Curl, $\vec\nabla \times (\hat{a}\times \vec{b})$ EDIT: FIXED TYPOS & Deleted most of my wrong work pointed out by others.
Calculate the curl of $f(\vec r,t)$ where the function is given by
$$
f(\vec r,t)=- (\hat{a}\times \vec{b}) \frac{e^{i(c r- d t)}}{r}
$$where this is a spherical coordinate system.
where $\hat{a}$ is a unit vector $\hat a=\frac{\vec r}{r}$ and $\vec b$ is a constant vector.  The curl of f is given by
$$
\vec \nabla \times f(\vec r,t)=-\vec \nabla \times\left( (\hat{a}\times \vec{b}) \frac{e^{i(c r- d t)}}{r}\right).
$$
I prefer $\epsilon_{ijk}$ notation to compute things, thanks! 
I am stuck here
$$
\vec \nabla \times \vec f=\partial_j(r_iu_{oj}-r_ju_{oi})g(r)=-2u_{oi}g(r)+(r_iu_{oj}-r_ju_{oi})\bigg(  \frac{ik}{r^2}-\frac{2}{r^3}  \bigg)e^{i(cr-dt)}.
$$
where $g(r)$ is a scalar function and  is given by
$$
g(r)=\frac{e^{i(cr-dt)}}{r^2}.
$$
So I am stuck on how to proceed, and write everything back in terms of vector notation.  Thanks
 A: If $b$ is a constant vector, you can just
$$\partial_j(a_i b_j-a_j b_i)=(b_j \partial_j)a_i-b_i(\partial_j a_j)$$
$$=(\vec{b}\cdot\nabla)\hat{a}-\vec{b}(\nabla\cdot\hat{a})$$
But it's easier to process things in the index form:
$$\partial_j\frac{r_i}{r}=\frac{\partial_j r_i}{r}-r_i\frac{1}{r^2}\partial_j r=
\frac{\delta_{ij}}{r}-\frac{r_i r_j}{r^3}
$$
Now a simple trace gives you 
$$\partial_i \frac{r_i}{r}=\nabla\cdot\hat{a}=\frac{2}{r}$$
And
$$(\vec{b}\cdot\nabla)\hat{a}=\frac{\vec{b}}{r}-\frac{\vec{r}(\vec{b}\cdot\vec{r})}{r^3}=\frac{1}{r}\hat{a}\times(\vec{b}\times\hat{a})$$
where I recognized the formula for the double cross product.
Putting things together:
$$\nabla\times(\hat{a}\times\vec{b})=-\frac{\vec{b}}{r}-\frac{\vec{r}(\vec{b}\cdot\vec{r})}{r^3}=\frac{1}{r}\hat{a}\times(\vec{b}\times\hat{a})-\frac{2\vec{b}}{r}$$

This is a part of the solution, you need to differentiate your original thing as a product:
$$\nabla\times(\vec{v}f)=f\nabla\times\vec{v}+(\nabla f)\times\vec{v}$$
where $\vec{v}=\hat{a}\times\vec{b}$ and $f=\frac{e^{i(cr-dt)}}{r}$. The first term we now have, but we still need the second:
$$\nabla\frac{e^{i(cr-dt)}}{r}=e^{i(cr-dt)}\nabla\frac{1}{r}+\frac{1}{r}\nabla e^{i(cr-dt)}$$
$$=-e^{i(cr-dt)}\frac{\vec{r}}{r^3}+ic\frac{\vec{r}}{r^2}e^{i(cr-dt)}$$
$$=\hat{a}\frac{e^{i(cr-dt)}}{r}\left(ic-\frac{1}{r}\right)$$

Put things together:
$$\nabla\times \vec{f}=-\left(\frac{e^{i(cr-dt)}}{r}\nabla\times (\hat{a}\times\vec{b})+(\nabla\frac{e^{i(cr-dt)}}{r})\times\left(\hat{a}\times\vec{b}\right)\right)$$
$$=-\frac{e^{i(cr-dt)}}{r}\left(-\left(\frac{1}{r}\hat{a}\times(\vec{b}\times\hat{a})-\frac{2\vec{b}}{r}\right)+\left(ic-\frac{1}{r}\right)\hat{a}\times\left(\hat{a}\times\vec{b}\right)\right)$$
$$=-\frac{e^{i(cr-dt)}}{r}\left(\frac{2\vec{b}}{r}+\left(ic\right)\hat{a}\times\left(\hat{a}\times\vec{b}\right)\right)$$
It doesn't seem the same to the thing you list as the solution.
It is very possible that I made many mistakes because there are a lot of signs to be careful about, but you get the general idea.
Use http://en.wikipedia.org/wiki/Vector_calculus_identities in the future.
p.s. optics?
A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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$\ds{\large\tt\mbox{There is a list of}}$ $\quad\large\mbox{Vector Identities in Wikipedia}$ !!!.

