Earth’s Atmospheric Window and Surface Temperature

 A E Banner Dec 2019

The Atmospheric Window has a vital role in governing the temperature of the Earth’s surface.  Without it, the temperature would be too great for life.  Radiative energy emitted from the surface can pass through the window to space with only moderate absorption because the are few greenhouse gas wavelengths within the wavelength range of the window. 
This range is generally taken to be from 8 to 14 microns.

The following hypothesis gives the required values for the temperature of the Earth’s surface both for an atmosphere with, and also without, greenhouse gases and water vapour, and hence quantitatively explains the “33 deg C rise”.  It also shows that with the current value of window power, increased greenhouse gas concentrations have only a limited effect on surface temperature, but any significant reduction in the window power can have serious results.

The Atmospheric Window
Reference (1) shows the transmission of radiative energy emitted from the surface of the Earth through the window.  The transmittance is between 80% and 70%, but reducing to zero through the 13 and 14 micron sections.  A computer program for the Planck distribution at 288 K has enabled the transmitted power to be calculated, and allowance has been made for the amount absorbed.  It provides a value of 90.2 Wm^-2 for the range of microns. 

The Planck curve for a surface temperature of 288K, and this window power of 90.2 Wm^-2 is given in Fig 1.  It must be pointed out that this data is the total energy in Joules per square metre per second, radiated across all wavelengths.  This is in line with the Stefan-Boltzmann equation. 

The Planck figures, however, apply in terms of the steradian, and must be multipied by ‘pi’ to achieve agreement with the S-B figures.

A new approach
In the following treatment, the term “Greenhouse Gases” includes water vapour and clouds, in addition to carbon dioxide, methane and all radiative energy absorbing molecules.

The greenhouse gases have very little effect within the window and so photons with wavelengths within the window pass through to space with only a little absorption.  Radiation from the Earth’s surface is absorbed and re-emitted by the greenhouse gases with wavelengths outside the window.

Fig 2 shows schematically the basis of this new approach.

The hypothesis starts with the emission of radiant energy from the Earth’s surface in line with the equation of Stefan-Boltzmann for a “black body”.  This is acceptable for Earth with an emissivity taken to be 0.98

Let   P = output power from the surface in Wm^-2
e = emissivity of the surface
s = Stefan-Boltzmann constant, 5.67*10^-8 Wm^-2K^-4
T = surface temperature in K
w = power emitted through the window to space, Wm^–2
Take T = 288 K and e = 0.98
P = e.s.(T^4)  …………………..(1)
= 0.98*5.67*(10^–8)*(T^4)
= 382.28 Wm^–2
This is the power emitted as photons from the surface of the Earth into the atmosphere.
Some of this power, w, escapes directly into space through the window.  ( little greenhouse absorption in the window)
Let   P1 be the power remaining in the atmosphere.
So    P1 = P – w
But there are GHGs effective in the wavelengths outside the window, and so absorption and emission occurs here.
Now, it may be that not all of the energy (P – w) is absorbed/emitted.  This might be due to insufficient greenhouse gases in the atmosphere, or too small a molecular cross section for absorption.
So let  f  be the energy absorption factor combining these effects.
If all the radiative power is being absorbed, then  f = 1.0
If none of the radiative power is being absorbed, then  f = 0.0
Therefore, the power absorbed and then re-emitted is (P – w)f.
Since greenhouse gas molecules emit photons equally in all directions, the power radiated upwards is 0.5(P – w)f , and this is equal to the power radiated downwards, 0.5(P – w)f.
However, if the energy absorption factor f is less than 1.0, there is energy still left unaccounted for in the atmosphere.  Let this remainder be R.
Therefore, it follows that  R = (P – w)(1.0 – f)
So the total power into space  = w + R + 0.5(P – w)f
And for Earth’s energy balance this must equal 239 Wm^-2.
So           w + R + 0.5(P – w)f  =  239
Hence,   P(1.0 – 0.5f) + 0.5wf  =  239
So           P = (239 0.5wf ) / (1.0 – 0.5f )
Substituting for P from eqn (1), this gives
T^4 = (239.0 –­ 0.5wf ) / (e.s.(1.0 – 0.5f ))
T^4 = 0.179966*(239 – 0.5wf )*10^8 / (1.0 – 0.5f )

The value of  T  has been determined for a range of energy absorption factors f,  and for specified values of window power w; the results are given in Fig 3.

