This is a simple explanation why an essential claim of greenhouse gas theory or the catastrophic man-made global warming hypothesis is very wrong. That theory believes that back-radiation from the atmosphere to the surface is about the same as the total solar radiation incident upon the top of the Earth's atmosphere. For a much longer exposition of many of the problems with this theory you might examine my earlier and much-updated post
This post is comparatively very short and will only discuss a simple explanation of the cornerstone argument of huge back-radiation to the surface. Indeed, I will show why that back-radiation is virtually zero. Because it is nearly zero, the catastrophic man-made global warming hypothesis is shown to be nonsense. There is no excuse for governments acting to force individuals and companies to reduce their carbon dioxide emissions. Doing so is very expensive, causes energy reliability problems, and puts people out of work. Causing these miseries in the name of a failed hypothesis is a very serious injustice to mankind.
Let us examine a current NASA depiction of the huge role the catastrophic AGW hypothesis gives to back-radiation:
Now notice that the emitted surface radiation is 117% of the top of the atmosphere average radiation from the sun and back-radiation is 100% of the top of the atmosphere incident solar radiation. Notice also that only 48% of the total solar radiation was ever absorbed by the Earth's surface. Of the 48% of the total solar incident energy absorbed by the surface, 25% is lost by evaporation of water and 5% by convection loss, leaving only 18% to be emitted by radiation. Now the proponents of the large greenhouse warming effect claim that the 18% is just a net difference between a larger surface radiation emission rate and a large back-radiation rate. In this case the difference of 117% - 100% = 17%, which is about 18%. We are going to evaluate this claim.
Let us examine a current NASA depiction of the huge role the catastrophic AGW hypothesis gives to back-radiation:
Now notice that the emitted surface radiation is 117% of the top of the atmosphere average radiation from the sun and back-radiation is 100% of the top of the atmosphere incident solar radiation. Notice also that only 48% of the total solar radiation was ever absorbed by the Earth's surface. Of the 48% of the total solar incident energy absorbed by the surface, 25% is lost by evaporation of water and 5% by convection loss, leaving only 18% to be emitted by radiation. Now the proponents of the large greenhouse warming effect claim that the 18% is just a net difference between a larger surface radiation emission rate and a large back-radiation rate. In this case the difference of 117% - 100% = 17%, which is about 18%. We are going to evaluate this claim.
Consider the surface of the Earth for a moment. The density of atoms per cubic meter in the surface is about 1 gram per cubic centimeter for the water that covers about 71% of the Earth's surface and even greater for land materials or if considering the salts in the oceans. Expressed as a density per cubic meter of water, this is 1000 kg per cubic meter. A cubic meter of water has 3.34 x 1028 water molecules. Infra-red emission and absorption in a surface occurs in the outer 2 micrometers of the surface. Consequently, there are about 6.7 x 1022 water molecules emitting infra-red at most from the surface. When infra-red radiation is incident upon the surface, these are the same water molecules that would be able to absorb that radiation.
The density of the atmosphere near the surface at sea level is 1.225 kg per cubic meter in the U.S. Standard Atmosphere at a temperature of 288.15K. Adding water vapor very slightly decreases that density. The number of molecules per cubic meter at sea level in the U.S. Standard Atmosphere is 2.55 x 1025/m3. The important infra-red active gas near the surface is water vapor and its density per cubic meter is commonly between 10 g/kg to 14 g/kg of air as shown in Fig. 1. below. At a specific humidity of 12 g/kg, the number of water molecules/m3 of air is about 4.9 x 1023. Therefore, there are more water molecules in the first cubic meter of air above 1 m2 of water surface emitters by a factor of 7.3. This should mean that radiation that can be absorbed by water vapor will be absorbed in the first meter of air above the surface at a humidity near the Earth average humidity. There is a report that the average mean free length for water vapor absorption in the atmosphere is as long as 8 m. If it actually is that long, then the result I get below on the amount of back-radiation can be multiplied by a factor of 8. That number will still be most unimpressive.
