# A controversial look at Blackbody radiation and Earth minus GHG's

Image via Wikipedia

Guest Post by Reed Coray

On Dec. 6, 2011 12:12 am Lord Monckton posted a comment on a thread entitled Monckton on sensitivity training at Durban that appeared on this blog on Dec. 5, 2011. In that comment he wrote:

“First, it is not difficult to calculate that the Earth’s characteristic-emission temperature is 255 K. That is the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere. Since today’s surface temperature is 288 K, the presence as opposed to absence of all the greenhouse gases causes a warming of 33 K”.

Since I’m not sure what the definition of the “Earth’s characteristic-emission temperature” is, I can’t disagree with his claim that its value is 255 K. However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.

When computing the Earth’s surface temperature difference in “the presence as opposed to the absence of all greenhouse gases”, (i) two temperatures (A and B) must be measured/estimated and (ii) the difference in those temperatures computed. The first temperature, A, is the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases. The second temperature, B, is the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only–i.e., an atmosphere that contains non-greenhouse gases but is devoid of greenhouse gases.

For temperature A almost everyone uses a “measured average” of temperatures over the surface of the Earth. Although issues may exist regarding the algorithm used to compute a “measured average” Earth surface temperature, for the purposes of this discussion I’ll ignore all such issues and accept the value of 288 K as the value of temperature A (the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases).

Thus, we are left with coming up with a way to measure/estimate temperature B (the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only). We can’t directly measure B because we can’t remove greenhouse gases from the Earth’s atmosphere. This means we must use an algorithm (a model) to estimate B. I believe the algorithm most commonly used to compute the 255 K temperature estimate of B does NOT correspond to a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only”. As will be evident by my description (see below) of the commonly used algorithm, if anything that algorithm is more representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases” than it is representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only.”

If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.

Although there are many algorithms that can potentially lead to a 255 K temperature estimate of B, I now present the algorithm that I believe is most commonly used, and discuss why that algorithm does NOT represent “the temperature of the Earth’s surface in the presence of an atmosphere that is devoid of greenhouse gases”. I believe the algorithm described below represents the fundamental equation of radiative transfer for the Earth/Sun system assuming (a) an Earth absorption albedo of 0.3, and (b) an Earth emissivity of 1.

(1) The “effective temperature” of the Sun [i.e., the temperature of a sun-size spherical blackbody for which the radiated electromagnetic power (a) is representative of the total solar radiated power, and (b) has a power spectral density similar to the solar power spectral density] is approximately 5,778 K.

(2) For a spherical blackbody of radius 6.96×10^8 meters (the approximate radius of the sun) at a uniform surface temperature of 5,778 K, (a) the total radiated power is approximately 3.85×10^26 Watts, and (b) the radiated power density at a distance of 1.5×10^11 meters from the center of the blackbody (the approximate distance between the center of the Sun and the center of the Earth) is approximately 1,367 Watts per square meter.

(3) If the center of a sphere of radius 6.44×10^6 meters (the approximate radius of the Earth) is placed at a distance of 1.5×10^11 meters from the center of the Sun, to a good approximation the “effective absorbing area” of that sphere for blackbody radiation from the Sun is 1.3×10^14 square meters; and hence the solar power incident on the effective absorbing area of the sphere of radius 6.44×10^6 meters is approximately 1.78×10^17 Watts (1.3×10^14 square meters x 1,367 Watts per square meter).

(4) If the sphere of radius 6.44×10^6 meters absorbs electromagnetic energy with an “effective absorption albedo” of 0.3, then the solar power absorbed by the sphere is 1.25×10^17 Watts [1.78×10^17 Watts x (1 – 0.3)].

(5) A spherical blackbody (i.e., a spherical body whose surface radiates like a surface having an emissivity of 1) of radius 6.44×10^6 meters and at a temperature 254.87 K (hereafter rounded to 255 K) will radiate energy at the approximate rate of 1.25×10^17 Watts.

(6) If independent of the direction of energy incident on a sphere, the surface temperature of the sphere at any instant in time is everywhere the same, then the sphere possesses the property of perfect-thermal-conduction. Thus, for (a) an inert (no internal thermal energy source) perfect-thermal-conduction spherical body of radius 6.44×10^6 meters and uniform surface temperature 255 K whose center is placed at a distance of 1.5×10^11 meters from the center of an active (internal thermal energy source) spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K, and (b) the inert perfect-thermal-conduction spherical body (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates electromagnetic energy with an emissivity of 1 then the perfect-thermal-conduction inert spherical body at temperature 255 K will be in radiation rate equilibrium with the active spherical blackbody at temperature 5,778 K.  If the phrase “inert perfect-thermal-conduction spherical body of radius 6.44×10^6 meters” is replaced with the word “Earth,” and the phrase ” active spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K” is replaced with the word “Sun”, it can be concluded that: If (a) an “Earth” at temperature 255 K is placed at a distance of 1.5×10^11 meters from the “Sun” and (b) the “Earth” (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates energy with an emissivity of 1, then the “Earth” will be in radiation rate equilibrium with the “Sun.” For the above conditions, the temperature of the “Earth” in radiation rate equilibrium with the “Sun” will be 255 K.

This completes the algorithm that I believe is commonly used to arrive at an “Earth’s characteristic-emission temperature” of 255 K, and hence is used to compute the 33 K temperature difference.

Even ignoring the facts that (1) it is incorrect to use the “average surface temperature” when computing radiation energy loss from a surface, and (2) in the presence of an atmosphere, (a) the blackbody radiation formula may not apply, and (b) blackbody radiation from the surface of the Earth is not the only mechanism for Earth energy loss to space (the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space), the problem with using the 255 K temperature computed above to determine the difference between the Earth’s temperature with and without greenhouse gases is that the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.

Thus an effective absorption albedo of 0.3 is based on the presence of a greenhouse gas–water vapor. It is illogical to compute a difference between two temperatures both of whose values are based on the presence of greenhouse gases and then claim that temperature difference represents the temperature difference with and without greenhouse gases. Without water vapor, there won’t be any clouds as we know them. Without clouds, the effective absorption albedo of the Earth will likely not be 0.3, and hence without the greenhouse gas water vapor, the Earth’s surface temperature in the absence of greenhouse gases is likely to be something other than 255 K. Thus, the 255 K “Earth characteristic-emission temperature” as computed using the algorithm above is NOT relevant to a discussion of the Earth surface temperature difference for an atmosphere that does and an atmosphere that does not contain greenhouse gases.  Only if 0.3 is the effective absorption albedo of the Earth in the presence of an atmosphere devoid of all greenhouse gases is it fair to claim the presence of greenhouse gases increases the temperature of the Earth by 33 K.

Because clouds reflect a significant amount of incoming solar power, without water vapor I believe the effective absorption albedo of the Earth will be less than 0.3. If true, then more of the Sun’s energy will be absorbed by an Earth whose atmosphere is devoid of greenhouse gases than by an Earth whose atmosphere contains clouds formed from the greenhouse gas water vapor. This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.

For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).

In summation, using the simplified arguments that I believe are also used to arrive at the 33 K temperature difference (i.e., assumed perfect-thermal-conduction Earth, blackbody Earth emission, greybody Earth absorption with an effective absorption albedo between 0 and 0.3, and ignoring atmospheric radiation to space for an Earth atmosphere devoid of greenhouse gases), I conclude the presence of greenhouse gases in the Earth’s atmosphere increases the Earth’s temperature by somewhere between 9 K and 33 K. Thus, I believe the claim that the presence of atmospheric greenhouse gases increases the temperature of the Earth by 33 K is based on an argument that has little relevance to the Earth’s temperature in the presence of an atmosphere devoid of greenhouse gases; and hence at best is misleading and at worst incorrect.

Note: Upon first publication – the guest author Reed Coray was accidentally and unintentionally omitted.

## 239 thoughts on “A controversial look at Blackbody radiation and Earth minus GHG's”

1. Bryan says:

The albedo of the Earth surface is likely to be even less than the 0.12 value for the Moon.
Reason, the 70% of Earth surface covered by water.
Without the greenhouse gases the extra radiation reaching the surface will largely be short wave infra red(slightly longer than visible red down to 5um).
For this EM region the albedo must be getting near zero.

2. D. Cohen says:

“…the sphere possesses the property of perfect-thermal-conduction….”
This assumption of perfect thermal conduction for the earth without greenhouse gasses is at least as bad as the albedo assumption. I just looked up the average surface temperature of the moon. Without any atmosphere at all the average lunar surface temperature found after a 2 minute search of the web (so take this value for what it’s worth) is -9 degrees Fahrenheit. The moon is about the same distance from the sun as the earth, so this value could be regarded as a ballpark estimate for what you want to know.

3. Lars P. says:

There are many omission in the +33°K warming calculation.
The direct warming of the atmosphere from direct sunlight is ignored as well as the ocean warming from direct sunrise and the way how the seas get their average temperature, through enthalpy & radiation.
The snowball earth theory without CO2 makes no sense as it ignores the 1365 W/m2 direct solar radiation on a spherical surface in rotation not 1/4 of it on a hypothetical flat world.
Greenhouse is only a fraction of the 33° and no earth energy budget ignoring the oceans has any scientific value.
It is used as it is a useful tool for the CAGW scare.

4. richard verney says:

There are strong arguments to suggest that the atmosphere if anything cools us rather than heats us. Certainly, if we did not have an atmosphere day time temperatures would be a lot hotter than we experience. Night time temperatures may be cooler but the extent to which this would be the case depends upon the heat capacity of the liquid storage reservoirs which cover approximately 3/4 of the surface area of the planet..
The average temperatrure of the globe may be out by several degrees, We do not accurately know its albedo nor its emissivity (which are not constants) and the oceans (which themselves are in effect a greenhouse gas albeit in liquid form) is a vast heat storage reservoir. The earth is not a blackbody and I consider that these fundamental problems make the entire BB assessment unreliable.
Further, I consider that it is misconceived to view the earth’s surface temperature as measured today as the base against which the comparison should be made. We are in the midle of an interglacial period and the earth’s surface is untypical warm. The oceans have an average temperature of about 4 degC and this low temperature will come back to bite. Some weighting should be given to take account that over geological time the average surface temperature of the earth is far less than we are privileged to enjoy today.
What if your calculation (or the accepted BB calculation) was performed 20,000 years ago? Many of the fundamentals would be the same and yet in this scenario, the cAGW crowd would have one believe that GHGs were raising the temperature by only about 23K (not 33K).
In posing that question I am aware that ice extent will have changed the albedo, but may be it was less cloudy elsewhere which would tend to offset the change in albedo made by ice.
I would not be surprised if in the real world GHGs add only a few degrees. It is the presence of water and its properties that keep us warm.

5. From the last few sentences one has therefore to conclude that the climate sensitivity to an additional greenhouse gas must be less than “previously thought”. Or in other words, CO2 adds only 9 K instead of 33 K. Am I right?

6. John Brookes says:

OK, that seems like a pretty silly argument. Take a look at Mars. I think Mars would be pretty close to earth without ghgs. No oceans, no vegetation, no ice no animals, just lots of red dirt. But Mars does not have an albedo of 0, it has an albedo of 0.25 – not too disimilar to that of Earth, and certainly a value which will get you closer to 255K than 279K.
As for the assumption of a perfectly thermally conducting Earth, if you abandon this, you will lower the average temperature. Because of the T^4 dependancy of outgoing radiation, having one part of earth at (say) 245 and another of equal area at 265 will result in more outgoing radiation than having it all at 255. Therefore to achieve the same amount of outgoing radiation, the average temperature needs to be lower than 255.
Anyway, well done for writing this, but make a bit more effort to think things through.

7. Stephen Wilde says:

This is yet another example of a set of calculations that simply ignores the role of sunlight into the oceans
Since most of the Greenhouse Gases are in fact water vapour the whole exercise is meaningless unless one removes the oceans and all surface water too.
Some warming proponents have suggested that without non condensing GHGs (all those other than water but especially CO2) then in that situation all the water on Earth would freeze. That is their way of attributing ultimate importance to the non condensing GHGs.
However I don’t see that as likely. Sunlight would still penetrate the oceans and water would still evaporate. The water cycle might be slower but it would still be present would it not ?
To my mind this issue goes to the heart of AGW theory which assumes that the Earth is about 33C warmer than it ‘should’ be as a result of GHGs.
What I would like to know is how they feel able to exclude the oceans in setting that temperature. The liquid in the oceans is a vastly more effective system for slowing down the exit of solar energy from the Earth system than are all the GHGs present in the air even including water vapour.
As far as I know Arrhenius et al only considered the atmosphere. They did not consider the role of the oceans so in order to maintain AGW theory it is essential to propose that it is only the non condensing GHGs that keep the oceans liquid.
But that is not the case. It is solar shortwave penetrating the ocean surface that keeps the oceans liquid, creating water vapour in the process and thus making the Earth’s atmosphere warmer than it otherwise would have been.
That solar shortwave into the oceans and the retention of much of that energy for a lengthy period of time is the primary mechanism for slowing down the loss of that energy back to space. Vastly greater than the slowing down caused by GHGs.
It is the atmospheric pressure that dictates the baseline energy content that the oceans can achieve because that pressure determines the energy cost of evaporation by fixing the energy value of the latent heat of evaporation. Namely the difference between the amount of energy required to break the bonds between water molecules relative to the energy required by the breaking process. At current atmospheric pressure the latter is about 5 times the former which is why evaporation is a very powerful net cooling process.
GHGs do NOT affect that baseline energy content for the oceans because they do not significantly affect atmospheric pressure and therefore they can only affect the rate of energy flow from surface to space. Therefore GHGs can only result in a faster water cycle and NOT a change to the baseline energy content for the system as a whole.
The rate of energy flow from surface to space at any given moment is reflected in the relative sizes, intensities and latitudinal positions of the permanent climate zones. It is the shifting of those zones that we perceive as climate change and NOT any significant change in total system energy content.

8. steveta_uk says:

Surely if you want to determine the effects of the sun on a body at the distance of the earth with no atmosphere, just look to the moon.

9. Richard M says:

I’ve mentioned before that I thought this computation was missing from the literature. I believe you’ve made a great start. However, there is more to the question than just albedo.
The surface of a bare Earth would still have some albedo, but even more important, an atmosphere of O2 and N2 would still absorb and radiate some energy. There would also be some conduction from the surface and 100% of the surface area would be land. Since this kind of atmosphere would NOT radiate very effectively due to its low emissivity, it’s important to determine how much heat would get trapped and what the equilibrium temperature would be.

10. Alan Statham says:

This is painful to read. You seem ignorant of many things. You even start by saying you don’t know what is meant by a characteristic emission temperature. I mean, that’s not particularly clear language but anyone with a basic physics education knows what is meant by that. You wrongly claim that you can’t use an average temperature to calculate radiation losses. And you have used nearly 2000 words simply to say that you don’t agree with the albedo commonly used in this simplified calculation.
The calculation neglects the greenhouse effect; it doesn’t neglect the reflective properties of the atmosphere. This is because we wish to estimate the magnitude of the greenhouse effect, so we only take that out.

11. And not forgetting that our EARTH HAS NO LID.
Then, the question arises: Where is heat saved?
Alternatives:
1) Atmosphere: Air: Volumetric heat capacity: 0.00192 joules /cu-cm.
2) Oceans: Water: Volumetric heat capacity:4.186 joules/cu-cm, i.e., 3227 times than that of Air.
3) Soil: Ground: volumetric heat capacity: About 2.0 joules /cu-cm.
Green House Effect = Confined Heat Effect
No confinement = No effect.

Remember: How soon atmosphere cools down during an eclipse.

12. richard verney says:

Stephen Wilde says:
December 26, 2011 at 4:23 am
//////////////////////////////////////
I am with you on much of what you have said.
On several occassions I have had arguments with Willis regarding the oceans freezing. He considers that without GHGs the oceans would freeze. I say to him, he is basing that claim on average conditions and one needs to look not at the average condition but rather to look at the considtions prevailing over the tropics. In the tropics, there is enough sloar energy to ensure that the oceans do not freeze at the tropics. The heat in the tropical ocean is then circulated poleward. May be the circulation pattern and extent would be different to that observed today but the material point is that the tropical ocean would not freeze and some of its heat would be distributed elsewhere.
The key to Earth’s climate is the oceans, the properties of water and the water cycle.

13. davidmhoffer says:

Stephen Wilde nails it when he says:
“Since most of the Greenhouse Gases are in fact water vapour the whole exercise is meaningless unless one removes the oceans and all surface water too.”
Further, I don’t think the 255K number is supposed to represent the surface temperature of the earth without greenhouse gases. It is supposed to represent the temperature of the earth without the greenhouse EFFECT. If that is the case, then the logic presented to arrive at 255K seems quite reasonable.
(I’d rather eat bark than admit that the IPCC et al got anything right, but in this case, I think they did)

14. BenAW says:

Try this for simplicity:
incoming total radiation 1364 W/m^2 on a disk with a radius the same as earth.
Since a sphere has a four times larger area than a disk divide by 4.
30% albedo: 0,7 x 1364 / 4 = 239 W/m^2 average incoming radiation.
Stefan Boltzmann (SB) then gives the infamous 255K
More realistic seems to divide the incoming radiation over half the sphere, as on the real earth.
The fourth power in SB then has an interesting effect:
0,7 x 1364 / 2 = 477 W/m^2
SB now gives 303K !!!!
So half the earth receives on average enough energie to reach 303K (30 C)
Since the average temp is lower 288K ? (15 C), this surplus is stored and released at the nightside of the earth. Seems very plausible that this results in an average temp of 288K.
No greenhouse effect needed.

15. Sionn5 says:

One must also understand that there is in inherent temperature due to gravity. Gravity pulls a non greenhouse atmosphere down and this produces a pressure. The ideal gas law can be used to approximate this inherent temperature. This effect occurs on any gravitating body with any type atmosphere and is already neglected in the calculations.

16. davidmhoffer says:

who is the author of this? the article only says “news staff” and it isn’t signed either?

17. It surprises me when people believe 5x10ee18kg of cold, rarefied atmosphere can heat 1350x10ee18kg of water. You can heat a material with a large thermal mass (like water) with a material with a small thermal mass (like a gas), but never if the gas is colder than the liquid. This is an intelligence test.
Imagine a gallon of water on a table in front of you. There are lots of methods that can be used to increase its temperature by 10% (33C). However, which of these methods are at work in the atmosphere?

18. A thought experiment. If the earth had an albedo of 0 and you could put whatever imaginary layer you wanted around it (with physical properties, i.e. you can’t use one-way glass that doesn’t exist in the real world). What is the maximum temperature?
Could you get the earth to have a higher temperature than a theoretical perfect black-body? If so, does that seem logical?

19. steveta_uk says:

John Brookes (December 26, 2011 at 4:18 am)
(245+265) / 2 = 255
sqrt( sqrt( ((245^4)+(265^4)) / 2 ) ) = 255.5864
Not what I was expecting.

20. wayne says:

Not only is the Earth not a black body but since oceans cover about 7/10th of the surface, a much smaller area than half of a sphere centered directly under the sun can even have the effect of absorption of incoming radiation as clearly shown in this simple experiment in relation to the specific critical angle:

I can find no reference of this in “climate science” and would appreciate any links someone might have, where this is taken into account or, whether this is calculated within the albedo ~0.30 figure used within such calculations used in this post.

21. PeterF says:

“… and the heated atmosphere will also radiate energy to space)”
No, it won’t. When you start with the assumption that there are no greenhouse gases in the atmosphere, then the remaining gases (oxygen and nitrogen in this case) will neither absorb nor radiate in the relevant micrometer range, and hence can’t radiate energy to space. They are not a blackbody.
I am puzzled as to what exactly will happen in such an atmosphere: we will get an adiabatic temperature profile, we will get convection and redistribution of heat through circulation (“wind”), but all heat removal from earth must come from radiative emission from the earth itself, not from its atmosphere. The albedo will be determined solely from the earth’s surface, i.e. its various soils, as the atmosphere will not be radiating.
At the end, this heat redistribution on the globe will possibly result in a justification of the assumption of a single, global temperature which determines the radiative heat loss of the earth.
Or not?

22. richard verney says:

steveta_uk says:
December 26, 2011 at 4:24 am
Surely if you want to determine the effects of the sun on a body at the distance of the earth with no atmosphere, just look to the moon.
///////////////////////////////////////////////////////////////////////////////////
Two problems. First, the length of the lunar day. Eg., what would be the average temperature of the moon if a lunar day was only one hour rather than about 27;5 days? This becomes very important when considering the second problem. Second, the moon is a rocky world without the heat storage reservoirs found on Earth (ie., it does not have the equivalent thermal ocean reservoirs).
On Earth, the oceans absorb solar irradiance during the day and then slowly release this at night. The Earth’s rotation of 24 hours playing an important role in this. There is long enough daylight for the oceans to respond to the warmth of the sun, but not too much darkness to permit them to fully convcet/conduct and radiate away the heat so absorbed. If the Earth rotated on its axis once every 100 years, the conditions on Earth would be radically different even though we would still be receiving the same amount of solar energy input.
The Earth is a water world and cannot be likened to a rocky world. As far as I understand matters BB calculations do not work properly for the moon nor for any of the planets in the solar system and the Earth (due to its watery nature) is far more complicated.
I consider the entire rational behind the BB calculations for the Earth to be misconceived . .

23. Richard M says:

I think were on the right track here. First, we need to consider the Earth with no oceans and an atmosphere just like the current one without GHGs. Then, we need to add back in the oceans but still keep the GHGs out of the atmosphere. Yes, I realize this is unphysical, but I think in doing those calculations we could learn a lot.
I think it’s very possible the first situation would have a surface temperature greater than what we currently have and what it would be in the second situation. I suspect, as others have indicated, that water cools the planet.

24. Dave Springer says:

“If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.”
Yes of course it is. It’s the earth without an atmosphere or ocean and an albedo of zero. The average temperature of the moon is 250K. It has an albedo of 0.15.
Ask the fraudsters sometime to take the ocean out of their climate models and see what happens. The global ocean is where most of the greenhouse effect happens. Water is quite transparent to visible light and quite opaque to infrared. It’s more opaque than water vapor and far more opaque than CO2. Transparency to visible light and opacity to infrared are the properties which distinguish greenhouse gases from non-greenhouse gases. Since gases are technically fluids and liquid water has the requisite properties for greenhouse warming one can then quickly realize that greenhouse heating of the ocean is what does most of the surface warming above blackbody temperature. Sunlight penetrates the ocean to some 100 meters where it absorbed along the way by impurities. All the energy in the sunlight is transferred to the water at the speed of light. However because water is opaque at all far infrared frequencies, unlike CO2 which is opaque only in narrow bands, none of that energy absorbed at depth can escape radiatively – it must be mechanically transported to the surface. That delay in transport, whether it’s water or CO2 doing the delaying, is what causes the so-called greenhouse effect. The ocean, you see, has many times the mass of the atmosphere and thus many times the greenhouse effect. Adding insult to injury the ocean doesn’t absorb any significant energy from downwelling infrared from greenhouse gases because the LWIR energy is completely absorbed in a skin layer just a few micrometers deep which doesn’t mix downwards but rather evaporates and carries the LWIR energy immediately away as latent heat of vaporization.

