The Compton shift, derived

Classical Thomson scattering has the incident electric field drive the electron into oscillation at the incident frequency, and an oscillating charge radiates at the frequency at which it oscillates. So the scattered light must return at exactly the incident wavelength, at every angle, from every target — the wave account has no free knob with which to change a colour. What Compton measured in 1923, sending 71.10 pm molybdenum Kα X-rays onto graphite, was a second line beside the unshifted one, displaced towards longer wavelengths, with the displacement growing as the detector swung from forward to backward angles and taking the same value whatever the scatterer was made of. No repair to the wave account produces a wavelength change without a moving source. A photon–electron collision does, with nothing left free to adjust.

Give the photon energy E = hc/λ and momentum p = h/λ, read off the m = 0 branch of the energy–momentum invariant. Put one electron at rest at the origin and let it be free: no lattice, no binding, nothing else available to absorb momentum. Three quantities are then conserved. Energy: hc/λ + mₑ c² = hc/λ′ + Eₑ. Momentum along the beam: h/λ = (h/λ′)cos θ + pₑ cos φ. Momentum across it: 0 = (h/λ′)sin θ − pₑ sin φ. The electron has to be handled relativistically even for modest X-rays, so Eₑ and pₑ are tied together by Eₑ² = (pₑ c)² + (mₑ c²)². Four unknowns, four relations, and the physics stops here — the assumption of a free electron at rest is the entire physical content, and everything that follows is bookkeeping.

The recoil angle φ is in the equations but not in the question. Isolate the electron terms: pₑ cos φ = h/λ − (h/λ′)cos θ and pₑ sin φ = (h/λ′)sin θ. Square both and add. On the left cos²φ + sin²φ = 1 collapses everything to pₑ², and multiplying through by c² gives (pₑ c)² = E² + E′² − 2EE′cos θ. Now take the energy equation in the form Eₑ = E − E′ + mₑ c², square it, and use the invariant to replace Eₑ² − (mₑ c²)² by (pₑ c)²: that route gives (pₑ c)² = (E − E′)² + 2mₑ c²(E − E′). Set the two expressions for (pₑ c)² equal. The E² and E′² terms appear on both sides and cancel, and what survives is 2EE′(1 − cos θ) = 2mₑ c²(E − E′).

Divide 2EE′(1 − cos θ) = 2mₑ c²(E − E′) through by 2EE′ and the right-hand side becomes a difference of reciprocals: 1 − cos θ = mₑ c²(1/E′ − 1/E). Since E = hc/λ, the reciprocal 1/E is just λ/hc, so mₑ c²(λ′ − λ)/hc = 1 − cos θ, and λ′ − λ = (h/mₑ c)(1 − cos θ). Now read what has vanished. λ is gone from the right-hand side, so 71 pm and 500 nm shift by the same amount. The target material is gone, because only one mass entered and it was the electron's. The intensity never appeared at all. What remains is the angle and a single constant with the dimensions of length, h/mₑ c = 2.4263 pm, multiplied by (1 − cos θ), which runs from 0 forward, through 1 at 90°, to 2 straight back — so no Compton shift can ever exceed 4.8526 pm.

A spectrometer never measures Δλ on its own; it measures Δλ against the smallest separation λ/R it can resolve at that wavelength, so visibility is set by the ratio Δλ/λ. At 90° that ratio is 2.4263 pm divided by λ. For Compton's 71.10 pm line it is 3.4% — with a crystal spectrometer of resolving power 2000 that is 68 resolution elements, an unmissable displacement. For 500 nm green light the same absolute 2.4263 pm is 4.9 parts per million, so you would need R above 2 × 10⁵ merely to watch the line move, and at a photon energy of 2.5 eV no electron in the target is remotely free anyway. Every optical scattering experiment before 1923 was therefore consistent with no shift at all. The shift is universal; its detectability is not.

Nothing in the derivation is approximate — granted the assumption, the result is exact. The assumption is the weak point. A real target electron is bound by anything from a few eV to tens of keV and is moving inside the atom. Binding matters once it stops being negligible beside the photon energy: molybdenum Kα at 17.44 keV is sixty-one times carbon's 284 eV K-shell binding, so the free-electron ledger holds and the shifted line is sharp. Push the photon energy down towards the binding energy and the electron can no longer be treated as free; the recoiling body effectively becomes the whole atom, and since the shift goes as 1/m it collapses — that is the origin of the unshifted line, taken up in the next topic. The initial motion of the bound electrons smears the shifted line into a Compton profile. And conservation says nothing at all about how many photons scatter into a given angle: that needs the Klein–Nishina cross-section.