Radiative_Transfer_Part2

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David Straus
George Mason University / COLA
The Physical Climate System
Atmospheric Radiative Transfer and Climate Part 2
 David Straus
George Mason University / COLA
Atmospheric Radiative Transfer and Climate - Part 2 
General Description of RadiativeTransfer
(Salby Section 8.2 )
Consider the figure on the next page (Fig 8.6 from Salby)
Here the unit vector describes the direction of a “pencil” of radiation.
The monochromatic intensity or radiance is defined by:
which describes the rate that energy of wavelength λ flows per unit area 
“upward” in the direction
Then
is the energy passing through area dA from the lower half space to the upper 
half space with wavelength between λ and λ + dλ, in a range of directions 
Ω to Ω + dΩ, in time dt.
has the units of (power)/(wavelength * area * steradian)
εω=
!̂
!I! = I!"̂
!̂
!I! dAd!d"dt
!I! =!I!(!x,!")
 David Straus
George Mason University / COLA
 David Straus
George Mason University / COLA
Atmospheric Radiative Transfer and Climate - Part 2
()ikrtEAeω⋅−=
kckcω==
Note that is positive definite by definition. 
(What happens if energy is going from the upper half to the lower half 
plane? - we will take that into account shortly. 
The component of intensity normal to the plane is just: 
 
where the unit vector n points up. 
Integrating this over the whole upper half space defines the 
monochromatic flux or irradiance crossing the surface: 
I!
!I! · n̂= I!cos(")
F
+
! =
Z 2"
0
!I! · n̂ d#+
=
Z 2"
0
I!cos($)d#+ (1)
=
Z 2"
0
d%
Z "/2
0
I!(%,$)cos($)sin($)d$
 David Straus
George Mason University / COLA
Atmospheric Radiative Transfer and Climate - Part 2
The flux or irradiance has units of power / (wavelength * area)
The total flux follows by integrating over wavelength: 
Then (F+) dA dt is the total energy passing through area dA in time dt from 
below the plane to above the plane
Similar to F+λ, the flux F-λ describes the rate of energy per unit 
wavelength per unit area which passes from above to below the plane. 
As with F+λ, the flux F-λ is positive definite.
The net flux is then given by:
1/(1)neβω=−
F
+ =
Z
!
0
F
+
"
d" (2)
F! = F+! −F−! (3)
 David Straus
George Mason University / COLA
Atmospheric Radiative Transfer and Climate - Part 2
For isotropic radiation, the intensity Iλ (Φ,θ) is independent of direction, 
and the integral over the solid angle can be readily done:
Extinction and Lambert’s Law:
Lambert’s Law: Extinction process is linear, both in the intensity of 
radiation and the amount of matter:
where dIλ is the change in intensity for an increment of path length ds 
along the direction of the radiation, ρ = mass density, kλ is the 
monochromatic cress section (units of area/mass), and the extinction 
coefficient βe = ρ kλ. Both scattering and absorption in general 
contribute to extinction, and their effects are additive:
F
+
!
= " I! (4)
dI! =−"I!k! ds=−#eI! ds (5)
 David Straus
George Mason University / COLA
Atmospheric Radiative Transfer and Climate
the extinction cross section is thus the sum of the absorption cross 
section σaλ and the scattering cross section σsλ:
Now equation 5 can be re-expressed as:
which can easily be integrated to: 
where we have defined the optical path length u(s) as:
k! = "a!+"s!
d (log I!) =−"k! ds (6)
I!(s) = I!(0)e−
R
s
0 "k! ds
′
= I!(0)e−u(s) (7)
u(s) =
Z
s
0
!k" ds
′ (8)
 David Straus
George Mason University / COLA
T!(s) = I!(s)/I!(0) = e−u(s) (9)
A!(s) = 1− e−u(s) (10)
Atmospheric Radiative Transfer and Climate - Part 2 
Monochromatic transmissivity Tλ(s) describes the fraction of radiation 
remaining in the pencil of radiation after distance s along the path: 
Monochromatic absorptivity Aλ(s) is the fraction of radiation which has 
been absorbed along the path. Since we must have Tλ(s) + Aλ(s) = 1, we 
have:
Black-Body Radiation (using wavelength λ as fundamental variable)
We had derived the emissivity as a function of frequency ω, in the 
section on Black Body Radiation:
9
Pe(!)d! = a(!)
h̄
4"2c2
!3
(e#h̄!−1)d!
where this held for each direction of polarization. We now set the 
absorptivity a = 1 for all ω. Note that we can relate ω to λ and the 
frequency ν as follows:
Further note that:
Writing Iω = Bω for Blackbody radiation flux as a function of ω, we 
have:
! = kc=
(
2"
#
)
c
!=2"$
$=
(
c
#
)
!h̄" =
h̄"
kT
=
h
2#
2#$
kT
=
h$
kT
=
hc
%kT
I! =
h
3
c2
(
e
hc
"kT −1
)
10
To convert from frequency w to l as the fundamental variable, we use:
Iλ dλ = Iω dω
which means that:
so that we obtain:
This particular form for the emissivity Iλ is written as Bλ to denote Black 
Body approximation.
