Rayleigh scattering
From Wikipedia, the free encyclopedia
Rayleigh scattering (named after Lord Rayleigh) is the scattering of light, or other electromagnetic radiation, by particles much 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 by the atmosphere is the main reason light from the sky is blue.
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 articles on optics and scattering.
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[edit] Derivations using particle theory
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 spherical particles larger than about a tenth of the illuminating wavelength is explained by the Mie theory.
The intensity I of light scattered by a single small particle from a beam of unpolarized light of wavelength λ and intensity I0 is given by:
where R is the distance to the particle, θ is the scattering angle, n is the refractive index of the particle, and d is the diameter of the particle.
The angular distribution of Rayleigh scattering, governed by the (1+cos2 θ) term, is symmetric in the plane normal to the incident direction of the light, and so the forward scatter equals the backwards scatter. Integrating over the sphere surrounding the particle gives the Rayleigh scattering cross section σs:
The Rayleigh scattering coefficient for a group of scattering particles is the number of particles per unit volume N times the cross-section. As with all wave effects, in incoherent scattering the scattered powers add arithmetically, while in coherent scattering, such as if the particles are very near each-other, the fields add arithmetically and the sum must be squared to obtain the total scattered power.
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. It should be noted that, despite the use of the term photon, Rayleigh scattering was developed prior to the invention of quantum mechanics and is not based fundamentally in modern theory about the interaction of light with matter. Nevertheless, Rayleigh scattering is a fair approximation to the manner in which light scattering occurs within various media.
[edit] An explanation of Rayleigh scattering using the S-matrix
The S-matrix is used in describing mathematical representation of particles scattering off each other in quantum mechanics. One starts with a set of incoming particles (the whole spectrum of photons in sunlight, and essentially stationary oxygen/nitrogen molecules in the atmosphere) and wants 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 square of the S-matrix, plus a few details regarding cross sections, contains a description of the forces or interactions between the particles. It is a matrix because the initial and final states of quantum objects (these initial and final states are also called scattering channels) 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 quantum spin.
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.
- 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.
- 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.
- 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 did not 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.
[edit] Why is the sky blue?
During sunrise and sunset 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.
Because of the strong wavelength dependence (inverse fourth power) of light scattering according to Rayleigh's Law, one would expect that the sky would appear more violet than blue, the former having a shorter wavelength than the latter. There is a simple physiological explanation for this apparent conundrum. It turns out that the human eye's high resolution color-detection system is made of proteins and chromophores (which together make up photoreceptor cells or "Cone" structures in the eye's fovea) that are sensitive to different wavelengths in the visible spectrum (400 nm–700 nm). In fact, there are three major protein-chromophore sensors that have peak sensitivities to yellowish-green (564 nm), bluish-green (534 nm), and blue-violet (420 nm) light. The brain uses the different responses of these chromophores to interpret the spectrum of the light that reaches the retina.
When one experimentally plots the sensitivity curves for the L, M, and S color sensors, three "bell-curve" distributions with peaks at each of these wavelengths are seen to overlap one another and cover the visible spectrum. We depend on this overlap for color sensing to detect the entire spectrum of visible light. However, when one looks at the bell-curve sensitivity of the short wavelength (S) color sensor, there is a narrow detection band with a rapid fall-off in sensitivity around 450 nm. This means that our eyes are many times less sensitive to violet light (400 nm) than to blue light. As a result, although the scattered light contains more violet photons than blue ones, the physiological makeup of the eye gives a more pronounced contribution to the blue-ness of the sky.
- See also: Why is the sun yellow?
[edit] See also
[edit] References
- Ditchburn, R.W. (1963). Light, 2nd ed., London: Blackie & Sons, pp582–585.
