 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/36a/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Assume that

v

is constant. Set

τ = t_{ret}, n = \frac{R}{R}, β = \frac{v}{c}, κ = 1 - n \cdot β, D = R - β \cdot R = R κ .

t = τ + R (τ) / c

gives

\partial_{t} τ = \frac{1}{κ}, \nabla τ = - \frac{n}{c κ} .

Since

d R / d τ = - v

, direct differentiation of

D

yields

\partial_{t} D = \frac{c ( β ^{2} - n \cdot β )}{κ}, v \cdot \nabla D = - \frac{c ( β ^{2} - n \cdot β )}{κ} .

Therefore

(\partial_{t} + v \cdot \nabla) D = 0.

ϕ = q / (4 π ϵ_{0} D)

and

A = v ϕ / c^{2}

. The preceding identity consequently implies

\partial_{t} ϕ + v \cdot \nabla ϕ = 0.

v

is constant,

\frac{1}{c ^{2}} \frac{\partial ϕ}{\partial t} + \nabla \cdot A = \frac{1}{c ^{2}} (\partial_{t} ϕ + v \cdot \nabla ϕ) = 0.

Thus the explicit uniformly moving potentials satisfy the Lorenz gauge.

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