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My syllabus of electromagnetism defines the complex wavenumber as: $$k = \omega\sqrt{\epsilon\mu}$$ with $\epsilon$ the complex permittivity and $\mu$ the complex permeability. Thus $\epsilon$ and $\mu$ are complex numbers, which makes the square root ambiguous. They state that this issue can be solved, by defining $k$ as a complex number with a negative imaginary part and a positive real part: $$k = \alpha-i\beta$$ with $\alpha \ge 0$ and $\beta \ge 0$.

I don't see though how this can be defined this way, because it seems to me that it is possible that $\epsilon\mu$ has no square roots in the fourth quadrant. I hope someone can clarify this.

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    $\begingroup$ I don't really understand why this question is downvoted. It would be constructive if the downvoter could explain why this isn't a good question. $\endgroup$
    – Rayman
    Oct 29, 2013 at 12:56

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Mathematically speaking, the square roots of $\epsilon\mu$ can be anywhere in the complex plane. But then the wavevector $k$ has to satisfy some physical requirements: the fact that the dielectric or magnetic permittivities are complex functions means absorption/dispersion in the material, which translates into an attenuation of the propagating wave only if $\beta > 0$: $$ e^{ikx} = e^{i(\alpha + i\beta)x} = e^{i\alpha x}\times e^{-\beta x} $$

Together with this the wave has to propagate to the right, say for $x>0$, otherwise the real exponential blows up, therefore also $\alpha > 0$. In general, if you define $k = \alpha + i\beta$, if you don't want explosive behaviours, you need $\alpha$ and $\beta$ of the same sign: $\alpha\beta > 0$.

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  • $\begingroup$ Thank you for your answer. There exist active media though, in which energy is delivered to the electromagnetic fields. So it seems to me that in such a medium the amplitude of the plane wave would in fact "blow up". I don't really see why this is non-physical, because the material doesn't need to have infinite extent. Am I right if I say that the sign convention for $k$ is only meant for passive mediums? $\endgroup$
    – Rayman
    Oct 29, 2013 at 12:54
  • $\begingroup$ Well, unless you can create energy out of nowhere, I guess such media will have a saturation threshold, above which the physics switches from linear to non-linear. If the material is of finite thickness, of course also the "exploding" regime is well acceptable. But no, the convention has to be the same for both the passive and active medium, because the sign of $\alpha\beta$ has to be specified by the dielectric constant of the medium! Otherwise you could turn an active into a passive medium... $\endgroup$
    – Mattia
    Oct 31, 2013 at 8:17

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