"Man, being the servant and interpreter of Nature, can do and understand so much and so much only as he has observed in fact or in thought of the course of nature. Beyond this he neither knows anything nor can do anything." Francis Bacon 1620 "The New Organon"
The most crucial issue in current AGW debate is of course the greenhouse warming effect of CO2 gas.Laboratory and satellite spectra show CO2 gas has three main absorption bands at wavelengths of 2.7, 4.3 and 15 µm, respectively.The 15 µm absorption band absorbs significant amount of thermal radiation emitted from the earth ground surface.We all know that absorption of radiation waves leads to warming up of an object.Water vapor and CO2 molecules absorb, N2 and O2 do not.It thus appears logical and straightforward to label water vapor and CO2 greenhouse gases that warm up the atmosphere.While the warming effect of absorption has been well addressed, the cooling effect of emission for CO2 has not.
There are two simple physical laws: one is the absorption law describing how much
energy, I, an object absorbs; the other is the Stefan-Boltzmann law expressing how much energy, J, the same object with surface temperature, T, emits per unit area and unit time.
(1) I = aI0
(2)J = ε σT4
where, α and ε are surface absorptivity and emissivity of the object, σ is the Stefan-Boltzmann constant equal to 5.670373 x 10-8 (W/m2K4), and I0 represents the radiation source.I is the energy in, and J is the energy out of the object.If I > J, the object gains heat resulting in warming; if I < J, the object loses heat leading to cooling.When I = J, the object reaches its radiative equilibrium.For a given radiation source, I0, there is a corresponding radiative equilibrium temperature, T, for an object.
Absorptivity and emissivity are the intrinsic material properties of an object.According to the Kirchhoff’s law, an object with absorptivity α = 0 (or α = 1), emissivity of the object must be equal to ε = 0 (or ε = 1).Visa verse, if α¹ 0 (or α¹ 1), there must be ε ¹ 0 (or ε ¹ 1).Put it in words: an object that absorbs emits, or an object that emits absorbs.Indeed, absorption and emission are two inseparable equivalent identities of the same physical essence.
An important message from Eqs (1) and (2) is that absorption of heat energy relies on and only on EXTERNAL factor - radiation source, I0, whilst emission is determined by and only by INTERNAL factor - surface temperature, T, of the object.
Bearing these concepts in mind, we are now ready for a thought experiment:
1) Cover the Earth with literally white cloth so that there is no radiation wave for CO2 molecules to absorb.The temperature of CO2 molecules shall approach 0 K (-273.15°C).This is because:
If I0 = 0, there must be I = 0 according to Eq. (1);
But J > 0 because T > 0 according to Eq. (2); therefore CO2 keeps losing heat and dropping its temperature until T = 0.
Figure 1.What will be the CO2 temperature if the Earth ground surface is covered with literally white cloth.
2) Remove the white cloth to expose CO2 molecules to the earth ground surface radiation, I0, which will raise the temperature of CO2 molecules at lowest altitude from -273.15°C to -78°C (math omitted here) because I0 is not strong enough.CO2 temperatures at higher altitude would be even lower because of damped intensities of the radiation waves.
3) Now take heat transfer by molecular collision into account.One may argue that CO2 will not be frozen to 0 K in 1) or -78°C in 2) because of constant heat transfer due to molecular collisions with neighboring N2 and O2 molecules.In fact, CO2 will be only slightly cooler or even in the practically same temperature as N2 and O2 depending on how fast heat transfer by molecular collisions as compared with radiative emission.
When CO2 is warmer than its radiative equilibrium temperature, it emits more heat energy than that it absorbs.In other words, CO2 emits not only the heat energy gained from absorption, but also the heat energy gained from N2 and O2 by molecular collisions.A heat transfer route is shown below:
N2 and O2 do not emit heat but pass heat to CO2 by molecular collisions;
CO2 dissipates heat by thermal radiation to space.
With this alternative interpretation, we have a better explanation of the temperature-altitude profile in the thermosphere.A CO2 molecule is heavier than N2 and O2 due to higher molecular weight; so is water vapour but due to aggregation of molecular clusters.Neither water vapour nor CO2reaches the high altitude thermosphere.Even if there are still residual greenhouse gas molecules in the thermosphere, there would be no effective heat transfer by molecular collisions any more because of extremely low air pressure.The temperature in the thermosphere is well above 100°C, increases steadily and exceeds 1000°K with increasing altitude.Without so-called greenhouse gases in the thermosphere, N2 and O2 have no mechanism for heat dissipation.
