When light interacts with an atom or molecule, it may be scattered. That is, its direction of travel after the encounter is different from that before the encounter.
Rayleigh scattering is an important process affecting the travel of light through the atmosphere. This is particularly true in the ultraviolet region, since the amount of light that is scattered is much greater at shorter wavelengths than at longer wavelengths.
For simplicity, let's think of a single atom. As long as they stay within a small spherical region around the atom's nucleus, an atom's electrons are fairly free to move around.
Along comes a photon, with its electric field oscillating, tracing out a sine wave in the "plane of polarization." Because the electrons have an electrical charge, they respond to the photon's electric field, feeling a force that, at any instant of time, is proportional to the magnitude of the electric field. So as the field oscillates, the electrons are forced up and down. Now, it is a law of electrodynamics that an electrically charged particle that accelerates produces an electromagnetic wave that propagates outward from the charged particle, like water waves radiate from where a pebble is thrown into a pond. At one instant the electron is going upward, and then it slows down, stops, and starts going downward. Thus, during this phase of its motion, it is accelerating downward. At some later time, it slows down, stops, and starts going upward, and so it must be accelerating upward. The frequency of this up and down oscillation is just the frequency of the photon, so the electromagnetic wave it produces has just the same frequency (and energy and wavelength) as the original photon. Also, the polarization of the outgoing wave is in the same direction as the original photon.
This is the outgoing scattered electromagnetic wave after the interaction of the electron with the incoming electromagnetic wave. The atom is at the center of the figure.
Now we have to think about energy conservation. If a photon, having some definite energy, encounters an electron bound in a molecule, the response of the electron produces an outgoing electromagnetic wave of the same energy. The electron, and the molecule or atom it is in, are not sources of energy. In fact, once this photon-electron encounter is over, the molecule is in no physical way different from what it was before the encounter. That means that this process must make the initial photon disappear! It also means that, from a single photon-electron encounter, only a single photon may leave. But we have the notion that a photon is a particle that travels in some direction, and this would seem to contradict the notion that the oscillation of the electron produces circular outgoing electromagnetic waves. Here we run headlong into one of the deep mysteries of the interaction of light with matter: Some phenomena of light must be explained in terms of the photon concept, the photon traveling in a definite direction. Other phenomena must be explained in terms of the wave concept, where a wave may be curved, just like water waves.
Here's how we resolve this apparent contradiction. In any single photon-electron encounter, the original photon disappears, and is replaced by a single outgoing photon, which proceeds away from the location of the encounter in a straight-line path whose direction lies more or less in the plane perpendicular to the direction of the polarization of the original photon. The next photon that encounters the same electron will elicit the same response, producing yet another single outgoing photon, which heads off in a straight line path, in the same plane, but in some other direction. Another photon comes by, and another, and another. Each is scattered in some essentially random direction in this plane. Actually, choose any two directions in this plane, and an outgoing photon is equally likely to scatter into one as into the other (even right back in the opposite direction of the original photon). So if you look at a single photon-electron encounter, you will see single photons leaving the encounter in definite directions, but if you look at the total of all the photons that leave, they form the circularly symmetric distribution of outgoing electromagnetic radiation suggested by the analogy with the pond water waves.
Let's do a thought experiment. If you sit in the plane of the outgoing photons, and look at the electron as it is responding to an incoming photon, you "see" the electron moving up and down. And when an outgoing photon comes past you, its plane of polarization is in that same direction. If you now move upward, out of the plane, just a bit, and look again, you can still see the electron moving up and down. But now, because of the fact you are looking at the electron from an angle, it does not seem to move as much as it did when seen from in the plane. The higher up you go, the less it appears to move. If you look down at the electron from the top, you can not tell it's moving at all. What does this signify?
If, when looking at the moving electron from a point at some distance away from it, you see it moving up and down, then some of the electromagnetic radiation its motion is producing is coming toward you. Because the amplitude of the motion is less the further out of the scattering plane you sit, the intensity of that radiation is less. In fact, it is less by a factor that's the cosine of the angle the line from you to the electron makes with the plane. When you are 90o out of the plane (i.e. looking down on the electron), you see no motion. The cosine of 90o is 0, so this says that no radiation is scattered in that direction.
