# How can the refractive index be below 1 in a dielectric?

Upon checking the optical properties of different dielectrics, I found the interesting case of $$Al_2O_3$$. It seems to be reported with a refractive index below 1 in the infrared range of $$10 - 12~\mu m$$, for example by Kischkat, Boidin and Querry. Also the absolute value of the complex index of refraction $$\bar n = \sqrt{n^2 + \kappa^2}$$ is below 1 at certain wavelengths.

(red: $$n$$, blue: $$\kappa$$)

Basic physics tells me that the phase velocity of a EM wave in a dielectric is related to the real part of the refractive index by $$v_{ph} = \frac{c_0}{n}$$

Now obviously, the phase velocity can be above $$c_0$$, but mostly this happens only in 'strange' cases: in plasmas, active media, metamaterials or magnetic substances. $$Al_2O_3$$ seems to be neither of those.

Another clue is that this happens around the region where $$n$$ and $$\kappa$$ cross - that is also the point where $$\varepsilon = 0$$ in a Drude metal, and hence the wavelength of the bulk plasmon. Could this therefore just be an effect of the $$Al_2O_3$$ becoming metallic, as mentioned by @boyfarrell?

What's going on here?

• Lots of free charge charges i.e. material behaving as a metal will cause this – boyfarrell Oct 23 '18 at 11:01
• @boyfarrell Ah, right, I absolutely didn't think of this. So Al2O3 just behaves in this range more like a metal. Feel free to add it as an answer. – ahemmetter Oct 23 '18 at 11:04
• Yeah could correspond to a spectral region near some sort of resonance. Not really an expert on metals. Would be interested in a good answer too. – boyfarrell Oct 23 '18 at 11:06

In these circumstances the phase and group velocities are different and indeed the group velocity can be greater than $$c$$. This isn't a problem since the group velocity is no longer the speed at which information travels.
• It seems the maximum of $\kappa$ does not coincide with the region where $n < 1$, no I'm not sure in how far absorption is applicable here. Clearly $n$ decreases with increasing wavelength, but the question is rather about the value of it, not its derivative. – ahemmetter Oct 23 '18 at 11:12