# Confusion over real and complex versions of the same quantities

1. Jul 22, 2010

### Squeegee

Some authors speak of the index of refraction'' and the relative permittivity'' (or dielectric constant'') as real numbers, as if they do not warrant any discussion of complex numbers. Other authors speak of the complex index of refraction'' and the complex permittivity.'' But they all seem to agree that the index of refraction squared equals the permittivity.

Say the index of refraction is $n'+in''$ and the relative permittivity is $\epsilon'+i\epsilon''$. According to the authors who are only talking about real numbers, I have $n'^2=\epsilon'$. According to those who talk about complex numbers, I have $(n'+in'')^2=\epsilon'+i\epsilon''$. Together, those two equations seem extremely weird, so I can only believe that I am mistaken about something. I think that at least one of the statements I have made so far is incorrect.

Can someone please clear up my confusion? Thank you!

2. Jul 23, 2010

### K^2

You have to go back to wave equation to work it all out. For a linear medium, the wave equation becomes this.

$$\nabla^2 E = \mu \epsilon \frac{\partial^2}{\partial t^2}E$$

This works even if μ and ε are complex. So does this general solution.

$$E = A e^{i(k \cdot x - \omega t)}$$

Where A is arbitrary complex amplitude. Substitution into the above yields the next line.

$$k^2 = \mu \epsilon \omega^2$$

Square of group velocity of the wave is then easy to deduce.

$$v_g^2 = \left(\frac{\partial \omega}{\partial k}\right)^2 = \frac{1}{\mu \epsilon}$$

And we know the definition of index of refraction.

$$n^2 = \frac{c^2}{v_g^2} = \mu \epsilon c^2$$

Why work with squares all the way to this point? Well, otherwise, in an earlier step, I would have had to take square root of 1/(με) to find group velocity, and that may be problematic if the quantity is complex. But keeping everything nice and squared, we can still work with it.

If you need to then work out real and complex components, there isn't really too much trouble.

$$Re(n^2) + i Im(n^2) = Re(\mu \epsilon c^2) + i Im(\mu \epsilon c^2)$$

Obviously that gives you two equations.

$$Re(n)^2 - Im(n)^2 = c^2 ( Re(\mu)Re(\epsilon) - Im(\mu)Im(\epsilon) )$$
$$2 Re(n)Im(n) = c^2 ( Re(\mu)Im(\epsilon) + Im(\mu)Re(\epsilon) )$$

If you have complete knowledge of μ and ε, you can easily solve this to find both Re(n) and Im(n).

3. Jul 24, 2010

### Antiphon

K^2 is right of course.

What happens is that in optics most materials are nonmagnetic and have a real mu equal to one. In that case the simple root relation works. In general though epsilon and mu are both complex.

The bottom line is that it takes two complex numbers to characterize an isotropic material at a given frequency. Using the index of refraction is insufficient in the general case but is a useful simplification for most optical materials.