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A photon of wavelength

λ

has energy

E = h ν = h c / λ

and momentum

p = (h / λ) \hat{n}

along its propagation direction.

The specified Compton scattering formula requires the electron to be initially at rest in the frame where the wavelengths and angle are measured. This initial-rest assumption is implicit in the printed request; it is not valid for arbitrary initial electron momentum. Use four-momentum with Minkowski metric signature

(+ - - -)

: initially

P = (m c, 0)

, while

K = (E / c, (E / c) \hat{n})

and

K^{'} = (E^{'} / c, (E^{'} / c) \hat{n}^{'})

are null. Conservation gives

P^{'} = P + K - K^{'}

, and the outgoing electron satisfies the same mass-shell condition

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

. Hence

P \cdot (K - K^{'}) = K \cdot K^{'}, m (E - E^{'}) = \frac{E E ^{'}}{c ^{2}} (1 - cos θ) .

After dividing by

E E^{'}

and substituting the photon energy relation,

\frac{1}{E ^{'}} - \frac{1}{E} = \frac{1 - c o s θ}{m c ^{2}}, λ^{'} - λ = \frac{h}{m c} (1 - cos θ) .

The scale

h / (m c)

is the electron Compton wavelength. For a moving initial electron, the earlier invariant identity remains valid, but

P \cdot K

includes its initial velocity and the wavelength shift is different.

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 44 / 4 / Solution Created 2026-10-03 Updated 2026-10-14

Take incoming electron momentum

p

and positron momentum

q

, with outgoing photon momenta

k_{1}, k_{2}

, so

p + q = k_{1} + k_{2}

. The leading electron-positron annihilation into two photons has two tree-level Feynman diagrams, differing by the order in which the two outgoing photons attach to the charged-fermion line.

Figure 1.
The two photon orders in electron-positron annihilation and the s and u channels of Compton scattering
. **Details**.

The arrows indicate fermion flow. An incoming positron has the opposite flow direction to its physical momentum; the lower row uses an outgoing electron instead.

The required QED Feynman rules, with interaction

- e \overset{ˉ}{ψ} γ^{μ} ψ A_{μ}

, are a vertex

- i e γ^{μ}

, a Dirac propagator

i (\neq r + m) / (r^{2} - m^{2} + i 0)

, an incoming electron spinor

u (p)

, an incoming positron adjoint

\overset{v}{ˉ} (q)

, and an outgoing photon polarization

ϵ_{μ}^{*} (k)

. Here Feynman slash notation means

\neq a = γ^{μ} a_{μ}

, and the external Dirac equations are

(\neq p - m) u (p) = 0

and

\overset{v}{ˉ} (q) (\neq q + m) = 0

Put

S (r) = (\neq r + m) / (r^{2} - m^{2} + i 0)

, omitting only the propagator's overall factor

i

. Multiplying the two vertices and the propagator gives

M = - e^{2} \overset{v}{ˉ} (q) [\neq ϵ_{2}^{*} S (p - k_{1}) \neq ϵ_{1}^{*} + \neq ϵ_{1}^{*} S (p - k_{2}) \neq ϵ_{2}^{*}] u (p) .

The diagrammatic expression is

i M

. The relative sign is plus because the two photons are bosons; exchanging their attachments does not exchange external fermions. The amplitude is of order

e^{2}

A physical photon polarization vector obeys

k^{2} = 0

and

k \cdot ϵ = 0

, with the gauge equivalence

ϵ \sim ϵ + α k

. One can choose a transverse representative with

ϵ^{0} = 0

and normalize

ϵ^{*} \cdot ϵ = - 1

. There are two physical transverse polarizations.

To prove the requested two-photon fermion Ward identity, replace

ϵ_{1}^{*}

k_{1}

. Away from an internal pole,

S (r)

is the inverse of

\neq r - m

, with the boundary prescription understood. The first term simplifies because

\neq k_{1} = (\neq p - m) - (\neq p - \neq k_{1} - m), S (p - k_{1}) \neq k_{1} u (p) = - u (p) .