For the current window power  w = 90.2 Wm^-2 , it shows that the surface temperature of 288K is obtained with an energy absorption factor  f = 0.981

If there were no greenhouse gases in the atmosphere, or indeed, no atmosphere, this would be equivalent to zero energy absorption factor, f = 0.0.  This gives a temperature of 256 K, which is correct for an emissivity of 0.98
This provides the temperature rise of 32 deg C.

The energy flux returning to the surface from the atmosphere is 0.5(P ‒ w)f.
For T = 288 K, P = 382.28 Wm^-2, and the value of  w = 90.2 Wm^-2, the downward power to the surface is 143.27 Wm^-2.  In addition, there is 239 from the Sun, making a total of 382.27 Wm^-2, as required for energy balance.

Fig 4 shows the critical role of the window.

For any value of w, the temperature cannot exceed that given by the curve for f = 1.0, because with f  = 1.0 all of the radiant energy in the atmosphere is already being absorbed and emitted by the greenhouse gases.  Further increases in greenhouse gas concentrations will, therefore, have no effect. 
With the current window of 90.2 Wm^-2, the temperature of 288 K is obtained with f = 0.981
It is clear that an increase of 1 K could occur, or has already occurred, if greenhouse gas concentrations increased f to 1.0   But no further rise in temperature would happen. Perhaps this could explain the “hiatus”.
However, if the window power were to be reduced, the results would be serious. 
Fig 5 shows the temperature increases for convenience. Even without any more carbon dioxide, the temperature rise with w = 0 could be 15 deg C.

The potential problem is due to the increasing use of compounds of fluorine; particularly the CFCs and the HCFCs.  Also, sulphur hexafluoride.  These substances have very significant wavelengths within the window, and so are very dangerous.   Fortunately, these are man-made substances, and so in principle it should be possible to exert some control on their use, in accordance with the Montreal Protocol.
However, these ozone destroying substances are being superseded by HFCs which also have high radiative absorption wavelengths within the window. And so the problem continues.

Ref (1)

A Really Simple Alternative to the GHG Theory for Global Warming

Eddie Banner    29/7/19

As a physicist, I have long had doubts about the Greenhouse Gas Theory for global warming, and I recently came across a post which reinforces these doubts.  Although the post itself supports the GHG theory, nevertheless it provides a very interesting diagram, shown below,

which is based upon excellent work by R A Hanel et al with real measurements by satellite taken over the Niger valley in North Africa. It shows the infrared radiative flux emitted to space at the top of the atmosphere.     

Fig 1.  Satellite measurements
Credit: Data from R. A. Hanel, et al., J. Geophys. Res., 1972, 77, 2829-2841 

This diagram gives the results of real, practical observations by the Nimbus satellite, and provides ideas which simply are not forthcoming with the theoretical computer models such as Nasa’s GISS model.

I should like to offer a completely new (as far as I know) treatment, based entirely on these practical results. 

The Planck curve for any particular surface temperature of the Earth gives the maximum radiation flux emitted by the surface of the Earth and transmitted to space, assuming the Earth to be a “black body”.  Accordingly, the spectrum trace in the diagram would seem to indicate that the radiation emanates from a surface temperature of 320 K.  The atmospheric window, through which radiation to space passes largely unimpeded by absorption, is clearly shown between infrared wavelengths of 8 microns and 14 microns. 

The figure also demonstrates the absorption of the radiative flux in the 15 micron region by carbon dioxide in the atmosphere. 

Working with wavenumbers, because they are a measure of the infrared frequency, and therefore of energy, the wavenumber of a 15 micron photon is 667 cm^-1. 

The abscissa is given in terms of cm^-1, which is a measure of frequency, and the ordinate is in terms of milliW.m^-2.(cm^-1)^-1.  Therefore, for a 1 second period, the area of any particular portion of the graph has units of the product of abscissa and ordinate.

That is   ( milliJoules.m^-2.(cm^-1)^-1) * (cm^-1)

i.e.          milliJoules per m^2

So the energy of a part of the graph is given by the area of that part. 

