Curl, $\vec\nabla \times (\hat{a}\times \vec{b})$

Question

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

Answer 1

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: $$\nabla\times(\hat{a}\times\vec{b})=-\frac{\vec{b}}{r}-\frac{\vec{r}(\vec{b}\cdot\vec{r})}{r^3}=\frac{1}{r}\hat{a}\times(\vec{b}\times\hat{a})-\frac{2\vec{b}}{r}$$

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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:

$$\nabla\frac{e^{i(cr-dt)}}{r}=e^{i(cr-dt)}\nabla\frac{1}{r}+\frac{1}{r}\nabla e^{i(cr-dt)}$$ $$=-e^{i(cr-dt)}\frac{\vec{r}}{r^3}+ic\frac{\vec{r}}{r^2}e^{i(cr-dt)}$$ $$=\hat{a}\frac{e^{i(cr-dt)}}{r}\left(ic-\frac{1}{r}\right)$$

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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?

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\large\tt\mbox{There is a list of}}$ $\quad\large\mbox{Vector Identities in Wikipedia}$ !!!.

\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}