Why doesn't a dielectric material having relative permittivity $\epsilon_{r}<1$ violate special relativity? If the material has its relative permittivity less than one, $$\epsilon_r < 1,$$ and since the wave speed of an EM-wave in that material is given by $$v = \dfrac{1}{\sqrt{\epsilon \mu_0}},$$ wouldn't this result in
$$\dfrac{v^2}{c^2} = \dfrac{\epsilon_0 \mu_0}{\epsilon \mu_0} = \dfrac{1}{\epsilon_r} > 1,$$ and therefore a Lorentz factor
$$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}}$$
that becomes complex and unphysical? Where did I do wrong?
 A: The dielectric constant $\epsilon_{r}=n^{2}$ (nor a nonmagnetic material) is always a function of frequency, $\epsilon_{r}(\omega)$. While it is possible to have $\epsilon_{r}(\omega)<1$ in some frequency ranges, for any material $\epsilon_{r}(\omega)\rightarrow1$ as $\omega\rightarrow\infty$, and this is responsible for preserving relativity.
First, I would like to explain why $\epsilon_{r}(\omega)\rightarrow1$ always as $\omega\rightarrow\infty$. The dielectric constant is related to the material polarization created an applied electric field. The DC dielectric constant $\epsilon_{r}(0)$ is determined by the polarization of the constituent atoms when they are exposed to a constant external $\vec{E}$. The frequency-dependent generalization $\epsilon_{r}(\omega)$ is determined by the oscillating polarization that occurs when the applied electric field is oscillating harmonically with angular frequency $\omega$ (i.e. period $2\pi/\omega$).  If the oscillation frequency $\omega$ is too high, then the electrons in the material do not have enough time to move and form small dipoles before the direction of $\vec{E}(t)=\hat{\varepsilon}E_{0}\cos\omega t$ has reversed itself. So for a large enough $\omega$, any material becomes essentially unpolarizable, meaning $\epsilon_{r}(\omega\rightarrow\infty)\rightarrow1$.
Now, we also need to understand why having $\epsilon_{r}(\omega\rightarrow\infty)\rightarrow1$ ensures that there are no violations of special relativity. The key restriction imposed by the special theory relativity is that no signal can travel faster than the speed of light $c$. To send an electromagnetic signal, you need to transmit a wave packet with a sharply defined front edge—something like a square wave, for example.  (There is no faster-than-light communication if you send a single monochromatic plane wave, because such a wave has no beginning or end, in space or time.  In other words, the plane wave started being sent infinitely far in the past, so there is no possibility of a signal arriving too early.) The speed at which the sharp edge of the electromagnetic signal moves is the "signal speed," (sometimes denoted $c_{s}$) and relativity requires that $c_{s}\leq c$.  To form a sharp leading edge to the signal, the wave packet needs to contain waves of very short wavelength (equivalent to very high frequency); to form a perfect square wave requires that the packet contain components all the way up to $\omega\rightarrow\infty$.  What's more, the rate at which the sharp disturbance carrying the signal moves is determined by the phase speed of those highest-frequency components.  The maximal signal speed is thus actually
$$c_{s}\leq\left.\frac{\omega}{k}\right|_{\omega\rightarrow\infty}=\left.\frac{c}{\sqrt{\epsilon_{r}(\omega)}}\right|_{\omega\rightarrow\infty}=c,$$
and no signal can travel faster than light.
A: In such cases the phase velocity $\omega/k$ is larger than c but the group velocity $d\omega/dk$ is not.
