Guest Post by Willis Eschenbach
OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?
a) 33°C (59°F) cooler
b) 20°C (36°F) cooler
c) 8° C (15°F) cooler
The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.
Why is the temperature rise so small? Here’s a diagram of what is happening.
Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.
As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.
The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.
This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.
Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.
As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.
The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:
1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.
2. The climate models underestimate the increase in evaporation with temperature.
3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .
4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.
5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.
6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.
Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:
a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.
b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.
c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.
d) Sensible heat losses increase with the surface temperature.
e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.
so joel shore, what is the specific sources and amounts of the additional w/m^2 after a 3.7 w/m^2 increase due to doubling the co2?
since the total w/m^2, including cloud cover, is around 150 w/m^2 and the total rise in temperature is 33 deg C, it should take a new total of about 13.6 w/m^2 to achieve a 3 deg C rise and your co2 is only going to contribute 3.7 so effectively, there must be 10 w/m^2 additional coming from somewhere.
cba: First, I might quibble with your estimate of 13.6 W/m^2 for a 3 C rise…I think that since the accepted value for the no-feedback case of doubling CO2 is ~1.0-1.1 C, then you need about 11 W/m^2 total to get to 3 C.
But, that is a small point of contention, and in answer to your basic question, these other amounts presumably come from the water vapor feedback, the ice-albedo feedback, and…with the most variation from model-to-model…the cloud feedback.
My basic point is just this: Willis believes that he has shown that the climate sensitivity is small. What I think he has actually shown is that the climate sensitivity in the absence of feedbacks is fairly small and hence it is the estimated positive feedbacks that make the sensitivity large, which is something that we already knew. (Although, like I said, I am still confused as to why his calculation got a climate sensitivity of about half the accepted “absence-of-feedbacks” value when I think, if correctly done, it should get something close to the “absence-of-feedbacks” value. One thing I have been wondering about is how the lapse rate feedback fits into this scenario, but I don’t think it is large enough to account for the discrepancy.)
Believe me, Joe, I’m not doing half a good a job at dispelling the notions of the World’s Leading Climate Scientists as the World’s Leading Climate Scientists are doing for themselves:
The Trenbeth “sulfate aerosol” negative feedback modulation of IR that are not observed from satellites in the IR; or the purloined “heat” from GHG warming unobserved in sea or air
The unobserved Santer predicted tropopause altitude response or tropopause warming magnification in the Tropics
The GISS climate modeling managed by Hansen that, by force fitting feedback over a thirty year period, can match observed temperatures but the predictions of climate bear no relation to natural climate particularly in the Tropics
At some point, these folks might step back and say to themselves, “the assumptions behind all this might not be right.”
Nope. They continue to press Government for drastic changes that will send the Free World back to the stone age.
That’s where my squeaky little barely heard voice comes in to say: stop
Re: Colin Davidson (Mar 22 01:42),
“Anna V(22:13:06, 21MAR10) wrote:
“One, imo, should be saying: the heat capacity of the atmosphere is increased by the existence of H20 and CO2 etc, from x to y, and thus heat transport to space is delayed by z , instead of handwaving watts/m^2.””
I think that is only partially correct. The “heat capacity” is presumably the amount of energy it takes to heat a certain mass of air by a certain amount. The concentrations of Water vapour and CO2 in the air make almost no difference to the heat capacity.
There is no presumption. Heat capacity is the way the heat content of any material is parametrized. From Wiki:
The equation relating heat energy to thermal mass is:
Q = C_\mathrm{th} \Delta T\,
where Q is the heat energy transferred, Cth is the thermal mass of the body, and ΔT is the change in temperature.
In thermodynamic terms this is the way to describe a mass system in a heat environment. Heat is energy and temperature its proxy. When we say that the atmosphere’s temperature has increased by deltaT because of the introduction of extra mass it means the heat capacity has changed by this delta Mass (CO2 in this case).