The Planck radiation curves are very lengthy to compute without a suitable computer program, so I have printed a copy of the diagram and simply interpolated, and sketched on the print, within the given curves for suitable temperatures.

It is instructive to determine the energies of the various parts of the diagram, and this has been done simply by measuring the areas on the print by pencil and ruler, estimating triangles and parallelograms as appropriate.

The “Planck” temperature values, except for the 320K curve, are not relevant because the radiative flux at all wavenumbers outside the window arises from atmospheric absorption and re-emission, at a lower atmospheric temperature.  This is the emission temperature.  Since all the radiation emanates from the Earth’s surface, in the absence of absorption the emission curve will follow the shape of the Planck curves.  The position of the appropriate curve on the diagram can be found from Fig 4(a) in the paper by W Zhong and J D Haigh (1).  This shows that the value of the transmitted radiation at 15 microns, if there were no absorption by CO2, would be 1.42 times the transmittance in the absorbed case.  This places the “non-absorbed” position of the trace for 667 cm^-1 at 80 mW.m^-2.(cm^-1)^-1 on the ordinate axis.  The corresponding Planck curve without absorption can be sketched in, as above. 

At wavenumbers less than about 800, there is clearly absorption by carbon dioxide; this is the area below the ‘sketched” emission temperature curve, between it and the trace of the measured radiation flux.

There is also radiation transmitted to space, shown above the curve at wavenumbers less than 600 cm^-1.  So where does this “extra” energy come from?  Not directly from the surface because the limit is set by the Planck curve being considered.  So it would seem to  come from the atmosphere, and this is supported as follows.

Suppose we measure the areas (and therefore the energies) above and also below the emission temperature curve, between 400 and 800 cm^-1.   (Several sets of measurements might be made for slightly different emission curves for comparison, if necessary.)

It is found that the two energies are equal, and this suggests that the energy absorbed by the carbon dioxide has been re-emitted to space by photon emission from water molecules below 600 wavenumber.

A temperature is found for which the areas, and therefore the energies, are equal.

Therefore, it appears that, in this case, the emission temperature is about 256, simply using this number to position the curve on the graph.

This emission temperature is readily stabilized.  If some perturbation caused this temperature to start to rise, then there would be more energy absorbed by CO2 below 800 cm^-1, and so more feedback to the transmission below 600 cm^-1, which means more cooling by radiation to space.  So the emission temperature would fall in response.  The converse argument also applies.  The system has negative feedback, and so the emission temperature is stabilized. 

Now, suppose that the carbon dioxide concentration is doubled.  This means that there is more absorption, and so equally more feedback, which means more radiation to space.  This means that more carbon dioxide causes more energy to be emitted, so causing a cooling effect.

So, extra carbon dioxide produces more cooling.

Therefore, if there is any increase in atmospheric temperature, it is not due to carbon dioxide.

It follows, therefore, that the GHG theory is NOT valid.

Ref. (1)

Energy causes global warming

Energy causes global warming

Energy Theory for Global Warming

Aubrey E Banner

11th May 2019

Global warming is certainly happening.
This has been proved by many measurements of the Earth’s temperature, and the resulting consequences are manifest.  A popular explanation is that this effect is due to the carbon dioxide in the Earth’s atmosphere, and it is known as the “Greenhouse Gas Effect.”  It claims that anthropogenic burning of fossil fuels continually increases the amount of carbon dioxide, and so causes the temperature of the atmosphere to increase.  However, many physicists disagree with this theory because they say that it violates the Second Law of Thermodynamics.

I should like to offer a new, simple idea which shows that the rise in temperature of the Earth’s atmosphere is due to two combined causes which have nothing to do with the Greenhouse Gas effect.  They are :-
(1) Primary Energy
(2) Energy from the Oceans 
Primary Energy
The larger of these effects is due to the vast amount of energy we generate and consume. The energy we get from burning fossil fuels, or from nuclear power, eventually ends up as heat put into the Earth’s system, and this heat energy causes the temperature to rise.  It’s fairly obvious, really.  Most of the energy goes into the oceans, but a small proportion remains in the air, and is enough to cause the effect.  It should be realized that although the amount of energy involved on a daily or yearly basis is relatively small, nevertheless the total amount accumulated over a period of years is sufficient to explain the temperature increase.  Moreover, once the energy has arrived in the atmosphere, it does not escape to space as many might think.  Please refer to the Appendix for the explanation. 