25. coldlynx says:

The classic mistake or “trick” is to calculate the outgoing radiation from earth surface.
As if GHG and cloud does not exist.
Do the math from middle of atmosphere: 50% of atmosphere weight is 500 Mb equal to 5500 meter altitude. Lapse rate for humid atmosphere are about 6 C/1000 meter or for 5500m …… “Ta da”……..33 C colder than on earth surface. Such a coincidence. Not.
The atmospheric window is just that, a small wavelength window where just some radiation escape from earth surface direct to space. Most of outgoing radiation are radiated from the atmosphere.
For an average temperature are an average altitude and by that an average radiation balance just making sense.
By the way, change in UV radiation from the sun alter atmosphere height, that is changing the 500mb altitude, which will change the average surface temperature.

26. Dave Springer says:

“the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.”
A sticky wicket indeed. The earth and moon are made of the same stuff. The moon has an albedo of ~0.15 which is what the earth would be without an atmosphere. The ocean has an albedo of ~0.00 which makes it a lot darker than the moon. But the ocean can’t exist without an atmosphere to keep it from boiling away. And you can’t have a liquid ocean and an atmosphere without clouds. Clouds have an albedo of ~0.80 and then the combination of all these things gives the planet an albedo in the range of ~0.35 which varies interannually by an unknowm but presumably small amount in the range of +-0.03.
No one knows exactly what the earth’s average albedo really is. Climate models use values which differ by as much as 0.07 from one to another. Speaking in W/m^2 that 0.07 uncertainty is 7% of some 200 Watts per square meter or about 14 W/m2. The net effect of all anthropogenic forcings is claimed to be around 2.5W/m2 so in essence this means that uncertainty in natural albedo forcing is about 6 times greater than anthropogenic forcing. In other words, the models are operating in the dark (so to speak) with uncertainties far greater than the anthropogenic factor they are trying to calculate. Albedo in the climate models is a proverbial fudge factor that is adjusted by the model maker to whatever value gives the best result in conjuction with all the other assumptions and variables. If these models weren’t being considered with more credibility than instruments (if observations disagree with models it is presumed the observations must be wrong!) this would be material for late night talk show comedy skits.

27. Leonard Weinstein says:

The writeup is wrong on the two major points. PeterF made the first one, that if there is no greenhouse effect, the atmosphere will not radiate much to space (there would be a small amount from aerosols and very weak diatomic gas effects, but not significant). The second point is the albedo effect. Oceans and clouds have effects, but if no greenhouse effect were present, the clouds (and water vapor) would not be present. The albedo would then be due to ground level absorption. It does not matter if there are oceans or not if the assumption of equivalent albedo is assumed to be the same, as long as no clouds or water vapor is present. The only effect of oceans is storage and distribution, and long term averages would be the same as if no oceans were present. Lord Monckton is correct as far as his statements are made. Get over it.

28. What about a “leaking condenser”,where the atmosphere, being the dielectric, short circuits during lightning storms (and also during tornadoes, etc.)?

29. Dave Springer says:

This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.
“For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).”
Bingo. The ocean has an albedo of effectively 0 and it covers 70% of the surface. The rocks have an effective albedo of 0.15, same as the moon, which is (believe it or not) about the same as weathered asphalt. The moon is a much darker color than one might guess by looking at it because it is contrasted against the inky blackness of space. If it had the albedo of a cloud or snow it would be blinding to look at it.
Anyhow, the albedo of the earth sans greenhouse gases would be around 0.05 instead of .035 which would make it quite a bit warmer than 255K on average.
Technically however, clouds aren’t greenhouse gases. They are composed of liquid water droplets. Water is highly reflective at high angles of incidence so unlike the surface of the ocean with presents a low angle of incidence to sunlight except near the poles and near sundown &* sunset when it doesn’t really matter because solar power is very weak then anyhow, clouds are liquid water in a highly reflective configuration.
It’s tough to imagine how clouds would form without atmospheric water vapor but to be fair one must point out that clouds aren’t composed of water vapor.

30. Don’t we already have a close by object with no atmosphere? What’s the surface temp on the moon? Can that not be used an an example?

31. “The Earth is a water world and cannot be likened to a rocky world. As far as I understand matters BB calculations do not work properly for the moon nor for any of the planets in the solar system and the Earth (due to its watery nature) is far more complicated.”
That makes sense. Then it begs the question, that as the earth rotates and continents interupt the water absorbtion, then less heat is captured. Correct? Then that begs the question, was then the differences in climate (ie no winters prior to this ice age) caused by the configuration of the continents? Less exposed land the warmer the planet became? Shallow seas means more life (our oil today), which changes the climate as well.
I don’t think the climate modelers are even close to how the planet works.

32. Dave Springer says:

Complicating all this even more is rotation rate. The *measured* average temperature of the moon’s surface is 250K. Its albedo is much lower than the earth’s so by that empirical yardstick the earth should be colder than the moon. However, the blackbody calculation for the earth presumes that it is rotating fast enough for day/night temperature difference to be erased. But it does rotate and does have a diurnal temperature difference. Sans atmosphere the difference would be nearly as great as on the moon (which is a couple of hundred degrees K). Because heat is lost faster when temperature is higher this makes the real-life rotating body colder than the blackbody calculation would produce. Thus we have the circumstance where the earth’s moon with an albedo of 0.15 and a 28 day rotation period has a measured average temperature of 250K and a theoretical earth at albedo 0.30 spinning infinitely fast would be 5K warmer even though the theoretical earth absorbs 15% less energy.
There are so many confounding factors that the 33K figure bandied about for an earth without greenhouse gases is completely untrustworthy and without merit. I use that figure solely for the sake of argument because it isn’t necessary to dispute it in order to show the hideous flaws in computer climate models.
But this is still a great OP as it does point out a very flawed assumption bandied about by CAGW fraudsters as some kind of basic fact that is beyond dispute. Little of what they claim is beyond dispute.

33. DirkH says:

PeterF says:
December 26, 2011 at 5:26 am
““… and the heated atmosphere will also radiate energy to space)”
No, it won’t. When you start with the assumption that there are no greenhouse gases in the atmosphere, then the remaining gases (oxygen and nitrogen in this case) will neither absorb nor radiate in the relevant micrometer range, and hence can’t radiate energy to space. They are not a blackbody.”
So O2 and N2 will keep their energy forever even if in the vacuum of space, as they can’t radiate anything? I don’t think that’s how it works. Would be an interesting energy storage if it worked.
http://www.americanthinker.com/2010/02/the_hidden_flaw_in_greenhouse.html
Now, of course O2 and N2 don’t have the absorption/emission bands in the LWIR range that CO2 or H2O have but that surely doesn’t stop them from having SOME emissivity.
Is there a substance with zero emissivity? Only at 0 K.
“real substances have an emissivity between zero and one (0 <ε< 1)."
http://www.learnthermo.com/T1-tutorial/ch04/lesson-B/pg22.php

34. Thermo says:

Infrared radiation can’t heat gases (atmosphere), so where is warming effect?

35. Luther Wu says:

jrwakefield says:
December 26, 2011 at 6:57 am
Don’t we already have a close by object with no atmosphere? What’s the surface temp on the moon? Can that not be used an an example?
__________________________
250K degrees.

36. Thierry says:

Everything concerning the fictitious 33°C greenhouse gases effect has been definitively explained and rebutted by Gerlich & Tscheuschner (2009) in “Falsi fication Of The Atmospheric CO2 Greenhouse E ffects Within The Frame Of Physics” (http://arxiv.org/abs/0707.1161)
This value comes from a wrong use of the Stefan-Boltzmann law (flux imposing local temperature, which is wrong) involving an over simplified planet model (no atmosphere, infinite conductivity, no rotation). The real temperature would be -129°C.
Therefore, this -18°C has no meaning at all.
I do not understand why there is still a debate here.

37. Dave Springer says:

Leonard Weinstein says:
December 26, 2011 at 6:32 am
“The writeup is wrong on the two major points. PeterF made the first one, that if there is no greenhouse effect, the atmosphere will not radiate much to space (there would be a small amount from aerosols and very weak diatomic gas effects, but not significant). The second point is the albedo effect. Oceans and clouds have effects, but if no greenhouse effect were present, the clouds (and water vapor) would not be present. The albedo would then be due to ground level absorption. It does not matter if there are oceans or not if the assumption of equivalent albedo is assumed to be the same, as long as no clouds or water vapor is present. The only effect of oceans is storage and distribution, and long term averages would be the same as if no oceans were present. Lord Monckton is correct as far as his statements are made. Get over it.”
No, you’re wrong. The ocean has an effective albedo near zero and it radiates very little. Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation. Land surfaces radiate. Water evaporates. Write that down. Land has an albedo of 0.15. Water has an albedo of 0.00. Write that down too. The go figure out your errors. Monckton is full of crap which is no surprise. He’s a legend in his own mind and has a goodly number of shallow-thinking sycophants such as yourself who worship him simply because he’s on the same side of the debate as you are.

38. This subject has been pretty well beaten to death by “Science of Doom”. Leonard Weinstein and DeWitt Payne were two of the folks who made the most sense there.

39. wayne says:

PeterF says:
December 26, 2011 at 5:26 am
“… and the heated atmosphere will also radiate energy to space)”
No, it won’t. When you start with the assumption that there are no greenhouse gases in the atmosphere, then the remaining gases (oxygen and nitrogen in this case) will neither absorb nor radiate in the relevant micrometer range, and hence can’t radiate energy to space. They are not a blackbody.
>>>
Sorry PeterF but your statement is not strictly correct. Have you never seen the oxygen radiation in lower frequency spectra? You are saying there are no vibrational lines in oxygen and nitrogen in mid-IR but all matter above zero Kelvin, even gases, though weakly, and primarily at microwave frequencies, do radiate electromagnetic wave energy away due to electron accelerations in the gas molecules interactions with each other.
However, a lone isolated molecule of oxygen or nitrogen in the void of space will do exactly as you said, never radiate at all, zero.
One paper I read about a year ago, listed on HITRAN site of I believe nitrogen-argon interaction continuum at about wavenumber 400/cm and lower.
If you have problems with this, give me a link to a radiometry scientific paper proving this otherwise.

40. davidmhoffer says:

PeterF says:
December 26, 2011 at 5:26 am
“… and the heated atmosphere will also radiate energy to space)”
No, it won’t. When you start with the assumption that there are no greenhouse gases in the atmosphere, then the remaining gases (oxygen and nitrogen in this case) will neither absorb nor radiate in the relevant micrometer range, and hence can’t radiate energy to space.>>>
Yes it will. The absorptive/radiative spectrum for Oxygen and NO2 is right in the graphic in the article. It may not be as pronounced as water vapour and CO2, but it isn’t zero either. Further, the atmosphere contains plenty of dust which will absorb and radiate depending on the composition of the dust, but more than likely in a very broad spectrum.

41. Joel Shore says:

To the author of this post: It is strange that with all that Monckton got wrong in that post (as was well-explained by some of us in the comments therein) you have chosen to attack the one thing that he got right. As davidmhoffer points out, the calculation of the 255 K number is for an atmosphere without a greenhouse effect, not an atmosphere without greenhouse gases. So, the correct calculation to do is to keep the Earth system albedo constant … Or, in other words, to ask, “Given that the Earth system (earth + atmosphere) absorbs 240 W/m^2, how warm could its surface possibly be if there were not a greenhouse effect?” And, the answer is that it could only be about 255 K at most.
BenAW: Your notions don’t even obey energy conservation. You are running yourself in circles getting confused about averages. I suggest you work in terms of total power in and out and then you will hopefully understand why your argument is nonsense.

42. Dave Springer says:

Richard M says:
December 26, 2011 at 5:52 am
” think were on the right track here. First, we need to consider the Earth with no oceans and an atmosphere just like the current one without GHGs. Then, we need to add back in the oceans but still keep the GHGs out of the atmosphere. Yes, I realize this is unphysical, but I think in doing those calculations we could learn a lot.”
What you get is a snowball earth. We already have something very similar to that condition to look at in the geologic column. The vast majority of the atmospheric greenhouse effect is from water vapor. On a snowball earth most of the water vapor is frozen out of the atmosphere and the ocean is effectively removed from the equation be being covered with permanent sea ice.
The earth with an atmosphere but no ocean would be a frozen wasteland. Without greenhouse gases it would be colder but not much colder so long as a liquid ocean covered most of the surface. Where non-condensing greenhouse gases are very important is in a snowball earth episode. On a snowball earth volcanoes continue to spew CO2 and dark ashes. On a snowball there are no CO2 sinks so CO2 keeps building up in the atmosphere. Ashes also accumulate as some float so when ice surfaces partially melt in the summer it concentrates the ash on the surface making it darker and darker. After millions of years of that a tipping point is reached. So non-condensing greenhouse gases are the kindling which ignites the water cycle. Once the water cycle is ignited it keeps the earth unfrozen (most of the time) and is self-regulating through cloud formation which caps the maximum temperature preventing a runaway greenhouse. It’s quite elegant and works so well that for most of the earth’s history, even though the sun has increased its output 10% over geologic time, the earth stays warm and green from pole to pole. Episodes where there is permanent ice cover at the poles are rare. If anything right now the big climate danger lurking around the corner is the end of the Holocene Interglacial period and unless one finds hunting wooly mammoths over frozen tundra to be a good way to earn a living then we should welcome what little additional warming that anthropogenic greenhouse gases can give us and be concerned about what happens when we no longer have a way of fluffing up the earth’s temperature enough to maybe avoid the return of mile-thick glaciers as far south as New York City.

43. davidmhoffer says:

snip – both you and Dave Springer need to take your arguments elsewhere – I’m sick of this moderating this war between you two – Anthony

44. Richard M says:

Why do you only consider radiation? The GHG-less atmosphere will receive energy by conduction. Since the atmosphere has low emissivity that heat will not be radiated away quickly as it can be with GHGs. However, once it gets warm enough it will start conducting heat back to the surface. Where is this equilibrium point? Is it 255 K and if so, why?
I’ll admit I don’t have any training in this area so maybe there is a simple explanation. I’ve never seen it though and can’t help feeling this could a positive feedback situation where the Earth would heat up.

45. Stephen Wilde says:

Dave Springer said:
“On a snowball earth most of the water vapor is frozen out of the atmosphere and the ocean is effectively removed from the equation by being covered with permanent sea ice.”
How could that happen when solar shortwave continues to be received into the oceans from equator poleward ? It is that which maintains our liquid oceans and not non condensing GHGs.
Removing non condensing GHGs would actually INCREASE such solar input to the oceans because it would no longer be absorbed by those gases on the way down through the atmosphere and thereby be converted from solar shortwave which DOES penetrate the ocean surface to infrared longwavewhich DOES NOT penetrate the ocean surface.
On that basis non condensing GHGS result in a COLDER oceanic equilibrium temperature for a given level of solar input to the system than would otherwise have been the case.
Only the energy content of the air is raised by non condensing GHGs but that extra energy is all in latent form and serves only to speed up the water cycle.
In contrast the energy content of the oceans is actually lowered.
However the total system energy content for the system as a whole in response to a given solar energy input remains pretty much unchanged because the atmospheric Greenhouse Effect is many magnitudes smaller than the Oceanic Hot Water Bottle Effect.

46. Dave Springer says:

@Stephen Wilde
http://climaterealists.com/attachments/ftp/TheSettingAndMaintainingOfEarth.pdf
I had no idea you were a member of the Royal Meteorlogical Society!
I swear by all that is Holy and my mother’s eyes too that I hadn’t read your treatise before arriving at the same conclusions.
The only caveat I would add to it is that the earth isn’t entirely covered by ocean so there is still some validity in the atmospheric greenhouse effect particularly at the surface of the rocks where we live and work. On average the ocean is the big kahuna but on average we live in continental interiors far enough removed from the ocean so that we experience some warming due to greenhouse gases. That warming though simply isn’t enough to throw the entire system into a catastrophic tailspin. Perhaps it might be enough, given the greatest effect is in higher latitudes over continental interiors, that it can extend the Holocene Interglacial through the peak of the Milankovitch cycle without glaciation getting the upper hand this time. In point of fact I’m very hard pressed to come up with anything at all in the way of negative consequences from much higher atmospheric CO2 level. The earth is pretty much green from pole to pole when it is far higher and unless one actually likes barren rocks and ice one should ardently adore the prospect of more CO2.

47. davidmhoffer says:

Dave Springer;
Adding insult to injury the ocean doesn’t absorb any significant energy from downwelling infrared from greenhouse gases because the LWIR energy is completely absorbed in a skin layer just a few micrometers deep which doesn’t mix downwards but rather evaporates and carries the LWIR energy immediately away as latent heat of vaporization.>>>
This would be true of still water with no impurities. In windy or rainy conditions, the surface tension layer doesn’t exist for the most part as the surface must be relatively still for a period of 5 to 7 seconds before the skin layer reforms. Ocean surfaces subjected to wind are covered about 15% in sea foam or more. Sea water also containe silt, algae and other impurities which are brought to the surface by turbulence and which are good absorbers in the IR spectrum.

First, I think the entire model is flat wrong. (From American Thinker) Certified Consulting Meteorologist Joseph D’Aleo and computer expert E. Michael Smith appeared together on KUSI TV [Video] to discuss the Climategate — American Style scandal they had discovered. This time out, the alleged perpetrators are the National Oceanic and Atmospheric Administration (NOAA) and the NASA Goddard Institute for Space Studies (GISS).
Perhaps the key point discovered by Smith was that by 1990, NOAA had deleted from its datasets
all but 1,500 of the 6,000 thermometers in service around the globe. For example, Canada’s reporting stations dropped from 496 in 1989 to 44 in 1991, with the percentage of stations at lower elevations tripling while the numbers of those at higher elevations dropped to one. That’s right: As Smith wrote in his blog, they left “one thermometer for everything north of LAT 65.” And that one resides in a place called Eureka, which has been described as “The Garden Spot of the Arctic” due to its unusually moderate summers.
Overall, U.S. online stations have dropped from a peak of 1,850 in 1963 to a low of 136 as of 2007. In his blog, Smith wittily observed that “the Thermometer Langoliers have eaten 9/10 of the thermometers in the USA[,] including all the cold ones in California.” But he was deadly serious after comparing current to previous versions of USHCN data and discovering that this “selection bias” creates a +0.6°C warming in U.S. temperature history.
Willis Eschenbach’s WUWT essay, “The smoking gun at Darwin Zero,” and it plots GHCN Raw versus homogeneity-adjusted temperature data at Darwin International Airport in Australia is another excellent article outlining huge problems with “averaging” temperatures with mostly fake data. Furthermore, Smith cited boatloads of problems and errors he found in the Fortran code “adjustments” written to accomplish this task, ranging from hot airport stations being mismarked as “rural” to the “correction” having the wrong sign (+/-) and therefore increasing when it meant to decrease or vice-versa.
The real chicanery begins in the next phase, wherein the planet is flattened and stretched onto an 8,000-box grid, into which the time series are converted to a series of anomalies (degree variances from the baseline). Now, you might wonder just how one manages to fill 8,000 boxes using 1,500 stations.
Here’s NASA’s solution:
For each grid box, the stations within that grid box and also any station within 1200km of the center of that box are combined using the reference station method.
Even on paper, the design flaws inherent in such a process should be glaringly obvious.

49. Is there a reason why the author, who refers to him/herself as “I”, continues to remain unknown?

50. davidmhoffer says:

[Language .. Robt]
As I was quoting Mr Springer, and you allowed his language through in the first place, I saw no reason to edit the foul language. However, upon reflection, Mr Springer’s foul mouth should be no excuse for my own, even in the circumstance where I am quoting him. My apologies, and I shall endeavour to be more circumspect in the future.

51. son of mulder says:

If we start with earth at 255 deg K and introduce CO2 eventually water vapour will be added and increasing CO2 to today’s level will warm the earth to qbout 288 deg C. Then start to remove CO2, is the process reversible, will all water vapour eventually drain from the atmoshere or will removal of all CO2 leave a residual amount of water vapour so that the temperature does not revert to 255 deg K? In that case the real contribution of CO2 becomes clear. Has anyone worked this out?

52. Ed Scott says:

enneagram says:
December 26, 2011 at 4:52 am
“Remember: How soon atmosphere cools down during an eclipse.”
—————
Or after dark in a desert.

53. Stephen Wilde says:

“Sea water also containe silt, algae and other impurities which are brought to the surface by turbulence and which are good absorbers in the IR spectrum.”
Sure they do but then they warm up the water molecules around them to speed up the rate of evaporation.
And it isn’t a surface tension issue. The ocean skin is the topmost 1 to 3 mm which is always cooler than the ocean bulk below.

54. Francois says:

Let’s call it planet Earth, forget the water, the atmosphere, the animals -us- , I still do not understand how the whole systeme works according to your calculations. If you think we should not be here, please tell us, somehow, we’ll manage.

55. Jim D says:

son of mulder, there was a Schmidt et al. paper (the control knob paper) that did this (removed CO2 starting with the current atmosphere) and found that the temperature dropped about 30 degrees, and not all the water vapor goes, but most does, while the albedo goes up due to the extra sea ice that covers a large area when you cool it that much, and this helps with the positive cooling feedback too.

56. David says:

Richard Verney and Steven Wilde, some questions for you. I think the earth conducts a yearly experiment, which, if looked closely at, can provide some underdstanding of the different effects a world with more water has verses a world with less.
This is seen on a hemispheric bases bi-annually, as the earth’s seasonal energy pulse can reveal on a hemispheric scale some of what happens with the ocean / atmosphere on a monthly daily and hourly basis. Sunlight, falling on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. This is close to 90 W/M2 more TSI. Because the Northern Hemisphere has more land which heats easier then water most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the stating that the ATMOSPHERE is 4 degrees higher in July. In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There are also other factors, such as the Northern hemispheres winter increase in albedo exceeds the southern hemisphere’s winter albedo due to the far larger northern hemisphere land mass.
So at perihelion we have a permanent loss to space of ? W/2m SWR, due to increased NH albedo, and, in the SH a temporary loss of SWR to the atmosphere, as at perihelion the SWR is falling on far more ocean, where it is absorbed into the oceans for far longer then if that SWR fell on land. Do these balance (unlikely) or is the earth gaining or losing energy during perihelion???
The TOA seasonal flux should tell us and climate models should accurately predict the observation. Does anyone know if the climate models accurately predict what happens seasonaly on earth?

57. Dave Springer says:

Stephen Wilde says:
December 26, 2011 at 8:09 am
“On a snowball earth most of the water vapor is frozen out of the atmosphere and the ocean is effectively removed from the equation by being covered with permanent sea ice.”
How could that happen when solar shortwave continues to be received into the oceans from equator poleward ? It is that which maintains our liquid oceans and not non condensing GHGs.
——————————————————————————–
I’m not sure how it happens, Stephen, but it does. There is no essential argument over whether the earth has experienced snowball episodes in the past. This is indisputable testimony taken from the geologic column. There is some dispute over whether the freezing was complete and perennial even at the equator but there is little dispute other than that.
What exactly triggers it is probably random catastrophe. Perhaps the equivalent of nuclear winter caused by an asteroid impact or the eruption of a supervolcano in synchronicity with a solar grand minimum and the optimal orbital configuration in the Milankovitch cycle. A perfect storm of some sort I suspect.
While I can’t say exactly how a snowball earth happens to start I cannot deny that the earth has entered into such episodes in the past. Obviously such espisodes are not quite permanent in nature so that raises the question of what snaps the earth out of it. Most people who’ve thought about it reach the same conclusion about volcanic activity.
In the meantime the snowball earth gives us a perfect opportunity to see what happens when the water cycle is, for all practical purposes, shut down by freezing temperatures over most of the globe and with it nothing left to provide a greenhouse effect but non-condensing greenhouse gases. The result is evidently a snowball for many millions of years until something happens to trigger a melt.
We would probably be in a snowball episode right now if it weren’t for solar output being a few percentage points higher than in past ice ages. Models (hate to refer to one but they aren’t all bogus) indicate that whenever continental ice sheets reach past 25 degrees latitude toward the equator the feedback from higher albedo over such a large area makes them unstoppable and results global freeze. Indisputable evidence of glaciation in the tropics well inside the 25 degree tipping point is the evidentiary basis for a complete freeze although as I said there is some controversy over whether the freeze was indeed complete and perennial.