For the second term, four-momentum conservation gives

k_{1} = (p - k_{2}) + q

, so

\overset{v}{ˉ} (q) \neq k_{1} S (p - k_{2}) = \overset{v}{ˉ} (q) [(\neq p - \neq k_{2} - m) + (\neq q + m)] S (p - k_{2}) = \overset{v}{ˉ} (q) .

The two contributions are therefore

- \overset{v}{ˉ} \neq ϵ_{2}^{*} u

and

+ \overset{v}{ˉ} \neq ϵ_{2}^{*} u

, which cancel. Interchanging the photon labels proves the other contraction vanishes. By linearity, adding a multiple of either photon momentum to its polarization leaves the summed amplitude unchanged. This is gauge invariance: longitudinal gauge representatives do not describe an additional physical photon state. An individual diagram does not generally have this property.

For Compton scattering, let

p, k

be incoming electron and photon momenta and

p^{'}, k^{'}

the outgoing ones. The diagrams in the lower row have intermediate electron momenta

p + k

and

p - k^{'}

respectively. The same QED Feynman rules give

M_{C} = - e^{2} \overset{u}{ˉ} (p^{'}) [\neq ϵ^{' *} S (p + k) \neq ϵ + \neq ϵ S (p - k^{'}) \neq ϵ^{' *}] u (p) .

Their denominators are

(p + k)^{2} - m^{2} + i 0

and

(p - k^{'})^{2} - m^{2} + i 0

, the

s

- and

u

-channel denominators. In annihilation the two exchanged-electron channels instead carry

p - k_{1}

and

p - k_{2}

. Crossing symmetry relates the two expressions by moving the positron to an outgoing electron and one outgoing photon to an incoming photon, with the corresponding momentum and polarization replacements. There is no two-photon contact vertex for the minimally coupled Dirac electron. Both processes require the sum of the two photon orders to satisfy the Ward identity.

Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 46 / 4 / Solution Created 2026-10-03 Updated 2026-10-14

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
. **Details**.

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},

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} .

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.

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}) .

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.

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

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) .

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},

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

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) .

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

Past exam of the mathematics course of the University of Cambridge / 2017 / ia / Paper 4 / 9A / a / Solution Created 2026-09-24 Updated 2026-10-20

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 .

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 )} .

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.

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

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 .

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 .

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],

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.

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}

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 301 / 3 / iv / b / Solution Created 2026-10-03 Updated 2026-10-14

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. **Details**.

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

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 .

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.

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.

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} .

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.

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

Let the incident and final photon momentum magnitudes be

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

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} .

Substitute

P = p + p^{'}

, isolate the square root, and square. Cancellation leaves

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

Dividing by

p p^{'}

and using the wavelength definitions gives the Compton scattering shift

λ^{'} = λ + \frac{2 h}{m c} .

The other collinear possibility is the trivial forward solution

λ^{'} = λ

, in which the electron remains at rest.

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

Label positron Compton scattering as

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

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^{'}) .

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

Past exam of the nuclear and particle physics course of the University of Oxford / 2022 / 2 / e / Solution 2026-10-25

The following are the leading quantum electrodynamics diagrams. Solid arrows indicate fermion-number flow, and wavy lines indicate photons. The heavy source line represents a charged nucleus; its recoil allows four-momentum conservation.

Figure 1.
Leading Compton, electron-line bremsstrahlung and nuclear-field pair-production diagrams
. **Details**.

For Compton scattering,

e^{-} γ \to e^{-} γ

, there are two diagrams: absorption of the incident photon followed along the electron line by emission of the outgoing photon, and the opposite ordering. These are the

s

- and

u

-channel electron propagators. Both scattering amplitudes are order

e^{2}

For bremsstrahlung,

e^{-} Z \to e^{-} Z γ

, a virtual photon transfers momentum between the electron and the source. The real photon is emitted before or after this interaction along the electron line. These two dominant external-field diagrams have scattering amplitudes of order

Z e^{3}

. A full finite-mass charged-target calculation also includes radiation from the target; it is suppressed in the heavy stationary-source approximation used here.

For nuclear-field pair production,

γ Z \to e^{-} e^{+} Z

, crossing the preceding diagrams gives two order-

Z e^{3}

contributions, with the incoming real photon and the exchanged virtual photon in either order along the pair's fermion line. An isolated real photon cannot produce a massive pair in vacuum: its invariant squared mass is zero, whereas the pair has invariant squared mass at least

4 m_{e}^{2}

. The source absorbs recoil momentum.

