Fusion and the Gamow peak

Two protons feel the strong force only when their centres are a couple of femtometres apart, and getting there costs Coulomb energy: with e²/4πε₀ = 1.44 MeV⋅fm, pushing them to 2.0 fm costs 0.72 MeV. The Sun's core sits at 1.5×10⁷ K, so kT = 1.29 keV, and the barrier is 557 kT high. A classical activation estimate — the fraction of pairs with enough energy to go over — gives e⁽⁻⁵⁵⁷⁾ ≈ 10⁽⁻²⁴²⁾. Multiply by every proton–proton collision the Sun has had in 4.6 billion years, of order 10⁸², and you still predict 10⁽⁻¹⁶⁰⁾ reactions. The Sun shines, so this is not a picture that needs a correction factor; it is the wrong picture. Whatever supplies the rate must send particles through the barrier, not over it.

Run WKB across the Coulomb barrier: the transmission is exp(−(2/ħ)∫√(2μ(V−E)) dr) between the classical turning points. For E well below the barrier top that integral collapses to 2πη, with η = Z₁Z₂αc/v the Sommerfeld parameter. Substituting E = ½μv² converts it to exp(−b/√E), where b = √E_G and E_G = 2μc²(παZ₁Z₂)². Everything about the pair — charges and reduced mass — is compressed into b, which scales as Z₁Z₂√μ: 22.2 keV^½ for p+p, 25.6 for p+d, 34.4 for D+T, 212 for p+¹⁴N. The function is savage at low energy. At E = 1.29 keV it is 3.3×10⁻⁹; at 5.9 keV, 1.1×10⁻⁴; at 30 keV, 1.7×10⁻². Merely doubling the energy from 5.9 to 11.8 keV multiplies it by about 15.

Multiply the population factor e^(−E/kT), which collapses as E rises, by the penetration factor e^(−b/√E), which explodes. The combined exponent is f(E) = E/kT + b/√E; setting df/dE = 1/kT − b/(2E³⁄²) to zero gives the Gamow peak at E₀ = (b⋅kT/2)²⁄³ = (E_G(kT)²/4)¹⁄³. For p+p in the Sun that is (22.2 × 1.29/2)²⁄³ = 5.9 keV — 4.6 kT, and still 122 times below the barrier. Expanding f about E₀ gives a Gaussian of full 1/e width Δ = 4√(E₀kT/3) = 6.4 keV, with peak value e^(−τ) where τ = f(E₀) = 3E₀/kT = 13.7. Note that Δ/E₀ = 4/√τ, so the solar pp 'peak' is about as wide as its own centre energy: a broad, lopsided hump, not a spike, yet the Gaussian integral still lands within about 3% of the exact one. Only when τ is large — CNO's 62 — is it genuinely narrow.

Write the cross-section as σ(E) = S(E)e^(−b/√E)/E. The 1/E is the geometric πƛ² factor and the exponential is the barrier; whatever remains, S(E), is nuclear structure, and for a non-resonant reaction it varies slowly enough to extrapolate across decades. The numbers show why that matters. S(0) for p+p is 4.0×10⁻²⁵ MeV⋅b, absurdly small because a proton must weak-decay into a neutron during the collision; for ³He+³He it is 5.4 MeV⋅b, twenty-five orders of magnitude larger. At the solar Gamow peak the p+p cross-section is around 10⁻²⁶ barn — it has never been measured and never will be, and the rate is computed from weak-interaction theory instead. Accelerators reach 0.1–1 MeV, underground laboratories such as LUNA a few tens of keV, and S bridges the gap. The trap is resonance: D+T has a broad ⁵He level at 64 keV, so its S is anything but flat and the smooth Gamow formula misleads.

Doing the Gaussian integral gives ⟨σv⟩ ∝ (kT)^(−2/3)e^(−τ), and since τ ∝ (kT)^(−1/3) the local logarithmic slope is n = d ln⟨σv⟩/d ln T = (τ−2)/3. For solar p+p, τ = 13.7 gives n = 3.9. The CNO cycle's slowest link, ¹⁴N(p, γ)¹⁵O, has Z₁Z₂ = 7 and b = 212 keV^½, so at the same temperature E₀ = 26.6 keV, τ = 61.7 and n = 19.9. Both routes do identical book-keeping — 4¹H → ⁴He + 2e⁺ + 2νₑ, releasing 26.73 MeV once the positrons annihilate, of which neutrinos carry off about 0.6 MeV in the pp chain and 1.7 MeV in CNO — but they answer temperature completely differently: a core 10% hotter multiplies CNO by 6.7 and pp by only 1.45. That is why the two cross near 1.7×10⁷ K, why stars above roughly 1.3 M☉ burn by CNO, and why the Sun at 1.5×10⁷ K draws only about 1% of its luminosity from CNO, a fraction Borexino has now measured. Remember n is a local slope, not a law: it falls as T rises.

The rate per unit volume is r₁₂ = n₁n₂⟨σv⟩/(1 + δ₁₂), the Kronecker δ halving the identical-particle cases p+p and D+D so no pair is counted twice. With nₚ ≈ 3×10³¹ m⁻³ in the solar core this yields a central power density near 280 W m⁻³ — less than a compost heap. The Sun is bright because it is enormous, not because its core is fierce. A reactor has no such option: it runs D–T at kT ≈ 15 keV, where ⟨σv⟩ is some 10²⁷ times the solar p+p value, and the problem becomes holding the energy in, which is the Lawson condition nTτ_E ≳ 3×10²¹ m⁻³ keV s. Two assumptions sit underneath all of this. The plasma must be non-degenerate and Maxwellian — in a white dwarf it is neither, and reactions become pycnonuclear, driven by density rather than temperature — and the barrier must be the bare Coulomb one, though electrons screen it: in the solar plasma they lift the p+p rate by a few per cent, and in the laboratory the target's bound electrons lift measured rates by up to tens of per cent at the lowest energies, biasing exactly the low-energy data you most want.