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

In summary: 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.
  • #36
drmalawi said:
I usually say in class "f*ck photons, they don't exist!"
This is also not true for some decades. Nowadays the quantum opticians have reliable true-single-photon sources!
 
Physics news on Phys.org
  • #37
vanhees71 said:
This is also not true for some decades. Nowadays the quantum opticians have reliable true-single-photon sources!
If I taught quantum optics I would probably say "f*ck waves, they don't exist" :)
 
  • Haha
Likes vanhees71
  • #38
Which is not true either. All there is are quantum fields, and they are following more the intuition of fields (as all relativistic physics is in fact only really consistent as a classical of quantum field theory; the point particle has been a stranger since the very beginning, when Lorentz introduced the "classical electron theory" into electrodynamics). Since "wave equations" fulfilling the causality constraints of relativity are hyperbolic, there's no way to avoid waves!
 
  • #39
vanhees71 said:
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"),
These stuffs are too much difficult for me, I just want to know what "shape" of radiation is reemitted after scattering like: if I look at the scattered light from any point around the molecule that scatters it, do I always see the scattered light? Is there always some part of the scattered EM radiation reaching my eye? If I look at pictures of the radiation emitted by an oscillating dipole, which is what we have in Rayleigh scattering, it looks like the "shape" of the scattered radiation I see in Rayleigh scattering pictures and that made me think the shape of the radiation emitted by an oscillating dipole is the same we see in Rayleigh scattering, this would mean that part of the scattered radiation, as well as part of the radiation emitted by a dipole, always reaches my eye but I can't understand if this is right or not. In the image, that's the radiation emitted by an oscillating dipole, and the eyes (circles) are reached by it in every point I think. Does this happen even in Rayleigh scattering and what about unpolarized light like the sunlight? I apologize for being repetitive but I just can't understand.

Cattura.PNG
 
Last edited:
  • #40
The light reaching your eye from even a small section of sky has been scattered by air molecules with all orientations, so much of the directionality is averaged out. The light, particularly from a direction 90deg from the sun will, however, be strongly polarized.
To understand in detail, you need to understand the details
 
  • Like
Likes vanhees71 and Salmone
  • #41
hutchphd said:
The light reaching your eye from even a small section of sky has been scattered by air molecules with all orientations, so much of the directionality is averaged out. The light, particularly from a direction 90deg from the sun will, however, be strongly polarized.
To understand in detail, you need to understand the details
Maybe is better for me to understand if we assume there's only one molecule which scatters the unpolarized sunlight, the beam of light arrives on this molecule, part of it proceeds straight forward and a component is scattered, now the scattered radiation should be the typical radiation emitted by an oscillating dipole, that is, radiation that has the "shape" of the one in the figure and will therefore have the same intensity given by Rayleigh's formula for intensity, plus having that shape some of the scattered radiation will reach the eye of any observer. Let's see if this is right, otherwise I will go back and study the theory behind Rayleigh for good because I hurt myself in my confusion and I do not want to lengthen this discussion unnecessarily.
 
  • #42
You need to understand both and then know how to fit them together. Scattering from an individual molecule is a solvable (approximately) problem and gives you much information. But when you ask "what will I see in the sky" you will need to include the fact that you are seeing many scattering events from various configurations. You cannot use one or the other. Both are required.
 
  • Like
Likes Salmone and vanhees71
  • #43
Salmone said:
Sorry I need to reopen this question, how can a photon be scattered downward or in another specific direction?
We are discussing Scattering and not Absorption. They are two entirely different mechanisms. This effect is not narrowly frequency selective (just a slope over the band).
Salmone said:
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.
There are two aspects to the radiation pattern; there's the directivity for inducing the oscillation in the dipole and there's the directivity for radiating the wave for each of a random selection of dipole orientations. This 'fills in' the null of the ideal donut shape.
hutchphd said:
The portion you see certainly has a direction.
I'm not sure that statement makes it clear that the radiation from the dipole is in all directions? The energy is shared out all over the 'donut'.
It's no wonder that the scattering effect is very slight; firstly, most of the energy passes through, unscattered and secondly the scattered light is spread out over the pattern. - just to amplify your point.
 
  • Like
Likes hutchphd, Salmone and vanhees71
  • #44
sophiecentaur said:
There are two aspects to the radiation pattern; there's the directivity for inducing the oscillation in the dipole and there's the directivity for radiating the wave for each of a random selection of dipole orientations. This 'fills in' the null of the ideal donut shape.
I did not understand what you mean
sophiecentaur said:
I'm not sure that statement makes it clear that the radiation from the dipole is in all directions?
So is it a yes or a not? The radiation from the dipole is in all directions? I guess yes of course.
 
  • #45
The intensity distribution is depicted nicely in a polar diagram. You get the characteristic two "blobs" with the maximum perpendicular to the direction of the dipole and zero in its direction. In the same way you can also depict the higher harmonics (quadrupole, octupole,...):

https://en.wikipedia.org/wiki/Dipole#Dipole_radiation
 
  • Like
Likes Salmone
  • #46
Salmone said:
I did not understand what you mean
Back to Radio; it's more familiar. Take a 1m long dipole. It will have a current induced in it by a passing VHF radio wave and the value of the current will depend on its directivity pattern and the angle of arrival. That current will radiate power in all directions, modified by the pattern. The signal received by a receiver out in some direction will depend on where it is on that pattern AND on the level of the induced current. So that donut pattern is actually involved twice - once for receiving and once for transmitting
Salmone said:
The radiation from the dipole is in all directions?
Yes, of course. The molecule can't 'know' where the observer is.
 
