Rayleigh scattering -- What is the true reason for the color of the sky?

The answer is yes, typically only one of the wavelengths that make up the incident light is scattered. This happens most often for shorter waves, like blue and violet. The intensity of the scattered light is proportional to the inverse of the fourth power of the wavelenght of the incident light, so even the intensity of the scattered beam of light is greater for blue and violet light.f
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I have a little question about Rayleigh scattering: I know that Rayleigh cross section is proportional to the inverse of the fourth power of the wavelenght of the incident light and that it is so even the intensity of the scattered beam of light, now:

1. What is the color of the sky due to? Both the things? The fact that intensity scattered is greater for blue and violet light and that more scattering processes occur for these two colors than for others?
2. How can the intensity of the scattered radiation be less than the incident one if the energy must be conserved? Since the intensity of a wave is the amount of energy the wave carries per unit of square and time if the energy remain the same, how can the intensity be smaller?
 
  • #2
It depends on the medium. If there is significant absorption, i.e., dissipation, the total energy of the scattered light is smaller than that of the source. The energy is of course dissipated into heat within the medium.
 
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1. What is the color of the sky due to? Both the things? The fact that intensity scattered is greater for blue and violet light and that more scattering processes occur for these two colors than for others?
I'm not quite sure if I understand your question correctly. The incident radiation from the Sun is a mixture of wavelengths that we call white. Typically it is scattered only once, but for shorter wavelengths with much higher probability. The scattered light is a different mixture of wavelengths with a preponderance of short wavelenghts (blue), but it also contains green, yellow, and even red light. But we cannot see these colours separately; the prevalence of the shorter wavelengths makes us perceive it as a kind of blue.

2. How can the intensity of the scattered radiation be less than the incident one if the energy must be conserved? Since the intensity of a wave is the amount of energy the wave carries per unit of square and time if the energy remain the same, how can the intensity be smaller?
On average, about one fifth of the incident visible light is scattered in the Earth's atmosphere (when the sky is blue). The intensity is the amount of energy not only per unit area and time, but also per unit solid angle (lumens per square meter per steradian). The incident light is concentrated within less than a square degree, whereas the blue of the sky is literally scattered over the whole of the sky, and therefore less intense than the direct light from the sun.
 
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  • #4
@WernerQH
I need to ask you some things: about answer no.1, what do you mean by "typically it is scattered only once"? That only one of the wavelenghts that compose the incident light is typically scattered? So for example we have blu+violet+green+orange+yellow+red and typically just one of these colours is scattered?And why if so, the scattered light is a different mixture of wavelengths?

About the answer no.2 I still can't understand, if the intensity of the scattered radiation is smaller, is this due to the fact that now the energy that passes through the unit solid angle is less than the one that passed through it before scattering since it was more "concentrate" in the space?
Like in the image I've posted, if I integrate the intensity over the whole solid angle before and after the scattering, what I get is the same result:it is the energy carried per unit time but if I measure this quantity without integrating, so per unit time and unit solid angle, before scattering I'd obtain a greater number than after scattering since in the second case the wave is more "spread" in the space, so if I only "concentrate" over a small portion of the solid angle what I measure is just one of the many rays the incident wave splits in, is that right?
r9L9RGzRRcKXQBBcmQSOoJR0ZK8JoqpPh7IHNI66I&usqp=CAU.png


Last thing: the intensity after the scattering is greater for shorter wavelenghts, what does it mean in terms of energy per unit of solid angle? In other words, if the blue goes into scattering or the red goes into scattering, talking about the definition of intensity, what are the physical differences we can note after the scattering? Is red light less "spread out" after the scattering? If I wanted to draw it, how should I illustrate it? How can I imagine the intensity of red light be smaller? How the energy per unit of solid angle and time could be smaller than the one we get from blue scattering? Hope this is clear.
 
