# Obtaining both relative permittivity and permeability from refractive index

The complex refractive index $$\tilde{n}$$ is related to the relative electric permittivity and the magnetic permeability with the relation

$$\begin{equation} \tilde{n} = n + \mathrm{i} k= \sqrt{\varepsilon_r \mu_r} \end{equation}$$

In a previous great answer available at the link Contradiction on the behavior of refractive index, the user @ahemmetter explains the relation between complex refractive index $$\tilde{n}(\omega)$$ and the electric permittivity $$\varepsilon$$ of a material under the assumption that is non-magnetic ($$\mu_r = 1$$).

I quote here the relevant part of the answer.

Permittivity and permeability are not just constants, but instead are complex functions that depend on a number of other quantities, including the wavelength of the light. In fact, the refractive index $$n$$ is only half the story - there is also a related quantity, the extinction coefficient $$k$$, that describes the absorption in a medium. To put it more accurately, the refractive index $$\tilde n$$ is a complex function that depends among other things on the wavelength.

$$\tilde n (\omega) = n(\omega) + i k(\omega)$$

As stated in the question, the refractive index is related to the permittivity and permeability. In fact, they contain the same information about the material, and which one is chosen largely depends on convention and convenience. Also the permittivity ("dielectric function") and permeability have real and imaginary parts. For a non-magnetic material ($$\mu_r = 1$$, valid for most common materials), dielectric function and refractive index are related as follows: $$\varepsilon' + i \varepsilon'' = (n + ik)^2$$

With the individual components being: $$\varepsilon' = n^2 - k^2$$ $$\varepsilon'' = 2nk$$

I would like to know if there is a way to compute both the permittivity and the permeability from the complex refractive index when we cannot assume that $$\mu_r = 1$$.

Given $$\tilde \mu_r = \mu' + \mathfrak j \mu''$$ and $$\tilde \epsilon_r = \epsilon' + \mathfrak j \epsilon''$$ you can calculate $$\tilde n = \sqrt {\tilde \mu_r \tilde \epsilon_r } = n(\omega)+\mathfrak j k(\omega)$$, you just have to make sure of selecting the correct branch of the $$sqrt$$ function. Below 100GHz you can even measure the real and imag components separately with reasonable easiness by using straight line electric dipole and circular loop magnetic dipole antennas to measure the $$E$$ and $$H$$ fields, resp., but that method is probably not very practical at much higher frequencies.
• Your answer is not completely clear to me. I was asking if there was a way to compute $\mu_r$ and $\epsilon_r$ knowing $\tilde{n}$. Jun 25, 2021 at 20:36
• If that is your question then you have only two (2) "knowns" $\Re { \tilde n }$ and $\Im {\tilde n}$ but four (4) "unknowns"; you will need two more relationships to get $\epsilon', \epsilon'', \mu',\mu''$ but those are not given. Jun 25, 2021 at 21:04
• Yeah, the Kramers-Kronig (ie., causality) relationship would be true for both complex $\epsilon$ and $\mu$ separately so then you have only "two" unknowns but the two "knowns", say, $\epsilon'$ and $\mu'$ you would have to know for every frequency and then use the Hilbert transform on them to get $\epsilon''$ and $\mu''$ but this has nothing to do with $\tilde n$. Jun 25, 2021 at 21:54