Evolution of Earth S Atmosphere (Greek - Vapor Ball)

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 109 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 36Ar/38Ar (Warneck)  present atmosphere derived from earth’s mantle.

3) Mantle 3.8 bya (during space bombardment) releases, through geologic activity (volcanoes), N, S, H2O, Cl, CO2 (based on weathering of pre-Cambrian (> 0.6 bya) rocks.

CO2 ~ 100 x > today. This much CO2 coupled with CH4 may have compensated for a lower solar constant (25%) to keep the earth’s overall temperature warm enough to preserve liquid water.

3.7 bya – oldest known sediments, well rounded pebbles  liquid water present.

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 CO2 from atmosphere, whereas Earth can, capturing CO2 in carbonates.

S, Cl also highly water soluble  trapped in oceans or rained out on surface.

As CO2 is removed and water condenses N2 builds up as there are few sinks.

But no O2, - O2 also absent from all other planetary atmospheres.

Evidence of no O2

1) Pyrite (FeS2) and Uraninite (large UO2) do not form in O2 atmosphere, yet are present in Archaean) sediments which show long transport  ample exposure to atmosphere.

2) Banded iron formations

Comprise majority of world’s iron reserves – Lake Superior source of US iron/steel produced. – Kiruna, Sweden, …

Present in 3.8 Ga rocks from Greenland and persisted until ~ 1.8 Ga, then no longer found.

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.

Formation from sedimentation more evidence of oceans present at 3.8 Ga.

Chemical nature of iron – primarily ferrous (reduced) iron (Fe2+), which is relatively soluble in water. This appears in the sediment as magnetite (Fe3O4).

The layering, lack of detrital, and widespread distribution → iron and silica was in solution → the Fe was ferrous iron since ferric (oxidized) iron (Fe3+) has very low solubility.

Sources
Weathering on land and transport in water to oceans. This could only occur if there was no oxygen present in atmosphere.

Volcanic or hydrothermal activity, also requires lack of O2.

What caused precipitation from solution? Silica continuous precip w/o micro-organisms to take the silica for shells, but the iron. Possibilities are:
Evaporation so fluctuations in temperature - daily/seasonal.

Primitive algae provide oxygen to precipitate the iron by conversion to ferric iron. If true this implies very early photosynthetic organisms.

End of banded iron formations at 1.8 Ga points to the rise of O2, 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 (Fe2O3), iron oxide.

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:

very low atmospheric oxygen,

sufficient sulphur in the atmosphere, and

substantial concentrations of reducing gases
redox – oxidation/ reduction
oxidation is decrease of electrons → increase in oxidation state
reduction is increase of electrons → decrease in oxidation state

4) Reducing gas provided by CH4. 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 O2?

Period associated with stable continents

Cyanobacteria production of O2 prior to 2.5 Ga but consumed faster than production?

Cyanobacteria produces oxygen, but organic matter decays and O2 → CO2 and oxygen lost. For O2 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):

0.001% of PAL, MIF of sulfur disappears
1% of PAL - iron retained in soils – red beds appear
< 40% of PAL required for oceanic anoxia in Proterozoic (1.5 – 0.8 Ga).
> 60% PAL required for fire.
Charcoal present for last 0.45 Ga sets lower limit for O2.
Charcoal gap at 0.38 Ga (middle late Devonian) coincident with evidence of marine anoxia.

Implications of no O2

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

a) Autotrophs – can synthesize organic carbon from inorganic carbon, e.g. CO2

b) Heterotrophs – depend on a source of ogranic carbon in the environment, all modern heterotrophs depend on autotrophs, because of oxidizing atmosphere. Prior to O2 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. C6H12O6  2C2H5OH + 2CO2 (Sugar  aclohol + CO2)
Causes no change in overall oxidation
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) CO2 + 4H2  CH4 + 2H2O (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 O2 production.

Anoxygenic photosynthesis

Does not produce oxygen
Reduced CO2 using H, H2S, 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.
This activity would have reduced hydrogen, allowing a buildup of O2 from the photolysis of water.
2H2O + h (UV)  2H2 + O2 H escape to space (escape velocity), O builds up. This reaction alone would require 26 by to create the earth’s O2 layer.
Buildup of O2 from photolysis of water  ozone (O3) 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.

