The Forcing Conundrum

Guest Post by Willis Eschenbach.

For all of its faults, the IPCC (Intergovernmental Panel on Climate Change) lays out their idea of the climate paradigm pretty clearly. A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. Today I found myself contemplating the concept of radiative forcing, usually referred to just as “forcing”.

So … what is radiative forcing when it’s at home? Well, that gets a bit complex … in the history chapter of the Fourth Assessment Report (AR4), the IPCC says of the origination of the concept (emphasis mine):

The concept of radiative forcing (RF) as the radiative imbalance (W m–2) in the climate system at the top of the atmosphere caused by the addition of a greenhouse gas (or other change) was established at the time and summarised in Chapter 2 of the WGI FAR [First Assessment Report].

tropopause temperature by latitude

Figure 1. A graph of temperature versus altitude, showing how the tropopause is higher in the tropics and lower at the poles. The tropopause marks the boundary between the troposphere (the lowest atmospheric layer) and the stratosphere. SOURCE 

The concept of radiative forcing was clearly stated in the Third Assessment Report (TAR), which defined radiative forcing as follows:

 The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values.

In the context of climate change, the term forcing is restricted to changes in the radiation balance of the surface-troposphere system imposed by external factors, with no changes in stratospheric dynamics, without any surface and tropospheric feedbacks in operation (i.e., no secondary effects induced because of changes in tropospheric motions or its thermodynamic state), and with no dynamically-induced changes in the amount and distribution of atmospheric water (vapour, liquid, and solid forms).

So what’s not to like about that definition of forcing?

Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.

We can measure the average surface temperature, or at least estimate it in a consistent fashion from a number of measurements. But we can never measure the change in the radiation balance at the troposphere AFTER the stratosphere has readjusted, but with the surface and tropospheric temperatures held fixed. You can’t hold any part of the climate fixed. It simply can not be done. This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement.

The problem is that the surface and tropospheric temperatures respond to changes in radiation with a time scale on the order of seconds. The instant that the sun hits the surface, it starts affecting the surface temperature. Even hourly measurements of radiative imbalances reflect the changing temperatures of the surface and the troposphere during that hour. There is no way that we can have the “surface and tropospheric temperatures and state held fixed at the unperturbed values” as is required by the IPCC formulation.

There is a second difficulty with the IPCC definition of radiative forcing, a practical problem. This is that the forcing is defined by the IPCC as being measured at the tropopause. The tropopause is the boundary between the troposphere (the lowest atmospheric layer, where weather occurs), and the stratosphere above it. Unfortunately, the tropopause varies in height from the tropics to the poles, from day to night, and from summer to winter. The tropopause is a most vaguely located, vagrant, and ill-mannered creature that is neither stratosphere nor troposphere. One authority defines it as:

The boundary between the troposphere and the stratosphere, where an abrupt change in lapse rate usually occurs. It is defined as the lowest level at which the lapse rate decreases to 2 °C/km or less, provided that the average lapse rate between this level and all higher levels within 2 km does not exceed 2 °C/km.

This is an interesting definition. It highlights that there can be two or more layers that look like the tropopause (little temperature change with altitude), and if there is more than one, this definition always chooses the one at the higher altitude.

In any case, the issue arises because under the IPCC definition the radiation balance is measured at the tropopause. But it is very difficult to measure the radiation, either upwelling or downwelling, at the tropopause. You can’t do it from the ground, and you can’t do it from a satellite. You have to do it from a balloon or an airplane, while taking continuous temperature measurements so you can identify the altitude of the tropopause at that particular place and time. As a result, we will never be able to measure it on a global basis.

So even if we were not already talking about an unmeasurable quantity (radiative change with stratosphere reacting and surface and tropospheric temperatures held fixed), because of practical difficulties we still wouldn’t be able to measure the radiation at the tropopause in any global, regional, or even local sense. All we have is scattered point measurements, far from enough to establish a global average.

This is very unfortunate. It means that “radiative forcing” as defined by the IPCC is not measurable for two separate reasons, one practical, the other that the definition involves an imaginary and physically impossible situation.

In my experience, this is unusual in theories of physical phenomena. I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable. Climate science is already strange enough, because it studies averages rather than observations. But this definition of forcing pushes the field into unreality.

Here is the main problem. Under the IPCC’s definition, radiative forcing cannot ever be measured. This makes it impossible to falsify the central idea that the change in surface temperature is a linear function of the change in forcing. Since we cannot measure the forcing, how can that be falsified (or proven)?

It is for this reason that I use a slightly different definition of the forcing. This is the net radiative change, not at the troposphere, but at the TOA (top of atmosphere, often taken to mean 20 km for practical purposes).

And rather than some imaginary measurement after some but not all parts of the climate have reacted, I use the forcing AFTER all parts of the climate have readjusted to the change. Any measurement we can take already must include whatever readjustments of the surface and tropospheric temperatures that have taken place since the last measurement. It is this definition of “radiative forcing” that I used in my recent post, An Interim Look at Intermediate Sensitivity.

I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.

w.

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GlynnMhor
December 12, 2012 8:24 pm

“… imaginary values that can never be measured or verified.”
That about says it all, doesn’t it?

Truthseeker
December 12, 2012 8:26 pm

Willis, you conclude with …
“I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.”
I think that is precisely the point. Since you cannot measure of verify these critical factors, you cannot “prove” the IPCC wrong. If you cannot prove them wrong, they must be right (at least in IPCC logic world).
A bureaucrat never lets real science get in the way of a policy for control over others.

noaaprogrammer
December 12, 2012 9:01 pm

Imaginary by design or by accident?

NZ Willy
December 12, 2012 9:11 pm

Radiative forcing has a familiar analog in albedo, which is reflectivity of sunlight. But albedo restricts sunlight downwards, “radiative forcing” restricts heat upwards. There are times when cloud cover does indeed restrict sunlight from reaching us and the sky is darkish — so, conversely, “radiative forcing” would make the Earth look cool from space, i.e. Earth would be retaining heat. But the fly in this ointment is that Earth, as a black box, must radiate more as it heats. There is no glass skin as with a greenhouse. When the clouds clear, the excess heat radiates away into space.
My car’s windshield still freezes in wintertime when parked out in the open, while the side windows remain thawed. This is because the side windows are warmed by the ground’s infrared and the windshield, which points to the sky and does not see the ground, is not. My car’s windshield tells me that there is no radiative forcing to prevent it from freezing — it shows that heat is radiating into space just like it ever has, IPCC Chicken Littles notwithstanding.

Bill H
December 12, 2012 9:18 pm

Just like the hidden agenda in the emails of climate-gate the “you cant prove me wrong” fantasies of the IPCC and Alarmists make it impossible to prove or disprove the theroy… Its kind of like hiding your work, methods, and answers then telling people you will just have to trust us..

john robertson
December 12, 2012 9:31 pm

By design, every action of the UN points to this choice of the IPCC to use imaginary values that can neither be measured, verified or discarded.
It leaves people who question the scientific claims of the IPCC in a fog, its like arguing with the mist.
It turned out that critiquing the science, with all its flaws, was just a means of tying critic up in knots.
Of course the science was settled, there is none and none required.
And after Climate gate and Copenhagen they are exposed for what they be.
Bandits and Con artists, they never had science to support the plan and never felt they needed any .
This scheme needs to be rewarded, by sentences proportional to those of Bernie Madoff.
Who, in hindsight, was just following the lead of his government.

AndyG55
December 12, 2012 9:32 pm

Willis, the tropical tropopause goes up and down.
Does it at any time go above 20km?
Also the mid lat tropopause looks like it has slight change point at 20km
I ask this because using a point just at or only slightly above a change-point seems a strange thing to do.. ……. or did you mean 20 miles (30km) ?

AndyG55
December 12, 2012 9:33 pm

Darn it ,, I am referring to your TOA height. !

Johnny
December 12, 2012 10:04 pm

It seems to me that climate science is not a science in the Newtonian sense. Results take 30 years to be disproven and even if all experimental evidence is contrary to the results of the science it doesn’t shake people’s beliefs. We have a science where the people who create the theory also get to play with the raw data and “adjust it” constantly to reflect their view of the perturbations they believe are most important so that the data conforms better to their model predictions. This is a science where repeatable experiments are impossible. Where the central physical laws proposed don’t have to be proven by any actual experiments or measurements to be “settled”. It sounds more like a middle ages religion than a science like physics or chemistry.

December 12, 2012 10:28 pm

My son, a surveyor, heard me mention watt per sq km at the troposphere and immediately interjected “At what altitude?”
Unless light passing through a horizonal sq m is parallel, like some laser light, then because there are more sq m of conceptual surface at a higher altitude quasi-sphere around the globe, the flux through each is less when there is a steady source. How does one define the flux when the “number of sq m” on a surface is forever changing rapidly with altitude, but by definition requires measurement to be taken when steady state is approached?
I have no idea about what models do about the variation in numbers of sq m shared by a watt as altitude is changed. Can someone assure me that the effect is built into the math?
If it is not, then a watt per sq m is ‘more powerful’ at low altitude than one at high and it would be hard to construct a Willis fig 1.

December 12, 2012 10:42 pm

But, Willis did you back that up with some fancy talk and show a chart?

Spector
December 12, 2012 10:48 pm

I have been assuming that they are assuming that the outgoing radiation at the top of the troposphere is presumed to be driving or ‘forcing’ the state of the troposphere as a whole.
Radiative Forcing
From The Wikipedia:
“In climate science, radiative forcing is defined as the difference between radiant energy received by the earth and energy re-radiated to space. Typically, radiative forcing is quantified at the tropopause in units of watts per square meter of earth’s surface. A positive forcing (more incoming energy) warms the system, while negative forcing (more outgoing energy) cools it. Causes of radiative forcing include changes in insolation (incident solar radiation) and in concentrations of radiatively active gases and aerosols.”
http://en.wikipedia.org/wiki/Radiative_forcing
I think that ‘Top of Troposphere, Net Radiation Input’ might be a better term, being less presumptive. I see that the MODTRAN plots they have, show only thermal outgoing radiation as a positive value.

December 12, 2012 10:50 pm

So the thing is noting but a large fudge factor that is made up to meet the needs or desires of the modelers.

E.M.Smith
Editor
December 12, 2012 11:21 pm

“Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.”
Hip Hip Hooooray!!! Willis Sees the Light!
@AndyG55:
See the chart at the bottom of here:
http://chiefio.wordpress.com/2012/12/12/tropopause-rules/
The one with density and temperature and height.
http://chiefio.wordpress.com/2012/12/12/tropopause-rules/atmosphere-temp-press-profile-1076/
Puts 20 km as top of the tropoPause and just at the bottom of the stratosphere, but not near the Top Of The Atmosphere at all, as the Mesosphere goes to 80 km…
I ‘eyeball it’ at a bit under 1% of sea level pressure though, so well above the bulk of the air mass.
It also ought to be about at the point where molecules are far enough apart that a radiative regime can begin. (Part of what I find funny about the IPCC definition is that they define a radiative based metric at the point where the air below it is, by definition, in a non-radiative driven mode. It is only above the tropopause that radiation matters to IR outbound.)

