Question about conductivity/reflexion

[fluidistic]

I'm all mixed. I've read passively through a tiny part of Jackson's book on electrodynamics and some Hecht on Optics.

Why are copper and gold excellent conductors while iron conducts less considering that copper and gold absorb somehow greatly light in the visible spectra, particularly in wavelengths corresponding to blue/green and hence their color? Wouldn't that mean that copper and gold doesn't reflect as well light as iron, since they do absorb? And a not so good reflector should be a not so good conductor? Or I'm wrong?
Maybe I should considering the absorption over the whole EM spectra, not only visible and I'd see that copper and gold absorb less than iron. In other words, their emissivity would be less than the one of iron.
Doing a quick search (http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html) I found out that indeed iron has a greater emissivity than copper and gold though it's hard to compare since surfaces aren't necessarily the same.

I'd like some clarifications.
Thanks.

 

[DrDu]

As you say, it is a wavelength dependent phenomenon. In the case of copper and gold, the color is due to transitions from the d band to the conduction band. And certainly, at frequencies corresponding to the blue-green part of the spectrum gold and copper will be not so good conductors as they are e.g. in the red part of the spectrum.

 

[fluidistic]

I appreciate your help. I have a question, what do you mean exactly by

And certainly, at frequencies corresponding to the blue-green part of the spectrum gold and copper will be not so good conductors as they are e.g. in the red part of the spectrum.

?
Do you mean if I put an alternate current into a gold/copper wire at those high frequencies?

 

[Born2bwire]

I appreciate your help. I have a question, what do you mean exactly by ?
Do you mean if I put an alternate current into a gold/copper wire at those high frequencies?

Well... you really can't do that. How do you excite currents at optical frequencies? The classical picture starts to break down around the terahertz regime. At around terahertz some metals, like gold and silver, start to behave like plasmas. An incident wave will excite a surface plasma. But at higher frequencies the electrons have too much inertia to be able to oscillate at the same frequency as the incident radiation. This corresponds to the metal becoming completely opague. Even further up in frequency you need to start taking into account the quantum nature of the material to be able to properly model it (like the atomic transitions that give gold its color). Where all this happens and starts to breakdown will depend on the material though. Suffice to say though that when you are in the optical regime you need to start considering some quantum effects.

There isn't a good way to passively predict the frequency response of a material. It is a very complicated property and you will find numerous resonances and behaviors across the spectrum. It is true that we can relate the real and imaginary parts of the permittivity/permeablity via the Kramer-Kronig relationship (Hilbert transform) and thus glean some insight from that. However, we still need to deduce the complete behavior of one of the components. We can provide some basic models that are effective over small bandwidths. For example there is the conductivity, plasma, or inhomogeneous mixed materials (Debye is one such model) models. In the end though, we generally fit these models to experimental data and this allows us to have limited picture of the materials over a certain bandwidth.

 

[fluidistic]

Good to know born2bwire. It also surprised me that the meaning of

And certainly, at frequencies corresponding to the blue-green part of the spectrum gold and copper will be not so good conductors as they are e.g. in the red part of the spectrum.

would be

Do you mean if I put an alternate current into a gold/copper wire at those high frequencies?

.
Do you have an idea about what DrDu meant?

 

[DrDu]

I meant that the absorption $$\kappa$$, i.e. the imaginary part of the index of refraction of a material can be expressed in terms of it's optical conductivity $$\sigma$$ as $$\kappa=\frac{i}{2n\omega} \sigma$$; See: http://en.wikipedia.org/wiki/Refractive_index "relation to dielectric constant" and http://en.wikipedia.org/wiki/Dielectric_constant "Lossy medium".
I don't think that the explanations of born2bwire really contradict an explanation of the damping of an electromagnetic wave to be due to the limited conductivity of the electrons moving at that frequency. After all, even the DC conductivity of a metal is essentially a quantum feature of the degenerate Fermi gas.