Here is an analogy that I like to use: (even though it is not really a correct physical explanation)

Imagine that you are out riding your segway over some strange surfaces, that each have a number $n_i$ that controls the speed that a segway wheel travels over it according to the formula $v_i=v_0/n_i$. Now imagine that you cross a straight boundary between two surfaces at an angle. Because of the angle, one wheel will cross the boundary before the other. If $n_i$ is higher for the entered surface this wheel will go slower than the other until it too crosses the boundary, which will cause the segway to turn towards the normal of the boundary. Similarly, if $n_i$ is lower for the entered surface, the first wheel to enter will go faster, and the segway will turn from the normal.

If you do the calculations for the segway you will get the the same results as for the wavefront explanation (basically Snell's law), but I really like how this analogy works with your intuition.

Lorentz came with a nice model for light matter interaction that describes dispersion quite effectively. If we assume that an electron oscillates around some equilibrium position and is driven by an external electric field $\mathbf{E}$ (i.e., light), its movement can be described by the equation
$$
m\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^2}+m\gamma\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}+k\mathbf{x} = e\mathbf{E}.
$$
The first and third terms on the LHS describe a classical harmonic oscillator, the second term adds damping, and the RHS gives the driving force.

If we assume that the incoming light is monochromatic, $\mathbf{E} = \mathbf{E}_0e^{-i\omega t}$ and we assume a similar response $\xi$, we get
$$
\xi = \frac{e}{m}\mathbf{E}_0\frac{e^{-i\omega t}}{\Omega^2-\omega^2-i\gamma\omega},
$$
where $\Omega^2 = k/m$.
Now we can play with this a bit, using the fact that for dielectric polarization we have $\mathbf{P} = \epsilon_0\chi\mathbf{E} = Ne\xi$ and for index of refraction we have $n^2 = 1+\chi$ to find out that
$$
n^2 = 1+\frac{Ne^2}{\epsilon_0 m}\frac{\Omega^2-\omega^2+i\gamma\omega}{(\Omega^2-\omega^2)^2+\gamma^2\omega^2}.
$$
Clearly, the refractive index is frequency dependent. Moreover, this dependence comes from the friction in the electron movement; if we assumed that there is no damping of the electron movement, $\gamma = 0$, there would be no frequency dependence.

There is another possible approach to this, using impulse method, that assumes that the dielectric polarization is given by convolution
$$
\mathbf{P}(t) = \epsilon_0\int_{-\infty}^t\chi(t-t')\mathbf{E}(t')\mathrm{d}t'.
$$
Using Fourier transform, we have $\mathbf{P}(\omega) = \epsilon_0\chi(\omega)\mathbf{E}(\omega)$. If the susceptibility $\chi$ is given by a Dirac-$\delta$-function, its Fourier transform is constant and does not depend on frequency. In reality, however, the medium has a finite response time and the susceptibility has a finite width. Therefore, its Fourier transform is not a constant but depends on frequency.

## Best Answer

In fact your statement is not quite right: air

doesdisperse light - its refractive index does depend on frequency, albeit very weakly. If you replace "air" with "vacuum" in your question, then the statement you are asking about is a correct one.Dispersion arises from the interaction of light (an electromagnetic field) with the electrical charge present in the atoms of the dispersing materials.

In air, there are far, far fewer atoms per unit volume than do materials we think of as dispersive. A vacuum, by definition, has no electric charge at all for the light to interact with. Therefore, its refractive does not depend on frequency. This last statement is true because (1) light is mediated by a massless field and (2) special relativity requires that all such fields propagate at exactly $c$.