58. cal says:

Richard M says:
December 26, 2011 at 8:07 am
Why do you only consider radiation? The GHG-less atmosphere will receive energy by conduction. Since the atmosphere has low emissivity that heat will not be radiated away quickly as it can be with GHGs. However, once it gets warm enough it will start conducting heat back to the surface. Where is this equilibrium point? Is it 255 K and if so, why?
I’ll admit I don’t have any training in this area so maybe there is a simple explanation. I’ve never seen it though and can’t help feeling this could a positive feedback situation where the Earth would heat up.
The conductivity of air is 0.024 W/mK. That means that a cube of air with a side of length 1metre and a temperature difference of 1 degree C from top to bottom would dissipate 0.024 watts. Now the adiabatic lapse rate is 6C per kilometre so the actual temperature difference across a cube of air at the earth’s surface would be only 6 thousandths of a degree. So the actual loss by conduction is only about 120 microwatts. This is miniscule compared with radiation. Convection and evaporation are the next largest with even these being many thousand times more important than conduction. In addition you mke the statement “when it gets warm enough”. However there is no mechanism whereby the air temperature could exceed the surface temperature and so “start conducting back to the surface”

59. Bryan says:

Joel Shore
Would like to fool you by saying that the Earth without greenhouse gases would still have the same albedo(i.e. 0.3)
Yet he advances no evidence to support this unphysical assumption!
This is the fiction that supports the figure of 33K as the greenhouse effect value
Using figures from an IPCC approved source such as the KT97 energy distribution diagram we can show this figure is nonsense.
The Earth without the greenhouse gases and clouds.
Examine the diagram KT97
Of the fraction 4 X 77W/m2 most would fail to be reflected by the now non existent clouds.
Of the fraction 4 X 67W/m2 absorbed by atmosphere all would pass through since there is now no absorption.
This indicates that almost all of the 1368W/m2 will reach the surface.
KT work out for us that 30W/m2 of 198W/m2 is reflected by surface giving an albedo of 0.15.
However since much more IR is now reaching the surface the figure of 0.15 is too high and a figure of 0.1 or less might be more realistic.
It appears (using the IPCC figures from KT 97) that the effect is now 16K rather than 33K

60. Dave Springer says:

Jim D says:
December 26, 2011 at 8:36 am
“son of mulder, there was a Schmidt et al. paper (the control knob paper) that did this (removed CO2 starting with the current atmosphere) and found that the temperature dropped about 30 degrees, and not all the water vapor goes, but most does, while the albedo goes up due to the extra sea ice that covers a large area when you cool it that much, and this helps with the positive cooling feedback too.”
I have no argument with that. The earth is in an ice age right now and it evidently requires nothing more than a slight cyclical shift in orbital alignment which makes northern hemisphere winters warmer and summers cooler to kick off a process at the end of which London and New York are beneath mile of ice.
The thing of it is though that CO2’s greenhouse effect is logarithymic in nature and the first 100ppm does most of the heavy lifting thus an increase from 280ppm to 400ppm will have only a very modest effect and due to other factors at work in heat distribution the effect is felt mostly on land and mostly at higher latitudes. Given that we’re in an ice age right the f now and given that anthropogenic emission helps warm up the joint mostly where the glaciers have their footholds and not where it’s already comfortably warm all year round it’s difficult to escape the conclusion that rising atmospheric CO2 is a good thing.

61. R. Gates says:

This is an interesting theoretical exercise, but one which seems to have little application to anything in the real world. Without greenhouse gases, earth would of course have no oceans and no ice, and simply be a lifeless rock, much like the moon. It really wouldn’t matter what the “average” surface temperature was, as one side of the planet would be boiling hot and one side freezing cold, as there’d be no greenhouse gases or oceans to modulate that “average”. So whatever the average would be, it would swing so wildly across a 24 hour period, that the average would be quite irrelevant. Bottom line is: whatever the “average” would be without greenhouse gases, we enjoy a much warmer planet, and more importantly, one that varies less severely, because of greenhouse gases. Also, as some or apt to do, the discussion of course goes into the relative importance of each type of greenhouse gas, and everyone knows that water vapor is more potent, but that does not indicate their overall effect. Here the discussion must turn to condensing versus non-condensing gases, and the difference between water vapor and CO2. As water vapor will condense with changes in pressure and temperature, the amount varies wildly over the earth’s surface, CO2 on the other hand, remains relatively well mixed, but more importantly, it does not condense out with the temperatures found on earth. Thus, were one to suggest we remove all the CO2 from the atmosphere (in addition to killing all the plant life) we’d also see all the water vapor condense out over time as the earth continued to cool, and we’d return to a snowball earth in fairly short order.

62. Kevin Kilty says:

Alan Statham says:
December 26, 2011 at 4:51 am
… You seem ignorant of many things. You even start by saying you don’t know what is meant by a characteristic emission temperature. I mean, that’s not particularly clear language but anyone with a basic physics education knows what is meant by that. You wrongly claim that you can’t use an average temperature to calculate radiation losses. And you have used nearly 2000 words simply to say that you don’t agree with the albedo commonly used in this simplified calculation.

Most people with a basic physics education can’t remember any of the basic physics….
With regard to the calculation of Earth average temperature, the equating of outgoing IR over a sphere and incoming over a disk is simply a calculation of some temperature without any useful application. What is it we are trying to illustrate? Are we trying to find the actual earth temperature in the absence of a “greenhouse effect”?
Even on a planet that rotates as rapidly as Earth, without a “greenhouse effect” and without a means to transfer heat poleward rapidly, temperature at the subsolar point is going to be quite high relative to the poles. If one assumes the subsolar point radiates all that it receives then T=T(sun)*(R(sun)/R(sun->earth))^(1/2)*(1-A)^(1/4). If Earth’s albedo (Bond’s Albedo) is 0.3, which was used to estimate the “average” surface temperature of 255K, then T=360K in the subsolar region and everywhere else is PDC (pretty damned cold), or as BenAW suggests, that the sunlit hemisphere radiates preferentially, T would be 303K, with the other hemisphere being PDC. These are very different temperatures and distributions.
On a related note, I’ve often wondered why we use average temperature rather than the M-th root of integrated temperature to the M-th power, with 4<M<5, considering the Planck function and filtering effect of the atmosphere. I mean, average temperature really doesn't correspond to anything important, does it?

63. davidmhoffer says:

Dave Springer;
The emissivity of water is dependent on temperature and can be overwhelmed by the coefficient of evaporation.>>>
In physics, particularly as it applies to undestanding climate processes, we are not as concerned about what “can” happen as we are about what does happen in general over a period of time. The following image from ERBE is instructive:
http://eos.atmos.washington.edu/cgi-bin/erbe/disp.pl?olr.ann.d
As can be seen in this image, the outgoing LW radiation from the ocean surfaces over a period of one year in general exceeds that of the land masses by a considerable amount. I will leave it to you to determine if you are wrong due to the reasons I stated, or if you are wrong due to other reasons. Data trumps theory however, and your theory appears to be incorrect.
Dave Springer;
Aside from those facts of physics which you and Monckton are evidently unaware of is the fact that water is totally opaque to long wave infrared and its emissivity is limited to a surface layer just a few microns deep. The solar heated water below that depth must first be somehow transported to the surface before it can emit any radiation. Then when it reaches the surface it must compete with the evaporation coefficient.>>>
As I explained earlier, still water is opaque to IR, turbulent water is not. As for mechanisms to promote mixing, please note that rain is pretty much a daily occurance over the open ocean and the rain is in general not only a source of turbulence, but also causes mixing because it is cooler than the ocean itself. As for the warmer water below, you are neglecting conduction. If the radiance emitted was overwhelmed by evaporation, then the ERBE data would not show such high amounts of outbound LW.
Dave Springer;
The net result is reflected in many ocean heat budget studies. Approximately 70% of ocean heat loss is evaporative, 25% emissive, and 5% conductive. These are the facts. You may deny them but you cannot change them and neither can Lord Monckton.>>>
I do not deny those facts. Your contention was that escape of LW to space is negligible from the ocean surface. I’ve demonstrated that this is incorrect.
Dave Springer;
With all due respect to your expertise as an IT salesman>>>
A rather snide remark coming from someone who spent most of his career working in the computer industry designing hardware for gaming consoles and desk top computers. BTW, you may want to consider that what I did in my career for one period of time does not apply to my entire work experience, nor is it indicative of my educational background.

64. Stephen Wilde says:

Dave,
I’m not convinced that there has ever been a snowball Earth. As I understand it there has always been an ice fee area either side of the equator.
So all we have is an extreme form of ice age with the ice moving down across more of each hemisphere than is usual.
Over the 4 billion year history of the Earth I am unsurprised that such things happened from time to time. The most likely cause would be large changes in the level of solar activity for whatever reason with a very negative polar oscillation spreading polar air equatorward and cool oceans failing to resist. Your ‘perfect storm’ suggestions are plausible too.
So the water cycle never did stop despite vast areas of ice cover and in due course the water cycle regained its power once more.

65. gnomish says:

to springer, wilde and the author – in this vast psychodrama that was never about science, you guys have found some. and to think the entire circus could have been avoided if you’d had a say about it…lol
i think you have the correct view on how the planet is heated and refrigerated.
arthur clarke once wondered why it was called ‘earth’ when it was mostly water…
i like the meme ‘hot water bottle effect’. it might be supplemented with the ‘sauna effect’.
the greenhouse meme is precisely as useful as the notion of phlogiston – it is a useful indicator of how well a person’s mind is navigating the catastrophist diorama.

66. Stephen Wilde says:

“we’d also see all the water vapor condense out over time as the earth continued to cool, and we’d return to a snowball earth in fairly short order.”
How could that ever happen with solar shortwave still entering the equatorial oceans and raising vast quantities of water vapour with or without the aid of non condensing GHGs?

67. Paul Marko says:

Dave Springer notes: “Sunlight penetrates the ocean to some 100 meters where it is absorbed along the way by impurities.”
Why can you see the mid-ocean ridges thousands of feet below the surface as viewed from space?

68. Lars P. says:

Dave Springer says:
December 26, 2011 at 7:45 am
“What you get is a snowball earth.”
Only in a flat world with 1/4 of solar radiation.
I would be very interested to see calculation that leads to snowball earth in a rotating sphere-world with 1365 W/m2 not a 1/4 of it, including day & night. This would not let water freeze in the tropics – no way, sorry.
With ocean circulation this expands north and south so snowball earth needs additional conditions to happen, cannot be caused by missing carbon dioxide from the atmosphere.
R. Gates says:
December 26, 2011 at 9:03 am
“one side of the planet would be boiling hot and one side freezing cold, as there’d be no greenhouse gases or oceans to modulate that “average””
The earth is a sphere that is spinning no way to have one side boiling hot and one side freezing cold – what you say does not make sense R. Gates, sorry.

69. malcolm says:

Bomber_the_Cat says:
December 26, 2011 at 8:35 am
David, what you should be aware of is that NO2 is not nitrogen, it is Nitrous Oxide, which is a greenhouse gas. The graphic for Oxygen is actually for Oxygen and Ozone shown on the same plot, for some reason. The little blip in the Long Wave Infrared (LWIR) is caused exclusively by the Ozone; Oxygen does not absorb nor emit significantly in the LWIR, as the graphic shows. Neither does Nitrogen, which isn’t shown at all in the graphic.
I feel a spasm of chemistry pedantry coming on. So:
Nitrous oxide is N2O, laughing gas, as used in whipped cream aerosols, and not yet banned, even though it’s a greenhouse gas. NO2 is nitrogen dioxide.
(NO, on the other hand, is Nitric oxide, responsible for the vasodilation effect of such useful drugs as trinitroglycerine and sildenafil)
Sorry. Back to the discussion.

70. Stephen Wilde says:

Harry Dale Huffman says:
December 26, 2011 at 9:16 am
Harry, the problem I have with your work is that you base the whole thing on the pressure related (adiabatic) lapse rate which is of course the same for any planet such that all other things being equal the ‘temperature’ (I prefer energy content) of the planet will be related directly to atmospheric pressure at the surface and distance from the sun.
The trouble is that the actual (so called ‘environmental’) lapse rate is very different from planet to planet so Venus CANNOT be directly compared to Earth.
On Earth we have a fully functioning and highly effective water cycle which Venus does not have so the outturn on each planet is NOT comparable.
Earth has a powerful Hot Water Bottle Effect and a very weak Greenhouse Effect. Venus has a very powerful Greenhouse Effect and NO Hot Water Bottle Efffect.
Chalk and cheese come to mind.

71. Mydogsgotnonose says:

The albedo of the earth without clouds and ice is 0.07. This gives an effective no-atmosphere temperature of 273K or ~0°C. However, have a non-GHG atmosphere with aerosols and you will have lapse rate cooling. In that case the present GHG warming is ~9K.
So, IPCC climate science overestimates CO2 climate sensitivity by at least a factor of 33/9 = 3.66. But there’s also the -0.7W/m^2 ‘cloud albedo effect’ cooling in AR4. Being a figment of Sagan’s incorrect aerosol physics, it doesn’t exist and could be warming. So reduce CO2 climate sensitivity by a further 44%.
The maximum CO2 climate sensitivity is 0.45. in reality, it’s probably much smaller because the IR physics is wrong as well, not taking account of self-absorption at IR band saturation.
The other thing wrong with the IPCC farrago is no ‘back radiation’, it’s really Prevost Exchange Energy so can’t do thermodynamic work. All in all, this is probably the biggest scientific fraud in History: time it was ended.

72. gnomish says:

@ wilde:
good job. now if you only had willis’ awesome jimmy.buffetness… nah- you’ll do without… lol

73. R. Gates says:

To those discussing how the earth may have broke out of the snowball epoch, this fairly current research is most germain, and not at all irrelevant to the topic of greenhouse gases and earth’s average temperature:

74. Dave Springer says:

Thierry says:
December 26, 2011 at 7:22 am
“Everything concerning the fictitious 33°C greenhouse gases effect has been definitively explained and rebutted by Gerlich & Tscheuschner (2009) in “Falsi fication Of The Atmospheric CO2 Greenhouse E ffects Within The Frame Of Physics” (http://arxiv.org/abs/0707.1161)
This value comes from a wrong use of the Stefan-Boltzmann law (flux imposing local temperature, which is wrong) involving an over simplified planet model (no atmosphere, infinite conductivity, no rotation). The real temperature would be -129°C.
Therefore, this -18°C has no meaning at all.”
On must explain why temperature probes buried a meter or more deep in lunar regolith at 45 degrees latitude record a year-round constant temperature of -23C while the same probe buried a meter deep at the latitude on the earth records a year-round temperature of 11C.
Something is making the earth ~33C warmer than the moon. This is not only a theoretical calculation but an observation as well since the early 1970’s when two separate Apollo missions buried temperature probes in lunar regolith from surface to 3 meters deep that radioed the temperatures back to NASA for the next several years. Gerlich et al are certainly entitled to an opinion on why the earth is so much warmer than the moon but they aren’t entitled to their own facts where their “fact” is that the earth isn’t really warmer than an approximate black body lacking a dynamic ocean/atmosphere at the same distance from the sun.

75. Kevin Kilty says:

Dave Springer says:
December 26, 2011 at 7:24 am
Leonard Weinstein says:
December 26, 2011 at 6:32 am
… .

I agree that the oceans are black at normal incidence, but at lower angles water reflects a large fraction of SW radiation. This is just a part of the complications involved in this whole debate–which ends up with numbers that have little utility as far as I can see. The temperature of 255K depends on such things as whole-earth equal intensity of outgoing LW radiation, but that is far from the actual case.
If one would like an “average” radiation temperature, then what I suggested earlier,

…the M-th root of integrated temperature to the M-th power, with 4<M<5, considering the Planck function and filtering effect of the atmosphere….

makes more sense.

Joel Shore says:
December 26, 2011 at 7:40 am …
BenAW: Your notions don’t even obey energy conservation. You are running yourself in circles getting confused about averages. I suggest you work in terms of total power in and out and then you will hopefully understand why your argument is nonsense.

In what manner do his notions violate energy conservation? There are many ways to sum or integrate to a given number. For instance, the Moon, which many people have used as example here, the subsolar point is quite hot, and the rest is pretty damned cold. The average temperature of 250K achieves energy conservation, but is not reflective of the actual distribution, which also obeys energy conservation, at all.

Thierry says:
December 26, 2011 at 7:22 am
Everything concerning the fictitious 33°C greenhouse gases effect has been definitively explained and rebutted by Gerlich & Tscheuschner (2009) in “Falsi fication Of The Atmospheric CO2 Greenhouse E ffects Within The Frame Of Physics” (http://arxiv.org/abs/0707.1161)
This value comes from a wrong use of the Stefan-Boltzmann law (flux imposing local temperature, which is wrong) involving an over simplified planet model (no atmosphere, infinite conductivity, no rotation). The real temperature would be -129°C.
Therefore, this -18°C has no meaning at all.
I do not understand why there is still a debate here.

I agree generally with you, but occasionally Gerlich & Tscheuschner frame points in such a narrow manner that they fall prey to the same flaws they critique.

76. ferd berple says:

DirkH says:
December 26, 2011 at 7:13 am
So O2 and N2 will keep their energy forever even if in the vacuum of space, as they can’t radiate anything?
Bingo!

77. XCapglider says:

Quote from the article: “However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.
Maybe this website sheds somewhat more light on this topic:
It is really not hard to understand, IF you think the right way. If there was not something like the ‘greenhouse effect’ we all would be dead, unless we all could live at 255K. Much confusion arises from the term ‘Greenhouse effect’, which sadly is a misnomer.

78. John M. Chenosky, PE says:

SLAY THE SKY DRAGON——END OF ARGUMENT!!!

79. Stephen Wilde says:

David says:
December 26, 2011 at 8:50 am
So at perihelion we have a permanent loss to space of ? W/2m SWR, due to increased NH albedo, and, in the SH a temporary loss of SWR to the atmosphere, as at perihelion the SWR is falling on far more ocean, where it is absorbed into the oceans for far longer then if that SWR fell on land. Do these balance (unlikely) or is the earth gaining or losing energy during perihelion???
The TOA seasonal flux should tell us and climate models should accurately predict the observation. Does anyone know if the climate models accurately predict what happens seasonaly on earth?”
i) and ii) I don’t think they ever balance precisely. There is always a net gain or net loss and the system adjusts accordingly to try to maintain balance. Always a negative system response to ANY forcing whether that be solar, oceanic, volcanic, aerosols or CO2.
iii) The climate models do include seasonal variations. Where they collapse is in identifying and quantifying trends beyond normal seasonal variability.

80. Vince Causey says:

Alan Stratham,
you said: “You wrongly claim that you can’t use an average temperature to calculate radiation losses.”
I must disagree. Whenever you have a non linear relationship between two variables, you have to be very careful about averaging the independent variables.
Suppose for sake of argument, you have half the surface area with temperature T1, and half with temperature T2. What would be the radiative flux from each area? Ignoring Stefan-Boltzmann’s constant, you would have F1 = T1^4 and F2 = T2^4.
Total flux would be Area1 x T1^4 plus Area2 x T2^4. Since each area is half of the total area A, total radiative flux = 1/2A(T1^4 + T2^4).
Now let’s find the average temperature Tav. Tav = 1/2(T1 + T2).
Flux density = (Tav)^4 = 1/16(T1 + T2)^4.
Multiplying this by the total area A, total radiative flux = 1/16A(T1 + T2)^4.
You can expand that if you like, but I think you’ll find it is not equal to 1/2A(T1^4 + T2^4).
Based on the mathematics, I suggest that the author is correct in claiming that you can’t use average temperature to calculate radiation loss.

81. Stephen Wilde says:

“those discussing how the earth may have broke out of the snowball epoch, ”
Where is the evidence that it actually happened ?
That is, with NO open water at the equator.

82. Dave Springer says:

R. Gates says:
December 26, 2011 at 9:54 am
“To those discussing how the earth may have broke out of the snowball epoch, this fairly current research is most germain, and not at all irrelevant to the topic of greenhouse gases and earth’s average temperature:”
Thanks Gates. You did good. That research confirms my own conclusion which I arrived at all by my lonesome, about the effect of volcanic ash on deglaciation.
I would also remind you that James Hansen himself, earlier in his career, pointed out that accumulation of black soot from dirty smokestacks and unfiltered old diesel engines travels up to a few thousand kilometers from its source then comes to rest on NH permanent snow cover accelerating the melt of sea ice and glaciers as it gets darker and darker every single year. In fact back when he wasn’t so afraid of the facts getting in the way of his agenda he assigned black soot up to half the blame for global warming.
This was, of course, before it was determined that the United States needed to be the scapegoat for global warming because, you see, since the Clean Air Act of 1964 the U.S. has not been emitting much black soot. But Europe with its abiding love for antique diesels in its transportation fleet (ya gotta love how those old diesel engines just keep going and going and going like the energizer bunny) and poor countries who cling to slash & burn agriculture and cook with anything combustable, continue to emit copious amounts of black soot which then goes on to melt things like the Himlayan Glacier. But we can’t go blaming Europe and poor undeveloped countries for these things… that’s just not politically correct. So the CO2 bogeyman and attendant water vapor amplification were invented so the US could be made to appear to be the bad guy.
Isn’t that just precious?

83. Vince Causey says:

John Brookes,
“OK, that seems like a pretty silly argument. Take a look at Mars. I think Mars would be pretty close to earth without ghgs. No oceans, no vegetation, no ice no animals, just lots of red dirt. But Mars does not have an albedo of 0, it has an albedo of 0.25 – not too disimilar to that of Earth, and certainly a value which will get you closer to 255K than 279K..”
Mars is nothing like Earth. Earth is covered by 70% water which lowers albedo and would lead to an equilibrium temperature closer to 279k than 255k.

84. wayne says:

Lars P., thoroughly enjoyed reading you recent comment in tips&notes. You should re-post it here, not link, for general consideration by the many who are digging to bury the pseudo-science surrounding this very subject. I found yours very relevant.

85. Leonard Weinstein says:

David Springer,
You and most of the other responders do not “get it”. The ASSUMPTION WAS MADE by Monckton that if there were no greenhouse gases, AND THAT THE ALBEDO WAS 0.3, what the result would be. This is not the case for real oceans, since with oceans there always is present clouds and greenhouse gases. The model Monckton described is an idealized one that can’t have real oceans, or you could not reach the albedo stated. You are confusing the assumptions stated with the actual surface, which violates the simplified model stated. When doing simplifying analysis, you have to pay attention to what is said.

