Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 1 / i

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- Solution i

Solution

Using the Levi-Civita symbol and the Einstein summation convention in Cartesian coordinates, the product rule gives

\nabla \cdot (A \times B) = \partial_{i} (ϵ_{ijk} A_{j} B_{k}) = ϵ_{ijk} (\partial_{i} A_{j}) B_{k} + ϵ_{ijk} A_{j} \partial_{i} B_{k} = B_{k} ϵ_{kij} \partial_{i} A_{j} - A_{j} ϵ_{jik} \partial_{i} B_{k} .

The cyclic permutation in the first term preserves the sign, whereas exchanging the first two indices in the second reverses it. Recognizing the two curls proves

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

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