Compton Scattering

When a beam of monochromatic X-rays scatters off a block of graphite, the scattered radiation contains not only the original wavelength but a second, longer wavelength that grows with scattering angle. Classical Thomson scattering can describe elastic re-radiation at the incident frequency, but it cannot account for the observed wavelength shift. Compton's result requires treating the radiation as photons carrying definite energy and momentum. In 1923 Arthur Holly Compton showed that the shift is inevitable the moment you treat the X-ray as a particle — a photon of energy $E=h\nu$ and momentum $p=h/\lambda$ — that collides with a nearly free electron and recoils, handing over some of its energy. The scattered photon, poorer in energy, is redder. The whole effect is fixed by two conservation laws applied to a relativistic two-body collision. Energy and momentum conservation, together with the photon dispersion $E=pc$ and the electron's relativistic energy $E^2=(pc)^2+(mc^2)^2$ , force a single clean prediction: the wavelength increases by $\Delta\lambda=\lambda_C(1-\cos\theta)$ , where $\lambda_C=h/m_ec$ is the electron's Compton wavelength, $2.426$ pm. Remarkably, this shift depends only on the scattering angle and a universal constant — not on the incident wavelength, the material, or the intensity. That angle-only, material-independent, intensity-independent signature is what made Compton scattering a decisive piece of evidence for the photon. This article derives the shift from four-momentum conservation, extracts the scattered-photon energy and the electron recoil, and locates the classical (Thomson) limit as the low-energy edge of the same formula. Downstream refinements — the angular intensity distribution (the Klein–Nishina formula) and inverse Compton scattering — are named where they belong and are not derived here.

Key takeaways

A photon scattering off a free electron loses energy, so its wavelength increases by $\Delta\lambda=\lambda_C(1-\cos\theta)$ with $\lambda_C=h/m_ec=2.426$ pm.
The shift follows from relativistic energy–momentum conservation for a photon scattering from a free electron initially at rest, using $E_\gamma=pc$ for the photon and $E_e^2=(p_e c)^2+(m_e c^2)^2$ for the electron.
The shift is independent of the incident wavelength, the target material, and the beam intensity — it depends only on $\theta$ and the universal $\lambda_C$ .
The scattered-photon energy is $E'=E/[1+(E/m_ec^2)(1-\cos\theta)]$ ; the electron carries away the difference as recoil kinetic energy.
In the low-energy limit $E\ll m_ec^2$ the fractional shift vanishes and one recovers classical Thomson scattering, where wavelength is unchanged.
Key idea: treating light as a particle that carries momentum $h/\lambda$ turns wavelength change into an ordinary billiard-ball recoil, and the answer is dictated entirely by kinematics.

The core insight

A photon is a relativistic particle with energy $h\nu$ and momentum $h/\lambda$ . Requiring energy and momentum to balance in its collision with a resting electron forces the scattered wavelength to grow by $\Delta\lambda=(h/m_ec)(1-\cos\theta)$ — an angle-only shift set by a single universal constant.

Where this lives today

Compton scattering is important whenever photons in the keV-to-MeV range interact with matter. In medical physics it is a major interaction in soft tissue over much of the diagnostic and therapeutic energy range, contributing both to dose deposition in radiotherapy and to the scatter that degrades X-ray images. Gamma-ray detectors and spectrometers show the "Compton continuum" and "Compton edge" arising from Compton scattering, with the edge corresponding to the maximum energy transferred to the recoil electron at backscatter. Compton telescopes exploit the same scattering kinematics to reconstruct the direction of astrophysical $\gamma$ -rays. The inverse process — a low-energy photon gaining energy from an ultra-relativistic electron — powers inverse-Compton emission in astrophysical jets and is engineered in laboratory Compton light sources to produce tunable high-energy photon beams. The angular intensity of the scattering, beyond the pure kinematics derived here, is given by the Klein–Nishina formula, computed from quantum electrodynamics; at low energy it reduces to the classical Thomson cross section $\sigma_T=6.65\times10^{-29}\,\mathrm{m}^2$ . The kinematic shift itself remains a standard undergraduate demonstration of the reality of the photon.