86. ferd berple says:

Does CO2 increase the temperature inside real greenhouses due to radiation?
Isn’t in point of fact the greenhouse effect in greenhouses a result of convection?
Doesn’t CO2 in point of fact play little or no role in the average temperature of real greenhouse?
Doesn’t the EPA and IPCC claim that atmospheric CO2 warms due to the exact same effect that warms real greenhouses?
Why if CO2 has a warming effect, is climate change not called global warming?
So, where is the experimental evidence that an atmosphere in which convection is not restricted by some sort of artificial cover will in fact behave like a real greenhouse in which convection is artificially restricted?
Where is the controlled experiment that demonstrates that a doubling of a trace amount of CO2 will in point of fact significantly increase average temperatures over a 24 hour period in any container? Not simply when the sun is shining, but also when the sun is not shining, to remove the thermal difference between CO2 and air.
If such an experiment has been done under controlled, double blind conditions, where are the results?

87. Robert of Ottawa says:

Has anyone done a vaguely realistic model of the earth representative of land/water spacial distribution and daily rotation a) without an atmosphere b) with a nitrogen atmsophere c) with water vapor and independantly variable albedo – given the contraversial nature of cloud.
That would be a better to start a model on. Starting a model based upon a flat static disc cannot evolve beyond its innacurate assumptions; rather like assuming a spherical horse.

88. Alan D McIntire says:

More on a greenhouse gassless atmosphere. I’m reiterating what other’s have pointed out earlier:
The atmosphere would pick up heat by conduction with the earth’s surface.
The heat would drop off the higher you get in the atmosphere due to conservation of energy.
When the planet reached equilibrium, the energy lost by conduction from earth’s surface to the atmosphere would equal the net energy picked up by conduction from earth’s atmosphere.
Net radiation from the sun to earth’s surface would equal net radiation from earth’s surface back to space.
In the real universe there’s no such thing as a completely non-greenhouse atmosphere- all gases radiate in SOME frequencies. Depending on the densities of the different gases making up an atmosphere, these different gases would have different “effective radiation heights”, in the atmosphere, giving differernt greenhouse effects.
In the real universe there’s no such thing as a perfect black body.. All surface temperatures are
hotter than a black body giving off the same amount of radiation.
In the real universe, there’d be no free oxygen in an atmosphere without life, There’d be no life on a planet without water, so the argument about the effective temperature without greenhouse gases would be moot..

89. R. Gates says:

Lars P. says:
December 26, 2011 at 9:39 am
R. Gates says:
December 26, 2011 at 9:03 am
“one side of the planet would be boiling hot and one side freezing cold, as there’d be no greenhouse gases or oceans to modulate that “average””
The earth is a sphere that is spinning no way to have one side boiling hot and one side freezing cold – what you say does not make sense R. Gates, sorry.
______
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does. The essential point is that the oceans and greenhouse atmosphere prevent a strong diurnal temperature variation. The oceans provide the primary heat “sink” of the planet and the greenhouse atmosphere provides a transport mechanism for the heat.

90. mkelly says:

There are two things wrong with the normal formula used. 1. The idea of an average temperature using the 255 W/m^2. It can be shown that much of the NH up To 45N in the summer gets near 700 W/m^2. The troplcs near 1000 all year. 2. IR in the range CO2 emits (15 micro) back to the earth has no ability to do work so cannot heat anything.
Further there are no examples of gases doing anything other than dissipating heat. Our lives depend on this fact.
Sent from my new Christmas gift tablet. Please forgive spelling errors etc. as I get used to it.

91. Dave Springer says:

Stephen Wilde says:
December 26, 2011 at 9:28 am
“I’m not convinced that there has ever been a snowball Earth. As I understand it there has always been an ice fee area either side of the equator.”
I’m not convinced either but the fact remains that it was close enough so that the distinction between slushball and snowball is academic for what we’re concerned about. Ice ages happen. We must somehow incorporate that into the hot water bottle theory. The way I do that is by keeping in mind that the hot water bottle only covers 70% of the earth’s surface.
Also keep in mind that as ice grows that 70% shrinks. The ocean’s shorelines retreat as more and more surface water is locked up in ice. That’s yet another positive feedback that works to take the ocean further and further out of the equation.
“So all we have is an extreme form of ice age with the ice moving down across more of each hemisphere than is usual.”
Glaciers at sea level in the tropics doesn’t seem fairly characterized by “more than usual”.
“Over the 4 billion year history of the Earth I am unsurprised that such things happened from time to time. The most likely cause would be large changes in the level of solar activity”
There is zero evidence for large changes in the level of solar activity. There are a zillion class M stars at all stages of their life cycles for us to look at and relatively speaking a class-M star like ours is before it becomes a red giant in its cups is one of the most constant dependable things in the universe. Large changes like that would leave fingerprints in the geologic column as well and they aren’t there. Unless there is strong evidence otherwise we must assume the sun is no different than other star like it and its constancy is dependable to a fault.
“for whatever reason with a very negative polar oscillation spreading polar air equatorward and cool oceans failing to resist. Your ‘perfect storm’ suggestions are plausible too.”
You’re just guessing about vague somethings. I cited events that are know to happen with frequencies that are known on average as well. Milankovitch cycles are calculable. Supervolcanoes happen every several million years. Asteroid strikes likewise. The latter two are not predictable with current knowledge as individual events but over geologic spans of time a predictable number of events occurs. It’s almost inevitable that a perfect storm of such events happens now and then. In fact I’m wondering right now how many Pinatubos it would take in succession to end the Holocene Interglacial. The Milankovitch alignment is just getting more favorable for glaciation over the next few thousand years and it appears the so-called Modern Maximum in solar magnetic activity has closed out and if Svensmark is right that’s going to help cool things off. Maybe one more Pinatubo is all it would take right now. I’m just sayin’ we don’t know what causes ice ages with certainty but we are certain they do happen.
So the water cycle never did stop despite vast areas of ice cover and in due course the water cycle regained its power once more.

92. Louis says:

R. Gates says:
“…Thus, were one to suggest we remove all the CO2 from the atmosphere (in addition to killing all the plant life) we’d also see all the water vapor condense out over time as the earth continued to cool, and we’d return to a snowball earth in fairly short order.”
Thanks for pointing out how ridiculous and dangerous it is for the U.S. EPA to treat CO2 as a pollutant and for the U.S Supreme Court to uphold their argument. CO2 sustains life on this planet. The idea that it is a “pollutant” and thus doesn’t belong in our atmosphere at all is suicidal.

93. Vince Causey says:

R. Gates,
Thanks for the link to the snowball earth paper. The main points to summarise are:
1) CO2 levels of even 0.2 bar (200,000 ppm) are not sufficient to end glaciation.
2) Dark particles from super volcanoes would be needed in addition to 0.2 bar of CO2.
3) Models diverge at high levels of CO2.
4) The models themselves already contain the assumption of high sensitivity to CO2.
5) Nobody knows what actually ended snowball earth.

94. Luther Wu says:

R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does.
_______________________
What?

95. Myrrh says:

Dave Springer says:
December 26, 2011 at 6:06 am
“If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.”
Yes of course it is. It’s the earth without an atmosphere or ocean and an albedo of zero. The average temperature of the moon is 250K. It has an albedo of 0.15.
=============================
Sigh, you’re saying it like this because the AGW meme is that ‘greenhouse gases’ excludes oxygen and nitrogen. Another discussion with everyone talking at cross purposes..
Firstly, before AGWScience Fiction began playing around with words and changing definitions, the Earth’s Greenhouse WAS THE WHOLE OF THE EARTH’S ATMOSPHERE, therefore, all the gases within it are greenhouse gases, and, that means some 96% oxygen and nitrogen, around 4% water, around 1% argon and the rest trace…
Earth without an atmosphere, without its real greenhouse, which is real greenhouse with windows.., the temp is around -18°C.
With the real Greenhouse, the whole of the Earth’s atmosphere including oxygen and nitrogen, the temp is 15°C.
So, the usual +plus 33°C of the AGWSF meme, is NOT from warmth added by AGWGreenhouse gases, but a straight comparison between Earth with and without its RealGreenhouseAtmosphere which includes oxygen and nitrogen as greenhouses gases.
How does that temp arrive?
Without the Water Cycle, but with the rest of the atmosphere in place, the Earth would be 67°C.
In other words, the greenhouse gas Water vapour cools the Earth by 52°C to get the temp back down to 15°C.
I do hope you can all see what the AGWSF memes have done here. Before going into arguments about albedos and comparisons with the Moon, first get out of the muddle created by AGWSF messing around with names and definitions.

96. Dave Springer says:

ferd berple says:
December 26, 2011 at 10:13 am
“Doesn’t the EPA and IPCC claim that atmospheric CO2 warms due to the exact same effect that warms real greenhouses?”
No. Nobody is claiming that there’s a sheet of glass blocking convection of heat away from the surface.

97. Important question here is:
What drives ‘global’ temperatures (Land and Land & Oceans) ?
There are two major drivers:
1. The AMO with principal the ‘9 year’ oscillation, which is not related to the sunspot cycle (but not totally immune to it) , for more details see:
http://www.vukcevic.talktalk.net/theAMO-NAO.htm
for the ‘9 year’ and importance of the AMO to global temps.
2. the Solar magnetic (Hale) cycle, not the sunspot one.
On this one I am writing more (on lines of the above linked as ‘theAMO-NAO’)
Hope everyone had a good Xmas

98. Stephen Wilde says:

Actually, Dave , I should have said ‘solar influence’ rather than ‘solar activity’ which would have brought in astronomical changes such as the Milankovitch cycles.
I think we are all guessing at vague somethings at this distance in time. The snowball Earth idea is a concept rather than a fact.
The continents were in very different positions back then so that there are many bits of continental land mass bearing signs of glaciation that are now near the equator.

99. R. Gates says:

Luther Wu says:
December 26, 2011 at 10:41 am
R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does.
_______________________
What?
______
A shocking revelation isn’t it. At the moon’s equator on the sunlit side the temperature is around 120C whereas on the opposite dark side, the temperature is about -230C. Amazing what an ocean and greenhouse atmosphere can do!

100. Reed Coray says:

John Brookes says [December 26, 2011 at 4:18 am]: “Anyway, well done for writing this, but make a bit more effort to think things through.”
John, I have given more thought to the problem. I don’t claim to know either (a) how to calculate or (b) what the temperature of the Earth’s surface would be in the absence of greenhouse gases. I believe such a computation is extremely complex and is might be meaningless except as an exercise. What I argue is that the 33 K temperature difference arrived at by (a) measuring an average Earth surface temperature and (b) the temperature of the Earth based on a measured absorptivity of 0.7 (albedo of 0.3, which is in large part due to clouds) and an emissivity of 1 is not relevant to a computation of the Earth with and without greenhouse gases. I, like others who have commented on my post, believe (and I admit I may be wrong) CAGW proponents want the temperature difference associated with greenhouse gases to be as large as possible because to do so increases the importance of greenhouse gases.
Richard M says [December 26, 2011 at 4:37 am]: “I’ve mentioned before that I thought this computation was missing from the literature. I believe you’ve made a great start. However, there is more to the question than just albedo.”
I agree, there is much more to the problem. I’ll outline the problem I would like to see someone attack. Assume you have a spherical solid (non-liquid) body that exists in the vacuum of space isolated from all other objects. Assume the spherical body has a radius, “R”, and uniformly-distributed mass equal to the mass of the Earth–this assumption will approximate the gravitational effects of the Earth at the spherical body’s surface. Assume the spherical body is NOT rotating–I know this does not represent the Earth, but I want to create spherical symmetry which in part requires the removal of all non-radial forces (primarily coriolis). Assume the surface of the sphere radiates like a graybody (blackbody with emissivity greater than 0 and less than 1) with the same emissivity, “e”, over the entire surface. Assume at the center of the sphere a source of energy exists (e.g., radioactive decay) at a constant but known energy rate “W” Watts. For these conditions, I believe it is relatively simple to compute the spherical body’s surface temperature, “T” when radiation-rate equilibrium is reached–i.e., when the power radiated to space from the spherical body’s surface is equal to the power generated internal to the sphere. Specifically, I argue that by symmetry the surface temperature of the spherical body will be everywhere the same and equal to
T = [W/(sigma*e*A)]^0.25
where “sigma” is the Stefan-Boltzmann constant, and “A = 4*pi*R^2” is the surface area of the sphere.
Okay, so much for the easy problem. Now, to the spherical body characterized above (a) uniformly over the surface of the sphere add oxygen and nitrogen where the masses of the added oxygen and added nitrogen are equal, respectively, to the mass of the oxygen and nitrogen in the Earth’s atmosphere, and (b) in addition to the atmosphere of (a), add elements/compounds of greenhouse gases uniformly over the sphere’s surface. I don’t care what the greenhouse gases are, but it would be nice to consider only adding CO2 as well as adding both CO2 and water vapor. If the latter (both CO2 and water vapor), let the masses of the added gases be equal to the “average” masses of CO2 and water vapor in the Earth’s atmosphere. Then for both (a) and (b) compute in radiation-rate equilibrium (1) the surface temperature of the sphere and (2) the temperature as a function of altitude above the sphere’s surface.
davidmhoffer says [December 26, 2011 at 5:01 am]: “Further, I don’t think the 255K number is supposed to represent the surface temperature of the earth without greenhouse gases. It is supposed to represent the temperature of the earth without the greenhouse EFFECT. If that is the case, then the logic presented to arrive at 255K seems quite reasonable.”
Lord Monckton specifically used the words “greenhouse gases” when discussing the 33 K temperature difference. I believe most people when discussing the source of the 33 K temperature difference either directly or indirectly also treat the source to primarily be “greenhouse gases.” If as you say “the 255 K number is supposed to represent the temperature of the earth without the greenhouse EFFECT“, then why mention greenhouse “GASES” at all? If you remove “greenhouse gases” from the “greenhouse effect”, any mention of “greenhouse gases” in a discussion of the 33 K temperature difference becomes irrelevant.

101. wayne says:

R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side?
>>>
The moon… spinning huh, like Venus you must mean, right? R. Gates , that sure is a cheap shot to Lars P. Or, I guess using your logic, that even with a greenhouse gas-less Earth spinning at 1 RPM, it would have the sun lit side boiling hot and the dark side freezing cold. Why don’t you stop quoting from your AGW manual every now and then and just think of what you are saying. Any intelligent person could understand Lars point.

102. Thierry says:

Once again, only Gerlich & Tscheuschner have explained where the 255 K comes from : a wrong calculation (that more scientists use in their models, which is ridiculus) meaning nothing of a non rotating earth without any atmosphere using Stefan-Boltzmann law the other around. The only reason we are not burning or freezing to death on earth is because of our 1 bar atmosphere temperates the climate.They also explain why calculating the real temperature of a rotation earth year along (still without an atmosphere though) is impossible, even with the most powerful computers.

103. R. Gates says:

Short of slight variations in mineral content affecting actual albedo, using the moon’s temperature on the sunlit and then the dark side should give you a pretty good idea of what the “average” temperature would be on earth. The daytime side of earth would be approximately 120C and the nighttime side would fall to a frosty -230C (but of course no frost would form as there would be no water vapor). The moons albedo is about .12, and you could make a reasonable assumption that the earth’s would be somewhere in the region without atmosphere, oceans, or clouds. But of course the earth has a different mineral mix than the moon, perhaps making a bit darker, so a slightly lower albedo is probably in order. But as the above extreme dirurnal temperature extremes the earth would experience indicate– the notion of an average temperature in a world devoid of ocean and atmosphere becomes meaningless.

104. Neo says:

The Earth does warm itself with heat from the mantle.
The oil man’s rule of thumb is that the ambient temperature goes up about 1 degree C for ever 1000 ft deeper you go down, so obviously the surface is cooler than the ground below it.

105. Agile Aspect says:

It’s trivial to calculate the temperature of the Earth without the conjecture of the “greenhouse effect” using Stefan-Boltzmann Law and the equation of state of gases in the atmosphere.
Simply calculate the temperature using the Sun’s flux, locate the altitude of the temperature playing with the albedo, then use the equation of state of the atmosphere to reach the surface.
The temperature will be to high since the calculation assumes a static atmosphere, i.e., no rotation, no slipping at the boundary and no wind.
The Stefan-Boltzmann Law can not be applied to gases since they do not have a continuous frequency distribution.
In short, your calculation is nonsense since you ignored gravity and what you refer to as the “greenhouse” effect is nothing more than the heat capacity of gases.

106. Mydogsgotnonose says:

Dave Springer says:December 26, 2011 at 9:56 am
‘On must explain why temperature probes buried a meter or more deep in lunar regolith at 45 degrees latitude record a year-round constant temperature of -23C while the same probe buried a meter deep at the latitude on the earth records a year-round temperature of 11C.
Something is making the earth ~33C warmer than the moon’

Yes, correct. It’s the combination of GHG warming and lapse rate heating. The IPCC fails to take into account the lapse rate heating.

107. LaMaisonDieu says:

Next problem of the Kindergarten-Model:
The “surface temperature” of 288K which is measured in a Stevenson screen in ABSENCE of radiation and in 2m height is a “convective” air temperature.
Thus one can not compare an air temperature of 2 m height in absence of radiation with a radiation temperature in 0 m height.

108. Richard M says:

Thanks cal for your response. However, the conduction of air itself is not what I was referring to. I was only referring to conduction from the surface to the air right above it. After that we would still have convection and the Coriolis effect to move the heat around. And, don’t forget time. Even a small effect can grow large over millions of years (unless something else counters it). I think the response by Alan D McIntire covers more closely what I am wondering about.
In addition, I believe the point has been made that O2 and N2 still radiate a small amount. Hence, we would still have some cooling at higher altitudes and a lapse rate.
I believe Myrrh attempted to answer my question by asserting a temperature of 69C. Ok fine. How is that determined? I’d really like to understand in more detail how such a system would behave. I really think we’d learn a lot if we could understand this GHG-less Earth.
Maybe the effect would be minimal and the answer is 255.x or maybe it would be large as Myrrh stated. I still don’t feel comfortable with the various responses.

109. Neo says:

Away from tectonic plate boundaries, the geothermal gradient is 22.1°C per km of depth (1°F per 70 feet of depth) in most of the world.

110. Luther Wu says:

R. Gates says:
December 26, 2011 at 11:00 am
A shocking revelation isn’t it.
____________________
What’s shocking is that you would try to sidetrack this thread with nonsense.

111. richard verney says:
December 26, 2011 at 3:48 am

. Some weighting should be given to take account that over geological time the average surface temperature of the earth is far less than we are privileged to enjoy today.

Only if you don’t count the temps in deep geological time, which were much longer at much higher temps, ~24°C. The “Hot House” state is more common than the “Ice Box”.

112. It would be advisable for the “carbon fanatics” to start walking naked and eating stones. Cotton is made by plants from CO2, Water and Sunlight,and making sugar, floor, food,etc too, in a process you ignore called photosynthesis, …and you know what? plants breath the CO2 you exhale, so you will have to stop breathing at all, …but rejoice! so you will cooperate with the world sustainability. Let Gaia take care of you once and for all!

113. Bill Treuren says:

The concept of temperature is only possible with an atmosphere.
This is a very good discussion but it’s clear that the very complexity of the human generated models from all sides is dwarfed by natures complexity.
Albedo changes with latitude and time of year when you consider the incident angle of incoming solar radiation, however outgoing radition must vary on a different schedule. I know the models are very complex and they have big computers, but nature would Demand many many thousands of variables which are full of discontinuous functions.
Good stuff and I would agree with the view that sensitivity is over estimated.
Bill

114. jorgekafkazar says:

Too much eggnog. Far too many half-truths I’ll just comment on one:
Dave Springer says: “…The ocean has an albedo of effectively 0 and it covers 70% of the surface.”
Although these statements are independently true, juxtaposing them creates a false impression. The zenith angle of the sun varies across the globe, and reflectance (albedo) varies with zenith angle. To put it in simple terms for the noggists, have you ever stood on a cliff above the ocean and seen the sun reflecting off the water? It’s often too bright to look at, corresponding to an albedo close to 1.00. Albedo of the ocean is a function of zenith angle, clouds, temperature, waves, wind, ice, tide, flotsam, foam, and plankton (!). The range of ocean albedo overlaps that of ice.
Dave Springer also says: “.No one knows exactly what the earth’s average albedo really is. Climate models use values which differ by as much as 0.07 from one to another.”
Correct. You can see why variability in ocean albedo makes computation of average albedo problematic. It’s constantly changing. Worse yet, the ocean absorptivity-emissivity ratio is not 1.00, just another example of how the complexity of the earth will forestall development of anything close to a valid climate model within our lifetimes.

115. michael hammer says:

Why make the analysis so needlessly long and complex. Surely its much simpler to say, the solar energy in space near the earth is 1367 watts/sqM. The earth presents a disc to the sun while the actual surface is that of a sphere. The ratio of areas is pi *R^2/4*pi*R^2 or 1:4 hence the energy averaged over the surface of the Earth is 1367/4 = 342 watts/sqM. Now the Earth has an albedo of 0.3 which means 30% of this energy is reflected back out to space so the Earth only absorbs 243 watts/sqM. For a black body to emit 243 watts/sqM it must have a temperature of 255K. However without any green hosue gases in the atmosphere there would be no water vapour, hence no clouds and clouds are responsible for most of Earth’s albedo. If without clouds the albedo dropped to say 0 Earth woudl be receiving 342 watts/sqM and a black body radiating 342 watts/sqM would be at a temperature of 279K. In fact the Earth would still have an albedo greater than 0 but so the temperature would be somewhat less than 279K but non the less your argument is very sound. However it does not go far enough!!!!
What also needs to be recognised is that the green house warming of water vapour is logarithmic like all green house gases while the cloud cooling is close to linear. At very low concentrations of water vapour green hosue warming dominates greatly and the presence of water vapour warms the Earth, however as water vapour levels rise the INCREMENTAL warming impact of water vapour falls but as total concentration rises the impact of cloud cooling becomes far more significant and eventually dominates INCREMENTALLY over the warming impact. Water vapour content is determined by temperature (exponential relationship) so put another way, at very low temperatures water vapour warms the Earth but as the temperature rises the warming is increasingly offset by cloud cooling and eventually cloud cooling dominates so at high temperatures water vapour cooling the Earth,
In short, water vapour is a powerful stabilising influence that sets an operating point for Earth’s temperature and maintains it by strong negative feedback.
How strong; consider water vapour content of the atmosphere doubles for each 10C rise in temperature. Doubling water vapour at a first approximation doubles cloud mass (either denser clouds or more of them both of which increase albedo) and we knwo that clouds at present contribute about 80 watts/sqM of cooling. So potentially a 10C rise would incrementally double that for an incremental 80 watts/sqM of cooling. The green house warming due to water vapour is generally taken to be about 80-100 watts/sqM but this is the result of about 10 doublings so the incremental warming per doubling is about 8-10 watts/sqM. Thus the net impact of doubling water vapour is +10 – 80 or 70 watts/sqM of cooling. Or about 7 watts.sqM per degree C. Compare that with the claimed impact of doubling CO2 of 3 watts/sqM. A rise of 3/7 of a degree C would cancel it out and yes a net impact of 0.4-0.5C per doubling of CO2 sounds about right.
Clearly the anaysls above is very crude and approximate and could not be relied on for a good estimate but non the less it gives an idea of just how strong the negative feedeback from water vapour is.

