Figure 1: Electromagnetic wave

Electromagnetic waves

An electric generator rotates a magnet to create electricity. An magnetic resonance imager foundin a hospital creates a magnetic field by passing a large electric current in a circle. A changingelectric field induces a magnetic field and a changing magnetic field induces an electric field. Themagnetic field is in each case normal to the electric field.

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� i!t

⌘

~H =

~H0 exp

⇣i~k · ~x� i!t

⌘

where E is the electric field and H is the magnetic field, the real component of ~k is the wavenumber2⇡/�, ~x is the direction of wave propagation and ! is the angular frequency of the radition 2⇡⌫.

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 ✏ 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 vacuumhave the symbols ✏0 and µ0. The instantaneous flux density of the electric field in the direction of

1

the wave is defined by the Poynting vector

~S =

~E ⇥ ~H

In a vacuum this can be simplified to

S =

E20

µ0ccos

2(!t)

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

Sav =

1

2

Re(

~E0 ⇥ ~H0)

orSav =

1

2

c✏0E20

Sav has units of W m�2 (or J s�1 m�2) and is a function of wavelength. This is units of flux (quantityper length squared per second). In atmospheric sciences, we more commonly use the symbol F .The point here is that the flux of energy F is proportional to the square of the magnitude of anelectromagnetic wave. It is the flux density of electromagnetic waves that is of primary interest toclimate studies.

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

~E =

~E0 exp

⇣i~k · ~x� i!t

⌘

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

~k =

~k0 + i ~k”

Therefore, the consequence of k being complex is the following

~E = E0 exp

⇣� ~k” · ~x

⌘exp

⇣i⇣~k0 · ~x� !t

⌘⌘

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

� =

~k0 · ~x� !t

At constant phased� = 0 =

~k0 · d~x� !dt

2

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

v =

d~x

dt=

!

~k0

or alternativelyv = ⌫� =

!

2⇡�

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⇥ 10

8 m/s, the “speed of light”, where

c = ⌫� =

!

2⇡� =

!

k

For something that is not a vacuum, we can introduce a “complex refractive index” N for describ-ing a medium.

~k =

~k0 + i ~k” =

!

cN =

!

c(n0

+ in”)

where N is the (complex) refractive index of the material with real and imaginary componentsn

0 and n00 , respectively. It can be shown that the real component of the refractive index is just a

measure of how much the phase speed of light is slowed by a medium

n0=

c

v

We showed previously that~F / ~E2

Therefore~F =

~F0 exp

⇣�2

~k” · ~x⌘

Thus, the flux decays exponentially with distance through an absorbing medium. Now, since

��� ~k”

��� =

!

cIm (N) =

!

cn” =

2⇡n”

�

we can define an absorption coefficient

�a = 4⇡n”/�

such that~F =

~F0 exp (��ax)

3

This is known as Beer’s Law

4

Figure 1: Electron two slit experiment, showing that photons exhibit a duality in their behaviouras both waves and particles.

Light as quanta

A paradox of modern physics is that particles sometimes appear to be waves, and sometimes appearto be particles, depending on how you look at them. These totally disparate views of EM radiationhave yet to be reconciled. No one understands it. It is just plain weird. From the point of viewof this course, just accept it. Sometimes we’ll treat radiation as waves and sometimes as particles.Neither is any more wrong or right.

The frequency is related to wavelength � through the speed of light c

⌫ =c

�

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.

⌫̃ =1

�

Strangely though, energy comes in discrete, although very small packets or particles, but their

1

energy is related to the frequency of a wave!

E = h⌫

where h is Planck’s constant. 6.626⇥ 10�34 J s. Einstein figured this one out too, also in 1905, inhis explanation of what is known as the photoelectric effect. It won Einstein the Nobel prize, andis immortalized in the following song:

http://www.scientainment.com/ptsongbook.htm

Example

In 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 = h⌫ = hc/� = 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 =F

h⌫=

F�

hc= 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.