CO2 + H2O + h (visible)  O2 + CH2O (organic matter) – Photosynthesis. At present photosynthesis creates 20 bTons/year of O2

2) The dissociation of water requires intermediate compounds such as HO2, H2O2 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 CO2 to organic matter and release O2 to atmosphere and oceans O2 converts dead organic matter back to CO2

b) Some CH2O gets buried  carbon lost to the system, at least for a while, and O2 increases

c) CO2 stored in air, oceans, organic matter (living and dead), limestones, humus, peat, coal, oil.

d) O2 released stored in air/oceans, oxidized soil and minerals. Source of O2 must supply atmospheric O2 and satisfy all surface sinks, e.g. FeO, CO2, SO2, H2O.

CO2 – thermodynamics of CO2 geochemistry.

1) Chemical reactions, activities, rate constants for A + B ↔ C [8.2]

Forward reaction = k+[A][B]
[ ] ≡ activity = probability of collision • probability of reaction, generally α concentration.

Reverse reaction = k-[C]

Rate constants
d[A]/dt = d[B]/dt = - k+[A][B] if there is no C
With C present, d[C]/dt = - k-[C], then
d[A]/dt = - k+[A][B] + k-[C]. In steady state, d[A]/dt = 0 →
[C]/[A][B] = k+/k- ≡ keq

pH, H2O ↔ H+ + OH- →
[H+][OH-]/[H2O] = Kw, but [H2O] = 1 →
[H+][OH-] = Kw = 10-14 (T=288), then
–log[H+] – log[OH-] = 14 → pH = -log[H+] and neutrality → pH=7

2) Gas – solid equilibria – Ebelmen-Urey reactions [8.3]

MgSiO3(s) + CO2(g) ↔ MgCO3(s) + SiO2(s)

CaSiO3(s) + CO2(g) ↔ CaCO3(s) + SiO2(s)

FeSiO3(s) + CO2(g) ↔ FeCO3(s) + SiO2(s)

As long as all solid reactants are available their activity is unity so only one activity [CO2] can vary and this = pCO2 = the equilibrium constant.

pCO2(T) = p1 exp(- ∆H/R*T) (Arrhenius law), p1≡ constant
Same form as Clausius Clapeyron Eqn
vp(T) = vp1(T1) • exp[∆H/R* (1/T-1/T1)]
from dP/dT=L/T/∆V).
R* ≡ universal gas constant
∆H ≡ difference in enthalpy of formation between product and reactant.
Enthalpy ≡ total energy of a thermodynamic system.

pCO2(T=300 K) Mg (<19 ppmv) and for Ca (< 0.1 ppmv) → weathering always trying to reduce CO2 but can’t keep up with production, outgassing. Whereas for Fe (6000 ppmv) implies no significant formation of iron carbonates in today’s world.
In the past presence of iron carbonates a signature of high CO2 and significant greenhouse warming.
Figure 8.2 shows CO2 is released in magma as T > 700 K
CO2 then released through volcanic activity.

3) Gas – liquid equilibria [8.4]

pA = KH(T) • cA or cA = pA/ KH(T) -- Henry’s law, pA ≡ partial pressure of A in gas, cA ≡ concentration of A in liquid, KH ≡ Henry’s law constant. Note then as KH decreases cA increases. The form of Henry’s law constant is sometimes written pA = KH(T) • cA and sometimes cA = KH*(T) • pA, where KH* = 1/KH. Pierrehumbert uses pA = KH(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.

Temp dependence of KH - Arrhenius law

KH(T) = KH(To) exp[ -CH • (1/T – 1/To), CH > 0 →

KH decreases as T decreases → Slide

Solubility increases as T decreases. Gases more soluble in cold water. Air bubbles in heated water.

Comes from the van’t Hoff equation, d ln KH /dT = ∆ KH / (R T2)

KH(To) exp[ -CH • (1/T – 1/To), CH = ∆ KH/R

∆ KH ≡ reaction enthalpy at constant temperature and pressure.

