Atmospheric Radiation

January 3, 2018

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Contents1 Electromagnetic waves 3

2 Light as quanta 62.1 Atmospheric application: Oxygen dissociation in the stratosphere . . . . . . . . . 7

3 Flux and Intensity 73.1 Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.2 Intensity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Atmospheric application: Solar intensity . . . . . . . . . . . . . . . . . . . . . . . 93.4 Relationship between flux and intensity . . . . . . . . . . . . . . . . . . . . . . . 10

4 Blackbody radiation 104.1 Atmospheric example: Earth’s energy balance . . . . . . . . . . . . . . . . . . . . 15

5 Local Thermodynamic Equilibrium 155.1 Kirchoffs Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

6 Why is the sky blue? 18

7 Why are clouds white? 19

8 The mathematics of scattering and absorption and emission 258.1 Absorptance and transmittance . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.2 Optical depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258.3 Transmittance in a plane parallel atmosphere . . . . . . . . . . . . . . . . . . . . . 278.4 Size parameter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.5 Single-scattering albedo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288.6 Phase function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

9 Energy transitions of molecular absorption 309.1 The simplest case: The Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . 309.2 Polyatomic molecules and rotational transitions . . . . . . . . . . . . . . . . . . . 319.3 Polyatomic molecules and vibrational transitions . . . . . . . . . . . . . . . . . . 339.4 Vibrational-rotational spectra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

10 Molecular absorption profiles 3510.1 Line strengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3510.2 Line profiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

11 Line absorption 3711.1 Weak and strong lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

12 Molecular absorption, the greenhouse effect, and climate change 39

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1 Electromagnetic waves

Figure 1.1: Electromagnetic wave

The fundamental equations that describe electromagnetic radiation are Maxwell’s equations.The solutions to the equations are sinusoidal of form

~E = ~E0 exp⇣

i~k ·~x� iwt⌘

(1.1)

~H = ~H0 exp⇣

i~k ·~x� iwt⌘

(1.2)

where E is the electric field and H is the magnetic field, the real component of~k is the wavenumber2p/l ,~x is the direction of wave propagation and w is the angular frequency of the radiation 2pn .

Some fundamental properties of the waves are that they do not diverge, that H and E are normalto each other, and that radiation of one frequency or wavelength does not interact with radiationof another frequency. The last point is particularly useful because it means we can analyse theseparate contributions of a range in frequency or wavelength to the total energy input from radiationas it accumulates over time.

The symbols e and µ stand for the electric permittivity and magnetic permeability, respectively.They are merely properties of matter. Curiously, unlike say sound or water waves, E-M waves donot require a medium and can travel in a vacuum. The permittivity and permeability of a vacuum

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have the symbols e0 and µ0. The instantaneous flux density of the electric field in the direction ofthe wave is defined by the Poynting vector

~S = ~E ⇥ ~H (1.3)

In a vacuum this can be simplified to

S =E2

0µ0c

cos2 (wt)

Although averaged over time it is equal to the average Poynting Flux

Sav =12

Re(~E0 ⇥ ~H0)

orSav =

12

ce0E20 (1.4)

Sav has units of W m�2 (or J s�1 m�2) and is a function of wavelength. This is units of flux(quantity per length squared per second). In atmospheric sciences, we more commonly use thesymbol F . The point here is that the flux of energy F is proportional to the square of the magnitudeof an electromagnetic wave. It is the flux density of electromagnetic waves that is of primaryinterest to climate studies.

Figure 1.2: Electromagnetic wave quantities

Okay, but let’s look at interactions of radiation with something that is not a vacuum. Rememberfrom Eq. 1.1

~E = ~E0 exp⇣

i~k ·~x� iwt⌘

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Assume that the wavenumber vector~k is a complex number with a real and imaginary component,i.e.

~k = ~k0 + i~k” (1.5)

Therefore, the consequence of k being complex is the following

~E = E0 exp⇣�~k” ·~x

⌘exp⇣

i⇣~k0 ·~x�wt

⌘⌘(1.6)

Note that if ~k” = 0 then the amplitude of the wave is constant.

Figure 1.3: Atmospheric radiation and transmission in the atmosphere as a function of wavelength

Thus the phase of the wave is given by

f = ~k0 ·~x�wt (1.7)

At constant phasedf = 0 = ~k0 ·d~x�wdt (1.8)

implying that the phase speed of radiation in the direction the waves propagate is given by

v =d~xdt

=w~k0

(1.9)

or alternativelyv = nl =

w2p

l (1.10)

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The point here is that frequency and wavelength are linked through the phase speed of the wave.In a vacuum the phase speed is c = 3.0⇥108 m/s, the “speed of light”, where

c = nl =w2p

l =wk

(1.11)

It is common practice in the atmospheric sciences to divide n by the speed of light to obtain thewavenumber

n =nc

which is most commonly expressed in units of cm�1. The only reason for dividing by c is thatwe end up with numbers around 1000 that our poor little brains can handle more easily than 1011.So, for example, we can express visible light with a wavelength of l = 0.3 µm (0.3⇥10�4 cm or0.3⇥10�6 m) as having a frequency of n = c/l = 1015 Hz or n = 1/l = 3⇥106 cm�1.

2 Light as quantaA paradox of modern physics is that particles sometimes appear to be waves, and sometimes appearto be particles, depending on how you look at them. Sometimes we’ll treat radiation as waves andsometimes as particles. As far as we know, neither is any more wrong or right. But sometimes oneof the two is more useful.

Figure 2.1: Electron two slit experiment, showing that photons exhibit a duality in their behaviouras both waves and particles.

From Eq. 1.11, the frequency of light is related to its wavelength l through the speed of light c

n =cl

(2.1)

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The speed of light in a vacuum is about 2.998⇥108 m s�1, regardless of how fast you are movingrelative to another frame of reference (this is the principle behind Einstein’s special theory ofrelativity, published in 1905, which led to the famous E = mc2). Another definition often used byatmospheric scientists is the wavenumber, usually described in terms of cm�1. The wavenumberis related to the frequency through a factor of c.

n =1l

(2.2)

Strangely though, energy comes in discrete, although very small, packets or particles, whose en-ergy is related to the frequency of a wave!

E = hn (2.3)

where h is Planck’s constant, 6.626⇥10�34 J s. While Planck was the first to postulate, as much asa mathematical trick as anything, the quantization of energy, Einstein was the first to make a com-pelling theoretical argument, also in 1905, in his explanation of what is known as the photoelectriceffect. It won Einstein the Nobel prize, and is immortalized in this song.

2.1 Atmospheric application: Oxygen dissociation in the stratosphereIn the formation of ozone, only radiation with wavelengths smaller than 0.2424 µm is capable ofdissociating molecular oxygen into atomic oxygen, according to the reaction

O2 +photon ! O+O

Based on this information, how much energy is apparently required to break the molecular bondof a single molecule of O2?

E = hn = hc/l = 6.6⇥10�34 ⇥3.0⇥108/0.2424⇥10�6 = 8⇥10�19 J

A very small number, but in the atmosphere these short wavelength photons are the only ones withsufficiently high energy to break molecular bonds and thereby play a role in chemical reactions.

The number of photons raining down on us is huge. Assuming the flux density of sunlightreaching the surface on a cloudy day is 100 W m�2 with a mean wavelength of 0.5 µm the flux ofphotons is

N =Fhn

=Flhc

= 2.3⇥1020 s�1m�2

Huge!Since this huge number will wash out any sense of photons being discretized events, in atmo-

spheric sciences we usually treat radiation as a wave, except in the case of chemical photolysis, orin certain computational approaches to radiative transfer.

3 Flux and Intensity

3.1 FluxRadiative flux has units of W m�2, and is perhaps more appropriately called the flux density,although atmospheric scientists never call it this. Radiative flux may be coming from one direction

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(the sun), or on a cloudy day at the surface from all directions. The only restriction is that the fluxrefers to radiation incident on a surface normal to a hemisphere enclosing the surface.

Electromagnetic waves typically represent a superposition of many many wavelengths l or fre-quencies n (remember the two are related by c), each with its own amplitude F0. By superpositionwe mean that we can describe the electromagnetic wave as a time series F (t) (think of rapidlypulsed light striking you with varying intensity)

A Fourier transform can convert this time signature to frequency space:

F (n) =Z •

0F (t)e�2pintdt (3.1)

This function contains all the original information present in the original time series but displaysit in the frequency domain. The original time series can be restored exactly by taking the inverseFourier transform

F (t) =1

2p

Z •

0F (n)e2pintdn (3.2)

We now define a spectral flux Fn or Fl with units W m�2 s or W m�2 µm�1.

Fl = limDl!0

F(l ,l +Dl )

Dl(3.3)

andFn = lim

Dn!0

F(n ,n +Dn)

Dn(3.4)

The broadband flux is the integrated flux over a range of wavelengths

F (l1,l2) =Z l2

l1Fl dl (3.5)

So, as a function of time, a sensor receives an oscillating intensity associated with the sinusoidalnature of the electromagnetic waves, e.g. F (t). Applying a Fourier transform to this signal yieldsthe intensity of the waves as a function of frequency F (w). In real life we sense this as an objectlooking “red” or “green”, even though our eyes are only receiving F (t). Our brain is converting atemporal signal to information in frequency space.

Note that the broadband flux could also be calculated as

F (l1,l2) = F (n1,n2) =Z n2

n1Fndn (3.6)

provided that ni = c/li. That is Fl dl = Fndn = dF even if Fl 6= Fn . The flux of energy F mustbe the same whether we express it within a band of n or l . However, the spectral flux Fl and Fnis dependent on the co-ordinate system considered. More on this in an assignment.

