Guest Post by Willis Eschenbach
[Update: I have found the problems in my calculations. The main one was I was measuring a different system than Kiehl et al. My thanks to all who wrote in, much appreciated.]
The IPCC puts the central value for the climate sensitivity at 3°C per doubling of CO2, with lower and upper limits of 2° and 4.5°.
I’ve been investigating the implications of the canonical climate equation illustrated in Figure 1. I find it much easier to understand an equation describing the real world if I can draw a picture of it, so I made Figure 1 below.
Be clear that Figure 1 is not representing my equation. It is representing the central climate equation of mainstream climate science (see e.g. Kiehl ). Let us accept, for the purpose of this discussion, that the canonical equation shown at the bottom left of Figure 1 is a true representation of the average system over some suitably long period of time. If it is true, then what can we deduce from it?
Figure 1. A diagram of the energy flowing through the climate system, as per the current climate paradigm. I is insolation, the incoming solar radiation, and it is equal to the outgoing energy. L, the system loss, is shown symbolically as lifting over the greenhouse gases and on to space. Q is the total downwelling radiation at the top of the atmosphere. It is composed of what is a constant (in a long-term sense) amount of solar energy I plus T/S, the amount of radiation coming from the sadly misnamed “greenhouse effect”. T ≈ 288 K, I ≈ 342 W m-2. Units of energy are watts per square metre (W m-2) or zetta-joules (10^21 joules) per year (ZJ yr-1). These two units are directly inter-convertible, with one watt per square metre of constant forcing = 16.13 ZJ per year.
In the process of looking into the implications this equation, I’ve discovered something interesting that bears on this question of sensitivity.
Let me reiterate something first. There are a host of losses and feedbacks that are not individually represented in Figure 1. Per the assumptions made by Kiehl and the other scientists he cites, these losses and feedbacks average out over time, and thus they are all subsumed into the “climate sensitivity” factor. That is the assumption made by the mainstream climate scientists for this situation. So please, no comments about how I’ve forgotten the biosphere or something. This is their equation, I haven’t forgotten those kind of things. I’m simply exploring the implications of their equation.
This equation is the basis of the oft-repeated claim that if the TOA energy goes out of balance, the only way to re-establish the balance is to change the temperature. And indeed, for the system described in Figure 1, that is the only way to re-establish the balance.
What I had never realized until I drew up Figure 1 was that L, the system loss, is equal to the incoming solar I minus T/S. And it took even longer to realize the significance of my find. Why is this relationship so important?
First, it’s important because (I – Losses)/ I is the system efficiency E. Efficiency measures how much bang for the buck the greenhouse system is giving us. Figure 1 lets us relate efficiency and sensitivity as E = (T/I) / S, where T/I is a constant equal to 0.84. This means that as sensitivity increases, efficiency decreases proportionately. I had never realized they were related that way, that the efficiency E of the whole system varies as 0.84 / S, the sensitivity. I’m quite sure I don’t yet understand all the implications of that relationship.
And more to the point of this essay, what happens to the system loss L is important because the system loss can never be less than zero. As Bob Dylan said, “When you got nothin’, you got nothin’ to lose.”
And this leads to a crucial mathematical inequality. This is that T/S, temperature divided by sensitivity, can never be greater than the incoming solar I. When T/S equals I, the system is running with no losses at all, and you can’t do better than that. This is an important and, as far as I know, unremarked inequality:
I > T/S
or
Incoming Solar I (W m-2) > Temperature T (K) / Sensitivity S (K (W m-2)-1)
Rearranging terms, we see that
S > T/I
or
Sensitivity > Temperature / Incoming Solar
Now, here is the interesting part. We know the temperature T, 288 K. We know the incoming solar I, 342 W m-2. This means that to make Figure 1 system above physically possible on Earth, the climate sensitivity S must be greater than T/I = 288/342 = 0.84 degrees C temperature rise for each additional watt per square metre of forcing.
And in more familiar units, this inequality is saying that the sensitivity must be greater than 3° per doubling of CO2. This is a very curious result. This canonical climate science equation says that given Earth’s insolation I and surface temperature T, climate sensitivity could be more, but it cannot be less than three degrees C for a doubling of CO2 … but the IPCC gives the range as 2°C to 4.5°C for a doubling.
