In metals, below a certain frequency the real part of permittivity is negative. This is the reason for reflection of light. If the imaginary part were zero, the reflectivity would be 1.  There would not be any absorption. Just because e^1/2 would be imaginary. Then, you'd have to take absolute value of your expression for reflection and obtain 1.

Imaginary part of permittivity allows for absorption and thus reduces reflectivity from 1. R = 1- A. It is determined  by the scattering. More scattering, more absorption,  lesser reflection. 

Thanks for your answer. I'm not sure if I've fully got what you said. So since n= h+ki, and e = h^2-k^2+2hki, and k become large below a certain frequency, e1=h^2-k2 will be negative. Now if there were no absorption, the e would be equal to e1, and e1^1/2  would be imagenary, which is a representative of absorption. When there is no absorption, e1^1/2 would be zero, thus, R = 1? Does that mean at that particular small frequency there is no real permittivity?

But I'm so confused. First why does absorption increase at low frequencies? is there any physical explanation for this? I mean conceptual explanation not a mathematical. Plus, my main question still remains that why more scattering leads to more absorption? What is the interaction between light and scattered electrons?

No, h is negative. Even if k=0. That is metal.

Thanks again Igor, 

Could you please give me some simple explanation that why h is negative, and why k becomes so important at small frequency in metals? Again I am more interested in physical meaning of these parameters than their effects in a mathematical model. 

And my main question is still the effect of scattering on reflectivity and absoption (the physical reasons). 

Thanks so much. 

Ahmad

Electrons in metals are free. A free electron reacts to the oscillations of the electric field of the electromagnetic wave by moving forth and back. This movement with respect to the ionic cores creates dipoles, which create an electric field opposite to the field of the incident wave. The field created by one electron cannot compensate the field of the incident wave, but hte field of many electrons can, so, up to a certain frequency, which depends on the number of electrons, the incident wave is completely cancelled in the metal, so it is reflected. mathematically this is described with negative permittivity.

There is a rub, the motion of electrons is hampered by scattering, effectively friction. Due to this friction a part of energy of the incident wave is transformed into heat, absorbed and in not reflected. This friction is the reason also for the finite conductivity of metals, id est resistance. At very low frequencies the effect of the resistance is stronger than the effect of electrons inertia (overdamped mode) and the permittivity is complec. Above some frequency (Maxwell limit) the inertia is stronger than  the friction. The permittivity becomes almost real but negative. Above some other frequency that depends on the electron density, electrons are too slow to react to the field of the incident wave and cannot compensate it. Then the permittivity is real and the metal is transparent to light. 

So, how scattering effects the response of the metal to light? By absorbing a part of its energy converting it to heat.

Let me try, I really love the question!

1. As the electric field in metals is zero, most of the physics plays on the metal surface.

2. The less light is reflected, the more light is absorbed, and vice versa (as the transmission is zero).

According to the Wiedemann-Franz law, the ratio of thermal conductivity vs electrical conductivity increases with temperature. This somehow makes sense, as the more phonons are around, the more likely an electron can pass its energy on to phonons. Consequently, if we now heat a metal up, the incident light will be much easier converted to heat (absorption), as a cold metal surface populates less phonons. Therefore, less energy can be reflected by a hot metal surface (still assuming that the metal is not burning hot and emitting lights...)

What do you think? Does this make any sense?

Thanks Igor for your detailed explanation and also Christian for simplifying the matter. The more I read and study about this topic, the more confused I get.

I learnt that metals in general are kind of lossy materials, cause valance electrons can easily absorb photons and transfer to conduction band (Is this correct?) At the same time below a certain frequency their refractive indexes increases dramatically so the wave gets reflected before even has a chance to pass through the material. This is related to the big, but opposite electric field created by dipoles in metals, as Igor explained. So although a metal is so lossy naturally, it is reflective at low frequencies, true?.

Now first question is if the wave does not get inside the metal, how does the wave produce dipoles inside the metal? I mean if the wave is outside, how does it influence the bulk? Or does that mean that the wave just affect the surface and the atoms on the surface reflect the wave back? 

At plasma frequency wave can pass through and a big portion of that is absorb, metal becomes too lossy. This is again because of photon-electon reaction, and the friction that Igor mentioned. Is this all about electron scattering and free path of electrons? Does that mean that at plasma frequency photons lead to more electron scattering by producing phonons, making the electron motion harder? (is this correct?) If so, where is the role played by the electrons transferred to conduction band by absorbing photons? how are these two different and how does every phenomenon affect an electron motion?

At very high frequency the whole wave pass through without interacting with electron, just like x-ray that can go through the metal, for which metal is transparent. Igor said that electrons are too slow for the wave, does that mean that electrons have slower velocity relative to the wave that cannot influence the wave (e.g. x-ray)? 

You've got confused and need some time to arrange information in your head.

1. I learnt that metals in general are kind of lossy materials, cause valance electrons can easily absorb photons and transfer to conduction band (Is this correct?)

No, this is not correct. The valance band in metals is half filled with electrons, therefore it is also the conduction band. Call it whatever you like, but it is the same band.  The electrons may be considered free. Intraband (that is inside one band) transitions are forbidden by energy and momentum conservation laws. 

2. first question is if the wave does not get inside the metal, how does the wave produce dipoles inside the metal?

It does penetrate slightly. Just decays exponentially. The penetration depth depends on frequency. So, you may say that the surface reflects light, but actually it is a thin layer of metal near the surface.

3. Is this all about electron scattering and free path of electrons? Does that mean that at plasma frequency photons lead to more electron scattering by producing phonons, making the electron motion harder? (is this correct?) If so, where is the role played by the electrons transferred to conduction band by absorbing photons? how are these two different and how does every phenomenon affect an electron motion?

Nothing particularly interesting happens with light at plasma frequency. Due to symmetry of the electromagnetic waves. They are transverse and plasma oscillations are longitudinal, therefore they do not interact. At plasma frequency real part of permittivity is zero. Below plasma frequency light is reflected, above plasma frequency it penetrates. That's all. No increased absorption.

4. Igor said that electrons are too slow for the wave, does that mean that electrons have slower velocity relative to the wave that cannot influence the wave (e.g. x-ray)? 

Slow means just that the amplitude of forced oscillations of electrons is small, the dipole moments (with respect to cores) is not sufficient to compensate for the electric field of incident light, so the light penetrates into metal. The transition happens at the plasma frequency. When it is approached from below the penetration depth becomes larger (more electrons are needed to compensate for the field of the incident wave. When the light's frequency becomes larger than the plasma frequency electrons kind of give up resistance. This is not related to the electron velocities but to the amplitude of their oscillation.

Look up Drude model. This is the simplest model of metals but it contains equations that are useful for understanding.