Atmospheric Structure, Blackbody Radiation, the Radiation Budget and the Greenhouse Effect
2. Blackbody radiation: Planck, Wien, Stefan–Boltzmann and Kirchhoff
A blackbody absorbs all radiation falling on it and, at its temperature, emits the maximum possible radiation at every wavelength. Its spectral radiance is the Planck function, B_λ(T) = (2hc²/λ⁵) / [exp(hc/λkT) − 1], with h = 6.626×10⁻³⁴ J s, c = 3.00×10⁸ m/s and k = 1.381×10⁻²³ J/K. Three working laws follow from it or sit beside it.
- Wien displacement law: setting dB_λ/dλ = 0 gives λ_max T = 2.898×10⁻³ m·K (2898 μm·K). A hotter body peaks at a shorter wavelength.
- Stefan–Boltzmann law: integrating over all wavelengths and the hemisphere gives the flux F = σT⁴, with σ = 5.67×10⁻⁸ W m⁻² K⁻⁴. A real grey surface emits F = εσT⁴ with emissivity 0 < ε ≤ 1.
- Kirchhoff's law: in thermodynamic equilibrium, at each wavelength the emissivity equals the absorptivity, ε_λ = a_λ. A good absorber at a wavelength is a good emitter at that wavelength. The equality is wavelength by wavelength: snow reflects most sunlight yet is nearly black (ε close to 1) in the thermal infrared.
| Quantity | Sun | Earth |
|---|---|---|
| Temperature | about 5800 K (photosphere) | 288 K surface; 255 K effective |
| Peak wavelength λ_max | 0.50 μm (visible) | 10.1 μm at 288 K; 11.4 μm at 255 K (infrared) |
| Emitted flux σT⁴ | 6.4×10⁷ W/m² | 390 W/m² at 288 K |
| Name of the band | Shortwave, about 0.2 to 4 μm | Longwave, about 4 to 100 μm |
Worked example. Take the 288 K surface as a blackbody. Peak wavelength: λ_max = 2898 μm·K / 288 K = 10.06 μm. Flux: F = 5.67×10⁻⁸ × 288⁴ = 5.67×10⁻⁸ × 6.88×10⁹ = 390 W/m². Doubling T would halve λ_max and raise F by 2⁴ = 16. Because the solar and terrestrial spectra hardly overlap, atmospheric radiation is treated as separate shortwave (solar) and longwave (terrestrial) streams.
3. Radiation balance, effective temperature and the global budget
The solar constant S₀ ≈ 1361 W/m² is the solar flux on a surface square to the beam at Earth's mean distance (older books give 1367 to 1370). Earth intercepts the beam over its disc, πR², but radiates from its whole sphere, 4πR², so the global-mean incident flux is S₀/4 ≈ 340 W/m². A fraction α, the planetary albedo (≈ 0.30), is reflected by clouds, air, aerosol and the surface. The absorbed flux is S₀(1 − α)/4 ≈ 238 W/m².
Effective temperature. In steady state, absorbed solar equals emitted longwave: S₀(1 − α)/4 = σT_e⁴. So T_e = [S₀(1 − α)/(4σ)]^¼ = [238.2 / 5.67×10⁻⁸]^¼ = 254.6 K ≈ 255 K (−18 °C). T_e is the blackbody temperature that would emit the observed outgoing flux, not any real air temperature. The mean surface temperature is about 288 K, so the greenhouse effect adds about 33 K. With a 6.5 K/km lapse rate, 255 K is found (288 − 255)/6.5 ≈ 5 km up, the typical level from which longwave radiation escapes.
