According to this federal government story, infrared radiation emitted by our atmosphere provides a surface warming power flux of 340.3 W/m2 compared to 163.3 W/m2 of radiation (UV, visible light, and infrared) directly absorbed from the sun. In the official pronouncements of the United States of America federal government, so-called back radiation from our cooler atmosphere provides 2.084 times as much heat to the warmer surface as does the sun directly.
In this viewpoint, commonly claimed to be the consensus viewpoint of 97% of all scientists, the Earth does not have a gravitational field which acts upon air to to provide a temperature gradient in accordance with simple physics which has been well-known for a very long time. This article will explain this very simple, well-known, yet now completely ignored physics in relatively simple terms. If one is to understand the equilibrium climate of the Earth's surface and its lower atmosphere, the troposphere, that understanding requires that we understand how gravity acts upon our atmosphere.
Yet we cannot ignore the role of radiation in developing this understanding either. Let me reprise some considerations from my earlier article The Simple Physics Explaining the Earth's Surface Temperature by way of introduction.
NASA says that the Earth emits 239.9 W/m2 of longwave infrared radiation into space. This implies an effective Earth system radiative temperature T:
P = 239.9 W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) T4,
Now we can go a step further, since NASA says that 40.1 W/m2 of longwave infra-red radiation emitted from the surface passes through the atmospheric window without absorption by the atmosphere directly into space. This allows us to calculate the effective radiative temperature of the atmosphere alone. Consequently, the effective radiative temperature of the atmosphere as a black body is found to be:
This effective atmospheric temperature is important because it is at an altitude at which the atmosphere is in equilibrium with radiation into it and out of it with respect to space. This is an effectively pinned temperature in our atmosphere from the standpoint of radiation. Changes in the water vapor concentration and those of other infrared-active gases may move this point somewhat, but there is such a point given any such changes in infrared-active molecule concentrations which serves as a reference point for an equilibrium temperature of the atmosphere as seen from space. Where this altitude is is mostly determined by water vapor in the atmosphere.
According to the U.S. Standard Atmosphere Table of 1976, the average air molecule has a mass of 0.028964 kg/mol. So the kinetic energy difference of one mole of air molecules at the surface and at an altitude of 6846 m is
KE (h=0) - KE (h = 6846m) = (0.028964 kg/mol) (66992.2 m2/s2 ) = 1940.36 J/mol
The heat capacity at constant pressure for a mole of the average air molecule is 29.07 J/K mol. So we have
T(h=0) - T(h=6846 m) = (1940.36 J/mol) / (29.07 J/K mol) = 66.75K
This is a temperature gradient per km of altitude of 9.75 K/km, which is a bit less than the more frequently given 9.8 K/km because we took into consideration the fact that the gravitational constant is not really quite constant and decreases slightly with altitude.
Now comes the important lesson of this exercise. We can now calculate the equilibrium surface temperature of the Earth due to the temperature gradient in the troposphere resulting from gravity acting on the atmosphere. Recall that at 6846 m altitude, the temperature interpolated from the US Standard Atmosphere Table of 1976 is 243.7 K. With a gravitational temperature gradient of 9.75 K/km, the surface temperature is
T (h=0) = 243.7K + (6.846 km) ( 9.75 K/km) = 310.4 K
The air temperature at the bottom of the atmosphere, where the molecules of the atmosphere are bombarding the surface, should be about 310 K were it not for the existence of cooling effects. This is much higher than the 288 K average temperature of the Earth's surface.
In fact, given that the surface is said in the Earth Energy Budget above to be absorbing 163.3 W/m2 of direct radiation from the sun, the surface should be much hotter than 310 K. The problem is still greater if the surface is also warmed by back radiation from the atmosphere of 340.3 W/m2 , despite the exaggerated surface infrared emissions of 398.2 W/m2 of that energy budget. I discussed why the surface infrared emissions are greatly exaggerated recently in Infrared Radiation from the Earth's Surface and the So-Called Scientific Consensus. The claim of a large back radiation from the atmosphere was needed because even without an exaggerated cooling by radiation of the surface, there was no way to achieve an average temperature of 288 K at the surface without the gravitational effect being taken into account.
