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I Difference between Scattering and Emission of Photons

  1. Jan 10, 2017 #1
    Hello,

    Ok, so I’ve been searching about this for a while and there seems to be a difference in explanations that confuses me. What concerns me is the exact detailed physical mechanism that makes particles cause a scatter of a photon and an emission of a photon.

    The Wiki page about scattering says the following about one type of scattering, Rayleigh Scattering:

    “The particles may be individual atoms or molecules. It can occur when light travels through transparent solids and liquids, but is most prominently seen in gases. Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation we see as scattered light.”

    To me, this explanation seems to be the same as when a particle meets a photon that has the same energy/frequency as a potential energy/frequency state of the particle, with the result of the particle absorbing this photon and emitting it again. This is the description found here: http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html#c3

    However, several sources say that that absorption and emission is different from scattering. My question is then, in what way exactly? I am aware that, for example, Rayleigh scattering is elastic which means that the photon that left the particle has the same energy level as the one that hit the particle, but what happened exactly in between? Did it get absorbed and emitted in the same energy quantity?

    Here’s a good discussion about the difference between emission (fluorescence) and Raman scattering:
    http://physics.stackexchange.com/questions/38459/what-is-the-difference-between-raman-scattering-and-fluorescence

    There seems to be 2 different opinions on this in that discussion:
    1. That they’re the same in the sense of the photon being absorbed and then emitted again, with the only difference being the time between absorption and emission if you compare fluorescence with Raman scattering. So there is a time in which the particle is being excited by the photon.
    2. That, in the case of Raman scattering, there is no absorption of the photon but a partial interaction of the photon with the particle which results in the particle taking a bit of energy away from the photon.So there is no excitation of the particle whatsoever by the photon.

    Point 2 bothers me because I’m taught that a photon cannot be partially absorbed or interacted with, since it’s a discrete energy state.

    So what is it exactly that separates the mechanism of emission from scattering? Preferably comparing it with every type of scattering.
     
    Last edited: Jan 10, 2017
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  3. Jan 10, 2017 #2

    jambaugh

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    The distinctions being made are distinct paradigms but physically you cannot distinguish between a photon interacting with another system and the photon being absorbed and re-emitted. In fact the free propagation can be seen as repeated "absorption and re-emissions" in the sense of Huygen's principle.

    One can however distinguish types of absorption/re-emission and you should re-read the discussion with this in mind.
     
  4. Jan 11, 2017 #3

    Saw

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    I think the question is very good (I had the same doubt!) and well deserves some elaborated answer, which unfortunately I don't have...

    When one reads about transitions, you see a clear distinction between:

    Case A, where the E of the incident photon matches a gap between E levels of the atom, so the atom is excited into a higher state (electron transition). This is often called "resonance absorption".
    Case B, where the E of the incident photon is lower than such gap, so there is no such excitation / transition, and the photon goes out as it came in. This is called "Rayleigh scattering". When the E of the incident photon is higher than the gap, the photon loses energy to the particle but we don't say that there is partial absorption (?) and this is called "Raman scattering" (also Stokes scattering, but there is also an anti-Stokes effect, let us leave that aside...).

    But then you learn that in case A the excitation lasts for an incredibly short time after which the photon is re-emitted with the same E as it came in, just as in case B. May it be that the duration of "excitation" in A is longer than "whatever happens" in B? (but it doesn't look like a great difference, given that the time is very short).

    Well, yes, given the difficulties, I would also prefer to call everything simply absorption / re-emission. But still we need to distinguish cases A and B, by both the causes and the effects. The causes are clearly different (case A, excitation/transition; case B, no such thing). But what about the effects, if in both cases the photon is emitted after a very short time with same E as it came in? Well, I am aware that if a second photon impinges upon an already excited atom, the phenomenon of stimulated emission arises, but any other difference? Well, it can be that in A, before re-emission, the E may be absorbed by the material via collisions as thermal E, but that looks more like a collective property of the material, rather than an individual feature of the atom.

    On another note, I am reading here that Raman scattering is not a phenomenon of re-radiation due to an oscillating dipole, just to complicate things a little more.
     
