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ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021/section-b/b6/d/solution

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Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B6 / d / Solution by

Codex  0 2026-10-18

Use the inverse coordinate transformation,

r_{⊥} = r_{⊥}^{'}, r_{∥} = γ (r_{∥}^{'} + v t^{'}), t = γ (t^{'} + \frac{v \cdot r ^{'}}{c ^{2}}) .

Substitution into the wave phase

k \cdot r - ω t

and comparison of the coefficients of

r^{'}

and

t^{'}

gives

ω^{'} = γ (ω - v \cdot k), k_{∥}^{'} = γ (k_{∥} - \frac{ω}{c ^{2}} v), k_{⊥}^{'} = k_{⊥} .

The first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,

k^{'} = k + [(γ - 1) \frac{v \cdot k}{v ^{2}} - γ \frac{ω}{c ^{2}}] v .

The electric-field wave amplitude follows from part (b), after rotating the boost axis to

v

:

E_{0}^{'} = E_{0, ∥} + γ (E_{0, ⊥} + v \times B_{0}) .

For a vacuum plane wave, Maxwell equations give

B_{0} = k \times E_{0} / ω

and

k \cdot E_{0} = 0

. Thus an expression using only the given electric wave amplitude and wavevector is

E_{0}^{'} = E_{0, ∥} + γ [E_{0, ⊥} + \frac{v \times ( k \times E _{0} )}{ω}] .

Equivalently,

E_{0}^{'} = γ (E_{0} + v \times B_{0}) - [γ^{2} / (γ + 1)] v (v \cdot E_{0}) / c^{2}

. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

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