116. jorgekafkazar said @ December 26, 2011 at 1:00 pm
“You can see why variability in ocean albedo makes computation of average albedo problematic. It’s constantly changing. Worse yet, the ocean absorptivity-emissivity ratio is not 1.00, just another example of how the complexity of the earth will forestall development of anything close to a valid climate model within our lifetimes.”
The modelers admit to needing computers at least a couple of orders of magnitude greater computational power. Certainly not going to happen in my lifetime. One wonders if they will require orders of magnitude more energy to run…

117. In any complex system, with many variables, the complexity overwhelms simple analysis. Our minds try to extract simple cause & effect, but the system isn’t amenable to such analysis. While a living cell is even more complex, it is a suitable analogy, In the cell we have a complex mix of feedback, adaption and adjustment. While we can examine components of the cell and understand them, we cannot grasp the total complexity of what happens. We CAN describe the overall function of a cell, and we can come up with a simplified model. But we cannot remove any major part without destroying the whole. The complexity of climate involves chemical cycles and the biosphere, not just simple physics and climatology.
We can’t just remove all the green house gases and see what would happen. It makes as much sense as a cell sans mitochondria.
Both life and climate appeared to pre-modern man as evidence of the hand of God.
Maybe there was some truth in that!

118. Jimmy Haigh says:

Don’t forget that the core of the Earth is very hot – millions of degrees acording to the Goreacle….

119. Richard M says:

kiwistonewall says:
December 26, 2011 at 1:24 pm
We can’t just remove all the green house gases and see what would happen. It makes as much sense as a cell sans mitochondria.

That is true if you wanted to do something with that model, but it doesn’t mean you can’t learn something interesting. If we want to understand the temperature impact of the gases called GHGs then it would be nice to understand what might happen if they were removed. Of course we have to be careful, but we might just learn a few things we didn’t know before. If the answer is 69C as Myrrh indicated that should set off alarm bells all over the place since the basic assumption of climate science is 15C.

120. Richard M says:

Oops, sorry … I meant -18C for the last temperature.

121. michael hammer says:

David Springer at 9:56 stated
On must explain why temperature probes buried a meter or more deep in lunar regolith at 45 degrees latitude record a year-round constant temperature of -23C while the same probe buried a meter deep at the latitude on the earth records a year-round temperature of 11C.
The reason is much simpler than any claims of lapse rate or green house warming. The moon in effect rotates once every 28 days relative to the sun (28 times slower than Earth), it has no atmosphere to distribute heat and no water with high thermal mass to store it. Thus the surface has a very short time constant and tracks solar iput. However the relationship between energy input and temperature is a fourth power law which is extremely non linear. Thus at lunar noon the solar input is the full 1367 watts/sqM which yields a temperature of 121C yet during the long lunar night with no energy input the temeprature falls to -233C or just 40K. Burying a probe 3 M down under the surface does not remove this non linearity because the 3M only serves to average out the surface temperature fluctuation not the non linear relationship between energy input and surface temperature. When you average a T^4 relationsip over a large range of T the average will come out too low.

122. Rosco says:

This “effective temperature” nonsense is the biggest scam I have ever seen portrayed as “science”.
The calculation of the “effective temperature” relies on the ratio of the area of a disk to a sphere – one quarter. No matter how the explanationis presented it boils down to a disk irradiated by 957 W/sq m (1367 W/sq m X (1 – 0.3 )) is in radiative balance with a sphere radiating ~ 239 W/sq m at an “effective temperature of 255 K.
Here’s where the con begins. We know the surface is warmer so the difference must be the “greenhouse effect”.
I say rubbish – the Sun heats the Earth during the day and the heating is variable – not constant.
Further, it is obvious that using the geometry of the “radiative balance” has no bearing on the solar insolation – if it did then does it mean that my back being turned away from the radiator means the power of the radiator is halved because of “radiative balance” ?
I know that sounds silly but if this “radiative balance” calculation was correct then as Wikipedia state:-
“If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C”
Well this is verifiable as there is a body in space the same distance from the Sun – the Moon. And any search of the net will reveal the Moon reaches up to 396 K or 123 degrees C during the lunar day.
Let’s verify the theory shall we :-
1367 W/sq m solar constant – divide by four to get solar insolation as required by “radiative balance” gives solar insolation of ~342 W/sq m.
Stefan-Boltzmann says 342 W/sq m yields a maximum temperature of ~278.7 K or 5.3 degrees C.
DAMN IT, all my theory goes up in smoke – the maximum temperature on the moon is actually ~5.3 degrees C.
So NASA is simply wrong when it says the Moon reaches 123 degrees C – the “radiative balance” calculations as used for Earth’s “effective temperature” prove the data is, as usual, simply wrong and theory and models are right.
If you believe this nonsense you really are gullible.
The other nonsense is the implication Nitrogen and Oxygen don’t play a part in warming and cooling of the atmosphere.
If anyone is stupid enough to believe that 99% of the atmosphere doesn’t become heated and radiate IR in accordance with the temperature then they have either been conned or lack the cognitive ability to recognize such an impossibility.
Nitrogen and Oxygen may not absorb much IR but they MUST become heated and MUST radiate.
The amount of IR from water vapour is at best a trace amount and the amount of IR from CO2 is tiny – unless CO2 possess some hidden magical power that science has not detected yet.

123. Pat Frank says:

Here (pdf download) is a NASA archive document, titled, “Earth Albedo and Emitted Radiation — NASA SP-8067” showing how the 255 K effective radiant temperature of Earth was derived — by experimental measurements using satellites. The document is dated from 1971, which means it was written — and the data were collected — prior to the politicization of NASA. Which means we can safely assume those who wrote it are trustworthy, and the data are not fudged. This is a very sad commentary about the times we live in now, but we’ve all been burned too many times to ignore that truth.
In any case, after a long analysis they report the average TOA emitted radiant power is 237(+/-)7 W/m^2. This is an empirically measured value, not calculated from a model. The word “average” is very important because, as the document points out, the locally radiated power varies greatly.
Here’s what the document says: “As a first approximation, suitable for estimating the effects of Earth emission and albedo on spacecraft subsystems and elements having relatively long thermal time constants, the Earth may be treated as a uniform and diffuse (Lambertian) emitter and reflector. The mean annual values to be used are:
Emission: 237(+/-)7 Watts m^-2 (0.34(+/-)0.01 cal cm^-2 min^-1)
Albedo: 0.30(+/-)0.02
These values were derived fron analyses of data acquired by spacecraft (ref. 38). It should be recognized that they were derived from albedo values that range from about 0.10 to 0.80 and from long wave radiation values ranging from about 105 to greater than 350 Watts m^-2 (0.15 to 0.50 cal cm^-2 min^-1) over limited geographical regions.
(my bold)”
This 237 W/m^2 translates directly to an average Stefan-Boltzmann blackbody TOA temperature of 254 K. A distant spacecraft looking back at Earth radiating as a thermal point-source would measure that temperature.
The greenhouse warming is just the difference between the bottom-of-the-atmosphere (BOA) surface temperature and the TOA radiant temperature. As Earth is at overall energy equilibrium — energy in = energy out — and outgoing energy is by radiation into space, then Earth radiates as though it is at equilibrium with an energy source equivalent to a radiant temperature of 254 K. So, the surface is warmer than would be expected by the observed thermal equilibrium, and that is the GH difference. Note: observed, not deduced.
GH average temperature just a kind of convenient fiction to illustrate how the water-rich atmosphere impacts the surface, making Earth habitable down here for us all. Nevertheless, the atmosphere does make the BOA average ~33 K warmer than the TOA average.

124. Reed Coray said in part:
“(the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space)”
Non-“greenhouse” gases do not significantly radiate thermal energy to space. A planet with an atmosphere free of GHGs, such as an atmosphere made of neon and argon, has thermal radiation to space only from its surface.
However, I do see some of the points on theoretical temperature of Earth’s surface if it had an atmosphere free of GHGs and its current albedo, as in 255 K not being the “likely correct average”. For example, 255 K would not be the average, but the “root mean 4th”, or 4th root of global-average 4th power of surface temperature. The average would be less. This means existing GHGs are boosting the global average surface temperature by more than the mentioned 32 K.

125. Dave Springer says:
December 26, 2011 at 6:48 am
“The ocean has an albedo of effectively 0 and it covers 70% of the surface.”
True but in a very limited set of circumstances that are not part of the real world: Yes, at very low degrees incidence angle with no waves, hardly the case all over the Earth at any one time. Water is mostly a mirror. Albedo equations are not proper for water, Fresnel equations are. Nevertheless, water is considered to have a very low albedo, I believe that to be wrong.
Best,
J.

127. To ALL:
It is amazing to see how much confusion can be created out of a simple problem. Here are a few points for all of you to consider since no one has addressed the issue in a physically correct way including Reed Coray:
1) When assessing the strength of the GH effect, one needs to compare two temperatures – (a) current mean temperature of Earth (Ts) with atmosphere, and (b) the estimated ‘gray-body’ temperature of Earth (Tgb) WITHOUT atmosphere. Discussions involving a hypothetical atmosphere with no ‘greenhouse’ gases is physically meaningless, because the GH effect measured as a surface thermal enhancement, i.e. GHE = Ts/Tgb may not be due to heat-absorbing gases at all …
2) Calculating Earth’s mean gray-body temperature without atmosphere requites an explicit spherical integration of the temperature field. In other words, temperature at each point on Earth’s sphere needs be calculated first by taking the 4th root of the absorbed solar radiation at that point, and then the resulting temperature field must be averaged. This approach is different from the conventional method attempting to estimate the mean planetary temperature from the spherically AVERAGE absorbed solar radiation using the S-B law. Due to the non-linear relationship between temperature and radiative flux, and a highly non-uniform distribution of temperatures on an airless planet, one CANNOT obtain the true (arithmetic) mean temperature from an AVERAGE absorbed radiation using S-B equation! This is explained by the Hoelder’s inequality between non-linear integrals in math. According to this inequality, the effective emission temperature calculated from the spherically average radiation flux will ALWAYS be significantly HIGHER than the true mean temperature of the planet. That’s is why effective emission temperatures are not physically compatible (and should not be equated) with palatable mean temperatures!
3) When one computes the MEAN surface temperature of an airless Earth through EXPLICIT integration of the temperature field resulting from a differential solar heating of the sphere and assuming an average surface albedo of 0.12, one arrives at 154.3K. Since the actual near-surface temperature of Earth is currently 287.6K, the true ‘Greenhouse Effect’ turns out to be 133.3K. In terms of the proposed thermal enhancement ratio above, this means GHE = 287.6/154.3 = 1.86. This raises the question: Can a handful of trace gases, which amount to less than 0.5% of atmospheric mass trap enough radiant heat to boost the planet surface temperature some 86% above that of a moon-like airless gray body? According to classical thermodynamics, this not possible. Hence, the atmospheric GH effect is not caused by heat-absorbing gases, but is likely due to an entirely different mechanism … The nature of this mechanism is explained in a special posting to occur soon on this website … Stay tuned!

128. Ian S says:
December 26, 2011 at 5:13 am
“A thought experiment. If the earth had an albedo of 0 and you could put whatever imaginary layer you wanted around it (with physical properties, i.e. you can’t use one-way glass that doesn’t exist in the real world). What is the maximum temperature?
Could you get the earth to have a higher temperature than a theoretical perfect black-body? If so, does that seem logical?”
The answer is yes. There is such a thing as a “hot mirror” – an object that is largely transparent to wavelengths shorter than 700 nm or so, and largely reflecting of wavelengths longer than that, up to maybe 2,000 nm. This is an example of a “dichroic filter”. It is not fundamentally physically impossible to create spherical ones whose infrared-specific reflection extends through “low temperature thermal IR” out to 20,000 or 50,000 nm.
Put a black rock inside one of these “oversimplified” (but not fundamentally physically impossible) dichroic puppies with a vacuum in between, put it in sunlight, and then the black rock will get hotter until its visible outgoing radiation is in balance with its visible incoming radiation. At that point, the black rock will warm up to theoretically 1,476 degrees K, and appear orange in daylight and a brightish, possibly-slightly-whitish, and quite yellowish shade of orange in someone’s living room or bedroom. The rock may be lava at that point.
As for greenhouse gases: For an oversimplified case, consider a blackbody planet surrounded by an “oversimplified” sheet of GHGs. Incoming radiation comes in, through the GHGs, unimpeded The outgoing radiation from the planet is absorbed by the GHG layer. Half of the GHG-absorbed radiation gets re-radiated outwards, and half gets re-radiated inwards towards the planet. So, the planet radiates twice the radiation that its sun delivers to it in order to achieve
thermal equilibrium. Its GHG atmosphere layer, if oversimplifyable to being a perfect full-performance GHG one of unifrorm temperature and not running into convection complicating things, would have the temperature that achieves outgoing radiation equal to incoming radiation to the planet.

129. 9 to 33C? More like 9 to 16 C. Latent cooling alone drops the surface average flux from approximately 393 to about 313 Wm-2 (273K) minus 240 Wm-2 TOA OLR is 73 Wm-2 combined for greenhouse radiant and conductive/convective. The greenhouse is only the amount over 255C. We live on a water world and Trenberth in a fantasy world.

130. Bryan says:

Pat Frank says that by experimental evidence and calculation the Earth with an albedo of 0.3
results in;
” Earth radiates as though it is at equilibrium with an energy source equivalent to a radiant temperature of 254 K.”
Nobody disputes these points.
What is in dispute is the claim that if the so called greenhouse gases did not emit/absorb in the IR the Earths albedo would stay fixed at 0.3.
It should be fairly obvious that much more radiation(particularly in the IR) would reach the surface.
The surface has an albedo of 0.15 for the existing solar radiation.
With a large increase of IR radiation the surface albedo is bound to fall much further.
It is an abuse of any kind of logic to use an albedo of 0.3 for a calculation of a mythical greenhouse effect of 33K.
Any theory to be taken seriously must be internally consistent.
By using IPCC accepted values the claimed 33K turns out to be nearer 16K.
The greenhouse theory as presented in popular expositions is internally inconsistent.

131. Leonard Weinstein says in part:
December 26, 2011 at 6:32 am
“Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation”.
This may be the case on a planet with evaporative cooling and precipitating-cloud-forming convection, and these items involve a major greenhouse gas (water vapor).
I thought the discussion was for comparison of Earth to a most-Earth-like-possible planet with an atmosphere free of GHGs. Even though such a GHG-free planet would be one without bodies of water, as in a dry, deserted rock!

132. R. Gates says:

wayne says:
December 26, 2011 at 11:10 am
R. Gates says:
December 26, 2011 at 10:27 am
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side?
>>>
The moon… spinning huh, like Venus you must mean, right? R. Gates , that sure is a cheap shot to Lars P. Or, I guess using your logic, that even with a greenhouse gas-less Earth spinning at 1 RPM, it would have the sun lit side boiling hot and the dark side freezing cold.
——
Why would it be any different? It would not. Just as we get large diurnal temperature swings in the desert areas of earth because there is less water vapor in the atmosphere above the desert, if you took away all greenhouse gases planet wide you would get a diurnal temperature swing that would be even more extreme. It might not be as dramatic as that of the moon (-230C at night to 120C during the day) because of some heat coming from earth’s interior, but it would still be extreme enough to make Earth a very inhospitable place.

133. Diatomic and mono-atomic molecules have very very low emission in the thermal (IR) bands. Only at higher pressure and temperature does interactions between molecules create the dipoles allowing such thermal emission. But thermodynamics will not be denied! What happens is that such gases (if warmer than their surroundings) will cool by expansion.
http://www.heliosat3.de/e-learning/remote-sensing/Lec7.pdf

134. To Donald L. Klipstein
I hope I can interest you further to consider this comparison.
A black body is often approximated by a spherical cavity with a small hole in it. The energy gets in at one angle, but bounces around ‘trapped’ at many other angles with a small percentage chance of finding the hole again.
I believe that band-pass, or high-pass filters are just another type of ‘hole’. The energy gets in at one frequency and is converted (through absorption and emission) to different frequencies with a small percentage chance of hitting the escape frequency again.
I like to think of things in terms of energy densities. It helps get rid of the confusion of temperature which is material dependent and adds complexity.
Whether we are talking about angles or frequencies, the end result remains the same (I believe) and you can not create an average energy density inside the spherical cavity that is higher than the energy density outside the cavity. [Of course if our energy source is coherent (i.e. radiation in a single direction, or at a single frequency) then we can be focus the energy to obtain extremely high energy densities, however the average energy density will remain the same.] Average energy density conservation is basically just a different way of looking at second law of thermodynamics and how you can not transfer heat from a cold object to a hot object and special filters of any kind we can imagine are not a form of Maxwell’s demon.

135. Lars P. says:

R. Gates says:
December 26, 2011 at 10:27 am
______
I guess then that the moon, which is also a spinning sphere, would not have a boiling side and freezing side? But it does. The essential point is that the oceans and greenhouse atmosphere prevent a strong diurnal temperature variation. The oceans provide the primary heat “sink” of the planet and the greenhouse atmosphere provides a transport mechanism for the heat.
————–
Dear R. Gates, yes the oceans are the primary “sink” of the planet as you can also see from my other comments in this thread – we agree on this point, quelle surprise, this is exactly the main point that is forgotten in greenhouse. But please don’t try to defend greenhouse theory based on the oceans, as it was already said many times the downwelling infrared does not even reach under the first millimetre of water. Greenhouse does not add to the warming of the oceans. And the 24 hours versus 28 days to make one rotation is a fundamental difference that allows for only 12 hours cooling or 14 days. And of course the oceans with their heat capacity, the way how they warm and how they lose heat play the main role – not greenhouse R. Gates.

136. Dave Springer says on December 26, 2011 at 7:24 am:
“Land surfaces radiate. Water evaporates. Write that down. Land has an albedo of 0.15. Water has an albedo of 0.00. Write that down too.”
=====
What you are saying there seems, at first glance, to be correct but unfortunately it ignores one very important fact, which is that the soil anywhere on this planet of ours does have a fairly high ‘water-content’. Many of the Earth’s desert-soils, or sands, have very low water contents – which is the – or at leas one, very important reason for rapid cooling after “Sun-down”
I come from “Way up North” (60° 57′ 06″ N) and in that district, when I was a youngster, the “Winter Ground Frost” used to reach down to a depth of – roughly 2 + meters (6′ 6″) and I am convinced it is the water-, and not the soil molecules that freeze solid . If the snow was gone by mid-April, the ground frost would also all be gone by mid. to late May. —
And therein lays another “We cannot mention that point” that “climate scientists” treasure, which, of course shows the importance of “Conduction” and by that I mean heat/energy transferred from the top of the surface – downward, or inwards, as well as from the surface to the Atmosphere. And it must mainly happen by evaporation from anywhere where water is present

137. I see that some people think N2 and O2 are incapable of emitting radiation. If that is a fact, wouldn’t that make them into very powerful GHGs?

138. Scott says:

This “what if” analysis of removing all green house gasses but keeping the earth’s albedo unchanged sure promotes discussion. It kind of reminds me of a Saturday Night Live skit they did years ago where they would ask a question such as “What would have happened if Genghis Khan had a machine gun” and had a panel discuss it. It seems that the writers for that program understood the ridiculous nature of such proposals better than climate scientists!

139. wayne says:

@ Ned Nikolov:
December 26, 2011 at 3:36 pm
Absolutely fascinating poster & contents Dr. Nikolov. I wholly agree with the path you and your colleagues are taking, without a doubt. Congratulations and looking forward to your paper.
There is one aspect I found was not mentioned within that poster and text that may, only possibly, help account for some of the remaining small variances between each of the bodies and their individual atmosphere chemical composition. I will drop an e-mail via Dr. Zeller if you would be willing to consider a bit of input even though it may have already been considered. — Wayne Jackson

140. Michael Hammer said:
“Doubling water vapour at a first approximation doubles cloud mass (either denser clouds or more of them both of which increase albedo)”
As I see it, increasing global temperature increases density of clouds. So, albedo of many of them may increase from 99.9% to 99.95% or something-like-that should global warming achieve a 10-11 degree C/K temperature rise.
Meanwhile, I see vertical convection with clouds getting more efficient at circulating heat as the oceans warm from AGW. While, need to circulate heat decreases because the Arctic warms more than the tropics when the globe warms. So, I expect warming of the globe by increase of “greenhouse gases” would result in a slight *decrease* of cloud cover.
However, I suspect IPCC and its favored scientists do over-estimating of this effect, and neglect to consider that a warmed atmosphere with more-efficient more-compact clouds has its global relative humidity decreased – partially negating the “water vapor positive feedback”.
As a result, I like to consider “real-world” combined water vapor and cloud albedo feedbacks
equal to the IPCC-favored value or the MODTRAN-derivable figure for water-vapor positive
feedback. As in, for oversimplification, use some published or available figure for the water vapor feedback, next consider cloud albedo feedback to be zero, Plug this into GCMs (global circulation models), and I hope honest scientists see what I like to see – a “less-partial snowball” Earth has global climate sensitivity decreasing as CO2 increases, towards “regulating” global
average surface temperature of 23 C (9 degree rise) if CO2 increases by a factor of 20 and
Antarctica repeats a previous latitude warmer than its current one that gets it largely covered by 2 miles of ice.

141. Jim Masterson says:

>>
Dave Springer says:
December 26, 2011 at 10:29 am
There are a zillion class M stars at all stages of their life cycles for us to look at and relatively speaking a class-M star like ours is before it becomes a red giant in its cups is one of the most constant dependable things in the universe.
<<
There may be a “zillion” class M stars, but the Sun is not one of them. I’m assuming you mean spectral class. The Sun is a main sequence, spectral class G2, dwarf star (some say G2.5). That would give it a somewhat yellow color. M class stars are red.
Jim

142. wayne says:

@ Ned Nikolov:
December 26, 2011 at 3:36 pm
This approach is different from the conventional method attempting to estimate the mean planetary temperature from the spherically AVERAGE absorbed solar radiation using the S-B law. Due to the non-linear relationship between temperature and radiative flux, and a highly non-uniform distribution of temperatures on an airless planet, one CANNOT obtain the true (arithmetic) mean temperature from an AVERAGE absorbed radiation using S-B equation!
>>>
Yes Ned, a fellow commentator here, I believe ‘anna v’, was pressing that very point over two years ago of the impossibility to the non-linear distribution when jumping from temperatures to radiative power and vice versa, and I have never forgotten that very aspect, knowing all “consensus” climate science calculations were off by at least this factor. But you have the deep math training that is one of my weaknesses. I would have been immediately be jumping into numeric integration.
T.gb = 2/5 • √√( (1362.0 + 13.25e-05) • (1 – 0.12) / (0.955 * ‹kSB›) )
= 154.28 K
TrueGHE = 287.6 K – T.gb
= 133.32 K
ends at your starting conclusions in this aspect, I see it now. Thanks to you both to swim against the collective current, everyone does seem to know simple Stefan-Boltzmann application is not correct in this arena, but you both are the first to understandably frame it to near completeness.

