0
$\begingroup$

Short question on electromagnetic waves. I am interested to know why the $z$ component of the complex monochromatic plane wave of the electric part of an electromagnetic wave is zero. If we consider the monochromatic planar electric wave $$\mathbf{\tilde{E}}(z,t) = \mathbf{\tilde{E}}_0e^{i(kz- \omega t)},$$ where $\mathbf{\tilde{E}_0} = \vec{\mathbf{E}}e^{i \delta}$ is the complex amplitude.

Is the reason that $(\tilde{E})_{z} = 0$ because Gauss's law $\nabla \cdot \vec{\mathbf{E}} = 0$ implies $$\nabla \cdot \mathbf{\tilde{E}} = \nabla \cdot[\vec{E}e^{i(kz-\omega t + \delta)}] = E_{z}(ik)e^{i(kz- \omega t + \delta)}=0\implies(E_{z}) = 0\implies(\tilde{E}_{z}) = 0?$$

Is this the correct reasoning? Thanks.

$\endgroup$

1 Answer 1

1
$\begingroup$

Yeah, that's pretty much it. That's why that happens.

Just to make this longer than a one-liner: note that an identification of the form $\widetilde{\mathbf E}_0=\vec{\mathbf E}e^{i\delta}$ sort of misses the point of complex amplitudes, since $\delta$ is ill-defined, and there's no guarantee that you can make $\vec{\mathbf E}$ real-valued. How would you deal, for example, with $\widetilde{\mathbf E}=|E_0|e^{i\delta}(\hat{\mathbf e}_x+ i\hat{\mathbf e}_y)$?

$\endgroup$
10
  • $\begingroup$ In the book I am using (which is introductory), it seems to suggest that the electric field $\mathbf{\vec{E}}$ is a real-valued vector field, are the components more generally complex valued? Also, how is $\delta$ ill-defined exactly? $\endgroup$
    – user100411
    Jan 24, 2017 at 10:49
  • $\begingroup$ Your book is wrong (or you are misinterpreting it). There is generally no way to make all the components of $\vec {\mathbf E}$ real-valued simultaneously; the counterexample is the circularly polarized wave at the end of my answer. This is why $\delta$ is ill-defined. $\endgroup$ Jan 24, 2017 at 10:58
  • $\begingroup$ Just to be clear though, when dealing with electric fields which originate from stationary charges, we consider the electric field as being real valued? $\endgroup$
    – user100411
    Jan 24, 2017 at 11:11
  • $\begingroup$ The physical fields are always real-valued. The field you wrote down, $\mathbf{\tilde{E}}(z,t) = \mathbf{\tilde{E}}_0e^{i(kz- \omega t)},$ is not physical - you get to the physical field by taking the real part. For further small questions along this line, though, it's better to ask on chat. $\endgroup$ Jan 24, 2017 at 11:13
  • $\begingroup$ I sent a question on chat, not sure if you received it. So real part of what I wrote $Re[\mathbf{\tilde{E}}] = Re[\mathbf{\vec{E}}e^{i(kz-\omega t + \delta)}] = \mathbf{\vec{E}}\cos(kz-\omega t + \delta)$ where $\mathbf{\vec{E}}$ is real valued since we got it by taking the real part of $\mathbf{\tilde{E}}$. But you are saying that $\mathbf{\vec{E}}$ is not necessarily real, what is wrong the with what I stated? Thanks. I think you have a typo in your answer. Did you mean "there's no guarantee you can make $\mathbf{\tilde{E}}$ real"? $\endgroup$
    – user100411
    Jan 26, 2017 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.