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.
http://www.junkscience.com/ipcc_tar/wg1/044.html
“If the amount of carbon dioxide were doubled instantaneously, with everything else remaining the same, the outgoing infrared radiation would be reduced by about 4 Wm-2. In other words, the radiative forcing corresponding to a doubling of the CO2 concentration would be 4 Wm-2. To counteract this imbalance, the temperature of the surface-troposphere system would have to increase by 1.2°C (with an accuracy of ±10%), in the absence of other changes. In reality, due to feedbacks, the response of the climate system is much more complex. It is believed that the overall effect of the feedbacks amplifies the temperature increase to 1.5 to 4.5°C”
Joel Shore (19:12:16) :
“Thus, it is the greenhouse effect that continuously de-stabilizes the atmosphere, ‘trying’ to create a temperature profile that the atmosphere cannot sustain, which then causes all different kinds of weather as the atmosphere convectively overturns. Thus, the greenhouse effect is actually required to explain why weather occurs. ”
No. You are ignoring thermalization, which is facilitated by CO2 and HOH and which would produce “weather,” even without any “greenhouse effect.”
JAE says:
I’m not sure I understand your point, but the main thing I wanted to note is that you are not quoting me directly; you are quoting Roy Spencer (as quoted by me).
anna v says:
It doesn’t. However, the 2nd Law applies to the NET flow of heat…which, as I noted, is clear once you realize that all bodies at non-zero temperature radiate heat and a colder body next to a hotter body is not going to magically stop radiating heat to the hotter body. Rather, the 2nd Law just tells you that the hotter object will absorb less heat from the colder object than the colder object will absorb from the hotter object.
So, what your statement should say is, “The atmosphere cannot IN NET transfer heat to the ground without violating the second law.” And, the answer is: It doesn’t. The net flow is from ground to atmosphere as the 2nd Law requires. However, this still leaves the ground warmer than the comparison case of an IR-transparent atmosphere, in which case ALL of the heat that the earth radiates escapes into space and NONE of it makes it back to the earth.
Willis Eschenbach says:
Yes, but these numbers are talking about stratospheric adjustment. They are not talking about feedbacks. (And, at any rate, what Hansen et al. are concerned with is the forcing, so they don’t talk about the effect of the feedbacks on the TOA radiation balance.)
What Bill is talking about is the “effective radiating layer”. Of course, in reality what you have is a distribution function, i.e., if you looked at the emission that escapes into space and at what height in the atmosphere it originated, you would have a continuous function but it would have a peak and the location of that peak is, roughly speaking, the effective radiating layer. And, Bill is right that the effect of an increase in greenhouse gases is to raise that effective radiating layer to higher levels where the atmosphere is colder, which then means the amount of radiation emitted is less. See here for more discussion: http://www.aip.org/history/climate/simple.htm#L_0623
No…That is not what Hansen is talking about. He is talking only about the forcing due to CO2 after stratospheric adjustment. That is not something you measure after all feedbacks because, by definition, it doesn’t include the effect of feedbacks.
Think of it this way: How do you imagine that the water vapor feedback works? The way it works is by creating an additional imbalance in the TOA radiation balance. However, that change, by definition, is not considered part of the forcing…It is considered part of the feedbacks.
Willis Eschenbach says:
Actually, that is not how you calculate the effect of the feedback. I’ve always wondered what a feedback parameter expressed in (W/m^2) per °C means and I have finally figured it out. You have to use the formula that is in footnote 6 of the IPCC AR4 WG1 Chapter 8 on p. 631. Basically, the amplification of the temperature change relative to that predicted in the absence of feedbacks is 1 / (1 + lambda/lambda_0) where lambda is the total feedback and lambda_0 is approximately –3.2 (W /m^2) per °C. Note that the lambda_0 is -1/S where S = the sensitivity in the absence of feedbacks (i.e., basically that derived from the Stefan-Boltzmann Equation). Hence, a 2 W/m^2 per °C feedback yields about a 2.67 magnification of the no-feedback temperature response. And, in the limit that the feedbacks approach 3.2 W/m^2 per °C, the magnification diverges because this means the feedbacks are sufficiently strong to cause a runaway effect.
It is actually not too hard to derive that formula in footnote 6, by the way. I just did it on a piece of paper, although I am not sure I yet understand what I did quite well enough to explain it…at least without writing down the equations.