But wait, there’s more. Remember, I just calculated the minimum sensitivity (3°C per doubling of CO2). As such, it represents a system running at 100% efficiency (no losses at all). But we know that there are lots of losses in the whole natural system. For starters there is about 100 W m-2 lost to albedo reflection from clouds and the surface. Then there is the 40 W m-2 loss through the “atmospheric window”. Then there are the losses through sensible and latent heat, they total another 50 W m-2 net loss. Losses through absorption of incoming sunlight about 35 W m-2. That totals 225 W m-2 of losses. So we’re at an efficiency of E = (I – L) / I = (342-225)/342 = 33%. (This is not an atypical efficiency for a natural heat engine). Using the formula above that relates efficiency and sensitivity S = 0.84/E, if we reduce efficiency to one-third of its value, the sensitivity triples. That gives us 9°C as a reasonable climate sensitivity figure for the doubling of CO2. And that’s way out of the ballpark as far as other estimates go.
So that’s the puzzle, and I certainly don’t have the answer. As far as I can understand it, Figure 1 is an accurate representation of the canonical equation Q = T/S + ∆H. It leads to the mathematically demonstrable conclusion that given the amount of solar energy entering the system and the temperature attained by the system, the climate sensitivity must be greater than 3°C for a doubling of CO2, and is likely on the order of 9°C per doubling. This is far above the overwhelming majority of scientific studies and climate model results.
So, what’s wrong with this picture? Problems with the equation? It seems to be working fine, all necessary energy balances are satisfied, as is the canonical equation — Q does indeed equal T/S plus ∆H. It’s just that, because of this heretofore un-noticed inequality, it gives unreasonable results in the real world. Am I leaving something out? Problems with the diagram? If so, I don’t see them. What am I missing?
All answers gratefully considered. Once again, all other effects are assumed to equal out, please don’t say it’s plankton or volcanoes.
Best wishes for the New Year,
w.
I’m sorry, I am not a climate scientist and the terms you use here I’m only familiar with by seeing them here and in other blogs. But I thought that as S increases the number of degrees of temperature increase per doubling of CO2 goes down. Surely if S was so high that T/S is practically zero then we have I = F-(a very small amount), and thus there would be almost no temperature increase per doubling of CO2. This would mean that your bound of S>.84 would indicate that the temperature increase per doubling of CO2 is less than 3 degrees not more.
Well Willis, I don’t like your energy cartoon any more than I like Dr Trenberth’s; but yours is at least understandab;le.
The problem is that the TSI is 1366 W/m^2, and not 342; and the thermal conductivity of the earth is not infinite; so the earth is not isothermal at 288 K.
A practical result of a 1366 w/m^2 blow torch, suitably attenuated by atmospheric absorption, scattering (Raleigh and clouds) impinging on a 70% near black body oceanic absorber, and a somewhat less absorbtive solid land surface, is that local temperatures rise far above 288 K; as high as 333 K or more fro the hottest land surfaces; and as a result those surfaces cool (radiate) at a much higher rate than your 288 K isothermal earth; so the result is nothing like oyur staic infinite conductivity, non rotating earth model.
Otherwise your coolors are prettier than Trenberth’s; but still just pink elephants.
Willis,
I do not see or understand how the derivation of E=(T/I) / S from the system efficiency E which equals (I – Losses) / I is done.
Dave Springer, Jan. 4, 5:17 am says:
“The fly in the ointment is that thermal IR cannot heat the ocean. It only heats the land. IR is absorbed by water in the first few micrometers at the surface.”
That’s a good point, but it also extends to land surfaces which are moist or covered with vegetation. So maybe 85% of the Earth’s surface is not heated by thermal IR.
_Jim says:
January 4, 2011 at 8:40 am
Zero hits searching this web page for Boltzmann or Wein …
One hit for Planck.
Two hits for Atmospheric Window however!
Until consideration is given to these facets of LWIR from the earth discussion on this topic is pretty well separated from that physical processes that take place in reality.
But, maybe you have to start somewhere before graduating to the stage where one comprehends radiative energy flow from a ‘black body’ is proportional to temperature to the fourth power …
Perhaps if you actually read through the article and comments, rather than lazily searching for terms you think may appear in precise terms, you’d appreciate that your snide, patronising tone is unwarranted.
There are no shortcuts to knowledge, my friend.
Dave Springer, Jan 4, 6:16 says:
“Technically correct. What back radiation does is slows down the rate of cooling. Just like a layer of clothing doesn’t actually warm your body but rather slows down how fast your body loses warmth. Greenhouse gases are insulators. Insulation doesn’t supply heat it just slows down the loss of heat.”
Well, ALL the gases in the atmosphere function as “insulation,” not just the GHGs (remember, the N2 and O2 are heated by collisions/thermalization). That’s why the there is way too much emphasis on radiative effects, alone. The real reason the Earth stays warmer than it “should be” is simply heat storage of the oceans and atmosphere–and insulation by those gases.
