Why does the vector Laplacian involve the double curl of the vector field?

Question

The scalar Laplacian is defined as $\Delta A =\nabla\cdot\nabla A $. This makes conceptual sense to me as the divergence of the gradient... but I'm having trouble connecting this concept to a vector Laplacian because it introduces a double curl as $\Delta \mathbf{A}=\nabla(\nabla\cdot\mathbf{A}) - \nabla\times(\nabla\times \mathbf{A})$. I understand what curl is but I don't understand why it's introduced in the vector Laplacian.

Answer 1

The definition of Laplacian operator for either scalar or vector is almost the same. You can see it by noting the vector identity $$\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-(\nabla\cdot\nabla)A$$ Plugging it into your definition produces still $$\Delta A=(\nabla\cdot\nabla)A$$