interval with no inverse interconversion or nuclear capture; τ is a mean lifetime, not a measured value
u = Δt/τ	N_n	N_p	Total nucleons	N_n/N_p
0	24	96	120	1/4
ln(4/3) ≈ 0.287682	18	102	120	3/17 ≈ 0.176471
ln 2 ≈ 0.693147	12	108	120	1/9 ≈ 0.111111

A lifetime definition changes the calculation

After one mean lifetime τ, a freely decaying expected population retains e⁻¹ of its initial count. One half remains after the half-life τ ln 2. These two time definitions should not be interchanged.

For the supplied initial counts, the half-life stage leaves 12 neutrons and produces 12 additional protons, giving 108 protons. The neutron/proton ratio becomes 1/9, rather than half of the original 1/4.

The smooth decay law describes an expected count or a large population scaled to these units. It does not require that an individual finite set of 24 actual neutrons decays in exact deterministic fractions.

Once nuclear capture becomes important, free-neutron decay alone is no longer the complete account. Bound-nucleus reaction and decay channels need their own equations; the free interval deliberately excludes them.

Two separate supplied mechanisms are shown; they are not successive panels of a solved thermal history. Panel A uses the dimensionless temperature ratio θ=T/T_star from 1/2 to 2, Γ=H_star θ^5 and H=H_star θ² with a common unspecified positive inverse-time H_star and constant effective radiation degrees of freedom. Its linear θ axis and logarithmic base-two Γ/H axis show θ³: 1/8, 1, 8 at θ=1/2, 1, 2. Equality is a timescale comparison, not an exact freeze-out surface or abundance. Cooling goes toward smaller θ. Panel B is an independent isolated free-neutron decay interval after inverse interconversion and nuclear capture are switched off. Its horizontal axis is u=Δt/τ from 0 to ln 2, where τ is a supplied mean lifetime rather than a measured value or half-life. Its linear vertical axis gives comoving or proportional expected counts. Solid blue circles show N_n=24exp(−u); dashed orange squares show N_p=120−N_n; dotted green diamonds show their constant total 120. At u=0 the counts are 24 and 96 with n/p=1/4. At u=ln(4/3) they are 18 and 102 with ratio 3/17. At u=ln 2 they are 12 and 108 with ratio 1/9. Daughter protons must be included rather than freezing the denominator. One half remains at τ ln 2, while exp(−1) remains after one mean lifetime τ; the latter time lies beyond the displayed interval. The smooth curves refer to expected/scaled counts, not deterministic decay times of 24 individual particles. Comoving counts remove common expansion dilution. The lepton products and their energy are not represented, and no nucleon-only energy-closure claim is made. No actual neutron lifetime, numerical reference temperature, measured rate or precision BBN yield is supplied. Original Ark teaching model.

An initial balance is one input to a reaction network

Neutrons can later be incorporated into light nuclei. The timing of that incorporation depends on whether intermediates form and survive, on baryon densities and reaction coefficients, and on expansion and reverse reactions.

A full calculation therefore combines weak interconversion, neutron decay and a connected nuclear network. Changes in a rate or an expansion history can propagate into several abundances rather than adjusting only one count.

The two chosen mechanism panels establish useful checks: rate ratios need domains, and a decay account must retain the daughter particles and its time convention. They do not supply an observed neutron lifetime, a measured freeze-out temperature or a precision primordial yield.

The next lesson examines why a bound deuterium nucleus can be repeatedly broken apart by photons even when typical thermal energy is below its binding scale. That bottleneck helps connect the nucleon balance to the later nuclear sequence.

Worked reasoning

Follow the daughter protons

Starting from 24 neutrons and 96 protons, what happens after Δt = τ ln 2 if only free beta decay is allowed?

The expected surviving neutron count is 24e^(−ln 2) = 12.
The 12 decays add 12 protons, so N_p = 96 + 12 = 108.
Total nucleons remain 120 and N_n/N_p = 12/108 = 1/9.

12 neutrons, 108 protons and neutron/proton ratio 1/9 in the supplied expected-count account.

Try the reasoning

Does Γ/H = 1 alone determine an exact final neutron/proton abundance?
Show the answer and reasoning
No.
It compares timescales; the actual coupled population evolution still needs to be solved.
After one mean lifetime, is exactly one half of a freely decaying expected population left?
Show the answer and reasoning
No; the remaining fraction is e⁻¹.
One half remains after τ ln 2.
May the proton count remain fixed at 96 in the supplied isolated decay interval?
Show the answer and reasoning
No.
Each neutron beta decay produces a proton, so the proton count rises as the neutron count falls.

What to carry forward

An equilibrium target, a rate crossing and an evolved abundance are different results.
A free-neutron decay account must include the daughter protons.
Mean lifetime and half-life give different survival fractions.

References

Sources supporting the explanations, physical distinctions and examples in this lesson.