Robert Clemenzi
December 12, 2012 11:36 pm

At equilibrium, for every altitude, the flux in equals the flux out. It does not matter if it is the tropopause or not. (That is the definition of *equilibrium*.) That is why the IPCC definition has to be so strange.
As for using a linear equation – what ever the shape of the actual function is, the first derivative will provide the slope of a straight line that can be used to estimate values over some range. Since CO2 is such a minor player, it is highly likely that the error between the straight line approximation and the actual function is less than we can measure.

Larry in Texas
December 12, 2012 11:44 pm

Thanks very much Willis, this is one of the most informative pieces you have ever written (not that the others weren’t informative, but this one was well written for a layman like me). Now, if you could define for me what the “lapse rate” is, I might get a better handle on a very complicated subject for me.

December 12, 2012 11:52 pm

“This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement” This is quite consistent with all other aspects of CAGW theory. If the theory is threatened, the theory shifts ground to new unmeasureables grounds, until in turn these are threatened and the Caravan is forced to move on to Climate Change, Climate Disrution, Extreme Climate or whatever ground is still available. Slowly, very slowly, people are noticing this and quietly distancing themselves to solid ground..

December 13, 2012 12:25 am

“Top Of the Atmosphere” and “Troposphere” are useful concepts to develop climate models. They are moving targets. No exact physical definition of such boundaries will ever be possible.
“Forcing” before, during, or after allowance for thermal equilibrium to be re-established is also just a concept. Any one can chose a definition that will be adequate for his modelling purpose.
Models are no scientific hypothesis. They are speculations based on more or less plausible scenarii and mathematical formulas about the behaviour of physical parameters within a complex non-linear multi-variable system.
When model calculations get too complicated, then variable reduction and linearizations are sought to simplify them. This doesn’t make the model better but at least it becomes computable.
A model delivering a fair or a wrong prediction (into the future or as reconstruction of the past) does not prove or falsify a scientific theory. It is just a successful or a failed speculation.

December 13, 2012 12:26 am

Could this factor be similar to what we call a “heat transfer coefficient” in engineering? This is a factor that is used to calculate the flow of heat in, say, a heat exchanger where you have fluids on each side of a metallic barrier. The simplified equation is Q=UAdT, where Q= btu/hr, A=area of heat transfer surface, dT=temperature difference between the two fluids, and U=overall heat transfer coefficient. U can usually be broken down into different components to account for resistance from the hot fluid to the metal surface, resistance to conduction in the metal, and then heat transfer from the surface of the metal to the cooler fluid. And each of those heat transfer coefficients can be affected by he fluid state (liquid, gas, 2-phase), the degree of superheat of the fluid relative to the metal temperature, etc.
These coefficients are calculated/inferred from temperature measurements of the fluid and the solid metallic surface, and once calculated, are used to determine the performance of the equipment. They can be very hard to measure/calculate, so engineers usually incorporate safety factors into their overall equipment performance calculations. One would normally include allowances for fouling and corrosion in U, and you might even include additional area to the heat exchanger to make sure that the equipment performs as expected.

peterg
December 13, 2012 12:34 am

I suppose the forcing is a control theory abstraction. Add a GHG gas. Model its impact as if it were an extra radiative input. Deduce the increase in surface temperature. So it only has value as an imaginary construct for a particular way of modelling the “control” system so as to get the answers you want your model to give.

AlecM
December 13, 2012 12:37 am

Agreed, it’s a stupid parameter. However. the level of stupidity is far worse than Willy claims.
Radiative forcing is the temperature radiation field from GHG thermal emission. Except for some water vapour side bands, it annihilates the surface radiation field. There is no net ‘forcing’ so no CO2-AGW. The GHE is from surface temperature rising above the level set by lapse rate.
This is basic heat transfer physics between two emitters at nearly the same temperature that all process engineers know but far too may physicists buy into ‘back radiation’, the artefact of a pyrgeometer measurement.
This is the biggest scientific cock up in History compounded by the Houghton mistakes of assuming lapse rate is caused by GHGs and using the two – stream approximation when only net radiative flux can do thermodynamic work. A truly massive scam by con artists even now trying to justify non-science.

Liberal Skeptic
December 13, 2012 12:52 am

Untestable hypothesis is the reason people outside of the theory don’t take string theory seriously yet. This doesn’t look any different.
On a related note, I pointed out an appeal to authority logical fallacy in someone famous’s tweet regarding climate change last night and now i’ve got a bunch of people tweeting me who clearly have no idea what a logical fallacy is. No wonder they can’t see that Catastrophic climate change is just alarmism. This is basic critical thinking.

Spector
December 13, 2012 12:57 am

RE: Willis Eschenbach says: (December 12, 2012 at 11:51 pm)
“… For example, in radiation budget, TOA is considered 20 km because above that altitude the optical mass of the atmosphere is negligible.
Again, I see a standard definition is not quite what I would expect it to be, as this definition neatly avoids the strong CO2 radiation from the mesosphere. The default MODTRAN setting of 70 km would make much more sense for me. 20 km may be good enough for optical calculations, but I think it’s too low for good LWIR estimates.

richardscourtney
December 13, 2012 1:02 am

Willis:
I write to make two observations.
Firstly, Vincent Grey made your point about the non-physical nature of ‘radiative forcing’ in his peer review comments of the first IPCC Report. And he has made it in his review comments on each of the subsequent IPCC Reports, too.
I suspect it may be to your mutual benefit if the two of you were to converse on this matter.
Secondly, you rightly say that an important issue with radiative forcing is that its calculation holds surface temperature constant. However, some changes to global temperature result from surface effects which spread from the surface through the climate system; for example, Lindzen has pointed out that redistribution of heat by ocean currents could account for all twentieth century global temperature rise.
Considering radiative forcing to be THE driver of global temperature change is nonsense when there are effects whereby some changes to global temperature drive radiative forcing.
Richard

Meynard François
December 13, 2012 1:10 am

The physical problem of the coupling between radiation convection heat exchange is in general difficult to calculate exactly. Going back to the early papers of Manabe & Wetherhald, you see that an “ad hoc” algorithm has been used in order to calculate something. It consists in discoupling the calculation in two steps.
1) pure radiative heat exchange due to a so called radiative forcing.
2) then the corrective effect of the other heat flows convective.
Apparently this “CO2 green house effect algorithm” is still used today in the large 3-D climate models. But, from a physical point of view this is incorrect, unless it is proven that this algorithm reflects the real coupled heat exchange. To my knowledge this has not been scientifically established yet. Does somebody know more about this point ?

Stephen Richards
December 13, 2012 1:15 am

i’ve got a bunch of people tweeting me who clearly have no idea what a logical fallacy is. No wonder they can’t see that Catastrophic climate change is just alarmism.
You must have got one from my daughter ; )) She can talk me through her logical sequences for 10 mins and end it with me completely confused.

Roger Longstaff
December 13, 2012 1:43 am

To me, the IPCC definition says it all: “….at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium…”.
There is never radiative equilibrium in the atmosphere as a consequence of planetary rotation, so the differences between equilibrium and non-equilibrium thermodynamics make the IPCC analysis fatally flawed. Add to this the incorrect use of mathematics (they do not use Holder’s Inequality, as far as I am aware) and the whole thing becomes completely meaningless.

peter Miller
December 13, 2012 1:45 am

This reminds me of a discussion I had many years ago with the late Prof. Krige, when he was supervising my Master’s thesis.
The statistical method of kriging is well known in the mining world for resource and reserve calculation.
I was telling him I could not duplicate the results of two companied’ official reserves from drill results.
He looked at my workings and commented: “but you haven’t made the necessary adjustments.”
So I asked: “how do I know what ‘adjustments’ to make?”
His response was: “from observations.”
Ever since that day, I have treated statistical models and interpretations with a sack of salt.
So, in the case of radiative forcings, we need to make a huge number of observations to ascertain some form of average unrepresentative value and this, from a practical point of view, is virtually impossible. ,
Makes perfect clumate science sense to me.

E.M.Smith
Editor
December 13, 2012 1:56 am

Oh what a Humpty Dumpty world… where even “Top Of Atmosphere” doesn’t mean the actual top of the actual atmosphere but varies based on what you want it to mean… Sigh…

richard verney
December 13, 2012 2:11 am

NZ Willy says:
December 12, 2012 at 9:11 pm
/////////////////////////////////////////////////////////////////////////
Willy, I made a very similar observation about 6 or 8 montha ago on another article written by Willis. Save that I was talking about dew, and save that I would suggest that it is warm air being convected from the ground that keeps the car doors/door windows ice free.
Is it not the case that dew would be a very rare event if DWLWIR truly had sensible energy? After all it is claimed that DWLWIR can warm the oceans (although I am very sceptical of that claim due to the absorption characteristics of water and since the top micron layer of the ocean is frequently little more than airborn windswept spray and spume such that it is divorced from the ocean below).
Without checking, I believe that 40% of all DWLWIR is a absorbed within 1 micron of water and 60% within 2 or 3 microns. With that absorption and the DWLWIR figure given by Trenbeth, there is enough energy to evaporate water such that dew (which is often fractions of a millimetre)would quickly evaporate.
Whilst the point you raise is slightly off topic, I think that you are right to look at the practical application of the theory to the real world as we experience every day in our lives.

TimTheToolMan
December 13, 2012 2:11 am

Willis notes “This means that the area of the TOA layer is about 0.6% larger than the surface of the earth. As a result, for any but the most detailed of analyses, the difference can be (and usually is) safely ignored.”
Is it really that large? That would make the difference between the TOA and the surface averaged over the entire planet about 2 W/m2. Considering the net absorbed energy is thought to be less than 1 W/m2, I dont think that could be considered too small to worry about!
http://spark.ucar.edu/longcontent/energy-budget

Dumbed Down Science
December 13, 2012 2:26 am

The IPCC are the true deniers, scientific charlatans degrading climate, one false hypothesis after another.

AndyG55
December 13, 2012 2:54 am

E.M.Smith says:
Puts 20 km as top of the tropoPause and just at the bottom of the stratosphere, but not near the Top Of The Atmosphere at all,
Seems it is a NASA definition.. seems pretty silly to me to use a point so close to the change of the troposphere. but i guess if it makes it easier for them !!!!
Just saying 😉

AndyG55
December 13, 2012 3:00 am

What I really am trying to say is that its always better to do your calculation “away” from areas where things could be changing. and at 20km , we are very close the point where deltaT changes from +ve to -ve. Not clever.

