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 Incoming links: Compton scattering

Baryon Jeans length across recombination 2026-10-05

Before cosmological recombination, photon-baryon sound speed controls the baryonic comoving Jeans length. The tightly coupled photon-baryon fluid has

c_{s}^{2} = c^{2} / [3 (1 + R_{b})]

, where

R_{b} = 3 \overset{ρ}{ˉ}_{b} / (4 \overset{ρ}{ˉ}_{γ}) \propto a

. Using total background density in the instantaneous dynamical Jeans estimate, the scale grows as

a

during radiation domination, approaches a plateau in strongly baryon-loaded matter domination, and drops when photon decoupling removes photon pressure support. Subsequently, adiabatic monatomic-gas cooling gives

T_{b} \propto a^{- 2}

and a comoving Jeans length proportional to

a^{- 1/2}

. Residual Compton scattering can delay that cooling law.

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Bremsstrahlung 2026-10-06

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Bremsstrahlung is electromagnetic radiation emitted when a charged particle accelerates during interaction with another charged particle or field. The microscopic process

e^{-} + Z \to e^{-} + Z + γ

creates a photon, while its inverse absorbs one. Thus it contributes to a photon-number collision source in the photon number current, unlike ordinary Compton scattering. Thermal bremsstrahlung averages this emission over a thermal distribution of charged particles.

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Compton cooling by the cosmic microwave background 2026-10-07

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Hot electrons can transfer energy to Cosmic microwave background photons through Compton scattering. For a fixed electron fraction

x_{e}

and electron temperature well above the radiation temperature, the optically thin gas cooling time scales approximately as

(1 + z)^{- 4} / x_{e}

. Its volume loss scales with electron density, not with

n_{H}^{2}

, because the radiation density is independent of local gas density. The process tends toward the radiation temperature; it heats rather than cools electrons colder than that radiation.

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Double Compton scattering Created 2026-10-06 Updated 2026-10-07

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Double Compton scattering is

e^{-} + γ \to e^{-} + γ + γ

, together with its inverse. The forward reaction has one incoming and two outgoing photons, so it changes their number and can source the divergence of the photon number current. Ordinary Compton scattering has one incoming and one outgoing photon and conserves photon number. Forward and inverse rates can balance without either reaction individually conserving it.

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Inverse Compton cooling 2026-10-07

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Thermal electrons can transfer energy to lower-energy radiation photons through Compton scattering. In the nonrelativistic Thomson regime, the net cooling power per electron is

4 σ_{T} c U_{γ} k_{B} (T_{e} - T_{γ}) / (m_{e} c^{2})

for a blackbody photon bath. A sufficiently hot electron population therefore loses energy; a colder population is heated instead. The Cosmic microwave background radiation energy density grows as

(1 + z)^{4}

, making this channel more important for high-redshift diffuse ionized gas.

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Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 3 / 10A / Solution Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 44 / 4 / Solution 2026-10-03

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Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 46 / 4 / Solution Created 2026-10-03 Updated 2026-10-07

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Take the muon mass to be

m_{μ}

, and use the quantum electrodynamics convention

L_{int} = - e \overset{ˉ}{ψ} γ^{μ} ψ A_{μ}

with

e > 0

. Let

p_{1}, p_{2}

be the incoming muon and antimuon momenta and

k_{1}, k_{2}

the outgoing photon momenta, so

p_{1} + p_{2} = k_{1} + k_{2}

. External particles are on shell:

p_{1}^{2} = p_{2}^{2} = m_{μ}^{2}

and

k_{1}^{2} = k_{2}^{2} = 0

. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

Figure 1.
Two photon orders for muon annihilation and Compton scattering
.

