double curl and vector laplacian
Andrew Henderson 
I am currently reading "Classical Electrodynamics" by Jackson and I couldn't reproduce a calculation made there.
I got $$\vec{A} = - ik \frac{e^{ikr}}{r} \vec{d}$$ and I want to calculate the following two things $$\vec{B} = \nabla \times \vec{A} \\ \vec{E} = \frac{i}{k} \nabla \times \left( \nabla \times \vec{A} \right)$$Calculating $\vec{B}$ was pretty easy and straightforward by using $\nabla \times \left( f(r)\vec{d} \right) = f'(r) \frac{\vec{r}}{r} \times \vec{d}$ for a constant $\vec{d}$. I get $$\vec{B} = e^{ikr} \left( \frac{k^2}{r^2} + \frac{ik}{r^3} \right) \vec{r} \times \vec{d}$$which is identical to the calculation in the book.
When I calculate $\vec{E}$ I don't get the book's solution, which is $$\vec{E} = k^2 \left(\vec{n} \times \vec{d} \right) \times \vec{n} \frac{e^{ikr}}{r} + \left[ 3 \vec{n} \left( \vec{n} \cdot \vec{d} \right) - \vec{d} \right] \left[ \frac{1}{r^3} - \frac{ik}{r^2} \right] e^{ikr} $$
I tried using the identity $$\nabla \times \left( \nabla \times \vec{A} \right) = \nabla \left( \nabla \cdot \vec{A} \right) - \Delta \vec{A},$$ but I couldn't find an elegant way of calculating $\Delta \vec{A}$. I think it should be possible by using the vector laplacian as described here and just calculating it but isn't there a better way to do this?
$\endgroup$ 21 Answer
$\begingroup$Yes. The basic identity for the Laplacian acting on purely radial functions is $$ \Delta f(r)= \frac{1}{r} \partial_r^2~ (rf(r)), $$on top of the action of the gradient on such, which you know, above.
Thus,$$ \vec{E}=\frac{i}{k} \nabla \times \left( \nabla \times \vec{A} \right) =\frac{i}{k} \nabla \left( \nabla \cdot \vec{A} \right) - \frac{i}{k}\Delta \vec{A}, $$and$$ \frac{i}{k}\Delta \vec{A} =\frac{1}{r}\vec{d} ~\partial_r^2 e^{ikr}= \frac{-k^2}{r}\vec{d} ~ e^{ikr}. $$
Furthermore, $$ \frac{i}{k}\nabla \cdot \vec{A}=\vec{d}\cdot \nabla ~\frac{e^{ikr}}{r}= \vec{d}\cdot \vec{r} ~ e^{ikr}\left (\frac{ik}{r^2}-\frac{1}{r^3}\right ), $$so that $$ \frac{i}{k}\nabla (\nabla \cdot \vec{A})= \vec{d}~e^{ikr}\left (\frac{ik}{r^2}-\frac{1}{r^3}\right )+\hat{r} (\hat{r}\cdot\vec{d}) r\partial_r~ e^{ikr}\left (\frac{ik}{r^2}-\frac{1}{r^3}\right )\\ = \vec{d}~e^{ikr}\left (\frac{ik}{r^2}-\frac{1}{r^3}\right )+\hat{r} (\hat{r}\cdot\vec{d}) ~ \frac{e^{ikr}}{r}\left (-k^2 -\frac{3ik}{r}+\frac{3}{r^2}\right ). $$
Consequently,$$ \vec{E}= \vec{d}~\frac{e^{ikr}}{r}\left (\frac{ik}{r}-\frac{1}{r^2} +k^2\right )+\hat{r} (\hat{r}\cdot\vec{d}) ~ \frac{e^{ikr}}{r}\left (-k^2 -\frac{3ik}{r}+\frac{3}{r^2}\right )\\ =\frac{e^{ikr}}{r} \left ( \vec{d}~\left (\frac{ik}{r}-\frac{1}{r^2} +k^2\right )+\hat{r} (\hat{r}\cdot\vec{d}) ~ \left (-k^2 -\frac{3ik}{r}+\frac{3}{r^2}\right )\right ) . $$
- Needless to say, this is the very expression you are targetting, once you evaluate its double cross product, $ \frac{e^{ikr}}{r}\left ( k^2 (\vec{d} -\hat{r} ~\hat{r}\cdot \vec{d} ) + ( 3 \hat{r} ~ \hat{r} \cdot \vec{d} - \vec{d}) \left[ \frac{1}{r^2} - \frac{ik}{r} \right] \right )$ , and reduce it to a canonical form such as the one found.