3.2 IntensityIntensity I (or commonly radiance) has units of W m�2 per steradian, where a steradian is andefinition of a solid angle dw . An infinitesimal increment of solid angle is

dw = sinqdqdf = A/r2 (3.7)

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Figure 3.1: Spectrum as a function of wavelength

where in the atmosphere, the convention is that q is the zenith angle (0� if directly overhead),and f is the azimuthal angle (0� in the direction of an incoming beam). A hemisphere covers 2psteradians and a full sphere 4p steradians. A is the cross-sectional area normal to radius vector r.

3.3 Atmospheric application: Solar intensityThe intensity of radiation coming from any direction is a function of angle W = (q ,f) and isdefined as

I (W) =dFdw

(3.8)

For example, the sun - earth distance is Ds = 1.496⇥ 108 km and the sun radius is Rs = 6.96⇥105km. Compute the angular diameter subtended by the sun, its solid angle, and the intensity ofsolar radiation if S0 = 1370 W m�2.

The solution requires recognizing that

dw = A/D2s = pR2

s/D2s = 6.8⇥10�5 sr

Thus,I = dF/dw = 1370/6.8⇥10�5 = 2⇥107W m�2 sr�1

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4.2 Quantitative Description of Radiation 115

The monochromatic flux density (or monochro-matic irradiance) F! is a measure of the rate of energytransfer per unit area by radiation with a given wave-length through a plane surface with a specified orien-tation in three-dimensional space. If the radiationimpinges on a plane surface from one direction(e.g., upon the horizontal plane from above) the fluxdensity is said to be incident upon that surface, inwhich case

(4.5)

The limit on the bottom of the integral operator indi-cates that the integration extends over the entirehemisphere of solid angles lying above the plane, d"represents an elemental arc of solid angle, and # is theangle between the incident radiation and the directionnormal to dA. The factor cos # represents the spread-ing and resulting dilution of radiation with a slantedorientation relative to the surface. Monochromaticflux density F! has units of W m$2 %m$1. Analogousquantities can be defined for the wave number andfrequency spectra.

Exercise 4.1 By means of a formal integration oversolid angle, calculate the arc of solid angle subtendedby the sky when viewed from a point on a horizontalsurface.

Solution: The required integration is performedusing a spherical coordinate system centered on apoint on the surface, with the pole pointing straightupward toward the zenith, where # is called the

F! & !2'

I! cos # d"

zenith angle and ( the azimuth angle, as defined inFig. 4.3. The required arc of solid angle is given by

■

Combining (4.3) and (4.5), we obtain an expres-sion for the flux density (or irradiance) of radiationincident upon a plane surface

(4.6)

Flux density, the rate at which radiant energy passesthrough a unit area on a horizontal surface, isexpressed in units of watts per square meter. The fol-lowing two exercises illustrate the relation betweenintensity and flux density.

Exercise 4.2 The flux density Fs of solar radiationincident upon a horizontal surface at the top ofthe Earth’s atmosphere at zero zenith angle is 1368 Wm$2. Estimate the intensity of solar radiation. Assumethat solar radiation is isotropic (i.e., that every pointon the “surface” of the sun emits radiation with thesame intensity in all directions, as indicated in Fig. 4.4).For reference, the radius of the sun Rs is 7.00 ) 108 mand the Earth–sun distance d is 1.50 ) 1011 m.

Solution: Let Is be the intensity of solar radiation. Ifthe solar radiation is isotropic and the sun is directly

F & !2'

I cos # d" & !!2

!1

!2'

I! cos #d" d!

!2'

d" & !2'

(&0!'"2

#&0sin #d#d( & 2'!'"2

#&0sin #d# & 2'

Fig. 4.2 The curve represents a hypothetical spectrum ofmonochromatic intensity I! or monochromatic flux density F!

as a function of wavelength !. The shaded area represents theintensity I or flux density F of radiation with wavelengths rang-ing from !1 to !2.

! 1

I ! o

r F

!

! 2

The“spectrum”

Wavelength !

Fig. 4.3 Relationship between intensity and flux density. # isthe angle between the incident radiation and the normal tothe surface. For the case of radiation incident upon a hori-zontal surface from above, # is called the zenith angle. ( isreferred to as the azimuth angle.

δθ

δφ

θ

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Figure 3.2: Illustration of a solid angle

3.4 Relationship between flux and intensityFrom any given level in the atmosphere, we often talk about the upwelling and downwelling flux.This is because when studying radiative transfer in the atmosphere we are primarily concernedwith the amount of energy entering or leaving a system through its upper and lower boundaries.

• The upwelling flux is simply the sum of the upward component of all intensities integratedover the upward hemisphere

F" =Z 2p

0

Z p/2

0I" (q ,f)cosq sinqdqdf (3.9)

Similarly for the downwelling flux

F# = �Z 2p

0

Z p

p/2I# (q ,f)cosq sinqdqdf (3.10)

4 Blackbody radiationThe basic principles of thermal emission are as follows:

• An object of temperature T radiates energy at all wavelengths (or equivalently frequencies).The amount of energy radiated at a specific frequency follows a relation known as the Planckfunction.

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116 Radiative Transfer

overhead, then from (4.5) the flux density of solarradiation at the top of the Earth’s atmosphere is

where !" is the arc of solid angle subtended by thesun in the sky. Because !" is very small, we canignore the variations in cos# in the integration. Withthis so-called parallel beam approximation, the inte-gral reduces to

and because the zenith angle, in this case, is zero,

The fraction of the hemisphere of solid angle (i.e.,“the sky”) that is occupied by the sun is the same asthe fraction of the area of the hemisphere of radius d,centered on the Earth, i.e., occupied by the sun, i.e.,

from which

!" $ % !Rs

d "2

$ % ! 7.00 & 108

1.50 & 1011"2$ 6.84 & 10'5 sr

!"

2%$

%R2s

2%d2

Fs $ Is & !"

Fs $ Is & cos # & !"

Fs $ #!"

Is cos # d"

and

$ 2.00 & 107 W m'2 sr'1 ■

The intensity of radiation is constant along raypaths through space and is thus independent of dis-tance from its source, in this case, the sun. The corre-sponding flux density is directly proportional to thearc solid angle subtended by the sun, which isinversely proportional to the square of the distancefrom the sun. It follows that flux density variesinversely with the square of the distance from thesun, i.e.,

(4.7)

This so-called inverse square law also follows fromthe fact that the flux of solar radiation Es (i.e., theflux density Fs multiplied by the area of spheres, con-centric with the sun, through which it passes as itradiates outward) is independent of distance fromthe Sun, i.e.,

Exercise 4.3 Radiation is emitted from a planesurface with a uniform intensity in all directions.What is the flux density of the emitted radiation?

Solution:

(4.8)

Although the geometrical setting of this exerciseis quite specific, the result applies generally toisotropic radiation, as illustrated, for example, inExercise 4.31. ■

Performing the integrations over wavelengthand solid angle in reverse order (with solid anglefirst) yields the monochromatic flux density F(

as an intermediate by-product. The relationships

$ %I $ %I [(sin2(%$2) ' sin2(0)]

$ 2%I #%$2

0cos # sin # d#

F $ #2%

I cos # d" $ #2%

)$0 #%$2

#$0I cos # sin # d#d)

Es $ Fs & 4%d2 $ const.

F * d'2

Is $Fs

!"$

1368 W m'2

6.84 & 10'5 sr

Ir

r

R

Er

Fr

ER

FR

R

δ ω

Fig. 4.4 Relationships involving intensity I, flux density F, andflux E of isotropic radiation emitted from a spherical sourcewith radius r, indicated by the blue shading, and incident upona much larger sphere of radius R, concentric with the source.Thin arrows denote intensity and thick arrows denote flux den-sity. Fluxes ER $ Er and intensities IR $ Ir. Flux density Fdecreases with the square of the distance from the source.

P732951-Ch04.qxd 9/12/05 7:41 PM Page 116

Figure 3.3: Relationships between intensity I, flux density F and flux E of isotropic radiationemitted from a spherical source with radius r, indicated by the blue shading, and incident upona much larger sphere of radius R , concentric with the source. Thin arrows denote intensity andthick arrows denote flux density. Fluxes ER = Er and intensities IR = Ir. Flux density F decreaseswith the square of the distance from the source.

• The wavelength of peak radiation is inversely proportional to the object temperature. Equiv-alently, the peak frequency is proportional to the temperature.

• The total amount of energy radiated, summed over all wavelengths, is proportional to thetemperature to the fourth power F µ T 4

• With certain caveats (described later) an object absorbs as effectively as it emits

By definition, a blackbody absorbs all radiation incident upon it. An example is the sun. ThePlanck function, which describes the intensity of radiation emitted from a blackbody is (W m�2

sr�1 frequency�1)

Bn (T ) =2hn3

c2�ehn/kT �1

� (4.1)

which, in terms of wavelength is

Bl (T ) =2hc2

l 5�ehc/klT �1

� (4.2)

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4.3 Blackbody Radiation 117

discussed in this section can be summarized interms of an expression for the flux of radiationemitted by, incident upon, or passing through asurface !A

(4.9)

which is expressed in units of watts (W).

4.3 Blackbody RadiationA blackbody4 is a surface that completely absorbsall incident radiation. Examples include certainsubstances such as coal and a small aperture of amuch larger cavity. The entrances of most cavesappear nearly black, even though the interior wallsmay be quite reflective, because only a very smallfraction of the sunlight that enters is reflected backthrough the entrance: most of the light that entersthe cave is absorbed in multiple reflections off thewalls. The narrower the entrance and the morecomplex the interior geometry of the cave, thesmaller the fraction of the incident light that isreturned back through it, and the blacker theappearance of the cave when viewed from outside(Fig. 4.5).