The physical picture

Picture a single X-ray photon travelling to the right and striking an electron that is initially at rest and only weakly bound compared with the photon energy (typical X-ray energies are in the keV range, much larger than ordinary atomic binding energies). The photon scatters into a new direction making angle $\theta$ with its original path; the electron recoils into the lower half-plane at angle $\varphi$ . Because the photon transfers momentum to the electron, the scattered photon carries less energy and momentum than the incident photon; for a photon, the reduced momentum corresponds to a longer wavelength. The redshift is the direct fingerprint of that recoil. The size of the effect is set by comparing the photon's momentum to the electron's rest momentum scale $m_ec$ . If the photon's momentum $h/\lambda$ is tiny compared with $m_ec$ (long wavelengths, visible light), the electron barely budges and the wavelength is essentially unchanged — this is the classical regime. Only when $\lambda$ approaches the Compton wavelength $\lambda_C=h/m_ec$ does the fractional shift become appreciable. That is why the effect was first seen with X-rays, whose wavelengths of order tens of pm are only about an order of magnitude larger than $\lambda_C=2.426\,\mathrm{pm}$ .

Figure

Compton collision geometry: photon in, scattered photon and recoil electron out

There is a single decisive prediction. In the classical Thomson-scattering picture for a free electron, the scattered radiation retains the incident frequency, independent of the scattering angle. Compton's data instead showed a wavelength that grew with angle by an amount matching $\lambda_C(1-\cos\theta)$ to within experimental error, using the known electron mass. No free parameter, no material dependence — just kinematics. That rigidity is the whole point of the second figure.

Figure

The wavelength shift depends only on angle, never on incident λ

The core equations

The whole effect lives in relativistic four-momentum conservation. We treat the photon as a massless relativistic particle and the electron by its exact relativistic energy relation; the algebra collapses to a single formula for the wavelength change.

E_\gamma = h\nu = \frac{hc}{\lambda}, \qquad p_\gamma = \frac{E_\gamma}{c} = \frac{h}{\lambda} \qquad (1)

where: $E_\gamma$ is the photon energy, $\nu$ its frequency, $\lambda$ its wavelength, $p_\gamma$ its momentum, $h$ Planck's constant, and $c$ the speed of light. A photon is massless, so its energy and momentum are locked together by $E_\gamma=p_\gamma c$ ; combined with $E_\gamma=h\nu$ this gives $p_\gamma=h/\lambda$ , the de Broglie relation read for light. The recoiling electron obeys the relativistic energy–momentum relation, the mass-shell condition that any free particle must satisfy.

E_e^2 = (p_e c)^2 + (m_e c^2)^2 \qquad (2)

where: $E_e$ is the electron's total energy after the collision, $p_e$ the magnitude of its momentum, and $m_e$ its rest mass. Before the collision the electron is at rest, so its energy is $m_ec^2$ and its momentum is zero. Together with conservation of total energy and momentum, this relativistic energy--momentum relation allows the electron variables to be eliminated and yields the central Compton-shift formula.

\Delta\lambda = \lambda' - \lambda = \frac{h}{m_e c}\,(1-\cos\theta) = \lambda_C\,(1-\cos\theta) \qquad (3)

where: $\lambda'$ is the scattered wavelength, $\lambda$ the incident wavelength, $\theta$ the photon scattering angle, and $\lambda_C=\frac{h}{m_ec}=2.426\times10^{-12}\,\mathrm{m}$ is the Compton wavelength of the electron. The shift is additive in wavelength, angle-dependent through $(1-\cos\theta)$ , and set in size by one universal constant. Written in terms of energy rather than wavelength, the same relation gives the scattered photon's energy directly, which is the more useful form at $\gamma$ -ray energies.

E' = \frac{E}{\,1 + \dfrac{E}{m_e c^2}\,(1-\cos\theta)\,} \qquad (4)

where: $E=E_\gamma$ is the incident photon energy, $E'$ the scattered photon energy, and $m_ec^2=0.511\,\mathrm{MeV}$ the electron rest energy. The dimensionless ratio $E/m_ec^2$ controls how hard the photon hits: when it is small the photon barely loses energy; when it is of order one, energy loss is dramatic. By energy conservation, whatever the photon loses the electron gains as kinetic energy.

T_e = E - E' = E\,\frac{\dfrac{E}{m_ec^2}(1-\cos\theta)}{1+\dfrac{E}{m_ec^2}(1-\cos\theta)} \qquad (5)

where: $T_e$ is the recoil electron's kinetic energy. This is maximal at backscatter ( $\theta=180^\circ$ ), where the photon reverses direction and transfers the most momentum — the origin of the "Compton edge" seen in $\gamma$ -ray detector spectra.

