Zdunkowski W., Trautmann T., Bott A. Radiation in the atmosphere: A course in Theoretical Meteorology

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7.7 Problems 275

radiation use

B =

κσT

with κ = 0.25, σ = 5.6697 × 10

−8

−2

−4

7.14: Consider two idealized lines in triangular form as shown in the ﬁgure.

,max

αα

The center lines are located at ν

and ν

. The distance between the lines

exceeds the half-width. (Note that here the half-width is twice as large

as in the usual deﬁnition.) Find an analytic representation of the spectral

absorption coefﬁcient, the equivalent width W

, W

of each line and the

equivalent width W of the line spectrum. Moreover, consider the limiting

cases u → 0 and u →∞.

7.15: In an isothermal atmospheric layer an absorbing gas is homogeneously

distributed, that is the concentration c

gas

is constant. Calculate the equivalent

width of a spectral line for a vertical path in the region of the weak line

approximation. Assume the Lorentz line shape.

Absorption by gases

8.1 Introduction

In this chapter we are going to discuss some of the more elementary ideas in

connection with the absorption spectra of gases. The energy E of a molecule may

be expressed as the sum of the rotational energy E

rot

, vibrational energy E

vib

and

electronic energy E

. Of these three types of energy, E

rot

is generally the smallest,

typically a few hundredths of an electron Volt.

Vibrational energies are of the order

of a few tenths of one electron Volt, while the largest energies are of the electronic

type which generally amount to a few electron Volts.

The absorption (emission) spectrum arising from the rotational and vibrational

motion of a molecule which is not electronically excited will be located in the

infrared region. In infrared absorption experiments light from a suitable source

penetrates an absorption chamber containing the gas to be studied and then enters a

spectrograph. If the instrument is of low resolving power, a series of wide bands is

observed which correspond to the vibrational transitions. If an instrument of high

resolving power is used, these bands are seen to consist of numerous spectral lines

resulting from the energy levels of rotation.

In the next section we are going to discuss the vibrational motion of two relatively

simple molecules (CO

and H

O) which are particularly important in the study

of radiative transfer in the atmosphere. The forces between the atoms making

up the molecule may be crudely approximated by forces exerted by weightless

springs which hold the atoms relative to each other in the neighborhood of certain

conﬁgurations. The forces due to stretching or compression of the springs are

assumed to follow Hooke’s law.

If a molecule contains n atoms, there are 3n modes of motion. Of these, three

correspond to translation, and three to rotation (or two for a linear molecule). The

If an electron falls through a potential difference of one Volt it attains a kinetic energy of 1.602 × 10

−19

joule

which is used as the deﬁnition of one electron volt, i.e. 1 electron volt = 1.602 ×10

−19

joule.

276

8.2 Molecular vibrations 277

Fig. 8.1 Two coupled harmonic oscillators with equilibrium positions at x

= 0,

= 0.

remaining 3n − 6 (or 3n − 5) correspond to the normal vibrational modes of the

molecule.

8.2 Molecular vibrations

8.2.1 Two coupled harmonic oscillators

Before studying the vibrational motion of the carbon dioxide and water molecules,

we are going to discuss a fairly simple example by considering two particles, each

of mass m, connected by light springs of stiffness k, see Figure 8.1. The particles

are constrained to move in a straight line. The distances x

and x

stand for the

displacements of particles 1 and 2 from their equilibrium positions.

The equation of motion of each particle can be obtained quite easily by using

Lagrange’s equation of motion. The Lagrangian function L is deﬁned as the dif-

ference of the kinetic energy K and the potential energy V of the system, that is

(a) L = K − V

(b) K =

k(x

− x

)

(8.1)

The second term on the right-hand side of (8.1c) refers to the potential energy

stored in the spring connecting the two masses. If q

and

stand for the generalized

coordinate and the generalized velocity of the particles, then Lagrange’s equation

of motion of a conservative system is given by

∂ L

∂

∂ L

∂q

(8.2)

Substituting for q

= x

, x

equation (8.1) into (8.2) yields the equations of motion

for the two particles

∂ L

∂

∂ L

∂x

or m

=−kx

+ k(x

− x

)

∂ L

∂

∂ L

∂x

or m

=−kx

+ k(x

− x

)

(8.3)

278 Absorption by gases

Let us ﬁnd the possible common frequencies of vibration of the two particles.

These frequencies are known as eigenfrequencies. The associated vibrational states

are the corresponding eigenvibrations or normal modes of vibrations. We are going

to solve the system (8.3) by means of the trial solutions

= A

cos ωt, x

= A

cos ωt (8.4)

which requires that both particles vibrate with the frequency ω. Had we used a sine

function or the combination of a cosine and a sine function we would still obtain the

same equations expressing the conditions for the frequency. Substitution of (8.4)

into (8.3) gives two linear homogeneous equations for the amplitudes A

and A

(−mω

+ 2k)A

− kA

= 0

−kA

+ (−mω

+ 2k)A

= 0

(8.5)

Nontrivial solutions for the amplitudes exist only if the determinant of the coefﬁ-

cients vanishes. The expansion of the determinant results in the frequency equation



−mω

+ 2k −k

−k −mω

+ 2k



= (−mω

+ 2k)

− k

= 0 (8.6)

The positive roots

, ω

(8.7)

are the eigenfrequencies of the system.