CO2 alone is not very soluble in water, but CO2 will disassociate in water, particularly seawater to form carbonic acid (H2CO3) which further disassociates into bicarbonate (HCO3-) and carbonate (CO32-). This will occur in water which is somewhat basic, since the disassociation increases the pH or log [H+], but diminishes significantly when pH ≤ 7.

CO2(aq) + H2O ↔ H2CO3, and H2CO3 ↔ HCO3- + H+

CO2(aq) + H2O ↔ HCO3- + H+
CO2(aq) = cCO2 = pCO2/KH

HCO3- ↔ CO32- + H+

Since [H2O] is an inexhaustible reservoir it is absorbed into K1 and K2 below.
c.i → [HCO3-][H+]/[CO2(aq)] = K1(T)

c.ii →[CO32-][H+]/[ HCO3-] = K2(T)
K1/2 ≡ equilibria constants
Here [ ] = no. of moles liter-1
cCO2 depends on pH which controls [ HCO3-].
At:
pH=6.2, [ HCO3-] = cCO2
pH=8.2, [ HCO3-] = 100 • cCO2

To find the total aqueous CO2 = [CO2(aq)T] = c CO2T
c CO2T = c CO2 + [HCO3-] + [CO32-]

c CO2T = c CO2 + c CO2 K1 /[H+] + cCO2 K2 [ HCO3-] / [H+]

c CO2T = c CO2 • (1 + K1 /[H+] + K2 K1/ [H+]2)

c CO2T = pCO2/KH • (1 + K1 /[H+] + K2 K1/ [H+]2)
c CO2T = pCO2 / KH*

Where KH* = KH / (1 + K1 /[H+] + K2 K1/ [H+]2)

From these eq’ns and Henry’s law we can derive the ratio of atmspheric to oceanic carbon.

Henry’s law constant gives pCO2 as partial pressure of CO2

Then pCO2 = KH*/No • cCO2, No ≡ moles of ocean water. Note Pierre-Humbert’s value for No is off by three orders of magnitue. It should be 7.68 x 1022 moles.

But also p=weight of the atmospheric column above. Thus
dp = g ρ dz, now ρ dz = Mass/Area, g ≡ gravity.
M/A = Nco2 • Mco2/A, with
Nco2 ≡ number of moles of CO2
Mco2≡ molecular weight of CO2
Thus, pCO2 = g • Nco2 • Mco2/A

fatm = KH*/ No • A / (Mco2 g), where fatm ≡ ratio of atmospheric to oceanic carbon, in this case mole to mole. A ≡ surface area of planet. → Exercise to derive this.

To solve this we need the pH of the ocean = -log[H+] since KH* = fn([H+].
Ocean, however, has an additional source of carbonate ion, limestone, CaCO3. Now summing up the concentration of all the ions involved we have.

2[Ca2+] + [H+] = [OH-] + [HCO3-] + 2[CO32-]

Using this and c.iv.1 and c.iv.2 and [H+][OH-] = Kw for the disassociation of water leads to

2[Ca2+] + [H+] = Kw/[H+] + pCO2/KH •K1/[H+] + 2 K2/[H+] • (pCO2/KH •K1/[H+])

[Ca2+] = 0.5 * {Kw/[H+] + pCO2/KH •K1/[H+] + 2 pCO2/KH • K1 K2/[H+]2 - [H+]}

[Ca++] = ½ • { Kw/[H+] + pCO2/KH •K1/[H+] • (1 + 2 K2/[H+]) – [H+] }

For today’s ocean, pH=8.17 → [Ca2+] = 2.2 x 10-5 as a mole fraction.

Now if we assume that [Ca2+] remains fixed and can not keep up with a rapidly changing pCO2, such as happening now, then we can use c.ix.4 to find a new [H+] and thus pH for pure water reservoir in equilibria with some pCO2. → Exercise

Thermodynamics of atmosphere, briefly [2.3]

1) Ideal gas law, p V = N k T, N ≡ number of molecules = n NA = no. of moles • Avagadro’s number. k ≡ Botlzmann’s constant.

