The proton–proton chain and stellar fusion

Write the three links of pp-I with the interaction that drives each. p + p → d + e⁺ + ν is weak: a proton becomes a neutron while the pair sits inside the barrier. d + p → ³He + γ is electromagnetic. ³He + ³He → ⁴He + 2p is strong. The first two links face the same barrier, Z₁Z₂ = 1, so the barrier is not what separates them; the S-factors are. S₁₁ = 4.0 × 10⁻²² keV b, S₁₂ = 2.1 × 10⁻⁴ keV b, S₃₃ = 5.2 × 10³ keV b: eighteen and then seven more orders of magnitude. The third link gives much of its advantage back: Z₁Z₂ = 4 and three times the reduced mass make E_G ≈ 24 MeV, so τ ≈ 49 at 15.7 MK against 13.5 for pp, and ³He is scarce. At the solar centre that makes a proton wait about 7 × 10⁹ yr for a partner, a deuteron about 2 s, and a ³He nucleus of order 10⁵ yr. The slowest link is the rate-limiting step, so the solar luminosity is a weak-interaction rate, and deuterium and ³He settle to steady-state abundances at which each is destroyed as fast as it is made. Everything downstream of the first step is fast bookkeeping; everything about the chain's pace is in S₁₁.

Mass tables list atomic masses, each carrying Z electrons. For p + p → d + e⁺ + ν the nuclear balance is 2mₚ − m_d − mₑ, while 2M(¹H) − M(²H) = 2mₚ + 2mₑ − m_d − mₑ = 2mₚ − m_d + mₑ. The difference is 2mₑc² = 1.022 MeV: one electron because the tables gave you one too many, one because the positron is new mass. So Qₚₚ = 1.442 − 1.022 = 0.420 MeV, shared by the positron and the neutrino, whose spectrum ends at 0.420 MeV and averages 0.265 MeV. The rare capture route p + e⁻ + p → d + ν has no positron and no spare electron, so its Q is the full 1.442 MeV, the monoenergetic pep line. Now do the whole chain: 4M(¹H) − M(⁴He) = 0.02870 u = 26.73 MeV, and here nothing is subtracted, because the two positrons annihilate with two of the four electrons and their 2 × 1.022 MeV comes back as photons. The only energy that leaves the star is neutrino energy: 0.53 MeV for pp-I, 1.08 for pp-II (mostly the 0.862 MeV ⁷Be line), 7.0 for pp-III (the ⁸B spectrum) and 1.70 for CNO.

At the Gamow peak, E₀ ≈ 6 keV, σₚₚ = (S/E₀) exp(−√(E_G/E₀)) = (4.0 × 10⁻²² keV b / 6.1 keV) × e^(−9.0) ≈ 8 × 10⁻²⁷ b ≈ 10⁻⁵⁰ cm². Even at 1 MeV, where the barrier factor has risen to 0.5, S(0) gives only 2 × 10⁻²⁵ b: a milliamp of protons stopping in a thick hydrogen target would make one deuteron per 10³ to 10⁴ years, far too few to count against any beam-induced background. So S₁₁(0) comes from theory. Two low-energy protons in a relative s-wave must be spin singlet, ¹S₀ with T = 1, by the Pauli principle; the deuteron is ³S₁ with T = 0. The Fermi operator Σ τ₊ changes neither S nor T, so the transition is pure Gamow–Teller, driven by Σ σ τ₊ with coupling g_A = 1.27, and its matrix element is the overlap Λ of the pp scattering wavefunction with the deuteron wavefunction, Λ² ≈ 7.0. Normalising to the superallowed 0⁺ → 0⁺ ft value of 3071 s removes G_F and the radiative corrections, and the phase-space integral fₚₚ = 0.144 fixes the lepton kinematics. The result, S₁₁(0) = 4.01 × 10⁻²² keV b to about 1%, is the best-known number in solar fusion precisely because it is the one nobody can measure.

