Some useful identities
For a scalar field \(\phi\),
(1)\[ \nabla \times \nabla \phi
\rightarrow
\epsilon_{ijk} \frac{\partial }{\partial x_{j}} \frac{\partial \phi}{\partial x_{k}} = \epsilon_{ijk} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}\]
Since \(\epsilon_{ijk} = - \epsilon_{ikj}\),
(2)\[ \epsilon_{ijk} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}
= - \epsilon_{ikj} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}\]
However, the differentiation with respect to \(x\) is free to exchange, so that
(3)\[ - \epsilon_{ikj} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}
= - \epsilon_{ikj} \frac{\partial^{2} \phi}{\partial x_{k} \partial x_{j}}\]
The indices \(j\) and \(k\) in the last equation are dummy; therefore rewriting \(j \rightarrow k\) and \(k \rightarrow j\) gives
(4)\[ - \epsilon_{ikj} \frac{\partial^{2} \phi}{\partial x_{k} \partial x_{j}}
= - \epsilon_{ijk} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}\]
Adding this result to the first equation yields
(5)\[ 2 \nabla \times \nabla \phi
\rightarrow
\epsilon_{ijk} \frac{\partial }{\partial x_{j}} \frac{\partial \phi}{\partial x_{k}} = \epsilon_{ijk} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}
- \epsilon_{ijk} \frac{\partial^{2} \phi}{\partial x_{j} \partial x_{k}}
= 0\]
Therefore, the rotation of the gradient of a scalar field is identically zero:
(6)\[ \nabla \times \nabla \phi = 0\]
For a vector field \(\mathbf{f}\),
(7)\[ \nabla \cdot \nabla \times \mathbf{f}
\rightarrow \frac{\partial }{\partial x_{i}} \epsilon_{ijk} \frac{\partial f_{k}}{\partial x_{j}}
= \epsilon_{ijk} \frac{\partial^{2} f_{k}}{\partial x_{i} \partial x_{j}} \]
With the same manner we used above, it can be shown that
(8)\[ \nabla \cdot \nabla \times \mathbf{f} = 0 \]
When we have \(\times\) twice, we often use
(9)\[ \epsilon_{kij} \epsilon_{kmn}
= \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm} \]
We often meet \textit{rotation of rotation}, \(\nabla \times \nabla \times \mathbf{f}\), in vector calculus for fluid mechanics. This can be rewritten in a form expressed in terms of \(grad\) and \(div\) as follows:
(10)\[\begin{split}\begin{split}
\nabla \times \nabla \times \mathbf{f}
&\rightarrow \epsilon_{ijk} \frac{\partial }{\partial x_{j}} \epsilon_{kmn} \frac{\partial f_{n}}{\partial x_{m}}
= \epsilon_{ijk} \epsilon_{kmn} \frac{\partial }{\partial x_{j}} \frac{\partial f_{n}}{\partial x_{m}}
= ( \delta_{im} \delta_{jn} - \delta_{in} \delta_{jm} ) \frac{\partial }{\partial x_{j}} \frac{\partial f_{n}}{\partial x_{m}}
= \frac{\partial }{\partial x_{j}} \frac{\partial f_{j}}{\partial x_{i}} - \frac{\partial }{\partial x_{j}} \frac{\partial f_{i}}{\partial x_{j}} \\
&= \frac{\partial }{\partial x_{i}} \frac{\partial f_{j}}{\partial x_{j}} - \frac{\partial^{2} f_{i}}{\partial x_{j} \partial x_{j}}
\end{split}\end{split}\]
Therefore,
(11)\[ \nabla \times \nabla \times \mathbf{f}
= \nabla \nabla \cdot \mathbf{f} - \nabla^{2} \mathbf{f}\]