 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2009/paper-1/1/a/solution

\nabla \cdot F = \partial_{i} F_{i}, (\nabla \times F)_{i} = ϵ_{ijk} \partial_{j} F_{k},

with summation over repeated indices. In components the curl is

(\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})

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

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

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