2) Define universal gas constant, R* = NA 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

R≡ particular gas constant. R(dry air) = 287 J mol-1 K-1.

Gas law holds for each gas making up a mixture of gases → def of partial pressure. Thus pi = ρi Ri T, Ri = R*/Mi

Mass mixing ratio = ρA / ρB = MA pA /(MB pB)

4) First law of thermodynamics, Conservation of energy. When a gas expands or contracts it does work by pushing against its environment.

Work = force • distance = p dV = p d(1/ρ), if we write it as work per unit mass.
Energy can also change by changing kinetic energy of molecules = cv dT, cv ≡ specific heat at constant volume.
Then δQ = cv dT + p d(1/ρ)
Now p d(1/ρ) = d(p/ρ) – 1/ρ dp = d(RT) – 1/ρ dp
So δQ = (cv + R) dT – 1/ρ dp = cp dT – 1/ρ dp, cp ≡ cv + R

5) Entropy = ds = δQ/T = (cp) dT/T – 1/(ρT) dp = (cp) dT/T – R dp/p

6) For adiabatic process, entropy is conserved → ds=0

d ln(T) = R/cp d ln(P) → T(p) = To (p/po)R/cp
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/cp ≡ dry adiabatic lapse rate. Explains relative atmospheric stability, and why temperature cools with height in the atmosphere.
Potential temperature, θ = T (p/po)-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

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 Bυ(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 = ρr(z).
Now use the hydrostatic equation, dp/dz = -ρ g, which is valid for any species and for the whole atmosphere. Thus
dz = dpatm/(-ρatm g) and dz = dpr / (-ρr g), where pr, and ρr are the density and pressure of species r, then
dτ* = -κ ρr (z) dz = κ dpr(z) / g. Now if r is well mixed with mixing ratio q then
dpr(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 (m2/kg), κ(υ,p,T,qi) = ∑i=0n κ(υ,p,T) qi(p), where qi is mass specific concentration of greenhouse gas i. If i is well mixed then qi = 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(po) • p/po →

τυ*(p1, p2) = - 1/g κ(po) ∫ q(p) p/po dp, where po ≡ surface pressure

For a well mixed gas → q ≠ q(p) = mixing ratio, then

τυ*(p1, p2) = - 1/g κ(po) q/po • ∫p1p2 p dp = - 1/g κ(po) q/po (p22 – p12)/2

Optical depth = τυ*(p1, p2) = - 1/g ∫p1p2 κ(p) q(p) dp, for a single greenhouse gas.

Then we obtain Schwartzchild’s equation:

dI = dτ (Iυ - Bυ(T)) → -dI/dτ = - (Iυ - Bυ(T))

-dI/dτ* = -1/cosθ • ( Iυ(τ*,n) - Bυ(T(τ*))

Now { -dI/dτ = - Iυ + Bυ(T) } • e-τ(s’,s)
-dI e-τ + I e-τ dτ = B e-τ dτ
-∫0s’ d (I e-τ) = ∫0s’ B e-τ dτ
-I(s’) e-τ(s’,s’) + I(0) e-τ(s’,0) = ∫0s’ B e-τ(s’,s) dτ
I(s’) = I(0) e-τ(s’,0) - ∫0s’ 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 ∫0τ 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 ∫ττ∞ B(υ,T(τ’)) e-(τ’-τ) dτ’
τ∞ ≡ optical thickness of entire atmosphere = τ*(ps,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.

Transmission f’n Tυ(p,p’) = exp(-(τ(p)-τ(p’)) → T(p,p’)dτ = ± dT (+,p>p’, -,p<p’)
If the optical depth is calculated between p and p’, then
Tυ(p,p’) = exp(-τ)

I+(τ, υ) = I+(ps,υ) Tυ(p,ps) - ∫p’=pps πB(υ,T(p’)) dT(p,p’)
I-(τ, υ) = I-(0,υ) Tυ(0,p) + ∫p’=0p πB(υ,T(p’)) dT(p,p’)

