Proving vector identities using suffix notation

Question

Show, using suffix notation, the following identity $$\nabla\times\nabla\phi = \mathbf{0}.$$

So we have:

\begin{align}\nabla\times\nabla\phi &= \Big[\nabla\times\nabla\phi\Big]_{i}\\&= \varepsilon_{ijk}\dfrac{\partial(\nabla\phi)_{k}}{\partial x_{j}}\\&= \varepsilon_{ijk}\dfrac{\partial}{\partial x_{j}}\left[\dfrac{\partial\phi}{\partial x_{k}}\right]\end{align}

But I'm not really sure what to do next with this. Any hints?

Answer 1

$[\nabla\times\nabla\phi]_i = \epsilon_{ijk}\partial_j[\nabla\phi]_k=\epsilon_{ijk}\partial_j\partial_k\phi = -\epsilon_{ikj}\partial_j\partial_k\phi = -\epsilon_{ikj}\partial_k\partial_j\phi$

The last part is due to the equality of mixed partials. Now do the change of notation $j \leftrightarrow k$ on that last term (you can always change dummy indices) to get $\epsilon_{ijk}\partial_j\partial_k\phi = -\epsilon_{ijk}\partial_j\partial_k\phi$. The only function equal to its negative is the zero function. Thus $[\nabla\times\nabla\phi]_i = 0$.

Answer 2

$$\nabla_{k}(\nabla \times A)_{k} = \nabla_{k}\epsilon_{ijk}\nabla_{ j}A_{k}$$ Which is identically zero by the nablas but antisymmetric in the $\epsilon$

\begin{eqnarray} \nabla_{i}(A \times B)_{i} &=& \nabla_{i}(\epsilon_{ijk}A_{j}B_ {k}) \\ &=& \epsilon_{ijk}\nabla_{i}(A_{j}B_ {k}) \\ &=& \epsilon_{ijk}\nabla_{i}(A_{j})B _{k} + \epsilon_{ijk}A_{j}\nabla_{i}B_{ k} = \epsilon_{ijk}\nabla_{i}(A_{j})B _{k} - \epsilon_{jik}A_{j}\nabla_{i}B_{ k} \\ &=& 0 \end{eqnarray}