# Complex refractive index and absorption

I have a quite plain and simple question: Why is the imaginary part of the refractive index negative? I read it is to allow a positive sign in case of absorption. But how/why? I don't see why it should be positive when absorbing? What does the sign change?

## 1 Answer

The refractive index of a material tells you how to get the wavevector $k$ from the angular frequency $\omega$, via the dispersion relation $$k=n\frac{\omega}{c},$$ which directly determines the relationship between the spatial and temporal dependence of a plane wave $$f(x,t)=e^{i(kx-\omega t)}.$$

If the refractive index is complex, then the wavevector will be complex, which means that what used to be an oscillatory imaginary exponential over position now contains, additionally, a decaying real exponential: \begin{align} f(x,t) &=\exp(i((k_\mathrm{re}+ik_\mathrm{im})x-\omega t)) \\&=e^{-k_\mathrm{im}x}\exp(i(k_\mathrm{re}x-\omega t)) . \end{align} Here the plane wave $f(x,t)=e^{i(kx-\omega t)}$ represents a wave that is travelling to the right, with power flowing from negative $x$ to positive $x$ (and some power getting absorbed), so we expect there to be less signal the more positive that $x$ gets, and this requires the exponential dependence $e^{-k_\mathrm{im}x}$ to be decaying, i.e. it requires $k_\mathrm{im}=\frac\omega c\mathrm{Im}(n)$ to be positive (or at least nonnegative).