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FriedPhoton
2009-Jun-13, 05:24 PM
When light is reflected or refracted, does it really bend and bounce or is it absorbed by the material's atoms and emitted in a different direction?

This is the "tip of the iceberg" question. I have more that will follow depending on the answer to this one.

PetersCreek
2009-Jun-13, 06:02 PM
I've moved your question to the Q&A forum.

While it's not strictly a question about space or astronomy, I think it's related closely enough to merit putting it here where you're likely to get more responses.

grant hutchison
2009-Jun-13, 07:05 PM
I think you might need to define "really". :)
The behaviour of light is described by a wavefunction, which interacts with the wavefunction of matter in interesting ways that lead to what we experience as reflection and refraction.
Is that way really happens? How could we tell what really happens?

Grant Hutchison

Ken G
2009-Jun-14, 02:38 AM
Grant's question is certainly a valid one to ask, but I'm going to choose a particular interpretation of what you are asking, and say that the answer is "it could be either". Basically, it sounds to me like you are asking, is there ever a time when there is clearly no photon, and then a later time when it is back, or is there always a photon there undergoing some kind of bounce? It turns out there are actually two kinds of scattering, coherent and incoherent. Coherent scattering keeps the phase of the photon the same, so what this means is that there is a period of time when the wave function of the incident photon and the reflected photon exist at the same time-- one cannot say if the photon has bounced yet or not, it is in a superposition of not having bounced and having bounced. But incoherent scattering is not like that, the photon is clearly gone, it's energy has gone into exciting some atom, and then later on it reappears, out of phase with the original photon. I may be oversimplifying some of the details, but these are the kinds of issues that would appear if you have a "wave packet" coming in and a "wave packet" coming out from the interaction. The degree of overlap can be a lot or a little.

grant hutchison
2009-Jun-14, 12:15 PM
It turns out there are actually two kinds of scattering, coherent and incoherent.Is the latter applicable to the "reflection and refraction" of the OP?

Grant Hutchison

PraedSt
2009-Jun-14, 01:01 PM
I hope you two aren't going to start fighting again. :lol:

G O R T
2009-Jun-14, 01:06 PM
When light is reflected or refracted, does it really bend and bounce or is it absorbed by the material's atoms and emitted in a different direction?

For simplicity, the light is indeed absorbed and reemitted. Light cannot travel far through a dense material without being absorbed even if it is transparent. Typically photon absorbtion occurs statistically in the first half wavelength of material. Anti-reflection coatings make use of this by surface depositing layers of various refractive index materials that are thinner than 1/2 wavelength of the target light frequency.

The wavelength of light, the material, and the permittivity of the material are important. The reflectivity and transparancy of materials are dependant on the wavelength (energy) of the photon.


Some characteristics of optical interfaces are best described by wave theory. Answers may refer to wave of particle theory depending on the question.

Ken G
2009-Jun-14, 04:44 PM
Is the latter applicable to the "reflection and refraction" of the OP?
Them's fighting words! Just kidding. I see what you are saying, the technical meaning of reflection and refraction are phase-coherent processes, so only apply to the former case I mentioned. The latter case is actually just "another thing light can do", but may not be what the OP was interested in. We'd have to ask the OPer!

Ken G
2009-Jun-14, 04:56 PM
Light cannot travel far through a dense material without being absorbed even if it is transparent.This actually depends on the specific meaning one is using for "absorbed", as there are really two possibilities, related to the two cases I distinguished above. The meaning you take is absorbed=interacts, and it is true that even a pane of glass, which is transparent even when many many wavelengths thick, still slows the passage of the light we see through it. So clearly the molecules are doing something to the light, and that is what you mean by absorbing and re-emitting it. Then it is the phase relationships that cause the light to not be scattered (Huygens' principle), but only if the interaction is phase-coherent, which is why the light does not scatter and we see a relatively undisturbed image through the glass. To make that distinction, sometimes the term "absorb" is reserved for incoherent processes, so those that would make objects opaque. Other times, it is even reserved for processes that thermalize (destroy) the photon completely. So that's why your statement may sound strange, though is completely correct given the meaning of absorb that you are using.

FriedPhoton
2009-Jun-15, 02:33 AM
Gort's reply came closest to a) what I understand, and b) what I thought.

I was trying to understand if photons actually impacted a surface and changed direction, like a rubber ball off a wall, or if they were absorbed and re-emitted.

So, the answer is, if they are reflected they have been absorbed and re-emitted back in some direction. If they are refracted, they are absorbed, re-emitted in a forward direction (although possibly at some different angle), then re-absorbed, and so on until they pass out of the material. Essentially passing from atom to atom until free again.

