1. Radiative Transfer Virtually all the exchanges of energy between the earth-atmosphere system and the rest of the universe take place by radiative transfer. The earth and its atmosphere are constantly absorbing solar radiation and emitting their own radiation to space. Over a long period of time, the rates of absorption and emission are very nearly equal, thus the earth-atmosphere system is very nearly in equilibrium with the sun. Radiative transfer also serves as a mechanism for exchanging energy between the atmosphere and the underlying surface, and among different layers of the atmosphere. Radiative transfer plays an important role in a number of chemical reactions in the upper atmosphere and in the formation of photochemical smogs. The transfer properties of visible radiation determine the visibility, the color of the sky and the appearance of clouds. Radiation emitted by the earth and atmosphere and intercepted by satellites is the basis for remote sensing of the atmospheric temperature structure, water vapor amounts, ozone and other trace gases. 2. Spectrum of Radiation Electromagnetic radiation may be viewed as an ensemble of waves propagating at the speed of light (c = 2.998 10 8 m/s through vacum). We characterize radiation in terms of: frequency ν = c /λ wavelength λ = c /ν Radiative transfer in planetary atmosphere involves an ensemble of waves with a continuum of wavelengths and frequencies. We partition them into bands : shortwave (λ < 4µm) carries most of the energy associated with solar radiation or longwave (λ > 4µm) which refers to the band that encompasses most of the terrestrial. The visible region 0.39 0.79µm is defined by the range of wavelengths that the human eye is capable of sensing, and subranges of the visible are discernible as colors. 3. Definitions Solid Angle ω Consider a cone with its vertex at the origin of a concentric spherical surface. The solid angle is defined as the ratio of the area of the sphere 1
Figure 1: Ahrens, Chapter 2 2
intercepted by the cone to the square of the radius. ω = A r 2 (1) In spherical coordinates dω = da r 2 (2) da = r 2 sinθdθdφ (3) The unit of the solid angle is the steradian. The area cut out of a sphere by one steradian is equal to the square of the radius. Integration over the entire spherical surface then gives ω = 4πsteradians Monochromatic flux density F λ Amount of radiant energy with a given wavelength passing through a unit area per unit time. Expressed in [W m 2 ]. Total flux density is calculated as the integral over all wavelengths F = λ2 λ 1 F λ dλ = ν 2 ν 1 F ν dν. Monochromatic Intensity (or Monochromatic Radiance) I λ Radiant energy in a specific wavelength per unit time coming from a specific direction and passing through a unit area perpendicular to that direction. Total intensity is calculated as the integral over all wavelengths I = λ 2 λ 1 I λ dλ = ν 2 ν 1 I ν dν. The units are [W m 2 steradian 1 ].The intensity I λ and flux density F λ are related by I λ = df (4) dωcosθ We can integrate over the solid angle subtended by a hemisphere to determine the monochromatic flux density coming from all directions F λ = 2πsteradians 4. Blackbody Radiation 0 I λ cosθdω = 2π 0 dφ π/2 A blackbody is a surface that absorbs all incident radiation. 0 I λ cosθsinφdφ (5) 3
The Planck Function Determined experimentally, the intensity of radiation emitted by a blackbody is c 1 λ 5 B λ = (6) π(e c 2/λT 1) where c 1 = 3.74 10 16 W m 2 and c 2 = 1.45 10 2 mk. Theoretical justification of this empirical relationship led to the development of the theory of quantum physics. Wien s Displacement Law Differentiating Equation 6 and setting the derivative equal to zero, gives the wavelength of peak emission for a blackbody at temperature T (HW6). λ m = 2897 (7) T where T in K and λ m in µm. An important consequence of Wien displacement law is the fact that solar radiation is concentrated in the visible and near-infrared parts of the spectrum, while radiation emitted by the planets and their atmospheres is largely confined to the infrared. The nearly complete absence of overlap between the curves justifies dealing with solar and planetary radiation separately in many problems of radiative transfer. Stefan-Boltzmann Law The black body flux density obtained by integrating the Planck function πb λ over all wavelengths. F = σt 4 (8) Where σ is the Stefan-Boltzmann constant equal to 5.67 10 8 W m 2 K 4 4