The way water is heated by sunlight is complicated. Visible light passes through the surface layers and is absorbed deeper down, in the top few tens of metres. It then rises mainly by convection, but with some conduction (and a tiny amount of radiation) to the surface. (Assuming the water is well above freezing point – cold water does strange things because of water’s odd temperature-density properties.) The water then cools by radiation, evaporation, and conduction at the surface.
The role of downwelling IR is to slow the rate at which the surface cools, which in turn affects the temperature differential that drives convection and conduction of heat to the surface. If you have water pouring into a tank from a tap and out again through a hole in the bottom, you can increase the water level in the tank by partially blocking the hole. Putting your finger over the hole at the bottom does not pour any additional water into the tank above it, but it can still have an effect on the level there.
Interestingly, the role of convection can be studied by turning it off – there is an arrangement called a solar pond in which salt is dissolved in water in layers, with more dense/concentrated solution at the bottom and less dense/fresher water at the top. The heavier water at the bottom does not rise when heated – it does not convect. So sunlight shines through to the bottom of the pond, and can then only escape to the surface by conduction. Solar ponds can easily reach temperatures over 90 C at the bottom with only a few metres of water.
Water acts like concentric shells to radiation, but with millions of shells micrometres apart. If pure radiation was all there was to it, the greenhouse effect in water would result in boiling below the surface with only a few centimetres of water. (Water molecules inside the bulk of a body of water still radiate.) Conversely, the surface would radiate all its heat away faster than it could be replenished by radiation from below and (if we ignore conduction from the air above) would freeze! A radiation-only world would be strange indeed! Fortunately, conduction can convey heat a lot faster, reducing the gradient to mere tens of degrees C per metre, and convection reduces it further still. As with air, you cannot understand the temperature of the water surface without understanding the role of convection.
I’ve been pondering the “thermostat” hypothesis for a while and looking up at the cold clear night sky it struck me. On cold days, cloud cover warms — on warm days, it cools. On warm days T > Tcrit and the atomsphere convects, on cold days it doesn’t (even to the point of inversion layers… a cold weather phenomenon.
Thoughts all.
Nullius in Verba says:
January 4, 2011 at 1:23 pm
“Assuming the water is well above freezing point – cold water does strange things because of water’s odd temperature-density properties”
Seawater does nothing odd. Maximum density occurs at the freezing point which at 3.5% salinity is -1.8C.
Dear Willis,
the problem is easy to solve.
The definition of the sensitivity in your simple model would be:
1/Sensitivity=dU(Greenhouse)/dT=Differential of Greenhouse Back Radiation and Temperature.
Assuming Stefan-Boltzmann Law this yields in your model (which indeed is very simple)
1/Sensitivity=4*U(Greenhouse)/T
Or in your consideration
Sensitivity>1/4*T/I=1/4*288/341K/W/m2=0.211K/W/m2
If the Radiative Forcing for 2*CO2 would be 3.7W/m2 a temperature of T>0.78 would result.
The values for the sensitivity and temperature for 2*CO2 of course only valid in this simple model.
It is just a factor 1/4 which was missing
Best regards
Rainer
George E. Smith says:
January 4, 2011 at 12:45 pm
“The problem is that the TSI is 1366 W/m^2, and not 342”
TSI is 1366 W/m^2 if the earth is flat and orientated towards the sun.
The earth isn’t flat and only a small portion is orientated towards the sun at any given time.
To get average TSI one divides by 4.
JAE says:
January 4, 2011 at 1:03 pm
“That’s a good point, but it also extends to land surfaces which are moist or covered with vegetation. So maybe 85% of the Earth’s surface is not heated by thermal IR.”
Point taken. I considered it but didn’t think the complication of trying to figure out what percentage of land surface is wet and for what percentage of the time was needed to make the point.
What fascinates me about all the learned exchanges and calculation here is that most of it shows just how little understanding of what’s going on in the region of temperature/energy input and output and such we really have! No wonder the AGW is still alive and kicking – it’s proponents are CERTAIN that they’re right and need no discussion or argument – for them there is consensus and ‘the science is settled’!
charlie says:
Moritz Petersen says:
I agree. This is what I see as a very serious problem. Not sure it is the only problem (see for example the 1st part of Moritz’s post) but this confusion of T and deltaT is a big one.
@Scott Brim,
sorry, wrong wording. I meant “reference situation”, around which linear relations can hopefully be derived for small enough variation.
Nothing special about pre-industrial CO2, besides the fact that I am inclined to believe the mainstream climate view there, i.e. that CO2 was relatively stable during historic era (at least after the last glaciation), and that the current CO2 increase is due to the burning of fossil fuels.
