It is well known that a prism can "split light" by separating different frequencies of light:
Many sources state that the reason this happens is that the index of refraction is different for different frequencies. This is known as dispersion.
My question is about why dispersion exists. Is frequency dependence for refraction a property fundamental to all waves? Is the effect the result of some sort of non-linearity in response by the refracting material to electromagnetic fields? Are there (theoretically) any materials that have an essentially constant, non-unity index of refraction (at least for the visible spectrum)?
Best Answer
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.