4.3.1 The Planck Function

It has been determined experimentally that the inten-sity of radiation emitted by a blackbody is given by

(4.10)

where c1 " 3.74 # 10$16 W m2 and c2 " 1.44 #10$2 m K. This mathematical relationship, known asthe Planck5 function, was subsequently verified onthe basis of the theory of quantum physics. It is alsoobserved and has been verified theoretically thatblackbody radiation is isotropic. When B%(T) is

B%(T) "c1%$5

& (ec2!%T $ 1)

E " "!A

"2&

"%2

%1

I% ( ', () d% cos ( d) dA

4 The term body in this context refers to a coherent mass of material with a uniform temperature and composition. A body may be agaseous medium, as long as it has well-defined interfaces with the surrounding objects, media, or vacuum, across which the intensity of theincident and emitted radiation can be defined. For example, it could be a layer of gas of a specified thickness or the surface of a mass ofsolid material.

5 Max Planck (1858–1947) German physicist. Professor of physics at the University of Kiel and University of Berlin. Studied underHelmholtz and Kirchhoff. Played an important role in the development of quantum theory. Awarded the Nobel Prize in 1918.

Fig. 4.5 Radiation entering a cavity with a very small aper-ture and reflecting off the interior walls. [Adapted fromK. N. Liou, An Introduction to Atmospheric Radiation, AcademicPress, p. 10 (2002).]

0

7000 K

6000 K

5000 K

0.5 1.0 1.5 2.00

10

20

30

40

50

60

70

B! (

MW

m–2

µ m

–1 s

r –1 )

! (µ m)

Fig. 4.6 Emission spectra for blackbodies with absolute tem-peratures as indicated, plotted as a function of wavelength ona linear scale. The three-dimensional surface formed by theensemble of such spectra is the Planck function. [Adapted fromR. G Fleagle and J. A Businger, An Introduction to AtmosphericPhysics, Academic Press, p. 137 (1965).]

plotted as a function of wavelength on a linearscale, the resulting spectrum of monochromaticintensity exhibits the shape shown in Fig. 4.6, witha sharp short wavelength cutoff, a steep rise to a

P732951-Ch04.qxd 9/12/05 7:41 PM Page 117

Figure 4.1: A blackbody defined by a cavity where emission and absorption are in equilibrium soas to maintain a constant temperature

What is notable about this function is that it has a peak. We can derive the location of the peak(Wien’s Law) by taking the first derivative of Bl with respect to l and setting to zero to get

lmax = 2902/T (4.3)

Note that this is the peak for Bl not for Bn . If we were to find the maximum of Bn with respect ton and convert back to l the formula would be

lmax = 5107/T (4.4)

Quite different! But why is there a peak? Where does this function come from? A first statementhere is that the derivation of the Planck function is not straightforward, and was approached ini-tially through combinations of classical, statistical and quantum mechanics. But it is an extremelyimportant result, perhaps the first important result of the newly developed quantum theory. Itsingredients have much to say about other topics in atmospheric physics. Here we will hand-wavesome of the more central concepts.

The way this problem was first approached was to think of the radiating body as a bunchof individual oscillators. The total internal energy of the oscillators would be distributed amongtranslational, vibrational, rotational, and electronic energy

utot = utrans + urot + uvib + uelec (4.5)

12

4.3 Blackbody Radiation 117

discussed in this section can be summarized interms of an expression for the flux of radiationemitted by, incident upon, or passing through asurface !A

(4.9)

which is expressed in units of watts (W).

4.3 Blackbody RadiationA blackbody4 is a surface that completely absorbsall incident radiation. Examples include certainsubstances such as coal and a small aperture of amuch larger cavity. The entrances of most cavesappear nearly black, even though the interior wallsmay be quite reflective, because only a very smallfraction of the sunlight that enters is reflected backthrough the entrance: most of the light that entersthe cave is absorbed in multiple reflections off thewalls. The narrower the entrance and the morecomplex the interior geometry of the cave, thesmaller the fraction of the incident light that isreturned back through it, and the blacker theappearance of the cave when viewed from outside(Fig. 4.5).

4.3.1 The Planck Function

It has been determined experimentally that the inten-sity of radiation emitted by a blackbody is given by

(4.10)

where c1 " 3.74 # 10$16 W m2 and c2 " 1.44 #10$2 m K. This mathematical relationship, known asthe Planck5 function, was subsequently verified onthe basis of the theory of quantum physics. It is alsoobserved and has been verified theoretically thatblackbody radiation is isotropic. When B%(T) is

B%(T) "c1%$5

& (ec2!%T $ 1)

E " "!A

"2&

"%2

%1

I% ( ', () d% cos ( d) dA

4 The term body in this context refers to a coherent mass of material with a uniform temperature and composition. A body may be agaseous medium, as long as it has well-defined interfaces with the surrounding objects, media, or vacuum, across which the intensity of theincident and emitted radiation can be defined. For example, it could be a layer of gas of a specified thickness or the surface of a mass ofsolid material.

5 Max Planck (1858–1947) German physicist. Professor of physics at the University of Kiel and University of Berlin. Studied underHelmholtz and Kirchhoff. Played an important role in the development of quantum theory. Awarded the Nobel Prize in 1918.

Fig. 4.5 Radiation entering a cavity with a very small aper-ture and reflecting off the interior walls. [Adapted fromK. N. Liou, An Introduction to Atmospheric Radiation, AcademicPress, p. 10 (2002).]

0

7000 K

6000 K

5000 K

0.5 1.0 1.5 2.00

10

20

30

40

50

60

70B

! (

MW

m–2

µ m

–1 s

r –1 )

! (µ m)

Fig. 4.6 Emission spectra for blackbodies with absolute tem-peratures as indicated, plotted as a function of wavelength ona linear scale. The three-dimensional surface formed by theensemble of such spectra is the Planck function. [Adapted fromR. G Fleagle and J. A Businger, An Introduction to AtmosphericPhysics, Academic Press, p. 137 (1965).]

plotted as a function of wavelength on a linearscale, the resulting spectrum of monochromaticintensity exhibits the shape shown in Fig. 4.6, witha sharp short wavelength cutoff, a steep rise to a

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Figure 4.2: The blackbody spectrum. Note that colder temperatures correspond to lower energiesof emission and longer wavelengths.

Each of these types of energy could be further subdivided into distinct independent modes, eachhaving total energy kT , where k is the Boltzmann constant (1.381 ⇥ 10�23 J/K). Now all thisjiggling associated with the temperature corresponds to displacements of electric charge. As dis-cussed previously this generates an induced radiation field – the dipole radiation – whose mag-nitude varies as frequency n (or w) squared. Intuitively, the more jiggles that happen, the moreenergy that is lost and the cooler the object must get. Absent external inputs, a stove doesn’t stayhot.

In a so-called black-body, we don’t let dipole radiation just radiate away into space. Ratherit is trapped inside an enclosed “black” (non-reflecting) box. Thus, any radiation that is emittedand lost also acts as a source for creating more jiggles. Accordingly, we can imagine an idealizedequilibrium state in which the amount of energy that enters the system (say through a small aperturein the box) is the same as the amount of energy that escapes it.

What is this equilibrium state? Well first, we can think of this as being like any linear system ina condition of steady-state: the rate of change of a substance is proportional to coefficient a timesthe substance (the source) minus coefficient b times the substance (the sink), e.g., for temperature

dTdt

= aT �bT

In equilibrium, a = b. The final step is to determine the absorption of incoming radiation, becausethe absorbed incoming radiation must be equivalent to the loss rate for there to be an equilibrium.

13

In our case, the energy sink is proportional to the amount of energy of the system (kT ) and to thepower of absorption and emission. Things get more complicated here but classical theory suggeststhat a = b µ w2. Ultimately, the central feature is that the radiant intensity is a product of thethermal energy kT and its frequency squared w2 µ n2. This “classical” approach gives us theso-called Rayleigh-Jean’s law

BnRJ =8pn2

c3 kT (4.6)

which looks like the long tail of the Planck function, and where, of course w = 2pn .The problem here is that the Rayleigh-Jeans law just goes to infinity for high frequencies and

short wavelengths. We don’t get fried by X-rays, thankfully. Here’s the safety catch: remember,energy is “quantized”, and the amount of energy in a oscillator mode kT can never be less thanhn . When kT ⌧ hn , the Rayleigh-Jeans assumption that total oscillator energy is kT is no longercorrect. Now, a “quantum” description must take over. It accounts for the rapid dropoff in energyat short wavelengths, high frequencies in B. If energies are sufficiently high that hn � kT , quan-tum mechanics must take over from the classical mechanics perspective because it is impossiblethat anything is smaller than hn . From this perspective, radiant energy decreases with increasingfrequency.

Combining the Rayleigh-Jeans and Quantum mechanical approaches, the desired expressionfor Black-body radiation (Eq. 4.1) is

Bn (T ) =2hn3

c2 (exp(hn/kT )�1)

Which implies a peak in the spectrum as prescribed by Wien’s law (Eq. 4.3). To get the totalintensity averaged over all frequencies we integrate Bn to get (W m�2 sr�1)

B(T ) =Z •

0Bn (T )dn = bT 4

whereb = 2p4k4/

�15c2h3�

Since blackbody radiation is isotropic the blackbody flux is (W m�2)

FBB (T ) = pbT 4 = sT 4 (4.7)

where s = 5.67⇥10�8 J m�2 sec�1 deg�4 is the Stefan-Boltzman constant.Ultimately there are three methods for obtaining F µ T 4:

1. Stefan showed that F = sT 4 based on experimental measurements

2. Boltzmann showed that F µ T 4 based on thermodynamic arguments

3. Planck showed that F = sT 4 by assuming that energy is quantized, obtaining the Planckfunction, and integrating this to obtain s , where s = pb is a mix of fundamental constants.

14

F F

Figure 4.3: A planetary blackbody where emitted thermal radiation from the sphere is in balancewith absorbed solar radiation.

