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# Intensity Wave Length Module 9 Radiative Transfer in the Atmosphere

9.1 Introduction It has been emphasized in earlier lectures that energy from Sun drives the circulation of the atmosphere and ocean. Earth receives energy from Sun as ultraviolet, visible and near- radiation. This radiation band is called the short radiation or radiation with wavelengths λ < 4 µm . An equal amount of energy is re-emitted by earth to maintain an overall energy balance. Earth emits energy in the form of infrared thermal radiation or longwave radiation with wavelengths λ > 4 µm . These two bands shown in Fig. 9.1 pertain to the wavelength intervals given as, Band 1: Solar 0.1− 4 µm (1 µm = 10−6 m) Band 2: Infrared region 4 −100 µm

Sun Earth

Fig. 9.1 Blackbody emission at the

temperatures of Sun and Earth. 6000 K 288 K

Intensity Overlap region Overlap

0.1 0.5 1 5 10 50 100 Wave ( µ m )

Microwave radiation is not important for energy balance of the earth, but it finds wide use in remote sensing of the earth system because it is capable of penetrating through clouds. The study of radiative transfer in atmosphere and ocean is important because it leads to (i) a better understanding of energy transfer in these two components of the climate system; and (ii) in interpreting remote sensing from satellites. Radiation is described by , intensity and ; and there are certain terms also that are frequently used while discussing the mathematics of the radiative transfer. First of all, it is required to calculate a differential amount of dEλ ( J ) in a wavelength ( µm ) interval λ and λ + dλ that crosses an ( m2 ) element dA in time () interval dt in a direction confined by an arc dω of the ( abbreviated as sr ). Some frequently used terms in radiative transfer are defined below and their dimensions (given in brackets) indicate how they are interrelated.

Radiant Energy: The energy of radiation beam at a given wavelength (J) . Flux: Radiant energy crossing unit area per unit time ( Js−1m−2 or Wm−2 ). Intensity: Flux in unit wavelength and unit solid angle (W m−2µm−1 sr−1 ) Flux density or : Normal component of intensity integrated over the hemispheric solid angle (W m−2µm−1 ). It is the emittance of an emitting surface. Extinction: Loss of energy from a due to absorption and scattering.

1 Radiative decay: Electron falls back to the original state by re-emitting photon of same energy and same . Photodissociation: Absorption of solar leading to breakdown of molecules initiating photochemical reactions and photo-ionization (i.e. outer electrons are stripped from atoms). Absorption: At sufficiently high pressures, molecular collisions are likely to occur and as a result of molecular collisions energy of excitation will be transferred to other forms of energy before re-emission could take place. In such situations, the photon is said to have been absorbed. Due to absorption of energy of excitation, the is produced, which is shared between molecules by collisional interaction. Since thermal energy is the macroscopic expression of molecular kinetic energy, the increased kinetic energy of molecules leads to local heating. In this process the photon energy has been transferred to heat. This process is also called “thermalization” or “quenching”. Scattering: Scattering of a photon of given energy and frequency by an atmospheric molecule implies that electron falls back to ground state re-emitting photon of same energy and frequency as the original one but in random direction. There are two kinds of scattering that occur in the atmosphere. Scattering by air molecules (Rayleigh) gives the blue colour to the sky; and scattering by aerosol particles (such as dust), called the Mie scattering, produces most scenic sunsets or sunrises. Thermal (infra-red) cooling: Thermal photons are also absorbed or scattered much like solar photons. When emitted, the energy is drawn from molecular kinetic energy leading to cooling of the atmosphere.

Thus, radiant energy dEλ and the intensity Iλ are related as follows:

dEλ = Iλ dAdλ dω dt (9.1)

9.2 Absorbing constituents in the atmosphere The main constituents that absorb radiation in different wavelength regions (or bands) are shown in Fig. 9.2. One may conspicuously note in this figure two atmospheric windows where the absorption is weak: one in the solar spectrum of the wavelengths 0.3 < λ < 1 µm (visible) and the other in the infrared region of wavelengths 8 < λ < 12 µm . Ozone (O3), water vapour (H2O) and carbon dioxide (CO2) are the dominant absorbers that primarily heat the atmosphere though oxygen (O2) also absorbs radiation in the ultraviolet and the visible region of the . In the solar spectrum ( λ < 4 µm ), O3 absorbs radiation in the ultraviolet and visible region; both CO2 and water vapour are greenhouse gases and absorb radiation in the near infrared region 1 < λ < 3.5 µm . In the infrared region, radiation is weakly absorbed by CO, CH4 and N2O in the troposphere, but CO2 and water vapour are the strong absorbers. In the atmospheric window region, 8 < λ < 12 µm , absorption is weak by these gases except for a band near λ = 9.6 µm which is associated with ozone (O3). The atmospheric window regions in the visible and infrared wavelength bands are important for remote sensing of atmosphere; that is why satellites are regularly launched by different countries in the space and their constellation forms an important part of the earth observing system. Electromagnetic radiation, a packet of different , is characterized by wavelength, frequency and wave number. Wavelength is denoted by λ; wave number is the reciprocal of wavelength; frequency is the product of wave number and the speed of

2 with which all electromagnetic waves propagate. Wavelength (λ), wave number (ν~) and frequency (ν) are thus related as: 1 ν c ν = = and ν = cν~ = , c is the . λ c λ

Fig. 9.2: The electromagnetic spectrum and the main atmospheric absorbents

Small variations in the speed of light in air produce mirage and distortions, which indeed limit the resolution of ground based telescopes. Moreover, differences in the speed of light in air and water produce beautiful rainbows. Wavelength is expressed in the units of micrometre ( µm ). Based on wavelength (or frequency or wave number), energy can be partitioned according to wavelength bands. However, for radiative transfer in the atmosphere, radiation is categorized into two wave bands: Shortwave radiation: λ < 4 µm energy associated with solar (downwelling) radiation Longwave radiation: λ > 4 µm energy associated with terrestrial (upwelling) radiation

Monochromatic radiation: refers to radiation of one single frequency or wavelength (i.e. single colour radiation). Similar meaning is also ascribed to monochromatic radiance. Visible Region: 0.39 ≤ λ ≤ 0.76 µm ; that is, the wavelengths sensed by the eye. Photon: Discrete packets of radiation; each photon contains energy E = hν .