December 13, 2012 3:13 am

E.M.Smith says:
December 13, 2012 at 1:56 am
Shut up and obey you’re masters

AndyG55
December 13, 2012 3:20 am

Hey Willis, in figure 1, what lapse rate is being used ??
I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

Mack
December 13, 2012 4:00 am

Tim the Tool Man,
Good show putting up that link to the Earth’s Energy budget…..none of the figures you see in that cartoon are actual measurements but may well have been pulled from Trenberth’s ass.

December 13, 2012 4:00 am

I was convinced Willis’ logic, integrity and intellect would eventually lead him to discard the added-CO2-causes-warming theme. Anyone with engineering skill would know what happens when you add additional agents of conduction or convection to a system. Agents of conduction are agents of cooling. Anyone puzzled by what radiation can do should simply imagine the radiative path replaced by a conductive path. Conduction, convection and radiation always work in the same direction. The rate and effectiveness are different, but the outcome is always in the same direction.

Bob Ryan
December 13, 2012 4:01 am

Willis: the problem you point too has been a long running issue in the philosophy of science. You may be interested in the work of David Papineau and in particular his book ‘Theory and Meaning’. He would argue that much of modern science is made up of theoretical language and constructs which are not directly observable or measurable. He has a very interesting talk on this issue at http://philosophybites.com/2009/01/david-papineau-on-scientific-realism.html.

Bob Ryan
December 13, 2012 4:04 am

Sorry, hasty read through. My comment at 4.01am: in the first sentence it should be ‘to’ not ‘too’.

December 13, 2012 4:07 am

The thing I find most troubling about these discussions, is that the warmists claim that they can accurately estimate the change in radiative forcing which occurs when CO2 concentrations change, by using radiative transfer models. Now my understanding is that radiative transfer models are engineering type models which have been fully validated. I cannot understand why radiative transfer models are suitable to estimate radiative forcing. They were never designed to estimate radiative forcing, and no-one seems to have shown that they are capable of estimating radiative forcing. Myhr et al 1998 merely states that they have used three such radiative transfer models, with not a word as to why they are suitable to do the job. Any comments, Willis?

Dirck
December 13, 2012 4:12 am

“A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. ”
That is not a fundamental part of anything except your catastrophic misunderstandings of climate science.

richardscourtney
December 13, 2012 4:22 am

Dirck:
Your post at December 13, 2012 at 4:12 am says in total

“A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. ”

That is not a fundamental part of anything except your catastrophic misunderstandings of climate science.

You are wrong.
See? I have just repeated your silly game. It’s not helpful, is it?
And, incidentally, if pressed I can show that the IPCC says you are wrong, but I won’t until you have justified your untrue assertion that Willis’ statement is wrong.
Richard

phi
December 13, 2012 5:02 am

Willis Eschenbach,
“I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.”
It’s even worse than that. GHG forcing is in fact undefined and indefinable. The need to calculate an instantaneous balance just after the addition of material created ex nihilo does not allow this material to adapt his temperature to the environment. As the temperature of material created ex nihilo is not defined, the GHG forcing is not defined either.

Ryan
December 13, 2012 6:40 am

I have been thinking of a way to demonstrate to the layman the basic problems with the greenhouse theory and I came up with this. It assumes you have a so-called “radiator” in your house and it is working hard this winter.
Put your hand on the “radiator”. Is it hot? Well that’s because it has hot water inside (about 60Celsius so not THAT hot but still pretty warm) and the well-known phenomena of conduction is carrying that heat to the surface. Now put your face in front of the radiator – feel the heat? No? Well what your feeling (or rather not feeling) is the radiation coming off the “radiator”. Now put your face just above the hot “radiator”. Do you feel the heat now? Sure you do. You can actually feel a slight breeze of warm air past your face due to the well-known phenomena of convection. The fact is the phenomena of conduction and convection are far more powerful means of removing heat energy from warm surfaces than radiation is. The hotter the surface the more conduction and convection you will get.
For the atmosphere we get the same thing. During the day the land and sea gets warmed by the sun. Replacing oyxgen by CO2 has almost no impact on this so peak daytime temperatures of land and sea cannot be impacted by greenhouse gases. The air gets warm too – partly from the suns rays directly but in great part due to the land and sea causing convection currents in the air. These convection currents go up a long way – right up to the troposphere at about 30,000 ft where the jet planes and the high cirrus clouds are. The closer you are to the ground the stronger the convection currents are – on a hot day you can often see the effect of strong convection as the heating effect is enough to bend the light through the air due to changes in density. Close to the ground the air temperature is determined almost entirely by these strong convection currents.
The ground and sea itself maybe at a completely different temperature to the air just above it. Why is that? During the day the suns rays penetrate the atmosphere. It doesn’t matter if we think of the visible sunlight or infra-red – if it didn’t penetrate the atmosphere the Earth would be a cold dark place. It isn’t. Those photons ejected by the sun come hurtling across space, through the air and then meet the immovable object known as the land. Those photons might on occassion give up their energy to the air but most simply crash into land or sea where they are completely absorbed. The suns rays hitting the atmosphere may cause it to heat up quite readily because the air isn’t very dense – it is a different story for land and sea because the high density of atoms in both makes heating the land and sea hard work for those incoming photons. CO2 plays absolutely no part in this at all.
It is during the night that CO2 has a part to play. It acts like a nightime insulating blanket has appeared over the Earth, but it is lightly distributed throughout the atmosphere at low density. However, it isn’t a proper insulating blanket because in such a blanket the air is in pockets and is not free to move – in the lower atmosphere there are strong air currents always conducting heat away from the warm land and sea. At ground level CO2 cannot stop much of the heat radiating from planet Earth (and indeed if it did then we would already be at saturation point and adding more CO2 wouldn’t make any difference) and as far as conduction and convection is concerned CO2 is just as capable of conducting heat away as oxygen. Frankly CO2 can make no difference at ground level. Even just above ground level the air itself may be warmer but there is little energy in warm air itself because it has low density and that extra energy is distributed throughout the atmosphere. You cannot melt the ice of the Arctic by simply making the nightime air above it very slightly warmer – it takes far too much energy to melt water and there just isn’t ever enough trapped in the air. You can only melt the ice by making the Arctic oceans warmer – and that cannot be done by making the air a slightly better insulator.
In summary the greenhouse effect largely ignores the fact that where life exists on Earth the temperature is almost entirely dependent on absorption, conduction and convection.

BillC
December 13, 2012 6:53 am

Willis,
Apologies for a shorter than necessary comment to explain what I’m thinking but I’d like your reaction if any. Isaac Held has blogged about a radiative forcing term defined where you let not only the stratospheric temperatures adjust, but also the immediate surface and atmosphere. This forcing is used to constrain the heat flux into the ocean. See the last (long) paragraph of this post.
While this doesn’t satisfy several of your criteria, it at least provides a break point on the flux/temperature diagram that is more physically meaningful, and in the absence of feedbacks should yield a fairly linear flux-to-temperature relationship at a given, single location on the globe.
thoughts?

JP Miller
December 13, 2012 6:58 am

I wonder what would happen if Willis, E.M. Smith, and Stephen Wilde were to put their ideas, data, and analytical approaches together and present a new formulation for climate processes including the likely impact of CO2 and water vapor. I imagine them getting it vetted by thoughtful professional climate heavy-weights and it getting published in Nature. Nah, never happen. The world’s poorer for it…..

phi
December 13, 2012 7:57 am

François Meynard,
(December 13, 2012 at 1:10 am)
“Apparently this “CO2 green house effect algorithm” is still used today in the large 3-D climate models.”
I ask myself this question for a long time. The fact is that the narrative always considers convection as feedback. This is obviously unacceptable in a potentially unstable system.

Chris B
December 13, 2012 8:03 am

AlecM says:
December 13, 2012 at 12:37 am
======================
Here’s what I found in “Lapse Rate” on Wiki (I don’t pretend to understand it):
Thermodynamic SS/Radiative GHG lapse rate
Robert H. Essenhigh developed a comprehensive thermodynamic model of the lapse rate based on the Schuster-Schwarzschild integral (S-S) Equations of Transfer that govern radiation through the atmosphere including absorption and radiation by greenhouse gases.,.[11][12] “The solution predicts, in agreement with the Standard Atmosphere experimental data, a linear decline of the fourth power of the temperature, T^4, with pressure, P, and, at a first approximation, a linear decline of T with altitude, h, up to the tropopause at about 10 km (the lower atmosphere).” The predicted normalized density ratio and pressure ratio differ and fit the experimental data well. Sreekanth Kolan extended Essenhigh’s model to include the energy balance for the lower and upper atmospheres.[13]

December 13, 2012 8:12 am

As a son of a rocket engineer, I find all this talk about “top of the atmosphere” at 20 km to be rather misguided.

If a space vehicle comes within 120 to 160 km of the Earth’s surface, atmospheric drag will bring it down in a few days, with final disintegration occurring at an altitude of about 80 km. http://www.braeunig.us/space/orbmech.htm

OK, difference science, different criteria.
But why is 20 km considered to be TOA for climate science?
20 km is only 65,617 feet. The anvils of thunderheads reach that high. The U-2 and SR-71 are air-breathing planes that aerodynamically fly above that altitude.
20 km might be “Top of the Weather”, but it is misguided to think of it as a real Top of the Atmosphere.

Jeff Alberts
December 13, 2012 8:23 am

My car’s windshield still freezes in wintertime when parked out in the open, while the side windows remain thawed. This is because the side windows are warmed by the ground’s infrared and the windshield, which points to the sky and does not see the ground, is not. My car’s windshield tells me that there is no radiative forcing to prevent it from freezing — it shows that heat is radiating into space just like it ever has, IPCC Chicken Littles notwithstanding.

My experience is different. My car’s windows won’t frost over when exposed to sunlight, but any in the shade will, whether the windshield or side windows.

Jpatrick
December 13, 2012 8:42 am

Yes. When you do “science” without operational definitions the results can be interpreted as the beholders wishe.

Roger Longstaff
December 13, 2012 8:43 am

Stephen Rasey says: December 13, 2012 at 8:12 am:
“As a son of a rocket engineer, I find all this talk about “top of the atmosphere” at 20 km to be rather misguided”
I completely agree. The system that I work on (SKYLON) breathes air up to 26 km, in a lifting, climbing, accelerating trajectory. To define TOA @ 20 km is moronic.
In my view, TOA should be defined as a notional boundary at which only radiative energy transfer is possible – 300km, anyone?

john robertson
December 13, 2012 8:44 am

@JP Miller 6:58am. Just stay tuned we are watching that happen on WUWT, the future of honest science? And a forum of empirically based speculation.As science is never absolute, informed speculation is the next best thing.
With so much specialization in the sciences now, it is necessary for outside points of view.
The field description of expert(Drip under pressure) has never been so accurate.
Open access and source can provide exposure of the “minor error” inherent in those crucial misconceptions that seem to ooze out of our universities.
As for climate heavy weights… probably more on the blogs than present in the IPCC team .

mkelly
December 13, 2012 8:50 am

http://climaterealists.com/index.php?id=9004
The link above shows what power CO2 has. The photo is of someone back yard. It shows what NZWILLY spoke of concerning his wind shield. CO2 has no power to warm the earth as it cannot even melt alittle frost.