I’m still waiting for some indication that the radiation-fixated discoursers on this thread have any grasp of the enthalpy of gases and of the fact that convection and the creation of pressure gradients (which drive the geaostrophic winds) involve WORK done by the atmosphere. It is such lapses, along with neglect of biological dissipation of radiation that render the usual “energy budgets” an unrealistic exercise in number juggling.
“JAE says:
No. You are ignoring thermalization, which is facilitated by CO2 and HOH and which would produce “weather,” even without any “greenhouse effect.”
I’m not sure I understand your point, but the main thing I wanted to note is that you are not quoting me directly; you are quoting Roy Spencer (as quoted by me).”
Wow! If you don’t understand this point, then I don’t think you should not be discussing this subject and you should read a physics book.
Here is the story: when the GHG molecules (HOH and CO2) absorb IR, they get all excited (vibrational, rotational, translational) and they strike the O2 and N2, which are the bulk of the atmosphere, thereby energizing (heating) them. This “heats them,” and spreads the “wealth.” This is LTE, which you should look up at Wicki. This is the real function of GHGs, IMHO (not radiation). There are far too many people that completely ignore this inconvienient physical fact and try to explain everything with radiation (including Willis), again IMHO.
Why will NOBODY try to explain to me why it is only 85 F where Willis lives, when that area has 6-7 times as much greenhouse gases as Phoenix, Arizona, which routinely has 110 F and higher in the summer.
Where is the empiricism, which is the FUNDAMENTAL pillar of science?
Re: Joel Shore (Mar 24 17:40), Re: Joel Shore (Mar 24 17:40),
Joel, I am not being confrontational on this, I want to clear it up, and your statement does not clear it.
So, what your statement should say is, “The atmosphere cannot IN NET transfer heat to the ground without violating the second law.” And, the answer is: It doesn’t. The net flow is from ground to atmosphere as the 2nd Law requires. However, this still leaves the ground warmer than the comparison case of an IR-transparent atmosphere, in which case ALL of the heat that the earth radiates escapes into space and NONE of it makes it back to the earth.
All the terminology of back radiation and multiplication talks of heat transfer from the atmosphere to the ground. What is NET? a million years? a microsecond?
Reservoir1 + Reservori2 is NET ? Then the classic Clausius formulation has no meaning, because from the first law NET does not change or decreases by radiation anyway.
I would also like to see you point the finger where the image by Peden,
http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html
which creates a paradoxical device from back radiation, is getting off track.
Joel Shore (17:55:41)
Not true at all. From Hansen et al. cited above:
Note that this says “this approach allows both stratospheric, tropospheric, and land surface feedback mechanisms to operate.” So your claim above that “They are not talking about feedbacks.” is falsified.
JAE (20:04:31)
Because they are in different climate zones, and because of the high humidity in the South Pacific.
The South Pacific is tempered by, and stays at about the temperature of, the surrounding ocean. A combination of evaporation, cumulus clouds, and thunderstorms caps the high temperatures, preventing them from rising too far. As a result, the average temperature rarely exceeds 30° C (86°).
Hot moist air near the equator rises in the thunderstorms and drives the “Hadley circulation”. This air has the moisture stripped out of it as it rises and moves towards the poles. The air, now dry, descends at about 30°N and 30°S. As it descends, it warms. This creates the great bands of desert which circle the earth at ~ 30°N and 30°S.
Finally, because the air is dry, there is much less “greenhouse effect” and little cloud in the desert belts. As a result, there are much greater day/night swings. For example, tomorrow in Honiara, Solomon Islands, the forecast is for a high of 90°F and a low of 74°F (average 82°F), a 16° swing. In Phoenix, on the other hand, tomorrow is slated to have a high of 80°F and a low of 53°F, a 27° swing.
Hope that answers your question.
Joel and Willis
This is the fourth time I have asked this question with no reply even though I have rephrased it
What would be regarded as a lower measured value for atmospheric backradiation and under what conditions would this occur?
Given that the backradiation shown in KT diagram is twice the incoming solar EMR to the surface this should be easy to measure and a rough estimate is all I’m looking for.
Joel
I must say I agree with anna v ‘s comments above regarding the 2nd Law of TD.
I did not want to stray too far away from the main points of the thread namely the science behind the KT diagram and whether Willis’s attempts to tweak it are justified.
Regarding G&T (and refutations thereof) I have simply referred people to the Arthur P Smith site where exactly the difficulty for Arthur in the use of the word “heat” is well illustrated.
