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How is light absorbed/reflected?

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webdesignmad
#1
Dec30-11, 09:42 AM
P: 3
Light particles, or photons, are "created" when an electron moves down 1 or more orbitals and the energy of the photon equals the energy difference between the rings. Light is then absorbed by an electron receiving the energy and moving up the proportional amount of orbitals. But electrons always return to their original orbitals by releasing this gained energy, so how then can light disappear, as the cycle starts again?
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Bill_K
#2
Dec30-11, 12:01 PM
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Thanks
Bill_K's Avatar
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Excitation and deexcitation of atoms is only one small way that light interacts with matter, corresponding for example to the spectral lines in solar radiation. Most of the interaction is in the form of scattering and thermal motion.
aimforclarity
#3
Dec30-11, 05:36 PM
P: 34
but scattering is absorption and random re-emtition no?
You can look at it classically with Maxwell's equations and look at the refractive index on two sides of an interface of say air and metal, but on a microscopic level, quantum i thought light scatters by absorption and reemition

Cthugha
#4
Dec31-11, 03:46 AM
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P: 1,626
How is light absorbed/reflected?

Quote Quote by aimforclarity View Post
but scattering is absorption and random re-emtition no?
Not really. Reemission usually takes some time, while scattering is usually considered instantaneous. Therefore real absorption and reemission is associated with fluorescence. However, elastic or inelastic scattering can be described by absorption to and reemission from a virtual energy state.
webdesignmad
#5
Dec31-11, 05:36 AM
P: 3
What I'm asking is that, for example, if I turn off a light, the photons are immediately dispersed. Where have they gone? Or what has destroyed the quanta of energy?
DrDu
#6
Dec31-11, 06:01 AM
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P: 3,555
most of the photons are converted into vibrations of the molecules that absorbed the photons.
webdesignmad
#7
Dec31-11, 06:03 AM
P: 3
Thanks, thats just what i was looking for
morrobay
#8
Dec31-11, 11:57 PM
P: 374
Quote Quote by Cthugha View Post
Not really. Reemission usually takes some time, while scattering is usually considered instantaneous. Therefore real absorption and reemission is associated with fluorescence. However, elastic or inelastic scattering can be described by absorption to and reemission from a virtual energy state.
Can you give an example of instantaneous scattering ?
Every reference I have seen defines the mechanism of scattering , as well as reflection and refraction , as absorption / reemission
Cthugha
#9
Jan1-12, 10:30 AM
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An example? Pretty much any instance of scattering works that way. If scattering is described in terms of absorption and reemission, these are virtual processes and therefore also "instantaneous" compared to spontaneous reemission from a real state.

There is a review article called "Laser Rayleigh scattering" by R. B. Miles et. al. (Meas. Sci. Technol. 12 (2001) R33–R51) in case you are interested in that topic in depth. It should be freely available, if I remember correctly.
aimforclarity
#10
Jan4-12, 07:31 PM
P: 34
Quote Quote by Cthugha View Post
An example? Pretty much any instance of scattering works that way. If scattering is described in terms of absorption and reemission, these are virtual processes and therefore also "instantaneous" compared to spontaneous reemission from a real state.

There is a review article called "Laser Rayleigh scattering" by R. B. Miles et. al. (Meas. Sci. Technol. 12 (2001) R33R51) in case you are interested in that topic in depth. It should be freely available, if I remember correctly.
to be honest im very confused now, is there any way you can describe the process to someone unfamilar with virtual photons but in a physics phd program
Cthugha
#11
Jan5-12, 04:26 AM
Sci Advisor
P: 1,626
This is somewhat hard without actually doing the math, so I suggest you pick up a book where the interaction between an atom and light is discussed in detail. One understandable approach is given e.g. in the introduction to quantum optics book by Grynberg, Aspect, Fabre and Cohen-Tannoudji.

In a nutshell you find that the interaction Hamiltonian in the long-wavelength approximation has two terms. One is linear and one is quadratic in the vector potential. The linear term contains one operator acting on the light field and one acting on the atom and is responsible for emission and absorption processes. The quadratic term contains two operators acting on photons, but none acting on the atom. You have four possible combinations of two operators acting on photons. One is creating two photons. One is destroying two photons. These processes do not conserve energy and are very unlikely to occur. The other two processes destroy one photon and create one photon without affecting the atom at all. These are the processes leading to elastic scattering.
aimforclarity
#12
Jan8-12, 03:04 PM
P: 34
ok that makes a bit more sense Cthugha, thank you.

so there are 2 scattering processes for an atom in free space:
1. the linear abs/emit
2. quadratic a, a-dagger combinations for the light field that emit a photon of the same energy but in a random direction? (should be quasi-random right, only on average would it appear random)

my followup thoughts would be:
A. phase relation b/w incident & scattered photon -> 'coherent?'

B. does the incident photon 'slow down' or how much 'time' does this process take

C. the anhialated photon is the incident one and the scattered one is the outgoing one, so where is the virtual one? (sorry about my lack of qed understanding)

D. just to be clear when you say long wavelength limit that means semiclassical derivation - eg photons are not quantized? or what is the importance?

E. but this is not the same process that would occur for light scattering in a piece of plastic or a crystalline structure?

sorry for all the question, i really hope you could shed some light on my ignorance :)
Cthugha
#13
Jan8-12, 05:18 PM
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Quote Quote by aimforclarity View Post
A. phase relation b/w incident & scattered photon -> 'coherent?'
Elastic scattering with a single scatterer is usually coherent, yes.