December 13, 2012 8:54 am

Jeff
Now I wish I had taken a photo of neighbour’s car at 3pm, just before sunset. Both the windscreen and all the side windows were still frosted over. They were like that at dawn. We had sunshine but a temperature of around freezing did not let the ice melt. Also, my other neighbour has a garage with a flat roof. Some rain fell on it a few weeks ago and did not clear before temps fell below freezing. The glacier is still there and does not appear to be receding. I live close to London.
And it is brass monkeys in my barrel.

Robert Clemenzi
December 13, 2012 9:08 am

Larry in Texas asks:
December 12, 2012 at 11:44 pm

could define for me what the “lapse rate” is?

In this context, it is the change in temperature with a change in height.
AndyG55 says:
December 13, 2012 at 3:20 am

in figure 1, what lapse rate is being used ??
I’m really tired, but for say the polar regions, a drop of 50C in just under 9km doesn’t make sense.

That drawing is obviously wrong. The polar lapse rate is drawn as about 7.2 K/km, the correct value is about 6.5 K/km. Also, the polar tropopause is not drawn correctly. Over Antarctica, in winter, the tropopause is higher and colder than the tropical tropopause. Figure 1 shows a shape more like what is found over Alaska, not over the pole.
All the books say that the tropopause is about 17 km at the equator and 7 km at the poles. However, on 08-16-2008, the Antarctic polar tropopause was over 20 km.
If you want to see what the tropopause really looks like, you can plot real radiosonde data. I have a program that will show a full year’s data for several different locations.
http://mc-computing.com/Science_Facts/Lapse_Rate/Lapse_Rate_Animations.html

NZ Willy
December 13, 2012 10:16 am

richard verney said: “Willy, I made a very similar observation … save that I would suggest that it is warm air being convected from the ground that keeps the car doors/door windows ice free.”
Warm air cannot be the agent which keeps cars’ side windows ice free, because that would equally well keep the windshield ice free — but the windshield freezes. Furthermore, the grass tops in nearby fields, and rooftops, freeze too. This is a common phenomenon in New Zealand in winter. What they all have in common is that they do not see the ground, and so don’t get the infrared radiation from the ground.
If there was “radiative forcing”, then these sky-facing surfaces would not freeze as they do because radiative forcing equates to infrared from the sky. But they do freeze, so there is no infrared from the sky, ergo, no radiative forcing. So my point is totally on-topic, despite your further demur.

Curious George
December 13, 2012 10:40 am

The argument seems to be based on average temperatures. However the total (blackbody) radiation is proportional to the fourth power of the absolute temperature, so we are “averaging” a highly nonlinear relationship. For example, the total blackbody radiation more than doubles when temperature rises from -23C (250K) to 37C (310K).
A related question: how good a blackbody the Earth surface is? Do oceans, deserts, forests, cloud tops behave the same way? Do IPCC calculations involve the heat capacity, which varies wildly in these examples? Do they consider daily or seasonal variations?

Robert Clemenzi
December 13, 2012 10:50 am

NZ Willy says:
December 13, 2012 at 10:16 am

If there was “radiative forcing”, then these sky-facing surfaces would not freeze as they do because radiative forcing equates to infrared from the sky. But they do freeze, so there is no infrared from the sky, ergo, no radiative forcing.

It is my understanding that a freezer makes ice even though the inside walls of the freezer are emitting infrared radiation. This is because everything above absolute zero emits infrared radiation. Therefore, the fact that something freezes does NOT prove that there is “no infrared from the sky“. It only demonstrates that the amount of IR from the sky is less than the amount of IR released in forming frost.

Gail Combs
December 13, 2012 11:06 am

Johnny says:
December 12, 2012 at 10:04 pm
…. It sounds more like a middle ages religion than a science like physics or chemistry.
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
Actually my husband pointed out that IPCC ‘science’ most closely resembles Alchemy.
It looks like the problem does not reside in just climate science either.

This … interview with Eric S. Raymond I think summarizes the Alchemist position prevalent in much of clinical medicine with regard to its software today: ‘Page 8:
The occult school of alchemy didn’t turn into the science of chemistry until alchemists abandoned the practice of secrecy and instead started sharing results with each other and checking each other’s experiments. And that was a very early stage in the development of modern science and engineering. It happened about 400 years ago. And the thing that we’ve discovered over the last 400 years, as we’ve pursued experimental science and developed engineering from an art into a craft into a repeatable discipline, is that human beings doing complex, creative work, doing design work, make mistakes. There is no way to mechanically check the results of creative work. If you could do that, it wouldn’t be creative work; it would be something you could do with a machine. So the only way to check complex, creative work for correctness is by the critical judgment of peer experts…’
http://linuxmednews.com/1078854224/addPostingForm

So it looks like we are taking a GIANT STEP BACKWARDS 400 yrs to the middle ages. Of course this is where the Neo-luddites want to drag us back to in all technologies.

Stephen Wilde
December 13, 2012 11:35 am

“But when a cloud passes over in that winter night, it is immediately much warmer. That is the effect of the downwelling longwave radiation from the cloud.”
I don’t think so.
Descending air is adiabatically warmed by conversion of PE to KE as air descends from above,
With a clear sky at night, radiation from the surface is faster than the warming effect of descending air and an inversion forms.
If the descending air is moist, clouds then form at the inversion height because the surface cooling reduces the temperature of the descending air to below its dew point.
Those clouds are then at the temperature of the descending, adiabatically warmed, air from above and thus warmer than the air at the ground so they start to conduct energy downward and that offsets radiative cooling from the surface, often warming the lower layer in the process.
So, why should you believe that?
Consider the difference between a low cloud base and a medium or high cloud base.
The nearer the cloud base to the surface the greater the warming effect which points to conduction as the cause.
If downward IR were the cause it would make no difference whether the cloud base were high or low.
We know full well from observations that low clouds break down an inversion more effectively than high clouds.
So, conduction it must be.

Frank
December 13, 2012 11:36 am

Willis: Here are some points your presentation missed:
We can measure the radiative imbalance (not radiative forcing) from space by starting with incoming solar and then subtracting reflected solar and OLR. The errors in these measurements apparently are too large to make a useful comparison between observed radiative imbalance and theoretical radiative forcing.
The upward and downward radiative flux through the atmosphere can only be calculated if one specifies the composition and temperature (which controls emission) everywhere along the path. In instantaneous change in composition will produce and instantaneous change in flux. In locations where convection is not important (at the tropopause and higher), one can use the power absorbed and emitted by the changed flux to calculate how the temperature will change WITH TIME when assumptions about the composition of the atmosphere are changed. The radiation flux is then recalculated with the new temperature information and a new equilibrium state will be reached. This is what the IPCC does with its defines radiative forcing after the tropopause and the stratosphere reach radiative equilibrium. Since temperature below the tropopause is controlled mostly by convection and the lapse rate, calculations based on radiative equilibrium make no sense. Instead, the IPCC assumes that (to a first approximation) the lapse rate remains constant so that a 1 degC temperature increase at the tropopause will be linked with a 1 degC rise at the surface. The tropopause is special because it allows one to realistically calculate temperature change by radiative forcing/equilibrium and have some rational – a fixed lapse rate – for expecting that temperature change to be relevant to what happens at the surface.
One can calculate radiative forcing at the surface from 2X CO2 (about 1 W/m2). Since the average photon reaching the earth will have been emitted from GHG’s close surface where it is warmer, 2X CO2 will increasing DLR. At the surface, however, you can’t predict whether the additional energy will leave the surface by convection or increased OLR (after surface warming).
The IPCC’s definition of radiation forcing has the limitations you describe. However, it is the best system we have for coming up with one number that compares and integrates the effects of changing atmospheric gases, solar output, aerosols, surface albedo, clouds, etc. When small percent changes in radiative forcing and temperature are involved, a linear relationship is a reasonable approximation.
We sent radiosondes into the atmosphere twice a day from about a hundred locations on the planet. We have excellent information about what is happening in the vicinity of the tropopause.

anticlimactic
December 13, 2012 11:37 am

We are expected to believe without question that GHGs warm the Earth in various ways, but the more I think about the less I can see how they can.
All atoms in the universe are constantly trying to shed their heat and fall to absolute zero [-273C] and the only reason they don’t is that they are usually receiving about as much energy as they are losing.
Assuming a GHG molecule absorbs some heat from the Earth, it will be immediately reradiated, roughly 50% back to Earth and 50% out in to space. When the heat returns to Earth it will again be reradiated and again only 50% of any captured heat will be returned – and so on.
The warming effect of GHGs will approach zero in less than a minute!

richard verney
December 13, 2012 11:49 am

NZ Willy says:
December 13, 2012 at 10:16 am
///////////////////////////////////////////////////////////////////////////////////////
I agree with the general thrust of your final paragraph, although never freeze may be a bit too strong.
As regards the second para, you overlook that radiative heating is the least efficient form of heating. I have gas central heating with radiators. Ther gas boiler heats water which is circulated throgh the radiators. One has to place one’s hand within about 1/2 inch of the horizontal panel which sits parallel with the wall before one can feel noticeable heat. On the other hand, one can feel noticable heat when one’s hand is three feet above the top of the radiator. This is not withstanding that the surface area of the top is a fraction of the surface ares of the horizontal panel. Although they are called radoators, it is convection that performs the bulk of the heating. The same is so with a BBQ. One cannot cook meat 4 inches from the side of a BBQ whereas one can cook meat a foot above it. Radiative heating soon gets swamped by convection.
As regards the car, the windscreen, the bonnet and the roof are in the shaddow of the convective heat being given off by the ground on top of which the car is parked, and the adjacenet ground surrounding the car. Unless there is a lot of wind, the convective heat cannot heat the windshield because it is rising vertically.
But consider my point with dew. Consider Spring or Autumn. You have a hollow. When the sun rises, it heats one side of the hollow, the other side of the hollow remains in shaddow. Relatively quickly the dew is burnt off on the sunny side. However, dew on the shaddow side can linger most of the day.
Now Trenbeth suggests that DWLWIR has almost double the power of solar irradiance. In my example, with early morning sun the angle of incidence is low so solar energy in this example is very weak (much below the Trenbeth average). Not withstanding that it is very weak, it can relatively quickly burn off the dew on the sunny side of the hollow. Yet even in hours, DWLWIR cannot burn of the dew in the shaddow part of the hollow. Why is that if DWLWIR has real/sensible energy?
One may be able to measure a signal from DWLWIR (Willis I do not doubt that one can measure a signal). But the issue is, can it do any work? Can it actually heat anything? (Willis is there any published paper showing that DWLWIR actually heats something, rather than mere speculation?) If DWLWIR cannot burn off dew in the shaddow area of a hollow notwithstanding exposure of the dew to DWLWIR of many many hours, I would suggest that it appears that DWLWIR lacks significant sensible energy.
PS> My example is a deep hollow, or perhaps a valley, one side is sunny and the other is not.

