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 × 108 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
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Figure 1: Ahrens, Chapter 2
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intercepted by the cone to the square of the radius.
ω=
A
r2
(1)
dω =
dA
r2
(2)
In spherical coordinates
dA = r2 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 Fλ Amount of radiant energy with a given wavelength passing through a unit area per unit time. Expressed in [W m−2 ].
flux density
R ν2 is calculated as the integral over all wavelengths F =
RTotal
λ2
Fλ dλ = ν1 Fν dν.
λ1
Monochromatic Intensity (or Monochromatic Radiance) Iλ Radiant energy in
a specific wavelength per unit time coming from a specific direction and
passing through a unit area perpendicular to that direction.
TotalRintensity
R λ2
ν
is calculated as the integral over all wavelengths I = λ1 Iλ dλ = ν12 Iν dν.
The units are [W m−2 steradian−1 ].The intensity Iλ and flux density Fλ are
related by
dF
(4)
Iλ =
dωcosθ
We can integrate over the solid angle subtended by a hemisphere to determine the monochromatic flux density coming from all directions
Z
Fλ =
2πsteradians
Z
Iλ cosθdω =
0
2π
Z
π/2
dφ
0
Iλ cosθsinφdφ
0
4. Blackbody Radiation
A blackbody is a surface that absorbs all incident radiation.
3
(5)
The Planck Function Determined experimentally, the intensity of radiation emitted by a blackbody is
c1 λ−5
(6)
Bλ =
π(ec2 /λT − 1)
where c1 = 3.74 × 10−16 W m2 and c2 = 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).
2897
(7)
λm =
T
where T in K and λm 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 πBλ 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
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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 Tλ = IλI(transmitted)
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 Rλ = IλI(transmitted)
λ (incident)
3. Absorbed Increase in internal energy of the object. We define the monochromatic fractional absorptivity as αλ = IIλλ(absorbed)
(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
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N2 and O2 , 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 αλ + Rλ + Tλ = 1. For a black body αλ = 1.
5a. Other Definitions
Total Mass Extinction Coefficient If a beam of intensity Iλ becomes Iλ + dIλ
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
dIλ = −kλT ρIλ ds
(9)
where kλT is the total mass extinction coefficient and has units [m2 kg −1 ].
We define the optical depth
R s τλ in terms of the total mass extinction coefficient as follows: τλ = 0 kλT ρds
Mass Absorpion Coefficient If a beam of intensity Iλ becomes Iλ + dIλ 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:
dIλ = −kλa ρIλ ds
(10)
2
−1
where kλa is the mass absorption coefficient and
R shas units [m kg ]. We
define the absorption optical depth τλa as τλa = 0 kλa ρds
Mass Scattering Coefficient If a beam of intensity Iλ becomes Iλ + dIλ 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:
dIλ = −kλs ρIλ ds
(11)
2
−1
where kλs is the mass scattering coefficient and
R s has units [m kg ]. We
define the scattering optical depth τλs as τλs = 0 kλs ρds
It follows that kλT = kλa + kλs
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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:
Iλ − Rλ Iλ = Tλ Iλ + Eλ
(12)
Where Eλ is the emitted radiance in the same direction as Iλ . But Tλ Iλ = Iλ (1 −
αλ − Rλ ) since αλ + Rλ + Tλ = 1. Therefore,
Iλ (1 − Rλ ) = Iλ (1 − αλ − Rλ ) + Eλ
(13)
Thus, Eλ − αλ Iλ = 0 or Eλ = αλ Iλ
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:
Eλ = αλ Iλ0
(14)
Where Iλ0 is the incident radiation on the slab in the new enclosure, thus it
follows that Iλ = Iλ0 .Thus, the radiation within an opaque, hollow enclosure is
independent of the material from which the walls are made. Re-writing the above
Eλ
= Iλb = f (T, λ) only and Iλb is the radiance inside an
equation we see αl ambda
opaque hollow enclosure at temperature T and wavelength λ.
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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 λ as the ratio of the radiation
emitted at the wavelength λ to that within a hollow enclosure at the same temperature or:
Eλ
(15)
Iλb
From this definition, we see that Eλ = λ Iλb . But from Kirchoff’s Law it then
follows:
λ =
λ = αλ
(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
Iλb = Eλ
(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.
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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 Fe 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 7×108 m. Calculate the equivalent blackbody temperature or effective temperature of this layer. The mean
distance between the earth and sun is 1.5 × 1011 m.
3. Calculate the equivalent blackbody temperature of the earth assuming a
planetary albedo αp = 0.3 where αp 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.
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