And this would account for the misnomer that light "slows" as it passes through a material. It simply takes more time due to a) repeated absorption and re-emission, b) a change in angle that causes it to travel further.

I'm hoping to understand that light always travels at light speed and never travels "slower". Because if it did travel slower it would experience time, and I wonder if it would convert to a particle if it actually slowed down.

I understand these things at a pop-science level but I find it all fascinating to contemplate so please bear with me if my terminology is vague. I appreciate the quantum physics replies but I need to work my way up to those. Although, feel free to hit me with them if you think it'll help prevent my misunderstanding the true nature of the behaviors I'm asking about.

Thanks!

FriedPhoton
2009-Jun-15, 02:39 AM
This statement by Ken, "Other times, it is even reserved for processes that thermalize (destroy) the photon completely." is interesting.

I have wondered about how energy is "used". From what little I understand about entropy I gather that at some point all the energy in the universe will eventually be useless. But it will still exist, because it can not be destroyed.

So if I start with a bunch of cosmic rays it is possible to use up that energy and watch the resulting photons drop in frequency? Would those photon's wavelengths grow longer until they spanned the universe, or is there some threshold below which there is no matter that can interact with it at which point it is functionally useless?

mugaliens
2009-Jun-15, 05:08 AM
Cosmic rays are particles streaming through space at less than c. They are not light rays.

Ken G
2009-Jun-15, 05:24 AM
Would those photon's wavelengths grow longer until they spanned the universe, or is there some threshold below which there is no matter that can interact with it at which point it is functionally useless?The "functional uselessness" of thermalized light (which would be isotropic and a "Planck spectrum", like the CMB) does not come from the fact that it cannot interact, it comes from the fact that it is in thermodynamic equilibrium-- a state wherein every single process is balanced by its inverse process. So nothing ever "happens" in the net in thermodynamic equilibrium. There could be an enormous amount of energy, and it could be interacting all over the place, but it is only deviations from equilibrium that end up doing anything-- in the net.

astromark
2009-Jun-15, 06:59 AM
My understanding is that light is both a wave form and a particle stream.. As such it can exhibat interesting charactoristics when said to being refracted or reflected.
If light is absorbed and has zero emissions then its been converted into heat or disipatated. That part of the spectrum that is visable light is indeed interesting in its manor of reflection. As Ken has so quintessentially said. What is really happening...

FriedPhoton
2009-Jun-16, 03:58 AM
The "functional uselessness" of thermalized light (which would be isotropic and a "Planck spectrum", like the CMB) does not come from the fact that it cannot interact, it comes from the fact that it is in thermodynamic equilibrium-- a state wherein every single process is balanced by its inverse process. So nothing ever "happens" in the net in thermodynamic equilibrium. There could be an enormous amount of energy, and it could be interacting all over the place, but it is only deviations from equilibrium that end up doing anything-- in the net.

OK, so if I have two x-ray photons (ignoring the cosmic ray issue for now) that are 180 degrees out of phase with each other and they cancel each other out, then the work the photons can do is nil. Is that correct? Then the energy that exists is two x-ray photons worth, but it's useless?

Can photon's collide like that? Could the energy be made useful again by the addition of a third photon? How do photon's interact? Do they have to meet each other at some desired angle to cancel each other or do the waves simply combine?

I don't want to go to much further with this line of questioning in case I've misunderstood something, so I'll leave it at this for now.

FriedPhoton
2009-Jun-16, 04:00 AM
Cosmic rays are particles streaming through space at less than c. They are not light rays.

My apologies! I meant gamma rays.

Ken G
2009-Jun-16, 07:06 PM
OK, so if I have two x-ray photons (ignoring the cosmic ray issue for now) that are 180 degrees out of phase with each other and they cancel each other out, then the work the photons can do is nil. Is that correct? Then the energy that exists is two x-ray photons worth, but it's useless?That's a different issue, you are talking about coherence now. The uselessness of energy in thermodynamic equilibrium is incoherent, i.e., there's no need to worry about coherence effects. Coherence has more to do with telling photons what to do and where to go, in a situation of complete destructive interference everywhere there isn't any photon to begin with, no photon is allowed to do anything. Usually destructive interference doesn't work that way, it tells the photon it cannot do certain things, but then the photon just does other things instead-- there isn't total destructive interference everywhere. But we're really not talking about coherence here.


Can photon's collide like that? Could the energy be made useful again by the addition of a third photon? How do photon's interact? Do they have to meet each other at some desired angle to cancel each other or do the waves simply combine? Photons can do interesting things when they have enough energy to make particles, but the key point in all this is no matter what interesting things can happen, they will always happen in balance with the inverse reaction when we are in thermodynamic equilibrium. That's the "equilibrium" part, nothing changes, and if nothing is changing, nothing is really "happening" in the net. So all the energy is kind of "useless" in that situation.