mkelly
December 13, 2012 11:50 am

Willis says: “Thanks, mkelly. I see you and other folks above making the argument that since the downwelling longwave radiation doesn’t do what y’all expect it to do, it must not exist,…”
Willis you often get annoyed with folks who don’t quote you or put words in your mouth. I have never said that it does not exist I have said and maintain that it has not power to do anything. ie work. So please don’t think or say that I said it does not exist. The question is can it do anything and I say no.

phi
December 13, 2012 12:02 pm

Frank,
“…the IPCC assumes that (to a first approximation) the lapse rate remains constant so that…”
Yes. But the added CO2 is not constrained by the lapse rate because this gas is created ex nihilo. Emulate its temperature on the lapse rate is an arbitrary operation. The value obtained (radiative forcing) is therefore also arbitrary.

richard verney
December 13, 2012 12:05 pm

Stephen Wilde says:
December 13, 2012 at 11:35 am
////////////////////////////////////////////////////////////////////
The cloud example is a difficult example and proves next to nothing because one cannot separate the various mechanisms that may be at play, and thereby correctly ascribe cause and effect.
The cloud is an example of the classic greenhouse, ie., it really is akin to the glass roof in that it hinders convection thereby trapping warm(er) air. So whilst one may feel less cold on a cloudy night, one does not know whether this is due to the points you raise to which I would add the obstruction of convection (which would otherwise be more efficiently cooling the body of air immediately above the ground), or whether as Willis argues there is more DWLWIR.

george e. smith
December 13, 2012 12:40 pm

Well nowhere in the earth climate system, is anything IN EQUILIBRIUM.
“””””…..The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values……””””
In particular, given that the earth rotates, so that no point on earth ever is in thermal equilibrium. or radiative equilibrium, so this definition is BS.
Moreover, there is no known way to hold either surface or tropospheric temperatures or “state” fixed at any value let alone any unperturbed value.
Who are these people who think it is ok for them to simply make up their own physical laws, and make comlex systems obey their silly ideas ?

AndyG55
December 13, 2012 12:56 pm

Its morning here now, and even though I haven’t had my coffee yet…
It appears Figure 1 is based on wet lapse rate.
If you take the tropical (pink) line.. its surface value is 20C (ignore where the numbers are, they are badly printed) That line crosses the grid again at -30C at 9km so the slope is 5.5C/km
Same slope with the polar (blue line) but isn’t polar air mostly dry ?
Still.. the point sits that using 20km as TOA is really pretty ridiculous.

Mack
December 13, 2012 1:05 pm

Comment from Ryan Dec 13th 2012 6.40am …..Good comment Ryan . Well said.

Stephen Wilde
December 13, 2012 1:12 pm

richard verney said:
“So whilst one may feel less cold on a cloudy night, one does not know whether this is due to the points you raise to which I would add the obstruction of convection”
I agree, Richard.
Suppression of convection facilitates downward conduction.
It cannot be anything to do with DLR because if it were the effect would be the same from both high and low clouds yet we know it is not.
I don’t see it as anything to do with DLR in any event because the solar throughput is unhindered at equilibrium. At equilibrium solar shortwave in must equal longwave out at top of atmosphere.
Solar shortwave comes in and planetary longwave goes out and at equilibrium they are always equal.
What matters is how the atmosphere intervenes and that is a constant related to mass. gravity and insolation and nothing else.
PE + KE = Constant.
Only mass, gravity and insolation set that constant.
The atmosphere shuffles KE and PE through the speed of adiabatic processes as necessary to maintain that constant.
If the radiative characteristics of individual atmospheric components vary so as to disrupt the balance between solar shortwave in and longwave out then the atmospheric heights change to redistribute KE and PE to negate the imbalance.
The constant being set by mass, gravity and insolation, any other factor such as the radiative characteristics of individual constituents of the atmosphere can only affect the proportion of KE relative to PE.
The more the atmosphere expands from more GHGs the more PE there can be and the less KE (cooling) and the more the atmosphere contracts from less GHGs the less PE and the more KE (warming).
GHGs might cause the atmosphere to expand but if they do that then PE increases relative to KE and there is an equal and opposite cooling effect due to the reduction in KE.
If they cause the atmosphere to contract then KE increases relative to PE and there is an equal and opposite warming effect due to the increase in KE.
That is the true Thermostat Hypothesis and Willis’s version is correct as far as it goes but it does not extend to the entire globe which is where it fails.

Stephen Wilde
December 13, 2012 1:15 pm

“Heck, many clouds are created by convection, they are areas of increased rather than decreased convection, and despite that, they are sources of increased downwelling radiation ”
Really ?
Clouds created by convection as compared to clouds formed beneath an inversion layer result in air flowing in from surrounding sunlit regions to feed the uplift.
No increase in DLR there.

Curious George
December 13, 2012 1:22 pm

The argument seems to be based on average temperatures.
Willis: What argument are you talking about? Details, friends, the devil is in the detail.
The red, blue, and green lines in Fig. 1 are all averages – I may be wrong. Has IPCC ever considered temperature variations in space or time? I thought Willis might know.

mkelly
December 13, 2012 1:32 pm

Willis Eschenbach says:
December 13, 2012 at 12:52 pm
“So where does the energy go when such radiation hits the earth?”
First if you did not indend for me to be included in the group of folks that say the radiation for CO2 does not exist then I with draw my comment.
It is called reflection. You assume it is absorbed.
Just because radiation strikes a body does not mean it is absorbed and turned into thermal energy (heat).
I have said before if you or others can write a radiative heat transfer equation showing how CO2 warms the surface of the earth I’ll happily accept that. ie makes the surfaced go from 255K to 288K.

AndyG55
December 13, 2012 1:34 pm

Ummm.. clouds are where the convective flow stops.
If you stop the convective flow, the convective flow below is forced to slow down.
Upward energy transfer is thus slowed down.
Think of a big heavily laden truck going up a hill on a one lane road, with many cars behind that want to go faster.

richard verney
December 13, 2012 1:37 pm

(since heat cannot be effectively conducted down

richard verney
December 13, 2012 1:40 pm

Some posts have been a little off topic, but that is no doubt because one can hardly add to Willis’s closing comment. It is only too true that one can’t have any particular conclusions given the use of imaginary values that can never be measured or verified.
I am in the mkelly camp that the issue is not whether DWLWIR exists, but rather it is one of whether DWLWIR has sensible energy/the ability to do real work in the Earth environment.
The key to AGW is the behavoir of DWLWIR and water. The reason for this is threefold. First, it is the oceans that control the weather. Second (and this is related),the latent heat content of the oceans dwarfs the latent heat content of the atmosphere. Third, the absorption of DWLWIR in water is very different to that in dry air (such that the behavoir of DWLWIR over the oceans is different to that over land). IR cannot penetrate water to any significant degree such that water acts as an IR block. Almost all IR is absorbed within 10 microns, with about 40% of all DWLWIR being absorbed within 1 micron and 60% within 3 microns. Sunlight can penetrate water to great depths. This is fundamental.
The power from sunlight can warm water; it does not immediately cause the top micron layers to burn off/evaporate but instead the energy penetrates to depth slowly heating a vast volume of water at depth. IR is very different since approximately 60% of all IR is absorbed within 3 microns. This means that energy in IR does not heat a vast volume of water at depth, but instead it is concentrated and heats a relatively small volume at the very top 1 to 3 micron layer. Since the energy is not dissipated, but concentrated, the very top microns surely must burn off.
If one considers the amount of DWLWIR that is, according to Trenbeth/AGW proponents, hitting the oceans, as a matter of the absorption characteristics of water, the energy in this (assuming that DWLWIR can perform sensible work) would lead to the top micron layer quickly evaporating. If this was happening, then there would be 10s of metres of rainfall annually, which there is not. The fact that there is not such an amount of rainfall suggests: (i) the characteristics of absorption of IR in water is wrong (extremely unlikely) such that IR is not completely absorbed in the first microns but can infact penetrate many metres, (ii) DWLWIR does not exist or not in the quantity postulated by AGW proponents, or (iii) DWLWIR lacks sensible energy. It is a signal but cannot perform real work in the conditions encountered in Earth’s environ.
Until one can witness DWLWIR performing real work such as burning off dew, or melting ice, questions quite obviously arise as to whether it does (or does not) possess sensible energy.
PS. If one looks at the temperature profile of the oceans, the top few microns are cooler than the top 15 microns such that the direction of heat flow is up towards the surface, not down towards the deeper ocean.

richard verney
December 13, 2012 2:17 pm

PS. In my last post where I use IR, I mean LWIR. The wavelength is very important to the characteristics of absorption.

December 13, 2012 2:26 pm

One can generally state the Forcings Theory as,
The Earth’s climate warms/cools from changes in the net radiation balance at a hypothetical contiguous shell that surrounds the climate.
TOA would fit the contiguous shell surrounding the climate criteria. And the tropopause would, as you point out, seem to fail the contiguous criteria.
Ignoring heat from the Earth’s interior, the Forcings Theory, as stated above, would appear to be a priori true. But being a priori true, doesn’t make it science.
Firstly, we cannot measure (at our hypothetical shell) the radiative source for most radiative forcing changes. Therefore cannot measure the change in a particular forcing.
Secondly, the theory gives us no criteria to determine whether something is a radiative forcing or not. Which means there may be yet undiscovered forcings, as GCRs were a few years ago.
All in all, I have trouble considering the Forcings Theory as properly scientific.

December 13, 2012 2:42 pm

The graphic showing a smooth variation in troposphere height from equator to poles is wrong.
There are discontinuities between the equatorial, mid latitude and polar tropopauses. The jet streams occur there.

Matthew R Marler
December 13, 2012 5:00 pm

I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable.
Historically, most of them have done that aplenty. Conceptual and definitional clarity have usually followed decades of confusion. Consider the fundamental concept of “ergodic”, described in Lawrence Sklar’s book “Physics and Chance”, or just about any history of atomic theory.
Possibly the most famous imaginary is the “body in uniform motion”.

TimTheToolMan
December 13, 2012 5:08 pm

Willis writes a strawman argument “Go buy an infrared lamp, put it over a pan of water, and see what happens.”
You have just made DLR > ULR which isn’t the case for the ocean. You’ve fallen into the precise trap that you’ve argued all along doesn’t matter regarding whether something is cooling more slowly or actually “warming”.

Mack
December 13, 2012 6:07 pm

Willis,
Maybe I can get a bit of traction over DLR “can’t warm the ocean” I suppose anything in nature is possible but what you are contending here is that a trace gas wafts around in the air above the ocean waves and warms the water beneath more than otherwise by the sun. Call me a stickler for not introducing amounts and figures but when thinking about all this I get a feeling of insignificance and impossibility, and so should you.