In order to get some idea about the type of the normal vibrations, we substi-

tute (8.7) into the system (8.5) and ﬁnd the conditions for the symmetric and the

antisymmetric modes

antisymmetric mode: A

=−A

for ω

=⇒ x

=−x

symmetric mode: A

= A

for ω

=⇒ x

= x

(8.8)

The number of normal vibrations is equal to the number of coordinates which are

required for a complete description of the system. The equally large amplitudes

of this example result from the assumption that the two masses are equally large.

The general motion of the mass points will be given by superimposing the normal

vibrations with different amplitudes and phases.

8.2.2 Review of physical principles

Theoretical calculations of molecular vibrations often are carried out in a coordinate

system in which the center of mass of the particles is at rest. To prevent the molecule

8.2 Molecular vibrations 279

Fig. 8.2 Center of mass system.

from rotating we must require that the angular momentum of the molecule vanishes.

A brief review of the physical principles involved in the solution technique now

follows.

We ignore the translational motion of the system as a whole since for spectro-

scopic considerations only the motion of the atoms relative to the center of mass is

of importance. As shown in Figure 8.2, the origin of the primed coordinate system

is the center of mass whose position in the unprimed system is given by the vector

The position of the particle m

in the primed and unprimed (laboratory) system

is r



and r

, respectively. In general, the center of mass is deﬁned by



i=1

with M =



i=1

(8.9)

where M is the total mass. Replacing r

by the vector sum r

+ r



results in



i=1



+ r





= Mr



i=1



(8.10)

From this equation it follows that



i=1



= 0,



i=1



= 0 (8.11)

whereby the second equation results from the time differentiation of the ﬁrst one.

Hence in the center of mass system the sum of the mass moments m



as well as

the sum of the linear moments m



vanish.

280 Absorption by gases

321

Fig. 8.3 Linear symmetric triatomic molecule in equilibrium position.

The total angular momentum of the system is deﬁned by

J =



i=1

(

× v

)



i=1



+ r





× (v

+ v



)

= M

(

× v

)



i=1





× v





= J



i=1



(8.12)

showing that the angular momentum J can be expressed as the sum of two terms. The

ﬁrst term represents the angular momentum of the center of mass J

having mass

M, while the second term gives the sum of the angular momenta of the individual

particles about the center of mass.

8.2.3 Linear triatomic molecules

In the following we calculate the normal vibrations of the linear symmetric CO

molecule whose equilibrium position is shown in Figure 8.3. The atoms m

are

separated from the atom m

by the equilibrium distance l. We assume that the

potential energy of the molecule depends only on the distances between the atoms

and on the angle of bending. The motion of the atoms takes place in the (x, y)-plane.

First we are going to discuss longitudinal vibrations.

If the vector x

with components (x

, y

) stands for the displacement of atom i

from its equilibrium position r

0,i

then the momentary position of this atom is given

= r

0,i

+ x

(8.13)

The forces holding the atoms together, in ﬁrst approximation, follow Hooke’s law.

For longitudinal vibrations the Lagrangian function L of the system is expressed by

L = K − V =





−

[(x

− x

)

+ (x

− x

)

] (8.14)

where k

is the spring constant for the longitudinal motion. To eliminate one

coordinate, say x

, we make use of



i=1



i=1

0,i

(8.15)

8.2 Molecular vibrations 281

which is a statement for the conservation of the center of mass. Application of

(8.13) and (8.15) yields

+ x

) + m

= 0 (8.16)

so that L is given by

L =





(

)

−





+ x



(

+ x

)

+ x

)



(8.17)

We introduce the new set of coordinates (η, ξ ) by means of

η = x

− x

, ξ = x

+ x

=⇒

ξ + η

, x

ξ − η

(8.18)

Utilizing the new coordinates the Lagrangian function may be written as

L =

˙η

−

(8.19)

with m

= 2m

+ m

. In contrast to (8.17), in this form L does not contain any

cross terms.

Differentiation of the Lagrange function (8.19) according to (8.2) yields

∂ L

∂

ξ,

∂ L

∂ξ

=−

ξ (8.20)

so that the equations of motion for the (η, ξ) coordinates are given by

ξ +

ξ = 0, ¨η +

η = 0 (8.21)

From these equations it is seen that due to the introduction of the coordinates (η, ξ)

the differential equations of motion are decoupled which was not the case in the

example of two coupled harmonic oscillators, see (8.3). The coordinates (η, ξ ) are

known as the normal coordinates. It is a characteristic feature of normal coordinates

that the differential equations of motion are automatically separated, there being

one differential equation for each normal coordinate.

