17

I'm asking a question that has bothered me for years and years. First of all, let me give some context. I'm a layman in physics (college educated, math major). I've read Feynman's QED cover to cover, and watched his messenger lectures. My hope is to get an answer at the "Feynman level", ideally in the crystal clear terms that feynman himself uses :)

Okay, so with that out of the way, my question is about transparency. My understanding from Feynman and other materials is that when a photon hits an atom, the photon can either be absorbed, or absorbed and released.

What I don't understand is how transparency could arise from this. If the light were absorbed and then a new photon emitted, it seems incredible that the new photon would have the exact trajectory as the old one. Why wouldn't it go off randomly in a random direction?

Here's some of the homework I've done trying to find out the answer on my own:

1) Transparency of materials (one commentor alludes to scattering not being explained here)

2) http://alemassociates.com/mambo/content/view/33/1/ (this seems to address it, but I will admit it's mostly over my head)

3) http://www.av8n.com/physics/white.htm (seems to talk about it, but doesn't explain why the photon would ever come out the same direction as it came in)

4) Discover magazine posted this: "when a packet of light energy, or photon, hits a solid object, three things can happen. Light can disappear: If the photon has the same vibrational frequency as the electrons in the material it strikes, those electrons absorb its energy, changing the photon from light into heat. Light can also be scattered: the surface electrons can grab the photon's energy and then eject a photon of the same wavelength which is how you see pretty much everything that doesn't emit light on its own. But if the photon doesn't have the right vibrational energy for absorption, and if the atoms in the material are arranged in patterns that discourage reflection (such as the random jumble of molecules in glass or air) then the photon's energy passes from atom to atom, emerging on the other side still bright and shiny. Then you have transparency."

I'm posting here for help. I realize that this forum is geared for graduate level and beyond physics, but I'm hoping that this transgression will be forgiven. I would really, truly, like to understand why anything is transparent at all. From my understanding, nothing should ever be transparent! At best we might get translucency (a bunch of photons coming out in random directions), but I cannot understand transparency. Why would a photon interacting with an atom come out the exact way it came from? In other words, as per that Discover magazine article, why would the photon be passed from atom to atom always at the exact same vector that it came from?

Thank you!

edit: If possible, I'd like to keep it in terms of photons. What's going on at the photon level, and avoid the wave framework. Let's talk about photons and probability amplitudes a-la Feynman.

Community
  • 1
RefiAgain
  • 171

3 Answers3

6

I believe your puzzlement comes from confusing two frameworks: the quantum mechanical (photons) and the classical mechanics one, waves.

When one is calculating in terms of classical electromagnetic waves there are classical considerations : refraction, absorption, reflection with their corresponding constants .

When one is zooming in the microcosm and talking of photons, a wave is composed of zillions of photons which go through, each at the velocity of light.

The bulk of the target material is in effect the electric and magnetic fields holding the atoms together to form it: the nuclei are tiny targets and the electrons are small zooming targets. The probability of a single photon to scatter on a nucleus or an electron is miniscule. It interacts/scatters out of its optical ray path with the electric/magnetic fields that are holding the glass or crystal together. The scattering angles are very small in transparent materials thus preserving the optical path, enormous in opaque ones . It is those fields that one has to worry about, not the individual atoms and their excitations.

The photons scatter mostly elastically with the fields holding the solids together with tiny or high cross sections depending on the frequency of light and spacing of the materials. In crystals and glasses the optical frequencies have small probability of interaction.

x rays find most materials transparent because the photons' energy is much larger than the energies available by the fields holding the crystals together, and the scattering angles with the fields are very small, except when they hit the atoms, and then we get x ray crystallography.

Edit after comment

Here is the sequence as I see it:

A classical electromagnetic wave is made up of photons in phase according to the wave description.

There is an enormous number of photons in the wave per second making it up. Here is a useful article which explains how a classical wave is built up from a quantum substrate.

Each photon does not change the atomic or crystalline energy levels going through a transparent material in the quantum mechanical way by emitting a softer photon. It scatters quantum mechanically elastically, through the medium, changing the direction infinitesimally so that it keeps the quantum mechanical phase with its companions and displays transparency. Since a medium has a composite collective electric and magnetic field it is not a simple "electron photon going to electron photon" QED diagram. In the case of a crystal one could have a model of "photon crystal photon crystal" scattering amplitude for example.

The higher the sequential probability of scattering going through a medium the larger the final deflection through it will be, and the higher the over all probability of losing the phase with its companions in the wave.( the thicker the glass the less transparency and image coherence).