Jeff Root
2009-Jun-16, 07:27 PM
sometimes the term "absorb" is reserved for incoherent processes, so
those that would make objects opaque. Other times, it is even reserved
for processes that thermalize (destroy) the photon completely.
How is thermalizing the photons different from just absorbing them
so that the material is opaque? A black surface is thermalizing while
a white one is not? And a fuschia surface is thermalizing for photons
of some wavelengths but not others? (Ignoring everything outside
the visual range!)

-- Jeff, in Minneapolis

Ken G
2009-Jun-16, 09:45 PM
How is thermalizing the photons different from just absorbing them
so that the material is opaque? A black surface is thermalizing while
a white one is not? Right. A white surface is rough, so it scatters incoherently (unlike a mirror), but it doesn't change the wavelengths that are incident, whereas the emission coming from a black surface is coming in an entirely different wavelength regime (the infrared, typically). Then there is also "fluorescence", where you start with ultraviolet light (say, though for lights you just use some electrical excitation), and light comes out in the visible, but not because it is thermalized (fluorescent lights aren't hot). If you start with UV light and fluoresce it, then you are absorbing the UV and it is undergoing atomic transitions that destroy the UV photons and replace them with visible ones, etc. This can also be called "Raman scattering", but there you usually start with visible light and end up with IR, due to molecules not atoms. The terms can get very confused, and may mean different things in different places.

And a fuschia surface is thermalizing for photons
of some wavelengths but not others?Yes, though I have no idea what color fuschia is. Your question is related to the "albedo", and when that depends on wavelength.

grant hutchison
2009-Jun-16, 09:58 PM
Yes, though I have no idea what color fuschia is. Turns out to be spelled fuchsia, despite the pronunciation. A shrub with purply-red flowers, named after Leonhard Fuchs, who would cringe at how English speakers have mangled his name.
So: it's a sort of purply-red colour. :)

Grant Hutchison

FriedPhoton
2009-Jun-16, 10:02 PM
That's a different issue, you are talking about coherence now. The uselessness of energy in thermodynamic equilibrium is incoherent, i.e., there's no need to worry about coherence effects. Coherence has more to do with telling photons what to do and where to go, in a situation of complete destructive interference everywhere there isn't any photon to begin with, no photon is allowed to do anything. Usually destructive interference doesn't work that way, it tells the photon it cannot do certain things, but then the photon just does other things instead-- there isn't total destructive interference everywhere. But we're really not talking about coherence here.
Photons can do interesting things when they have enough energy to make particles, but the key point in all this is no matter what interesting things can happen, they will always happen in balance with the inverse reaction when we are in thermodynamic equilibrium. That's the "equilibrium" part, nothing changes, and if nothing is changing, nothing is really "happening" in the net. So all the energy is kind of "useless" in that situation.

So, I think you're saying that in a situation like the double slit experiment where the waves cancel each other out is destructive interference and where the destructive interference is is a point at which the energy is useless? Is that energy simply balanced by the constructive interference so that you essentially haven't really lost anything? Or does it work some other way?

I think I've been making the mistake of thinking of photons as little wave packets, meaning that they would appear to have a frequency of their own. Are wave effects seen with photons or must there be many photons to see a waveform?

Jeff Root
2009-Jun-17, 12:14 AM
I tried to spell 'fuchsia' from memory. I read about the origin of the name
and the alteration of the pronunciation a couple of years ago, but that
seems to be long enough for me to forget, nowadays.

-- Jeff, in Chartreuse

Jeff Root
2009-Jun-17, 12:29 AM
I think I've been making the mistake of thinking of photons as little
wave packets, meaning that they would appear to have a frequency
of their own. Are wave effects seen with photons or must there be
many photons to see a waveform?
That is a fabulously good question. I've wondered about it a lot.
My personal expectaton is that each photon has a frequency of
its own -- relative to an observer -- but that measurement of the
frequency is essentially impossible. An individual photon can come
off a spectroscope grating at any angle, but statistically it is most
likely to come off at about the angle characteristic of its frequency.
(or wavelength...)

-- Jeff, in Minneapolis

Ken G
2009-Jun-17, 01:18 AM
Is that energy simply balanced by the constructive interference so that you essentially haven't really lost anything? I would say you don't really even know how much energy is involved until you add up all the coherences and decide what the photons are "told" to do, and how many are told to do it. That's why lasers are so bright and involve such high energies-- their coherence allows a lot more photons to take part.