4.1 Atmospheric example: Earth’s energy balanceExercise 4.6 in Wallace and Hobbs

Calculate the equivalanet blackbody temperature of the Earth, assuming a planetary albedo of0.30. Assume that the Earth is in radiative equilibrium.

FE = sT 4E =

(1�A)FS

4=

(1�0.30)⇥13684

= 239.4W m�2

Solving for TE we get

TE =

✓FE

s

◆1/4=

✓239.4

5.67⇥10�8

◆1/4= 255K (4.8)

5 Local Thermodynamic EquilibriumA key atmospheric concept is Local Thermodynamic Equilibrium. If the atmosphere was not inLTE we couldn’t define temperature. Here we describe what this means.

Local thermodynamic equilibrium means that over time scales of interest fluctuations in theBoltzmann distribution between energetic modes are so fast that effectively all modes are in equi-librium with each other. This sounds very abstract but it is critically important because it meansthat we can describe a volume of air as having a temperature and pressure.

With respect to the atmosphere, the energy associated with a molecule has two major compo-nents: the kinetic energy associated with translational motions (molecular movement in space),and the kinetic energy associated with molecular scale energy transitions

utot = utrans + umol

15

The molecular scale energy can in turn be broken up into rotational, vibrational, vibrational-rotational and electronic transition components (plus a few others that we won’t concern ourselveswith here)

utot = utrans + urot + uvib + uelec + ...

Each of these components, as we discussed, has several, modes or degrees of freedom. Transla-tional energy has three degrees of freedom: one each in the x, y, and z directions. For internalenergy, CO2 for example has two rotational degrees of freedom, three distinct vibrational degreesof freedom, and many, many vibrational-rotational degrees of freedom.

Translational energy is what we sense as temperature. A classical treatment works fine here.Each degree of translational freedom has kinetic energy

utransx =12

kT

for a total in the x, y, and z direction (3 DOFs)

ux,trans =32

kT

This energy is what we interpret as “temperature” in daily life (more on this later). It is the kineticenergy of the molecules that causes the pressure on our skin that we interpret as heat.

Molecular energy is more complicated. If we look at the exponential expression that is part ofthe Planck radiation equation

Bn =2hn3

c2�ehn/kT �1

�

you will notice that this radiation depends on temperature also, but in a way that depends not atall on the translational motions of the molecules but rather only the energy transitions (or jiggles)within the molecules themselves. So really, we have two completely different temperatures

Thermal Temperature 32kTtherm

Planck Temperature Bn(TPlanck)

Local Thermodynamic Equilibrium, the precondition for Kirchoff’s Law, and what applies to thebottom 60-70 km of our atmosphere, requires that

Ttherm ⌘ TPlanck (5.1)

How does this happen?Imagine the following sequence of events

1. A molecule maintains a Boltzmann distribution that is a function of its radiating temperatureTPlanck

2. The molecule is bombarded by electromagnetic radiation that, if it is at frequencies corre-sponding to molecular modes, disturbs the Boltzmann distribution, raising molecular ener-gies over all to a higher state.

16

3. Two things can happen here. If left to themselves, the molecule will reestablish a Boltzmanndistribution by releasing photons (energy). This is what causes the Northern Lights. How-ever, if the pressure of the gas is high enough, the molecules collide before this release ofenergy can happen. Instead, through collisions, the molecular energy gets passed betweenmolecules and gets turned into kinetic energy.

4. Through continual absorption of radiative energy, and redistribution as kinetic energy, anequilibrium is maintained between utrans and umol such that

Ttherm ⌘ TPlanck

For atmospheric pressure above 0.05 mb (i.e. 99.5 % of the atmosphere) this conditionapplies.

Maintenance of LTE is usually glossed over in introductory texts because it applies to most of theatmosphere. However, understanding why it occurs is integral to understanding why our planet islivable, i.e. how it is that radiative energy is converted to the thermal energy that keeps our planetwarm. The concept of temperature really is conditional on an assumption that a system is in LTE.

5.1 Kirchoffs LawUnder conditions of Local Thermodynamic Equilibrium, we often hear that Kirchoff’s Law ap-plies, which states that the emissivity of a layer is equivalent to the absorptivity

el = al (5.2)

The premise here is that if an object has a temperature T and an absorptivity al < 1 then it is nota blackbody. Something may have an absorptivity less than unity at a particular wavelength if it istransparent or shiny. A blackbody absorbs all radiation incident upon it at a particular wavelength,hence its emissivity is unity. To see why Kirchoff’s law holds consider that at LTE there must beequilibrium between total absorption and emission. Otherwise temperature would be changing.Thus absorption equals emission and

Z •

0al Bl dl =

Z •

0el Bl dl

implying thatel = al

Note that an object can still absorb more radiation than it emits at a particular wavelength - if,for example, the radiation it absorbs is coming from another object with a higher temperature.

For example, consider object a and object b with Tb ⌧ Ta. Object a radiates energy

Fa = sT 4a

Object b absorbs a fraction of energy from a at a particular wavelength according to its absorptivity:

Fbl (absorbed) = al sT 4a

17

It reemits the energy with emissivityel = al

but at temperature Tb. ThusFbl (emitted) = el sT 4

b

Notice that Fbl (emitted) ⌧ Fbl (absorbed), even though the emissity and absorptivity are thesame, because Tb ⌧ Ta.

6 Why is the sky blue?The atmosphere is made up of oxygen and nitrogen mostly. Oxygen absorbs some solar radiation,but mostly these two molecules only scatter light to make the sky blue. How do they do this? Tounderstand how, we need to understand what is known as dipole radiation.

Consider an electric field incident on an atom represented by an electron surrounding a nucleus,i.e. an electromagnetic “dipole”. From Eq. 1.1, the incident electric field at a fixed point can bedescribed by

~E = ~E0eiwt (6.1)

The incident electric field makes the electric field of the atom oscillate. There’s a fair bit of hand-waving here, but it makes sense that the amplitude of displacement of electric charge within themolecule should be proportional to the incident electric field, i.e.

~x(t) =~x0eiwt µ⇣~E = ~E0eiwt

⌘(6.2)

But, now the molecule is vibrating and moving its electric charge back and forth, it generatesits own electric field, distinct from the incident electric field. This is new electromagnetic fieldrepresents what we term the scattered radiation. Remember that force required to accelerate thecharge qe of an electron, which has mass me, is

F = qeE = mea

An electric field is created by the acceleration of the electric charge created by the incident radia-tion. Therefore the scattered field, or “dipole radiation”, is determined by:

~Edipole (t) µ a(t)

But we know that acceleration of the electron is simply the second derivative of the displacement~x(t), so from Eq. 6.2

a(t) = x(t) = �w2~x0eiwt

Therefore the scattered electric field - the dipole radiation - is given by

~Edipole µ w2~E0 (6.3)

Both the intensity and flux of an electromagnetic wave are proportional to the square of the electricfield. Therefore, because I µ E2, we can write

~Idipole µ w4~I0 (6.4)

18

In other words, the intensity of scattered dipole radiation is proportional to the intensity of theincoming radiation, and the fourth power of the frequency of the radiation.

Now, noting that Nitrogen and Oxygen molecules respond as dipoles to incoming solar radia-tion, we can explain why the sky is blue.

If~Idipole µ w4

then, because w/2p = n = c/l , then

~Idipole µ 1/l 4

Blue has a wavelength of 0.4 µm, and red has wavelength of 0.7 µm. Therefore

Iblue/Ired = 0.74/0.44 = 9

All wavelengths of light are scattered, but blue light is scattered 9 times more effectively than redlight.

Of course, this then begs the question, if short wavelengths are what are favored, than why is thesky not violet? To answer this question, we must consider human physiology and the diminishedsensitivity of the eye to the color violet.

7 Why are clouds white?The concentration of water molecules in clear and cloudy air is about the same. Why then, can wesee clouds, and why do they look white?

To address why we can see clouds, consider the following. The amount of dipole scattering wesee in clear air is proportional to the amount of light scattered and the concentration of moleculesN

~Itotal =~Idipole1 +~Idipole2 +~Idipole3 + . . . = N~Idipole (7.1)

This is why the sky gets darker and darker the higher we go in the atmosphere, until it becomescompletely black. N is getting exponentially smaller with height.

When molecules are in the condensed phase, as they are in water droplets, the molecules areso close together (less than 1 nm) that any electrical vibrations are no longer independent of eachother. Molecules that are close to each other instead prefer to vibrate sympathetically, or “inphase”. This amplifies the response of the dipoles such that

~Itotal =⇣

N~Idipole

⌘·⇣

N~Idipole

⌘= N2~Idipole (7.2)

The number of water molecules in even a tenuous cirrus cloud is probably about N = 1030, soobviously N2

o N, and the cloud is visible where clear is not.But this still doesn’t tell us why clouds are white. We are still looking at dipole radiation in a

cloud, even if the dipoles are all very close together and vibrating in phase.

19

Figure 7.1: Interference pattern created by a plane wave with wavelength l passing through twoslits separated by distance 2r. The phase difference between the two wavefronts Df is a functionof angle from the forward direction q .

Radiation scattered by a single dipoles only always vibrates in phase in the forward scatteringdirection. Off the forward scattering direction, there is positive and negative interference, andwhich it is is a function of particle size and the incident wavelength. The phase difference Df is afunction of the angle between the scattered waves q

Df =2prl

(1� cosq) (7.3)

The relationship between the intensity of the signal for two interfering light sources I1 and I2 andDf is

I1+2 = I1 + I2 +2p

I1I2 cosDf (7.4)

Let’s consider two examples and the special case that I1 = I2. If q ' 0,2p,4p..., or if 2pr ⌧ l ,or 2pr (1� cosq)/l ' 0,2p,4p..., then Df ' 0,2p,4p... and I1+2 ' 4I. This is the case ofconstructive interference.