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  • #5
I need to ask you some things: about answer no.1, what do you mean by "typically it is scattered only once"? That only one of the wavelenghts that compose the incident light is typically scattered?
What I meant is that each photon of the diffuse light reaching our eyes has been scattered by only one molecule. But of course there are many molecules everywhere in the atmosphere. Looking straight up there are fewer scattering molecules, so the sky appears somewhat darker in that direction than when looking more towards the horizon. But each molecule produces a similar spectrum, with a higher percentage of blue than red light, compared to the incident white light from the Sun. As determined by the Rayleigh cross section. Perhaps it helps if you look at the Wikipedia page on Diffuse sky radiation.
So for example we have blu+violet+green+orange+yellow+red and typically just one of these colours is scattered?And why if so, the scattered light is a different mixture of wavelengths?
Light is almost always a mixture of different wavelengths. Rayleigh scattering increases the percentages of blue and green light compared to yellow and red. On the other hand, the tiny droplets of water in clouds scatter all wavelengths with equal probability, which is why they appear white (like the incident radiation).
About the answer no.2 I still can't understand, if the intensity of the scattered radiation is smaller, is this due to the fact that now the energy that passes through the unit solid angle is less than the one that passed through it before scattering since it was more "concentrate" in the space?
Yes.
Like in the image I've posted, if I integrate the intensity over the whole solid angle before and after the scattering, what I get is the same result:it is the energy carried per unit time but if I measure this quantity without integrating, so per unit time and unit solid angle, before scattering I'd obtain a greater number than after scattering since in the second case the wave is more "spread" in the space, so if I only "concentrate" over a small portion of the solid angle what I measure is just one of the many rays the incident wave splits in, is that right?View attachment 304340
If you consider scattering by a single molecule, you are talking about a different quantity under the same name "intensity". In this case the intensity decreases as the inverse square of the distance from the scatterer. When looking at the blue sky, this decrease is compensated by an increase in the number of scatterers (molecules).
Last thing: the intensity after the scattering is greater for shorter wavelenghts, what does it mean in terms of energy per unit of solid angle? In other words, if the blue goes into scattering or the red goes into scattering, talking about the definition of intensity, what are the physical differences we can note after the scattering? Is red light less "spread out" after the scattering? If I wanted to draw it, how should I illustrate it? How can I imagine the intensity of red light be smaller? How the energy per unit of solid angle and time could be smaller than the one we get from blue scattering?
No, the red light is spread out in the same way as the blue light. The angular distribution emerging from a single molecule is exactly the same. It is only the total number of photons that is less (relative to the number in the incident beam). You could draw shorter arrows if you wanted to illustrate it.

Hope it's become a little bit clearer. :smile:
 
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Unfortunately, it is all much less clear.

What I meant is that each photon of the diffuse light reaching our eyes has been scattered by only one molecule. But of course there are many molecules everywhere in the atmosphere. Looking straight up there are fewer scattering molecules, so the sky appears somewhat darker in that direction than when looking more towards the horizon.

I think my problem is I completely don't know what happens when a beam of light "encounters" a molecule, when you say "each photon of the diffuse light reaching our eyes has been scattered by only one molecule" do you mean each time light is scattered but then since there are many molecules, the same photon could be scattered again?

each molecule produces a similar spectrum, with a higher percentage of blue than red light, compared to the incident white light from the Sun

I don't understand the first part of the sentence, each molecule produces a similar spectrum means that even if a monochromatic wave goes into Rayleigh scattering the diffused wave contains all the colors? I thought that when a white incident beam goes into scattering, all the colors except the scattered one continue straight on their directions and the scattered color is transformed into a spherical EM wave.

If you consider scattering by a single molecule, you are talking about a different quantity under the same name "intensity". In this case the intensity decreases as the inverse square of the distance from the scatterer. When looking at the blue sky, this decrease is compensated by an increase in the number of scatterers (molecules).
I can't understand, you answered yes to my question but now I have to consider a different intensity?

No, the red light is spread out in the same way as the blue light. The angular distribution emerging from a single molecule is exactly the same. It is only the total number of photons that is less (relative to the number in the incident beam)

Here, if the total number of photons is less, how could the energy conservate? Less photons=less energy I think so where does the energy go?
Also, is it possible to have Rayleigh scattering for a single photon? How can we justify the less diffused intensity for red photon if just one photon goes into scattering?
 
