Compton scattering and photon momentum

A classical electromagnetic wave of frequency ν drives a free electron into oscillation at ν, and an accelerating charge radiates at its own frequency. So Thomson scattering returns light at exactly the incident wavelength, at every angle, with intensity going as 1 + cos²θ and a total cross-section σ_T = 6.65 × 10⁻²⁹ m² that does not depend on λ at all. In that picture Δλ = 0 is not an approximation, it is a theorem. In 1923 Compton sent 17.4 keV molybdenum Kα X-rays into graphite and measured the scattered spectrum with a calcite crystal. At each angle he measured — 45°, 90° and 135° — he found two lines: one at the incident 71.1 pm, and a second displaced to longer wavelength by an amount that grew with the angle. Nothing you can do to a classical wave produces the second line. Giving the light momentum does.

Once the light has momentum, the rest is bookkeeping. Before: a photon along +x with energy hc/λ and momentum h/λ, and an electron at rest with energy mₑc² and no momentum. After: a photon at angle θ with hc/λ′ and h/λ′, and an electron with momentum pₑ at angle φ on the other side of the beam, carrying total energy √(pₑ²c² + mₑ²c⁴). Three conservation statements follow. Energy: hc/λ + mₑc² = hc/λ′ + √(pₑ²c² + mₑ²c⁴). Momentum along x: h/λ = (h/λ′)cos θ + pₑ cos φ. Momentum along y: 0 = (h/λ′)sin θ − pₑ sin φ. That is three equations in the four unknowns λ′, pₑ, φ and θ, so fixing one of them — in practice θ, by where the detector sits — determines the other three. Beyond the photon relations themselves, exactly two physical assumptions are buried in the setup: the electron is free, and it is initially at rest.

Neither φ nor pₑ is wanted, so eliminate them. Write A = hc/λ and B = hc/λ′. Move the photon terms of the two momentum equations to one side and square: (pₑc cos φ)² = (A − B cos θ)² and (pₑc sin φ)² = (B sin θ)². Add them, and sin²φ + cos²φ = 1 gives pₑ²c² = A² + B² − 2AB cos θ. Now the energy equation: it says Eₑ = A − B + mₑc², and squaring against Eₑ² = pₑ²c² + mₑ²c⁴ gives pₑ²c² = (A − B)² + 2(A − B)mₑc². Set the two expressions equal. The A² and B² terms cancel on both sides, leaving 2(A − B)mₑc² = 2AB(1 − cos θ), which rearranges to 1/B − 1/A = (1 − cos θ)/mₑc². That step is the whole result, because 1/A and 1/B are λ/hc and λ′/hc — the answer is naturally linear in wavelength, not in frequency. Multiply through by hc and λ′ − λ = (h/mₑc)(1 − cos θ). Every h and every c has collapsed into one length.

Read what survived. Only θ and mₑ remain; the incident wavelength, the beam intensity and the target material have all cancelled out. The shift runs from nothing in the forward direction, through λ_C = 2.4263 pm at 90°, to 2λ_C = 4.853 pm straight back, and those are the same picometres for every incident wavelength, so what changes between experiments is not the shift but the fraction it represents. Green light at 550 nm shifts by 2.43 pm at 90°: a relative change of 4.4 × 10⁻⁶, invisible to any optical spectrometer of the day. Compton's 71.1 pm X-rays shift by the identical 2.43 pm, which is 3.4% and easily resolved on a crystal spectrometer. A 1.87 pm gamma from ¹³⁷Cs shifts by that same 2.43 pm, now 130% of its own wavelength. Energy behaves quite differently: ΔE/E = α(1 − cos θ)/[1 + α(1 − cos θ)] with α = λ_C/λ, so at 90° the X-ray loses 3.3% of its energy while the gamma loses 56%. The shift is universal; the damage is not.

mₑ entered the formula as the mass of whatever recoils. An electron bound more tightly than the collision can pay cannot recoil alone; the momentum goes to the whole atom, and mₑ becomes M. For carbon, M = 12 u = 1.99 × 10⁻²⁶ kg, so h/Mc = 1.11 × 10⁻⁴ pm — 22 000 times smaller than λ_C and far below any spectrometer's resolution. That light returns at the incident wavelength, and it is the unshifted line sitting beside the shifted one in every one of Compton's spectra. Which electrons count as free is settled by comparing binding energy with the recoil energy on offer. Graphite's four valence electrons per atom sit in bands no more than about 20 eV deep; carbon's two K-shell electrons are bound by 284 eV. A 17.4 keV photon scattering at 30° hands the electron only 79 eV, so the K electrons scatter without shifting while the valence electrons recoil freely. At 90° the offer rises to 576 eV, above the K-shell binding, and the unshifted line weakens. Heavier elements, holding their electrons more tightly, show a stronger unshifted line — exactly as observed.

The same three equations fix the electron once θ is known. Divide the y-momentum equation by the x-momentum equation: cot φ = (λ′/λ − cos θ)/sin θ, and since λ′/λ = 1 + α(1 − cos θ) with α = E/mₑc², the numerator is (1 + α)(1 − cos θ), so cot φ = (1 + α) tan(θ/2). The electron never goes backward and never leaves at 90° with any momentum worth having; it is thrown forward, more sharply the harder the photon is scattered. Its energy is whatever the photon lost, K = E α(1 − cos θ)/[1 + α(1 − cos θ)], which is largest at θ = 180°: Kₘₐₓ = 2E²/(mₑc² + 2E), the Compton edge that a γ spectrometer records as a cliff — 477 keV for the 662 keV line of ¹³⁷Cs. The parameter α is the whole story of the regime: for Compton's X-rays it is 0.034 and the electron takes a few hundred electronvolts, for a 662 keV gamma it is 1.29 and the electron takes most of the energy. The lock between θ and φ is itself testable, and Compton and Simon tested it in a cloud chamber in 1925: the recoil track and the direction of the next scattering event lay where the two-body formula said they must, event by event, and Bothe and Geiger caught the scattered X-ray and its electron in time coincidence. Whether that record also forces the field itself to be quantised is a separate question, held over to the next topic.