The QED Feynman rules needed here are the vertex

- i e γ^{μ}

, the Dirac propagator

\frac{i ( \neq q + m _{μ} )}{q ^{2} - m _{μ}^{2} + i 0}, \neq q = γ^{μ} q_{μ},

(1)

an incoming muon spinor

u (p_{1})

, an incoming antimuon adjoint

\overset{v}{ˉ} (p_{2})

, and an outgoing photon polarization

ϵ_{μ}^{*} (k)

. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define

R (q) = (\neq q + m_{μ}) / (q^{2} - m_{μ}^{2} + i 0)

and

\neq ϵ = γ^{μ} ϵ_{μ}

With the convention that a diagram contributes

i M

, the two muon-antimuon annihilation amplitudes are

M_{t} M_{u} M_{ann} = - e^{2} \overset{v}{ˉ} (p_{2}) \neq ϵ_{2}^{*} R (p_{1} - k_{1}) \neq ϵ_{1}^{*} u (p_{1}), = - e^{2} \overset{v}{ˉ} (p_{2}) \neq ϵ_{1}^{*} R (p_{1} - k_{2}) \neq ϵ_{2}^{*} u (p_{1}), = M_{t} + M_{u} .

(2)

Their denominators are respectively

t - m_{μ}^{2} + i 0

and

u - m_{μ}^{2} + i 0

, where

t = (p_{1} - k_{1})^{2}

u = (p_{1} - k_{2})^{2}

. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k \cdot ϵ = 0$ , and represents an equivalence class

ϵ \sim ϵ + α k

. For a real null momentum, adding

α k

preserves transversality. One can additionally choose transverse spatial vectors with

ϵ^{0} = 0

and

ϵ^{*} \cdot ϵ = - 1

; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace

ϵ_{1}^{*}

k_{1}

and put

q_{t} = p_{1} - k_{1}

q_{u} = p_{1} - k_{2} = k_{1} - p_{2}

. The external Dirac equations give

(\neq p_{1} - m_{μ}) u (p_{1}) = 0, \overset{v}{ˉ} (p_{2}) (\neq p_{2} + m_{μ}) = 0.

(3)

Away from propagator poles,

R (q) (\neq q - m_{μ}) = (\neq q - m_{μ}) R (q) = I

; the identity extends with the common Feynman prescription. Therefore

R (q_{t}) \neq k_{1} u (p_{1}) \overset{v}{ˉ} (p_{2}) \neq k_{1} R (q_{u}) = R (q_{t}) (\neq p_{1} - \neq q_{t}) u (p_{1}) = - u (p_{1}), = \overset{v}{ˉ} (p_{2}) (\neq q_{u} + \neq p_{2}) R (q_{u}) = \overset{v}{ˉ} (p_{2}) .

(4)

The two terms in the bracket then give

- \overset{v}{ˉ} \neq ϵ_{2}^{*} u

and

+ \overset{v}{ˉ} \neq ϵ_{2}^{*} u

, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under $ϵ_{i} \mapsto ϵ_{i} + α k_{i}$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as

p, k

and outgoing momenta as

p^{'}, k^{'}

, with

p + k = p^{'} + k^{'}

. The outgoing muon contributes

\overset{u}{ˉ} (p^{'})

, the incoming photon contributes

ϵ (k)

without conjugation, and the outgoing photon contributes

ϵ^{' *} (k^{'})

. The two tree-level Compton amplitudes are

M_{s} M_{u} M_{C} = - e^{2} \overset{u}{ˉ} (p^{'}) \neq ϵ^{' *} R (p + k) \neq ϵ u (p), = - e^{2} \overset{u}{ˉ} (p^{'}) \neq ϵ R (p - k^{'}) \neq ϵ^{' *} u (p), = M_{s} + M_{u} .