Past exam of the optics and electrodynamics course of the University of Cambridge / 2019 / Section B / B5 / Binding scale and recoil 2026-10-18

A distinct classical Thomson scattering window requires two separated scales:

ω_{0} ≪ ω ≪ ω_{C}, ω_{C} = \frac{m c ^{2}}{ℏ} .

To remove that window, the binding frequency must be comparable with or greater than the Compton scattering recoil scale. Thus the formal condition for bound-electron scattering without a Thomson frequency window is

\frac{k}{m} ≳ \frac{m c ^{2}}{ℏ}, k ≳ \frac{m ^{3} c ^{4}}{ℏ ^{2}} .

Using the stipulated approximate electron mass

m = 1 0^{- 30} kg

c = 2.998 \times 1 0^{8} m s^{- 1}

and

ℏ = 1.055 \times 1 0^{- 34} J s

gives

k_{scale} ≃ 7.3 \times 1 0^{11} N m^{- 1}, ω_{C} ≃ 8.5 \times 1 0^{20} s^{- 1} .

The dimensions are

[m^{3} c^{4} / ℏ^{2}] = kg s^{- 2} = N m^{- 1}

. The inequality denotes a scale comparison, not a uniquely specified spring constant or a sharp transition.

There is also a physical limitation to the question's premise. Compton scattering does not begin discontinuously at

ℏ ω = m c^{2}

; recoil corrections grow as photon energy approaches the rest energy. If

ℏ ω_{0} ≳ m c^{2}

, the assumed nonrelativistic binding model is itself no longer reliable. The calculation therefore locates where no well-separated classical Thomson scattering range exists; it cannot establish a literal direct transition of a real strongly bound system using this oscillator equation alone.

Past exam of the particle physics course of the University of Oxford / 2015 / 8 / a / Solution 2026-10-25

The Goldhaber experiment inferred neutrino helicity from the circular polarization of a nuclear gamma ray. A spin-zero metastable europium-152 nucleus captures an inner-shell electron, producing a spin-one excited samarium-152 daughter and an electron neutrino; the daughter promptly emits a roughly

960 keV

gamma ray to its spin-zero ground state. For a nearly stationary initial atom, the daughter recoils opposite the neutrino. Selecting photons emitted along that recoil relates the daughter spin projection, the photon helicity and the neutrino helicity.

The selection used resonant absorption and scattering in samarium. Nuclear recoil normally shifts an emitted gamma ray below the absorption resonance; the daughter recoil from electron capture supplies a compensating forward Doppler shift. Resonant scattering therefore preferentially selects forward photons, fixing their direction relative to the otherwise invisible neutrino. A samarium scatterer and scintillation counting identify this resonant component. Transmission through magnetized iron, whose Compton scattering depends on the relative electron and photon spins, then analyzes the circular polarization. Reversing the magnetization reverses the analyzing asymmetry and helps control instrumental offsets.

The selected photon and neutrino normalized helicities have the same sign: their momenta are opposite, while the daughter angular-momentum accounting gives opposite spin projections along the recoil axis. The observed negative photon helicity therefore establishes predominantly left-helicity neutrinos, consistent with the V-A interaction. It measures a helicity correlation, not a tiny neutrino mass or a literal observation of the neutral particle's trajectory.^[ref]

Past exam of the particle physics course of the University of Oxford / 2016 / 2 / 3 / Solution 2026-10-25

The nuclear gamma ray can be absorbed by ground-state samarium-152, excite the same level and be re-emitted: nuclear nuclear resonance fluorescence. Its energy in the absorber rest frame must overlap the nuclear absorption line within the relevant natural and thermal broadening. Atomic radiation from refilling the K shell does not have the required transition energy.

Let the excitation energy be

E_{0}

and the nuclear mass be

M

. A free stationary emitter loses recoil energy while a free stationary absorber requires recoil energy:

E_{em} ≃ E_{0} - \frac{E _{0}^{2}}{2 M c ^{2}}, E_{abs} ≃ E_{0} + \frac{E _{0}^{2}}{2 M c ^{2}}, E_{abs} - E_{em} ≃ \frac{E _{0}^{2}}{M c ^{2}} .

