9

Wikipedia says that "in general, the refractive index of a glass increases with its density." And the refraction index of water vapor is less than ice, and even less than liquid water. Is there any simple explanation to that?

Qmechanic
  • 201,751
arax
  • 1,168

2 Answers2

11

The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time.
The more this happens, the slower the effective average speed.
The denser the material, the more atoms there are in the way.

Martin Beckett
  • 30,766
  • Great explanation. – boyfarrell Mar 29 '14 at 06:04
  • 5
    The explanation is only apparent. In classical EM theory there is no time lag between interaction of external EM wave with charges and emission of secondary EM wave, both occur simultaneously. – Ján Lalinský Mar 29 '14 at 08:02
  • 4
    By what mechanism is the light absorbed? Not an atomic electron transition, as surely re-emission would change frequency and hence color. – xxyzzy Jul 21 '18 at 11:22
  • 1
    This explanation is wrong. See the Ewald–Oseen light extinction theorem for the correct explanation. – Mihai Danila Aug 27 '23 at 02:12
  • No, this answer is incorrect. In fact, the field of spectroscopy fundamentally relies on the principle that different materials interact narrowly with only specific wavelengths of light. Certain materials only interact with certain wavelengths, yet dense mediums like water slows light in a continuous, not discrete, fashion depending on wavelength. – Zamicol Dec 31 '23 at 12:47
5

This is quite a subtle issue. The charges in the medium produce secondary spherical expanding EM waves when hit by the primary wave (external forces). There is immense number of these secondary waves. At any point of space, each secondary wave has slightly different wave vector. In a medium dense enough, these secondary waves add to the primary wave in such a way that the resulting wave has behaviour that is well described by a single macroscopic wave of the same frequency and (usually) same direction but (for most frequencies) with a reduced wavelength.

A common picture backed by successes of dispersion theory is that the relation $\mathbf j(t) = c\mathbf E(t-\Delta t)$ is valid, where $c, \Delta t$ are some medium property constants that depend on frequency of the wave, $\mathbf j$ is current density and $\mathbf E$ is total macroscopic electric field. With this assumption, Maxwell's equations imply that the resulting wave in the medium will have modified (in usual cases shorter) wavelength hence lower velocity (for a certain limited interval of frequencies it can have a longer wavelength and higher velocity).

Ján Lalinský
  • 37,229
  • 1
    This explanation cannot be correct. You state that "the resulting wave in the medium will have shorter wavelength hence lower velocity" but this is incorrect. If you change the wavelength (or the frequency, because frequency = c / wavelength) you only change the energy of the photon / light wave (and thus colour) but not the speed, which is fixed as long as it remains in the same type of medium. Indeed though as @arax notes, this fixed speed actually depends on the type of medium and can become smaller than c, but I do not know the correct explanation. – PDiracDelta Aug 11 '17 at 22:38
  • 1
    @PDiracDelta, the question is about index of refraction. Index of refraction is ratio of phase velocities of a wave in two media. Since frequency is the same in both media (linear media) and since $v = f \lambda$, decrease in wavelength implies decrease in phase velocity $v$. The energy of a photon or light wave has nothing to do with this explanation. – Ján Lalinský Aug 12 '17 at 23:22
  • I'm trying to understand what you are saying, but if not light, then what kind of wave are you talking about? AFAIK a light and EM wave are the same thing. Also, to me your claim that the frequency remains the same is non-trivial. – PDiracDelta Aug 14 '17 at 11:03
  • I think I've found a good explanation here: https://physics.stackexchange.com/a/476/51901 (this question now apparently has been marked as a duplicate) – PDiracDelta Aug 14 '17 at 11:18
  • I am talking about an EM wave in material medium, which is a model of a light wave, but the important part in this explanation is relation of electric field to current density, not energy of the wave or the concept of a photon. – Ján Lalinský Aug 14 '17 at 23:07
  • @JánLalinský But how is it possible that secondary waves to exist? Doesn't this violate the conservation of energy? – Antonios Sarikas Nov 20 '19 at 20:53
  • @ado sar That is a good question. We know that secondary radiation exists (from experience) and we believe energy is locally conserved (an extrapolation of definition of energy allowed by experience). Sometimes it seems as if presence of secondary wave near accelerating charged object means there is some additional EM energy that appeared out of nowhere (not from the primary wave). But this is just confusion about the concept of EM energy and how it can be assigned to regions or particles. – Ján Lalinský Nov 20 '19 at 21:14
  • @adosar EM energy can be attributed either to a region of space (as often done with Poynting's expressions), or to some subset (at least two) of charged particles and some region of space (as in Tetrode/Fokker/Frenkel/Feynman-Wheeler theories of point particles), but it cannot be, in general, assigned to some given component of EM field that obeys Maxwell's equations (which secondary EM wave is). Whatever the secondary wave looks like, EM energy + matter energy is locally conserved. – Ján Lalinský Nov 20 '19 at 21:16
  • @JánLalinský Thanks for the answer. But I still don't get it. Lets consider an em wave interacting with a single charge in space. What bothers me is if we treat light as photons then the total number of photons that hit the charge must be the same that come out after interact with it (conservation of energy). But if we treat light as an em wave then as the (electric) field oscillates when it meets the charge the latter start to oscillate so it emitts radiation. But does the initial em wave continue to propagate like a wave in a lake (e.g. when it meets a ball in the surface of the lake) ? – Antonios Sarikas Nov 23 '19 at 22:42
  • In classical EM theory, EM field obeys linear equations, so net field is just a sum of primary and secondary field. The primary field is unchanged by the secondary field, they both contribute to total EM field. If primary field is plane wave, then it continues on its path, undisturbed by the charge or its field. The only thing the charge changes it produced additional field component. – Ján Lalinský Nov 24 '19 at 04:05