University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.6

The Klein-Nishina formula derived in the laboratory frame

Topic 10.05 left Compton scattering as a Lorentz scalar in p⋅k and p⋅k′. In a target the electrons sit, near enough, at rest, so here that scalar is rewritten in laboratory energies and angles, given its flux and its recoil phase space, and turned into the formula that γ-ray detectors, shielding and dosimetry still run on.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 10.05's averaged Compton |M|² is a Lorentz scalar, but a measurement happens in a frame: photons of energy ω strike electrons at rest, and a detector at angle θ counts photons of energy ω′. In that frame p⋅k = mω and p⋅k′ = mω′, the Compton relation ties ω′ to θ, and the four terms of the trace collapse to 2e⁴[ω′/ω + ω/ω′ − sin²θ]. The dynamics is then finished. What remains is the bookkeeping of a target experiment, and it carries the physics of recoil.

The flux is 4mω. The phase space is not the centre-of-mass one: because the struck electron recoils, ω′ falls as θ grows, and the energy delta function returns a Jacobian mω/(ω′E′) that leaves ω′²dΩ/(16π²mω). Multiplying out gives Klein and Nishina's 1929 formula, with the length rₑ = α/m = 2.818 fm setting its scale. Its content is a competition.

The squared amplitude never falls below its Thomson value, but the kinematic factor (ω′/ω)² does, so forward scattering keeps Thomson's 0.499 b per unit cos θ at every energy while the large angles empty as ω/m grows, and the total, integrated in closed form, is 0.287 b at ω = m. The model stops at a free electron at rest, at tree level: atomic binding, the electrons' motion that Doppler-broadens the scattered line, and radiative corrections all lie outside it.

Simple definition: The Klein-Nishina formula is the tree-level QED cross-section for an unpolarised photon of energy ω to scatter through angle θ off a free electron at rest: dσ/dcos θ = πrₑ²(ω′/ω)²[ω′/ω + ω/ω′ − sin²θ], with ω′ fixed by the Compton relation.
Example: For ω = m = 511 keV at θ = 90°, ω′ = 255.5 keV and dσ/dcos θ = πrₑ² × (1/2)² × (1/2 + 2 − 1) = 0.0936 b, only 37.5 per cent of Thomson's 0.2495 b at that angle; integrated over every angle the total is 0.287 b.

Compton |M|² with the electron at rest (derived)¼Σ|M|² = 2e⁴[ω′/ω + ω/ω′ − sin²θ]

Topic 10.05's trace with p⋅k = mω and p⋅k′ = mω′; the Compton relation folds its m² and m⁴ terms into −sin²θ. The bracket is never below 1 + cos²θ.

ω, ω′ incident and scattered photon energies (J, quoted in keV); θ the lab scattering angle (rad); e² = 4πα; |M|² dimensionless

Flux factor with the electron at rest (derived)F = 4√((p⋅k)² − m²m_γ²) = 4mω

Invariant, so 4p⋅k serves in any frame; in the laboratory it is the product of the two energies with a relative speed of exactly c.

m the electron's rest energy mc² = 0.511 MeV (8.19 × 10⁻¹⁴ J); ω the photon energy (J); m_γ = 0; F has dimension energy²

Phase space with a recoiling electron (derived)dΦ₂ = ω′² dΩ / (16π² mω)

Not topic 5.04's centre-of-mass |p_f|/(16π²√s): recoil makes ω′ fall with θ, and the delta function charges a Jacobian for it. dΦ₂ is dimensionless.

dΩ the photon's solid angle (sr); includes the slope mω/(ω′E′) of the energy delta function, E′ = m + ω − ω′ the recoiling electron's total energy (J)

Klein-Nishina cross-section (derived)dσ/dcos θ = (πα²/m²)(ω′/ω)²[ω′/ω + ω/ω′ − sin²θ]

Forward, ω′ = ω and it equals Thomson's 2πrₑ² = 0.499 b at every energy; backward the factor (ω′/ω)² = (1 + 2ω/m)⁻² empties it.

α²/m² = rₑ², rₑ = αħc/mc² = 2.818 fm, so πrₑ² = 0.2495 b (1 b = 10⁻²⁸ m²); dσ/dΩ is this divided by 2π

Trading cos θ for the invariant t (derived)dt/dcos θ = 2ω′², where t = (k − k′)² = −2ωω′(1 − cos θ)

Multiplying the frame-free dσ/dt = ¼Σ|M|²/(16π(s − m²)²), s − m² = 2mω, by it rebuilds the formula without lab phase space: an independent check.

t the squared four-momentum transfer (energy², J² in SI); differentiated at fixed ω using the Compton relation

Total cross-section in closed form (derived)σ = 2πrₑ²[((1 + κ)/κ²)(2(1 + κ)/(1 + 2κ) − ln(1 + 2κ)/κ) + ln(1 + 2κ)/(2κ) − (1 + 3κ)/(1 + 2κ)²]

From x = ω/ω′ as the integration variable. It tends to (8π/3)rₑ² = 0.665 b as κ → 0, the limit topic 10.07 develops; at κ = 1 it is 43 per cent of that.

