 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/ia/paper-3/4a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Use suffix notation, with repeated indices summed from one to three. The cross product, divergence, and curl have components

(A \times B)_{i} = ϵ_{ijk} A_{j} B_{k}

\nabla \cdot A = \partial_{i} A_{i}

, and

(\nabla \times A)_{i} = ϵ_{ijk} \partial_{j} A_{k}

. Here

ϵ

\partial_{i} (ϵ_{ijk} A_{j} B_{k}) = ϵ_{ijk} (\partial_{i} A_{j}) B_{k} + ϵ_{ijk} A_{j} \partial_{i} B_{k} .

Using the cyclic identity

ϵ_{ijk} = ϵ_{kij}

in the first term and

ϵ_{ijk} = - ϵ_{jik}

in the second identifies these terms as

B_{k} (\nabla \times A)_{k}

and

- A_{j} (\nabla \times B)_{j}

. Therefore

\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B) .

For the second identity, contract two Levi-Civita symbols, using

ϵ_{ijk} ϵ_{k l m} = δ_{i l} δ_{jm} - δ_{im} δ_{j l}

[\nabla \times (A \times B)]_{i} = ϵ_{ijk} ϵ_{k l m} \partial_{j} (A_{l} B_{m}) = \partial_{j} (A_{i} B_{j} - A_{j} B_{i}) = B_{j} \partial_{j} A_{i} + A_{i} \partial_{j} B_{j} - A_{j} \partial_{j} B_{i} - B_{i} \partial_{j} A_{j} .

The four component expressions translate into

\nabla \times (A \times B) = (B \cdot \nabla) A - B (\nabla \cdot A) - (A \cdot \nabla) B + A (\nabla \cdot B) .

Both derivations require only continuously differentiable vector fields; no additional divergence-free assumptions are used.

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