It is important to understand that under equilibrium conditions, the gravitational field is not transferring heat through the atmosphere or to the surface. The gravitational field effect on the temperature of the atmosphere is trying to stabilize the temperature gradient in the atmosphere and at the surface and heat transfer only occurs when heat transferring effects such as radiation, water evaporation and condensation, and thermals upset the balance. In this sense, the gravitational field effect does not belong in the Earth Energy Budget. But, the fact that it does not belong there points out that the Earth Energy Budget is inadequate to understand equilibrium temperatures and is actually misleading.
While it is wrong to think of the gravitational effect as a power input in the system, it is still instructive for the sake of comparison to those factors that do put power into the Earth system to calculate its effective power input. Using the Stefan-Boltzmann equation this would be given as
P = σ T4 = (5.6697 x 10-8 W/m2K4) (310.4 K)4 = 526.3 W/m2
If one adds to this the 163.3 W/m2 of direct solar radiation and subtracts the 18.4 W/m2 of cooling thermals and the 86.4 W/m2 of cooling water evaporation power, the Earth's surface temperature would be given by
The net radiation from the surface is another cooling mechanism, which according to NASA is (398.2 - 340.3)W/m2 = 57.9 W/m2 . If we subtract this power from the 584.8 W/m2 , we get 526.9 W/m2 for the effective surface power emission given these inputs and outputs. So,
P = 526.9 W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) T4,
and T = 310.5 K.
This temperature is almost exactly the same as the temperature calculated from the effect of gravity alone, which is to be expected because the heat flows into and out of the surface in the NASA scheme were equal.
I do not trust the NASA value for the net cooling by radiation of the surface, which is given as 57.9 W/m2 in their scheme at all. I do believe that they can measure the 40.1 W/m2 of surface radiation through the atmospheric window into space. I also believe that this implies an actual surface emission about three times that amount. Furthermore, there is no back radiation in the equilibrium condition, though given that the atmosphere is often not in equilibrium, there are times when the air is warmer than the surface, so there is an average low value of back radiation. I do not have a decent number for the back radiation, but it is much smaller than the heat loss by thermals, given by NASA as 18.4 W/m2 . Therefore, I will ignore back radiation as being in the noise. Taking net surface radiation as about 120 W/m2 then instead of 57.9W/m2 , we get a surface temperature of
P = 464.8 W/m2 = σ T4 = (5.6697 x 10-8 W/m2K4) T4,
and T = 300.9 K.
Consequently, with a more realistic surface radiation cooling, the surface temperature is still about 12.9 K too warm. Another 74.7 W/m2 of cooling is still missing. Most of this is certainly due to underestimating the cooling effects of thermals and water evaporation.
Already, one should be able to see that the only way to support the idea that infrared-active gases are alone warming the surface is to believe in the reality of the 340.3 W/m2 of back radiation shown in the diagram. This back radiation requires that a black body radiator be at a temperature of 278.34 K, which in the U.S. Standard Atmosphere Table of 1976 occurs at an altitude of 1510 m. The infrared radiation from such a black body would have to travel 1510 m to reach the surface without any absorption by the infrared-active molecules in that path. In reality, because water vapor and carbon dioxide only emit a portion of the spectrum of a black body radiator, to emit so much radiation they would have to be at a much cooler temperature than 278.34 K, which means that the radiation they emitted exclusively at the wavelengths that they also want to absorb would be emitted at a very much higher altitude than 1510 m. This would require a much longer mean free path for their emitted radiation than 1510 m. In reality, the mean free path is very much shorter than 1510 m. This makes this large back radiation from the atmosphere a fiction.
Note that adding still more CO2 to the atmosphere will actually further reduce the mean free path for infrared radiation and reduce the temperature differentials that are required to transport much heat between areas at different temperatures by radiation. Because of the T to the fourth power dependence of radiative emissions, smaller temperature differentials yield rapidly decreasing radiative heat transport back from the atmosphere to the surface. Most of the time, when an infrared-active molecule absorbs a photon, it transfers almost all of that energy to other infrared-inactive molecules such as nitrogen, oxygen, or argon during many collisions with them.
The large back radiation of the NASA Earth Energy Budget is a fiction inserted because they have failed to acknowledge the critical role of gravity in providing us with a warm surface.
There is much more that is very odd about the NASA notion of back radiation here. Let us do a little accounting. Note that the atmosphere absorbs energy at the following rates:
77.1 W/m2 from incoming solar radiation
358.2 W/m2 of radiation from the surface
18.1 W/m2 from thermals rising from the surface
86.4 W/m2 from the condensation of water evaporated at the surface
Total atmospheric energy flux input is 540.1 W/m2 .