κ = ω/m dimensionless; σ in m², quoted in barns: 0.640 b at 10 keV, 0.287 b at κ = 1, 0.211 b at 1.00 MeV

Put the electron at rest and the invariants become energies

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 10.05 left the averaged Compton trace in invariant form, quoted here: ¼Σ|M|² = 2e⁴[p⋅k′/p⋅k + p⋅k/p⋅k′ + 2m²(1/p⋅k − 1/p⋅k′) + m⁴(1/p⋅k − 1/p⋅k′)²]. A target supplies a frame: p = (m, 0), k = (ω, ωn̂), k′ = (ω′, ω′n̂′), with n̂⋅n̂′ = cos θ. Then p⋅k = mω and p⋅k′ = mω′, and the first two terms read ω′/ω + ω/ω′. The Compton relation 1/ω′ − 1/ω = (1 − cos θ)/m, quoted from UQMI-01.04, makes 1/p⋅k − 1/p⋅k′ = −(1 − cos θ)/m², so the last two terms give −2(1 − cos θ) + (1 − cos θ)² = −sin²θ. Derived: ¼Σ|M|² = 2e⁴[ω′/ω + ω/ω′ − sin²θ]. It is dimensionless, as a 2 → 2 squared amplitude must be. And because ω′/ω + ω/ω′ ≥ 2, the bracket never drops below 1 + cos²θ, its value when ω′ = ω. Whatever makes the cross-section fall at high energy, it is not a shrinking amplitude.

Flux and phase space when the electron recoils

The cross-section is dσ = ¼Σ|M|² dΦ₂/F. The flux factor F = 4√((p⋅k)² − m²m_γ²) is invariant, and with m_γ = 0 it is 4p⋅k = 4mω. Phase space is where the laboratory parts company with topic 5.04's centre of mass. Start from dΦ₂ = (2π)⁴δ⁴(p + k − p′ − k′) d³k′/((2π)³2ω′) d³p′/((2π)³2E′). The three-momentum delta function gives the electron momentum ωn̂ − ω′n̂′, so E′ = √(m² + ω² + ω′² − 2ωω′ cos θ). Write d³k′ = ω′²dω′dΩ; the energy delta function δ(m + ω − ω′ − E′) then divides by its slope, 1 + (ω′ − ω cos θ)/E′. With E′ = m + ω − ω′ and the Compton relation, E′ + ω′ − ω cos θ = m + ω(1 − cos θ) = mω/ω′, so the slope is mω/(ω′E′) and, derived, dΦ₂ = ω′²dΩ/(16π²mω), dimensionless as two-body phase space must be. At ω = m and θ = 90°, ω′ = 255.5 keV, E′ = 766.5 keV and the slope is 4/3: a calculation that forgets it overstates the phase space by a third.

Assemble the formula and check it two ways

Divide by the flux: dσ/dΩ = ¼Σ|M|² ω′²/(64π²m²ω²). With e² = 4πα and the azimuthal symmetry of an unpolarised beam, dσ/dcos θ = 2π dσ/dΩ = (πα²/m²)(ω′/ω)²[ω′/ω + ω/ω′ − sin²θ], derived: the formula Klein and Nishina obtained in 1929 from Dirac's equation, reached here as tree-level QED. Dimensions: α²/m² = 203.9 GeV⁻² is an area, and (ħc)² = 0.3894 GeV² mb turns it into 79.41 mb = rₑ², with rₑ = αħc/mc² = 2.818 fm, so πrₑ² = 0.2495 b. Invariant check: dσ/dt = ¼Σ|M|²/(16π(s − m²)²) holds in every frame. Here s − m² = 2mω and t = −2ωω′(1 − cos θ), and differentiating with the Compton relation gives dt/dcos θ = 2ω′²; the product is the same formula, reached without the lab phase space. Low-energy check: as ω/m → 0, ω′ → ω and dσ/dcos θ → πrₑ²(1 + cos²θ), Thomson's result from UAQM-12.09.

