Zdunkowski W., Trautmann T., Bott A. Radiation in the atmosphere: A course in Theoretical Meteorology
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7.6 Appendix 265
Further inspection of Figure 7.18 shows that infrared cooling decreases in the
lower stratosphere and then increases to about −10 K day
−1
just below 50 km. Even
at a height of 60 km, cooling rates still amount to about −5 K day
−1
. The strong
radiative cooling in the layer extending from 30 to 60 km is mainly caused by CO
2
.
In the height range from 30–50 km radiative cooling is intensified by the presence
of ozone and to a smaller extent by water vapor. The strongest cooling should take
place at the top of the stratospheric O
3
layer located between 20 and 40 km since
shielding effects of still higher absorbing layers are absent. For a height of about
40 km Plass (1956) calculated an ozone cooling rate of about −2.5 K day
−1
. This
result was verified by more recent investigations.
Finally, we will investigate the net radiative temperature change. Within the
entire troposphere the net radiative heating is almost zero. Above the troposphere
radiation causes a net warming of the atmosphere. The strongest radiative heating is
found in the height range from 25–50 km which roughly coincides with the strato-
spheric temperature increase in this height interval. The fact that the atmospheric
temperature decreases above 50 km indicates that physical processes other than
radiation cause atmospheric cooling in this region.
7.6 Appendix
7.6.1 Maxwell’s velocity distribution and the mean molecular velocity
Let v = (v
x
,v
y
,v
z
) be the vector of the thermal velocity of the air molecules.
Temperature is just another measure for the average kinetic energy of the molecules.
If we limit our discussion to translational energy, the relation between temperature
and mean kinetic energy is given by
1
2
m
v
2
=
3
2
kT (7.242)
In total there are three degrees of freedom for the translational motion. Along each
of the three Cartesian axes the average kinetic energy contributes
1
2
kT to the total
thermal energy. Hence, we obtain
1
2
m
v
2
x
=
1
2
m
v
2
y
=
1
2
m
v
2
z
=
1
2
kT (7.243)
Maxwell’s velocity distribution F(v
x
,v
y
,v
z
)isgivenby
F(v
x
,v
y
,v
z
) =
m
2πkT
3/2
exp
−
m
v
2
x
+ v
2
y
+ v
2
z
2kT
(7.244)
266 Transmission in spectral lines and bands of lines
v
y
v
x
v
v +
dv
Fig. 7.19 Two-dimensional illustration of a spherical shell volume for the deriva-
tion of Maxwell’s velocity distribution f (v).
Maxwell’s distribution has spherical symmetry and depends only on the magnitude
of v, but not on the direction of v. To be a proper probability distribution F must
be normalized
∞
−∞
Fdv
x
dv
y
dv
z
= 1 (7.245)
Furthermore, each translational degree of freedom must contribute
1
2
kT to its aver-
age kinetic energy, see (7.243).
Due to the fact that F only depends on v
2
=|v|
2
= v
2
x
+ v
2
y
+ v
2
z
, integration of
F over the Cartesian volume element dv
x
dv
y
dv
z
can be replaced by an integration
over spherical shells having the volume 4πv
2
dv, see Figure 7.19. Hence, we may
write
f (v)dv = 4π nF(v)v
2
dv = 4πn
m
2πkT
3/2
exp
−
mv
2
2kT
v
2
dv
(7.246)
The function f (v) is Maxwell’s velocity distribution for the magnitude v of the
velocity. Note that the factor 4π stems from the integration over azimuth from 0 to
2π and over the cosine of the polar angle from −1 to 1. Furthermore, the factor n
is necessary to account for the contributions of each molecule contained in a unit
volume of space.
The most probable velocity ˆv is defined by that velocity for which f (v) attains
its maximum, i.e.