Derivation AThe wavelength shift from four-momentum conservation

The goal is to derive equation (3) with no approximation, starting only from conservation of relativistic energy and momentum and the two dispersion relations (1) and (2). The strategy is to isolate the electron's four-momentum, then square it to invoke the mass-shell condition $E_e^2-(p_ec)^2=(m_ec^2)^2$ , which conveniently eliminates the electron's recoil angle. The one thing to watch is bookkeeping of the dot product between the incoming and outgoing photon momenta, which is where the scattering angle $\theta$ enters through a cosine. Everything is exact; no term is dropped.

Step 1. Write conservation. Energy: $E + m_ec^2 = E' + E_e$ , so $E_e = E - E' + m_ec^2$ . Momentum (a vector equation): $\mathbf{p}_\gamma = \mathbf{p}'_\gamma + \mathbf{p}_e$ , so $\mathbf{p}_e = \mathbf{p}_\gamma - \mathbf{p}'_\gamma$ .

Step 2. Square the electron momentum. Using $|\mathbf{p}_\gamma|=E/c$ and $|\mathbf{p}'_\gamma|=E'/c$ and the angle $\theta$ between them, $p_e^2 = |\mathbf{p}_\gamma-\mathbf{p}'_\gamma|^2 = \dfrac{E^2}{c^2}+\dfrac{E'^2}{c^2}-\dfrac{2EE'}{c^2}\cos\theta$ .

Step 3. Apply the electron mass shell (2): $E_e^2-(p_ec)^2=(m_ec^2)^2$ . Substitute $E_e$ from Step 1 and $p_e^2c^2$ from Step 2. Expanding $E_e^2=(E-E'+m_ec^2)^2$ gives $E^2+E'^2+(m_ec^2)^2-2EE'+2m_ec^2(E-E')$ .

Step 4. Combine. $E_e^2-(p_ec)^2 = \big[E^2+E'^2-2EE'+2m_ec^2(E-E')+(m_ec^2)^2\big] - \big[E^2+E'^2-2EE'\cos\theta\big]$ . The $E^2$ and $E'^2$ terms cancel, leaving $(m_ec^2)^2 = -2EE'+2EE'\cos\theta+2m_ec^2(E-E')+(m_ec^2)^2$ .

Step 5. Cancel $(m_ec^2)^2$ and solve. $0 = -2EE'(1-\cos\theta)+2m_ec^2(E-E')$ , i.e. $m_ec^2(E-E')=EE'(1-\cos\theta)$ . Divide by $EE'\,m_ec^2$ : $\dfrac{1}{E'}-\dfrac{1}{E}=\dfrac{1-\cos\theta}{m_ec^2}$ . Since $E=hc/\lambda$ , $1/E=\lambda/hc$ , so $\dfrac{\lambda'-\lambda}{hc}=\dfrac{1-\cos\theta}{m_ec^2}$ , giving $\Delta\lambda=\dfrac{h}{m_ec}(1-\cos\theta)$ , equation (3). Squaring $\mathbf{p}_e=\mathbf{p}_\gamma-\mathbf{p}'_\gamma$ and imposing the mass shell yields exactly $1/E'-1/E=(1-\cos\theta)/m_ec^2$ and hence $\Delta\lambda=\lambda_C(1-\cos\theta)$ .

Go deeper · Derivation B · 1 minThe shift is independent of the incident wavelength

Derivation BThe shift is independent of the incident wavelength

Here the aim is to prove the article's owned structural claim: the absolute wavelength shift $\Delta\lambda$ depends only on the scattering angle and the universal constant $\lambda_C$ , and carries no dependence on the incoming wavelength $\lambda$ (equivalently, on the incoming energy $E$ ). This is what makes Compton scattering a clean, material- and beam-independent test of the photon hypothesis, and it is not obvious — the fractional shift and the scattered energy both depend strongly on $E$ . The route is simply to inspect equation (3) and contrast it with the fractional and energy forms, being careful to distinguish "the shift $\Delta\lambda$ " from "the fractional shift $\Delta\lambda/\lambda$ ."

Step 1. Read equation (3): $\Delta\lambda=\lambda_C(1-\cos\theta)$ . The right-hand side contains only $\lambda_C=h/m_ec$ (fixed by nature) and $\theta$ (chosen by the detector). Neither $\lambda$ nor $E$ appears. Hence $\partial(\Delta\lambda)/\partial\lambda=0$ identically.