Jeff Alberts
December 13, 2012 6:33 pm

Diogenes says:
December 13, 2012 at 8:54 am
Jeff
Now I wish I had taken a photo of neighbour’s car at 3pm, just before sunset. Both the windscreen and all the side windows were still frosted over. They were like that at dawn. We had sunshine but a temperature of around freezing did not let the ice melt. Also, my other neighbour has a garage with a flat roof. Some rain fell on it a few weeks ago and did not clear before temps fell below freezing. The glacier is still there and does not appear to be receding. I live close to London.
And it is brass monkeys in my barrel.

I’m not doubting what you saw, just doubting that it’s due to radiation from the ground.
I live north of Seattle, on the US northwest Pacific coast. When the ground frosts over here, any areas not hit by sun will usually stay frosted over throughout the day if the ambient temp doesn’t get too far above freezing. One road I drive to/from work has about a mile stretch that never gets direct sun. It will stay frosty all day long, even though all the other roads are dry and clear. It’s pretty dangerous.
I can often also see frost and non-frost areas side by side where the shade and sunlight meet.
Sorry, not getting the brass monkeys reference 🙁

TimTheToolMan
December 13, 2012 7:03 pm

Willis write “Not sure what your point is here, Tim.”
I know. Its because you dont actually understand the issue yet otherwise you wouldn’t have put up the strawman argument.
Willis writes “If you don’t, as I mentioned above, you should read “Radiating the Ocean“. ”
Actually I think it is you who should read that thread. Specifically my comments within it 😉

jae
December 13, 2012 7:41 pm

The problem, Willis, is that there is absolutely no empirical evidence that the backradiation does anything, as some of the commenters say. Of course the backradiation exists, but It is simply a fact of physics, a property of matter. It does no necessarily have anything to do with the temperature of the atmomsphere. The construct of “backradiation” looks right, because the backraidiation does exist. But IT IS ONLY A PROPERTY OF MATTER. Only the STORAGE OF ENERGY matters. Again, again, again, again, I keep comparing Atlanta and Phoenix, same latitude and elevation——AND Phoenix on a clear day ALWAYS has a higher temperature, day and night, than Atlanta, DESPITE THE FACT THAT ATLANTA HAS AT LEAST 4 TIMES AS MUCH GREEHNOUSE GASSES AS PHOENIX! OH, ORACLE, PLEASE EXPLAIN THIS PHENOMENON! THERE IS SIMPLY NO PROOF OF THE PUTATIVE GHE! AND WITHOUT EMPIRICAL PROOF, THERE IS NO SCIENCE THERE. WILLIS, SORRY, BUT THEORY IS NOT SCIENCE!

TimTheToolMan
December 13, 2012 7:58 pm

Willis attacks me with “I have no time for people who don’t play well with others, Tim, especially jerks that attack me rather than answer a question.”
But didn’t read posts in the other thread like I asked him to. I didn’t attack you Willis, I accused you of not understanding the issue and I stand by that statement because you made a strawman argument on warming water with an IR lamp vs understanding how DLR interacts with the ocean’s cool skin.

TimTheToolMan
December 13, 2012 8:29 pm

Willis writes in a reply to tallbloke the other thread “You’re looking at this backwards. In a system in which the surface is maintained at a slightly lower temperature than the bulk … what happens if you forcibly warm the surface? Think it through all the way, Roger. The very surface warms … until it’s slightly warmer than the bulk … which then heats up slightly, until the surface is slightly cooler than the bulk, and the previous condition (cooler surface) is restored.
Unless your claim is that such a system can’t be heated from the top … but I don’t think that’s your claim.”
And this indicates to me that Willis gets confused by DLR “warming”. At no time is the surface temperature warmer than the bulk, heating it up. This would be a new result that Willis would have to justify by reference. Willis doesn’t understand the process of warming yet. Thats not an attack, its an observation.
The ocean is always losing energy. The relation is DLR < ULR. It doesn't matter if clouds increase the DLR by a seemingly whopping 100 W/m2 if the ocean surface is radiating 150 W/m2…its losing energy and the surface isn't heating and transferring its energy to the bulk.
When Willis writes about heat lamps its a totally different situation. In that case there is sufficient energy to actually heat the water and there is no cool skin. The water is warming and not cooling. Hence there is no reason why the energy cant conduct down into the bulk.

Matthew R Marler
December 13, 2012 9:19 pm

Willis, just chatting. I admire your work.
You are conflating an unmeasurable variable (the radiation imbalance with the troposphere and surface fixed) and an abstract concept (“body in uniform motion”). The difference is that the radiation imbalance is a number, while a body in motion is not a number.
Hmm. How about “black body radiation”? Is that an unmeasurable variable or an imaginary concept? How would you classify Avogadro’s number?

Mack
December 13, 2012 10:09 pm

Willis,
Medical conditions caused by small amounts of anything be it ricin or ozone will not wash with me either. We are talking about heat and the physics of gases and their interaction with the sea, not the cellular biology of humans.
However, addressing your strawman arguement in the medical line……….as an example of ignorance and worry over small amounts, lets take the example of the flouridation of a towns water supply. There are people,(shall we call them ignorant cranks),who believe that increasing the F’ ion in a towns water supply from a naturally occuring say 0.02ppm.to about 0.8ppm will cause all sorts of disease,Alzhiemers, you name it, Yet there are towns in the US where there is (or was?) naturally occuring F’ in the water of 100s of ppm and the only ill effect is a mottling of the teeth.
I hope you are not among the “ignorant cranks” Willis.
But yes, you’re right about the “heavy lifting” of water,be it in the clouds,water vapour, fog, snow, ice,ocean, ie water everywhere. The Earth’s temperature is purely hydrological.
Well the heavy lifting could be hydraulic Willis 🙂 whatever.

Bernie McCune
December 13, 2012 11:31 pm

I am not sure this is the place to post some observations that have been made by a small group of high altitude balloon flyers but I will make an attempt to say something about it which might be of value in trying to understand the strange environment of the troposphere and stratosphere. I was involved with a team that supported the NASA high altitude balloon program and also briefly the Air Force high altitude balloon program. Figure 1 is certainly an idealized view of this environment. These balloons fly for hours and sometimes days in this very strange environment. These very large balloons have been launched from all over the world with a very active Antarctic flight schedule as part of the program. Thermal effects relative to the balloon payloads (mostly radiative) are unusual since there is very little air and convective thermal exchange at altitude. Our experience of this environment can only be considered the blink of an eye in terms of months and years that go on when we are not there to observe these high altitude phenomenon. It will be very difficult to probe this dynamic environment because we just cannot position instrumentation there for very long. In situ measurement is, at this time, the only way to gather this data. The use of white paint and aluminized foil coverings can keep the balloon instruments at an acceptable temperature rather than freezing or cooking them. The “trop” is generally very cold (often -60 to -80 C) but as you pass up above it, temperatures can “warm up” to a toasty -30 C. Those boundaries are often dramatic. The short 90 minute flight of a radiosonde balloon can quickly probe this environment but most of these small “weather” balloon flights do not ever get above 100,000 feet. High altitude balloon flights are much rarer than sonde flights but they have reached altitudes of 160Kft and regularly fly at 120 to 140 Kft. The balloons fly at all hours and night flights have a really big problem of staying warm enough to stay alive. When internal temperatures in electronic packages drop much below -10 C there is a good chance they will stop functioning. The same group I worked with developed Space Shuttle and satellite payloads and we found that the thermal design for the orbiting space payloads was easy compared to balloon payloads. Being completely in the vacuum of space rather than in the last remnants of the atmosphere makes things much easier thermally. Both have some strange radiative thermal characteristics but remnants of gases complicates things for the high altitude balloon flights. Proximity to the earth’s surface for balloon flights is a thermal factor. Daytime albedo radiative effects on balloon payloads from clouds we were flying over can rapidly heat the electronics packages if we hadn’t designed the thermal shell properly. There are direct sun effects and then the reflected cloud effects from below. Antarctic flights occur when the sun never sets and there are serious snow and ice albedo effects from below. Those Antarctic flights have lasted for weeks and in at least one case the balloon did a double spiral twice around the continent (may have been about a 6 week flight)
The winds at these high altitudes give an indication that there is still plenty of “gas” up there. Winds on the way up can vary in speed and direction with speeds of 0 to 50 knots being common. Contrary to some popular ideas, winds at 120 to 140Kft can reach these same speeds (50 to 60 knots). Winds if we happen to get into the jet stream as we climb into the “trop” can reach well over 100 knots (I have seen cases of 150 knots). Winds as we reach above the trop can be as low as 5 knots and sometimes drop even lower to almost 0. Mid latitude northern hemispheric winds at altitude tend to travel from west to east during the fall months and then there is a “turn around” that occurs over a period of days and weeks in the spring where winds at altitude travel from east to west until late summer where another “turn around” occurs. This turn around effect can be seen occurring over several months through a whole series of high to lower altitudes and latitudes. It is a very complex pattern but does come in a very seasonal clockwork fashion. Some interesting box winds occur during this changeable period. A balloon traveling to altitude can “see” east west winds through a several thousand foot climb and then get into a west east wind and so forth. The balloon can travel 10s of thousands of feet and remain mostly overhead during these unusual conditions. High altitude winds in the southern hemisphere are a mirror image of the NH winds and follow the mirror image seasonal pattern.
I wanted to post these observations so that I could show you that we must be very careful about making assumptions about the environment that we think is there. There are some very dramatic differences from our surface environment that will make it hard for us to visualize what is happening there. Things are occurring up there mostly beyond our ability to “see” and since there are no manned vehicles ever flying through that area on a regular basis we really have no idea what is really happening over long periods of time except from some of these tiny snapshots of that near space overhead environment.
Bernie

Cliff Huston
December 14, 2012 12:46 am

TimTheToolMan says:
“And this indicates to me that Willis gets confused by DLR “warming”. At no time is the surface temperature warmer than the bulk, heating it up. This would be a new result that Willis would have to justify by reference. Willis doesn’t understand the process of warming yet. Thats not an attack, its an observation.”
When a water molecule is impacted by DLR it becomes warmer than other surface molecules and the bulk molecules below. The fact that the surface water molecules on average are cooler than the bulk water molecules below, has nothing to do with the fact that DLR caused the water to be heated. The warmed molecule does not know it belongs to the cool layer and will be happy to transfer its energy to a cooler molecule, that belongs to the warm group below it.
Feynman: Jiggling Atoms

Think about it.
Cliff

phlogiston
December 14, 2012 1:00 am

Bernie McCune says:
December 13, 2012 at 11:31 pm
Important observations, the turbulent and chaotic nature of high atmosphere winds will present a much larger surface area for thermal convective exchange and could profoundly affect vertical thermal dynamics in the atmosphere. As modellers of supernovae discovered, incorporating turbulent gas mixing can mean the difference between success and failure of effective simulation.