I also believe that CO2 increase is good: it makes plants grow faster, put the biosphere back in the safety zone while previous levels were dangerously low. The warming effect is probably there (good theoretical reasons to believe it’s there), but it seems small compared to natural variability, is certainly not catastrophic, maybe positive, and basically nothing to worry about. Peak oil is worrying me (a little), CAGW, not at all.
This place me firmly in the skeptic camp, although not among the rabid skeptic. I quite like the theories of Willis, as I said his analysis of tropical effects is original, well explained, and looks probable (I think there should be something like his regulating effects to maintain T within tight bounds at least at the tropics, it would explain the faint sun paradox). But this do not detract me to point glaring errors when I see them, and here the error is quite simple and would be embarassing if highlighted on other blogs…
kcrucible says:
January 4, 2011 at 11:29 am
“That seems to be nonsense to me. Shallow pools of water heated by the sun are demonstrably warmer than deeper pools with the same surface area. For particularly deep bodies, with not a lot of turbulance, you can end up with a scenario where the top couple of feet are much warmer than the lower areas.”
The sun isn’t heating them with thermal IR.
“The IR doesn’t vaporize water. IR increases molecule speed (heat.) With more heat, more water molecules evaporate certainly, _but not all particles that gain energy evaporate_. The additional heat increases molecular motion and a large amount gets distributed throughout the water body via conduction and convection.”
Convection only occurs when you heat water from the bottom or cool it from the top.
JAE says:
The reason the emphasis is on radiative effects is because the only way that the earth / atmosphere system communicates with the sun or space is via radiation! Oceans can’t really keep the earth warmer overall. In the absence of greenhouse gases, energy balance shows that in steady-state the earth’s average temperature (or more precisely, the average of T^4 over the earth’s surface) is determined completely by its distance from the sun and its albedo. The ocean mainly serves as a heat sink that reduces the variations in temperature (e.g., during the diurnal cycle). [Although technically, the reduction of this cycle can lead to a somewhat warmer average temperature, , because the behavior of vs is such that if you have two temperature distributions with the same then the one with a larger variation will have a somewhat lower value of .]
“Charles Duncan says:
January 4, 2011 at 11:37 am
The circumference of the Earth is about 40,000km, so the diameter is about 12,730km. In comparison the atmosphere is but a thin veneer, and the additional area at TOA is trivial, I believe.”
Is it trivial? Climate theory assumes that all of the heating occurs in this thin veneer. Before you can dismiss the extra heat loss through radiation from a less dense, expanded atmosphere, you must first quantify its value. Skylab crashed to the earth at least two years before NASA’s planned rescue, because unanticipated atmospheric expansion caused increased drag on the lab. Calling it trivial doesn’t make it so.
“”””” Nullius in Verba says:
January 4, 2011 at 1:23 pm
The way water is heated by sunlight is complicated. Visible light passes through the surface layers and is absorbed deeper down, in the top few tens of metres. It then rises mainly by convection, but with some conduction (and a tiny amount of radiation) to the surface. (Assuming the water is well above freezing point – cold water does strange things because of water’s odd temperature-density properties.) The water then cools by radiation, evaporation, and conduction at the surface. “””””
Well unfortunately, that odd density temperature function ONLY applies to fresh water, and low salinity salt water. Sea water with more than 2.47% of dissolved salts, exhibits NO MAXIMUM DENSITY before it freezes. Average sea water has 3.5% salinity, so it keeps getting denser right down to its freezing point which is of the order of a couple of degrees C below zero. So the turnover phenomenon of fresh water lakes, does not happen in the ocean.
The optical absorption coefficient of sea water is well documented fro all wavelengths from the Atmospheric UV limit near 180 nm and all the way out to wavelengths of about 1 metre; but with good data to around 100 microns wavelenght which is the end of the climatically useful far infra-red region.
Specifically for sea water at around 460 nm which is close to the solar spectrum peak, the absoption coefficient is between 1 and 2 x 10^-4 cm^-1 Even if it was as high as 2 E-4, that would put the 1/e transmission dept at 50 metres, or 250 metres for a 99% extinction.
So solar energy proceeds much deeper in the deep ocean than just a few tens of metres.
I agree that the local expansion caused by the heating of deeper waters should create an upward convection gradient, that would cause heated waters to slowly carry much of that energy back towards the surface; which is warmed not so much by direct absorption in the shallows; but by upward convection.
The warming of shallow water bodies, simply reflects the fact that the bottom reflects much of the light back towards the surface. In the deeper oceans, the water behaves like a grey body with an absorptance of about 97%, which is not a bad stand in for a black body absorber, and radiator.