ˆv =
%
2kT
m
(7.247)
7.6 Appendix 267
Likewise, the average velocity ¯v can be computed from
¯v =
1
n
∞
0
v f (v)dv =
%
8kT
πm
(7.248)
and the root-mean-square velocity is given by
(
v
2
)
1/2
=
1
n
∞
0
v
2
f (v)dv
1/2
=
%
3kT
m
(7.249)
Note that the factor 1/n in the last two expressions is required for normalization.
7.6.2 Original derivation of the Ladenburg and Reiche function
We start with the following expression for the average absorption of an isolated
Lorentz line
A(
¯
u) =
∞
−∞
1 − exp
−
2
¯
uy
2
x
2
+ y
2
dx
with
¯
u =
Su
2πα
L
, x =
ν
ν
, y =
α
L
ν
(7.250)
It is convenient to introduce the substitution
x = y tan
(
s/2
)
=⇒ dx = y
d
ds
tan
(
s/2
)
ds (7.251)
Using this substitution, the argument of the exponential function in (7.250) can be
transformed to
2
¯
uy
2
x
2
+ y
2
=
2
¯
uy
2
y
2
[1 + tan
2
(
s/2
)
]
= 2
¯
u cos
2
(
s/2
)
=
¯
u(1 + cos s) (7.252)
so that the average absorption may be rewritten as
A(
¯
u) = y
π
−π
1 − exp
[
−
¯
u(1 + cos s)
]
d
ds
tan
(
s/2
)
ds (7.253)
This expression can be further simplified by means of integration by parts
A(
¯
u) = y
1 − exp
[
−
¯
u(1 + cos s)
]
cot
(
s/2
)
π
−π
+ y
¯
u
π
−π
exp
[
−
¯
u(1 + cos s)
]
tan
(
s/2
)
sin sds (7.254)
268 Transmission in spectral lines and bands of lines
The first term on the right-hand side is of the form 0/0. Utilizing l’Hˆopital’s rule
one can easily verify that this term vanishes. Using
tan
(
s/2
)
=
1 − cos s
sin s
(7.255)
the second term on the right-hand side of (7.254) can be rewritten yielding
A(
¯
u) = y
¯
u exp
(
−
¯
u
)
π
−π
exp
(
−
¯
u cos s
)
ds −
π
−π
cos s exp
(
−
¯
u cos s
)
ds
(7.256)
The two integrals appearing in (7.256) are proportional to the Bessel functions of
the first kind of order 0 and 1. From mathematical tables (Abramowitz and Stegun,
1972) we find
J
n
(ρ) =
(−i)
n
2π
π
−π
exp
(
iρ cos s
)
cos(ns)ds (7.257)
Hence, by setting ρ = i
¯
u, we can write for (7.256)
A(
¯
u) = 2π y
¯
u exp
(
−
¯
u
)
[J
0
(i
¯
u) −iJ
1
(i
¯
u)] (7.258)
The modified Bessel functions I
n
are real functions and are related to the Bessel
functions by
J
n
(ix) = i
n
I
n
(x) (7.259)
This completes the alternative derivation of the Ladenburg and Reiche function,
that is
A(
¯
u) = 2π y
¯
u exp
(
−
¯
u
)
[I
0
(
¯
u) + I
1
(
¯
u)] = 2π yL(
¯
u)
(7.260)
7.6.3 The Mittag–Leffler theorem and the Elsasser model
Before stating the Mittag–Leffler theorem it might be useful for the reader to review
some terminology from complex variable theory. Undoubtedly, the student was
exposed to the theory of residues which is an integral part of any course on the
mathematics of physics and engineering. By necessity our review will be very
brief.
Consider the complex function f (z) where z is the complex variable z = x + iy.
