# Why phase velocity of light is slower in a dielectric than vacuum?

Why does light speed suffer a reduction when it passes through a homogeneous dielectric medium? I know my math says so, i.e.- a highly polarizable ($\chi$) medium is associated with high $\epsilon$ (=$1+\chi$), and according to Helmholtz equation $\nabla^{2}E+\omega^2\mu\epsilon E=0$, the wavelength $\left(\lambda=\dfrac{2\pi}{\omega\sqrt{\mu\epsilon}}\right)$ should shrink down with larger $\epsilon$, leading to slower phase velocity. But is there any intuitive explanation behind that? Why should a highly polarizable medium (with no scattering) lead to slow speed (or smaller wavelength)?

Let me put my question in some other way. Lets assume, we have a group of identical charged harmonic oscillator arranged along a straight line in air, with a finite spacing between them. No damping is there. This should essentially mimic a $1 D$ lossless dielectric medium. Once the rightmost oscillator is set in motion, it will begin radiation. It radiates in all direction, but for the time being, let's just care about the forward radiation only. This propagating radiation will eventually cause all other oscillator to vibrate and re-radiate. If we can explain how the phase difference between any two neighboring oscillators depends on the polarization amplitude $(P=q\times d)$, $d$ being the distance from equilibrium and $q$ is charge), it should suffice to answer my original question, i.e.- why light speed is slower in dielectric.

## 1 Answer

A key to understanding this is realizing that it's not always true. In fact, at x-ray frequencies, refractive indices are typically less than 1, so that the phase velocity is faster than the vacuum speed of light. The key difference is that x-ray frequencies are well above the natural frequencies of most of the electronic excitations that are involved in the polarizability, while visible light frequencies are generally well below many of those electronic excitation frequencies.

And the reason that that matters is because the sign of the phase shift changes. Recall the theory of the damped driven harmonic oscillator: The response of the oscillator is out of phase with the driving frequency, so the light that the oscillator re-emits is also out of phase with the driving frequency. Superposing the original light with the phase-shifted re-emitted light results in a progressive phase shift that's linear with the propagation distance--which is exactly the same thing as a change in the phase velocity. The phase shift depends on the frequency, and in particular it reverse sign as the driving frequency passes through the natural frequency of the oscillator, which explains the refractive index being greater than 1 (typically) at visible frequencies and less than 1 at x-ray frequencies.

Weird things happen in between, for example at the plasma frequency where the real part of the dielectric function goes to zero. And if you structure the material on the mesoscopic scale then even weirder things happen (see metamaterials).

• Thanks a lot elifino. However, I appreciate your statement as an annexed information, instead of a direct answer to my question. Its true that, for different frequencies, $\epsilon$ can be smaller than 1 (leading phase velocity greater than lightspeed), equal to zero (longitudinal surface plasmon wave), or even negative. It all depends on how atoms are polarized by electric field. But how will you intuitively explain the phenomenon of reduced phase-velocity of visible light in a typical dielectric, say glass? – Joy Nov 25 '15 at 6:26
• Also, phase-shift of radiated field due to damping/scattering is accounted as loss in dielectric, and has nothing to do with the imaginary part of propagation constant ($\gamma=\alpha +i\beta$). So it does not explain why we have reduced light speed in a lossless medium. – Joy Nov 25 '15 at 6:38