University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.7
Thomson's limit and the high-energy Compton fall
Every practical number for Compton scattering sits somewhere between two ends of the Klein-Nishina curve. This topic works both: the low-energy plateau at 0.665 b, which QED protects to all orders and uses to define α, and the high-energy fall as m/ω times a logarithm, which sets how far gamma rays travel in matter.
Build the model
Connect the measurement to the mechanism.
The Klein-Nishina formula has two ends, and each says something the middle does not. As ω/m → 0 the photon can no longer move the electron, ω′ → ω, and the formula collapses to Thomson's (rₑ²/2)(1 + cos²θ), (8π/3)rₑ² = 0.665 b in all with rₑ = α/m = 2.82 fm, the number UAQM-12.09 took from the A² term of the non-relativistic Hamiltonian. The Dirac electron has no A² term.
In the Feynman propagator the pair states, whose energy gap is 2m, supply e²/m instead, while the one-electron states contribute only at order ω/m. Nor is the limit a tree-level accident: the low-energy theorem of Thirring, Low, Gell-Mann and Goldberger makes it exact to all orders when e and m are the physical charge and mass, which is why QED's charge is the Thomson-limit one, α = 1/137.036. At the other end, ω ≫ m, the photon hands most of its energy to the electron, the u-channel electron line comes nearly on shell, and the total falls as (πrₑ²m/ω)[ln(2ω/m) + ½]: the 1/s that dimensions force on a point-like process, times a logarithm the electron mass cuts off.
Between the ends the full formula holds at tree level for a free electron. In matter it is most of the γ-ray attenuation once photoabsorption has faded and before pair production takes over, in water from a few tens of keV to a few tens of MeV.
- Simple definition
- The Thomson limit is the ω ≪ m end of Compton scattering, where recoil vanishes and dσ/dΩ = (rₑ²/2)(1 + cos²θ), (8π/3)rₑ² in all with rₑ = α/m; for ω ≫ m the total falls instead as (πrₑ²m/ω)[ln(2ω/m) + ½].
- Example
- For 10 keV X-rays, κ = ω/m = 0.0196 and a free electron's cross-section is 0.640 b, 3.7% under Thomson's 0.665 b. At 100 MeV the high-energy form gives 0.00825 b against the exact 0.00820 b.
An average of rₑ² for polarisation normal to the scattering plane and rₑ²cos²θ for polarisation in it, so light scattered at 90° comes out fully polarised.
θ: lab angle between k⃗ and k⃗′; rₑ = α/m = 2.818 fm; dσ/dΩ in m² sr⁻¹, 39.7 mb sr⁻¹ at 90°. Derived as the ω/m → 0 limit.
In SI, rₑ = e²/(4πε₀mc²) contains no ħ, so the result is classical: the ħ in α cancels the ħ in the Compton wavelength 1/m = 386 fm.
rₑ = αħc/(mc²) = 2.818 fm; σT = 0.6652 b = 6.652 × 10⁻²⁹ m². Derived: 1 + cos²θ integrates to 16π/3 over the sphere.
The first correction is recoil: at 5.11 keV, κ = 0.01, the cross-section is already 2% under Thomson. The series converges only for κ below ½.
κ dimensionless, ω and m in MeV. Derived by expanding topic 10.06's closed-form total; to κ² it is good to 1 × 10⁻⁴ at κ = 0.02.
The first term is a one-electron intermediate state, the second the pair state of time-ordered theory. As ω → 0 the second, with q⁰ + E ≈ 2m, carries the whole Thomson amplitude.
Sums over spin; E = √(q⃗² + m²) in MeV; u at (E, q⃗), v at (E, −q⃗); q off shell. Shown here by partial fractions in q⁰.
The order-ω term is fixed by the magnetic moment and structure enters at ω². A proton gives 0.197 μb; a neutral, spinless atom gives zero, leaving Rayleigh's ω⁴.
Ze, M: the target's physical charge and mass (MeV); α = 1/137.036, at q² = 0; ε⃗, ε⃗′: photon polarisations. Stated, not derived.
1/s is forced by dimensions; the logarithm comes from backscatter, where ω′ falls to m/2. It is 4% above the exact total at 10 MeV and 0.6% above at 100 MeV.
ω: lab photon energy; m = 0.511 MeV; σ in m² or b. Invariantly (2πα²/s)[ln(s/m²) + ½]. Computed here from the y = ω′/ω integral.
