The velocity 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 velocity 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.


Although I think that the actual explanation is possible by using quantum electrodynamics, I just want to provide a basic argument to show that it must slow down. Let $\vec E_0$ and $\vec B_0$ denote the electric and magnetic fields of the electromagnetic wave in vacuum. Now suppose this enters a medium(which is not conducting free charges). The total current density is now given by :

$ J_t = \frac{\partial \vec P}{\partial t} + \nabla \times \vec M + J_d$

where $J_d$ = displacement current(this obviously exists, because of the alternating field of the EM wave)

$\vec P$ = polarisation vector(which changes with time as the wave passes through)

$\vec M$ = magnetisation vector(which also changes as the atomic dipoles oscillate as the wave passes through)

It can be seen that this $\vec J_t$ will produce a magnetic field(maxwell's 4th eqn) which in turn will affect the electric field and so on.

The key observation here, is to note that the current produced by $\vec J_t$ will oppose its source(the EM wave) by Lenz's law. In other words, it will decrease the amplitude of the electric and magnetic fields(of the wave). Thus the electric and magnetic parts of the wave in the medium will satisfy:

$|\vec E_m| < |\vec E_0|$ and $|\vec B_m| < |\vec B_0|$

So the poynting vector in the medium $\vec S_m = \frac{1}{\mu_o} \vec E_m \times \vec B_m$

This will be less that than the magnitude of the Poynting vector in vacuum.

Since the Poyting vector is a measure of rate of transmission of energy through a medium, it is a measure of the speed of light itself in the medium.

PS: I cannot provide detailed mathematics due to my limited knowledge, but I happened to find some reason in this explanation. Sorry, if you were looking for rigorous proofs(wait for the other answers). Please point our any mistakes in my explanation if possible.


Your premise is incorrect. The speed of light in a medium is not necessarily less than in vacuum. For a theoretical prediction of light propagation in a medium you need a model of the medium, which will vary from case to case. One model that is often used, e.g. for dilute gases, is the Drude model, which models the electrons attached to molecules as driven-damped harmonic oscillators, driven by the incident light wave, each with a natural oscillation frequency, with damping attributed vaguely to radiation, etc. Once you have the model of wave-matter interaction you can derive a dispersion relation, which is the relationship between frequency $f$ and wavelength $\lambda$ of a wave. Then the phase velocity is $v=f\lambda$. In some cases it can come out greater than $c$. You should be able to find more information on it in any junior level E&M textbook, e.g. Griffiths ch. 9.


Refractive index of thin metal films or, in general, in plasma, is less than 1, that is, velocity of light in plasma and metals is greater than in vacuum, but only phase velocity, of course.

  • $\begingroup$ I don't know why someone down-voted this. You are correct! (I wish they didn't allow anonymous down-votes - it encourages cowardice.) $\endgroup$ – pwf Dec 8 '16 at 18:54

protected by Qmechanic Dec 4 '16 at 0:27

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