Why do photons continue in same direction after interaction with transparent media

Phalanx

Ok, i have recently been trying to learn a lot about light, and I have found that the reason the speed of light is slower in dense materials is because of the photon-atom interactions.

However, I have also been taught that when a photon is absorbed by an atom that an electron in that atom jumps from its current energy shell to a higher one, then jumps back down emitting a pulse of light that is the same energy as the original. My question is how does that pulse of light "know" to go in the same direction as the original as opposed to scattering randomly. I feel like my understanding of the photon/particle interaction needs to be expanded upon.

Physics Monkey

This is a very good question that is only properly answered in a nice quantum mechanical treatment of the atom-light interaction. Here is the basic idea: imagine you have a beam of light incident upon an atom sitting somewhere. The atom can absorb a photon and change its internal energy and the rate of this absorption is proportional to the intensity of the light beam. Now, this excited atom likes to decay back down to the ground state and re-emit a photon. Where does this photon go? Well, the atom can decay in two ways: spontaneous emission or stimulated emission. Spontaneous emission is caused by vacuum fluctuations of the electromagnetic field and every excited state of the atom has a lifetime due to spontaneous emission. When the atom emits by spontaneous emission, the photon can go pretty much anywhere i.e. the light is scattered. Now, the more interesting process for your question is stimulated emission. If the incident light is intense enough, the atom really likes to emit that photon back into the beam. This is basically because photons are bosons and they like to be near their friends. In precise language, the rate of stimulated emission is proportional to incident intensity, so the more intense the beam the more likely that the photon will be emitted right back into the beam. Now, if you put a bunch of these atoms together in a solid lots of things can happen and the story will change some. However, the very basic idea remains the same, if the intensity is high enough then the photon almost always decays by stimulated emission and continues on its way. Of course, some photons will always be emitted spontaneously, this is the scattered light you can see from any transparent material.

Hope this helps.

Danger

It sure helped me. That's something that had me wondering too. Thanks.

pervect

This explanation (stimulated emission vs spontaneous emission) would seem to imply that very weak light (for instance, sometimes beams are created that are expected to have only one photon at a time) would always be scatterered. I don't think this happens.

Spin_Network

Take the explanation of PM to be correct, one can ask of photons inside the universe: At any one moment, are there more photons "inside" or external to atoms?

Is there a scattering amplitude to the Whole Universe, and is this related to the CC?

DaTario

photons also have a cross section for the interaction process, so photons may pass through the material without suffering any disturbance. Perhaps in this case, neither spontaneous nor stimulated process has happened.

DaTario

Tide

The classical argument is that the medium is essentially a continuum of electromagnetic dipoles that radiate due to interaction with the incident wave. Interference between the incident wave and the radiated field results in a combined field propagating in the forward direction.

ZapperZ

Again, as Tide as alluded to, optical transmission in solids, as in this case, does not involve any "atomic transition". The band structure of the solid does not look anything close to the atomic energy levels of atoms - the atoms have, for most practical purposes, lost their individuality due to the overlapping of their valence shell.

What is more important in the optical properties of such solids is the phonon structure, and if the optical wavelength being sent can be sustained by the phonons. These "dipoles" that Tide referred to can produce optical phonons that can dictate the tranparency of a material at a particular wavelength.

Zz.

zoobyshoe

This is how Claude Bile explained it a few months ago, in more detail, though. Someone had asked if the photon that emerges from a piece of transparent material is the same one that went in. He explained how the energy and momentum of a photon propagates through the material by turning the first atom it hits into a dipole, which causes the next to respond by becoming a dipole, and so on through the thickness of the glass. This explanation made it clear to me why it continues in the same direction.

ZapperZ

Ah, but there is a subtle but crucial difference between the two explanation. An isolated atom turning into a "dipole" due to such interaction does not explain why there are opaque and transparent solids. Tide stated of a continuum of dipoles. I have, in another thread, tried to explain this as a chain of ions separated by negative charges and how optical modes from such a chain can be obtain.

I think there is a very crucial, moral-of-the-story point to be brought across here, and that is that in a solid, "atoms" or the properties of individual atoms are no longer that significant in a number of common physical characteristics of a solid. You can have the identical atom content, such as carbon, but arrange them in different ways and you get two very different material (graphite and diamond). The individuality of the atoms (i.e. the atomic carbon energy levels) alone doesn't tell you the if you're going to get graphite or diamond characteristics.

Zz.

Physics Monkey

I see I must respond to some critics of my answer, but before I do, thank you all for your valid points.

