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

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

All you need to know is the definitions of (just taking $\mathbb{R}^3$ for simplicity) $$ \nabla=\left[\matrix{\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}}\right]\quad\text{(a symbolic vector)},\qquad \vec A=\left[\matrix{A_1(x,y,z)\\A_2(x,y,z)\\A_3(x,y,z)}\right]\qquad\text{(a vector)} $$ the scalar product and the vector product of two vectors. Well, also $\Delta=\nabla\cdot\nabla$. Then just apply the products formally to the vector coordinates and keep the order so that the derivatives affect functions.

Answer 2

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

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

Answer 3

Similar to @wltrup with notational changes, we have

$$\begin{align} \nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\ &=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\ &=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\ &=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\ &=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\ &=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5 \end{align}$$

In going from $(1)$ to $(2)$ we made use of the vector triple product. Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.

In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.

In going from $(3)$ to $(4)$, we rearranged terms.

In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.