15
$\begingroup$

The speed of light in a material is defined as $c = \frac{1}{\sqrt{\epsilon \mu}}$. There are metamaterials with negative permittivity $\epsilon < 0$ and permeability $\mu < 0$ at the same time. This leads to a negative refractive index of these materials.

But do (meta-) materials exist with only negative $\epsilon < 0$ and positive $\mu > 0$ or vice versa? This would lead to a complex speed of light inside such materials.

What would be the consequences of a complex speed of light? Could particles reach unlimited speed inside these materials? Would there still be Cherenkov radiation?

$\endgroup$
4
  • 14
    $\begingroup$ I don't know about metamaterials, but it seems to me that if $\epsilon\mu<0$, it would mean that the phase velocity was purely imaginary. That would mean that waves in the material would die away exponentially rather than oscillating. The solutions to the wave equation are of the form $e^{i(kx-\omega t)}$ with $\omega=ck$. If $c$ is imaginary, then $k=i\kappa$ is imaginary for real $\omega$, and the solution looks like $e^{\pm\kappa x-i\omega t}$. The physically useful solution in these cases is the decaying one. $\endgroup$
    – Ted Bunn
    Commented May 21, 2011 at 21:25
  • $\begingroup$ Keep in mind that the maximum speed of a particle, even in a material, is 299792458 m/s, regardless of how light behaves in the material. So particles wouldn't be able to reach unlimited speed even if the speed of light did become complex. $\endgroup$
    – David Z
    Commented May 22, 2011 at 2:51
  • $\begingroup$ @Ted Bunn Within/close to absoprtion bands/lines n can become < 1, but difficult to observe due to absorption. "Anomalous Dispersion" In the far IR or Microwave band paramagnetic substances have "bands", eg ruby (+ magnetic field) is used in circulators and similar devices. $\endgroup$
    – Georg
    Commented May 22, 2011 at 11:06
  • 1
    $\begingroup$ @Ted Bunn If waves in the material die away exponentially, would that imply that also Cherenkov radiation is damped exponentially? So I guess one couldn't measure the speed of a fast charged particle in such a material based on the emitted Cherenkov radiation? $\endgroup$
    – asmaier
    Commented May 29, 2011 at 10:26

3 Answers 3

8
$\begingroup$

Complex quantities always denote loss. So if the velocity is imaginary, it is impossible for a wave to travel from one point to another. If you look at the Drude model, for some certain frequency the signal will pass so it behaves like a dielectric at that time, but for frequencies lower than the Plasma frequency it will behave like a metal where no transmission is possible and at that time permittivity is less than zero, so at that time the velocity of the wave is imaginary.

So, in my opinion, imaginary velocity means no transmission.

$\endgroup$
5
$\begingroup$

Technically, all of these materials will have (effective) complex dielectric and/or magnetic properties. Thus, you'll be dealing with complex wave speeds.

Complex speed means lossy transmission. The amplitude of the wave will decay with a decay constant that is inversely proportional to the imaginary part of the square root of the speed.

Note that if this term is small, or the material is thin enough, then you can have transmission.

$\endgroup$
0
$\begingroup$

Since there's only one square root, one would end up with an imaginary velocity rather than the more general case of a complex velocity.

In physics, one occasionally deals with complex energies. I don't know what imaginary velocities will mean and there's not a lot of references in the literature. One paper is "Velocities" in Quantum Mechanics

$\endgroup$
2
  • 1
    $\begingroup$ Dear @Carl Brannen: A comment to the answer (v1): The case $\epsilon\mu<0$ is not excluded, cf. the same wikipedia article en.wikipedia.org/wiki/… $\endgroup$
    – Qmechanic
    Commented Jun 8, 2011 at 11:47
  • $\begingroup$ Yes, I'll modify my answer. (Embarrassingly, I took plasma classes in college.) $\endgroup$ Commented Jun 8, 2011 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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