First, let me point out that I indicated that the story would change in solids. I am perfectly aware of the fact that atoms tend to loose their identities in a solid. That being said, I don't think the OP has any idea what phonons are, let alone optical phonon modes. I was trying to give a simple yet realistic explanation in keeping with the level of the OP, and in doing so I ended up giving something like an Einstein model for the transition structure. Now, it is true that in insulators (and almost all transparent materials are insulators) much of the action happens with the phonon modes, but all you have to do is replace "atomic transition" with "excitation of phonon mode" to make the simple picture fit, at least roughly. Either way, a phonon mode that emits light will do so with a rate proportional to a "phase space" factor of [tex] n+1 [/tex] (the number of photons already present plus one), this is just Fermi's Golden Rule. Now, there are plenty of complications, details, even other factors that can be just as important, and so on and so forth. For practical purposes, one often doesn't even need to quantize the field, but I think this was the flavor and level of explanation that the OP was looking for.

As an aside, I should note that phonons are also not the most important player for a number of very interesting processes. In particular, many solid state lasers use the discrete levels of color centers in their operation (like the very first laser built by Maiman). These same color centers, defects in the crystal lattice, are also largely responsible for the coloration of ionic crystals. So the individuality of defects can sometimes be important too.

ZapperZ

I was more concerned with the OP's question in reference to the thread topic. The inference to "transparent media" indicated to me that this was a question of optical transmission in something like glass, for instance, and not a "pumped" solid to produce or amplify light. This is a very common question here on PF - it appears every couple of weeks, it seems, and have been repeatedly addressed. At the simplest level, I tried to emphasize the idea that most of the optical characteristics of solids do not resemble the atomic structure of the the atoms that make up the solid. If such take-away message gets across, I consider it a success.

As you have said, beyond the simple "transparent" issue, the phenomena and explanation can become involved and complicated (example: UV absorption in glass, but not for quartz, would require a very involved explanation).

Zz.

zoobyshoe

Yes, I see there is a whole other level to go into as to why some solids are transparent and others not.

DaleSpam

Really? I didn't know that. If a photon does pass through without suffering any disturbance does it pass through at c or at the "c" of the material? I would think that it would go at the lower "c", but then it would seem like it was disturbed, so I am probably missing something.

-Dale

Hans de Vries

Good question indeed.

This is related to issues like the "Single Photon Interference".

A picture taken with an 8 meter telescope of an extremely faint
astronomical object is typically a "one-photon-a-time" process,
nevertheless, the single photon's trajectory after the lens is
extremely well defined.

In this case, even with a single photon there is still a spread wave-
function associated to it. According to Huygens principle (cq. Feynman's
path integral principle) the sum of all paths defines a well defined wave-
front in the direction of motion of the single photon.

All the paths of the wave function are slowed down in the transparent
material. The question then of how this works leads you directly to the
problematic issue of the Interpretation of Quantum Mechanics.

There is still a lot of active research going on and photon recoil
measurements have been made only recently:

Photon recoil momentum in dispersive media:
Gretchen K. Campbell, and al. PRL 94, 170403 (2005)

http://www.aip.org/pnu/2005/split/732-1.html
http://cua.mit.edu/ketterle_group/Pr..._05/camp05.pdf



A photon inside a dielectric media, however dilute, like the BEC in the
paper above has a momentum modified by the refraction index n of the
medium. With [itex]\lambda[/itex] as the wavelength in vacuum we get:

[tex]p\ =\ \frac{nh}{\lambda}[/tex]

Thus when a photon hits an atom, even in a very diluted gas, the
interaction of the light with all the atoms has to be taken into account,
even if the specific interaction being measured, in effect, is that of
a single atom.



Regards, Hans

Spin_Network

Which then leads to the puzzling Luminocity Function associated with ..say far away faint QSO's ?

The amount of outpouring light, has a specific "pressure", defined as Luminocity. Take a single far away faint object, this is sitting within a field of Radiative energy of its own making (outpouring photons), but it is also recieving a VAST amount of Photon energy that is "opposite"..it is infalling from every Luminous object from the rest of the Universe?

As mentioned elsewhere in thread, the infalling faint light "weak-field" from the rest of the Universe, will still impose an effect that is, or can be equated to that of a 'material-like', transparent field, or Dark Energy.

This will in effect impose a 'Grip' and restrain the outpouring light source, as percieved to have an extra "restraining" halo?

Just as the Electron is WRT an Atom (electron=shroud) that interacts with incoming photons, coming from the source atoms, and from a specific direction, and via a connecting medium as explained by QM or QCD, then on a Universal scale, Galaxies as source's, if are isolated enough, will have a greater number of infalling photons, across a great light spectrum , by fact that all other Galaxies that are providing light that is infalling upon an isolated single observed object?

In a simplistic way (and my interpretation may be challenged), a far away QSO may be thought as emmiting a single source signal, whilst recieving a vast, multi-number of incedent photons?

Hans de Vries

This all has very little to do with the original question.

Are you suggesting Light Pressure as Dark Energy?

Can't be, because it would accelerate differently sized objects in different
ways (acceleration = exposed-surface/mass)

Also, why should in-falling light restrain the emission of light?


Regards, Hans