Joel,
Can you educate me? What exactly is “Lapse Rate Feedback”?
joel shore
well joel, the numbers I showed are quite simple but they use the entire actual Earth values. 150w/m^2 = difference between the 239 w/m^2 actual mean radiated amount from the Earth that balances the incoming (after albedo) value for solar. divide that into the actual 33 deg C rise in T provides the mean sensitivity.
the h2o vapor feedback can be estimated as well. with only a small increment in T – like 5 deg c along the entire atmospheric column, if one held the relative humidity constant – as is usually assumed, one finds one can have only about a 30% increase in the h2o vapor. Running a 1-d model, one can find that this amounts to a little over 3 w/m^2 increase in total forcing which is less than that of the co2 doubling in the first place, making the combined value equivalent to a mere 1 degree or so.
I note there is a big fat PRESUMABLY in the area of your post talking about other factors. You use ice – albedo as one. Note that although there is a big problem with Khiel and Trenberth’s estimates, they call for 0.08 total surface fraction and 0.22 cloud (and atmospheric) fraction of the 0.3+ total albedo. Considering that ocean makes for around 70% of the surface total for a surface albedo of somewhat under 0.08, land accounts for a bit under 0.2 albedo, including all that snow & ice, at least for that part of the surface that isn’t ocean that is visible under about a 62% average cloud cover. That should provide an excellent hint that the small area currently covered in ice and snow on average accounts for very little of the actual albedo. Now an ice age with glaciation is going to make a bit of a difference but that isn’t what we have currently.
That leaves clouds. And you’re trying to tell me that somehow or another, adding an additional 30% h2o vapor to the mix is going to result in a decrease in cloud cover and hence, in albedo and also, presumably in a decrease in convection via the water vapor cycle.
just off hand, I can’t think of a portion of the h2o vapor cycle that is going to offer positive feedback when taken in the context of that cycle. h2o vapor is a lighter molecule than any other predominant molecule in the atmosphere. That alone creates less density and adds buoyancy to moist parcels. In general, lifting air to higher altitudes and lowering air parcels to lower altitudes is rather the zero sum game until one adds in h2o vapor which caries with it heat of evaporation (and sometimes also heat of fusion) and this is carried from lower to upper atmosphere. As the altitude rises, the moist air eventually gives up most of the h2o vapor in the form of liquid h2o and solid h2o particles – giving up the hts of evaporation and fusion to the upper atm. These either precipitate down or form clouds as having supersaturated air is just not acceptable for any length of time. In either case, more surface heat gets transferred up. While it appears that clouds forming is a poorly understood subject, there’s plenty of evidence that pollution, volcanic ejecta, and even cosmic rays contribute to the formation of clouds and provide a mechanism for external involvement in the climate for such things, albeit in a round about fashion.
it makes far more sense for this than does the presumption that heat gets trapped in the oceans. IR is blocked quickly by liquid h2o. Except in very limited circumstances, warm h2o liquid rises. While solar radiation can penetrate well below the surface, even many dozens of meters, IR is trapped at the surface – and that is what increases with increase ghgs. Like trying to boil a pot of water by putting a heating element on top, one has all of the facets working in opposition to bringing heat downward in water.
perhaps one should rethink the status quo when it comes to the actual mechanisms presumed to raise temperature rather than lowering temperature. After all, you’re still a far cry from explaining just how something that should further reduce waming with negative feedback is somehow or another being assumed to increase it with postive feedback and that it’s in an area that is very poorly understood.
Colin,
I see what they are doing now. A CO2 doubling gives 3.7 W/m2 at TOA (or actually at the 255 deg C layer) which is amplified by the existing GH effect by 2.45 to get 9 W/m2 at the surface. This will increase the surface temp 1 C and produce enough water vapor to give another 3.7 W forcing at TOA, which of course is also amplified to 9 W to the surface. These 18 W will lead to reduced snow and ice cover ++ to produce the third 3.7 W at TOA. Then we have 27 W at the surface and can subtract a few W for evaporation.
Joel
Regarding the mistake of Arthur P Smith
I presume you attended classes where the Kinetic Theory was explained to you!