9.3 Black-body radiation It is the radiation emitted from a small hole that has been cut in an isothermal cavity whose walls are maintained at a uniform temperature. The radiation is isotropic and the spectral energy density (the energy per unit volume per unit frequency interval) depends only on one frequency and cavity wall temperature. Thus, a blackbody surface completely absorbs all incident radiation to it. Most caves appear nearly black as all the sunlight rays that enter the cave are absorbed in multiple reflections and only a small fraction of sunlight (incident) emerges out from the entrance.

The Planck function: According to the Planck’s Law (the first fundamental law of radiation), spectral energy density of the blackbody radiation at temperature T is given by 8π hν 3 u (T ) = (9.2) ν c3 {exp[hν / (kT )]−1}

3 34 Here u (T ) is the spectral energy density of blackbody radiation; h 6.6262 ×10− J s ν = 23 1 (), k = 1.38062 ×10− J K − (Boltzmann constant) and c is the speed of 8 1 light (c = 2.99792458 ×10 ms− ). The energy density of a group of photons moving within a small arc of the solid angle Δω is given by uν Δω / (4π ) ; 4π is the solid angle of the sphere. Then the energy flow at speed c by this group of photons is given as

⎛ uν Δω ⎞ c uν c ⎜ ⎟ = = Bν (T ) (9.3) ⎝ 4π ⎠ Δω 4π

Using the expression of uν (T ) , we obtain an expression for the Planck function Bν (T ) as 2hν 3 B (T ) = (Planck Function) (9.4) ν c2 {exp[hν / (kT )]−1}

Bν (T ) is the per unit area, per unit solid angle, per unit frequency interval (spectral radiance) of blackbody radiation. The Planck function may also be expressed as a function of wavelength λ and temperature T, and it reads as: 2hc2 −3 −1 −2 −1 −1 . (9.5) Bλ (T ) = 5 (Wm sr or Wm sr µm ) λ {exp[hc / (λkT )]−1} Since h, c and k are constant, the intensity of radiation Bλ(T) may be written as: −5 c1λ 2 −16 2 Bλ (T ) = ; c1 = 2πhc = 3.74 ×10 W m ; c2 π exp −1 c = hc / k= 1.45 ×10−2 mK. (9.6) { λT } 2

Thus Bλ (T ) is the blackbody spectral radiance written in terms of power (W ) per unit area (m−2 ) , per unit solid angle (sr−1) , per unit wavelength interval (m−1 or µm−1) . Whenever radiation and matter interact; photons are absorbed, scattered or emitted by molecules of the optically active gas

If Bλ (T ) is integrated over all wavelengths, then one obtains the following expression for the blackbody radiance, ∞ σ 2k 4π 5 B (T )dλ = T 4 , σ = = 5.67 ×10−8 Wm−2K −4 (9.7) ∫ λ 3 2 0 π 15h c ∞ x3 π 4 In evaluating the above integral we use, dx = . Write the expression (9.7) as ∫ x 0 e −1 15

∞ ∞ dλ σ ∞ σ λB d(lnλ) = λB (T ) = T 4 ⇒ T − 4 λB (T )d(lnλ) = (9.8) ∫ λ ∫ λ λ π ∫ λ π 0 0 o

Both σ and π are constants; σ is called the Stefan-Boltzmann constant. Thus, the integral on the left in eq. (9.8) multiplied by T − 4 is a constant. With π=3.14159625 and 5.66961 10−8 W m−2K − 4 , the plotting of T − 4 B (T ) against ln suggests that the σ = × λ λ λ area under the resulting curve is independent of T . The curves of this kind are shown in Fig. 9.1, which have been produced for T = 6000 K (Sun) and T = 288 K (Earth).

4 The monochromatic spectral intensity I (or I ) is also expressed in units of λ v per square , per unit arc of solid angle, per unit wavelength in the electromagnetic spectrum. The intensity or the radiance I , in units of Watt per square metre per steradian (W m−2sr−1 ), is written as

λ2 v2 I = I dλ = I dv (9.9) ∫ λ ∫ v λ 1 v1 For energy balance of the Earth, the entire electromagnetic spectrum is considered. Separate integrations are performed for the shortwave part over wavelengths corresponding to incoming solar radiation (λ < 4 µm); likewise, over the wavelength corresponding to outgoing terrestrial radiation (λ > 4 µm), the integration is performed for the longwave part. Thus, integral I in (9.9) is proportional to the area under the spectrum plotted as a linear function of λ or v~ . From this fact, one may establish that

2 ⎛ 1 ⎞ Iv = λ Iλ or vIv = λ Iλ ⎜ v = ⎟ (9.10) ⎝ λ ⎠

The monochromatic flux density Fλ is a measure of the rate of energy transfer per unit area by radiation of given wavelength λ through a plane surface with specified orientation in the three-dimensional space. If the radiation is assumed to impinge upon a plane surface, then the flux density incident upon the surface (Fig. 9.3) is given by

F = I cosθ dω (Wm−2µm−1 ) (9.11) λ ∫ λ 2π In (9.11), dω is the elemental arc of the solid angle; θ is the angle between the incoming beam and the direction of the normal n to the area dA ; cosθ accounts for the slanted orientation of the beam relative to the surface. In radiative transfer computations, a spherical coordinate system (θ , ϕ) is used in which the axis of rotation of the earth points upwards and coincides with the zenith. The angle θ is the zenith angle and ϕ is azimuth angle. Thus, θ varies between 0 to π /2 , and ϕ varies between 0 to 2π . When Sun is at the zenith then it is at 900 to the surface dA . The amount of radiant energy falling on the area dA depends on the arc of the solid angle.

