 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-3/11b/i/solution

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

In components,

[\nabla \times \nabla ϕ]_{i} = ϵ_{ijk} \partial_{j} \partial_{k} ϕ = 0,

because the Levi-Civita symbol is antisymmetric in

j, k

while equality of mixed partial derivatives makes

\partial_{j} \partial_{k} ϕ

symmetric. Hence

\nabla \times \nabla ϕ = 0 .

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