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EMR's Interaction with the Earth & Atmosphere ∝ ∝ ∝ EMR’s interaction with the earth & atmosphere HANDOUT’s OBJECTIVES: • familiarize student with EMR scattering basics in environment • develop basic principles of radiative transfer in environment • develop Lambert’s law & Schwarzschild equation, plus introduce remote-sensing applications of same Lord Rayleigh & sky blueness “Why is the sky blue?” is a good (and hardly trivial) starting point for studying EMR’s interactions with the earth’s atmosphere and oceans. Water (whether vapor or liquid) is often invoked to explain sky blueness. After all, if ocean water is blue, shouldn’t that explain the sky’s color? Unfortunately, several meters of liquid water are required for water’s blue-green absorption minimum to be visible. If we condensed all atmospheric water vapor into a uniform shell around the earth, the shell would only be a few centimeters thick. Visible scattering by water molecules differs little from that of other atmospheric gases. (At the molecular level, we distinguish between the scattering, or redirecting, of EMR and its absorption, or conversion to other wavelengths (or kinds) of energy.) We retrace the ideas of 19th-century British scientist Lord Rayleigh, who showed that atmospheric molecules themselves are responsible for clear-sky color. Rayleigh Rp reasoned that if a spherical particle’s radius Rp << than that of λlight ( < 0.1, the λlight Rayleigh scattering criterion) its total scattering is ∝ its volume V. So when an EMR wave “illuminates” a molecule, scattering by its atoms is in phase because they are within λlight of each other. Thus the amplitude of the scattered electric field Es ∝ V. Clearly Es also depends on the strength of the incident electric field Ei. Rayleigh knew that energy conservation required the intensity (AKA irradiance) of the scattering to decrease with 1 distance r from the particle as . In turn, this means that the amplitude E r2 s decreases as 1/r. What else describes Es/Ei? Dimensional balance requires a variable with the units of length2, and λ2 is a plausible choice. So 2 Es V Es 2 V ∝ and intensity [ ] ∝ 2 4 . Ei r λ 2 Ei r λ 3 Because each spherical particle’s V = (4/3) πRp , relative scattering per particle is 3 2 6 Es 2 (Rp ) Rp [ ] ∝ 2 4 ∝ 2 4 . Ei r λ r λ -2- Rayleigh’s arguments here are simply those of dimensional consistency and plausible (but not rigorous) use of EMR variables. Is it possible to get a similar answer from another tack? Remember earlier that we said Ei = E0 sin(ωt − δ) describes the time and space dependence of an electric vector’s amplitude (ω = 2πν, the angular frequency). We take the second time derivative of Ei to be a measure of the acceleration (Es) that the EMR wave exerts on a charged particle (a scatterer). Thus: 2 ∂ Ei 2πc 2 E ∝ = -ω2 E = -[ ] E , s ∂t2 i λ i and the relative scattered intensity is given by: Es 2 2πc 4 [ ] ∝ [ ] . Ei λ 6 Es 2 Rp Rayleigh’s relationship [ ] ∝ 4 has two important consequences for Ei λ remote sensing: 1) It describes the λ-dependence of visible scattering by molecules in the atmosphere as ∝ λ−4, meaning that the atmosphere’s gaseous constituents scatter blue light (small λ) more efficiently than red light (larger λ). At the crudest level, this explains the blue sky. 2) It describes the raindrop size-dependence of radar scattering by rain. In 6 other words, the reflected radiance of radar beamed at spherical raindrops is ∝ Rp , meaning that N large raindrops will scatter much more than N smaller raindrops. Although there are fewer large raindrops than small drops per volume of rainy air, the 6 Rp factor usually means that the larger drops will return a stronger radar signal. Note that we've jumped over an enormous range of particle sizes above. If an average atom’s nuclear radius is ~ 10-15 m and a typical raindrop radius is 10-3 m, our examples of Rayleigh scatterers span 12 orders of magnitude. In describing scattering, the ratio of particle size to wavelength is more important than absolute sizes. A measure of relative size is the size parameter x: 2πR x = p (Eq. 1). λ SO431 — EMR’s interaction with the earth & atmosphere (4-2-2010) -3- R Rayleigh’s theory assumes that p < 0.1. When this Rayleigh scattering λlight criterion isn’t met, Rayleigh theory is inadequate and we must use the more general (and much more complicated) Mie theory. Below are Mie theory’s predictions about spectral scattering by a wide range of water-drop sizes. As a subset of Mie theory, Rayleigh theory holds only for the smallest drops (see bottom edge of nearest plane). Because of its power, Mie theory is used routinely in remote sensing applications. 