I(θ) n θ Fig. 9.3 Radiation beam at angle θ to an area dA; n is normal to the area dA. dA

Solid angle: A solid angle can be defined as the ratio of the area σ on the spherical surface at a point P at the centre of the sphere to the square of the radius (r2 ) . The calculation of the solid angle is illustrated in Fig. 9.4 where point P is the centre of the sphere. The point P is, for example, a point on the earth then we can calculate the solid angle subtended by whole sky (the semi spherical cap) by formal integration. Thus, solid angle subtended by an area element dσ is given as

5 dσ (rsinθdϕ)(rdθ) dω = = = sinθ dϕ dθ (9.12) r2 r2 The integration of the arc of the solid angle dω = sinθ dθ dϕ with given limits for θ and φ, will give the solid angle subtended by the whole sky (forming a semi spherical cap) when viewed from the point P on the horizontal surface as π /2 2π π /2 ∫ dω = ∫ ∫ sinθ dϕ dθ = 2π ∫ sinθ dθ = 2π (9.13) sky θ=0 ϕ=0 0

Fig. 9.4 The celestial sphere (sky blue colour) of radius r having centre at point P. The Earth occupies the position P in this celestial sphere. The arc of the solid angle dω subtended at point P by an area dσ is illustrated in the figure. The figure also shows the sides of the quadrilateral as rsinθ dϕ and rdθ ; thus the area dσ is given as dσ = r 2 sinθ dϕ dθ The plane containing the point P is a horizontal surface, which intersects with the celestial sphere in a circular disc (white colour) of radius r as shown in the figure.

The flux of radiation incident upon a plane surface is obtained by integrating the expression of Fλ in (9.11) over all the wavelengths, and we get the expression

λ2 F = F dλ = I cosθ dω dλ = I cosθ dω W m−2 (9.14) ∫ λ ∫ ∫ λ ∫ ( ) λ λ1 2π 2π The expression (9.14) is general and gives the integrated flux density. The expression is used to obtain the Solar constant, meaning the radiation flux reaching the Top-of-the- Atmosphere (TOA) from the Sun, and to calculate the infrared radiation flux leaving the earth-atmosphere system. Note difference in the units of F and F . λ

NP Fig.9.5 The parallel beam approximation: r Sundays falling on the earth surface are all . parallel.

P Sunrays falling on Earth on falling Sunrays

Parallel beam approximation: The radiation beam falling on the earth could be imagined as a solid cylinder of radiation (as illustrated in Fig. 9.5) of radius equal to that of the earth. The parallel beam approximation (PBA) therefore assumes that the sunrays falling on the spherical earth are unidirectional and parallel, i.e., zenith angle is zero. This

6 approximation is used for two-stream (upwelling and downwelling) solutions of the radiative transfer equation.

Relation between intensity and flux density of radiation: This is illustrated for the solar radiation reaching the ground. The basic data for establishing the relation between intensity and flux density of solar radiation are:

(i) Flux density Fs of solar radiation at TOA incident upon a horizontal surface at a zero zenith angle; it shows some variability as its values are: ⎧ −2 ⎪ 1366 Wm Fs = ⎨ or 1370Wm−2 ⎩⎪ (9.15) (ii) Radius of the sun and its distance from the earth R = 6.96 ×108 m Radius of the Sun s (9.16) d = 1.49598 ×1011 m Distance between the Sun and the earth

Let us assume that Is be the intensity of solar radiation which is isotropic; that is, the surface of sun radiates an amount Is from every point in all directions; then F = I cosθ dω (9.17) s ∫ s δω On applying the parallel beam approximation (PBA) the variations in cosθ may be ignored because the zenith angle under PBA is zero. Hence we obtain Fs as F F = I δω ⇒ I = s (9.18) s s s δω Now it remains to compute δω, the fraction of the solid angle ( 2π ) of the hemispherical dome subtended by sun’s disc at a point on the earth. The sun’s disc with its radius Rs 2 2 occupies an area π Rs on the surface area 2π d of the hemispherical dome of radius d. Thus, the ratio of these two is equal to the ratio of the corresponding sold angles that are subtended at the point P; hence from the simple arithmetic, we have

2 2 δω π Rs ⎛ Rs ⎞ −5 = 2 ⇒ δω = π ⎜ ⎟ = 6.84 ×10 sr (9.19) 2π 2πd ⎝ d ⎠ With this unique value of the solid angle δω , the intensity of the solar radiation is calculated from (9.18) as F 1370 I = s = = 2.003×107Wm−2 sr−1 (9.20) s δω 6.84 ×10−5 Intensity of radiation is constant along ray paths and it is independent of the distance from its source, i.e. the sun, in this case. From (9.18), is proportional to (because Fs δω I = Is is constant) at any point in the solar system, that is, −2 Fs c× δω ⇒ Fs c× d (inverse square law as R s is const.) (9.21) The inverse square law also follows from the fact that flux of solar radiation (Es) is independent of the distance from the sun, i.e. 2 − 2 Es = Fs × 4πd = const. ⇒ Fs c× d (Flux of solar radiation) (9.22)

7 Ex. 9.1: Radiation of uniform intensity is emitted from a plane surface in all directions. Determine the flux of radiation. Solution: Flux density integrated over the entire hemispheric solid angle gives the flux: 2π π /2 F = ∫ I cosθ dω = ∫ ∫ I cosθ sinθ dθ dϕ 2π ϕ=0 θ=0 π /2 π /2 or F = 2π I cosθ sinθ dθ = π I ⎡sin2 θ ⎤ ∫ ⎣ ⎦ 0 θ=0 or F = π I (Wm−2 )

−2 −1 That is, the flux of radiation is π times the intensity (Wm sr ) of radiation.

What is important in radiative transfer calculations? No doubt, it is amount of radiation that is emitted by, incident upon or passing through a surface area δA. Hence, the radiant energy E is expressed as λ2 E = I (ϕ,θ)dλ cosθ dω dA (9.23) ∫ ∫ ∫ λ δ A 2π λ1 Area solid wave- angle length The quantity E () is a hemispheric average value of radiant energy in a given wavelength interval ( λ1, λ2 ) that makes the plane parallel approximation valid in weather prediction and climate studies.