Sea water has strong absorption bands at around 1, 2, and 3 microns, reaching a maximum absorption coefficient of about 9,000 cm^-1 at 3.0 microns, which gives a 1/e depth of around 1.1 microns or less than 6 microns for 99% absorption.
For the longer wavelength regions of interest to atmospheric thermal radiation (surface too), from around 6.0 to around 100 microns sea water has an absorption coefficient between about 200 and 6,000 cm^-1 which gives 1/e depths of around 2-60 microns; so the downward LWIR gets absorbed in the top few hundred microns of the ocean; where it mainly promotes prompt evaporation.
You cannot simply add downward LWIR radiative fluxes to incoming solar insolation, as if they both do the same thing; they don’t, and that is one of the failings of the climatists to accept that.
Joel Shore
I think that you are quite familiar with the climate models that the IPCC rely on.
Perhaps you have one of the basic models yourself?
What would happen if one of the models was left to run with the IR radiative effects of CO2 shut down?
H2O still operating with radiative properties in the IR and the usual phase change effects still intact.
I guess that the Earth climate would be pretty much as we experience it at present.
“”””” harrywr2 says:
January 4, 2011 at 2:03 pm
George E. Smith says:
January 4, 2011 at 12:45 pm
“The problem is that the TSI is 1366 W/m^2, and not 342″
TSI is 1366 W/m^2 if the earth is flat and orientated towards the sun.
The earth isn’t flat and only a small portion is orientated towards the sun at any given time.
To get average TSI one divides by 4. “””””
You obviously didn’t read what I said. TSI averages around 1366 W/m^2, and varies over an eleven year cycle by about 0.1% and of course seasonally due to the small changing sun earth distance change.
Your figure of 342 W/m^2 has nothing to do with earth flatness or roundness; it has to do with a completely fictional and quite absurd assumptiuon that the sun shines on the entire earth surface 24/7/365; from pole to pole and even on Antarctica in the depths of Winter midnight.
It doesn’t do that, on either this planet or any other planet in the universe. the earth rotates, and the sun that reaches the ground (clear air) packs about 1000 W/m^2 of projected area (normal to the sun-earth line), everywhere on the sunlit half of the planet.
The resulting thermal response of the planet, is nothing at all like your fictitious non-rotating isothermal earth receiving 342 W/m^2 over its entire surfqace at all times.
The system is quite non-linear; so you simply can’t average the real Physical effects; the heating of the earth to well above 288 K during a single daytime pass of the sun, persists well beyond the following night time period; so the temperature rise during daylight hours does not simply revert to some average value over night
TSI is NOT 342 W/m^2 it is 1366 give or take a smidgeon; but it illuminates only half the earth at any one time instant; well actually slightly more than half, by at least one degree so 181/360ths of the earth.
Last time I checked 181/36 of the whole would not normally be referred to as “only a small portion”.
This is not rocket science; maybe a third grade science question for “Are You Smarter than a Fifth Grader ?” What fraction of the earth surface is in daylight at any instant of time ? Answer; about half !
Willis
Missing from the cartoon is geothermal energy from radioactive decay in the earth. Energy in from space must be less than energy radiated to space. Otherwise the oceans would have boiled off long ago.
Joel:
It is true that virtually all the energy in the Earth system is gained and lost only through radiation, but it seems to me that you need to think beyond radiation and also consider that the atmosphere, land and (especially) water stores a lot of heat. In order to have a troposphere that is many kilometers in depth, there has to be a LOT of stored energy in the atmosphere (kinetic AND potential). And I think that that stored energy is all you need to consider to explain the surface temperature on Earth. For gases (the atmosphere), PV = RT; so T = (the constant, R)x(PV). Both P and V of a quantity of gas depend only upon the amount of stored energy in that gas, so T must also depends on this.
Bryan says:
This recent paper addresses that issue: http://www.sciencemag.org/content/330/6002/356.abstract
The equation is wrong – firstly it assumes a linear system in equilibrium when its a non-linear chaotic system.
Surface temperature is linked to ocean temperatire, the energy traded back and forth between the two varies with albedo / cloud cover / aerosols etc… and these are related to the sea temperature and surface temperature – the result is a climate sensitvity that changes size and sign depending on the systems state at a given point i.e. the deviation from the “ideal climate”) – but this results in an unsolvable equation that results in randon chaotic fluctuations, not a nice linear system in equilibrium, so it is instead simplified to the point where the equation is of very limited use for projecting temperatures for future states where the resulting climate sensitivity is unknown (see IPCC equation above).