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This is also not true for some decades. Nowadays the quantum opticians have reliable true-single-photon sources!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!drmalawi said:I usually say in class "f*ck photons, they don't exist!"
If I taught quantum optics I would probably say "f*ck waves, they don't exist" :)vanhees71 said:This is also not true for some decades. Nowadays the quantum opticians have reliable true-single-photon sources!
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.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"),
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.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
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:Sorry I need to reopen this question, how can a photon be scattered downward or in another specific direction?
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.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.
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'.hutchphd said:The portion you see certainly has a direction.
I did not understand what you meansophiecentaur 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.
So is it a yes or a not? The radiation from the dipole is in all directions? I guess yes of course.sophiecentaur said:I'm not sure that statement makes it clear that the radiation from the dipole is in all directions?
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 transmittingSalmone said:I did not understand what you mean
Yes, of course. The molecule can't 'know' where the observer is.Salmone said:The radiation from the dipole is in all directions?
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?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
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?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.
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: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?
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.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 = A_{0} X (1+cos^{2}θ) . (1+cos^{2}Φ)
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.
Yes - that makes sense but are diatomic molecules spherically symmetrical? Maybe the random orientations takes care of that.tech99 said:for a spherical scatterer the induced dipole is always normal to the Sun's ray.
Good point, so I presume these molecules will degrade the polarisation purity, because they can be canted at random angles.sophiecentaur said:Yes - that makes sense but are diatomic molecules spherically symmetrical? Maybe the random orientations takes care of that.
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.Salmone said:Don't we have to consider the unpolarized sunlight as two linearly polarized radiations