RADIATIVE transfer is an incredibly important topic when it comes to the earth’s climate system. It is the only way for the earth to either heat up, via absorption, or cool down, via emission. The temperature of the earth is set by these two counter acting mechanisms. To solve this balance for the earth’s dynamic systems is incredibly difficult. But to understand the basic principles involved is fairly straight forward and central to understanding the theory of anthropogenic global warming (AGW).
Some caveats first:
Black Bodies(BB). BB’s are perfect absorbers of all incident radiation. BB’s are also the ideal or most efficient emitters of radiation. No object will emit more than a BB when at the same temperature. In the real world no object is a BB, but many objects approach BB’s over certain spectrums. For instance the sun is a good approximation of a BB.
Grey Bodies(GB). GB’s are bodies that have uniform emissivity across all wavelengths. As with BB’s no real world bodies are perfect GB’s. As such it is a useful approximation.
For example, the earth can be approximated by a GB with emissivity around 0.7 for solar wavelengths(λ < 4μm) and an emissivity near 1.0 for terrestrial wavelengths(λ > 4μm). Emissivity is a wavelength dependant quantity but approximations like the one used for the earth work very well.
Kirchhoff’s Law. This law is an equality between the absorptivity and emissivity. It holds under local thermodynamic equilibrium(LTE). The vast majority of the atmosphere is in LTE, only at high altitudes where the atmosphere is tenuous does this break down. For practical purposes we are always in LTE.
Equilibrium. Equilibrium is the state where temperatures are no longer changing. At this point a body must be emitting the same amount of energy that it is absorbing. This is a consequence of the Conservation of Energy.
For ease of calculations all bodies are assumed to be isothermal and in isolation with no conduction or convection(ie: isolated in a vacuum). Also all bodies are all solid surfaces, there is no transmission only absorption and emission. These simplifications aid in understanding the underlying pattern of radiative transfer. They are not a requirement of actually solving problems but are necessary to simplify the equations.
Radiative exchanges in a nutshell:
Kirchhoff’s equates absorptivity α and emissivity ε. Generally only ε is used when speaking of both absorptivity and emissivity.
1) α = ε
Absorption by an object is simply a function of ε. The temperature of the absorber is irrelevant to absorption. The amount absorbed(q) is a function of emissivity and the incident power(Q).
2) q = εQ
Emission from an object is a function of both ε and its temperature. The emitted power(p) is a function of emissivity, the Stefan-Boltzmann constant(σ), the temperature(T) in Kelvin raised to the fourth power and the area(A) of the emitting surface. This is the Stefan-Boltzmann Law. This law is the integration over all wavelengths of the Planck function. The Plank function describes the intensity of emitted radiation per wavelength for a given temperature.
3) p = εσT^4.A
A body will absorb radiation at a rate given by 2) and will emit radiation at a rate given by 3). For a body to be in equilibrium 2) must equal 3). If the temperature of the object is to low to allow equilibrium then the body will heat up until it reaches a temperature that satisfies this equality.
To say that a body is not emitting as much as it is absorbing is just another way as saying that the body is heating up.
4) p = q
Those are the basics as both I understand them and how they are recorded in literature. Getting these four simple concepts wrong will cause all manners of erroneous conclusions and beliefs.
This thread is opened with the intent to discuss the above compared to the understanding of other readers.
A good primer on the subject is a series of pages provided by http://hyperphysics.phy-astr.gsu.edu/hbase/bbcon.html#c1
Example of using above basic radiative transfer equations to calculate the temperature of an object.
Body A is illuminating Body B. Body B is at a distance D from body A.
Source, body A, is represented as:
T(a) = 5778K
R(a) = 6.955*10^8m
ε = 1.0
Target, body B.
T(b) = ?
R(b) = 6.371*10^6m
D(ab) = 1.496*10^11m
ε = 0.7
Body A emits:
5) p = εσT^4.A
6) p = 1 * 5.67*10^-8W/m^2K^4 * 5778K^4 * (4 * π * 6.955*10^8m^2)
7) p = 3.84*10^26W
Power Body B intercepts:
8) Q = (πR(b)^2 / 4πD(ab)^2) * p
9) Q = (π * 6.371*10^6m^2 / 4 * π * 1.50*10^11m^2) * 3.84*10^26W
10) Q = 4.53*10^-10 * 3.84*10^26W
11) Q = 1.74*10^17W
Body B absorbs:
12) q = εQ
13) q = 0.7 * 1.74*10^17W
14) q = 1.22*10^17W
Body B emits:
15) p = q
16) p = 1.22*10^17W
Body B temperature:
17) p = εσT^4.A
18) T = (p/εσA)^.25
19) T = (1.24*10^17W / (0.7 * 5.67*10^-8W/m^2K^4 * (4 * π * 6.371*10^6m^2))^.25
20) T = 278.6K (or 5.4C)
This calculation is for the Sun/Earth system. It contains one error, the emissivity for the earth was held constant at 0.7 for both absorption of solar and emission of terrestrial radiation. The better approximation of the earth described at the top of the article would result in T = 254.8k(or -18.3C).
Of course the presence of an atmosphere would change this calculation. That is why in the real world we have a temperature of 15C, the greenhouse effect explains the difference.
This post follows on from discussions at the thread http://jennifermarohasy.com/blog/2011/04/determining-the-total-emissivity-of-a-mixture-of-gases-containing-overlapping-absorption-bands/