Joel you should think long and hard about whether this is your settled position
….”applies to net flows; it does not mean that no heat can be radiated by a colder body and absorbed by a hotter one.”
If so, define heat in the sense that Physicists use when communicating with one another.
Somewhat off topic…. A P Smith , Roy Spencer and adiabatic expansion.
What they refer to is free adiabatic expansion which is isothermal
In a Gravitational Field the expansion is NOT FREE.
Any molecule moving up has to overcome gravity.
The system( gas particles) in moving up has to supply work against gravity.
Where does this work come from?
It can only come from one place the internal energy of the system(gas particles).
They lose heat and their temperature drops.
And as we all know this is exactly what happens.
Joel Shore (19:36:51) : edit
Yes, I know that. It has to do with the fact that if there is a feedback, the feedback is also amplified.
Now, back to what I had said:
The total of the water vapor and the albedo forcings given in the citation I gave above is about 1.6 Wm-2 °C-1. Using the formula you reference gives a total forcing of 2 Wm-2 °C-1, which is what I quoted.
However, I still don’t see what this has to do with my underlying calculation, which is that a change of about 150 W/m2 leads to a warming of about 20°C. This includes all of the feedbacks. And as I understand it, the IPCC warming expected from a doubling of CO2 (climate sensitivity) includes all of the feedbacks as well.
The IPCC figure is 3°C for an increase of 3.7 W/m2, or about 3°C / 3.7W/m2 = ~ 0.8°C/Wm-2.
My numbers give 20°C / 150Wm2 = ~ 0.1°C/Wm2.
Since both numbers include all possible feedbacks, I don’t see why feedback calculations are relevant to the discussion.
Willis Eschenbach (00:42:35) :
“Hope that answers your question.”
Thank you, but no. What you say is correct, but it all ignores the question. If the GHE is a determinant of temp., then more GHE amounts = more back-radiation = more heat. Instantly. Now, if evaporation and convection keep this heat in check, fine. Then let’s stop worrying about adding more GHEs to the atmosphere.
anna v says:
I don’t understand the question. NET means you take the difference between the heat transferred from A to B and that transferred from B to A. It doesn’t matter over what time you consider and, in fact, if you measure in Watts, you are doing it in terms of power (i.e., energy per second). No matter what time period you measure it over, the net heat flow will be from hot to cold.
It gets off track because in the case of the chicken, you don’t have an input source of energy, as you do from the sun in the case of the earth. However, consider if there was an input heat source for the chicken. Do you think it might get hotter if it were in a well-insulated oven that reflected some of the heat back than it would if the insulation were not there?
Also, do you think that some hot liquid in a thermos might stay hotter for longer in the thermos than not in the thermos?
Joel I just read this
http://www.physicsforums.com/showthread.php?t=385923
Apparently Chris does not think that you or Arthur are prominent Physicists.
I see it it required six of you to comment on what you regard as rubbish.
I just hope the committee approach does not lead to lack of clarity.
I look forward to reading it and no doubt it will be commented on.
…………………Well, word has just come through so I’m going to announce it here. I’m now officially a published physicist! If you really go into the details it isn’t quite as impressive as might appear at first glance, but hey. I’ll accept congratulations in any case!
I’m very pleased, but to keep this in perspective, it is in a comparatively minor journal, and the accepted paper is a “comment” on a preceding paper, and the paper we comment on is an oddity which has no real scientific influence, and the content of our paper is easily within what is accessible to an undergraduate studying the relevant aspects of thermodynamics. Originally I had felt it wasn’t even worth writing a paper on it, but I was persuaded to join in the project. None of my co-authors are prominent as physicists.
The reference is:
Joshua Halpern, Christopher M. Colose, Chris Ho-Stuart, Joel D. Shore, Arthur P. Smith, Jörg Zimmermann (2010) Comment On “Falsification of the Atmospheric CO2 Greenhouse Effects within the Frame of Physics”, (to appear in) International Journal of Modern Physics (B), Vol 24, Iss 10, March 30 2010.
Re: Joel Shore (Mar 25 06:32),
It gets off track because in the case of the chicken, you don’t have an input source of energy, as you do from the sun in the case of the earth.
But you agreed that in my example of turning off the radiation from the sun the two reservoirs R1 R2 , T1>T2 , turning it on would make no difference in the thermodynamic picture?
InRe: Joel Shore (Mar 24 17:40),
The chicken would radiate as a black body, the radiation would be reflected by the reflector ( not by a complicated CO2 logic) etc. and with the logic of adding the back radiation that is used in the CO2 problem, one would get a runaway heating problem!