As long as T > 0, CO2emits, regardless whether its heat is gained by absorption of radiation or molecular collisions.The downward radiation waves emitted by CO2are absorbed by the earth’s ground surface, resulting in a small temperature increase so that the outgoing radiation blocked by CO2goes to space via other wavelength bands.CO2functions as a half-mirror hanging on the sky to the earth’s ground surface.This theme, however, will be addressed in a separate article.
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It is true that absorption of radiation waves leads to an object warmer than otherwise.The key question is: what is the temperature of the GHGs without absorption of radiation waves?It shall not be difficult to find that we have counted, without knowing, twice or more of the warming effect of CO2absorption.This perhaps is the most important underlying conceptual issue that must be resolved to advance climate science.
Earth energy budget is of fundamental importance for understanding what factors determine the climateand climate trends of the earth-atmosphere system.Figure 1 shows a diagram in the IPCC fourth report, AR4, for estimate of the Earth’s annual and global mean energy balance .This diagram is sometimes known as the K & T diagram.
Figure 1Earth energy budget diagram of IPCC report AR4 2007
The diagram basically shows two giant energy trees: one is on the left hand side describing the incoming solar radiation energy and its components; and the other on the right hand side portraying the outgoing radiation and its components with respect to the earth-atmosphere system.In the middle are two lovely bunches of grass showing energy transfers due to evaporation and other thermal events.
This diagram has been widely discussed by, in particular, sceptical scientists.This article, however, will point out that there is a fundamental flaw in the diagram due to misuse of the Stefan-Boltzmann equation.
Examine first how 390 W/m2 surface radiation was obtained.The original paper of the diagram  states “for the outgoing fluxes, the surface infrared radiation of 390 W/m2 corresponds to a blackbody emission at 15°C.”Assuming the surface radiation being p, the statement can be best translated in a mathematical form:
This is to say if ε is not 1, but 0.99, 0.95, 0.9 or 0.8, there will be an error for the surface radiation of 3.9, 19.5, 39 or 78 W/m2 respectively.Obviously, as long as the emissivity of the ground surface of the Earth, ε, is not accurately determined, the figure for the surface radiation is intrinsically uncertain; it is not a minor issue at all.
The error does not stop here.There is a further error that easily escapes from scrutiny, i.e.what temperature T stands for in the Stefan-Boltzmann equation - is 15°C correct?
It is true that the Earth’s mean near-surface air temperature, as measured by global weather stations using thermometers, is around 15°C.As such it is largely of the temperature of nitrogen and oxygen gases that consist of 99% dry air.Thus this temperature is not what one should use in the Stefan-Boltzmann equation.This is because nitrogen and oxygen are non-radiative (literally ε = 0 for transparent and white bodies); 0 multiplying anything leads to 0.
The T represents the temperature of the ground surface of the Earth, which is measured by infrared spectroscopy.Figure 2 shows a ModTran spectrum of the Earth outgoing radiation detected at 20 km high altitude with detector facing down under standard US air conditions.The spectrum was fitted with curves of the Planck’s law. The top envelop 285.04 K (11.89°C) is the temperature of the earth’s ground surface.One obtains:
The above shows what is the proper methodology to apply the Stefan-Boltzmann law to quantify the surface radiation, though how to determine the emissivity and the mean ground surface temperature may need further investigations.
Figure 2 ModTran earth outgoing radiation spectrum for detector facing down at 20 km altitude under standard US air conditions.
A number of values shown in the IPCC AR4 Earth Energy Budget have been revised in a paper published in Bulletin of the American Meteorological Society in 2009 [3, 4].With respect to the 390 W/m2 surface radiation, effect of temperature variation from mean value was taken into consideration.This leads to a replacement of the 390 W/m2 with a new value of 396 W/m2.However, setting ε to 1 remains unchanged.
This revision may be an improvement, but is minor as compared with the uncertainty of emissivity and temperature of the ground surface of the Earth.As this figure is uncertain, many of the other figures on the outgoing radiation tree would be in doubt as well.