2

Flux

Radiative 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(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 � or fre-quencies ⌫ (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 (⌫) =

Z 1

0F (t) e�2⇡i⌫tdt

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 inverse

Fourier transform

F (t) =

1

2⇡

Z 1

0F (⌫) e2⇡i⌫td⌫

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

F� = lim

��!0

F (�, � + ��)

��

andF⌫ = lim

�⌫!0

F (⌫, ⌫ + �⌫)

�⌫

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

F (�1, �2) =

Z �2

�1

F�d�

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 (!). 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.

1

Figure 1: Spectrum as a function of wavelength

Intensity

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

d! = sin ✓d✓d� = A/r2

where in the atmosphere, the convention is that ✓ is the zenith angle (0� if directly overhead),and � is the azimuthal angle (0� in the direction of an incoming beam). A hemisphere covers 2⇡

steradians and a full sphere 4⇡ steradians. A is the cross-sectional area normal to radius vector r.

Example

The intensity of radiation coming from any direction is a function of angle ⌦ = (✓, �) and isdefined as

I (⌦) =

�F

�!

For example, the sun - earth distance is Ds = 1.496 ⇥ 10

8 km and the sun radius is Rs = 6.96 ⇥10

5km. Compute the angular diameter subtended by the sun, its solid angle, and the intensity of

2

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

P732951-Ch04.qxd 9/12/05 7:41 PM Page 115

Figure 2: Illustration of a solid angle

3

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: Relationships between intensity I , flux density F and flux E of isotropic radiation emitted

from a spherical source with radius r, indicated by the blue shading, and incident upon a much

larger sphere of radius R , concentric with the source. Thin arrows denote intensity and thick

arrows denote flux density. Fluxes ER = Er and intensities IR = Ir. Flux density F decreases

with the square of the distance from the source.

solar radiation if S0 = 1370 W m�2.

The solution requires recognizing that

�! = A/D2s = ⇡R2

s/D2s = 6.8⇥ 10

�5 sr

Thus,I = �F/�! = 1370/6.8⇥ 10

�5= 2⇥ 10

7 W m�2 sr�1

Relationship between flux and intensity

From 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 integrated over

4

the upward hemisphere.

F "=

Z 2⇡

0

Z ⇡/2

0I" (✓, �) cos ✓ sin ✓d✓d�

Similarly for the downwelling flux

F #= �

Z 2⇡

0

Z ⇡

⇡/2I# (✓, �) cos ✓ sin ✓d✓d�

5

Dipole radiation

The atmosphere is made up of oxygen and nitrogen mostly, but also greenhouse gases. Greenhousegases absorb radiation. Oxygen absorbs some solar radiation, but mostly nitrogen and oxygenmolecules only scatter light to make the sky blue. How do they do this? To understand how, weneed 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”. The incident electric field can be described by

~

E =

~

E0e�i!t

An electric field exerts a force. Remember that force required to accelerate the charge q of a dipole,which has mass m, is

F = qE = ma = m

d

2~x

dt

2

By doing thermodynamic work by displacing a charge through a force field, the incident electricfield makes the electric field of the atom oscillate. The full expression for the displacement of thecharge is the equation for a damped simple harmonic oscillator

m

d

2~x

dt

2= �K~x� b

d~x

dt

+ w

d

3~x

dt

3+ qE0 exp (�i!t)

= restoring + damping + recoil + forcing

You might be unfamiliar with the recoil or “jerk” term. It is a recoil associated with the radiativeemission. There must be radiative emission to balance the radiative forcing and maintain equi-

librium. How? The molecule is vibrating and moving its electric charge back and forth, so itgenerates its own electric field, distinct from the incident electric field. This new electromagneticfield represents what we term the scattered radiation.

It makes sense that the amplitude of displacement of electric charge within the molecule shouldbe proportional to the incident electric field, i.e.

