Prove that $\nabla \times (\nabla \times \vec A) = \nabla(\nabla \cdot \vec A) - \nabla^2\vec A$. For any vector function of $\vec A$.

Asked Apr 14 '21 at 12:50

Active Apr 14 '21 at 13:05

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Show that $\nabla \times (\nabla \times \vec A) = \nabla(\nabla \cdot \vec A) - \nabla^2\vec A$. For any vector function of $\vec A$.

edited Apr 14 '21 at 13:03

asked Apr 14 '21 at 12:50

DEEPANSHU ARYA

Doesn't this just come from the definition of the vector Laplacian? https://en.wikipedia.org/wiki/Laplace_operator#Vector_Laplacian – Stephen Donovan Apr 14 '21 at 12:56
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Apr 14 '21 at 13:05
The question has been previously asked, and a detailed answer is available there: https://math.stackexchange.com/questions/1381381/showing-that-nabla-times-nabla-times-veca-nabla-nabla-cdot-veca-de?rq=1 – Jack Pote Apr 14 '21 at 13:05

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