Sensitivity. Since T_e ∝ [S₀(1 − α)]^¼, a 1% change in S₀ changes T_e by 0.25%, about 0.6 K. Raising α from 0.30 to 0.32 lowers T_e from 254.6 K to 252.7 K, a fall of 1.8 K, before any feedback. An order-of-magnitude check: forgetting the factor 4 gives (952.7/5.67×10⁻⁸)^¼ ≈ 360 K, far above any observed mean, which exposes the slip.
| Component | Value | Remark |
|---|---|---|
| Incoming solar at top of atmosphere | 340 | S₀/4 |
| Reflected solar | ≈ 100 | mostly by clouds and air; about a quarter by the surface |
| Solar absorbed by atmosphere / by surface | ≈ 80 / ≈ 160 | sum ≈ 240 = 340 − 100 |
| Surface longwave emission / downward longwave from air | ≈ 396 / ≈ 340 | the large back-radiation is the greenhouse effect |
| Latent / sensible heat flux from surface | ≈ 80 / ≈ 20 | non-radiative loss from the surface |
| Outgoing longwave radiation at top | ≈ 240 | balances absorbed solar |
Check the surface: it gains 160 + 340 = 500 and loses about 396 + 80 + 20 = 496 W/m², balanced to within the few W/m² by which published budgets differ (each figure in the table is rounded and varies by a few W/m² between estimates). Latitude by latitude the balance fails: low latitudes absorb more than they emit, high latitudes less, and the surplus is carried poleward by atmosphere and ocean (of order 5×10¹⁵ W at its peak). That transport drives the general circulation.
4. Modes of heat transfer in the atmosphere
- Conduction is molecular transfer. Air is a poor conductor (k ≈ 0.025 W m⁻¹ K⁻¹), so it matters only in the millimetre-thick film touching the ground and in the soil.
- Convection, in meteorology, is vertical bulk transport: buoyant thermals and turbulent eddies carry sensible heat up from the heated surface. It is the main way the lowest kilometres of air are warmed.
- Advection is horizontal transport of heat by the mean wind, as in a warm front, a sea breeze or the hot loo wind.
- Radiation needs no medium. It is the only way the Earth system exchanges energy with space, and the main way air cools in the free atmosphere.
Latent heat transport deserves its own mention. Evaporation takes L_v = 2.5×10⁶ J/kg from the surface; condensation in cloud returns it to the air at height. Energy thus moves upward, and with the winds poleward, carried by moisture. The ratio of sensible to latent heat flux at the surface is the Bowen ratio: well below 1 over tropical oceans, well above 1 over deserts.
Worked example. Evaporating 1 mm of water from 1 m² per day is 1 kg per day, so the latent flux is 2.5×10⁶ J / 86 400 s = 28.9 W/m². A global-mean precipitation near 2.7 mm/day therefore corresponds to 2.7 × 28.9 ≈ 78 W/m², matching the budget's latent flux of about 80 W/m². The handy conversion is 1 mm/day ≈ 29 W/m².
5. The greenhouse effect, greenhouse gases and radiative forcing
The atmosphere is nearly transparent to the Sun's shortwave radiation but absorbs much of the surface's longwave radiation and re-emits it in all directions at its own, lower temperature (Kirchhoff). Radiation to space therefore leaves from a high, cold level (about 5 km, near 255 K) instead of from the 288 K surface. The surface must be warmer than T_e so that the cold emission level still radiates the 238 W/m² that the planet absorbs. Adding a greenhouse gas raises the emission level to a colder height, the outgoing flux drops, and the planet warms until balance returns.
| Gas | Main band(s) | Remark |
|---|---|---|
| Water vapour | 6.3 μm vibration; pure rotation beyond about 15 μm | strongest absorber; amount set by temperature, so a feedback |
| Carbon dioxide | 15 μm bending band; 4.3 μm | long-lived and well mixed; 15 μm lies near Earth's emission peak |
| Ozone | 9.6 μm, inside the window | also absorbs ultraviolet in the stratosphere |
| Methane, nitrous oxide | about 7.7 μm and 7.8 μm (N₂O also 17 μm) | small amounts, large effect per molecule |
A molecule absorbs infrared only if its vibration or rotation changes its electric dipole moment. N₂ and O₂, homonuclear diatomics making up 99% of air, have no such change and are transparent; H₂O, CO₂, O₃, CH₄ and N₂O are not. The atmospheric window, about 8 to 12 μm, is the spectral gap through which surface radiation escapes directly.
