 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/36b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\partial_{μ} A^{μ} = 0

. Substituting the electromagnetic four-potential into the covariant Maxwell equations then removes the mixed derivative and gives

□ A^{μ} = μ_{0} J^{μ}, □ = \frac{1}{c ^{2}} \partial_{t}^{2} - \nabla^{2} .

Equivalently,

(\nabla^{2} - \frac{1}{c ^{2}} \partial_{t}^{2}) A^{μ} = - μ_{0} J^{μ} .

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