Evolution of Earth’s atmosphere
(Greek - vapor ball)
For reference see: Evolution of the atmosphere (Walker, 1975)
Chemistry of Atmospheres (Wayne, 1991)
Chemistry of the Natural Atmosphere (Warneck, 2000)
Primordial
1) If the atmosphere formed from the solar nebula then its noble gases should be similar to solar noble gases.

Neon would be 10-100 times more abundant in earth’s atmosphere than it is. (Fig 9.1, Wayne, 1991)
2) Assumption is original solar nebula atmosphere blown away by solar wind.
Present atmosphere
1) Began formation ~ 4 x 10 9 years ago (bya) based on isotopic ratios of Ar/Xe.
2) Noble gases in ocean island basalts (origin from deep mantle) have same isotopic ratio as in atmosphere today for 36 Ar/ 38 Ar (Warneck)  present atmosphere derived from earth’s mantle.
3) Mantle 3.8 bya (during space bombardment) releases, through geologic activity
(volcanoes), N, S, H
2
O, Cl, CO
2
(based on weathering of pre-Cambrian (> 0.6 bya) rocks. a.
CO
2
~ 100 x > today. This much CO
2
coupled with CH
4
may have compensated for a lower solar constant (25%) to keep the earth’s overall temperature warm enough to preserve liquid water. b.
3.7 bya – oldest known sediments, well rounded pebbles

liquid water present. c.
Liquid water (Goldilocks effect) – initial temperatures of Venus (too warm), Mars (too cold), and Earth (just right) for liquid water to form.
With no liquid water Venus and Mars cannot remove CO
2
from atmosphere, whereas Earth can, capturing CO
2
in carbonates. d.
S, Cl also highly water soluble

trapped in oceans or rained out on surface. e.
As CO2 is removed and water condenses N2 builds up as there are few sinks. f.
But no O
2
, - O
2
also absent from all other planetary atmospheres.
Evidence of no O
2
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 1

1) Pyrite (FeS
2
) and Uraninite (large UO
2
) do not form in O
2
atmosphere, yet are present in Archaean) sediments which show long transport

ample exposure to atmosphere.
2) Banded iron formations a.
Comprise majority of world’s iron reserves – Lake Superior source of US iron/steel produced. – Kiruna, Sweden, … b.
Present in 3.8 Ga rocks from Greenland and persisted until ~ 1.8 Ga, then no longer found. c.
Sediment precipitated from solution, thinly bedded, interleaved with chert
(silica), free of detrital (rock pieces derived from other rock). → water was saturated in both iron and silica. Little organic carbon. Layers extend hundreds of km. d.
Formation from sedimentation more evidence of oceans present at 3.8 Ga. e.
Chemical nature of iron – primarily ferrous (reduced) iron (Fe
2+
), which is relatively soluble in water. This appears in the sediment as magnetite
(Fe
3
O
4
). f.
The layering, lack of detrital, and widespread distribution → iron and silica was in solution → the Fe was ferrous iron since ferric (oxidized) iron (Fe
3+
) has very low solubility. g.
Sources i.
Weathering on land and transport in water to oceans. This could only occur if there was no oxygen present in atmosphere. ii.
Volcanic or hydrothermal activity, also requires lack of O2. h.
What caused precipitation from solution? Silica continuous precip w/o micro-organisms to take the silica for shells, but the iron. Possibilities are: i.
Evaporation so fluctuations in temperature - daily/seasonal. ii.
Primitive algae provide oxygen to precipitate the iron by conversion to ferric iron. If true this implies very early photosynthetic organisms. i.
End of banded iron formations at 1.8 Ga points to the rise of O
2
, and the oxygenation of the oceans. After this time iron is in the form of red beds
→ sandstone, shale. These indicative of ferric iron, not soluble in water, and the deposits contain haematite (Fe
2
O
3
), iron oxide.
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 2

3) Mass independent fractionation of sulfur (Kump, Nature, 2008; Farquhar, et al., sci, 2000) points to oxygenation events at 3.2 and 2.45 Ga. To preserve the MIF signature, three conditions needed: a.
very low atmospheric oxygen, b.
sufficient sulphur in the atmosphere, and c.
substantial concentrations of reducing gases i.
redox – oxidation/ reduction ii.
oxidation is decrease of electrons → increase in oxidation state iii.
reduction is increase of electrons → decrease in oxidation state
4) Reducing gas provided by CH
4
. Oxygenation result from collapse of methane? If so this could also explain first major glaciation as Earth lost a significant greenhouse gas.
5) But why the rise in O
2
? a.
Period associated with stable continents b.
Cyanobacteria production of O
2
prior to 2.5 Ga but consumed faster than production? c.
Cyanobacteria produces oxygen, but organic matter decays and O
2
→ CO
2 and oxygen lost. For O
2
to build up the organic matter must be buried → plate tectonics, subduction zones.
6) Impact of oxygen at various levels wrt present atmospheric level (PAL): a.
0.001% of PAL, MIF of sulfur disappears b.
1% of PAL - iron retained in soils – red beds appear c.
< 40% of PAL required for oceanic anoxia in Proterozoic (1.5 – 0.8 Ga). d.
> 60% PAL required for fire. i.
Charcoal present for last 0.45 Ga sets lower limit for O2. ii.
Charcoal gap at 0.38 Ga (middle late Devonian) coincident with evidence of marine anoxia.
Implications of no O
2
All early life forms (earliest ~ 3.6 Ga) were anoxic.
Over first 1.5 by after first life formed life evolved in its ability to capture energy.
All life must metabolize (derive energy from environment) which has two functions:
1) Biosynthesis – the synthesis of organic molecules to make cells – building cell material from carbon. Two types of biosynthesis
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 3

a) Autotrophs – can synthesize organic carbon from inorganic carbon, e.g.
CO
2 b) Heterotrophs – depend on a source of ogranic carbon in the environment, all modern heterotrophs depend on autotrophs, because of oxidizing atmosphere. Prior to O
2
in atmosphere there was a larger source of abiotic organic carbon.
2) Provide energy to support biologic function, including biosynthesis a) Radiant energy – phototrophs - absorbed energy is converted into chemical energy. Some of these organism use this energy to synthesize organic moleculse

photosynthetic organisms (derive carbon from inorganic compounds). b) Chemical energy - chemoheterotrophs – extract energy from chemical reactions of primarily organic compounds extracted from environment.
This energy source requires the oxidation of one element (electron donor) and reduction (electron acceptor) of another element.
Energy sources for early life forms
Fermentation –
 e.g. C
6
H
12
O
6

2C
2
H
5
OH + 2CO
2
(Sugar

Causes no change in overall oxidation

aclohol + CO
2
)

Can derive energy from organic compounds which are not highly reduced or oxidized

Relatively inefficient energy source < 10% as efficient as aerobic respiration

Life (heterotrophs) slow growing, required organic molecules created abiotically

not a geologic force.
First autotrophs

Anoxic micro-organisms – e.g. methane bacteria (methanogens, archea) CO
2
+
4H
2
 CH
4
+ 2H
2
O (cannot tolerate oxygen –wetlands, guts of mammals – responsible for methane production of digestion.)