3.0 m blue µ red 0.15 µm visible-λ scattering by 0.15-3.0 µm radius water drops SO431 — EMR’s interaction with the earth & atmosphere (4-2-2010) -4- Rayleigh’s equation for scattering by atmospheric molecules says their scattering −4 efficiency is ∝ λ . Since λviolet < λblue, why isn’t the sky violet rather than blue? Our answer refers to human vision, but can easily be extended to satellite sensors. 1) Our visual systems are not equally sensitive to all wavelengths of light. In particular, note the low response at the violet end of the spectrum. 1 1 y t 0.1 i 0.1 v i t i s n e 0.01 0.01 s peak response l a at 555 nm u s i 0.001 0.001 v n a -4 m 10 10-4 u h -5 10 10-5 380 420 460 500 540 580 620 660 700 wavelength (nm) 2) As we have seen earlier, the light source for skylight — sunlight — has less power in the short wavelength end of the visible spectrum. 3) Skylight does not consist of light of a single wavelength, but rather is a mixture of many Lλ. Our visual system converts this mixture of monochromatic radiances into three different signals, each averaged over a part of the visible spectrum. 1 1 y c n e i For visible-λ molecular (i . e . , c 0.8 0.8 i f f -4 e Rayleigh) scattering, E /E ∝ λ sct inc g n 0.6 0.6 i r e t t a c 0.4 0.4 s r a l u 0.2 c 0.2 e l o m 0 0 380 420 460 500 540 580 620 660 700 wavelength (nm) SO431 — EMR’s interaction with the earth & atmosphere (4-2-2010) -5- 4) We have neglected multiple scattering’s effect on sky color (i.e., repeated scattering of a photon before it reaches us). By describing scattering at the molecular level, we have tacitly implied that scattering by many molecules is the same as one molecule’s single scattering. Multiple scattering is crucial to correctly interpreting both our ground-based views of the atmosphere and satellite images taken through it. In the atmosphere (or any other suitably dense medium), not only direct sunlight illuminates each molecule; so does sunlight scattered by other molecules. As shown below, an observer receives light scattered by molecules and particles all along a given line of sight. parallel sunlight “top” of atmosphere d directi ire on A cti B´ o A´ n B observer’s eye surface Note that the pathlength in direction A > pathlength in direction B. This simple fact explains a common observation: if molecular scattering ∝ λ-4, why is the clear sky’s horizon white rather than blue? The answer begins with the fact that atmospheric pathlengths increase as you look toward the horizon.. What you see along paths A and B above is the sum of all scatterings along those paths. Consider how this works to produce the white horizon: -4 (1) Preferential blue scattering (∝ λ ) is identical at points A´ and B´. (2) But each subsequent scattering along A and B (i.e., closer to the eye) preferentially removes blue light first scattered at A´ and B´. The net result is that nearby scatterers contribute a larger amount of blue skylight to the total reaching your eye, while more distant scatterers contribute a smaller amount of reddened skylight. (3) Because A > B, in direction A you see more scattered photons (i.e., brighter skylight) that are less dominated by blue (i.e., slightly reddened skylight). (4) For the longest pathlengths (at the horizon), this produces a brighter, whiter sky. SO431 — EMR’s interaction with the earth & atmosphere (4-2-2010) -6- To analyze the multiple-scattering problem of horizon whiteness quantitatively, we begin by distinguishing between single- and multiple-scattering media. Our analysis will also tell us how satellites are affected by multiple scattering in the atmosphere. We assume that all scatterers and absorbers in an idealized homogeneous medium are identical spheres of radius Rp that are uniformly distributed. Consider a slab of thickness h and area A that’s lit by light of intensity I0. The number of scatterers per unit volume in the slab is the particle number density N. scatterer of I0 I0 {horizontal surface radius R area = A} p h I I Every photon encounter with a particle in the slab results either in absorption or scattering (which together we call extinction), and except when the photon is scattered exactly in the direction it was originally headed (the forward direction), every encounter causes a reduction in I0. Over area A and infinitesimal thickness dh in the slab there are N A dh particles (i.e., {number/volume}*area*thickness = {number/volume}*volume = number of particles).
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