Frank
December 14, 2012 1:40 am

Willis: Thanks for your reply. I generally like your posts, but rebel when I think you overstated the skeptical case.
About the surface radiative forcing, you wrote: “Sure, you can calculate it … but is it correct?” Later, about radiative forcing in general, you wrote: “If you can’t measure it, you can’t falsify it. If you can’t falsify it, it’s not science.” These are noble sentiments, but it think they are misapplied here. Like you, I’d like to see an experiment done with the atmosphere: Forget about MODTRAN or HITRAN; just point a giant variable-wavelength IR laser at a spaceship with a photodetector and take the absorption spectrum of the whole %*&^* atmosphere! Unfortunately, we can’t instantaneously double the amount of CO2 and take a second set of measurements. We also need to know about emission.
I think our current information is a reasonable substitute for impractical experiments on the whole atmosphere. The absorption spectra of gases have been carefully studied by scientists since at least the 1930’s, long before politicization of climate science. The width of absorption lines of gases is determined by broadened by Doppler broadening (solid theoretical basis) and collision/pressure broadening (characterized imperfectly by experiment). Each GHG has been studied carefully in the laboratory and parameters characterizing the central frequency, intensity and pressure- and temperature-dependent half-width are entered in the HITRAN and MODTRAN databases (originally created for the aerospace industry). It’s trivial to pick a temperature, pressure and frequency range for a gas like CO2 and experimentally determine if the calculated and observed spectra agree. One can determine if the absorption spectrum of pure CO2 is changed by dilution with a mixture of nitrogen and oxygen. Characterizing the optical properties of GHG’s is clearly science and it probably has been done with a high degree of accuracy and reproducibility because the measurements are made in a laboratory.
The second step is to use parameters obtained from laboratory studies to calculate what should happen as radiation passes upward or downward through pre-defined atmospheres meant to be representation of various locations on earth. The radiative forcing for 2X CO2 at the tropopause has been calculated from such models independently from MODTRAN or HITRAN data by several groups and by GCM’s (which don’t use line-by-line methods). The least reliable part of this process is dealing with clouds, because we don’t have a very good handle of the temperature and emission by cloud tops. We can’t measure radiative forcing in the atmosphere itself, but the results for dozens of models have been compared. While these calculations are not “falsifiable by experiment”, they aren’t equivalent to Shakespeare’s Glendower calling the spirits from the deep. Radiative forcing isn’t anywhere near as bad as the unfalsifiable temperature predictions made for 2100 by GCMs that can’t properly model convection and clouds and that contain a dozen or more adjustable parameters. The no-feedbacks climate sensitivity of about 1 degC for 2X CO2 appears far more reliable than any estimates of its amplification by various feedbacks; another creation of climate science that haven’t been confirmed by observation. There’s so much poor science to criticize that I personally wouldn’t invest my time complaining about radiative forcing. 3.7 W/m2 translates to a temperature rise of only 1 degK (without feedbacks), so I think radiative forcing is a good issue for skeptics.
Radiative forcing is calculated at the tropopause because this is the lowest spot in the atmosphere where the temperature is controlled by radiative equilibrium. Where temperature is controlled by radiative equilibrium, radiative forcing can be converted to temperature change. Radiative forcing calculated at the surface can’t be converted to a temperature change, because some of the energy delivered by increased DLR can escape by convection rather than being used to warm the surface until radiative balance at the surface has been restored. If most of the energy delivered by increased DLR leaves by increased convection, 2X CO2 won’t produce much warming.
As for your comments about large and small changes, an average of 239 W/m2 of solar radiation reaches the surface and troposphere and an average of 333 W/m2 of DLR reaches the surface. A radiative forcing of 3.7 W/m2 for 2X CO2 is small compared with these values. You correctly point out that there can be a 1000 W/m2 difference in solar radiation between day and night, but no one is worried about the temperature difference between day and night. Even if they were, DLR is still present during the night, and radiative forcing is only a small fraction of DLR. As for temperature change, you need to work in terms of degrees Kelvin (W = oT^4), not Celsius. Global warming and climate change aren’t concerned with the temperature the difference between the Sahara and Antarctic; just a rise of a few degK on a planet where the mean temperature is 288 degK. That is a small change.
If you take the derivation of W = oT^4 with respect to temperature and do a little algebra, you get dW/W = 4*dT/T. For small changes (with no feedbacks), the percent change in T is one-fourth the percent change in W. The IPCC’s feedbacks allegedly increase this by a factor of 2 to 4.5.

TimTheToolMan
December 14, 2012 3:49 am

Cliff writes ” The warmed molecule does not know it belongs to the cool layer and will be happy to transfer its energy to a cooler molecule, that belongs to the warm group below it.”
That would be true if there were cooler molecules below to transfer the heat to. But that’s not the case. Here is a diagram of how the very surface of the ocean looks
http://en.wikipedia.org/wiki/File:MODIS_and_AIRS_SST_comp_fig2.i.jpg
Its a logarithmic scale. Take notice of where the DLR interacts with the ocean. Thats from the dot at the very top down to about the second dot. Then below that, the ocean continues to warm for one mm or so. Thermodynamics tells us that if molecules are heated at the surface then that energy wont travel downwards because its going from a colder place to a warmer one.
Its a very interesting area and is far from intuitive.

richardscourtney
December 14, 2012 3:53 am

Jeff Alberts:
At December 13, 2012 at 6:33 pm you say to Diogenes

Sorry, not getting the brass monkeys reference 🙁

I reply with this off-topic explanation to assist non-UK readers of WUWT who may also not understand the phrase but may be amused by its explanation.
Cold weather in the UK is often said to be,
“Cold enough to freeze the balls off a brass monkey”.
And this is often shortened to become a phrase such as “It’s brass monkeys outside” when describing cold weather.
The origin of “brass monkeys” is not documented but is often said to have a naval history. Many similar phrases do have a naval origin (e.g. ‘show a leg’). And these phrases derive from the British humour of using an innocuous phrase in a manner which could be thought to be rude (i.e. double entendre).
The story about “brass monkeys” goes like this.
Warships prepared for battle by – among other things – stacking cannon balls alongside their cannons. The lower layer of balls needed to be constrained or the entire stack would collapse and roll away. Initially, this was achieved by putting the balls in flat trays with low walls around their edges, but lifting balls from the bottom layer of the trays was difficult (fingers could not get under a heavy iron ball). This was solved by replacing each tray with a thick, wooden board which had round depressions in its top surface. The iron balls would key into the depressions which were called ‘moons’ because their appearance resembled the Moon when they were seen in the dim side-lighting within a ship.
The Royal Navy liked to place cannons beside the entrances to its shore installations and to stack pyramids of cannon balls beside the cannons. The wooden moon-keys for the balls soon degraded, so they were replaced by brass ones (and sailors were required to polish them daily).
However, brass has a much higher coefficient of thermal expansion than iron. So, as temperature drops the brass moons contract relative to the iron balls. This squeezes the balls up so they are less well keyed into place. Indeed, on very cold nights it could become cold enough to ‘freeze the balls off a brass moon-key’ and the balls would roll down the street.
Slurred pronunciation changes ‘moon-key’ to ‘monkey’.
Richard

phi
December 14, 2012 4:35 am

Frank,
“There’s so much poor science to criticize that I personally wouldn’t invest my time complaining about radiative forcing. 3.7 W/m2 translates to a temperature rise of only 1 degK (without feedbacks), so I think radiative forcing is a good issue for skeptics.”
Well, science is not the stock market! Either radiative forcing has a meaning, or it does not.
Adding CO2 to the atmosphere modifies the system and in particular the lapse rate. The surface Warming expected is a result of a change in lapse rate and not an effect of extra energy. There is no reason to think that we can characterize an increase of CO2 in the unit used to describe a change in albedo or solar activity.

Ryan
December 14, 2012 4:49 am

Willis, you are quite wrong to compare clouds to GHGs. That is an entriely different mechanism. Clouds are made up of dispersed droplets of liquid water (not gases) and they absorb a great deal of incoming radiation during the day as water has a very high specifi heat capacity. We can see that because not only do they make the ground darker, but they do so evenly across the entire visible spectrum. They do so into the infra-red as well. Consequently clouds contain a lot of heat energy. During the night they will emit that heat energy as radiation. That is like heating up a tank of hot water during the day and sitting it just above someones head at night. Yes, doing that will make you feel slightly warmer (but actually only slightly warmer because the effect due to emitting photons is actually tiny). This has nothing to do with the greenhouse effect, however, which claims that heat emitted from the ground will be absorbed and re-emitted by CO2. Even the much stronger cloud effect you have mentioned will noyl have a very small impact.
It is very difficult to get atoms to emit photons. Every method we have tried in the lab tends to end up with massive amounts of conducted heat energy. A lightbulb emits photons which are visible – but you can’t feel the heat of them when they hit your face – the energy they have is tiny (thankfully our eyes are sensitive to this tiny amount of energy). You can feel the heat of the highly energetic atoms in the filament as they cause the glass envelope of the bulb to heat and the air above it by conduction. Getting photons out of a wire is like getting blood out of a stone. Much easier to get conducted heat out.
The sun heats the Earth only because it emits such huge numbers of photons that travel to the equator of the Earth more or less completely unimpeded. Here they warm the oceans which then keep the northern lattitudes warm by a series of rather convenient ocean currents (the people of Britain should thank the people of Panama because their country keeps Britain warm). The sea temperature around the UK is about 10Celsius in the Winter and 16Celsius in the Summer. Doesn’t bear much relation to the air temperature above it (which can vary from -12 to +38) so why should CO2 make a difference?

December 14, 2012 6:22 am

Wills
I find all of this quite interesting and am throwing my 0.5 cents in.
The amount of radiative forcing delta between the 1950’s and today should be easily measurable. The U.S. Air Force through their “upper atmospheric research” program pushed the state of the art in IR spectrometers to the limit of measurement, or a fraction of an individual wavelength. These measurements were done in order to design the detectors for the Sidewinder and other IR heat seeking missiles. It was done at all altitudes up through at least 50-70,000 feet.
These instruments measured the extinction coefficients for CO2, N2O, H2O, CH4, and other trace gas absorbers in the atmosphere. In theory the additional CO2 and CH4 and N2O in the atmosphere should show up as a broadening of the spectrum of absorption at ALL altitudes up to an including the altitudes where the lines desaturate due to reduced pressure and temperature.
In theory additional CO2 (to use this as the example) should do two things. It should broaden the absorption spectrum (just compare the 1950’s values to today, easily measurable) AND it should increase the altitude where the lines desaturate (the altitude where the absorption lines are no longer fully saturated i.e absorbing all the radiation at that particular wavelength).
Both of these quantities are measurable and then you can back out that the effects on local temperature. Everything I have ever seen on this subject relates to classical physics and this is not a classical physics phenomenon but a Quantum Mechanical phenomenon. I could blast a terawatt per square meter at 0.7 microns but since nothing at this wavelength absorbs that energy it is as if it does not exist. Energy is only transferred between radiation to thermal motion of a molecule through absorption and emission and the time constant between the two. That does not change. If the absorption spectrum is wider, more energy is absorbed (and emitted) but during the time between the two the molecular excitation can lead to a temperature rise. These quantities are strictly governed by QM rules and the equations state that absorption and emission are themselves variable, dependent on temperature and pressure of the ENTIRE atmosphere, not just the relative increase in concentration. This is what is called collision broadening of the absorption lines, which is a gaussian to lorentz transformation of the QM spectrum.
Where is the basic QM physics in all of this? I rarely if ever see it discussed and these quantities are calcuable and measurable AND we have a huge baseline from the USAF work stretching from the early 1950’s on the subject.