Note reduction of carbon and oxidation of hydrogen.

This process limited by supply of hydrogen: Source - volcanoes, hot springs, hydrothermal vents, occupied by extremophiles. Sinks – escape to space, incorporation in organic matter not susceptible to further dissociation, fermentation.

No O
2
production.
Anoxygenic photosynthesis

Does not produce oxygen

Reduced CO
2
using H, H
2
S, and organic molecules (purple and green bacteria examples).

Reduced organic compounds resulting from metabolism are lost to the system in sediments and thus volcanoes still the only source of new reduced compounds which have been lost to system.
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 4


This activity would have reduced hydrogen, allowing a buildup of O
2 from the
 photolysis of water.
2H
2
O + h

(UV)

2H
2
+ O
2
H escape to space (escape velocity), O builds up.
This reaction alone would require 26 by to create the earth’s O
2
layer.

Buildup of O
2
from photolysis of water

ozone (O
3
) also appears and this shuts off the photolysis of water since UV now absorbed by ozone.
Oxygenic photosynthesis – 2 problems to overcome
1) Considerable input of energy required to dissociate water into hydrogen and oxygen. This process essentially uses water as the electron donor instead of hydrogen. Sunlight can provide this energy.

CO
2
+ H
2
O + h

(visible)

O
2
+ CH
2
O (organic matter) –
Photosynthesis. At present photosynthesis creates 20 bTons/year of O
2
2) The dissociation of water requires intermediate compounds such as HO
2
,
H
2
O
2
and OH. These compounds, particularly OH are very reactive and would attack organic matter. Thus organisms had to develop methods to suppress the concentrations of these molecules, modern organisms use enzymes to do this.

Once these problems were overcome life was freed from dependence on volcanoes and became a geologic force in modifying the atmosphere, releasing abundant oxygen.

Leads to carbon cycle a) Plants covert CO
2
to organic matter and release O
2
to atmosphere and oceans
O
2
converts dead organic matter back to CO
2 b) Some CH
2
O gets buried

carbon lost to the system, at least for a while, and
O
2
increases c) CO
2
stored in air, oceans, organic matter (living and dead), limestones, humus, peat, coal, oil. d) O
2
released stored in air/oceans, oxidized soil and minerals. Source of O
2
must supply atmospheric O
2
and satisfy all surface sinks, e.g. FeO, CO
2
, SO
2
, H
2
O.
CO
2
– thermodynamics of CO2 geochemistry.
1) Chemical reactions, activities, rate constants for A + B ↔ C [8.2] a.
Forward reaction = k
+
[A][B] i.
[ ] ≡ activity = probability of collision • probability of reaction, generally α concentration. b.
Reverse reaction = k
-
[C] c.
Rate constants i.
d[A]/dt = d[B]/dt = - k
+
[A][B] if there is no C ii.
With C present, d[C]/dt = - k
-
[C], then iii.
d[A]/dt = - k
+
[A][B] + k
-
[C]. In steady state, d[A]/dt = 0 →
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 5

iv.
[C]/[A][B] = k
+
/k
-
≡ k eq d.
pH, H
2
O ↔ H + + OH → i.
[H
+
][OH
-
]/[H
2
O] = K w
, but [H
2
O] = 1 → ii.
[H
+
][OH
-
] = K w
= 10
-14
(T=288), then iii.
–log[H + ] – log[OH ] = 14 → pH = -log[H + ] and neutrality → pH=7
2) Gas – solid equilibria – Ebelmen-Urey reactions [8.3] a.
MgSiO
3
(s) + CO
2
(g) ↔ MgCO
3
(s) + SiO
2
(s) b.
CaSiO
3
(s) + CO
2
(g) ↔ CaCO
3
(s) + SiO
2
(s) c.
FeSiO
3
(s) + CO
2
(g) ↔ FeCO
3
(s) + SiO
2
(s) d.
As long as all solid reactants are available their activity is unity so only one activity [CO
2
] can vary and this = pCO
2
= the equilibrium constant. e.
pCO
2
(T) = p
1
exp(- ∆H/R*T) (Arrhenius law), p
1
≡ constant i.
Same form as Clausius Clapeyron Eqn
1.
vp(T) = vp
1
(T
1
) • exp[∆H/R* (1/T-1/T
1
)]
2.
from dP/dT=L/T/∆V). ii.
R* ≡ universal gas constant iii.
∆H ≡ difference in enthalpy of formation between product and reactant. iv.
Enthalpy ≡ total energy of a thermodynamic system. f.
pCO
2
(T=300 K) Mg (<19 ppmv) and for Ca (< 0.1 ppmv) → weathering always trying to reduce CO
2
but can’t keep up with production, outgassing. Whereas for Fe (6000 ppmv) implies no significant formation of iron carbonates in today’s world. i.
In the past presence of iron carbonates a signature of high CO2 and significant greenhouse warming. ii.
Figure 8.2 shows CO
2
is released in magma as T > 700 K iii.
CO2 then released through volcanic activity.
3) Gas – liquid equilibria [8.4] a.
pA = K
H
(T) • cA or cA = pA/ K
H
(T) -- Henry’s law, pA ≡ partial pressure of A in gas, cA ≡ concentration of A in liquid, K
H
≡ Henry’s law constant.
Note then as K
H
decreases cA increases. The form of Henry’s law constant is sometimes written pA = K
H
(T) • cA and sometimes cA =
K
H
*(T) • pA, where K
H
* = 1/K
H
. Pierrehumbert uses pA = K
H
(T) • cA, but be careful when looking up Henry’s law constants to check which form is used. The units on the Henry’s law constant will tell which form is used. b.
Temp dependence of K
H
- Arrhenius law
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 6

