Douce A.P. Thermodynamics of the Earth and Planets

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628 Topics in atmospheric thermodynamics

=(r

)

Fig. 13.4

Radiant energy exchange between a cavity at temperature T

and a body at temperature T

. Both are assumed to

behave as black bodies, and the intervening medium has unit transmittivity.

of body and cavity, the total rates of energy radiation are a

σ T

and a

σ T

, respectively.

There is an important difference between the two, though: the cavity radiates onto itself,

but the body does not. Heat transfer by radiation between the two is thus not symmetrical.

In particular, the total amount of energy radiated by the body is absorbed by the cavity, but

the converse is not true.

The radiation emitted by the body gives rise to an incident energy flux F

b→c

on the

surface of the cavity given by:

b→c

σT





σT

(13.33)

which is the inverse square law of radiation, a.k.a. the equation of energy conservation. The

flux of electromagnetic radiation emitted by the cavity, σ T

, bathes its interior uniformly.

This can be demonstrated formally (Winterton, 1997) or you can accept it intuitively on

the basis of symmetry. The body interposes a surface area a

to this energy flux, so that the

total amount of energy emitted by the cavity that is absorbed by the body is a

σ T

, and

the flux of radiation incident on the body F

c→b

, is, therefore:

c→b

σT

=σT

. (13.34)

629 13.3 Radiative energy transfer

Because the body is a black body it absorbs this energy, so that the net flux of electromagnetic

energy that leaves the body, F

b,net

, equals the energy emitted minus the energy absorbed, i.e.:

b, net

−F

c→b

=σ



−T



. (13.35)

If T

> T

then the body is losing internal energy, and must either be cooling down, if

this energy is not being replenished, or there is an internal source that supplies energy at

this rate. Conversely, if T

> T

radiant energy is being transformed to internal energy in

the body. Think of this conversion in terms of photons colliding with particles of matter,

whereupon the kinetic energy of the photons is added to the translational, vibrational or

rotational energies of the particles. As an aside, photons carry not only kinetic energy but

also momentum, and momentum conservation must be obeyed too. Transfer of momentum

from photons to particles of matter gives rise to radiation pressure.

Equation (13.35) shows that radiative heat flux varies as the difference between the fourth

powers of temperature, in contrast to diffusive and convective heat flux, which are linear

functions of temperature difference, or nearly so (e.g. equations (3.5) and (3.89)). It justifies

equation (2.16): the planet at temperature T is immersed in the solar nebula, which we can

think of as the “cavity”, at temperature T

From (13.35) we can define the equilibrium temperature for the body, T

b,eq

, as the tem-

perature at which there is no net flux of electromagnetic radiation from the body and hence

no change in its internal energy content with time. This is simply:

b, eq

. (13.36)

A body at the same temperature as its environment does not exchange electromagnetic

energy with it.

Worked Example 13.1 Radiative energy balance at a planet’s surface

The Sun can be approximated as a spherical black body with an emission temperature of

∼6000 K, and planetary orbits can be thought of as circumferences on spherical cavities

centered on the Sun. Setting r

= r

= solar radius, r

= r

= orbital radius and T

= T

= Sun’s emission temperature, equation (13.33) gives the flux of solar radiation across a

planet’s orbit, which is called the solar constant (Fig. 13.5). The solar constant for the Earth

is ∼1368 W m

−2

. This is the absolute maximum rate at which energy can be extracted

from sunlight at the Earth’s surface. Even if all of this radiant energy could be converted to

mechanical energy, it represents somewhat less than 2 horse-power per square meter. The

energy density of solar power is quite low, which is a reality that economic development

of this energy source must cope with. In practice, moreover, only a fraction of the solar

energy flux can be converted to usable energy at the Earth’s surface, partly because for

the atmosphere A > 0 (see below), and also because conversion to electrical or mechanical

energy is never 100% efficient (Chapter 4).