  5. Jan 11, 2017 #4
    I tried rereading it again, but there are people who say that in case of scattering, there isn't any absorption at all. That's where the problem lies. Is there or isn't there any absorption in the case of scattering?

    @Saw : Thanks for your well extensive explanation. I got some questions when reading your post.

    I've also read another interaction in this scenario. That if an incident photon has a lower energy than the energy gap in a particle, the particle would let it just through instead of scattering it. I assume that's what happens in transparent objects if a material let's all the visible light through?

    Won't an incident photon with a higher energy than the gap lead to shooting the electron away and ionizing the atom? Or do you mean a photon with an energy level between 2 energy gaps of the particle?

    Regarding interaction with the rest of the material, someone in the discussion link said the following about Raman scattering and fluorescence;

    "Maybe the difference is really just terminology, but we normally think of Raman spectroscopy as the excited state decaying to a vibrationally excited state by emitting a photon of reduced frequency, and the vibrationally excited state then decays by interaction with the lattice. In fluorescence it's the other way around. The excited state decays to a lower energy state by interaction with the lattice and this lower excited state then decays by emission of a photon."

    I'm not really sure what to conclude from that, other than there is absorption in both cases and that there's a time difference. But if this quote is true, then one could say that Raman scattering and fluorescence also depend on the collective property of the material.


    There's something else important that I noticed about scattering types. Rayleigh scattering, Mie scattering and geometrical scattering all depend on the size of the particle with respect to the wavelength, the particle being smaller (Rayleigh), equally large (Mie) or larger (geometrical scattering) than the wavelength. If scattering really involves absorption and re-emission of photons, then that means that the energy gaps of particles depend somehow on the size of the particle (and/or perhaps the mass as well?) which would seem weird to me. Are there partciles that are equally large/massive with different energy gaps?

    Second to that, if we extend scattering to reflection and call that absorption and re-emission of photons as well, then how would one explain the same emission angle for every particle since reflection is characterized by an emission angle that is the same as the incident angle of a photon? If absorption and emission can't do that, then surely it must be nothing more than a "bounce" of the photon against the particle. Is there even a slight energy reduction of the photon in the case of reflection?
     
  6. Jan 11, 2017 #5

    Saw

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    The usual description does assume that the “lattice” (because it is a collective behavior, see this FAQ) briefly vibrates with a non-resonant frequency and then re-emits it. That brief stop and ensuing delay is what would explain that light travels more slowly in the medium. Whether that is "scattering" or not is another question. I would say, yes, the light is scattered/re-emitted in random directions. How is it that it keeps its original direction? What I read (classical explanation) is that the network of atoms acts cooperatively so that there is destructive interference except in the original direction, but someone objected once to that in a discussion...

    Yes, I meant the second thing. Ionization would happen when the incident photon has enough E to make the electron jump beyond the highest level.

    Interesting. What I had noted down is that fluorescence is for instance that the particle absorbs UV radiation through several E jumps but relaxes in steps emitting photons of several frequencies (jointly white light). But your quote implies that part of the E is dissipated (through interaction with the lattice) and then a single photon of lower frequency is emitted, which would be the same outcome as Raman scattering, although here the process would follow the inverted order…

    I tend to believe that is true and see this site for a similar description of fluorescence.

    Yes, good question…

    In terms of reflection, scattering in random directions would be what happens when there are rugosities (bigger than light’s wavelength BTW…) and so there is diffuse reflection. If the principle angle of incidence = angle of emission is to be respected, then it will be because the surface is polished and so we have specular reflection, but I don’t know how to describe that at quantum level. We may have to read Feynman's description.
     
    Last edited: Jan 12, 2017 at 2:11 AM
  7. Jan 11, 2017 #6

    PeterDonis

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    I don't know where you were taught this, but it's not correct. What is correct is that EM radiation of a given frequency is quantized--heuristically, it comes in "packets" of energy ##h \nu##, where ##\nu## is the frequency. But that does not mean it is impossible for EM radiation to undergo interactions that change its frequency. That is perfectly possible, and point 2 is just an example of it.

