Another Look at Climate Sensitivity

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.

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dp
March 16, 2010 8:46 pm

Hmmm – Isn’t the moon pretty much earth without air? The temperature there is entirely radiative to surface dwellers (the few that have dwelt), and the soil temperature is dependent upon it’s exposure to the sun. In shadows it is damned cold – I’d wager colder than 40 something degrees.
I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.

March 16, 2010 8:51 pm

For a real chart showing thermal balance go look up figure 2.3 Principles of Atmospheric Physics and Chemistry. Goody 1996
The back radiation term is pure bs. It is tied in with the nonsense argument of the atmospheric reradiation of 1/2 up /1/2 down. Absolute rubish

March 16, 2010 9:03 pm

Wiki Answer to: “What is the temperature on the moon?”
The average daytime temperature on the Moon is around 107°C (225°F), but can be as high as 123°C (253°F).
When an area rotates out of the sun, the “nighttime” temperature falls to an average of -153°C (-243°F).
253 F
-243 F
_______
10 F
Don’t know if that helps. 8 degrees is pretty close.

Jan
March 16, 2010 9:10 pm

I find this analysis be very important. Thanx

March 16, 2010 9:15 pm

Willis Eschenbach (21:07:38) :
enough (20:51:43) : edit
For a real chart showing thermal balance go look up figure 2.3 Principles of Atmospheric Physics and Chemistry. Goody 1996
The back radiation term is pure bs. It is tied in with the nonsense argument of the atmospheric reradiation of 1/2 up /1/2 down. Absolute rubish
Sorry, you have not given enough information for me to respond … plus you misspelled “rubbish”. If you post and link the figure you refer to, we can discuss it. In the interim, you might read the Kiehl/Trenberth paper referred to above.

Willis, I think he is talking about a figure numbered 2.3 in this book:

Principles of Atmospheric Physics and Chemistry (Hardcover)
~ Richard Goody (Author) “In the longest view, the earth is passing through a series of disequilibrium conditions between the origin of the universe, 15 billion years (By) ago,…”

http://www.amazon.com/Principles-Atmospheric-Physics-Chemistry-Richard/dp/0195093623
You can get it used for only $50.00!

B. Smith
March 16, 2010 9:16 pm

Without an atmosphere there would be no oceans, ice, snow or plant life to affect albedo, so the albedo of the Earth should be more similar to that of the barren Moon. It would seem reasonable and logical to expect the high and low temperature variations on an atmosphere-less, barren, waterless Earth to more closely resemble those of the Moon. Am I missing something here?
“Lunar Surface Temperatures
Temperatures on the Lunar surface vary widely on location. Although beyond the first few centimeters of the regolith the temperature is a nearly constant -35 C (at a depth of 1 meter), the surface is influenced widely by the day-night cycle. The average temperature on the surface is about 40-45 C lower than it is just below the surface.
In the day, the temperature of the Moon averages 107 C, although it rises as high as 123 C. The night cools the surface to an average of -153 C, or -233 C in the permanently shaded south polar basin. A typical non-polar minimum temperature is -181 C (at the Apollo 15 site).
The Lunar temperature increases about 280 C from just before dawn to Lunar noon. Average temperature also changes about 6 C between aphelion and perihelion.
Marvin Ostrega
Reference: Heiken et al. Lunar Sourcebook: A User’s Guide to the Moon. Cambridge: University of Cambridge Press, 1991.”

March 16, 2010 9:19 pm

Willis Eschenbach (21:04:42) :
My calculations are based on a blackbody using the Stefan-Bolzmann equation. This equation allows us to convert between radiation intensity and temperature.
I would have said 20C, not 8C…

Matt
March 16, 2010 9:31 pm

The link about the manifold ways in which thunderstorms cool the earth appears not to be working. Pity – I really wanted to read it.

Jim Masterson
March 16, 2010 9:31 pm

>>
In the interim, you might read the Kiehl/Trenberth paper referred to above.
<<
The link isn’t working (for me). Which paper? KT1997? Or the update, TFK2009? I made an infinite layer model and the back radiation doesn’t match KT1997, but it comes close. That 40 W/m^2 for the atmospheric window may be bogus. KT1997 doesn’t calculate it correctly. It may be 80 W/m^2 or 87 W/m^2, depending on how you define “cloudy.”
Jim

March 16, 2010 9:35 pm

I keep reading a number of .05-.09C of CO2 warming and that may indeed be right.
less than .5C or a factor of TEN! decade vs century?
Thanks for the post, id bet is was a lot of work!
You will note that the closer we come to the TRUTH the more they attack, without sources, or logic.
Tim L

March 16, 2010 9:46 pm

How of the incoming energy from the Sun is converted into atmospheric and oceanic convection?
None, according to Kielh and Trenberth.
Which is amazing.
Climate physics is a fascinating subject.

jorgekafkazar
March 16, 2010 9:47 pm

dp (20:46:32) : “Hmmm – Isn’t the moon pretty much earth without air? The temperature there is entirely radiative to surface dwellers (the few that have dwelt), and the soil temperature is dependent upon it’s [sic] exposure to the sun. In shadows it is damned cold – I’d wager colder than 40 something degrees.
“I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.”
Your understanding is flawed across the board, and your prose is impenetrable. What do you mean by “radiative to surface dwellers?” Where did you come up with “our celestial toroid?”

dp
March 16, 2010 9:52 pm

A thing you’ve raised to ponder is does the moon radiate solar heat as fast as it absorbs it? Well, obviously yes, at some temperature it does, and necessarily into that part of the sky not filled with sunlight. (And I’m perfectly happy to accept regional variances in temperature owing to maria vs disrupted lunar surfaces.) As would an airless earth. Actually, so too would an atmospheric earth as Venus demonstrates.
Since we’re talking about energy balance here – clearly the moon is now radiating back to space as much energy as it receives after all these billions of years, there being no mechanism for storage and transport as there is on earth, the current energy level, the black body temperature of the moon, allowing for the rotational gradient, is it really that close to earth’s temperature which is moderated by atmospheric and oceanic energy distribution? My problem is I don’t see how black body temperature translates to an embracing thermal experience as provided to living things by the atmosphere and oceans.
If what you say is true it is really quite profound as it suggests our global temperature is a matter of albedo, no?

March 16, 2010 10:09 pm

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.

I think this first part is questionable. Nothing wrong with your division of 20/150 mathematically, just whether it means anything.
If you take a fixed albedo and increase the radiative forcing at the surface you can calculate the increase in temperature without feedbacks. You obviously know how to do this..
(For people who don’t, you can see more at The Earth’s Energy Budget – Part One and at the end of CO2 – An Insignificant Trace Gas – Part Seven )
So your calculation is basically saying that the response of the climate system from no atmosphere to current status is a linear model.
I know that of course, we can’t assume the same albedo. I see your comments on the various aspects of climate response, don’t know what to think about those yet.
Rightly or wrongly to assume no climate through to current climate is linear seems way way out there!
Perhaps you are aiming at a parody of GCMs producing a linear response?
I find their linear response hard to believe but perhaps you could say that if you look at one small part of a big curve, the slope is linear or can be approximated linear close to that area..

savethesharks
March 16, 2010 10:11 pm

Great post, Willis.
Maybe I missed it, but it should also be stated that, as a new point number 7, the general circulation models are also woefully lacking in accounting for the multi-decadal oscillations of the oceans, including, but not limited to the PDO and the AMO, among many others.
But that point on the importance of cooling with tropical thunderstorms, poorly understood, can not be overemphasized!
Dr. Spencer and Lindzen and others are cracking that block….and so little is known. But you could not be more right. Thunderstorms…are magnificent (albeit ephemeral) players in the earth’s climate.
They boil in atomic-bomb-scale local and mesoscale fury…to heights up to the tropopause…and we wonder how they can bring down airlines (Brazil to France, last year).
Then they fade away in cirrus wisps the next day….all of that heat escaping through an open “window” in the atmosphere…
Thanks again for this post.
Chris
Norfolk, VA, USA

Brian W
March 16, 2010 10:14 pm

enough (20:51:43)
Bingo! Spot On.
Willis Eschenbach
It’s pretty late here and I would like to debate your so called diagram. So a couple of things now and more tomorrow. Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation. As enough(my feelings exactly) stated it’s nonsense and simply not true.
The holy grail of the black/grey body is inadequate to analyse the earth’s weather and climate. Does the earth look like a black/grey body? The warmers have concocted a heady brew from this as the diagram shows. Forget Kiehl/Trenberth they are climate quacks. Take clouds for example, according to the diagram 76w/m2 is not only reflected by the clouds, but also by aerosols and the atmosphere. Really? How is the number 76 arrived at? Clouds most certainly absorb the suns energy. Where is the figure for that? Another thing, with clouds constantly forming and dissipating across the planet, is this an averaged figure for the entire surface? Are all the numbers in the diagram static and unchanging or subject to WIDE variations?
It all looks like an add up the numbers exercise. My BS detector goes off the charts whenever I look at that diagram. Its a mess. So please explain to me your understanding of backradiation and how it occurs and I’ll debunk it.
By the way, in your post at 21:04:42 you misspelled Boltzman. Pretty easy to leave a letter out, eh.

JDN
March 16, 2010 10:15 pm

Let’s try it and find out. 🙂

stan stendera
March 16, 2010 10:21 pm

Sometimes my birdfeeder has hawks circling above it looking for easy pray. One is them is Eschenbach’s hawk. a subspecies of Cooper’s hawk; both of which are far too uncommen

JinOH
March 16, 2010 10:31 pm

‘Sometimes my birdfeeder has hawks circling above it looking for easy pray. One is them is Eschenbach’s hawk. a subspecies of Cooper’s hawk; both of which are far too uncommen’
LOL – I have hawks visiting my feeders too at times.
Nature is a bummer for science.

Ray Boorman
March 16, 2010 10:33 pm

The Moon is very different to Earth, which makes the simple comparisons of pundits above worthless.
1) It is 3,743 km in diameter versus earth’s 12,756 km;
2) Earth is over 1000 deg C at the top of the mantle, which is a mere 5km below the seabed in the deep ocean, & about 50km under the land surface.
3) Moon rotates once each 27 & a bit days, Earth rotates every 24 hours.
4) Earth has lots of volcanic activity, which brings heat to the surface. None has been detected on the Moon, yet.

Allan M R MacRae
March 16, 2010 10:41 pm

Good stuff Willis!
Posted on June 28, 2009
http://wattsupwiththat.com/2009/06/27/new-paper-global-dimming-and-brightening-a-review/#comments
The sensitivity of global temperature to increased atmospheric CO2 is so small as to be inconsequential – much less than 1 degree C for a doubling of atmospheric CO2. CO2 feedbacks are negative, not positive. Climate model hindcasting fails unless false aerosol data is used to “cook” the model.

March 16, 2010 11:02 pm

Brian W:

Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation

How about measurement of it?
We can measure longwave radiation (>4um) downward at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation 4um coming from?
You can see one example of some specific measurements at CO2 – An Insignificant Gas? Part Six – Visualization
Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.

Brian Johnson uk
March 16, 2010 11:27 pm

dp….
“I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.”
dp, Homer Simpson rules – we live on a celestial donut!
“Donuts. Is there anything they can’t do?”

stan stendera
March 16, 2010 11:32 pm

Way, way, way off topic!
for JinOH
I will never forget this: I put birdseed on my deckrail. And of course the squirrels get into it. I was playing on the compute , came to a stopping point,and glanced out the sliding door to the deck. The squirrel, stealing birdseed, suddently started running back and forth then leapt off the deck. Out of nowhere came a Cooper’s hawk and snatched it out of the air. That is why Altamonte is such a crime.
MODERATORS! I know this is off topic. You have my complete permission to forward it to jinOH including my E-mail!!!!

March 16, 2010 11:43 pm

(Something funny happened to my last post, I think because I used a less than sign..)
Brian W:

Backradiation as well as feedbacks are pseudoscience. There is no real physical basis for back radiation

How about measurement of it?
We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.
Where is this radiation coming from?
You can see one example of some specific measurements at CO2 – An Insignificant Gas? Part Six – Visualization
Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.

Steve Goddard
March 16, 2010 11:47 pm

Venus is the same size as earth and has a fairly uniform temperature of over 400C across the entire planet (day or night equator or pole) This is because of their atmosphere.
Sunlight on Mercury is four times stronger than Venus, yet temperatures average much cooler, because of the lack of an atmosphere.
My plane flight on Saturday descended 35,000 ft. and temperatures warmed by 150 degrees F as we descended – because of the blanket generated by the atmosphere.

anna v
March 16, 2010 11:48 pm

Willis,
All the temperatures we have are in the air , and there is a large difference between ground and air. I do not know of any measured ground temperatures. Where do you take the 14C from?
SSTs might be OK, so maybe you should take the SST average temperature as the 0 atmosphere temperature. Earth is 75% ocean after all. the AMSU plots give a scale from -2 to 30, that would give 14, but I would think a true average would come higher.
The albedo of the moon is around 0.1 so it is not a good analogue for your purposes.

anna v
March 16, 2010 11:49 pm
Spence_UK
March 16, 2010 11:59 pm

The figure of 0.13 K per W/m^2 is interesting.
Spencer and Christy found that the LT temperature varies by 2.3K over the year due to the eccentricity of the orbit of the earth. There is little or no perceptible phase lag in this variation wrt the orbital pattern, suggesting this is a full figure and there is no “warming in the pipeline”.
The orbital eccentricity results in a variation of solar radiance of 90 W/m^2. Of course, this arrives over a pi*r disc and is spread over 4*pi*r surface of the sphere, resulting in an equivalent of 17 W/m^2 averaged over the surface of the earth.
This yields a climate sensitivity of 2.3/17 = 0.135 K per W/m^2, entirely independently of Willis’ calculation.
Coincidence?

Sou
March 17, 2010 12:09 am

Is this post meant to be parody, or is the humour unintentional?

Spence_UK
March 17, 2010 12:09 am

re: my 23:59
Good grief, why don’t I check my posts before pressing submit 🙂
Firstly, the area of a circle is pi*r^2 and the surface of a sphere is 4*pi*r^2. Secondly, 90/4 is 22.5 not 17. I’m sure I got that number from somewhere, just can’t remember where…
This yields a climate sensitivity *lower* than Willis at 0.1 K per W/m^2. Not such a coincidence, but at least in the same ball park…

B. Smith
March 17, 2010 12:14 am

Thank you for clarifying, WE.

wayne
March 17, 2010 12:17 am

Willis :
enough (20:51:43) :
I follow you Willis. Good point and well done!
And as ‘enough’ pointed out about the 1/2 up and 1/2 down being rubbish, he is right. All of these energy balance charts are basically wrong, misleading, or incomplete but that doesn’t take one iota away from what you have just have shown us on the sensitivity. The one thing they all do agree on is output equal input.
You see, none of them take into account the dip of the horizon correction or just “the dip”. Any good sailor who knows celestial navigation is well aware of that factor and it applies whether radiation is inbound from stars or outbound from CO2 molecules. Half up and half down can only occur exactly at the sea’s surface and ALL radiation leaving randomly from atoms or molecules in the atmosphere must account for the dip factor.
If you even stand up and are 5 feet above sea level it is no longer half up and half down however tiny. At 20,000 feet it is sizeable. There is now much more going up to space and much less going down back to warm the surface and this factor increases with the height above sea level quickly at first and slower near satellite altitudes. At high tropic cloud top levels this imbalance can get rather large, shunting heat directly to space. If you want the equations, holler. I learned celestial navigation from a captain right out of college (had a dream of circumnavigating when young).
That reinforces your statements about the mention of wind speed on evaporation rates which feed the clouds which shunt the heat to space, but you can’t if “half is up and half is down” and that often used term IS rubbish.
This is only my opinion, wink.

kevoka
March 17, 2010 12:24 am

I am missing something. Everywhere I have read the calculations result in a blackbody temp of 255K.
No atmosphere , no blanket, the temp would be 255K. About 30K less than it is. 1C = 1K.

March 17, 2010 12:35 am

anna v:
You have a point when you say:

All the temperatures we have are in the air , and there is a large difference between ground and air. I do not know of any measured ground temperatures.

The air temperature is taken as the closest proxy to the land surface temperature. During the day it is a reasonably close value because the land heats up, heats the air directly over it and this air then expands and rises. The result is convective mixing which means that the air temperature six feet off the ground is a reasonable estimate.
At night it’s a different story. The temperature can vary significantly in the first few feet because the bottom layer is now colder and there is no convective mixing. “No wind” conditions give very different results from wind conditions. For this reason and UHI issues we should just give up on trying to measure land temperatures
But given what we do have our best estimate of annual average global surface temperature is around 15’C.

tallbloke
March 17, 2010 12:35 am

Hi Willis, great post for opening up discussion. The ‘atmospheric window’ is only 13W/M^2 according to Trenberth. You commented that thunderstorms would increase with increasing temperature. Presumably that would widen the window quite a bit. Would it also stick quite a lot more water vapour into the higher part of the troposphere? What effect would that have?

Nylo
March 17, 2010 12:35 am

You should not compare to a black body of an homogeneus temperature. Because of the emisivity dependence on t^4, two black bodies with the same average temperature but different distribution of it along their surface radiate different ammounts of energy. For a given average temperature, the more temperature differences you have along the surface, the more you radiate to space.
To continue with the moon example, if the average of the moon’s daily temperature is 107C and the average night temperature is -153C, you can be sure that the emisivity total is quite higher than that of a black body with a uniform average temperature of -23C, i.e. the same average temperature. That’s because the emisivity you gain with the +130C of the illuminated side is much bigger than the emisivity you lose with the -130C of the dark side. Actually, it would be equivalent to a black body with a uniform temperature of 320K (47C). Quite hot huh? But this is considering a moon with one half at some uniform 107C and the other half at some uniform -153C. In reality, there are also differences of temperature along both surfaces leading to those averages. And remember: any inhomogeneity in the temperature leads to more emisivity than that of a black body of the same average temperature. So the equivalence of the emisivity would be to a black body of even more than 47C average temperature.
Calculations:
Daily temperature = 107C = 380K
Night temperature = -153C = 120K
The black body with equivalent emisivity would have a temperature equal to the 4th root of ((380^4+120^4)/2) = 320K = 47C
(The day side, with 3.16 times the temperature of the night side, radiates more than 3.16^4= 100.5 times the energy radiated by the night side.

Brian W
March 17, 2010 12:44 am

Willis Eschenbach (22:57:51)
Well, lets see, you used the diagram to show us that “this is what’s happening”.
I and many others do not agree. If the diagram is wrong then using w/m2 and blackbody physics will not give you the right answer. If the numbers are wrong then how do you calculate anything? Yes indeed a very impressive link but since incoming sunlight is comprised of 10% ultraviolet, 44.8% visible (shortwave) and 45.2% INFRARED how is it these wonderful scientists can accurately separate upwelling from downwelling without proper hard experiments in the atmosphere? Sensitivity based on what, fancy algorithms with built in beliefs. Explaining part or all of the diagram shows me that you know what you are talking about, and you avoided it so I’m dubious.
As to the sensitivity. CO2 with a specific heat less than N2, O2 and even aluminum, poor absorption compared to air, fast emission and most importantly a concentration of .038% by volume contributes NO sensible (usable) heat to our atmosphere. A doubling of CO2 will not give even .1 degree increase. I wish people would stop fixating on CO2. So the models do overestimate sensitivity by a huge amount. The whole AGW/CO2 thing is a stinking scientific fraud. Period.

stan stendera
March 17, 2010 12:46 am

It was not a Cooper’s hawk. it was an Eschwnbach hawl!

stan stendera
March 17, 2010 12:47 am

It was not a Cooper’s hawk. it was an Eschenbach hawk!

March 17, 2010 12:58 am

Willis, you are only looking at the high frequency end of an amplifying process, there is definetely a frequency dependency in climate sensititivity noticable, although not as high on a century timespan as IPCC models make of it..
http://members.casema.nl/errenwijlens/co2/howmuch.htm
See also Pelletier for a temperature power spectrum
Jon D. Pelletier, 2002, Natural variability of atmospheric temperatures and geomagnetic intensity over a wide range of time scales, doi:10.1073/pnas.022582599 PNAS February 19, 2002 vol. 99 no. Suppl 1 2546-2553 Link

March 17, 2010 1:34 am

Brian W (00:44:44) :

but since incoming sunlight is comprised of 10% ultraviolet, 44.8% visible (shortwave) and 45.2% INFRARED how is it these wonderful scientists can accurately separate upwelling from downwelling without proper hard experiments in the atmosphere?

It’s easy.
Incoming sunlight is 99% less than 4um in wavelength. Climate science calls this conventionally “shortwave”.
Radiation from the surface of the earth is 99.9% greater than 4um in wavelength. Climate science calls this conventionally “longwave”.
Which means solar and terrestrial radiation can be easily distinguished and lots can be learnt even without “fancy algorithms with built in beliefs.
You can see some explanation about longwave and shortwave

March 17, 2010 1:34 am

Surprisingly, new NASA diagram shows no back-radiation..
http://eosweb.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
If Kiehl-Trenberth diagram is correct and backradiation of 321W/m2 is responsible for the hypothetical +33K effect, it means additional 3.7W/m2 will cause 0.38°C net warming.

March 17, 2010 1:42 am

kevoka (00:24:44) :
Everywhere I have read the calculations result in a blackbody temp of 255K.

This calculation is wrong, since it is calculated with present albedo of 0.3. 2/3 of this albedo is made by clouds, which should actually not be there if there are no GH gases or atmosphere.
More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.

Nylo
March 17, 2010 1:45 am

Re: Willis Eschenbach (00:44:35) :
“Regarding the moon, I’m not the one who brought it up”.
I know. I have used the moon as an example only because it is easy to see the implications as the temperaure swings are so huge. But the same idea could be applied to the Earth. The energy radiated by the Earth is not the same as the energy radiated by a black body of the same average but homogeneous temperature. It is quite bigger, because also the Earth has noticeable swings in the temperature between night and day (especially in the deserts), or the poles and the ecuator, or the summer and the winter hemispheres. The hottest parts radiate sooo much more.

Lindsay H
March 17, 2010 1:49 am

interesting:::
I have a problem with your black body figures for the earth
http://en.wikipedia.org/wiki/Black_body
[edit] Temperature of Earth
If we substitute in the measured values for the Sun and Earth:
If we set the average emissivity to unity, we calculate the “effective temperature” of the Earth to be:
TE = 254.356 K or -18.8 C.
This is the temperature that the Earth would be at if it radiated as a perfect black body in the infrared, ignoring greenhouse effects, and assuming an unchanging albedo. The Earth in fact radiates almost as a perfect black body in the infrared which will raise the estimated temperature a few degrees above the effective temperature. If we wish to estimate what the temperature of the Earth would be if it had no atmosphere, then we could take the albedo and emissivity of the moon as a good estimate. The albedo and emissivity of the moon are about 0.1054[20] and 0.95[21] respectively, yielding an estimated temperature of about 1.36 C.
Estimates of the Earth’s average albedo vary in the range 0.3–0.4, resulting in different estimated effective temperatures. Estimates are often based on the solar constant (total insolation power density) rather than the temperature, size, and distance of the sun. For example, using 0.4 for albedo, and an insolation of 1400 W m−2), one obtains an effective temperature of about 245 K.[22] Similarly using albedo 0.3 and solar constant of 1372 W m−2), one obtains an effective temperature of 255 K.[23][24]
this is substantially different from your estimates is the author of the wiki misleading us?
the earth atmosphere albido is in the order of 30% based on satelliteMODIS values.
so the greengouse effect plus .3 albido gives an average temp of 14 deg c , a difference of 32deg c from the theoretical black body temp.
I’m scratching my ancient head a bit ! applying the albido of 30% should give the planet a temp of 279 deg k = 5.85 deg C meaning the greenhouse effect adds 8 deg c.
confusion abounds!

Brian W
March 17, 2010 1:59 am

Steve Goddard (23:47:55)
The shell of your airplane heated due to air friction! Blankets indeed!
Science of Doom (23:43:15)
“We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.”
The sun is a “broadband radiator” it radiates at ALL frequencies. Since everyone likes w/m2 here is some watts for you. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is INFRARED light, 445 watts is visible light, and 32 watts is ultraviolet light(wiki). And these are only three tiny bands of frequencies in a much larger spectrum. So good luck with your pseudoscience of doom. You are just another warmist site promulgating calculated nonsense.

Ronaldo
March 17, 2010 2:00 am

scienceofdoom (23:43:15) :
“Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.”
Isn’t this because the absorption of IR at specific wavelengths raises the internal (vibrational) energy of the molecule (of CO2, H2O), which then promptly re-emits energy at the same wavelength but in a random direction, hence the notches in the upwelling radiation? The absorption is an internal process, not a thermal process.

Frozen man
March 17, 2010 2:08 am

Hi Willis,
good post i think. I’ve been studying some time this kind of radiation balances, So some things can be added for further knowledge:
1. The graph is old in this authors (Trenberth et altea, 2008), the new one is clearly unbalanced to warm ( i couldn’t see clear what was based on)
2. I think they don’t have applied some basic laws of energy and thermodynamics like: internal energy of the system (U), all the kinds of energy that the earth also receipt, f. e., magnetic, gravitational, etc…. Also thety didnt take into account the effect of winds as cinetic energy, etc… this graph suffers from too much simplicity (only radiation).

Dermot O'Logical
March 17, 2010 2:29 am

I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?
Thanks
Dermot

Louis Hissink
March 17, 2010 2:39 am

Willis
You might have fun adding a graph showing the Earth’s temperature profile up to 500km above surface – it might surprise quite a few that the temperature does not continue dropping the further from Earth but actually rises significantly into the thermosphere.
So which temperature would be representative of the Earth’s temperature, and does the temperature at the thermosphere change diurnally etc etc etc.
The climate people are waxing thermally over the bottom part of the plot – but presumably remain ignorant of what causes the rest of the profile up to 500 km etc.
http://www.windows.ucar.edu/tour/link=/earth/images/profile_jpg_image.html
Have fun folks – I am bush till Easter drilling holes in the ground testing scientific hypotheses in situ.

John A
March 17, 2010 2:40 am

Steve Goddard:

My plane flight on Saturday descended 35,000 ft. and temperatures warmed by 150 degrees F as we descended – because of the blanket generated by the atmosphere.

But blankets don’t warm by suppressing radiation.

March 17, 2010 2:44 am

I think that in time the Kiehl/Trenberth diagram will be considered an even bigger joke than the Mann hockey stick.
Proof
The hockey stick had at least a possibility of being correct, if the predictions turned out to true.
Its a travesty for him that reality dealt a massive raspberry.
The Kiehl/Trenberth diagram is wrong from the start
Just two points
1. They seem at great pains to show that the radiation intensity is conserved.
There is no such conservation law in Physics.
2. The magnitude of so called back radiation from the trace gasses co2 and water vapour is shown as almost twice the total solar radiation must be physical nonsense.

Toho
March 17, 2010 2:55 am

Juraj V. (01:42:28) :
“More, much colder Earth with (frozen) oceans would turn partially to white snowball and its albedo would be mightily increased.”
There can’t be exposed ice if there is no atmosphere. Ice without an atmosphere vaporizes. That is why there is no exposed ice on Mars or the Moon.

Cement a friend
March 17, 2010 3:00 am

Willis,
Another consideration of the energy budget is in figure 2 of http://physicsworld.com/cws/article/print/17402 . You will need to log in to physicsworld.com (IOP website)but it is free. As far as I know has discredited this assessment and the work of Ahilleas Maurellis.
In my assessment (from texts books on heat transfer and personal experience and measurement) if there are two molecules of CO2 near each other in the atmosphere and they are at the same temperature there will be no heat transfer them because both molecules will be saturated at the wavelengths of absorption and emission. If the temperature is different, heat flux will be from the higher to the lower temperature and never the reverse. The same thing applies to applies to the earth’s surface which is about 70% water with an emissivity of about 0.95 (almost a black body). If the surface temperature is above that of the atmosphere, which is the case due to due the the lapse rate and radiation from H2O and CO2 heat will only from the surface.
In my assessment the water vapor (and water and ice in clouds) and to a minor extent CO2 can be compared to a perforated wide spaced sieve. A large quantity of heat flows straight through while some is absorbed and re-radiated upwards reducing the overall heat flux (or heat flow rate). The heat flux will vary considerable around the globe and at different times of the various cycles which affect the earth.
I have no respect for the Kiehl and Trenberth paper. However, I do not dispute (more correctly agree) some of the other points you make such as about evaporation and forced (wind) convective heat transfer.

Nylo
March 17, 2010 3:01 am

Dermot O’Logical (02:29:29) :
“I thought incoming solar radiation was about 1400W/m^2 i.e. the TSI figure.
Why is it 342 W/m^2 in the diagram? Or is it a different physical quantity?”
1400W/m2 is what you would get in space, in a surface perpendicular to the incident light. 342W/m2 represents an average in the surface of the Earth between night and day zones as well as the different angles of the incident light wrt the Earth’s surface. That is, if you multiply 342 times the area of the surface of the Earth you would get the total energy we receive from the Sun at any given time.

rbateman
March 17, 2010 3:20 am

So, Willis, if a station is located where it is naturally near maximum for it’s latitude (due to phsyical geography and circulation), then UHI from any C02 increase will be negligible.
Give me a day or so, I am almost done with just such a station.

Dermot O'Logical
March 17, 2010 3:21 am

Willis Eschenbach (02:51:43)
Much obliged. Kind of obvious when you…. think… about…. it. I must learn to do that.
But why divide by the _whole_ surface area of the Earth? We are only getting sunlight on half of the surface area i.e. the daylight side. It seems invalid to average the incoming insolation over the whole surface area, as that would imply complete absorption of the energy, then being evenly distributed across the whole surface, _before_ being re-radiated away. Stefans law would suggest the daylight side radiates away significantly more energy that the night side, and that’s non-linear, at T^4.
(Please forgive amateur analysis of a physics graduate 20 years out of University – the day job doesn’t require me to keep up to speed with thermodynamics and statistics)
Dermot.

anna v
March 17, 2010 3:24 am

Re: scienceofdoom (Mar 17 00:35),
It is not only in the night that the ground temperature is different than the air.
I live in Greece, in the summer the ground in sunlight is more than 50C, the air temperature may be anywhere from 38 to 42, not higher. One could fry eggs on the rocks.
Searching for ground temperatures, I found one study in Puerto Rico where they measured in the ground in the sun for planting studies, and the temperature 20cm in the ground was the same as the air temperature ( in that particular study 22C). This means that the surface, which is the radiator, would be much higher, as soil is not a good conductor.
That is why I think that average SST temperatures would be more realistic, since they are being measured.

March 17, 2010 3:24 am

Willis: Do the models take account of the endothermic photosynthesis process? The vegetation on the earth is cooling the surrounding air because conversion of CO2 and water through photosynthesis is an endothermic reaction. When biomass is burned, liberating CO2 and water again, this energy is released. As CO2 concentration increases the rate of photosynthesis of plants using the C3 photosynthetic pathway (trees, most grasses, most plants) increases much more rapidly than the increase in CO2 concentration.

March 17, 2010 3:30 am

Ronaldo asked:
On my comment about downward longwave radiation in the bands of various “greenhouse” gases and outgoing longwave radiation with an equivalent “notch” in it

Isn’t this because the absorption of IR at specific wavelengths raises the internal (vibrational) energy of the molecule (of CO2, H2O), which then promptly re-emits energy at the same wavelength but in a random direction, hence the notches in the upwelling radiation? The absorption is an internal process, not a thermal process.

It’s both. When a “greenhouse” gas absorbs energy there is the possibility that it will reradiate this same energy in a very short space of time in a random direction.
But especially in the troposphere and generally in the first 100km of the atmosphere, the gas in the excited state is rapidly deactivated by collisions and the energy absorbed from the original photon is distributed thermally.
This layer of the atmosphere therefore warms up and radiates energy both up and down. This process is the dominant one.

dr.bill
March 17, 2010 3:34 am

Willis is just “testing the class” by changing horses a little bit with the answers to his pop quiz.
Using 342 W/m&sup2; and 14°C, here’s the “point spread”:
The 8°C cooler is correct if comparing “now” to a blackbody (albedo = 0), but if comparing “now” to a denuded Earth with its current average surface albedo (≈ 0.16), the 20°C is correct. If we turn the Earth into a “big Moon” (albedo ≈ 0.12), we get 17°C. In any case, it’s nice to have our cosy blanket, and even nicer for it to be so stable. 🙂
/dr.bill

Recipy
March 17, 2010 3:39 am

Willis . You are confusing things. Is it deliberate?
1) You do not define which temperature you are talking about. From the 14degC average you’d think it was surface temperature, but then you stumble by looking at TOA fluxes. Try looking at the total non-reflected downwelling fluxes at the surface instead and apply stefan-boltzman to that. In that case you’d get an earth which is ~50 degC warmer than a blackbody. This is reduced to ~33C, if you account for non-radiative heat loss terms accounting for the mixing in the atmosphere (i.e. sensible and latent).
2) Wikipedia says: the surface temperature of Earth as a blackbody would be -19C. That is 33C colder than you 14C with an atmosphere. See T_E calculation here: http://en.wikipedia.org/wiki/Black_body.

March 17, 2010 3:58 am

Brian W said:

The sun is a “broadband radiator” it radiates at ALL frequencies. Since everyone likes w/m2 here is some watts for you. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is INFRARED light, 445 watts is visible light, and 32 watts is ultraviolet light(wiki). And these are only three tiny bands of frequencies in a much larger spectrum. So good luck with your pseudoscience of doom. You are just another warmist site promulgating calculated nonsense.

I don’t expect to convince Brian W.
Max Planck was the original promulgator of this “nonsense”. Born 1858, won the Nobel prize for physics in 1918. Many people know his name.
He found the equation that described radiation from a body according to its temperature, and the shape of the waveform. You can see the formula here:
http://en.wikipedia.org/wiki/Planck's_law – along with typical radiation vs wavelength curves.
The energy from the sun, as measured by satellite, amazingly follows this curve. So misguided, that sun..
It’s sad when so little is understood of the absolute basics and yet the passion and the belief is so strong.
Of course, Brian W. can win the Nobel prize by showing that Max Planck – and the sun – are wrong.

jmrSudbury
March 17, 2010 4:10 am

The http://www.worldscinet.com/cgi-bin/details.cgi?id=pii:S021797920904984X&type=html paper disagrees with the idea of back radiation too.
The net is not equivalent to 50/50. The atmosphere is not made up of a single atom re-emitting half of what it absorbed back to earth. For air particles that are 1km up, the earth would be about 170 degrees of the field of view. For those that are 10km up, the earth would be something like 80 degrees of the field of view. This is just an example. I made up the numbers, but the fact there is a big difference is what is important.
John M Reynolds

March 17, 2010 4:19 am

My quick 2 cents on this is I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.
I’m not going to call Trenberth a quack, as someone else did, but I will say it’s not clear to me that we have the technology needed to build an accurate energy balance model at this time.
Which means it’s not clear to me that anything build upon a foundation of our current energy models is accurate.

Allen63
March 17, 2010 4:21 am

I seem to remember reading that Earth’s atmosphere causes 33C heat up vs no atmosphere. Sometimes writers say the “greenhouse effect” for earth is 33C.
Was 33C wrong?
Guess this echos a post just above.

wayne
March 17, 2010 4:27 am

Willis, I have a question for you. Want to see if this seems real to you. You say they measure the down dwelling LW and I assume you are speaking of 321 W/m2 but I can’t get the math to jive. Using their numbers I always come up with the surface being too hot.
If you take all radiation absorbed by the ground, 169 for SW Absorbed By Surface and 321 for LW Back Radiation Absorbed By Surface, you get 490 W/m2. That is the radiation from the sun the chart says you would need to warm the earth to 14 C as if there was no atmosphere at all, right?
Take (490/SB)^0.25 = 304.9 K or 89 F for surface temperature according to the chart. Does that seem correct to you? Am I viewing it wrong? To me it seems too hot and the back radiation is way too high.
To correct that would recalculate to SB*(14+273.15)^4 – 169 = 216.5 for the LW back radiation absorbed by the surface. That is where I don’t connect right to K&T math. Seems 216 not 321, or 105 Wm-2 too high.
Don’t think this would affect your sensitivity logic at all, just the internal numbers.

March 17, 2010 4:36 am

jmrSudbury:
Which one are you going with?
No “back-radiation” according to the entertaining Gerlich and Tscheuschner or back-radiation but modified by the volume integral?
Equally divided up and down is just a useful approximation to explain the basic principles to people who are starting out. Mathematic treatments take this effect into account.
The Planck formula for radiance is in units of W/m^2/sr. “sr” is steradians, or solid angle, i.e. the energy is radiated in all directions.

Basil
Editor
March 17, 2010 4:38 am

I’m confused. At the beginning, the answer to the Quiz is said to be (c), 8° C cooler. But latter on, I read
“If the earth had no atmosphere and had an average albedo like the present earth of 0.16, it would be about 20°C cooler than it is at present.”
And Leif chose (b). So which is it? (b) or (c)? While I don’t see eye to eye with Leif on some things, the “smart money” would bet (b).

1DandyTroll
March 17, 2010 4:39 am

Has anyone actually measured the incoming wattage per square meter?

March 17, 2010 4:43 am

[quote anna v (03:24:15) :]
Searching for ground temperatures, I found one study in Puerto Rico where they measured in the ground in the sun for planting studies, and the temperature 20cm in the ground was the same as the air temperature ( in that particular study 22C). This means that the surface, which is the radiator, would be much higher, as soil is not a good conductor.
That is why I think that average SST temperatures would be more realistic, since they are being measured.
[/quote]

I’m not sure if this is what you’re looking for, but the ISCCP does produce surface skin temperatures (as opposed to surface air temperatures).
Just follow the link below, select “Mean Surface Skin Temperatures” from the “Select a variable:” dropdown, select a time period, and click the “View” button to see a picture.
http://isccp.giss.nasa.gov/products/browsesurf1.html

March 17, 2010 4:48 am

magicjava:

I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.

Which page on Kiehl and Trenberth’s 1997 paper are you referring to?

March 17, 2010 4:56 am

[quote Allen63 (04:21:29) :]
I seem to remember reading that Earth’s atmosphere causes 33C heat up vs no atmosphere. Sometimes writers say the “greenhouse effect” for earth is 33C.
Was 33C wrong?
[/quote]

The “greenhouse effect” isn’t going to be the same as “effect of greenhouse gasses”. The greenhouse effect is always a warming effect. The effect of greenhouse gasses can be warming or cooling.
An example of a cooling effect created by greenhouse gasses is water vapor becoming clouds. Overall, clouds cool the Earth by about 20 degrees, even though they’re made from what started out as a greenhouse gas.

March 17, 2010 5:03 am

[quote: scienceofdoom (04:48:54) :]
magicjava:
I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.
Which page on Kiehl and Trenberth’s 1997 paper are you referring to?
[/quote]

It’s not the 1997 paper. It’s the updated paper (2005 I believe).
The exact quote is:
[quote: Trenberth :]
The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 8.5 +/-0.15 Wm-2 by Hansen, et. al.
[/quote]

I’d consider any earlier paper to be even less accurate than this one.

dr.bill
March 17, 2010 5:09 am

Recipy (03:39:01) :
Allen63 (04:21:29) :
Regarding the 33°C issue:
That value comes from using an albedo of ≈ 0.31, which is roughly what you get with the atmosphere, clouds, etc, left intact. In that case, the 342 W/m² available for absorption by the surface becomes 236 W/m² (= 0.69×342), Willis’ values (and mine a little further up the page) are found by elimininating the atmosphere altogether. If you don’t have an atmosphere, you can’t have clouds, and you wouldn’t have such a high albedo, so the comparison is between “everything in place as now” versus “bare planet like the Moon”. You can’t logically use the albedo that one gets with the clouds and such if you don’t have an atmosphere to produce them.
/dr.bill

March 17, 2010 5:10 am

What about internal warming due to radioactivity and tidal forces (although, I guess those are mitigated by the fact that the oceans can move in response to the forces)? My understanding is that Lord Kelvin’s thermodynamic calculation of the age of the earth, which was based on a black-body irradiated by the Sun, failed to produce the correct age because he was unaware of the concept of radioactivity.

March 17, 2010 5:17 am

[quote 1DandyTroll (04:39:37) :]
Has anyone actually measured the incoming wattage per square meter?
[/quote]

Yes. The CERES satellite measures longwave and shortwave radiation. Their web site is here:
http://science.larc.nasa.gov/ceres/index.html
And an overview of the meaning of it all is here:
http://asd-www.larc.nasa.gov/ceres/brochure/intro.html

March 17, 2010 5:23 am

scienceofdoom:
Here’s the link to Trenberth’s paper. See page 3 for the quote:
http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/BAMSmarTrenberth.pdf
1DandyTroll:
On page 3 of that paper you’ll see Trenberth saying he ignores data from the CERES satellite when building his energy balance model.

Kay
March 17, 2010 5:30 am

@ Bill Parsons (21:03:27) :
Wiki Answer to: “What is the temperature on the moon?”
The average daytime temperature on the Moon is around 107°C (225°F), but can be as high as 123°C (253°F).
When an area rotates out of the sun, the “nighttime” temperature falls to an average of -153°C (-243°F).
253 F
-243 F
_______
10 F”
=============================
For a range of 496 degrees between the light and dark sides. That’s pretty extreme. What is the range on Earth? Thanks to the atmosphere, its nowhere near close to that.