We can calculate the equilibrium temperature of the planet, T

, that we referred to in

Section 2.1. This is a temperature such that the planet emits electromagnetic radiation at the

same rate as it absorbs it from the Sun. The total amount of solar radiation that reaches a

planet is equal to the solar constant multiplied by the cross section of the planet, πr

, where

is the planet’s radius. Note that this is not the surface area of the hemisphere facing the

sun, but the cross section that intersects the solar energy flux, i.e. the stream of solar photons

(Fig. 13.5). Think of the planet as the “body” in Fig. 13.4, inside a cavity of infinite extent.

630 Topics in atmospheric thermodynamics

4πr

σT

πr

(

)

σT

Fig. 13.5 Radiant energy exchange between the Sun (radius r

) and a planet of radius r

in a circular orbit of radius r

. The

planet absorbs the solar ﬂux contained in the light grey solid angle, deﬁned by the cross section area πr

at the

planet’s orbital distance. Thermalized radiation is radiated back to space (a cavity of inﬁnite extent symbolized by the

dark grey circle) over the entire surface area of the planet, 4π r

We see from equation (13.35) that, because during the day T

, the planet absorbs

sunlight, whereas at night T

> T

and the planet emits radiation. This radiation is what

we called thermalized radiation, or thermalized sunlight, in Chapter 2. If the planet rotates

fast enough we can assume that it radiates at the same equilibrium temperature, T

, over

its entire surface area, 4πr

. We will ignore the planet’s internal energy flux. The Earth’s

internal energy flux is ∼10

−5

times the solar constant, so it cannot have any noticeable effect

on its surface equilibrium temperature. The same is true of the other terrestrial planets but,

as we saw in Chapter 2, it is not the case for the fluid planets nor for Io. Assuming that

the planet behaves like a black body, T

must satisfy the following energy conservation

condition:

4πr





=πr





σT

(13.37)

631 13.3 Radiative energy transfer

or:





1/4





1/2

. (13.38)

At this temperature the planet radiates energy at the same (average) rate as it receives it

from the Sun. For the Earth we calculate an equilibrium temperature of ∼278 K. We can use

Planck’s law (equation (13.29)) to compare the spectrum of sunlight with that of thermalized

sunlight (Fig. 13.3). The emission peak shifts from 0.5 µm, in the visible range, to 10 µm,

in the mid infrared. Because of the very strong dependency of emitted flux on temperature

there is a difference of seven orders of magnitude between the height of the peaks at 6000 K

and 278 K.

The Earth’s equilibrium temperature is about 10 K lower than the actual mean terrestrial

surface temperature (∼288 K). Why the discrepancy? In short, because the Earth has an

atmosphere, oceans and a climate system. For example, some trace components in the tro-

posphere such as H

O, CO

and CH

absorb in the infrared part of the spectrum, causing

warming of the lower atmosphere (the greenhouse effect, see below). Additional complica-

tions arise from variations in solar radiation with latitude and seasons (that drive convective

heat transport) and with changes in albedo related to ground cover and cloud cover.

I stated above that at night T

> T

, but what is the “cavity temperature” at night?

Disregarding for now the planet’s atmosphere (and heterogeneities that may arise from

interplanetary and galactic energy sources) this is the temperature of the microwave back-

ground radiation, ∼2.7 K (Fig. 13.3). In terms of heat transfer, you can think of this radiation

as being emitted by an infinitely distant cavity at this uniform temperature, although this is

of course not its true physical nature (it is strongly red-shifted radiation from the early Uni-

verse). If the Earth were to be ejected from the Solar System into the depths of intergalactic

space, would it still radiate electromagnetic energy to space? Yes, at its current average

internal energy output (∼87 mW m

−2

) it would radiate as a black body at a temperature

of ∼35K (equation (13.31)) with a peak at a wavelength of ∼82 µm (equation (13.30)),

corresponding to the far infrared part of the spectrum (Fig. 13.3). Emission flux originating

from the Earth’s internal heat (at 35 K) is some five orders of magnitude lower than the flux

of thermalized solar radiation at 278 K (Fig. 13.3). This explains why it is generally not

possible to use remote sensing to determine heat flux from terrestrial planets. It is interest-

ing (and pointless) to speculate that the low surface temperature that would result from the

Earth being ejected to intergalactic space would increase (slightly) the mantle’s Rayleigh

number and hence the rate of plate tectonics.