(5)

These are the lower two diagrams. Their intermediate muon momenta are

p + k

and

p - k^{'}

, with denominators

s - m_{μ}^{2} + i 0

and

u - m_{μ}^{2} + i 0

. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order

e^{2}

in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing

p_{2} \mapsto - p^{'}

and one outgoing photon momentum

k_{1} \mapsto - k

, with the corresponding external wavefunction replacements. The annihilation

t

channel becomes the Compton

s

channel and the other channel becomes its

u

channel. Identical final photons require a factor

1/2!

in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

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Past exam of the mathematics course of the University of Cambridge / 2015 / ia / Paper 4 / 4C / Solution Created 2026-09-24 Updated 2026-10-06

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Use the Minkowski metric convention

+ - - -

. A particle's four-momentum is

P = (E / c, p)

, with

P \cdot P = m^{2} c^{2}

for rest mass

m

. For a photon, its energy and wavelength obey

E = h c / λ

, and its four-momentum is null. Here

h

is the Planck constant.

The initially stationary electron has

P = (m c, 0, 0, 0)

. Choose the scattering plane as the

x y

plane. The incoming and outgoing photon four-momenta are

K_{1} = \frac{h}{λ _{1}} (1, 1, 0, 0), K_{2} = \frac{h}{λ _{2}} (1, cos θ, sin θ, 0) .

(1)

By four-momentum conservation, the final electron has

P^{'} = P + K_{1} - K_{2}

. Its rest mass is unchanged, so the relativistic energy-momentum relation gives

P^{'2} = P^{2} = m^{2} c^{2}

. Since

K_{1}^{2} = K_{2}^{2} = 0

, expansion yields

P \cdot (K_{1} - K_{2}) = K_{1} \cdot K_{2},

(2)

m c h (\frac{1}{λ _{1}} - \frac{1}{λ _{2}}) = \frac{h ^{2}}{λ _{1} λ _{2}} (1 - cos θ) .

(3)

Multiplying through gives the Compton scattering shift:

λ_{2} - λ_{1} = \frac{h}{m c} (1 - cos θ) = \frac{2 h}{m c} sin^{2} (θ /2) .

(4)

It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.

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Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 9A / a / Solution Created 2026-09-24 Updated 2026-10-05

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Use momentum conservation and conservation of energy in the laboratory inertial frame. The photon has momentum magnitude

E / c

, while the outgoing Electron satisfies the relativistic energy-momentum relation. Its total energy and momentum obey

E_{e} = m c^{2} + E_{1} - E_{2}, c^{2} ∣ p_{e} ∣^{2} = E_{1}^{2} + E_{2}^{2} - 2 E_{1} E_{2} cos θ .

(1)

Substitute these into

E_{e}^{2} = m^{2} c^{4} + c^{2} ∣ p_{e} ∣^{2}

. Cancelling

m^{2} c^{4}

and

(E_{1} - E_{2})^{2}

gives

m c^{2} (E_{1} - E_{2}) = E_{1} E_{2} (1 - cos θ)

. Both photon energies are positive, so division yields

\frac{1}{E _{2}} - \frac{1}{E _{1}} = \frac{1 - cos θ}{m c ^{2}}, E_{2} = \frac{E _{1}}{1 + ( E _{1} / m c ^{2} ) ( 1 - cos θ )} .

(2)

This is Compton scattering. The energies include the Electron's rest energy; using only its kinetic energy in the mass-shell equation would give the wrong result. The outgoing photon has unchanged energy at

θ = 0

and smaller energy for a nonzero scattering angle.

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Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 312 / 2 / iii / Solution Created 2026-10-03 Updated 2026-10-06

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In the local orthonormal tetrad, the components of the photon number current are

J^{\hat{0}} = \int f d^{3} p = n_{γ}, J^{\hat{i}} = \int f e^{i} d^{3} p .

(1)

They are respectively the photon number density and the photon number flux. Thus $J^{μ}$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,

\overset{n}{ˉ}_{γ} = \frac{4 π}{a ^{3}} \int_{0}^{\infty} ϵ^{2} \overset{ˉ}{f} (ϵ) d ϵ .

