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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.

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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 $\epsilon_{kjk} = 0$

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  • $\begingroup$ Wow! It's very amazing and cool!) it never occurred to me to rewrite the derivative in tensor form! This this conclusion is correct up to the choice of a variable (if we differentiate by $r$, $x=|r-R|$, then the expression from here will be fair: $\mathbf{div}\ (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot \mathbf{rot}\ \mathbf{A} - \mathbf{A} \cdot \mathbf{rot}\ \mathbf{B}$ ). Very thanks! $\endgroup$ – DJNZ Apr 16 at 17:09
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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.

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  • $\begingroup$ That's a general result. Whenever you contract something symmetric with something antisymmetric you'll get 0. $\endgroup$ – Jackozee Hakkiuz Apr 28 at 23:43
  • $\begingroup$ Yes, thank you, you are absolutely right, this is the main thing. $\endgroup$ – DJNZ Apr 29 at 17:05

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