One familiar equation should be
Average KE of molecule =1/2 mv^2 =3/2kT
v is the RMS speed, T is the Kelvin Temperature the other quantities should be familiar to you if you attended an elementary KT class.
What happens if you throw a ball up in the air in a gravitational field?
Yes that’s right the speed decreases until it stops.
If the molecule moves up its speed and its KE decreases .
The KE changes into GPEnergy(mgh)
If the KE decreases then by equation above its Temperature DECREASES.
This gives rise to the lapse rate.
This elementary derivation does not involve the question of whether or not the molecule is radiative or not.
That someone who considers himself to be a Physicist does not seem to be aware of this is unusual.
I thought it was a momentary lapse and he would correct the record but he has not.
This is most disappointing as apparently he is a co author with you in the attempt to refute G&T.
I was hoping that G&T were going to be given a more rigorous testing than has been the case up till now.
It looks that this is now unlikely to be the case.
lgl (01:32:52) :
The water vapor forcing is a feedback loop so it was a bit silly of me describing it the way I did. It is of course the resulting 3 deg C that produces enough vapor to give the additional 3.7 W.
Re: HankHenry (Mar 20 16:58),
I had to look up R value. It seems it is thermal resistance, and it is a measure of insulation
here is a calculation of the atmosphere,
http://www.energyadvocate.com/fw79.htm
though I have not checked it. It says that in the units used in the US the value for the atmosphere is 0.8, where fiber glass is 11.
anna,
the R value also roughly approximates thickness of plywood equivalent. 0.8 is around 3/4″ of plywood. The fiberglass reduces convection and radiation making it an R-11 for a 3 1/2 inch void, roughly equivalent to 11 inches of plywood – which is conduction only. It’s a totally ‘engineering’ type of number and applied to the atmosphere essentially gives the equivalent thickness of plywood necessary to achieve the same heat flow per area – which should be ft^2 rather than m^2. Not sure why anyone would do that other than to just compare the lack of insulation factor is present in the atmosphere. BTW R-12 is rather pathetic for walls too,
Note that buoyancy forces of air are directly proportional to the volume coefficient of expansion (which, for ideal gases, is the reciprocal of the absolute temperature). This is well known to helicopter pilots, who are well aware that colder air has a lot more lift than warm air.
There is a lot we don’t know that we could all study together, but “global warming” has become the Civil War amongst those who seek to increase knowledge of atmosphere and climate.
Moving away from global warming, and climate modeling is still very weak, and we don’t even know the origins of phenomena such as El Nino.
We don’t know how the historical climate is influencing the present climate via movement of heat in the oceans, and I think we have to move away from GHG modeling someday.
We have a lot more in common than we do differences, and it starts with the motivation to be of use and benefit to humanity in some way
I’m moving away from this discussion, salutations to Joel
bgvavlentine@verizon.net
Brian – I might add *to all pilots*…Lift of rotary wings, fixed wings, even displacement of hot air balloons is proportional to density of the gas they are in, which is inversely proportional to temperature of that gas. I’m not sure most pilots think of the “coefficient of expansion” of the gas, and least not most pilots.
cba (23:44:32, 22MAR10) wrote, in an interesting and thought provoking post (much of the thrust of which I agreed with):
“In either case, more surface heat gets transferred up.”
I don’t agree with that statement. The surface is powered by the sun. It has to get rid of all the energy it receives from the sun, any way it can – otherwise it adjusts its temperature until it does.
What happens if the back-radiation increases is that the surface temperature rises, evaporation increases and surface radiation increases to bbalance (in sum) the change. As far as the atmosphere is concerned there is the same amount of net flux from the surface.
However cba’s comment is partly correct. If the surface temperature rises, the balance of flux transfer is tipped even more heavily in favour of the latent heat carried by water vapour. So less flux enters the atmosphere near its base and more enters at the clouds.