Let f (z) be single-valued and analytic inside and on a circle C except at the point
z = a which is taken as the center of C. Then f (z) can be expanded as a Laurent
7.6 Appendix 269
series about z = a as given by
f (z) =
∞
n=−∞
b
n
(z − a)
n
(7.261)
The part
∞
n=0
b
n
(z − a)
n
of f (z) is called the analytic part while the remainder of
the Laurent series
−1
n=−∞
b
n
(z − a)
n
is known as the principal part. The expansion
coefficients can be calculated with the help of
b
n
=
1
i2π
0
C
f (z)
(z − a)
n+1
dz, n = 0, ±1, ±2,... (7.262)
For the special case that n =−1wefind
0
f (z)dz = i2πb
−1
(7.263)
The latter equation can be evaluated by termwise integrating (7.261), utilizing
0
1
(z − a)
p
dz =
i2π for p = 1
0 for p = 1, p an integer number
(7.264)
Since (7.263) only involves the expansion coefficient b
−1
it is called the residue
of f (z) at the point z = a. If the principal part of f (z) has only a finite number of
terms
f (z)
principal part
=
b
−1
z − a
+
b
−2
(z − a)
2
+···+
b
−n
(z − a)
n
(7.265)
where b
−n
= 0, then z = a is a pole of order n. In case that n = 1 we have a pole
of order 1 or a simple pole. If z = a is a pole of order k then the residue b
−1
can be
obtained from
b
−1
= lim
z→a
1
(k − 1)!
d
k−1
dz
k−1
[(z − a)
k
f (z)] (7.266)
In case of a simple pole (7.266) simplifies to
b
−1
= lim
z→a
(z − a) f (z) (7.267)
We are now ready to state the Mittag–Leffler theorem. Suppose the only singu-
larities of f (z) in the finite z-plane are the simple poles a
1
, a
2
,.... Let the residues
of f (z) at these poles be given by b
1
, b
2
,.... Then the Mittag–Leffler theorem can
be expressed as
f (z) = f (0) +
∞
n=1
b
n
1
z − a
n
+
1
a
n
(7.268)
270 Transmission in spectral lines and bands of lines
The special function f (z) = cot z − 1/z is fundamental in showing that the reg-
ular Elsasser spectral band model (7.119) can be written in the form (7.120). We
will now show this by performing two major steps.
Step 1
By writing
f (z) = cot z −
1
z
=
z cos z − sin z
z sin z
(7.269)
we recognize that f (z) has simple poles at z = nπ with n =±1, ±2,... so that
the residues at the poles are
b
n
= lim
z→nπ
(z − nπ)
z cos z − sin z
z sin z
= lim
z→nπ
z − nπ
sin z
lim
z→nπ
z cos z − sin z
z
= 1
(7.270)
In this equation we have made use of the fact that the limit of a product is equal to
the product of the limits. Using l’Hˆopital’s rule we find that at z = 0 the function
has a removable singularity since
lim
z→0
z cos z − sin z
z sin z
= 0 (7.271)
Defining f (0) = 0 we find from (7.268) together with (7.270)
f (z) = cot z −
1
z
=
n
1
z − nπ
+
1
nπ
, n =±1, ±2,...
= lim
N →∞
−1
n=−N
1
z − nπ
+
1
nπ
+
N
n=1
1
z − nπ
+
1
nπ
(7.272)
To accept this result we would still have to show that f (z) is bounded on circles
C
N
having their center at the origin and radius R
N
= (N + 1/2)π which is part of
the requirements used to derive the Mittag–Leffler theorem. We omit this part of
the proof.
Step 2
We split cot z into the real and imaginary part. According to (7.269)
cot z =
1
z
+ f (z) =
1
x + iy
+
n
1
x + iy − nπ
+
1
nπ
, n =±1, ±2,...
(7.273)
7.7 Problems 271
After a few simple steps we find
cot z =
x
x
2
+ y
2
+
n
x
2
+ y
2
− nπ x
nπ[(x − nπ )
2
+ y
2
]
−i
y
x
2
+ y
2
+
n
y
(x − nπ)
2
+ y
2
=(cot z) + i(cot z) (7.274)
It is a simple matter to show that we may also write
cot z =
cos(x + iy)
sin(x + iy)
=
sin(2x)
cosh(2y) −cos(2x)
− i
sinh(2y)
cosh(2y) −cos(2x)
(7.275)
A comparison of the latter two equations shows that for the imaginary part we must
have
sinh(2y)
cosh(2y) −cos(2x)
=
y
x
2
+ y
2
+
n
y
(x − nπ)
2
+ y
2
, n =±1, ±2,...