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do you mean each time light is scattered but then since there are many molecules, the same photon could be scattered again?
Right. A photon comes from the sun. It is scattered by a molecule in the sky directly overhead. It happens to scatter downward and strikes your eye. That photon has been scattered once.

Another photon comes from the sun. It is scattered once somewhere else and then scattered again by a molecule directly overhead. It happens to scatter downward and strikes your eye. That photon has been scattered twice.

The assertion is that blue light you are seeing when looking directly upward consists mostly of photons that have been scattered only once.

That makes sense since the illumination in the atmosphere above your head directly from the sun is much brighter than the diffuse illumination up there from once-scattered, twice-scattered and many-times-scattered photons.
 
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Here, if the total number of photons is less, how could the energy conservate? Less photons=less energy I think so where does the energy go?
Where does the energy go when you shine white light on a dark surface? Or on something blue?

That was answered in post #2.
Also, is it possible to have Rayleigh scattering for a single photon? How can we justify the less diffused intensity for red photon
Rayleigh scattering can be explained by classical EM- which is where you posted this thread. For the purposes of this discussion, therefore, there is no such thing a photon. The photon is part of the QM theory of light.

Rayleigh scattering is significantly more effective for shorter wavelength light, so red light is scattered much less than blue. To the extent that the scattered blue light is the only one we are aware of.
 
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  • #9
To the extent that the scattered blue light is the only one we are aware of.
During the daytime anyway. At sunrise and sunset, scattering of reds can be seen since the blue has been scattered to near-extinction.

[On second thought, I do not know if that is un-scattered red being scattered by non-Rayleigh means... On third thought, if it's a fourth power effect and the reds are twice the wavelength then the effect should be 6.25% intensity for red scattering, so Rayleigh should still figure in]
 
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  • #10
Unfortunately, it is all much less clear.
Oh. I'll try again. Hopefully I'm not creating yet more confusion.
I think my problem is I completely don't know what happens when a beam of light "encounters" a molecule, when you say "each photon of the diffuse light reaching our eyes has been scattered by only one molecule" do you mean each time light is scattered but then since there are many molecules, the same photon could be scattered again?
Yes, jbriggs444 has explained it nicely.
[...] that even if a monochromatic wave goes into Rayleigh scattering the diffused wave contains all the colors? I thought that when a white incident beam goes into scattering, all the colors except the scattered one continue straight on their directions and the scattered color is transformed into a spherical EM wave.
No, if the incident wave is monochromatic, the scattered wave has the same wavelength. But in the sky you have to consider the direct light from the Sun with all its wavelengths plus all the scattered waves from all the molecules. But since the molecules are far apart, the scattered waves are incoherent and you can treat the scatterings as independent and just add up the intensities. The spectrum of the diffuse light is then just the initial solar spectrum multiplied by the Rayleigh cross section. (At least over the wavelength range of visible light.)
Here, if the total number of photons is less, how could the energy conservate? Less photons=less energy I think so where does the energy go?
Energy is conserved: The total number of photons, scattered plus those remaining in the direct beam, remains the same. For the red photons the fraction remaining in the direct beam is higher. (As you can observe during sunset.)
 
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Much of the light going through the atmosphere goes right on through without scattering at all (that is why the sun still looks really bright when overhead). The probability of Rayleigh scatter of light by air at STP is roughly 10-5 per meter of air traversed. So the scattering to the side is quite selectively blue but a only a few percent of the beam.
 
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  • #12
Is some of this problem explainable by the fact that all the colours we see in the sky are very desaturated (just slight, very broad dips or peaks in the spectrum)? Our intuition tends to magnify what is, in effect, a very subtle effect. After all, the perceived colour of an object under morning, evening or mid day conditions is much the same. We could only do that if there were a broad spectrum of illuminant (and a lot of clever colour processing in our brains).

This is unlike what we see in a rainbow, where there are very narrow bands of colour (but still very desaturated). They are very far from being spectral colours because of the presence of light from many directions from the sky.
 
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Of course this is all against a black background and our vision is logarithmic so the subtle change is exaggerated. The Rayleigh form is radiometric and the photometric outcome will be (seem to be) smaller...how does one quantify "blueness". Lots of ways.
 