In summary, there are two technical errors in calculation of the surface radiation: 1) false assumption that the ground surface of the Earth is a black-body surface; 2) confusion of the near surface air temperature with the ground surface temperature in application of the Stefan-Boltzmann law.As long as ε can not be determined, the outgoing radiation from the ground surface of the Earth is not certain, casting doubts over the value of the diagram. (By J. Cao)
Figure 3Revised earth energy budget diagram 2009
Le Treut, H. et al. 2007: Historical Overview of Climate Change. In: Climate Change 2007: The Physical Science Basis. Contribution of Working Group I to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change [Solomon, S., D. Qin, M. Manning, Z. Chen, M. Marquis, K.B. Averyt, M. Tignor and H.L. Miller (eds.)]. CambridgeUniversity Press, Cambridge, United Kingdom and New York, NY, USA.
J.T. Kiehl and K.E. Trenberth; “Earth’s Annual Global Mean Energy Budget,” Bulletin of the American Meteorological Society, 78(2) pp197-208 (February 1997)
K. E. Trenberth, J. T. Fasullo, and J. Kiehl, 2009: “Earth’s Global Energy Budget,” Bulletin of the American Meteorological Society, pp311-324 (March 2009)
K. E. Trenberth and J. T. Fasullo; “Tracking Earth’s Energy: From El Nin˜o to Global Warming,” Surv. Geophys., DOI 10.1007/s10712-011-9150-2 (October 2011)
Late Prof. Richard Schwartz was an astrophysicist, and had his article entitled “An Astrophysicist Looks at Global Warming” published posthumously by the Geophysical Society of America in GSA Today, 22(1), 44-45 (January 2012).Schwartz demanded in the article “most important, contrarians must show why the scientific basis of greenhouse gas warming is incorrect.”
This question does deserve an answer.Let’s answer the question by addressing what Schwartz considered the scientific basis of greenhouse gas warming was.
First of all, Schwartz employed the planetary mean temperature for the Venus, the Mars and the Earth to interpret the greenhouse gas warming effect; the total atmospheric greenhouse gas warming raised the temperature by 33 °C for the Earth, 6°C for the Mars and 460°C for the Venus.
To explain why this interpretation is incorrect, we need to examine how the 33°C greenhouse effect for the Earth is obtained.33°C = 15°C – (-18°C).The -18°C is obtained by radiative equilibrium between incoming absorbing radiant flux from the Sun and outgoing emitting radiant flux from the Earth:
(1) pr2 (1-a) S0 = 4 pr2 ε σ T4
where, r, is radius of the Earth, and T earth’s mean surface temperature; α is albedo and S0 is the solar constant representing the incoming solar radiation energy per unit area and unit time with its value being around 1368 W/m2; σ is the Stefan-Boltzmann constant equal to 5.670373 x 10-8 (W/m2K4), and ε is the emissivity of the earth surface.
In current climate research, ε is either missing in the equation or is assumed to be unit.Inserting the value of α = 0.3 and ε = 1 into and rearranging Eq. (1) leads to:
(2) T = 254.9 (K) @ 255 (K) @ -18°C
However, by adopting ε = 1, one has assumed that the earth surface is a black-body surface, which of course can not be true.If ε is not 1, but 0.9, 0.8, 0.7 and 0.6, T would be -11.5°C, -3.6°C, 5.5°C or 16.5°C respectively.This -18°C is simply a result of technical error.
On the other hand, the Earth’s mean near-surface air temperature, as measured by global weather stations, is around 15°C (@ 288K).Another widely spread technical error is to use this 15°C to subtract the -18°C.To explain why, it is essential to decode highly symbolised notions “surface” and “surface temperature T” of the Stefan-Boltzmann law to extract true physical meanings for the case of earth-atmosphere system.
If there is no atmosphere, the surface means the land and water ground surface of the Earth, and T represents the mean temperature of the ground surface.If there is atmosphere that are all of nitrogen and oxygen, the surface is still the ground surface, and T still the mean temperature of the ground surface, regardless what the temperature of nitrogen and oxygen may be.This is because nitrogen and oxygen are non-radiative (literally ε = 0 for transparent and white bodies).0 multiplying anything leads to 0.
The real earth-atmosphere system consists of the ground surface, non-radiative gases as well as radiative gases such as water vapour and carbon dioxide.In this case the earth surface is not straightforward any more: over the absorption bands of water vapour and carbon dioxide (e.g. the absorption band 15 μm for CO2), the surface is a layer of atmosphere starting from the top of atmosphere (TOA) with thickness equal to absorption depth, and the temperature is the mean temperature of CO2 in this air layer, Tco2(h).One can similarly find out the surface and surface temperature for any other absorbing bands of radiative gases.For the rest of infrared bands, the surface and surface temperature are the ground surface and its mean temperature, TGSurf.What T stands for in Equation (1) is the mean temperature averaged over all the infrared bands.Figure 1 shows that over the 15 μm infrared band, the earth surface and surface temperature are the top layer of the atmosphere and the temperature of CO2 in this layer respectively.
