Multivariable Calculus – Showing Curl of Curl of A Equals Gradient of Divergence of A Minus Laplacian of A

Question

I have faced difficulties while trying to prove that

$$\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$$

I don't have any clue how can I start to work with it. Any hint will be helpful.

Answer 1

For some details missing from the derivation below, see my answer to this question.

First, we write the LHS in terms of its components, using the Kronecker delta and Levi-Civita symbols. (Note: I'll drop the vector arrow on $\vec{A}$ but it's a vector)

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\partial_{j}\,(\nabla \times A)_{k} = \epsilon_{ijk}\,\partial_{j}\,(\epsilon_{krs}\,\partial_{r}\,A_{s}) $$

The $\epsilon_{krs}$ are constants, so

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\epsilon_{krs}\,\partial_{j}\,\partial_{r}\,A_{s} $$

But

$$\epsilon_{ijk}\,\epsilon_{krs} = \delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr} $$

so

$$[\,\nabla \times (\nabla \times A)\,]_{i} = (\delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr})\, \partial_{j}\,\partial_{r}\,A_{s} = \delta_{ir}\,\delta_{js}\,\partial_{j}\,\partial_{r}\,A_{s} - \delta_{is}\,\delta_{jr}\,\partial_{j}\,\partial_{r}\,A_{s} $$

Simplifying,

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \partial_{j}\,\partial_{i}\,A_{j} - \partial_{j}\,\partial_{j}\,A_{i} = \partial_{i}\,(\partial_{j}\,A_{j}) - (\partial_{j}\,\partial_{j})\,A_{i} = \partial_{i}\,(\nabla \cdot A) - \nabla^2A_i $$

so

$$\nabla \times (\nabla \times A) = \nabla(\nabla \cdot A) - \nabla^2A $$