(7.276)
Setting in (7.276) x = πν/δ and y = πα
L
/δ we obtain
S
δ
sinh
(
2πα
L
/δ
)
cosh
(
2πα
L
/δ
)
− cos
(
2πν/δ
)
=
∞
n=−∞
S
π
α
L
(ν − nδ)
2
+ α
2
L
= k
ν,E
(7.277)
Finally, by introducing in (7.277) the abbreviations β = 2πα
L
/δ and s = 2πν/δ
we find
k
ν,E
=
S
δ
sinh β
cosh β − cos s
= k(s) (7.278)
in accordance with (7.120).
The proof of the Mittag–Leffler theorem can be found in various textbooks such
as Whittaker and Watson (1915). The theory of residues is nicely discussed, for
example, in Wylie (1966) and Spiegel (1964). The latter author also proves the
Mittag–Leffler theorem.
7.7 Problems
7.1: Show by direct integration that the Lorentz line-shape factor f
L
(ν − ν
0
)is
normalized.
7.2: Estimate the pressure levels for which the half-width α
D
of the Doppler line
equals the half-width α
L
of the Lorentz line for a
272 Transmission in spectral lines and bands of lines
(a) CO
2
line located at 667 cm
−1
,
(b) H
2
O line at 1600 cm
−1
. Assume that α
L
(p
0
= 1013.25 hPa) = 0.1cm
−1
.You
may ignore the temperature dependence of the Lorentz half-width. Use the
following model atmosphere. At the Earth’s surface: z = 0 km, p
0
= 1013.25
hPa, T = 288.15 K,
Height level (km) Lapse rate (K km
−1
)
0–11 6.5
11–25 0.0
25–47 −3.0
47–53 0.0
53–80 4.4
80–90 0.0
7.3: The centers of two identical Lorentz lines are separated by a distance of eight
half-widths. Calculate the transmission between the lines at a wave number
of three half-widths away from the line center. S = π cm
−2
, α
L
= 0.1cm
−1
,
u = 1 cm.
7.4: Carry out the integration (7.149) of the Schnaidt model.
Hint: First assume that the line stretches to infinity on both sides of the line
center and then subtract the remaining parts. Make use of the approximation
that in the atmosphere y = α
L
/δ 1.
7.5: Consider an isothermal atmospheric layer so that the line intensity is con-
stant. Also assume that the specific humidity q is constant within this layer.
Show that for an Elsasser band the monochromatic transmission of this
layer is given by
T =
cosh β
2
− cos s
cosh β
1
− cos s
η
, η =
qSp
1
sec ϑ
2πgα
L,1
where ϑ is the zenith angle of the radiation, s = 2πν/δ, and β
i
=
2πα
i
/δ, i = 1, 2 are taken at the lower and upper boundary of the layer
where the pressure is p
1
and p
2
, respectively.
7.6: Determine the vertical transmission of radiation within the spectral range
of a Lorentz line originating at the surface of the Earth where the pressure
is p
0
= 1000 hPa. Ignore the temperature dependence of the line intensity
and of the half-width α
L
. The specific humidity is distributed according
to q( p) = q( p
0
)(p/ p
0
)
2.5
. Assume that S = π cm
−2
,α
L,0
= 0.08 cm
−1
,
δ = 20α
L,0
, and q(p
0
) = 5gkg
−1
.
Hint: Use the Curtis–Godson approximation.
7.7: Consider a vertical transmission path between pressure levels p
1
and
p
2
< p
1
. The concentration c
gas
of an absorbing gas of density ρ
gas
is homo-
geneously distributed.
7.7 Problems 273
(a) Assuming that the line intensity is height independent, by using the Curtis–
Godson approximation, show that
˜
p = ( p
1
+ p
2
)/2.
