 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/ia/paper-3/11c/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-07

[F \times (\nabla \times G)]_{i} = ϵ_{ijk} F_{j} ϵ_{k ℓ m} \partial_{ℓ} G_{m} = (δ_{i ℓ} δ_{jm} - δ_{im} δ_{j ℓ}) F_{j} \partial_{ℓ} G_{m} = F_{j} \partial_{i} G_{j} - F_{j} \partial_{j} G_{i} .

The last term cancels

(F \cdot \nabla) G_{i}

. Interchanging

F, G

gives the analogous cancellation with

(G \cdot \nabla) F_{i}

. The surviving terms are

F_{j} \partial_{i} G_{j} + G_{j} \partial_{i} F_{j} = \partial_{i} (F_{j} G_{j})

\nabla (F \cdot G) = (F \cdot \nabla) G + (G \cdot \nabla) F + F \times (\nabla \times G) + G \times (\nabla \times F) .

The proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.

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