~x (t) = ~x0ei!t /

⇣~

E =

~

E0ei!t

⌘

and if we try this as a solution then we obtain for the amplitude of the displacement

~x0 =

q

m

E0

!

20 � !

2 � i�!

where !0 =

qK/m and

� = �a + �s!2/!

20

where�a =

b

m

and�s = !0

w

m

1

The instantaneous rate at which the incoming electric field does work on the dipole oscillator isthe force times the rate of displacement

P = q

~

E

d~x

dt

The time averaged power hP i is distributed between absorption and scattering so that

hP i = hPai+ hPsi

wherehPai =

e

2

2m

E

20

!

2�a

(!

20 � !

2) + �

2!

2

andhPsi =

e

2

2m

E

20

!

2�s

(!

20 � !

2) + �

2!

2

!

2

!

20

Energy is conserved. �s and �a determine how it is partitioned. Note that the relationship of powerdissipated to the incident flux ~

F is through the absorption and scattering cross-sections �

hPai = �a~

F

hPsi = �s~

F

2

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 1: A blackbody defined by a cavity where emission and absorption are in equilibrium so as

to maintain a constant temperature

Blackbody radiation

The 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.

• 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 / 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

1

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

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

Figure 2: The blackbody spectrum. Note that colder temperatures correspond to lower energies of

emission and longer wavelengths.

sr�1 frequency�1)

B⌫

(T ) =

2h⌫3

c2(eh⌫/kT � 1)

which, in terms of wavelength is

B�

(T ) =

2hc2

�5(eh⌫/k�T � 1)

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 dervitation of B with respect to � and setting to zero to get

�max

= 2897/T

Why is there a peak? Where does this function come from? A first statement here is that thederivation of the Planck function is not straightforward, and was approached initially throughcombinations of classical, statistical and quantum mechanics. But it is an extremely importantresult, perhaps the first important result of the newly developed quantum theory. Its ingredients

2

have much to say about other topics in atmospheric physics. Here we will hand-wave some of themore central concepts.

The way this problem was first approached was to think of the radiating body as a bunch ofindividual oscillators. The total energy of the oscillators would be distributed among translational,vibrational, rotational, and electronic energy

etot

= etrans

+ erot

+ evib

+ eelec

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 ⌫ (or !) 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. Rather it istrapped inside an enclosed “black” (non-reflecting) box. Thus, any dipole radiation that is emitted(our E0 in dipole radiation) and lost also acts as a source for creating more jiggles. Accordingly,we can imagine an idealized equilibrium state in which the amount of energy that enters the system(say through a small aperture in 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 systemin a condition of steady-state: the rate of change of a substance is proportional to coefficient a

times the substance (the source) minus coefficient b times the substance (the sink). In equilibrium,a = b. In our case, the energy sink is proportional to the amount of energy of the system (kT ) andto the loss rate of energy through emission (/ !2). Contrast this with dipole radiation where theincoming energy was external leading to a power that is proportional to !2⇥!2 : so in a blackbodyone of the !2 is now replaced by kT . The final step is to determine the absorption of incomingradiation, because the absorbed incoming radiation must be equivalent to the loss rate for there tobe an equilibrium. Things get more complicated here, but ultimately, we do not lose the centralfeature that the radiant intensity is a product of the thermal energy and its frequency squared. This“classical” approach gives us the so-called Rayleigh-Jean’s law

B⌫RJ

=

8⇡⌫2

c3kT

which looks like the long tail of the Planck function. The problem here is that the Rayleigh-Jeanslaw just goes to infinity for high frequencies and short wavelengths. We don’t get fried by X-rays,

3

thankfully.Here’s the safety catch: remember, energy is “quantized”, and the amount of energy in a os-

cillator mode kT can never be less than h⌫. When kT ⌧ h⌫, the Rayleigh-Jeans assumption thattotal oscillator energy is kT is no longer correct. Now, a “quantum” description must take over. Itaccounts for the rapid dropoff in energy at short wavelengths, high frequencies in B.