Taking ω/m → 0 in the Klein-Nishina formula
Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 10.06 derived, for an electron at rest, dσ/dΩ = (rₑ²/2)(ω′/ω)²[ω′/ω + ω/ω′ − sin²θ] with ω′/ω = 1/(1 + κ(1 − cosθ)) and κ = ω/m, quoted. As κ → 0 the photon cannot move the electron, ω′ → ω, and the bracket becomes 2 − sin²θ = 1 + cos²θ, so dσ/dΩ → (rₑ²/2)(1 + cos²θ), derived. The angular factor is a polarisation average. The amplitude tends to 2e²ε⋅ε′*, so dσ/dΩ = rₑ²|ε⃗⋅ε⃗′*|²: rₑ² for polarisation normal to the scattering plane, rₑ²cos²θ for polarisation in it. Integrating, ∫(1 + cos²θ)dΩ = 2π(2 + 2/3) = 16π/3 and σT = (8π/3)rₑ². Restoring units, rₑ = αħc/(mc²) = 197.327 MeV fm/(137.036 × 0.511 MeV) = 2.818 fm and σT = 0.6652 b, UAQM-12.09's value. Dimensions: α is a pure number and 1/m a length, so rₑ² is an area. Expanding topic 10.06's closed-form total gives σ/σT = 1 − 2κ + (26/5)κ² − …, derived: at 10 keV the free-electron value is 0.640 b, 3.7% under the plateau.
No seagull: the pair states supply e²/m
The coupling −eψ̄γμψAμ is linear in A, so Thomson's amplitude must come from topic 10.04's diagrams. Split each propagator in q⁰ into Σ u ū/(2E(q⁰ − E)), a one-electron state, plus Σ v v̄/(2E(q⁰ + E)), the pair state of time-ordered theory, shown: in the s-channel diagram the outgoing photon leaves with a new e⁺e⁻ pair whose positron then meets the first electron and the incoming photon, and in the u-channel diagram the incoming photon makes the pair and its positron meets the first electron to emit the outgoing photon. For an electron at rest and ω → 0 the one-electron denominators are ±ω, but each transverse vertex between slow electron spinors costs the recoil velocity, so these terms are of order e²ω/m. The pair denominators are q⁰ + E ≈ 2m, and the pair part of each propagator tends to (γ⁰ − 1)/(4m). Moving γ⁰ through the transverse ε̸ and using ū(γ⁰ + 1)u = 4m at rest, the two pair terms give exactly 2e²ε⋅ε′*, shown. At 5.11 keV and 90°, polarisations normal to the plane, the pair terms are −0.995e² and −1.005e², the Thomson −2e², while no one-electron term exceeds 0.02e² for any polarisations or spins, computed. The 1/m in rₑ is the 1/2m of the pair gap; the Pauli Hamiltonian's A²/2m is these pairs integrated out, stated.
The low-energy theorem makes the limit exact
Thirring showed in 1950 that radiative corrections to Thomson scattering vanish as ω → 0. In 1954 Low, and independently Gell-Mann and Goldberger, proved from gauge invariance, Lorentz invariance and crossing alone that for any target of charge Ze and mass M the ω → 0 amplitude is Thomson's, built from the physical charge and mass; the term linear in ω is fixed by the magnetic moment, and structure first enters at order ω². The theorem is stated here, not derived. It does three jobs. It defines the charge: whatever loops do to the vertex elsewhere, the e of Compton scattering at q² = 0 is the one topic 12.02 imposes through its vertex condition, α⁻¹ = 137.036, and inserting α(mZ) ≈ 1/128 instead would overestimate σT by 15%. It holds for composite targets: a proton gives (8π/3)(α/mₚ)² = 0.197 μb whatever its quarks do. And a neutral atom with no magnetic moment, helium say, has no term below order ω², its polarisability's, so σ ∝ ω⁴: Rayleigh's law, stated. Atomic electrons scatter like free ones only when ω far exceeds their binding.
Where the high-energy logarithm comes from
Change variable from θ to y = ω′/ω. The Compton relation gives 1/y = 1 + κ(1 − cosθ), so d(cosθ) = dy/(κy²), and topic 10.06's dσ/d(cosθ) = πrₑ²y²[y + 1/y − sin²θ] becomes dσ/dy = (πrₑ²/κ)[y + 1/y − sin²θ], with y from 1/(1 + 2κ) at backscatter up to 1, derived. Its y + 1/y is topic 10.05's massless −2e⁴(u/s + s/u) divided by 2e⁴, because y = (m² − u)/(s − m²) → −u/s. Integrate term by term: 1/y gives ln(1 + 2κ), y gives ½[1 − (1 + 2κ)⁻²] → ½, and sin²θ, appreciable only where y is below about 1/κ, gives about (2/κ)(ln 2κ − 2) → 0. So σ → (πrₑ²m/ω)[ln(2ω/m) + ½], computed, or (2πα²/s)[ln(s/m²) + ½] with s ≈ 2mω. The 1/s is dimensions: above m no other scale remains. The logarithm is the u-channel pole: massless, the 1/y term would diverge at backscatter, and the mass stops it by keeping ω′ above ω/(1 + 2κ) → m/2. At 10 MeV the form gives 0.0531 b against the exact 0.0510 b; at 100 MeV, 0.00825 b against 0.00820 b.
