The speed of electromagnetic waves in a medium is smaller than its value in the vacuum: $$v=\frac{1}{\sqrt{\mu\epsilon}}=c/n<c$$ with the refractive index $n=\sqrt{\frac{\mu\epsilon}{\mu_0\epsilon_0}}\approx \sqrt{\epsilon_r}>1$ always. Why is this the case?
Naively and qualitatively, I think, when the wave falls on a medium, it is absorbed by the medium particles, which then oscillate and re-emit the radiation, and this might cause a delay in the propagation. However, I'm looking for a classical mathematical model (in terms of microscopic interaction between atoms and fields similar in spirit to the Lorentz theory of dispersion) of the propagation of electromagnetic wave in a medium that explains physically why does the speed decrease and enables one to derive the relation $v=c/n$.
EDIT: In this question, the OP talks about photon absorption-(re)emission theory, and qualitatively explains how it changes the "drift velocity". I want a quantitative version of this model/theory that enables me to define $v$, and show that $v<c$. The answer here is nice, but still qualitative.
Best Answer
This question has led to at least one totally incorrect statement, namely that the speed of light in a medium can exceed the speed of light in vacuum. Although the phase velocity, $\omega / k$, can be larger than $c$, this is not true for the group velocity, $d \omega / dk$. Information and energy are travelling at group speed.
Secondly, there is no absorption and remission. Instead there is scattering off electronic excitations, which have lower than light speed. This has the effect to lower the speed in the medium. For very short wavelengths a phase advance may result do that $n \lt 1$ is possible. However in such a case the group velocity is still lower than $c$.