March 17, 2010 5:42 am

Trenberth is far less certain than he used to be about the energy budget. Remember he said it was a ‘travesty’ that he couldn’t account for the lack of warming in the last decade. One of his recent papers talks about ‘missing energy’ that he simply can’t account for: ‘Tracking Earth’s energy: From El Niño to global warming’ (February 8, 2010, submitted to ‘Science’)
http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/Tracking%20Energyv5.pdf
“Increasing concentrations of carbon dioxide and other greenhouse gases have led to a post-2000 imbalance at the top-of–atmosphere (TOA) of 0.9±0.5 W m-2 that produces “global warming”. Tracking how much extra energy has gone back to space and where this energy has accumulated is possible, with reasonable agreement between model and observational results for 1993 to 2003…Because carbon dioxide concentrations have further increased since 2003 the amount of heat subsequently being accumulated should be even greater.”
“…although some heat has gone into the record breaking loss of Arctic sea ice, and some has undoubtedly contributed to unprecedented melting of Greenland and Antarctica, it does not add up to be anywhere near enough to account for the measured TOA energy. Closure of the energy budget over the past 5 years is elusive. Thus state-of-the-art observations are unable to fully account for recent energy variability implying error bars too large to make the products useful.”
[“not…anywhere near enough”, “elusive”, “unable to fully account” “too large to make the products useful” etc: read: we’re at a loss to know what’s going on – it’s a travesty.]
“After 2000, observations from TOA…referenced to the 2000 values, show an increasing discrepancy…relative to the total warming observed…The quiet sun changes in total solar irradiance reduce the net heating slightly …but a large energy component is missing…”
“A large energy component is missing” – you can say that again! The discrepancy is huge and increasing by the year – this is the ‘travesty’ because it is a mysterious component that Trenberth knows nothing about, so for five years he’s been unable to ‘close’ the energy budget. Big problem! The contributions from melting glaciers, ice caps, Greenland, Antarctica and Arctic sea ice plus contributions from land and atmosphere warming and changes in solar irradiance all taken together are completely swamped by the “missing energy” due to an energy flux that is not in any way understood.
Of course, it’s only ‘missing’ on the assumption that the greenhouse effect is working as it is alleged to, with positive feedbacks. Lindzen and Choi (though rubbished by Trenberth) showed that, far from positive feedback, there may even be negative feedback – the surface temperature increased AND the escaping IR was increasing just as fast (this is probably Trenberth’s TOA discrepancy, and where the ‘missing energy flux’ is going). This seems to suggest that surface warming through the 1990s was NOT largely due to greenhouse warming, and with increasing CO2 concentration year by year the effect is considerably less than assumed by IPCC. Wind out the positive feedbacks and this ‘missing energy’ component disappears and things start to look sensible again.

March 17, 2010 5:43 am

The label “back- radiation “Is causing needless confusion.
I take it to mean that all of the 321w/m2 has come “back” from somewhere.
The somewhere must be the atmosphere.
This label seems to separate it from all other IR and long wave from scattering and direct to surface by whatever means EM radiation.
“No “back-radiation” according to the entertaining Gerlich and Tscheuschner”
In fact what G&T say is no HEAT from back radiation which they further state is so small that it can be ignored.
Wayne
….If you take all radiation absorbed by the ground, 169 for SW Absorbed By Surface and 321 for LW Back Radiation Absorbed By Surface,…….
I agree with Wayne does all the 321 come back from somewhere?
This diagram was meant to educate but instead it seems full of obvious ambiguities

March 17, 2010 5:47 am

[quote Kay (05:30:20) :]
For a range of 496 degrees between the light and dark sides. That’s pretty extreme. What is the range on Earth? Thanks to the atmosphere, its nowhere near close to that.
[/quote]

Looking at the raw data for the aqua satellite from the last 13 months, it gives a high of 274.18 K and a low of 197.06 K, for a difference of 77.12 K.
Obviously, the Earth’s experienced greater extremes than what it’s seen in the last 13 month, but 77.12 K is a decent enough rough estimate.

David L
March 17, 2010 5:56 am

“What is the temperature on the moon?”
First, remember that heat is the thermal unit of measure that “flows”, not temperature. Temperature is a result of heat content, defined as the average translational kinetic energy of atoms and molecules, and is dependent on the mass and specific heat of the object (such as air). There is less specific heat (and less mass) for a planet with no atmosphere than one with atmosphere. So if the amount of energy were the same for both cases, the temperature would be higher for the planet with no atmosphere. But heat is lost radiatively (black body radiation) and the atmosphere will prevent some of this loss, but itself also radiates it’s heat back to the earth and out to space. Quite a complicated system to figure out exactly as Willis nicely shows in his graphic.
There is no atmosphere on the moon, so there’s no air temperature (there’s no air to absorb heat), only surface temperature (the surface absorbs heat). When the sun is hitting the surface, it’s very hot. When the sun isn’t hitting the surface, it’s very cold because, having lost whatever heat it had by radiation. One side is hot, the other cold, and the average works out to be about -10F.
This illustrates the problem I have with a single term like average global temperature. Statistically if I put one foot in boiling water and one foot in ice water, the average is luke warm and I should feel fine. Some stations show an increase in temperture, others show a decrease. What’s the average? If North America becomes an ice sheet but Antarctica melts and becomes a jungle, and the average global temperature is still zero, should we be concerned? Is this global climate change? It’s certainly local climate change!!!!
Scientists should be focussed on the thermal budget and not a meaningless average global temperature that is dependant on so many other factors. I guess thermometers are cheap and everyone can read them…it’s harder to track the energy. But that’s the core of the problem. Is the earth retaining too much of the solar energy it receives and not allowing the proper balance to radiate back into space? When a climate scientist shows me these fundamental calulations and experiments on the heat budget then I’ll believe we have a problem.
In closing, a guy who lived in my neighborhood in 1902 wrote in his diary that yesterday (in 1902) was 58F….Today (2010) the high is 47F……not warmer for you climate scientists keeping score at home.

March 17, 2010 5:59 am

Willis Eschenbach (01:08:38) :
On a blackbody earth the current distance from the sun, it would only be 8°C cooler.
You are misunderstanding this. There is a difference between a ‘blackbody’ [20C cooler] and a ‘black body’ [8C cooler]. The correct answer to your question as stated is 20C cooler.

March 17, 2010 6:02 am

[quote ScientistForTruth (05:42:08) :]
[Quoting Trenberth]
“Increasing concentrations of carbon dioxide and other greenhouse gases have led to a post-2000 imbalance at the top-of–atmosphere (TOA) of 0.9±0.5 W m-2 that produces “global warming”
[/quote]

I’ll just add that the 0.9 Wm-2 Trenberth uses _does not_ come from measurements. It comes from Hansen’s climate model. Actual measurements give a higher or lower number than 0.9, depending on what measurement you use.
So, Trenberth is basically saying “Global Warming is 0.9 Wm-2 because Hensen’s climate models say it is so.”

David L
March 17, 2010 6:05 am

This is an excellent website that explains the fundamental concepts of Physics for those of you new or “rusty” on these concepts.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
Click on “Heat and Thermodynamics” for links to those concepts, under that link click on “Radiation” to read about Stefan-Boltzman.

Bill Illis
March 17, 2010 6:15 am

I think you are right Willis.
They have actually calculated the 33K greenhouse effect wrong.
The Albedo figure used in the 33K calculation is 0.3 which includes the effects of solar reflection by clouds. But then the calculation excludes the positive greenhouse contribution of those same clouds. You can’t include one without the other.
So everything should start over at the beginning and say an atmosphere without any greenhouse gases would be 20K cooler.
Put water vapour/clouds back in, and you have -13K from cloud Albedo and something around +17K to +25K from the water vapour/cloud greenhouse effect so you do get something like +16K to +8K greenhouse effect from CO2/non-water vapour GHGs.
After that, the Stefan Boltzmann equations are logarithmic so one needs to calculate the K/W/m2 impact at each individual differential rather than take the total temperature K / total forcing change [which is one of the problems with the greenhouse theory – they stop using the differential].
Once you get close to the 390 watts/m2 total forcing, the differential is 0.18K / W/m2.

Basil
Editor
March 17, 2010 6:33 am

“The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.16 (16% of the solar radiation hitting the ground is reflected by the surface back to space). ”
Another question. 29/.16 = 181.25. I don’t see ~181 in the diagram. I do see 29+169=198 (168 absorbed, and 29 reflected).
Is this CBO scoring, or .gov accounting? I.e. 181 is close enough to 198 for government work?

March 17, 2010 6:36 am

Bill Illis (06:15:41) :
I think you are right Willis.
I think he is not.

Anders L.
March 17, 2010 6:43 am

“If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?

March 17, 2010 6:57 am

“The Wikipedia article is using the amount of sun after albedo losses, which is about 235 W/m2. That gives the colder value you quote above.”
Just a simple question. The wikipedia equations regarding radiating in all directions and receiving only in one direction seem to suggest the calculation assumes the sun is a point source, and the earth a hard disk. In actual fact, the sun has a finite size and slightly more than half the earth’s surface is being irradiated at any time. In addition, the atmosphere itself tends to increase the size of the cross sectional area – which could both increase and decrease the temperature of the earth.
Finally, I really wish people would stop reproducing these rubbish static models of the atmosphere. The atmosphere is a dynamic system with convectional currents taking heat from the surface up above otherwise blocking layers. The hysteria about global warming, is essentially a static view of the atmosphere (both in terms of time and space), in contrast I would suggest the sceptic view is a dynamic view whereby the climate not only varies in time, but it doesn’t stick around the surface but in contrast there are dynamic convective currents taking hot air high into the atmosphere where gases with a high emissivity (CO2!) enable the heat to escape.
Think about it this way
Why does hot air rise to create thermals …. it is because cold air descends! (Ha ha got you!) OK, a bit flippant, but if cold air didn’t descend, then we’d be left with a vacuum at the surface much like the space between Al Gore’s ears.
Hot air rises because cold air descends and so we should stop focussing on what is heating the air, and start asking the question: What is cooling the air?
Now, the answer is pretty obvious: the heated air is loosing its heat by the emission of IR, and therefore anything that helps to increase the emission of IR … like e.g. a complex molecule with a high IR interaction … what could that be? CO2? helps to cool the planet! That is to say, any molecule able to absorb IR is also (at the same temperatures) able to emit CO2 and so is just as capable of being a global cooler as being a global warmer!
The warmists see the atmosphere as a big blanket keeping us warm. In contrast I think the sceptical view should be to see the atmosphere as a huge cooling system taking heat from the surface of the earth up into space. The more CO2, the better the cooling system, and if like the warmists, we were to ignore the other effects of CO2, we also could create a hysteria that adding CO2 will dramatically increase the effectiveness of the world’s cooling system and lead to runaway global cooling.
OK, that’s far too simplistic but no less simplistic than the rubbish about CO2 causing runaway warming!

David L
March 17, 2010 7:05 am

Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Understanding if the earth is warming due to greenhouse gases has everything to do with the thermodynamics of the system. If you can’t explain the heat gain and the heat loss of this system then you don’t understand the system and shouldn’t be scaring people into doom and gloom scenarios.
Consider this, satellites orbiting the earth are in “vacuum” and being irradiated by the sun. Plus they contain electronics that generate heat. Here on earth my Pentium stays cool because air circualtes around it. How do you cool a pentium in a vaccum??? I’ll give you a hint “Black Body Radiation”. You ever see those giant sails on the space station? Some are solar panels, some are cooling radiators.
Now if scientists and engineers can use the basic concepts of physics to allow satellites to work in orbit, the Rover drive all over Mars, and the hubble space telescope develop great images, and in every case without freezing or melting down in a vacuum, then those same concepts should be able to explain the temperature on earth and it’s response to solar flux, greenhouse gases, and radiative cooling to name just a few.
Comparing the system with an atmosphere and without leads to a basic understanding of the maximum and minimum of the system. In calculus it’s understanding the limits of the equation. This helps in framing what’s going on in the middle.

March 17, 2010 7:06 am

[quote Leif Svalgaard (06:36:25) :]
Bill Illis (06:15:41) :
I think you are right Willis.
I think he is not.
[/quote]

I think the skeptic community would be well served if we built out own energy balance model, rather than using one based on numbers James Hansen pulled out of his ass, which is all Trenberth’s model is.
Starting with data from the CERES satellite would be a good idea, IMHO. Building a model that changes with conditions rather than just giving a hard-coded answer would be a good idea too. And being honest about things we don’t know, like where extra energy is going, would also be a plus.

Claude Harvey
March 17, 2010 7:16 am

Willis:
According to the chart of the past 450,000 years of reconstructed global temperatures for planet earth (the one The Goracle likes to wave about), the last Ice Age saw temperatures of at least 9 to 10 degrees C below today’s. I believe we had an atmosphere back then.
I think your “black body” calculation is off by a country mile. I admire much of your work, but suspect you got wrapped around your own axle on this one.

Frank Schroeder
March 17, 2010 7:19 am

>Leif Svalgaard (05:59:34) :
>>Willis Eschenbach (01:08:38) :
>>On a blackbody earth the current distance from the sun, it would only be
>>8°C cooler.
>You are misunderstanding this. There is a difference between a ‘blackbody’
>[20C cooler] and a ‘black body’ [8C cooler]. The correct answer to your
>question as stated is 20C cooler.
Yes, this distinction is important – as I understand it, a “black” blackbody earth would be 8 degrees colder, whereas a blackbody earth would be 20 degrees colder. However, does this matter for the energy flux diagram and Willis’ conclusions?

harrywr2
March 17, 2010 7:21 am

Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Because it’s a discussion on how sensitive the earths climate is.
One set of scientists says that if you modify one little part of the atmosphere there are going to be gigantic changes the climate.
I believe Willis is attempting to show if you just completely got rid of the atmosphere altogether the climate wouldn’t change that much.
Of course it would be hard to breath and the polar bears would surely die in Willis’s example.

Enneagram
March 17, 2010 7:23 am

Basil (06:33:41) :
This means that the surface albedo is about 0.16 (16% of the solar radiation hitting the ground is reflected by the surface back to space). ”

Should we thank the chinese for so kindly lowering that albedo?..with Carbon soot they lower it but with SO2 they increase it. But all this is nonsense as just one humble volcano will surpass any amount of anthropogenic “forcings”.

Ryan Stephenson
March 17, 2010 7:28 am

Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius. That underground heat energy must be percolating up to the surface eventually and contributing to the background temperature at sea level.
Interestingly, the bottom of the ocean cannot be lower than 4Celsius. If it fell below this it would freeze, and then the ice, being less dense than water, would float to the surface, where it would then melt in the sun. So the deepest ocean trenches cannot be lower than 4Celsius, even though in principle they can be getting almost no energy directly from the sun. Complicated isn’t it?
The Moon doesn’t have a lot of geothermal energy or ocean’s that refuse to freeze. It is relatively easy to understand why it gets close to absolute zero at night and above boiling point during the day. Fortunately we are well insulated from such extremes.

March 17, 2010 7:28 am

Why is it that PV=nRT is never considered when figuring a surface temperature of the earth. It is a valid scientific formula. It explains the normal use of STP when saying under what conditions you did your test etc. As such 0 deg C is what the earth with an atmosphere has as a temperature and the 33 deg C warm up from GHG is really only 15 deg. at max.
Also if IR goes from 4-60 micro and the band for CO2 is 14-15&16 accounting for any over absorbtion then all CO2 can only get 5.4% of total reradiation. Of that 5.4% human input to CO2 is roughly 3% ergo 3% of 5.4% is .0016. The best that human CO2 can do is .0016 of any longwave radiation. If you use TOA of 247 W/m2 then .4 W/m2 is max for human input. It is probably less than that.

March 17, 2010 7:43 am

Frank Schroeder (07:19:28) :
Yes, this distinction is important – as I understand it, a “black” blackbody earth would be 8 degrees colder, whereas a blackbody earth would be 20 degrees colder. However, does this matter for the energy flux diagram and Willis’ conclusions?
To be frank, I can’t tell what his conclusions are.

March 17, 2010 7:45 am

Your theory seems to be Swiss cheese on this one Willis.

David L
March 17, 2010 7:50 am

mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.
In the earth’s atmoshere there is no change of volume (it’s the boundless sky..bounded only by gravity), no change in pressure (except small barometric changes when fronts move around) and there’s really no change in moles of gas (unless you consider the tiny fractional increase in CO2). Plus the ideal gas law only applies in the ideal cases…in this marginal increases in moles of CO2 it wouldn’t be an accurate equation anyhow.

Jim
March 17, 2010 8:19 am

******************
David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.
In the earth’s atmoshere there is no change of volume (it’s the boundless sky..bounded only by gravity), no change in pressure (except small barometric changes when fronts move around) and there’s really no change in moles of gas (unless you consider the tiny fractional increase in CO2). Plus the ideal gas law only applies in the ideal cases…in this marginal increases in moles of CO2 it wouldn’t be an accurate equation anyhow.
**************
Isn’t the ideal gas law used to calculate the adiabatic lapse rate?

March 17, 2010 8:33 am

David L (07:50:25) :
Are you saying that the 5.3million giga tons of atmosphere has no effect on the earth’s temperature? If I leave T as an unknown in the formula and use standard pressure for P and figure for a volume up to say 100km I get a temperature of 0 deg C.
The term STP means standard temperture and pressure. If it is standard to have 0 deg C at one atm then that should apply to the surface of the earth also.
I realize air is not an ideal gas but then again we use radiation formula that were meant for cavities and black bodies with no depth.

March 17, 2010 8:46 am

Is anyone going to consider the second law of thermodynamics and “back radiation” supposedly “warming” the earth’s surface. ?
Quite simply anything based off the bunkum IR budgets,
is bunkum itself.
That is a shame Willis because your recent post(s) regarding thunderstorms (atmospheric heat pipes) is damned excellent, and a great leap forward in our understanding. I particularly have to mention the illuminating “sun’s eye view” you used, brilliant.
Excellant posts by Brian W, Ryan Stephenson, David Haseler and David L. to mention only a few.
The rest, this place is becoming a warmists site.

Richard M
March 17, 2010 8:47 am

As a corollary it would be interesting to look at historic situations. For example, when the Earth had no ice caps what was the sensitivity? At the height of glaciation what was the sensitivity? That might put some bounds on the problem.

Bill DiPuccio
March 17, 2010 8:48 am

“1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.”
I would agree with this statement as it applies to a global scale. But positive water vapor feedbacks do occur on smaller spatial and temporal scales (Dessler, Lindzen, Pielke).
The error occurs in trying to generalize these situations to the entire atmosphere. As you point out, all the negative feedbacks come into play canceling or even reversing the potential effect.

OceanTwo
March 17, 2010 8:53 am

harrywr2 (07:21:49) :
Anders L. (06:43:13) :
““If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?”
Because it’s a discussion on how sensitive the earths climate is.
One set of scientists says that if you modify one little part of the atmosphere there are going to be gigantic changes the climate.
I believe Willis is attempting to show if you just completely got rid of the atmosphere altogether the climate wouldn’t change that much.
Of course it would be hard to breath and the polar bears would surely die in Willis’s example.

That’s essentially what I got out of it, also.
The actual energy flows in the atmosphere in this case are (mostly) irrelevant. The linearity is irrelevant. It seems we are trying to determine what the (theoretical) temperature of the earth is without an atmosphere.
And honestly, even at 20 degrees, it seems like you’d need a radical atmospheric change to significantly alter the atmospheric contribution. (Of course some people claim that the current and predicted CO2 change is a ‘radical’ change).
Maybe Willis *does* hate polar bears, who knows? they aren’t particularly friendly creatures, after all. 😛

Charles Higley
March 17, 2010 8:56 am

Tempest in a teapot when you realize that there is not enough carbon out there for us to burn to double the CO2 in the atmosphere.
Why? Good question.
CO2 partitions 50 to 1 between water and air. It would take 50 times the CO2 that the atmosphere contains to double the CO2 in the atmosphere. One estimate is that, if we really tried, we might be able to raise the CO2 by 20%. We are fighting against a partition coefficient.

Alan D McIntire
March 17, 2010 9:00 am

The temperature for a blackbody earth should be
T = (1368 watts/square meter)/(4* 5.67*10^-8 watts/square meter K^4)
gives a temperature of 278.683 K- Willis Eschenbach’s figure. What you’re all arguing about is whether or not you should throw the factor (1 – albedo) into the numerator. If you throw
in the clouds, you throw a (1-.3) into the numerator, you decrease the prior figure of 278.683 to 278.683 * (0.7^.25) = 254.909 -blackbody earth without greenhouse gases but with clouds

March 17, 2010 9:08 am

Derek (08:46:04) : No warmist I.
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Someone should patent this.

Allen63
March 17, 2010 9:09 am

dr.bill,
Thanks for the clear response. Choice of albedo it is, then.

March 17, 2010 9:18 am

Alan D McIntire (09:00:18) :
The temperature for a blackbody earth should be […] a temperature of 278.683 K- Willis Eschenbach’s figure.
No, this is for a black blackbody, but the Earth is not black. It is very nearly a blackbody, but the albedo of the Earth without atmosphere [and clouds] is not zero, it is [as Willis points out] more like 0.15, thus a real Earth with atmosphere removed would be 20C cooler, not 8C.

March 17, 2010 9:19 am

Alan D McIntire (09:00:18) :
The temperature for a blackbody earth should be […] a temperature of 278.683 K- Willis Eschenbach’s figure.
No, this is for a black blackbody, but the Earth is not black. It is very nearly a blackbody, but the albedo of the Earth without atmosphere [and clouds] is not zero, it is [as Willis points out] more like 0.15, thus a real Earth with atmosphere removed would be 20C cooler, not 8C.

Steve Goddard
March 17, 2010 9:19 am

Using this logic, Mt. Everest should be only four or five degrees cooler than Saudi Arabia. Another example of the misuse of mathematics.

March 17, 2010 9:23 am

David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I’m afraid mkelly is quite correct.
The atmosphere is subject to the Earths Gravitational Field
This causes gravitational compression of the gases.
High school physics student would be expected to answer question below.
Typical HSPQuestion is” using the kinetic theory of gases
a/calculate the rms speed of a N2 molecule at STP
b/If this molecule moved vertically upward and had no further interactions how high would it get
c/ What would happen to the temperature as it went higher”
(Answers 517m/s,13.6Km,drops constantly)

lgl
March 17, 2010 9:30 am

Close enough for the black-body sensitivity which is 0.18 °C per W/m2, but we want to know the ‘climate-body’ sensitivity, and then you have to include the feedbacks.
Trenberth’s diagram shows that the system amplifies the external input (sun) by 2.4 so the resulting sensitivity is somewhere around 0.43 °C per W/m2, or 1,6 C per CO2 doubling.
This assumes todays gain of 2.4 will not change if the forcing increases, but it probably will, probably not linear. But since a 3.7 W/m2 increase is <1% of the initial, setting a 2.2-2.6 boundary should be safe (+/- 8% of initial) and then the sensitivity range becomes 0.40-0.47 or 1.48-1.73 C per CO2 doubling.
calcs:
Solar input to surface: 169 + (58+10)/2=203 W/m2 (/2 because only 1/2 downward)
Total to surface: 321+169=490
Gain: 490/203=2.41
Black-body sensitivity: (289^4-288^4)*5.678*10^-8=5.45 W/C=0.18 C/W/m2
This holds if we do not reach a Tipping Point, but the good thing is we already reached the cold-to-warm TP 10000 years ago, so the next TP will be a warm-to-cold TP.

brent
March 17, 2010 9:34 am

Willis,
Thanks for your superb efforts and excellent way to get open discussion
I’m a little wary should we ourselves draw overly firm conclusions/convictions in the sense of taking the Kiehl/Trenberth graph figures as a given, and then drawing some overly firm conclusions therefrom
(All sorts of conjecture is absolutely warranted )
Isn’t this the same Kevin Trenberth referred to below?
I.E. Who admits in a private email we are not close to balancing the earths energy budget (and agreed by Tom Wigley) and therefore geoengineering would be hopeless, as we would not be in a position to evaluate it’s effect
If so what is the level of confidence in all the various fluxes shown in that diagram in the first place.?
It may well be that it’s just that I’ve not taken the time myself to study these particular aspects in depth required to get a personal comfort level
The Warmers are absolutely open to highly warranted criticism on multiple levels. I thought your robust 🙂 comments to Judith Curry and Ravetz were spot on. They reflected my views exactly.
Just a word of caution (very slight) that we not get ahead ourselves in term conviction/confidence
Warm Regards
brent
http://www.eastangliaemails.com/emails.php?eid=1056&filename=1255550975.txt
Kevin,
I didn’t mean to offend you. But what you said was “we can’t account
for the lack of warming at the moment”. Now you say “we are no where
close to knowing where energy is going”. In my eyes these are two
different things — the second relates to our level of understanding,
and I agree that this is still lacking.
Tom.
++++++++++++++++++
Kevin Trenberth wrote:
> Hi Tom
> How come you do not agree with a statement that says we are no where
> close to knowing where energy is going or whether clouds are changing to
> make the planet brighter. We are not close to balancing the energy
> budget. The fact that we can not account for what is happening in the
> climate system makes any consideration of geoengineering quite hopeless
> as we will never be able to tell if it is successful or not! It is a
> travesty!
> Kevin
>

March 17, 2010 9:44 am

An argument that is made on the basis of a Kiehl-Trenberth diagram makes me nervous for this type of diagram makes a muddle of thermodynamics. In the language of thermodynamics, there is no such thing as an “energy flow.” The only energy that “flows” is heat. In a Kiehl-Trenberth diagram, some of the “flows” are heat. Others are radiation intensities. It is clear that the “back radiation” is not heat for it “flows” from cold to hot matter; if it were heat, it could not flow in this manner, under the second law of thermodynamics. However, while we have a conservation principle for heat flows, we do not have one for radiation intensities. Thus, the proposition that the K-T diagram portrays some kind of “balance” is false.

KDK
March 17, 2010 9:48 am

Amazing to see so many using ‘wiki’… LOL
What is truly amazing is this: So many of you consider yourself scientists but, as seen in many, if not all posts, is that you cannot agree on actual numbers. I thought sciences, like mathematics, uses numbers that have been accepted by measurements that can be validated.
How is it that there are so many theories and only a few datasets to use? How can we almost NEVER agree on an answer? Why so many theories? Can’t someone just factually prove what is being presented… ever? Certainly, I am not saying we shouldn’t continue to push forward as new and enlightening evidence is brought forth via discovery, but, damn, I didn’t know science was so grey… shouldn’t it be either black or white, one or the other? I realize that some of these are just theories and can never be proven, but like AGW, the one’s claiming to KNOW MUST prove it via facts, or it should just be academia speaking for the sake of spurring on innovation/discovery and NOT be presented like it is the end all–unless it truly is.
Again, I am not saying that because Joe, Jim, Janet, and Jomojo all say it is fact, that one should stop looking, but let’s find out which measurements are the best, accurate and then all use the same info after analysis. How are there so many theories surrounding supposed fact-based research?

March 17, 2010 9:49 am

Steve Goddard (09:19:41) :
Another example of the misuse of mathematics.
Nonsense, it is of misuse of the physics.

renminbi
March 17, 2010 9:58 am

If the earth’s atmosphere were only N2 and O2 would either of these gases absorb or emit radiation from the sun or earth. Sorry to expose my ignorance,but it is better to ask than to remain ignorant. This is a great blog,with great comments.

Richard Sharpe
March 17, 2010 10:01 am

mkelly (09:08:39) said:

Derek (08:46:04) : No warmist I.
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Someone should patent this.

I don’t think this is a good argument. The devil is in the details.
1. Coffee is normally at a temperature that is outside the frequency range that CO2 efficiently absorbs.
2. Evacuating the thermos shell is to prevent conduction between the inner part of the shell and the outer part of the shell. CO2 will conduct much better than a vacuum.
So, it really depends on whether the losses due to conduction are higher or lower than the reduction in losses due to radiation.

March 17, 2010 10:20 am

Willis Eschenbach (10:09:49) :
8°C if the no-atmosphere earth is considered to be a blackbody.
You do not understand what a blackbody is. The 8C is for a black blackbody. The Earth is not black, but is nevertheless very close to a blackbody. Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate the same amount. If not, we would violate the thermodynamics, where we could get work out of the radiative difference [use it to power a steam engine] between two bodies at the same temperature.

March 17, 2010 10:25 am

Willis Eschenbach –
Do you think you’re up to creating an energy balance diagram from scratch using actual data rather than James Hansen’s climate model the way Trenberth did?
Somebody with motivation has to do it and my plate is full at the moment with the Aqua satellite.

Eric (skeptic)
March 17, 2010 10:26 am

Enneagram states ”just one humble volcano will surpass any amount of anthropogenic “forcings””
Pinatubo released 42 Mt of CO2 (see http://vulcan.wr.usgs.gov/Projects/Emissions/Reports/Pinatubo/pinatubo_abs.html). Mankind releases 6 or 7 Gt of carbon per year (multiply by 44/12 to get CO2)

March 17, 2010 10:41 am

[quote Willis Eschenbach (10:35:46) :]
Are there any substantive claims that I have not disputed?
[/quote]

Well, um, yes. Mine. Trenberth’s model uses made up numbers and ignores real life measurements. Using his work to check your results means you’re comparing your results to made up numbers.
That’s not a flaw in your line of reasoning, but it is a flaw with what you’re using for a sanity check on your results.
Which is why I was wondering if you feel up to the task of creating a more realistic energy balance. One that uses real-world numbers.

March 17, 2010 10:42 am

Willis Eschenbach (10:34:11) :
Is my writing really that opaque? My conclusion is that the climate sensitivity is on the order of half a degree C per doubling of CO2, and the equilibrium sensitivity is much less than that.
As far as I can see you concluded that based on the 20C figure. You never used the 8C figure for anything, which is good because it is wrong.

March 17, 2010 10:47 am

scienceofdoom (23:02:15)
Thanks for the response to brian W. I find your blog a wonderful resource for people get an understanding of things like back radiation.
There is enough detail there for those with technical backgrounds and your way of explaining things to those with “not so technical” backgrounds is very good. I often wonder what the blog wars would have been like if reasonable voices like yours had been in charge of real climate.
Scienceofdoom: WRT Willis contention. It seems a first good guess that
the first order effect of albedo is linear, it’s effectively a TSI knob, but one thing that bothers me about the diagram is what happens to the 29W/sq meter that get reflected from the surface. Shouldnt some portion of that be absorbed by trop and lower strat ( as is the case with that same radition when it is incoming)? And shouldnt some be relected back down by clouds and aerosals? is the 29 a net net net. Let me put it this way. Short wave reflected at the surface has to travel back through the same phyiscal medium it took to get to the earth.. probably should not read too much into the cartoon I suppose.

Steve Goddard
March 17, 2010 10:50 am

Willis,
There is no need for laboratory experiments. The earth itself is the laboratory, with some locations having much less atmosphere than others. Temperatures in the upper elevations of the Himalayas average 40-50C cooler than similar latitudes at lower elevations, because of the lack of atmosphere. Your claim below is simply incorrect.

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.

Kevin
March 17, 2010 10:55 am

WTF? Environment Canada says that this winter was 4.0°C above average?
http://www.msc.ec.gc.ca/ccrm/bulletin/national_e.cfm
The national average temperature for the winter 2009/2010 was 4.0°C above normal, based on preliminary data, which makes this the warmest winter on record since nationwide records began in 1948. The previous record was 2005/2006 which was 3.9°C above normal. At 3.2°C below normal, the winter of 1971/1972 remains the coolest. As the mean temperature departures map shows all of the country, but for a small area over the southern Prairies, was above normal, with some areas of the arctic and northern Quebec more than 6°C above normal. Southern Saskatchewan did have a cooler than normal winter, with temperatures more than 1°C below normal. This past spring was the first season in the past 5 years with temperatures below normal, as shown in the consecutive seasons graph.

March 17, 2010 11:00 am

Willis Eschenbach (10:34:11) :
Is my writing really that opaque?
The initial quiz was a bit opaque. I would have agreed completely if the premises were ‘no atmosphere and totally black’. The ‘blackbody’ bit is somewhat off, because in the infrared, the Earth is an almost perfect blackbody and that is where it radiates.

March 17, 2010 11:08 am

Willis Eschenbach (10:59:56) :
Can’t have it both ways … the 8C is right for a black blackbody. Since I was not talking about a white blackbody, the 8C figure is not wrong.
But you were not explicitly talking about a ‘black’ blackbody either so the built-in assumption is that you were talking about a blackbody with an albedo [0.15] as that the surface actually has…
But perhaps we have generated enough heat by now without much new light. 🙂

David L
March 17, 2010 11:09 am

mkelly (08:33:01) :
David L (07:50:25) :
“Are you saying that the 5.3million giga tons of atmosphere has no effect on the earth’s temperature? If I leave T as an unknown in the formula and use standard pressure for P and figure for a volume up to say 100km I get a temperature of 0 deg C. ”
Not due to PV=nRT realistically. Are you saying that because there’s pressure and gas molecules and a volume (what’s the volume of the earth’s atmosphere by the way?) then we have a temperature? Turn off the sun, shut down the geothermal heat, and see how long it takes for the atmosphere to become pools of liquid N2 on a frozen ocean. Pressure is not driving temperature here. It’s the other way around. Classic example of mixing up Causation with Correlation. Temperature and Pressure are correlated…but which causes which? If your highschool student compresses air, the compressed air temp changes….but that temperature is soon absorbed by the surroundings. Ever touch a high pressure gas cylinder? How hot is it? It’s room temperature…But it’s under high pressure? how can that be? Now put that tank in the sunlight….guess what…temp goes up AND pressure goes up.
So why would the amount of air, the earths atmospheric volume, and pressure have anything to do with global temperatures according to PV=nRT? Look up R….what are the terms? J/K mol. It’s the amount of energy per temperture per mole….it’s all about the energy folks Know the energy flow and you’ll know all the temperatures!!!!!! If you’re not doing work on the system you aren’t moving energy around!! Just having an atmosphere is not doing work!!!! You have to do something to it…like heat it up with sunlight!!!!!!

March 17, 2010 11:15 am

“Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate the same amount. If not, we would violate the thermodynamics, where we could get work out of the radiative difference [use it to power a steam engine] between two bodies at the same temperature.”
This warmist logic! The body with the lower emissivity will radiate less, that is what emissivity means: it radiates less at the same temperature.
As for breaking fundamental laws of thermodynamics, if that were true, then thermos flasks and space blankets both break your “fundamental” laws as the low emissivity material prevents irradiation of heat compared to similar “blackbody” materials at the same temperature.

March 17, 2010 11:18 am

Richard Sharpe (10:01:57) : I don’t have “smiley” faces to put at the end of sentences, but you it the nail on the head. It was a rhetorical question.
Most of the people on the earth live in areas that get above the temperature of where CO2 radiates. That’s part of the point.
If CO2 would not warm up coffee or keep gespacho luke warm how is it able to warm up an entire global. The atmosphere has conduction and convection. Gases dissipate heat. My frozen turkey will warm to room temperature but no more. No amount of CO2 in my house will heat the turkey higher than what the room temperature is.
Is there a ratio of convection/conduction by CO2 versus radiation by CO2 when dissipating heat?
If I remember my heat transfer class when two discs at different temperatures radiate at each other the lower will rise to the higher then it becomes a push/pull and nothing further happens. So the best that could happen with CO2, is it could rise to the temperature of the earth and then all stops. The pot does not warm the burner.

Micky C
March 17, 2010 11:21 am

Willis, I think inadvertently you have stumbled upon the key wrong assumption in this whole AGW debacle. The value for Trenberth back radiation is very wrong. In the paper they show the upward radiation from the surface and then from the top of the atmosphere. At the TOA we can see the absorption peaks and how there is a reduction in the spectra. The difference in W/m2 is the difference in the integrals of these curves i.e. what does not come out of the TOA is the difference in the integrals. If those curves are integrated an eyeball estimate is about 10 to 15% difference. If we have 339 going up how can we then have 321 coming back?
Secondly using the same curves, and using the measured CO2 15 µm band, which is heavily saturated for the current ppm of CO2. Any increase in ‘the wings’ is 4 orders of magnitude different than the main peak. This puts any decrease in transmissivity of that order of magnitude. Now considering that the difference in back radiation is around 30 to 50 W, how exactly is a doubling of CO2 going to produce 3W/m2 extra forcing. The simple answer is it can’t hence the assumption that water vapour increases; an assumption that as yet has not been tested.

March 17, 2010 11:22 am

Just realised, when I mentioned thermos flasks, I forgot to mention I was referring to the silver coating used to reduce heat loss.

anna v
March 17, 2010 11:34 am

groundRe: magicjava (Mar 17 04:43),
Thanks for the link. One can see the difference between air and skin surface. in the maps.

March 17, 2010 11:39 am

mkelly (09:08:39)
For anyone a question. If CO2 can cause an increase in temperature based on reradiation of IR then why are thermos bottles not filled with 100% CO2 instead of using a vacuum?
By the IPCC formula of 5.35ln(C/CO) you could get up to 75 W/m2 of extra “heat” going back into your coffee. OUCH!
Let’s see if I can explain. Heat is transferred by Convection, Conduction and radiation. See the sun? It’s in a vacuum. There is a lot of nothing around it.
So the heat of the sun get here by radiation. Take a metal rod in your hand hold it in a fire. That’s conduction.
In a thermos you build a vacuum chamber. You try to get as few as particles in there as possible to prevent conduction. Filling that chamber with anything would defeat the purpose of reducing the heat transfered by CONDUCTION. How do you prevent radiation through this vacuum? Easy, you put a SILVERED LINING on the metal.
That shiny metal “blocks” or reflects the IR. It does this much better than C02 does. C02 only “blocks” certain regions of Longwave.
http://www.howstuffworks.com/thermos.htm/printable
When you look at how radiation move through a material you have to look
at all the ‘windows’ in the material. .
This may help:
http://www.crisp.nus.edu.sg/~research/tutorial/atmoseff.htm
Then go google radiative transfer equations.

Bill Illis
March 17, 2010 11:45 am

Technically, if we have now reduced the greenhouse effect/atmospheric pressure impact to 20K, the TOA forcing relative to that figure is not 150 W/m2 anymore – it would only be about 95 W/m2.
The equilibrium emission temperature would now be 268K rather than 255K and 95 W/m2 gets you up to the surface temperature of 287K or 14C.
So the average over the entire 95 W/m2 would be 0.21K /W/m2. [As I said the first part of the 95 watts would have more impact than the last one which is just 0.18K /W/m2].
Technically, it also wouldn’t be the TOA forcing either but just the 3.1 km high forcing.

March 17, 2010 11:51 am

[quote Willis Eschenbach (10:57:20) :]
I greatly regret putting up the global energy budget diagram, as people are overlooking the subject of the thread, which is climate sensitivity.
[/quote.

My apologies for being one of those people. When I see Trenberth’s diagram, it just sets off a giant red flag for me.
There are so many topics covered in the post that it’s difficult for me to find the core substance of it all. I _think_ the point to the entire thing is:
[quote Willis Eschenbach:
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase.
[/quote]
… but that was saved until the next-to-last sentence.
Also, you’re tossing around a lot of unsubstantiated claims. Examples include almost everything you say about the IPCC. For example, does the IPCC really calculate the effects of Evaporation incorrectly? And no one ever noticed this? With all the things coming out about the IPCC lately, this may very well be the case, but statements like this need some supporting evidence.
The use of 20°/8° is also very confusing to me. It’s not clear when one or the other applies to what you’re saying.
I’d suggest stripping out as many non-essential points as possible, so folks like me don’t get tripped up over Trenbreth, and folks like Lief don’t nit-pick you to death on black blackbody and white blackbody stuff.
If your main point is the increase in temperature due to increase in W/m-2 is logarithmic, start with that and make it clear.

March 17, 2010 12:02 pm

Willis Eschenbach (10:35:46) :
Are there any substantive claims that I have not disputed?
Derek says – Well, yes actually, for instance see the below.
AND my own post elsewhere.
http://www.globalwarmingskeptics.info/forums/thread-609.html
Terry Oldberg (09:44:22) :
In the language of thermodynamics, there is no such thing as an “energy flow.” The only energy that “flows” is heat. In a Kiehl-Trenberth diagram, some of the “flows” are heat. Others are radiation intensities. It is clear that the “back radiation” is not heat for it “flows” from cold to hot matter; if it were heat, it could not flow in this manner, under the second law of thermodynamics. However, while we have a conservation principle for heat flows, we do not have one for radiation intensities. Thus, the proposition that the K-T diagram portrays some kind of “balance” is false.

Frank
March 17, 2010 12:12 pm

John A wrote (21:46:51):
How of the incoming energy from the Sun is converted into atmospheric and oceanic convection? None, according to Kielh and Trenberth. Which is amazing. Climate physics is a fascinating subject.
What causes atmospheric and oceanic convection to slow down (the only reason solar energy would be needed to keep them going)? Friction. Where does the energy consumed in overcoming friction appear? As heat. Physics is a fascinating subject.

March 17, 2010 12:30 pm

magicjava

I’m not sure Trenberth’s energy balance is accurate. It ignores readings we get from satellites and thermometers in favor of equations taken from Hansen’s climate model.

And in response to my question..

It’s not the 1997 paper. It’s the updated paper (2005 I believe).
The exact quote is:
[quote: Trenberth :]
The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 8.5 +/-0.15 Wm-2 by Hansen, et. al.
[/quote]
I’d consider any earlier paper to be even less accurate than this one.

The updated paper is 2008.
In fact, Kiehl and Trenberth don’t throw out any measurements in favor of climate models. In the 1997 paper they assume that incoming and outgoing energy balances. Why?
Because the instrument error means the measurement results are not accurate enough to be able to say whether energy in = energy out. In a non-warming and non-cooling world this would be true.
And there is a small challenge of whether the measurements of solar energy in is more accurate than the measurement of terrestrial energy leaving and reflected solar energy.
So there is an uncertainty around 5-10W/m^2.
In the updated 2008 paper, being good scientists they attempt to provide better numbers all around. In the case of the instrument error, there are still the same unknowns. So rather than fix energy in=energy out, they say, “well, the earth is warming up, how much by?”.
Anyhow, for those for whom the word “model” causes outrage.. look at the statement which was inaccurately quoted:

The TOA energy imbalance can probably be most accurately determined from climate models and is estimated to be 0.85±0.15 W m-2 by Hansen et al. (2005) and is supported by estimated recent changes in ocean heat content (Willis et al. 2004; Hansen et al. 2005).

As you can see -first the number is pretty small, and second, the calculation is supported by the measurement of the increase in ocean heat (as you can see in The Real Measure of Global Warming ).
1W/m^2 out of 240W/m^2 – and you want to throw out the results?
For anyone wanting to learn a little about climate basics, read the whole original 1997 paper.

Frank
March 17, 2010 12:32 pm

Willis: Technically speaking, climate sensitivity is the partial derivative of temperature with respect to energy (probably both incoming energy from the sun and energy radiated downwards by the atmosphere and clouds). Your post may be assuming that temperature is a linear function with respect to incoming energy. The relationship between T and W is certainly a very complicated one if conditions are dramatically different from today – a snowball earth or even an ice-age where the relative proportions of land and ocean change. For small changes in W and T, the relative is approximately linear and climate sensitivity can be treated as a constant.
The IPCC also worries about the time-scale associated with estimates of climate sensitivity. For a given dW, how long does it take for T to come to equilibrium? Does equilibrium include changes in the deep oceans and ice-caps? For models with the highest climate sensitivity, equilibrium of the atmosphere and upper ocean apparently requires decades.