Information for primary energy consumption has been obtained from the British Petroleum website at
and the following total figures have been calculated for the Northern and Southern Hemispheres separately, over the 50 years from 1965 to 2015.

The BP figures are given in units of million tons of oil equivalent (mto equiv), but are expressed below in Joules, where the conversion factor 1 mto equals 4.187*1016 Joules has been used.  This data shows that the annual energy consumed in the Northern Hemisphere increases each year in an approximately linear way. 

The graph below shows the increase in the aggregate (accumulated)  Primary Energy in the Earth’s system from 1965 to 2015, calculated from the BP data.

Primary Energy over 50 years 1965 -2015   
Most of the energy is consumed in the Northern Hemisphere. 
The energy in the atmosphere stays in its hemisphere of origin because of the well-known circulation of the currents in the atmosphere.  BP data, after conversion. 
            Northern Hemisphere    8.8510*1021 Joules
            Southern Hemisphere    5.2513*1020 Joules
This is the primary energy consumption over 50 years. 
Initially all the energy enters the atmosphere.

I have had some difficulty in obtaining a definitive figure for the proportion of this energy which remains in the atmosphere.  Most of it enters the oceans; some enters the land mass and the ice.  However, from the IPCC report
and scroll to we find for the period 1961 to 2003 that the total energy entering the oceans was 89.3% of the total energy from all sources, and the energy remaining in the atmosphere was 3.14%, this latter figure being subject an error of +or – 40%.  This means that the proportion of the total energy remaining in the atmosphere was between 1.89% and 4.40% of the total.  The following calculations use the value 3.14%.   
Oxygen and nitrogen molecules do not radiate, so the increase in energy of the atmosphere is
Northern hemisphere
0.0314 * 8.8510 * 1021 Joules,   that is  2.77921 * 1020 Joules, over 50 years.
Southern hemisphere
0.0314 * 5.2513 * 1020 Joules,   that is  1.64891 * 1019 Joules, over 50 years.
Now, the surface area of the Earth is 5.1*1014 m2
So, the area of each hemisphere = 2.55 * 1014 m2
Northern Hemisphere

Therefore energy consumption  =  (2.7792*1020) / (2.55*1014
= 1.08988 * 106 Joules per square metre, over 50 years.
From  we find that the number of molecules in the Earth’s atmosphere is 1.09*1044   
So, the number of “air” molecules in the standard column based on 1m2 of the surface =  (total number of molecules in the atmosphere) / (surface area of Earth)
  =  (1.09*1044) / 5.1*1014  
  =  2.137 * 1029 molecules in column (on 1 m2
It has been assumed that the added energy is confined to half of the troposphere, which contains 75% of the atmosphere. ( Refer to 
So the number of participating molecules in the column is 0.5*0.75*2.137*1029 
= 8.0138*1028  molecules per m2  
The increase in energy is shared between these molecules in the column. 
So the increase in energy per molecule = (1.08988*106) / (8.0138*1028)   
1.36*10-23  Joules per molecule  
From the kinetic theory of gases,
the kinetic energy of a molecule moving in a gas is (3/2)*k*T  where  k is the Boltzmann constant 1.38*10-23 J/K, and T is the Absolute temperature of the molecule. 
So,                 increase in energy = 1.5*k*(increase in temp)
So,   increase in temperature = (increase in energy) / (1.5*1.38*10-23)   K  

Therefore, for the Northern Hemisphere      
increase in temperature = (1.36*10-23  ) / (2.07*10-23)  K  = 0.657 K 
So, increase in atmospheric temperature =  0.657  K  over the 50 year period.

From we find measured values for atmosphere temperature increases from 1965 to 2015, and these are shown in blue in the graph below.  The graph for the calculated values is in red.  The equation for this curve is
Temp.Anomaly = (7.423*10-23 )*(Primary Energy Aggregate (gross) in Joules)  o C

From this graph, it is clear that the effect of the Primary Energy is not sufficient to explain the measured increase in temperature of the atmosphere in the Northern Hemisphere.  But the effect of warming by the Oceans has still to be considered.  For this we must first deal with the temperature rise in the Southern Hemisphere. 