If, however, q ' p,3p,5p..., or 2pr (1� cosq)/l ' p,3p,5p..., then Df ' p,3p,5p..., andI1+2 ' 0. This is the case of destructive interference..

20

Now let’s consider a given separation between sheets of dipoles r, so 2pr/l is fixed. Thenthere is no cancellation of forward scattered light (1� cosq = 0) and maximum cancellation atback angles (1� cosq = 2), as shown in Fig. 7.1.

Construc)veInterference

Destruc)veInterference

��

Figure 7.2: Interference by sheets of dipoles.

21

Figure 7.3: Interference pattern created by a droplet.

A similar argument can be applied to stacked sheets of dipoles. If two sheets of dipoles areseparated by a distance 2r such that 2pr/l ⌧ 1, then then sheets vibrate close to in phase, actingeffectively as a single sheet of dipoles. But if the sheets are much farther apart and 2pr/l � 1,then there are interference effects and cancellation of dipole radiation from one sheet by anothersheet.

22

Figure 7.4: Interference pattern created by a cloud of droplets, the Corona. Can you explain thecolor separation?

Since the wavelength of visible light is about 0.5 µm, and the distance between water moleculesis about 1000th of that, so, very roughly, we can get about 10003 = 1 trillion water molecules tovibrate in phase. If a particle gets very much bigger than 0.05 µm across, then dipole radiationby the molecules is no longer “coherent” or all in phase with constructive interference. As aparticle grows, some combination of constructive and destructive interference occurs as any twosheets of dipoles span a larger range of separations. Supposing that q = p , then from Eq. 7.3,Df = 4pr/l . Suppose too that there is destructive interference so that Df = p = 4pr/l . As theparticle grows, destructive interference affects the shortest (blue) wavelengths first before it affectsthe longer wavelengths of light. Normally, blue light is scattered preferentially. But now, thescattered color becomes less and less blue, and more and more red. Individual dipoles still scatterblue preferentially, but this is offset by interference effects.

Of course, as the particle grows further, blue shifts back to constructive interference and thered undergoes destructive interference, so this explanation is not enough for why clouds are white.If the particle is large enough compared to the wavelength, then all colors are scattered roughlyequally for particles about 2 µm across (for sunlight). This is difficult to explain, but the bestapproach is to think that for large particles, there are an similarly large number of possible sep-arations of all dipoles with all other dipoles. Ultimately, there is a roughly even combination ofconstructive and destructive interference for all colors and for all combinations of dipoles. Thisinterference scattering signature dominates the blue scattering signature for single dipoles. Clouddroplets are typically at least 10 µm across so clouds scatter all wavelengths roughly equally, i.e.,white.

The words we use for this phenomenon are Rayleigh scattering I µ I0/l 4, which applies to

23

particles that vibrate entirely in phase. In the atmosphere this works for gases, and very verysmall aerosols we can’t see. Mie scattering goes roughly as I µ I0/l 2, and applies to most aerosolparticles that are still small enough to preferentially scatter blue but less effectively so. Geometricscattering applies to particles where a wave treatment is no longer necessary, and which scatterlight of all wavelengths equally (i.e. I µ I0). Clouds particles fall into this regime.

A nice demonstration of various scattering regimes is to make a solution of sodium thiosulfate,into which you titrate sulfuric acid. A precipitate will form that should initially scatter blue andtransmit red, but scatter white and transmit little as the particles grow.

Figure 7.5: Milk is blue if you dilute it with enough water, meaning it is not white due to interfer-ence phenomena.

24

But the reason why clouds are white in fact actually has nothing to do with the fact that allcolors are scattered nearly equally by cloud droplets. Rather it is due to multiple scattering. Toprove this to yourself, milk is white even though really dilute milk scatters blue preferentially.

8 The mathematics of scattering and absorption and emission

8.1 Absorptance and transmittanceThe transmittance of an layer depends on its optical depth t , which in turn depends on how muchof the substance the radiation has to pass through, and how dark the substance is to the radiat-ing wavelength. A stout (cola) transmits less light from a direct beam of radiation than a lager(gingerale), even if in the same sized glass. The direct intensity I of the radiation is attenuatedexponentially, so that the brightness of an object seen through the medium drops of in proportionaccording to the most simple of first-order differential equations

dIldtl

= �Il

This has the standard exponential decay solution

Il (tl ) = Il (0)exp(�tl )

Thus the transmittance isTl =

I (tl )

I (0)= exp(�tl ) (8.1)

and the absorptance isAl = 1� exp(�tl ) (8.2)

Note that as tl ! •, the transmittance goes to 0 and the absorptance goes to 1.Note that Beer’s Law expresses only extinction from a direct beam of radiation. Milk, which

scatters rather than absorbs light, might have the same optical depth as a cola, and attenuate thedirect radiation equally. Beer’s Law says nothing about whether a direct beam of radiation isattenuated due to scattering (as in the case of milk) or absorption (as in the case of a cola). However,the milk looks much brighter because it is scattering radiation in all directions, and this radiationbecomes multiply scattered. Multiple scattering mean that the radiation is just re-directed andhomogenized (made more isotropic). This is a more advanced topic.

8.2 Optical depthHow do we express t in terms of physically meaningful parameters? There are a number of differ-ent ways that are used, all related. At the most fundamental level, every molecule has an effectiveabsorption cross-section sa or scattering cross-section ss that has units of area. What this repre-sents is the effective cross-sectional area of the molecule that absorbs all radiation incident uponit. For example a glass of cola is black, and probably has an absorption cross-section close to it’sactual cross-sectional area normal to the plane of incident radiation. The same glass filled withginger-ale would have a much smaller absorption cross-section, even though the glass itself has

25

the same physical cross-sectional area. Ginger-ale is lighter in color. The absorption and scatter-ing cross-sections are related to the geometric cross-section s through a scattering or absorptionefficiency Kl , which is dimensionless. Thus

ssl = Ksl s (8.3)

sal = Kal s (8.4)

so, for exampleKs =

ss

s(8.5)

The volume absorption (or scattering) coefficient ba is the product of sa and the concentrationof molecules or particles N in units of #/m3

bal = Nsal = NKal s (8.6)

bsl = Nssl = NKsl s (8.7)

Thus ba has units of 1/m. Since the optical depth t is dimensionless it follows that

dtal = bal ds = Nsal ds = NKal sds (8.8)

dtsl = bsl ds = Nssl ds = NKsl sds (8.9)

where ds is the pathlength of the radiation. The more soda or beer you peer through, the harder itis to see to the other side.

It is also common to express ba in terms of a mass absorption coefficient ka in units of m2/kgsuch that

dtal = karwds (8.10)

where r is the density of air and w is the mixing ratio of the gas in the air. This is very close to afinal, very commonly used formulation that defines the ta in terms of the mass path u with units ofg m�2 such that

du = rwds (8.11)

anddta = kadu (8.12)

Of course, it is important to recognize that in the atmosphere u = u(z) and therefore, if the masspath is defined with respect to a vertical column.

ta (z) = ka

Z •

z

du(z)dz

dz (8.13)

where • refers to the top of the atmosphere. Note that dz = cosqds where ds is the slant path.

26

Figure 8.1: Illustration of the slantpath through a vertical layer.

8.3 Transmittance in a plane parallel atmosphereThus according to Eq. 8.1

T (z) = exp(�kau)

or, coming back to the expression for exponential decay of intensity, we must modify

dIldtl

= �Il

Il (tl ) = Il (0)exp(�tl )

toµ dIl

dtl= �Il

Il (tl ) = Il (0)exp(�tl /µ) (8.14)

where, µ = cosq , where q = 0 and µ = 1 for the up direction, q = 90 and µ = 0 for the sidedirection. and What this shows is that for a particular atmospheric optical depth, radiation is

27

attenuated most when it is coming it at side angles, because it must pass through much moreatmosphere before coming out the other side than if it is passing straight through. There are afew other important “single-scattering” parameters that are necessary to describe the full radiationfield:

8.4 Size parameterThe size parameter x is a dimensionless number that express the size of the droplets relative to thewavelength of incident radiation

x = r |k| = 2pr/l (8.15)

where a is the particle radius, and l is the wavelength of the light. Remember that we have shownthat the size of a particle relative to incident radiation that is quite important in determining howlight is scattered.

4.4 Physics of Scattering and Absorption and Emission 123

convenient to express the rate of scattering orabsorption in the form

(4.17)

where ! is the density of the air, r is the mass of theabsorbing gas per unit mass of air, and k" is the massabsorption coefficient, which has units of m2 kg#1.

In the aforementioned expressions the productsN$K" and !rk" are volume scattering, absorption, orextinction coefficients, depending on the context,and have units of m#1. The contributions of thevarious species of gases and particles are additive(i.e., K"N$ % (K")1 N1$1 & (K")2 N2$2 & . . . .), asare the contributions of scattering and absorptionto the extinction of the incident beam of radiation; i.e.,

(4.18)

4.4.1 Scattering by Air Moleculesand Particles

At any given place and time, particles includingaerosols with a wide variety of shapes and sizes, aswell as cloud droplets and ice crystals, may be pres-ent. Nonetheless it is instructive to consider thecase of scattering by a spherical particle of radius r,for which the scattering, absorption, or extinctionefficiency K" in (4.16) can be prescribed on the

& K"(absorption)K"(extinction) % K"(scattering)

dI" % #I"!rk"ds

basis of theory, as a function of a dimensionless sizeparameter

(4.19)

and a complex index of refraction of the particles(m % mr & imi), whose real part mr is the ratio ofthe speed of light in a vacuum to the speed atwhich light travels when it is passing through theparticle. Figure 4.11 shows the range of size param-eters for various kinds of particles in the atmos-phere and radiation in various wavelength ranges.For the scattering of radiation in the visible part ofthe spectrum, x ranges from much less than 1 forair molecules to !1 for haze and smoke particles to''1 for raindrops.