Joel Shore
March 17, 2010 12:34 pm

Willis:
I am confused as to where you have gotten the 150 W/m^2 figure that you quote. I also think, in line with what scienceofdoom said, that what you have essentially calculated here is not the climate sensitivity but the sensitivity in the absence of feedbacks. In particular, what you have to recognize is that the distinction between forcings and feedbacks is somewhat arbitrary. For example, CO2 change is, in some sense, a feedback in the glacial – interglacial cycles but is a forcing in our current predicament. The thing is that you have to be consistent. What you have done is not consistent because you have presumably counted everything, e.g., including the water vapor, as a forcing in getting that 150 W/m^2 figure but then you have not counted any change in water vapor as a forcing in computing the resulting climate sensitivity.
Another way of putting it is this: If the warming due to the change in CO2 causes additional water vapor to go into the atmosphere, this water vapor will produce an additional radiative forcing…and if the additional warming causes ice melt that changes the earth’s surface albedo, this will also produce an additional radiative forcing. (And, the change in clouds would also produce an additional radiative forcing whose magnitude, and even admittingly, sign are uncertain.) Now, we usually call all of these things feedbacks instead of forcings and that is fine if we do so consistently. The problem is that I don’t think you have done so consistently…i.e., as near as I can tell, your “150 W/m^2” (which I don’t really understand anyway) counts everything as a forcing, not a feedback.
By the way, Jim Hansen has made a similar point recently about being consistent with what you call a feedback and a forcing…And, in fact, he has argued that what he calls the Charney sensitivity derived from the glacial – interglacial cycles considers changes in albedo due to changes in ice sheets as a forcing whereas any such changes in our current discussions are considered to be a feedback. Hence, he argues that the 3 C climate sensitivity for doubling CO2 is probably too small for our current “experiment”…and says it is more like 6 C when you properly consider the effects of the albedo changes due to changes in ice sheets. (Others question whether the ice sheet albedo effects would really be that large in our current climate and also how fast the ice sheet changes can occur…so, I am not saying his 6 C estimate is correct, but I think that his basic point about being careful what you consider to be forcings and feedbacks is.)

March 17, 2010 12:51 pm

steven mosher (11:39:03) : Steve I guess you missed reading the part about patenting the idea. It was a rhetorical question. I cannot put smiley face at the end of the sentences.
However, it is possible to construct an idealized thermos floating in space, with an inside chamber transparent to IR but infinite R value, and an outside surface that blocks incoming radiation whose inside surface is a perfect reflector of IR. So no conduction no convection only IR from the soup, coffee whatever to the next chamber. Fill the space where the vacuum would normally be with CO2.
The IPCC formula makes no difference as to anything except the forcing of the log of (C/CO). So with that in mind using the IPCC 5.35ln(C/CO) we get
ln 1000000 is 13.81 times 5.35 for 73.91 W/m2. According to IPCC whatever you have in the thermos will heat by 74 W/m2 indefinitly to vaporization and explosion.
Please don’t miss the point Steve I don’t think CO2 can do what is claimed and none of the information I have read demonstrates via physics, math, chemisty etc or whatever, over turns my understanding of the laws of thermodymanics, heat transfer etc.
My ultimate point is that if CO2 could increase heat energy through radiation then someone somewhere would have found a way to make money off the idea. But as of yet nothing even though we are told this has been know for a 100 years. Hog wash.

OceanTwo
March 17, 2010 12:58 pm

Honestly, I can’t find anything about a ‘colored’ black body. A Black body is a black body (or blackbody if you are so inclined). Either way, anything which as an emissivity (not equal to 1) is not a black body.
Unless there’s something changed in the last 20 years. Maybe: they are teaching all sorts of funky stuff these days (perhaps an object with a specific emissivity is classified as a specific ‘black body’ color? ). Even so, it’s an irrelevant argument: as I understand it, based on what had been demonstrated, if earth was a black body it [surface] would be 8 degrees cooler; as it has an emissivity, it is theorized to be 20 degrees cooler than what we measure today.
It seems like the prosecution is stating that the perpetrator accelerated from a stand still to 100mph when they breezed through the red light. The defense [Willis] is arguing that it’s only 50 yards from where the perpetrator started to the red light, so how can they be going 100 mph? Weather it’s 50 yards, or 60, or 20, or 70.235 is quite irrelevant; or what the shade of the red light is, or even the decimal accuracy of the radar gun used for measurement.
Does the theory hold? Are the results of Willis calculation so far off ? Is the principle employed sound? If not, why not? Can it be demonstrated that his theory is false? Are there any significant factors not accounted for?
In other words…

Mattias, Sweden
March 17, 2010 1:09 pm

8 degrees cooler? That contradicts what can be found using Stefan-Boltzmann law, 255K or -18C. Or have I missed something?
http://answers.yahoo.com/question/index?qid=20090408183916AASBK1D
http://en.wikipedia.org/wiki/Black_body#Temperature_of_Earth

March 17, 2010 1:23 pm

OceanTwo (12:58:04) :
if earth was a black body it [surface] would be 8 degrees cooler; as it has an emissivity, it is theorized to be 20 degrees cooler than what we measure today.
albedo, perhaps, rather than emissivity.
The 8C was not really used by Willis for anything, so it is perhaps not worth harping too much on it. There is also a possible confusion because you can talk of blackbody radiation as that which has a blackbody spectrum without the body be required to be black [the Sun is not black, for example]. More confusion comes from the statement: “That is to say, the net gain from our entire complete system, including clouds, surface albedo, etc..” where the clouds and surface albedo are not gains but deficits. Anyway, I may just have been too pedantic about this, but I was disappointed that the lead paragraphs were on something not actually used further on. The important things was the 20C difference.

wayne
March 17, 2010 1:32 pm

Willis:
Re: wayne (04:27:21) :
I may have answered my own question to you.
Seems the 105 W/m2 might very well come from the 22 for Sensible Heat and 76 for Latent Heat. Those total 98, close but not exact to the 105.
Seems the K/T chart is organized in a confusing manner. Their fault, not yours. It should have been broken in three sections. The Input, the internal flows, and the output. That would tie the back radiation to the back radiation mathematically.
On the left, the SW radiation from the sun, the input, and it is pretty clear, then a dashed light gray vertical division.
Next is the internal energy flows. That is the atmosphere, clouds, the two upper pointing heats, sensible and latent, and the down dwelling back radiation on the far right. Then another vertical dashed light gray line.
What remains on the far right is the output section. That is all of the up dwelling radiation that gathers from the ground and from the center internal section.
That would de-mystify it a bit, mathematically anyway.
Willis, somewhat OT:
Some of your graphs are coming out vary fuzzy on this end. If you are getting these via screen captures, this might help. It works on xp anyway. Before capturing, turn the screen font edge smoothing off. On xp it’s standard or ClearType. Capture the picture then turn the smoothing back on. It does wonders on my machine, no more fuzziness! You get am exact pixel for pixel copy.

Anders L.
March 17, 2010 1:37 pm

Willis Eschenbach (10:25:22) :
Anders L. (06:43:13)
“If the earth had no atmosphere, and if it were a blackbody …”
Yes, but it does have an atmosphere, and it most certainly is not a blackbody, so why is this entire exercise relevant?
“Thought experiments” have a long and proud history in science. They are widely used to examine conditions that we cannot replicate in a laboratory.
I agree wholeheartedly, but I am not convinced that anything meaningful
about the detailed properties of our climate system can be inferred by
thinking about a planet completely devoid of a climate system.

March 17, 2010 1:55 pm

[quote scienceofdoom (12:30:53) :]
In fact, Kiehl and Trenberth don’t throw out any measurements in favor of climate models.
[/quote]

They do, readings that are more than 6 times higher than their estimate of 0.9. The CERES satellite shows an imbalance of 6.4 W/m-2. It’s right there in the paper. The sentence before the one I quoted. So is the reason why they threw it out: because it’s not expected by their models.
And yes, I think Trenberth’s model needs to be redone. I’ve given reasons why on this thread. Willis Eschenbach has given reasons why he believes the model can’t even work. In a nutshell, we need to start looking at reality rather than models.
Anyway, Willis has stated that Trenberth’s energy budget is not central to what he was trying to say, so I don’t want to clog up this thread with debates over it. If you’d like to respond to what I’ve posted here, I’ll let you have the last word on the matter.

DavidB
March 17, 2010 2:03 pm

Just curious (really) about one point. If the average temperature of the moon in sunlight is about 225F, how come the average temp of the earth in sunlight is so much lower? As far as I can see from the energy budget diagram, about half of the incoming radiation is either reflected by the atmosphere or the earth’s surface itself, or absorbed by the atmosphere on the way in (and some of the latter would be radiated down to the surface). So it seems that we should expect the average surface temp of the earth in sunlight to be not less than about 110F, i.e. about half that of the moon. True, this is reached or exceeded in the middle of the Sahara or Death Valley, but it seems very high for a global average. Am I wrong about this, or is there something wrong in my reasoning?

DavidB
March 17, 2010 2:05 pm

….maybe cooling by evaporation of water accounts for the apparent shortfall?

Richard Sharpe
March 17, 2010 2:08 pm

steven mosher (11:39:03) said:
In a thermos you build a vacuum chamber. You try to get as few as particles in there as possible to prevent conduction. Filling that chamber with anything would defeat the purpose of reducing the heat transfered by CONDUCTION. How do you prevent radiation through this vacuum? Easy, you put a SILVERED LINING on the metal.
That shiny metal “blocks” or reflects the IR. It does this much better than C02 does. C02 only “blocks” certain regions of Longwave.
Ahhh, that’s a much better explanation than I managed to produce. I forgot about the silver coating on the inside side of the outer sleeve … and yes, almost complete reflection is much better than 50% reflection.
Looks like I didn’t get my invite from CTM. Was hoping to meet you.

Joel Shore
March 17, 2010 2:16 pm

mkelly says:

However, it is possible to construct an idealized thermos floating in space, with an inside chamber transparent to IR but infinite R value, and an outside surface that blocks incoming radiation whose inside surface is a perfect reflector of IR. So no conduction no convection only IR from the soup, coffee whatever to the next chamber. Fill the space where the vacuum would normally be with CO2.
The IPCC formula makes no difference as to anything except the forcing of the log of (C/CO). So with that in mind using the IPCC 5.35ln(C/CO) we get
ln 1000000 is 13.81 times 5.35 for 73.91 W/m2. According to IPCC whatever you have in the thermos will heat by 74 W/m2 indefinitly to vaporization and explosion.

There’s an awful lot of confusion in these couple of paragraphs but I’ll try to point out some of the major problems:
(1) In fact, if you postulate a system where incoming radiation gets in and (at least some of it) gets absorbed but any outgoing radiation cannot escape then indeed that system will keep heating up indefinitely. That’s what energy balance (essentially the 1st Law of Thermodynamics) says it has to do!
(2) In reality, in the case that you described, the system could never get hotter than the surface of the sun because, as the temperature rises, its emission of radiation will shift toward the shorter wavelengths and eventually it won’t be emitting mainly in the IR but rather in the visible where, by the assumptions of the problem (e.g, such radiation is able to get into the system), the radiation is able to escape.
(3) The IPCC formula is not a magical formula that holds for any container with a concentration of CO2 in it. It is a formula obtained from detailed line-by-line radiative (or even radiative-convective) transfer calculations in the actual earth’s atmosphere. Among other things, the actual temperature profile of the atmosphere with height and the other IR-active constituents like water vapor play an important role. See here for further discussion: http://www.aip.org/history/climate/simple.htm#L_0623

daniel
March 17, 2010 2:22 pm

I have one question regarding the Kiehl/Trenbeth radiation budget scheme : I have the understanding that since some quite extraordinary papers
Gerlich & Treuschchner (1989), Robitaille, etc. this kind of ‘budget’ was heavily falsified for basic physics reasons ?
I’m wondering whether these papers had eventually to face rebuttals in peer reviewed papers…
Ar this stage I’m not aware of any, and would just conclude that they represent the current stage of physics, which mean that next IPCC report’s science basis would be interesting to draft

Joel Shore
March 17, 2010 2:25 pm

DavidB says:

Just curious (really) about one point. If the average temperature of the moon in sunlight is about 225F, how come the average temp of the earth in sunlight is so much lower?

It is the global radiation in and out that half to balance, not the local one. The heat transport in the earth’s atmosphere and the thermal inertia relative to the diurnal variation means that the earth’s temperature goes through less extreme ranges. If the earth rotated much slower and/or if there was less thermal inertia (e.g., not so much ocean) and/or if there was less convective processes in the atmosphere to mix things then there would be more extreme temperature ranges.

It's always Marcia, Marcia
March 17, 2010 2:38 pm

Roy Spencer on clouds and negative feedback. May be evidence for Lindzens’ Ifrared Iris
Part 1

It's always Marcia, Marcia
March 17, 2010 2:38 pm

Part 2

Brian G Valentine
March 17, 2010 3:03 pm

Thank you, Willis, this is interesting.
It’s a common calculation in the senior-level course of radiation heat transfer to show the Earth’s radiation exchange temperature with the Sun is (about) 280K, then the extra 8K are observed from satellites
In other words, isn’t this pretty common knowledge to people with a BS in mechanical engineering etc?

pochas
March 17, 2010 3:07 pm

Also recommended:
Nir Shaviv: “On Climate Sensitivity and Why it is Probably Small”
http://www.sciencebits.com/OnClimateSensitivity

Brian P
March 17, 2010 3:29 pm

Willis
A bit old but I really liked your thunderstorm post from some months ago

DavidB
March 17, 2010 3:30 pm

To Joel Shore: Thanks. Those are good points. The influence of the oceans as a heat store is probably the key factor. In the interior of the continents there is a much larger diurnal range than over or near the oceans, and even there convection currents must carry the extremes of daytime heat away from the surface. I’m just a curious newbie to these subjects, so please forgive my errors and oversights.

Alan D McIntire
March 17, 2010 3:31 pm

You’ve also got to adjust for emissivity. The earth’s albedo would be maybe 0.15 without clouds, but it would also would not emit as a black body- what would the emissivity be? I’ve read estimates that the oceans’ emissivity is about 0.94, what would it be for land only, about 0.9?
in that case, you’d figure in a correction of ( 1 – 085/.9)^0.25

DocMartyn
March 17, 2010 3:35 pm

‘David L (07:50:25) :
mkelly (07:28:59) :
“Why is it that PV=nRT is never considered when figuring a surface ”
I don’t believe it applies in any significant way. Gas temperature by this equation changes given a change in volume, moles of gas, or pressure. but it says nothing of surface temperature.’
One liter of liquid water contains 55.56 moles of water molecules and occupies 1000 mls or one litre. At STP, as a gas, this would generate a volume of 1244.544 liters; giving a ratio of 1,244.544:1.
The globally-averaged annual precipitation is 990 millimetres (39 in), or 0.27 mls/cm2 per day; so that during the day/night cycle, on average, each cm2 of the planet generates and then collapses one third of a liter of volume, or an ingot 336 cm or 3.36 meters.

Colin Davidson
March 17, 2010 3:36 pm

I think Willis’s estimate makes sense on other grounds.
Using the diagram the atmosphere receives its energy primarily from the Surface.
If you look at the Surface energy flux balance, you get a sensitivity between 0.095 and 0.15 DegC/W/m^2, depending on what assumption is made on evaporation.
(This number makes no allowance for “feedbacks” eg greater back-radiation due to increased CO2 and water vapour concentrations, or lesser solar radiation absorbed into the surface due to increased cloud cover. However it is interesting that if the surface temperature rises by the median IPCC estimate of 3 DegC for a doubling of CO2, the 3.7W/m^2 “radiative forcing” at the top of the atmosphere would necessarily need to be a huge “surface forcing” of between 22 and 32 W/m^2 at the surface. )
The usual IPCC calculations, without feedbacks, use a TOA sensitivity of around 0.3. This is then assumed to apply throughout the troposphere, right down to ground level. But that sensitivity is between twice and thrice the surface sensitivity, and I reckon that the assumption that the temperature change at TOA translates to the same temperature change at the surface cannot be sustained without invoking magic, or unless the sensitivities are in fact similar.
Willis’s calculation brings the TOA sensitivity in line with the surface.

Alex Heyworth
March 17, 2010 4:13 pm

magicjava (07:06:00) :
[quote]
I think the skeptic community would be well served if we built out own energy balance model, rather than using one based on numbers James Hansen pulled out of his ass, which is all Trenberth’s model is.
Starting with data from the CERES satellite would be a good idea, IMHO. Building a model that changes with conditions rather than just giving a hard-coded answer would be a good idea too. And being honest about things we don’t know, like where extra energy is going, would also be a plus.
[/quote]
You mean something like this? http://www.palisad.com/co2/eb/eb.html

Alan D McIntire
March 17, 2010 4:21 pm

Sorry, that should be a correction factor of roughly
(albedo/emissivity) ^0.25 = (.85/,9)^.25 times the original temperature.
Another point, Using Trenbeth ballpark figures, if we get 240 watts from the sun, and the greenhouse effect makes that an effective 390 watts and a 33 C increase, isn’t the increase in temperature per delta watt increase in the ballpark of
33 C/(390-240) = about 0.22C per watt increase?
A thrid point using Trenbeth ballpark figures. The surface flux is about 490 rather than 390 watts, with 100 watts going into conduction and convection rather than sensible heat. Figuring that half of that is radiated to space, half back to earth, about half of 100 watts or 50 watts is removed from sensible heat by convection and conduction. 50 watts latent heat + 150 watt sensible heat implies that 25% of any past wattage increase has gone into latent heat. Shouldn’t that apply to the 3.8 watt increase for a doubling of CO2? Shouldn’t at least 0.95 watts go into latent heat, only 2.85 watts or less go into sensible heat?
A fourth point, since emissivity is less than 1, assuming it’s about 0.9, isn’t the actual greenhouse effect something like 100 latent+ 150*.9 = 135 watts sensible heat?

Brian G Valentine
March 17, 2010 5:07 pm

On a “small” climate sensitivity:
Estimates from Lindzen, Shaviv, Monckton, Schwarz, others, put the value BELOW what is known to be “natural” variability – partly stochastic, partly determinate.
My question is: if there is no way to observe a physical quantity, does it have meaning at all?
I can construct a thousand quantities that I can calculate in some way (e.g, change in Earth’s rotational angular momentum from buildings, transportation, etc) but the construct is entirely meaningless because there is no way to physically measure it

cba
March 17, 2010 5:25 pm

Willis,
there are some significant problems with the albedo numbers. If you go by K&T, it would seem the 0.3 total albedo is 0.22 for clouds and 0.08 for surface with about 0.62 cloud cover fraction. In essence, the 0.08 is the average albedo where there are no clouds present.
Your number of 0.15 would be fairly reasonable based upon the Moon and Mars. It is not reasonable for Earth as over 70% of the surface is covered by oceans and the oceans have less than 0.04 albedo for incoming solar radiation, at least for high angles of incidence (wrt horizon where the incoming flux is most significant).
Try these numbers:
surface (ocean – use 0.7 ocean fraction) 0.04
surface( land) 0.19 (includes snow cover, forests, sand etc)
average surface albedo 0.085 (corresponds to 29/342 on chart )
the land albedo is calculated from the 0.085 total surface and the 0.04 ocean.
clouds (62% cover, averaged value) 0.36
this corresponds to the assumption of 0.22 for cloud albedo and fractional contribution.
NOTE that these numbers do not add up.
The weighted average for 0.62 cloud cover leaves 0.38 surface contribution.
Using the 0.085 overall surface albedo and fraction visible, one gets
0.38*0.085 = 0.032 contribution to the overall albedo which conflicts with the chart.
Since the overall albedo listed is 0.307, the cloud contribution would actually be 0.307 – 0.032 = 0.275 at a 0.62 fractional cover.
That puts the cloud albedo at 0.275/0.62 = 0.44
Note that this leaves the clouds with what would seem a more realistic albedo value than the original 0.38 albedo value.
Your question of temperature drop really depends upon the conditions.
if the surface is stripped of ocean and ice and vegetation, then it’s going to have an albedo of around 0.17 or 0.15. That’s around 265K or -8 deg C or 22 C less (20 C less if 0.15). If the surface stayed the same, the drop would be 15 deg C. If the cloud albedo somehow stayed along with the surface albedo so that the total was 0.307 the temperature drop would be 34 deg. lower.

Joel Shore
March 17, 2010 5:26 pm

daniel says:

I have one question regarding the Kiehl/Trenbeth radiation budget scheme : I have the understanding that since some quite extraordinary papers
Gerlich & Treuschchner (1989), Robitaille, etc. this kind of ‘budget’ was heavily falsified for basic physics reasons ?
I’m wondering whether these papers had eventually to face rebuttals in peer reviewed papers…

That Gerlich & Treuschchner saw the light of day in even a second-rate physics journal is extremely embarrassing, as its fundamental error regarding the 2nd Law of Thermodynamics is basic enough that one could lead a group of bright first-year physics students through the logic. (And, some of its other arguments are just plain bizarre.) A comment on G&T that I am a party to and that will essentially trash it completely has been accepted by the journal and should appear within the next few months.

Brian G Valentine
March 17, 2010 5:36 pm

Thank you, Mister Troll Shore, for your erudite analysis of Gerlich and Tscheuschner’s paper, whether you believe it or not, the analysis is correct, and if you think Phys B is second rate, don’t look at it, read some first rate stuff like “Scientific American” instead, the favored bedtime reading material of the first-rate watermelon

Joel Shore
March 17, 2010 5:38 pm

Alan DMcIntyre says:

Shouldn’t that apply to the 3.8 watt increase for a doubling of CO2? Shouldn’t at least 0.95 watts go into latent heat, only 2.85 watts or less go into sensible heat?

Well, the issue of latent heat is really one of distribution of the temperature change with height in the atmosphere since what evaporates at the surface has to condense higher in the atmosphere, and this in turn can affect the radiative balance since most of the emission back out into space escapes from the upper troposphere. But, now the punchline: What you are describing is, I believe, just another way of talking about the (negative) lapse rate feedback, i.e., the notion that because of the condensation of water vapor, the upper troposphere tends to warm faster than the surface. This is a feedback that is already included in all the climate models…and, in fact, because much of the physics governing this feedback also governs the (positive) water vapor feedback, models that have a higher magnitude for the water vapor feedback tend to have a higher magnitude for the lapse rate feedback…and hence the sum of these two feedbacks tends to vary less from model-to-model than each of them individually.

March 17, 2010 5:40 pm

Willis Eschenbach (17:36:52) :
G&T is bad, bad, bad.
I’ll agree with that.

Joel Shore
March 17, 2010 5:51 pm

Willis says:

Dang, here I am agreeing with Joel again … will wonders never cease?

Watch out…It could be addicting! 😉

Francisco
March 17, 2010 5:52 pm

yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
—————–
Geothermal is supposed to play its biggest role in warming the bottom of the oceans, as the crust under the ocean floor is much thinner than under most land. The amounts of geothermal energy and its variations and distributions are largely unknown. Considering how close it is from us, it’s amazing how little we really know about what is going on at the earth’s core. Some interesting new theories are emerging:
http://tinyurl.com/y8pguwr
Beginning in 1969, astronomers discovered that three of the giant planets, Jupiter, Saturn, and Neptune, each radiate about twice as much energy as they receive from the Sun. Those planets each contain a powerful energy source which was inexplicable until J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source for those planets
http://tinyurl.com/y8pguwr

Harry Lu
March 17, 2010 5:59 pm

Leif Svalgaard (13:23:50) :
A questions perhaps you could answer please?
Albedo refers to reflectivity.
The normally referred albedo is to the visible light reflectance. But surely the reflectance in the LW IR will be different. Is this known?
Emissivity of a surface is not he same as albedo. Having used an IR camera a few times I know that the IR temperature of an object changes with emissivity. and corrections can be incorporated in the IR to temp conversion (this is why Mr Watts IR photos in his surface station project are not very valid – he need to correct for surface emissivity.) When using IR cameras one usually coats the objects with a thin layer of mat paint to equalise (approx) emissivity.
Surely albedo controls the absorbed solar radiation, but emissivity controls the radiated LW energy.
If so then one needs to know the emissivity of the earth in the 4u range not its albedo.
/harry

March 17, 2010 6:00 pm


Leif Svalgaard (10:20:53) :
You do not understand what a blackbody is. … Take a white sphere and heat it to 400C, then take a black sphere and heat it to 400C, they both radiate …

“Surface emissivity” must be considered; both may not have equal emissivity figures … which will affect radiation.
An Emissivity Primer
Emissivity Coefficients of some common Materials
Examples, emissivity:
Aluminum Foil – 0.04
Beryllium —- 0.18
Granite —– 0.45
Sand —- 0.76
Wrought Iron – 0.94
Plaster —– 0.98
.
.

March 17, 2010 6:02 pm

Francisco (17:52:28) :
J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source
Recent measurements of geo-neutrinos rule out the hypothesis that most of the planet’s internal heat is generated by a uranium-fuelled nuclear geo-reactor in the Earth’s core.:
http://www.physorg.com/news187946006.html

Francisco
March 17, 2010 6:08 pm

yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
———————-
See also letter by Herndon in Current Science, titled “Variables unaccounted for…”
http://tinyurl.com/yfx4dax

sky
March 17, 2010 6:15 pm

Nothing impedes clear physical reasoning than a wrong conceptual model. A blackbody Earth without an atmosphere might be a good starting point for teaching students, but no realistic analysis of actual temperatures can be made without fully mastering the implications of enthalpy. That atmospheric pressure is a vital variable seems to have been forgotten.

Harry Lu
March 17, 2010 6:15 pm

Thermal imaging and emissivity here
http://en.wikipedia.org/wiki/Thermal_imaging
“Since there is no such thing as a perfect black body, the infrared radiation of normal objects will appear to be less than the contact temperature. The rate (percentage) of emission of infrared radiation will thus be a fraction of the true contact temperature. This fraction is called emissivity.
Some objects have different emissivities in long wave as compared to mid wave emissions. ”
/harry

March 17, 2010 6:53 pm

Update, pls:


Joel Shore (17:38:05) :

Well, the issue of latent heat is really one of distribution of the temperature change with height in the atmosphere since what evaporates at the surface has to condense higher in the atmosphere, and this in turn can affect the radiative balance since most of the emission back out into space escapes from the upper troposphere.

Hmmm … which space-based sensor (and what wavelength) can allow measurement, can allow this to be actively seen during convective T-storm activity?
I’m going to posit the theory that a largish amount of thermal energy is _not_ radiated directly from the troposphere (from the gases constituting the air e.g. CO2 which are *capable* of EM radiation) into space, but rather becomes part of Hadley Cell circulation and the bulk of that sensible heat is released back into space is at latitudes greater than ~45 degrees due to *surface cooling* …
Also note that thermal budgets show a ‘deficit’ of incoming solar energy at those latitudes (for the average temps seen), so it is circulation from the lower latitudes that keeps those temperatures elevated relative to what might otherwise be experienced over ~50 latitude.
.
.

HankHenry
March 17, 2010 7:12 pm

Willis tells us “The average temperature of the planet is about 14°C.” Of course that’s disregarding the massive cold of the ocean which is more like 4°C – averaged out. What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths. Also, do I remember my grade school science correctly when I say the total mass of the atmosphere is represented by the mass of just 33 feet of water – the theoretical lifting limit of a suction pump?

March 17, 2010 7:24 pm

Willis Eschenbach (19:16:37) :
a warming from a no-atmosphere situation of ~ 18°C rather than the ~ 20°C I used above.
I like your number now.

Brian G Valentine
March 17, 2010 7:25 pm

Willis and Leif think Gerlich & Tscheuschner’s analysis is bad.
Sorry folks, it’s good. Two legitimate criticisms leveled at it: it does not account for convection heat transfer in the atmosphere (considers conduction only), and it does not consider clouds.
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer), and all radiation in the atmosphere is eventually lost via degradation to 20+ micrometer wavelengths, for which everything in the lower atmosphere is transparent
– or it increases the internal energy of the atmosphere, the contribution (via increasing the average k.e. of molecules) is negligible.
Ladies and Gentlemen: Either Gerlich and Tschneuschner are correct, or the atmosphere is a perpetual motion machine.
If it is, then copy the principle with a scale model and make yourself wealthy. Power your invention up a little bit with cold fusion.

Joel Shore
March 17, 2010 7:59 pm

Brian G Valentine says:

Ladies and Gentlemen: Either Gerlich and Tschneuschner are correct, or the atmosphere is a perpetual motion machine.

That’s a “head-I-win, tails-you-lose” style of argumentation. Actually, either Gerlich and Tschneuschner are correct and the greenhouse effect describes a perpetual motion machine or they are wrong and it doesn’t. It is very easy to show that the second of these possibilities is in fact the case…And, I think Brian that you are intelligent enough to comprehend this.

Joel Shore
March 17, 2010 8:05 pm

Willis Eschenbach says:

But none of these change my point. Use whichever TOA and albedo figures you want, you won’t get anything near the UN IPCC canonical sensitivity of 3°C per doubling.

But, I’d like to see you address the more fundamental objections that I raised in my post of (12:34:08), namely that I don’t see how your calculation includes feedback effects. I think you are just calculating the sensitivity before feedbacks and somehow getting the wrong number (as I don’t understand where the 150 W/m^2 value you use comes from).

March 17, 2010 8:08 pm

Brian G Valentine (19:25:34) :
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer)
It is a lot more. Conduction is mediated by micro-scale vibration without any bulk transfer of matter, while convection is macro-scale movements of matter.

Tsk Tsk
March 17, 2010 8:11 pm

Leif,
I don’t buy the white 400C sphere and the black 400C sphere radiating the same. The fact that the white sphere is white has altered its emissivity and its radiation. It’s no longer a black body but a grey body. You can’t extract energy from the two spheres initially as the white sphere will also have lower absorption to go with the lower emissivity and so you won’t get a net flow of heat from the black sphere to the white. Maxwell’s demon still naps even though the total radiation from the two spheres initially is not the same.
Willis,
Yes, your post was a bit confusing because you didn’t show the math for the two cases until your follow up comments. I originally agreed with Leif until I re-read your post a couple of times to get your point. And, as others have pointed out, why didn’t you use the derivative with respect to power of S-B at mean global temp to get your sensitivity? The error is relatively small (0.18 vs. 0.13 K(C)/W) but it’s still a bug, or were you claiming that the sensitivity is something other than black body?

cba
March 17, 2010 8:14 pm

Willis,
to measure the actual sensitivity of the atmospheric absorption you’re going to have to have similar albedo, despite its unreal conditions. That means before and after albedo of 0.307 despite a lack of atmosphere or of atmospheric absorption. The numbers are an average temperature of 288K with the associated surface radiation of 390 w/m^2. Balance occurs with about 239w/m^2 making it out of the atmosphere – leaving 150 w/m^2 to be absorbed in the atmosphere or blocked in the atmosphere. Note that some of this is due to ghgs and some due to cloud blocking.
doing the comparisons for various warming values:
33 deg C = 34/ 150 = 0.23
18 deg C = 18/150 = 0.12
8 deg C = 8/150 = 0.05 deg C rise per W/m^2 power increase
keeping albedo constant means we have 342 W/m^2 – 107 = 235 w/m^2 which corresponds to a bb temperature of 254 or about 34 deg C.
Note that even the 34 deg C rise due to the current atmosphere corresponds to only 0.8 deg C rise for a 3.7 w/m^2 forcing increase due to a co2 doubling.
Applying an increase in absolute humidity based on constant RH and a 2 deg C increase in temperature, one finds that h2o vapor (ignoring cloud formation) results in only about 1/3 of the power increase due to the co2 doubling (it’s about 3.1 w/m^2 for an increase of 30% based on a 5 deg C rise)which means that most of such a 2 deg C rise doesn’t have a mechanism to cause the increase. That is even assuming the rise of 5 deg. C, the co2 + h2o vapor increase doesn’t have enough forcing to generate even a 2 deg. C rise. Less warming results in even lower forcing from h2o vapor.

Glenn Tamblyn
March 17, 2010 8:44 pm

Willis
A few points
Your TOA figure of 147 seems wrong, from the diagram. You seem to be using the radiation that ORIGINATES at the TOA and not including the radiation making it THROUGH FROM LOWER LEVELS – 237 w/M^2 on the diagram
“1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.”
Well I would say rather estimated from calculations. Just because it may not have been adequately demonstrated does not automatically mean it isn’t valid
“2. The climate models underestimate the increase in evaporation with temperature.”
I can’t comment on whether they do but the question is what impact does this have on energy balance. What altitude does this additional heat get transported to before it condenses out?
“3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways . “.
Yes, conceivably they do, but what is the quantitative impact of that? What percentage of the energy transport in the atmosphere could be attributed to Thunderstorms. Is it a significant percentage or small enough that it can be ignored as a 2nd or 3rd order effect. Or simply subsumed into the general circulation within a cell in the model. A qualitative idea about a possible factor is only useful if you can get a handle on how important that factor is quantitatively.
“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.”.
How can the effect of CO2 be ‘tuned to the temperature record’? Surely the effect of CO2 is used to estimate the radiative balance consequences. Any tuning to the temperature record is then about relating radiative balance to temperature change. And the UHI effect won’t be that big. 70% of the surface temperature is oceans – no UHI. Most of the worlds land area isn’t in cities, no UHI. If it is the historical record being used, even in cities, UHI was much lower in the past.
“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.”
See my comments at 4 about UHI. Also isn’t aerosol effects also analysed by modelling and measuring the optical properties aerosols?
“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”
See my comments at 3 about quantifying such statements. How much do winds speeds increase relative to the current speed of the Jet Stream for example. Also, what if any impact does any of this have on vertical mixing and energy transport to higher altitudes which could increase outgoing radiation. Greater mixing at one altitude does not necesarily result in changing the vertical heat balance for that level, just the horizontal distribution of that heat.
You don’t mention increasing contributions from Methane, Ozone, Nox. You don’t discuss long term albedo change due to decline of ice coverage or vegetation change. You don’t discuss the consequences if cooling due to aerosols declines due to humanity trying to clean up our emissions. You don’t mention that the capacity of the air to hold water vapour does not go up exponentially with temperature, so increased evaporation will be balanced by increased precipitation.
There is a lot you don’t mention Willis

March 17, 2010 9:00 pm

HankHenry:

What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths.

The sun warms the bottom of the atmosphere and the top of the oceans. When a gas or liquid warms it expands and so rises because it is less dense.
So the oceans are very stratified in temperature – because the top is warmer and gets further warmed by the sun.
But the atmosphere experiences a lot of convective overturning as the warmed air at the bottom rises up. So the atmosphere is more mixed.
You can see more about this at Why Global Mean Surface Temperature Should be Relegated, Or Mostly Ignored
PS More technical note, yes the top of the stratosphere gets warmed as well due to the O2-O3 process. But most of the sun’s energy then travels through the “transparent” atmosphere and so warms the surface of the earth and the surface of the oceans.

March 17, 2010 9:20 pm

Tsk Tsk (20:11:34) :
I don’t buy the white 400C sphere and the black 400C sphere radiating the same.
My argument goes like this:
Consider a solid white sphere with an opaque interior in empty space far from everything. In its interior it has a nuclear reactor that heats the sphere [by conduction] to a temperature of 400C. This means that the electrons just inside the surface due to thermal jiggling will be accelerated constantly [in random directions]. Accelerated charges radiate determined by their acceleration. Since we stipulate that the sphere stays at the same temperature it must radiate corresponding to the thermal jiggling determined by that temperature.
The same must hold for a black sphere, or a green sphere, etc. The radiating, accelerated electrons don’t know what color the sphere has. The energy emitted must equal that generated by the reactor.

gbaikie
March 17, 2010 9:53 pm

“Temperature of a body at the distance of the earth from the sun, with an albedo equal to that of the earth: about 20°C cooler than the present earth.
TOA radiation from the present earth: about 150 W/m2
Since 150 W/m2 TOA radiation gives a warming of about 20°C, this gives a climate sensitivity of about half a degree for a doubling of CO2. This is only one-sixth of the UN IPCC canonical value of 3°C.
I don’t care if I’m off by 10%, this is a first-order analysis. My answer is way below the IPCC answer, that’s the issue, not the exact details.”
Couple questions.
First, it’s well known that earth’s average temperature is about 15 C. Is that suppose to be air temperature or temperature of the ground or surface of oceans. I assume it’s air temperature in the shade and on sea level elevation. Or simply the average air temperature of the ocean [since a large part of the surface of the Earth is covered by oceans- and therefore one can basically ignore all the land with it’s all of varying elevation.
Now you talking about an imagined airless world with the same albedo as earth. It seems then that you talking about the surface temperature- as you have removed the air.
Take Mars [a nearly airless world] it’s “air temperature” varies widely depending upon whether it is a meter or two above the surface:
“The daytime SURFACE temperature is about 80 F during rare summer days, to -200 F at the poles in winter. The AIR temperature, however, rarely gets much above 32 F. ”
http://www.astronomycafe.net/qadir/q2681.html
“The problem with measuring the temperatures on Mars is that they change in time and space in radical ways,” he said. “On Mars Pathfinder, for example, over a distance of only 40 centimeters (about 16 inches) up and down a little mast, we had three temperature sensors. They often measured temperatures that differed by about 15 degrees Celsius — that’s about 25 degrees Fahrenheit.”
http://www.space.com/news/mplmet_991201.html

anna v
March 17, 2010 10:14 pm

Re: DavidB (Mar 17 15:30),
A contributor posted the following link
http://isccp.giss.nasa.gov/products/browsesurf1.html
where one can find the “surface skin temperature”.
It is higher than the air at 2meters temperature that is used in all the calculations.
Mind you, it is the surface skin temperature that should be used in black body formulae, because it is the solids/liquids that radiate as T^4.( I think the atmosphere is something like a T^6 and in any case has small heat capacity with respect to the surface). Nobody uses the “surface skin temperature” in all these radiation budgets.
In any case, my physicist trained brain gives up with all the hand waving and double counting that goes on in “climate science”.

LightRain
March 17, 2010 11:16 pm

Why are we worried about a black body earth when there is, and never will be such a thing. I see the point, saying a perfect black body would only be 8°C colder doesn’t allow the tiny CO2 increase much latitude to increase temperatures. But in reality we have a 30°C difference caused by our atmospheric blanket, therefore a 3-4°C increase is less impossible, and at the same time quite significant.

anna v
March 17, 2010 11:29 pm

Re: Willis Eschenbach (Mar 17 23:19),
They are repulsive in any case, and totally unmerited in Joel’s case. I disagree with him a lot, but he is absolutely not a troll.
Truly said. I have had interactions with Joel and I confirm this.

HankHenry
March 18, 2010 12:15 am

scienceofdoom, ok but my question still stands. On the continents temperatures at a depth of a few feet below the frost line will represent something pretty close to the annual average surface temperature (but no colder) while the temperature in the depths of the ocean is *colder* than the average surface temperature. (Remember that the geothermal gradient of the earth is something like 25-30 C per kilometer so by this extrapolation the bottoms of the oceans with an average depth of 3.7 km should be very very warm). This is the main point I am trying to drive home, namely, that on the continents the earth below the frost line (or at comparable depths for tropical latitudes) is no cooler than the average surface temperature while in the oceans most of the ocean is cooler than the average surface temperature. Once one starts thinking about temperatures a short distance beneath the surface of the earth it strikes one as quite surprising. I believe that so much cold ocean water (3.9 C overall average according to my encyclopedia) suggests that the earth is really quite different than a simple black body model. Rather there must be substantial movement by advection of heat along and across the surface of the globe; otherwise there would not be so much of the ocean’s water at the same average temperatures as polar or sub polar regions. My guess is that on the earth there is a general motion of heat poleward (at the surface but not at oceanic depths) and further that because air is dryer in cold polar regions, heat is radiated away more effectively in those regions. The Earth is much different than the idealized black body radiator of our imaginations. I am suggesting that the earth with its oceans is probably better at radiating than a true black body with a strictly solid stable surface and that the prime evidence of this is the great coldness of the subsurface ocean.

anna v
March 18, 2010 12:47 am

Re: Willis Eschenbach (Mar 18 00:01),
It is infrared that is important at these temperatures.
various emissivities:
http://www.icess.ucsb.edu/modis/EMIS/html/em.html
0.95 is closer for soils. .985 for oceans. Sand can go as low as o.75 ( think deserts).
There is still the assumption that average air temperature at 2 meters is the same as average surface skin temperature. It is not.
http://isccp.giss.nasa.gov/products/browsesurf1.html
It is obvious in Africa, and also the changes when one blinks from surface to air.
look at the antarctic. These are the average mean annual. they have options per month also.
They have the data there also but I do not have the tools to use it ( ftp, Fortran)

Recipy
March 18, 2010 2:20 am

Dr.Bill (&Willis): Agreed. The earth would not have clouds without an atmosphere. But the surface albedo would be higher than that of the moon. Presumably we would have more snow and ice in a colder world. Since Earth is ~70% covered by water at the moment i would argue that the albedo would be closer to that of Jupiters moon Europa than that of the moon. Perhaps an albedo of 0.5 would be a nice round guess (that is actually higher than it is today.). In that case i get a difference 54degC.

HelmutU
March 18, 2010 2:40 am

Dear Dr.Eschenbach,
as Prof. Dr. G. Gerlich at the Technical Univerity of Braunschweig has shown in his Paper “Falosificartion of the Atmospheric CO2 Greenhouse Effects wirthin the Frame of Physics” that diagrams like the one of Kiehl and Trenberth are contradict to physics for the following reasons:
1. they connot represent radiation intensities,
2. they connot represent sourceless fluxes, i.e. divergence free vector fields in three dimensions, sice avanishing three-dimensional divergence still allows that aportion of the field goes sidewards,
3. thes do not fit in the framework of Feynman diagrams, which represent mathematical expressions clearly defined in quantum field theory,
they do not fir in the standard language of system theory.
Further if the calculation of the earth without atmosphere is done in the correct way (first take the 4th root of every temperature and than calculated the mean temperature), the mean temperature is -129° Celsius for 0,7 as usally assumed.