    You are assuming a binary distinction that does not exist. "Absorption" and "scattering" are not names for two distinct fundamental processes. They are just two different approximate descriptions of processes involving EM radiation, and the cases for which they are reasonable approximations overlap.
     
  8. Jan 12, 2017 at 11:54 AM #7
    Impossible for EM radiation to undergo interactions that change its frequency, is not what I meant. What I'm saying is that in order to change a photons energy level, the incident photon has to be fully absorbed/destructed first. In the discussion link it is said that the incident photon is interacted with partially without the incident photon being wholly destructed/absorbed first. Someone in the link mentioned the Feynmann Diagtam and according to that diagram, Compton scattering for example involves total destruction of the incident photon first.

    So they both involve the same mechanism? What is the exact process that I'm wrongly making a distinction in? For example, when it is mentioned that they both involve absorption, it is said that they differ in time duration between absorption and emission.
     
  9. Jan 12, 2017 at 12:05 PM #8

    PeterDonis

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    And that is not correct. The photon has to interact, but "fully absorbed/destructed" is not the only possible interaction. At least, not if you're talking about the level of approximation you were talking about before, where we have EM radiation and electrons interacting with each other in a basically classical manner.

    Yes, but this is a different approximation, the one in which we use Feynman diagrams to describe what is going on. In this approximation, there is no such thing as "scattering" in the sense of two particles interacting "at a distance" as in the classical scattering model. There is just one interaction, a vertex with one photon "leg" and two electron/positron "legs". Whether you call this vertex "photon emission" or "photon absorption" or just "electron-photon interaction" is a matter of convenience of terminology.

    The reason this is still just an approximation is that it depends on perturbation theory, i.e., on trying to solve the equations of quantum electrodynamics approximately, by computing successive terms in a series expansion, because we don't know how to solve them exactly.

    Whatever the exact solution (which we don't know, see above) of the equations of quantum electrodynamics is. But even in the Feynman diagram approximation described above, you can see that there is only one interaction, not two.
     
  10. Jan 12, 2017 at 12:26 PM #9
    If that's not the only possible interaction, then there may be a possible distinction between scattering and absorption like it is said. For example, giving the fully absorption/destruction of a photon the name "absorption" and if it's not fully absorbed/destructed the name "scatter". The discussion between people being that if it is possible for a photon to partially interact or not. If it is possible like you said then the distinction is that the photon isn't absorbed in the case of scatter. If it isn't possible, then perhaps they're in both cases absorbed but there's a time difference until re-emission. That's where my question lies.

    So there's no exact mechanism described? Only that scatter and absorption are overlapping approximations of the same mechanism? But there is apparently another way of photon interaction like you said (see above).
     
  11. Jan 12, 2017 at 1:18 PM #10

    PeterDonis

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    And the answer to that is "it depends on which approximate model you are using". There is no single answer. If you use one approximate model, photons can either be "scattered" or "absorbed". If you use another approximate model, there is only one interaction, the vertex with one photon leg and two electron legs. Both models are approximations.

    The closest thing we have to an "exact mechanism" is the description in terms of quantum fields--there is an electron/positron field and a photon field. But we can't exactly solve the equations for this field. That's why we have to use approximations, and why different approximations are used in different scenarios.

    You have to pay more careful attention. I was talking about one particular approximation. See above.
     
  12. Jan 12, 2017 at 2:06 PM #11
    Ok, this clears things up a bit for me. Is there a limit to where these different approximations give different results? Like for example, seeing that the type of scattering is dependent on the particle size (Rayleigh, Mie, geometrical like reflection), can the approximation theory of total absorption/destruction with re-emission of a photon predict such a particles size/geometry-dependent behavior?

    I'm not sure if you've edited your last post but when I replied, your post didn't contain the explanation of other possible approximations.
     
  13. Jan 12, 2017 at 2:11 PM #12

    PeterDonis

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    In principle, yes. In practice, nobody does this because the math is too complicated. That's one of the main reasons we have different approximations for different scenarios--because using the most fundamental one (which at the current state of our knowledge is the Feynman diagram one) all the time would be beyond our practical capabilities.

    Read my post #8.
     