Now, we take the same step we did before to get from the electromagnetic wave concept to the photon concept. We observe that the electromagnetic wave is propagating away from the moving electron not only in the plane, but in almost all other directions as well. As for the photons, an individual photon can only be going in one direction away from the moving electron. If we watch lots and lots of photons undergoing Rayleigh scattering from the electron, we will see some leave pretty much in the scattering plane, we will see others (somewhat fewer) that are leaving along lines that make a 20 degree angle with the scattering plane. We will see somewhat fewer still leaving along lines that make 80 degree angles with the scattering plane. The relative numbers that leave at some angle are actually proportional to the square of the cosine of that angle. The following figure shows the relative amounts of scattering in the various directions.
Amplitude of Rayleigh scattering by an atom or molecule (at the center of the "doughnut"). The electromagnetic waves are coming in from the left, and are polarized with their electric fields in the vertical direction. There is no scattering into the vertical direction. Choose a point on this surface, and draw a line from that point to the center. The distance between the points is proportional to the intensity of the radiation scattered in the direction of the chosen point. The intensity is largest in any direction in the plane perpendicular to the plane of polarization.
In the discussion above, we have described the relationship of the polarization of the outgoing photon to that of the original photon. The sunlight falling on the top of the Earth's atmosphere is almost completely unpolarized. That is, if you choose any two lines perpendicular to the light's direction of travel, there will be just as many photons aligned with one of the lines as with the other. However, due to the Rayleigh scattering in the atmosphere, the light that reaches the Earth's surface, or that are scattered back into space, is partially polarized. That is, if we look at a large number of photons that are scattered into some direction, not all the photons will have exactly the same polarization. To quantify this, we discuss the degree of polarization in the Rayleigh scattered light.
Suppose we have a sample of a Rayleigh-scattering material, and we introduce an incoming beam of light. If the scattering angle (i.e. the angle between the incoming and outgoing photon directions) is , then the degree of polarization is proportional to sin 2 θ. This means that the degree of polarization is greatest when the angle between the incoming and outgoing photon directions is 90o.
As mentioned elsewhere, the short answer to the everlasting question, Why is the sky blue? is that there is Rayleigh scattering of the Sun's light by the molecules in the air. You can see the polarization effect for yourself. Find a pair of polaroid sunglasses or a polarizing filter such as is used in photography. On a clear day, when the Sun is low in the sky, and when the sky does not look very hazy (whitish glow all around the horizon), look at the sky through the polarizing filter or one of the lenses of the sunglasses. First, look straight up. Most of the light you will see has been scattered through an angle of roughly 90o. Now rotate the polarizer (the lens or filter) slowly, and notice what happens to the light. Note the orientation of the polarizer when the greatest amount of light is seen. Now look at different regions of the horizon, relatively near the Sun (don't look right at the Sun!), at right angles to the Sun, and in the opposite direction. Every time, choose a direction, look in that direction through the polarizer, and rotate the polarizer slowly, noting any brightening or darkening effect.
It is important to realize that the light that arrives at the top of the atmosphere from the Sun is unpolarized: Just as many photons have their planes of polarization in any direction as in any other direction. The degree of polarization of the light reaching the surface of the Earth, or reaching some satellite instrument in space, is due to the Rayleigh scattering process in the atmosphere.
Some creatures, notably the hymenoptera family of insects (bees, wasps, and hornets), have eyes with built-in polarizers, and they use the polarization of the sky light to navigate. Because these creatures' eyes are responsive further into the ultraviolet than our own, and because the ultraviolet light is even more strongly Rayleigh-scattered than the blue-violet light we can see, they actually see a greater contrast in sky brightness as they rotate than we see when we rotate the polarizer. (Their ultraviolet vision also helps them see flower pigments.)
Rayleigh Scattering Cross Sections
Not every photon that passes by a molecule, with its semi-mobile electrons, gets Rayleigh scattered. The mathematical theory of Rayleigh scattering must take into account the wavelength of the light, the size and shape of the region the semi-mobile electrons can move around in, how tightly bound they are to the atom's nucleus, and how many of them there are in a molecule. The development of this theory is beyond the level of this lecture. However, the result is essentially this: The cross section for scattering is most sensitive to the wavelength, and goes as
Visible light has wavelengths in the range of about 400 nm to 700 nm. So, according to this law, blue light (at 400 nm) is roughly 9 times as strongly scattered as red light. Of course, infrared light is very weakly Rayleigh-scattered, while ultraviolet is even more strongly scattered than any visible light.
The Rayleigh scattering cross section as a function of wavelength, through the UVA, UVB, and visible regions of the spectrum. The greater the cross section, the more efficient the molecules are at scattering the light. The scattering cross section is about 9 times as great at the blue-violet end of the visible spectrum than at the red end. Over the UV region, the scattering cross section increases by a futher factor of about 3.