Particles with x (( 1 are relatively ineffective atscattering radiation. Within this so-called Rayleighscattering regime the expression for the scatteringefficiency is of the form

(4.20)

and the scattering is divided evenly between theforward and backward hemispheres, as indicated inFig. 4.12a. For values of the size parameter compara-ble to or greater than 1 the scattered radiation isdirected mainly into the forward hemisphere, as indi-cated in subsequent panels.

Figure 4.13 shows K" as a function of size parame-ter for particles with mr % 1.5 and a range of valuesof mi. Consider just the top curve that corresponds

K" ) "#4

x %2*r

"

I! – dI!

dz

ds = sec θ dz

I !

θ

Fig. 4.10 Extinction of incident parallel beam solar radia-tion as it passes through an infinitesimally thin atmosphericlayer containing absorbing gases and/or aerosols.

Wea

ther

rada

r

Sol

arra

diat

ion

Ter

rest

rial

radi

atio

n

1 10 102 103 104 105

104

103

102

10

1

10–1

10–2

10–3

Geometric optics

Mie scatterin

g

Rayleigh scatterin

g

Negligible scatterin

g

Raindrops

Drizzle

Clouddroplets

Smoke,dust, haze

Airmolecules

r (µ

m) x = 1

! (µ m)

Fig. 4.11 Size parameter x as a function of wavelength (")of the incident radiation and particle radius r.

P732951-Ch04.qxd 9/12/05 7:41 PM Page 123

Figure 8.2: In this plot, the size parameter increases as one moves from the Rayleigh Scatteringregime to the geometric optics regime.

8.5 Single-scattering albedoThe single-scattering albedo w0 ranges from 0 to 1 and represents the fraction of energy incidenton a particle that is scattered rather than absorbed. i.e.

w0 =Ks

Ks +Ka=

ss

ss +sa=

ts

ts + ta(8.16)

28

124 Radiative Transfer

to mi ! 0 (no absorption). For 1 " x # 50, referredto as the Mie10 scattering regime, K$ exhibits adamped oscillatory behavior, with a mean around avalue of 2, and for x % 50, the range referred to as

the geometric optics regime, the oscillatory behavioris less prominent and K$ ! 2.

Exercise 4.9 Estimate the relative efficiencies withwhich red light ($ " 0.64&m) and blue light($ " 0.47&m) are scattered by air molecules.

Solution: From (4.20)

■

Hence, the preponderance of blue in light scattered byair molecules, as evidenced by the blueness of the skyon days when the air is relatively free from aerosols.

Figure 4.14 shows an example of the coloring ofthe sky and sunlit objects imparted by Rayleigh scat-tering. The photograph was taken just after sunrise.Blue sky is visible overhead, while objects in theforeground, including the aerosol layer, are illumi-nated by sunlight in which the shorter wavelengths(bluer colors) have been depleted by scattering alongits long, oblique path through the atmosphere.

Ground-based weather radars and remote sensingof rainfall from instruments carried aboard satellitesexploit the size strong dependence of scatteringefficiency K upon size parameter x for microwaveradiation in the 1- to 10-cm wavelength range inci-dent upon clouds with droplet radii on the order ofmillimeters. In contrast to infrared radiation, which

K(blue)K(red)

! #0.640.47$

4! 3.45

(a) (b)

(c)

Incident Beam

Forward

Fig. 4.12 Schematic showing the angular distribution of theradiation at visible (0.5 &m) wavelength scattered by sphericalparticles with radii of (a) 10'4 &m, (b) 0.1 &m, and (c) 1 &m.The forward scattering for the 1-&m aerosol is extremely largeand is scaled for presentation purposes. [Adapted fromK. N. Liou, An Introduction to Atmospheric Radiation, AcademicPress, p. 7 (2002).]

10 Gustav Mie (1868–1957) German physicist. Carried out fundamental studies on the theory of electromagnetic scattering and kinetictheory.

0

Sca

tterin

g ef

ficie

ncy

K!

Size parameter x1 5 10 50 100

1

2

3

4

5

mi = 1

mi = 0.1

mi = 0.01

mi = 0

Fig. 4.13 Scattering efficiency K$ as a function of sizeparameter x, plotted on a logarithmic scale, for four differentrefractive indices with mr ! 1.5 and mi ranging from 0 to 1, asindicated. [From K. N. Liou, An Introduction to AtmosphericRadiation, Academic Press, p. 191 (2002).]

Fig. 4.14 Photograph of the Great Wall of China, taken justafter sunrise.

P732951-Ch04.qxd 9/12/05 7:42 PM Page 124

Figure 8.3: The extinction efficiency as a function of size parameter for a range of values of thecomplex index of refraction. More absorption brings the extinction efficiency closer to unity and asingle scattering albedo of 0.5.

In the visible, clouds and Rayleigh scattering gases like N2 and O2 have a single scattering albedounity (i.e. no absorption). In their IR absorption bands, gases have a single-scattering albedo ofzero (i.e no scattering).

8.6 Phase functionIf light is scattered (w0 > 1), the probability it is scattered in any given direction is expressed by thephase function p(µ), where µ = cosq , where q is relative to the direction of the incoming beam.µ = 1 corresponds to forward scattering (the same direction as the incoming beam), and µ =�1 corresponds to back-scattering (towards the light source). With the exception of such thingsas mirrors, light is scattered over a distribution of angles. Cloud particles are strongly forwardscattering of visible light, but they have more isotropic Rayleigh scattering of radar wavelengths.Gases have isotropic Rayleigh scattering in the visible also.

In general p(µ) is normalized such that

12

Z 1

�1p(µ)dµ = 1 (8.17)

29

124 Radiative Transfer

to mi ! 0 (no absorption). For 1 " x # 50, referredto as the Mie10 scattering regime, K$ exhibits adamped oscillatory behavior, with a mean around avalue of 2, and for x % 50, the range referred to as

the geometric optics regime, the oscillatory behavioris less prominent and K$ ! 2.

Exercise 4.9 Estimate the relative efficiencies withwhich red light ($ " 0.64&m) and blue light($ " 0.47&m) are scattered by air molecules.

Solution: From (4.20)

■

Hence, the preponderance of blue in light scattered byair molecules, as evidenced by the blueness of the skyon days when the air is relatively free from aerosols.

Figure 4.14 shows an example of the coloring ofthe sky and sunlit objects imparted by Rayleigh scat-tering. The photograph was taken just after sunrise.Blue sky is visible overhead, while objects in theforeground, including the aerosol layer, are illumi-nated by sunlight in which the shorter wavelengths(bluer colors) have been depleted by scattering alongits long, oblique path through the atmosphere.

Ground-based weather radars and remote sensingof rainfall from instruments carried aboard satellitesexploit the size strong dependence of scatteringefficiency K upon size parameter x for microwaveradiation in the 1- to 10-cm wavelength range inci-dent upon clouds with droplet radii on the order ofmillimeters. In contrast to infrared radiation, which

K(blue)K(red)

! #0.640.47$

4! 3.45

(a) (b)

(c)

Incident Beam

Forward

Fig. 4.12 Schematic showing the angular distribution of theradiation at visible (0.5 &m) wavelength scattered by sphericalparticles with radii of (a) 10'4 &m, (b) 0.1 &m, and (c) 1 &m.The forward scattering for the 1-&m aerosol is extremely largeand is scaled for presentation purposes. [Adapted fromK. N. Liou, An Introduction to Atmospheric Radiation, AcademicPress, p. 7 (2002).]

10 Gustav Mie (1868–1957) German physicist. Carried out fundamental studies on the theory of electromagnetic scattering and kinetictheory.

0

Sca

tterin

g ef

ficie

ncy

K!

Size parameter x1 5 10 50 100

1

2

3

4

5

mi = 1

mi = 0.1

mi = 0.01

mi = 0

Fig. 4.13 Scattering efficiency K$ as a function of sizeparameter x, plotted on a logarithmic scale, for four differentrefractive indices with mr ! 1.5 and mi ranging from 0 to 1, asindicated. [From K. N. Liou, An Introduction to AtmosphericRadiation, Academic Press, p. 191 (2002).]

Fig. 4.14 Photograph of the Great Wall of China, taken justafter sunrise.

P732951-Ch04.qxd 9/12/05 7:42 PM Page 124

Figure 8.4: A graphical representation of the phase function with forward scattering at zero degreesto the right for a) Rayleigh Scattering b) Mie Scattering and c) the Geometric Optics regime. Notethat the larger the size parameter, the more strongly forward scattering is the particle.

9 Energy transitions of molecular absorptionWe have discussed how the internal energy of a molecule can be separated into translational, ro-tational and vibrational modes. Energy can be added or subtracted from each of these modes bythe absorption/emission of a photon of electromagnetic radiation. The energy can also change thecharge distribution within the molecule, or even break apart the molecule if the photon is suffi-ciently high energy (as in the case of O2 and UV radiation).

9.1 The simplest case: The Hydrogen atomFor example the absorption of a photon by a hydrogen atom will raise the energy level of theelectron from its ground state to an excited state. The frequency of the photon that is associatedwith this energy transition is

DE = hn (9.1)

Now, at the molecular level energy is quantized. Without going into details, what this means is thatonly certain discrete frequencies can be absorbed in order to excite a transition in energy level. Fora hydrogen atom these are

n = RH

✓1j2 �

1k2

◆

where j and k are integers associated with the energy level of the electron (ground state = 1).

30

Figure 8.5: The phase function of water droplets with dominant optical features (M. King et al.).