  14. Jan 12, 2017 at 3:18 PM #13
    Great. Thanks for this enlightment.

    Yes, that's the post I was referring to being extended after I replied. I understand now what you mean.
     
  15. Jan 14, 2017 at 3:20 PM #14
    @Saw : Just have been searching a bit about the quantum descriptions/approximations of different types of scattering and its dependence on the size of particles.

    Regarding the dependency on the particle size, there are for example semiconductor atoms called "quantum dots". It seems that the widths between the available energy gaps in such atoms do depend on the size of the atom. This dependency is called quantum confinement effect. This effect explains that an electron in very large atoms usually have continuous available energy levels but as the atom size gets smaller, it would possess increasingly more discrete steps of energy levels. The smaller an atom size gets, the larger the gaps between these available discrete energy states which would mean that a small atom would need a higher energy photon to get excited/(partially) absorb it and would therefore emit blue light instead of red for example. I'm not sure to what extent this theory would be applicable to non-semiconductor atoms but this might explain why very small particles scatter blue light much more strongly in the case of Rayleigh scattering. I wonder if fluoresence is also dependent on particle size?

    Another quantum description regarding the difference between scattering and fluorescence is found here . It shows that scattering involves, not really the excitation of electrons to the higher available energy states, but more of a virtual energy state between these energy states that makes them go back immediately to their original state again. I think you gave a similar description on this in your post. This could explain the shorter time delay until emission compared to fluorescence. This description could be called an absorption but I think people don't prefer it to call it that since that term is reserved for if electrons are truly excited to higher energy levels. Reaching a virtual energy state and going back immediately to the original state is considered more of a "bounce" instead. The article further shows that there is a difference between scattering and fluorescence regarding how much the energy of the emitted photon differ from the absorbed one. An exception is resonant Raman scattering which seem to be the same as resonant fluorescence.

    As for a quantum approximation of reflection, the atoms do indeed scatter/emit photons in all directions after absorption but in the case of reflection where the atoms are more or less at the same level, interference destructs the emission/scatter of photons at any angle except for the angle that is the same as the angle of incident. You've given a similar description on transparency. A good explanation and discussion about reflection and transparency is found here .

    This is what I have so far. I'm planning to delve into the QED Lectures of Feynman about reflection and scatter based on a quantum description to get more details on all this. There are videos available.
     
    Last edited: Jan 14, 2017 at 3:38 PM
  16. Jan 15, 2017 at 3:15 AM #15

    Saw

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    You've been doing excellent work, this "quantum dots" issue is really interesting.

    I will study and comment later the rest of your post but wanted to share now the following, a sort of mechanical analogy with your findings on the above issue. In Conceptual Physics by Paul Hewitt, a simple but very enlightening book that I am lately enjoying, the author makes this comparison when discussing about Rayleigh scattering: "this is similar to the way that small bells ring with higher notes than larger bells".

    See also this quote from here:

    I don't know to what extent the analogy can be pushed further, but it is worth noting at least. Actually for a complete analogy with Rayleigh scattering we would need two things: 1/ the bell or fork should be much smaller than the wavelength of sound (diameter not larger than 1/10th of the wavelength) and 2/ then the bell or fork tends to vibrate more with wavelengths with regard to which it is not too small (higher notes...).

    This is what I gather that your quantum confinement example guarantees: 1/ first you need an atom small enough to have discrete E gaps and 2/ once we are in the discrete domain, it appears that the small size of the atom entails small vibration radius, which calls for higher energy photons.

    As to the reason for 1/ a hypothesis could be that the atom is simpler this way and suffers fewer interactions; instead a big one has many constituents, hence more interactions and thus becomes more like, ultimately, a network of atoms, which give off continuous spectra.
     
  17. Jan 17, 2017 at 1:34 PM #16

    jambaugh

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    My apologies for not getting back to this discussion earlier and it appears you're in good hands with the other posts. I would add one qualifier to your thinking and that is that you should not think of photon absorption, emission, or scattering as instantaneous events. Especially when photons have highly resolved momenta (frequencies) the scattering process takes an extensive length of time and the "emission" isn't per se after the "absorption" but rather they happen together. This is true both classically and quantum mechanically.
     
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