Francisco
March 18, 2010 3:22 am

Leif Svalgaard (18:02:49) :
Francisco (17:52:28) :
J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source
Recent measurements of geo-neutrinos rule out the hypothesis that most of the planet’s internal heat is generated by a uranium-fuelled nuclear geo-reactor in the Earth’s core.:
http://www.physorg.com/news187946006.html
========================
I’ve asked Marvin Herndon how he interprets those measurements, and below is his reply. Herndon continues to be adamant about the physical impossibility of convection in the earth’s core and mantle as a source of the earth’s magnetic field in the current standard earth model, as explained by him here: http://tinyurl.com/ylnjzur
Home page with links to much of his work is here: http://www.nuclearplanet.com/
This was his reply to my inquiry:
“No, those measurements do not [rule out the georeactor]. The last measurements said it ruled out georeactor energy output above 3 terawatts. From KAMland data, the limit was 6 terawatts. They only detected a small number of events (signals in their detector), from which they had to subtract estimates of background and signals caused by man-made nuclear reactors. There will have to be many, many more measurements before any statement can be made with confidence.”
“There is much wrong with textbook geoscience. I have now shown conclusively that convection in the Earth’s core and in the Earth’s mantle is physically impossible. If fact I have a new paper coming out soon.”

March 18, 2010 3:25 am

Willis Eschenbach
What would you expect a thermopile or some other radiation detector pointed vertically upwards to read(average ball park figure) given these conditions.
a. Radiatively insulated to exclude upward radiation.
b. In the middle of a desert.
c. Two hours before dawn.
d. Gentle or stronger wind.

Brian G Valentine
March 18, 2010 4:22 am

I have always referred to it as the “fallacy of the false dichotomy,” Willis.
The “CO2 in the air from people is causing something to happen” concept is abysmally poor science, for which there is no evidence (and evidently impossible); anybody who believes otherwise has absolutely no comprehension of “geological time”

Ryan Stephenson
March 18, 2010 4:33 am

Willis: It is interesting that one branch of taxpayer funded science woul dhave us believe that geothermel heating has no impact on a planetary body’s surface temperature, whilst another would have you believe that it is sufficient to maintain life on distant moons.
http://www.sciencedaily.com/releases/2009/12/091215141510.htm
Excuses, excuses…. to spend other people’s money.
By the way, the mid-ocean ridges are 40,000km long, are spewing out lava at 1300Celsius along their entire length and have an efficient heat-exchange system called water to distribute the heat evenly over the planet.

Glenn Tamblyn
March 18, 2010 5:08 am

Willis
“The UN IPCC says that the climate sensitivity is the temperature change divided by the change in downwelling TOA longwave radiation. That is to say, radiation that ORIGINATES at the TOA. So that’s what I have used. If you want to use the upwelling radiation as well you are free to do so … but then you are not measuring climate sensitivity as defined by the IPCC.
What is your source for this. The only definition I am aware of for Climate Sensitivity. whether it be the Charney Sensitivity or others is Temperature change per Doubling of CO2. Basing seisitivity on downwelling radiation levels seems strange. Source please.
If there are differing definitions then how are these differing definitions related so that we can compare Apples with Apples. From the diagram above, total downwelling is 321 w/M^2

Glenn Tamblyn
March 18, 2010 5:13 am

Willis
“I don’t know if you’ve ever lived in the tropics, but if you, you would not say they were a third order effect “. Yes, I have lived in a semi tropical environment and locally a Thunderstorm is impressive, visceral, quite emotional. Which is all irrelevent. The question is what percentage transports due to thunderstorms contribute to the total transport over the entire globe. Again, quantitative assessments rather than emotive, local or qualitative ones.

gbaikie
March 18, 2010 6:03 am

“You raise an interesting point. The average surface air temperature is generally considered to be a reasonable approximation of the actual surface temperature. In general the surface is warmer than the air during the day, and cooler than the air during the night. So for the purpose of this first-order analysis, the difference in the averages doesn’t make a difference.”
I would say loosely, that the top surface of ground at nite should be about the same temperature as the air. And on the top surface of the ground [or tree top] in sunlight is much warmer than the air.
But as I said before, since about 70% of the surface of earth is covered by water*, one could almost not even bother considering land temperatures.
BUT if we are essentially measuring ocean temperature instead of land temperatures, there obvious a limits to how warm the water can get in terms of higher temperatures [and that will affect this whole blackbody type calculations].
E.g. The top surface of the sand of a desert could reach somewhere around 160 F during the day- such temperature aren’t possible in a ocean. One factor being that even though it’s not at the boiling point such higher temperatures have significant increases in the evaporation rate of water.
So in regard to a ocean the surface temperature I would say that it is normally about the same temperature as air temperature regardless of whether it’s day or day.
But then again, what is meant by ocean surface temperature- are we talking about to first 2 mm of ocean OR the top layer of the ocean thermocline.
Because when I say there are about the same temperature I mean the top 2mm of water rather than the top say, 10 or 20 meters of water.
* It should noted that as far the average temperature of earth the tropical region is the most important “in getting an average global temperature”- and I don’t have an reference or know off hand, what the percent of land vs ocean is in tropical region.

Francisco
March 18, 2010 6:10 am

Ryan Stephenson (04:33:23) :
Willis: It is interesting that one branch of taxpayer funded science would have us believe that geothermal heating has no impact on a planetary body’s surface temperature, whilst another would have you believe that it is sufficient to maintain life on distant moons.
http://www.sciencedaily.com/releases/2009/12/091215141510.htm
Excuses, excuses…. to spend other people’s money.
By the way, the mid-ocean ridges are 40,000km long, are spewing out lava at 1300Celsius along their entire length and have an efficient heat-exchange system called water to distribute the heat evenly over the planet.
=======================
I agree. Never understood how this is so trivial.
http://en.wikipedia.org/wiki/Geothermal_power
“The Earth’s internal heat naturally flows to the surface by conduction at a rate of 44.2 terawatts, (TW,)[20] and is replenished by radioactive decay of minerals at a rate of 30 TW.[21] These power rates are more than double humanity’s current energy consumption from all primary sources”

yusuke
March 18, 2010 6:30 am

Your argument is wrong!!
You can never assume that the downward radiation at TOA and temerparature rise are proportional. It is proven that they are not!!
Well, IPCC says that (radiative forcing)*(climate sensitivity) = (temperature rise),
but this is because this equation is a first-order equation, which is applicable only when radiative forcing and temperature rise are both small compared to their base values (150W/m^2 and 300 kelvin, in this case).
“Well, my body temperature decreased by 2 degrees after taking 2 tablets of medicine, so my body temp. should reduce by 10 after taking 10 tablets”.
You made exactly same mistake as this statement!!
Please do not decieve innocent people!!

anna v
March 18, 2010 6:31 am

Re: Francisco (Mar 18 06:10),
Terra watts sound grand, but in the context of the earths area, and the language of climatology, geothermal only amounts to something like 0.075/m^2. I checked it when I first got involved in this AGW stuff.
http://geophysics.ou.edu/geomechanics/notes/heatflow/global_heat_flow.htm
Not a player in the game.
CO2 though coming out of vents etc is another story.

yusuke
March 18, 2010 6:32 am

In addition, all of the effects you mentioned are actually “included” in models referred to by IPCC.

anna v
March 18, 2010 6:53 am

Re: gbaikie (Mar 18 06:03),
E.g. The top surface of the sand of a desert could reach somewhere around 160 F during the day- such temperature aren’t possible in a ocean. One factor being that even though it’s not at the boiling point such higher temperatures have significant increases in the evaporation rate of water.
I live by the sea. In the summer air temperatures can be for a month from 37C to 40C the water is below 25C.
At night it is the opposite, particularly in autumn, when the water may be a steady20C while the air drops to 12 and 14C. In spring, the air may be 23C and the water 13C. Even in the oceans, winds from the land blow hot or cold and are not reflected in the surface skin temperature by several degrees.
Again I refer to the surface skin temperatures in
http://isccp.giss.nasa.gov/products/browsesurf1.html
they are different than the air temperatures. Unless one does the calculation, i.e. integrate over the provided data, one cannot assert that there is no global difference by handwaving.
But then again, what is meant by ocean surface temperature- are we talking about to first 2 mm of ocean OR the top layer of the ocean thermocline.
It is infrared at these temperatures that are being radiated, and the first mms of ocean or soil or vegetation is the important one.

March 18, 2010 7:26 am

Francisco (03:22:49) :
I’ve asked Marvin Herndon how he interprets those measurements, […] Herndon continues to be adamant There will have to be many, many more measurements before any statement can be made with confidence.
Fair enough, such measurements are continuing. I wonder what he would say if the continue to give the same result…
His statement: “I have now shown conclusively that convection in the Earth’s core and in the Earth’s mantle is physically impossible” rubs me the wrong way.

Francisco
March 18, 2010 8:02 am

anna v (06:31:47) :
Re: Francisco (Mar 18 06:10),
Terra watts sound grand, but in the context of the earths area, and the language of climatology, geothermal only amounts to something like 0.075/m^2. I checked it when I first got involved in this AGW stuff.
==============
yes, I am aware of those figures. An often-quoted one is 0.9 (at realclimate). I have also seen much larger figures given at physics forums. I suppose it is small, but at a purely intuitive level, the notion that more than twice the total human energy consumption is coming at us from underneath sounds significant at least at first sight.
I find a lot of the approaches puzzling. For example, the following paper. They observe an increase of 24 mW/m2 in ground heat flux from Canada’s geothermal database over the last 200 years, and they attribute all that to changes in surface temperature, and apparently none of it to changes in geothermal heat flux coming up.
http://esrc.stfx.ca/pdf/halifaxtalk.pdf
From the abstract:
“Results from the analysis of Canada’s geothermal database indicate that the ground heat flux has increased an average of 24 mW/m2 over the last 200 years. Application of this method to the global geothermal data base allowed for a quantification of the global ground energy balance at the Earth’s surface for the
past few centuries. Preliminary global ground surface emperature and surface heat flux histories indicate that the Earth’s continents have warmed by about 0.5 K and
received an additional 26 mW/m2 of energy in the last 100 years.”

Ryan Stephenson
March 18, 2010 8:05 am

“geothermal only amounts to something like 0.075/m^2. I checked it when I first got involved in this AGW stuff.”
This calculation is clearly wrong. If it were true then there would be no possibility of lakes under Antartica, but there are. These lakes are created by geothermal heat counteracting the freezing effect of the overlying super-cooled ice.
Here’s another one for you. Every day massive volumes of water are hauled from one end of the earth to the other by the tides. This massive kinetic energy – where does it go? The laws of entropy tell you. Did you know that tidal forces due to the moon also cause the land to be pulled upward by as much as a meter too? It is these forces that cause heating on moons like Europa which orbits Jupiter. It causes heating on the earth too, but your article only mentions nuclear fission.

anna v
March 18, 2010 8:38 am

Re: Ryan Stephenson (Mar 18 08:05),
Tides also within the vocabulary of climate science give fractions of a Watt/m^2.
The link I found by googling, I had not saved my calculations.
Certainly over direct flues the heating will be important, but that is another story. Averaged over the area of the earth it is small.

March 18, 2010 8:42 am

Francisco (08:02:39) :
they attribute all that to changes in surface temperature, and apparently none of it to changes in geothermal heat flux coming up.
And that is correct. If you heat the surface, the heat will propagate downwards too. In fact, one can see the MWP in borehole temperatures as well: http://www.leif.org/research/T-Boreholes.png

March 18, 2010 8:45 am


Brian G Valentine (19:25:34) :

But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer), and all radiation in the atmosphere is eventually lost via degradation to 20+ micrometer wavelengths, for which everything in the lower atmosphere is transparent …

(Emphasis mine)
Can you point to an article or source where this is actually measured (the quantitative measurement of atmospheric radiation of ‘heat’ energy ostensibly removed from the surface via convection)?
Joel Shore could not …
.
.

March 18, 2010 8:51 am

Willis Eschenbach (23:21:59)Sir, with all respect a thought process of constructing an idealized thermos is possible since you can imagine a material with an infinite R value, etc.
I know it won’t work that is the point. The earth and the thermos are in someways the same. A heated body surrounded by gas surrounded by a vacuum, as I tried to describe. If CO2 cannot heat or keep something warm via a thermos then heating an entire globe is utter nonsense.
Again, as I pointed out if CO2 could increase heat in some object then (and we have known this for over a hundred years) why has no body found a way to make money off the idea. A CO2 coat, a wrap for your house etc. Because it is a false idea. CO2 will not increase the heat in anything. As I said my thawed turkey will not get warmer than room temperature no matter how much CO2 is in my house.
I don’t know the patent process but if someone made a claim such as a product that with the aid of CO2 you got extra heat it would sure bring this to a head.
I wish I could put a smiley face on things so folks would know when I am pointing out what I think is a false hood in the CO2 causes warming theory. Or I could just say that I guess.
Enjoy your work very much. Thank you for it.
Regards, MKelly

Brian W
March 18, 2010 9:20 am

Brian G. Valentine (17:07:48) (17:36:44) (19:25:34) (04:22:43)
Willis Eschenbach & others
I concur 100%. Gerlich & Tscheuschner are correct in their analysis. One of the best overviews of anything I have seen. A “must” read. Physics rooted in real testable observations. There is a lot in this paper there for anyone in any discipline. I am a Ham Radio Operator. I have been one for 37 years and the first at age 13 to get the basic licence here in Canada and Willis I will tell you that this paper is good I’d give it a 5 star rating with ease. The bits about electromagnetic radiation and the atmosphere are spot on the money. Above roughly the 10 meter band or 30 megahertz the atmosphere becomes transparent to radio waves. This is called “line of sight” communications.The wavelength becomes short enough that the ionisphere allows them to pass through unobstructed. Below 30 mhz you enter shortwave radio territory, the wavelength becomes long enough that the ionisphere reflects radio waves. IR’s wavelength is above 30mhz and since IR is electromagnetic radiation it to will also pass through the atmosphere unobstructed. There is no reflection. So called radiating layers are only implied and cannot be physically observed(two eyes). Gases by their nature disperse and diffuse. Whatever the scientists are measuring is not what they think it is. Backradiation is falsified and thus falls into the realm of pseudoscience. I’m callin ’em as I see ’em. Now back to my family.

anna v
March 18, 2010 9:23 am

Re: mkelly (Mar 18 08:51),
You will enjoy this
http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html
Double glaze glasses for windows with CO2 instead of inert gases should be patented.

Francisco
March 18, 2010 9:31 am

mkelly
I wonder if someone can point to references to specific lab experiments attempting to measure the amount of surface warming caused by a given increase in CO2 in a column of air that is receiving infrared radiation from below. Do the IPCC reports give references to these kinds of experiments?
If no such experiments have been made, how feasible would they be? If they are feasible, how expensive would they be in comparison with the total amount of money dedicated to fund AGW science?
Are any such tests being made at all?

JAE
March 18, 2010 9:31 am

Hmmm.
Bill Parsons commented (third comment):
“Wiki Answer to: “What is the temperature on the moon?”
The average daytime temperature on the Moon is around 107°C (225°F), but can be as high as 123°C (253°F).
When an area rotates out of the sun, the “nighttime” temperature falls to an average of -153°C (-243°F).”
Now, note that the average daytime temperature on the moon is not that far from what is reached by the soil in a low-altitude desert on Earth. Even a greenhouse easily attains 170 F.
It is interesting that that even with the slow rotation of the moon, “nighttime” temperatures don’t go anywhere near 3 K (-270 C), which is the temp of space. Ergo, the moon stores a lot of heat “overnight.” Without the aid of any “greenhouse gases.” Or without any water. This heat is stored in rocks and dust.
“Moon-Day” length is 29 earth days (14.5 earth days in light; 14.5 earth days in darkness). http://www.solarviews.com/eng/moon.htm
As indicated above, there is a 260 degree change between the moon’s “night” and “daytime” average temperatures. Assuming a linear loss of heat (bad assumption, but I don’t know how to do the proper T^4 math), the moon loses 260/14.5 = 17.9 deg/earthday. Or only 17.9/24 = 0.75 deg/hr. Therefore, crudely, if the moon had only a 24 hour day like earth, its temperature would only vary something like 17.9 degrees. So, it might be 123 C during the day and 105 C at night. Without any greenhouse effect.
The temperature of the earth could easily be a function of heat storage and have nothing to do with “greenhouse gases,” except that they aid in thermalization.
??

Brian W
March 18, 2010 9:32 am

mkelly
Sorry dude, almost forgot about you. You also understand physics.

Ryan Stephenson
March 18, 2010 9:51 am

“Tides also within the vocabulary of climate science give fractions of a Watt/m^2.”
You are thinking only of tides in terms of ocean movements. When the moon and the earth rotate around each other they both get deformed like a squash ball and the area of most deformity travels around the earth at 1000 miles per hour. This happens to the whole of planet earth, not just the oceans. Thus the planet warms, just like it has been discovered that Europa is warmer than it should be for a body with no obvious source of external heating and too small for internal fission.
Meanwhile, the earth is 4000Celsius at its core which is similar to the surface temperature of the sun, only its not 93million miles away.
“The link I found by googling, I had not saved my calculations.
Certainly over direct flues the heating will be important, but that is another story. Averaged over the area of the earth it is small.”
Flues on land pump heat into the air and most of that is lost into space. Flues under the sea do not lose their heat to space. The specific heat capacity of water if very high, so the ocean simply stores the heat, then convection ensures fresh colder water gets a chance of further heating. The flues can be over 1000Celsius.
By the way, at some points the earths crust is very thin, and at others huge magma chambers are very close to the surface. There is much more of this volcanic activity under the oceans than on land. The average heat flux is not important if you have an efficient heat exchange mechanism in an area of very high heat flux – which you do.

Dave Springer
March 18, 2010 10:06 am

Willis,
I believe you need to explain lunar regolith temperatures a bit more.
The temperature of lunar regolith at the Apollo 15 site was measured at an almost constant -23C (day or night) at a depth of 50cm (1). Apollo 15 landed at 26 degrees latitude (2) so it is roughly equidistant from polar and equatorial extremes of solar irradiance. Internal heating is almost nil so the regolith temperature is almost solely reflective (pun intended) of the intensity of solar irradiance and surface albedo.
Lunar albedo averages 12% (3) which is lower than the figure you give for the earth of 15% so we should expect the regolith temperature at depth would be even cooler if the surface albedo was the same as the earth.
Could you please explain the large discrepancy between average temperatures actually measured on the earth’s airless moon and your theoretical prediction of the earth’s temperature sans atmosphere?
Disclosures: I believe the experimental evidence is overwhelmingly against anthropogenic global warming of any adverse significance and what actual AGW there is has a net beneficial effect which in combination with increased CO2 makes for longer growing seasons and higher crop yields. Sure, there are winners and losers on a slightly warmer earth as climate patterns change the winners far outnumber the losers. What is really deserving of alarm is a cooling earth – crops don’t grow well in ice and snow.
references:
(1) http://education.ksc.nasa.gov/esmdspacegrant/LunarRegolithExcavatorCourse/Chapter5.htm#OtherPhysicalProperties
(2) https://nasm.si.edu/collections/imagery/apollo/AS15/a15landsite.htm
(3) http://www.geo.lsa.umich.edu/~shaopeng/Huang08ASR.pdf

Steve Goddard
March 18, 2010 10:35 am

A more realistic view of the amount of warming produced by the atmosphere is closer to 70C – not the 20C claimed in this article. Temperatures in the atmosphere drop linearly from about 280K to 210K as we rise from sea level to 10km, and then remain constant at higher elevations.
http://mtp.jpl.nasa.gov/missions/cirex/results.html
This gives a lapse rate of 2.1C/1000ft.
I remember one summer day in Arizona where I was hiking on dark colored rock in a snowstorm at the top of Humphrey’s peak at 12,000 ft, and later in the day was in 105 F weather in Phoenix. Temperatures varied by about 40C between the two locations, mainly because of the amount of atmosphere above the ground surface at the two locations. Cloud cover also played a role. Planes flying overhead at 35,000 feet would have been in temperatures close to minus 70 F.

gbaikie
March 18, 2010 10:43 am

“I live by the sea. In the summer air temperatures can be for a month from 37C to 40C the water is below 25C.”
I assume you mean the summer high temperatures for a month can reach 37 to 40 C.
I live a few miles from the beach- near LA.
I used to live in British Columbia on Vancouver Island- there ocean water was well below 25 C- probably something like 10 C. One couldn’t really swim in the ocean [though my uncle and others did a lot of scuba diving in this water]- and there were some beaches where was warm enough- basically the water would warm up when the tide came in.
In Canada I lived right on the beach and few times did swim briefly in the ocean at low tide between the shore and the kelp bed- I could feel a sharp difference between the very top layer of water and the rest of the water. Of course having the very top part of the water warm didn’t change the fact that the water was too cold to swim in for any amount of time- unless you worn a wet suit [though I think dry suits were mostly used for scuba diving in this area].
Though swimming in lakes was a different matter- depending the lake but in the summer a lot of them were warm enough.
Though with most coastal water the wave action will mix the water temperature- I lived on the east coast of Vancouver Island and if it wasn’t stormy weather the Georgia Straits could be fairly calm water.
“At night it is the opposite, particularly in autumn, when the water may be a steady20C while the air drops to 12 and 14C.”
There is no doubt there is a difference between ocean temperature and land temperature. Most of Europe and the place I grew up is milder and warmer because of the influence of the Ocean temperatures. Or the place where live now is cooler then say 10 or 20 miles inland- because of the ocean’s influence. And it can get considerably warmer here when there are what’s called Santa Anna wind- a constant air flow lasting for days coming from a land area instead of normal pattern daily ebbing back and forth from ocean to land.
“Again I refer to the surface skin temperatures in
http://isccp.giss.nasa.gov/products/browsesurf1.html
I went over there, but I don’t know what exactly you referring to. I selected annual and both surface skin and air temperature, but it doesn’t seem to tell exactly what temperature each of the color represent [obviously it’s in Kelvin- that’s not a problem- but they have 12 colors and 4 numbers- I suppose you can sort of get that the global average temperature per that map seemed to be somewhere around 289 K? [[16 C]] but it shows up on my computer as a small map and at pretty coarse scale.

March 18, 2010 11:33 am

Francisco (09:31:15) : I know of no experiment done under STP conditions (0 deg C at 1 atm) or standard conditions (59 deg F at 1 atm) that has demonstrated CO2 could increase the temperature of anything.

March 18, 2010 11:42 am

anna v (09:23:56) : Thank you. I got a chuckle from it. That goes along with my CO2 thermos bottle. We should be able to come up with all sorts of CO2 heat increasing items.
“Save electricity throw out your old electric blanket and purchase a new CO2 filled comforter. Depending on per cent of fill you can have up to 74 additional watts of heating power per square meter. With a fleece cover that is washable this blanket will be the one you want when winter chill sets in. Purchase now for $29.99 for one but if you order now get two for $45.00. State tax not included.”
There is big opportunity out there.

James F. Evans
March 18, 2010 12:05 pm

Brian W (00:44:44) wrote:
“As to the sensitivity. CO2 with a specific heat less than N2, O2 and even aluminum, poor absorption compared to air, fast emission and most importantly a concentration of .038% by volume contributes NO sensible (usable) heat to our atmosphere. A doubling of CO2 will not give even .1 degree increase. I wish people would stop fixating on CO2. So the models do overestimate sensitivity by a huge amount. The whole AGW/CO2 thing is a stinking scientific fraud. Period.”
Brian W, thank you, in my opinion this is the salient analysis with regards to CO2 in the atmosphere (I’ve raised it here before).
There is only one problem with it. It’s too simple and straight-forward: It ends the argument and obviates any reason for further study of the issue. All that is left to study is fixing an upper limit of CO2 animals can properly and healthily respirate over the long term (which in all likelyhood would be rather simple and not require public funding, the life-blood of “modern science”).
Even scientists on the sceptical side of the argument seemingly don’t want to acknowledge this undeniable physical reality because all their angles of disputation (and funding) become unnecessary.
Needless to say, AGW scientists don’t want to acknowledge this physical constraint, either, as it defeats their arguments in total — no passing “GO”, no collecting $200 (government funding).
Problem: Until, heavy-hitter sceptical scientists write papers outlining this physical reality, AGW scientist won’t bargain against themselves and raise the issue on their own accord.
I’ve stated this issue before here on previous posts, asking to be corrected if I’m wrong, so far no-takers (that I’m aware of).
And, as I’ve stated before, hair-splitting on this parts per million or that parts per million CO2 concentration, is conceding over half the scientific battle and is akin to arguments about the arrangement of deck chairs on the Titanic.
(Just where the AGW proponents want the “game” to be played.)
But that’s where we find most sceptical scientists — arguing about deck chairs on the Titanic.
And, frankly, beyond my suspicions about funding dollars, it’s perplexing.
An old lawyer joke: “One lawyer in the town starves, two lawyers do a thriving business.”
Sure, I maybe wrong — but I’m from Missouri, “show me”.
And, until it is explained why I’m wrong, I’ll keep banging the drum and encourage others to do the same.

March 18, 2010 12:14 pm

Willis Eschenbach
….This is far from having enough information to answer. What is your point?
Thanks for the reply.
The conditions add up to eliminating any direct to surface from Sun radiation.
That should then mean that any radiation being measured would be back radiation from the atmosphere.
I’m not looking for an accurate figure just a very rough estimate.

Francisco
March 18, 2010 1:32 pm

mkelly (11:33:35) : I know of no experiment done under STP conditions (0 deg C at 1 atm) or standard conditions (59 deg F at 1 atm) that has demonstrated CO2 could increase the temperature of anything.
==================
For such a controversial matter as this, the lack of such experiments is appalling. Some may argue they are impossible, but I don’t think so. You don’t need to reproduce the entire atmosphere, or anything remotely resembling that, to get an idea of how doubling or tripling or quadrupling CO2 concentrations in an isolated, open-top column of sunlit (or man-lit) air inside a controlled environment, affects its temperature near the bottom. Or any number of variations along that kind of setup. It cannot be that hard. It seems to me that infinitely more complicated experiments are routinely carried out in other fields.
It also seems to me that if we cannot detect any clear effect on air temperature under such easily measurable conditions, then what on earth are we talking about when we presume that little fluctuations in a quantity called “global mean surface temperature” (which is measured in grotesquely primitive fashion and using impenetrable methods) are the result of significantly smaller increases in CO2 concentration than the ones I mentioned??? None of this makes any sense.
I imagine those kinds of experiments must have been attempted, and if they had yielded any clear results in the desired direction, they would be routinely mentioned by the climate establishment as the empirical anchor to its fantasies. But we don’t see any of that. Nor do we see any attempt to set up such experiments. I wonder why that is.

Nemesis
March 18, 2010 1:34 pm

O/T but Ive just been watching a re-run of Jeremy Clarkson’s ‘Inventions that changed the world’. In the one about jet engines, he claims that for 3 days after 9/11 when there were no areoplanes in the sky, the earth warmed 1degree during the day and cooled one degree at night. Does this mean that areoplanes rather than cause global warming actually causes cooling ?

daniel
March 18, 2010 1:51 pm

This tread is amazing as most interveants are spending so much time making some more or less warranted maths in order to make the so-called consensus model work, whereas this model a) appear flawed from a theretical point of view (Gerlich & T, Robitaille, ) b) is no longer ‘alone’ (Miszolczi, the russian school…), c) so many predictions according to the model are just not to be found in the true life; it’s sad to see so little time spent on the respective merits of Miskolczi or the Russian
Anothe interesting point for the near future is how the contributions of Gerlich & T, Robitaille, Miszolczi, the russian school, the Galactic Cosmic rays assumption (CERN’s CLOUD experience which should fascinate anybody seriously interested in doing physics through acual experiences) will be treated in the frame of IPCC’s AR5, as it’s no longer possible to just ignore all these alternatives to the IPCC’s affirmation that CO2 is responsible because volcanoes and the sun’s irradiance are not
As to Joel Shore announcement of a peer reviewed “comment on G&T” paper, will this be more successful that Smith tentative ? anyway is the paper already available through Arxiv ?

anna v
March 18, 2010 1:59 pm

Re: Willis Eschenbach (Mar 18 12:57),
in http://isccp.giss.nasa.gov/products/browsefmt.html they describe the various data
and on the main page next to the download data button they have choice of format, so choosing ASCII one gets a table .

March 18, 2010 2:24 pm

Willis Eschenbach (13:38:27) :
where I show boreholes a few miles apart that show radically different temperature histories.
I know, I know, just like temperatures…
But when you average over thousands of them, the statistics gets a bit better.

Spector
March 18, 2010 3:40 pm

I do not know how valid the 30 percent reflectivity figure is — I would assume it would be measured from space as a long-term average of that fraction of the radiation from the Earth that had the same signature as direct solar radiation without regard for the source of that reflected energy. As such, I presume this is a measure of that fraction the incoming solar radiation that is returned directly to outer space and plays no role in determining the Earth’s climate.
Thus, after reflection, we have an average level of about 239 W/m2 that must be emitted after becoming involved with the thermodynamics of the Earth. Perhaps this could be further reduced by including energy that is captured and released by the upper atmosphere without affecting conditions on the ground. In any case, the Stefan-Boltzmann equation yields a temperature of 255 K or -18 C for this energy flow per unit area. As I see it, this is a reference temperature only — it does not imply what conditions might produce it.
In the presentation of Dr. Miskolczi’s new greenhouse theory, Dr. Zagoni says the difference between this reference temperature and the Earth’s average surface temperature (+15C) is called the Greenhouse Temperature G (33 degrees C.) As such, he appears to be using this as a measure of the current effective thermal resistance or R-factor of the atmosphere.
Dr. Miskolczi has developed a new theory that includes a maximum limit to the greenhouse effect. He is reported to have discovered that the original work on this theory used equations only applied to planets with infinitely thick atmospheres.

James F. Evans
March 18, 2010 4:14 pm

In this thread there has been reference to IPCC definitions like ‘radiative forcing’ and ‘climate sensitivity’ and some values attached to those definitions, therefore, those definitions and values have been incorporated in various papers.
But there is a problem with that: Namely, that a presupposition exists that these are accurate and useful definitions and values.
Given the track record of the IPCC and it’s agenda & outcome driven processes, one is left to question that accuracy of such standards.
Sure, don’t use or abide by what the IPCC sets up and your paper won’t get into the IPCC report, but if they are flawed requirements then being constrained by these requirements garantees the papers, while jumping through the IPCC’s hoops, won’t reflect actual physical reality:
And, isn’t reporting on what is actually happening more important than being properly constrained by arbitrary IPCC standards?
Or is it simply about playing the game as the rule master dictates…damn the reality?

Francisco
March 18, 2010 4:25 pm

Willis Eschenbach (12:26:45) :
44.2 terawatts = 4.4 * 10^13 watts
Earth’s surface = 5.1 * 10^14 square metres
Heat flux = ~ 0.1 W/m2
================
Yes, assuming the 44 terawats are right. I have no idea how that is measured.
However that may be, the above conversion has lead me to another surprising perspective.
If total human energy consumption is about 15 terawatts, then this would translate to ~ 0.3 W/m2.
So let’s think about the meaning of this. If we dedicated ALL the energy we consume to the exclusive purpose of heating the earth’s surface *permanently* from underneath using the most efficient means at our disposal — the warming effect would be barely noticeable.
On the other hand, if we just take a gas that happens to be released as a byproduct of our energy use — the same gas that is exhaled by the biosphere and and the oceans in quantities that are orders of magnitude larger than our contribution — and let it go in the air, the result of the concentration increase in this gas putatively attributed to our modest exhalations, is supposed to be a surface warming many, many, many times larger than what we could ever achieve by using all that energy permanently and directly for heating the surface as hard as we could.
When I look at it in this perspective, I find that extremely bizarre. Which is why I would like to see some piece of empirical evidence, no matter how primitive, of this marvel at work.

Francisco
March 18, 2010 4:28 pm

Francisco (16:25:37)
Of course I meant 0.03 (not 0.3) W/m2.
Sorry.

brent
March 18, 2010 4:37 pm

The science of climate change
The clouds of unknowing
There are lots of uncertainties in climate science. But that does not mean it is fundamentally wrong
http://www.economist.com/displayStory.cfm?story_id=15719298
brent comments:
The Economist should reflect on the Wegman equation:
“Method Wrong + Answer Correct = Bad Science.”
Of course “answer correct” in this context means “politically correct”
brent’s amendment to the Wegman Equation follows”
Method Wrong + Answer (Politically) Correct =”Post Normal Science”
Post Normal Science = Sceance
cheers
brent

Alan D McIntire
March 18, 2010 5:12 pm

“Another point, Using Trenbeth ballpark figures, if we get 240 watts from the sun, and the greenhouse effect makes that an effective 390 watts and a 33 C increase, isn’t the increase in temperature per delta watt increase in the ballpark of
33 C/(390-240) = about 0.22C per watt increase?”- AMC
“No, no, no. I’ll keep going over this until everyone gets it. According to the UN IPCC, climate sensitivity is the change in surface temperature divided by the change in TOA forcing … yes, I know, it’s a really dumb measurement, but there it is. I suspect it was specifically chosen by AGW supporters because downwelling TOA radiation can’t be measured directly … but perhaps that’s just my paranoia speaking.”- Willis Eschenbach
I REALIZE climate sensitivity would be delta surface temperature divided by delta watts at surface, but I believe the 33c/150 watts gives an upper limit ballpark figure.
With NO forcing, the earth’s temperature would be 33 C lower, radiating at 240 watts. Those additonal 150 watts have been pumped into earth’s atmosphere, gradually warming the earth at a lower and lower rate. The total warming was 33C, the total additional sensible wattage was 150 watts, so the average sensitivity over the full range of 240 to 390 watts was
33C/150 watts = 0.22C. It stands to reason that ACTUAL climate sensitivity should be less.
The Stefan-Boltzmann equation for a blackbody goes
T(degrees Kelvin) = S(constant)*(watts/square meter)^0.25. Our first step is to find that S constant.
Doing a google search, I find 1000 K implies a blackbody flux of 56790 watts/square meter.
1000K = S* (56790 watts/square meter)^ 0.25. Click on your calculator and use the scientific view. Plug in 56790
X^Y
0.25
=
and you get 15.43718 Divide 1000K by 15.43718 and you get S = 64.77867
We now know
T(kelvin) = 64.77867 ( watts/square meter)^0.25.
Let’s plug in some numbers.
At 100 watts per square meter flux,
T = 64.77867 * 100^0.25 = 204.848 K, at 400 watts per square meter,
T = 64.77867 * 400^0.25 = 289.699 K
T = 64.77867 W^0.25. From elementary calculus, the derivative of that
equation will give you the sensitivity.
dT/dW = 64.77867* 1/4 * (W^(-3/4)) = 64.77867*1/4 * (1/(W^0.75)).
At 100 watts/sq meter, you get a sensitivity of
64.77867*1/4*(100^-0.75) = 64.77867*0.25* 0.0316228= 0.512121 K/watt per meter squared
At 400 watts/sq meter, you get a sensitivity of
64.77867*1/4 *0.01118034 = 0.181062 degrees K per watt per meter squared.
My above example was equivalent to
taking (289.699-204.848)/(400-100) and getting an “AVERAGE” sensitivity of 0.283 K/watt- note that this is higher than the “actual” sensitivity of 0.181 once we’ve reached 400 watts. The earth is not a blackbody, but the reasoning should be the same, that “AVERAGE” sensitivity of 0.22 C must be larger than the current sensitivity with a surface 390 watts.

sky
March 18, 2010 5:22 pm

Willis Eschenbach (00:03:04):
You claim to have accounted for atmoshperic pressure. I’d like to see where! All the cartoonish schematics (not just yours) show only power fluxes–not enthalpy calculations–presented in an disconnected way, contradicting Kirchhoff’s current law. Furthermore, the figures for the flux of latent heat seem to be land-based, rather than accounting for the oceans, where evaporation exceeds upwelling IR radiation and conduction combined. Sorry, but that’s just not a physically cogent exposition of the “budget” of the Earth’s climate sytem.

Joel Shore
March 18, 2010 5:27 pm

daniel says:

As to Joel Shore announcement of a peer reviewed “comment on G&T” paper, will this be more successful that Smith tentative ? anyway is the paper already available through Arxiv ?

I am not sure what you mean about “more successful” in regards to Arthur Smith’s paper. I think that his paper was very nice, but because at the time that he wrote it G&T had not yet gotten their paper published in a peer-reviewed journal, Arthur did not even try to submit his. (It is very hard to get a paper published in a peer reviewed journal that just rebuts nonsense floating around on the internet. Arthur pretty much admits in the introduction that his paper does not really break significant new ground but just clarifies well-understood science in light of G&T’s confusions.)
Also, Arthur focused on one aspect of G&T’s claims (the supposed “fictitiousness” of the calculation that the greenhouse effect warms the earth’s surface by about 33 C), whereas our comment on G&T addresses a broader array of aspects of their paper, including the central claim that the atmospheric greenhouse effect somehow violates the 2nd Law of Thermodynamics. Our paper is not yet available on the Arxiv although it may be up there before too long.

Alan D McIntire
March 18, 2010 5:37 pm

Sorry, I meant to agree with W.E. and state, “I realize forcing shold be delta T at surface divided by delta watts at TOP of atmosphere.

March 18, 2010 6:03 pm

Willis Eschenbach (16:16:56) :
1. Contrary to your claim, temperatures are well correlated, particularly over short distances. It is very unusual to find two sites a mile apart where one is cooling and one is warming.
That’s where careful cherry picking comes in.
Boreholes: I saw this guy’s presentation: http://adsabs.harvard.edu/abs/2009AGUFMPP53A..10C
And couldn’t put a finger on where it was wrong, so he convinced me.
This study looks solid too: http://home.badc.rl.ac.uk/mjuckes/mitrie_files/docs/mitrie_borehole.pdf
Unfortunately it shows global warming so many would reject it just on that, regardless of the merit of the method.

March 18, 2010 6:31 pm


Brian W (09:20:12) :

The bits about electromagnetic radiation and the atmosphere are spot on the money. Above roughly the 10 meter band or 30 megahertz the atmosphere becomes transparent to radio waves.

To a point; even energy in the 52 or 146 or 432 MHz ranges can become affected or susceptible to ‘action’ by the atmosphere *and* the ionosphere. I won’t go into other effects e.g. Faraday rotation: rotation of the plane of polarization of a linearly polarized wave propagating through a magnetized dielectric medium – that can be seen at GPS frequencies in the L-band (1.2 and 1.5 GHz) .
THEN there is an ‘opaqueness’ in the atmosphere that sets in at wavelengths shorter than 1 cm (above about 30 GHz) .
Click my name to see a chart.
.
.

Tsk Tsk
March 18, 2010 6:34 pm

Leif,
You’ve over constrained the system or at least implied that the emissivity of the two objects is identical even though one of them clearly isn’t a black body. Take out the nuclear reactor and start with identical temperature spheres but with different emissivities because one of the spheres is not a black body. We’ll return to the case where there’s internal heat generation, but for now simply rely on the internal energy of the two spheres giving them equal temperatures. BTW, I don’t really understand the electron acceleration comment. That sounds like bremsstrahlung to me which isn’t black body radiation at all. Anyway…
Here’s a simple way to look at the problem. Let’s turn the spheres into two parallel semi-infinite walls separated by a vacuum. The only way these walls can lose heat is to radiate it to each other. Let’s also say that the white wall has an emissivity and therefore absorptivity of 0.5 and the black wall is a perfect black body with emissivity of 1. And yes, let’s make the emissivity a wavelength independent quantity because I’m lazy. To avoid messing around with constants and T^4 let’s also say that the black wall is of such a temperature as to radiate 10 W/m^2 at its initial state. Since the white wall has the same temperature but half the emissivity, it emits 5W/m^2. Now let’s look at the energy balance for the two walls. The white wall is exposed to 10W/m^2 incoming radiation of which it absorbs half, 5W/m^2. The other half is reflected back toward the black wall It also emits 5W/m^2 of its own radiation for a net heat transfer of: 5-5=0W/m^2. The black wall emits 10W/m^2. It receives 5W/m^2 generated from the white wall and 5W/m^2 of its own radiation reflected back for a net heat transfer of: 5+5-10=0W/m^2. So, two bodies at the same temperature and different emissivities do not have to radiate the same amount and can still be in radiative equilibrium at the same temperature. No work can be done. Entropy is constant. Carnot is happy. Maxwell is happy. And his demon sleeps.
To return to your example, unless the emissivities are the same, then the problem you framed is not in steady state. If the same amount of energy is being produced in both spheres and they have different emissivities, which they will since they are different colors and hence absorb/emit differently, then the white sphere is only temporarily at 400C on its way to its final temperature. Put another way your white/green/blue sphere can’t be a black body, but it will still emit according to Planck’s law albeit with a sub-black body emissivity.

Joel Shore
March 18, 2010 6:35 pm

Francisco says:

So let’s think about the meaning of this. If we dedicated ALL the energy we consume to the exclusive purpose of heating the earth’s surface *permanently* from underneath using the most efficient means at our disposal — the warming effect would be barely noticeable.
On the other hand, if we just take a gas that happens to be released as a byproduct of our energy use — the same gas that is exhaled by the biosphere and and the oceans in quantities that are orders of magnitude larger than our contribution — and let it go in the air, the result of the concentration increase in this gas putatively attributed to our modest exhalations, is supposed to be a surface warming many, many, many times larger than what we could ever achieve by using all that energy permanently and directly for heating the surface as hard as we could.

Why is that all that surprising? As many people have noted, the amount of energy that we use in miniscule compared to that supplied by the sun. Hence, a much more efficient way of causing heating of the earth is to change the atmospheric chemistry so that more of that energy from the sun is retained, rather than simply heating it directly ourselves.
By the way, when you say, “the same gas that is exhaled by the biosphere and and the oceans in quantities that are orders of magnitude larger than our contribution”, you are neglecting the other side of the equation – that it is also inhaled by the biosphere and oceans in these large quantities. And, the exchange is only large with the mixed layer of the ocean. The exchange between the mixed layer and the deep ocean is quite a bit smaller…and that is the rate-limiting step. The mixed layer + biosphere + atmosphere are just passing the same carbon around and around. We are taking a quantity of carbon that has long been locked away from the atmosphere and rapidly liberating it into this system.

cba
March 18, 2010 7:15 pm

willis,
sorry but i’m out of town again and have limitations on access to things. I tried to go through the erbe etc. paper you referenced looking for the 0.123 clear ssky albedo. It occurred to me that perhaps there is more clear sky over land than ocean or something to that effect. Putting the number 0.123 in to the spread sheet indicates essentially a requirement of 0.32 land surface albedo and a 0.38 cloud albedo. 0.32 seems rather high, only snow and sand produce this and there’s not that much of those around. the 0.38 (or even 0.44 assuming surface is 0.085 or so) seems rather realistic. Hopefully, I can get some time to go through that erbe paper in detail.

Joel Shore
March 18, 2010 7:30 pm

Much as I’d like to do this all the time, I do have a life. Anyhow, 11 PM here, I’m back in the saddle.