Step 2. Contrast with the fractional shift: $\dfrac{\Delta\lambda}{\lambda}=\dfrac{\lambda_C}{\lambda}(1-\cos\theta)$ , which does depend on $\lambda$ , growing as $\lambda$ shrinks. This is why the effect is invisible for visible light ( $\lambda\sim500$ nm gives $\Delta\lambda/\lambda\sim10^{-5}$ ) yet obvious for X-rays ( $\lambda\sim70$ pm gives $\Delta\lambda/\lambda\sim3\%$ ).

Step 3. Contrast with the energy form (4): $E'=E/[1+(E/m_ec^2)(1-\cos\theta)]$ , whose energy loss $E-E'$ depends strongly and nonlinearly on $E$ . So the "no incident-wavelength dependence" statement is special to the absolute wavelength variable $\Delta\lambda$ ; it is a happy consequence of writing the kinematics in $1/E$ , which is linear in $\lambda$ .

Step 4. State the physical upshot. Because $\Delta\lambda$ is angle-only, plotting scattered wavelength against angle for any incident line, on any target, must fall on the single universal curve $\lambda+\lambda_C(1-\cos\theta)$ . Compton's confirmation of exactly this — with the coefficient equal to $h/m_ec$ using the independently known electron mass — is what settled the photon debate. Evaluating $\Delta\lambda$ at fixed $\theta=90^\circ$ for three very different incident wavelengths — visible light, Mo Kα, and a hard $\gamma$ -ray — gives $\Delta\lambda=2.426$ pm identically in all three cases, while the fractional shift and the scattered energy differ by orders of magnitude.

Limiting cases

Each limit of the exact formulas is a physical check on the construction.

Case	Result	Meaning
Forward scattering $\theta\to0$	$\Delta\lambda\to0$ , $E'\to E$	A photon that continues straight ahead transfers no momentum; nothing happens.
Backscatter $\theta=180^\circ$	$\Delta\lambda=2\lambda_C$ , $T_e$ maximal	The photon reverses; maximum momentum transfer and the largest energy loss — the Compton edge.
Low energy $E\ll m_ec^2$	$E'\to E$ , $\Delta\lambda/\lambda\to0$	The photon is too soft to move the electron appreciably; wavelength is essentially unchanged — the classical Thomson limit.
High energy $E\gg m_ec^2$	$E'\to \dfrac{m_ec^2}{1-\cos\theta}$ (for $\theta\neq0$ )	the scattered-photon energy approaches an angle-dependent limit set by the electron rest energy, while most of the incident energy is transferred to the recoil electron.

Common trapThinking the classical wave picture "almost" explains the shift

It does not — classical Thomson scattering from a free electron predicts no change in the photon frequency. In the linear classical picture, the electron oscillates at the incident frequency and radiates at that same frequency. There is no continuous "classical Compton effect" that becomes the quantum one as intensity rises: the Compton shift is intrinsically associated with a photon carrying momentum $h/\lambda$ and transferring energy and momentum to the electron. Raising the intensity increases the number of scattered photons but does not change the kinematics of an individual photon, so $\Delta\lambda$ is intensity-independent. An unshifted component can also arise from coherent scattering by bound electrons or the atom as a whole; because the effective recoiling mass is much larger than $m_e$ , the corresponding wavelength shift is negligible.

Worked examples

Example 1 — The shift for molybdenum Kα X-rays. Compton's canonical measurement used Mo K $\alpha$ radiation, $\lambda=71.1$ pm, scattered through $\theta=90^\circ$ from graphite. Find the scattered wavelength and the fractional shift.

Step 1. Compute the Compton wavelength: $\lambda_C=h/m_ec=(6.626\times10^{-34})/[(9.109\times10^{-31})(2.998\times10^8)]\,\mathrm{m}=2.426\times10^{-12}\,\mathrm{m}=2.426\,\mathrm{pm}$ .

Step 2. Apply equation (3) at $\theta=90^\circ$ : $1-\cos90^\circ=1$ , so $\Delta\lambda=\lambda_C\cdot1=\mathbf{2.426\ pm}$ . The scattered wavelength is $\lambda'=71.1+2.43=\mathbf{73.5\ pm}$ .

