 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/1/a/solution

\partial_{i} = \partial / \partial x_{i}

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

Here the Kronecker delta selects the indicated indices, and continuous second partial derivatives allow their order to be interchanged. Identifying the gradient, divergence and componentwise Laplacian proves the curl of the curl identity:

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

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