Southern Hemisphere
we find the measured temperature anomalies for the Southern atmosphere, and these are shown in blue in the graph below.  The effect of the Primary Energy in the Southern Hemisphere is shown ingreen. The difference between the two is shown in red.

The increase in atmospheric temperature in the Southern hemisphere over the 50 years 1965 to 2015 was 0.43 oC.
The energy theory gives a value of only 0.039 oC.  For 2015, the difference, 0.391 oC, is explained by means of the energy emitted from the oceans.
This difference provides crucial information about the amount of energy transferred from the oceans to the atmosphere.

Assuming that the warming occurs in a similar way in the Northern Hemisphere, but allowing a factor of 0.75 for the smaller area of the Northern Oceans, the contribution of the oceans to the total temperature increase in the North can then be readily calculated, as shown in the graph below.  This is the “Ocean Effect”.

Northern Hemisphere, revisited, including the Ocean Effect

It will be seen that the sum of the temperature anomalies for the Primary Energy and the Ocean effects gives a value slightly in excess of the measured temperature anomaly.

Projection for the Northern Hemisphere until 2065
If annual energy consumption and ocean warming were to increase at the same rate as the average value over the last 20 years, the increase in atmospheric temperature from 1965 to 2065 levels would be 2.37o C, as shown in the graph below

It would seem that energy considerations alone could account for global warming, so the problem is far more serious than the enhanced greenhouse gas effect attributed mainly to carbon dioxide.  Even if continuing increases in concentrations of greenhouse gas levels were to be reduced to zero, temperature rises would still continue unabated because of the continuing energy increases here considered.

It follows that in order to limit temperatures to present levels, it is vital to avoid adding any further energy to the system.  This means that all present methods of energy generation must be discontinued, except for the renewables such as solar, wind and wave.
The renewable energy is derived from the Sun, and is ultimately absorbed into the Earth’s system.  In using this energy, we are simply diverting it first for our own use, so it does not cause any increase in temperature.

To estimate the possible loss of energy from the atmosphere
Consider a standard column of the atmosphere based on an area of 1square metre of the Earth’s surface.
As before, the number of nitrogen molecules in the column = 2.15*1029 Therefore, for carbon dioxide at 400ppm, number of CO2
molecules = 8.6*1025
From the PNNL spectra at we find that the absorption cross section of a CO2 molecule for a 15 micron photon is  5*10-22  m2 per molecule.
So, in a standard column on 1 m2 surface, the number of photons absorbed = (8.6*1025 )*(5*10-22 )
= 4.3*10photons / m2
Now, the energy per 15 micron photon = 1.3252*10-20 Joule.
Therefore, the sum total of the absorbed energy
= (4.3*104 )*(1.3252*10-20 )Joule
= 5.7*10-16 Joule
Let us suppose that this energy can be transported to space, and so lost.
But, the collision relaxation time of the carbon dioxide molecule is taken to be 2*10-10 second, so the absorption process can be repeated 5*109 times per second.

So the total energy lost to space by carbon dioxide per square metre of the Earth’s surface could be  (5.7*10-16 ) * 2*109   Joule per second
=  1.14*10-6 Joule per second per m2
=  1.14*10-6 Watts per m2

Similarly, for the same repetition rate, it can be shown that the total energy lost to space by water per square metre of the Earth’s surface could be  1.14*10-9  Joule per second
=  1.14*10-9 Watts per m2
These figures must be compared with the rate of supply of energy to the atmosphere from anthropogenic sources.
In the Northern Hemisphere, the energy entering the atmosphere
= 5.148*1020 Joules over 50 years
=  3.265*1011 Joule per second
The surface area of the hemisphere = 2.55*1014 m2
Therefore, the input rate = (3.265*1011) / (2.55*1014)  Joule/sec/m2
= 1.28*10-3  Watts per m2 
Therefore, although a very small proportion of energy might be lost to space, the re-supply rate at average consumption over the 50 years period is much greater, by a factor of 2.2*10