Step 3. Fractional shift: $\Delta\lambda/\lambda = 2.426/71.1 = \mathbf{3.4\%}$ — small but readily measurable with a crystal spectrometer, which is exactly why the effect was found with X-rays and not visible light. The takeaway: the $90^\circ$ shift equals one Compton wavelength exactly, $2.426$ pm, regardless of which X-ray line you use.

Example 2 — Backscattering a 0.511 MeV photon and the Compton edge. A photon with energy equal to the electron rest energy, $E=m_ec^2=0.511$ MeV (an annihilation photon), backscatters at $\theta=180^\circ$ . Find the scattered energy and the recoil electron energy.

Step 1. Set $u\equiv E/m_ec^2=1$ and $1-\cos180^\circ=2$ . From equation (4), $E'=E/[1+u(1-\cos\theta)]=0.511/[1+1\cdot2]=0.511/3=\mathbf{0.1704\ MeV}$ .

Step 2. Recoil kinetic energy by conservation (5): $T_e=E-E'=0.511-0.1704=\mathbf{0.3407\ MeV}$ . This is the maximum energy the electron can acquire from a $0.511$ MeV photon — the position of the Compton edge in a scintillator's pulse-height spectrum.

Step 3. Sanity check with the wavelength form. $\lambda=hc/E$ . For $E=0.511\,\mathrm{MeV}$ , $\lambda=2.426\,\mathrm{pm}=\lambda_C$ (the photon energy equals $m_ec^2$ exactly when $\lambda=\lambda_C$ ). Then $\lambda'=\lambda+2\lambda_C=3\lambda_C$ , so $E'=hc/\lambda'=E/3=0.1704$ MeV — consistent. The takeaways: at $E=m_ec^2$ and $180^\circ$ the photon keeps one-third of its energy ( $0.170\,\mathrm{MeV}$ ) and the electron carries two-thirds ( $0.341\,\mathrm{MeV}$ ).

Origins · 1 minArthur Holly Compton and how this was discovered

Arthur Holly Compton

1892–1962 · American physicist whose X-ray scattering experiments established the particle nature of light

Arthur Holly Compton performed his decisive X-ray scattering measurements at Washington University in St. Louis in 1922–1923, publishing "A Quantum Theory of the Scattering of X-rays by Light Elements" in the Physical Review in 1923. By treating the X-ray as a photon carrying momentum $h/\lambda$ and applying relativistic collision kinematics, he predicted and confirmed the angle-dependent wavelength shift now bearing his name, quantitatively vindicating Einstein's light-quantum hypothesis. He shared the 1927 Nobel Prize in Physics with C. T. R. Wilson, whose cloud chamber made the recoil electrons visible. Compton later led the University of Chicago's plutonium work during the Manhattan Project.

History

1905

— Einstein proposes the light quantum to explain the photoelectric effect

1916

— Einstein shows a light quantum must also carry momentum

h\nu/c

1923

— Compton measures the X-ray wavelength shift and derives

\Delta\lambda=\lambda_C(1-\cos\theta)

, publishing in Phys. Rev. 21, 483

1923

— Debye independently publishes the same kinematic analysis

1925

— Bothe and Geiger confirm photon and electron are emitted simultaneously, ruling out the BKS statistical alternative

1927

— Compton shares the Nobel Prize in Physics with C. T. R. Wilson

1929

— Klein and Nishina compute the relativistic scattering cross section using the Dirac equation

Going further

References. Compton, "A Quantum Theory of the Scattering of X-rays by Light Elements," Phys. Rev. 21, 483 (1923) · Eisberg & Resnick, Quantum Physics, 2nd ed., §2.4 · Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., Ch. 1 (photon momentum) · Klein & Nishina, Z. Phys. 52, 853 (1929) · Krane, Modern Physics, Ch. 3. Key numerical benchmarks: $\lambda_C=h/m_ec=2.4263$ pm; the reciprocal form $1/E'-1/E=(1-\cos\theta)/m_ec^2$ follows from four-momentum conservation; the $90^\circ$ shift $=2.426$ pm and the $180^\circ$ shift $=4.853$ pm; $\Delta\lambda$ is independent of the incident wavelength at fixed angle; at $E=m_ec^2$ and $\theta=180^\circ$ , $E'=0.1703$ MeV and $T_e=0.3407$ MeV; and in the Thomson low-energy limit $E'/E\to1$ .