Fig. 1. The average specific humidity of air is shown as a function of latitude. The specific humidity is the weight of water in grams in one kilogram of air.
In comparison, at 400 ppm of carbon
dioxide, there are 1.0 x 1022 molecules of CO2 /m3. There are 49 water vapor molecules for every
carbon dioxide molecule at a specific humidity of 12 g/kg. In addition, the emissivity of CO2
molecules is less than 40% that of water vapor molecules. Since many of the carbon dioxide
molecule absorption peaks are largely or mostly already absorbed by
overlapping water vapor absorption lines, the effect of CO2 near the surface is very minimal compared to
that of water vapor. This is less true
at altitudes above 4 km where the concentration of water vapor drops greatly and the carbon dioxide remains in proportion to nitrogen and oxygen, but such altitudes have nothing to do with the
back-radiation to the surface issue.
Returning to the back-radiation caused by water vapor, we find that the temperature difference over the 1 meter absorption range according to the U.S. Standard Atmosphere is only about 0.0065 K/m. For humid air, the temperature gradient is even less. So if the water surface and the water vapor in the first meter of air above the surface are treated as gray bodies, we have a power transfer from the surface to the water vapor in the atmosphere of PW
PW = σ (εs Ts4 – εa Ta4) ,
for the surface s and the atmosphere water vapor a and if we take the emissivities to be a high value of 0.95 often used by the catastrophic man-made global warming promoters, this is equal to 0.033 W/m2.
This is actually an over-estimate because we have ignored the fact that some of the heat energy in the surface is used to evaporate water and some is lost to the air by means of air molecule collisions with the surface. Much of the solar radiation absorbed by the surface is actually transferred out of the surface by evaporation and by conduction. Of the 48% of solar radiation absorbed by the surface according to the above NASA schematic, a total of 30% is removed from the Earth's surface by evaporation and convection/conduction. This leaves only about 18% to be radiated away from the surface. So if the Earth's surface was an interface directly with vacuum, the 48% of solar radiation absorbed would be radiated in its entirety. But other processes, evaporation and conduction, remove most of this energy, leaving a fraction of only 18%/48% = 0.375 of the total energy flux to be removed by radiation.
But water vapor does not absorb infra-red across the entire surface emission spectrum. It actually absorbs only about 65% of it. At least 28% of the surface emission is into the atmospheric window where no atmospheric gases absorb the infra-red radiation. See the transmission spectrum of the Earth's infra-red surface emissions taken by Nimbus IV in 1970 near Guam below:
This points out one of many flaws in the NASA depiction of the Earth Energy Budget above. Taking the two atmospheric windows at higher wave numbers than the peak of the Earth emission spectrum, the infra-red radiated into them is about 28%. 28% of the 117% surface infra-red emission claimed by NASA is 33%, not the 12% quantity their schematic assigns to the atmospheric window radiation lost directly into space from the surface. The direct loss to space of surface emission with the NASA depicted 117% surface emission should be 2.75 times what NASA puts in its Earth Energy Budget schematic diagram at the start of this article. This inconsistency suggests that the surface emission radiation is actually much less than the 117% claimed. If that emission is much less, then the back-radiation would also have to be much less.
So the actual water vapor absorbed infra-red radiation is only about
PW = (0.375) (0.65) (0.033 W/m2) = 0.0080 W/m2.
But because of the 6.9 x 109 collisions/s in air near sea level, most of this energy is transferred to non-radiating nitrogen, oxygen, and argon. Only about 20% is re-radiated and half of that is radiated toward space. Consequently, the total back-radiation which can be absorbed by the surface, PB, is about
PB = (0.2) (.5) (0.0080 W/m2) = 0.00080 W/m2
Thus, the absorbed back-radiation has an upper limit of about 0.00023% of the average solar insolation at the top of the atmosphere (342 W/m2)! For all intents and purposes, the absorbed back-radiation is zero. This value should be multiplied by the mean free path for absorption by water vapor, which may be as large as 8 m on average.
