divergence of gradient of scalar function in tensor form I found simple expression in tensor notation for a divergence of product vector and gradient of scalar function:
$$\operatorname{div}(\mathbf{j}) = 0 \text{, where } \mathbf{j} = \mathbf{m}\times \nabla f(|\mathbf{r} - \mathbf{R}|)$$
and I want to write this expression in tensor form.
So, I have scalar function of scalar argument:
$f(|\mathbf{r} - \mathbf{R}|)$;
now, in tensor form: $\nabla f = f_{,i}$, $\mathbf{m}\times\nabla f = \varepsilon_{ijk} m_j f_{,i}$, right? 
Also, $\operatorname{div}f = f_{i,i}$, then $\operatorname{div}(\mathbf{m}\times\nabla f) = \varepsilon_{ijk} m_j f_{,ki}$?
How to write this expression in tensor form, so that equality to zero follows from the property of the Levi-Civita symbol?
Thanks.
 A: You will need to use the functional form of $f$ at some point. That is $\nabla_x f(|x|) = f'(|x|) \frac{x}{|x|}$ so in tensor notation $(\nabla_x f(|x|))_i = f' \frac{x_i}{|x|} $, then
$$\nabla \cdot (m\times \nabla f) = \partial_i \left(\epsilon_{ijk}m_j f'\frac{x_k}{|x|}\right)$$
Assuming you meant $m$ is a constant vector you have
$$\nabla \cdot (m\times \nabla f) = \epsilon_{ijk}m_j \left[f''\frac{x_ix_k}{|x|^2} + f' \frac{\delta_k^i}{|x|} - f'\frac{x_kx_i}{|x|^3}\right]$$
$$ = \frac{xf''}{|x|^2}\cdot (m\times x) + \epsilon_{kjk}m_jf'\frac{1}{|x|} - \frac{xf'}{|x|^3}\cdot (m\times x) = 0$$
The 1st and 3rd follow from $x\perp (m\times x)$ and the second is zeros as $\epsilon_{kjk} = 0$
A: Also, there is a beautiful way to get an answer: we write the derivatives in tensor form (by comma term):
$$(m\times \nabla f) = \varepsilon_{ijk} m_j  f_{\color{magenta},k}$$
Hence, 
$$\nabla \cdot (m\times \nabla f) = \varepsilon_{ijk} m_j f_{\color{magenta}{,i}k}$$
We can rename $i$ and $k$: 
$$\varepsilon_{ijk} m_j f_{,ik} = \varepsilon_{ijk} m_j f_{,ki}$$
Also note that $\varepsilon_{kji} = - \varepsilon_{ijk}; ~~f_{,ki}=f_{,ik}~$, then $~~\varepsilon_{ijk}f_{,ik} = \varepsilon_{kji}f_{,ki} = - \varepsilon_{ijk}f_{,ik}~$. 
Therefore, since $\varepsilon_{ijk}f_{,ik} = -\varepsilon_{ijk}f_{,ik}~$, then $~~\varepsilon_{ijk}f_{,ik} = 0$
Thank you all very much.