It is all those additions of watts/m^2 that are double counting in my opinion of course in the greenhouse expositions. A delay of cooling is fine, but it cannot be represented by adding watts/m^2 at imaginary boundaries.
I think what happens in this case solved correctly is that the chicken stays at its temperature, if the oven is isolated. The thermos effect as you observe.
Let me rephrase it. If you consider as “net” the combined R1 R2 energy balance , the Clausius formulation of the second law becomes moot and void, in my opinion.
Thanks for your answers.
Willis Eschenbach says:
I would have to read the Hansen paper in more detail (and perhaps the Gregory paper too) to understand what they are getting at. A lot of different things are called feedbacks and it is not clear which are included. What I would call your attention to is this sentence that you quoted:
Note that by looking at the flux in the limit of zero temperature change, you can’t be seeing any feedbacks, such as the water vapor feedback, that depend on the temperature…because you are looking at the zero temperature change limit.
I am still confused about what you are even claiming: Are you claiming that the ~4 W/m^2 forcing due to doubling CO2 means a 4 W/m^2 change in the TOA flux because it already includes the water vapor feedback. Or, are you arguing that the water vapor feedback somehow doesn’t affect the TOA flux? I don’t think either of these claims is correct but I am not really sure at this point which one you believe to be correct.
anna v says:
No…What I said is that having the sun in the picture does not then make it possible for there to be net radiation transfer from the (cooler) atmosphere to warmer earth. That is very different than saying the sun makes no difference to what would happen to the whole system…It makes a huge difference obviously. You turn off the sun and the earth will cool to a ridiculously low temperature (probably until internal processes producing heat in the earth, e.g., due to radioactive decay balanced the radiant loss out into space).
The chicken would radiate as a black body, the radiation would be reflected by the reflector ( not by a complicated CO2 logic) etc. and with the logic of adding the back radiation that is used in the CO2 problem, one would get a runaway heating problem!
You are basically running yourself around in circles by choosing analogies that are nowhere near analogous (e.g., no heat source like the sun present). I suggest you go and solve the problem that I described in a previous post:
You have a blackbody sphere representing the sun that is at a constant temperature T_s and then you have it surrounded by one or two blackbody concentric spherical shells (A and B, which are analogs…in a very abstract way…of the earth and an IR-active atmosphere, respectively, with the radius of B slightly larger than A). There is empty space (assumed to be at zero temperature) beyond that. The question is then: What is the steady-state temperature of spherical shell A when shell B is absent and what is the temperature of shell A when shell B is present? Also, when Shell B is present, which way is the net radiative heat flow, from A to B or B to A? (These questions can be answered exactly by a simple calculation if you assume that the radii of the sphere and the two shells are close enough that one can neglect any “self-view” that the inner surface of shell A or B has of itself.)
I’ll even give you the answers:
T_A = [(1/2)^(1/4)] T_s ~= 0.84 T_s when shell B is absent.
T_A = [(2/3)^(1/4)] T_s ~= 0.90 T_s and
T_B = [(1/3)^(1/4)] T_s ~= 0.76 T_s when shell B is present.
With B present, the net flow of radiation between B and A is (1/3)*sigma*(T_s^4) from A to B.
This simple example illustrates all the features of the greenhouse effect: The presence of Shell B causes Shell A to be at a higher temperature than when Shell B is absent. However, Shell B is at a lower temperature than Shell A…and the net flow of radiation is from the warmer A to the cooler B. If you understand this example, then you will basically understand the greenhouse effect and also why the claim that it violates the 2nd Law is WRONG, WRONG, WRONG.
Bryan says:
Well, I would tend to agree with Chris, depending on what is meant by the word “prominent”. Certainly, if you surveyed 100 random physicists, I doubt many (maybe even any) of them would have heard of us. However, by that standard, G&T aren’t very prominent either. And, in terms of publication record, from what I’ve seen, I think that I would stack up very favorably with them.
I’ll tell you a little story. I once was part of a comment on a paper in Physical Review Letters (PRL), where papers are limited to 4 pages in length. I was subsequently told by someone that the paper’s main author had bragged about how wonderful his paper was and said that he really could have gotten several papers in PRL out of that work but had chosen instead to put it all into one. When I heard this, my reaction was, “Well…In a bizarre way I sort of agree with him because comments on papers in PRL are limited to one page and it was very difficult to explain all of the errors that he had made in that one paper in just a one-page comment. So, he really did have at least several PRL-length-paper’s worth of errors crammed into that one paper.”
