evaluating curl of $\vec r/r^2$

Question

how do I calculate curl of : $\vec r/r^2$ I don't know how to solve this problem can someone help me please

Answer 1

Note that $\nabla \times (\phi \vec A)=\nabla (\phi) \times \vec A+\phi \nabla \times \vec A$. Then, with $\phi=\frac{1}{r^2}$ and $\vec A=\vec r$, we have for $r\ne 0$

$$\begin{align} \nabla \times \left(\frac{\vec r}{r^2}\right)&=\nabla \left(\frac{1}{r^2}\right)\times \vec r+\frac{1}{r^2}\nabla \times \vec r\\\\ &=0 \end{align}$$

since $\nabla \left(\frac{1}{r^2}\right)$ has only a radial component and $\nabla \times \vec r=0$ since the position vector is irrotational.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\nabla\times\pars{\vec{r} \over r^{2}}} & = \nabla\times\bracks{\nabla\ln\pars{r}} = \color{#f00}{\vec{0}} \end{align}

Note that \begin{align} \nabla\times\nabla\phi &= \sum_{i}\hat{e}_{i} \partiald{}{x_{i}}\times\sum_{j}\hat{e}_{j}\partiald{\phi}{x_{j}} = \sum_{i,j}{\partial^{2}\phi \over \partial x_{i}\partial x_{j}}\,\hat{e}_{i}\times\hat{e}_{j} = \sum_{i,j}{\partial^{2}\phi \over \partial x_{i}\partial x_{j}}\, \sum_{k}\epsilon_{ijk}\,\hat{e}_{k} \\[3mm] & = \sum_{k}\hat{e}_{k}\pars{\sum_{i,j} \epsilon_{ijk}\,{\partial^{2}\phi \over \partial x_{i}\partial x_{j}}} = \half\sum_{k}\hat{e}_{k}\ \underbrace{\pars{\sum_{i,j} \epsilon_{ijk}\,{\partial^{2}\phi \over \partial x_{i}\partial x_{j}} + \epsilon_{jik}\,{\partial^{2}\phi \over \partial x_{j}\partial x_{i}}}} _{\ds{=\ 0}} \end{align} whenever $\ds{{\partial^{2}\phi \over \partial x_{i}\partial x_{j}} = {\partial^{2}\phi \over \partial x_{j}\partial x_{i}}\,,\quad \forall\ i,j = x,y,z}$.