I just proceeded down the true path of analysis dictated by the use of Occam's Razor. Let us assume that NASA is right that 117% of the solar incident radiation at the top of the atmosphere actually is emitted from the surface due to some unknown extra input energy that lifts the 48% of the solar absorbed energy flux at the surface by 117% + 25% + 5% - 48% = 99% so that one can have 117% radiation, 25% evaporation, and 5% conduction cooling mechanisms at work. Is it possible for that 99% warming source to be back-radiation?
The 117% radiation is greater than the 18% upon which I performed the Occam's Razor version of the calculation. In fact, the back-radiation per meter of mean free path is then 117% / 18% = 6.5 times greater than the value I calculated. So, if we assume that the average water vapor mean free path is 8m, we get 8 (6.5) (0.00023%) = 0.012%, which does not look at all like the 100% claimed by NASA.
Because back-radiation to the surface is insignificant compared to the claims made by the proponents of catastrophic man-made global warming, the mechanism upon which that theory stands is wrong. Indeed, adding carbon dioxide to the atmosphere will actually cause more incoming solar radiation to be absorbed by the atmosphere before it reaches the surface. This results in a cooling of the surface. In addition, more CO2 in the atmosphere near the surface will also cause the temperature gradient in the atmosphere to become slightly smaller, just as infra-red absorbing water vapor makes it smaller. This is because radiation transport effects operate at the speed of light, which is faster than evaporation/condensation transport or conduction/convection transport of energy. Since all remove energy from the surface, they are all cooling effects. As a result, adding CO2 to the atmosphere actually causes a very slight cooling of the surface, contrary to the claims of a substantial warming effect.
The primary sources of CO2 in the atmosphere are warming oceans, decaying plant life, and heat vents and volcanic emissions. Since CO2 in the atmosphere creates a slight cooling of the surface, it acts as a negative feedback to the warming oceans that cause it to increase and it slightly cools the decaying plants to slow down the further generation of CO2 from that source. Of course volcanic and heat vent sources of CO2 are also providing heating, so carbon dioxide as an surface coolant acts to stabilize the Earth's temperature much as water vapor does. It has negative feedback effects. It is no more subject to the sort of tipping point catastrophes that global warming alarmists put out than is water vapor, though its effects are much, much weaker.
This article was last updated on 29 July 2014.
[The fact that back-radiation itself is insignificant in explaining the 33C warming of the surface claimed by the proponents of catastrophic AGW is not at all to say that the effects of water vapor on the surface temperature are without significance. During the day the evaporation of water is an important surface cooling effect and during the night its condensation can be an important warming effect. During the day, the infra-red active gases absorb a portion of the incoming solar radiation and prevent it from being absorbed by the surface. At night, fog and clouds scatter, reflect, and absorb the infra-red radiation emitted from the surface. Also at night, the small true back-radiation effect is not off-set by the absorption of incoming solar radiation by the same infra-red active molecules in the atmosphere. There is a very important moderation of the surface temperature between day and night due to these effects. This note was added on 13 November 2014 to remind people of the context in which my criticism of the usual greenhouse gas theory is framed.]
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4 comments:
Hi Charles. I believe you are entirely correct in stating that the radiation from the surface is less than the 117% shown, as this figure is not in the atmosphere's energy budget. I have a different way to describe why. Tell me what you think.
For me the problem arises from the notion of photon fluxes which leads to an inevitable mis-use of the SB equation.
If we consider the SB thought experiment of two, infinite, high emissivity parallel plates in a vacuum at equilibrium, and view them with photon fluxes we find the following;
"As photons carry energy away from each plate we find that at any point between the plates we have twice the emission of each plate. So if both emit a photon flux of 'x' then at all points 2x can be found as energy cannot cancel energy. You can't have negative energy. This leads to the notion that energy IS available at all points between two surfaces at equilibrium with no thermal gradient (!!!!)"