i.
K
H
(T) = K
H
(T o
) exp[ -C
H
• (1/T – 1/T o
), C
H
> 0 → ii.
K
H
decreases as T decreases → Slide iii.
Solubility increases as T decreases. Gases more soluble in cold water. Air bubbles in heated water.
iv.
Comes from the van’t Hoff equation, d ln K
H
/dT = ∆ K
H
/ (R T
2
) v.
K
H
(T o
) exp[ -C
H
• (1/T – 1/T o
), C
H
= ∆ K
H
/R vi.
∆ K
H
≡ reaction enthalpy at constant temperature and pressure.
c.
CO
2
alone is not very soluble in water, but CO
2
will disassociate in water, particularly seawater to form carbonic acid (H
2
CO
3
) which further disassociates into bicarbonate (HCO
3
) and carbonate (CO
3
2). This will occur in water which is somewhat basic, since the disassociation increases the pH or log [H
+
], but diminishes significantly when pH ≤ 7. i.
CO
2
(aq) + H
2
O ↔ H
2
CO
3
, and H
2
CO
3
↔ HCO
3
-
+ H
+ ii.
CO
2
(aq) + H
2
O ↔ HCO
3
-
+ H
+
1.
CO
2
(aq) = cCO
2
= pCO
2
/K
H iii.
HCO
3
-
↔ CO
3
2-
+ H
+ iv.
Since [H
2
O] is an inexhaustible reservoir it is absorbed into K
1 and K
2
below.
1.
c.i → [HCO
3
-
][H
+
]/[CO
2
(aq)] = K
1
(T)
2.
c.ii →[CO
3
2-
][H
+
]/[ HCO
3
-
] = K
2
(T) a.
K
1/2
≡ equilibria constants b.
Here [ ] = no. of moles liter -1
3.
cCO
2
depends on pH which controls [ HCO
3
-
].
4.
At: i.
pH=6.2, [ HCO
3
ii.
pH=8.2, [ HCO
3
-
] = cCO
2
] = 100 • cCO
2 v.
To find the total aqueous CO
2
= [CO
2
(aq)
T
] = c CO
2
T
1.
c CO
2
T = c CO
2
+ [HCO
3
] + [CO
3
2] a.
c CO
2
T
= c CO
2
+ c CO
2
K
1
/[H
+
] + cCO
2
K
2
[ HCO
3
-
] / [H
+
] b.
c CO
2
T
= c CO
2
• (1 + K
1
/[H
+
] + K
2
K
1
/ [H
+
]
2
) c.
c CO
2
T
= pCO
2
/K
H
• (1 + K
1
/[H
+
] + K
2
K
1
/ [H
+
]
2
)
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 7

d.
c CO
2
T
= pCO
2
/ K
H
* e.
Where K
H
* = K
H
/ (1 + K
1
/[H + ] + K
2
K
1
/ [H + ] 2 ) vi.
From these eq’ns and Henry’s law we can derive the ratio of atmspheric to oceanic carbon.
1.
Henry’s law constant gives pCO
2
as partial pressure of CO
2
2.
Then pCO
2
= K
H
*/N o
• cCO
2
, N o
≡ moles of ocean water.
Note Pierre-Humbert’s value for N o
is off by three orders of magnitue. It should be 7.68 x 10
22
moles.
3.
But also p=weight of the atmospheric column above. Thus a.
dp = g ρ dz, now ρ dz = Mass/Area, g ≡ gravity. b.
M/A = N co2
• M co2
/A, with c.
N co2
≡ number of moles of CO
2 d.
M co2
≡ molecular weight of CO
2 e.
Thus, pCO
2
= g • N co2
• M co2
/A
4.
f atm
= K
H
*/ N o
• A / (M co2
g), where f atm
≡ ratio of atmospheric to oceanic carbon, in this case mole to mole. A
≡ surface area of planet. → Exercise to derive this. vii.
To solve this we need the pH of the ocean = -log[H
+
] since K
H
*
= fn([H
+
]. viii.
Ocean, however, has an additional source of carbonate ion, limestone, CaCO
3
. Now summing up the concentration of all the ions involved we have.
1.
2[Ca
2+
] + [H
+
] = [OH
-
] + [HCO
3
-
] + 2[CO
3
2-
]
2.
Using this and c.iv.1 and c.iv.2 and [H + ][OH ] = K w
for the disassociation of water leads to
3.
2[Ca
2+
] + [H
+
] = K w
/[H
+
] + pCO
2
/K
H
•K
1
/[H
+
] + 2 K
2
/[H
+ ] • (pCO
2
/K
H
•K
1
/[H
+
])
4.
[Ca
2+
] = 0.5 * {K w
/[H
+
] + pCO
2
/K
H
•K
1
/[H
+
] + 2 pCO
2
/K
H
• K
1
K
2
/[H
+
]
2
- [H
+
]}
5.
[Ca ++ ] = ½ • { K w
/[H + ] + pCO
2
/K
H
•K
1
/[H + ] • (1 + 2 K
2
/[H + ]) – [H + ] }
6.
For today’s ocean, pH=8.17 → [Ca 2+
] = 2.2 x 10
-5
as a mole fraction.
7.
Now if we assume that [Ca
2+
] remains fixed and can not keep up with a rapidly changing pCO
2
, such as happening now, then we can use c.ix.4 to find a new
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 8