Forward stays Thomson; recoil empties the large angles

At θ = 0 the photon is undeflected, ω′ = ω, t = 0 and the bracket is 2, so dσ/dcos θ = 2πrₑ² = 0.499 b at every photon energy: at tree level the forward value never moves. Backward, ω′/ω = 1/(1 + 2κ) with κ = ω/m, and backward over forward is [(1 + 2κ)⁻³ + (1 + 2κ)⁻¹]/2: 0.925 at κ = 0.02 (10.2 keV), 0.185 at κ = 1 and 0.024 at κ = 10. The squared amplitude is not responsible, since backward at κ = 10 its bracket is 21.05 against Thomson's 2. The fall is all in (ω′/ω)², which phase space supplied as ω′² and the flux and slope as 1/ω². Writing x = ω/ω′ = 1 + κ(1 − cos θ), that factor is 1/x², already halved once κ(1 − cos θ) = 0.41, so the scattering crowds into a forward cone where κ(1 − cos θ) ≲ 1: at κ = 10, θ ≲ 26°. Turning a hard photon through a large angle hands most of its energy to the electron: backward at κ = 10 the photon keeps only 1/21 of it.

Integrate over angle in closed form

Change variable to x = ω/ω′ = 1 + κ(1 − cos θ), which runs from 1 forward to 1 + 2κ backward, with dcos θ = −dx/κ and sin²θ = u(2 − u), u = (x − 1)/κ. Then σ is πrₑ²/κ times the integral from x = 1 to 1 + 2κ of [x⁻³ + x⁻¹ − u(2 − u)x⁻²] dx, and every piece is a power or a logarithm. At κ = 1 the three integrals are 4/9 = 0.4444, ln 3 = 1.0986 and 4 ln 3 − 4 = 0.3944, so σ = πrₑ² × 1.1486 = 0.287 b, computed: 43 per cent of Thomson's 0.665 b. In general the closed form on the formula card gives 0.640 b at 10 keV, 0.493 b at 100 keV, 0.287 b at 511 keV, 0.211 b at 1.00 MeV, 0.146 b at 2.00 MeV and 0.0510 b at 10 MeV. At small κ its bracket is a difference of nearly equal numbers, so keep digits or use the series σ = (8π/3)rₑ²(1 − 2κ + 26κ²/5 − …). Both ends of the curve, Thomson's constant and the high-energy logarithm, are topic 10.07's.

Measured against matter, and what a free electron omits

Water holds 3.343 × 10²³ electrons per gram, so at 1.00 MeV, with σ = 0.2112 b, the formula predicts a mass attenuation coefficient of 0.0706 cm² g⁻¹. The tabulated value, quoted here, is 0.0707 cm² g⁻¹, almost all of it Compton scattering, and narrow-beam transmission measurements reproduce such tables to about a per cent. The model's limits are named here, not developed. Atomic electrons are bound, by 0.54 keV in oxygen's K shell and 88 keV in lead's (quoted), and inner electrons move with momenta of order Zαm. When the momentum transfer is not large compared with those momenta, binding suppresses small-angle scattering below the free value, and the electrons' motion Doppler-broadens the scattered line about ω′ into the Compton profile. Radiative corrections, of relative order α/π ≈ 0.2 per cent with logarithms, begin in Unit 12. And everything here is unpolarised: a polarised beam or target needs the unsummed trace.

Change one variable at a time

Make the relationship visible.

Interactive model

Photon energy κ = ω/m1.00 × 511 keV

Scattering angle θ90 °

Set θ to 0° and sweep κ from 0.02 to 10: the forward dot never leaves 0.499 b. Then set θ to 180° and watch it fall to 2.4 per cent of the forward value at κ = 10, while the total drops from 0.640 b to 0.082 b.

SCATTERED ENERGY ω′255 keV

dσ/dcos θ AT θ0.094 b

KN ÷ THOMSON AT θ0.38

TOTAL σ0.287 b

Live interpretationSCATTERED ENERGY ω′: 255 keV. dσ/dcos θ AT θ: 0.094 b. KN ÷ THOMSON AT θ: 0.38. TOTAL σ: 0.287 b

Catch the common trap

Explain before calculating.

A 511 keV photon, ω = m, scatters through θ = 60° off a free electron at rest. Taking πrₑ² = 0.2495 b, what is the laboratory dσ/dcos θ?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 200 keV gamma ray scatters through 60° off a free electron at rest. Find the scattered photon's energy, the Klein-Nishina dσ/dΩ in mb sr⁻¹, and its ratio to Thomson's value at the same angle.