Builds on: Photoelectric Effect and the Photon · Wave–Particle Duality · de Broglie Matter Waves. Leads to: The Klein–Nishina Cross Section · Inverse Compton Scattering · Pair Production and Annihilation.

Why does the electron have to be treated as free when it is bound in an atom?

For X-rays the photon energy (thousands of eV) is far larger than the outer-electron binding energy (a few eV), so on the collision timescale the electron responds as if free — the binding is a negligible perturbation. This is why the shifted "Compton line" appears. Photons that scatter off tightly bound inner electrons, whose binding is comparable to the photon energy, cause the whole atom to recoil; then the relevant mass in $\lambda_C=h/Mc$ is the atomic mass, thousands of times larger, so the shift is negligible and you see an unshifted line at the incident wavelength. Real spectra show both lines, and their relative strength depends on the target and the geometry.

Is Compton scattering the same thing as the photoelectric effect?

No. In the photoelectric effect the photon is absorbed entirely and the electron is ejected, which requires a bound electron and a nearby nucleus to conserve momentum — a free electron cannot absorb a photon while conserving both energy and momentum. In Compton scattering the photon is not absorbed; it survives with reduced energy and a new direction, and the electron recoils. The two dominate in different energy ranges: the photoelectric effect at low photon energies and high atomic number, Compton scattering at intermediate energies, and pair production above $2m_ec^2$ .

Can a free electron simply absorb a photon instead of scattering it?

No — it is kinematically forbidden. If a free electron at rest absorbed a photon, conservation of energy and momentum could not both be satisfied: in the electron's final rest frame the absorbed photon's momentum would have nowhere to go. Working through the four-momentum algebra gives a contradiction ( $m_ec^2$ would have to equal $\sqrt{(m_ec^2)^2+(pc)^2}$ with $p\neq0$ ). That is precisely why a photon interacting with a free electron must scatter (Compton), whereas absorption (the photoelectric effect) needs a third body — a bound nucleus — to take up the recoil.

Why does the shift depend on cosine of the angle and not the angle itself?

Because the angle enters through the dot product of the incoming and outgoing photon momenta. When you square the electron's momentum $\mathbf{p}_e=\mathbf{p}_\gamma-\mathbf{p}'_\gamma$ , the cross term is $-2\,\mathbf{p}_\gamma\!\cdot\!\mathbf{p}'_\gamma=-2|\mathbf{p}_\gamma||\mathbf{p}'_\gamma|\cos\theta$ . The cosine is not a modelling choice; it is the geometric content of the momentum-conservation vector equation. This is also why forward scattering ( $\cos\theta=1$ ) gives zero shift and backscatter ( $\cos\theta=-1$ ) gives the maximum: the momentum transfer is smallest and largest there, respectively.

How big is the effect for visible light — could I see it?

The absolute shift is always $\lambda_C(1-\cos\theta)\le4.85$ pm. For visible light, $\lambda\approx500{,}000$ pm, so the fractional shift is at most about $10^{-5}$ — utterly negligible and far below what an optical spectrometer can resolve. The effect only becomes fractionally significant when $\lambda$ approaches $\lambda_C=2.43$ pm, i.e. for X-rays and $\gamma$ -rays. So no, you cannot see Compton scattering with visible light; the wavelength change is real but immeasurably small.

Where does the classical Thomson cross section fit in?

Thomson scattering is the low-energy limit of the full quantum (Klein–Nishina) treatment. The kinematics derived in this article tell you the wavelength of the scattered photon; they say nothing about how much scatters in each direction. That angular intensity is the cross section, computed in QED as the Klein–Nishina formula, which reduces at $E\ll m_ec^2$ to the classical Thomson cross section $\sigma_T=6.65\times10^{-29}\,\mathrm{m}^2$ , independent of energy. So in the low-energy limit both the wavelength shift and the intensity go over to their classical values — the photon is too soft to reveal its particle nature.

What is inverse Compton scattering?

It is the same process viewed in a different frame. If a low-energy photon collides with an ultra-relativistic electron, the electron can give energy to the photon, blueshifting it to high energy — the reverse of the ordinary case where the photon loses energy to a slow electron. The kinematics are identical four-momentum conservation; only the initial conditions differ. Inverse Compton scattering is a major mechanism in high-energy astrophysics (upscattering starlight or the cosmic microwave background off relativistic electrons in jets and clusters) and is used in laboratory light sources to produce tunable X- and $\gamma$ -ray beams.