Calculating $\nabla\left(\nabla \cdot \boldsymbol{r}_{0} e^{i\boldsymbol{k} \cdot \boldsymbol{r}}\right)$

Question

• In a longer derivation I ran into the following quantity: $$ \nabla\left[\nabla\cdot\left(% {\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right) \right] $$ ( i.e., the gradient of the divergence ) where ${\bf k}$ is a vector of constants and ${\bf r}$ is a position vector.
• Can someone help explaining how to calculate this? I am hoping it gives: $$ \nabla\left[\nabla\cdot\left(% {\bf r}_{0}\,{\rm e}^{{\rm i}{\bf k} \cdot {\bf r}}\,\right) \right] = -{\bf k}\left({\bf k}\cdot{\bf r}_{0}\right) $$ ( because then the rest of my equations add up ).

Answer 1

Summing over repeated indices, the divergence is $r_{0i}\partial_ie^{\text{i}k_jr_j}=r_{0i}\text{i}k_ie^{\text{i}k_jr_j}=\text{i}(k\cdot r_0)e^{\text{i}k_jr_j}$. Applying $\partial_l$ pulls down another $\text{i}k_l$ factor, so the gradient is $-k(k\cdot r_0)e^{\text{i}k\cdot r}$. Your desired result drops the exponential, which I suspect is a typo.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\nabla\bracks{% \nabla\cdot\pars{{\bf r}_{0}\expo{\ic{\bf k}\cdot{\bf r}}}}} = \nabla\bracks{% \overbrace{\pars{\nabla\cdot{\bf r}_{0}}}^{\ds{=\ 0}}\ \expo{\ic{\bf k}\cdot{\bf r}} + {\bf r}_{0}\cdot\nabla\pars{\expo{\ic{\bf k}\cdot{\bf r}}}} = \ic\nabla\pars{{\bf r}_{0}\cdot{\bf k}\,\expo{\ic{\bf k}\cdot{\bf r}}} \\[5mm] = &\ \ic\,{\bf r}_{0}\cdot{\bf k}\,\nabla\expo{\ic{\bf k}\cdot{\bf r}} = \ic\,{\bf r}_{0}\cdot{\bf k}\,\pars{\expo{\ic{\bf k}\cdot{\bf r}} \,\ic{\bf k}} = \bbx{-{\bf k}\pars{{\bf k}\cdot{\bf r}_{0}}\expo{\ic{\bf k}\cdot{\bf r}}} \\ & \end{align} There are a lot of vectorial identities over here.