[H
+
] and thus pH for pure water reservoir in equilibria with some pCO
2
. →
Exercise
Thermodynamics of atmosphere, briefly [2.3]
1) Ideal gas law, p V = N k T, N ≡ number of molecules = n N
A
= no. of moles •
Avagadro’s number. k ≡ Botlzmann’s constant.
2) Define universal gas constant, R* = N
A
k → p V = n R* T, n ≡ no. of moles
3) Now m ≡ mass of the molecules = n M, M ≡ molecular weight of one mole → p V = m R*/M T = m R T and p = m/V R T = ρ R T → p= ρ R T a.
R≡ particular gas constant. R(dry air) = 287 J mol -1 K -1 . b.
Gas law holds for each gas making up a mixture of gases → def of partial pressure. Thus p i
= ρ i
R i
T, R i
= R*/M i c.
Mass mixing ratio = ρ
A
/ ρ
B
= M
A
p
A
/(M
B
p
B
)
4) First law of thermodynamics, Conservation of energy. When a gas expands or contracts it does work by pushing against its environment. a.
Work = force • distance = p dV = p d(1/ρ), if we write it as work per unit mass. b.
Energy can also change by changing kinetic energy of molecules = c v
dT, c v
≡ specific heat at constant volume. c.
Then δQ = c v
dT + p d(1/ρ) d.
Now p d(1/ρ) = d(p/ρ) – 1/ρ dp = d(RT) – 1/ρ dp e.
So δQ = (c v
+ R) dT – 1/ρ dp = c p
dT – 1/ρ dp, c p
≡ c v
+ R
5)
Entropy = ds = δQ/T = (c p
) dT/T – 1/(ρT) dp = (c p
) dT/T – R dp/p
6)
For adiabatic process, entropy is conserved → ds=0 a.
d ln(T) = R/c p
d ln(P) → T(p) = T o
(p/p o
) R/cp b.
Thus temperature at any pressure can be estimated from temperature and pressure at some starting point. Note that T decreases as pressure decreases and dlnT/dlnp = R/c p
≡ dry adiabatic lapse rate. Explains
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 9

relative atmospheric stability, and why temperature cools with height in the atmosphere. c.
Potential temperature, θ = T (p/p o
)
-R/cp
7) Hydrostatic equation. Pressure at the base of an air column = weight of the air above. Thus dp = - g ρ dz, ρ dz = mass in the column/area.
8) If there is a change of phase on an air parcel which is lifted then latent heat of condensation/freezing is added to the system and the temperature lapse rate decreases → clouds can generate their own lift through producing latent heat.
Will not go into details.
9) Examples of temperature profiles [2.2] - slides
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 10

Radiative transfer
4.2 Plane-parallel radiative transfer
4.2.1 Optical thickness and Schwarzchild eq’ns
I(p,n,υ) ≡ spectral irradiance = flux density of electromagnetic radiation, at pressure, p, in direction n, at frequency, υ, or more generally at position s.

I propagates through a layer of atmosphere, ds. I is absorbed in the layer and the number of absorbers are α ds, if ds is small.

Then dI = - κ ρ ds I + ρ ds j, where j ≡ source function in the layer = κ B
υ
(T)
 dI = -κ ρ ds (I – B)

Define dτ ≡ - κ ρ(s) ds ≡ optical thickness = - κ ρ(z) dz/cosθ = dτ*/cosθ
 Where dτ* = -κ ρ(z) dz ≡ optical depth
 Notes in this expression ρ(z) is the density of the absorbing gas = ρ r
(z).

Now use the hydrostatic equation, dp/dz = -ρ g, which is valid for any species and for the whole atmosphere. Thus
 dz = dp atm
/(-ρ atm
g) and dz = dp r
/ (-ρ r
g), where p r
, and ρ r
are the density and pressure of species r, then
 dτ* = -κ ρ r
(z) dz = κ dp r
(z) / g. Now if r is well mixed with mixing ratio q then
 dp r
(z) = d(q p) = q dp since q ≠ q(p) and dτ* = κ q/g • dp
 θ ≡ angle between n and vertical. dτ* ≡ optical depth
 κ ≡ absorption coefficient = absorption cross section per unit mass (m 2 /kg),
κ(υ,p,T,q i
) = ∑ i=0 n
κ(υ,p,T) q i
(p), where q i
is mass specific concentration of greenhouse gas i. If i is well mixed then q i
= constant. The dependence of κ on υ, p, T depends on quantum mechanical nature of the specific absorber.
 For pressure broadening of absorption lines we can write κ(p) = κ(p o
) • p/p o
→
 τ
υ
*
(p
1
, p
2
) = - 1/g κ(p o
) ∫ q(p) p/p o
dp, where p o
≡ surface pressure

For a well mixed gas → q ≠ q(p) = mixing ratio, then
 τ
υ
*
(p
1
, p
2
) = - 1/g κ(p o
) q/p o
• ∫ p1 p2
p dp = - 1/g κ(p o
) q/p o
(p
2
2
– p
1
2
)/2

Optical depth = τ
υ
*
(p
1
, p
2
) = - 1/g ∫ p1 p2
κ(p) q(p) dp, for a single greenhouse gas.

Then we obtain Schwartzchild’s equation:
 dI = dτ (I
υ
- B
υ
(T)) → -dI/dτ = - (I
υ
- B
υ
(T))

-dI/dτ* = -1/cosθ • ( I
υ
(τ*,n) - B
υ
(T(τ*))
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 11


Now { -dI/dτ = - I
υ
+ B
υ
(T) } • e -τ(s’,s)

-dI e -τ + I e -τ dτ = B e -τ dτ

-∫
0 s’
d (I e -τ ) = ∫
0 s’
B e -τ dτ

-I(s’) e -τ(s’,s’) + I(0) e -τ(s’,0) = ∫
0 s’
B e -τ(s’,s) dτ
 I(s’) = I(0) e -τ(s’,0) - ∫
0 s’
B e -τ(s’,s) dτ
 I(s’) = I(0) e -τ(s’,0) + ∫ s’
0 B e -τ(s’,s) dτ

Radiation at the top of the layer = Incident radiation from the bottom attenuated by absorption and added to by emission attenuated.

Plane parallel atmosphere – two stream solution for I
+
and I
-
, ignoring scattering which is appropriate for infrared radiation..

I
+
(τ, υ) = I
+
(0,υ) e -(τ-0) + π ∫
0
τ
B(υ,T(τ’) e -(τ-τ’) dτ’
 π arises from switch from radiance to irradiance.

Upwelling radiation arises from radiation from the surface attenuated exponentially by absorption plus radiation from the atmosphere similarly attenuated.

Most important emission to space arises near the top of the atmosphere where radiation is least attenuated, when the atmosphere is opaque to surface emission.

I
-
(τ, υ) = I
-
(τ
∞
,υ) e
-(τ∞-τ)
+ π ∫
τ
τ∞
B(υ,T(τ’)) e
-(τ’-τ)
dτ’
 τ
∞
≡ optical thickness of entire atmosphere = τ*(p s
,0)/cosθ.
 Downwelling radiation at any level, τ, arises from radiation at the top of the atmosphere attenuated by absorption plus emission from the atmosphere above the level τ also attenuated.

