[Physics] How to interpret the complex index of refraction

complex numberselectromagnetismopticsrefractionwaves

The index of refraction which represents how much light gets refracted when entering a medium is defined as

$$n = \frac{c}{v}$$

I have seen it stated in several places, such as here, that we can replace the index of refraction with a complex index of refraction

$$\tilde{n} = n + ik$$

I don't understand how we can do this and what it actually represents. Either the index of refraction is complex valued or not, I don't see how you can just arbitrarily replace a real valued variable with a complex valued variable. So how is this valid?

Also, in the link above it states that $k$ represents the absorption loss. Why is a complex number needed to represent this?

Best Answer

Electromagnetic theory combined with optics gives an electromagnetic interpretation for the refractive index: $$ v = \frac{1}{\sqrt{\epsilon_0\mu_0}\sqrt{\epsilon_r\mu_r}} = \frac{c}{\sqrt{\epsilon_r\mu_r}} = \frac{c}{n} \quad\Longrightarrow\quad n^2 = \epsilon_r\mu_r $$

If we ignore magnetic properties, ie $\mu_r\approx 1$, then we have: $n^2 = \epsilon_r$. If you solve maxwell equations for an waving electromagnetic field in vacuum, you will get this result for plane waves: $\mathbf k\times\mathbf B = -\omega/c^2\epsilon_r\mathbf E$. The form of the $\mathbf{E}$ plane wave is: $$\mathbf E = \mathbf E_0 e^{i(\omega t - \mathbf k\cdot\mathbf r)}$$

But, then, when you plug an electromagnetic plane wave into the Maxwell Equations for a non-magnetic material with conductivity $\sigma$ and relative complex permittivity $ \hat{\epsilon}_r = \epsilon'+i\epsilon''$, the current is $\mathbf J_{\text{tot}} = J_{\text{displacement}} + J_{\text{conduction}} = i\omega\epsilon\mathbf E_0 + \sigma\mathbf E_0 = i\omega\epsilon_0(\epsilon' - \frac{i\sigma}{\epsilon_0 \omega})$ and the Ampere Equation reduces to: $$ -i\mathbf k\times\mathbf B_0 = \frac{i\omega}{c^2}\left(\epsilon' - \frac{i\sigma}{\epsilon_0\omega}\right)\mathbf E_0 $$

Then, in comparison with previous result, we defined a complex dielectric constant: $$ \hat\epsilon_r = \epsilon' + i\epsilon'' = \epsilon' - \frac{i\sigma}{\epsilon_0\omega} $$

And then, we define a complex refraction index: $\hat n = n +i\kappa = \sqrt{\hat\epsilon_r}$, $\frac{kc}{\omega} = \hat{n}$ with the branch chosen so that the real exponential is decaying. Assuming a plane wave propagating in the $+x$ direction, you now get a damped wave, where the complex part of the refraction index $\epsilon''$ makes the wave decrease its electric field exponentially: $$ \mathbf E(x, t) = \mathbf E_0 e^{i \left( \omega t - kx \right)} = \mathbf E_0 e^{-\frac{\omega}{c}\kappa x} e^{i\left(\omega t - \frac{\omega n}{c}x\right)} $$

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