9.4 Radiation laws In the preceding section we have presented the first and most fundamental radiation law, the Planck’s law. We now present the other three fundamental laws of radiation that are at the foundation of the radiative transfer theory and allow us to formulate the relevant equations of radiative transfer in a grey atmosphere.

(i) Wien’s displacement law: The Wien’s displacement law states that the wavelength of peak emission for a blackbody is inversely proportional to temperature. Thus, the wavelength of peak emission for a blackbody at temperature T is given by 2897 λ = (9.24) m T where λm is the wavelength of maximum emission and T is the absolute temperature (K) of the blackbody. To obtain expression (9.24), differentiate the Planck function Bλ (T ) with respect to λ and then set the resulting expression to zero, ∂B (T ) λ = 0 (9.25) ∂λ Further, from the emission spectrum of the source, it is possible to determine the temperature of the radiating body using the Wien’s displacement law. For example, the wavelength of maximum solar radiation is observed at λm ~ 0.475 µm, then the Sun’s temperature can be easily calculated as 2897 2897 Tsun = = = 6099 K (Sun’s temperature) λm 0.475

8 The solar radiation is divided in two bands: (a) Visible band, 0.4 – 0.7 µm; (b) near infrared band, 0.7 – 4 µm. The Wien’s displacement law explains why solar radiation is concentrated in the visible and near infrared regions of the spectrum (λ < 4 µm) whereas radiation emitted by planets and their atmosphere is largely confined in the infrared region (λ > 4 µm). For earth with T = 300 K, the wavelength of maximum radiation is λm = 10 µm, which lies well into the infrared region of the electromagnetic spectrum.

(ii) Stefan–Boltzmann Law: The Stefan-Boltzmann law states that the flux density of radiation emitted by a blackbody is proportional to the fourth power of the absolute temperature (K). Thus, if T is the temperature of the blackbody then the flux density of the radiation is expressed as: 4 4 F c× T or F = σT (9.26) 2k 4π 5 In (9.26), σ = = 5.67 ×10−8 W m−2 K −4 is the Stefan-Boltzmann constant. The 15h3 c2 ( ) Stefan-Boltzmann law can be derived by integrating the Planck function over all the wavelengths of the electromagnetic spectrum as shown by relation (9.7) and (9.8) in the preceding section. From this law, one can find the equivalent black body temperature, which is the effective emission temperature (TE ) of a black or a grey (nonblack) body if measurements of flux density (F) are available. To fix the idea, this is illustrated by the following example.

Ex. 9.2 Calculate the equivalent blackbody temperature TE of the solar photosphere (i.e., the outermost visible layer of the sun), if the flux density (or irradiance) of solar radiation reaching the earth is 1370 Wm-2.

Solution: The flux density of solar radiation Fs at the top of the earth’s atmosphere is -2 given as 1370 Wm , and Fs = Isδω . The intensity of radiation ( Is ) is constant along ray paths and is independent of the distance from its source; δω is the arc of the solid angle 2 ⎛ Rs ⎞ subtended by sun at the earth. So, δω = π⎜ ⎟ and RS = radius of Sun. Hence, ⎝ d ⎠ 2 2 2 2 ⎛ Rs ⎞ ⎛ Rs ⎞ ⎛ d ⎞ ⎛ d ⎞ 7 −2 Fs = π Is ⎜ ⎟ = Fphoto ⎜ ⎟ ⇒ Fphoto = π Is = Fs ⎜ ⎟ = 1370 × ⎜ ⎟ = 6.28 ×10 Wm ⎝ d ⎠ ⎝ d ⎠ ⎝ Rs ⎠ ⎝ Rs ⎠

Note that π Is is the flux emitted by the photosphere of the Sun which is equal to the flux 4 4 calculated from the Stefan–Boltzmann law: Fphoto = σTE . That is,π Is = σTE ; and from this relation, we can now calculate the equivalent blackbody temperature TE of the Sun’s π I 6.28×107 photosphere as: T 4 = S = = (5769)4 E σ 5.67 ×10−8

That is, TE = 5769 K = Tsun (Sun’s temperature).

This value of Tsun is slightly lower than its earlier calculation from Wien’s law where we obtained the value . It is thus evident that Sun’s emission spectrum slightly Tsun = 6099 K differs from blackbody spectrum prescribed by the Planck’s law.

9 (iii) Kirchhoff’s Law: Blackbodies absorb all incident radiation, but all surfaces are not blackbodies. This implies that nonblack (grey) bodies such as gaseous media (atmosphere) besides absorption can also reflect and transmit radiation which introduce further complexity to the radiative transfer theory. Nevertheless, their behaviour can be understood from the laws of radiation derived for blackbodies. An important quantity in this regard is the wavelength dependent (ελ) of a medium or a body. Emissivity is defined as the ratio of the monochromatic intensity of the radiation emitted by a given body to the corresponding blackbody radiation. Thus,

Iλ (emitted) ελ = (9.27) Bλ (T ) The other important quantities, which need to be defined, are the fractions of incident radiation that are (i) absorbed, (ii) reflected or (iii) transmitted in different spectral ranges by a body or a medium. Hence, we have the following definitions for absorptivity, reflectivity and transmissivity.