Sorry…I didn’t mean to be impatient. I just wanted to make sure that you saw my post amidst everything else.

Joel, the 150 W/m2 is what I estimate as the TOA downwelling forcing. See Figure 1 above. Kiehl/Trenberth give a higher figure, but their diagram is not physically possible. But use either number, it doesn’t matter.

Okay…I will have to think about this more.

This number perforce includes everything. It includes whatever water vapor feedback (or forcing) exists at the current earth temperature. It includes the cloud feedbacks (or forcings) that exist at the current earth temperature. By definition, it includes each and every feedback that exists, known and unknown, measured and unmeasured.

Since the sensitivity for a doubling of CO2 is 3.7*?T/?F, that gives on the order of half a degree for a doubling of CO2.

Ah…but therein lies the problem. All those enter into what you are saying the forcing is…but then when it comes time to calculate how this relates to CO2, you say that doubling CO2 produces a forcing of 3.7 W/m^2. That is not correct. In your first calculation (the 20 C divided by 150 W/m^2), you have called everything a forcing whereas in the second calculation you have called everything a feedback except for the change in CO2.
Let me give you a concrete example: Suppose that the increase in CO2 directly is responsible for 30 W/m^2 of forcing in the current climate due to CO2’s IR absorption and the water vapor is responsible for the other 120 W/m^2 due to its IR absorption (and, for simplicity, we just ignore everything else). Now, suppose that the water vapor feedback works in such a way that if we removed all the CO2, the feedback processes would be such that all of the water vapor also disappears (again, I am not claiming this is true…but just supposing so).
In that case, then to get the climate sensitivity due to changing CO2, I think you would agree that you should divide the 20 C change by 30 W/m^2, not the full 150 W/m^2. But, alas, that is not what you have done.
To put it another way, when the IPCC says that the central estimate for climate sensitivity is 3.0 C (with an assumption that doubling CO2 produces a TOA forcing of 3.7 W/m^2), they are not predicting that the TOA downwelling radiation will be increased by only 3.7 W/m^2 if you double CO2. What they are saying is that the direct effect of the IR absorption by CO2 will be to increase the TOA downwelling radiation by 3.7 W/m^2 but that the total increase, once you consider all the feedback effects will be an increase in the TOA downwelling radiation by almost 3 times this (where I am using the result that a 3.7 W/m^2 increase in TOA downwelling radiation produces a temperature rise of about 1 C, as calculated by the S-B Equation).
It would be a violation of radiative balance to claim that a 3.7 W/m^2 increase in downwelling radiation would produce a 3 C increase in temperature! We know from the S-B Equation that this is not the case.
The one thing that admittedly remains a mystery to me is why your calculation of the climate sensitivity in the way that you did it does not just come up with the value for climate sensitivity determined by the S-B Equation in the absence of any feedbacks, which is what I think it ought to get. So, I am still confused about whether the 150 W/m^2 is what you actually wanted to use, but I will have to think more about that.

PS – I note that someone above called you “Troll Shore”. Folks, I’ve got good news and bad news. Joel is a physicist, understands the math, and is one of the few AGW supporting trained scientists with the balls to post here. So cut out the personal attacks. They are repulsive in any case, and totally unmerited in Joel’s case. I disagree with him a lot, but he is absolutely not a troll.

Thanks, Willis (and anna v). I appreciate it.

March 18, 2010 7:42 pm

Tsk Tsk (18:34:25) :
then the white sphere is only temporarily at 400C on its way to its final temperature.
The 400C is its final temperature, once the white and the black sphere have reached a final same temperature, they both have the same thermal spectrum. Every opaque body radiates a thermal spectrum determined by its temperature. The balls are in empty space far VERY from each other and from anything else. And do not exchange [measurable] radiation with each other or with anything else.

March 18, 2010 7:55 pm

Tsk Tsk (18:34:25) :
I don’t really understand the electron acceleration comment
From a different view [more modern and more ‘correct’ view], electrons in atoms are ‘excited’ by thermal motions; when the atom relaxes [excited states have finite – short – life times], the electrons return to a lower state emitting photons in the process.

Brian G Valentine
March 18, 2010 8:15 pm

Jim writes, “Can you point to an article or source where this is actually measured (the quantitative measurement of atmospheric radiation of ‘heat’ energy ostensibly removed from the surface via convection)?”
Thank you for pointing that out, Jim, I have been trying myself to address that, I am sorry I don’t have a reference for you. That’s a good observation
Brian W, that’s a good comparison. Note that the ionosphere (Kenelly-Heaviside layer) acts as a deus ex machina to overcome the curvature of the Earth so far as short wave radio transmission is described by the laws of electromagnetic propagation; there’s no such associated “outside influence” as far as thermal IR transmission is concerned (1-30 micrometer radiation), and Gerlich and Tscheushner must be correct. (Look at the “wrong” description of how the “greenhouse” effect supposedly “works”.)
The Trenberth “analysis” is nothing more than some Enron-type bookkeeping to give an extra 1 w/m**2 or so (on the disc) that he wants to present as “overheating” the world from Greenhouse gases. It isn’t observed by satellites measuring incoming and outgoing radiation (which balance), but what the hell. It serves the purpose of overheating some temperaments and some egos, which the what the construct was really ever designed for

Dave Springer
March 18, 2010 8:32 pm

The moon couldn’t possibly be a better experimental setup for isolating the atmosphere as a variable in average surface temperature, we put temperature sensing instruments on it decades ago, have years of data from them, and yet almost everyone here is ignoring this hard data.
Would someone please put forward an hypothesis which explains subsurface lunar regolith temperature. Any hypothesis must explain these observed temperatures to be at all credible. From what I can determine from experimental evidence the earth’s atmosphere accounts for a minimum of 30C warmer surface temperature.

Brian G Valentine
March 18, 2010 9:04 pm

The Moon’s reflectivity is known to be partly specular in the visible, since the Moon appears uniformly bright from center to limb as viewed from the Earth. Is it this reflectivity in the visible is the same reflectivity through the thermal IR? I don’t know.
I can’t answer your question about the Moon, Dave, but I can give you a calculation that shows the thermal exchange temperature of the Earth with the Sun is about 279.5K, and and we also know that the Earth may be viewed as a radiator at a temperature of about 288K, other than that, I don’t know what to say about the 30K warmer figure

Tsk Tsk
March 18, 2010 9:25 pm

Leif,
Again, you’ve over constrained the problem. The ball cannot be white and a black body. What makes it white? Clearly it does not absorb all incident light and therefore it cannot emit perfect black body radiation. You’re right that every body radiates a spectrum based on its temperature and based on its emissivity. You keep leaving that part out. By your definition every object is a perfect black body (emissivity of 1) and the only thing that matters is its temperature. That’s simply not correct. Fluke has some pretty graphs to demonstrate the impact of emissivity on measured temperatures using an IR thermometer: http://www.hartscientific.com/publications/pdfs/3187781_A_w.pdf
I placed the two spheres/walls in proximity to counter your claim that different amounts of radiation from the two objects would allow you to extract work and therefore that proved that the two objects must be emitting the same amount of radiation. What I showed was that there is no net heat transfer and no way to extract work from the transfer of heat from two bodies at the same temperature but different emissivities emitting different amounts of radiation.
I understand the emission of photons comes from electrons transitioning between different states. I’ve just never heard acceleration used to describe the process, but then again quantum was a long time ago and I spend most of my time on magnons and not phonons these days.

Bill Illis
March 18, 2010 9:46 pm

The 3.0C per doubling sensitivity comes from the following:
– doubled CO2/GHGs add 3.7 to 4.0 W/m2 of forcing at the tropopause emission layer where it is 255K (sometimes they call this the TOA forcing but the top of the atmosphere is much warmer than this level – it is really the tropopause at about 5 kms up but they often mix this up to keep people off balance);
– then there is feedbacks from increased water vapour and albedo which add another 6.5 to 7.4 W/m^2 after this initial forcing;
– the layer now increases in temperature by 3.0C (11.5 extra watts on top of the original 240 watts).
– the layer where the equilibrium emission temperature of 255K occurs is now 461 Metres higher.
– the adiabatic lapse rate of 6.5C/km stays intact and the surface warms by the same 3.0C.
So, the sensivitiy is calculated as 4.0 extra watts of GHG forcing eventually results in an increase of 11.5 watts (once all feedbacks occur) and temperatures at the surface rise 3.0C.
So, the theory calculates 0.75C/W/m2. It takes the Planck Response of 0.265C/W/m2 (the Stefan-Boltzmann Dtemp/Dforcing for 255K or 240 W/m2) at the tropopause and multiplies it by another 134% in feedbacks to arrive at the final sensitivity.
Its a theory based on some principles. But theories need to have factual empirical data behind them.

anna v
March 18, 2010 10:01 pm

Re: Willis Eschenbach (Mar 18 16:02),
The two links connecting directly to the images I gave in my first post give this message you quote. I did not put a correction because the main page was linked also in the same post.
anyway:
I use Firefox with windows XP
I click on
http://isccp.giss.nasa.gov/products/browsesurf1.html
sometimes it might give a “not found”, but it is from traffic or bad width.
“select variable” example: mean surface skin
select a time period : Mean annual
select a format: EQ grid; Ascii text
and “download” .
I find the file in my download area labeled with sequential numbers.
No other secret except in persisting if I get a “not found”.
the differences. between skin and air are large in the cold and hot places. I do not know if they would integrate out. I have not learned to use EXCEL for plot making 🙁 . Maybe I should start.

March 18, 2010 10:09 pm

Tsk Tsk (21:25:20) :
By your definition every object is a perfect black body (emissivity of 1) and the only thing that matters is its temperature.
Did not say that, I said that maintained at the same temperature the radiations would be the same.
I’ve just never heard acceleration used to describe the process, but then again quantum was a long time ago
Since we were talking blackbody and S-B laws in a 19th Century [pre-quantum] setting I stayed in that framework. You can even say that electron transitioning [to stay in a Bohr model] from one orbital to another is a change of velocity and hence an acceleration, but hose details are not really necessary and at a deeper level that is not how it works anyway [but we often use pictures based on incomplete models if they are useful]

Brian G Valentine
March 18, 2010 10:12 pm

I understand where this is coming from, Bill, but where I (and apparently Gerlich and Tscheuscher and a lot of others) have the worst difficulty, is accepting the idea that the “atmosphere” is “warming” the “surface”.
Either that idea violates the Second Law or it doesn’t. If someone could just present ONE parallel example to show this is at least possible, then I would not say that Gerlich and Tscheusher’s analysis is as good as anyother application of the Second Law

F. Ross
March 18, 2010 10:39 pm

Very interesting post Mr. Eschenbach; it is great to see the varied, open air give and take between you and and the variety of posters who question several of your points.
I enjoyed reading, if not fully understanding, nearly all the posts and your responses to them.
Infinitely more interesting than anything on the boob tube. Thanks.

March 18, 2010 11:19 pm

Brian W (01:59:09) :
Steve Goddard (23:47:55)
The shell of your airplane heated due to air friction! Blankets indeed!

Blankets. The air friction at 35000 ft is the same as 5000 ft is the airplane is flying at the same indicated airspeed, and airliners do stay in a fairly small range of indicated airspeeds. The thinner air up high requires they fly a much faster true airspeed for the wings to generate the lift, but the dynamic pressure the airplane feels is about the same.

anna v
March 19, 2010 12:49 am

Re: Brian G Valentine (Mar 18 22:12),
I understand where this is coming from, Bill, but where I (and apparently Gerlich and Tscheuscher and a lot of others) have the worst difficulty, is accepting the idea that the “atmosphere” is “warming” the “surface”.
Yes.
http://discover.itsc.uah.edu/amsutemps/execute.csh?amsutemps+001
At 36000 feet (15km) temperatures play between -47.5 to -46.5.
at 14000 feet (5km) -21.5 to -19.5
near surface 20.8 to 21.5
Now basic thermodynamics says heat is not transferred from a cold body to a hot body without work being done. That is elementary . CO2 is not doing any work ( running a compressor circuit, at least it has not been shown as such).
The earth is a macroscopic body. There is absolutely no reason not to calculate everything within classical thermodynamics : there are no coherent phenomena and radiation is included in classical thermodynamics. If one wants to calculate the effect of a change in the composition of the gas, one has to calculate the effect of the change in the heat capacity of the medium, to see what happens in the heat transports.
The confusion with the quantum mechanical guts of the molecules of the atmosphere is what , in my opinion, makes for double counting.
One can work with quantum statistical mechanics, but then one should use that formulation throughout, making statistical ensembles and studying their evolution etc. Mixing two techniques over the same material creates the contradiction.

Spector
March 19, 2010 1:18 am

Perhaps one method of calculating climate sensitivity to some parameter, like increasing CO2 concentration, would be to calculate how much a very small step change in the controlling parameter would change the effective Stefan-Boltzmann temperature of the total energy that escapes from the top of the atmosphere. I suspect the negative of this change might give the surface temperature change required to make up the difference.
Of course this would all assume that we knew how to make such calculations in the first place.

cal
March 19, 2010 1:26 am

Lief
Tsk Tsk
Tsk Tsk has my vote on this one. Lief, you define the two spheres as having a temperature of 400 but you also imply that the spheres as have equal power generation. If these two are defined then they must have identical heat losses equal to the power generated, for the temperature to remain constant. But you also define them as having different emmissivities (one is black and one is white). You cannot define all these three as being true. As Tsk Tsk says the system is overdefined. Tsk Tsk also clarified the point that good absorbers are good emmitters and vice versa so that energy laws are not violated.

March 19, 2010 4:13 am

Joel Shore (17:27:26) :
daniel says:
As to Joel Shore announcement of a peer reviewed “comment on G&T” paper, will this be more successful that Smith tentative ? anyway is the paper already available through Arxiv ?……
Looking forward to reading your paper but I have to agree with Daniel that the previous so called refutations of G&T have backfired.
I would invite anyone who has any doubts on the matter to go to Arthur P Smith website and the thread “the arrogance of Physicists”.
Arthur has great difficulty coping with Fred Staples and Terry Oldberg never mind G&T

March 19, 2010 5:01 am

Willis Eschenbach (01:56:43) :
“That’s where careful cherry picking comes in.”
I don’t know if that’s humor or not … I hope it is.

It was not. You can [and did] pick two close boreholes that give different results. That is the cherry. For it not to be cherry picking you will have to show that any two close boreholes are always wildly different, which they are not.
Finally, I note that your borehole reconstruction shows the Medieval Warm Period, and theirs shows nothing of the sort … which one is wrong?
One only went back to 1500, so cannot show the MWP.
The error bar is probably calculated from the spread of the 837 individual reconstructions. Let that be, typically [go look at several] 3 degrees, then the usual 1-sigma error of the mean would be 3/sqrt(836) = 0.10 degrees. You can find the individual reconstructions here http://www.ncdc.noaa.gov/paleo/borehole/core.html
I do believe [Red Queen-like] the borehole technique is sound. This does not guarantee that all applications of it are.

beng
March 19, 2010 5:04 am

********
HankHenry (19:12:32) :
Willis tells us “The average temperature of the planet is about 14°C.” Of course that’s disregarding the massive cold of the ocean which is more like 4°C – averaged out. What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths. Also, do I remember my grade school science correctly when I say the total mass of the atmosphere is represented by the mass of just 33 feet of water – the theoretical lifting limit of a suction pump?
********
No, it’s rather simple physics — the earth’s crust (in this case, the sea bottom) is a very good insulator because of its thickness. And, surprisingly, the warmer water above the cold-water layer is also a good insulator because of its thickness and also the water’s general stratification (little mixing).
The cold water is there because cold meltwater from glaciers and exposure to arctic air causes it to sink, and is then preserved below by the insulation above and below it. The cold water would eventually warm up, but it’s being replaced by more cold water continuously, so a equilibrium situation arises such as we observe. The cold deep-water is a reminent of the previous ice-ages.
Heat conduction is a feeble way to transfer heat compared to convection and even radiation, especially when there is little mixing (stratified).

Dave Springer
March 19, 2010 5:37 am

Brian Valentine
I don’t understand why you responded since you had nothing to offer in way of explanation why the constant subsurface temperature of lunar regolith in the middle latitudes agrees very closely with the oft-cited figure of the earth being heated approximately 30 degrees celsius through greenhouse gases.
There are many studies of the optical properties of lunar regolith available. It isn’t exotic material of any sort and that shouldn’t be surprising given the moon is made of the same material as the earth. Lunar regolith is generally darker than earth surfaces because “space weathering” of the soil darkens it. Fresh impact craters have bright rays coming out from them because it is newly exposed regolith which has yet to be darkened by space weathering.
I believe the data from the Apollo experiments thoroughly disproves Willis’ hypothesis of only 8 degrees celsius heating from GHG. His continued silence on it I’m taking as a tacit admission that he can’t account for it and rather than admit an error is choosing to simply ignore the inconvenient data. I had thought hiding contrary data that doesn’t agree with theoretical model predictions was a tactic used by the AGW alarmists. I’m very disappointed to see it being employed by the anti-alarmists.

Francisco
March 19, 2010 6:03 am

Joel Shore (18:35:55) :
Well I should have emphasized that I find it “surprising” only if I force myself to look at things in a certain innocent way.
Regarding our release of carbon that had been locked away… just like you point out that our energy use is tiny compared with what we get from the Sun, I shall point out that our release of carbon is also tiny compared with the vast transactions between the different parts of the system. This NASA chart of the carbon cycle is illustrative for piece of mind, if viewed with proper calm:
http://nasascience.nasa.gov/earth-science/oceanography/ocean-earth-system/ocean-\carbon-cycle
From that chart, you see that total amount of carbon in the extant fossil fuels still in the ground is roughly 1/8 the amount of carbon in the active system. The amount we are recycling annually back into the system is about 0.015% of the total carbon in the system.
The cause for alarm, it seems to me, comes only if you adopt the assumption that our modest recycling program represents a very grave disturbance to a static and balanced system, whose balance coincides with its state around 1850. I consider this assumption to have no rational or scientific basis whatsoever. It is merely ideological.
To begin with, the carbon we are releasing was once part of that dynamic system. We are recycling it back into life at what appears like a reasonable rate.
A few years ago there was some talk of a “missing sink” to refer to the apparently surprising fact that about half the CO2 we release does not accumulate in the atmosphere. This nonsensical phrase has now been sensibly dropped. How can sinks be missing? The surprise came only from the unwarranted assumption that the other entities that hold CO2 should refuse to take any extra amounts, on some kind of unwritten principle of carbon austerity by plants and oceans.
But why on earth would it be surprising that the plants take notice of variations in what is available for them to take – and proceed take it – as they are indeed taking it. Didn’t they once hold much of the very carbon we are now putting back in circulation? And why would it be surprising that the oceans also “take notice” of variations in the partial pressure of this gas in the atmosphere, and open their huge arms accordingly?
Alternatively, for illustrative purposes, you could look at the atmosphere as a giant pool with huge drains and pipes leading in and out of two other giant pools represented by the oceans and the plants/soil. So now we add our own little dripping faucet contribution to this giant global commerce, and we take ourselves so seriously that we assume the huge drains of this pool will be overwhelmed by our contribution, which will thus accumulate indefinitely. There are ribald jokes about ants having grandiose perceptions of their sanitary relations with elephants, or ants standing on a railway track and worried sick that they may cause a catastrophic derailment of the next train. I find them appropriate, sometimes, to the AGW discussions.
I think it could probably be demonstrated that even a doubling of CO2 concentrations from current levels is extremely unlikely, if not impossible, as a strict consequence of our contribution.

Brian G Valentine
March 19, 2010 6:06 am

Thank you Anna. Somebody tried to tell me that the “work” expended to accomplish this is the pV work done “by” the atmosphere; that certainly cannot be. One could in principle write down quantum statistical conservation and evolution conditions I suppose; don’t know how much meaning that would have on getting some numbers out of it unless some dramatic simplifications could be made (like if it was near zero Kelvin).
Thank you Willis for your most enlightening construct. I might have normalized to radiance quantities energy/unit area/unit if solid angle to possibly avoid some difficulty of interpretation of conserved quantities.
Wonderful weather here on the Mid Atlantic region this past week! I just can’t WAIT for some watermelons to start howling that it’s “too warm” and can’t we pass an energy and climate bill to fix this problem?
(You think I’m kidding? I’m not kidding one bit.)

March 19, 2010 6:49 am

Bryan (04:13:36) :
I would invite anyone who has any doubts on the matter to go to Arthur P Smith website and the thread “the arrogance of Physicists”.
And to compare with: http://arxiv.org/ftp/arxiv/papers/0904/0904.2767.pdf

Brian G Valentine
March 19, 2010 8:06 am

Professor Kramm sent to me that response, I would hope this appears someplace else other than good old arXiv
Anyway I still believe Gerlich and Tscheuschner,
… they oughta take me outside, and shoot me …

JAE
March 19, 2010 9:17 am

Can anyone tell me why it is not hotter in the desert on a humid day (when there is a lot of the best GHG) than on a very dry day?

Steve Goddard
March 19, 2010 9:37 am

The greenhouse effect is due to the fact that the atmosphere has a finite temperature and therefore emits radiation towards the surface. The surface receives more radiation from the atmosphere than it does from the sun.
The amount of warming generated by the atmosphere is about 7C per kilometer or 70-80C. This is well established, and as Feynman said :
It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.

beng
March 19, 2010 9:45 am

********
Brian G Valentine (06:06:09) :
Thank you Anna. Somebody tried to tell me that the “work” expended to accomplish this is the pV work done “by” the atmosphere; that certainly cannot be.
********
I had alittle trouble w/this issue until I thought alittle.
Work compressing a gas does cause it to rise in temp. But once the compression is finished, the gas will cool back to ambient. So a gas under static conditions (even at an extremely high pressure) w/no further work being done on it , doesn’t produce any heat & will return to ambient temp. Standard gas cylinders of compressed H2 are @ 2000 psi, but they’re at ambient temp once the heat of filling it (work) transfers away.
The earth does in fact do “work” of compression on descending air masses (high-pressure systems) & Chinook-type winds, but since the atmosphere is of a fixed mass & volume, air will have to rise somewhere else & then “cooling” occurs, as in low-pressure systems. The temp effects balance out & for the whole earth are zero. The earth as a whole isn’t doing any net “work” on the atmosphere.

JAE
March 19, 2010 10:02 am

“The amount of warming generated by the atmosphere is about 7C per kilometer or 70-80C. This is well established, and as Feynman said :
It doesn’t matter how beautiful your theory is, it doesn’t matter how smart you are. If it doesn’t agree with experiment, it’s wrong.”
For Feynman’s sake where is the empirical evidence that “well-establishes” this?

JAE
March 19, 2010 10:23 am

beng: what you say is accurate. But don’t forget that a gas at, say 20 C, possesses a lot of potential and kinetic energy. This is stored energy, which manifests itself as heat and causes the thermometer to read 20 C. A column of air several miles high has to have a lot of stored heat in it to exist. In fact, the amount of such stored heat is probably equivalent to the “greenhouse effect.”

Steve Goddard
March 19, 2010 11:42 am

JAE,
Nights in the desert are much warmer when it is humid, due to the increased GHG.

March 19, 2010 12:02 pm

Steve Goddard
How can you separate out the radiation from the Sun from the radiation from the atmosphere?

Steve Goddard
March 19, 2010 12:24 pm

Bryan,
Good question. Solar radiation comes at much shorter wavelengths than radiation from the atmosphere.
http://www.globalwarmingart.com/wiki/File:Atmospheric_Transmission_png

March 19, 2010 12:31 pm

Willis Eschenbach (12:01:17) :
why is your reconstruction a degree warmer than theirs 500 years ago if their error bar is only a tenth of a degree?
There are random errors and systematic errors [e.g. depending on site distribution]. The systematic error is unknown and can [and probably does] explain the difference.
We’re not dealing with borehole scientists. We’re dealing with borehole acolytes.
I think you have a bad case of ‘confirmation basis’ [actually anti-confirmation], but to each his own bias 🙂
I would separate the technique [I think it will be useful on the Moon too – also no water 🙂 ] from current practitioners [or at least the ones you have had a run-in with].

HankHenry
March 19, 2010 12:54 pm

Beng-
“The cold deep-water is a remnant of the previous ice-ages. ”
This occurred to me but I find it surprising. If heat were able to conduct through seawater at just 1 meter per year and the average ocean depth is just 3.7 km that would only take us back to a couple thousand BC, and I usually think of the last ice age as ending something like 10,000 BC. I wonder if there could be another phenomenon having to do with a pressure gradient and/or density at work “fighting” against downward heat conduction in the ocean?
If it is true that deep ocean coldness is an ice age remnant, it raises more questions. Now I wonder in maintaining the equilibrium you mention how much heat is extracted in a year and whether it is something worth putting into the models…. or is it all a subsystem that can be disregarded?

JAE
March 19, 2010 1:45 pm

Steve Goddard:
“Nights in the desert are much warmer when it is humid, due to the increased GHG.”
Some data would be nice here. In summertime, it is much warmer at night in the dry desert (say, Phoenix) than it is at humid areas at the same lat. and long. (say Atlanta). Why?

JAE
March 19, 2010 1:48 pm

Steve Goddard (10:57:07) :
?? I fail to see how the lapse rate supports your position. I don’t understand the other reference, so if it proves anything, I cannot tell.

JAE
March 19, 2010 2:35 pm

Willis:
“If you have data on this question I’d love to see it …”
It’s mostly from experience that I say it. I don’t know where to get long-term daily temp., humidity, and cloud-cover records. However, if you look at the Phoenix weather for February and the first 18 days of this month, selecting only those days which have a cloud cover of 0 or 0.1 (11 days during the period), you will find that the high daily temperature varies betwee 85 F to 46 F, and the daily low relative humidity readings vary from 13-29 percent. There is a slight NEGATIVE correlation between the high temp. and humidity (r^2 = 0.33). At night there is also a negative correlation–it is generally warmer when the r.h. is LOWER (r^2 = 0.46). Not a lot of data, I know, but something to start with….
Data here: http://www.nws.noaa.gov/climate/index.php?wfo=psr

JAE
March 19, 2010 2:55 pm

Nuts, I used relative humidity, rather than absolute humidity. When abs. humidity values are used there is essentially no correlation between daytime high temperatures varying between 66-85 and daytime low abs. humidities varying from 2.9-5.9 (essentially a doubling of the amount of water in the air had no discernable effect on temperatures). Same with nighttime.

kwik
March 19, 2010 2:56 pm

Doesnt the earth surface, and all the layers upwards obey Zeroth Law of Thermodynamics?
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/thereq.html
And doesnt that mean CO2 cannot increase the temperature?
Except, say if Co2 increased to the double amount overnight, the temperature would increase for a short time, and then fall back to the original value?
Or have I totally misunderstood the situation?

JAE
March 19, 2010 5:16 pm

Willis:
“However, as you point out, at those times we also get evaporation and sometimes clouds … which makes it very hard to separate out the effects.”
I hope you noticed that these were all virtually cloudless days (less than 10% cloud cover). And most of the days were in sequential periods, indicating no rainy periods.
“Thanks, JAE. It’s very hard to prove that way. The difficulty is that deserts get rain. When the residual moisture in the soil evaporates, we get water vapour, which tends to hold the heat in.”
Within this small amount of data, there is certainly no indication that more water vapor meant more heat, either in the day or the night. I’ll try to get more data somewhere.
Note, also that the amount of water vapor in the air down where you are is about 6-7 times what it is in Phoenix this time of year (during the day). Yet your temperatures are about the same… The negative feedbacks evidently keep any “greenhouse effects” well in check.

Colin Davidson
March 19, 2010 6:18 pm

Apologies if this has already been covered.
Joel Shore (19:30:38, 18MAR10) wrote:
“What they are saying is that the direct effect of the IR absorption by CO2 will be to increase the TOA downwelling radiation by 3.7 W/m^2 but that the total increase, once you consider all the feedback effects will be an increase in the TOA downwelling radiation by almost 3 times this (where I am using the result that a 3.7 W/m^2 increase in TOA downwelling radiation produces a temperature rise of about 1 C, as calculated by the S-B Equation). ”
and Bill Illis (21:46:42, 18MAR10) wrote:
“- then there is feedbacks from increased water vapour and albedo which add another 6.5 to 7.4 W/m^2 after this initial forcing;
– the layer now increases in temperature by 3.0C (11.5 extra watts on top of the original 240 watts).
– the layer where the equilibrium emission temperature of 255K occurs is now 461 Metres higher.
– the adiabatic lapse rate of 6.5C/km stays intact and the surface warms by the same 3.0C.
So, the sensivitiy is calculated as 4.0 extra watts of GHG forcing eventually results in an increase of 11.5 watts (once all feedbacks occur) and temperatures rise 3.0DegC”
I don’t agree with either of these statements.
The Surface Balance Equation at Equilibrium (I’m not claiming we are ever at equilibrium, this is the simple KT model) is:
Absorbed Solar Radiation + Back IR Radiation from the Atmosphere = IR Radiation from the surface + Latent Heat in Evaporated Water Vapour + Direct Conduction to the Air from the Surface.
Differentiating, again at equilibrium,
Change in Forcings (absorbed solar +back Radiation) = Change in Radiation from the surface + Change in Evaporation + Change in conduction (assumrd to be nil, as at equilibrium the relationship between the surface and the boundary layer is likely to be unaltered for any small change in temperature).
Cranking in the numbers:
Change in Forcing = 4sT^3dT + 78xdT. (s is 5.67×10^-8, dT is change in temperature, x is percentage increase in evaporation per DegC).
The big unknown in this equation is x, the increase in ebaporation. The Clausius-Clapyeron equation gives an upper limit of 6.5%, and some scientists think it is as low as 2%. It’s not 0%.
For x=2% the surface sensitivity is 0.15DegC/W/m^2, and the Change in Forcing is 22W/m^2 if the temperature rises by 3DegC.
For x=6.5% the surface sensitivity is 0.095DegC/W/m^2 and the Change in Forcing is 32W/m^2 for a temperature rise of 3DegC.
I don’t know what the relationship of the forcing at the tropopause (11.5 W/m^2 according to Bill) is to Surface Forcing. But to get 3DegC at the surface you need to double or triple that already trebled Radiative Forcing. (The assumption that the Lapse Rate stays constant is no help – the Surface is massively imbalanced. It cannot maintain the 3DegC increase without additional flux, and it can’t get that from an inflexible Lapse Rate.)
I don’t know what forcing was assumed by Joel. If it’s say 12W/m^2, then you get around 1DegC at the surface. But again you need to double or treble that to get 3DegC.
What am I missing here? I get a really insensitive surface and a large mismatch in surface forcing to tropopausal “Radiative Forcing”. Is there a problem with the no-change-in-conduction assuption (It would have to decrease by 50%)?

Alan D McIntire
March 19, 2010 6:35 pm

Regarding Pollock and his boreholes:
Here’s a 1997 paper reporting boreholes showing an early holocene climate optimum, medieval warm period, and little ice age
http://www-personal.umich.edu/~shaopeng/97GL01846.pdf
a summary from 1998 addressing only the 20th century
http://www.sciencemag.org/cgi/content/abstract/282/5387/279
and a summary of a paper from 1999 addressing only the warming since the Little ice age
http://www.nature.com/nature/journal/v403/n6771/abs/403756a0.html
It looks like an example of “playing to the global warmers”, papers since 1998 only address the period since the little ice age
– evidently these warming results more easily make it iinto the approved journals

Bill Illis
March 19, 2010 6:46 pm

“HankHenry (12:54:15) :
Beng-
“The cold deep-water is a remnant of the previous ice-ages. ” ”
The deep ocean temperature is a reflection of the past several hundred years of the coldest, densest ocean water there is on the planet.
The coldest densest ocean water occurs in the thermohaline ocean sinking regions near the sea ice pack and under the sea ice pack in the Arctic and around Antarctica.
Ocean water is most dense at about 1.0C with high saline content. Fresh water is most dense at 4.0C but with the introduction of very high salt content, the density can maximize close to 0.0C.
If ocean water gets colder than the maximum density combination of temperature and saltiness, it becomes less dense and rises. If it gets warmer than this or less salty than this, it also becomes less dense and rises. If it freezes, it becomes the least dense water of all and it floats to the top or it only occurs at the surface in the first place. The saline content will then fall out of it as well and it becomes even less dense.
Even in the ice ages, the deep ocean temperature probably did not go much lower, maybe 0.0C, and it has never, ever in the history of the planet, been lower than this. The entire ocean has to warm up, especially in the polar regions for the deep ocean temperature to go higher than today. In other words, there can be no sea ice, even in the winter, to increase much from today’s level.
In the paleoclimate, there are estimates of the deep ocean temperature being as high as 10C above today’s level. That means, even in the winter, the poles did not go below 10C.

Steve Goddard
March 19, 2010 6:48 pm

JAE,
One reason Phoenix is warm at night is because of the humidity from lawns, golf courses, etc. I have seen dew points in excess of 70 degrees there.
Also, why do you think it is extremely cold on the top of Mt. Everest? Could it have anything to do with the lack of atmosphere?

Joel Shore
March 19, 2010 7:04 pm

Willis Eschenbach says:

If I am following your argument, you misunderstand the “per CO2 doubling”. This is just how the IPCC tends to measure climate sensitivity. It has nothing to do with how much of any given forcing is from CO2. It is merely a unit of measurement.
For example, we could say that climate sensitivity is 1°C/W-m2, or we could say that climate sensitivity is 3.7°C / CO2 doubling. There is absolutely no difference between those statements. They are just expressed in different units. The conversion factor is
1 CO2 doubling = 3.7 W/m2
or
1 W/m2 = 0.27 CO2 doublings
So we can convert from one unit to the other, and it means nothing about whether we are talking about cloud forcing or H2O forcing or CO2 forcing. It’s just different units.

I suggest you read what Bill Illis wrote in his comment on 18/03/2010 (21:46:42). At the end of the day, the original forcing of ~4 W/m^2 from doubling CO2 results in another ~7 W/m^2 of “forcing” (in the sense of change in downwelling radiation at the TOA).
The IPCC is not saying that ~4 W/m^2 increase in downwelling radiation at the TOA leads to a 3 C temperature rise…They are saying that ~4 W/m^2 of forcing results in feedback that increase the downwelling radiation by considerably more than that.

Joel Shore
March 19, 2010 7:14 pm

Francisco says:

From that chart, you see that total amount of carbon in the extant fossil fuels still in the ground is roughly 1/8 the amount of carbon in the active system. The amount we are recycling annually back into the system is about 0.015% of the total carbon in the system.

That is not really relevant as the rate of transfer to the deep ocean reservoir is rather small. So, you have to consider the amount we are adding to the other reservoirs (that exchange fast enough that they can be thought of as a rapidly-equilibrating subsystem that exchanges carbon only very slowly with the deep oceans. Also, note that if you add only 1% annually to a reservoir that is not able to get rid of it, you will double the amount in that reservoir in 100 years.

The cause for alarm, it seems to me, comes only if you adopt the assumption that our modest recycling program represents a very grave disturbance to a static and balanced system, whose balance coincides with its state around 1850. I consider this assumption to have no rational or scientific basis whatsoever. It is merely ideological.

Ah…Do you mean no basis beside the overwhelming empirical evidence that we have increased the levels of CO2 in the atmosphere from the pre-industrial baseline of 280ppm to the current ~390ppm?

To begin with, the carbon we are releasing was once part of that dynamic system. We are recycling it back into life at what appears like a reasonable rate.

Reasonable compared to what? And, the fact is that CO2 levels on geological timescales has been much higher than they are now, with the correlated effects of warmer temperatures, higher sea levels and everything else.

A few years ago there was some talk of a “missing sink” to refer to the apparently surprising fact that about half the CO2 we release does not accumulate in the atmosphere.

I think you are woefully misunderstanding the science. There was some confusion regarding exactly what fraction of that carbon that is not accumulating is going into the ocean and what part into the biosphere (and soils). There was not confusion about the basic fact that about half of it was not remaining in the atmosphere.

I think it could probably be demonstrated that even a doubling of CO2 concentrations from current levels is extremely unlikely, if not impossible, as a strict consequence of our contribution.

Well, you are welcome to believe what you want to believe but it certainly isn’t a belief based on the scientific evidence.

Joel Shore
March 19, 2010 7:32 pm

Brian G Valentine says:

I understand where this is coming from, Bill, but where I (and apparently Gerlich and Tscheuscher and a lot of others) have the worst difficulty, is accepting the idea that the “atmosphere” is “warming” the “surface”.
Either that idea violates the Second Law or it doesn’t. If someone could just present ONE parallel example to show this is at least possible, then I would not say that Gerlich and Tscheusher’s analysis is as good as anyother application of the Second Law

Debunking of G&T’s Second Law argument in 3 easy steps:
(1) The Second Law statement about heat flowing from hotter to colder applies to net flows; it does not mean that no heat can be radiated by a colder body and absorbed by a hotter one. This should be obvious from an understanding of the basic statistical physics from which the Second Law derives, but even if one is ignorant of that, we know that all bodies radiate when they are at a nonzero temperature; if you have a hotter body near a colder body, the colder body is not going to magically detect the hotter body and stop radiating. The Second Law is not about magic. The Second Law means simply that the colder body will absorb more radiation emitted by the hotter body than the hotter body will absorb radiation emitted by the colder body.
(2) All reasonable models of the atmospheric greenhouse effect, whether toy models like Willis’s steel greenhouse or Trenberth and Kiehl’s energy budget diagrams or line-by-line radiative transfer calculations, have the net flow of heat being from the warmer earth to the colder atmosphere, as the Second Law requires. If you believe otherwise, show us.
(3) Despite the fact that the net flow is from earth to atmosphere, the effect of an IR-active atmosphere is still to produce warming relative to the case of an IR-transparent atmosphere. This may be the point that trips the most people up….It seems like an almost unstated assumption in many arguments of a Second Law violation that in order for the greenhouse effect to occur, the net flow must be from atmosphere to Earth. ****THIS IS WRONG.**** The reason that this is wrong is because in the comparison case, of an IR-transparent atmosphere, all the radiation emitted by the earth escapes into space. Hence, anything that causes some of that radiation to be returned to the earth will cause warming relative to that case. The greenhouse effect just says that SOME of the radiation that would otherwise escape from the Earth into space finds its way back to the Earth; it makes no claim that the amount that finds its way back is greater than the amount that the atmosphere absorbs from the earth, as such a claim would be a violation of the Second Law (and just plain silly to boot).

Joel Shore
March 19, 2010 7:36 pm

Bryan:

I would invite anyone who has any doubts on the matter to go to Arthur P Smith website and the thread “the arrogance of Physicists”.
Arthur has great difficulty coping with Fred Staples and Terry Oldberg never mind G&T

Arthur has troubles with them in the same way that I would have troubles with a particularly dense and opinionated student who had all sorts of wrong-headed ideas about physics and did not have the necessary combination of knowledge, intellect, and ability to listen to things that go against what they want to believe to understand when I tried to explain things to them. I suppose that someone reading the exchange who does not have the necessary background to evaluate the scientific arguments might find Fred Staples or Terry Oldberg’s arguments compelling; those of us who do, do not find them at all compelling and are actually somewhat amused to hear that people take them at all seriously.

Brian G Valentine
March 19, 2010 8:09 pm

Joel, I appreciate very much your understanding and interest of the forcings at the tropopause and the associated feedbacks. I really do, I know you have labored and contributed extensively. Your understanding of the climate sensitivity corollaries is as extensive as anyone I know.
But if we come back to the same old problem above, Joel, this trying to move heat from cold to hot without compensating work … we know that space is something of a homogeneous radiation source or sink at a temperature of like 2.9K – one looks at that and thinks, the heat has to be going the other way …
The bottom like to all the feedback stuff, Joel, is that it doesn’t look right, what more can anyone say?

JAE
March 19, 2010 8:35 pm

Steve Goddard (18:48:19) :
“One reason Phoenix is warm at night is because of the humidity from lawns, golf courses, etc. I have seen dew points in excess of 70 degrees there.
Also, why do you think it is extremely cold on the top of Mt. Everest? Could it have anything to do with the lack of atmosphere?”
Well, forget Phoenix. and look at Daggett, California, then: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/23161.txt
You won’t find many lawns there.
I don’t understand the comment about Mt. Everest.

Steve Goddard
March 19, 2010 8:39 pm

JAE,
The reason it is cold at high elevation is because there is less atmosphere above to keep it warm.

JAE
March 19, 2010 9:04 pm

Folks, MY bottom line is that I have been looking for an empirical demonstration of the “atmospheric greenhouse effect” for a long time, and I still have found absolutely NO empirical demonstration of the phenomenon, despite the fact that many learned scholars have expounded upon the theory for over 100 years. I see a lot of really cool ideas, including the K&T diagram that Willis provided above. The theory looks good and offers a nice tidy way to explain some things. The problem is that it does NOT explain many other things, which I mentioned in my earlier posts. It is simply not any hotter in areas that have copious amounts of greenhouse gases than it is in areas which have very minor amounts of greenhouse gases. WHY? Ahrrenius provided a THEORY, not a FACT. And nobody to my knowledge has done better.

anna v
March 19, 2010 10:18 pm

Re: Dave Springer (Mar 19 05:37),
You say elsewhere that the temperature on the moon is measured at -23C at a depth of 50cm.
All black body calculations have to do with skin surface temperatures. Even a few mm down from the skin, be it water or ground, infrared is blocked and only convection/contact works.
Thus one should go from the average skin temperature and calculate how long it takes to heat a body at near absolute zero up to -23C by conduction from a surface heated by the sun with an average ( day night year) temperature of “xx” which you do not quote.
This is not a simple problem, for all I know it might take a billion years, as regolith is a good insulator. I was not able to find a global skin surface temperature for the moon, only calculations.
My point is that the -23C at 50cm argument is not relevant to the discussion.

anna v
March 19, 2010 11:20 pm

Re: Steve Goddard (Mar 19 09:37),
The amount of warming generated by the atmosphere is about 7C per kilometer or 70-80C. This is well established,
I am puzzled. Not by the temperature gradients, but of the claim it is the atmosphere that is generating the heat to produce the gradient. Link please?

gbaikie
March 19, 2010 11:49 pm

“Once again, see my post, “The Steel Greenhouse”. It shows that the greenhouse effect has nothing to do with gas, or lapse rate, or any of that. If you have problems with the concepts there, let me know.”
I don’t think it would be it would be twice the temperature on surface planet as compared to the temperature of the outer shell. As you said:
“Figure 1. Building a steel greenhouse. (A) Planet without greenhouse. Surface temperature is 235 W/m2 heated from the interior. (B) Planet surrounded by steel greenhouse shell. Shell radiates the same amount to space as the planet without the shell, 235 W/m2. It radiates the same inward, which warms the planet to 470 W/m2 (29 C, 83°F, 302 K).
I agree the that outer shell radiates the same amount of energy. And we will disregard the rather insignificant increase of the area of the outer shell.
Instead the surface temperature could depend upon how the energy is conducted to the outer shell- conduction, convection, and/or radiation.
It make simple; let’s assume it’s a complete vacuum- and so only allowing radiation.
In that case the surface of the planet should be the same temperature as the shell.
Now, let’s return to the Moon- which is a nearly perfect vacuum.
What temperature would the moon surface have if it was a light year or two from the Sun?
I would say the surface temperature would be close to the present temperature in it’s deep polar craters which are never exposed to sunlight. This surface temperature is well above the temperature of 4 kelvin which is said the be the universe’s temperature. I don’t want to argue about what the temperature exactly is- say somewhere around 20 to 100 kelvin and say that at least 20 Kelvin of that heat is from the Moon’s core heat.
So if we put the Moon a couple light years from the Sun and say it’s surface is somewhere around 20 to 30 K. And then we put a steel shell around it- the shell would be around 20 to 30 K and the surface would remain at 20 to 30 K. And it doesn’t matter how many shells you put around it, it will still be 20 or 30 K at the surface. And you hang a mile of fiberglass insulation from the sphere- it of course wouldn’t make any difference. Or if you made a perfect mirror inside the sphere it would not make any difference.
But if you increase the surface area on the outside of the sphere- put spikes on it, or simply have bumpy surface- and that could make a difference [you could double or triple the surface area]- so you could lower the outside temperature of the sphere. Though of course it can not get below 4 Kelvin.
Though, since a sphere can be made into a smooth exterior surface and the Moon [or any planet] would be rougher surface [and therefore more surface area], such reduction in surface area would reduce it’s ability to radiate heat and therefore increase the temperature on the lunar surface.