Checking the curve against γ-ray attenuation
In matter each atom offers Z electrons, and where ω far exceeds their binding, Compton scattering removes photons at n σKN per unit length, n the electron density. Water has Z/A = 10/18.015, so 3.343 × 10²³ electrons per gram. At 1.00 MeV, κ = 1.957 and σKN = 0.2112 b, so the Compton mass attenuation coefficient is 3.343 × 10²³ g⁻¹ × 0.2112 × 10⁻²⁴ cm² = 0.0706 cm² g⁻¹. The tabulated total (NIST XCOM, quoted) is 0.0707 cm² g⁻¹: a tree-level formula for a free electron accounts for 99.8% of what water does to a 1 MeV gamma ray. At 10.0 MeV the same count gives 0.0170 cm² g⁻¹ against a tabulated 0.0222, because the high-energy fall has handed almost a quarter of the attenuation to pair production in the nuclear field, which grows with ω and roughly as Z², named here. In lead at 1 MeV the Compton share is 0.0503 of the tabulated 0.0710 cm² g⁻¹; photoabsorption, rising steeply with Z, supplies most of the rest. Between those two intruders lies the Compton window, in water from a few tens of keV to a few tens of MeV.
Change one variable at a time
Make the relationship visible.
Start at log₁₀(ω/m) = −2, where the solid curve sits on Thomson's and σ/σT = 0.98. Raise ω past m: large angles empty while θ = 0 keeps rₑ², and above ω/m ≈ 10 the total runs along the dashed high-energy form as the lowest ω′ settles at m/2.
PHOTON ENERGY ω0.511 MeV
ω′/ω AT θ0.500
TOTAL σ (KLEIN-NISHINA)0.2865 b
LOWEST ω′, AT θ = 180°0.1703 MeV
Live interpretationPHOTON ENERGY ω: 0.511 MeV. ω′/ω AT θ: 0.500. TOTAL σ (KLEIN-NISHINA): 0.2865 b. LOWEST ω′, AT θ = 180°: 0.1703 MeV
Catch the common trap
Explain before calculating.
An unpolarised 5.11 keV photon (ω/m = 0.01) scatters off a free electron at rest. The Dirac coupling −eψ̄γμψAμ is linear in A, so there is no seagull. Splitting each propagator into one-electron and pair intermediate states, which part of the two tree diagrams supplies the Thomson amplitude 2e²ε⋅ε′*?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA linearly polarised 1.5 J Nd:YAG pulse (λ = 1064 nm) crosses a hydrogen plasma with nₑ = 5.0 × 10¹⁹ m⁻³. Optics view a 20 mm length of the beam at θ = 90°, along a direction perpendicular to the polarisation, and collect a solid angle of 2.0 × 10⁻² sr. How many scattered photons reach them, and how many come from the plasma's protons?
- Photon energy: ħω = hc/λ = 1239.84 eV nm/1064 nm = 1.165 eV, so κ = ω/m = 1.165 eV/511.0 keV = 2.28 × 10⁻⁶. The recoil correction 2κ is 5 × 10⁻⁶, so the Thomson limit is exact here, and the electrons of an ionised plasma are free.
- Photons per pulse: N = 1.5 J/(1.165 eV × 1.602 × 10⁻¹⁹ J eV⁻¹) = 8.03 × 10¹⁸.
- Summed over the scattered polarisation, dσ/dΩ = rₑ²[1 − (ε̂⋅k̂′)²]. Viewing perpendicular to ε̂ makes the bracket 1, so dσ/dΩ = rₑ² = (2.818 × 10⁻¹⁵ m)² = 7.94 × 10⁻³⁰ m² sr⁻¹, twice the unpolarised value at 90°.
- Probability per photon of scattering into the optics: nₑL(dσ/dΩ)ΔΩ = (5.0 × 10¹⁹ m⁻³)(0.020 m)(7.94 × 10⁻³⁰ m² sr⁻¹)(2.0 × 10⁻² sr) = 1.59 × 10⁻¹³, dimensionless as it must be.
- Collected: 8.03 × 10¹⁸ × 1.59 × 10⁻¹³ = 1.28 × 10⁶ photons. Protons have the same density and charge, and the low-energy theorem sets their Thomson cross-section by their physical mass: (mₑ/mₚ)² = (1/1836.15)² = 2.97 × 10⁻⁷, so they contribute 0.38 photon.