5. Radiative properties of Nonblack materials Unlike blackbodies, which absorb all incident radiation, nonblack bodies such as gaseous media can also reflect and transmit radiation. We will give a brief description of the radiative processes in nonblack bodies. The fate of radiation depends on wavelength. 1. Transmitted Radiation passes undisturbed. We define the monochromatic fractional transmissivity as T λ = I λ(transmitted) I λ (incident) 2. Reflected Radiation. Reflectivity is depicted by albedo. Albedo usually represents all wavelengths and refers to the earth-atmosphere reflection. We define the monochromatic fractional reflectivity as R λ = I λ(transmitted) I λ (incident) 3. Absorbed Increase in internal energy of the object. We define the monochromatic fractional absorptivity as α λ = I λ(absorbed) I λ (incident) (a) Ionization-Dissociation Interactions. In these interactions, an electron is stripped from an atom or molecule, or a molecule is torn apart. These interactions occur primarily at ultraviolet and shorter wavelengths. All solar radiation shorter than about 0.1 µm in wavelength is absorbed in the upper atmosphere by ionizing atmospheric gases, particularly atomic oxygen. Between 0.1 and 0.2 µm molecular oxygen dissociates into atomic oxygen. Radiation between 0.2 and 0.3 µm is absorbed by dissociation of ozone. These bands are important for preventing the radiation from reaching the ground and in satellite meteorology for measuring ozone concentrations. (b) Electronic Transitions Orbital electron jumps between quantized energy levels. These occur mostly in the UV and visible. Ozone, and molecular oxygen. (c) Vibrational transitions A molecule changes vibrational energy states. These transitions occur mostly in the infrared portion of the spectrum and are extremely important for satellite meteorology. The two chief absorbers in the infrared region of the spectrum are carbon dioxide and water vapor. Symmetric stretching has neither a static or dynamic electric dipole moment because the symmetry of the molecule is maintained. If a molecule has no electric dipole moment, the electric field of incident radiation cannot interact with the molecule. (This is why 5
N 2 and O 2, the two most abundant gases in the atmosphere, are transparent in the infrared. (d) Rotational transitions a molecule changes rotational states. These occur in the far infrared and microwave portion of the spectrum. They can occur at the same time as vibrational transitions. Figure 4.7. The three are related by α λ + R λ + T λ = 1. For a black body α λ = 1. 5a. Other Definitions Total Mass Extinction Coefficient If a beam of intensity I λ becomes I λ + di λ upon traversing a distance ds in its direction of propagation through a medium of density ρ, then the reduction of intensity due to extinction (which could be absorption, reflection, scattering, diffraction, refraction etc.) is di λ = k λt ρi λ ds (9) where k λt is the total mass extinction coefficient and has units [m 2 kg 1 ]. We define the optical depth τ λ in terms of the total mass extinction coefficient as follows: τ λ = s 0 k λt ρds Mass Absorpion Coefficient If a beam of intensity I λ becomes I λ + di λ upon traversing a distance ds in its direction of propagation through a medium of density ρ, then the reduction of intensity due to absorption is: di λ = k λa ρi λ ds (10) where k λa is the mass absorption coefficient and has units [m 2 kg 1 ]. We define the absorption optical depth τ λa as τ λa = s 0 k λaρds Mass Scattering Coefficient If a beam of intensity I λ becomes I λ + di λ upon traversing a distance ds in its direction of propagation through a medium of density ρ, then the reduction of intensity due to scattering is: di λ = k λs ρi λ ds (11) where k λs is the mass scattering coefficient and has units [m 2 kg 1 ]. We define the scattering optical depth τ λs as τ λs = s 0 k λsρds It follows that k λt = k λa + k λs 6
5b. Kirchoff s Law It can be shown that the radiation emitted by a given material is a function of temperature and wavelength only. Consider an opaque, hollow enclosure with zero transmissivity into which is placed a slab of finite thickness. In general, this slab will reflect, absorb and transmit parts of the incident radiation. In addition, it will emit radiation itself. We now allow the enclosure and the slab to reach thermodynamic equilibrium, such that the slab and the enclosure walls are the same temperature. Under this condition, the flow of energy in all directions must be the same. In thermodynamic equilibrium, the amount entering the slab must exactly equal the amount leaving, or there would be a net flow of heat to or from the walls, into or out of the slab. Since the slab and the walls are in thermodynamic equilibrium, this would constitute a violation of the Second Law of