Let the phasor electric field of a uniform plane wave be $\vec{E_{0}}e^{-i\vec{k} .\vec{r}}$ where $\vec{k}$ is the wavenumber vector and $\vec{r}$ the position vector.

why is then, $\frac{1}{-i\omega \mu} \nabla \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}} = \frac{1}{\eta}\hat{k} \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}}$

I have seen this statements in Cheng's book Field and wave electromagnetics 2nd ed. page 367 for those who are curious.

  • 2
    $\begingroup$ hint: $\textrm{curl}(f\textbf{F})=f\textrm{curl}(\textbf{F}) + \textrm{grad}(f) \times \textbf{F}$ $\endgroup$
    – hyportnex
    Jan 4, 2018 at 13:30
  • 1
    $\begingroup$ ... and remembering (after @hyportnex) that $\vec E_0$ is a constant vector. $\endgroup$ Jan 4, 2018 at 13:33

1 Answer 1


This is because $k=\omega/c=\omega\sqrt{\epsilon \mu}$, $\eta=\sqrt{\mu/\epsilon}$, $\nabla e^{-i\vec{k}. \vec{r}}=-i\vec ke^{-i\vec{k}. \vec{r}}$ and thus $$\frac{1}{-i\omega \mu} \nabla \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}} =\frac{-i}{-i\omega \mu}\vec{k} \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}}= \frac{-ik}{-i\omega \mu}\hat{k} \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}}=\frac{1}{\eta}\hat{k} \times \vec{E_{0}}e^{-i\vec{k}. \vec{r}}$$ Added note: The curl is the cross product of the vectorial nabla operator and this electric vector field phasor. The cross product of vectors is bilinear.


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