Radiative forcing is the change in net downward flux at the tropopause, in W/m², caused by an imposed change (a gas, aerosol, solar output or albedo), after the stratosphere adjusts but with surface temperature held fixed; positive forcing warms. For CO₂ a widely used simplified fit is ΔF = 5.35 ln(C/C₀) W/m². The logarithm arises because band centres are already saturated and only the wings keep absorbing. Doubling CO₂ gives 5.35 × 0.693 = 3.7 W/m². The no-feedback (Planck) response is ΔT = ΔF / (4σT_e³) = 3.7 / 3.76 ≈ 1.0 K; feedbacks, chiefly water vapour, amplify it, and the assessed best estimate of the full equilibrium response to doubling is about 3 K.
One-layer model. Let the atmosphere be one isothermal layer at T_a that is transparent to sunlight and has longwave emissivity (= absorptivity) ε. At the top: ε σT_a⁴ + (1 − ε)σT_s⁴ = σT_e⁴. In the layer, absorbed = emitted: ε σT_s⁴ = 2ε σT_a⁴, since it radiates both up and down. Hence σT_a⁴ = σT_s⁴/2 and T_s⁴ = 2T_e⁴/(2 − ε). For a fully absorbing layer (ε = 1), T_a = T_e = 255 K and T_s = 2^¼ × 255 = 303 K, too warm. To obtain T_s = 288 K the layer needs ε = 2[1 − (255/288)⁴] = 0.77. The real atmosphere is cooler at the surface because convection and latent heat also remove energy from the surface.
Key takeaways
- The atmosphere is layered by the sign of dT/dz: troposphere (falls, 6.5 K/km mean), stratosphere (ozone heating), mesosphere (falls) and thermosphere (rises). Dry air is 78.08% N₂, 20.95% O₂, 0.93% Ar and about 0.04% CO₂; water vapour is the variable gas and the tropopause is higher at the equator.
- Planck gives the spectrum, Wien gives λ_max T = 2898 μm·K, Stefan–Boltzmann gives F = σT⁴ (σ = 5.67×10⁻⁸ W m⁻² K⁻⁴), and Kirchhoff gives ε_λ = a_λ wavelength by wavelength. The Sun peaks near 0.5 μm and the Earth near 10 μm, so shortwave and longwave are treated separately.
- Absorbed solar is S₀(1 − α)/4 ≈ 238 W/m² (the 4 is sphere area over disc area), so T_e = [S₀(1 − α)/(4σ)]^¼ ≈ 255 K. The surface is about 33 K warmer than this at 288 K; T_e is the blackbody temperature of the outgoing radiation, found near 5 km height.
- Heat moves by conduction (negligible except at the ground), convection (vertical), advection (horizontal) and radiation, plus latent heat carried by moisture: 1 mm/day of evaporation is about 29 W/m².
- Greenhouse gases need a changing dipole moment, so H₂O, CO₂, O₃, CH₄ and N₂O act and N₂ and O₂ do not. They raise the emission level to a colder height, reducing outgoing flux and warming the surface. A one-layer model gives T_s⁴ = 2T_e⁴/(2 − ε).
- Radiative forcing for CO₂ follows ΔF = 5.35 ln(C/C₀) W/m², about 3.7 W/m² per doubling, logarithmic because the band centres are saturated; the no-feedback warming is about 1 K, and feedbacks amplify it.
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In which layer of the atmosphere does temperature rise with height mainly because ozone absorbs solar ultraviolet?
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Answer: C — Stratosphere, because the ozone layer lies in it and heats it from within
The ozone layer (about 15 to 35 km) lies in the stratosphere and absorbs ultraviolet, so temperature rises with height there and the layer is stable. The troposphere is heated from below and its temperature falls with height. The thermosphere is heated by extreme ultraviolet absorbed by O₂ and N₂, not by visible light.Why is the tropical tropopause higher and colder than the polar tropopause?