Spector
March 20, 2010 12:01 am

I would think that one possible indicator of changes in the ground-level greenhouse effect, if any, would be a year-by-year listing of the coldest near sea-level temperature recorded for each of those years in the Arctic dark zones. I believe it would be the clear cold quiet periods in this region that would be most affected by changes in the classical greenhouse effect.

gbaikie
March 20, 2010 12:01 am

Said differently make a box a meter cubed. Have it have a sandwich of vacuum and thin sheets of metal. Say have sheet 2 mm thick separated by 2 mm of vacuum- so that gives you 250 “shells”. Put this on the ground on the moon in the shade- and it will not increase the temperature of the ground it is on.

Steve Goddard
March 20, 2010 12:59 am

anna,
The only significant heat source in the climate system is the sun. A coat keeps you warm without generating any heat.
Why are you puzzled by my statement, but apparently not by the premise of this article?

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.

March 20, 2010 1:09 am

Joel Shore (19:32:16) :
we know that all bodies radiate when they are at a nonzero temperature
I was being lectured that a white body [emissivity = 0] does not radiate at any temperature [100C, 400C or millions of degrees] …

March 20, 2010 3:00 am

Joel Shore (19:36:57) :
Bryan:
I would invite anyone who has any doubts on the matter to go to Arthur P Smith website and the thread “the arrogance of Physicists”.
Joel, Arthur was asked by Fred what difference would be made to the thermodynamics of the atmosphere if it was composed entirely of N2 and O2 and he replied that he thought it would be ISOTHERMAL from the surface up to a height of 30Km.
Not only does Arthur seek to contradict the 2nd law of THD but now also the Ist law of THD and the slight matter of abolishing gravity thrown in for good measure.
Yet you find this type of Physics perfectly OK!
Fred Staples and Terry Oldfield have a background in heat transfer thermodynamics.
I would suggest that before Arthur writes more nonsense he should attend some thermodynamics classes

March 20, 2010 4:27 am

Correction
Name above should read Terry Oldberg not Terry Oldfield

beng
March 20, 2010 6:26 am

********
Bill Illis (18:46:57) :
“HankHenry (12:54:15) :
Beng-
“The cold deep-water is a remnant of the previous ice-ages. ” ”
The deep ocean temperature is a reflection of the past several hundred years of the coldest, densest ocean water there is on the planet.

*******
That’s prb’ly right, Bill, tho I’ve read it takes a couple thousand yrs for the deep ocean to recycle. So the cold deep water isn’t exactly a remanent of the last ice-age (10000 yrs ago), but it is of the last thousand or so yrs of cold polar weather and sinking cold water there, which continues today. Some study I read said that before Antarctica was glaciated (before 35 mil yrs ago), the deep water was as warm as 60F (15C).
*******
HankHenry (12:54:15) :
Beng-
“The cold deep-water is a remnant of the previous ice-ages. ”
This occurred to me but I find it surprising. If heat were able to conduct through seawater at just 1 meter per year and the average ocean depth is just 3.7 km that would only take us back to a couple thousand BC, and I usually think of the last ice age as ending something like 10,000 BC. I wonder if there could be another phenomenon having to do with a pressure gradient and/or density at work “fighting” against downward heat conduction in the ocean?
If it is true that deep ocean coldness is an ice age remnant, it raises more questions. Now I wonder in maintaining the equilibrium you mention how much heat is extracted in a year and whether it is something worth putting into the models…. or is it all a subsystem that can be disregarded?

*******
Hank, read my response to Bill above — the cold deep-water is prb’ly no more than a thousand or so yrs old, but still is an accumulated remanent of cold polar regions & resulting sinking water. So your thermal-conduction rate is prb’ly pretty close. But more cold water is sinking & replacing it in polar regions even now in the interglacial.
As far as “fighting” conduction — no such action is necessary, just relatively simple heat-transfer laws. Thickness of the insulating layer is the key — kilometer-thick, static layers slow down conduction to a crawl. That’s the same reason the ground isn’t hot — a few kilometers of insulating crust keep us shielded from 2000C molten rock.
And as far as I know, the deep cold-water isn’t included in GCM models — even ocean-heat content (OHC) measurements only go down several hundred meters (700 meters?), so basically water below that is ignored. Not sure if this is reasonable or not.

Joel Shore
March 20, 2010 6:58 am

Brian G Valentine says:

But if we come back to the same old problem above, Joel, this trying to move heat from cold to hot without compensating work

No, it is not, as I explained in my post of (19:32:16). You can’t just assert that something violates the Second Law because you just feel that it does. You have to demonstrate that it does. I have explained why it doesn’t.
Leif Svalgaard says:

I was being lectured that a white body [emissivity = 0] does not radiate at any temperature [100C, 400C or millions of degrees] …

Do you know of any object in nature that is an ideal white body? At any rate, I don’t think it is vital to my argument that such an ideal white body (or something very close to it) exists does not exist. I was just interested in the transfers of radiation between two bodies at different temperatures.
Bryan says:

Joel, Arthur was asked by Fred what difference would be made to the thermodynamics of the atmosphere if it was composed entirely of N2 and O2 and he replied that he thought it would be ISOTHERMAL from the surface up to a height of 30Km.

Well, I would have to think more about whether I agree with Arthur on it being isothermal (if that is really what he said). However, the important point is that some people seem to think that one can create higher temperatures at the earth’s surface simply by virtue of the pressure, even if the atmosphere is transparent to IR radiation. And, the point is that this is wrong. The earth still has to satisfy the first law of thermodynamics, which means that in steady-state the amount it radiates away has to be equal to the amount of energy it receives and this is what would determine the temperature at the surface in an IR-transparent atmosphere where all of the radiation from the surface will escape into space.

Brian G Valentine
March 20, 2010 7:19 am

“I suppose that someone reading the exchange who does not have the necessary background to evaluate the scientific arguments might find Fred Staples or Terry Oldberg’s arguments compelling; those of us who do, do not find them at all compelling”
The same old worn out thing, thrown at people who disagree, I hate that, damn it

Joel Shore
March 20, 2010 7:32 am

Brian G Valentine:

The same old worn out thing, thrown at people who disagree, I hate that, damn it

Well, look at it from our point of view: There are some people who, by virtue of their lack of background combined with their own preconceptions based on strongly-held beliefs, one seems unable to reach by scientific arguments, no matter how hard we try. When they say, “Convince me that…” they are asking us to do the impossible, which essentially means they are just wasting our time.
In the end, many people will believe what they want to believe…and there is only so much one can do to change minds.

anna v
March 20, 2010 7:45 am

Re: Steve Goddard (Mar 20 00:59),
I have expressed my questions with respect with assumptions in posts up ways.
The coat keeps me warm because I am the heat source and it is an insulating body, i.e. its heat capacity and conductivity of heat and permeability to air currents is such that it delays the cooling of my body by radiation, conduction and convection (lets not enter into the biological system that keeps me at 37C playing havoc with entropy).
Generally insulation delays equilibrium with ambient temperature, but if there is no heat source, ambient wins.
The difference between moon conditions and earth conditions makes it self evident that the atmosphere plays such a function for the surface of the earth, miss named “a greenhouse effect”, and misused with CO2 runaway imaginations.
In the same way that the coat does not heat me, the atmosphere does not heat nor create the temperature gradients. It is gravity and the thermodynamic and dynamical properties of atmospheric gases (including convection precipitation evaporation etc) that define the temperature gradients that the input sun energy and radiating earth generate in the atmosphere.

March 20, 2010 8:51 am

Gerlich and Tscheuschner have further developed their ideas regarding atmospheric physics. March 2010
Readers will require a background in Physics to follow the article completely. Consensus advocates are now getting a bit reluctant to fully engage G&T on matters such as heat transfer and radiative balance.
arxiv4.library.cornell.edu/pdf/1003.1508

HankHenry
March 20, 2010 9:16 am

Beng – Thank you for the knowledgeable and thoughtful response. I started my comments in this thread with the remark, “Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold.” I think that is a true statement based on things I read about boreholes, the thermal gradient, and deep mines in South Africa. My comment probably doesn’t bear directly on the article at the top of the thread about radiative heat. It just strikes me as a peculiar phenomenon that nature presents us. Perhaps something of as much interest as the much vaunted and misnamed “greenhouse effect.” I also want to say there must be more at play than insulation. There seems to be some process actively removing heat from the deep ocean. I think that process is probably the “sinking and replacing” that you mention.

March 20, 2010 9:20 am

Joel Shore (06:58:15) :
“I was being lectured that a white body [emissivity = 0] does not radiate at any temperature [100C, 400C or millions of degrees] …”
Do you know of any object in nature that is an ideal white body?

A black hole is perhaps the closest to a perfect white body [as its emissivity is zero – quantum effects aside], although the concepts of temperature and entropy begin to lose their meaning here. There are also no ideal blackbodies.
At any rate, I don’t think it is vital to my argument
You are quite correct.

March 20, 2010 10:05 am

Joel
……The earth still has to satisfy the first law of thermodynamics, which means that in steady-state the amount it radiates away has to be equal to the amount of energy it receives and this is what would determine the temperature at the surface in an IR-transparent atmosphere where all of the radiation from the surface will escape into space…….
There is no such law as the conservation of radiation.
The energy that the Earth receives from the Sun can be accounted for in a number of ways for instance Chemical Potential Energy (Plants included), stored as internal energy in the oceans etc
The Planet can warm up and cool down over periods lasting several centuries and still be in harmony with the Ist law of THD if the Suns energy is stored as other energy forms on Earth.

Brian G Valentine
March 20, 2010 10:06 am

Well, look at it from my point of view: There are some people who, by virtue of their lack of background combined with their own preconceptions based on a strongly-held need to believe in AGW, one seems unable to reach by scientific arguments, no matter how hard I try. When they say, “Convince me that…” they are asking me to do the impossible, which essentially means they are just wasting my time.
In the end, many people will believe what they want to believe…and there is only so much I can do to change minds. But I’m not going to let nonsense and paranoia about AGW cause other people to suffer if I can help it.

Steve Goddard
March 20, 2010 10:36 am

anna,
If there was no atmosphere, there would be no temperature gradient through the atmosphere. Without a blanket you are cold. The sun warms the earth, and the atmosphere keeps the temperature up. Both are required to keep the earth livable.

Dave Springer
March 20, 2010 11:50 am

@anna v
Of course the lunar regolith temperature at 50cm is relevant. The surface temperature in full sunlight fluctuates by hundreds of degrees C from day to night. Every centimeter you drop below the surface the fluctuation decreases. By the time you get to 50cm there is no more measurable diurnal fluctuation and so what you’re measuring is predominantly the average surface temperature.
Lunar regolith is, by the way, a much better insulator than typical earth soil. You have to go much deeper into the earth before diurnal temperature swings become too small to measure.

anna v
March 20, 2010 12:44 pm

Re: Dave Springer (Mar 20 11:50),
Relevant to what?
The moon radiates heat in the infrared according to the formula flux= CxT^4
and that T must be the temperature of the first mm of surface . Because infrared does not penetrate either way more than the skin.
The temperature at 50cm is irrelevant to the heat radiated out. It is heat that is important, not a steady temperature. It fluctuates widely on the skin surface? Too bad for the calculations, but those are the numbers that take part in the problem.
The shell at 50cm, is isolated from the skin surface because the regolith is such a good insulator.

Dave Springer
March 20, 2010 12:51 pm

willis
The accusation of ignoring evidence was a device to goad you into responding. It worked and I’m not apologizing for it.
The main problem with your post is that you begin it with the impossible situation of the earth being a perfect black body with no atmosphere and say it would be “only” 8C cooler.
Yeah, but so what? The earth isn’t a perfect black body. Albedo plays a huge role in surface temperature. The heat flow experiments in Apollo 15 and 17 are excellent points of reference for what temperature realistic chunks of airless rock with non-zero albedos will attain at approximately 93 million miles from the sun.
Now you’re reduced to nit-picking by saying that two experiments deployed at mid-latitudes in full sun on the lunar surface aren’t enough to warrant consideration.
Well, here’s the deal with that. That’s still a heck of lot more warrant than your completely imaginary situation of a perfect black body. So there.

Dave Springer
March 20, 2010 1:44 pm

willis
To address your question of what’s up with the average between max daytime and nighttime surface temperatures being markedly different than subsurface average…
Someone else posted that and I’m not at all familiar with the reference it came from. The Apollo 15 and 17 heat flux experiments showed no such thing.
In fact there was no attempt to measure surface temperature in the heat flow rigs. It was only inadvertantly measured in the A15 setup because the astronaut couldn’t drill deep enough to bury all the thermocouples in one of the boreholes. As a result there were thermocouples (encased in a fiberglass rod) riding up a meter above the surface. Those above ground thermocouples yielded interesting information but it wasn’t regolith temperature, it was fiberglass temperature. In fact the most interesting thing they measured was a very small but gradual increase in nighttime minimum temperature. Several researches have speculated that the observed increase is due to increasing intensity of earthshine over the 5 years of data acquisition (1972 to 1977).
One might be interested in that as I believe back here on terra firma we had global cooling going on between 1972 and 1977. Increased earth albedo would give us global cooling and at the same time increase the intensity of earthshine on those inadvertantly exposed Apollo 15 ALSEP thermocouples.
Albedo in my opinion is what is really playing the big role in so-called climate change and a lot of the albedo change is in fact man-made by the deposition of black carbon (soot). Black carbon fits all the observations:
– it generates more heating at the surface than at altitude
– it has a limited range of effect of up to a few thousand kilometers from the source
– the sources are almost all in the northern hemisphere
– its effect is amplified by permanent snow cover as even during partial melts it floats to the surface getting darker and darker with each passing year
– recent experiments measuring earthshine on the moon show it decreasing
Even more interesting is the political ramifications of black carbon. The United States cleaned up its act with regard to black carbon decades ago. The primary offenders are Europe with their love of dirty deisel engines and poor countries that utilize slash/burn agriculture, heat their homes and cook their food by burning anything combustable in individual stoves and furnaces, and things of that nature. The United States then can’t be used for a scapegoat for global warming and in fact we would be the heroes instead of the villians. But NOooooooooooo. CO2 has to be the bogeyman so that the big wealthy United States gets the blame for it.

Colin Davidson
March 20, 2010 2:10 pm

Is The Surface Unbalanced?
In my previous offering (Colin Davidson 18:18:13, 19MAR10) I used the Surface Balance equation to derive the sensitivity of the surface to “Surface Forcing”, and also calculated what that “Surface Forcing” must be to maintain a surface temperature which is 3DegC warmer.
The numbers turn out to be: a sensitivity between 0.095 and 0.15 DegC/W/m2, and an increase in “Surface Forcing” of between 22 and 32W/m2.
Joel Shore (19:04:26, 19MAR10) said:
“At the end of the day, the original forcing of ~4 W/m^2 from doubling CO2 results in another ~7 W/m^2 of “forcing” (in the sense of change in downwelling radiation at the TOA).The IPCC is not saying that ~4 W/m^2 increase in downwelling radiation at the TOA leads to a 3 C temperature rise…They are saying that ~4 W/m^2 of forcing results in feedback that increase the downwelling radiation by considerably more than that.”
As I understand this statement, the additional TOA “Radiative Forcing” at equilibrium for a 3DegC warmer planet is in the region of 12W/m2.
But this is nowhere near enough to keep the surface at an additional 3DegC warmer. Even if there is no additional evaporation, there needs to be 16W/m2 of additional downwelling radiation at the surface to cover the increased Stephan Boltzmann IR radiation.
And it is unrealistic to assume no increase in evaporation if the surface temperature increases. After all, my limited understanding is that the feedbacks invoked by the IPCC rely on increased water vapour.
So the additional amount of downwelling radiation, 12W/m2 is completely inadequate to maintain a 3DegC increase in surface temperature. It is between half and a third of what is required.
If you don’t get 22-32W/m2 you don’t get 3DegC. If you only have 12W/m2 you get between 1 to 1.5DegC.
Are all the models perpetually unbalanced at the surface? This seems quite likely based on the numbers cited by Joel.

Dave Springer
March 20, 2010 2:21 pm

anna v
“relevant to what?”
Relevant to the average surface temperature of an airless rock with an albedo around 10% orbiting around 93 million miles from the sun.
Why would you think that NOT relevant to a discussion of what the temperature of the earth’s surface would be absent an atmosphere?
I can’t help you out if you don’t understand why a thermocouple buried just deep enough in the moon so that there are no variations in it from day to day or year to year is not indicating the average surface temperature.

Joel Shore
March 20, 2010 4:46 pm

Brian G. Valentine says:

Well, look at it from my point of view: There are some people who, by virtue of their lack of background combined with their own preconceptions based on a strongly-held need to believe in AGW, one seems unable to reach by scientific arguments, no matter how hard I try.

Well, I understand that in your state, you might feel that way. But, there are accepted ways that have worked well for our societies to distinguish between science and nonsense…and those scientific authorities have clearly come down on one side of the argument. And, in fact, you are so far out in the weeds on issues like G&T that even many who share your general view about AGW can’t stomach such nonsense.
Bryan says:

There is no such law as the conservation of radiation.
The energy that the Earth receives from the Sun can be accounted for in a number of ways for instance Chemical Potential Energy (Plants included), stored as internal energy in the oceans etc
The Planet can warm up and cool down over periods lasting several centuries and still be in harmony with the Ist law of THD if the Suns energy is stored as other energy forms on Earth.

Ah…240 W/m^2 is an awfully lot of energy to store away by the methods that you are proposing. Besides which, the various energy flows can be measured, maybe not down to the fractions of a W/m^2 that one might ideally like, but certainly well enough to dismiss any notions that the earth’s temperature can grossly violate the constraints of radiative balance at will.

HankHenry
March 20, 2010 4:58 pm

anna v
I worked my way through the video presentation on stefan boltzmann for undergraduates of Dr. David Archer over at realclimate. In those lectures he throws in this discussion about how much time it takes for energy (I’m not sure if you’d call it heat or radiation) to work its way through the atmosphere. I gathered those considerations were something apart and supplementary to just a discussion of stefan boltzmann proper. Is this as much as saying the earth’s atmosphere has an R value? I wonder what that value would be like – on an averaged out basis?

Joel Shore
March 20, 2010 5:12 pm

Colin Davidson: There is a reason why climate scientists like to talk in terms of the top-of-the-atmosphere and that is because the temperature at the surface is really governed by the top-of-the-atmosphere radiative balance and the processes that set the thermal structure of the atmosphere with height.
For example, in the book “Global Warming: The Hard Science” by L.D. Danny Harvey, he does a calculations where he shows that if you increase the downward infrared flux from the atmosphere at the earth’s surface by 10 W/m^2, you would only warm the surface by ~0.1 C (while cooling the atmosphere by 0.3 C). However, an externally-imposed 10 W/m^2 change in the greenhouse gas forcing produces (in his estimate, using an ~2 C per CO2 doubling climate sensitivity) about a 5 C surface temperature increase, i.e., a 50X larger response. The reason has to do with the atmospheric response: In the former case, the response is governed by turbulent heat exchange, which is very efficient in neutralizing the effect. However, in the latter case, the response is governed by the radiative damping to space, which is much weaker.

Brian G Valentine
March 20, 2010 7:53 pm

A thought experiment:
Joel, there was a day that you knew nothing about AGW. You never heard of it, but you had rational thought.
At that time, I held up a picture of the Earth and atmosphere in outer space. I pointed out temperatures to you at tropopause, Earth, and outer space. Then I asked, “which way is heat at the tropopause going?”
When you gave me the answer, “to space,” I corrected you, saying, “the downwelling radiation is an exchange with the warmer Earth, and increases the average surface temperature due to forcing at the tropopause”
… you would have taken me for a fool, trying to beguile you with misinformation
But then, Joel, people who you admired and evidently sympathized with, have sold you on an argument that you (I feel sure) would have taken for complete and utter nonsense.
Right out in the weeds, Joel

anna v
March 20, 2010 10:51 pm

Re: Dave Springer (Mar 20 14:21),
I do not know about thermocouples, what about a diode?
Lets try again:
Temperature is what enters the radiation flow equations.
The temperature of the skin.
Temperature is not conserved. Energy is conserved. The -23- degrees 50cm in the ground have a deformed and unknown connection with whatever the average of the skin is: conduction equations and constants.
Take the average numbers in
http://www.asi.org/adb/m/03/05/average-temperatures.html
107C daytime
-153 night average
add and divide by two, as is done in the GISS records of earth temperature, and you find: -28C.
The actual average will be different, integrated correctly, but not the -23C 50cm in the ground.

Colin Davidson
March 21, 2010 12:21 am

Joel wrote (17:12:14, 20MAR10):
“Colin Davidson: There is a reason why climate scientists like to talk in terms of the top-of-the-atmosphere and that is because the temperature at the surface is really governed by the top-of-the-atmosphere radiative balance and the processes that set the thermal structure of the atmosphere with height. ”
Don’t the energy fluxes at both boundaries have to be in balance? There is no mechanism for the TOA to warm the Surface. It has no thermal mass to speak of, and is at a very feeble temperature. Declaring that the Surface is warmed by “the processes that set the thermal structure of the atmosphere with height” is akin to saying that the surface temperature is created by magic.
The Atmosphere is mostly warmed from the surface upwards. It is not warmed from the TOA downwards. At the surface (in a Kiehl & Trenberth Diagram, and at equilibrium – both these things are simplifications so that the Physics does not get bogged down in all the other variables) only two forcings are available – absorption of solar radiation, and back-radiation from the atmosphere. If you increase the temperature of the surface, how does one balance the new KT diagram at the elevated temperature?
Only by increasing the sum of the two forcings by a massive 22 to 32 W/m^2 can the surface remain balanced. If it is not balanced then it will cool down or heat up until it is balanced, regardless of what the TOA is doing. Allowing for the (very disputed) IPCC positive feedbacks, Joel has indicated that the additional downwelling radiation is 12W/m^2 or so. That’s not enough to maintain the surface at an additional 3DegC. The most you get is 1.5DegC.
There may be something I have missed here (most likely) but I don’t think Joel has identified what it is. At present I am forced to conclude that anyone who predicts more than 1.5DegC for a doubling of CO2 has a surface imbalance hidden in his model.

lgl
March 21, 2010 3:59 am

Joel Shore (17:12:14) :
“if you increase the downward infrared flux from the atmosphere at the earth’s surface by 10 W/m^2, you would only warm the surface by ~0.1 C”
This sounds very strange. Has he developed a service pack for the S-B law?
Colin Davidson (00:21:13) :
How do you arrive at “a massive 22 to 32 W/m^2” ?

Joel Shore
March 21, 2010 6:46 am

Brian G Valentine says:

At that time, I held up a picture of the Earth and atmosphere in outer space. I pointed out temperatures to you at tropopause, Earth, and outer space. Then I asked, “which way is heat at the tropopause going?”
When you gave me the answer, “to space,” I corrected you, saying, “the downwelling radiation is an exchange with the warmer Earth, and increases the average surface temperature due to forcing at the tropopause”
… you would have taken me for a fool, trying to beguile you with misinformation

Yes, and if by heat you mean net heat (which is really the only logical thing you could talk about as flowing only in one direction) then I would to this day say that you are wrong. The net heat flow is in fact to space. It is in Willis’s steel greenhouse, it is in the Trenberth & Kiehl diagram, and it is in any other explanation of the greenhouse effect that I have seen.
It is just that the heat flow into space is less than it would be if the atmosphere were transparent to IR radiation (and the Earth’s surface were still at its current temperature).

Dave Springer
March 21, 2010 6:48 am

anna v
You don’t know what a thermocouple is? That is taught in high school physics. Very basic stuff. What grade are you in?

Joel Shore
March 21, 2010 6:50 am

Igl says:

This sounds very strange. Has he developed a service pack for the S-B law?

No…It is just that the earth’s surface can exchange heat with the atmosphere by other processes besides radiation. However, the earth system as a whole can only exchange radiation with space by radiation.

March 21, 2010 6:54 am

Joel
I take it you agree with me that there is no law of conservation of radiation!
……..but certainly well enough to dismiss any notions that the earth’s temperature can grossly violate the constraints of radiative balance at will….
But then you try to bend it back to some kind of law that does not hold unless you try to grossly violate it!
My boat uses a solar panel to keep my batteries charged up until I need to convert the chemical potential energy into some other form of energy of my choosing.
The coal that I use in my fire releases energy that was stored over 400 million years.
You could yourself think of countless other examples of EM radiation being converted into other energy forms.
The preoccupation by “consensus”advocates on radiation balance is leading them into serious errors.
Also on the topic of the millions of scientists beavering away fully united in support of a co2 greenhouse effect is as bogus as the effect itself.
What any sceptic finds is this theory is like an inverted triangle balanced on a tiny base.
The three sets of so called independent data supporting the theory is reduced to two because of the Phil Jones affair.
We then find that the other two were not as independent as claimed.
The sceptics choose a variety of methods to test the AGW theory.
Some do it by actually checking the data to see it it supports the theory.
Some do it by providing an alternative theory such as solar flares.
Some(mainly geologists) take a historical approach to find out if the present time is highly unusual.
Others like G&T check to see if the Physics behind AGW theory stacks up.
To expect all sceptics to be in total agreement with one another is really stupid.
I am from a Physics background and I must say I am much more impressed by G&T than any counter argument.
In fact some of those attacking G&T show serious signs of scientific ignorance.
If say a Physics Professor specialising in Heat Transfer Thermodynamics was to criticise their paper, I’m sure they would regard that as a serious challenge, but as yet this has not happened.

Brian G Valentine
March 21, 2010 7:28 am

Dave Springer, I don’t exactly care for that kind of tone directed at Anna because you’ve got no basis for it

Dave Springer
March 21, 2010 7:42 am

anna v
You made a mistake in your calculation:

Take the average numbers in
http://www.asi.org/adb/m/03/05/average-temperatures.html
107C daytime
-153 night average
add and divide by two, as is done in the GISS records of earth temperature, and you find: -28C

diurnal mean temperature swing = 107+153 = 260
half of diurnal change = 260 / 2 = 130
half of diurnal change added to nightime mean = -153 + 130 = -23
half of diurnal change subtracted from daytime mean = 107 – 130 = -23
Actual measured regolith temperature at 50cm depth was, unsurprisingly, a constant -23C.
QED

Joel Shore
March 21, 2010 11:05 am

Bryan says:

I take it you agree with me that there is no law of conservation of radiation!

My boat uses a solar panel to keep my batteries charged up until I need to convert the chemical potential energy into some other form of energy of my choosing.
The coal that I use in my fire releases energy that was stored over 400 million years.
You could yourself think of countless other examples of EM radiation being converted into other energy forms.
The preoccupation by “consensus”advocates on radiation balance is leading them into serious errors.

Look, you could also complain that climate scientists don’t include the effects of quantum gravity or the relativistic corrections to the speeds of winds and ocean currents. The question is whether this is at all relevant. I haven’t done the calculation myself but I’ll give you a hint from what I do know: All those fossil fuels that accumulated over hundreds of millions of years that we are now burning very rapidly (relative to the time that it took them to form) amounts to an energy usage averaged over the earth’s surface of ~0.02 W/m^2.
And, by the way, your solar panel only successfully converts a small fraction of that energy and what it stores gets converted eventually into heat when you run your boat. In the end, the net amount of energy you store in that battery averaged over the course of year will be very small (and will be either negative or positive depending on whether the battery is in a more or less charged state at the beginning of the year as at the end).
We are also measuring the various energy fluxes and, while we can’t yet measure them to the fraction of a W/m^2 that would be necessary to directly measure what radiative balance there is, I believe that they are measured accurately enough to say empirically that there is radiative balance within at least a few W/m^2.

I am from a Physics background and I must say I am much more impressed by G&T than any counter argument.
In fact some of those attacking G&T show serious signs of scientific ignorance.
If say a Physics Professor specialising in Heat Transfer Thermodynamics was to criticise their paper, I’m sure they would regard that as a serious challenge, but as yet this has not happened.

There are so many things I could say to this that I don’t know where to start:
(1) One doesn’t need to be a specialist in heat transfer thermodynamics to understand that G&T is nonsense.
(2) Most “skeptics” like to say that it is not the credentials that matter but rather the argument. So, why should G&T only take seriously those with such credentials?
(3) G&T are hardly the ones in the position to be deciding what constitutes a serious challenge to their paper. The fact that it was published at all is an extreme embarrassment and nobody in the field is taking it seriously. In fact, one of my motivations for participating in the comment debunking it was the notion that it was a public service so that the climate scientists themselves could focus on real science rather than wasting their time dealing with such nonsense. (I also felt a sense of duty as a physicist since the paper appeared in a physics journal and G&T make much of the fact that their physics viewpoint somehow allows them to see what others have missed.)
(4) As it happens, I am a physicist who has published papers in the area of statistical physics (which really provides the underlying foundations for thermodynamics) in some of the top physics journals in the world (e.g., Physical Review Letters). I have also done some heat transfer calculations in a more practically-oriented corporate research and development environment.

anna v
March 21, 2010 12:17 pm

Re: Dave Springer (Mar 21 07:42),
Well, I did make an arithmetic mistake
(-153+107)/2 makes -23.
(I was hurrying to keep an appointment, if that is an excuse for bad arithmetic)
I consider it serendipitous, or maybe the choice of 50cms was made because of the coincidence of the numbers ?
Temperatures on the Lunar surface vary widely on location. Although beyond the first few centimeters of the regolith the temperature is a nearly constant -35 C (at a depth of 1 meter), the surface is influenced widely by the day-night cycle.

so -23 is a choice, why not -35 as this reference says?
http://www.asi.org/adb/m/03/05/average-temperatures.html
BTW, I am a retired particle physicist doctorate and all.

lgl
March 21, 2010 2:10 pm

Joel Shore (06:50:50) :
But only 20% of the energy transport from the surface is non-radiative. Why would suddenly most of the added 10 W be returned as non-radiative? Still sounds very strange.

lgl
March 21, 2010 2:23 pm

… and 5 C after a CO2 doubling?? That’s 27.8 W/m2 more from the surface + the non-radiation, or a 7.5 amplification of the 3.7 W. I really don’t understand any of this …

March 21, 2010 4:37 pm

Joel
I can hardly believe some of the things you are writing.
You claim to be a Physicist yet you think there is such a thing as conservation of radiation.
Point me in the direction of one Physics book to back up your claim
….We are also measuring the various energy fluxes and, while we can’t yet measure them to the fraction of a W/m^2 that would be necessary to directly measure what radiative balance there is, I believe that they are measured accurately enough to say empirically that there is radiative balance within at least a few W/m^2………..
The total energy stored in fossil fuels over millions of years can be ignored apparently
… I haven’t done the calculation myself but I’ll give you a hint from what I do know: All those fossil fuels that accumulated over hundreds of millions of years that we are now burning very rapidly (relative to the time that it took them to form) amounts to an energy usage averaged over the earth’s surface of ~0.02 W/m^2…….
Please send this information to the IPCC
I have pointed you to elementary mistakes by A. P. Smith and you just brush them aside
You are completely contemptuous of G&T yet all you produce is a string of insults with no serious attempt to deal with any of the many points they make.
You claim to have produced more papers than Rupert Murdoch, so we will just have to take that on trust.
When and if your refutation is published we can then check just what the qualities of the papers were.
On present evidence of yourself, A.P.Smith, Eli Rabett and so on I think that G&T will not be losing any sleep.
I hope your paper will be read and given constructive criticism.
The usual way to resolve matters of scientific fact is to point to some experimental proof rather than a computerised projection.
The ability to hurl insults is inversely proportional to scientific proof.

Joel Shore
March 21, 2010 4:46 pm

Igl: No…In that calculation, Harvey assumed a 5 C increase from an imagined 10 W/m^2 forcing. Since a CO2 doubling is ~4 W/m^2 forcing, he is assuming a sensitivity of ~2 C per doubling.

Brian G Valentine
March 21, 2010 4:53 pm

Thank you, Joel.
So then at the tropopause, to make the surface temperature higher, the radiant heat of the Earth is
– reflected back to Earth at the tropopause (nope)
– absorbed by some molecules there and re-radiated …
… Which way. You tell me which way the radiation is going – up, or down.
The answer is: The radiation at the tropopause is going the opposite direction that AGW theory is going

March 21, 2010 6:46 pm

Willis –
I have to say I have read through this entire thread (I’m home sick with a bad cold), and I am impressed at how calm and professional you remain in the face of many reasons not to.
Great job.
In the interest of transparency, I will say that I have a PhDEE, I love Physics, have read several books on the subject of “Global Warming” including the CRUTAPE letters and very much appreciate this website. Hats off to all of you guys who work to stay objective.
Thanks,
Mike

Joel Shore
March 21, 2010 7:16 pm

Brian: I suggest that you re-read my post of (19:32:16) 19 March. It is actually not that difficult to understand if your mind is at all open to new things even if they go against your ideological predispositions.
To summarize: The net flow of terrestrial radiation is out. This still makes the temperature higher due to the greenhouse effect than it would be in an IR-transparent atmosphere because the net flow is less than it would be in that case.
I don’t know how I can explain it any simpler. There are also various toy models that you can play with to get the picture. Willis’s steel greenhouse is one. Another one, which is admittedly more abstracted from the actual system in question but is kind of fun is where 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 the 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.)

Joel Shore
March 21, 2010 7:32 pm

You claim to be a Physicist yet you think there is such a thing as conservation of radiation.

That’s not what I said; I suggest you might try reading it again.

The total energy stored in fossil fuels over millions of years can be ignored apparently

The point is that such energy storage happens only very slowly. Even though we are burning those fuels at a rate that far exceeds the rate at which they were formed, we are still producing only about 0.02 W/m^2 of direct heat by doing so.

Please send this information to the IPCC

The IPCC already knows this, which is why the current concern is over the changes that we are making to the chemistry of the atmosphere by burning fossil fuels that cause it to retain more of the sun’s energy rather than concern about the effect of the waste heat that we are producing by burning fossil fuels.

I have pointed you to elementary mistakes by A. P. Smith and you just brush them aside

I don’t see where you pointed to any such mistakes. The one specific thing that you questioned is whether Arthur is correct when he asserted that he thought an IR-transparent atmosphere would be isothermal. I would have to think more about that, but it is irrelevant to the main issue because, whether or not it is isothermal, we know in the case of an IR-transparent atmosphere what the surface temperature has to be.

You are completely contemptuous of G&T yet all you produce is a string of insults with no serious attempt to deal with any of the many points they make.

I guess you too missed my post of (19:32:16) 19 March where I explained very clearly where G&T go wrong in their claim of a violation of the Second Law. It seems strange to me that neither you nor Brian Valentine have been interested in engaging me on those points, seeming to prefer instead to go around attacking strawmen.

You claim to have produced more papers than Rupert Murdoch, so we will just have to take that on trust.

I don’t see where I gave you a specific number. But, if you want to see my papers, all you have to do is a simple google scholar search since I post under my real name here. (Additional fact: Middle initial is D.)

Colin Davidson
March 21, 2010 8:21 pm

Igl (03:59:11) asked: “How do you arrive at “a massive 22 to 32 W/m^2″ ?

Please see my post at 18:18:13, 19MAR10. In part:
“Cranking in the numbers:
Change in Forcing = 4sT^3dT + 78xdT. (s is 5.67×10^-8, dT is change in temperature, x is percentage increase in evaporation per DegC).
The big unknown in this equation is x, the increase in ebaporation. The Clausius-Clapyeron equation gives an upper limit of 6.5%, and some scientists think it is as low as 2%. It’s not 0%.
For x=2% the surface sensitivity is 0.15DegC/W/m^2, and the Change in Forcing is 22W/m^2 if the temperature rises by 3DegC.
For x=6.5% the surface sensitivity is 0.095DegC/W/m^2 and the Change in Forcing is 32W/m^2 for a temperature rise of 3DegC.”
The sensitivity at the hot, massive Surface is very different to the sensitivity of the thin cold atmosphere at the Tropopause. The system has to work much harder to heat the surface and keep it hot than it does at the Tropopause.

Colin Davidson
March 21, 2010 8:35 pm

Joel Shore (06:46:04, 21MAR10) said:
“It is just that the heat flow into space is less than it would be if the atmosphere were transparent to IR radiation (and the Earth’s surface were still at its current temperature).”
I think that statement may be wrong. At equilibrium (planetary and atmospheric temps stable) the energy flux radiated to space equals the energy flux received from space (neglecting any flux from the centre of the planet.), regardless of the state of the greenhouse.
My understanding is: Some of the radiated energy from the Earth (according to KT around 40W/m^2) is at the “window” frequencies and passes directly out to space. The rest of the energy which is going into the atmosphere (a small amount of net radiated energy, about 25W/m^2, plus the directly conducted energy about 25W/m^2 plus the latent heat in evaporated water about 80W/m^2, plus the absorbed sunlight about 75W/m^2) can only leave the planet by radiation. It gets radiated from two primary bands:
1. The band of water vapour. The effective level is somewhere near the top of the clouds. Most of the radiation (at least two thirds) is from here.
2. The CO2 in the mid Tropopause to Stratosphere. About 20-25% comes from here.

Brian G Valentine
March 21, 2010 9:19 pm

Here’s an even simpler toy model from about p200 of An Inconvenient Truth (and about all he has to say about how it works):
“CO2 acts like a blanket that keeps the Earth too warm.”
Tell me about it, Joel.

Brian G Valentine
March 21, 2010 9:21 pm

Folks, if some fish that Al Gore or Joel want to sell you don’t smell too good,
DON’T BUY

March 21, 2010 9:35 pm

Brian –
I know I am new to this thread – and to this site (I have been lurking for awhile while studying), but you cannot be seriously quoting Al Gore on thermodynamics or physics, rght? I mean really?
This is a guy who shows a glass of water with Ice in it, that does not protrude above the rim of the glass, and then shows it melting with the result that the water flows out of the glass onto the table the glass sits on…really? Hardly Mr. Wizard, I don’t think he is worth a quote. A Nobel Prize maybe, but that’s politics, not what is being discussed here.
Mike

anna v
March 21, 2010 10:13 pm

Re: Joel Shore (Mar 19 19:32),
Joel I agree with 1) and 2)
For
(3) Despite the fact that the net flow is from earth to atmosphere, the effect of an IR-active atmosphere is still to produce warming relative to the case of an IR-transparent atmosphere. This may be the point that trips the most people up….It seems like an almost unstated assumption in many arguments of a Second Law violation that in order for the greenhouse effect to occur, the net flow must be from atmosphere to Earth. ****THIS IS WRONG.**** The reason that this is wrong is because in the comparison case, of an IR-transparent atmosphere, all the radiation emitted by the earth escapes into space. Hence, anything that causes some of that radiation to be returned to the earth will cause warming relative to that case.
I come to think that it is the semantics that causes the confusion at this point.
The active energy role given to the atmosphere . Instead of “returned” one should state “retained” .
This misuse, as standard thermodynamics goes, is enhanced by the terminology of “forcings”.
Radiation is the dominant way energy leaves this ball and enters this ball, and can be used instead of total energy balance with small errors.
Within this ball, earth, the equivalence given and imposed to think in terms of “radiations” and “forcings” when a multitude of other heat transport involving processes exist, leads to errors, ( at least in temperature estimates) imo . It is wrong to state the problem in terms of Watts/m^2 as if all processes are perpendicular to the surface of the earth.
It is true that when one integrates to solve a problem, one can turn it into shells dx and consider the forces acting on this , but one does not go around making a terminology : the central part of the earth give A contribution to gravity, the magma adds B, the mantle C, in order to describe to the layman scientist how gravity works on the surface. ( remember the mechanics problem of a marble falling in hole that goes completely through the center of the earth to the other side?) It would be confusing and not useful for calculations involving gravity on the surface where people need a formula.
I believe this sort of confusion has been introduced by the climate terminology and makes it difficult to see the forest for the trees and introduces errors in actual calculations.
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.
So it may only be semantics, but thermodynamics says that a cooler body cannot warm a hotter body unless work is done. I can imagine the gigantic system of convection precipitation evaporation etc in the atmosphere/ocean system to be described as an equivalent heat pump where the energy is supplied by the sun and the the work is done by the winds gravity tides and the chemistry. A passive ball as is described with forcings and watts/m^2 and an atmosphere actively warming violates the second law , imo always of course.
CO2 has maybe a tiny role in this heat engine, by changing the heat capacity.