For

E_{0} ≃ 0.96 MeV

the gap is about

6.5 eV

. A photon emitted along the daughter recoil receives a positive Doppler shift of approximately

E_{0} p_{ν} / (M c)

, whereas one emitted opposite the recoil is redshifted. Emission before the daughter slows down therefore selects photons roughly opposite to the neutrino. Comparable neutrino and photon energies nearly compensate the recoil mismatch; finite line widths and thermal motion permit overlap. Exact compensation need not hold for every atom, and this is not an assumption of recoil-free emission.

Magnetized iron supplies polarized electrons. Their Compton scattering cross section depends on the relative electron spin and photon helicity, so the iron attenuates the two circular polarizations differently. Reverse the magnetization and measure the resonance-selected count asymmetry

A = (N_{+} - N_{-}) / (N_{+} + N_{-})

. With a calibrated analysing power

a

and background correction,

A = a P_{γ}

determines the sign and magnitude of the circular polarization. A scintillation counter such as NaI(Tl), coupled to a photomultiplier, determines the gamma ray energy from calibrated pulse height; a semiconductor detector could perform the same spectroscopic function.

Past exam of the particle physics course of the University of Oxford / 2017 / 4 / c / Solution 2026-10-25

At low charged-lepton energy, collisional ionization and excitation dominate over radiation. A slowing Positron also eventually annihilates, usually yielding photons. Low-energy photons lose energy through Compton scattering and, at sufficiently low energy, photoelectric absorption; pair production is kinematically unavailable below its threshold.

The electron critical energy

E_{c}

is the energy where mean bremsstrahlung and collisional losses are equal. With

- (d E / d x)_{rad} ≃ E / X_{0}

and a slowly varying collisional loss,

\frac{d E}{d x}_{rad, E_{c}} = \frac{d E}{d x}_{coll, E_{c}} .

Above this energy, radiation and pair creation multiply shower particles efficiently. Below it, charged particles mainly deposit their energy by collisions and the shower dies out. Critical energy is material dependent; it is not a new photon-production threshold.

Past exam of the physics 1 paper of the University of Oxford / 2022 / 7 / c / Solution 2026-10-25

Write the incident and scattered photon energies as

ε = h c / λ

and

ε^{'} = h c / λ^{'}

. Conservation of relativistic energy gives the recoiling electron energy

E_{e} = m c^{2} + ε - ε^{'}

. By momentum conservation, its momentum is the difference of the two photon momenta; hence

p_{e}^{2} c^{2} = ε^{2} + ε^{'2} - 2 ε ε^{'} cos θ .

Insert these into the relativistic energy-momentum relation for the electron:

(m c^{2} + ε - ε^{'})^{2} = m^{2} c^{4} + ε^{2} + ε^{'2} - 2 ε ε^{'} cos θ .

Expanding and cancelling terms gives

m c^{2} (ε - ε^{'}) = ε ε^{'} (1 - cos θ), \frac{1}{ε ^{'}} - \frac{1}{ε} = \frac{1 - c o s θ}{m c ^{2}} .

Since

1/ ε = λ / (h c)

, the Compton scattering formula follows:

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

The scale

h / (m c)

is the electron Compton wavelength. At

θ = 0

λ^{'} = λ

: there is no energy or momentum transfer in this ideal forward limit, so the electron remains at rest. At

θ = π

, the wavelength shift is maximal:

Δ λ_{m a x} = \frac{2 h}{m c}, ε_{m i n}^{'} = \frac{ε}{1 + 2 ε / ( m c ^{2} )} .

(5)

The electron gains its largest kinetic energy,

ε - ε_{m i n}^{'}

, and recoils forward while the photon reverses direction.

Past exam of the physics A course of the University of Cambridge / 2010 / Paper 2 / Section C / C10 / b / Solution 2026-10-20

In Compton scattering, a photon transfers energy and momentum to an initially stationary, approximately free electron. With incident and scattered wavelengths

λ, λ^{'}

and photon deflection angle

θ

, momentum conservation gives

p_{e}^{2} = h^{2} (λ^{- 2} + λ^{' - 2} - \frac{2 c o s θ}{λ λ ^{'}}) .