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Answer: B — Deep tropical convection carries the layer of falling temperature to a greater height
Strong surface heating and deep moist convection in the tropics keep temperature falling with height up to 16 to 17 km, where it reaches about 190 to 200 K. At the poles convection is weak and the tropopause is 8 to 9 km. The lower-pressure explanation is wrong: the tropical troposphere is deeper, not thinner.According to Kirchhoff's law, at a given wavelength and in thermodynamic equilibrium, a surface that absorbs strongly will
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Answer: D — emit strongly at that same wavelength, because emissivity equals absorptivity there
Kirchhoff's law is ε_λ = a_λ at each wavelength. A good absorber at a wavelength is therefore a good emitter at it. The tempting inverse (weak emission) confuses the energy absorbed with energy reflected; absorption is the part that is not reflected.A blackbody has its Planck spectral peak (per unit wavelength) at 0.50 μm. Its temperature is raised so that the total emitted flux becomes 16 times larger. The new peak wavelength is
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Answer: B — 0.25 μm, since flux ∝ T⁴ so T doubles and λ_max ∝ 1/T halves
Flux is F = σT⁴, so 16 times the flux means T is multiplied by 16^(1/4) = 2. Wien's law λ_max T = constant then halves the peak wavelength to 0.25 μm. Using 1/T⁴ for the peak applies the Stefan–Boltzmann exponent to the wrong law.The global-mean solar flux incident at the top of the atmosphere is S₀/4, not S₀, because
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Answer: D — the Earth intercepts the beam over its disc πR² but its surface area is 4πR²
Intercepted power is S₀πR², spread over a sphere of area 4πR², giving S₀/4 per unit area. The albedo is applied separately afterwards. The other statements invent absorption above the top, a mean cosine and a seasonal fraction, none of which sets the factor.A student computes the effective temperature of a planet using S₀(1 − α) = σT_e⁴, forgetting the factor 4. For S₀ = 1361 W/m², α = 0.30 and σ = 5.67×10⁻⁸ W m⁻² K⁻⁴, the answer is approximately 360 K instead of 255 K. The ratio of the wrong to the correct value is
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Answer: C — 4^(1/4) ≈ 1.41, since T_e is the fourth root of the flux
T_e ∝ (flux)^(1/4), so omitting a factor 4 in the flux multiplies T_e by 4^(1/4) = √2 = 1.414. Check: 255 × 1.414 = 360 K. A temperature error of a factor 4 or 2 would be wildly larger than observed, which is how the slip is caught.Which pair of gases makes up about 99% of dry air yet has essentially no thermal-infrared absorption?
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Answer: B — Nitrogen and oxygen, because their vibrations and rotations do not change the dipole moment
N₂ (78%) and O₂ (21%) are homonuclear diatomics: no vibration or rotation produces a changing dipole, so they do not absorb or emit in the thermal infrared. CO₂ and O₃ do absorb, so the pairs containing them fail; being linear or being made of one element is not the criterion.In a one-layer greenhouse model the atmosphere is isothermal, transparent to sunlight, with longwave emissivity ε = 0.5 and T_e = 255 K. Using T_s⁴ = 2T_e⁴/(2 − ε), the surface temperature is closest to
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Answer: B — 274 K, since T_s = 255 × (4/3)^(1/4)
T_s⁴ = 2T_e⁴/1.5 = (4/3)T_e⁴, so T_s = 255 × 1.0746 = 274 K. The 303 K value needs ε = 1 and 255 K needs ε = 0. The 312 K value takes a square root instead of a fourth root and has no physical basis.Radiative forcing by CO₂ is approximated by ΔF = 5.35 ln(C/C₀) W/m². Compared with the first 100 ppm added to a 280 ppm atmosphere, the next 100 ppm added gives
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Answer: A — a smaller forcing, because the dependence is logarithmic in concentration
ln(380/280) = 0.305 gives 1.63 W/m² and ln(480/380) = 0.234 gives 1.25 W/m², so the second increment forces less. The band centres are saturated but the wings keep absorbing, so the forcing does not fall to zero.Select ALL correct statements about the atmosphere.