9.2 Polyatomic molecules and rotational transitionsBut we don’t care about monatomic molecules, since the only monatomic molecules in the atmo-sphere are Nobel gases like argon and krypton that require really really high energies to be excited(as seen in neon signs). In addition to the three translational degrees of freedom (whose energydepends only on temperature not radiation), there is rotational and vibrational energy.

For example monatomic molecules have zero degrees of rotational freedom. Symmetric lineartriatomic molecules like CO2 have one degree of rotational freedom (two have the same energy),and asymmetric molecules such as H2O have three independent modes.

In quantum mechanics the angular momentum associated with molecular rotation is

L =h

2pp

J (J +1) (9.2)

where J is an integer rotational quantum number.In classical mechanics and quantum mechanics, the energy associated with rotation is

E = L2/2I (9.3)

where I is the moment of inertia. Therefore, the discrete energy associated with a particularmolecular mode of rotation is

EJ (J) =12I

✓h

2p

◆2J (J +1) (9.4)

31

Figure 8.6: The phase function of ice crystals with dominant optical features (M. King et al.).

The frequencies associated with molecular transitions are

DE = EJ0 �EJ” = hn (9.5)

wheren = BnJ0

�J0 +1

��BnJ”(J”+1) (9.6)

where Bn = h/�8p2cI

�. It turns out that in the rules of quantum mechanics energy transtions that

might be associated with absorption of EM radiation are only associated with

DJ = J”� J0 = ±1

. The frequencies associated with pure rotational transitions tend to be low energy in the 10 to 100cm�1 range.

An important note here is that the molecule must have a dipole moment. In other words, oneend of the molecule must be more positive than the other. If this is not the case, there is no potentialfor an incoming electromagnetic wave to exert “torque” on the molecule. As examples, nitrogen,oxygen and carbon dioxide are linearly symmetric, so they have no dipole moment. Consequentlythey have no purely rotational molecular transitions.

The greenhouse gas N2O on the other hand, with structure N-N-O, does have a dipole moment.It is linear but asymmetric. It has two modes of rotation, one along an axis lying vertically alongthis page, and a second lying along an axis pointing into the page. These two modes are called“degenerate” because they differ not in the amount of energy required but only in what directionyou are looking at the molecule. A third mode, with axis lying horizontal to this page, goes straightthround the length of the molecule, so it has zero energy.

32

126 Radiative Transfer

• coronas produced by the diffraction of light inwater droplets in low or (sometimes) middlecloud decks. They consist of colored rings atan angular radius of less than 15° from thelight sources, e.g., the sun or moon. If thecloud droplets are fairly uniform in size,several sequences of rings may be seen: thespacing of the rings depends upon dropletsize. In each sequence the inside ring is violetor blue and the outside ring is red. An exampleis shown in Fig. 4.19.

4.4.2 Absorption by Particles

The absorption of radiation by particles is ofinterest in its own right, and affects scattering aswell. The theoretical framework discussed in theprevious subsection (commonly referred to asMie theory) predicts both the scattering and theabsorption of radiation by homogeneous sphericalparticles, where the real part of the refractiveindex relates to the scattering and the imaginarypart relates to the absorption. This brief sectionmentions just a few of the consequences of absorp-tion. In the plot of K versus x (Fig. 4.13) the pres-ence of absorption tends to damp the oscillatorybehavior in the Mie regime. In the limit of x !! 1,

the extinction coefficient always approaches 2 but,in accordance with (4.18) the scattering coefficient,ranges from as low as 1 in the case of strong absorp-tion to as high as 2 for negligible absorption. In thelongwave part of the electromagnetic spectrum,cloud droplets absorb radiation so strongly thateven a relatively thin cloud layer of clouds behavesas a blackbody, absorbing virtually all the radiationincident from above and below.

Refractingangle 90°

Minimum angle of deviation 46°

Minimum angle of deviation 22°Refracting

angle 60°

Light from sun

Cloud of hexagonal ice columns

Observer

46° halo

22° halo22°

46°

Fig. 4.18 Refraction of light in hexagonal ice crystals to produce the 22° and 46° haloes.

Fig. 4.19 A corona around the sun produced by the diffrac-tion of light in cloud droplets. [Photograph courtesy of HaraldEdens.]

P732951-Ch04.qxd 9/12/05 7:42 PM Page 126

Figure 8.7: Professor Peter V. Hobbs with his ice crystals halos

Water, however, which has structure H/O\H has a different rotational mode along all threeaxes. In the atmosphere, purely rotational water modes matter most at very long wavelengthsgreater than 20 µm, where water has a very complex rotational structure. This is because severalmodes of rotation can be excited simultaneously.

9.3 Polyatomic molecules and vibrational transitionsVibrational transitions are possible provided there is a molecular bond. A molecular bond is like aspring. The classical analogy is Hooke’s law where

F = �k�r0 � r

�(9.7)

where r0 is the equilibrium separation, and the spring resonates like a simple harmonic oscillatorwith resonant frequency

n 0 =1

2pp

k/m0

(9.8)

where m0 is the reduced mass and the energy at each frequency is E = hn 0. In the quantum world,allowable vibrational transitions have energy

DEn = v0hn 0 � v”hn 0

where the quantum mechanical rule is

Dv = v0 � v” = ±1

33

v is ±1. Thus, the frequency of photons that excite a transition is given by

DEn = hn = |Dv|hn 0

orn = |Dv|n 0

With a simple molecule like O-O, there is only one vibrational mode. With triatomic moleculesthere are normally three: the v1 mode, or “symmetric stretch” mode, the v2 or “bending” modeand the v3 or “asymmetric stretch” mode. Vibrational transitions are higher energy than rotationaltransitions and are in the 1000 cm�1 range. A nice illustration of vibrational motions is here.

9.4 Vibrational-rotational spectra

Figure 9.1: Vibrational - rotational transitions for Dv = +1 and DJ = +1

It gets more complicated. CO2, as stated previously, has no purely rotational modes. But clearlyit can have vibrational modes due to vibrational stretching of the molecular bonds. For example, thegreenhouse effect associated with CO2 is due to the n2 bending mode. If the stretching is asymmet-ric, than it can induce a dipole moment in the molecule, which then makes the molecule susceptibleto rotational transitions. There are huge numbers of options for such vibrational-rotation transi-tions, and they account for much of the complexity in the absorption bands seen by various gasesin the atmosphere. A simple illustration of these transitions is shown in Fig. 9.1. The important15 µm (670 cm�1) absorption band for CO2 and the 6.7 µm band for water vapor (used for watervapor satellite imagery) are both vibrational-rotational bands The CO2 band is shown in Fig. 9.2.

34

Figure 9.2: Vibrational - rotational transitions for CO2

10 Molecular absorption profiles

10.1 Line strengthsThe strength of any particular absorption line (i.e. the amount of energy associated with any par-ticular molecular transition) can be derived from quantum theory or measured in a lab. Databasesexist for these quantities. The amount of energy absorbed in the atmosphere is going to dependon the number of molecular transitions that occur, which in turn will depend on the concentra-tion of molecules. The symbol used for the line strength is S. We’ll show later how this affectsatmospheric absorption.

10.2 Line profilesBut the line is not infinitely narrow, rather it has a characteristic shape or profile that means that arange of frequencies are absorbed for any individual molecular transition.

Imagine Hooke’s law again (Eq. 9.7).

F (x) = �kx = mx

Solving the ode, the oscillator has a single natural frequency that it can be forced at

w0 =p

k/m0 (10.1)

However there is always a frictional damping force which is usually proportional to the velocityv. In the atmosphere this frictional damping force is associated with collisions between molecules.

35

This induces pressure broadening of the absorption line, smearing the range of frequencies atwhich the molecule will undergo a molecular transition. Note that w0 = 2pn0 or w0 = 2pcn0.

4.4 Physics of Scattering and Absorption and Emission 129

and the points at which the amplitude is equal to halfthe peak amplitude) is , m is the mass of themolecule and k is the Boltzmann’s constant (1.381 !10"23 J K"1 molecule"1).

The shape factor for pressure broadening, com-monly referred to as the Lorentz13 line shape, isgiven by

(4.28)

In this expression the half-width of the line is deter-mined by

(4.29)

which is proportional to the frequency of molecularcollisions. The exponent N ranges from 1!2 to 1depending on the molecular species.

Shapes of absorption lines of the same strengthand half-width, but broadened by these two dis-tinctly different processes, are contrasted inFig. 4.21. The “wings” of the absorption lines shapedby pressure broadening extend out farther from thecenter of the line than those shaped by Dopplerbroadening. For a water vapor line at 400 cm"1 anda temperature of 300 K, the Doppler line width is7 ! 10"4 cm"1. A typical water vapor line width forair at the same temperature at the Earth’s surface is"100 times wider due to the presence of pressurebroadening.14 Below "20 km, pressure broadeningis the dominant factor in determining the width ofabsorption lines, whereas above 50 km, wheremolecular collisions are much less frequent,Doppler broadening is the dominant factor. In theintermediate layer between 20 and 50 km, the lineshape is a convolution of the Doppler and Lorentzshapes.

# $ pTN

f %#

& [(' " '0)2 ( #2]

#D√ln 2

Laboratory measurements of absorption spectraexist for only a very limited sampling of pressuresand temperatures. However, through the use of theo-retically derived spectra, adjusted empirically toimprove the fit with existing measurements, atmos-pheric physicists and climate modelers are able tocalculate the absorption spectra for each of theradiatively important atmospheric gases for anyspecified thermodynamic conditions.15 An exampleshowing the excellent agreement between observedand theoretically derived absorption spectra is shownin Fig. 4.22. Note the narrowness of the lines, evenwhen the effects of Doppler and pressure broadeningare taken into account. The greatest uncertainties intheoretically derived absorption spectra are in theso-called “continua,” where the superposition of theoutermost parts of the wings of many different linesin nearby line clusters produces weak but in somecases significant absorption.