A final consideration is that in equilibrium systems follow a distribution of states in proportionto their potential energy. This is called the Boltzmann distribution, which follows

n = n0 exp (�P.E./kT )

The most familiar example of this is that in a gaseous medium (e.g. the atmosphere) densitydecreases with height. Making an isothermal approximation

⇢ = ⇢0 exp (�mgh/kT )

An added ingredient for blackbody radiation is that quantum mechanical energy is quantized (i.e.e

n

= nh⌫, where n = 0, 1, 2...). But we get something similar anyway which is that the energydistribution follows

eqm

=

h⌫

exp (h⌫/kT )� 1

Notice that this is much like e�h⌫/kT and actually would be if energies weren’t quantized. The mainpoint here is that if energies are sufficiently high that h⌫ � kT , quantum mechanics must takeover from the classical mechanics perspective. And in this perspective, radiant energy decreases

with increasing frequency.Combining the Rayleigh-Jeans and Quantum mechanical approches, we end up with our de-

sired expression for Black-body radiation

B⌫

(T ) =

2h⌫3

c2(eh⌫/kT � 1)

Which implies a peak in the spectrum as prescribed by Wien’s law. So, as you can see, the deriva-tion is involved, even in our hand-waved approach. However, it touches on a number of conceptsthat we will get back to repeatedly, so are worth keeping in mind.

To get the total intensity averaged over all frequencies we integrate B⌫

to get (W m�2 sr�1)

B (T ) =

Z 1

0B

⌫

(T ) d⌫ = bT 4

4

F F

Figure 3: A planetary blackbody where emitted thermal radiation from the sphere is in balance

with absorbed solar radiation.

whereb = 2⇡4k4/

⇣15c2h3

⌘

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

FBB

(T ) = ⇡bT 4= �T 4

where � = 5.67⇥ 10

�8 J m�2 sec�1 deg�4 is the Stefan-Boltzman constant.Note that, we can show that F = �T 4 using three methods

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

2. Boltzmann showed that F / T 4 based on thermodynamic arguments (for an assignment)

3. Planck showed that F = �T 4, based on quantum mechanical arguments, where � = ⇡b is amix of fundamental constants.

Example

Exercise 4.6 in Wallace and HobbsCalculate the equivalanet blackbody temperature of the Earth, assuming a planetary albedo of

0.30. Assume that the Earth is in radiative equilibrium.

FE

= �T 4E

=

(1� A) FS

4

=

(1� 0.30)⇥ 1368

4

= 239.4 W m�2

5

Solving for TE

we get

TE

=

✓F

E

�

◆1/4

=

✓239.4

5.67⇥ 10

�8

◆1/4

= 255 K

6

Local Thermodynamic Equilibrium

A 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

e

Total

= e

trans

+ e

mol

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)

e

total

= e

trans

+ e

rot

+ e

vib

+ e

elec

+ . . .

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

e

trans x

=

1

2

kT

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

x

trans

=

3

2

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. Every DOF is associated with an energy transition.At an equilibrium we find that, averaged over a large number of molecules, the probability thatwe will find a molecule with a particular energy level is determined by what is known as theBoltzmann distribution. Ignoring that some modes are degenerate (different physically but withthe same energy), consider a hypothetical molecule with two energy levels 0 and 1 (0 being theground). Then the relative population of state 1 to state 0 is given by

n1

n0= exp (�h⌫/kT )

i.e. there’s exponentially fewer in a particular energy state, the higher the energy level. In fact wesee this exponential expression as part of the Planck radiation equation

B

⌫

=

2h⌫

3

c

2(e

h⌫/kT � 1)

1

In fact, maintenance of this distribution is what creates the Planck blackbody radiation. Notice thatthis radiation depends on temperature also, but in a way that depends not at all on the translationalmotions of the molecules but rather only the energy transitions within the molecules themselves.So really, we have two completely different temperatures

Thermal Temperature 32kT

therm

Planck Temperature B

⌫

(T

Planck

)

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

T

therm

⌘ T

Planck

How does this happen?Imagine the following sequence of events

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

Planck

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.