Colin Davidson
March 22, 2010 1:01 am

Igl wrote(06:50:50, 21MAR10):
“But only 20% of the energy transport from the surface is non-radiative. Why would suddenly most of the added 10 W be returned as non-radiative? Still sounds very strange.”
It is important to note that it is the NET radiative transfer (absorbed – radiated) which heats the air, not the absorbed radiation. So the actual transfer of energy from the Surface into the Atmosphere is:
Net Absorbed Radiation (= IR Radiated from the Surface(390) – The portion which is not absorbed and escapes straight to space (40) – The Back Radiation (324), the figures come from K&T 1997 from the IPCC AR4, Chapter 1) = 26W/m^2
+ Direct Conduction into the air = 24W/m^2
+ Latent Heat in Evaporated Water = 78W/m^2
So about one fifth of the energy transferred from the surface is radiated,
one fifth conducted,
and three fifths is in water vapour as latent heat.
When the surface warms further, evaportion increases, and the NET surface radiation absorbed by the atmosphere falls by exactly the same amount. At a 3 DegC increase, the evaporation increases by between 6 and 16 W/m^2 (scientists are unsure) to 84 to 94W/m^2, and the NET radiation drops by that amount to between 10 and 20W/m^2.

Colin Davidson
March 22, 2010 1:22 am

Joel Shore wrote (19:16:13, 21MAR10)
“The net flow of terrestrial radiation is out. This still makes the temperature higher due to the greenhouse effect than it would be in an IR-transparent atmosphere because the net flow is less than it would be in that case”
That is incorrect I think.
The net flow is identical in all cases of the atmosphere. Neglecting the flows into and out of the solids/liquids of the planet, (which for clarity only I ignore by claiming an equilibrium which I acknowledge never exists in practice) the incoming solar radiation is always balanced by the outgoing radiation.
This outgoing radiation comprises reflected sunlight, IR “through the window” direct from the surface, and IR from the gases in the atmosphere, primarily Water Vapour, and to a considerably less extent, CO2.
It is reasonably obvious that the higher one goes in the atmosphere the more chance a photon emitted by a water molecule will have of NOT being absorbed by a higher water molecule. If one is very low down then any radiation from water molecules is completely blocked by the overlying water molecules. It is only where the water vapour thins out (top of the clouds) that emissions from water vapour have a good chance of making it to space. (Note that the intensity from this layer is much higher than from the CO2, as it is relatively warm – around minus10DegC and water vapour has a much more exrensive spectrum).
Similarly for CO2, but in this case the extinction horizon is deep in the Tropopause, and the majority of emissions are coming from the stratosphere. (The gas is much colder, around -45DegC, and CO2 has less spectrum, so it emits much less energy than the water vapour).
If you increase the concentration of either gas, the level at which extinction occurs changes. For an increase in gas concentarition this is higher in the atmosphere. The resultant energy imbalance is called “Radiative Forcing” (or “Feedback” in the case of water vapour, which the IPCC doesn’t like very much). That imbalance causes local cooling or heating of the atmosphere until a balance is restored. This is called the “Greenhouse Effect”.
That’s my understanding, which means it is probably wrong!

March 22, 2010 1:40 am

Joel
Here is the direct link you require I hope you read the whole piece so as not to say I pulled this out of context.
Its easy to make a slip when responding in a blog but what alarms me is that Arthur did not take the chance to set the record straight.
It seems an elementary mistake – draw your own conclusions.
Try to cut down on the insult throwing.
If you think about it it is more the language of the football terraces than of science.
Rational debate and adherence to the scientific method is the only way to make progress.
…..In fact, I stated earlier that the atmosphere would be essentially isothermal, at the surface temperature. So your claim that I think the “temperature at top of the N2/O2 atmosphere would be about 30 degrees C below the surface” is wrong. Not that I see any significant logical conclusion you are drawing from the issue. http://arthur.shumwaysmith.com/life/content/the_arrogance_of_physicists

Colin Davidson
March 22, 2010 1:42 am

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.
But I said partially correct. Where I think the statement is sort of correct is that if water vapour concentration is increased then the air is actually carrying much more energy – due to the increased latent heat.
So I think it is untrue to say that if CO2 concentration is increased the heat capacity of the air is increased.
But if the surface temperature goes up and as a result there is more evaporation, then the air is holding more energy.
It’s quite interesting – as the surface temperature increases, less energy is being transported into the lowest levels of the atmosphere (by radiation and convection), and more is going to the middle levels (the clouds) in evaporated water vapour.

lgl
March 22, 2010 2:38 am

Colin Davidson (20:21:36) :
Sorry Colin. I did search your name… well, bad excuse
Then we pretty much agree. Correcting for the evaporation I’m ending in the area 1.3 C per CO2 doubling. So if K/T budget is right then IPCC is wrong. Cleary it can’t be us two being wrong 🙂
But I totally disagree with your (01:01:42) …

Dave Springer
March 22, 2010 4:10 am

anna v
I have no idea where they get the lunar temperate data in those quotes you keep using. The reference is not published by NASA and it contains no references to the actual data returned by the sensors in the boreholes.
The full data record is not public yet but NASA says it will be available for anonymous FTP download “soon”. If you had checked my original comment on this you’d find I got my data from the link below which has the data plotted at low resolution for Apollo 15 & 17. Apollo 17, by the way, recorded a constant regolith temperature of 18C. 23C was the constant temperature recorded at the Apollo 15 site. Apollo 17 was closer to the moon’s equator (20 degrees n. latitude vs. 26) and the regolith there was also darker, both of which combine to make the average surface temperature there 5C warmer. I used the Apollo 15 site as it’s more representative of the typical lunar surface in albedo and equidistant between pole and equator.
http://education.ksc.nasa.gov/esmdspacegrant/LunarRegolithExcavatorCourse/Chapter5.htm#OtherPhysicalProperties
That said, there should be no argument from any physicist that once the borehole depth is deep enough to smooth out the diurnal heating/cooling fluctuations the constant temperature obtained is the average of the surface temperature. This is pretty simple stuff. The calculations are not simple if you are trying to go backward from watts/meter^2 at the surface to the temperature at any particular depth. Then you need to know the thermal diffusion characteristics of the substrate and because any real world substrate is not going to be homogenous the calculated result will never be quite as good as a measured result. When you have a thermocouple buried at a certain depth you don’t need to do any calculations as the thermocouple is giving you the correct result of those calculations with the actual dirt doing the integration of the heat flow equations for you. The actual dirt does the calculation for every quantum particle in the whole system which is something you cannot hope to do on paper. The best you can do on paper is use statistical thermodynamics to predict bulk behaviors of systems with many many particles. This is how experimental science works. You produce hypothetical results based on your perception of how things work then you obtain actual results from the real world and compare to your hypothetical results. Experiments can yield bogus data of course but in this case (the Apollo heat flux experiment) the results are quite reliable as the instrumentation was nothing new, the deployment rather simple, and no uncontrolled variables of any significance to worry about (the interior of the moon is geologically dead, no atmosphere, no water transport of heat through the soil, no seasons, etc.)
Evidently you’ve never worked on the experimental side or it’s been so long you’ve forgotten what it’s like because a thermocouple is as basic a piece of lab equipment as test tubes, centrifuges, litmus paper, and bunsen burners. The principle of operation of thermocouples is basic high school physics and, if nowhere else, you had to have used thermocouples in high school and college science courses where there was any laboratory work included in the course. I was shocked when you didn’t know what one was.
Anyhow, if you want the equations which apply to soil physics and heat flow I suggest going here:
http://soilphysics.okstate.edu/software/SoilTemperature/index.html
The above is a downloadable software package that does the math for you, either calculating average surface air temperature taken from from soil temperature readings at depth or going the other way and predicting soil temperature at depth from average air temperature. The documentation with the software lays out the equations used by the software. This is important stuff when doing things like designing energy efficient earth-insulated buildings and using heat pumps with heat exchangers in boreholes. For instance, if you’re in upstate New York the constant soil temperature, when you have enough thermal mass to smooth out seasonal temperature change, is about 52 degrees. In south-central Texas where I’m at the average soil temperature is 72 degrees. That’s a big difference if you’re sinking a house in the ground or piling dirt around it to leverage the thermal inertia of the earth for heating/cooling or drilling a hole in the ground to drop an HVAC heat exchanger into. Same goes for the moon should we ever start having buildings there. It’s a very handy fact to know that burying a habitat one meter deep at 26 degrees north latitude on the moon with regolith similar to Apollo 15 landing site would produce a stable inside temperature of -23C absent any internal heating or cooling or -18C for a similar habitat at 20 degrees north latitude under regolith with an albedo like that at Apollo 17 landing site.
And back to the main point – if the earth didn’t have an atmosphere it would be quite similar to the moon as far as average surface temperature. I’m sure Willis knows this even if he won’t admit it and now he appears to be sulking. What the heck is it about the internet that makes people so unwilling to admit mistakes?

Joel Shore
March 22, 2010 5:15 am

Colin Davidson says:

That is incorrect I think.
The net flow is identical in all cases of the atmosphere. Neglecting the flows into and out of the solids/liquids of the planet, (which for clarity only I ignore by claiming an equilibrium which I acknowledge never exists in practice) the incoming solar radiation is always balanced by the outgoing radiation./blockquote>
Colin,
Yeah…There is always some confusion when talking about the greenhouse effect regarding at what point in time one is talking about. When I said that the net flow out is reduced, I am imagining instantaneously “turning on” the greenhouse effect and seeing the immediate result. But, of course, once the net flow out of the system is reduced, the result will be that the earth heats up until radiative balance is restored. And, once that occurs, what you have is, as you note, that the net terrestrial radiation that escapes is the same as before but the earth has to be warmer in order to produce this result.

That’s my understanding, which means it is probably wrong!

No, I think what you describe is (at least essentially) correct.
Bryan says:

Here is the direct link you require I hope you read the whole piece so as not to say I pulled this out of context.
Its easy to make a slip when responding in a blog but what alarms me is that Arthur did not take the chance to set the record straight.

At this point, I am confused about what in Arthur’s piece you feel is incorrect (since you seem to say that you don’t really disagree with the isothermal part).
Brian G Valentine says:

Here’s an even simpler toy model from about p200 of An Inconvenient Truth (and about all he has to say about how it works):
“CO2 acts like a blanket that keeps the Earth too warm.”

Brian, I don’t understand why you want to quote Al Gore rather than discuss the science. The question I thought we were talking about is whether the atmospheric greenhouse effect violates the Second Law not whether Al Gore’s analogy is perfect or not.

anna v
March 22, 2010 5:42 am

Re: Dave Springer (Mar 22 04:10),
I am tired of this one upmanship.
I was not questioning the term “thermocouple”. I was questioning your assigning it to the 50cm layer of the moon surface.
from wiki
A thermocouple is a junction between two different metals that produces a voltage related to a temperature difference.
Which is what I remembered and has nothing to do with your model of the ground integrating global temperatures of the moon.
It must be the “couple” that is confusing.
That there will be a functional connection, there will be. But until one integrates the skin surface temperatures ( which are the ones responsible for radiation and its balance) and has the average moon temperature I will not be convinced that the 50cm layer temperature gives the average moon temperature, and not the 1m or 2m. The steadiness in day night changes is a necessary condition if you are talking of an integrator, but not a sufficient one, always in my opinion. It is the differences that are eliminated but that does not fix the absolute value.
After all, if the regolith were a complete insulator, the 50cm layer would be at close to 0K while the average skin surface would still be -23 ( with the numbers above for the sake of argument ). Raise the heat conductivity incrementally and the temperature will rise correspondingly, but in no way it will be the surface average energy, though the change will be functionally connected ( if the average temperature became -20C correspondingly the 50cm level would rise by a calculable increment).

anna v
March 22, 2010 6:13 am

Re: Dave Springer (Mar 22 04:10),
That said, there should be no argument from any physicist that once the borehole depth is deep enough to smooth out the diurnal heating/cooling fluctuations the constant temperature obtained is the average of the surface temperature.
This is one physicist who doubts it, as I explained in my previous post.
There are more with me I guess.
If this were true, why go through the trouble of surface thermometers and not grid the land with boreholes?
Our disagreement lies in the absolute value of the temperature, when first the seasonal/time variation stops being evident. You are saying it is the average temperature over the seasons/times and I am doubting this identity.
For example, in:
http://www-personal.umich.edu/~shaopeng/science98.pdf
They are talking of deltaT. Not absolute T. I have no quarrel with that.

Brian G Valentine
March 22, 2010 2:08 pm

“The question I thought we were talking about is whether the atmospheric greenhouse effect violates the Second Law ”
The answer is, I can’t go any farther to convince you, and apparently, conversely.
If you really think about it this greenhouse gas idea causing global warming in the sense that it applied has to be completely erroneous. CO2 has been in the air since the Earth had an oxidizing atmosphere, probably less than a billion years after the first appearance of life on Earth, and and from all appearances of the geologic record, the rise and fall of CO2 in the air has never been associated with true climactic change. It has been the result of it, related to ocean solubility of it.
From a phenomenological standpoint, it MUST violate the second law Joel OR CLIMACTIC CHANGE FROM CO2 IN THE AIR WOULD HAVE HAPPENED IN THE PAST
can’t you see that? It’s Cold Fusion all over again. If Cold Fusion were possible at anything exceeding quantum tunneling there would be no such thing as an ocean.
If greenhouse CO2 were possible at ANY level the atmosphere would have saturated with water vapor long ago. There is no way it could be stopped
Don’t you have any appreciation for what three billion years of geologic history is?
Fleishman and Pons could be stopped because nobody could build their little cold fusion test tube thing (not that anybody need to try). This AGW thing cannot be so easily stopped because there is nothing but rational thought to counter it and that’s JUST NOT ENOUGH for some folks who for one reason or another actually like the idea and will go to the end of the Earth to defend it

Colin Davidson
March 22, 2010 3:18 pm

Igl wrote (02:38:39, 22MAR10):
“Then we pretty much agree. Correcting for the evaporation I’m ending in the area 1.3 C per CO2 doubling. So if K/T budget is right then IPCC is wrong. Cleary it can’t be us two being wrong 🙂
But I totally disagree with your (01:01:42) …”
I think what Igl may disagree with is my statement that the atmosphere is heated by NET absorbed surface radiation, rather than the actual amount of radiation absorbed.
You are standing in a room. It feels neither hot or cold. You are receiving around 325 W/m^2 of radiation from all directions. Why don’t you warm up? Because you are also radiating about the same out to your surroundings.
Replace the person with some air. looking to the side, the air absorbs about the same radiation as it emits, so it does not heat or cool from that source. Looking down at the ground however, it absorbs somewhat more than it emits, so it warms slightly. It is not the absolute anmount of radiation but the NET which counts.
In my post I calculated the relative strengths of the surface fluxes which heat the atmosphere. These are:
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
March 22, 2010 3:46 pm

I must point out that my posts are really only trying to explain and extend the physics of the ficticious world depicted by Kiehl &Trenberth, particularly their 1997 version which is the one used by the IPCC AR4 Chapter 1. The reason for doing this is to remove all of the variables (non-equilibrium, night/day, cloud/not, land/sea, latitude, season, wind, precipitation, altitude, sea currents, energy from the centre of the earth) from the discussion so that the basic physics can be explored without the complications that these other factors involve.
Going back to that ficticious world, I have a problem, enunciated in my posts at (14:10:34, 20MAR10) and (00:21:13, 21MAR10). That is that the additional downwelling radiation in a 3DegC higher temperature world, is much less than what is required to keep the surface at a 3DegC higher temperature. The surface is much less sensitive to downwelling radiation than the air at the tropopause for two major reasons:
1. It is much hotter than the Tropopause , so the increase in temperature yields a much greater increase in surface radiation.
2. Any temperature increase increases the rate of evaporation, so there is an additional flux required to supply the energy of latent heat in the evaporated water.
The IPCC claims that a doubling of CO2 prodduces an additional 4W/m^2 of “Radiative Forcing”, which through the controversial positive feedback due to water vapour and clouds turns into about 12W/m^2. It claims this warms the upper atmosphere by about 3DegC, and that this local warming gets transmitted to the surface by the magic of constant lapse rate. OK, suppose all that is true.
Where does the additional 22 to 32 W/m^2 downward radiation required to maintain the surface temperature at an additional 3DegC come from? (I note in passing that these are minimum, not maximum estimates, as the additional cloud produced due to the additional evaporation would tend to increase these figures).

March 22, 2010 3:48 pm

Joel
You either did not follow the link or you failed to understand it.
The link shows that Arthur Smith thinks that if CO2 and H20 vapour were removed from the atmosphere.
The atmosphere would then become isothermal from the surface to a height of 30Km.
I certainly do not believe in such nonsense.
Any person that had a clue about the Kinetic Theory (and/or ) had climbed a hill would be in no doubt that the temperature drops steadily.
I thought that he would correct the record once someone had pointed out how foolish it was.
There was no correction and perhaps he still believes it!
Since you are a friend of his perhaps you could have a word with him and explain why this is physical nonsense.

Joel Shore
March 22, 2010 4:30 pm

Brian G Valentine says:

From a phenomenological standpoint, it MUST violate the second law Joel OR CLIMACTIC CHANGE FROM CO2 IN THE AIR WOULD HAVE HAPPENED IN THE PAST

One doesn’t “prove” 2nd Law violations from a phenomenological standpoint (especially when one’s interpretation of the phenomenology disagrees with most of the the people who, you know, actually publish in the field).

Fleishman and Pons could be stopped because nobody could build their little cold fusion test tube thing (not that anybody need to try). This AGW thing cannot be so easily stopped because there is nothing but rational thought to counter it and that’s JUST NOT ENOUGH for some folks who for one reason or another actually like the idea and will go to the end of the Earth to defend it

Or maybe it is because those who who are arguing against it don’t really have very coherent scientific arguments and just put up a bunch of nonsense, like G&T did, that then gets swallowed by people who want to believe it even though it is easy to demonstrate what nonsense it is. If that is “rational thought”, I think you should try something else!

Brian G Valentine
March 22, 2010 5:32 pm

Joel, the only suggestion I have right now is to go “prove” your point with some more condescension and arrogance at people who write papers you don’t like and who question you on the web logs.
With some practice, maybe you can get as arrogant as Eli Rabbit!
Hey – maybe you and Eli can do some joint research together to answer the question
Which of the following is the stinkiest fraud?
– Gerlich and Tscheuschner
– Gerhard Kramm
– Brian Valentine
Publish your results in a reputable journal

Colin Davidson
March 22, 2010 6:29 pm

Joel Shore (05:15:45, 22MAR10) wrote:
“Yeah…There is always some confusion when talking about the greenhouse effect regarding at what point in time one is talking about. When I said that the net flow out is reduced, I am imagining instantaneously “turning on” the greenhouse effect and seeing the immediate result. But, of course, once the net flow out of the system is reduced, the result will be that the earth heats up until radiative balance is restored. And, once that occurs, what you have is, as you note, that the net terrestrial radiation that escapes is the same as before but the earth has to be warmer in order to produce this result.”
I would like to thank Joel for his answer.
I have a problem with”Radiative Forcing” which is described as a step change to a stable system (The imbalance at the Tropopause if all change in the Stratosphere is complete but no change has happened to the Troposphere). This is because we are imposing a step change on the static, imperfect model that is described by K&T.
I want to understand, instead, the final state, not the transition. I want to see the K&T diagram for a +3DegC world, and I want the surface balance explained in detail. For example, I want to know what the REDUCTION in sunlight absorbed by the Surface is (because it’s cloudier). I want to know how much extra back radiation there is due to a warmer atmosphere. I want to know how much surface radiation is predicted to escape direct to space. I want to know what the amount of direct conduction there is, and if it has changed from the +0DegC world, why. I want to know how much evaporation there is going to be, and why. I want to know how much extra back radiation there is due to the increase in concentration of the two main gases, with a justification.
I guess it’s not really important, because we are only talking about the really simplified, unrealistic model. But I’m curious. I can’t get a surface balance in that world – not one that’s driven by 2xCO2, unless the First or Second Laws are in abeyance.

Joel Shore
March 22, 2010 6:32 pm

Bryan:
I don’t see how the climbing a hill notion proves anything: Arthur is talking about a hypothetical case. He knows that the lapse rate exists in our current atmosphere. I also am not clear as to what kinetic theory dictates as to the thermal structure of the atmosphere. Care to enlighten me?
I honestly don’t have a “dog in this fight”. What the thermal structure of the atmosphere would be in the absence of IR-absorbing gases is not obvious to me. However, the one thing that I do know is that the temperature of the surface would be dictated by radiative balance, which I think was Arthur’s main point, the entire thermal structure being somewhat of a side discussion.
Brian Valentine:
It is certainly a bizarre world we live in where those people who call others “arrogant” are those like you and G&T who somehow feel that they know climate science much better than most of the world’s climate scientists. I sort of associated arrogance with having a sense of the correctness and importance of one’s own ideas that is way out of balance relative to the ideas of those who have clearly studied these things much more closely and have the track record of accomplishments to show for this. But, perhaps you have a very different definition.

cba
March 22, 2010 7:42 pm

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.

Joel Shore
March 22, 2010 8:08 pm

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.)

Brian G Valentine
March 22, 2010 9:05 pm

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

anna v
March 22, 2010 10:36 pm

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).

Colin Davidson
March 22, 2010 11:37 pm

Joel,
Can you educate me? What exactly is “Lapse Rate Feedback”?

cba
March 22, 2010 11:44 pm

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.

lgl
March 23, 2010 1:32 am

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.

March 23, 2010 1:42 am

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
March 23, 2010 2:30 am

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.

anna v
March 23, 2010 4:50 am

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.

cba
March 23, 2010 5:34 am

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,

Brian G Valentine
March 23, 2010 5:42 am

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

March 23, 2010 11:59 am

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.

Colin Davidson
March 23, 2010 3:36 pm

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)

Joel Shore
March 23, 2010 6:19 pm

Colin Davidson says:

Joel,
Can you educate me? What exactly is “Lapse Rate Feedback”?

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.

Joel Shore
March 23, 2010 6:38 pm

Bryan says:

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.

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.

Joel Shore
March 23, 2010 7:12 pm

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:

So, let’s imagine an extremely cold Earth and atmosphere, without any water vapor, carbon dioxide, methane or any other greenhouse gases – and with no surface water to evaporate and create atmospheric water vapor, either. Next, imagine the sun starts to warm the surface of the Earth.

And what happens when there is a temperature difference in a material? Heat flows by thermal conduction, which would then gradually warm the upper atmosphere to reduce that temperature difference. The process would be slow, because the thermal conductivity of air is quite low. But eventually, the entire atmosphere would reach a constant temperature with height.
Only the surface and a shallow layer of air next to the surface would go through a day-night cycle of heating and cooling. The rest of the atmosphere would be at approximately the same temperature as the average surface temperature. And without a falloff of temperature with height in the atmosphere of at least 10 deg. C per kilometer, all atmospheric convection would stop.

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.

March 24, 2010 2:41 am

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?

Brian G Valentine
March 24, 2010 8:47 am

“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.

anna v
March 24, 2010 8:52 am

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.

lgl
March 24, 2010 11:06 am

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.

lgl
March 24, 2010 11:16 am

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”

JAE
March 24, 2010 11:55 am

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.”

Joel Shore
March 24, 2010 5:40 pm

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).
anna v says:

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.

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.

Joel Shore
March 24, 2010 5:55 pm

Willis Eschenbach says:

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

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.)

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.

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

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 …

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.

Joel Shore
March 24, 2010 7:36 pm

Willis Eschenbach says:

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.

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.

sky
March 24, 2010 7:57 pm

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
March 24, 2010 8:04 pm

“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?

anna v
March 24, 2010 9:52 pm

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.

March 25, 2010 2:11 am

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.

JAE
March 25, 2010 6:30 am

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.

Joel Shore
March 25, 2010 6:32 am

anna v says:

What is NET? a million years? a microsecond?

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.

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.

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?

March 25, 2010 10:11 am

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.

anna v
March 25, 2010 11:02 am

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.

Joel Shore
March 25, 2010 7:17 pm

Willis Eschenbach says:

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.

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:

An estimate of the forcing is obtained by regressing the flux at the top of the atmosphere against the change in surface air temperature, with the flux at zero temperature change being the estimated forcing.

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.

Joel Shore
March 25, 2010 7:42 pm

anna v says:

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),

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!

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.

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.

Joel Shore
March 25, 2010 8:03 pm

Bryan says:

Apparently Chris does not think that you or Arthur are prominent Physicists.

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 see it it required six of you to comment on what you regard as rubbish.

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.

March 26, 2010 1:49 am

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.”

Invariant
March 26, 2010 2:33 pm

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?

Invariant
March 26, 2010 5:59 pm

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?

March 27, 2010 1:20 am

Willis
…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……….
A cold object can radiate to a hotter object.
It cannot radiate HEAT.
Heat has the thermodynamic property of the capacity to do Work.
Ask yourself if this applies to the above statement.
Look at the definition of HEAT in any physics textbook.

Guenter Hess
March 27, 2010 2:05 am

Anna V,
I totally agree with you that there is a lot of confusion because of semantics.
In climate science heat and heat radiation are sometimes used as synonyms.
This is sloppy language and creates statements that apparently violate the 2nd law. If one applies the correct language everything is fine.
Per example one reads sentences like:
“all bodies at non-zero temperature radiate heat according to their temperature…”
Such sentences are wrong and incomplete. In physics heat is defined as energy transport from warm to cold. Therefore such sentences need to read:
“all bodies at non-zero temperature radiate heat radiation according to their temperature and emissivity”
If physics is phrased correctly and the word radiation is used, confusion is minimized.
I agree with your analysis.
Best regards
Guenter

lgl
March 27, 2010 2:51 am

Willis,
Where is TOA in a world without an atmosphere?

Joel Shore
March 27, 2010 7:13 am

Bryan says:

A cold object can radiate to a hotter object.
It cannot radiate HEAT.
Heat has the thermodynamic property of the capacity to do Work.
Ask yourself if this applies to the above statement.
Look at the definition of HEAT in any physics textbook.

You may be right about the technical definition. These sort of things are why I have never been a big fan of thermodynamics and always try to think in terms of the underlying statistical physics. However, as Guenter Hess notes, what you are then quibbling over is just semantics. It does not in any way change our argument.
For example, here is a discussion by retired meteorology professor who is really militant on these issues of semantics and pedogagy: http://www.ems.psu.edu/~fraser/Bad/BadGreenhouse.html However, note that unlike you and G&T, he doesn’t jump to the conclusion that because the semantics are arguably not perfect, therefore the phenomenon does not exist. He just argues for a better way to describe the phenomenon.
Invariant says:
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.
Well, if you erroneously get that impression, it is your own erroneous conclusion. Hansen et al. very well understand that there are forcings that occur from natural causes too. In fact, that is how they have arrived at one estimate of the climate sensitivity: They have looked at the estimated forcings during the Ice Age – interglacial cycles (triggered by the Milankovitch cycles of which you speak) and at the resulting temperature change. Forcings and temperature change due to volcanic eruptions like Mt. Pinatubo in the early 1990s provides another piece of empirical data to estimate the climate sensitivity.
You are confusing two different things here:
(1) Changes in temperature due to changes in natural forcings (such as the ones that I have listed above).
(2) Changes in temperature due to internal variability.
The latter are NOT assumed not to exist and, in fact, there has been quite a bit of study of such internal variability using climate models and comparing to the real world. Unfortunately, however, this can be hampered a bit by the fact that we don’t have very accurate estimates for forcings in the past, particularly during times when the forcings were fairly small (e.g., changes in solar forcing over the last couple thousand years) and, of course, our estimates of the temperature changes themselve is more difficult once we go back to before the instrumental record.

Possibly natural climate oscillations may be an order of magnitude larger than the climate changes due to increased CO2.

We know quite accurately (within about 10% or so) the radiative forcing due to a given increase in CO2. We can also estimate the radiative forcing due to past events such as the last glacial maximum and the Mt Pinatubo eruption. This allows us to calculate what the effect of CO2 should be. So, in fact, it is the past climate changes…particularly the significant ones for which we can get the best estimates of temperature change and forcings…that allow us to estimate how large a change in climate will be produced by the known radiative effect of increasing greenhouse gases.

Joel Shore
March 27, 2010 7:27 am

Willis Eschenbach says:

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.

Willis,
I don’t think you have really disposed of the question at all. In fact, you haven’t really addressed the questions from the last post: I.e., I still don’t understand what your argument is. Are you claiming (1) that the ~4 W/m^2 number includes the effects of feedbacks such as the water vapor feedback, or (2) that it doesn’t but somehow the water vapor feedback doesn’t change this number?
Also, how do you feel that an estimate of forcing discussed in the way that Gregory et al. define it (by this regression to zero temperature change) could possibly include the effects of the water vapor feedback, which only operates when there is a nonzero temperature change?
And, I still don’t understand the mechanism by which you imagine the IPCC would claim that a ~4 W/m^2 change in the TOA radiation would lead to such a large surface temperature change. The only mechanism that I could see for that happening is if there were a huge amplification of the surface temperature relative to that higher up in the troposphere and, in fact, we know that the prediction is for the amplification to go the other way (although on a global scale it is not that large).
When you find yourself arguing against what is clearly a fallacious argument…particularly one that you think has been embraced by a very large community of scientists, that is a sign that you may be arguing against a “strawman” that you have created rather than their actual argument. I am afraid that this is indeed true in this case.

March 27, 2010 9:10 am

Joel
You may be right about the technical definition. These sort of things are why I have never been a big fan of thermodynamics and always try to think in terms of the underlying statistical physics. However, as Guenter Hess notes, what you are then quibbling over is just semantics. It does not in any way change our argument…….
The matter of whether one is a big fan of one branch or physics or not we must use the language developed within the discipline or it will lead to all sorts of confusion.
For instance I have never been completely comfortable with Quantum Mechanics yet I use its methods and language until some better explanation comes along.
I hope that in your new paper you will use language that is consistent within the Framework of Physics after all that is what the article is meant to be addressing.

Invariant
March 27, 2010 9:19 am

> Unfortunately, however, this can be hampered a bit by the fact that we don’t have very accurate estimates for forcings in the past, particularly during times when the forcings were fairly small (e.g., changes in solar forcing over the last couple thousand years) and, of course, our estimates of the temperature changes themselve is more difficult once we go back to before the instrumental record.
Thanks Joel for comments. I am honestly trying to figure out what is your point of view – disagreements pave the road to progress!
Now, from what you state, do you argue that any change in global temperature need to be caused by an internal or external force?

Guenter Hess
March 27, 2010 9:54 am

Joel,
I agree with Bryan.
Semantics in this context is very important, since your statement implied to the reader a violation of the 2nd law of thermodynamics.
I made my statement because I find the confusion between heat and heat radiation or thermal radiation in popular books about climate change written by prominent physicists like Schellnhuber and Rahmstorf.
Such a confusion suggests to many scientists that the grasp of thermodynamics by the author, who uses the technical words wrongly, is not sufficient to talk about issues like global warming with expertise.
I do not assume that Rahmstorf or Schellnhuber do not know this. However, I do think one should strive to phrase the physics correctly.
G&T refuted wrong statements like yours and in that they were right in some particulars.
You in turn tried to refute G&T in turn with the same wrong statement they already refuted. This is a waste of time. Therefore, lets strive to phrase correct statements about physics.
By the way I am a big fan of thermodynamics classical and statistical. Both branches augment each other and are necessary to describe the physical world within any field of science.
Best regards
Guenter

Invariant
March 27, 2010 10:11 am

Joel,
I would argue that transitions from one climate state to another do not have to be caused by any particular change in an external driver. In driven, non-equilibrium systems such changes may occur even though a specific cause can not be identified. If the temperature is not in equilibrium, the right hand side of the heat balance is non-zero,
m c dT/dt = Qin – Qout
and the temperature may change even though Qin and Qout is constant. In fact the temperature may change both for constant and variable values of Qin and Qout, but not if the difference, Qin – Qout, is identically equal to zero.
Most of the arguments I have seen from the AGW community seems to assume that there is always equilibrium, in order for the temperature to change, something has to change it…. This is clearly nonsense, in particular when the AGW community argues that something has to have caused the warming from the little ice age. Well, do they argue that there was equilibrium during the little ice age in 1750 when the world was much colder? If there is also equilibrium today this clearly does not make sense since the earth is warmer today… We know from the heat balance that the last term Qout is temperature dependent (Stefan Boltzmann radiation), so it’s impossible that two different temperatures at the same time can be the equilibrium temperature for our planet! Then the question is, given that the temperature has changed a lot the last 1000 years,
http://pages.science-skeptical.de/MWP/MedievalWarmPeriod1024×768.html
would it be possible to argue that the climate behaves as an oscillator, with normal fluctuations in temperature around equilibrium?
m c dT/dt = Qin – Qout
Look at the heat balance; wouldn’t it be fantastic if the temperature fluctuations would be zero? A perhaps more realistic result would be, given the Milankovitch cycles, ocean cycles and other cycles, that the temperature would oscillate a couple of Kelvin?
What is your gut feeling Joel? That the temperature fluctuations are small or large?

Invariant
March 27, 2010 11:26 am

(The html link was corrupted!)
The question to Joel is, given that the temperature has changed a lot the last 1000 years,
http://pages.science-skeptical.de/MWP/MedievalWarmPeriod1024x768.html
would it be possible to argue that the climate behaves as an oscillator, with normal fluctuations in temperature around equilibrium?
m c dT/dt = Qin – Qout
Look at the heat balance; wouldn’t it be fantastic if the temperature fluctuations would be zero? A perhaps more realistic result would be, given the Milankovitch cycles, ocean cycles and other cycles, that the temperature would oscillate a couple of Kelvin?
What is your gut feeling Joel? That the temperature fluctuations are small or large?

Joel Shore
March 27, 2010 2:43 pm

Invariant: Yes, it is quite possible that there has been a negative internal fluctuation in temperatures over the last century that has partially masked the warming due to greenhouse gases which otherwise would have been larger.
Actually, I am being somewhat facetious in the above. However, my point is that internal variability doesn’t only occur in one direction. And, more important than that is the fact that saying that internal variability exists doesn’t repeal the fact that climate responds to forcings too…and that we have estimates for its sensitivity to such forcings.
And, by the way, there is quite a bit of work within the climate science community in characterizing internal variability…so it is not like it is something that they haven’t thought about, or that the climate models themselves don’t show. And, as I understand it, the estimates are that the internal variability is pretty small.

A perhaps more realistic result would be, given the Milankovitch cycles, ocean cycles and other cycles, that the temperature would oscillate a couple of Kelvin?

What evidence do you have for such numbers? And, why are you including Milankovitch cycles at all…Those provide a forcing; they are not internal variaibility.

Invariant
March 27, 2010 4:44 pm

Invariant: Yes, it is quite possible that there has been a negative internal fluctuation in temperatures over the last century that has partially masked the warming due to greenhouse gases which otherwise would have been larger.
OK. Agreed!
Actually, I am being somewhat facetious in the above. However, my point is that internal variability doesn’t only occur in one direction.
Sure! Oscillations go both ways, just like a pendulum.
And, more important than that is the fact that saying that internal variability exists doesn’t repeal the fact that climate responds to forcings too…and that we have estimates for its sensitivity to such forcings.
OK. Agreed!
And, by the way, there is quite a bit of work within the climate science community in characterizing internal variability…so it is not like it is something that they haven’t thought about, or that the climate models themselves don’t show
OK. I think I have to learn your language, it is internal variability I should state when I mean internal oscillations not caused by an (external) force.
And, as I understand it, the estimates are that the internal variability is pretty small.
Now, this is interesting. What is regarded as large and small interests me a lot. Compared to what? What is the evidence for this? And is it easy to tell internal variability from increased CO2 or other external variations? Sometimes I tend to listen more to what people have experienced in historical times than dubious climate reconstructions…
“Comparatively rapid variations of climate, of the order of a century, have presumably always occurred…”
Climate Through the Ages (1950) C E P Brooks. http://www.archive.org/download/climatethrouchth033039mbp/climatethrouchth033039mbp.pdf
(p. 379)
What evidence do you have for such numbers?
Well, I have no experience in climate science, however I have a Ph.D. in Physics with specialization in non-equilbrium thermodynamics, and I suspect that it is unlikely that internal fluctuations (as you like to call them) are negligible. I do not regard a couple of Kelvin as much since the climate never has a chance to reach equilbrium.
And, why are you including Milankovitch cycles at all…Those provide a forcing; they are not internal variaibility.
I know, but I think the internal fluctuations may be triggered by external fluctuations. Think of an harmonic dissipative oscillator RLC circuit with many resonsance frequencies that is driven by oscillations on many frequencies due to Milankovitch cycles and/or oscillations between
1. day and night,
2. summer and winter,
3. ice ages and warmer periods
The fact that the equilbrium temperature is not fluctuating more than 287 ± 1-2 K is quite impressive. A process engineer would be very happy with a regulator that manages to keep the temperature that stable…

anna v
March 27, 2010 11:40 pm

Joel and Willis,
I do not doubt that the calculations of shells and temperatures are correct. Of course there is a “greenhouse effect”, we are here, aren’t we?
The classical thermodynamic view of “reservoir1” and “reservoir2” and the second law should be working on a ball with an atmosphere whether there is an extra heat source or not. Otherwise the Clausius formulation of the second law is defied.
My point is that talking of “back radiation”, adding watts/m^2 all over the place is not within classical thermodynamics, it mixes in quantum statistical terms and leads to confusion and I suspect double counting in the famous energy budgets.
The chicken in the oven
http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html
being an illustration of double counting.
So it is semantics, but it is easy to make logical errors when the semantics is confused.

lgl
March 28, 2010 3:42 am

Willis Eschenbach (17:32:57) :
The relevance is that you put the 0 W at surface altitude and the 150 W at 6 km altitude in your equation.
And the 3.7 W is without feedbacks.
anna v
“I suspect double counting in the famous energy budgets.”
There is double counting. That’s how feedback loops work. If 59% of the LW from surface is back radiated, the resulting gain is 1/(1-0.59)=2.44 and the 200 W solar input becomes 490 W at the surface.

Joel Shore
March 28, 2010 7:02 am

By the way, Willis, the Hansen et al. paper is very clear that the forcing definition used by the IPCC does not include feedbacks other than those due only to the STRATOSPHERIC temperature change:

[17] The adjusted radiative forcing, Fa, might be expected to be a good measure of the radiative forcing acting on the climate system and relevant to long-term climate change. The reason to anticipate this is that the stratospheric temperature adjusts rapidly, in comparison with the response time of the troposphere, which is tightly coupled to the ocean, and most forcing agents are present longer than the stratospheric radiative relaxation time. Thus Fa, the flux at the top of the atmosphere and throughout the stratosphere after the stratospheric temperature has come to radiative equilibrium, is the principal measure of climate forcing employed in RFCR and by IPCC [2001].
[18] Ultrapurists may object to calling Fa a forcing, and object even more to forcings defined below, because they include feedbacks. Fa allows only one climate feedback, the stratospheric thermal response to the forcing agent, to operate before the flux is computed.

Dave F
March 28, 2010 7:52 am

@ Joel Shore:
Models could only simulate temperatures in the 1990s by adding forcing from GHGs. Why does Earth return to pre-Pinatubo temperature so quickly if lags and feedbacks are involved in GHG forcing? ~.4C swing both directions in a matter of 3 years. Does this not suggest climate seeks to stabilize?

Invariant
March 28, 2010 12:26 pm

And, as I understand it, the estimates are that the internal variability is pretty small.
Could you please be a little more specific. I would really appreciate to know what is the current state of the art knowledge here. Does it mean that internal variability may cause the temperature to fluctuate something like T = 287 ± 0.4 K?

cba
March 29, 2010 4:34 am

invariant,
the current state of knowledge? That’s been summed as “natural variation? we don’t need no stinkin’ internal variability!”
Of course, this is the same crew that presumes constant albedo in order to estimate sensitivities. Well, constant expect for the usual man-made surface changes.
seems like the closest to measuring albedo for any time period (about 20 yrs) came up with peak to peak variations of around 10%. That’s about 10w/m^2 average.

cba
March 29, 2010 4:37 am

ooops! sorrry – that’s 10w/m^2 averaged p-p power difference for that event.

Invariant
March 29, 2010 5:55 am

cba (04:34:41) :
Sure. However, I would very much like to know what they think, not necessary to determine if it makes sense, but merely to figure out what is going on inside their head. I do not think most of the climate scientists deliberately are cheating, and I think it is fortunate that Joel is contributing to WUWT.
So, with state of the art knowledge, I really mean what most of the scientists developing climate models think! Possibly Mojib Latif think internal variability is larger, however, that’s not what I intend to find out, what is the opinion of the main stream climate scientists?
I think this is relevant for the whole climate sensitivity discussion too; imagine that we knew that the internal variability was T = 287 ± 1.6 K with a characteristic cycle length of ~330 years. Then we could argue that the sensitivity of the climate would be less – it is harder to influence an oscillating system than a system at (or close to) equilibrium.

Pamela Gray
March 29, 2010 6:51 am

Internal variability is what destroys farmers into bankruptcy. The condition of the Pacific Ocean, combined with both the Arctic Oscillation and the Jet Stream can cause the inland northwest area of the US to freeze sewer tanks. Or allow us to grow something. Not a small percentage of crops are planted in the fall or once very few years, so a deep freeze will kill these kinds of plants. And if these weather pattern variables are oscillating in the cold mode and in tandem for a few years, you can kiss your farm subsidies, let alone your crops, goodby. It is a decidedly no small thing and exposes the fact that armchair and Ivory tower climate warming believers haven’t a clue about the strength of internal variability, in one season, and over several seasons.

Pamela Gray
March 29, 2010 6:55 am

Addendum, internal variability cannot be “averaged”. Internal variability is a climate zone specific entity. Some zones are stable to variability, and some are not. A calculated average lulls everyone into unpreparedness. And many deaths. Anyone here who is willing to state such a mathematical calculation as a single application to climate modeling is guilty by association to such nonsense and engages in dangerous propositions.

Invariant
March 29, 2010 7:59 am

Pamela Gray (06:55:38) :Addendum, internal variability cannot be “averaged”.
Then internal variability is not what I have in mind. What would you call PDO Pamela?

Joel Shore
March 29, 2010 11:08 am

Invariant says:

I do not think most of the climate scientists deliberately are cheating, and I think it is fortunate that Joel is contributing to WUWT.

Thanks. You should keep in mind though that I am not a climate scientist by any stretch of the imagination. I am a physicist who has been studying climate science as a “hobby”.
In terms of internal variability, I don’t really know although my general impression is that the belief is that completely unforced variability (i.e., not due to major volcanic eruptions, changes in solar forcing, …) is pretty small…probably a few tenths of a degree. But, if I get the chance, I’ll try to see where they talk about this in the IPCC report and point you in that direction.

