Curl of a Vector Product

Question

Let $f(r)$ be a smooth scalar-valued function of $r=|\mathbf{r}|$, and let $a\in\Bbb{R}^{3}$ be a constant vector. I am trying to use suffix/index notation to calculate the expressions $\nabla\times(\mathbf{r}\times\mathbf{a}f(r))$ and $\nabla\cdot(\mathbf{a}f(r))$. This is my progress on the first expression by considering the $i$th component so far:

\begin{align} [\nabla\times(\mathbf{r}\times\mathbf{a}f(r))]_{i} & = \epsilon_{ijk}\partial x_j[\mathbf{r}\times\mathbf{a}f(r)]_{k} \\[2ex] & = \epsilon_{ijk}\partial x_j\epsilon_{klm}r_{l}a_{m}f(r) \\[2ex] & = \epsilon_{ijk}\epsilon_{klm}((\dfrac{\partial r_{i}}{\partial x_{j}}a_{m}f(r)+x_{l}\dfrac{\partial}{\partial x_{j}}(a_{m}f(r))) \\ \end{align}

After that, I am not sure on how to further simplify it but I am aware that I will be using $\epsilon_{ijk}\epsilon_{klm}=\epsilon_{kij}\epsilon_{klm}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$ at one point. Appreciate any help I can get, thanks.

Answer 1

\begin{eqnarray} \nabla \times ({\bf r} \times {\bf a}f(r)) &=& {\bf e}_i\epsilon_{ijk} \partial_j \left[{\bf r} \times {\bf a}f(r))\right]_k \\ &=& {\bf e}_i \epsilon_{ijk}\partial_j \left[\epsilon_{kmn} x_m a_n f(r)\right] \\ &=& {\bf e}_i \epsilon_{kij} \epsilon_{kmn}[(\partial_j x_m)a_n f(r) + x_m a_n (\partial_j f(r))] \\ &=& {\bf e}_i (\delta_{im}\delta_{jn} - \delta_{in}\delta_{jm})\left[\delta_{jm}a_n f(r) + x_m a_n f'(r) \frac{x_j}{r}\right] \\ &=& {\bf e}_i \delta_{im}\delta_{jn} \delta_{jm}a_n f(r) + {\bf e}_i \delta_{im}\delta_{jn} x_m a_n f'(r) \frac{x_j}{r} \\ && - {\bf e}_i \delta_{in}\delta_{jm} \delta_{jm}a_n f(r) - {\bf e}_i\delta_{in}\delta_{jm} x_m a_n f'(r) \frac{x_j}{r} \\ &=& {\bf e}_i \delta_{ji}a_jf(r) + {\bf e}_i x_i a_jf'(r)\frac{x_j}{r} - {\bf e}_i \delta_{jj}a_if(r) - {\bf e}_{i} x_ja_if'(r)\frac{x_j}{r} \\ &=& ({\bf e}_i a_i) f(r) - n({\bf e}_i a_i) f(r) + ({\bf e}_i x_i)\frac{(a_j x_j)}{r}f'(r) - ({\bf e}_ia_i) \frac{x_j x_j}{r}f'(r) \\ &=& (1 - n){\bf a}f(r) + {\bf a}\frac{{\bf a}\cdot r}{r}f'(r) - {\bf a}\frac{r^2}{r}f'(r) \\ &=& -2{\bf a}f(r) + {\bf a}({\bf a}\cdot \hat{\bf r} - r)f'(r) \end{eqnarray}

I will leave the other one to you