The rough numbers from the K&T diagram are:
AT 15DEGC:
CONDUCTION (24W/m^2) which heats the air in direct contact with the surface
NET ABSORBED RADIATION (26W/m^2) which heats the air up to about 500m, but over half that radiation is absorbed within the first 50m
LATENT HEAT IN WATER VAPOUR (78W/m^2) which heats the air when the water vapour condenses into water droplets, ie in the clouds.
If the surface temperature increases by 3DegC, these values alter and become:
CONDUCTION (24W/m^2)
NET ABSORBED RADIATION (Between 10 and 20 W/m^2)
LATENT HEAT IN WATER VAPOUR (Between 84 and 94W/m^2)
Colin Davidson says:
Short answer: Because of what is caused the moist adiabatic lapse rate, the warmer the air at the surface, the less fast the temperature drops with height. What this means is that warming at the surface is expected to get magnified as you go up in the tropical troposphere. Since the layer from which most of the radiation escapes into space is well up in the troposphere, that is the layer that must rise in temperature by a certain amount in order to restore radiative balance after the GHGs cause their forcing. So, the air temperature at the surface does not have to rise as much as one predicts that the temperature at that level in the atmosphere has to rise. Hence, it is a negative feedback in the sense that it reduces the expected surface temperature rise relative to what one predicts by just applying the Stefan-Boltzmann Equation.
Bryan says:
Actually, Arthur addresses this quite directly. He points out that the assumption in this is that the system is adiabatic, i.e., a parcel of air is not exchanging energy with its surroundings. In some contexts, such as during convection, it is a reasonable approximation that a rising parcel of air does not have the time to have significant energy exchange with its surroundings; however, in other contexts, it is not. In particular, the equation for the lapse rate that is derived from that analysis gives a stability limit: If the lapse rate is greater than that, a parcel of air that starts to rise in the atmosphere will continue to rise and hence the air is unstable. The tendency in this situation is then to get convective mixing, which tends to reduce the lapse rate. (The convective processes happen quite quickly in comparison to heat conduction and thus the adiabatic approximation is pretty good here.) This is all well-known and well-understood in the meteorological and climate science communities.
However, if you have a lapse rate smaller than the stability limit, the air is stable and such convective processes are suppressed. In that case, and in the absence of an IR-active atmosphere, Arthur argues that the adiabatic approximation is not a good one and the atmosphere will actually tend towards being isothermal, as is true of a system in thermal equilibrium.
Having not thought about this that much myself, I can’t say that I am yet convinced that Arthur is right. But, his argument seems very reasonable to me…and Arthur is a very smart guy whose scientific intuition I respect a lot.
And, at any rate, the larger point is that some people seem to think that arguing that the lapse rate is determined by physics other than radiation somehow negates the argument that invoking the greenhouse effect is necessary to explain the high surface temperature of the Earth. It does not. Regardless of the temperature distribution of the atmosphere with height, if the atmosphere is transparent to infrared radiation, all of the radiation from the Earth will escape to space and radiative balance will determine the surface temperature. You cannot magically create a warmer surface of the earth by invoking this argument.
Bryan:
It might be instructive for you to read Roy Spencer here http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/ , who seems to agree with Arthur on this point about the temperature profile of the atmosphere in the absence of the greenhouse effect:
Joel Shore (19:04:26) : edit
Bill had said:
Bill Illis (21:46:42) : edit
TOA in the climate science CO2 forcing context almost always means the tropopause.
There is a clear distinction made in the literature between Fi, the instantaneous forcing at the tropopause, and Fs*, the same forcing after ten years of adjustment. For a doubling of CO2, these are given by Hansen et al. as
Fi = 4.52 W/m2
Fs* = 3.95 W/m2 ± 0.11
Not according to anyone I’ve read. I’d need a citation for that, especially since increased albedo will decrease the tropospheric forcing, not increase it. Hansen et al. (ibid) give a temperature change of ~ 2°C for a doubling of CO2. The IPCC says the models incorporate a feedback of 2 W/m2 per °C, which would give an increase in forcing of 4 W/m2.