To conclude this section we derive a fundamental relationship between emissivity and

absorptivity.Assume that the body in Fig. 13.4 is not a black body. Then, for each wavelength

λ it has an emissivity /

=1 and an absorptivity A

=1. The cavity is a black body that emits

an energy flux F

∗

c, λ

, given by equation (13.29). From equation (13.34) and the definition

of absorptivity, the energy flux absorbed by the body at a given wavelength is:

c→b,λ

∗

c, λ

. (13.39)

From the definition of emissivity, the emission from the body is given by:

b, λ

∗

b, λ

, (13.40)

632 Topics in atmospheric thermodynamics

where F

∗

b, λ

is the black body emission at the body’s temperature. The net radiant flux

leaving the body is then given by:

b, net, λ

b, λ

−F

c→b,λ

∗

b, λ

−A

∗

c, λ

. (13.41)

If the temperatures of the body and the cavity are the same then F

∗

c, λ

= F

∗

b, λ

and also,

from equation 13.35, F

b, net, λ

= 0. The following result, known as Kirchoff’s law, follows:

. (13.42)

In other words, absorptivity and emissivity at a given wavelength are equal. There are

some important caveats. First, if a body absorbs radiation that was emitted at a temperature

different from that of the body itself then the integrated absorptivity and emissivity are not

equal. For example (see Fig. 13.3), because sunlight peaks at ∼0.5 µm whereas thermalized

sunlight on Earth peaks at ∼10.4 µm, the absorptivity and emissivity of the Earth’s surface

integrated over all wavelengths are not equal. Second, it is in general the case that A = /

even if emitted and absorbed radiation correspond to the same black body temperature. The

reasons for this have to do with the fact that, in contrast with a black body, radiation from

real bodies is in general not isotropic and may also be polarized differently from absorbed

radiation. Discussion of these effects is beyond the scope of this book.

13.3.3 Molecular absorption and emission of electromagnetic radiation

Electromagnetic radiation transports heat because photons can interact and exchange energy

with particles of matter. Heat transport by radiation is relatively minor in liquids and gener-

ally negligible in solids, but absorption and emission of radiant energy by planetary liquids

and solids is important. In contrast, gases are important in terms of radiant energy trans-

ported as well as absorbed and emitted. We saw in Chapter 1 that molecules in gases carry

energy as translational, rotational and vibrational kinetic energy, and that, as temperature

increases, all three modes come to participate in the heat capacity of gases. Molecules

also store energy in their electronic configurations. This mode does not show up in the

heat capacity of gases at “normal” planetary conditions because at these conditions there

is not enough energy to bring about changes in the electronic configuration of a molecule

or atom, or, in the terminology of quantum mechanics, to excite the electronic energy

modes.

In order to discuss emission, absorption and transport of radiant energy in gases we must

take a closer look at the various modes in which energy is stored in molecules. The rotational,

vibrational and electronic modes are all quantized, which means that they can carry energy

only in specific levels that are separated by discrete energy differences. Molecules can

absorb or emit energy only in discrete packages that correspond to the differences between

quantum energy levels. Photons carry discrete amounts of energy too, given by (hc)/λ.A

photon can interact with a molecule only if its energy corresponds to some of the possible

energy transitions in the molecule. Only photons of the appropriate wavelength (= energy)

can be absorbed and when they are, one of the molecular energy modes is excited to a higher

energy level. Conversely, when the molecule “drops” from an excited state to the ground

state it emits a photon of a wavelength that corresponds to this energy difference.

It is perhaps intuitively apparent that it takes relatively little energy to change the state

of rotation of a molecule, more energy to change the vibrations of the atomic bonds, and

even more energy to alter the electronic configuration of the molecule. This is formally

633 13.3 Radiative energy transfer

proved in quantum mechanics and is important for our purposes. The energies required

to excite electronic modes are carried by photons with wavelengths of order 0.1–0.5 µm,

within the ultraviolet and visible part of the spectrum (see Fig. 13.3). Electronic excitation

corresponds to breaking or forming of chemical bonds, i.e. chemical reactions. Vibrational

energy transitions are associated with lower energies, corresponding to infrared photons

with wavelengths of 1–10 µm. Rotational transitions occur in response to even lower

energy photons, in the microwave region of the spectrum (10

− 10

µm). Excitations

of rotational modes are generally unimportant in planetary processes, as they correspond

to exceedingly low temperatures, but are important in astrophysics, where non-thermal

mechanisms for the emission of microwave electromagnetic radiation exist (they are also

what makes microwave ovens work).