The transparency of the medium depends on the ordering of the atoms and molecules composing it so that it allows to keep the coherence between individual photons of the beam. The lower the density the better chance to keep the transparency, viz water and air.

hope this helps conceptually.

anna v
  • 233,453
  • "In crystals and glasses the optical frequencies have small probability of interaction." Well, depends on what you call small here – they're no bigger in, say, paper, at any rate. Yet that is opaque, which has little to do with individual interaction probabilities but a lot with the less ordered structure, which causes the phases to be more or less randomized for every path, unlike in crystals. – leftaroundabout Jun 18 '12 at 21:28
  • 1
    Thank you for the answer. I do not believe I am confusing the two frameworks. I'd like to keep this strictly at the QED level -- light is made up of photons that interact with electrons, and the probabilty of an event is calculated by summing probability amplitudes. leftaroundabout summarizes my point well. If it's just a low probabilty of interaction, that doesn't explain paper vs glass. It also doesn't explain transparency vs translucency. – RefiAgain Jun 18 '12 at 22:00
  • @leftaroundabout Yes, I included it in my edit. I was concentrating on making clear that there is no absorption and emission but it is elastic scattering. Of course for the coherence to be kept the atomic ordering of the material is important. – anna v Jun 19 '12 at 04:38
  • How irregular materials as glasses or liguid and gazes may lead to an overall probability to make a coherent output beam of photons (what wee see as transparency)? – dan Mar 06 '15 at 09:54
  • @danielAzuelos glasses and liquids have a regularity short and medium range, they cannot be described by unit crystals and thus are much more complicated to study quantum mechanically.http://en.wikipedia.org/wiki/Structure_of_liquids_and_glasses . gases have large distances between molecules that allow for coherence to a through going electromagnetic wave. – anna v Mar 06 '15 at 11:49
  • Nonetheless, if I take the examples of air or water, a photon beam entering these materials won't go straight forward, but will be deviated in what macroscopically look like a coherent direction (Snell's law). How might this team behaviour be explained? – dan Mar 06 '15 at 13:21
  • Have a look at http://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html . Air is different than water. quantum mechanical boundary conditions solutions will be different – anna v Mar 06 '15 at 13:48
5

I certainly can't match the clarity of Feynman's explanations, but I'll try to stay along his lines. He actually talks about this very issue quite extensively in multiple places in the Lᴇᴄᴛᴜʀᴇꜱ Oɴ Pʜʏꜱɪᴄꜱ, firstly in chapter 26-6 in the first volume.

You're completely right: there is no reason why the emitted photon should have the same trajectory as the absorbed one. As a matter of fact, it doesn't! Actually, at the atomic level, it doesn't even really make sense to talk of an outgoing photon's direction. You must have heard about particle/wave duality, and in the wave picture, it's quite easy to explain: the light has a wavelength much larger than an atom, so at those scales the emitted photon is just the center bulge of a wave spreading in all directions rather equally. But if there are many such spherical waves superimposed (from all the atoms that absorb and emit photons in the glass) with some specific phase relation (which is "inherited" from the incoming light) then you get the good old classical Huygens principle: the waves form a single directed wave front, which is what makes the outgoing light beam.

Now, in quantum field theory, one tends to avoid this explicit wave picture – at least the classical one – but the equivalent path integral formulation of course amounts to the same thing: you have to consider the amplitude for the photon to be absorbed and subsequently emitted by any of the atoms in its way. These amplitudes will be shifted by some phase depending on the position of the atom, and when all the phases are different the amplitudes will cancel to almost 0. Only in the direction where the phases are all the same, you get constructive inteference, so that's where you'll actually measure the photon.

leftaroundabout
  • 14,903
  • Thank you for the reply. I'd like to keep it in terms of photons and probability amplitudes, as in your second paragraph. I get that the path integral formulation must add up in such a way as to cancel out in other paths, but WHY? Feynman explains a lens focusing light by showing us how the paths end up adding in the right way. He also shows us how a diffraction grating will make colors by adding up the little arrows – RefiAgain Jun 18 '12 at 21:58
  • I don't understand WHY the little arrows would add up that way for glass, but not for anything else. I understand that they must, but I'm not seeing the WHY. What is it about glass that's so special that would make it so all the arrows (probability amplitudes) would cancel out exactly that way? It seems like an impossible coincidence. What's the underlying thing that's going on? Does that make sense? – RefiAgain Jun 18 '12 at 21:58
  • It's not just for glass, but for pretty much anything that doesn't have any structure bigger than the wavelength of light (and of course, doesn't absorb "permanently"). Why this is so – well, even from a path integral standpoint you'd trace this back to the mathematics that are easily visualised with the wavefront picture. Of course you can also calculate it without this analogy, but there's not really any reason not to use it. – leftaroundabout Jun 18 '12 at 22:53
2

This is not using any QED or Feynman level physics but I recommend Chaikin and Lubensky Principles of Condensed Matter Physics section 1.2 that, I believe, answers difference between paper and transparency. A further question that may be considered on this direction is "Why metals are shiny?"

Xenomorph
  • 21