Photon momentum and Compton scattering

In first-year mechanics p = mv, which makes a massless particle look momentum-free. Relativity replaces that with the invariant E² = (pc)² + (mc²)², and for m = 0 it gives p = E/c exactly, with no slow-speed approximation anywhere. With E = hf = hc/λ the photon momentum is p = h/λ, directed along the propagation. Check the dimensions: [h] = J s = kg m² s⁻¹, and dividing by a length leaves kg m s⁻¹. The numbers are small. A 500 nm photon carries 1.33 × 10⁻²⁷ kg m s⁻¹; a 71.1 pm X-ray photon carries 9.32 × 10⁻²⁴ kg m s⁻¹, the momentum of an electron moving at 3.4% of the speed of light. Two consequences follow. One photon's momentum is noticeable only when it strikes something light enough to recoil measurably, which is why a nearly free electron is the target that reveals it. And a beam transfers momentum at a rate fixed by its power alone, which is radiation pressure, a result Maxwell's theory already contained and the photon picture must reproduce.

Take the electron free and at rest: no binding energy, no initial momentum. A photon of wavelength λ arrives along x, leaves at angle θ with wavelength λ′, and the electron recoils at angle φ on the far side of the beam with momentum pₑ. Write the three conservation laws. Energy: hc/λ + mₑc² = hc/λ′ + √((pₑc)² + (mₑc²)²). Momentum along x: h/λ = (h/λ′) cos θ + pₑ cos φ. Momentum along y: 0 = (h/λ′) sin θ − pₑ sin φ. The electron's energy is relativistic because nothing guarantees a slow recoil: a 1.00 MeV gamma ray scattered straight back hands the electron 796 keV, more than its 511 keV rest energy, and ½mv² would put its speed at 1.77c. Now count. λ′, θ, pₑ and φ are four unknowns and there are three laws, so the outcome is a one-parameter family: choose θ, the angle at which the detector sits, and everything else is fixed. These kinematics follow without knowing anything about the force between photon and electron; only how often each θ occurs needs the dynamics.

Square and add the two momentum equations and φ disappears: pₑ² = (h/λ)² + (h/λ′)² − 2(h/λ)(h/λ′) cos θ. Isolate the square root in the energy equation and square it: (pₑc)² = (hc/λ)² + (hc/λ′)² − 2(hc)²/(λλ′) + 2mₑc² hc(1/λ − 1/λ′). Set the two expressions for (pₑc)² equal. The squared photon terms cancel identically, leaving mₑc² hc(1/λ − 1/λ′) = (hc)²(1 − cos θ)/(λλ′), and multiplying through by λλ′/(mₑc² hc) gives λ′ − λ = (h/mₑc)(1 − cos θ). The incident wavelength has gone. The shift is set by h/mₑc = 6.626 × 10⁻³⁴ J s ÷ (9.109 × 10⁻³¹ kg × 2.998 × 10⁸ m s⁻¹) = 2.426 × 10⁻¹² m, the Compton wavelength of the electron: the wavelength of a photon whose energy equals mₑc². Limiting cases check it. θ = 0 gives no shift, as a photon that is not deflected must; θ = 180° gives the maximum, 4.85 pm. Molybdenum X-rays near 71 pm scattered from graphite show a line shifted by about 2.4 pm at 90°, as predicted.

Because Δλ is absolute, the fractional shift is Δλ/λ = (h/mₑc)(1 − cos θ)/λ = (E/mₑc²)(1 − cos θ), with E = hc/λ. For 500 nm light at 90° that is 2.43 pm in 500 000 pm, five parts per million: the photon loses 5 ppm of its 2.48 eV and the electron recoils with 12 μeV, hopelessly below any spectrometer's resolution. For 71.1 pm X-rays it is 3.4%. For a 662 keV gamma ray ε = E/mₑc² = 1.30, and a backscattered photon leaves with E/(1 + 2ε) = 184 keV, so the electron carries 478 keV. That maximum transfer, the Compton edge, is the sharp shoulder in a scintillation detector's gamma spectrum. So ε is the number that tells you which physics you are in. As ε → 0 the shift and the recoil vanish and the photon scatters with its frequency unchanged, which is Thomson scattering: a classical free electron, driven by the wave's electric field, re-radiates at the driving frequency. The quantum formula contains the classical one as its low-energy limit, the correspondence check every photon result must pass.

The ledger assumed an electron that is free and at rest, and a real target offers neither. A scattered X-ray line arrives in two parts: the shifted line, and a second at the original wavelength. The unshifted line is the free-electron assumption failing. An electron bound by more than the recoil on offer cannot take the momentum alone; the whole atom recoils, and the shift becomes (h/Mc)(1 − cos θ) with M the atomic mass. For carbon M = 12.01 u = 21 900 mₑ, so the 90° shift is 2.43 pm ÷ 21 900 = 1.1 × 10⁻⁴ pm, far below resolution. Molybdenum Kα photons at 17.4 keV offer a 0.58 keV recoil at 90°, and carbon's deepest electrons are bound by only 0.28 keV, so most of graphite's electrons scatter as free and the shifted line is the stronger. Copper binds its K-shell electrons by 8.98 keV, far beyond that recoil, and the unshifted line grows with atomic number just as the criterion predicts. Even the free electrons are not at rest: their orbital momenta Doppler-broaden the shifted line, and that width is now used to map electron momenta in solids.

Return to p = E/c and add up a beam. A black surface absorbing energy dE in time dt takes momentum dE/c, so the force is the absorbed power divided by c and the pressure at normal incidence is P = I/c, with I in W m⁻². A mirror reverses each photon's momentum and doubles it to 2I/c. At angle α from the normal, unit area intercepts only I cos α, and only the normal component of each photon's momentum, a second factor cos α, presses on the surface, so P = (1 + R) I cos²α / c for reflected fraction R. Where the intensity varies across the surface, integrate: F = ((1 + R)/c) ∫ I cos²α dA over the lit area. The photon count never appears, because every photon carries the same momentum per unit energy, 1/c, whatever its wavelength, and the result matches Maxwell's radiation pressure: another correspondence check passed. Sunlight at 1361 W m⁻² presses on a black surface with 4.54 μPa and on a mirror with 9.08 μPa, so an 800 m² reflecting sail near Earth feels 7.26 mN, roughly the weight of a paper clip, applied without pause.