Derivation of the Compton Shift Equation

Key idea: Derive the Compton wavelength shift from two-dimensional momentum conservation, relativistic energy conservation, and photon momentum.

  • Advanced Physics
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Learning objectives

  • Connect blackbody evidence, Planck quantisation, photon momentum, and Compton scattering.

This derivation treats the target electron as free and initially at rest. The incident photon has momentum magnitude p, the scattered photon has magnitude p', and the electron recoils with momentum magnitude P.

Incident and scattered photon momenta with photon angle theta and electron recoil momentum at angle phi
Choose the incident photon direction as positive x. The photon scatters through theta and the electron recoils through phi.

1. Governing relations

For each photon,

E = pc, p = h/λ

For the electron of rest mass mₑ and total energy Eₑ,

Eₑ² = P²c² + mₑ²c⁴

Both momentum components and total energy are conserved.

2. Momentum conservation

Along the initial photon direction,

p = p' cos θ + P cos φ

Perpendicular to it,

0 = p' sin θ-P sin φ

Hence

P cos φ = p-p' cos θ

and

P sin φ = p' sin θ

Squaring and adding eliminates φ:

P² = (p-p' cos θ)² + (p' sin θ)²

Therefore

P² = p² + p'²-2pp' cos θ

3. Energy conservation

Initially, the electron contributes only its rest energy:

pc + mₑc² = p'c + Eₑ

Thus

Eₑ = (p-p')c + mₑc²

Insert this into the electron energy–momentum relation and divide by c²:

(p-p' + mₑc)² = P² + mₑ²c²

Expanding and cancelling mₑ²c² gives

P² = (p-p')² + 2mₑc(p-p')

4. Eliminate the electron momentum

Equate equations (1) and (2):

p² + p'²-2pp' cos θ = (p-p')² + 2mₑc(p-p')

Since (p-p')² = p² + p'²-2pp', cancellation leaves

pp'(1- cos θ) = mₑc(p-p')

Divide by pp':

1- cos θ = mₑc(1/p'-1/p)

Using p = h/λ and p' = h/λ',

1- cos θ = (mₑc/h)(λ'-λ)

Therefore

λ'-λ = (h/mₑc)(1- cos θ)

5. Checks and interpretation

  • Dimensions: h/(mₑc) has units of length.
  • Forward limit: θ = 0 gives λ' = λ.
  • Back-scattering: θ = 180° gives the maximum shift 2h/(mₑc).
  • Energy direction: 1- cos θ ≥ 0, so λ' ≥ λ; the photon never gains energy from an electron initially at rest.

The electron recoil angle disappears from the result because its momentum and energy are eliminated using the conservation laws.

Mind stretcher 1: Check the derived equation independentlyExtension

Use Δλ = h(1- cos θ)/(mₑc) to state the forward-scattering shift, the maximum shift and the direction of photon-energy transfer.

Answer

At 0°, the shift is zero. At 180°, it is 2h/(mₑc). Since 1- cos θ ≥ 0, the outgoing wavelength cannot be shorter, so the photon cannot gain energy from an electron initially at rest.

5. Common mistakes

  • Conserving photon energy by itself. Energy transfers to the electron.
  • Using P²/(2mₑ) for electron kinetic energy inside an otherwise relativistic derivation.
  • Losing the rest-energy term mₑc².
  • Squaring momentum components and forgetting the cross term -2pp' cos θ.
  • Substituting p = h/λ with the primed wavelength paired to the wrong momentum.

6. Worked Examples

Modelled example 1

Recovering the maximum shift

Core

Problem

Use the derived equation to calculate the maximum shift for an electron. Take h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg and c = 3.00 × 10⁸ m s⁻¹.
Study the worked solution
  1. Maximise the angular factor

    Method

    1- cos θ = 2 at 180°.

    Reason

    Back-scattering gives the smallest possible cosine.

    Working

    Δλₘₐₓ = 2h/(mₑc)
  2. Evaluate

    Method

    Δλₘₐₓ = 4.85 × 10⁻¹² m = 4.85 pm.

    Reason

    Substitute the electron mass into the derived relation.

    Working

    Δλₘₐₓ = (2(6.63 × 10⁻³⁴))/((9.11 × 10⁻³¹)(3.00 × 10⁸)) = 4.85 × 10⁻¹² m

Next: return to Compton Shift or the Quantum Theory of Light Hub.

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Course and syllabus information
Course
Advanced Physics
Edition
Advanced Physics