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 e

trans

and e

mol

such that

T

therm

⌘ T

Planck

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.

Kirchoffs Law

Under 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

"

�

= ↵

�

2

The premise here is that if an object has a temperature T and an absorptivity ↵

�

< 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 1

0↵

�

B

�

d� =

Z 1

0"

�

B

�

d�

implying that"

�

= ↵

�

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 T

b

⌧ T

a

. Object a radiates energy

F

a

= �T

4a

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

F

b�

(absorbed) = ↵

�

�T

4a

It reemits the energy with emissivity"

�

= ↵

�

but at temperature T

b

. ThusF

b�

(emitted) = "

�

�T

4b

Notice that F

b�

(emitted) ⌧ F

b�

(absorbed), even though the emissity and absorptivity are thesame, because T

b

⌧ T

a

.

3

Why is the sky blue?We showed that the scattered power of dipole radiation is given by

hPs

i =e2

2mE2

0

!2�s

(!20 � !2) + �2!2

!2

!20

If ! ⌧ !0 then absorption is negligible and the scattered electric field - the dipole radiation - isgiven by

hPs

i ' e2

2mE2

0

!4�2s

!60

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

~Idipole

/ !4~I0

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

Noting that Nitrogen and Oxygen molecules respond as dipoles to incoming solar radiation,we can explain why the sky is blue. If

~Idipole

/ !4

then, because !/2⇡ = ⌫ = c/�, then~Idipole

/ 1/�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 favoured, than why isthe sky not violet? To answer this question, we must consider human physiology.

1

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

~

I

total

=

~

I

dipole1 +

~

I

dipole2 +

~

I

dipole3 + . . . = N

~

I

dipole

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

~

I

total

=

⇣N

~

I

dipole

⌘·⇣N

~

I

dipole

⌘= N

2~

I

dipole

The number of water molecules in even a tenuous cirrus cloud is probably about N = 10

30, soobviously N

2o 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 acloud, even if the dipoles are all very close together and vibrating in phase.

Radiation scattered by a single dipoles only actually vibrates precisely in phase in the forwardscattering direction. Off the forward scattering direction, there is positive and negative interference.The phase difference �� is a function of the angle between the scattered waves ✓

�� =

2⇡r

�

(1� cos ✓) = x (1� cos ✓)

so, for a given separation between sheets of dipoles r, there is no cancellation of forward scatteredlight (1� cos ✓ = 0) and maximum cancellation at back angles (1� cos ✓ = 2).

A similar argument can be applied to stacked sheets of dipoles. If two sheets of dipoles areseparated by a distance r such that 2⇡r/� ⌧ 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 2⇡r/� � 1,then there are interference effects and cancellation of dipole radiation from one sheet by anothersheet.

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 0.5 µm across, then water molecules inone part of the particle no longer “know” about water molecules in the other part - they do notvibrate “in synch”. As a particle grows, this affects the shortest (blue) wavelengths first before itaffects the longer wavelengths of light. Normally, blue light is scattered preferentially. But now,the scattered color becomes less and less blue, and more and more red. Individual dipoles stillscatter blue preferentially, but this is offset by interference effects, which affect blue wavelengths.first. If the particle is large enough compared to the wavelength, then all colours are scatteredroughly equally for particles about 2 µm across (for sunlight). Cloud droplets are typically at least10 µm across so clouds are white.

1

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

particles that vibrate entirely in phase. In the atmosphere this works for gases, and very very smallaerosols we can’t see. Mie scattering goes roughly as I / I0/�

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 / 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.

2