Thus downwelling radiation will be most sensitive to the lowest layers of the atmosphere.

For Earth there is no significant IR radiation at top of atmosphere, so I
-
(τ
∞
,υ) =
0.
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 12


Transmission f’n T
υ
(p,p’) = exp(-(τ(p)-τ(p’)) → T
(p,p’)dτ = ± d
T (+,p>p’, -,p<p’)
 If the optical depth is calculated between p and p’, then

T
υ
(p,p’) = exp(-τ)

I
+
(τ, υ) = I
+
(p s
,υ) T
υ
(p,p s
) - ∫ p’=p ps πB(υ,T(p’)) d T (p,p’)

I
-
(τ, υ) = I
-
(0,υ) T
υ
(0,p) + ∫ p’=0 p πB(υ,T(p’)) d T (p,p’)
 Integrate by parts, ∫ udv = uv - ∫ vdu

I
+
(p, υ) = I
+
(ps,υ) Tυ (p,ps) – [ πB(υ,T(p’)) T (p,p’)| p ps
- ∫
p ps
π T
(p,p’) dB(υ,T(p’)]

I
+
(p, υ) = πB(T p
) + [I
+
(ps,υ) – πB(υ,T sa
] T (p,ps) + ∫
p ps
π T
(p,p’) dB(υ,T(p’)

Similarly for downwelling radiation

I
-
(p, υ) = πB(T p
) + [I
+
(∞,υ) – π B(υ,T
∞
)] T (0,p) + ∫
0 p π T (p,p’) dB(υ,T(p’))

Finally expand dB = dB/dT
T(p’)
dT/dp’ dp’
(12 a) I
+
(p, υ) = πB(T p
) + [I
+
(ps,υ) – πB(υ,T sa
] T
(p,ps) + ∫
p ps
π T
(p,p’) dB/dT
T(p’) dT/dp’ dp’
(12b) I
-
(p, υ) = πB(T p
) + [I
-
(∞,υ) – π B(υ,T
∞
)] T (0,p) - ∫
0 p
π T
(p,p’) dB/dT
T(p’)
dT/dp’ dp’

Net heating rate = h
υ
= -d/dτ (I
+
(τ,υ) – I
-
(τ,υ)) = g d/dp(I
+
- I
-
), which must be integrated over all frequencies to obtain the net heating rate. A negative value → cooling.

Heating rate per unit mass H
 → H
υ
υ
= g d/dp(I
+
- I
-
) = g dτ/dp d/dτ(I
 dτ/dp = -κ/(g <cosθ>) or dp/dτ = - g <cosθ>/κ
= κ/<cosθ> (-d/dτ(I
+
- I
-
)
+
- I
-
)

4.2.2 Special solutions
 Beer’s law

If atmosphere is too cold to radiate significantly, e.g. Earth’s atmosphere in the visible, → B(υ,T) = 0. Then I
+
(τ, υ) = I
+
(0,υ) e
-τ
and I
-
(τ, υ) = I
-
(τ
∞
,υ) e
-(τ∞-τ)
, which is Beer’s law.
 Isothermal slab, τ = τ
∞ /2 at top, τ = -τ ∞ /2 at bottom. No incident flux top or bottom →

For the upwelling radiation
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 13


I
+
(τ, υ) = I
+
(0,υ) e -(τ-0) + π ∫-τ
∞/2
τ
B(υ,T(τ’) e -(τ-τ’) dτ’, where we integrate from the bottom of the layer to τ. Now I
+ constant = B, and then ∫-τ
∞/2
τ
(0,υ) = 0, and isothermal layer → B(υ,T) is
B(υ,T(τ’) e
-(τ-τ’)
dτ’ = B•e
-τ •∫-τ
∞/2
τ
e
τ’
dτ’
 B•e -τ •∫-τ
∞/2
τ
e
τ’
dτ’ = B•e -τ • [e τ
- e -τ∞/2 ] = B•[1- e -τ+τ∞/2] ]

Thus, I
+
(τ)= π B [1 – exp(-(τ-(-τ
∞
/2))]

For the downwelling radiation
→ I
+
(τ)= π B [1 – exp(-(τ + τ
∞
/2))]

I
-
(τ, υ) = I
-
(τ
∞
,υ) e
-(τ∞-τ)
+ π ∫
τ
τ∞/2
B(υ,T(τ’)) e
-(τ’-τ)
dτ’, where we integrate from the bottom of the layer τ
∞
/2 to τ. As before B(υ,T) is constant and
 B•e τ •∫ τ∞/2
τ
e
-τ’
dτ’ = B•e
τ • [-e -τ∞/2 –
(-e
-τ) ] = B•[1- e (τ-τ∞/2)
]

I
-

(τ)= π B [1 – exp(-(τ
∞
/2-τ)]
→ I
-
(τ)= π B [1 – exp(τ-τ
∞
/2)]

At the top τ = τ
∞
/2 → I+ = π B (1-exp(-τ
∞
)) = πB for τ
∞
>> 1. So all cooling is at the top of the layer and = the black body radiation. This is the case when the layer is optically thick.

At the bottom τ = -τ
∞
/2 → I
-
(τ)= π B [1 – exp(-τ
∞
/2-τ
∞
/2)] = π B [1 – exp(-τ
∞
)] =
π B

Thus same relation holds and all the cooling occurs at the bottom of the layer.

Heating rate =
 h
υ
= -d/dτ (I
+
- I
-
) = -d/dτ [π B(– exp(-(τ + τ
∞
/2) + exp(τ-τ
∞
/2))]
 h
υ
= -d/dτ (I
+
- I
-
) = - π B exp(-τ
∞
/2) d/dτ [– exp(-τ) + exp(τ)]
 h
υ
= -d/dτ (I
+
- I
-
) = - π B exp(-τ
∞
/2) [exp(-τ) + exp(τ)]
 In optically thick case heating rate is nearly zero, τ
∞
>> 1, in interior and only becomes significant when near the edges, i.e. when τ approaches τ
∞
/2 or –
τ
∞
/2, and then it is a cooling.

Example clouds, since water droplets are much greater absorber than water vapor. → Boundaries of clouds will cool significantly.

Radiation fog intensifies after the fog forms.

Strato cu form due to destabilization of top of stratus deck by cooling.
 Optically thin case τ
∞
<< 1 then cooling is uniform through the layer.