143. J.H. says:

Dave Springer says………..
…No one knows exactly what the earth’s average albedo really is. Climate models use values which differ by as much as 0.07 from one to another. Speaking in W/m^2 that 0.07 uncertainty is 7% of some 200 Watts per square meter or about 14 W/m2. The net effect of all anthropogenic forcings is claimed to be around 2.5W/m2 so in essence this means that uncertainty in natural albedo forcing is about 6 times greater than anthropogenic forcing. In other words, the models are operating in the dark (so to speak) with uncertainties far greater than the anthropogenic factor they are trying to calculate…
—————————————————————————————————————————
…. Yep, that has been my opinion for this whole debate on AGW and the purported catastrophic effects to the Earths climate. Like Dave I consider that the noise of natural climate variation to be wholly overwhelming of any anthropogenic effect on the climate system, and real observations seem to bear this out…… It is only within the unnatural computer models of the Earths climate system that the AGW hypothesis continues to exist.
The confounding effects of the natural world defeat any simplistic climate model or math equations

144. Eco-Geek says:

Greenhouse gasses increase the Earth’s outgoing radiation budget. Remove them from the atmosphere and the remaining theremal sources: black-body and thermally radiating gasses (NGHGs) must increase their radiative output to compensate for this loss. As thermally radiative sources they can only do this through an increase in temperature. Hence:
GREENHOUSE GASSES CAUSE ABSOLUTE GLOBAL COOLING.
And lots of it.
It is so mind cringingly obvious and yet the sceptics continually support the warmist myths by denying the obvious: If GHGs are better radiating gasses than other radiative sources they increase the radiative output of the Earth and therefore cool the planet. Likewise remove the atmosphere entirely and all thermally radiation would have to be emitted by the surface so the temperature would rise much higher.
To put it another way: RADIATORS COOL THE SOURCE.
Otherwise we would have no central heating.
Get cool. Think..

145. Andre says:

I think there are a few essential elements missing in this generation of the null hypothesis,
1: The standard no-atmosphere “black body” model uses temperature of the surface, whereas the global temperature is the atmospheric temperature at 1.5 meters above the surface. Think about that
2: Non radiating gasses cannot lose their heat by radiation, only conduction and convection and there is only the earth surface to conduct heat to. Nothing can be emitted to the atmosphere.
3: There is no negative convection.
This is how I wrapped the null hypothesis up in UKww:
“The null hypothesis is usually describing the situation which would it make different from the actual -or alternative- hypothesis. It’s used mostly in statistics, I think, but why not try and see what happens if we apply it to the GHG hypothesis.
We read all over internet that the black body temperature of the Earth would have been -18C, but the actual average temperature is +15C; consequently this 33 degrees difference is supposed to be the greenhouse effect. But is this true?
Is the blackbody situation the “null hypothesis”? I don’t think so. The black body calculation assumes a sphere with a constant flux of light energy, uniformely distributed over the surface, using the Stefan Boltzman equation to derive it’s temperature like this.
But the earth is nowhere near a blackbody and if we want to really look at the null hypothesis, we would have to look at an earth without greenhouse effect, but still with an (inert) atmosphere and still rotating in 24 hrs, with seasons and all.
Now instead of using an average steady state solar radiation, we need to realize that we have the diurnal cycle with max insolation radiation at noon and no radiation incoming when the sun is below the horizon. So during daytime the earth surface warms up and much more than the according the average radiation. Equilibrium temperature at the equator in a steady state with the sun in zenith, using the full incoming 1365 w/m2 (albedo 30%) would be 360K or 87C. This follows from applying the Stephan Boltzman equation for the spot directly under the sun, instead of a uniformely distributed radiation.
So this much higher temperature of the earth surface is transmitted via conduction to the lowermost boundary layer of the atmosphere. This heated air gets is less dense, and it becomes buoyant so it rises up; Convection, the very basics of meteorology. So at daytime the atmosphere receives thermal energy of the earth. How can it lose this energy again? Remember we are in the null hypothesis, no radiation, no greenhouse effect, so the inert atmosphere cannot lose the energy by radiation.
Now, at night time the Earth does not receive radiation energy from the sun but it radiates energy out and cools quickly, obviously much more quickly in the null hypothesis even than with the greenhouse effect, which would have directed (“reflects”) some radiation back to earth. Now the cooler earth also cools the boundary layer of the atmosphere by conduction again, however there is no negative convection as the cool air gets more dense and tends to stay put; the inversion; also very basic meteorology. So despite the cooling of the earth, the missing radiation from the atmosphere prevents it from cooling at night and the next day more conducted energy is convected into the atmosphere, that stays there again.
Obviously we have an unbalance. And equilibrium can only be reached, maybe after thousands of years, when the convection at daytime has reduced so much to balance heat loss at night time via conduction back to the surface. For that the lower level atmosphere needs to be at the same temperature / density than the boundary layer would reach due to the conduction of heat from the surface.
Conclusion, in the null hypothesis, without greenhouse effect, the average temperature of the lower atmosphere would be considerably higher than the black body temperature of the surface. How much I don’t know. But the main point is that a certain portian of the temperature difference between black body and actual atmospheric temperature is not due to greenhouse effect but to the inability of the inert atmosphere to cool down by radiation. “

146. @Andre says:
December 27, 2011 at 1:58 am
At last, somebody on the right track.

147. Tony Mach says:

I think there is at least one (potentially major) error in this whole atmospheric transmission mess-up. There not TWO (Transmission vs. “Absorption AND Scattering”) but THREE (wavelength dependent) components for each atmospheric component:
1. Transmission
2. Absorption (and Re-Emission!)
3. Reflectance (“Scattering”)
I think Climate Science has an oversimplified model of this mess-up by putting Absorption and Scattering together.

148. Myrrh says:

Baa Humbug says:
December 26, 2011 at 8:24 pm
Myrrh says:
December 26, 2011 at 5:50 pm
A precis of results would be appreciated..
Will this do?
http://www.gewex.org/BSRN/BSRN-11_presentations/Tues-wacker.pdf
============================
Many thanks! I wondered how the science had progressed from its beginnings:

MONTHLY WEATHER REVIEW AUGUST, 1928
FURTHER STUDIES IN TERRESTRIAL RADIATION ‘
By G. C. SIMPSON, C. B. F. H. S., F. R. Met. SOC.
The results obtained in this paper are so different from
those obtained in the previous paper, Some Studies in
that it is necessary to review the
whole position, to find the causes of the differences and
to estimate their significance.
http://docs.lib.noaa.gov/rescue/mwr/056/mwr-056-08-0322.pdf

149. John Marshall says:

I am sorry to disagree with the famous Lord but the theory of GHG’s violates the laws of thermodynamics.
The BB radiated temperature calculated is incorrect because the earth is not a black body. the nobel Lord has completely forgotten the small fact that the atmosphere generates heat through adiabatic compression, rather like a diesel engine increases its cylinder temperature to the diesel oil combustion temperature of 250C by compressing the internal gasses. Refrigerators work by the same process. Using the combined gas laws will produce the required temperature at the earth’s surface. Venus, with a surface pressure of 90 atmospheres generates 480C or 750K by the same process and the high surface pressure is due to the heave CO2 atmosphere. It is interesting to note that in the Jovian atmosphere, without any greenhouse gasses present, generates a similar temperature to earth at the 1000mb level.
The GHG theory relies on these gasses to be IR reactive. But are they? The rerun of Gore’s experiment by Anthony Watts in which two identical containers with air in one and a CO2 rich mixture in the second were heated by light. Anthony got no temperature rise in his CO2 enriched container above that of the air filled container. Both remained at similar temperatures. He repeated this experiment several times without any greenhouse effect. The rigerous way Anthony carried out this experiment should have got people thinking harder about this discredited GHG theory.
A moments thought will show that if a molecule of CO2 is preferentially heated it will immediately share this heat with its surroundings. It cannot do anything else. It cannot ‘store it’ because the 2nd law must hold true. Any heated volume of gas in the atmosphere will convect due to density changes and convecting air in the atmosphere must expand and loose heat due to adiabatic expansion. The gas will very soon be cooler than the surface so cannot re-radiate heat to the surface, heat can only flow by whatever means from hot to cold as the laws of thermodynamics dictate. Also as confirmation the so called heat anomaly in the mid troposphere as predicted by the theory of GHG’s cannot be found. Why? because the theory is WRONG.

150. kwik says:

Andre says:
December 27, 2011 at 1:58 am
Finally someone able to do some logical thinking.

151. Myrrh says:

For interest:

“NASA Abandoned Flawed Climate Calculations in 1960’s
Siddons, Hertzberg and Schreuder were astonished to find that “the principal method for predicting a planet’s temperature is surprisingly arbitrary and simplistic.” That was, they believe, why NASA needed to set aside the blackbody equations when doing their own calculations for the Moon landings.
To climate sceptic scientists it seems self-evident that the Earth’s surface should not be treated like a flat, two-dimensional blackbody. It is more properly a complex spinning sphere with large variability in reflectivity and absorption of the Sun’s light and energy. But, despite the U.S. government knowing since the 1960’s that the blackbody equations were of no use to real-world science, these facts don’t appear to have been passed on to climatologists.
Lunar Temperatures Cast Doubt on Climate Theory
NASA had found that daytime temperatures on the lunar surface were lower than expected because planetary bodies also conduct heat to their inside rather than radiating it all into space – an empirical fact that challenges the GHG theory. Computer models supporting GHG theory had predicted that such heat energy would be ‘blanketed’ above a planet’s surface.
In fact, the Apollo data proves the Moon’s surface temperatures throughout its two-week night were higher than predicted by the blackbody equations because the moon “feeds on” the heat it had previously absorbed.
Thus the success of NASA’s moon landings becomes evidence against the unreliability of the Stefan- Boltzmann equations in real world science.”
http://www.suite101.com/content/apollo-mission-a-giant-leap-to-discredit-greenhouse-gas-theory-a241363

152. JeffC says:

In the words of Dr. Evil … magma … the earth is hardly a cold rock in space …

153. richard verney says:

Ned Nikolov says:
December 26, 2011 at 3:36 pm
///////////////////////////////////////
Dr Nikolov
#
On the assumption that we are heading towards an ice age, what process according to your theory, explains such a change in temperature?

154. Richard Verney wrote: “Two problems. First, the length of the lunar day. Eg., what would be the average temperature of the moon if a lunar day was only one hour rather than about 27;5 days? ”
A while back, I wrote a simulation model of what the Earth surface temperature would look like if it had no atmosphere, and spun on its axis once an hour. The mean temperature is the same as with a 24 hour rotation period, but there’s less variation around it.

155. Jim Masterson says:

>>
D. Cohen says:
December 26, 2011 at 3:35 am
I just looked up the average surface temperature of the moon.
<<
All the comments about the Moon’s surface temperature are essentially comparing apples to oranges. The temperature usually quoted for the Earth’s surface is really the Surface Air Temperature or SAT. This is a value measured about two meters ABOVE the Earth’s surface–not the actual Earth’s surface. (Climatologists and climate modelers seem to confuse the two.) The Moon has no atmosphere, so there can be no SAT. The Moon’s surface temperature is the actual surface temperature.
I’ve been to beaches where the sand burns my feet. If I wear sandals to insulate my feet from the hot surface, I find that the corresponding air temperature is quite pleasant.
Jim

156. Richard M says:

I like Dr. Nikolov’s presentation. I believe it is the same concept as has been put forward by several others, most notably, Postma and Huffman. The argument against it has always been that the air pressure can do no additional work once it stabilizes. However, I think another comment above is important. Without gravity the atmosphere would expand and lose its thermal energy. Gravity prevents this and hence keeps the gas molecules in close contact. This continued kinetic action must have some affect on temperature.
However, the fact is GHGs do radiate and absorb thermal heat. So, it seems they still need to be looked at more formally than I saw in Nikolov’s presentation. I don’t think asserting they are “completely counterbalanced” is sufficient. Do they block heat (through absorption) or do they cool the planet (through emission). I’m starting to lean towards the latter.

157. beng says:

****
John Marshall says:
December 27, 2011 at 3:39 am
The BB radiated temperature calculated is incorrect because the earth is not a black body. the nobel Lord has completely forgotten the small fact that the atmosphere generates heat through adiabatic compression, rather like a diesel engine increases its cylinder temperature to the diesel oil combustion temperature of 250C by compressing the internal gasses.
****
Yes but since the atmospheric mass is fixed, any compression (downward movement) in a local area must be countered by an equal decompression somewhere else, so the net effect is zero. Obviously, some regional areas are warmed by compression, like Chinook winds just east of the Rockies in NA, but other regions must experience counteracting decompression (upward movement).

158. KR says:

The 33C temperature difference with/without greenhouse gases is simply a Gedankenexperiment (http://en.wikipedia.org/wiki/Thought_experiment), exploring the potential consequences of a change, what happens if a single variable is modified – and in this case _with all other things held to be unchanging_.
Now, if we could experimentally remove all greenhouse gases, the end resulting temperature of the Earth would not be 33C colder, because that would kick in all kinds of other changes: missing clouds reducing albedo, glaciers increasing albedo, vegetation changes, changes in the lapse rate due to modified atmospheric energy, etc. Some of these changes would make the Earth warmer, some would make it colder.
But – the greenhouse gas contribution to current temperatures is 33C. That’s just spectroscopy, shown both by computations from numerically integrating over the depth of the atmosphere and measurements of the outgoing IR spectrum. The IR emissivity of ground and water is ~0.98 (98% as efficient as a pure blackbody), while the IR emissivity of the Earth to space through our greenhouse-gas containing atmosphere is ~0.612. An IR emissivity of 0.98 would allow the Earth radiate energy equal to incoming sunlight at ~-18C, while the emissivity of 0.612 means a temperature of ~15C is required.
And again, that’s just spectroscopy.
Secondary changes which would occur in a “real-world” removal of greenhouse gases, positive or negative, don’t change the GHG contribution to current temperatures – because those secondary changes are additional contributions +/-.

159. Bomber_the_Cat says:

John Marshall says:
December 27, 2011 at 3:39 am
I am sorry to disagree with the famous Lord but the theory of GHG’s violates the laws of thermodynamics.
So you disagree with Lord Monkton? Well John, if you learn some basic physics your opinion may count for something, but until then your misconceptions of the 2nd Law of Thermodynamics and pressure causing heat don’t carry much weight.
For instance, the car in my garage has pressurised tyres. Do you think the tyres are hotter than the rest of the car?

Bomber_the_Cat says:
December 27, 2011 at 8:35 am:
“For instance, the car in my garage has pressurised tyres. Do you think the tyres are hotter than the rest of the car?”..
How about the charge temp in the cylinder at the top of the compression stroke as JM indicated?
Not sure what your argument is, but if it is re: compression and atmospheric temps check this:

161. Bryan says:

KR says:
“Now, if we could experimentally remove all greenhouse gases, the end resulting temperature of the Earth would not be 33C colder.”
In this he agrees with the vast majority of posters who notice that the bogus calculation requires an Earth system albedo of 0.3.
Any realistic estimate of the real Earth system without greenhouse gases would be less that 0.1
Making the greenhouse effect less than 16K
However he then goes on to say
“But – the greenhouse gas contribution to current temperatures is 33C.”
So even though the theory is bogus we refuse to change the value!!!
“That’s just spectroscopy, shown both by computations from numerically integrating over the depth of the atmosphere and measurements of the outgoing IR spectrum. ”
This is just another layer of subterfuge used by CGW advocates like KR.
Only by incorrectly calibrating instruments so as to arrive at the 33K figure.
Instrument makers acknowledge that there is circular reasoning involved in radiation measurement.
You can calibrate the instruments to agree with your initial assumptions – what does that prove?
By defending the indefensible the adherents of the greenhouse theory are increasingly being seen as rather odd.

162. Spector says:

I think the number being referenced may have been made by making a few very simple assumptions that avoid all complex details about the surface of the Earth.
1. The Earth is characterized by a single temperature.
2. The total black body energy radiated at that temperature over the entire spherical surface of the earth is equal to the energy received from the sun on its disk shaped cross-section.

beng says:
December 27, 2011 at 8:12 am
“Yes but since the atmospheric mass is fixed, any compression (downward movement) in a local area must be countered by an equal decompression somewhere else, so the net effect is zero. Obviously, some regional areas are warmed by compression, like Chinook winds just east of the Rockies in NA, but other regions must experience counteracting decompression (upward movement).”
How about as you move up and down vertically in the atmosphere? Is pressure constant? Is any decrease in pressure accompanied by an increase elsewhere? Think of getting in a submarine and descending in the sea – does the hull experience a change in pressure loading?…
It is the vertical pressure gradient which supports the temperature lapse rate. Check the link above for bomber_the_cat.

164. richard verney says:

Frank Davis says:
December 27, 2011 at 6:29 am
//////////////////////////////
Frank
Thanks for the link. I have not yet read your paper but will definitely do so. I am not surprised by your summary. The position with respect to the Earth would be different if the speed of rotation was very much slower, may be measured in decades rather than in a day. The thermal capacity of the oceans stores the energy received during the day and releases it during the night. It smoothes the energy budget. However, for the thermal reservoirs to work, obviously they must not freeze (since in that condition they reflect most of the incoming solar energy) and this requires that solar inpout is regularly received (as in the case of the Earth every 12 hours).
The point with the moon is that it has no such thermal reservoirs. If it did, the diurnal differences would not be as stark. Since the moon does not have these thermal reservoirs, there is little to smooth the diurnal range since the rocky surface does not get heated to a significant depth. Lacking the thermal reservoirs the only other way to obtain diurnal smoothing would be a fast rotation. In my post, I suggested a rotation of once an hour, I have not done the calculations and it may be that the day would need to be even shorter than that!

165. R. Gates says:

There are a few additional thoughts to add to this already “interesting” notion of removing GHGs, and that has to do with what exactly what you’re defining as the “temperature” of the planet, and the role that greenhouse gases play in providing a temperature in the atmosphere at various levels. If the Earth were simply a rock spinning in space (much like the moon), without an atmosphere at all, then what exactly would you mean by its temperature?
The temperature right at the ground (0m), or 1m above the ground, or 1m below? These would all be different. SW radiation striking the ground will either be absorbed by the ground and/or reflected back into space. Once absorbed by the ground, it will converted to LW. Some of that LW will be reemitted back toward space. Of course, without greenhouse gases it will transfer instantly to space as there would be intervening gases for it to be absorbed by. Additionally of course is the dramatic diurnal change that you’d see on an Earth devoid of GH gases, such that the very notion of an “average” becomes suspect.

166. richard verney says:

@Bomber_the_Cat says:
December 27, 2011 at 8:35 am
///////////////////////////////////////////////////
Bomber the Cat
I have had this argument for years with car tyres.
When you inflate the tyres, the air inside heats in relation to the pressure. After inflation, the air in the tyres is hot. However, although the rubber is a fairly good insulator eventually, the heat from the air inside the tyre is lost by conduction and radiation.
In a static condition there is nothing replenishing the heat so eventually the air inside the tyres cools to ambient garage temperature. If on the other hand, immediately following inflating the tyres whilst the air temperature inside the tyres was still hot from the pressure brought about by the inflation, you were to drive on the tyres, the very slight flexing of the side walls would be sufficient to maintain the heat inside the tyres. The flexing of the side walls does not create very much pressure but the work being done is sufficient to maintain the temperature of the air inside the tyres. If you could drive indefinitely you would maintain the temperature indefinitely,
The point is that you need some external input of energy to maintain the temperature.
Reverting to the Earth. The Earth has two such external sources of energy that maintain the temperature caused by the pressure of the Earth’s atmosphere. First, there is the sun. Some of the solar energy is absorbed by the atmosphere. Second, the atmosphere flexes just like the side walls of the tyre (when you are driving on the tyre). As you are no doubt aware there is an atmospheric/diurnal bulge which is constantly fluctuating. These inputs taken together are sufficient to maintain the temperature caused by the pressure of the atmosphere.

167. richard verney says:

@Bomber_the_Cat says:
December 27, 2011 at 8:35 am
///////////////////////////////////////////////////
Bomber the Cat
I have had this argument for years with car tyres.
When you inflate the tyres, the air inside heats in relation to the pressure. After inflation, the air in the tyres is hot. However, although the rubber is a fairly good insulator eventually, the heat from the air inside the tyre is lost by conduction and radiation.
In a static condition there is nothing replenishing the heat so eventually the air inside the tyres cools to ambient garage temperature. If on the other hand, immediately following inflating the tyres whilst the air temperature inside the tyres was still hot from the pressure brought about by the inflation, you were to drive on the tyres, the very slight flexing of the side walls would be sufficient to maintain the heat inside the tyres. The flexing of the side walls does not create very much pressure but the work being done is sufficient to maintain the temperature of the air inside the tyres. If you could drive indefinitely you would maintain the temperature indefinitely,
The point is that you need some external input of energy to maintain the temperature.
Reverting to the Earth. The Earth has two such external sources of energy that maintain the temperature caused by the pressure of the Earth’s atmosphere. First, there is the sun. Some of the solar energy is absorbed by the atmosphere. Second, the atmosphere flexes just like the side walls of the tyre (when you are driving on the tyre). As you are no doubt aware there is an atmospheric/diurnal bulge which is constantly fluctuating. These inputs taken together are sufficient to maintain the temperature caused by the pressure of the atmosphere.

168. richard verney says:

@Richard M says:
December 27, 2011 at 7:41 am
////////////////
The temperature in the first instance is created by the gravitaional pressure. Thereafter, there is heat loss from the system. You need some external input which need not be sufficient to create the remperature brought baout by the gravitional process (in our case circa 288K) but rather that which is sufficient to cover/make good the heat loss from the system.
Obviously, the sun is one such source. A second source is work done in the atmoshere itself. The atmosphere is dynamic and in constant flux. Work is being done and a by product of this is to heat the atmosphere at any rate to the extent necessary to cover/replenish the heat loss.
I would not definitely rule out some heat trapping effect caused by GHGs but there is in practice no need to resort to such a theory to explain the temperature of the Earth. Gas pressure alone is quite capable of explaining why the Earth is sufficiently warm for life to flourish.

169. Doug Proctor says:

Great post for showing how the most elemental “fact” is neither certain nor settled! The probable best response is to ask what the purpose of the question of the Earth’s non-GHG temperature is: to eliminate the oceans as part of the removal of GHG is to change what the Earth is, so the answer becomes suitable to a non-Earth. As with other variations on the calculation.
The question in the CAGW story is used to give a background to either the great warming of GHG gases, and hence their danger (the warmist side), or that total GHGs give 33C of heating, so how much change can only 300 ppm of CO2 bring (the skeptic side)?
As for the non-atmospheric temperature of the planet (any planet): I’m a geologist, and I’ve been struggling to figure that out, although I have access to almost 300,000 temperature readings (in Alberta alone) at various depths in the Western Canadian Sedimentary Basin. The loss of heat at the surface is sudden and rapid; fifty meters down most places (outside remnant glacially cooled areas) it is about 16*C. Another sudden change happens at the continental crust contact of 1000 or more meters. Both the air and the sedimentary covers have a strong cooling effect, i.e. have high thermal conductivity. Elsewhere, the Greenland ice cores seem to show that, at the base of the ice sheets, the rock is – 9*C, while in Antarctica there are buried liquid lakes of 0*C. So what is even the non-solar insolation temperature of the planet?
Mars has an apparent surface temperature averaging – 60*C. But what is it 200 m down? The moon is – 272*C on the dark side and +250*C (pardon the inexactitudes; the details are not the point here). Same question: the Apollo boys did a little digging and left some instruments for temperature monitoring, but I couldn’t figure out where the surficial rapid cooling equalled the deeper radiative heat loss. Which is why I started to try to figure out what temperature the planet – any planet – would be at in the absence of an atmosphere.
The question that in a strict physicist’s way needs answering is: for a rock surface average of Earth, where the only two ways of temperature are 1) the present cooling of a central core through a thermally conductive surrounding medium like the Earth, and 2) the different rates of warming and cooling of the average dry surface rock averaged through the day, night and orbital variations, what would the surface of the planet be?
I can’t figure out what the planetary rock should be, but I THINK, it might be 18*C. Which is higher than the 14.5*C it is supposed to be, which then says that the atmosphere COOLS the planet. What does the warming/cooling differential of solar insolation have, considering the thermal conductivity of rock? What if there were a ball of rock at 3K, a billion years ago, with a billion years of sunshine falling on it at Earth’s orbital distance, and it is now in equibilbrium with the energy pouring down on it? Would that be the blackbody radiation of 255K? So, you add the 18C, or 281K, to the 255K and you get … hell, I don’t know.
The more you think about what you think you know, the more you know you know less about what you think you know than you thought you know. And then you have to consider the question. What are we really asking?