(2)

Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, gives

\nabla_{μ} J^{μ} = \int \frac{d ^{3} p}{E} C [f],

(3)

where

C [f]

is the collision operator in the covariant streaming equation. Therefore

\nabla_{μ} J^{μ} = 0 ⟺ \int \frac{d ^{3} p}{E} C [f] = 0.

(4)

Collisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives

\overset{n}{ˉ}_{γ} \propto a^{- 3}

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 3 / iv / b / Solution Created 2026-10-03 Updated 2026-10-05

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There are two tree-level Feynman diagrams for Compton scattering: an

s

-channel ordering in which the electron first absorbs the incoming photon, with internal momentum

p + q

, and a crossed

u

-channel ordering in which it first emits the outgoing photon, with internal momentum

p - q^{'}

Figure 1.
Tree-level Compton-scattering diagrams
. The two orderings of photon absorption and emission give the electron-exchange s-channel and u-channel diagrams.

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Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 346 / 1 / a / Solution Created 2026-10-03 Updated 2026-10-05

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Write

x

for a comoving coordinate,

u

for the peculiar velocity,

H = \overset{a}{˙} / a

for the Hubble parameter, and

δ = δ ρ / \overset{ρ}{ˉ}

for the density contrast. For a homogeneous self-gravitating fluid, the linearized continuity equation, Euler equations for an inviscid fluid, and cosmological Poisson equation are

\dot{δ} = - \frac{1}{a} \nabla \cdot u, \dot{u} + H u = - \frac{c _{s}^{2}}{a} \nabla δ - \frac{1}{a} \nabla ϕ, \nabla^{2} ϕ = 4 π G a^{2} \overset{ρ}{ˉ} δ .

(1)

Here

c_{s}

is the physical adiabatic sound speed, gradients refer to comoving coordinates, and the homogeneous background force has already been subtracted. Differentiating the first equation and eliminating

u

and

ϕ

gives

\ddot{δ} + 2 H \dot{δ} - \frac{c _{s}^{2}}{a ^{2}} \nabla^{2} δ - 4 π G \overset{ρ}{ˉ} δ = 0.

(2)

A Fourier mode with comoving wavenumber

k

therefore obeys

\ddot{δ}_{k} + 2 H \dot{δ}_{k} + (\frac{c _{s}^{2} k ^{2}}{a ^{2}} - 4 π G \overset{ρ}{ˉ}) δ_{k} = 0.

(3)

The Jeans wavenumber occurs where the coefficient in parentheses vanishes. Since a comoving wavelength is

2 π / k

, the result is

k_{J} = \frac{a 4 π G ρ ˉ}{c _{s}}, λ_{J}^{com} = \frac{c _{s}}{a} \frac{π}{G ρ ˉ} .

(4)

Below this comoving Jeans length, pressure restores a displaced fluid element: there are acoustic waves rather than growing Jeans instability. In a static background they oscillate; expansion changes their amplitude and frequency, and dissipation can damp them. Pressure support alone does not erase them. In the pre-recombination photon-baryon fluid, Silk damping supplies a separate erasure mechanism.

For collisionless matter, replace the sound speed by a characteristic velocity dispersion. For an isotropic Maxwell-Boltzmann velocity distribution, defining

σ^{2} = ⟨ u_{x}^{2} ⟩ = ⟨ ∣ u ∣^{2} ⟩ /3

, the static collisionless marginal-stability calculation gives

λ_{J, collisionless}^{com} = \frac{σ}{a} \frac{π}{G ρ ˉ} = \frac{v _{rms}}{a} \frac{π}{3 G ρ ˉ} .

(5)

The coefficient depends on the distribution and the convention for velocity dispersion. Small-scale suppression is now collisionless free streaming and phase mixing, rather than collisional acoustic waves. An instantaneous collisionless Jeans length must be distinguished from the accumulated comoving distance

\int v d t / a

traveled since decoupling.