The TOA forcing is only on the order of 150 – 160 W/m2, not 240 W/m2. See Kiehl/Trenberth or my budget in the head post. The 240 W/m2 is the total forcing coming from the sun, and is thus the total amount being radiated to space from all levels of the climate system including the surface. In equilibrium, this will never increase or decrease. This statement doesn’t make sense.
Assuming blackbody conditions, 255K = 240W/m2. This is the total radiation from the entire planet, not the emission from any layer. There is no atmospheric layer where there is an “equilibrium emission temperature of 255K”. Again, this statement doesn’t make sense.
Not true, for all of the reasons given above. Hansen et al. (ibid) says that the ~ 2°C for a doubling of CO2 is after all feedbacks, and that the net change in the TOA forcing after all feedbacks is ~ 4 W/m2 …
Joel, you are usually more skeptical than this …
Joel Shore (19:14:09)
While the 1% = double in 100 years number is tempting, in fact the climate system sequesters a large amount of what is emitted. Simple calculations show that if we increase the amount emitted by 1% of the 2010 emissions, we will not double the atmospheric concentration for a long, long time. By 2100, the concentration will only be about 35% more than at present. And it won’t double until about 2275.
This is because the amount sequestered is a percentage of the “excess” carbon dioxide in the air. As airborne CO2 concentration increases, the amount sequestered also increases. This slows the rise of the atmospheric concentration.
Here, I agree with Joel. There is plenty of evidence, not computer model results but evidence, that the recent increase in CO2 is from human activities. I don’t agree that a change in CO2 concentration from 0.03% of the atmosphere to 0.04% of the atmosphere (as happened over the last 200 years) has measurably changed the temperature, but I do agree that the CO2 rise is from human industry.
Joel Shore (19:32:16)
Couldn’t have said it better. I agree 100%. See my Steel Greenhouse post for the illustrated version.
Bryan (16:37:47)
The IPCC, like almost everyone in the field, knows that the thermal energy from the burning of fossil fuels is quite small. In the most recent year for which I have data (2003), the global consumption of fossil fuels was about 8,500 million tons of oil equivalent (mTOE).
1 TOE is 42 gigajoules. That gives a yearly consumption of 7.6e20 joules, which is about 1.1e13 joules per second, which is 1.1E13 watts instantaneous.
The surface area of the earth is 5.1e14 square metres, so the thermal energy from fossil fuel burning is about 0.02 W/m2 …
…
Absolutely not true. Joel has given you a clear and comprehensive explanation of why G&T are wrong.
He made no such claim, that’s a straw man.
Nope. The only way to check what the qualities of his previous published works are is to look at them.
Sometimes yes, sometimes no. Sometimes I get angry enough at some blockhead to hurl insults, but that means nothing about whether my scientific work is any good.
Bryan, Joel has been even tempered here. He laid out a clear explanation of why he thinks G&T is wrong. You have not said a single word about that. Instead, you raise straw men and make accusations about his past work. You even imply that his statement about his past work is fictional, rather than simply google his name and see.
If you want to get some traction, show us where you think Joel’s explanation above is incorrect. Anything else is just wasting words.
Joel and Willis
A P Smith and it now appears Roy Spencer think that an atmosphere without CO2 and H2O would become isothermal to a height of 30km at the surface temperature.
Stop and think about the enormity of such nonsense.
The surface temperature would be much higher than now because of failure to absorb much of the Suns IR.
Lets say surface temperature rises to 300K on average.
Outside of the atmosphere the temperature drops to near absolute zero.
Gravity still exists, day and night still exist, surface temperatures will still vary widely.
Tidal currents will still flow,winds will still exist etc.
Look again at the proposition at the top and see how silly it is.
Joel you seem to think that quoting Roy Spencer is a clincher.
I pointed out earlier that sceptics will come at a theory from different directions and will not necessarily agree with each other on all things.
There is no Church of the Latter Day Sceptic.
However I am glad that you have distanced yourself from Arthur on the point above.
Willis
You seem to think I know who Joel is, your wrong.
G&T make 16 separate points of disagreement with Greenhouse Theory.
