[Physics] How is refractive index related to the density of a medium (for example, air)


I have a question regarding refractive index dependency on the density of a dielectric, specifically air.


Let us start from Newton's second law form of driven harmonic oscillators $$-kx-bv+F_e=ma.$$

Knowing that $F_e=eE$ the equation may be rewritten as $$m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}+b\frac{\mathrm{d}{x}}{\mathrm{d}t}+kx=eE.$$

The relevant solution is given by $$x(t)=\frac{eE}{\sqrt{(-m\omega^2+k)^2+(2b\omega)^2}}\cos({\omega}t+\phi)$$

Assume a dielectric material. Let incoming light be monochromatic. In this case, I have shown that refractive index $n$ is dependant on the frequency $\omega$ via


where $\omega_0=\sqrt{\frac{k}{m}}$, $\beta\equiv\frac{b}{2m}$, and $n_c$ is the number of charge dipoles per unit volume.


The next step is to show that refractive index varies as density changes. Intuitively, the dipole concentration $n_c$ is related to the density of a material. Any further than that, I am clueless.

  1. How is refractive index related to the density of the medium, e.g., density of air? Please give a short, simple derivation if possible.

  2. When we have the equation from question (1), am I allowed to make ideal gas substitution in order to directly derive how the refractive index depends on air temperature? In other words, I would like to substitute density $\rho$ for $\frac{MP}{RT}$.


I have glanced over the following links:

Nevertheless, these seem not to be exactly what I am looking for.

Best Answer

You have to account for the local fields and how they change as the density changes. Look up the Clausius–Mossotti Equation and its derivation.

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