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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.