\begin{align}
\nabla\times\vec{\fermi}\pars{\vec{r},t}&=
\nabla\times\bracks{%
{\expo{\ic\pars{cr- d t}} \over r}\,\vec{b}\times{\vec{r} \over r}}
\\[3mm]&=\nabla\bracks{{\expo{\ic\pars{cr- d t}} \over r^{2}}}\times
\pars{\vec{b}\times\vec{r}}
+{\expo{\ic\pars{cr- d t}} \over r^{2}}\nabla\times\pars{\vec{b}\times\vec{r}}
\\[3mm]&=\braces{{\vec{r} \over r}\,
\partiald{}{r}\bracks{{\expo{\ic\pars{cr- d t}} \over r^{2}}}}\times
\pars{\vec{b}\times\vec{r}}
+{\expo{\ic\pars{cr- d t}} \over r^{2}}\nabla\times\pars{\vec{b}\times\vec{r}}
\\[3mm]&={\phi_{\rm r} \over r}\,\,
\color{#44f}{\vec{r}\times\pars{\vec{b}\times\vec{r}}}
+ \phi\ \color{#c00000}{\nabla\times\pars{\vec{b}\times\vec{r}}}
\quad\mbox{where}\quad\phi\equiv{\expo{\ic\pars{cr- d t}} \over r^{2}}
\qquad\quad\pars{1}
\end{align}

$$
\color{#44f}{\vec{r}\times\pars{\vec{b}\times\vec{r}}}
=\vec{b}\pars{\vec{r}\cdot\vec{r}} - \vec{r}\pars{\vec{b}\cdot\vec{r}}
=r^{2}\,\vec{b} - \pars{\vec{b}\cdot\vec{r}}\vec{r}\tag{2}
$$

\begin{align}
&\color{#c00000}{\nabla\times\pars{\vec{b}\times\vec{r}}}
=\vec{b}\ \overbrace{\nabla\cdot\vec{r}}^{\ds{=\ 3}}\ -\
\vec{r}\ \overbrace{\nabla\cdot\vec{b}}^{\ds{=\ 0}}\ +\ \overbrace{\pars{\vec{r}\cdot\nabla}\vec{b}}^{\ds{=\ 0}} - \pars{\vec{b}\cdot\nabla}\vec{r}
\\[3mm]&\mbox{and}\quad\pars{\vec{b}\cdot\nabla}\vec{r}
=\sum_{i}b_{i}\,\partiald{}{x_{i}}\sum_{j}\hat{e}_{j}x_{j}
=\sum_{ij}b_{i}\hat{e}_{j}\delta_{ij}=\sum_{i}b_{i}\hat{e}_{i}=\vec{b}.
\\[3mm]&\mbox{Then,}\quad\color{#c00000}{\nabla\times\pars{\vec{b}\times\vec{r}}}
=2\vec{b}\tag{3}
\end{align}

With $\pars{2}$ and $\pars{3}$, $\pars{1}$ is reduced to:
\begin{align}
\nabla\times\vec{\fermi}\pars{\vec{r},t}&=
{\phi_{\rm r} \over r}\bracks{r^{2}\,\vec{b} - \pars{\vec{b}\cdot\vec{r}}\vec{r}}
+\phi\pars{2\vec{b}}
\end{align}

\begin{align}&\color{#44f}{\large%
\nabla\times\vec{\fermi}\pars{\vec{r},t}=
\pars{2\phi + r\phi_{\rm r}}\vec{b}
-{\vec{b}\cdot\vec{r} \over r}\,\phi_{\rm r}\,\vec{r}}
\\[3mm]&\mbox{with}\quad\phi\equiv{\expo{\ic\pars{cr- d t}} \over r^{2}}
\quad\mbox{and}\quad\phi_{\rm r} = \partiald{\phi}{r}.
\end{align}

