Reflectivity of Metals - real and imaginary part of k-vector and complex dielectric function

De electric field component of an electromagnetic wave that is traveling in the $x$-direction is given as $$E(x,t) = E_0 e^{i(kx - \omega t)}$$ with $E_0$ the amplitude and $k$ the wave vector.

From what I understand, this wave vector, since it is not the same as the wave number, has both a real and an imaginary part.

The imaginary part is responsible for the damping of the wave and dissipation of energy? And what does the real part mean physically? That, if it is large enough, the wave gets entirely transmitted by the medium, and nothing is reflected?

Also, I was wondering what's exactly the physical meaning of the complex dielectric function for metals. For metals it is given as : $$\epsilon_r (\omega) = 1 - \frac{\omega_p^2}{\omega^2 + i \gamma \omega}$$ with $\omega_p$ the plasmafrequency and $\gamma$ a damping factor.

I know a dielectric constant describes the behaviour of some material in the presence of an electric field, but why do we need a function here?

• The dielectric "constant" is always a function. It depends at least on material, temperature, frequency and pressure. – Sebastian Riese May 24 '15 at 22:28
• I've made an addition to my post, don't know if it helps you. – Constantine Black May 25 '15 at 16:08

First, note that the way you have written the electrical wave-field isn't anything more than exactly a way to write a wave function in general. This is because the term $e^{i(kx-ωt)}$ can be written as: $$e^{i(kx-ωt)} = cos(kx-ωt) +i sin(kx-ωt)$$, and from here you can keep in general the real or the imaginary part as you wish.

As for $κ$, the imaginary part is a term that expresses absorption(the amplitude decreases as we get inside the material).That is: $$E(x,t)=E_0e^{-k_{im} z}e^{i(kx-ωt)}$$ , with $$k_{im} =ω \sqrt{ {εμ \over 2}}[\sqrt{1+ {σ \over εω}^2}-1]^{1/2}$$. The distance at which the amplitude decreases by a factor of $1/e$ is defined as $$d={1 \over k_{im}}$$ Note that $σ$ is the special electrical conductivity.

The electric susceptibility $χ_e$ of a dielectric material is a measure of how easily it polarizes in response to an electric field. This, in turn, determines the electric permittivity of the material and thus influences many other phenomena in that medium, from the capacitance of capacitors to the speed of light.

It is defined as the constant of proportionality (which may be a tensor) relating an electric field E to the induced dielectric polarization density P such that: $$P=ε_0 χ_ε Ε$$

where $ε_0$ is the electric permittivity of free space.

The susceptibility of a medium is related to its relative permittivity $ε_0$ by $$χ_ε = ε_r -1$$.

Note, as commented at your question by Sebastian Riese, that in general $ε_r$ is a function of the properties of the material.

Hope this helps with your question.

EDIT

As for reflectivity, it depends on the angle of incident on the surfaco of the material and properties as $ε and μ$. You can prove, as the most simple example that that the Reflection and Transmission factors are: $$R={n_1 -n_2 \over n_1 +n_2}^2$$ $$T={4n_1 n_2 \over (n_1 +n_2)^2}$$ where n is the index of refraction defined as: $$n=\sqrt{εμ \over ε_0 μ_0}$$ and it is also: $$k=k_{re} +i k_{im}$$ with: $$k^2 =μεω^2 +ι μσω$$, so there is a connection between $k$ and the reflection factor.

That's all. Hope I helped.