Figure 1An illustration showing what surface and surface temperature should be over the CO2 absorption band 15 μm.For infrared bands transparent to the atmosphere, the surface and surface temperature are the ground surface and ground surface temperature, TGSurf
The global mean surface temperature 15°C is measured by weather stations using thermometers, and can be denoted Tair(h).Obviously, Tair(h) is the mean temperature of Tn2o2(h), Tco2(h) and temperature of water vapour etc., averaged in terms of heat capacity.As such the 15°C is largely of the temperature of nitrogen and oxygen gases that consist of 99% dry air.Therefore, it is not physically meaningful to subtract -18°C from this 15°C.
In calculation of Mars planetary mean temperature using the Stefan-Boltzmann equation, climate scientists made the same error as they did in calculation of the earth mean temperature, i.e. they falsely assumed the Mars has a black-body surface.If this error is corrected, one obtains the mean surface temperature for Mars -47.13°C, which is in agreement with measurements -47°C.http://www.ucar.edu/learn/1_1_2_1t.htm
When there is a net heating source on a planet, we’ll not be able to calculate its planetary mean temperature any more from radiative equilibrium.For example, the incoming radiation energy for the Sun is almost 0, while its outgoing radiation energy is still εσT4, with ε@ 1 and T = 5778 K, which is a result of its energy generation due to nuclear fusion of hydrogen nuclei into helium.
The Venus has almost same temperature day and night in spite of the fact that a venusian day is as long as 243 earthly days.This is an indication that there are heat generating sources, most likely magma covering generated by volcanoes on the Venus.
Clearly, how to apply and interpret the Stefan-Boltzmann’s law is the problem that has led to a misunderstanding of the greenhouse gas warming effect.
In explaining the molecular mechanism of greenhouse gases warming, Schwartz stated:
“When a greenhouse molecule absorbs an infrared photon, the molecule rotates or vibrates faster and is said to be in an “excited” state.At low gas densities, an excited greenhouse gas molecule will spontaneously (by the rules of quantum mechanics) reradiate an infrared photon, which may escape the atmosphere into space and produce no net warming.
“At the higher densities of Earth’s atmosphere, the excited molecule will bump into (collide with) another molecule (any molecule in the atmosphere). In the collision, the energized greenhouse gas molecule loses its rotational energy, which is transferred to the kinetic energy of the molecule it collides with (this is called collisional de-excitation). The increased kinetic energies of the colliding molecules means that the molecules are moving faster than they were prior to the collision, and the increased velocities of such molecules represents a direct measure of increased atmospheric temperature.
“’Greenhouse gas’ warming occurs because the collisional de-excitation time for greenhouse molecules in Earth’s lower atmosphere is much shorter than the radiation lifetime of excited molecular states.This is the basic science of greenhouse gas warming.”
A paper by Prof. Pierrehumbert (Pierrehumbert, R.T., 2011, Infrared radiation and planetary temperature: Physics Today, v.64, p.33–38.) specifies radiation lifetime ranging from a few milli-seconds to a few tenth of a second, and collisional time 10-7 s, implying that the thermal transfer process between N2O2 and CO2/H2O by molecular collisions is far faster than the heat loss/gain by radiation for CO2/H2O.
Unfortunately, wrong physics has been employed to explain the thermal absorption and in particular emission phenomena of radiative gases.De-excitation occurs only after excitation; however, emission occurs 24/7 to a non-white/transparent object as long as temperature of the object is not 0 K (-273.15°C), regardless whether it absorbs or not.
The kinetics of heat transfer by radiation is determined by the equation εσT4 (or I = aI0) that measures the heat energy gained/lost per unit area and unit time for an object.Assuming the total surface area, S, mass, M, and specific heat capacity, cp, for the object, one can readily convert absorption/emission energy to temperature rise/drop for the object.The emission rate will be 1012 times faster at 1000 K than that at 1 K.On the other hand, the collision time of 10-7 s, which is related to the mean free path of air molecules, does not mean that air homogenizes its temperature in 10-7 s time scale.In fact air is a thermal insulator because thermal transfer for air by molecular collision is slow.Industrial examples of taking advantage of air’s thermal insulation property include: double glass windows for trains and hollow synthetic fibres etc.
Hopefully, this article is comprehensive enough and does answer late Schwartz’s question.