Note that the atmosphere radiates the following energy fluxes:
(239.9 - 40.1) W/m2 = 199.8 W/m2 into space
340.3 W/m2 to the surface
Total atmospheric energy flux output is 540.1 W/m2 .
These sums are in balance, as one expects in a system without gravity. However, one has to note that of the total atmospheric infrared radiation, only 37% is radiated to space, while 63% is somehow radiated back to the surface. As each infra-red active molecule absorbs an infrared photon, it is equally likely to subsequently emit an infrared photon in any direction according to the ideas of photon emission to which the scientists who created this vision of the Earth's energy budget ascribe. If so, because the atmosphere becomes less dense with altitude and because beyond the altitude at which water vapor condenses there is much less absorbing water vapor, more infrared radiation should escape the atmosphere to space than should be returned to the surface. The reality is that in the lower troposphere, the transport of energy or heat as radiation is small compared to the transport by convection. The transport by convection is much greater because the rate of molecular collisions is much greater than is the rate of photon emission by the infrared-active molecules in the lower, very dense atmosphere. But even if radiative transport of energy were the dominant means of energy transport in the lower troposphere, one could not have more than half of the energy transported downward.
Here is still another unphysical oddity: According to NASA and the many similar energy diagrams adopted by many other governmental organizations such as the UN and the European Union, all but about 10% of the infrared radiation of the surface is absorbed by the atmosphere. Here, the surface emits 398.2 W/m2 and the atmosphere absorbs 358.2 W/m2 of that. If the surface temperature is Ts and the effective atmospheric temperature is Ta, then one has
σ (Ts)4 - σ (Ta)4 = 358.2 W/m2
and we know that for the NASA Earth energy budget the first term on the left is 398.2 W/m2 , so we have
Ta = 163.0 K
Now this is a very interestingly low temperature. There is no temperature this low in the U.S. Standard Atmosphere Table of 1976. The temperature with altitude drops in the troposphere, stabilizes at 216.65 K in the tropopause and then increases with altitude until there is essentially no atmosphere left to do any absorbing of infrared photons.
In order to get enough energy flux to return a large back radiation to the surface to replace the role of gravity, NASA has to find a way to absorb almost all of the infrared radiation emitted from the surface in the atmosphere and to do that they have to make the atmosphere very cold. They have to make it colder than it is. Even if it were cold enough in its upper reaches and dense enough to absorb the radiation, how then could it actually return that radiation to the surface and also appear to be a black body radiator with a temperature of 378.3 K, which we calculated above? And note that this problem only becomes worse when we recognize that none of the infrared-active gases are black body absorbers or emitters. They are much less efficient both as absorbers and emitters of radiation.
The infrared active molecules speed the upward transport of solar insolation which has warmed the surface by short hop radiation from a warmer layer of air to a cooler layer just above it. The mean free path for the reabsorption of emitted infrared is short, so the amount of energy transported this way is small. But, this effect is a cooling effect in that the energy transport at the speed of light is much faster than the velocity of air molecules traveling upward in convection currents.
The primary infrared active molecule, water vapor, has a mass to heat capacity ratio less than that of the average air molecule, so it reduces the temperature gradient created by gravity in accordance with the calculation performed above. Because the water molecule is lighter than the average air molecule, it also has a tendency to increase updrafts of air at higher humidity and thus speed up heat dissipation from the surface by carrying heat to the altitude at which water condenses. The Earth is also cooled by plant growth and the absorption of water and carbon dioxide by minerals, which increases with more carbon dioxide and water vapor in the atmosphere.
All of these factors are candidates for being underestimated by the NASA Earth Energy Budget and provide mechanisms in which infrared-active molecules actually cool the surface of the Earth and, ultimately, the atmosphere.
We are very fortunate to have a deep atmosphere in a substantial gravitational field composed mostly of infrared inactive molecules with enough water to dissipate heat adequately by evaporation at the surface and enough infrared active molecules to dissipate heat from the upper troposphere. It is also critically important that the mean free path length for the absorption of infrared radiation emitted by our infrared-active molecules is short so that heat near the surface is not immediately and directly dissipated to space, but must instead rise slowly through the troposphere to the top portion of the troposphere before it is lost to space. The infrared-active gases play a critical role, but they do so in conjunction with the powerful effect of our Earth's gravitational field acting on our air molecules.
Additions made on 21 October 2017 are in green.