Rayleigh scattering
|
|
Rayleigh scattering (named after Lord Rayleigh) is the scattering of light by particles smaller than the wavelength of the light. It occurs when light travels in transparent solids and liquids, but is most prominently seen in gases. Rayleigh scattering of sunlight from particles in the atmosphere is the reason why the light from the sky is blue.
The amount of Rayleigh scattering that occurs to a beam of light is dependent upon the size of the particles and the wavelength of the light; in particular, the scattering coefficient, and hence the intensity of the scattered light, varies inversely with the fourth power of the wavelength, a relation known as the Rayleigh law. Scattering from particles of about the same size as the wavelength is handled by Mie theory. The angular intensity polarization relationships for Rayleigh scattering are conveniently simple. For particles not larger than the Rayleigh limit, there is complete symmetry in the probability of scattering with respect to a plane normal to the direction of the incident radiation, so that the forward scatter equals the backward scatter. The Rayleigh scattering coefficient ks is
- <math> k_s = \frac {2 \pi^5} {3} n \left( \frac {m^2 - 1} {m^2 + 2} \right)^2 \frac {d^6} {\lambda^4} <math>
where n is the number of scatters of diameter d; m is the index of refraction; and λ is the wavelength of the radiation.
The strong wavelength dependence of the scattering (~λ-4) means that blue light is scattered much more than red light. In the atmosphere, this results in blue photons being scattered across the sky to a greater extent than photons of a longer wavelength, and so one sees blue light coming from all regions of the sky whereas the rest is still mainly coming directly from the Sun.
A notable exception occurs during sunrise and sunset, when the Sun's light must pass through a much greater thickness of the atmosphere to reach an observer on the ground. This extra distance causes multiple scatterings of blue light, but relatively little scattering of red light; this is seen as a pronounced red-hued sky in the direction towards the sun.
If the size of particles are larger than the wavelength of light, light is not separated and all wavelengths are scattered as by a cloud which appears white, as do salt and sugar. For scattering by particles similar to or larger than a wavelength, see the article on Mie scattering.
An explanation of Rayleigh scattering using the S-matrix
What it refers to is the mathematical representation of particles scattering off ("hitting", in a way) each other in quantum mechanics. If you start with a set of incoming particles (in the student's example, the whole spectrum of photons in sunlight, and essentially stationary oxygen/nitrogen molecules in the atmosphere) and want to know the probability of the particles being in a certain configuration after scattering (where the different wavelength photons go after hitting the molecules), the S-matrix is your guy...well, the square of it, plus a few details. It contains a description of the forces between the particles, and is a matrix because it turns out that in Einstein's relativity, quantum objects can easily be expressed as vectors, and matrices are operations performed on vectors - so, forces acting on particles, in a simple way to think of it. To use the S-matrix, we have to input information about the particles that will scatter, such as their speed and direction, polarization which is really spin (a quantum mechanical property; they're not really "spinning").
For the example of sunlight shining on the atmosphere, the S-matrix predicts that shorter-wavelength light (blue end of the spectrum) will scatter at larger angles than longer-wavelength light (red end of the spectrum). And this is exactly what we see! Let me go through it. It helps to have a globe handy, perhaps using a pencil or straight piece of wire to simulate an incoming ray of sunlight; imagine a very thin layer over the surface which is the atmosphere. A small scattering angle means the light continues on nearly in the direction it started out in, while a large angle means close to perpendicular to the incoming direction.
One thing to realize is that most of the time photons completely miss a molecule, passing only close by. This means that sunlight scatters very little unless it travels through a lot of atmosphere. This is where the hands-on visualization will be useful.
(1) For the sun high overhead, sunlight goes through very little atmosphere, so little scattering takes place, which is why the sky close to the overhead sun in midday appears mostly white, the sun's color. The ray of sunlight travels through just the thickness of the atmosphere relative to the surface - the smallest amount of travel possible before getting to the surface.
(2) Again for the sun high in the sky, or at least not near the horizon, if you look in the sky far away from the sun it appears blue. This is the short-wavelength light that scattered nearly perpendicular to the sunlight. Realize that for some direction in the sky other than the sun, a ray of sunlight went from the sun toward the horizon in that direction, so you're looking nearly perpendicular (at a large angle) to that ray. Aim the pencil from the sun to a horizon relative to where you are standing on the globe to visualize this.
(3) For the sun low on the horizon, at morning or evening, a ray of sunlight travels through a lot more atmosphere to get to you than when it's high overhead. Hence, a lot more light scattering takes place between the sun and you. Hold the pencil tangent to your position on the surface of the globe, and imagining a thin layer of atmosphere, this is easy to see. Since there is more scattering over this distance, the color enhancement is quite strong! Looking up, we see dark blue to purple,the very shortest visible wavelengths of light scattered perpendicular to the ray's incoming direction. Looking toward (but not at!) the sun, the sky appears very red. Out of the (nearly) white light coming from the sun, if the blue light is scattered perpendicular to the ray, then the rest of the ray must be the red component. This is indeed what we see!
The student has actually happened across one of the most beautiful problems in physics, scattering of sunlight. One can solve this problem with quantum mechanics of photons and electrons using the S-matrix, or using classical electrodynamics of light waves and molecules. The results are exactly the same! It is a particularly stunning example of mathematical beauty in physics: quantum mechanics, which one normally thinks of as applying only to objects too small to see, translates to classical physics at the macroscopic level.
This scattering of light even has names: Rayleigh scattering for blue light perpendicular to the incoming ray, and Mie scattering for the red light nearly along the incoming ray's direction. Actually they are the same thing, since quantum mechanics describes them both via the S-matrix, but this is an accident of history since Rayleigh and Mie didn't know about quantum mechanics or electrodynamics when these effects were first observed.
For the seriously curious student, another feature of this scattering is that the blue light is polarized, in the plane perpendicular to the incoming light ray. Try this out: with a sheet of polarizing film, hold it up against a very blue sky in a direction away from the sun. You may need to put it over the end of a shoebox and cut a hole in the other end of the shoebox, to get rid of the light in your peripheral vision. Rotate the film (shoebox) and notice how the light from that direction gets brighter and dimmer. It will be darkest when the polarizing direction of the sheet is aligned with the direction of that piece of sky toward the sun.
See also
External links
- HyperPhysics description of Rayleigh scattering (http://hyperphysics.phy-astr.gsu.edu/hbase/atmos/blusky.html#c2)
- Rayleigh scattering article from Eric Weisstein's World of Physics (http://scienceworld.wolfram.com/physics/RayleighScattering.html)de:Rayleigh-Streuung
fr:Diffusion Rayleigh nl:Rayleighverstrooiing pl:Rozpraszanie Rayleigha vi:Tán xạ Rayleigh