And, so it is with G&T…There is so much wrong in there that it is hard for just one person to tackle it all. I would also add that all of us were doing this basically in our “free time” so it was also a case of “many hands make light work”. Personally, I think that Arthur’s original unpublished paper in response to G&T is considerably more elegant than our 6-author comment although ours is a more comprehensive litany of the errors in G&T, whereas Arthur picked mainly one issue to focus on.
Joel Shore
Looking forward to reading it.
Will it be published in arXiv. form ?
Perhaps one of the main platforms such as WUWT or Realclimate will host a feature on it to facilitate an in depth discussion on its merits!
Will it contain gems such as
….”applies to net flows; it does not mean that no heat can be radiated by a colder body and absorbed by a hotter one.”
JAE (06:30:00)
I think that evaporation and convection do keep this heat in check. See here for the details.
Joel Shore (19:42:48)
Joel is right. You chould look at the diagrams and follow the math at my post, The Steel Greenhouse, for a simpler explanation of why he is right.
Again, folks seem to think that a colder object cannot radiate heat to a warmer object. It most definitely can. The constraint from the Second Law is only that the net flow has to be from the warmer to the colder object. It says nothing about the individual flows except that the flow from warm to cold has to be greater than the flow from cold to warm.
Joel Shore (19:17:27)
Joel, you are correct that the definitions are not always clear. Here is the IPCC FAR definition:
However, the differences between the various forcings Fi, Fa, Fg, Fs, and Fs* are not large. Hansen defines these forcings as follows:
The Hansen paper cited above gives the following values in W/m2 for the forcings resulting from a doubling of CO2:
Fi: 4.52
Fa: 4.12
Fg: 4.08
Fs: 4.11
Fs* 3.95 ± 0.11
Note that the final value, Fs* includes all possible feedbacks/forcings from a model which is run until such time as the surface temperature is no longer changing.
I used the IPCC canonical value of 3.7 W/m2 for the equilibrium value of the change in forcing from a doubling of CO2. However, if I use 3.95 W/m2 from Hansen’s paper it makes no difference. Here are the results using the IPCC midrange sensitivity of 3°C from a doubling of CO2:
With 3.95 W/m2 change in equilibrium TOA forcing per doubling, we get 3°C/3.95Wm-2 = 0.75°C/Wm-2.
With 3.7 W/m2, we get 3°C/3.95Wm-2 = 0.81°C/Wm-2.
With real world measurements, we get 20°C/150Wm-2 = 0.13°C/Wm-2.
That’s why I say this whole question of feedbacks is a red herring. The problem is the huge difference between theory and measurement, regardless of which exact forcing we use for a doubling of CO2.
Note also that sensitivity is dependent on temperature, because parasitic losses and tropical cloud albedo both increase with temperature. So the equilibrium sensitivity that I calculate above from the real world numbers is going to be higher than the equilibrium sensitivity.
So … having disposed of the feedback question, I am still very interested if you have other objections to my analysis, and what those might be.
Joel Shore,
One of the most annoying things with the climate models is the assumption of thermal equilibrium. Based on the “forcings” in GISS climate model,
http://data.giss.nasa.gov/modelforce/,
we may erroneously get the impression that the global temperature would be completely constant in a world without humans. This is particularly peculiar since we know that Milankovitch cycles, ocean cycles and other cycles may lead to severe fluctuations – it is usually not the case that the criterion of thermal equilibrium is satisfied. So what is the magnitude of the natural oscillations the last 10 000 years, in a world without humans, both from a model and empirical point of view?
A) T = 287 ± 0.1 K
B) T = 287 ± 1.0 K
C) T = 287 ± 2.0 K
D) T = 287 ± 3.0 K
What is your opinion here Joel?
Joel Shore,
Take a close look at this:
http://pages.science-skeptical.de/MWP/MedievalWarmPeriod1024x768.html
(click the curves to see papers)
It is quite evident that there is no thermal equilibrium, but that the temperature oscillates significantly. Then the question becomes, given that the temperature oscillates significantly, what is the climate sensitivity to CO2? Surely this is not the problem Trenberth is addressing, in his world the energy budget is nearly balanced all the time.
Possibly natural climate oscillations may be an order of magnitude larger than the climate changes due to increased CO2.
How can we be certain that natural climate oscillations are insignificant?