However, electromagnetic radiation requires a wave description to depict interference, diffraction and dispersion. ie in order to describe the superposition of radiation from different sources and the optical properties we observe in reality we are required to view radiation as electromagnetic waves.
As waves, electromagnetic radiation is thus rendered a vector quantity.
Returning to our infinite parallel plates we now find;
"At all points between the plates, the monochromatic addition of the opposing radiation vectors across the entire emission spectrum sum to zero!
By respecting the very nature of 'light' to our best ability we can then deduce the following;
At equilibrium between two equal radiators there is no spontaneous exchange of energy. There is NO energy available at points in between due to this 'lack' of exchange.
This fits in with thermodynamic arguments about heat transfer, the Carnot cycle, heat engines and energy available for work and power from thermal gradients"
If we apply this to the surface radiation and line by line subtract the calculated monochromatic 'sky emissions' from the calculated monochromatic 'surface emissions' the difference is the net flux in Watts, which is the only energy that leaves the surface. Every available emission from the sky downwards is cancelled completely by the larger upward flux.
This means 17% upwelling is the only energy that leaves the surface. Around 30% of this is absorbed by the atmosphere as part of its energy budget. The remaining 70% leaves through the atmospheric window.
The massive photon fluxes shown in Trenburth's energy budget are unrealisable. They are 'effective radiation potential' referenced to zero Kelvin, something that doesn't exist on Earth!
Regards.
Hello Dr Anderson and thank you for this web page to help laymen. Will need to read your essays many more times yet. I followed the link to this site from GREENIE WATCH. As a layman I have difficulty understanding many of the finer points. I have a problem understanding 'back-radiation' from the atmosphere warming the surface. If I explain my level of understanding you might point me in the correct direction. I will attempt to explain my understanding of the radiation fingerprint of CO2 and how it relates to the basic rules of radiative heat transfere.
In your essay you state...
""But because of the 6.9 x 10^9 collisions/s in air near sea level, most of this energy is transferred to non-radiating nitrogen, oxygen, and argon. Only about 20% is re-radiated and half of that is radiated toward space.""
My reading leads me to believe that any small volume of air will be in thermal equilibrium so the CO2 molecules will have the same average kinetic energy as all the other atmospheric molecules in our small volume. I also read this kinetic battering effects the VIBRATIONAL level of the CO2 molecules. There is a term 'translational energy', which seems to imply that the vibrational level of the individual CO2 molecules will be raised above the absorption level of the 15 micron band which has a peak temperature of 193.2K. In other words I believe the CO2 molecular vibrational level will be equivalent to the local air temperature, thermal equilibrium, assuming local air temperature is 15C or 288.15K.
I also read that CO2 can emitt/absorb over some 3,800 lines of IR from 13 microns down to 17 microns. This would constitute some 18% of black body radiation, at a temperature of 288K, but even the most energetic photons in the 13 micron band have a peak temperature of 223K (-50C). Thus the CO2 molecule would need to be at the tropopause to be this cold and 80% of the atmosphere is below the tropopause and much warmer.
My understanding of thermal energy transfer by IR leads me to believe that the source radiation must emitt photons of higher energy level, shorter wavelength, than the peak radiation temperature of the target. In this case the surface IR at 15C, assuming black body level, meets that requirement BUT, (a very big BUT!) CO2 cannot absorb radiation from just below 4.3 microns down to 13 microns and CO2 will be TOO warm to absorb any radiation from 13 to 17 microns.
Thus my layman belief is that CO2 CANNOT absorb radiation from the surface but that CO2 is very actively radiating, half up half down, over the whole 13 to 17 micron band, (cooling the atmosphere) and that radiation does not warm the surface because the surface is already radiating over that range. My very basic calculations make it look like CO2 CAN absorb some 4.3 micron radiation from the surface, especially the sea surface, but this constitutes less than 1% of the energy leaving the surface. And IF the CO2 molecule does absorb a 4.3 micron photon only a fraction of that energy will pass to the N2, O2 and Ar molecules and the remainder will leave via 13 to 17 micron emissions.