282 Absorption by gases

(a)

(b)

Fig. 8.4 (a) Antisymmetric mode of the longitudinal vibrations, η = 0, ν

vibration. (b) Symmetric mode of the longitudinal vibrations, ξ = 0, ν

-vibration.

Substituting the trial solution exp(iωt) into (8.21) we immediately ﬁnd the

frequencies

antisymmetric vibration: ω

symmetric vibration: ω

(8.22)

Let us examine the vibrations more closely and consider case (a) of Figure 8.4.

If x

= x

then η = 0 resulting in the antisymmetric vibrational mode. This is the

reason we have added the sufﬁx a to the circular frequency in equation (8.22). This

type of motion is usually called the ν

-vibration. In case (b) we set x

=−x

that ξ = 0 resulting in the symmetric ν

-vibration (sufﬁx s in (8.22)).

So far we have restricted the motion of the atoms to one direction. A nonrigid

triatomic molecule, such as CO

, vibrates not only longitudinally but also transver-

sally as shown in Figure 8.5. In this case the Lagrangian function is given by

L =





−

(lδ)

(8.23)

where the constant k

of the potential energy part of L refers to the transversal

displacement. The angle δ stands for the deviation from 180

◦

. The meaning of the

angles α

and α

follows from the ﬁgure. Since the angle δ is assumed to be very

small, we may replace it by the sine functions as shown in

δ =



− α





− α



≈ sin



− α



+ sin



− α



= cos α

+ cos α



− y





− y



(8.24)

8.2 Molecular vibrations 283

Fig. 8.5 Coordinates of the transversal vibration of a linear triatomic molecule.

As before, we eliminate one coordinate by using the conservation of the center of

mass. The result is

+ y

) + m

= 0 (8.25)

In order to exclude the rotation of the molecule, the total angular momentum

must vanish. The mathematical form of the angular momentum is given by (8.12).

Since the deviation from the equilibrium position is small, we may approximate

the vector r

by r

0,i

. Thus the total angular momentum is approximately given by

J =



i=1

× v

) ≈



i=1

0,i

× v

) =



i=1

0,i

× x

) = 0 (8.26)

which is satisﬁed by



i=1

0,i

× x

) = 0 (8.27)

This equation can be used to show that y

= y

so that the Lagrangian function

may be written as

L =

δ)

−

(lδ)

(8.28)

After a few steps we ﬁnd the eigenfrequency of the transversal vibration

(8.29)

Details of the derivations will be left to the exercises.

The transversal vibration shown in Figure 8.6(a) is called ν

-vibration. At the

beginning of the chapter we have stated that this molecule should have four vibra-

tional modes. Our calculations, however, provided only three of these. The fourth

vibrational mode of the CO

molecule results from a twofold degeneracy, i.e. the

direction of one vibrational mode is perpendicular to that of the other as shown in

284 Absorption by gases

(a)

(b)

+ −

Fig. 8.6 Transversal vibration of a linear triatomic molecule.

Figure 8.6(b). This does not provide any new information. The + attached to m

and the − attached to m

simply indicate that they vibrate in opposite directions in

a plane perpendicular to the plane of the paper.

As will be shown later, the interaction of an electromagnetic wave with the

molecule to produce absorption or emission results from the interaction of the ele-

ctric ﬁeld vector E with the (variable) dipole moment M of the system. The dipole

moment of an electrically neutral molecule is a vector whose direction is along the

line joining the center of charge of the negative charges to the center of charge of

the positive charges. The magnitude of the dipole moment is the length of that line

multiplied by the total negative or positive charge, these being equal. An atom or a

molecule is said to be polarized by an electric ﬁeld when the displacements of the

charges caused by the electric ﬁeld produce or alter the dipole moment.

In a Cartesian coordinate system the components of M are given by



, M



, M



(8.30)

where e

is the charge of the particle k at the position (x

, y

, z

). If the particles

are the atoms of a molecule, the charges e

must be considered as effective charges.

Some molecules have a permanent dipole moment such as the heteronuclear

diatomic molecule CO. The dipole moment results from the asymmetric charge

distribution. In contrast, homonuclear diatomic molecules such as N

have no elec-

tric dipole moment due to the symmetric charge distribution. Similarly, in the

equilibrium conﬁguration the CO

molecule has no permanent dipole moment due

to the symmetric distribution of charges. Further details may be found, for example,

in Wilson et al. (1955).

Let us re-examine Figure 8.4. The symmetric longitudinal stretching of the CO

molecule, usually called the ν

-vibration, does not produce any dipole moment

so that this type of vibration is inactive in the infrared spectrum. The remain-

ing vibrations are classiﬁed as parallel or perpendicular according as the change