Thus a thin cloud will display much different cooling rates than a similar layer of water vapor.
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 14


Optically thick limit when temperature varies, dτ = κ dp/g >> 1. Take dp=p s
= depth of atmosphere.

In this case T (p,ps) << 1, and take dB/dT and dT/dp outside integral, assuming dT/dp varies little over the range of p. This can be done since integrand is only significant when. T
(p,p’) is significant which is when p’ is near p, and dB/dT can then be evaluated at p.
 ∫
p ps π T (p’,p) dB/dT
T(p’)
dT/dp’ dp’ = dB/dT dT/dp ∫
p ps π T (p’,p) dp’
 ∫
p ps
π T
(p’,p) dp’ = π g <cosθ>/κ ∫
τ
∞
T
(τ’, τ) dτ’ = π g <cosθ>/κ (1)
 ∫
τ
∞
T
(τ’, τ) dτ’ = ∫
τ
∞
exp(-(τ’-τ) dτ’ = -exp (-(τ’-τ)|
τ
∞
= exp(0) – exp(-∞) = 1

Thus

I
+
(τ, υ) = πB(T p
) + π g <cosθ>/κ dB/dT dT/dp

I
-
(τ, υ) = πB(T p
) - π g <cosθ>/κ dB/dT dT/dp
 Second term disappears as κ increases, optically thick limit, so upwelling and downwelling radiation approach the blackbody emission at the local temperature. These values modified if dT/dp is appreciable then there is some input from the warmer/cooler temperature below/above.

At top of atmosphere I
+∞
≈ π B(T
∞
) at υ where atmosphere is optically thick, e.g. certain regions of IR. Thus planet cools at TOA temperature in optically thick regions.

Near surface Is
≈ π B(T s
) → lowest layers most important for heating sfc. If dT/dp > 0, i.e. T cools as p decreases (usual), then the radiation near sfc is reduced due to the cooler air above.

Heating rate = h
υ
= -d/dτ (I
+
- I
-
) = gd/dp(2 π g <cosθ>/κ dB/dT dT/dp)
 = d/dp [2 π g 2
<cosθ>/κ dB/dT dT/dp].

Optically thin limit
 τ
∞
<< 1, and τ < τ
∞
→ e
-(τ-0)
and e
-(τ∞-τ)
are near unity → they can be expanded in the boundary terms and set to 1 in the integrals for I
+
, I
-
, thus

I
+
(τ, υ) = I
+
(0,υ) e
-(τ-0)
+ ∫
0
τ
π B(υ,T(τ’)) e
-(τ-τ’)
dτ’

= I
+
(0,υ) (1-τ) + ∫
0
τ
π B(υ,T(τ’)) dτ’
 and

I
-
(τ, υ) = I
-
(τ
∞
,υ) e
-(τ∞-τ)
+ ∫
τ
τ∞
π B(υ,T(τ’)) e
-(τ’-τ)
dτ’

= I
-
(τ
∞
,υ) (1-(τ
∞
-τ)) + ∫
τ
τ∞
π B(υ,T(τ’)) dτ’
 Let ∫
τ
τ∞
B(υ,T(τ’)) dτ’ /τ
∞
= B(<T>) ≡ mean emission temperature. Then
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 15


I
+∞
= (1-τ
∞
) I
+s
+ τ
∞
π B(<T>)

I
-s
= (1-τ
∞
) I
-∞
+ τ
∞
π B(<T>)
 → Optically thin atmosphere behaves as an isothermal atmosphere with temperature <T> and small emissivity τ
∞
.

Radiative heat flux

H
υ
= κ/<cosθ> (-d/dτ(I
+
- I
-
))

H
υ
= κ/<cosθ> (-d/dτ{[(1-τ
∞
) I
+s
+ τ
∞
π B(<T>)] – [(1-τ
∞
) I
-∞
+ τ
∞
π B(<T>)]})

= κ/<cosθ> (1-τ
∞
) (I
-∞
- I
+s
) – doesn’t agree with text.

Pierrehumbert claims H
υ
= κ/<cosθ> [(I
+s
+ I
-∞
) – 2π B(<T>)]

H
υ
is small since κ is small and represents heating due to absorption of radiation from surface and space and cooling due to blackbody radiation from the atmospheric layer in question.

Greenhouse gases are optically thin in some regions and thick in others. The IR heating of atmosphere is dominated by regions where optical thickness is ≥ 1.
Thus the IR heating will be controlled by the regions where the optical thickness is the smallest since the optically thick regions could be saturated, allowing almost not IR cooling to space.

Similarly if a gas is optically thin its impact on the IR heating/cooling will be dominated by the thickest of the thin regions.
3.2 (Pierrehumbert) Blackbody radiation
1) Planck function - B(υ,T) dυ = 2hυ
3
/c
2
1/(exp(hυ/k
B
T) -1) dυ a.
h = Planck’s constant 6.32 x 10 -32
J s b.
k
B
≡ Boltzmann’s constant = 1.38 x 10 -23 J/K c.
Transform to wavelength space d.
dυ = d(c/λ) = -c/λ 2
dλ e.
B(λ,T) dλ = 2h(c/λ) 3
/c
2
1/(exp(hc/λk f.
= - 2 h c
2 /λ 5
1/(exp(hc/λk
B
T) -1) (-c/λ
2
dλ)
B
T) -1) dλ, negative sign → flip integration limits
2) Planck function variations a.
Radiance (B), irradiance (π B), Energy density (E = B 4π/c) b.
Radiance ≡ radiation crossing a sfc perpendicular to direction of radiation. c.
B = E (J/m3) • c (m/s) • 1/4π(sr) = all possible directions d.
Digress on solid angle i.
Ω = area of sphere/r 2
= ∫ ii.
Ω = 1/r 2
∫ ∫ r
2 s
sfc of sphere/r
sinθ dθ dφ = 1/r
2
∫
0
2π
r
2
2
dφ ∫
0
π
sinθ dθ = 4π e.
Irradiance ≡ Energy crossing a surface A oriented in space from radiance about some solid angle f.
I = ∫ B(Ω) cosθ dΩ = ∫ ∫ B(Ω) cosθ (sinθ dθ dφ) g.
I = B(υ,T) ∫
0
2π
dφ ∫
0
π/2
cosθ sinθ dθ = B(υ,T) π
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 16