Quote Quote by aimforclarity View Post
B. does the incident photon 'slow down' or how much 'time' does this process take
Well, the incoming wave or photon can experience a phase shift which could be interpreted as causing a tiny retardation. Apart from that, there is no slowing down.

Quote Quote by aimforclarity View Post
C. the anhialated photon is the incident one and the scattered one is the outgoing one, so where is the virtual one? (sorry about my lack of qed understanding)
There are no virtual photons, but virtual energy states involved. The annihilated photon should in principle cause some change before the other photon gets emitted in order to conserve energy. However, there are no allowed states to go to. However, due to uncertainty reasons, energy levels are not defined very well at short time intervals and the annihilated photon is modeled as going to an intermediate virtual state which is extremely short lived and therefore its energy is not well defined. Therefore you conserve energy.

Quote Quote by aimforclarity View Post
D. just to be clear when you say long wavelength limit that means semiclassical derivation - eg photons are not quantized? or what is the importance?
The long wavelength limit just means that the wavelength is so large compared to the size of the scattering particle that the shape of that particle does not matter. If the wavelength gets shorter, the actual shape does matter. That basically is the difference between Rayleigh scattering and Mie scattering.

Quote Quote by aimforclarity View Post
E. but this is not the same process that would occur for light scattering in a piece of plastic or a crystalline structure?
In crystalline structures you rarely have scattering from a single scatterer like an atom, but rather a large number of scatterers which may be more (ions) or less (electrons in metals) at well defined positions
aimforclarity
#14
Jan8-12, 07:15 PM
P: 34
Quote Quote by Cthugha View Post
Elastic scattering with a single scatterer is usually coherent, yes.
-usually? maybe you can be a bit more explicit about what you mean by coherence? i was imagining that the phase and polarization of the outgoing photon is the same as the ingoing one
Quote Quote by Cthugha View Post
There are no virtual photons, but virtual energy states involved. The annihilated photon should in principle cause some change before the other photon gets emitted in order to conserve energy. However, there are no allowed states to go to. However, due to uncertainty reasons, energy levels are not defined very well at short time intervals and the annihilated photon is modeled as going to an intermediate virtual state which is extremely short lived and therefore its energy is not well defined. Therefore you conserve energy.
- this makes sense to me if i believe the that for processes on short time scales the effective "linewidth" of the energy levels grow. but this seems strange to me?


i also found this interesting 'physical' explanation byBorn2bwire of photon scattering in terms of more classical concepts and the idea of fields, i think he is talking about the same process you are

Quote Quote by Born2bwire View Post
As for the scattering, that arises due to the coupling of the photon's fields with the electron cloud of the bulk's atoms. The electromagnetic field from the photon disturbs the electron cloud which in turn creates its own electromagnetic wave in response. You can think of the total sum of these waves as giving rise to the scattering of the photon. In this manner, you do not need to be at an absorption level to achieve scattering. I believe that you can look into Griffith's introductory quantum mechanics books and find a few sections regarding scattering.
Cthugha
#15
Jan9-12, 01:42 AM
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P: 1,626
Quote Quote by aimforclarity View Post
-usually? maybe you can be a bit more explicit about what you mean by coherence? i was imagining that the phase and polarization of the outgoing photon is the same as the ingoing one
Not necessarily the same phase, but at least some well defined relationship between initial and outgoing phase. Introduction of a well defined phase shift still counts as coherent.

The "usually" was aiming at the small probability of having inelastic scattering. For molecules, there is for example some small probability for scattering from an excited state of the molecule instead of from the ground state. So the initial and the final state of the molecule differ and the energy of the scattered photon is different from the incoming photon. This spontaneous scattering event (spontaneous Raman scattering) is inelastic and incoherent.

Quote Quote by aimforclarity View Post
- this makes sense to me if i believe the that for processes on short time scales the effective "linewidth" of the energy levels grow. but this seems strange to me?
It is strange, but in accordance with experiments.

Quote Quote by aimforclarity View Post
i also found this interesting 'physical' explanation byBorn2bwire of photon scattering in terms of more classical concepts and the idea of fields, i think he is talking about the same process you are
Yes, it is quite possible that he means the same situation.
aimforclarity
#16
Jan9-12, 08:20 PM
P: 34
Quote Quote by Cthugha View Post
The "usually" was aiming at the small probability of having inelastic scattering. For molecules, there is for example some small probability for scattering from an excited state of the molecule instead of from the ground state. So the initial and the final state of the molecule differ and the energy of the scattered photon is different from the incoming photon. This spontaneous scattering event (spontaneous Raman scattering) is inelastic and incoherent.

ok! so if you have a linewidth on your state (eg a non-ground state) then you can also `transition` within the linewidth with some small probabily, which would be incoherent and inelastic. the reason this is called spontaneous Raman scattering is because it only occurs sometimes?
Cthugha
#17
Jan10-12, 03:55 AM
Sci Advisor
P: 1,626
It is not necessarily within the linewidth. Mostly it is assumed to be some excited state, say the lowest vibrational excitation of the molecule.

It is called spontaneous as opposed to stimulated. The meaning is the same as in spontaneous versus stimulated emission. The emission or scattering rate to the vacuum state is given by the spontaneous rate, but if the final state is already populated with photons, the emission/scattering rate gets enhanced proportionally. These are stimulated processes.


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