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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

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  • $\begingroup$ What is the relationship between $x,y,z$ and $r,t$? These are cylindrical coordinates? $\endgroup$ – user7530 Apr 4 '14 at 4:57
  • $\begingroup$ Spherical @user7530 $\endgroup$ – Jeff Faraci Apr 4 '14 at 4:58
  • $\begingroup$ Curl obeys the product rule $\nabla \times fv = f\nabla \times v + \nabla f \cdot v$ for scalar function $f$ and vector-valued function $v$. Here instead of dealing with Levi-Civita symbols (shudder) it might be easiest to apply the product rule, compute the curl of $v$ in Cartesian coordinates, and the gradient of $f$ in spherical. $\endgroup$ – user7530 Apr 4 '14 at 5:00
  • $\begingroup$ I am looking for a solution, not a comment. Thanks though. $\endgroup$ – Jeff Faraci Apr 4 '14 at 5:04
  • $\begingroup$ No problem. Just a suggestion that might make your life easier ;) $\endgroup$ – user7530 Apr 4 '14 at 5:05
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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?

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  • $\begingroup$ well I re-wrote the problem, since my step 2 was wrong, therefore everything follow was also wrong. Can you answer it in this post? Thanks $\endgroup$ – Jeff Faraci Apr 4 '14 at 7:23
  • $\begingroup$ How come this solution is different than what I have in the post, which is the correct solution. The answer should be $$ \vec \nabla \times f(\vec r,t)=-\hat a \times (\hat a \times \vec b) \left(\frac{ik e^{i(cr-dt)}}{r}-\frac{e^{i(cr-dt)}}{r^2} \right). $$I would like to up vote your answer and as the correct one and all but the solution isn't correct...Thanks for your time...If you can solve it thanks. What you originally posted in your solution I think was closer to the correct answer. You had the correct term, except one additional one. . But now this solution looks further away.. $\endgroup$ – Jeff Faraci Apr 4 '14 at 8:01
  • $\begingroup$ That 2b/r bothers me too. Let's both go through the procedure again - if we can find any problems. $\endgroup$ – orion Apr 4 '14 at 8:18
  • $\begingroup$ It almost looks like we shouldn't even take the curl of $\hat{a}\times\vec{b}$, as if it was a constant. Is it possible that the $\vec{r}$ in this term is different from $r$ in the exponential term? For instance, could it be that one is the position of the source and the other position at an arbitrary point in space? $\endgroup$ – orion Apr 4 '14 at 8:29
  • $\begingroup$ The function $$ \frac{e^{i(c r- d t)}}{r} $$ is just a scalar function. It is the same r in the picture. We are essentially taking the curl of a cross product*scalar function. I cannot find mistakes in your solution. Look at what I just added to the posting. thanks a lot $\endgroup$ – Jeff Faraci Apr 4 '14 at 16:38
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\large\tt\mbox{There is a list of}}$ $\quad\large\mbox{Vector Identities in Wikipedia}$ !!!.

\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}

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  • $\begingroup$ Thank you Felix. I like the physicist approach ;) +1 $\endgroup$ – Jeff Faraci Jun 1 '14 at 17:21
  • $\begingroup$ @Integrals You're welcome. Thanks. $\endgroup$ – Felix Marin Jun 1 '14 at 18:46

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