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So for a complex dielectric constant $\epsilon = \epsilon_a + i\epsilon_b$, the wave vector and index of refraction are related to it through $k = \frac{\omega}{c}n$ and $n = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}}$. According to Jackson, the real part of the dielectric is related to polarization and anomalous dispersion, while the imaginary part is associated with dissipation of energy into the medium.

If you write the wavevector as $k = \beta + i \alpha/2$ and plug it in the general wave formula (just in 1D right now) of $e^{ikr} = e^{-\alpha r/2}e^{i\beta r}$, the intensity drops as $e^{-\alpha r}$, so $\alpha$ is the attenuation constant, which tells you how quickly the wave dies out in the medium.

But, if you plug that form of $k$ into the above equations to solve for $\alpha$ and $\beta$ as a function of $\epsilon_a$ and $\epsilon_b$, you find that $\alpha$ and $\beta$ are both a function of both $\epsilon_a$ and $\epsilon_b$.

This is counterintuitive to me, because intuitively I'd think that the attenuation constant $\alpha$ would only be based on $\epsilon_b$, due to dissipation, and the same with $\beta$ and $\epsilon_a$.

Can anyone give a good physical explanation for this "mixing"?

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  • $\begingroup$ to me there should be dependence of $\alpha$ on $\epsilon_a$ because the dissipation is a kind of friction on the motion of the polarized molecules and $\epsilon_a$ tells you how much of the matter is polarized. does this make sense to you? $\endgroup$ – hyportnex Feb 13 '14 at 21:19
  • $\begingroup$ The dissipation can be in frequency-space (corresponds to time-dependent growth/decay) or wavenumber-space (corresponds to spatial-dependent growth/decay). In the case you listed, the decay is spatially dependent. This is because in Jackson, he considers a plane EM wave incident on a dielectric medium. What if you had a medium that could grow/produce such waves, then the decay rate would be $\propto$ $e^{-\gamma t}$, where $\gamma$ = $\Im \left[ \omega \right]$. $\endgroup$ – honeste_vivere Oct 3 '14 at 18:16
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There isn't really a good physical explanation - this simply arises from the conventions we choose to represent our electromagnetic fields.

The electric constant $\epsilon_0$ was defined as the constant needed to make Gauss's law for electricity and Coulomb's law work for whatever units of length, charge and force you want to choose. When we add a medium, we find it is useful to define the electric displacement vector and a new effective electric constant $\epsilon$ for that medium: $\epsilon$ accounts for the shifting of charge wrought by the electric field and the consequent "reaction field" from the bound charge: so we need to put $\epsilon$ into Gauss's law if we want it to work for the nett charge.

However, when you get to studying waves, you're putting together the Faraday and Ampère Laws (with Maxwell's "displacement current"): two different equations describing a different phenomenon than simply force and flux from an electric field. You have two first order coupled differential equations, so the speed and propagation constants depend on $\sqrt{\epsilon}$ and $\sqrt{\mu}$, because when you decouple second order equations you get square roots of the constant co-efficients involved in the arguments of the $\exp(i (k z - \omega t))$ basic solutions. If we had discovered waves first and Gauss's law second, we'd probably have defined things differently so that $\sqrt{\epsilon}$ and $\sqrt{\mu}$ were the more fundamental quantities. When you square or square root a complex quantity, you mix the components - that's all there is to it. You might even imagine defining $\epsilon^{\frac{1}{4}}$, $\mu^{\frac{1}{4}}$ as the fundamental quantities: this would be quite acceptable and you'd have squares of the fundamental quantities in Maxwell's equations. You'd still have mixing of real and imaginary components when you wanted to find the attenuation co-efficient. It's simply a matter of what conventions are used.

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The issue here is how much the refractive index $n$ tells you about dissipation. As you rightly said, the imaginary part of $n$, which depends on both real and imaginary parts of $\epsilon$, leads to an imaginary part in k which describes an exponentially decaying electric field. However, this doesn't necessarily correspond to dissipation (i.e. a drop in energy). For example, if we consider an ideal metal with real $\epsilon<0$, $k$ is imaginary leading again to spatial decay. But in the time domain, an incident plane wave will be perfectly reflected by such a metal rather than absorbed. To calculate the dissipation we need to workout the time averaged power, over a single cycle or period. If we do this for an arbitrary dielectric medium with $\epsilon=\alpha+i\beta$, the dissipated power is found to be $\langle P_d\rangle=\frac{\omega}{2}\beta|E|^2$ (a proof can be found here http://web.mit.edu/6.013_book/www/chapter11/11.5.html). So finally, the dissipated power is found to only depend on the imaginary part of the permittivity $\beta$.

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This is a good question that I struggled with myself for some time. I believe that the correct answer is the following. The imaginary part of the index of refraction, i.e. $\kappa$, quantifies the dissipation of light through a medium. However, if one wants to quantify the dissipation due to nonretarded electric fields alone, the quantity that quantifies this is not $\epsilon_b$ but $\operatorname{Im}\frac{1}{\epsilon}$, where $\epsilon$ is the total dielectric function. These, the index of refraction and the inverse dielectric function are therefore the most appropriate quantities to measure. As a further reference, I recommend the book Electrodynamics of Solids by Dressel and Gruner.

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In fact you can also write $n$ as a complex number, $n=n_r+in_i$, perhaps you have known the refractive index can be derived from $\sqrt{\epsilon}$, writing $\epsilon=\epsilon_r+\epsilon_i$ Solved as $$ n_r=\sqrt{\sqrt\frac{\epsilon_i^2+\epsilon_r^2}{2}+\epsilon_r/2}$$ $$ n_i=\sqrt{\sqrt\frac{\epsilon_i^2+\epsilon_r^2}{2}-\epsilon_r/2}$$

These expressions are all equivalent

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    $\begingroup$ I know that... did you even read the question? I talked about this in the first sentence. I was asking about the physical meaning behind it, if there is any. $\endgroup$ – YungHummmma Jan 14 '14 at 21:44
  • $\begingroup$ Just want to warn the expression of nr and ni are incorrect. Note that "2" under (ei^2+ er^2) are not in sqrt. the correct version is ni=sqrt(sqrt((e1^2+e2^2))/2-e1/2); $\endgroup$ – touchsky Jun 15 '18 at 15:34

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