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@jbriggs444 sorry could you be more clear? What happens to a photon that undergoes scattering? Is it re-emitted in a single direction? The fact that we are talking about photons is even more confusing to me, according to the image I posted the scattered light is re-emitted in all directions, not just one. I can't understand, I see images in which there is one incident beam and one scattered beam and images in which there is one incident beam and lot of scattered "rays".
 
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@PeroK
Where does the energy go when you shine white light on a dark surface? Or on something blue?
Ok I understand this but why when red light undergoes scattering the scattered beam is composed of less photons? Could I imagine that since red is less energetic the oscillating induced dipole it creates oscillates less than the one blue light would have created?
 
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The fact that we are talking about photons is even more confusing to me
You never said a truer word. Photons almost never help anyone. with this sort of problem; we seldom, if ever see individual photons so why not talk about waves. Rayleigh did and photons are not mentioned in this link. The link also discusses Mie scattering which is less wavelength dependent.
 
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  • #17
@PeroK

Ok I understand this but why when red light undergoes scattering the scattered beam is composed of less photons? Could I imagine that since red is less energetic the oscillating induced dipole it creates oscillates less than the one blue light would have created?
In this post you are straying into Quantum Physics and can you really justify your conclusions? Rayleigh considers a classical wave being scattered by a classical scattering object. We aren't dealing with absorption and re-emission of photons. If we were, we would be getting line and band spectra but what we do see is continuous spectra. (See what I meant in my previous post?)
 
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@PeroK

Ok I understand this but why when red light undergoes scattering the scattered beam is composed of less photons? Could I imagine that since red is less energetic the oscillating induced dipole it creates oscillates less than the one blue light would have created?
Less of the red light is scattered.
 
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Forget photons. Here we deal with classical light, and also photons are better thought about in terms of em. waves rather than particles. Photons are the quanta which have the least to do with localized classical point particles of everything you encounter in QT!
 
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It is very difficult for me at this point to understand the meaning of image I've posted, as far as I understand the arrows represent the Poynting vector of the scattered light depending on where it is issued which can vary with the scattering angle, being maximum in the direction of the incident light and minimum in the orthogonal direction. Since the scattered light is emitted by an oscillating induced dipole I think it must be a spherical wave so, does it make sense to talk about direction of a spherical wave?
 
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Since the scattered light is emitted by an oscillating induced dipole I think it must be a spherical wave so, does it make sense to talk about direction of a spherical wave?
The direction of a spherical wave is radially outward. Dipole radiation is a spherical wave far field but it is not spherically symmetric : nothing is emitted along the z axis (conventionally chosen as the dipole direction), while most comes out symmetrically equatorially (x-y plane). So your picture of the scattering is incorrect. Also (again) most of the sunlight suffers no scattering at all.
 
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  • #22
The direction of a spherical wave is radially outward. Dipole radiation is a spherical wave far field but it is not spherically symmetric : nothing is emitted along the z axis (conventionally chosen as the dipole direction), while most comes out symmetrically equatorially (x-y plane). So your picture of the scattering is incorrect. Also (again) most of the sunlight suffers no scattering at all.
Thank you for your answer, maybe I'm getting it. Let me ask you a question, is there an analogy between these tow images? The one on the left is the radiation emitted by an oscillating dipole and the other one is the Rayleigh scattering intensity, is the scattered intensity that way because of the form of oscillating dipole radiation?

Cattura.PNG
scattering-of-light.jpg
 
  • #23
Right. A photon comes from the sun. It is scattered by a molecule in the sky directly overhead. It happens to scatter downward and strikes your eye. That photon has been scattered once.
Sorry I need to reopen this question, how can a photon be scattered downward or in another specific direction? Aren't photons scattered in all directions?
 
  • #24
Sorry I need to reopen this question, how can a photon be scattered downward or in another specific direction? Aren't photons scattered in all directions?
What part of "there is no such thing as a photon in classical electromagnetism" do you not understand?

Light is an electromagnetic wave of oscillating electric and magnetic fields that satisfy Maxwell's equations.
 
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  • #25
What part of "there is no such thing as a photon in classical electromagnetism" do you not understand?
If I read in an answer the word photon, in order for the question to be clear I have to quote the word that was used.
 