Iλ (absorbed) Iλ (reflected) Iλ (transmitted) α λ = ; Rλ = ; Tλ = Iλ (incident) Iλ (incident) Iλ (incident)

For a given wavelength, the (Rλ ) , the (Tλ ) and the (α λ ) must all add up to unity; that is

Rλ + Tλ +α λ = 1 (9.28) Besides these radiative properties of nonblack materials, we also need the concept of local thermodynamical equilibrium while dealing with the radiative transfer through the atmosphere. Because temperature is not uniform in the atmosphere so the best approach is to regard it in thermodynamical equilibrium locally. Indeed, the first fundamental law of radiation has been derived without considering the matter. Thus, the Planck function applies to an isothermal cavity radiation, but without any matter in the cavity. A rigorous treatment of the interaction of matter and radiation requires both the matter and radiation as a fully coupled, quantized assembly. In such a coupled assembly in equilibrium at temperature T, the number of molecules of a gas (or for that matter any material system) in different states of energy levels will be present in accordance to the Boltzmann distribution, when radiation is neglected. If η1 and η2 are the number of molecules in states of energy E1 and E2 respectively, then their ratio is given by

η1 g1 ⎪⎧ ⎛ (E1 − E2 )⎞ ⎪⎫ = exp ⎨−⎜ ⎟ ⎬ , k = Boltzmann constant . (9.29) η2 g2 ⎩⎪ ⎝ kT ⎠ ⎭⎪

In the expression (9.29), g1 and g2 are the statistical weights. The equilibrium ratio of molecules in gases is however maintained by collisions between the gas molecules. Now consider an isothermal cavity containing both matter and radiation. Further assume that the interaction between the matter and radiation is sufficiently weak so that thermodynamical equilibrium prevails. Under such conditions, the radiation will continue to satisfy the Planck’s law, while the matter (molecules of a gas) will continue to satisfy the Boltzmann distribution. Indeed, for an ideal gas, the interaction between the molecules must be weak. The interaction between the matter and radiation is essential for achieving the thermodynamical equilibrium of the system, but molecular collisions

10 must not be so strong as to cause any significant departure in radiation from that given by the Planck’s law or of the matter from the Boltzmann distribution. However, in order to maintain equilibrium, some levels of interaction or collisions are necessary between matter and the radiation. The collisional processes are of two types: elastic and inelastic. For the first, there is no net energy transfer from kinetic to excitation energy while in inelastic collisions energy is transferred from kinetic to internal excitation energy. That is, colliding molecules cause collisional quenching. However, at sufficiently high pressures, molecular collisions are rapid enough for the Boltzmann distribution to hold for each small portion of the atmosphere for a given local value of the temperature T . Such a region of the atmosphere is said to be in local thermodynamical equilibrium (LTE) with respect to the given energy states. LTE applies to translational modes (associated with molecular kinetic energy and macroscopic thermal energy) below 500 km in the atmosphere. But for vibrational and rotational modes involved in absorption and emission from radiatively active gases, LTE holds for pressures greater than 0.1 hPa (~ 60 km height). The 99.9% mass of the atmosphere however lies below 0.1 hPa pressure level therefore it may be assumed that atmosphere is always in local thermodynamic equilibrium.

The Kirchhoff’s law states that the equality of emissivity ( ελ ) to absorptivity (α λ ) holds for gases in LTE; that is

ελ = α λ (9.30) Because emittance and absorptance are intrinsic properties of matter, the equality given in (9.30) holds even for objects that are removed from cavity (isotropic radiation field) and placed in a field of radiation that is not isotropic. The Kirchhoff’s law also states that emission can only occur at wavelengths where absorption occurs. Thus, a molecule that absorbs radiation at a particular wavelength shall also emit radiation at the same wavelength. Hence, emission rate of grey body will depend on the matter and the wavelength of radiation; while the radiation from a blackbody, in effect, depends upon temperature and wavelength. The absorption is maximum for a blackbody so also is the emission, therefore, we have

ελ = α λ = 1 (9.31) That is, complete absorption and emission of radiation in the case of a blackbody. A grey body can then be characterized as the one for which the following relation holds.

ελ = α λ < 1 (9.32) Accordingly, grey atmosphere is characterized by (9.32) as a medium with incomplete absorption and emission of radiation.

9.5 Basic equations of radiative transfer After discussing the basic laws in radiative transfer theory, we now derive the basic equations for the radiative transfer of energy in a plane parallel atmosphere. The equations will be derived for emission and absorption processes only. Since clouds and aerosols are not considered in their derivation hence scattering is unimportant. However, scattering is important for the shortwave radiation, but in the infrared wavelengths that we are considering, scattering is anyway neglected. In the plane parallel approximation (PPA) for radiative transfer calculations, temperature and density of atmospheric constituents are considered uniform in the horizontal; that is, they are functions of z (height) only. In this geometry, downward radiation comes from the Sun and upward

11 radiation emanate from earth; and accordingly with the PPA, the two-stream approximation of the radiative transfer equation is evidently possible.

(i) Law of Absorption (Bouguet’s Law): This is also known as the Lambert’s law which states that absorption which occurs when radiation of intensity I passes through a slab of thickness dz in the atmosphere, is proportional to the incident radiation intensity I itself and mass ρ dz (ρ density of the absorber) of the absorber in a unit cross-section of the slab. Hence, Bouguet law is mathematically expressed as dI dI c× I × ρ dz or dI = I k ρ dz or = k ρ dz (9.33) a I a

In the above equation ka is the absorption coefficient of the medium through which the ray of radiation passes. One can integrate (9.33) to obtain the following expression for the intensity I , I = I exp{− k ρ dz} (9.34) 0 ∫ a Let us define the absorber mass, optical thickness and the fractional transmission of the path as u = ∫ ρ dz (Absorber mass) (9.35) χ = k ρ dz () (9.36) ∫ a τ = exp{− k ρ dz} (Fractional transmission) (9.37) ∫ a

In view of the above defined quantities, one may write (9.34) in the following form

− χ − χ I = I0 e = τ I0 (writing τ = e ) (9.38)

It can be inferred from (9.38) that the transmitted fraction of the intensity is and I0 τ I0 the absorbed amount of radiation is therefore given by

A = I0 − I = I0 −τ I0 = (1−τ )I0 (9.39)

If the mass of the absorber gas varies in the vertical, then by subdividing the path into n homogeneous layers the intensity of radiation at the bottom of the n−th layer is given by

n In = Π τ i I0 i=1

In is the intensity of the radiation transmitted through the layer. The optical depth is also referred to as the optical path and the law of absorption is also known as Beer’s law.