Molecular gases absorb and emit radiation of wavelengths extending from the ultraviolet

to the infrared, by exciting electronic and vibrational energy modes. The mechanisms are

different for the two types of energy transitions. Absorption of ultraviolet radiation by

electronic transitions is associated to photodissociation reactions. Oxygen–ozone reactions

in the Earth’s stratosphere are a good example (Worked Example 12.3). The first and third

reactions in the Chapman cycle (equation (12.144)) absorb photons with wavelengths in

the 0.2–0.25 µm range. In both cases absorption of a photon excites an electronic transition

which results in breakage of an atomic bond. Absorption of ultraviolet radiation by these

reactions in the Earth’s stratosphere has two important effects. First, because complex

organic molecules such as proteins and DNA can also break up by absorbing photons in this

energy range, photoactivated reactions that produce and destroy ozone allow us to be here

discussing these things. Second, absorption of ultraviolet radiation heats the stratosphere

and inverts the temperature gradient that drives convection in the troposphere.

Infrared photons are not energetic enough to break atomic bonds and facilitate chemical

reactions. They are absorbed by exciting vibrational modes. There is an additional restriction

in this case, that arises from quantum mechanics selection rules. The selection rule for

vibrational excitations is that they can only happen in molecules in which the electrostatic

dipole moment (i.e. the distribution of electric charge across the molecule) is asymmetric.

What this means is that homonuclear diatomic molecules such as O

and N

cannot absorb

infrared radiation, because the electric charge of both atoms is identical. In contrast, diatomic

molecules made up of different atoms (e.g. CO or HCl) and polyatomic molecules have

dipole moments that are not symmetric relative to the molecular structure. By the selection

rule they can absorb infrared photons as long as they have the correct energy to excite one

of the possible vibrational transitions. A well-known example of this is the capability of

molecules such as CO

O and CH

to absorb infrared radiation at wavelengths that are

close to the emission peak of thermalized solar radiation (see below).

Gases emit and absorb photons of specific wavelengths only, that correspond to allowed

energy transitions in the molecules. Emission and absorption of electromagnetic radia-

tion by molecular gases gives rise to line spectra. The molecules and atoms in solids

and liquids, in contrast, are close enough that the quantum states of individual atoms are

not independent of one another and the discrete energy transitions become smoothed out.

The result is that absorption and emission of electromagnetic radiation in solids and liq-

uids extend over continuous regions of the spectrum. This is also true of gases in which

electrons are free, because in such case electrons do not have set energy levels, and it

is the reason why the outer envelopes of stars, made up of ionized gas, radiate as black

bodies.

634 Topics in atmospheric thermodynamics

13.3.4 Absorption and emission of electromagnetic radiation:

the macroscopic description

Consider a layer of material of infinitesimal thickness, dx (Fig. 13.6) and a beam of electro-

magnetic radiation of wavelength λ incident on one of its sides. Recall that the irradiance,

, is the total flux of radiation traveling in all directions. We will consider only the simplest

case of changes in radiation intensity in a single direction. The intensity of the radiation,

, is defined as the flux of radiation traveling in a single direction, which in this case is

perpendicular to the layer of material. As in all radiation problems, a complete analysis

requires that we consider the full geometry of the problem and how radiation intensity

changes with direction, but the physical principles involved are easier to see if we ignore

these complications.