I think I've been making the mistake of thinking of photons as little wave packets, meaning that they would appear to have a frequency of their own. Are wave effects seen with photons or must there be many photons to see a waveform?You can see wave effects with a single photon. Basically, I wouldn't think of the wave packet as the photon, I would think of the wave packet as the instructions the photon obeys (or to not personify the photon, the information that we can know about the photon, or use to predict the photon). The photon is just the quantum of action, they are the discrete things that are going on that add up to the full complexity of what is happening. Note that individual electrons also obey wave mechanics, as in the two-slit experiment for electrons.

Above all, I'd say the quanta here, the discreteness, has more to do with how you need to count. Money works this way too-- if I buy a house for 200,000 dollars, each "dollar" in that 200,000 is pretty meaningless, the whole 200,000 is just a number on paper-- I'll never say, "200,000? OK, hold your your palm: 1, 2, 3, ...." So that's how classical physics treats the field intensities, they are just numbers. But when you get into processes that really do involve a small number, like 3 bucks, then there does have to be a "1, 2, 3" kind of phase to the counting, or you don't get the right transaction (unless you use a credit card!). So when you go shopping, the "wave packet" is like your shopping list, the instructions that control how you shop. The quanta are the dollars you have in your wallet, the things that get the counting right when you want to make a transaction. And if you write a check and buy something for 1,000 dollars, there never will be a "counting" phase, the quantization is irrelevant, and all you care about is the shopping list (the wave amplitude). But if you deal with smaller amounts, only then will you notice the discrete character of bills.

Note also that there is something called a photon mode, which has a definite wavelength, and is used as a kind of "basis" to talk about what real photons (described by packets of modes) actually do. Because of the superposition principle, you can imagine what a photon would do if it had a definite wavelength, the photon mode, and then piece together what the real photon actually does by combining all the imagined photons that are really photon modes. Sometimes the word "photon" gets confused with the term "photon mode", they end up being treated interchangeably and you hear about "the wavelength of a photon". This is unfortunate, it causes a lot of confusion.

FriedPhoton
2009-Jun-17, 02:32 AM
I would say you don't really even know how much energy is involved until you add up all the coherences and decide what the photons are "told" to do, and how many are told to do it. That's why lasers are so bright and involve such high energies-- their coherence allows a lot more photons to take part.

I understand what you are saying. I guess if you have photons creating interference patterns the sum of the energy is still the same as what was originally produced. But have those dark portions of the interference pattern reduced themselves to the non-usable energy state because they are canceled out? Are they doomed to an eternity of sad uselessness?

I have another question related to this. Say you do this experiment in a vacuum. Is the interference occurring a centimeter before the light hits the surface (as well as everywhere between source and target) or does it only interact when it hits the surface? I suppose it is impossible to really know, but are there mathematical models that describe the behavior?


You can see wave effects with a single photon. Basically, I wouldn't think of the wave packet as the photon, I would think of the wave packet as the instructions the photon obeys (or to not personify the photon, the information that we can know about the photon, or use to predict the photon). The photon is just the quantum of action, they are the discrete things that are going on that add up to the full complexity of what is happening. Note that individual electrons also obey wave mechanics, as in the two-slit experiment for electrons.

Above all, I'd say the quanta here, the discreteness, has more to do with how you need to count. Money works this way too-- if I buy a house for 200,000 dollars, each "dollar" in that 200,000 is pretty meaningless, the whole 200,000 is just a number on paper-- I'll never say, "200,000? OK, hold your your palm: 1, 2, 3, ...." So that's how classical physics treats the field intensities, they are just numbers. But when you get into processes that really do involve a small number, like 3 bucks, then there does have to be a "1, 2, 3" kind of phase to the counting, or you don't get the right transaction (unless you use a credit card!). So when you go shopping, the "wave packet" is like your shopping list, the instructions that control how you shop. The quanta are the dollars you have in your wallet, the things that get the counting right when you want to make a transaction. And if you write a check and buy something for 1,000 dollars, there never will be a "counting" phase, the quantization is irrelevant, and all you care about is the shopping list (the wave amplitude). But if you deal with smaller amounts, only then will you notice the discrete character of bills.

Note also that there is something called a photon mode, which has a definite wavelength, and is used as a kind of "basis" to talk about what real photons (described by packets of modes) actually do. Because of the superposition principle, you can imagine what a photon would do if it had a definite wavelength, the photon mode, and then piece together what the real photon actually does by combining all the imagined photons that are really photon modes. Sometimes the word "photon" gets confused with the term "photon mode", they end up being treated interchangeably and you hear about "the wavelength of a photon". This is unfortunate, it causes a lot of confusion.

I think I get this. The wave is "what" happens, and the photon is "where" it happens?

Ken G
2009-Jun-17, 02:40 AM
Yes, that seems like a good way to think about it, I'd say anyway. Or it could be said that the photon is where and what, and the wave is how and why. But it's all in the realm of how we like to think about things.