170. I don’t understand this entire argument about characteristic emission temperature. If you want to understand what the Earth would be like without any atmosphere…..
Apollo 17 heat flow experiment.
http://www.ehartwell.com/afj/Reports/Apollo_17/Mission_report/4.0_Lunar_Surface_Science#4.3_Heat_flow_experiment
Remember that compared to the Earth, the Moon has far less internal heat. Also remember that the surface of the Moon is essentially an insulator against heat, it retains very little of it after dark. This can be seen in this paper of a reanalysis of the Apollo surface temperature record from Apollo 15 and 17.
After dark the temperatures falls almost instantaneously to about 110 degrees k (Apollo 15 and 17 were at low lunar latitudes). The slow decline in temperatures over the next 720 hours down to the ~75k level is from residual heat radiated from the Moon. The statement that if the Moon had a 24 hour rotational period the temperatures would be much higher is shown to be false by the data shown in figure 3.
http://www.lpi.usra.edu/meetings/lpsc2006/pdf/1682.pdf
Thus ANY Atmosphere provides a “greenhouse” effect.
Fools and children in the climate change world simply don’t get out of their little box to find out what the world is actually like.
Oh, by the way, a major climate skeptic is one of the people (Dr. Harrison Schmidt), who installed one of these two experiments and who is the only professional geologist to visit the Moon.

171. Reed Coray says:

First, I want to thank Anthony for “Guest Posting” my thoughts.
Second, I want to thank all commenters–both pro and con. Many valid points have been raised. Some commenters have asked: What’s the issue? The answer is clear and it is “X”. The problem is, the commenters don’t agree on “X”. We each have our own personal experiences, educational backgrounds, and biases. What is “correct” to one person is an absolute “falsehood” to someone else. We (and I am especially including myself) should realize that nature (which includes mankind) will behave according to a set of laws that we may or may not understand. Those laws, not opinion pieces, will determine what happens. This is not to say we shouldn’t try to understand those laws and base our actions on our understanding of their implications. It does, however, say that our “understanding of their implications” may be incorrect, and hence our actions may worsen not help the situation. As Humphrey Bogart said in the movie The African Queen: “You pays your money and takes your chances.”
Third, it was never my intent to quantify the temperature effects of greenhouse gases on the Earth’s temperature (surface, subsurface, water, land, atmosphere at any altitude, etc.). Nor was it my intent to discuss the “Greenhouse EFFECT”, whatever that means. My intents were to (a) describe the algorithm I believed is widely used to claim a 33 K “Earth temperature difference with and without greenhouse GASES”, and (b) describe why I believed that algorithm’s treatment of greenhouse gases is internally inconsistent–specifically the algorithm computes two temperatures each of which is a function of and requires the presence of greenhouse gases, and thus to claim the difference of those two temperatures represents the temperature difference with and without greenhouse gases is internally inconsistent and hence illogical.
Thank you all.
Reed Coray

richard verney says:
December 27, 2011 at 9:52 am:
“These inputs taken together are sufficient to maintain the temperature caused by the pressure of the atmosphere.”
pV=nRT
Which comes first – the temperature of the atmosphere or its pressure?

173. richard verney says:

@Jim Masterson says:
December 27, 2011 at 7:03 am
////////////////////////////////////////////////////////////////////////////
Jim
Others have hinted at the point that you so clearly have made and you are right to point this out. Regrettably, seeking to compare apples and pears is a common meme in the world of climate science.
The majority of the data is significantly flawed and not fit for purpose. It should have huge error bars and yet proponents of AGW would have one believe that the position is certain and measurements can be made to one hundredth of a degree!
They always use averages despite the fact that in the real world, the average condition is rarely encountered and tells us little about what is in reality going on. If a process is unknown or complex, it is either ignored or some fudge factor applied.
The Earth is nothing like a Blackbody so why would one expect a BB calculation to be accurate? Indeed, one point often overlooked is that we have no idea as to the area of the Earth. We look at it as if it were a uniformly smooth sphere of constant radius when we know that is not the case! What is the surface area of the Earth with all its valleys and peaks? I do not know but to give insight, the coastal distance of Norway is approximately the same as the circumference of the Earth, (ie., approximately 25,000 miles). Norway is just a small country and occupies only a small part of the circumference of the Earth and yet its coast line is equal in distance to the circumference!
The point I make is that the surface area of the Earth is nothing like that calculated from a uniformly smooth sphere (it is far larger) and each point has differences in emissivity, absorption etc such that the Earth is nothing like a Blackbody.
There are fundamental issues with the entire concepts behind the GHG theory. Now that observational data and models have ‘decoupled’, there are reasons to suspect that these fundamental issues are material and that the GHG theory is unsound. Whether this unsoundness is merely a matter of degree or more significant will no doubt become apparent the longer the ‘divergence’ continues between observational data and model projections. The heat is on to find the ‘missing heat’ or perhaps it is merely an apparition and is not missing because it was never there.

174. richard verney says:

December 27, 2011 at 11:24 am
//////////////////////
Obviously, it is a matter of some conjecture but it is likely that the temperature comes first in that it is all part of the process of how the planet was formed (out gassing of gases built up in the cooling core/mantle which gases would have been very warm) . Thus, when the Earth was cool enough to sustain an atmosphere, the atmosphere probably in its initial state would have been very warm since although the sun may have been somewhat cooler than today the Earth was hot. At this early stage, the amount of heat being radiated was high and thus over time the atmosphere cooled to the temperatures that the Earth has experienced these past billion years.

175. Shelama says:

Anthony, your credibility and that of your site remain unchanged. Sincerely. Keep up the good work.

176. richard verney says:

December 27, 2011 at 11:13 am
Doug Proctor – do you have any data on how temps change with depth below the seabed? Presumably this is data that offshore drilling people will know?
John Eggert is asking an interesting question here:
http://johneggert.wordpress.com/2011/11/06/more-on-why-is-the-ocean-so-cold/
//////////////////////////////////////////
About a year ago, I posed similar questions on this site as to whether the seabed helped warm the deep oceans. I enquired what if the oceans were drained, what would be the temperature at the valley floor (ie., at the deepest ridges of the now dry seabed)? In this regard, the average depth is just over 12,000 ft and the deepest depth somewhat over 36,000 ft. First, there would be more air pressure and hence more adiabatic heat. Second, one would, because of the depth of the valley, be standing nearer the mantle/core such that one may expect more heat to be conducted from the surface of the now dry seabed (this second point being relevant to whether the seabed heats the deep ocean). Third, the ocean crust is very much thinner (approximately on average 4 miles thick whereas continental crust is on average 25 to 30 miles thick) again meaning that seabed is far nearer the mantle/core such that perhaps more heat is being conducted from the mantle/crust. The latter two points suggested to me that if the oceans were to be drained, the now dry seabed would be warmer to the touch than would continental land such that the seabed if ever so slightly warm may to limited extent be warming the deep oceans from below.
Unfortunately I never got a satisfactory answer but I when I did some research I came across some borehole data which suggested that the continental crust has a temperature profile of about 25 to 30 deg C per kilometer depth and the ocean crust has a higher temperature profile. Unfortunately, since I have moved house, I cannot locate the relevant paper and borehole data. I do come across it, I will provide a link.

177. Let’s forget those rotating plates, black bodies and insulation integrals from north to south.
Why is our day colder than Moon’s day? Simply because
– air cools the surface and goes up, allowing cold air to sink and cool the surface
– water (condensed “greenhouse gas”) cools the surface by evaporation
– clouds (condensed “greenhouse gas”) shield the surface against the sunlight
– ice/snow (solidified “greenhouse gas”) reflects almost all sunlight, so the iced surface does not warm the atmosphere.
Result: our day is COLDER than Moon or black body thanks to presence of bulk atmosphere and water with its hydrological cycle and three states. So the “+33K” difference, even much more because our day is colder, must occur during the night.
Why is our night much warmer than Moon’s night?
If I go outside during the night, I am warmed by direct contact between my skin and warm bulk atmosphere. Bulk atmosphere holds the daily heat. Bulk atmosphere is nitrogen and oxygen, which got warmed during the day by conduction and convection. If they radiate IR (they should, since every object with >0 K temperature radiates something), they warm us, and there are 999,900 N2 and O2 radiating molecules per 100 anthropogenic CO2 molecules. If they do not radiate (as some of you claim), then they just hold the daily heat, effectively warming the night to much higher temperature as without their presence. Simple as that.
“Greenhouse gas” does nothing, if there is not a bulk atmosphere. NOTHING. Mars blackbody and practical temperature is the same 210K, despite its thin atmosphere contains 95% CO2, or 6,000 ppm. The point is, it does not have the bulk atmosphere.
http://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html

178. mkelly says:

richard verney says:
“First, there would be more air pressure and hence more adiabatic heat.”
You need to decide whether we keep the same volume of gas or would you increase the volume to account for the volume of the oceans. If the former then some areas would see a drop in temperature as there would be a thinner layer over what are now continents. If the latter then yes, the Marianas Trench would be warm indeed.

179. R. Gates says:

Juraj V. said:
Why is our night much warmer than Moon’s night?
If I go outside during the night, I am warmed by direct contact between my skin and warm bulk atmosphere. Bulk atmosphere holds the daily heat. Bulk atmosphere is nitrogen and oxygen, which got warmed during the day by conduction and convection. If they radiate IR (they should, since every object with >0 K temperature radiates something), they warm us, and there are 999,900 N2 and O2 radiating molecules per 100 anthropogenic CO2 molecules. If they do not radiate (as some of you claim), then they just hold the daily heat, effectively warming the night to much higher temperature as without their presence. Simple as that.
“Greenhouse gas” does nothing, if there is not a bulk atmosphere. NOTHING. Mars blackbody and practical temperature is the same 210K, despite its thin atmosphere contains 95% CO2, or 6,000 ppm. The point is, it does not have the bulk atmosphere.
_____
N2 and O2 have no absorption or emission in the IR range at all. It is the greenhouse properties of water vapor, CO2, and to a lessor extent, N2O and methane that hold all the IR, or as you call it, the “daily heat” in and keep it warmer at night. Considering what a small fraction all these are of the bulk atmosphere, their effect is remarkable. If you’ve ever spent any time in the desert at night, you can appreciate the potency of water vapor as a greenhouse gas, for even though the bulk atmosphere over a desert is not significantly different (certainly N2 and O2 are the same) than over a more moist region, the little bit of extra water vapor in the air makes a huge difference. The diurnal temperature change in deserts is far more extreme than in more moist regions because of the water vapor. Take it to an extreme, and keep the N2 and O2, (the majority of the bulk atmosphere) but take away all the greenhouse gases and the diunral temperature change gets even more extreme. Greenhouse gases do the heavy lifiting in moderating the diurnal temperture changes.

Juraj V. says:
December 27, 2011 at 1:13 pm
“Why is our day colder than Moon’s day?”
“Why is our night much warmer than Moon’s night?”
////////////////////////////////////////////////////////////////////////////
Juraj is asking the right questions.
During the day it is clear that convective and evaporative processes keep the earth cooler than it would otherwise be.
During the night however there is a real greenhouse effect keeping the surface temperature warmer than it would be without an atmosphere. A real greenhouse or glass house works primarily by providing a barrier to convection. There is a very small additional effect as the glass also provides a barrier to outgoing IR. Earth’s atmosphere is the same. Restricted convection coupled with the limited ability of oxygen and nitrogen to radiate IR is what keeps earth warm at night. The so called greenhouse gases play a more limited role. In a real green house convection is restricted by glass panes. In Earth’s atmosphere the restriction to convection is provided by the inertia of millions of tonnes of air. The primary greenhouse gasses in the atmosphere are therefore nitrogen and oxygen. The condensing and non-condensing trace gasses such as water vapour and CO2 may actually play a greater role in cooling than they do in insulating.

181. Stephen Wilde says:

“In Earth’s atmosphere the restriction to convection is provided by the inertia of millions of tonnes of air.”
Not so.
The restriction to convection is provided by the temperature inversion at the tropopause which is a consequence of solar heating of ozone in the stratosphere.
In my view the height of the tropopause especially the slope of that height between pole and equator is critical because that dictates the entire surface air pressure distribution, the positions, sizes and intensities of the permanent climate zones and thus the rate of energy flow from surface to stratosphere.
There we have the key to cloudiness and global albedo changes with an effect on the quantity of solar energy able to get into the oceans to fuel the climate system.
Solar activity levels appear to have the ability to raise and lower the height of the tropopause especially at the poles.
The likely cause is variations in the mix of particles and wavelengths from the sun differentially affecting ozone quantities at different levels.

Stephen – I thought bulk surface air pressure distribution would be dictated by static equilibrium and gravity with the height of the troposphere being dictated by temperature which, in turn, is largely a function of latitude?

183. richard verney says:

@R. Gates says:
December 27, 2011 at 2:40 pm
////////////////////////////////////////////////////////////
Mr Gates
You are right to point out the effectiveness of water vapour. In pointing this out, one can see how little warming/heat retention CO2 plays. However, the position is very complicated and we do not know precisely how everything interplays.
Let us assume the following: (i) that N2 and O2 cannot radiate heat and therefore can only give up heat by way of convection or conduction; (ii) that there is no CO2 nor water vapour in the atmosphere and the atmosphere consists only of N2 and O2; (iii) at the end of the day the N2 and O2 molecules are at a temperature of about 40degC (this being the air temperature measured in the shade in the usual way say 1.5m above the ground); (iv) it is a windless night; and (v) there is very thick cloud cover at say 50ft extending miles high which has a temperature of say 39 deg C on the desert side and below freezing on the space side (an unlikely scenario over a desert but since we are looking at theoretical compositions of the atmosphere a legitimate exercise).
In this scenario, the night air temperature cannot quickly cool since the thick cloud cover acts like a lid and hinders convection. The warm air (consisting of only N2 and O2) is trapped between the desert surface and low lying cloud (likewise the low lying cloud acts as a barrier to the coldness of space). The N2 and O2 unable to radiate away heat such that the air remains warm all night long.
Now what happens if we add CO2 to the mix?
Is the night air cooled by the addition of CO2 in this scenario since the N2 and O2 molecules whether by collision, conduction or what have you can transfer/give up their heat to a CO2 molecule which can then radiate away the heat (some going up, some going side ways and some going down – materially some going up and through the cloud thereby escaping out to space)?
What happens as we add more and more CO2? Obviously the greater concentration of CO2 the easier it will be for the N2 and O2 molecules to impart their energy onto the CO2 molecules and the more CO2 molecules the more radiators that are present to dissipate the heat.

184. Stephen Wilde says:

slow to follow,
It would, unless the temperature of the stratosphere is affected by a top down process.
A change in troposphere height can be effected by temperature changes from above OR below.

185. richard verney says:

Mr Gates
Further to my last post in which I agreed with you that water vapour makes a significant difference, perhaps I should have observed that whether this is due to the greenhouse properties of water vapour, or whether it is due to the fact that in moist air there is much more latent heat such that it obviously takes longer to cool since there is much more energy to give up/dissipate, or whether dry air convects quicker than moist air, or whether it is combination of more than one or all of these factors is a moot point.
In my view, the moderating of the diurnal range is primarily brought about by the oceans and not by GHGs. Far away from the oceans such as in deserts, the middle of Antarctica large diurnal ranges are experienced. To a lesser extent it is moderated by cloud cover which hinders convection (clouds in turn being a factor of the oceans).

Stephen Wilde says:
December 27, 2011 at 3:25 pm
Stephen, I think you have misunderstood my meaning. I am not suggesting that inertia is providing an absolute barrier to convection as in the glass of a greenhouse or the temperature inversion at the tropopause. Rather I am saying that inertia restricts the speed at which convection can remove heat from the surface. It is not necessary to have a total restriction of convection to experience a greenhouse effect. A green house with several panes of glass removed from the roof will still be warmer than one in which all the panes are removed.

187. Tim Minchin says:

shouldn’t it be fairlyeasy to create a model that also reflects the earths distance from teh sun and it’s rotation on it’s axis (and wobble) and breaks the surface down into 10k x 10k chunks. We would see that using a general number for the total surface is way out when we start looking at the actual energy loss/gain on each sector throughout not just a 24 hour day but a 365 day year and longer.

Stephen Wilde says:
December 27, 2011 at 3:58 pm
“A change in troposphere height can be effected by temperature changes from above OR below.”
Thanks Stephen – do you have any references you can point me towards?

189. gnomish says:

” It is the greenhouse properties of water vapor, CO2, and to a lessor extent, N2O and methane that hold all the IR, or as you call it, the “daily heat” in and keep it warmer at night. ”
BEEP! wrong!
water vapor holds 50,000 times more HEAT than the CO2 on a parcel of atmosphere and it is perfectly capable of conduction without condensation. steam is used to cook things NONRADIATIVELY. it can also condense without radiating – dew you know?
nitrogen, oxygen, friday’s beans – all gasses of every description conduct heat.
you’re not heated by radiation when you swim in a sea of gasses and the fan on my heat sink isn’t moving IR rays, mmk?
and also, ffs – TEMPERATURE is not heat. it is also not a property of radiation physics.
temperature is kinetic energy. nitrogen and oxygen don’t need permission from radiation freaks to be hot or not. dip something in some lox and guess what happens. explain that with radiation – go ahead and make my day… lol

190. gnomish says:

review avogadro, too – all molecules in a small volume have essentially the same temperature as the energy is completely shared by collisions.
co2 does not sit there glowering with special intensity. it gets slapped around just like all the other molecules so its temp is the same as the rest.

191. Stephen Wilde says:

December 27, 2011 at 4:25
Try this:
“if the
temperature differential between the surface and the stratosphere increases the
tropopause must rise. If the differential decreases then the tropopause must fall.
“Suppose, for example, that the surface temperature and the tropospheric
temperature gradient are given and that the temperature of the stratosphere
varies. Then, a cold stratosphere will be associated with a high tropopause (low
tropopause pressure), and a warm stratosphere will correspond to a low
tropopause (high tropopause pressure).”
from here page 14:
http://journals.ametsoc.org/doi/pdf/10.1175/1520-
0442(2001)014%3C3117%3ATTITPR%3E2.0.CO%3B2pm
So if solar variability alters ozone quantities above the tropopause to warm or cool the stratosphere then the tropopause height will change accordingly.
In another thread it has been confirmed that (unexpectedly) an active sun cools the mesosphere by destroying ozone above 40/45km. It has been observed that the stratosphere also cooled when the sun was more active. I think the stratospheric temperature trend follows the mesosphere trend DESPITE an active sun creating MORE ozone below that height.
That makes my hypothesis set out here:
http://climaterealists.com/attachments/ftp/How%20The%20Sun%20Could%20Control%20Earths%20Temperature.pdf
likely to be correct and the only current climate description that fits the observations.

192. Spector says:

From the above, Lord Monckton used the term “Earth’s characteristic-emission temperature.” I believe that is the temperature of a body with a single approximate emission temperature that radiates the same power over its whole spherical surface as it receives as a disk from the sun, based on the albedo-reduced solar constant. This emission temperature should not be confused with any normal average temperature because, according to the Stefan-Boltzmann law, energy emission is proportional to the fourth power of the absolute temperature.
The solar constant is nominally 1368 W/m² and the ratio of the Earth’s surface area to its cross-sectional area is 4:1 so that the Earth only needs to radiate about 342 W/m² to be in stable equilibrium. The nominal reflectivity of the Earth is usually stated to be about 30 percent, which reduces the required IR energy flow to about 239 W/m². The Stefan-Boltzmann law says this is equivalent to a temperature of 254.9 degrees K. Once again, this is an equivalent uniform surface temperature, based on the simplifying assumption that the Earth radiates like a star.

193. beng says:

****
December 27, 2011 at 9:14 am
Think of getting in a submarine and descending in the sea – does the hull experience a change in pressure loading?…
****
Certainly. Just like the change in pressure loading that happens when it ascends, except the opposite. 🙂

194. Reed Coray says:

I’m getting a better understanding of how some CAGW alarmists treat the Anthony Watts, Willis Eschenbachs, Tallblokes, Roy Spencers, etc. of the world. I chanced upon a critique of this post. The URL for that critique is:
Although there are “nuts” on both side of the AGC/CAGW discussion, the case is sometimes made that on a per capita basis the AGC/CAGW side more than the skeptic side engages in name calling, ad hominem attacks, and shrill rhetoric. I believe it would be informative for the AGW and CAGW advocates who have commented on this thread to read that critique and assess its relevance to what I wrote. Everyone is entitled to his opinion; and since I entered the arena, criticism comes with the territory. That’s as it should be. However, if I were an AGC and/or CAGC advocate, I would try to suppress critiques such as the one given above. Then again, maybe the critique author knows more about my intentions, background, and beliefs than I do.

195. SGW says:

R. Gates says:
December 27, 2011 at 2:40 pm
“It is the greenhouse properties of water vapor, CO2, and to a lessor extent, N2O and methane that hold all the IR, or as you call it, the “daily heat” in and keep it warmer at night.”
If so, why diurnal temp difference in Mars is so big despite having about 9 times more CO2 in much smaller surface area in its atmosphere?

196. John Marshall says:

Beng thinks that the adiabatic heating is a one time only event and has long gone.
Wrong. The compression due to mass is there all the time with changes, which are happening all the time both locally and between night and day hemispheres, only replacing losses due to radiation to space.
How does Jupiter maintain its temperature of 280K at the 1000mb level without any greenhouse gasses?
The theory of GHG’s has yet to be shown to be true within the bounds laid down by the laws of thermodynamics. Sorry Beng.

197. John Brookes says:

As has been known since 1966,[58] Jupiter radiates much more heat than it receives from the Sun. It is estimated that the ratio between the power emitted by the planet and that absorbed from the Sun is 1.67 ± 0.09. The internal heat flux from Jupiter is 5.44 ± 0.43 W/m2, whereas the total emitted power is 335 ± 26 petawatts. The latter value is approximately equal to one billionth of the total power radiated by the Sun. This excess heat is mainly the primordial heat from the early phases of Jupiter’s formation, but may result in part from the precipitation of helium into the core.[59]

198. richard verney says:

Dennis Ray Wingo says:
December 27, 2011 at 10:54 am
///////////////////////////////////
Dennis
Very interesting papers.
In one of my earlier posts, I suggested a rotation of about an hour was necessary to maintain temps. Fig 3 shows that the temps drop off like a cliff (the rock has little stored heat capacity at the depths of the probes) and it may be that an even faster rotation would be required. One would need more probes covering a different range of depths and a different scale to Fig 3 to make a more realistic estimate of the required speed of rotation.

