5
$\begingroup$

There is a related question (Why glass is transparent?) but I am coming at it only from Maxwell's equations. One can determine the skin depth $δ$ for poor conductors like (pure) water and glass using (see Wikipedia)

$$δ =2ρ \sqrt{\frac{ϵ}{μ_0}}$$

If I ignore the frequency dependence of the permittivity (only to get a board range for the skin depth of glass), using appropriate values for the resistivity ρ (water = $2.5×10^5$ Ω∙m and glass = $10^{10}−10^{14}$ Ω∙m), electric permittivity ($ϵ=ϵ_0ϵ_r$) and magnetic permeability ($μ ≈ μ_0$), I calculate that

$$δ(water) =10^4m$$ $$δ(glass) =10^8-10^{12} m$$

Maxwell’s equations determine the behavior of electromagnetic waves in conductors (as well as poor conductors), so if glass and water have such larger skin depths, then this is the reason why light is transparent for these two mediums – right? If so, I then have two related questions:

  1. Mathematically, it’s fairly straight forward to show that the skin depth is independent of frequency. However, is there a physical explanation why the skin depth is independent of frequency for poor conductors but not for good conductors?

  2. At least at optical frequencies, the skin depth is mainly dependent on the resistivity of the material. Since glass has a higher resistivity (is a poorer conductor) than water, electromagnetic waves penetrate farther through glass. So the key to understanding why glass is more transparent than water is physically understanding why δ ∝ ρ?

I have looked through the books of Griffiths and Jackson for help on this, and found nothing. Thank you in advance for any help on these questions.

Correction and edit due to Johannes’s comment below for question 2

|cite|improve this question
$\endgroup$
  • $\begingroup$ not an answer, but have you considered the molecular structure of glass? It's a nice amorphous crystal. I think I once read somewhere that much of it's superb transparency comes from light being able to pass easily through it because of this crystal structure. I'll look around and see if I can find the reference. $\endgroup$ – Jim Jul 25 '13 at 13:09
  • 4
    $\begingroup$ so after doing some looking around, commercial plate glass has a transmittance of ~90-95% in the visible range. As you might expect, for water, the thicker it is, the more it would absorb. But it turns out that distilled water about the same thickness as a pane of glass (I looked for about 0.5cm but found for a number of samples from 1 to 100mm) has a transmittance of ~95-100%. That means glass is less transparent than water. Really, it's true when you think about it. $\endgroup$ – Jim Jul 25 '13 at 13:40
  • $\begingroup$ @Jim: If the figures you're quoting are for the percentage of the light that passes through, then they're probably dominated by partial reflection at the air-glass and glass-air interfaces. That's a different mechanism than the one described by the skin depth. $\endgroup$ – Ben Crowell Jul 25 '13 at 14:53
  • $\begingroup$ @Jim: A thin sheet of glass has transmittance of ~92% because 4% of light is reflected back at each air-glass interface. The glass itself is much more transparent - think about fiber cables that can be many kilometers long. $\endgroup$ – gigacyan Jul 25 '13 at 14:56
  • $\begingroup$ fair enough, I was just presenting data. I believe my first comment shows I really don't know the answer. But for short distances, my point does remain valid. $\endgroup$ – Jim Jul 25 '13 at 15:44
2
$\begingroup$

I disagree with the premise of this question. Using DC permittivity and DC resistivity is an awful starting point if you want to understand anything about visible-light response. [Update: I should say that it's not that bad a starting point for metals specifically. Much worse for other materials.] When electrons move back and forth at 60 Hz, they usually move in a totally different way than when they move back and forth at 1 quadrillion Hz.

For example, in an n-type semiconductor, at 60 Hz, the conductance comes from electrons in the conduction band getting shifted within the band and traveling and sometimes bumping into defects. The conductance at 1 quadrillion Hz comes from electrons in the valence band being pulled into a quantum superposition state between valence and conduction band states. The superposition state happens to jiggle back and forth (by atomic-scale distances) at 1 quadrillion Hz, because of the energy difference between the two states and the laws of quantum mechanics. Soon the superposition is disturbed and you get an electron-hole pair.

For example, rubber has a very high resistivity but is not transparent. Indium-tin-oxide has a low resistivity but is transparent.

To understand visible absorption, you need to be thinking about energy levels and modes, not DC resistivity.

