[Physics] Why is the phase velocity used in the definition of the refractive index

Question

I'm aware of the so-called group index but why is the phase velocity used in the standard definition of the index of refraction? What advantage does this offer?

Answer 1

Well at the most fundamental level, the index of refraction of a material is defined as

$ n = \sqrt {\epsilon \mu} $

where $\epsilon $ is the electric permittivity and $\mu$ is the magnetic permeability of the material.

This arises from the solution of Maxwell's equations in a medium.

Also arising from the wave equation, which can be derived from Maxwell's equations, is that the index of refraction is the speed of light in vacuum, $c$ divided by the speed of light in the material $c_m$.

$ n ={ c \over c_m} $

Worth noting is that since $c_m$ is always less than $c$ , the index of refraction is always greater than 1.

Now the phase velocity for an electromagnetic wave of angular frequency $\omega$ is given by

$v_p = {\omega \over k}$ where $k= {2\pi \over\lambda_m}$ is the magnitude of the wave vector and $\lambda_m$ is the wavelength in the medium.

So after a little algebra, we find that the wave vector and index of refraction are related by

$k ={ n\omega \over c}$

Where does all this come into play in refraction and Snell's Law? Well, it is the wave vector that comes into play in satisfying the boundary conditions on the electric (and magnetic) fields at the interface between two media.

To see this, let's look at the simple case of a plane wave of monochromatic light incident on the interface between two media with indices of refraction $n_1$ and $n_2$:

In this simple case, considering the boundary conditions on the electric field at the interface is sufficient to derive Snell's Law.

The boundary condition is given by equation (1) in the diagram, namely that the incident and reflected electric fields minus the transmitted electric field must be zero, or equivalently that the total electric field at the interface must be continuous.

Since we defined y=0 as the plane of the interface, this boundary condition must hold for any value of x. This leads after a little algebra to equation (2) in the diagram. This equation depends only on the angles of incidence $\phi_{inc}$ and refraction $\phi_{tr}$, the wave vector magnitudes in both media $k_1$ and $k_2$, and the transmission and reflection coefficients $T$ and $R$ ( the fraction of energy that is transmitted into the new medium and reflected into the old medium respectively). By symmetry, $\phi_{inc} = \phi_{refl}$.

Because the left side of Equation 2 is independent of angle, so must the right side be. This leads to the term in the exponential being zero, which leads directly to Snell's Law, using the relation between wave vector and index of refraction shown above.

The group velocity never comes into play in the boundary conditions of refraction. It does come into play in the propagation of energy in the media (as opposed to the fields), but that's another question.