The previous topic's Laplace expansion of the Maxwell-averaged tunnelling factor supplies the tools: E₀ = (E_G (kT)²/4)¹⁄³, τ = 3E₀/kT and d ln⟨σv⟩/d ln T = (τ − 2)/3, the −2/3 coming from Δ(kT)^(−3/2) ∝ T^(−2/3). The work here is to apply them to the two bottlenecks of hydrogen burning. At 15.7 MK, kT = 1.353 keV. For pp, E_G = 493 keV, so E₀ = (493 × 1.830/4)¹⁄³ = 6.09 keV, four and a half times kT, τ = 13.5 and n = (13.5 − 2)/3 = 3.8. For p + ¹⁴N, with Z₁Z₂ = 7 and μc² = 875 MeV, E_G = 45.1 MeV, E₀ = 27.4 keV, τ = 60.8 and n = 19.6. Because τ ∝ E_G¹⁄³, a Gamow energy ninety times larger makes τ 4.5 times larger and n five times larger: the same thermal average, a far steeper law. The exponents are local: at 25 MK they have fallen to 3.2 and 16.7, which is why textbooks quote T⁴ against anything from T¹⁶ to T²⁰.

Once ³He exists it can meet another ³He, closing pp-I with Q = 12.86 MeV, or meet the ⁴He already present to make ⁷Be. ⁷Be captures an electron to ⁷Li, which takes a proton and splits into two ⁴He (pp-II), or, about once in a thousand times, captures a proton to make ⁸B, whose β⁺ decay gives the 6.7 MeV neutrinos (pp-III). In the Sun the terminations run near 85%, 15% and 0.02%, the split set by S₃₃ = 5.2 MeV b against S₃₄ = 0.56 keV b times the ⁴He abundance. Carbon and nitrogen catalyse the same net reaction: ¹²C(p, γ)¹³N(e⁺ν)¹³C(p, γ)¹⁴N(p, γ)¹⁵O(e⁺ν)¹⁵N(p, α)¹²C. Its slowest link is ¹⁴N(p, γ)¹⁵O with S(0) = 1.66 keV b, so in equilibrium nearly all the catalyst waits as ¹⁴N and ε_CNO ∝ X X_CNO. With X = 0.35 and X_CNO = 0.004 the two rates match near 19 MK. At the solar centre this model gives CNO about 5% of the local heat; averaged over the cooler burning core the share falls to about 1% of the luminosity, the value Borexino measured in 2020, while stars above roughly 1.3 M☉ sit on the CNO side. A T¹⁸ law packs the burning into a tiny central sphere, so the radiative gradient there exceeds the adiabatic one and those stars carry convective cores the Sun lacks.

Two assumptions sit under every number here. The Maxwellian average needs a non-degenerate gas: at the solar centre nₑ ≈ 6 × 10²⁵ cm⁻³ gives an electron Fermi energy near 0.6 keV against kT = 1.35 keV, so the electrons are only mildly degenerate and the ions not at all. Hydrostatic equilibrium is what makes a T⁴ or T¹⁸ law safe. The virial theorem gives a self-gravitating ideal gas a negative heat capacity: if ε rises, the core expands, cools as it works against gravity, and the rate falls back. The exponent sets how tightly the temperature is pinned, not whether the star survives. Remove either assumption and the thermostat is gone: in a red giant's degenerate helium core the pressure ignores the temperature, so the onset of helium burning runs away as the helium flash until the degeneracy lifts. Same Gamow physics, no restoring force. The Sun's lifetime is the same energy budget seen from outside: L☉/M☉ = 1.9 erg g⁻¹ s⁻¹ against 26.2 MeV per four protons, with a tenth of the star's hydrogen (X = 0.7) available, gives about 7 × 10⁹ years. That number is not set by S₁₁: L☉ is fixed by how fast radiation diffuses out, and the thermostat lets the core settle at whatever temperature makes the weak rate supply it, so a larger S₁₁ would buy a cooler core, not a proportionally shorter life.