Opponents of G&T have used the tactic of smear and innuendo instead of reasoned argument and rarely address any of the 16 points.
I have no particular reason to support G&T other than looking at the case they present and looking at the alternatives presented by their critics.
I would urge anyone interested to go to the critics sites such as Arthur’s with an open mind and draw your own conclusions.
Joel is part author of a proposed paper covering the G&T topic so I think its a good idea to wait and focus that discussion there.
I followed your comments in Barts blog yesterday and found them interesting.
Earlier VS welcomed the intervention of G&T into the debate and hoped that other Physicists would make a contribution to this important debate.
He was particularly scornful of those who depart from reasoned debate.
Why am I on this particular thread?
I think that the K&T diagram is very ambiguous however that was not my original question still unanswered so I will rephrase it.
What would be regarded as a minimum value for atmospheric backradiation and under what conditions would this occur?
“I’m not sure most pilots think of the “coefficient of expansion” of the gas, and least not most pilots.”
You’re probably right there, I think of “coefficient of expansion” of the (fluid) because, that is proportional to the buoyancy (forces) responsible for natural convection and that is important to me in heat transfer.
I rode on helicopters all the time while in Iraq and those things are hell on fuel because of the heat there.
So far, from what I read here, the following people believe Gerlich and Tscheuschner are correct
– Gerlich and Tscheuschner
-Kramm
-Me
– Brian W
So, according to Joel, we’re “fringe.”
Tough luck.
Let me give another example of why I think talking of back radiation etc , even though it is a way of looking at the fine detail, is wrong thermodynamically.
Take the law of increasing entropy and disorder.
This law is violated by an individual living organism: it increases its order thus decreasing the entropy of the organism. Thus one is forced to talk of “closed systems” where the total disorder is increased even though locally the biological organism violates the law.
In a closed system, for every process that occurs the entropy of the system will either increase or remain constant .
We see then that the second law, in its entropy form above, can be violated locally in an open system, which a biological organism is, while it exchanges energy with its surroundings ( second law takes over at death).The whole closed system has to be taken into account.
The second law is also formulated as:
No process is possible whose sole result is the removal of heat from a reservoir at one temperature and the absorption of an equal quantity of heat by a reservoir at a higher temperature.
Which is the form that G&T, and I, are saying is being violated by this back radiation concept.
The crucial words in this Clausius statement of the law is “sole”, and “equal quantity of heat”. The closed system is implied.
So the atmosphere as a reservoir of heat is at temperature T1, and the surface is at temperature T2>T1.
I am now trying to think how, in analogy to the entropy violation of a biological organism, a local violation of the Clausius statement would be permissible .
Let us forget the constant radiation coming from the sun. Let us say that somehow the sun suddenly turns into dark matter giving no radiation, and think what governs the ground reservoir and the atmospheric reservoir isolated in the cold of space as it cools off.
Would not this be the classic illustration of the Clausius statement of the second law? The atmosphere cannot transfer heat to the ground without violating the second law.
I cannot see why the introduction of a radiant sun by enlarging the “closed system” would change this.
http://www.realclimate.org/index.php/archives/2007/08/the-co2-problem-in-6-easy-steps/
“[Response: Feedbacks work for everything. That’s why the ‘radiative forcing’ concept works – it doesn’t matter if the initial push is from greenhouse gases or the sun. The change in temperature you’d need to balance a forcing of 4 W/m2 with no feedbacks is around 1.2 ºC and the difference between that and the real sensitivity (around 3 ºC) is a measure of how strong the net feedbacks are. – gavin]”
http://www.realclimate.org/?comments_popup=142
“For the response to 2xCO2, (around 3 deg C) you would expect an increase of about 30% in water vapour amounts. -gavin]”
“An instantaneous forcing calculation for 1.4xH2O over the whole globe gives a forcing of 5.5 W/m2 – demonstrating that over the whole spectrum, water is not saturated. – gavin]”
and then there is supposed to be a third 3.7 W/m2 after slow feedbacks.