Combine this with

h c / λ + m c^{2} = h c / λ^{'} + m^{2} c^{4} + p_{e}^{2} c^{2}

to obtain

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

The photon loses energy to electron recoil; the shift is independent of incident wavelength in this free-electron model and is largest at backscattering. At energies comparable with atomic binding, initial electron motion, binding and coherent scattering from the atom require additional treatment. Classical elastic scattering from a fixed charge alone does not produce this recoil wavelength shift.

In Arthur Holly Compton's spectral test, molybdenum-target X-rays scattered from graphite at

9 0^{\circ}

were compared with the primary spectrum using the same slits, crystal and tube potential energy. A calcite crystal resolved the shifted line by Bragg's law; lines on both sides established angular zero. The crystal glancing angle on the plotted axis differs from the photon scattering angle. The compared curves had different intensity scales, and the paper did not directly observe the recoil electrons in this X-ray test. These controls support the wavelength comparison without converting it into a direct recoil-track measurement.^[ref]

Past exam of the short options paper of the physics Prelims of the University of Oxford / 2019 / S3 / 2 / a / Solution 2026-10-25

Compton scattering transfers energy and momentum from a photon to an electron. For an initially stationary, approximately free electron, a photon scattered through a nonzero angle generally has lower frequency and longer wavelength; the recoiling electron carries the difference. The effect demonstrates the photon momentum

p_{γ} = h ν / c = h / λ

, beyond a purely elastic classical scattering description.

The middle panel of the figure in solution shows the incident photon along

+ x

, the outgoing photon at angle

θ

above that direction, and the recoiling electron at deflection angle

ϕ

below it. Each arrow labels its particle and momentum; the initial electron is at rest at the interaction point. The scattered photon and electron have opposite transverse momentum components, while their forward components sum to the incident photon momentum.

Past exam of the short options paper of the physics Prelims of the University of Oxford / 2021 / S03 / 2 / e / Solution 2026-10-25

Work in the initial rest frame of the free electron. If it absorbed a photon of positive energy

E_{γ}

, conservation of momentum would give the final electron momentum magnitude

p_{f} = E_{γ} / c

, while conservation of energy would give

E_{f} = m_{e} c^{2} + E_{γ}

. But the final electron must obey the relativistic energy–momentum relation,

E_{f}^{2} = m_{e}^{2} c^{4} + p_{f}^{2} c^{2} = m_{e}^{2} c^{4} + E_{γ}^{2} .

Combining these statements gives

(m_{e} c^{2} + E_{γ})^{2} = m_{e}^{2} c^{4} + E_{γ}^{2} ⟹ 2 m_{e} c^{2} E_{γ} = 0,

which contradicts

E_{γ} > 0

. Thus

An isolated free electron cannot absorb a single real photon.

The initial rest frame always exists for an electron, so this excludes the process in every inertial frame. Compton scattering is allowed because an outgoing photon carries additional energy and momentum. Absorption by an electron bound in matter is also possible because the rest of the system can share recoil.

Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S10 / 1 / a / Solution 2026-10-25

After injection, the radiopharmaceutical circulates and accumulates according to its biological targeting and clearance. Technetium-99m undergoes an isomeric nuclear transition, producing approximately

140 keV

gamma rays; its roughly six-hour half-life permits useful counting with limited persistence. Some photons undergo absorption or Compton scattering in the patient. A lead collimator in the gamma camera accepts a narrow range of directions so the accepted event retains positional information.

An accepted gamma ray deposits energy, commonly in a NaI(Tl) scintillator, by photoelectric effect or Compton scattering. The resulting visible particle scintillation light is collected by photomultipliers. Their relative pulse sizes determine an event position and their summed pulse estimates its energy. A pulse-height analyser accepts an energy window around the photopeak to reject many scattered events. Position and uniformity corrections are applied; accepted events increment a two-dimensional pixel histogram. After acquisition the computer displays this count map as a planar projection of tracer activity. Finite counting statistics, attenuation, scattering and the collimator limit the image's quantitative accuracy and spatial resolution.^[ref]