For the requested thermal histories, an Einstein-de Sitter universe describes the matter-era limit: it cannot literally also have a radiation-dominated epoch. Interpret the question as a spatially flat universe with negligible cosmological constant, passing from radiation domination,

a \propto t^{1/2}

, to matter domination,

a \propto t^{2/3}

. For the usual schematic dynamical Jeans scale use

\overset{ρ}{ˉ}_{tot}

in the estimate, so

λ_{dyn}^{com} \propto {c_{s} a, c_{s} a^{1/2}, t ≪ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 4}, t ≫ t_{eq}, \overset{ρ}{ˉ}_{tot} \propto a^{- 3} .

(6)

Near the particle horizon a relativistic perturbation calculation replaces the Newtonian derivation; the scaling estimate still identifies the relevant sound-crossing or streaming scale.

Before cosmological recombination, adiabatic initial conditions keep the entropy per baryon fixed in the tightly coupled photon-baryon fluid. With

R_{b} = 3 \overset{ρ}{ˉ}_{b} / (4 \overset{ρ}{ˉ}_{γ}) \propto a

, its photon-baryon sound speed is

c_{s}^{2} = \frac{c ^{2}}{3 ( 1 + R _{b} )} .

(7)

For example, this follows from

δ ρ_{γ} / \overset{ρ}{ˉ}_{γ} = (4/3) δ ρ_{b} / \overset{ρ}{ˉ}_{b}

and

δ P = c^{2} δ ρ_{γ} /3

. In the radiation-dominated limit

R_{b} ≪ 1

c_{s} ≃ c / 3

; in the strongly baryon-loaded matter-era limit

R_{b} ≫ 1

c_{s} \propto a^{- 1/2}

. Thus the baryon Jeans length across recombination has the asymptotic history

λ_{J, b}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, approximately constant, a^{- 1/2} \propto t^{- 1/3}, t ≪ t_{eq}, t_{eq} ≪ t < t_{rec}, t > t_{rec} with adiabatic gas cooling .

(8)

The transition to the plateau is smooth: retaining

1 + R_{b}

gives

λ_{J, b}^{com} \propto a^{1/2} / 1 + R_{b}

in matter domination. If baryon loading is still small, this intermediate segment instead rises as

a^{1/2}

until loading becomes important.

At photon decoupling, radiation ceases to provide pressure support to the baryons, so the sound speed drops from the coupled-fluid value to that of a nonrelativistic ideal gas,

c_{s, b}^{2} = γ k_{B} T_{b} / (μ m_{p})

, with

γ = 5/3

. The drop in comoving Jeans length at fixed density is the same ratio of sound speeds, typically several orders of magnitude. Subsequently an isentropic process gives

T_{b} \propto \overset{ρ}{ˉ}_{b}^{γ - 1} \propto a^{- 2}

, so

c_{s, b} \propto a^{- 1}

and

λ_{J, b}^{com} \propto a^{- 1/2}

. Residual energy exchange through Compton scattering can initially keep

T_{b} \propto a^{- 1}

, producing a short approximately flat gas segment before the decreasing segment. The idealized sketch assumes immediate adiabatic cooling:

log(comoving baryon Jeans scale)
  ^
  |                   __________________
  |                 /                   |
  |               /                     |  loss of photon pressure
  |             /                       |
  |           /                         +---\
  |         /                                \
  +-------------------|-----------------|----------> log(time)
                    t_eq              t_rec
        slope 1/2          slope ~0        slope -1/3

For the cold-dark-matter kinetic Jeans scale, take the stated toy history to mean that collisions maintain a particle temperature equal to the photon temperature until kinetic decoupling. This is an assumption about thermal coupling; a negligible collision rate after decoupling makes the subsequent treatment collisionless. Assume the ordering

t_{NR} < t_{dec} < t_{eq} < t_{rec}

. While relativistic, the characteristic particle speed is constant; while nonrelativistic but thermally coupled,

σ \propto T \propto a^{- 1/2}

; after kinetic decoupling, particle momentum redshifts as

a^{- 1}

and

σ \propto a^{- 1}

. Combining these with the background dynamical density gives

λ_{dyn, CDM}^{com} \propto ⎩ ⎨ ⎧ a \propto t^{1/2}, a^{1/2} \propto t^{1/4}, constant, a^{- 1/2} \propto t^{- 1/3}, t < t_{NR}, t_{NR} < t < t_{dec}, t_{dec} < t < t_{eq}, t > t_{eq} .