h.
This is the origin of π when we went to two stream solution. i.
These forms of B are in radiance/intensity ≡ power/(area • sr) 1/∆υ/λ
3) Wien’s law – total power emitted = F = σ T 4 a.
F = ∫
0
∞
π B(υ,T) dυ = ∫
0
∞ 2πhυ 3
/c
2
1/(exp(hυ/kT) -1) dυ b.
u= hυ/kT, du= h/kT dυ → c.
υ = kT/h u and dυ = kT/h du → d.
F = ∫
0
∞ 2πh (kT/h u) 3
/c
2
1/(exp(u) -1) kT/h du e.
= 2πh/c
2
(kT/h)
4 f.
= 2π (kT) g.
= [2π k
4
4
/(h
/(h
3 c
2
3 c
2
∫
0
∞
) ∫
) ∫
0
∞
0
∞ u
3 u
3
/(e u u
3
/(e u
-1) du
-1) du
/(e u
-1) du] T
4
= σ T
4
3.3 Radiation balance of planets.
1) Recall we already derived a.
Solar constant, L o
= σT b.
T p
= √(r s
/2r) • T s
4 s
• (1-α)
(r s
/r sp
)
2
1/4 and so
σT p
4
= L o
• (1-α)/4, or
2) How an atmosphere affects OLR. a.
From thermodynamics we showed that dT/dp > 0, i.e. T decreases as pressure decreases. b.
At the surface T=T s
and p=p s
and if atmosphere were transparent to OLR then outgoing radiation = σ T s
4 . c.
Add a greenhouse gas transparent to solar (vis), but not to terrestrial (IR), with absorption cross section/mass κ d.
Over an atmospheric layer, dp the mass is q dp/g for gas with mixing ratio q leading to an absorption of κ q dp/g e.
κ q dp/g < 1 → optically thin → IR can escape from surface. f.
κ q dp/g > 1 → optically thick → layer acts like a black body radiating at layer temperature. g.
Choose dp’ such that κ q dp’/g = 1 and each layer radiates isotropically at
<T’> h.
But each layer is blackbody so also absorbs perfectly → only the topmost layer radiates its energy to space. i.
Thus OLR determined by temperature of this topmost layer and will radiate as
σ T(p rad
)
4
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 17

j.
Since dT/dp > 0 → T(p rad
) < T(p s
) since p rad
< p s
. A greenhouse gas warms a planet to the extent that the atmosphere is colder than the surface. k.
Reality – green house gases have large variations in absorptive properties as a function of υ, thus they are both optically thin and thick in differing frequency ranges. Still in sum a level can be identified as the effective radiative level for a planet, characterizing the mean depth from which IR photons escape to space.
3)
Earth’s radiative equilibrium temperature h.
Calculated to be 255 K i.
Assuming a dry adiabatic atmosphere we can calculate T s j.
T s
= (p s
/p rad
)
R/cp
T rad k.
If know T s
can estimate p rad
≈ 670 hPa l.
If know p rad
can estimate T s
. m.
Now add more greenhouse gas, e.g. CO
2
→ p rad
decreases, thus T s increases. n.
Current indirect estimates suggest the Earth is currently receiving 1 W/m2 more energy that it is radiating to space. This based on estimates of OLR and the solar constant. But we don’t have an effective measure of overall
OLR, since all satellites are too close to Earth and thus have a limited view of Earth. i.
The current excess energy thought to be going to increase in ocean temperature and melting cryosphere. Once these reservoirs consumed then temperature will increase enough to bring Earth’s radiative temperature in line with solar in put and GHG concentrations through increasing the temperature of the radiating level. At present GHGs block the emission at that temperature. ii.
Direct measurements of Earth’s OLR proposed by
Deep Space
Climate Observatory (DSCOVR) (formerly known as Triana ) is a NASA satellite proposed in 1998 by then-Vice President Al Gore for the purpose of Earth observation. It is intended to be positioned at the Earth's L
1
Lagrangian point (stationary gravitational point between sun and earth), at a distance of 1.5 million kilometers. At this location it will have a continuous view of the Sun-lit side of the Earth. Thus it can measure the OLR over the whole Earth disk, and at the same time measure the solar input.
3.6 Optically thin atmosphere - skin temperature
1) Kirchoff’s law – in same spectral ranges emissivity = absorptivity, measured by
Kirchoff and other spectroscopists over a number of materials, and always the emissivity was a function of temperature leading to definition of a black body.
2)
Skin layer ≡ layer of atmosphere with so few molecules that it has a low IR emissivity, that is the upwelling terrestrial IR is only slightly absorbed, and we can assume that the upwelling terrestrial IR = the outgoing longwaver radiation
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 18

(OLR). What is the temperature of this layer? The skin layer radiates both up and down. Thus assuming emission = absorption a.
2 e ir
σ T skin
4 = e ir
OLR, where OLR = σ T rad
4 b.
T skin
= (OLR/(2 σ))
1/4
= T rad
/2
1/4
3) Development of a stratosphere a.
Optically thin atmosphere isothermal low emissivity, → skin layer extends to the ground → T atmosphere
= T skin
= T rad
/2
1/4
= T sfc
/2
1/4
, T sfc
= T surface b.
Sfc heated by solar radiation → σ T sfc
4
= (1-α) S, S≡solar insolation, so p rad
=p s c.
But the sfc warmer than air → convection → temperature profile follows the adiabat and the sfc layer grows until temperature of the layers reaches the skin temperature. Atmosphere above will remain isothermal. d.
T skin
defines tropopause height through T skin
= T sfc
(p trop
/p sfc
)
(R/Cp)
= T sfc
/2
1/4 e.
→ p trop
= 2 (-Cp/4R) • p sfc
, → Tropopause ≠ f(sfc) – only depends on characteristics of the atmosphere.
4) Stratosphere contains strong absorber (ozone) with absorptivity a sw
. a.
Energy balance new skin layer is 2 e ir
σ T nsk
4
= e ir
OLR + a sw
S b.
2 e ir
σ T nsk
4
= 2 e ir
σ T skin
4
+ a sw
S c.
= 2 e ir
σ T skin
4
(1 + a sw
S/( e ir
OLR)), T skin
≡ skin temp no atmos. d.
T nsk
= T skin
• (1 + a sw
S/( e ir
OLR)) 1/4 e.
New skin temp, T nsk
is warmer than previous skin temperature with no absorber. f.
T nsk
will increase with height if a sw
/e ir
increases with height, which would occur for ozone increasing in stratosphere.
5) Earth’s atmosphere a.
T skin
= T= (OLR/(2 σ))
1/4
= T rad
/2
1/4
= 255 * 0.84 = 214 K b.
Assume we know T sfc
= 288, then using T sfc
= T skin
(p sfc
/p trop
)
R/Cp c.
p trops
= (T skin
/T sfc
) Cp/R • p sfc
= 360 hPa d.
Or if we take T skin
= T sfc
/2
1/4
, then e.
p trops
= 2
(-Cp/4R)
• p s
= 0.545 • p s
= 545 hPa f.
Skin/tropopause temperature too warm partly because it ignores clouds and partly because we assume dry adiabatic process. g.
Moist process then we need to use equivalent potential temperature to define lapse rate. h.
Θ e
= T e
(p s
/p)
R/Cp
,
T e
≈ T • exp[L v
w / (c p
T) ] , T e
≡ equivalent temperature,
L v
≡ latent heat of evaporation/condensation, w ≡ mixing ratio of water vapor. i.
So T ≈ T s
(p/p s
)
R/Cp
- L v
/c p
• r
6) OLR Examples, multi-year, February, relationship to clouds a.
Brightness temperature, recall b.
B(υ,T)=2hυ
3
/c
2
[1/(e hυ/kT
-1)]
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 19