(ii) Schwarzschild’s Equation A slab of atmosphere will also emit radiation depending on its temperature. It is already mentioned that local thermodynamic equilibrium assumption applies to the atmosphere up to 0.1 hPa (60 km height) and Kirchhoff’s law is applicable for an atmosphere in local thermodynamic equilibrium, which allows us to write the amount of radiation emitted per unit area of the atmospheric slab. Thus, the amount emitted by slab of thickness dz is given by Femit = ε ρ dz B(T ) = ka ρ dz B(T ) using ε = ka . The function B(T ) is the black body emission per unit solid angle per unit surface area of slab at

12 temperature T. Also, the integration of B(T ) over the hemisphere (solid angle=2π), from Stefan-Boltzmann law, is proportional to T 4 , so we have σT 4 ∫ B(T ) cosθ dAdω = σT 4dA ⇒ B = 2π π where dA is element of surface area, dω is the element of solid angle of a beam of radiation inclined at angle θ to the normal to the area element dA . Radiation absorbed by the slab is given by (9.33). The Schwarzschild’s radiative transfer equation includes both absorption and emission from an atmospheric slab. Thus, any change dI in intensity I of radiation as it traverses through the atmospheric slab of thickness dz is due to absorption and emission. The corresponding balance is written as,

dI = −I ka ρdz + B ka ρdz = −(I − B)ka ρdz (9.40) On rearranging, one obtains the basic equation of radiative transfer as dI − = (I − B) (dχ = ka ρdz) (9.41) dχ Under the plane parallel approximation, the problem reduces to one dimension, but we need to consider two streams of F ↑ and F ↓ when the beam of intensity I(θ) makes an angle θ with the vertical as shown in Fig. 9.3. Hence, we have

F ↑ = ∫ I(θ)cosθ dω (upward facing hemisphere) (9.42a) 2π F ↓ = ∫ I(θ)cosθ dω (downward facing hemisphere) (9.42b) 2π Now we can write the radiative transfer equation for a plane parallel atmosphere where fluxes are hemispherically integrated, and optical depth is scaled by factor 5/3. The optical depth χ is replaced by χ∗ which allows replacing I by F in the radiative transfer equation. From (9.41), the radiative transfer equations for upward and downward traversing radiation are written as,

dF↑ = F↑ − π B (upward traversing) (9.43a) dχ∗ ( ) dF ↓ − = F ↓ − π B (downward traversing) (9.43b) dχ∗ ( ) 5 In the above equations χ∗ = χ = 1.666 χ ; χ∗ is the scaled optical depth and π B 3 replaces B , because two-stream approximation considers hemispherical averages. The equations (9.43) are the most important set of equations with fluxes integrated hemispherically and over all wavelengths for a deep atmosphere. The origin ( χ∗ = 0 ) is χ ∗ χ ∗ fixed at the position of the sun in the space and position of the earth ( = 0 ) is the other extremity of the interval that is considered in the radiative transfer calculations.

(iii) Heating Rate of the Layer: The net difference between the incoming and the outgoing radiation, as shown in the Fig. 9.6, will change the temperature of the layer.

13 Thus, change in the net is equal to net flux at the top minus net flux at the bottom boundary of the layer. This net change in the net radiative flux is responsible for heating or cooling ( dQ ), which can be mathematically expressed as

dQ = {(F↓ + dF↓ ) − (F↑ + dF↑ )} − {F↓ − F↑ } = d (F↓ − F↑ ) Wm−2

If absorption of dQ amount of net radiative flux changes the temperature of the layer of thickness dz by dT ; then the change in heat content of the layer is given by dT (ρdz)C ; ρ = density of air and C = specific heat of air at constant pressure p dt p Since this change happens due to dQ , the two quantities must be equal; that is,

⎛ dT ⎞ dT d ↓ ↑ ρ Cp ⎜ ⎟ dz = dQ or ρ Cp = F − F (9.44) ⎝ dt ⎠ dt dz ( )

d The quantity F↓ − F↑ is called the divergence of net radiation. The eq. (9.44) dz ( ) together with (9.43) formulates the complete set of equations for the transfer problem and their solutions can be obtained when initial conditions for F↓ , F↑ and T are specified.

F↑ + dF↑ F↓ + dF↓ Fig. 9.6 Diagram showing the flux balance for an z Top atmospheric layer of thickness dz . Height z increases upwards and pressure increases in the downward direction. dz Bottom p F↑ F↓

Now assume atmosphere transparent to incoming solar radiation and perform radiative transfer calculations. In equilibrium conditions, the temperature of each layer in dT the atmosphere will attain a steady value, which implies, = 0 ; this assumption is the dt first simplification of (9.44) giving,

d F↓ − F↑ = 0 ⇒ F↑ − F↓ = φ (constant) (9.45) dz ( )

↑ ↓ χ ∗ The fluxes F and F are obtained by solving the equations (9.43) as functions of , an independent variable that is formulated above using mass and frequency independent absorption coefficient k . In order to solve (9.43), define a new variable ψ =F↓ + F↑ ,

14 then by adding and subtracting the two-steam transfer equations (9.43a) and (9.43b), we obtain the following equations.