In general, the material may both absorb and emit radiation of a given wavelength, so

that the total change in the intensity of the beam over the thickness dx is given by:

=dI

λ,absorbed

+dI

λ,emitted

. (13.43)

From the definition of absorptivity we see that:

λ,absorbed

=−A

. (13.44)

It is reasonable to assume that the absorptivity is a function of some intrinsic material

property, on how much matter there is, and of the thickness of the layer, so we write:

ρdx, (13.45)

where ρ is density and k

is called the mass absorption coefficient for radiation of wave-

length λ. Because absorptivity is a non-dimensional number, the mass absorption coefficient

must have dimensions [L]

[M]

−1

, e.g. m

−1

. In this derivation we will consider only

absorption and emission by molecular mechanisms such as those that we discussed in the

previous section. The intensity of electromagnetic radiation can also change as a result of

scattering, as when photons interact with solid or liquid particles suspended in the atmo-

sphere. This process will not be considered here, but it is straightforward to add the effects

dτ

= k

ρ dx

= (–l

+ S

) dτ

+ dl

Fig. 13.6

Interaction between a monochromatic beam of electromagnetic radiation of intensity I

and a slab of material of

thickness dx.

635 13.3 Radiative energy transfer

of scattering to the absorption coefficient, in which case the name changes to extinction

coefficient to account for the combined effects of absorption and scattering.

We define a non-dimensional variable, called the optical thickness, τ

, as follows:

ρx, (13.46)

and from equations (13.44), (13.45) and (13.46), we write the change in beam intensity due

to absorption as follows:

λ,absorbed

=−k

ρI

dx =−I

dτ

. (13.47)

We can also write an expression for dI

λ,emitted

similar to (13.47) by defining a variable S

called the source term, which, as I

, has units of energy flux. Thus:

λ,emitted

dτ

. (13.48)

The source term is a measure of how much radiation the layer emits per unit of non-

dimensional optical thickness. Using equations (13.47) and (13.48), (13.43) becomes:

=−I

dτ

(13.49)

The solution of this differential equation is straightforward if S

is a constant (see Exercise

Problem 13.12, or differentiate (13.50) and substitute in (13.49) to verify that it is a solution):



(

)

−S



−τ

, (13.50)

where I

λ(0)

is the intensity of the incident beam and I

is the intensity at optical

thickness τ

The meaning of the optical thickness becomes apparent if we consider the case in which S

=0. Equation (13.50) then simplifies to the following, which is known as the Beer–Lambert

law (see also equation (12.137)):

(

)

−τ

. (13.51)

The optical thickness is the non-dimensional absorption length. About 2/3 of the incident

radiation is absorbed at τ

=1, and at five optical thicknesses more than 99% of the incident

radiation has been absorbed.As an example, consider transmission of solar radiation through

the terrestrial atmosphere. Solar radiation peaks at visible and ultraviolet wavelengths. The

temperature of the atmosphere is of the order of 250 K. We can see from Planck’s and

Wien’s laws that the atmosphere does not emit any significant amount of radiation in this

region of the spectrum, so that S

≈0. From our discussion in Section 13.3.3 it follows that

atmospheric oxygen has a large value of k

for λ ∼0.2 µm (ultraviolet photons). For the

Earth’s atmosphere τ

! 1 for ultraviolet radiation. We say that the atmosphere is optically

thick in the ultraviolet, or, equivalently that the atmosphere is nearly opaque to ultraviolet

radiation.

Returning to the full absorption–emission equation, (13.50), we consider what happens

if the temperature of the medium is such that emission at the wavelength of the incident

radiation (the term S

) cannot be ignored (this can be the case for infrared radiation in

-rich planetary atmospheres). In this case, for small optical thicknesses the source term

cancels out and the radiation flux is dominated by the incident flux, whereas for large optical

thicknesses radiation flux is dominated by thermal emission in the layer.