c.
In microwave region hυ/kT << 1 → e hυ/kT
= 1 + hυ/kT d.
Thus, B(υ,T)=2hυ
3
/c
2
[1/hυ/kT] = 2υ
2
/c
2
• k T brightness
4.4 Real Gas Radiation (Absorption)
1) Spectroscopy ≡ Identification of materials (gases) by their absorption/emission properties. a.
Spectral tubes, diffraction gratings, hand held spectrometers.
2) Absorption – quantized – requires the appropriate energy to, in order of decreasing energy: a.
Ionize b.
Disassociate c.
Excite = an electron change in orbital shell d.
Change vibrational state e.
Change vibrational - rotational state f.
Change rotational state g.
Kinetic/Translational change h.
E molec
= E
I
+ E
D
+ E
E
+ E vib
+ E vib-rot
+ E rot
+ E trans
3) Classical E&M – only accelerated charges radiate → for a molecule to have a charge it must have a dipole moment. a.
Homonuclear – no dipole moment – O2, N2 – limited vibrational and rotational states b.
Polar molecule has a permanent dipole moment - H2O has rich vibrational and rotational states. i.
3756 cm-1 = 2.7 µm ii.
3657 cm-1 = 2.7 µm iii.
1595 cm-1 = 6.3 µm c.
Linear symmetric molecule – CO2 → O – C – O, no permanent dipole moment – no rotational states, but several vibrational states. i.
Bending, k=667 cm
-1 , λ=15 µm – culprit global warming.
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 20

ii.
Asymmetric stretching, k=2349 cm -1 , λ=4.3 µm
4) Absorption X-section a.
Radiation, I, passing through a box of area A, length ∆x, containing N molecules/vol, with absorption X-section, κ. b.
Exiting the box is I – κ N A ∆x I/A = I - ∆I c.
∆I = – κ N ∆x I d.
In limit, d I/I = d lnI = -κN dx → I = I o
e
-κNx e.
κ N x must be dimensionless i.
If N is (m
-3 ) then κ (m 2
) ii.
If N is (1/kg) then the absorption = κ ρ N x, and κ still (m 2
) f.
With many species in the box then κ = ∑ i
κ i
N i
(ρ) so g.
I = I o
exp (- x ∑ i
κ i
N i
(ρ))
5) Absorption line κ (υ, p, T) = S/γ f’n[(υ-υ c
)/γ] a.
υ c
≡ line center – is fixed by the gas in question b.
S (T) ≡ line strength = ∫ κ dυ → as line broadens the peak decreases, so the integral of κ over the relevant line remains fixed. c.
γ(p, T) ≡ line width – characterized by the time required for spontaneous decay of the excited state, broadened by collisions with other molecules which can add or subtract energy to the excited state just enough to cause either emission or absorption – known as collisional or pressure broadening. Thus this is a function of molecular density (p) and speed (T). d.
Form of f’n[(υ-υ c
)/γ] depends on assumptions, but shape is normalized such that ∫ f’n[(υ-υ c
)/γ] dυ = 1. A common one is to assume a Lorentz line shape → f(x) = 1/[π ( 1 + x
2 )] → i.
κ = S(T) / [γ • π • ( 1 +(∆υ/γ) 2 )], ∆υ = υ-υ c ii.
κ = S(T) / [γ • π • ( 1 +(∆υ/γ) 2 )] • (γ/γ) 2 iii.
κ = S(T) γ / [π • ( γ 2 +∆υ 2 )] e.
Then T
υ
(p,p’) = exp(-(dτ(p,p’)) f.
Now dp/dz = -ρg, ρq=N, ds=dz/cos(θ) = -dp/[ρgcos(θ)] g.
dτ ≡ - κ ρ q ds = κ(p,T) q dp/(g cosθ) → h.
T
υ
(p,p’) = ∫
υ
exp[- 1/(g cosθ) • ∫ κ(p,T) dp] dυ i.
= ∫
-∆/2
-∆/2
exp{- 1/(g cosθ) • ∫ S(T) γ(p) q / [π • ( γ
2
+∆υ
2 )] dp }dυ
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 21

6) Real gas absorption lines a.
Composition of atmosphere b.
Absorption cross sections c.
Brightness temperature ≡ temperature of a blackbody in equilibrium which would give the observed emission from a grey body.
-------------------------------
 4.3 Gray gas model. τ ≠ fn(υ). In this case ∫ B(υ,T) dυ ≈ σ T 4 . Then

I
+
(τ) = I
+
(0) e -τ + σ T(τ) 4 .

I
-
(τ, υ) = I
-
(τ
∞
) e -(τ∞-τ) + σ T(τ) 4


1.
interaction with long wave radiation - 3.5, 3.6
2.
radiative transfer, grey gas - 4.3
3.
radiative transfer, absorbing atmosphere - 4.4
Terry Deshler, University of Wyoming, Notes on Atmosphere, 4/15/2020 22