Absorption, inelastic channels and the optical theorem

Units are SI, with nuclear numbers in MeV and fm (ħc = 197.327 MeV fm, 1 b = 100 fm²); everything here is non-relativistic, so the course metric diag(1, −1, −1, −1) never enters. Work in the centre-of-mass frame with ħk = √(2μE). QM-II's partial waves give, far from the target, ψ → Σ (2l + 1) Pₗ(cos θ)[Sₗ e^(ikr) − (−1)ˡ e^(−ikr)]/(2ikr): the incoming spherical waves are fixed by the plane wave, and the target acts only on what comes back. With one open channel, flux conservation forces |Sₗ| = 1 and Sₗ = e^(2iδₗ). A target that can capture, excite or break up the projectile sends some flux into other channels, so write Sₗ = ηₗ e^(2iδₗ) with 0 ≤ ηₗ ≤ 1. The inelasticity ηₗ, not topic 2.07's Sommerfeld η, lumps every other channel into one number per wave; resolving them needs the multichannel S-matrix, and predicting ηₗ an optical potential V − iW, both only named here. Subtracting e^(ikz) leaves f(θ) e^(ikr)/r with f(θ) = Σ (2l + 1)(Sₗ − 1)Pₗ(cos θ)/(2ik), derived. QM-II's points Sₗ sat on the unit circle; absorption moves them inside it.

The elastic flux is carried by the scattered wave. Integrating |f|² over angles, with ∫Pₗ Pₗ′ dΩ equal to 4π/(2l + 1) for l = l′ and 0 otherwise, gives σₑₗ = (π/k²) Σ (2l + 1)|1 − Sₗ|², derived. The absorbed flux is the net current into a large sphere, and it must be computed from the whole wave, not from f alone. For ψ = (A e^(ikr) − B e^(−ikr))/r the radial current (ħ/μ) Im(ψ* ∂ψ/∂r) is (ħk/μ)(|A|² − |B|²)/r²: the in-out cross terms, oscillating as e^(±2ikr), become real once multiplied by ik and drop out of the imaginary part exactly. Orthogonality separates the l again, and the inward excess divided by the incident flux ħk/μ is σᵣ = (π/k²) Σ (2l + 1)(1 − ηₗ²), derived. Two checks: π/k² is an area, so both sums are areas; and at ηₗ = 1, σᵣ = 0 while σₑₗ becomes (4π/k²) Σ (2l + 1) sin²δₗ, QM-II's elastic result.

Add the two. Since |1 − S|² = 1 − 2 Re S + |S|², σₜₒₜ = (π/k²) Σ (2l + 1)[|1 − Sₗ|² + 1 − |Sₗ|²] = (2π/k²) Σ (2l + 1)(1 − ηₗ cos 2δₗ). Now set θ = 0, where every Pₗ(1) = 1: Im[(Sₗ − 1)/(2ik)] = (1 − Re Sₗ)/(2k), so Im f(0) = (1/2k) Σ (2l + 1)(1 − ηₗ cos 2δₗ) and σₜₒₜ = (4π/k) Im f(0), derived: the form QM-II quoted, now with inelastic channels included. Dimensions: a length times 4π/k is an area. The reason is interference. Flux removed from the beam, whether scattered or absorbed, appears as a shortfall directly behind the target, and the only outgoing wave that overlaps the incident one there is the forward elastic amplitude. With spin ½ on a spinless target, topic 2.02's h(θ) carries Pₗ¹(cos θ), which vanishes at θ = 0, so the forward amplitude is g(0) times the unit matrix and σₜₒₜ = (4π/k) Im g(0) for every beam polarisation, the weight 2l + 1 splitting into l + 1 for j = l + ½ and l for j = l − ½.

Plot Sₗ as the point (ηₗ cos 2δₗ, ηₗ sin 2δₗ): purely elastic waves live on the unit circle, absorbing ones inside it. In units of (π/k²)(2l + 1), the elastic share is |1 − Sₗ|², the squared distance from Sₗ to the point 1; the reaction share 1 − ηₗ² depends on the radius alone; and the total 2(1 − Re Sₗ) is twice the horizontal gap to the line Re S = 1. On the circle of radius ηₗ the point nearest 1 is ηₗ itself, so the elastic share is at least (1 − ηₗ)², derived: a wave cannot be absorbed while its elastic scattering is switched off. The elastic share runs from (1 − ηₗ)² to (1 + ηₗ)²; the reaction share reaches 1 only at ηₗ = 0, where the elastic share is exactly 1 too. For a pure absorber, δₗ = 0, elastic over reaction is (1 − ηₗ)/(1 + ηₗ), never above 1. So when 14 MeV neutrons on lead give about 2.9 b elastic against 2.5 b non-elastic (quoted), the excess says the nucleus also refracts the wave through the real part of its potential.

Let a sphere swallow every partial wave whose classical impact parameter l/k, from lħ = pb, lies inside R and leave the rest alone: Sₗ = 0 for l ≤ L ≈ kR, Sₗ = 1 beyond. Since Σ (2l + 1) from 0 to L is (L + 1)², σᵣ = σₑₗ = π(L + 1)²/k² ≈ π(R + ƛ)² with ƛ = 1/k, tending to πR² as kR grows, and σₜₒₜ → 2πR², derived. The ƛ is an edge term: drawing the cut at (l + ½)/k = R instead gives π(R + ƛ/2)², so only the leading πR² is free of that choice. The elastic half is shadow scattering: f(θ) = (i/2k) Σ (2l + 1)Pₗ(cos θ) up to L, purely imaginary, the wave that cancels the beam behind the disc. Its forward value i(L + 1)²/(2k) ≈ ik(R + ƛ)²/2 returns σₜₒₜ = 2π(R + ƛ)² through the optical theorem. With Pₗ(cos θ) ≈ J₀((l + ½)θ) (quoted) the sum becomes iR′ J₁(kR′θ)/θ with edge radius R′ = (L + 1)/k ≈ R + ƛ, the Fraunhofer pattern of a disc, whose first zero sits at kR′θ = 3.83. So the second πR² hides in a forward cone about 1/(kR) wide: a transmission measurement with a small detector counts it, a counter at large angles never sees it. For light the same doubling is the extinction paradox, which Babinet's principle explains.

At low energy only the s wave counts, and QM-II's scattering length turns complex, a = α − iβ with β positive, so S₀ ≈ e^(−2ika) and η₀ = e^(−2kβ). Then σᵣ = (π/k²)(1 − e^(−4kβ)) ≈ 4πβ/k, derived: absorption grows as 1/v while σₑₗ tends to the constant 4π|a|². Thermal neutrons on ¹⁰B are absorbed with σᵣ = 3840 b at 2200 m s⁻¹ (quoted). With the centre-of-mass k = 3.17 × 10¹⁰ m⁻¹ that gives 1 − η₀² = 1.23 × 10⁻⁴ and β = 0.970 fm, and the 1/v law predicts 611 b at 1.00 eV; the measured cross-section follows 1/v closely up to the keV region (quoted). At the other extreme, TOTEM measured σₜₒₜ = 110.6 mb and σₑₗ = 31.0 mb for proton-proton scattering at √s = 13 TeV (quoted). A black disc gives the ratio 1/2; the measured 0.280 says the proton is grey, and a uniform grey disc with δₗ = 0, whose ratio is (1 − ηₗ)/2, needs ηₗ = 0.44. That reading rests only on flux conservation, which relativity keeps (quoted).