Thermodynamics. Therefore, the balance equation is: I λ R λ I λ = T λ I λ + E λ (12) Where E λ is the emitted radiance in the same direction as I λ. But T λ I λ = I λ (1 α λ R λ ) since α λ + R λ + T λ = 1. Therefore, I λ (1 R λ ) = I λ (1 α λ R λ ) + E λ (13) Thus, E λ α λ I λ = 0 or E λ = α λ I λ Thus, inside of an opaque, hollow enclosure in thermodynamic equilibrium, the amount emitted by the slab equals the amount absorbed by the slab. We now imagine our enclosure to be replaced by a different one, constructed from a different material, and again allow it to come into thermodynamic equilibrium with the same slab and at the same temperature as before. Consequently, the slab emission will be the same as before, since it depends only on temperature and wavelength, neither of which has been changed. Similarly, the slab absorption will not change because the slab material is the same. Thus we have: E λ = α λ I λ (14) Where I λ is the incident radiation on the slab in the new enclosure, thus it follows that I λ = I λ.thus, the radiation within an opaque, hollow enclosure is independent of the material from which the walls are made. Re-writing the above E equation we see λ = I α l ambda λb = f(t, λ) only and I λb is the radiance inside an opaque hollow enclosure at temperature T and wavelength λ. 7
This result is known as Kirchhoff s Law, which states that The ratio of the emission to the fractional absorptivity of a slab of any material in a state of thermodynamic equilibrium and at wavelength λ is equal to a constant. We may now define the fractional emissivity ɛ λ as the ratio of the radiation emitted at the wavelength λ to that within a hollow enclosure at the same temperature or: ɛ λ = E λ I λb (15) From this definition, we see that E λ = ɛ λ I λb. But from Kirchoff s Law it then follows: ɛ λ = α λ (16) Or the fractional emissivity equals the fractional absorptivity Kirchoff s Law is fundamental to further development of the subject of radiative transfer, and is frequently applied in a variety of applications. Recalling that it is strictly valid only under conditions of thermodynamic equilibrium, it is nevertheless generally assumed to be valid for atmospheric problems even though the atmosphere is not strictly in thermodynamic equilibrium. We may now carry this thought experiment one step further. Let s replace this slab by an ideal black body such that, by definition, it completely absorbs all radiation falling on it. Inside the hollow enclosure then, the radiation leaving the black body slab consists entirely of radiation emitted by the slab. The equilibrium condition becomes I λb = E λ (17) leading to the important conclusion that the radiation flowing in any direction within the hollow enclosure in thermodynamic equilibrium is equal to the energy emitted in the same direction as an ideal black body. Such radiation is called black body radiation, and from our earlier arguments is isotropic or equal in all directions. 8
6. Examples 1. Prove that the intensity I of solar radiation is independent of distance from the sun, provided that the distance is large and that radiation emitted from each elemental area on the sun is independent of the zenith angle. 2. The average flux density F e of solar radiation reaching the earth s orbit is 1370W m 2. Nearly all the radiation is emitted from the outermost visible layer of the sun, which has a mean radius of 7 10 8 m. Calculate the equivalent blackbody temperature or effective temperature of this layer. The mean distance between the earth and sun is 1.5 10 11 m. 3. Calculate the equivalent blackbody temperature of the earth assuming a planetary albedo α p = 0.3 where α p is the fraction of the total incident solar radiation that is reflected and scattered back to space. Assume that the earth is in radiative equilibrium. 4. A completely gray flat surface on the moon with an absorptivity of 0.9 is exposed to direct overhead solar radiation. What is the radiative equilibrium temperature of the surface? If the actual temperature is 300K, what is the net flux density above the surface? 5. A flat surface is subject to overhead solar radiation as in the previous example. The absorptivity is 0.1 for solar radiation and 0.8 in the infrared part of the spectrum, where most of the emission takes place. Compute the radiative equilibrium temperature. 6. Calculate the radiative equilibrium temperature of the earth s surface and atmosphere assuming that the atmosphere can be regarded as a thin layer with absorptivity of 0.1 for solar radiation and 0.8 for terrestrial radiation. Assume that the earth s surface radiates as a black body. 9