4
$\begingroup$

In the optics world, when light (assumed as a plane electromagnetic wave having complex wavenumber $k_0$), $$ E = E_0 \cdot e^{i(kx-\omega t)} $$ travels from one medium (e.g. vacuum) into another (denoted by index $M$, e.g. some glass), its wavelength changes from $\lambda_0$ to $\lambda_M$ and it is damped. We describe that using the refractive index $\tilde{n}$ of the medium $M$: $$ \tilde{n} = n + i \kappa $$ The complex wavenumber inside the medium is then $k_M = k_0 \cdot \tilde{n}$. This gives $$ E=E_0 \cdot e^{-\frac{2 \pi}{\lambda_0}\cdot \kappa} \cdot e^{i(\frac{2\pi}{\lambda_M} x - \omega t)} $$ where the first e-function decribes the damping and we see the changed wavelength $\lambda_M$ inside the medium in the second exponent, so in total a damped wave of different wavelength.

The complex refractive index basically consists of the (complex) relative permittivity $\epsilon_r$: $$ \tilde{n} = \sqrt{\epsilon_r \mu_r} \approx \sqrt{\epsilon_r} $$ (for $\mu_r \approx 1$, which I think is true for most dielectric materials at frequencies in the visible spectral range).

Now, in the electromagnetic world, when an E-field enters a medium of permittivity $\epsilon_r$, then the displacement field $D$ is $$ D = \epsilon_0 \epsilon_r E $$ and as discussed in Wikipedia (https://en.wikipedia.org/wiki/Permittivity), the permittivity results in damping and a phase difference between $D$ and $E$, which I guess is the cause for the wavelength change described by the real part of the refractive index in the optics world (please correct me if that's wrong)?

In this equation, the permittivity is just a factor for $E$, while in the optics world, it is in the exponent as part of the refractive index. But both views must be consistent... So how does the permittivity get into the exponent?

$\endgroup$
0
6
$\begingroup$

As with so many aspects of electrodynamics, we can elucidate things by returning to Maxwell's equations.

In the absence of free charges, the fields in a medium obey \begin{align} \vec{\nabla}\cdot\vec{D} &= 0 \tag{1}\\ \vec{\nabla}\cdot\vec{B} &= 0 \tag{2} \\ \vec{\nabla}\times\vec{E} + \frac{\partial \vec{B}}{\partial t} &= 0 \tag{3}\\ \vec{\nabla} \times \vec{H} - \frac{\partial \vec{D}}{\partial t} &= 0 \tag{4} \end{align} For an isotropic, linear, non-magnetic, non-dispersive medium, Equations (1) and (4) are equivalent to (respectively) \begin{align} \vec{\nabla}\cdot\vec{E} &= 0 \tag{5}\\ \vec{\nabla} \times \vec{B} - \mu_0\epsilon_0 \epsilon_r \frac{\partial \vec{E}}{\partial t} &= 0. \tag{6} \end{align} Taking the time derivative of this last equation and combining it with (3) and (5) yields $$ \nabla^2 \vec{E} - \frac{\epsilon_r}{c^2} \frac{\partial^2 \vec{E}}{\partial t^2}= 0, $$ where we have identified $\mu_0 \epsilon_0 = 1/c^2$.

We can see that this is a wave equation with a speed of propagation of $c/\sqrt{\epsilon_r}$. This means that the speed at which waves propagate will be affected by the permittivity of the medium, and so you end up with an $\epsilon_r$ "in the exponent" when you look at a plane-wave solution.

On a physical level, what is happening here (roughly) is that the magnetic field $\vec{B}$ is responding both to the changing electric field and to the polarization currents ($\vec{J}_d \equiv \partial \vec{P}/\partial t$, which equals $\epsilon_0 \chi_e \partial \vec{E}/\partial t$ for a linear medium). From Eq. (6), we can see that the spatial scale of variation of $\vec{B}$ will decrease for a given temporal rate of change of $\vec{E}$. (Assuming $\chi_e > 0$, which implies that $\epsilon_r > 1$.) In other words, $k$ increases for a fixed $\omega$, leading to a phase speed $\omega/k$ that is lower.

$\endgroup$
2
  • $\begingroup$ many thanks for this nice answer! I had to think about the last paragraph, but think I got it now: In Eq. 6, dE/dt stays constant when the photon enters another optical medium because the frequency stays constant. However, if $\epsilon_r$ increase, the whole term with dE/dt will increase, and thus rot(B) must increase. As rot B contains spatial derivatives, the change of B with distance increases, i.e. B rotates faster, but the whirl is smaller. And thus the wavelength is smaller (rotating B in turn produces an E field, which is how light propagates...). Is that correct? $\endgroup$ Jul 13 '21 at 7:46
  • $\begingroup$ @CharlesTucker3: Yes, that's what I was trying to get at in the last paragraph in a hand-wavy sort of way. $\endgroup$ Jul 13 '21 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.