↑ ↓ d( F − F dφ Adding : ( ) = F↑ + F↓ − 2π B ⇒ =ψ − 2π B (9.46) dχ ∗ dχ ∗ ↑ ↓ d( F + F dψ Subtracting : ( ) = F↑ − F↓ ⇒ = φ (9.47) dχ ∗ dχ ∗ dφ The direct integration of (9.47) gives ψ = φχ ∗ + c ; also φ = constant implies = 0 ; dχ ∗ hence from (9.46) one obtains ψ = 2π B , which can be equated to the expression of ψ obtained from direct integration of eq. (9.47) ; thus, we have φ 2π B = φχ ∗ + c or B = χ ∗ + c (9.48) 2π 1

The constant c1 can be evaluated using the condition at top of the atmosphere; that is, at χ ∗ = 0, the infrared downward flux F↓ = 0 . In view of this condition, eq. (9.45) φ gives F↑ = φ = 2π B . Hence B = . Applying afore stated boundary condition in (9.48), 2π φ one evaluates c = B ⇒ c = . Substitute c in (9.48), and it gives, 1 1 2π 1 φ π B = (1+ χ ∗ ) (9.49) 2 ↑ ↓ Taking φ = F − F = F0 , the eq. (9.49) can be written as follows F π B(T ) = σT 4 = o (1+ χ*) (9.50) 2

It is now possible to determine the upward and downward in terms of F0 χ ∗ and the scaled optical depth as both φ and ψ are completely determined. From the definitions of ψ and φ we have, F↑ + F↓ =ψ and F↑ − F↓ = φ ; therefore, F↑ and F↓ at any level χ ∗ are calculated as ψ +φ F (1+ χ ∗ ) + F ψ −φ F (1+ χ ∗ ) − F F↑ = = 0 0 ; and F↓ = = 0 0 ; 2 2 2 2 F F F↑ = 0 (2 + χ ∗ ); F↓ = 0 χ ∗ and F↑ − F↓ = F (9.51) 2 2 0 It may be noted that F↑ , F↓ and B(T ) are all linear functions of scaled optical depth χ ∗ . Their expressions given by (9.50) and (9.51) have been derived under the assumption of an atmosphere transparent to solar radiation. Further, it is assumed that infrared radiation from sun is negligible which is set as the upper boundary for radiative calculations. This simple model calculates the infrared radiative transfer in a grey atmosphere. Hence, it provides the simplest way to compute heating arising from radiative properties of atmospheric constituents that absorb infrared radiation and emit at the same wavelengths. The net effect is that surface temperature is raised; and, therefore, it is a simple model to illustrate the so called greenhouse effect. For a better understanding of

Two cases immediately arise for surface temperatures that can be calculated from (9.52).

(a) All incident radiation absorbed at the surface: This is the case when albedo of the

earth surface α = 0 ; and for this case the surface temperature Tg = 279 K . (b) Incident radiation partly absorbed at the surface: This is the case when albedo of the earth surface is finite, α = 0.3 (global mean albedo of the earth). Since part of the incident radiation will be reflected back to the space, a lower surface temperature of the earth than to the latter case is expected. For this case the surface temperature of

the earth calculates to Tg = 255 K . The temperature at the bottom of the atmosphere can be calculated from the expression χ ∗ χ ∗ ↓ (9.51) for upward radiation that corresponds to = 0 and noting that Fg = 0 ; thus, F F↑ = 0 (2 + χ ∗ ) = π B (T ) = σT 4 (9.53) g 2 g g g

Bg is the Planck function at ground temperature Tg which is the equilibrium temperature as well of the earth surface.

(iv) The greenhouse effect: It has been shown that atmosphere plays an important role in altering the surface temperature. If there were no atmosphere, the ground temperature would be Tg=255 K with an albedo of 30% of the earth’s surface. The natural levels of greenhouse gases, viz., water vapour, carbon dioxide and methane give rise to an increase of surface temperature by about 30-35 K ; consequently, life on earth and the humans have depended on the greenhouse effect, for without it the earth would have freezing temperatures everywhere. However, the greenhouse effect which in turn affect the climate system, does not behave in a linear way. More precisely, the increase in the concentration of minor constituents in the atmosphere does not mean that surface temperatures would also increase in same proportions.

16 An Important Inference: The radiation model just presented treats atmosphere transparent to solar radiation but opaque to longer wavelengths where radiative (infrared) cooling happens. Now, assume that Tb is the air temperature at base of the atmosphere χ ∗ χ ∗ ↑ and Tg the ground temperature. At the base of the atmosphere, = 0 , and F = π Bg . A key question arises: what is the magnitude of the difference between ground temperature Tg and air temperature Tb ?

An answer to the above question requires computation of the difference Tg − Tb . The χ ∗ χ ∗ temperature Tb can be calculated by putting = 0 in the expression (9.50), 1 π B(T ) = F 1+ χ ∗ = σT 4 . (9.54) b 2 o ( 0 ) b

For computing Tg , the corresponding blackbody radiation B(Tg ) is obtained from (9.53) which reads as 1 π B(T ) = F 2 + χ ∗ = σT 4 . (9.55) g 2 0 ( 0 ) g The difference in the two temperatures can be calculated from (9.54) and (9.55) by calculating the difference in the fluxes as F π [B(T ) − B(T )] = 0 = σ (T 4 − T 4 ) . (9.56) g b 2 g b

Since F0 > 0 implies that B(Tg ) − B(Tb ) > 0 henceTg > Tb ; that is, the ground temperature

Tg is greater than the air temperature Tb at the bottom of the atmosphere. This means the existence of a temperature discontinuity at the surface which is an unrealistic feature of this radiation model. Development of such discontinuities in the temperature field is not uncommon in the real atmosphere, but they are immediately removed by initiation of convection. A problem, inspired from a television film featuring zebra, has been designed which the readers are invited to solve so as to find the difference of temperatures of the white and black strips.

Fig. 9.7 Plot of B(T ) with scaled optical depth χ ∗ , which increases downward along the ordinate; B(T ) increases along the abscissa. F0 Top of the atmosphere ↑ ↓ 0 − The difference F − F (= F ) is constant as 0 shown in this figure. However, a temperature discontinuity depicted by a jump of magnitude χ ∗ F↓ π B(T) F↑ 1 F at the earth’s surface may be noted in − 2 0 ↓ accordance to eq. (9.56), which makes this model rather unrealistic. Such a discontinuity will be removed by convection. Thus, a radiative- χ ∗ convective model is quite realistic. 0 π B(Tb) π B(Tg) Earth's surface