(b) For a Lorentz line show that the equivalent widths of the strong line and weak
line approximations are given by
strong line approximation: W =
&
2Sα
L,0
c
gp
0
p
2
1
− p
2
2
weak line approximation: W =
Sc
g
(p
1
− p
2
)
where p
0
is the normal pressure.
7.8: For a uniformly distributed absorbing gas the height dependence of the
absorption coefficient is assumed to be given by k
ν
= k
ν,0
-
p/p
0
where p
0
is the surface pressure. The incoming solar radiation at the zenith angle ϑ
0
at the top of the atmosphere is S
ν,0
(W m
−2
). If the Sun’s position remains
fixed, find the parallel solar energy density E
0,t
within 1 hour per square
meter at the Earth’s surface whose albedo is A
g
= 0.31.
7.9: Show that the absorption of a Doppler line can be expressed by
A
D
=
Su
δ
∞
n=0
(−1)
n
b
n
(n + 1)!
√
n + 1
with b = Su
&
ln 2
πα
2
D
For convenience use the wave number notation.
7.10: A cylindrical absorption cell of length l is filled with an absorbing gas.
The pressure within the cell can be varied. The cell is illuminated by a
parallel infrared light beam parallel to the axis of the cylinder. Show that
the absorption at the line center of a Lorentz line is independent of pressure.
7.11: A hypothetical spectral interval contains 100 spectral lines of equal half-
width α
L
. The line positions are statistically distributed. Each decade of
lines is represented by a line of mean intensity S
0
with
Number of lines S
0
(cm
−2
)
1–10 0.05
11–20 0.10
21–30 0.15
31–40 0.20
41–50 0.25
51–60 0.30
61–70 0.35
71–80 0.40
81–90 0.45
91–100 0.50
For α
L
= 0.1cm
−1
, δ = 10α
L
and u = 1 cm find the transmission according
to the exponential model.
274 Transmission in spectral lines and bands of lines
7.12: Schnaidt expresses the average absorption as
A =
1
δ
δ/2
−δ/2
[1 − exp
(
−k
ν
u
)
]dν
This expression ignores the contribution of neighboring lines outside the
spectral range
[
−δ/2,δ/2
]
. Calculate the distribution function for this
model.
7.13: Given is the following tropical model atmosphere.
Height Pressure Temperature Relative humidity Specific humidity
(km) (hPa) (
◦
C) (%) (g/kg)
0 1011 27.7 82 18.76
1.0 902 22.8 68 13.03
1.5 858 20.6 63 11.09
1.8 830 19.2 60 10.00
2.0 808 18.2 59 9.50
2.3 780 16.9 57 8.75
3.0 710 13.3 52 6.96
4.0 636 7.8 48 4.96
4.4 604 5.5 46 4.28
5.0 562 2.0 44 3.43
6.0 495 −3.5 41 2.36
7.0 435 −10.0 39 1.46
8.0 385 −16.6 38 0.885
9.0 340 −23.2 37 0.518
9.8 302 −28.5 38 0.351
10.0 291 −30.0 38 0.312
11.0 253 −38.6 40 0.148
12.0 217 −47.2 43 0.0679
13.0 187 −55.7 49 0.0309
14.0 156 −62.3 58 0.0180
15.0 131 −69.0 70 0.0100
16.0 112 −75.5 90 0.0054
Calculate the radiative temperature change at 0, 1 and 3 km height for the
water vapor window (8–13)
µm. The gray absorption coefficient k may
be traced back to the dimer-water vapor molecule. k is temperature and
pressure dependent according to
k(z) = k
0
[
1 + α
2
(T (z) − T
0
)
]
p
1
(z)
p
0
with k
0
= 12 cm
2
g
−1
, α
2
=−0.02 K
−1
, T
0
= 278 K, p
0
= 1013.25 hPa.
p
1
is the partial pressure of water vapor. For the calculation of the Planck