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I would like to know why evanescent waves are not possible in pure vacuum?

Indeed, we have $$ k^2 = \left(\frac{\omega}{c}\right)^2 $$

in vacuum, so we could have for example : $$ k_x^2+k_y^2+k_z^2=\left(\frac{\omega}{c}\right)^2 $$

with $ k_z=i k_z'' $ and $ k_x^2+k_y^2 > \left(\frac{ \omega } {c}\right)^2$.

I have read that we need local charges to do it but I don't understand why.

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Evanescent waves appear as boundary conditions; vacuum solutions lack boundaries, except at infinity.

S < https://en.m.wikipedia.org/wiki/Evanescent_field>

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  • $\begingroup$ I understand that if I have two medium with n1>n2 it is possible to create a wavector along the x axis in medium 1 that will be larger than the allowed wave vector in the medium 2. So the kz will for example be imaginary in the second medium => evanescent wave. But here, it explains that If I am at the interface between two dielectric I can have evanescent wave in the second if I properly choose my angles. But it doesn't explain why in vacuum it can't happen also. If we don't add any condition on $ k^2=(\frac{w}{c})^2$ in the vacuum nothing forbid to have evanescent wave in vacuum. $\endgroup$
    – StarBucK
    Commented Feb 20, 2016 at 12:48
  • $\begingroup$ When you work the problem you will find that either the evanescent field will vanish, or it will be a transient. Evanescent fields are very important in some applications, such as fiber optics. Working a few problems is always worthwhile. $\endgroup$
    – Peter Diehr
    Commented Feb 20, 2016 at 12:57
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When we study diffraction in optics, we use the angular spectrum of plane waves and indeed, the evanescent waves in a vacuum are components of this spectrum when the sum of the squares of the direction cosines is greater than 1. This is treated in Goodman's book about Fourier optics. Doing a quick search, I found the following link: Angular Spectrum Method

Hope it can help.

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