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  • #26
Sorry I need to reopen this question, how can a photon be scattered downward or in another specific direction? Aren't photons scattered in all directions?
We are playing a bit fast and loose with the idea of photons as little bullets with a well defined path, sure. But we also know that the photon originated at the Sun, wound up in our eye, was sensed at a position in the retina that is associated with a downward trajectory and would not have been observed were it not for the presence of an atmosphere.

That suggests scattering. If some other photons were scattered in some other directions and were not observed... so what?

It can be helpful to adopt the simplest model that is consistent with the experimental results.
 
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  • #27
@MrKindness If I read in an answer the word photon, in order for the question to be clear I have to quote the word that was used.
Whoever first mentioned photons in this thread was wrong to do so.

If you want a QED/QFT explanation for Rayleigh scattering, then post in the quantum physics forum.
 
  • #28
The one on the left is the radiation emitted by an oscillating dipole and the other one is the Rayleigh scattering intensity, is the scattered intensity that way because of the form of oscillating dipole radiation?
These are different representations of basically the same phenomenon. The second is more of "sketch" of energy flow of the process I guess (it is difficult for me know).
The first is a quantitative graph showing an intermediate range snapshot of the electric field associate with oscillating dipole radiation. To understand why they can be the same process requires firm knowledge of Maxwell's equations.
 
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These are different representations of basically the same phenomenon. The second is more of "sketch" of energy flow of the process I guess (it is difficult for me know).
The first is a quantitative graph showing an intermediate range snapshot of the electric field associate with oscillating dipole radiation. To understand why they can be the same process requires firm knowledge of Maxwell's equations.
I thought you could relate the scattered intensity to the shape of the electric field generated by an oscillating dipole since I find them very similar. As far as I understand the second image represents Poynting vector so it's like the representation of the magnitude of the intensity of the scattered wave and since this last one is produced by an oscillating dipole I thought that distribution of intensities descends from the first image.
 
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That's pretty much what I said. The stellate "graph", without further explanation, could be a sketch of a porcupine.
Precisely what do you wish to know? Your original questions have been thoroughly answered I believe.

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That's pretty much what I said. The stellate "graph", without further explanation, could be a sketch of a porcupine.
Precisely what do you wish to know? Your original questions have been thoroughly answered I believe.

/




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I want to know if it makes sense to say that the scattered wave is re-emitted in a certain direction since as far as I know the oscillating dipole radiates a spherical wave and it is not possible to say that wave travel in a certain direction AND if you had to draw the Poynting vector of the scattered wave, would you have drawn it as in the image on the right?
 
  • #32
want to know if it makes sense to say that the scattered wave is re-emitted in a certain direction since as far as I know the oscillating dipole radiates a spherical wave and it is not possible to say that wave travel in a certain direction
Classically most of the field amplitude is as if there were no scattering but that is not what you see. A small portion is scattered into a dipolar spherical radiation field. It has preferred directions of maximal intensity but the associated wave always has a direction from the scattering center to your eye, usually in a straight line. The "direction" is along your line of sight.

The portion you see certainly has a direction.
 
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  • #33
The point is that for em. waves you cannot have a spherically symmetric solution, because the photon field is a spin-1 field, i.e., the multipole expansion starts with the dipole contribution. A monopole distribution is necessarily static. That's known as the Birkhoff theorem of electrodynamics. The analogous theorem for the gravitational field (general relativity) is that the only spherically symmetric vacuum solution of Einstein's equations is the static Schwarzschild solution. In the case of the gravitational waves the multipole expansion starts with the quadrupole term, because the gravitational field is a 2nd-rank tensor field ("spin 2"),
 
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  • #34
Forget photons
I usually say in class "f*ck photons, they don't exist!"
 
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  • #35
Regarding spherical re-radiation, if the scatterer is a sphere and we illuminate it with one polarisation then we approximately see the dipole "doughnut" radiation pattern. If we illuminate it with two incoherent waves polarised at right angles to each other then we see scattering in all directions. The Sun radiates random polarisation - we can say it is equivalent to two noise sources polarised at right angles to each other. Therefore I think the blue light of the sky will be radiated in every direction from each particle.
 

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