# 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.

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 $$\epsilon_{kjk} = 0$$

• 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! – DJNZ Apr 16 at 17:09

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.

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