636 Topics in atmospheric thermodynamics

13.3.5 Absorption cross section and mean free path

The mass absorption coefficient,k

, is a macroscopic parameter that describesthe interaction

between electromagnetic radiation and matter. We can give it a microscopic interpretation,

as follows. From the definition of optical thickness, equation 13.46, we see that the product

ρ has dimension of length

−1

, so that (k

ρ)

−1

is the natural lengthscale for absorption

of radiation of wavelength λ. We now search for a product of two microscopic variables

with the same dimension. We define the absorption cross section at wavelength λ, σ

,as

the effective target area that each molecule offers to photons of this wavelength. Thus, if

a substance is inert to a particular wavelength, such as molecular oxygen to infrared, σ

vanishes. In contrast, if a substance absorbs radiation of a given wavelength (e.g. CO

the infrared) its absorption cross section for that wavelength is a finite and potentially large

number. The dimension of σ

is length

molecule

−1

. If we multiply the absorption cross

section by the number density of molecules, N , defined as the number of molecules per unit

volume, we get a product with dimension of length

−1

. This product is the total absorption

cross section per unit volume, so we can suggest the following relationship between the

macroscopic absorption coefficient and the microscopic absorption cross section:

ρ =σ

N. (13.52)

Imagine now that we have a slab of absorbing substance of cross section a

and thickness

, such that ι

is the characteristic distance that a photon of wavelength λ can penetrate into

the substance before it becomes certain that it will be absorbed. This distance is called the

photon’s mean free path. The mean free path must be of the same order as the distance into

the layer at which the sum of the absorption cross sections of all of the molecules equals the

actual cross section of the slab. Since σ

N is the absorption cross section per unit volume,

multiplying this by the volume of the slab we get the total absorption cross section of the

slab, which we require to be equal to the physical cross section of the slab, i.e.:

, (13.53)

which shows that the mean free path is given by:

. (13.54)

The mean free path thus equals the lengthscale for absorption, (k

ρ)

−1

. From equations

(13.46) and (13.54) we have:

. (13.55)

An optical thickness of 1 is reached, and most of the incident radiation is absorbed, when

the distance traveled by the electromagnetic radiation is of the order of the mean free path.

Other things being equal, the mean free path varies inversely with the number density, N

(equation (13.54)), explaining in part why solids tend to be more opaque (absorptivity = 1)

than gases.

In a mixture of ideal gases, the number density N of a gas species at constant temperature is

proportional to the partial pressure of the species. For dilute gases at low pressure, therefore,

mean free path varies inversely with the partial pressure of the absorbing species (equation

(13.54)) and optical thickness can be approximated with a linear function of partial pressure

(equations (13.54) and (13.55)).

637 13.3 Radiative energy transfer

13.3.6 A radiative toy model of greenhouse warming

We close our discussion of radiative heat transfer by constructing a “toy model” of green-

house warming. The goal of this section is not to develop a complete and quantitatively

accurate model of the process, which is far beyond the scope of this book. Rather, I sim-

ply want to highlight the aspects of radiative heat transport that underlie the planetary

greenhouse effect. With this caveat, we proceed as follows (see Fig. 13.7).

Consider a planet such that the solar constant at its orbit is F

. The planet has an atmo-

spheric layer in which radiative heat exchanges take place. The elevation and thickness of

this active layer are unspecified, except for the fact that it is close enough to the planet’s

solid surface, and thin enough, that the radius of the planet, r

, and the mean radius of the

active atmospheric layer can be considered to be equal. We will simplify the mathematical

treatment by assuming that we can describe the interactions between electromagnetic radi-

ation, the atmosphere and the surface on the basis of only two sets of A, Θ,Rand /.As

we saw, all of these parameters depend on wavelength but here we will assume that we can

=1 (downwards)

= 1 (upwards)

= 1 (both directions)

(both directions)

Active atmospheric

layer, temperature = T

Planetary surface

temperature = T

= 1

π r

a,λs

π r

(Θ

a,λs

)

π r

a,λs

4π r

a,λt

g,λs

π r

a,λs

g,λs

π r

a,λs

g,λs

4π r

g,λt

4π r

g,λt

a,λs

= 1

g,λt

a,λs

a,λt

σT

g,λt

σT

4π r

g,λt

a,λt

Fig. 13.7

Energy balance for a simple radiative model of greenhouse warming of a planet of radius r

and solar constant F

. The

planet has an atmospheric layer that interacts with solar and thermal radiation, and attains an equilibrium

temperature T

. The equilibrium temperature of the planet’s surface is T

. The arrows illustrate the direction of the

various energy ﬂuxes, but not their relative intensities.