13 Hendrick Antoon Lorentz (1853–1928) Dutch physicist. Won Nobel prize for physics in 1902 for his theory of electromagneticradiation, which gave rise to the special theory of relativity. Refined Maxwell’s theory of electromagnetic radiation so it better explainedthe reflection and refraction of light.

14 The dominance of pressure broadening is mainly due to the fact that, for typical temperatures and pressures in the lower atmosphere,# )) #D. The difference in line shape also contributes, but it is of secondary importance.

15 The most comprehensive archive of these theoretically derived absorption spectra, the high-resolution transmission molecularabsorption (HITRAN) data base, contains spectra for many different gases. This data base contains line intensities, the wave numbers onwhich the lines are centered, the pressure half-widths and lower energy levels at reference temperatures and pressures for over a millionabsorption lines.

–5 –4 –3 –2 –1

Doppler

Lorentz

(ν − ν0)/α1 2 3 4 50

k ν

Fig. 4.21 Contrasting absorption line shapes associatedwith Doppler broadening and pressure broadening. Areasunder the two profiles, indicative of the line intensity S, arethe same. [Courtesy of Qiang Fu.]

P732951-Ch04.qxd 9/12/05 7:42 PM Page 129

Figure 10.1: Lorentz and Doppler profile expressed in terms of the absorption coefficient kn wherethe absorption optical depth is t = knu where u is the column amount of the absorbing gas.

Expressed in terms of a frequency n , the line shape takes on a characteristic shape known asthe Lorentz profile defined by

f (n �n0) =aL/p

(n �n0)2 +a2

L(10.2)

where aL (which replaces g) is the half width of the spectrum at half-maximum and is proportionalto temperature and pressure

aL = a0

✓pp0

◆✓T0

T

◆1/2(10.3)

a0 comes from lab studies. Typical values of a0 lie in the range 0.01 to 0.1 cm�1. Note that thefactor of p in the numerator is there so that

Z •

�•f (n �n0)dn = 1 (10.4)

Essentially, this normalization means that adding up the energetic contributions to single line dueto all frequencies gives the total.

36

There is also something called Doppler broadening associated with the fast motions of themolecules. It spreads out frequencies the same way that the siren from an ambulance changespitch depending on whether it is coming towards you are moving a way from you. But this onlybegins to matter where atmospheric pressures are very low. For most of the atmosphere belowabout 20 km, pressure broadening dominates. Finally, there is inescapable natural broadening dueto the Heisenberg uncertainty principle. If you are familiar with this, the upshot is that quantummechanics dictates that you can’t narrow down your estimate of the amount of energy absorbedwithout broadening your uncertainty in just which frequency this energy is getting absorbed at. Inthe atmosphere this effect is generally small compared to either Doppler or pressure broadening.

11 Line absorptionAs we describe earlier, individual lines actually cover a range of frequencies, due in particular topressure broadening. There are a number of ways to deal with this particularly thorny problem.We will deal with just one here, because it illustrates a particularly interesting property of theatmospheric greenhouse effect.

First we need to recognize that for one single absorption line (one individual molecular tran-sition) the mass absorption coefficient is spread over a wide range of wavenumbers. From Eq.10.2

ka (n) = S f (n) =SaL/p

(n �n0)2 +a2

L(11.1)

where S is the strength of the line, and has the same units as ka (meters squared per kilogram). Notethat in the hypothetical instance that there were no broadening of any kind then S and ka would beequivalent because the integral of f (n) is 1.

Now if we want to find the transmittance associated with an individual line, it wouldn’t besensible just to limit our selves to the center of the profile. Rather we would want to calculate aband - averaged transmittance T representative of a range of frequencies Dn that covers at leastone or two line widths aL on either side of the centerline n0. From Eq. 8.1

T =1

Dn

Z

nexp(�tn)dn (11.2)

wheretn = kau = S f (n)u (11.3)

The band averaged absorptance isA =1�T (11.4)

11.1 Weak and strong linesAs u increases, the transmittance goes down and the absorptance goes up. If u is sufficiently large,the value of t at the centerline n0 becomes large enough that the transmittance is 0 and the absorp-tance is 1. That particular wavelength becomes saturated. As u continues to increase the wings

37

0

1

0.3 0.4 0.5 0.6 0.8 1 1.2 1.5 2 2.5 3 4 5 6 7 8 910 12 15 20 25 30 40 50

Tran

smitt

ance Total

0

1

Tran

smitt

anceH2O

0

1

Tran

smitt

ance CO2

0

1

Tran

smitt

anceO3

0

1

Tran

smitt

ance O2

0

1

Tran

smitt

anceCH4

0

1

Tran

smitt

ance N2O

0

1

Tran

smitt

ance

Wavelength [µm]

CO

ZENITH ATMOSPHERIC TRANSMITTANCE

UV VIS Near IR Thermal IR

Figure 11.1: Absorption profiles for various atmospheric gases.

of the Lorentz profile become saturated also, starting from the center and moving progressivelyfurther outward.

So it seems that after a certain point you don’t get as much bang for your buck as you increaseu. We define weak lines as those that are nowhere saturated and strong lines as those that aresaturated at the very least at the center line. How do we quantify this effect?

From Eqs. 11.2 and 11.3, the band averaged transmittance of a line, as described above is

T =1

Dn

Z

nexp(�S f (n)u)dn

If we substitute the Lorentz profile for f (n) (Eq. 10.2) we end up with a rather hairy integral

T =1

Dn

Z

nexp

�S

aL/p(n �n0)

2 +a2L

u

!dn

38

It has a solution called the Landenberg-Reiche function (named after the folks who solved it) thatinvolves Bessel functions of the first kind of order 0 and 1. You are probably just as happy notto deal with this. The important thing to recognize is that in the limit of small values of u (weaklines), the above integral has the solution

T = 1� SuDn

(11.5)

and for large values of u (strong lines)

T = 1� 2p

SaLuDn

(11.6)

Thus for weak lines absorptance increases linearly with u and for strong lines it increases as thesquare root of u. Changes to greenhouse gas concentrations are most potent if the lines are initiallyweak.

12 Molecular absorption, the greenhouse effect, and climatechange

The greenhouse effect is a measure of how much atmospheric gases trap terrestrial radiative flux,reducing the amount of energy that escapes to outer space, and increasing the equilibrium temper-ature of the planet.

The primary greenhouse gases are water vapor (75 % of the total), CO2 (32 %), ozone (10 %)and methane CH4 (8 %). Note that there is overlap between bands, which is why the total is greaterthan 100%.

Now lets compare CO2 and CH4, which are increasing rapidly in the atmosphere due to humanactivities. The major source of CO2 is fossil-fuel combustion, which powers the world economy.Its current concentration is about 380 ppm. Methane is produced by microbes of the archaea familyin the guts of livestock (particularly cattle), in landfills, rice paddies, and whenever some area isflooded. Its current concentration is 1.85 ppm.

A great many of the absorption lines in the atmosphere are already saturated at their centers.This is particularly true of CO2 in its 15 µm vibrational-rotational absorption band, where thebroadband emissivity is 0.761 (it is unity around the centerline). Thus CO2 is a strong line. CO2is well-mixed in the atmosphere, so the implication of this result is that by doubling CO2 concen-trations in the atmosphere - something we are well on our way to doing - we only increase thegreenhouse effect due to CO2 by a factor of

p2 or about 40%. An additional complication to this

calculation is that there are also water vapor rotational bands at 15 µm, so the increase is attenuatedeven further.

Other greenhouse gases in the atmosphere are not yet saturated. Perhaps most important ofthese is methane CH4. At 7.6 µm methane has a vibrational-rotational band that is not quiteyet saturated but is located at a region of the terrestrial blackbody spectrum where there is stillsignificant energy. The broadband emissivity of this wavelength band is 0.166 - much lower thanCO2. Since the lines are not quite yet saturated the greenhouse climate forcing by methane follows

39

Figure 12.1: The Keeling Curve showing the increase in CO2 concentrations in the atmosphere

the weak line rule, and it is linearly proportional to its concentration. A doubling of methaneconcentrations will approximately double the methane greenhouse forcing.

We need to note however that the above arguments have an oversimplification. The wings ofindividual lines overlap. This further reduces the impact of doubling u. The extent of this effect isbeyond the scope of this course.

The climate sensitivity expresses the temperature change per doubling of CO2 concentrations.Figure 12.3 summarizes some estimates of climate sensitivity based on numerical model and paleo-climate estimates. (From Skinner 2012): Symbols indicate estimates of past global average ra-diative forcing anomalies and the corresponding global average temperature change, based onpublished data. Radiative forcing estimates include land albedo and greenhouse gas effects (andsometimes the impacts of atmospheric dust). Anomalies are relative to pre-industrial values, ex-cept where indicated. Uncertainties are 1sv or conservative estimates where published uncertaintiesare unavailable. Yellow shading indicates a summary probability distribution for numerical model-derived climate sensitivity values (18); dashed gray lines indicate 1�C increments. The solid blackline is a linear regression on all the data, forced through the origin and with 95% confidence limits(dotted black lines). The “ball-park” paleoclimate sensitivity estimates shown here range widelybetween 0.6�C and 6.5�C, but taken together imply a climate sensitivity of ~3�C, which is in veryclose agreement with the best estimate derived from numerical models. However, this agreementmay mask evidence for nonlinear feedbacks and abrupt climatic transitions that are not capturedin the climate sensitivity as commonly defined. LGM, Last Glacial Maximum; PETM, Paleocene-Eocene Thermal Maximum.

40

Figure 12.2: Trends in methane in the atmosphere

41

Figure 12.3:

42