17 However, in the present context, the temperature discontinuity at the surface justifiably demonstrates that the radiation model discussed above is incomplete because in the real atmosphere the other physical processes, such as convection, would lead to smoothening out of the temperature discontinuity. In view of this observation, a radiative- convective model should be quite capable of addressing the changes in the atmospheric profile as a result of increase in the optically active gases due to growing anthropogenic activity. Indeed, such models have been developed and used to address the issue. Nevertheless, this simple model has resemblance to the real atmosphere which can be seen by plotting (Fig. 9.7) the blackbody function B(T ) against the scaled optical depth χ ∗ . The greenhouse effect arises mainly due to water vapour, carbon dioxide and other gases composed of molecules having dipole electric moments, and thus absorb strongly in the longwave part of the electromagnetic spectrum that is essentially occupied by the outgoing terrestrial radiation. A simple exercise given below explains warming of the surface of earth when the greenhouse gases are present in a stratified planetary atmosphere constituted of optically transparent isothermal layers to solar radiation. Note that as the number of isothermal layers increases, the greenhouse effect accentuates resulting in runaway increase in surface temperatures. This however does not happen on the earth, which is why it is revered as the Gaia in Hindu and Greek mythologies and also prominent in the solar system as the Living Planet.

Ex. 9.3 Imagine a planet in the solar system that has its albedo same as earth’s but its atmosphere consists of multiple isothermal layers, each of which is transparent to shortwave radiation but completely opaque to longwave radiation. The layers and the surface of the planet are in radiative equilibrium. Find the surface temperature of the planet under this kind of envelop.

4 1/4 σ Tg = 2F ⇒ Tg = 2 (255) = 303 K

18 Next, consider two isothermal layers in the planet’s atmosphere (Fig. 9.8), which are at temperatures T1 and T2 respectively. Since the layers absorb LW radiation but are transparent to SW radiation, the incident F units of SW radiation will be absorbed only at the planet’s surface. The top layer at temperature T1 will emit F units of LW radiation but layer at temperature T2 emits 2F units to remain at same temperature. Thus the surface receives F units of SW and 2F units of LW radiation. Thus the surface receives (downward) 3F units of radiation, which must be balanced by an equal amount of emission from the surface. Hence the surface temperature of the planet with two isothermal layers in its atmosphere will be determined as

σ T 4 = 3F ⇒ T = 31/4 (255) = 355 K g g

Fig. 9.8 The radiation balance in an atmosphere F F consisting of isothermal layers at temperatures Top T1 , T2 and T3 respectively. The temperature of F T1 2F each layer is maintained through the radiation balance at the top and bottom of the isothermal F T2 3F F layer. Note that net radiation at top and bottom of the isothermal layer is zero. Downward arrows F T3 4F 2F indicate incoming radiation from Sun, which is absorbed only at the surface. The radiation T Surface s 3F balance is worked out from the top isothermal layer downward to the lowest layer and finally at Solar radiation Base of Emitted radiation the surface which the base of atmosphere. Atmosphere

It is essential to note that the base of the atmosphere is hypothetically represented by a dotted line (…..) which indeed coincides with the surface of the earth indicated by a continuous line ( ______) at the bottom in Fig 9.8. The two reference levels are distinctly shown in the above figure just to show that surface receives downward flux from atmosphere and surface emits upward flux into the atmosphere. There is no air between these two reference levels. Now, suppose that the planetary atmosphere of the planet is composed of N isothermal layers, the emitted radiation from such an arrangement of layers is F, 2F, 3F, …. , (N +1)F respectively. That is, F units enter at top layer, and (N +1)F units reach the surface, which must be balanced by radiation from surface. Thus the surface temperature is calculated as 4 1/4 σ Tg = (N +1)F ⇒ Tg = (N +1) (255)K This completes the solution of the given problem.

However, the actual depth of atmospheric layers of absorbing medium decreases exponentially due to increase in air density downwards. Hence, radiative equilibrium lapse rate will increase steeply with decreasing layer depth as one moves from top of the atmosphere to the earth surface. This means that radiative transfer becomes increasingly inefficient in removing energy stored at surface due to increasing blocking effect of

19 greenhouse gases. It is noteworthy that once the radiative equilibrium lapse rate exceeds the adiabatic lapse rate, convection becomes the most efficient mode of energy transfer and its distribution (mixing) in the depth of the atmosphere. Troposphere is the lowest layer of the atmosphere where convective overturning is dominant and energy is constantly mixed there. However convective overturning occurs not throughout the entire depth of the atmosphere but it is limited only up to a certain height from the surface with an upper bound of the tropopause, though under strong and deep convection its level is lifted up. That is why the tropopause height is about 12-16 km in the tropics but it is just 8 km in the polar latitudes. In the preceding paragraphs, it is demonstrated that without the greenhouse effect of the atmosphere, temperatures on earth would have been very cool. The Mother Earth would have been without life and inhabitable as calculations suggest. Therefore, the atmospheric constituent gases that are present in parts one in one million or even less than these proportions and appropriately named “trace gases”, remarkably change the radiative balance and produce the “natural” greenhouse effect that is so important to existence of life. Any further rise in the levels of these trace gases, characterized as “greenhouse gases (GHGs)”, will increase the greenhouse effect of the atmosphere producing rise in surface temperature of the earth globally with possible adverse effects. In the minds of non-scientists, such a consequence of “global warming” that arises from such extremely small or non-discernible changes in the atmospheric composition does not sink well and seems to be the main cause of the controversies and the global warming debate. It is also true that rising levels of greenhouse gases above their “natural” levels (existing before the industrial revolution) do not increase surface temperature in respective proportions mainly due to the presence of sulphate aerosols in the atmosphere, which scatter direct sunlight and thereby produce “cooling”. There are also associated “indirect effects” of sulphate aerosols that also results in cooling. The sulphur containing gases such as SO2, and H2S released in coal and biomass burning and, DMS and DMSO released from ocean surface are oxidised by the hydroxyl radicals to form sulphate aerosols in the atmosphere. Titanium oxide (TiO2) suggested by Peter Davidson, a freelance chemical engineer in U.K., is the other new compound that has entered into this discussion, as it is being considered as one of the key players in geo-engineering climate.

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