(9)

There is no sudden change at cosmological recombination: cold dark matter has already decoupled. Its small initial velocity dispersion makes its suppression scale much smaller than the pre-recombination baryon scale.

log(comoving CDM dynamical Jeans scale)
  ^
  |                        _______________
  |                    ___/               \
  |                ___/                    \       no jump
  |              /                          \      at t_rec
  |            /                             \
  +-----------|------------|--------------|----|------> log(time)
            t_NR         t_dec          t_eq t_rec
   slope 1/2    slope 1/4       slope 0       slope -1/3

The density convention matters. If one instead defines a species' self-gravitating collisionless Jeans length using only its own nonrelativistic density

\overset{ρ}{ˉ}_{m} \propto a^{- 3}

, its thermally coupled segment is constant and its decoupled segment decreases as

a^{- 1/2}

even during radiation domination. The constant radiation-era segment in the second sketch is the usual instantaneous background dynamical streaming estimate, not an exact single-species self-gravity threshold. A complete multicomponent calculation uses the separate perturbation equations and their combined cosmological Poisson equation.

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 12C / c / Solution Created 2026-09-24 Updated 2026-09-25

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Let the incident and final photon momentum magnitudes be

p = \frac{h}{λ}, p^{'} = \frac{h}{λ ^{'}} .

(1)

In a nontrivial collinear collision, the photon is backscattered and the electron moves in the original photon direction, so momentum conservation gives electron momentum

P = p + p^{'}

. Conservation of relativistic energy gives

p c + m c^{2} = p^{'} c + m^{2} c^{4} + P^{2} c^{2} .

(2)

Substitute

P = p + p^{'}

, isolate the square root, and square. Cancellation leaves

m c (p - p^{'}) = 2 p p^{'} .

(3)

Dividing by

p p^{'}

and using the wavelength definitions gives the Compton scattering shift

λ^{'} = λ + \frac{2 h}{m c} .

(4)

The other collinear possibility is the trivial forward solution

λ^{'} = λ

, in which the electron remains at rest.

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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 301 / 4 / iii / Solution 2026-09-28

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Label positron Compton scattering as

γ (k, λ) + e^{+} (p, s) ⟶ γ (k^{'}, λ^{'}) + e^{+} (p^{'}, s^{'}) .

(1)

There are two positron-exchange diagrams: the

s

-channel ordering with internal momentum

- p - k

, and the

u

-channel ordering with internal momentum

- p + k^{'}

. Reading opposite to the fermion arrow gives

i M = (- i e)^{2} \overline{v} (p) [γ^{μ} \frac{i ( - \neq p - \neq k + m )}{( p + k ) ^{2} - m ^{2} + i ϵ} γ^{ν} + γ^{ν} \frac{i ( - \neq p + \neq k ^{'} + m )}{( p - k ^{'} ) ^{2} - m ^{2} + i ϵ} γ^{μ}] v (p^{'}) ϵ_{μ} (k, λ) ϵ_{ν}^{*} (k^{'}, λ^{'}) .

(2)

It is the crossed form of electron Compton scattering and again satisfies the Ward identity only after both orderings are included.

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Photon absorption 2026-10-06

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Photon absorption removes a photon and transfers its energy and momentum to matter. It contributes a negative photon-number collision source. It differs from Compton scattering, which redistributes a photon's energy and direction while preserving the count. The integrated source can vanish if photon emission and absorption balance.

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