AnswerAbout 1.3 × 10⁶ photons per pulse before optical losses; the protons add about 0.4, so laser Thomson scattering sees only the electrons.
MediumTabulated mass attenuation coefficients for water (NIST XCOM) are 0.0707 cm² g⁻¹ at 1.00 MeV and 0.0222 cm² g⁻¹ at 10.0 MeV, and the exact Klein-Nishina totals there are 0.2112 b and 0.0510 b. Find how much of each coefficient is Compton scattering on free electrons, test the high-energy form at 10.0 MeV, and say what supplies the remainder.
- Electrons per gram of water: (Z/A)NA = (10/18.015) × 6.022 × 10²³ = 3.343 × 10²³ g⁻¹.
- At 1.00 MeV: (μ/ρ)C = 3.343 × 10²³ g⁻¹ × 0.2112 × 10⁻²⁴ cm² = 0.0706 cm² g⁻¹, 99.8% of the tabulated 0.0707. At this energy water attenuates gamma rays almost entirely by Compton scattering.
- At 10.0 MeV, κ = 10.0/0.511 = 19.57, so the high-energy form gives (πrₑ²/κ)[ln 2κ + ½] = (0.2495 b/19.57)(3.667 + 0.500) = 0.0531 b, 4% above the exact 0.0510 b. The gap is mostly the dropped sin²θ piece, 0.19 against the bracket's 4.17, and it shrinks only slowly, to 0.6% at 100 MeV.
- Compton coefficient at 10.0 MeV: 3.343 × 10²³ g⁻¹ × 0.0510 × 10⁻²⁴ cm² = 0.0170 cm² g⁻¹, 77% of the tabulated 0.0222 cm² g⁻¹.
- The remaining 0.0052 cm² g⁻¹ is pair production, chiefly in the field of the oxygen nucleus, which rises with ω while Compton falls as (m/ω)ln(2ω/m). So between 1 and 10 MeV the tabulated coefficient falls by a factor of 3.2 while Klein-Nishina alone falls by 4.1.
AnswerCompton is 0.0706 of 0.0707 cm² g⁻¹ (99.8%) at 1.00 MeV and 0.0170 of 0.0222 cm² g⁻¹ (77%) at 10.0 MeV, the rest pair production; the high-energy form is 4% high at 10 MeV.
HardWrite the Klein-Nishina total as σ = (πrₑ²/κ)∫[y + 1/y − sin²θ]dy, with y = ω′/ω running from 1/(1 + 2κ) to 1. For a 100 MeV photon on an electron at rest, evaluate each of the three pieces, compare σ with the high-energy form, and find the fraction of Compton events in which the photon keeps less than half its energy.
- From 1/y = 1 + κ(1 − cosθ), d(cosθ) = dy/(κy²). Topic 10.06's dσ/d(cosθ) = πrₑ²y²[y + 1/y − sin²θ] then gives dσ/dy = (πrₑ²/κ)[y + 1/y − sin²θ], with sin²θ = c(2 − c) and c = 1 − cosθ = (1/y − 1)/κ.
- κ = 100/0.511 = 195.70 and y₀ = 1/(1 + 2κ) = 1/392.39 = 2.548 × 10⁻³, so ω′ never falls below 0.2548 MeV, close to m/2 = 0.2555 MeV.
- Pieces: ∫dy/y = ln 392.39 = 5.9723; ∫y dy = ½(1 − y₀²) = 0.5000; and ∫c(2 − c)dy = (2/κ)[(1 + 1/κ)ln(1 + 2κ) − 2] = 0.0409, close to the large-κ estimate (2/κ)(ln 2κ − 2) = 0.0406. The bracket is 5.9723 + 0.5000 − 0.0409 = 6.4313.
- σ = (πrₑ²/κ) × 6.4313 = (0.24947 b/195.70) × 6.4313 = 1.2748 × 10⁻³ b × 6.4313 = 8.199 × 10⁻³ b. The high-energy form has ln 2κ + ½ = 5.9697 + 0.5 = 6.4697 and gives 8.247 × 10⁻³ b, 0.6% high.
- Photon keeping less than half, y below ½: ln(0.5/y₀) = 5.2791, ∫y dy = 0.1250, and the sin²θ piece from y₀ to ½ is 0.0409 − 0.0020 = 0.0389, the part above ½ being about (2/κ)(ln 2 − ½) = 0.0020. The fraction is (5.2791 + 0.1250 − 0.0389)/6.4313 = 5.3652/6.4313 = 0.834.
Answerσ = 8.20 × 10⁻³ b (pieces 5.972, 0.500, −0.041), with the high-energy form 0.6% high at 8.25 × 10⁻³ b; in 83% of events the photon keeps less than half its 100 MeV.