199. richard verney says:

SGW says:
December 28, 2011 at 12:58 am
///////////////////
The argument that the AGW proponents raise is that the atmosphere is so thin. Obviously, it is true that the Marsian atmosphere is thin, but whether this is the reason why the GHE cannot be seen on Mars is moot.

200. beng says:

*****
John Marshall says:
December 28, 2011 at 2:36 am
Beng thinks that the adiabatic heating is a one time only event and has long gone.
Funny how I think, huh? No, adiabatic compression-heating is a constant process in a dynamic atmosphere. But so is adiabatic decompression-cooling.
Wrong. The compression due to mass is there all the time with changes, which are happening all the time both locally and between night and day hemispheres, only replacing losses due to radiation to space.
The act of air compression is work, so produces heat. An equal amount of air elsewhere, tho, HAS to move upward to replace the air that moved downward (nature abhors a vacuum). That air decompresses and cools by essentially the same amount. The net result is zero.
How does Jupiter maintain its temperature of 280K at the 1000mb level without any greenhouse gasses?
Leftover internal heat! Jupiter’s mass is huge, and preserves much of the initial compression heat from formation. Earth’s residual heat is tiny in comparison (& lost much more quickly), & the insulating crust slows heat loss down to a minimum (guessing a couple of W/m2). Insignificant compared to solar input (except at the mouth of a volcano).
The theory of GHG’s has yet to be shown to be true within the bounds laid down by the laws of thermodynamics. Sorry Beng.
Don’t be sorry, John. I’m not the one lacking a basic understanding of thermo. 🙂

201. R. Gates says:
December 26, 2011 at 11:00 am
A shocking revelation isn’t it. At the moon’s equator on the sunlit side the temperature is around 120C whereas on the opposite dark side, the temperature is about -230C. Amazing what an ocean and greenhouse atmosphere can do!
———————————————————————
That has almost nothing to do with oceans and atmosphere, and everything to do with rotation speed. The surface of the Earth near the equator with cloudy skies heats up at a typical maximum rate of just under 1 C per hour in the late morning. Luna near the equator also heats up at a maximum rate of around 1 C per hour in the late morning. The difference is that earth rotates about 29.5 times faster than Luna, so Luna receives 29.5 times as much gross energy per unit area per day cycle.
Plugging in 393K, 288K and 43K (Note: -230C = 43K is only correct for those parts of Luna in permanent shadow, so not germane to the current thread), one gets, using 288K as unitary, (393/288)^4 or 3.4673 times as much energy in Luna late afternoon equilibrium, and 0.0005 times as much energy in Luna permanent shadow. All this tells us, is that, absent atmosphere, Earth would be a pretty good insulator.
For typical daily ranges on Earth, we get a ~22C range in the desert, and a ~11C range where people typically live. The minimum to maximum time is around 8-9 hours. For Luna near the equator, the daily range is typically from 165K to 375K, a range of 210C, in a period of about 292 hours. If Luna had a 24 hour day, and an average temperature of 255K, then it would be about 251K shortly after dawn, and about 259K in the late afternoon. This gives a range of about 8C.
We can check that by calculating cooling rates for Luna. To cool from 259K to 251K would take about 13.5 hours. Since Luna would, in a 24 hour day, take just under 10 hours to go from minimum to maximum, these numbers are definitely in the ballpark. Note that this gives a daily range for Luna that is LESS than the daily range for Earth.

beng – check out variation of pressure with horizontal displacement vs. with variation with vertical displacement. Check out temperature and pressure readings at any weather station as we move from night to day and back again.
I think you have misunderstood what happens wrt to lapse rate in the atmosphere but I do not have the time to argue it through with you. Check your definition of “adiabatic”. Search for “dry adiabatic lapse rate”, “hydrostatic equilibrium” and “hypsometric equation”. Apologies if you are on top of this and I have misunderstood your pov.

Stephen Wilde says:
December 27, 2011 at 5:56 pm :
Thanks Stephen – references appreciated.

204. pochas says:

@John, Being
Neither of you has any cause to be sorry. You are both a lot further along than the typical climate scientist, who builds his AGW chimera in some parallel radiative universe were convection does not exist.

205. gnomish says:

” if I were an AGC and/or CAGC advocate, I would try to suppress critiques such as the one given above.”
if you think that article is nasty- you ain’t been around.
be a big boy. suppress your butthurt instead. nobody needs more censors. also- give up subjunctives- you are not a CAGW so stop with the fairy tales altogether.
on the other hand, encourage the fools to indulge their mistakes and fantasies – it helps them move further down the road to extinction.

206. Matter says:

Let’s wave the magic wand and take out all greenhouse gases, but leave the surface as it is.
THe surface albedo is about 0.125 (Trenberth, 2008). So you’re actually looking at an equilibrium temperature of ~270 K. Assuming your thermally conducting sphere, the whole thing would freeze. A planet entirely covered in sea ice has a much higher albedo. Enceladus is an ice ‘planet’ – and it has a Bond albedo close to 1.
So the temperature would fall much, much further below freezing. You’d end up calculating that the greenhouse effect is many tens of degrees more useful than we think. Using a simplistic approach, I could argue that without it, we’d be hundreds of degrees worse off.
Another way of looking at it is to keep the gases, but magically remove their greenhouse effect. You cool Earth, ice spreads, it cools, clouds eventually disappear (too cold), ice spreads, ice spreads, albedo is close to 1, temperature falls far, far below the 255 K given.
So you’re right, this quantificatin of the greenhouse effect is only a rough one, in reality the presence of the greenhouse effect keeps us much warmer than the typical ‘back of the envelope’ calculation you gave, which is common in introductory planetary physics textbooks.

207. iya says:

The question is, would the (surface air-) temperature change, if earth had a much denser atmosphere, of equal composition?
Can’t we test it approximately by putting two bottles in the sun; one with compressed air and one with standard pressure?

Stephen Wilde says:
December 27, 2011 at 5:56 pm
This one worked for me:
http://journals.ametsoc.org/doi/pdf/10.1175/1520-0442%282001%29014%3C3117%3ATTITPR%3E2.0.CO%3B2
“The Tropopause in the Polar Regions”
GUNTHER ZANGL
Meteorologisches Institut der Universitat Munchen, Munich, Germany
KLAUS P. HOINKA
Institut fur Physik der Atmosphare, DLR Oberpfaffenhofen, Wesseling, Germany
Journal of Climate 15 JULY 2001

209. SGW says:

Matter says:
December 28, 2011 at 9:32 am
“Let’s wave the magic wand and take out all greenhouse gases, but leave the surface as it is.”
I think your reasoning has several problems.
You can’t take water wapor out from the atmosphere and leave the sea in place. Even if you could, day time heat from the sun would keep the sea from freezing anyway. Without any clouds acting as shades it would be obvious. And 95% percent of the atmosphere would still be there (nitrogen and oxygen) and do their job via convection and conduction giving night time insulation.

210. Bomber_the_Cat says:

Reed Coray says:
December 27, 2011 at 8:38 pm
“It would be informative for the AGW and CAGW advocates who have commented on this thread to read that critique and assess its relevance to what I wrote”
Reed, you don’t really want to ask this do you?
I believe the author of the article described you as “an ill-informed but enthusiastic amateur”. Do you object to this description?
Now there are many commentators here to which the same could be applied, and most of those support you. Not because they understand what you wrote, but because as Judith Curry once said, they decide if the article is for us or against us. That is the limit of their reasoning.

211. Reed Coray says:

Bomber_the_Cat. You may be right–asking AGW/CAGW proponents to read what others on their side of the issue say is probably counterproductive. However, I believe not all of the pro-AGWCAGW commenters on Anthony’s blog respond based solely on the criterion that “the article is for us or against us.” As far as the “ill-informed but enthusiastic amateur” comment goes, I agree with the “enthusiastic amateur”, but not necessarily with the “uniformed” part. I’ve been following much of the discussion for four years. There is some truth in what “gnomish” (December 28, 2011 at 9:04 am) said. I should probably have kept my mouth shut. However, I thought at the time that some commenters, in particular Joel Shore, might be interested the blog.

212. R. Gates says:

Thomas L. said:
“To cool from 259K to 251K would take about 13.5 hours. Since Luna would, in a 24 hour day, take just under 10 hours to go from minimum to maximum, these numbers are definitely in the ballpark. ”
_____
Your numbers are way off. The temperature of the moon’s darkened surface has been measured during a total lunar eclipse, with a maximum cooling of about 30K per hour. The first few centimeters of rock and dust on the moon cool rapidly during a lunar eclipse.

213. Spector says:

The University of Chicago ‘MODTRAN radiation code’ utility can be used to calculate the *raw* effect of changing the amounts of greenhouse gases in the atmosphere. For my own purposes, I usually use the default conditions and tropical locality with no clouds or rain. This program, using atmospheric component, absorption data, normally calculates radiant energy flow at various levels in the atmosphere as a function of surface temperature entered as an offset from 299.70 deg K.
For greenhouse effect calculations, I usually search for the temperature offset entry required to obtain a nominal energy flow, seen looking down from 70 km up of 292.993 W/m²—a number the program likes to produce. This standard value was based on my estimate of the incoming average daily solar radiation.
Starting with default conditions, except for a CO2 concentration of 396 PPM, I find that a surface temperature of 301.18 deg K is required to produce the nominal radiant energy flow (offset: 1.48).
If I set the CO2 concentration to zero, the nominal flow surface temperature drops to 293.58 deg K (offset: -6.14), an apparent 7.62-degree decrease.
Next, if I change the nominal CH4 from 1.7 ppm to zero, the nominal flow surface temperature drops to 293.17 deg K (offset: -6.53), an apparent 0.39-degree decrease.
Then, if I change the tropospheric ozone from 28 to zero and the stratospheric ozone scale from one to zero, the surface temperature for nominal flow drops to 291.75 deg K (offset: -7.955), an apparent 1.42-degree net decrease.
Finally, if I change the water vapor scale from one to zero, the required surface temperature drops to 275.86 deg K (offset: -23.839) for a final 15.89-degree decrease. Thus a 25.32-degree total calculated temperature decrease is obtained when all program optional greenhouse gases are removed.
Note that 275.86 deg K is *not* a calculation of the “Earth’s characteristic-emission temperature,” a hypothetical construct.

214. Crispin in Waterloo says:

@ Dave Springer
“No, you’re wrong. The ocean has an effective albedo near zero and it radiates very little. Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation.”
+++++++++
You surprise me. It may be true that the ocean loses most of its heat by evaporation, however water has an emissivity of about the same as black oil. Look it up. Write it down.
Water radiates heat very efficiently, certainly better than concrete or brickwork which are both about 0.93. Water is about 0.99. You can show this by covering a boiling pot with a lid. The water will boil at a much higher rate not only because of the insulating layer of air+lid and the limiting of evaporation, but because of the retention of a couple of hundred watts of energy that normally radiates from the water surface of the open pot onto the ceiling above.
Painting the oceans black, for example, would barely increase the radiation of heat at all.

215. Myrrh says:

Crispin in Waterloo says:
December 29, 2011 at 12:08 am
@ Dave Springer
“No, you’re wrong. The ocean has an effective albedo near zero and it radiates very little. Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation.”
+++++++++
You surprise me. It may be true that the ocean loses most of its heat by evaporation, however water has an emissivity of about the same as black oil. Look it up. Write it down.
Water radiates heat very efficiently, certainly better than concrete or brickwork which are both about 0.93. Water is about 0.99. You can show this by covering a boiling pot with a lid. The water will boil at a much higher rate not only because of the insulating layer of air+lid and the limiting of evaporation, but because of the retention of a couple of hundred watts of energy that normally radiates from the water surface of the open pot onto the ceiling above.
Painting the oceans black, for example, would barely increase the radiation of heat at all.
====================
And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?

216. iya says:

“And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?”
Yes, it all get’s absorbed in the first centimeters, but there’s not much infrared from the sun, most is from the atmosphere. As it does not mix that well, it’s not very plausible that the “missing heat” is in the deep ocean. If you go swimming in the summer, you’ll notice the first centimeters are much warmer than even 1m below the surface.

217. beng says:

*****
December 28, 2011 at 7:36 am
<i.Apologies if you are on top of this and I have misunderstood your pov.
****
slow to follow, I was following Patrick Michaels’ (who at the time was the VA State Climatologist before they found a way to get rid of him) objections to AGW back in the early 80’s when the AGW industry was just getting started. I was following John Daly’s site, then Climate Audit when it first appeared (~2000 IIRC). I’m a mechanical engineer w/a solid foundation in thermo. And I’ve heard the argument about “compression heating of the atmosphere” for years.
Instead of replying w/a long post, I recommend a site I was reading extensively some months ago:
http://scienceofdoom.wordpress.com/
and search for “Venus”.
He’s listed here as “Pro AGW”, but I found him accurate and w/o an apparent agenda (unbelievable). Eventually, the specific discussion there about Venus & GH effects reached an inconclusive end when some aspects of Venus’ lapse rate didn’t make sense according to standard theory. So I’m quite open to the idea that there is indeed something we don’t understand about Venus (& Earth too). But it ISN’T simple adiabatic compression/decompression — that’s straightforward thermo.
Read away — it’s a good discussion.

218. Myrrh says:

iya says:
December 29, 2011 at 6:47 am
“And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?”
Yes, it all get’s absorbed in the first centimeters, but there’s not much infrared from the sun, most is from the atmosphere. As it does not mix that well, it’s not very plausible that the “missing heat” is in the deep ocean. If you go swimming in the summer, you’ll notice the first centimeters are much warmer than even 1m below the surface.
————————
Thermal infrared direct from the Sun is invisible, it is the heat we feel from the Sun, the Sun’s thermal energy on the move to us. It arrives on Earth at the same time as non-thermal visible light, you can feel thermal infrared heat is direct from the Sun for example when a cloud temporarily blocks the Sun on a bright coldish spring day. It is readily absorbed by the body, being mostly water, it moves water molecules into vibrational states which heats up water.
Water has also a very high heat capacity, unlike say carbon dioxide with its low heat capacity, (which means it heats up very quickly and as quickly, practically instantly, gives up that heat), water actually stores the energy. It takes a long time to heat up but so keeps it longer, it takes longer to cool down – why we have water in our radiators in central heating systems. The temperature of the ocean isn’t a complete picture of the amount of heat energy it is absorbing and storing because heat capacity is the amount of heat energy required to raise temp one degree per given volume and water readily absorbs vast amounts of it before you’d notice any temperature change. Water has to absorb about four times as much heat energy as the same amount of air to raise its temperature 1°C. That’s how lakes and oceans have little temperature change even while the air around them is changing.
Water is transparent to visible light, visible light can’t heat the oceans at all.
Heat radiation from the Sun heats oceans, light radiation from the Sun is simply transmitted through.
The ‘energy budget’ as it is depicted with visible light claimed to heat land and oceans, is junk physics. Regardless it is now taught everywhere, visible light isn’t able to do this.

219. Phil. says:

iya says:
December 28, 2011 at 9:35 am
The question is, would the (surface air-) temperature change, if earth had a much denser atmosphere, of equal composition?

Yes because of broadening of the spectral lines of the GHG spectra, this is why the CO2 isn’t as effective on Mars.
See here for part of the CO2 spectra on the two planets:
http://i302.photobucket.com/albums/nn107/Sprintstar400/Mars-Earth.gif

220. Phil. says:

richard verney says:
December 27, 2011 at 3:57 pm
@R. Gates says:
December 27, 2011 at 2:40 pm
////////////////////////////////////////////////////////////
Mr Gates
You are right to point out the effectiveness of water vapour. In pointing this out, one can see how little warming/heat retention CO2 plays. However, the position is very complicated and we do not know precisely how everything interplays.
Let us assume the following: (i) that N2 and O2 cannot radiate heat and therefore can only give up heat by way of convection or conduction; (ii) that there is no CO2 nor water vapour in the atmosphere and the atmosphere consists only of N2 and O2; (iii) at the end of the day the N2 and O2 molecules are at a temperature of about 40degC (this being the air temperature measured in the shade in the usual way say 1.5m above the ground); (iv) it is a windless night; and (v) there is very thick cloud cover at say 50ft extending miles high which has a temperature of say 39 deg C on the desert side and below freezing on the space side (an unlikely scenario over a desert but since we are looking at theoretical compositions of the atmosphere a legitimate exercise).

Without water vapor in the atmosphere where does the cloud cover come from?

221. Phil. says:

John Marshall says:
December 27, 2011 at 3:39 am
I am sorry to disagree with the famous Lord but the theory of GHG’s violates the laws of thermodynamics.

Certainly doesn’t, which suggests that you don’t understand the thermodynamics
A moments thought will show that if a molecule of CO2 is preferentially heated it will immediately share this heat with its surroundings. It cannot do anything else. It cannot ‘store it’ because the 2nd law must hold true. Any heated volume of gas in the atmosphere will convect due to density changes and convecting air in the atmosphere must expand and loose heat due to adiabatic expansion.
Indeed you don’t understand thermo, ‘adiabatic’ means with no heat transfer!
Adiabatic cooling involves a drop in temperature as the gas expands, not a loss in heat.
The gas will very soon be cooler than the surface so cannot re-radiate heat to the surface, heat can only flow by whatever means from hot to cold as the laws of thermodynamics dictate.
Not true, radiation occurs in all directions irrespective of what the temperature of its ultimate target, how would we receive the 3K background radiation from the ‘Big Bang’?

222. Phil. says:

Rosco says:
December 26, 2011 at 2:07 pm
Nitrogen and Oxygen may not absorb much IR but they MUST become heated and MUST radiate.

They not only mustn’t they radiate but it is experimentally observed that they don’t.
The amount of IR from water vapour is at best a trace amount and the amount of IR from CO2 is tiny – unless CO2 possess some hidden magical power that science has not detected yet.
On the contrary, the emission of H2O and CO2 is substantial and is experimentally observed.

223. Phil. says:

steveta_uk says:
December 26, 2011 at 5:15 am
John Brookes (December 26, 2011 at 4:18 am)
(245+265) / 2 = 255
sqrt( sqrt( ((245^4)+(265^4)) / 2 ) ) = 255.5864
Not what I was expecting.

Why ever not, it’s high school math?
If the mean temperature is T and the range is a then what you want is
∜( (T+a)^4+(T-a)^4))/2
all the odd terms in a cancel so you’re left with:
∜( T^4+6(Ta)^2+a^4)
in the case you quoted T = 255 and a =10 you get the result above, if a=20 it’s around 256, for small a the deviation from T will be small.

224. George E. Smith; says:

Well I don’t understand just where all of this methane flap comes from.
For starters, at a black body Temperature of 288K which Trenberth et al use for their earth surface LWIR emission, the spectral peak wavelength is 10.1 microns. elementary high school BB radiation theory says that only 1% of the total BB spectral energy falls below one half of the peak wavelength or 5.05 microns, so the shorter wavelength methane and nitrous oxide absorption peaks, are quite innocuous as is the CO2 4 micron band. Only 25% is emitted below the peak wavelength of 10.1 microns (obtainable from Wien’s Law), and only 1% is emitted above 8 times the peak wavelength or 80.8 microns.
And the peak spectral irradiance varies as the fifth power of the Temperature, so if the equivalent BB temperature is only 255K and not 288K then the peak spectral irradiance is only 54% as large, and that peak is moved to an even longer wavelength of about 11.4 microns moving it further into the clear atmospheric window, and even away from the 9.6 micron Ozone band.
Note that the Planck formula BB spectral calculations represent the limiting envelope of any real thermal radiation source emissions, so those numbers above are worst (or best) case.
I simply cannot see how GHGs sans H2O can possibly absorb the majority of the surface LWIR emissions.
And any radiation shorter than about 4.5 microns has to be from incoming solar spectrum energy; it can’t be earth generated thermal emissions, so any atmospheric absorption below that wavelength has to be solar energy trapping in the atmosphere, which ALWAYS results in LESS solar energy being absorbed by the earth, so that leads to cooling of the surface.

225. Spector says:

RE: Main Article
“Since I’m not sure what the definition of the ‘Earth’s characteristic-emission temperature’ is, I can’t disagree with his claim that its value is 255 K.”
That is the temperature indicated by the Stefan-Boltzmann law to be characteristic of the average rate of energy flow escaping to outer space per square meter of the Earth’s surface.
“For temperature A almost everyone uses a ‘measured average’ of temperatures over the surface of the Earth.”
Since this calculation is based on average energy and the Stefan-Boltzmann law says this energy is proportional to the fourth power of the *absolute* temperature, what we really have is a fourth root of the mean fourth powers average temperature in degrees K. This applies to any calculation based on average radiant energy.
The one potential weak link in this equation is the assumption of a 30 percent average optical reflection factor that reduces the incoming 1368 W/m² nominal solar radiation to 958 W/m² over the disk of solar radiation intercepted by the earth. This hypothesis assumes that the atmosphere has absolutely no effect on determining the temperature of the Earth. That, of course, is unlikely in reality but this is a hypothetical value, not a real prediction, except, perhaps, on an airless planet having an average surface optical reflection factor of 30 percent.
As atmospheres without greenhouse gases cannot cool themselves except by contact with the ground, I think such an atmosphere subject to solar heating might expand indefinitely as it continued to heat and eventually escape to outer space.

226. Matter says:

Spector, I don’t understand your comment:
“As atmospheres without greenhouse gases cannot cool themselves except by contact with the ground, I think such an atmosphere subject to solar heating might expand indefinitely as it continued to heat and eventually escape to outer space.”
The ground has a non-zero emissivity, so it will radiate the heat out. In Earth’s position, it would tend towards a global heat distribution that has a quartic-root-mean-to-the-power-fourth temperature of 255 K.
The atmosphere can still convect and conduct, if it got hotter than the surface then heat would flow throug convection and conduction back into the ground. The ground would cool it off and prevent it expanding indefinitely.

227. Thomas L says:

R. Gates says:
December 28, 2011 at 8:29 pm
Your numbers are way off. The temperature of the moon’s darkened surface has been measured during a total lunar eclipse, with a maximum cooling of about 30K per hour. The first few centimeters of rock and dust on the moon cool rapidly during a lunar eclipse.
——————————————————————-
Hmm. I was using a model – no, not mocking models. Seen many odd numbers for Moon’s temperature. Example: a cable on top of the surface heats from ca. 150K to ca. 310K in a very short period of time (graph appears almost vertical – don’t have table) after sunrise. Then it takes many hundreds of hours to go from 310K to 390K near noon. And drops from 390K down to ca. 310K at sunset. Then plummets to near 165K (again, graph near vertical) in a short period of time, and after that cools very slowly to ca. 150K near sunrise.
I can explain the rapid rise in temperature due to the cable’s cross section at sunrise of 2 * r * length * sin theta, where theta is the angle of the cable to the sun. So the cable acts as a massless object. The cable would cool based on its surface of 2 * pi * r * length. Some radiation is exchanged between the cable and the surface of the moon.
The moon should heat up much slower at sunrise than a cable, as the surface of the moon is at a 90 degree angle to the sun at sunrise. But obviously, I would need to know specific heat, conductivity, and other such fun numbers – like how bumpy the surface of the moon is – in order to have my model match observations. Flipside, a standard Stevenson screen on the moon might give noon temperature as low as 45K or as high as 280K.
What I don’t get is the ‘knee’ in the lunar temperature data. Perhaps it is when conduction from subsurface slows further drop in temperature. That isn’t in my model, and needs to be.