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An harmonic oscillating source will emit fields of the form

\begin{align} &\mathbf{H}=\frac{ck^2}{4\pi}(\mathbf{n}\times\mathbf{p})\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr} \right) \tag{1}\\ &\mathbf{E}=\frac{1}{4\pi\varepsilon_0}\left\{ k^2(\mathbf{n}\times\mathbf{p})\times\mathbf{n}\frac{e^{ikr}}{r}+[3\mathbf{n}(\mathbf{n}\cdot\mathbf{p})-\mathbf{p}]\left( \frac{1}{r^3}-\frac{ik}{r^2}\right)e^{ikr} \right\} \tag{2} \end{align}

where $\mathbf{n}=\frac{\mathbf r}{r}$ is the unit vector that points in the radiation direction and $\mathbf{p}$ is the electric dipole vector.

As Jackson says (Classical Electrodynamics, chapter 9) it's possible to find equation (2) from equation (1) using the formula

$$ \mathbf{E}=\frac{iZ_0}{k}\nabla\times\mathbf{H} \tag{3}$$

Applying equation (3) to equation (1) I obtain

$$ \mathbf{E}=\frac{ik}{4\pi\varepsilon_0}\nabla\times\left[\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)(\mathbf{n}\times\mathbf{p})\right]$$

Then

$$ [\nabla\times(f\mathbf v\times\mathbf w)]_k=\varepsilon_{ijk}\partial_i\varepsilon_{abj}fv_aw_b=\varepsilon_{ijk}\partial_if\varepsilon_{abj}v_aw_b=[\nabla f\times(\mathbf v\times\mathbf w)]_k$$

Identifying $f=\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)$ I calculate

$$\nabla\left[\frac{e^{ikr}}{r}\left( 1-\frac{1}{ikr}e^{ikr} \right)\right]=\frac{e^{ikr}}{r}\left( ik-\frac{2}{r}+\frac{2}{ikr^2} \right)\mathbf n$$

Then i get

$$ \mathbf{E}=\frac{ik}{4\pi\varepsilon_0}\frac{e^{ikr}}{r}\left( ik-\frac{2}{r}+\frac{2}{ikr^2} \right)\mathbf{n}\times(\mathbf{n}\times\mathbf{p}) \tag{4}$$

That is definitely different from equation (2). So where am I wrong? How can I obtain the correct equation (2)?

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Note that $\mathbf{n}$ is not a constant vector but dependent on position. You have not taken the derivative of this, which is the source of error.

In particular, you say $$\nabla \times (f\mathbf{v}\times\mathbf{w}) = \nabla f \times (\mathbf{v}\times\mathbf{w}).$$

This is incorrect, the correct expression would be $$\nabla \times (f\mathbf{v}\times\mathbf{w}) = \nabla f \times (\mathbf{v}\times\mathbf{w})+ f\,\nabla \times (\mathbf{v}\times\mathbf{w}).$$

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  • $\begingroup$ Well, I can always choose $\mathbf{n}$ to be parallel to, for example, the x axis, then $\mathbf{n}=(1,0,0)$. $\endgroup$ – Alessandro Zunino Mar 13 '17 at 11:10
  • $\begingroup$ Still, when you apply the gradient, you should consider how $\mathbf{n}$ changes. If you move from $(1,0,0)$ to $(1,dy,0)$, $d \mathbf{n}/dy = 1$ which cannot be neglected. $\endgroup$ – Raziman T V Mar 13 '17 at 11:17
  • $\begingroup$ Yes, of course. I realized now that it was a quite dumb question. Thank you, anyway! $\endgroup$ – Alessandro Zunino Mar 13 '17 at 14:05

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