Cauchy's equation is an empirical relation that defines the refractive index of a material in terms of wavelength: $$ n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + ...$$

While I understand that it is at best empirical and is limited only to a certain range of wavelengths, I haven't seen any discussion on the justification for the inclusion of only the even power terms, namely $ \frac{1}{\lambda^{2}} $, $ \frac{1}{\lambda^{4}} $ , etc. Is there a physical / historical reason for this, or is it just for the sake of fitting?


1 Answer 1


Historically, Cauchy derived his equation assuming that light propagated in an elastic aether. The aether theory of light may be wrong, but Cauchy's derivation was pretty much equivalent to simple models of light propagation in dielectrics. If the resonant absorption frequency of the electrons in the media is $\omega_0$, then the index of refraction $n$ in such a model is given by (e.g. see Feynman Eq. 32.27)

$$n^2=1+\frac{\omega_p^2}{\omega_0^2-\omega^2-i\gamma\omega} $$

where $\omega_p=\sqrt{n_e e^2/m \epsilon_0}$ is the plasma frequency. As noted in this answer to "Why do we neglect higher order terms in Cauchy's Equation?", in the limit of negligible absorption (i.e. $\gamma=0$), this reduces to


where $p=\lambda_0/\lambda_p$ and $x=\omega/\omega_0$. When $\omega<<\omega_0$, this can be expanded in a Taylor series in $x=\omega/\omega_0\approx\lambda_0/\lambda$, which as an even function of $x$ only has even terms in $1/\lambda$:

$$n(\lambda)=\sqrt{p^2 + 1} + \frac{p^2\lambda_0^2}{2 \sqrt{p^2 + 1}} \frac{1}{\lambda^2} + \frac{p^2 (3 p^2 + 4) \lambda_0^4}{8 (p^2 + 1)^{3/2}}\frac{1}{\lambda^4} + O\left(\frac{1}{\lambda^6}\right)$$

This model turns out to be a good enough approximation that empirical fits to $A,B,C$ in $$ n(\lambda) = A + \frac{B}{\lambda^2} + \frac{C}{\lambda^4} + ...$$ work reasonably well to parameterize $n(\lambda)$ for some common materials for some wavelengths. In particular, it doesn't work badly in the visible for some gases and glasses.

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    $\begingroup$ It's also worth noting that the reason that the Taylor series turns out to only have even terms is because $n$ is an even function of $x$ (assuming negligible absorption.) $\endgroup$ Dec 12, 2022 at 16:05

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