κ = ω/m = 200/511.0 = 0.3914. The Compton relation gives ω′ = ω/[1 + κ(1 − cos 60°)] = 200 keV/1.1957 = 167.3 keV, so ω′/ω = 0.8363.
The bracket is ω′/ω + ω/ω′ − sin²60° = 0.8363 + 1.1957 − 0.7500 = 1.2820.
dσ/dΩ = (rₑ²/2)(ω′/ω)²[…], with rₑ²/2 = 39.70 mb: 39.70 mb × 0.6995 × 1.2820 = 35.6 mb sr⁻¹.
Thomson at 60° is (rₑ²/2)(1 + cos²60°) = 39.70 mb × 1.25 = 49.6 mb sr⁻¹, so the ratio is 35.6/49.6 = 0.72. Check: the bracket, 1.282, exceeds Thomson's 1.25, so the whole reduction comes from (ω′/ω)² = 0.70.

Answerω′ = 167 keV and dσ/dΩ = 35.6 mb sr⁻¹, 0.72 of Thomson's 49.6 mb sr⁻¹ at 60°.

MediumA broad beam of 662 keV photons, flux 1.00 × 10⁵ cm⁻² s⁻¹, bathes a 10.0 g aluminium rod (Z = 13, molar mass 26.98 g mol⁻¹). A detector subtending 2.00 × 10⁻³ sr views the rod at θ = 90°. Treating the electrons as free and at rest and neglecting attenuation in the rod, predict the energy of the scattered line and its count rate.

Electrons in the rod: Nₑ = (10.0/26.98) × 13 × 6.022 × 10²³ = 2.902 × 10²⁴.
κ = 662/511.0 = 1.2955, and at 90° ω′ = ω/(1 + κ) = 662 keV/2.2955 = 288.4 keV, so ω′/ω = 0.4356.
The bracket is 0.4356 + 2.2955 − 1 = 1.7311, so dσ/dΩ = (rₑ²/2)(0.4356)²(1.7311) = 0.03970 b × 0.1898 × 1.7311 = 0.01304 b sr⁻¹ = 1.304 × 10⁻²⁶ cm² sr⁻¹.
Rate = Φ Nₑ (dσ/dΩ) ΔΩ = 1.00 × 10⁵ cm⁻² s⁻¹ × 2.902 × 10²⁴ × 1.304 × 10⁻²⁶ cm² sr⁻¹ × 2.00 × 10⁻³ sr = 7.57 s⁻¹. Units: cm⁻² s⁻¹ × cm² sr⁻¹ × sr = s⁻¹.
Checks: Thomson's 0.0397 b sr⁻¹ would predict 23.0 s⁻¹, three times more: recoil at 90° costs (ω′/ω)² = 0.19, which the bracket, 1.73 against Thomson's 1, only partly restores. Aluminium's K shell is bound by 1.56 keV, far below 662 keV, so the free-electron rate stands, though binding Doppler-broadens the 288 keV line.

AnswerA line at 288 keV, counted at about 7.6 s⁻¹.

HardIntegrate the Klein-Nishina formula term by term at ω = 1.00 MeV, using x = ω/ω′ as the variable, to find σ per electron. Then predict the mass attenuation coefficient of water, compare it with the tabulated 0.0707 cm² g⁻¹, and find the fraction of 1.00 MeV photons crossing 10.0 cm of water unscattered.

κ = 1.000/0.5110 = 1.9570 and X = 1 + 2κ = 4.9139. With dcos θ = −dx/κ, σ = (πrₑ²/κ)(A + B − C), where A, B and C integrate x⁻³, x⁻¹ and u(2 − u)x⁻² from 1 to X, with u = (x − 1)/κ.
A = ½(1 − X⁻²) = ½(1 − 1/24.146) = 0.47929, and B = ln X = 1.59207.
Expanding u(2 − u)x⁻² = (2/κ)(x⁻¹ − x⁻²) − κ⁻²(1 − 2x⁻¹ + x⁻²) gives C = (2/κ + 2/κ²) ln X − 2/κ − 4/X − 2/(κX) = 2.45854 − 1.02200 − 0.81402 − 0.20798 = 0.41454.
σ = 0.24947 b × (0.47929 + 1.59207 − 0.41454)/1.9570 = 0.24947 b × 1.65682/1.9570 = 0.2112 b, as the closed form gives.
Water has 10 electrons per 18.015 g mol⁻¹, so 3.343 × 10²³ electrons per gram, and μ/ρ = 2.112 × 10⁻²⁵ cm² × 3.343 × 10²³ g⁻¹ = 0.0706 cm² g⁻¹, about 0.1 per cent below the tabulated 0.0707 cm² g⁻¹.
At 1.00 g cm⁻³, μ = 0.0706 cm⁻¹ and the mean free path is 14.2 cm, so a fraction e^(−0.706) = 0.494 crosses 10.0 cm unscattered.

Answerσ = 0.211 b per electron; μ/ρ = 0.0706 cm² g⁻¹ against the tabulated 0.0707 cm² g⁻¹; 49 per cent cross 10.0 cm of water unscattered.