Integrate by parts, ∫ udv = uv - ∫ vdu

I+(p, υ) = I+(ps,υ) Tυ(p,ps) – [ πB(υ,T(p’)) T(p,p’)|pps - ∫ pps π T(p,p’) dB(υ,T(p’)]
I+(p, υ) = πB(Tp) + [I+(ps,υ) – πB(υ,Tsa] T(p,ps) + ∫ pps π T(p,p’) dB(υ,T(p’)

Similarly for downwelling radiation
I-(p, υ) = πB(Tp) + [I+(∞,υ) – π B(υ,T∞)] T(0,p) + ∫ 0p π T(p,p’) dB(υ,T(p’))

Finally expand dB = dB/dTT(p’) dT/dp’ dp’

(12 a) I+(p, υ) = πB(Tp) + [I+(ps,υ) – πB(υ,Tsa] T(p,ps) + ∫ pps π T(p,p’) dB/dTT(p’) dT/dp’ dp’

(12b) I-(p, υ) = πB(Tp) + [I-(∞,υ) – π B(υ,T∞)] T(0,p) - ∫ 0p π T(p,p’) dB/dTT(p’) dT/dp’ dp’

Net heating rate = hυ = -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 Hυ = g d/dp(I+ - I-) = g dτ/dp d/dτ(I+ - I-)
dτ/dp = -κ/(g <cosθ>) or dp/dτ = - g <cosθ>/κ
→ Hυ = κ/<cosθ> (-d/dτ(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

I+(τ, υ) = I+(0,υ) e-(τ-0) + π ∫-τ∞/2τ B(υ,T(τ’) e-(τ-τ’) dτ’, where we integrate from the bottom of the layer to τ. Now I+(0,υ) = 0, and isothermal layer → B(υ,T) is constant = B, and then ∫-τ∞/2τ B(υ,T(τ’) e-(τ-τ’) dτ’ = B•e-τ•∫-τ∞/2τ eτ’ dτ’

B•e-τ•∫-τ∞/2 τ eτ’ dτ’ = B•e-τ• [eτ - e-τ∞/2] = B•[1- e-τ+τ∞/2]]

Thus, I+ (τ)= π B [1 – exp(-(τ-(-τ∞/2))] → I+ (τ)= π B [1 – exp(-(τ + τ∞/2))]

For the downwelling radiation

I-(τ, υ) = I-(τ∞,υ) e-(τ∞-τ) + π ∫ττ∞/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•eτ•∫τ∞/2 τ e-τ’ dτ’ = B•eτ• [-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 =

hυ = -d/dτ (I+ - I-) = -d/dτ [π B(– exp(-(τ + τ∞/2) + exp(τ-τ∞/2))]
hυ = -d/dτ (I+ - I-) = - π B exp(-τ∞/2) d/dτ [– exp(-τ) + exp(τ)]
hυ = -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.

Optically thick limit when temperature varies, dτ = κ dp/g > 1. Take dp=ps = 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.

∫ pps π T(p’,p) dB/dTT(p’) dT/dp’ dp’ = dB/dT dT/dp ∫ pps π T(p’,p) dp’

∫ pps 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(Tp) + π g <cosθ>/κ dB/dT dT/dp
I-(τ, υ) = πB(Tp) - 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 I-s ≈ π B(Ts) → 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 = hυ = -d/dτ (I+ - I-) = gd/dp(2 π g <cosθ>/κ dB/dT dT/dp)

= d/dp [2 π g2 <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) + ∫0τ π B(υ,T(τ’)) e-(τ-τ’) dτ’
= I+(0,υ) (1-τ) + ∫0τ π B(υ,T(τ’)) dτ’
and
I-(τ, υ) = I-(τ∞,υ) e-(τ∞-τ) + ∫ττ∞ π B(υ,T(τ’)) e-(τ’-τ) dτ’
= I-(τ∞,υ) (1-(τ∞-τ)) + ∫ττ∞ π B(υ,T(τ’)) dτ’

Let ∫ττ∞ B(υ,T(τ’)) dτ’ /τ∞ = B(<T>) ≡ mean emission temperature. Then

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
Hυ = κ/<cosθ> (-d/dτ(I+ - I-))
Hυ = 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 Hυ = κ/<cosθ> [(I+s + I-∞) – 2π B(<T>)]
Hυ 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