Water absorbs visible light because of various weak (harmonic) vibrational modes. Normally, vibration modes are only in the infrared, but water has unusually high-frequency vibration modes that reach just a bit into the visible. (Because hydrogen is light and bonds very tightly to oxygen. Just like a taut thin string on a guitar will vibrate at a higher frequency than a loose thick string.) Glass does not have that property.

Glass can be much more transparent than water: For example, fiber optics are glass strands through which light can travel many kilometers with negligible absorption. Fiber optics are manufactured very carefully to reduce absorption; if you made ordinary window glass that was 1km thick, it would certainly be opaque.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I am no expert, only trying to learn a little optics. It seems reasonable to me that Maxwell’s equations, which generates an electromagnetic wave solution and leads directly to the skin depth for conductors (good and poor alike), would be good enough to explain the difference between glass and water transparency. However, your excellent answer (which I am still trying hard to fully grasp) is telling me that Maxwell’s equations are insufficient to explain these details. That I have to resort to quantum physics to understand physically the difference between glass and water. Is this correct? $\endgroup$ – Carlos Jul 26 '13 at 16:16
  • $\begingroup$ (A) Electric and magnetic fields push around atoms and electrons; (B) The atoms and electrons in turn alter the electric and magnetic fields. You use Maxwell's equations for part (B) but not (A). (A) depends on how easily the electrons can move at each frequency, etc., and includes things like the frequency-dependence of permittivity and resistivity. I'm glad you've seen the skin depth derivation, but it assumes frequency-independent resistivity. That's a bad assumption at optical frequencies. (It's OK for metals at lower frequencies like radio frequencies). $\endgroup$ – Steve Byrnes Jul 26 '13 at 19:09
  • $\begingroup$ It seems straight forward to speak of frequency dependent permittivity because it’s related to the index of refraction through $ϵ≈ϵ_0n^2(ω)$, and n(ω) doesn’t change very much at optical frequencies ($ϵ≈ϵ_0$). However, as soon as I look at frequency dependent resistivity (say the Drude Model), the only way that I’ve learned through electrodynamics is specifically dealing with conductors, not dielectrics like glass or water. Do you have any suggestions where I can learn about this? $\endgroup$ – Carlos Jul 27 '13 at 9:16
  • $\begingroup$ Honestly, I’m lost. I assumed Maxwell’s equations allowed me to understand which medium is more transparent because of skin depth (because my professors have said this), and I’ve found out differently. In hindsight, I should ask “how far can I push the skin depth relationship, which is derived for conductors, be applied to dielectrics like water and glass?” This is the key point I am struggling with. This should be asked as a separate question and if possible, I would greatly appreciate you expanding on your answer there. It was extremely useful for a student of physics like me. $\endgroup$ – Carlos Jul 27 '13 at 9:19
  • $\begingroup$ The skin depth relationship is fine for any material at any frequency ... if you use the AC permittivity and AC conductivity. And it's very nice for metals at low frequencies. People discussing visible light in glass could talk about glass's AC conductivity, and use the skin depth formula, if they wanted to. But they usually don't - they use other notations & parametrizations instead. See en.wikipedia.org/wiki/Mathematical_descriptions_of_opacity $\endgroup$ – Steve Byrnes Jul 27 '13 at 12:53
0
$\begingroup$

Well the absorption data I gave for seawater, can be found in any reliable text on the properties of terrestrial materials. I have a source of data that covers the wavelength range from 0.1 microns, in the vacuum UV, up to 3.0 meters, or 100 MHz, in a Woods Hole Mass. paper that dates from 1965. I recently saw a similar graph in a 1981 Masters thesis, and the two graphs are virtually identical over that entire range.

So sea water absorption increases by a factor of 10^8 for a 6:1 wavelength change. That's hardly a square root dependency. The usual skin effect calculations are based on the assumption of a passive material. Optical absorption processes, almost always have anomalous absorption and refractive index functions that result from atomic or molecular resonances. Water in particular has well known resonances at 3.0 and 6.0 microns, that are a consequence of the water molecule structure, so they are the same for fresh and salt water. When you get into the microwave and radio frequency region at longer than about 30 microns wavelength, sea water, starts to behave much more like a passive homogeneous conductor, similar to ordinary metals. Then the usual skin effect behavior starts to show up. At those frequencies, you are beyond the vibration and rotation modes of the molecules.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ George, please see my comments with Steve above. Apparently, you are correct in recognizing that I was treating the medium as passive, which I didn’t understand until now. However, I was concerned with optical frequencies and transparency; the wavelengths you’ve quoted are outside the visible region. $\endgroup$ – Carlos Jul 27 '13 at 9:49

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.