I! = I"
d"
d!
=−I"2#
!2
c=−I"2#$
2
c
I! =
2"h#5
c3
(
e
hc
!kT −1
) = 2"hc2
!5
(
e
hc
!kT −1
)
11
The absorptivity aλ is defined so that the absorption of energy per unit 
time, per unit wavelength, per unit area is given by aλ Bλ. Note that if we 
consider a pencil of radiation traversing an incremental path length of 
ds through a gas, the amount of radiation absorbed will also be small, 
which we will write as daλ Bλ.
The emissivity ελ is defined so that the thermal emission of energy per 
unit time, per unit wavelength and per unit area is given ελ Bλ. Note that 
if we consider a pencil of radiation traversing an incremental path 
length of ds through a gas, the amount of radiation thermally emitted 
into the beam will also be small, and we will write this as dελ Bλ.
Thermal equilibrium (of the matter doing the absorbing and emitting) 
demands that:
ελ = aλ
which is a form of Kirchhoff’s Law.
For greybody radiation, ελ and aλ are independent of λ, but not 
necessarily equal to 1. 
12
The amount of radiation thermally emitted into the beam as it traverses path 
length ds of matter is dελ Bλ so that for emission into the beam we have:
dIλ / Bλ = dελ = daλ 
where we have used Kirchhoff’s Law.
Now writing Lambert’s Law, the extinction (fractional loss of energy) in path 
length ds due to absorption is just:
dIλ / Bλ = -daλ = - ρ σaλ ds
(Absorption and emission by the matter in ds must balance because of local 
thermodynamic equilibrium). Using this last equality in the expression for 
emission into the beam above, we get:
dIλ / Bλ = ρ σaλ ds
or:
dIλ = ρ σaλJλ ds
where
Jλ = Bλ(T)
denotes the source function in the absence of scattering.
13
Now we must take into account the fact that while the gas within the path 
increment ds is in thermal equilibrium at a temperature T, we in general do 
not know the equivalent black-body T ( or brightness T) of the incoming 
beam - this beam may have originated at a very different atmospheric level, 
therefore a very different brightness T.
Thus for the extinction law, we must use the more general form given by 
Lambert’s law:
dIλ / Iλ = - ρ kλ ds
or:
Rewriting the emission into the beam from the previous page as:
we have the general expression for radiative transfer:
 
dI!
"k!ds
=−I!
dI!
"k!ds
= +J!
dI!
"k!ds
=−I!+ J!
14
Please note that we have used the general expression for extinction, 
including scattering. If we neglect scattering, the source function 
becomes the Bl(T), and we can use the absorption cross section σaλ:
Introducing the optical thickness:
which increases in the negative s direction:
We can now rewrite the expression at the top of the page as:
which is known as Schwartzchild’s equation.
dI!
"#a!ds
=−I!+B!(T )
!(s) =
Z "
s
#$a% ds
′ = −u(s)
d! =−"#a$ ds
dI!
d"
= I!−B!(T )
15
- Ignore earth’s curvature
- atmosphere is homogeneous in the horizontal direction
- radiation field is horizontally isotropic
LW
Only
16
The thickness of an atmospheric slab is given by 
dz = μ ds, where μ = cos (θ). 
Now define the optical depth τλ, which is measured downward from the 
top of the atmosphere: 
Working on the LHS of Schwarzchild’s equation:
and we get:
It is important to recognize that since the atmosphereis homogeneous, 
the T-dependence in I and B is replaced by dependencies on t and u: 
!" ≡
Z #
z
$%a" dz
′ =
Z
0
s
$%a"µds= µ&"
dI!
d"!
=
dI#
d#!
d#!
d"!
= µ
dI#
d#!
µ
dI!
d"!
= I!−B!
I!=I! ("!,µ)
B!=B! ("!,µ)
17
It turns out that much of the absorption and thermal emission takes 
place in specific ranges of frequencies (wavelengths) known as 
“bands”, corresponding to classes of vibrational and rotational state 
transition frequencies. Thus the absorption and emission can be 
approximately localized in τλ,.
