Compton Scattering

Roll a cue ball into a resting ball and the cue ball comes away slower — it hands over some of its punch. Light does the same trick. Fire a high-energy X-ray photon at a loosely-held electron and the photon bounces off "tired": it comes away stretched to a slightly longer wavelength (a little redder), and the electron gets kicked away. That stretch is Compton scattering, and the amazing part is how simple the rule for it turns out to be.

The everyday version

A photon's colour is its wavelength $\lambda$ — short wavelength means high energy, long wavelength means low energy. When the photon glances off the electron and changes direction by an angle $\theta$, its wavelength always grows by a fixed amount $\Delta\lambda$. It never shrinks. The bounce always costs the photon energy, never adds it.

The one equation

The whole effect lives in a single line:

Here $\lambda_C = h/(m_e c) = 2.43\ \text{pm}$ is the electron's Compton wavelength — a fixed number set only by the electron's mass, nothing else. Put in a worked number: bounce the photon straight back ($\theta = 180^\circ$, so $\cos\theta = -1$) and $\Delta\lambda = 2 \times 2.43 = 4.85\ \text{pm}$ — the biggest stretch possible. Glance it sideways ($\theta = 90^\circ$, $\cos\theta = 0$) and $\Delta\lambda = 2.43\ \text{pm}$. Let it sail straight past ($\theta = 0^\circ$) and $\Delta\lambda = 0$ — no turn, no stretch. Notice what is missing: the photon's starting energy, the material, the temperature. Only the angle counts.

Why it happens (AP / intro-college)

Treat the collision as relativistic billiards. The photon carries momentum $p = E/c = h/\lambda$ and meets an electron sitting still. Demand that both energy and momentum balance before and after — using $E_e = \sqrt{(p_e c)^2 + (m_e c^2)^2}$ for the recoiling electron — and the algebra collapses to the boxed formula. Classical wave theory predicts no shift at all, so the shift itself is the fingerprint of the photon as a particle. The scattered photon keeps energy $E' = E/\!\left[1 + (E/m_e c^2)(1-\cos\theta)\right]$ and the electron carries off the difference $KE = E - E'$, largest for a straight-back bounce (the "Compton edge"). How likely each angle is comes from the Klein–Nishina cross section — symmetric at low energy, folded sharply forward at high energy. On the panel, the "Incident Photon E" slider sets $E$ and the "Scattering Angle θ" slider sets $\theta$; the readouts track $\lambda$, $\lambda'$, $\Delta\lambda$, $E'$, $KE$, and the recoil angle $\phi$.

Try this in the sim above

First, set $\theta = 180^\circ$ and watch $\Delta\lambda$ lock near $4.85$ pm — then drag $E$ from 50 to 1000 keV and see $\Delta\lambda$ refuse to budge (only the fractional shift $\Delta\lambda/\lambda$ grows). Second, open Wavelength Shift mode and sweep $\theta$ from $0^\circ$ to $180^\circ$ to trace the gentle $1-\cos\theta$ curve. Third, switch to Klein–Nishina mode and raise $E$ toward 1 MeV — watch the scattering pattern fold forward, the signature that high-energy photons barely deflect.

Compton scattering is the inelastic scattering of a high-energy photon (X-ray or γ-ray) by a free or weakly-bound electron. Unlike classical Thomson scattering, the scattered photon's wavelength increases — proof that light carries momentum and behaves as a particle. The 1923 experiment by Arthur H. Compton was decisive evidence for the photon hypothesis.

1. The Photon Picture

A photon has energy $E = h\nu = hc/\lambda$ and momentum $p = E/c = h/\lambda$. Treat the collision as a relativistic two-body problem: photon $(E, p)$ + stationary electron $(m_ec^2, 0)$ → photon' $(E', p')$ + electron' $(E_e, p_e)$.

2. Conservation of Energy

where $E_e = \sqrt{(p_e c)^2 + (m_e c^2)^2}$ is the relativistic energy of the recoil electron.

3. Conservation of Momentum

Decompose into the scattering plane. With photon initially along $+x$, photon scattered at angle $\theta$, electron recoiling at angle $\phi$:

4. Eliminating Electron Variables

Square and add the momentum equations to eliminate $\phi$:

From energy conservation: $E_e = E - E' + m_e c^2$, so

Setting the two expressions for $p_e^2 c^2$ equal and using $E = hc/\lambda$, $E' = hc/\lambda'$:

The constant $\lambda_C = h/(m_e c) = 2.426\times 10^{-12}\,\text{m} = 2.426\,\text{pm}$ is the Compton wavelength of the electron. The shift depends only on $\theta$ — not on the incident wavelength, the material, or any external parameter.

5. Scattered-Photon Energy

For $\theta = 180°$ (back-scatter), $E'$ reaches a maximum loss; for $\theta = 0°$, $E' = E$ (no scattering). The maximum recoil electron KE is the Compton edge:

6. Klein–Nishina Differential Cross Section

Quantum electrodynamics gives the angular distribution per unit solid angle:

where $r_e = e^2/(4\pi\varepsilon_0 m_e c^2) = 2.818$ fm is the classical electron radius. At low $E$ this reduces to the symmetric Thomson cross section; at high $E$ it becomes strongly forward-peaked.

Symbol Table

Mapping to the simulation

The "Photon Energy" slider sets $E$ in keV (1 keV–1 MeV span — soft X-ray to MeV γ). The "Angle" slider sets $\theta$. The simulation enforces exact conservation: $\Delta\lambda$, $E'$, $KE_e$, $\phi$ are all computed from the closed-form formulas above. The Klein–Nishina mode samples scattering angles weighted by $d\sigma/d\Omega$ — at low $E$ the distribution is fore-aft symmetric (Thomson limit); at high $E$ it peaks sharply forward.