Deriving the Compton shift equation

Compton’s equation follows from nothing more than a photon’s momentum and the conservation of energy and momentum in a collision. This lesson derives it. The plan is to write the electron’s momentum in two ways, once from momentum conservation and once from energy conservation, and then eliminate it.

The set-up

Treat the electron as free and at rest before the collision. The incident photon has momentum p, the scattered photon p', and the recoiling electron P.

Scroll across the figure to read all labels.

Take the incident photon's direction as the positive x-direction. The photon scatters through θ and the electron recoils at φ.

Each photon has E = pc and p = h/λ. The electron, of mass mₑ, has total energy Eₑ given by

Eₑ² = P²c² + mₑ²c⁴

The electron may recoil fast, so its energy must be treated relativistically, not as P²/(2mₑ).

Conservation of momentum

Along the original direction of the photon, and at right angles to it:

p = p' cos θ + P cos φ, 0 = p' sin θ-P sin φ

P cos φ = p-p' cos θ, P sin φ = p' sin θ

Squaring and adding removes the electron’s angle φ:

P² = (p-p' cos θ)² + (p' sin θ)² = p² + p'²-2pp' cos θ

Conservation of energy

Before the collision, the electron has only its rest energy:

pc + mₑc² = p'c + Eₑ ⇒ Eₑ = (p-p')c + mₑc²

Substitute this into the electron’s energy–momentum relation and divide by c²:

(p-p' + mₑc)² = P² + mₑ²c²

Expanding and cancelling mₑ²c²:

P² = (p-p')² + 2mₑc(p-p')

Eliminating the electron’s momentum

Equations (1) and (2) are both equal to P²:

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

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

pp'(1- cos θ) = mₑc(p-p')

Dividing by pp':

1- cos θ = mₑc(1/p'-1/p)

Finally, with p = h/λ and p' = h/λ':

λ'-λ = (h/mₑc)(1- cos θ)

The electron’s recoil angle and momentum have disappeared, because the two conservation laws were used to eliminate them. The constant h/(mₑc) has the units of length.

Problem

Use the derived equation to calculate the largest possible shift for scattering by an electron. Take h = 6.63 × 10⁻³⁴ J s, mₑ = 9.11 × 10⁻³¹ kg and c = 3.00 × 10⁸ m s⁻¹.

Show full solution

Make the angle factor as large as possible
Method
Choose θ = 180°.
Reason
Scattering straight back gives the smallest cosine, -1, so 1- cos θ = 2.
Working
Answer
Δλ max = 2h/(mₑc)
Evaluate
Method
Substitute the constants.
Reason
The electron’s mass sets the size of the shift.
Working
Answer
Δλ max = (2 × 6.63 × 10⁻³⁴)/(9.11 × 10⁻³¹ × 3.00 × 10⁸) = 4.85 × 10⁻¹² m
State the answer
Working
Answer
Δλ max = 4.85 pm, twice the electron’s Compton wavelength.

Quick check

The answer has the units of length, as the boxed equation requires.

Check your understanding 1 Checking the derived equation

Use Δλ = (h/mₑc)(1- cos θ) to find the shift for forward scattering, θ = 0, and decide whether a photon can ever gain energy from an electron that starts at rest.

Show answer

At θ = 0, 1- cos θ = 0, so there is no shift: a photon that carries straight on has not transferred any momentum. Since 1- cos θ ≥ 0 for every angle, λ' ≥ λ, so the scattered photon can never have a shorter wavelength: it cannot gain energy from an electron initially at rest.

Common mistakes

Conserving the photon’s energy on its own. Energy is shared with the electron.
Using P²/(2mₑ) for the electron’s kinetic energy in a relativistic derivation.
Losing the electron’s rest energy, mₑc².
Forgetting the cross term -2pp' cos θ when squaring the momentum components.
Pairing p' with the wrong wavelength when substituting p = h/λ.

Spotted an error? Report a correction

Deriving the Compton shift equation

Reading settings

The set-up

Conservation of momentum

Conservation of energy

Eliminating the electron’s momentum

The largest shift from the derived equation

Make the angle factor as large as possible

Evaluate

State the answer

Check your understanding 1 Checking the derived equation

Common mistakes