Invariant
March 29, 2010 12:37 pm

Joel Shore (11:08:43) : I am a physicist who has been studying climate science as a “hobby”.
Thanks.
Me to, I am also a physicist who has been studying climate science as a “hobby”.
Looking forward to hear from you! Why cannot the climate equilibrium oscillate a couple of Kelvin? I’ve never heard a good reason! Since you are a physicist, you know that all fluxes are instantaneously given by the gradients, meaning that the reach for equilibrium one day may be pointless the next day. Although the forces point toward equilibrium, it is missed over and over again… And thinking of the yearly fluctuations does not make the issue easier either, I mean, a yearly variation of ~90 W/m² is quite astonishing, the forces discussed here are almost negligible in comparison.

March 29, 2010 1:23 pm

Invariant (12:37:49) :
a yearly variation of ~90 W/m² is quite astonishing, the forces discussed here are almost negligible in comparison.
A pet peewee of mine too. People fight over the [purported] effects of a 0.1 W/m2 solar cycle variation and ignore the 90 W/m2 beam in their eyes.

Invariant
March 29, 2010 1:59 pm

Leif Svalgaard (13:23:15) : A pet peewee of mine too. People fight over the [purported] effects of a 0.1 W/m2 solar cycle variation and ignore the 90 W/m2 beam in their eyes.
Well you told me this Leif! 🙂 (I did not know).
Do you have some emprical data showing this?
God påske! 🙂

March 29, 2010 2:42 pm

Invariant (13:59:55) :
Well you told me this Leif! 🙂 (I did not know).
Do you have some emprical data showing this?

Here is the past 15 [or so] years of actually measured TSI [normalized to SORCE]:
http://www.leif.org/research/Erl76.png
You can see that all curves fall just on top of one another [showing that the orbit is very stable – no barycenter nonsense] and that there are tiny wiggles from time to time. Those are the biggest solar active regions. The solar cycle variation is smaller than the thickness of the lines.
God påske! 🙂
Tak og til dig såvel.

cba
March 29, 2010 2:45 pm

gee leif – but that’s peak to peak, LOL. the RMS difference is less.
Invariant, the 90W/m^2 difference is due to the orbital distance – aphelion to perihelion distances. December is perihelion – meaning that the peak incoming power to the southern hemisphere summer is actually 90W/m^2 more than the peak power coming to the northern hemisphere during its summer.
What’s more interesting to me is that the SH temperatures tend to be slightly lower than the NH temperatures, despite this massive difference in forcing. The difference is evidently that the NH has most of the land surface while the SH has most of the ocean surface. Considering that ocean albedo is much lower than land albedo, something like less than 0.04 versus something in the realm of 0.15 to 0.19, not only does the SH summer have more power (90w/m^2 peak to peak) at the TOA than the NH, ignoring cloud cover, there would be another 34 w/m^2 of absorbed power ignoring also additional differences in TOA power.
Of course, if one insists on ignoring cloud cover albedo changes, one is then faced with massive increases in absorbed power resulting in lower temperatures. LOL.
What has to be going on is something in the line of LIndzen’s iris effect where the added forcing, especially over water, is resulting in more water vapor cycle and cloud cover albedo reduction of incoming power to the surface.

March 29, 2010 2:52 pm

cba (14:45:21) :
gee leif – but that’s peak to peak, LOL.
So is the solar cycle effect…
the RMS difference is less.
I don’t know what you mean, or rather it doesn’t make much sense to compute RMS over a smooth curve…
December is perihelion
January 4th.
What has to be going on is something in the line of LIndzen’s iris effect
And yet, in spite of all that, the 1 W/m2 solar cycle effect manages to sneak by the 90 W/m2 and make LIAs etc. 🙂

Joseph
March 29, 2010 3:53 pm

A very interesting plot Leif! Thank you for that. Now I have a question.
Do you have any idea what that plot might look like over the course of the glacial/interglacial cycles?
What I am getting at is this: Is the effect of the Milankovitch cycles limited to a change in the latitudinal distribution of solar radiation that reaches the surface, or does the TSI that reaches the TOA change as well?
This has been bugging me for a long time (I can’t find any info) and I would appreciate anything you could contribute.
Thanks in advance.

Invariant
March 29, 2010 4:38 pm

Leif Svalgaard (14:42:18) :The solar cycle variation is smaller than the thickness of the lines.
Thanks a lot Leif! Really appreciate it! 🙂 Regarding the 1 W/m² TSI variation, perhaps some people thinks that our climate can be charged like a capacitor over many cycles by this extra offset? You know, if you have a simple RLC oscillatory circuit, the capacitor and the inductance can be charged and discharged each cycle. However, increasing 1365 W/m² to 1366 W/m² is not a big increase (in magnitude) after all, and I doubt that the capacitor wil notice the difference.
Regarding my proposed climate variation of T = 287 ± 1.6 K with a characteristic cycle length of ~330 years, perhaps this is not so unlikely after all? I mean, we know that the climate is chaotic, and chaos almost always means scale invariance, implying that the rapid oscillations we see at a short timescale (years) also must be visible on a larger timescale (decades, centuries). Whether the magnitude of the variations also must be scale invariant is an interesting question, because if they are, then small oscillations (in amplitude) on a short timescale may be invariant with large oscillations (in amplitude) on longer timescales.

March 29, 2010 4:41 pm

Joseph (15:53:09) :
What I am getting at is this: Is the effect of the Milankovitch cycles limited to a change in the latitudinal distribution of solar radiation that reaches the surface, or does the TSI that reaches the TOA change as well?
The average TSI [and all that reaches the TOA] over the year will not change measurably [as the average distance to the Sun is nearly constant]. The eccentricity of the orbit will change so the peak-to-peak swing will change [smaller the more circular the orbit is]. Another important changes will be in the phase of the curve, that is: when does it peak? A glaciation may hinge on the insolation at 65 degrees North during Northern summer.

March 29, 2010 5:11 pm

Invariant (16:38:38) :
Regarding my proposed climate variation of T = 287 ± 1.6 K with a characteristic cycle length of ~330 years, perhaps this is not so unlikely after all?
If that 3.2 degree swing were due to TSI, it would mean a 60 W/m2 swing in TSI, which is not very likely. At least 100 times larger than what we would expect.

Invariant
March 29, 2010 6:03 pm

Leif Svalgaard (17:11:57) : If that 3.2 degree swing were due to TSI, it would mean a 60 W/m2 swing in TSI, which is not very likely. At least 100 times larger than what we would expect.
Wow! Then the interal variations must be powerful… What could it be? Monster PDO? ….

cba
March 29, 2010 8:02 pm

“Leif Svalgaard (14:52:12) :
cba (14:45:21) :
gee leif – but that’s peak to peak, LOL.
So is the solar cycle effect…
the RMS difference is less.
I don’t know what you mean, or rather it doesn’t make much sense to compute RMS over a smooth curve…
December is perihelion
January 4th.
What has to be going on is something in the line of LIndzen’s iris effect
And yet, in spite of all that, the 1 W/m2 solar cycle effect manages to sneak by the 90 W/m2 and make LIAs etc. 🙂

the RMS I’d use to get an average power difference over half a cycle so as to get a number related to how much more power goes to the SH in their summer versus to the NH in our summer time. This 90w/m^2 peak to peak annual variation looks like it corresponds to a sine function.
the solar cycle effect is a bit longer than the few months of seasonal change which means it should have more of an effect due to the total energy, but still, it’s and order of magnitude less than the seasonal.

March 29, 2010 8:46 pm

cba (20:02:25) :
the RMS I’d use to get an average power difference over half a cycle
Just integrate over the half-cycle to get a physically meaningful number. But, no matter what, the yearly variation completely swamps the solar cycle variation, which is why Milankovitch works so well, because it exploits that fact.

Richard Sharpe
March 29, 2010 9:30 pm

I think that Luboš Motl has a very good discussion of this.
http://motls.blogspot.com/2010/03/black-body-limits-climate-sensitivity.html
I think he is more likely to be correct.

cba
March 30, 2010 5:47 am

Richard,
Just went over Lubos article (5 minutes worth at 6am) and there’s much to be said for it – not that I fully gleaned all of it at this hour of the day in such a short time. I do have some problems with it though.
Clear skies with surface radiation only account for about half of the Earth’s surface. Clouds block the rest with substantial ability. Clouds are not water vapor but are particles of liquid and solids which are no longer behaving as single molecules with well defined spectrums but are rather continuum radiators like larger solid and liquid objects. These are blocking the surface outgoing radiation and absorbing it. They are also radiating back down and radiating outbound at their characteristic temperature. However, this radiation outbound is a continuum at a lower temperature and it is above most all of the h2o vapor in the atmosphere and above both a good bit of the co2 AND it’s starting at pressures that are lower than the surface value and pressure is an important requirement for ghg absorption as it is required to spread the spectral lines out so as to permit greater capture.
Another problem with Lubos’ presentation is that the atmosphere in clear sky conditions is going to block about 30% of the radiation emitted from the surface. One needs just a bit more T to compensate.
For clear skies, radiation out of the atmosphere calculates to be around 270w/m^2, significantly more than a BB radiator around 288k. For an overall balance with 235 or 239 w/m^2 absorbed incoming solar, that means the cloud cover portion must radiate less than that through the balance of the atmosphere above the clouds. This throws in a new parameter, cloud cover fraction, which has an effect both on incoming and on outgoing radiation.

lgl
March 30, 2010 11:12 am

Willis Eschenbach (21:16:47) :
“The point is that ∆W, the change in radiation between the two situations, is 150 W/m2”
But radiation isn’t the only change. You are also jumping from the surface to the tropopause (or whereever, I thought TOA in this context was the 255 K height)
I have already cited IPCC in
lgl (11:16:12) : (and Gavin before that)
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”
Similar from a old Hansen paper: http://epa.gov/climatechange/effects/downloads/Challenge_chapter2.pdf
“A simple radiative calculation shows that doubling atmospheric C02 would raise the mean level of emission to space, averaged over the thermal emission spectrum, by about 200m. (Cf. discussion in the section below on empirical evidence of climate sensitivity.) Since atmospheric temperature falls off with altitude by about 6C/km, the planet would have to warm by about 1.2 C to restore equilibrium if the tropospheric temperature gradient and other factors remained unchanged. In general, other factors would not remain unchanged, and thus the actual temperature change at equilibrium would differ from the one in this simple calculation by some “feedback” factor,f,”

Invariant
March 30, 2010 1:36 pm

cba (05:47:25) : This throws in a new parameter, cloud cover fraction, which has an effect both on incoming and on outgoing radiation.
Could cover fraction! A most useful concept it seems. Are we monitoring this with the required precision?

cba
March 30, 2010 3:21 pm

it’s undoubtedly tied up in measurements of albedo – which have been few and far between so to say. There are some direct measurements of that going on as well that are in better shape than direct albedo measurements but I doubt it’s being done to the precision and coverage required. I’m probably oversimplifying too as there are different types of clouds with different effects. Khiel and Trenberth used the assumption of three types of clouds. Looking at Lindzen’s Iris concepts, it would seem that there is different albedo based upon the make up of the particular cloud type – like size of particulates involved in seeding the cloud.
My cloud cover fraction is the ultra simplified concept presuming the existence of a ‘generic’ cloud with average effects and then combining it with the clear sky area to get a better understanding of a simplistic cloudy+clear model.
What one has is a pair of straight line functions. These are incoming power density which ranges from 341w/m^2 in clear sky to a much lower value, like 10% of that for total cloud cover. One then has outgoing power which for clear skies is around 270W/m^2 (390 surface radiation – actual absorption by ghgs – something around 100 to 120 w/m^2 absorption (note that the 150 w/m^2 from 390-239 w/m^2 is for the actual mixed atmosphere of cloudy & clear) . The outbound IR in full cloud coverage is virtually total – but the cloud tops are going to radiate a continuum from their intrinsic T so there is no total blockage, only a reduction to something more like 200 or 210 W/m^2 so that the average becomes the same as the average incoming – around 239 w/m^2. These two lines intersect at a point on the cloud cover fraction axis – around 0.62 or so which represents the cloud fraction where there is balance.
While simplistic, it is more sophisticated than a clear sky only analysis.
Like you and some of the others in this thread, I’m involved professionally in physics (& astronomy) and not climatology or planetary science. Considering my current sleep patterns and the rest of my schedule these days, I have very little time at the moment for either sleep or for ‘hobby’ armchair climatology studies I’ve been doing the last couple of years.

Invariant
March 30, 2010 3:47 pm

cba (15:21:04) : I’m probably oversimplifying too as there are different types of clouds with different effects. Khiel and Trenberth used the assumption of three types of clouds.
Thanks. I’ve read a couple of papers by Trenberth, and I must say that there is something fishy going on, it is not anything near being well founded physics with some intuition and arguments based on insight, to me it seems mostly like a collection of (unfounded) claims…

Joel Shore
March 30, 2010 5:23 pm

Willis Eschenbach says:

I understand that very well, Joel. My point is that the forcing Fs* is one of the “forcings defined below”, and it does include feedbacks of all types. That’s why I used the Fs* numbers …

No, Fs* is the one that Hansen et al. note Gregory et al. say “is obtained by
regressing the flux at the top of the atmosphere against the change in surface air temperature, with the flux at zero temperature change being the estimated forcing”. If you are looking at the flux at zero temperature change, there is no way that this flux includes the effects of the water vapor feedback because the water vapor feedback depends on there being a temperature change!
I admit that the Hansen et al. paper is a little bit unclear and you can mistakenly misinterpret it in the way that you have. However, the evidence that this interpretation is just plain wrong is all around you. I would suggest that you actually entertain this possibly rather than to continue to cling to your notions that make essentially no sense on the grounds that you can find language in the Hansen et al. paper that is ambiguous enough so that you could read as supporting it (and ignoring the language such as that I have pointed out that clearly demonstrate that such an interpretation is incorrect).

cba
March 30, 2010 6:58 pm

K&T 97 is a fair job at trying to estimate some balance based upon simple theory and compared with some measurements, like satellite. The update (2007?) seems to really show a bit of bias IMHO. Not all of the numbers seem to be correct for the ’97 paper but it seems a fairly reasonable attempt on average.
as for how much is real physics versus unfounded claims, well, it is climate science.
the problems overall – not including the confusion between politics, the politics and subversion of science, religion, and climatology – is basically that albdeo/clouds etc. are not known or understood well enough to predict hardly anything and the sensitivity is way off, way too high.

March 30, 2010 7:44 pm

cba:
Bravo. The size and nature of the discrepancies, the size and nature of the unmodled effects, the degree of uncertainty in almost every known significant parameter makes predictions from the GCM’s highly suspect at best. It seems to me that one could manipulate the parameters and assumptions within arguable bounds and achieve wildly different results. In this context, predicting the effect of 100ppm, or 200 ppm change in C02 seems a stretch, almost laughable. Not that we shouldn’t try, but it seems to me we need to know a heck of a lot more before we put lots of faith in these predictions.
Another point, positive feedback as is being discussed is extremely sensitive to errors in the feedback gain (call it “g”). A term like 1/(1-g) is the result of a simple feedback term with gain g, and the sensitivity of this overall gain to changes, or errors in g is 1/(1-g)^2. Thus a term which produces a gain of 3, has a sensitivity of 9 to changes in g; quite substantial.
Perhaps I am not used to way Physicists or climate scientists discuss things, but the term “feedback” is plural in most control system terminology. Perhaps this is just a convention you guys all use?
Mike

cba
March 31, 2010 4:55 am

mike,
I think what we have is a natural temperature PID setpoint or closed loop control system with internal oscillations, an ability to be shorted out by high surface albedo due to fresh snow and ice, and a sensitivity to all sorts of serious external effects like cosmic rays and volcanism. The control mechanism is the cloud cover, at least during warm periods of low or no glaciation.
Consequently, an ice age lasts quite a long time as so much of the surface has similar albedo to that of clouds which short circuits the control mechanism. That combined with the naturally lower h2o vapor availability and content lead to the situation of long glaciation periods. Ultimately though, clear ice has very low albedo and as the snow and ice age and become dirty with soot from fires, volcanic dust, the roughness from sublimation and the like, along with the inability to replace the ice from precipitation, and perhaps also an external (to climate) event, one starts to lose the glaciation and the PID control mechanism takes over again.
Overall though, it looks like the setpoint system does a fairly good job at maintaining the temperature. Andddd, kudos to big Momma Nature who always seems to be able to figure out the best way to redistribute thermal energy and maintain that heat flow, even when climatologists can’t. BTW, this stuff reminds me of the old time evolution dogma that required a nice unchanging Earth to ‘cook’ those molecules up to the point that catastrophic events were not even allowed to be considered.

Invariant
March 31, 2010 9:53 am

cba (04:55:46) : I think what we have is a natural temperature PID setpoint or closed loop control system with internal oscillations…
This is a useful analogy. In a PID controller the proportional value (P) determines the reaction to the current deviation from equilibrium, the integral value (I) determines the reaction based on the sum of recent current deviations from equilibrium, and the derivative value (D) determines the reaction based on the rate at which the current deviation from equilibrium has been changing.
Sometimes a PID or just PI controller can be unstable, leading to oscillations. A nice example is a drunken man adjusting the temperature in the shower – his delayed response may lead to oscillations between warm and cold water…
🙂
Maybe our climate is a little out of focus or has a delayed response to the deviation from equilibrium – then the result will be temperature oscillations…

March 31, 2010 11:21 am

Interesting analogy, but I am not sure I would refer to it as a “controller”. It seems clear that there are hundreds of feedback loops with many minor loops inside bigger ones in any reasonable model of the climate. With positive feedback in lots of these loops, the system could indeed be unstable, and in fact will be under certain circumstances. The evidence we have, from anyone’s data, shows that so far the system has exhibited a reasonable local stability under some pretty heavy internal and external disturbances for hundreds of millions of years (at least). We certainly haven’t ramped off into a Venutian state at any time recently, despite many huge pertubations from inside and out. So the question to me is, do we know enough about this system to even get to Willis’ basic sensitivity analysis? When the operating point shifts, do we know enough about say, albedo or cloud cover shifts from pole to pole that we can really esitimate the feedback terms properly? Really?
Mike

cba
March 31, 2010 2:09 pm

it’s fairly easy to do the sort of things as Willis has been doing for this thread. We do know what we have in the way of conditions now- at least time independent and averaged. Where Willis and I disagree on approach is that he is starting with the Earth as it is expected to be without an atmosphere or clouds while I’m afraid there is way too much being included in there that is going to be changing in a fashion that is not just a small amount. He is right in the aspect that our current conditions are a result of Earth’s response including cloud formation and the change in albedo from around 0.15 to 0.19 up to around 0.30. However, I doubt that either improves the accuracy or clarifies the nature of a very small response, like a few doublings of co2. He is going with 15deg C rise (?) rather than a 33 deg C rise, making the sensitivity even less.
We do have several definite factors that can give us some basic idea. Average temperature is right at 288K now and if we assume constant albedo over the range of ghg influence, there’s a 33 deg C rise due to ghgs. (if not – then one has Willis’ number 15 deg C). We have with 0.3 albedo 239 W/m^2 average incoming power absorbed and for a general balance, we need to average 239 w/m^2 outgoing power radiating into space. In the IR, where Earth radiates, we’re going to be very close to an emissivity of 1.0 so at a temperature of 288k, we’re going to emit from the surface around 390 w/m^2 and of that amount only the 239 w/m^2 can be allowed to escape on average. That means in the real atmosphere, there must be around 150 w/m^2 absorbed for the outgoing, some from clouds, some from ghgs. 150/390 amounts to about 38% being blocked.
Above we have some rough numbers for the real Earth where we can calculate some average values. Given a 3.7 w/m^2 forcing increase for co2 doubling, one can figure the surface T rise needed to compensate – assuming other major factors are basically unchanged. Since 38% of any additional outbound radiated power is going to be absorbed as well, we must have a BB temperature that radiates more than an additional 3.7. That’s actually a needed 6 w/m^2 from the surface. Working stefan’s law backwards to find a new T which turns out to be 1.1 deg C and it amounts to a 1.1/3.7 = 0.3 deg C rise for an additional 1 w/m^2 of power being radiated from the Earth system.
However, if we take the average T rise for all ghgs and cloud cover etc, we have our 33 deg C / 150 W/m^2 = 0.22 deg C rise per w/m^2 – over the entire range for the addition of ghgs. Note the most obvious thing is that the average sensitivity over all ghg forcing is actually less than that calculated by the CO2 doubling in the real atmosphere using stefan’s law. That means there is a strong net negative feedback overall which means there can be no net positive feedback. 3.7 * 0.22 = around 0.8 deg C rise for a doubling of CO2. Note that Willis’ numbers will be even lower in sensitivity but there’s so much more lumped in there – like how we got the initial cloud cover – which may or may not be applicable now and I think it requires some additional accounting for (modeling) and I’m not sure that is possible to do it.
Invariant, Mike
offhand I’m not sure I can claim a non zero D term but I leave it in for completeness and to acknowledge the fact that maybe there really is one.
that there are a tremendous number of feedback loops is not in question. Just how well it behaves as a setpoint controller is also questionable. I think it should be obvious that it does because there are just too many things that are varying significantly.
Perhaps better questions to ask is whether the system is totally chaotic or partially chaotic and whether there really can be a setpoint control system. It appears probable to me that one cannot do a predictive time iterative model. I do think one can do these sorts of time independent concept models that tell us about how things tend to behave – like will adding ghgs increase or decrease temperatures.
I also think that these sorts of concept models essentially trash the notion of high sensitivity and show that these GCMs which predict high sensitivity are no better than video games when it comes to describing the climate or determining the sensitivity.

March 31, 2010 3:39 pm

cba
Thanks for re-stating all that, it is hard to put that all together unless you read the entire thread – good summary I think.
You make a good case, along with Willis and others, that general energy balance and average temperatures should be predictable to a first order. I didn’t mean to imply that at that level they aren’t, just that some of the faith put in the accuracy around ppm of this that or the other thing added will produce a certain outcome seems a bit beyond the pale. But your first order approximations seem reasonable.
I do wonder about long term time constants and systems that give rise to them, like Ocean Currents and heat resevoirs and land mass and their ultimate feedback on key parameters affecting things like albedo, and water vapor. In other words, some of the parametrics you guys use to calculate the sensitivities can change over years to provide a negative feedback which reduces the long term sensitivity. I know you guys already know or suspect these things, I am just trying to catch up.
Regarding chaotic effects, I personally believe they are of major import. A simple second order system exhibits chaotic behavior if a simple quantization model is added into it. I always found this fascinating. There are many critical phenomena in our climatactic system which have chaotic behavior: Cloud formulation and shape, wind patterns and turbulence, lightning to name but a few.
The other point people keep mentioining that I cannot help believe must be of great significance is the variation in the Sun, orbit, precession, etc which seem to dwarf GHG in magnitude, all of which have been present all along.
Thanks again fo rthe summary, I appreciate it.
Mike

Invariant
March 31, 2010 5:39 pm

Mike Workman (15:39:42) : I do wonder about long term time constants and systems that give rise to them,
I wonder about this too! Certainly the static picture that Trenberth is painting in his energy budged is useful, but to me the dynamics has always been more interesting – I cannot see how it is possible to explain climate variations without a clear picture of natural variations including the time constants and the systems that give rise to them.

cba
March 31, 2010 6:23 pm

you might want to look at / for an aging physical meteorology text. So much of the stuff has been worked out back in the 60s and it isn’t contaminated by an over reliance on computer modeling or gw politics. It’s a fascinating application of thermo. However, the basic ones assume adiabatic when there’s radiative activity going on.

Wolfgang Jünger
April 4, 2010 5:47 am

Dear Wiilis Eschenbach
From my point of view there is a fundamental error at the beginning.
Instead :”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 …..” it must be corrected to
My corrected version:
“As shown in the diagramm above,the surface lw radiation is 392 W/m². TOA radiation is 237 W/m². The difference 392 – 237 = 155 W/m² is absorbed by the system earth + atmosphäre and this is the cause for warming the earth by atmosphere.”
By accident the following calculation brings nearly the same result 20/155 = 0,129 grad C/W/m². But with a different atmosphäre like Venus, the given calculation would bring a total wrong result. Earth atmosphere in lw absorbs
155/392 = o,395 but Venus atmosphere absorbs 16250/16900 = 0,96 of surface-radiation. That means, Venus atmospheare is only radiating in lw range on TOA roughly 650 W/m² (Albedo 0,75).
Wolfgang Jünger

Joel Shore
April 9, 2010 6:39 pm

Willis Eschenbach says:

The only other time in the model run when the temperature change is zero is after the modelled system has fully equilibrated. At that point, the temperature is no longer changing. So that has to be the time they are discussing. And that is after the action of all of the feedbacks.

I personally would call that long time a case where dT/dt is going to zero, not a case where ∆T is going to 0. However, on the face of it, from just reading the Hansen et al. paper, I suppose your interpretation is not impossible if they have been a little careless with terminology.
However, today I got hold of the Gregory et al. paper and, in fact, I think you will find if you read it that they are indeed talking about regressing to the time when ∆T = 0, i.e., when the climate has not yet changed from the initial state. I suggest that you get hold of that paper and have a look.

So far, you have given no evidence at all that Fs* does not contain tropospheric, stratospheric, and surface feedbacks. So you’ll have to do more than just say that the evidence is all around me … where is it?

I am talking about the fact that your whole interpretation is convoluted and non-sensical. You seem to somehow want to believe that the radiative effects of, say, the water vapor feedback and the ice albedo feedback are insignificant and so they do not much change the radiative balance from what it would be if they were absent. This then has you believing that climate scientists are somehow imagining that the clear predictions of the Steffan-Boltzmann Equation get altered by what exactly? Exactly how is the planet, which only communicates with space (in an energetic sense) to any significant degree via radiation, going to heat up by 3 degrees if the net effect of the radiative changes is such that they should only cause it to heat up by a little over 1 degree? (One could, I suppose, seek refuge in something involving the relative heating of the surface vs the mid- and upper-troposphere, but alas that is a feedback operating in the opposite direction.) It is usually a sign that you are arguing against a “strawman” when you have come to the conclusion that your opponents believe silly things, as you have here (particularly, when those opponents are the strong majority of the relevant scientific community).
At this point, I also have to admit that I am confused by your whole argument about looking to the radiative balance at the TOA because what actually happens in the climate system is this: A forcing is introduced due to increased greenhouse gases that alters the radiative balance by increasing the downward-relative to-upward radiation at the TOA. The earth then responds by heating up and reduces this radiative imbalance. (Gregory et al. actually have a nice graph of this as predicted by the HadSM3 climate model.) However, because of positive feedbacks, it has to heat up more than would be predicted simply by applying the Steffan-Boltzmann Equation to the original radiative imbalance because the process of heating up causes further radiative changes that work to oppose this reduction in the radiative imbalance. [Note, however, that the radiative imbalance does still reduce with time because we are not in a state where the feedbacks are so strong that we get an instability…i.e., in net, as temperatures rise the radiative imbalance is reduced. It is just not reduced as rapidly (as a function of temperature) as it would be if the positive feedbacks were not operating and hence it takes a larger temperature change to restore radiative balance.]

April 10, 2010 3:20 am

All the W/m2, line by line figures for gases and radiation flows within the atmosphere used and bandied about in this thread by so many,
are they all just “model” answers / guesstimates rather than ACTUAL MEASUREMENTS. ?
If so, then the whole basis of the thread and discussion is on the models answers / guesstimates (and how good they are),
NOT observations.
It would appear that where a W/m2 figure is plotted on the Y axis of the various radiation plots, the figures used here are taken from, that / those.
The actual line by line calculations (HITRAN, LOWTRAN, etc.) are by an algorithm, presumably based upon, or derived from the Schwarzchild equation.
These models spectra indicate high emissivity values for gases (0.97),
yet gases do not behave like black or grey bodies.
Gases should have very low emissivity values (0.09).
I would suggest that the whole basis of the thread / discussions is from the above mentioned (falsely modelled) basis,
but there again, I think the whole of present climate science is from a false basis.
This is just one of them.
It does look suspiciously like the old “IPCC trick” of moving the decimal point one place though.
This trick works very well, I have to admit, in conjunction with the “all radiation is positive” rubbish as well.
(This could explain the “differences” between radiation / energy flows, and what we observe with heat flows ACTUALLY being relative)
You could create the impression (falsely) of a greenhouse effect with them combined,
AND TALK W/m2 RUBBISH UNTILL THE COWS COME HOME…..

Joel Shore
April 10, 2010 7:53 pm

Willis Eschenbach says:

Since they repeated the claim several times, it seems doubtful that they were just careless.

In which case we are left with the conclusion that your interpretation is likely incorrect because what they said is that they were regressing to was ∆T = 0, not dT/dt = 0, which is what you are talking about.

You are saying that Fs*, the forcing after the system has equilibrated with all feedbacks, is really Fi, the instantaneous forcing. I confess to being totally mystified by that claim. Perhaps you could say some more about that, I don’t get it.

No…What I am saying is that these forcings are subtly different but not radically different in the way that you seem to propose. In particular, Fs* is determined by doing a regression to zero temperature change rather than by looking at the instantaneous forcing. A point of their paper is that this way of doing things seems to get a value for the forcing that is more in line with that found after stratospheric adjustment than that found before. However, they are not claiming that it includes effects of the water vapor feedback that would manifest themselves as the temperature rises. I think you should get the Gregory paper and read it carefully. I just gave it a cursory reading, enough to confirm that my interpretation of what Hansen et al. quoted from that paper is the correct interpretation of what they are actually saying.

It “has to heat up more”? Has to? Why does it “have to heat up more”? What happened to the negative feedbacks? Did they magically disappear? You assume that the net feedback is positive … where is the evidence for that assumption?

Willis, the point of my discussion is to describe how things happen under the hypothetical situation that positive feedbacks dominate. Your claim was that you can somehow determine from the top-of-the-atmosphere radiation balance (by some method that I still don’t understand) what the average climate sensitivity is. I am explaining how your understanding of the relation between the top-of-the-atmosphere radiation balance and the radiative forcing (your basically equating the two) is incorrect and that is leading you to make unjustified conclusions. If you refuse to entertain the possibility that the feedbacks are positive even long enough to understand how the scientists understand how things work in that case then we can’t get anywhere. If it makes you feel better, take what you quoted from me and replace it with the following:

However, in a world (or model) where positive feedbacks dominated, it has to heat up more than would be predicted simply by applying the Steffan-Boltzmann Equation to the original radiative imbalance because the process of heating up causes further radiative changes that work to oppose this reduction in the radiative imbalance.

Changing it to such a hypothetical in no way undermines my basic point of trying to get you to understand the understanding of how these positive feedbacks operate and that they increase the temperature change precisely because they make additional changes to the radiative imbalance.

Joel Shore
April 10, 2010 8:25 pm

Willis,
I realize that a picture would be worth a thousand words in explaining what I think Gregory et al. are doing. I have drawn a little Excel graph to illustrate it, so if there is a way that I could send it to you, that would help.

Joel Shore
April 11, 2010 7:00 am

Willis: Here is the picture that illustrates, as I understand it, the difference between the instantaneous forcing F_i and the forcing F_s* as defined by Gregory et al.:
http://www.frontiernet.net/~jshore/Illustration_of_forcing_definitions

Joel Shore
April 11, 2010 10:06 am

Sorry…Let’s try that link again: http://www.frontiernet.net/~jshore and then click on “Illustration_of_forcing_definitions”.

Joel Shore
April 11, 2010 8:01 pm

Willis says:

As I recall, the underlying issue was whether the feedbacks were included in the estimation of Fs. Since the climate model used to estimate the data for the regression line contains all of the feedbacks (water vapor, ice, cloud, etc), it seems clear that they are included. Are we in agreement on that?

Yes…The climate model will include all of these things. So, for example, when the temperature changes, the water vapor feedback will come into play. However, the ***radiative forcing*** itself is defined by this regression to zero temperature change, which means that it includes only that part of the water vapor feedback that acts independent of any (tropospheric) temperature change. As far as I know, that is not any of the water vapor feedback, at least as I understand it. (Given Hansen’s cryptic language, perhaps there is some small adjustment of water vapor that occurs with the addition of CO2 even in the absence of any temperature change, although it is beyond my imagination to know what that would be. Actually, I suppose it could be the part due to any adjustment in the stratospheric temperature.) Similarly for the ice – albedo feedback.
It’s late now, so I will answer your second post tomorrow!
Take care,
Joel

Joel Shore
April 12, 2010 8:48 am

Willis says:

You say I am equating top-of-the-atmosphere radiation balance and the radiative forcing. I don’t recall saying anything about TOA radiation balance.

But, that is the problem. The definition of what is a “forcing” and what is a “feedback” depends on context. By calling that 150 W/m^2 the forcing, you are making a decision that is not compatible with determining the climate sensitivity that we are interested in from the radiative forcing due to increased CO2.
I suggest that you read the excellent piece here by Chris Colose, especially the last part where he talks about what would happen if we removed CO2 from the atmosphere: http://chriscolose.wordpress.com/2010/02/18/greenhouse-effect-revisited/ Note that Chris claims that the result would be a dramatic reduction in water vapor too and that the end result would be a much colder earth, nearly removing the entire greenhouse effect.
Now, you may not believe in this positive feedback due to water vapor (+ clouds) but that still doesn’t mean that you are entitled to do your calculation for climate sensitivity under the ***assumption*** that it doesn’t happen and then claim that such a calculation shows that it (i.e., a high climate sensitivity due to positive feedbacks) doesn’t happen. And, that is exactly what you have done because, if things work as he describes, the reduction in radiative forcing due simply to removing the CO2 from the atmosphere would cause a dramatic temperature reduction (basically, the 20 C temperature change that you are talking about). However, this reduction in radiative forcing is not the 150 W/m^2 that you calculate because that value also includes the radiative effect of the change that occurs in water vapor, which is clearly a feedback and not a forcing in the scenario that Chris describes (and in most scenarios that one can think of).

Joel Shore
April 12, 2010 9:04 am

Willis,
Now that I understand where your estimate of 150 W/m^2 is coming from, I think I see another problem with your argument, a problem that explains the mystery of why your climate sensitivity calculation doesn’t lead to something close to the sensitivity in the absence of feedbacks (like I thought it should).
It seems to me that you have gotten the 150 W/m^2 by taking the emission at a temperature of 288 K (the surface temperature of the earth) and subtracting the emission at the effective radiative temperature, 255 K. However, you then take your temperature difference to be the difference between 288 K and the ~268 K temperature that you calculate the earth would be at if we remove both the greenhouse effect and the albedo effect of clouds.
I think that this is inconsistent, i.e., if you use the 150 W/m^2 then the correct temperature change to use is 33 K. If you use the 20 K temperature change, then the correct radiative forcing to talk about is ~100 W/m^2 (the difference between radiation at 288 K and 268 K). With this correction, you will then get a value for the climate sensitivity (~0.2 – 0.22 K per W/m^2) that is closer to the no-feedback value. (It is still a little low of it, but I think that is accounted for by the fact that this sensitivity depends on the radiating temperature and is usually computed at the 255 K effective radiating temperature whereas you are computing the average sensitivity over a temperature range that is higher than this.)
So, at this point I think I now understand pretty much everything about your calculation, i.e., what it computes and why it is wrong (or, more precisely, why, if done correctly it would just be a complicated way of computing approximately the no-feedback value of the climate sensitivity).

Joel Shore
April 14, 2010 6:16 pm

I’m sorry if my lack of clarity lead you to that conclusion. The ~ 150 K is from the Kiehl-Trenberth global energy budget, or from my reworking of it in the head post. Basically, K/T say that of the ~ 235 W/m2 emitted by the earth to keep it in radiation balance, 40 W/m2 is directly emitted by the surface (through the “atmospheric window”), and 30 W/m2 is from the clouds. This leaves 165 W/m2 from the TOA. My calculations show a lower number, 147 W/m2 of up/downwelling longwave radiation from the TOA. Pick either one, it makes little difference to the end result.

Okay…Well, I will have to think more about this. It is not vital to understanding what is wrong with your argument…but it is useful for understanding why your argument doesn’t just give something close to the no-feedback sensitivity, which is what I think it should give by its construction if calculated correctly.

Since the 150W/m2 is accompanied by a 20°C warming, I divide one by the other to get the sensitivity. As I pointed out, this temperature change of 20°C includes all of the possible feedbacks.
I am still very interested in any objections you might have to that reasoning, thanks for your ideas. What am I missing here?

I keep trying to tell you what you are missing: You are missing the fact that what you are calling “forcing” is a combination of forcing and feedbacks, so if the feedbacks are positive then you are overestimating the radiative forcing. You have just created a circular argument that essentially says “If the feedbacks are not positive then the climate sensitivity isn’t very high.” We already knew that.

The problem with this is that decreasing water vapor decreases clouds, as does decreasing temperature. With decreasing temperature and water vapor, very rapidly, few clouds would form in the tropics. Any reduction in tropical clouds makes a huge difference in the amount of solar radiation striking the earth, one that would swamp any reduction in GHG forcing….
The fact that Chris doesn’t mention any effect from that reduction in clouds speaks volumes. Clouds are the elephant in the room, and when someone starts making claims about “what would happen if” and doesn’t mention clouds, they’re not talking science, they are engaged in advocacy. Which is a shame, because the rest of the post was so good and so clear on the science.

Willis, you are losing focus. I don’t personally care whether or not you believe in Chris’s positive feedback scenario. The point is that even if you don’t, you can’t argue for a low climate sensitivity with a circular argument that shows that if positive feedbacks are not allowed then the climate sensitivity is not very large.
Chris’s proposed positive feedback mechanism is just useful because it makes the circularity of your argument clear: If Chris’s description of what happens is correct, it is clear that your argument would not diagnose the correct climate sensitivity. After all, both of us would agree that the majority of the radiative greenhouse effect is due to water vapor and clouds, not CO2. Yet, in Chris’s scenario, all you have to do is remove the radiative forcing due to CO2 and you lose pretty much all of the water vapor and (I think) clouds too and their radiative effects. (Furthermore, you will likely also significantly increase the earth’s surface albedo by freezing over large expanses of both land and water, so even though cloud albedo would be less, surface albedo would be higher.)
So, Chris is saying, “Reduce the radiative forcing just by removing CO2 and you will get rid of all this other radiative effects.” However, in your calculation, you take the radiative forcing to be the entire amount of the radiative effects, including all of that due to water vapor. Ergo, if Chris’s (or any other positive feedback scenario) is correct, your calculation of the climate sensitivity will not correctly diagnose the climate sensitivity in the system.
Having said this, and I hesitate to add this because I worry that it will distract you, I have to say that I don’t really buy your argument against Chris’s picture. For one thing, you talk about clouds having this huge effect in the tropics but you yourself admit that even if you remove all effects of cloud albedo, there would still be ~20 C of warming to explain by greenhouse gases. And, as I noted, this doesn’t even touch the issue of the additional earth surface albedo that you get as you start to ice over more of the planet. Furthermore, I think a lot of the dramatic effects you talk about for clouds cooling in the tropics are really redistributions of heat within the troposphere. This is not to dispute that the net effect of those clouds on the earth’s radiation budget is cooling, but some of the effect that seems so dramatic to you is redistribution of heat up into the troposphere…which the models already have in there by more-or-less pegging the temperature structure to the moist adiabatic lapse rate.
But again, let’s take one issue at a time: Independent of whether the net feedbacks are negative or positive, your argument claiming to show that you can estimate the climate sensitivity by looking at the radiative effects and the resulting temperature change is wrong. We know the relation between radiative effects and temperature change: It is governed by the Steffan-Boltzmann Equation. (The only slight hitch is that one does have to worry about the temperature structure in the atmosphere; e.g., I think the negative lapse rate feedback is not really a radiative effect between the earth / sun / space but is instead an issue of how the temperature rise at the surface changes for a given temperature rise at the effective radiating level.) What we don’t know is what part of the current radiative effects that we see is a radiative forcing and what is a feedback. Just assuming that it is all forcing makes for a circular argument.
By the way, just to let you know: I am not just hypothesizing that your argument is wrong in this way. I am quite sure of it…much more sure than I actually am regarding whether the feedbacks really are positive. Even if you turn out to be right on that issue, your argument in this post is still wrong, wrong, wrong. I really hope you will be willing to listen closely to what I am saying and think about. So far, you seem to have a block that is preventing you from comprehending what I am saying.

April 14, 2010 7:31 pm

Joel Shore (18:16:30),
Hate to be a buttinsky here [not really], but this is too funny:

Well, I will have to think more about this. It is not vital to understanding what is wrong with your argument… I am not just hypothesizing that your argument is wrong in this way. I am quite sure of it …much more sure than I actually am regarding whether the feedbacks really are positive. …Even if you turn out to be right on that issue, your argument in this post is still wrong, wrong, wrong.

I think maybe one more “wrong” would have been the tipping point, especially if Joel stamped his foot when he said it.
To be a little more serious, what is the unit of measurement for a “forcing?” How about for a “feedback”?
We have quantifiable units of ohms, newtons, volts, henrys, Watts, etc. Without a unit of forcing and feedback, climate science is winging it.
So I propose the symbols ⇥ for a forcing, and ↩ for a feedback. The unit of forcing can be known as a Willis, the unit of feedback as a Smokey.
Let’s let Joel assign the quantities, since he’s never wrong.

Joel Shore
April 14, 2010 8:07 pm

Smokey,
Forcings are measured in Watts per meter-squared. Feedbacks are measured in Watts per meter-squared per degree Kelvin. So, for example, Dessler and Sherwood in their 2009 Science perspectives piece (http://www.sciencemag.org/cgi/content/summary/323/5917/1020) say that most estimates of the water vapor feedback are in the range of 1.5 to 2.0 W/(m^2)/K, which means that each 1 K rise in temperature would result in 1.5 to 2.0 W/(m^2) effect on the radiative balance due to the increase in water vapor in the atmosphere.
The Stefan-Boltzmann Equation itself gives a negative feedback of about -3.8 W/(m^2) per degree Kelvin. So long as the feedbacks due to everything else besides Stefan-Boltzmann are positive, then you get a magnification of the climate sensitivity due to the S-B Equation alone (i.e., the approximately 1. 1 K per CO2 doubling). If they are negative, you get a diminishing of this sensitivity. [If you imagine “tuning” the feedbacks, as the magnitude of all the other feedbacks approach 3.8 W/(m^2) per degree Kelvin, the sensitivity gets larger and larger, with feedbacks at or above this value leading to a true “runaway” effect.]