 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-3/11c/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-06

(\nabla \times V)_{i} = ϵ_{ijk} \partial_{j} V_{k} .

(\nabla \times (\nabla \times V))_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} \partial_{l} V_{m} = (δ_{i l} δ_{jm} - δ_{im} δ_{j l}) \partial_{j} \partial_{l} V_{m} = \partial_{i} (\partial_{j} V_{j}) - \partial_{j} \partial_{j} V_{i} .

Smoothness allows the partial derivatives to commute. This proves the curl of the curl identity:

\nabla \times (\nabla \times V) = \nabla (\nabla \cdot V) - \nabla^{2} V .

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