This allows for the definition of a single “effective” inclination
and a new optical depth: 
It is possible to then recast the problem in terms of a new set of 
variables.
Note that only the the effective inclination appears. This is essentially 
a one-dimensional (vertical) problem now. 
(see Section 8.4 of Salby for details on how this is accomplished). 
µ
!
∗
"
= µ−1!"
F
↑
!
("∗
!
)=#I!("!,+µ)
F
↓
!
("∗
!
)=#I!("!,−µ)
J∗
!
("∗
!
)=#J!("!)
18
Since
is replaced with
we get:
Integrating over the radiative transfer functions F over the upward and 
downward half-spaces yields: 
µ
dI!
d"!
= I!−B!
µ
dI!
d"!
= I!−B!
dI!
d"∗
!
= I!−B!
dF
↑
!
d"∗
!
=F↑
!
−B∗
!
−dF
↓
!
d"∗
!
=F↓
!
−B∗
!
(1)
In the equation for up-welling radiation, τ* decreases with rising z, i.e. 
with increasing path length from the surface.
In the equation for down-welling radiation, -τ* increases with falling z, i.e. 
with decreasing path length from the top of the atmosphere.
19
Radiative Equilibrium in a Grey Atmosphere
Assume the atmosphere is transparent to SW incoming radiation. The 
emissivity and absorptivity are independent of λ The extinction 
(absorption) cross section is also independent of λ. Then the optical path 
lengths τ and τ* are also independent of λ. 
Integrating the equations on the previous page over λ, we obtain:
Now for thermodynamic equilibrium of the atmosphere under LW heating 
alone, we must have the net flux be a constant:
We also define the “average flux” as : 
dF
↑
d!∗
=F↑−B∗ (1)
−dF
↓
d!∗
=F↓−B∗ (2)
F ≡ F↑−F↓ = c (3)
F ≡ F↑+F↓ (4)
20
Adding equations (1) + (2):
Taking (1) - (2) :
which can be integrated to:
and (5) becomes: 
Equations (7) and (8) imply: 
Since at the top of the atmosphere τ* = 0, 
dF
d!∗
= F↑+F↓−2B∗ = F−2B∗ (5)
dF
d!∗
= F (6)
F = F!∗+C (7)
F = 2B∗ (8)
B
∗(!∗) =
1
2
F!
∗+
1
2
C (9)
B
∗(!∗ = 0) =
1
2
C ≡ B∗
0
21
Using this for value for C in equation (9) we obtain:
Now we introduce solar radiation into the problem!
Invoking equilibrium of the total earth-atmosphere system together, the 
balance at the top of the atmosphere is as before, incident short-wave flux 
balancing OLR at top:
where Fs is the incoming solar flux, α the albedo. Of course
(no incoming LW radiation from space). Thus
So since F is constant, we have the simple result that:
 
(1−!)Fs = F↑(0)≡ F0 (11)
F
↓(0) = 0
F(0)=F↑(0)−F↓(0) = F↑(0) = F0
F(0)=F↑(0)+F↓(0) = F↑(0) = F0 (12)
B(!∗) =
1
2
F!
∗+B∗0 (10)
F = F0 (13)
22
Since equation 8 holds for all values of τ*, it holds for τ* = 0, so that:
Using equations (10), (13) and (14): 
B
∗
0 =
1
2
F(0) =
1
2
F0 (14)
B
∗(!∗) =
F0
2
!
∗+
1
2
F0 =
F0
2
(!∗+1) (15)
23
24
Thermal Equilibrium at the Surface:
The black surface absorbs all the incident SW flux, plus the downwelling 
LW flux from the atmosphere:
From equation 8:
From equation 13:
Subtracting the second from the first, and dividing by 2:
At the surface: 
From equations (16) and (17): 
B
∗(Ts) = F0+F↓(!∗s) (16)
F = F↓+F↑ = 2B∗
F = F↑−F↓ = F0
F
↓ = B∗− 1
2
F0
F
↓(!∗
s
) = B∗(!∗
s
)− 1
2
F0 (17)
B
∗(Ts) = B∗(!∗s)+
1
2
F0 (18)
25
The Temperature at the surface is discontinuous, since the ground 
temperature T=Ts is warmer than that of the air just above (at τ* = τ*s).
The radiative equilibrium temperature profile is given by:
where: 
Since density decreases nearly exponentially with z, so does τ*. This rapid 
decrease in height translates into a fairly rapid decrease of T with z. 
B
∗(!∗) = "B [T (!∗)] = #T 4 =
F0
2
(!∗+1) (19)
!∗ =
1
µ
!" =
1
µ
k"
Z #
z
$dz′