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Answer: A — In the homosphere the proportions of N₂, O₂ and Ar are nearly constant with height; B — The mesopause is the coldest region of the atmosphere, near 180 to 190 K; D — The thermosphere has high temperature but very low heat content because of its low density
Turbulent mixing keeps the dry-air proportions fixed to about 80 to 100 km. The mesopause is the coldest region, the thermosphere is hot but nearly empty. Water vapour is concentrated in the lowest few kilometres of the troposphere, so the stratosphere statement is false.A blackbody at 5800 K and a blackbody at 290 K are compared. Select ALL correct statements.
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Answer: A — The hotter body's peak wavelength is 20 times shorter than the cooler one's; C — The hotter body's emitted flux is 20⁴ = 160 000 times larger than the cooler one's; D — The cooler body peaks in the thermal infrared near 10 μm
The temperature ratio is 20. Wien gives λ_max inversely proportional to T, so the peak is 20 times shorter (0.50 μm against 10.0 μm). Flux scales as T⁴, 160 000 times, so the statement of 20 times is false.Select ALL molecules that absorb strongly in the thermal infrared in the way relevant to the greenhouse effect.
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Answer: A — H₂O, through vibration-rotation bands and pure rotation; C — CO₂, through its 15 μm bending band
Water vapour and CO₂ have dipole-changing motions. The CO₂ bending mode near 15 μm lies at the Earth's emission peak. N₂ has no dipole change and Ar is monatomic, with no vibrations or rotations at all.A body behaves as a blackbody at 300 K. Using Wien's constant λ_max T = 2898 μm·K, find its peak emission wavelength in μm, to two decimal places.
Numerical answer — type the value.
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Answer: 9.66
λ_max = 2898 / 300 = 9.66 μm, in the thermal infrared.A planet receives S₀ = 1361 W/m² at its distance from its star and has planetary albedo α = 0.25. Taking σ = 5.67×10⁻⁸ W m⁻² K⁻⁴ and no atmosphere, find its effective temperature in kelvin, rounded to the nearest integer.
Numerical answer — type the value.
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Answer: 259
Absorbed = S₀(1 − α)/4 = 1361 × 0.75 / 4 = 255.2 W/m². T_e = (255.2 / 5.67×10⁻⁸)^(1/4) = (4.501×10⁹)^(1/4) = 259.0 K.In the one-layer model (isothermal layer, transparent to sunlight, longwave emissivity ε, T_s⁴ = 2T_e⁴/(2 − ε)), take T_e = 255 K and a surface temperature T_s = 288 K. Find ε to two decimal places.
Numerical answer — type the value.
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Answer: 0.77
Rearranging, ε = 2[1 − (T_e/T_s)⁴] = 2[1 − (255/288)⁴] = 2(1 − 0.6146) = 0.77.Using ΔF = 5.35 ln(C/C₀) W/m², find the radiative forcing in W/m² when CO₂ rises from 280 ppm to 420 ppm, to two decimal places.
Numerical answer — type the value.
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Answer: 2.17
ΔF = 5.35 × ln(420/280) = 5.35 × ln 1.5 = 5.35 × 0.4055 = 2.17 W/m².Evaporation removes 3.0 mm of liquid water per day from a surface. Taking the latent heat of vaporisation L_v = 2.5×10⁶ J/kg, the density of water 1000 kg/m³ and 1 day = 86 400 s, find the equivalent latent heat flux in W/m², to one decimal place.
Numerical answer — type the value.
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Answer: 86.8
Mass per m² per day = 0.003 m × 1000 = 3 kg. Flux = 3 × 2.5×10⁶ / 86 400 = 86.8 W/m².