  • Like
Likes Salmone and vanhees71
  • #47
vanhees71 said:
The intensity distribution is depicted nicely in a polar diagram. You get the characteristic two "blobs" with the maximum perpendicular to the direction of the dipole and zero in its direction. In the same way you can also depict the higher harmonics (quadrupole, octupole,...):

https://en.wikipedia.org/wiki/Dipole#Dipole_radiation
Thank you but I can't understand to which of the many questions you're answering, are you saying that there is an analogy between intensity distribution in a polar diagram of an oscillating dipole and the intensity distribuzione as a function of the scattering angle 1+cos^2(\theta) in Rayleigh scattering?
 
Last edited:
  • #48
What I'm trying to answer is the question, whether "it radiates in all directions": The point is that this is precisely answered quantitatively by the polar diagram.

You are right, that diagram only depicts the scattered wave but not the unscattered part in Rayleigh scattering.
 
  • Like
Likes Salmone
  • #49
vanhees71 said:
What I'm trying to answer is the question, whether "it radiates in all directions": The point is that this is precisely answered quantitatively by the polar diagram.

You are right, that diagram only depicts the scattered wave but not the unscattered part in Rayleigh scattering.
Ok but now I have another doubt, why in Poynting vector expression we have a sin^2(x) and in the Rayleigh intensity expression there is a 1+cos^2(x)? Shouldn't they represent the same thing?
 
  • #50
Salmone said:
an analogy between intensity distribution in a polar diagram of an oscillating dipole and the intensity distribuzione as a function of the scattering angle 1+cos^2(\theta) in Rayleigh scattering?
It's more than an "analogy". It's strongly related. I'd say that the resultant amplitude A of the ray going to the observer will be something like:
A = A0 X (1+cos2θ) . (1+cos2Φ)
where the angle θ is the angle between the incident ray direction and Φ is the angle between the scattered ray direction. (angles relative to the normal) and X is a 'small number', corresponding to the effective cross section of the molecule. I'm also doing this in 2D with the polarisation aligned with the dipole.
Not perfect because the size of the dipole is small relative to the wavelength and you don't get a perfect zeros for polarisation normal to the dipole.
 
Last edited:
  • Like
Likes Salmone
  • #51
With respect to the OP, understanding the details of Rayleiigh scattering at an Intermediate level requires use of intermediate level physics. One sets up the salient physical process and does the math. This involves EM theory in 3 dimensions and integral calculus. It is not surprising when one has difficulty using intuition and pictures from the internet.
Despite good intentions, There is no royal path
 
  • Like
Likes sophiecentaur and vanhees71
  • #52
sophiecentaur said:
It's more than an "analogy". It's strongly related. I'd say that the resultant amplitude A of the ray going to the observer will be something like:
A = A0 X (1+cos2θ) . (1+cos2Φ)
where the angle θ is the angle between the incident ray direction and Φ is the angle between the scattered ray direction. (angles relative to the normal) and X is a 'small number', corresponding to the effective cross section of the molecule. I'm also doing this in 2D with the polarisation aligned with the dipole.
Not perfect because the size of the dipole is small relative to the wavelength and you don't get a perfect zeros for polarisation normal to the dipole.
The scattering dipole always lies in a principle plane containing the Sun and the observer. So there is no effect due to radiation pattern of the dipole - for a spherical scatterer the induced dipole is always normal to the Sun's ray. If, alternatively, we assume little rods, having random orientation, the polarisation purity will be destroyed due to re-radiation taking place with random polarisation, and this is not what is observed. If we resolve the Sun's radiation into two incoherent polarisations, and we consider the scatterers as spheres having induced crossed dipoles, then it accounts for all-round scattering and retains the polarisation effect observed. As the scattering loss of each particle is very high, it seems unlikey that double scattering is occurring to a significant extent, and the all-round scattering can be explained by the "unpolarised" nature of the Sun's radiation.
 
  • #53
tech99 said:
for a spherical scatterer the induced dipole is always normal to the Sun's ray.
Yes - that makes sense but are diatomic molecules spherically symmetrical? Maybe the random orientations takes care of that.
 
  • Like
Likes vanhees71
  • #54
sophiecentaur said:
Yes - that makes sense but are diatomic molecules spherically symmetrical? Maybe the random orientations takes care of that.
Good point, so I presume these molecules will degrade the polarisation purity, because they can be canted at random angles.
 
  • Like
Likes sophiecentaur
  • #55
Don't we have to consider the unpolarized sunlight as two linearly polarized radiations with electric fields oscillating one orthogonal to the other in a plane perpendicular to the direction of the sunlight so that we can imagine two oscillating perpendicular dipoles which give the ##1+cos^2(\theta)## from the ##sin^2(\theta)## of the single oscillating dipole Poynting vector? This is what I've understood but I don't even know if you're talking about this.
 
  • #56
None of this has anything to do with color. If you have a polarization question please start a new thread.
 
  • #57
Salmone said:
Don't we have to consider the unpolarized sunlight as two linearly polarized radiations
We sometimes choose to do that but the description of an elliptically polarised wave can be just that, with polarisation angle changing as the cycle progresses. It's like resolving forces - which we do or do not, according to when it suits us to solve a problem.
 
  • Like
Likes Salmone
  • #58
We must be careful to distinguish between elliptically polarised radiation and that from the Sun, which is called unpolarised. It can be resolved into any two orthogonal polarisations which are not related in phase, they are incoherent. With an elliptically polarised wave the two waves are in quadrature. Then we can find an exact cross polarised twin, but we cannot do so with the case of two cross polarised incoherent waves.
 

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

Back
Top