When the sun is shining then CO2 will be absorbing energy over the 2.7 micron band (1,073.33K peak temperature) and the 4.3 micron band (674K peak temperature). Since that energy is not reaching the surface this is again a cooling effect for the surface even as the air above is being warmed. CO2 is unlikely to re-radiate over those two bands though the intensity of 13 to 17 micron radiation will increase to no effect as mentioned above. (I hope this all makes sense.)
I have to ask, just how is CO2 in the atmosphere responsible for any 'back-radiation'?
Richard in Wales
UK
Hi Geoff and thanks for your comment.
Two infinite plates at the same temperature would actually be effectively a single black body cavity, at least with a surface emissivity of 1. The error the AGW people most commonly make is to treat two such surfaces as black body emitters and assume that the energy density in the space between the two plates is the sum of that provided by either plate. They thus double the energy density found inside a black body cavity.
The fundamental characteristic of the black body cavity is that the energy density is proportional to T to the fourth power. This two infinite plate case cannot double that energy density because each plate is really only the opposite wall of a single black body cavity and the radiation is fixed by the fixed energy density.
To be sure, applying this case to calculating the surface emission and back-radiation makes one plate the Earth's surface and the other either cold space or a nearby IR-active molecule for those frequencies such molecules can absorb and emit. Then the key thing is that the mean free path for such absorbable IR is between 1 and 50 meters when considering water vapor and carbon dioxide. Thus, the two opposing plates have slightly different temperatures. This requires the energy density to be intermediate between those two plates for the relevant frequency range. The AGW people claim it is effectively the sum of the energy density of the black body cavity of the Earth's surface and of the black body cavity of the emitting molecule. Since the temperature difference is very small, this very nearly doubles the energy density between them, contrary to the characteristics of a black body cavity.
Modeling energy flows using their model works as a calculational approach in some circumstances, which has given them great confidence in this technique. The problem is that it actually violates energy conservation and for many types of problems leads to incorrect answers.
Richard in Wales,
"My reading leads me to believe that any small volume of air will be in thermal equilibrium so the CO2 molecules will have the same average kinetic energy as all the other atmospheric molecules in our small volume."
It is this statement that leads to our differences. If an IR-active molecule has absorbed a photon of IR emitted by the surface, about 4 out of 5 times it will have a collision with other gas molecules before it can re-emit that energy as IR radiation. When it has such a collision, you and I are in essential agreement. The energy absorbed from the surface cannot be returned to the surface when the absorbing molecule has come into equilibrium with the surrounding gas molecules at a lower temperature than the surface.
But, in about one out of five cases in which an IR-active molecule absorbs a photon emitted from the surface, that molecule may have been excited to a temperature as high as the surface temperature. Because the mean free path length for absorption by water vapor or by CO2 is in the range of 1 to 50 meters, the absorbing molecule when it was in equilibrium with the surrounding air was generally only very slightly cooler than the surface before the absorption of the surface IR emission occurred. This is why there can be some very limited back-radiation.
Of these one of five cases of re-emission before any collision, half the emitted energy is directed toward the surface, but some of that will be intercepted by other molecules of the same type between our emitter molecule and the surface. Some of the photons incident upon the surface are also going to be reflected. But, some fraction may also be absorbed.
Because the energy radiated and absorbed by these IR-active molecules is that due to a small temperature difference, the energy involved is small to begin with. It is much smaller than in the catastrophic AGW model. That small number is knocked down by a factor of five, by a further factor of two, then by intercepting IR-active molecules, and further by any surface reflection. The result is that true back-radiation is a very small fraction of the surface-emitted IR radiation and the fraction of surface-emitted radiation that goes into the atmospheric window is much higher than indicated in a Kiehl-Trenberth Earth Energy Budget.
True back-radiation exists, but it is very small compared to the claims of the Kiehl-Trenberth Energy Budget. Even with those hugely exaggerated claims, the UN IPCC has great difficulty creating anything approaching a catastrophe.
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