AlecM
December 14, 2012 7:30 am

denniswingo: you won’t get any sensible discussion of the IR physics from IPCC luminaries such as Pierrehumbert because their house of cards’ science is based on fake physics. Thus they assume that a pyrometer measures energy flux when all it does is to measure an assembly of Poynting Vectors in the viewing angle of the instrument. If you made that viewing angle 2pi, in zero temperature gradient. the signal would be zero. So, ‘back radiation’ isn’t real and the positive feedback doesn’t exist.
As for IR absorption, this article clearly states that it is by dribs and drabs in many collisions.:google “PhysTodayRT2011.pdf”. This is the false physics to which you refer: quantum exclusion means it can’t happen. The need for LTE means the excess energy, if there is any, pseudo-diffuses to be thermalised at heterogeneities, mainly clouds.
The caveat about excess energy is that there isn’t any because thermal emission from the atmosphere annihilates the same emission from the surface. If this didn’t happen at equal temperature, we’d be an expanding ball of plasma. These people have no shame for their actions.

Bernie McCune
December 14, 2012 8:03 am

@ denniswingo
I wonder what the payload weight on modern day experiment of this sort might be? Balloon payloads continue to shrink due to all the recent (past 15 years) miniaturization process (includes lower power requirements and of course much smaller and lighter [modern Li technology] battery packs). Large latex balloons could easily probe to 80 or 90 Kft and they are relatively cheap. Latex versus polyethylene balloons rise to a burst altitude and drop the payload. Probably a couple hour flight but data can be taken going up and coming down. They also make small polyethylene balloons though. Poly balloons, however, reach a float altitude and remain there until evening and night cooling lower the float altitude a little each day/night cycle (or a slow loss of He). A large balloon can carry several tons to altitude and NASA is presently doing work on long duration balloons. This type of platform could drop sondes on a very regular basis and probe the upper atmosphere clear to the ground. There could be thousands of these probes aboard a single balloon flight. Balloon flights tend to be in the $1M range so this experiment could be done for almost nothing by today’s standards. Time to get real about this stuff and stop trying to do it with computers!
Bernie
Bernie

mkelly
December 14, 2012 8:51 am

AlecM says:
December 14, 2012 at 7:30 am
“The caveat about excess energy is that there isn’t any because thermal emission from the atmosphere annihilates the same emission from the surface. If this didn’t happen at equal temperature, we’d be an expanding ball of plasma.”
AlecM that is roughly my point. A photon that left the surface and returned via a CO2 emission cannot make the surface warmer only take it back at most to where it was. (yes I know it is not the same photon.)
The simple heat transfer equation q/A= e * SB * (T1^4- T2^4) tells you that if T1 and T2 are equal then NO thermal energy is transfered so how can it be that if T2 is less than T1 there is heat transfer sufficient to raise the temperature of the surface 33K.

December 14, 2012 9:44 am

Bernie
The answer is that the payload you speak of is easily doable and NASA does do some of this with their jet flight program where they fly spectrometers. I wish that I had the time to fully delve into this subject but to do it right is a full time job and no one is paying me to do so!

Ryan
December 14, 2012 9:53 am

Just been looking at some YouTube clips of firefighters using thermal imaging cameras. Plenty of evidence of convection currents carrying very hot air up to the ceiling (1200F)but the warming of walls and floors due to direct radiation is pretty minimal (200F). Most of this is probably direct radiation from the fire itself rather than heat radiated by hot air. Anyone that has stood in front of a bonfire will know you need to stand pretty close to feel the radiant heat (wouldn’t want to be standing right on top of one though!).

Bernie McCune
December 14, 2012 10:26 am

Sorry to go a bit off topic but how big and what pointing requirements (or other reqmts) do these instruments have? I suspect this might be a project that could easily be funded and could make several careers if there is a pressing need for it. I have an inkling that there might be niche available for this – no?
Bernie

December 14, 2012 10:50 am

We know from measurements that the sun puts about 170 W/m2 of energy into the ocean on a 24/7 average basis.
Willis, first of all, we need to disentangle the 24/7 average basis because that term is a fiction.
The oceans absorb energy from the sun during the day and radiate it at night, depending on the temperature of the atmosphere. This is basic physics.
The temperature of the ocean at the water/air interface is a function of the temperature of the atmosphere at that interface. If the air temperature is less than the ocean temperature then the ocean radiates energy into the atmosphere. If the temperature of the atmosphere is greater than the temperature of the ocean then the ocean absorbs energy through molecular collisions (conduction) at the air/water interface.
That is basic physics.
This can be a weak effect (glassy ocean) or a strong effect (turbulent ocean).
If there are clouds then there is a significant component of reflected IR radiation, but that IR radiation heats the atmosphere before it heats the water.
To find out what the radiative absorption of the ocean is you have to know what the absorption spectrum of water in the ocean is and what the skin depth is to determine how deep that radiation penetrates. However, this only works if the atmosphere is warmer than the water. If there is a clear sky and no humidity, the ocean or any body of water loses heat rapidly. I used to live on a lake on the Tennessee river and had a temperature monitor near the water. On a clear, cold, low humidity night the temperature would show a drop after dark but about 2-3 hours after dark the temperature would climb a couple of degrees from the conduction/radiation of heat from the water.
This is not that hard to figure out.

richardscourtney
December 14, 2012 11:15 am

Willis:
Please keep going. You are winning, and you are winning for science.
Richard

Robert Clemenzi
December 14, 2012 11:21 am

phi says:
December 14, 2012 at 4:35 am

Adding CO2 to the atmosphere modifies the system and in particular the lapse rate. The surface Warming expected is a result of a change in lapse rate and not an effect of extra energy.

That is not how I understand it. Please provide a reference.

phi
December 14, 2012 12:58 pm

Robert Clemenzi,
Adition of GHG in the atmosphere reduces radiation losses. It is these losses with the adiabatic cooling that determine the lapse rate. Add GHG therefore causes the change of the lapse rate. The proportion in which the surface will be heated depends on the proportion of energy flux reported on convection.

phi
December 14, 2012 1:35 pm

Robert Clemenzi,
You asked me for a reference, I do not have. It is a constant in climatology : the simplest things and the most obvious are generally not discussed but they are often denied.

Robert Clemenzi
December 14, 2012 2:43 pm

phi says:
December 14, 2012 at 12:58 pm
Actually, adding GHGs to the atmosphere increases radiation losses from the atmosphere. Everyone thinks it just increases to amount captured, but they actually cause more to be emitted than they capture.
Adiabatic cooling determines the maximum lapse rate, not the actual (measured / typical) lapse rate.
To see actual data, you can try my application.
http://mc-computing.com/Science_Facts/Lapse_Rate/Lapse_Rate_Animations.html

December 14, 2012 11:14 pm

richardscourtney said @ December 14, 2012 at 11:15 am

Willis:
Please keep going. You are winning, and you are winning for science.
Richard

Willis, I’m sure Richard didn’t mean that, Please don’t go. We want you to stay! For the science.

phi
December 15, 2012 1:08 am

Robert Clemenzi,
“Actually, adding GHGs to the atmosphere increases radiation losses from the atmosphere.”
What you say is indeed true, but you talk about the global level. Less energy is radiated directly from the ground. The flux through the atmosphere is increased by the same and therefore more energy is lost by the atmosphere.
But I was not talking about this global phenomenon. The potential of linear radiative losses in the column is reduced by the increase in opacity. The actual lapse rate is a consequence of both adiabatic (moist) cooling and radiative cooling. Increased GHG cause an increase in opacity, thus decreasing the potential of linear radiatives losses and therefore a decrease in lapse rate.

phi
December 15, 2012 1:31 am

Robert Clemenzi,
I would add that the amount of energy that the ground can no longer evacuate directly due to the increase in greenhouse gases can be effectively modeled by a forcing. But this part is certainly not dominant in what is supposed to represent the radiative forcing.

Brian H
December 15, 2012 2:49 am

typo: “often taken to me 20 km” [Thanks, fixed. -w.]
This handy-dandy ‘forcing’ power of the preferred variable is a limb of the “hidden variable fraud” written of on WUWT.

richardscourtney
December 15, 2012 3:11 am

The Pompous Git:
re your post at December 14, 2012 at 11:14 pm.
Thankyou for that. Yes, my choice of wording was unfortunate: of course I want more here from Willis.
And I did laugh. 😎
Richard

Mack
December 15, 2012 3:37 am

Willis,
You say “we know from measurement that the sun puts about 170w/sq.m into the ocean”
How is that measurement obtained Willis?

AlecM
December 15, 2012 5:02 am

phi: your ‘consensus’ view is incorrect.
Radiative equilibrium between equal temperature [100 m] lower atmosphere and Earth’s surface is by the annihilation of most surface IR in GHG bands by thermal IR. Otherwise we would be an expanding ball of gas. Because there can be no CO2-AGW, there can be no effect of increasing CO2 concentration on water vapour concentration so the Houghton/Hansen moist lapse rate GHE is disproved.

phi
December 15, 2012 5:53 am

AlecM,
I’m sorry but I do not quite understand the meaning of your message. That said, I do not think my views are consensual, unfortunately!

AlecM
December 15, 2012 9:25 am

phi; let me explain.
You are claiming the GHE is ~33 K, the level claimed by Hansen et al 1981. This is adapted from Houghton 1977 who made some serious mistakes.
The climate models use the two stream approximation to claim IR warming of the atmosphere is ~7 times reality. That warming is mostly water vapour side bands. There is no CO2 IR absorption of surface IR. There is no positive feedback.
This is because that CO2 band surface emission is annihilated by the thermal GHG IR from the atmosphere, standard radiative physics.
So, there is no mechanism by which increase of CO2 can cause increase of water vapour. Therefore the Houghton/Hansen idea of the GHE being lapse rate warming is debunked.
The real GHE is ~9 K with 24 K from lapse rate/gravitational potential energy.
There is no CAGW risk, ever, not even from methane. The windmills and carbon trading are from evil corporations on the inside tack manipulating bent politicians like Obama.

phi
December 15, 2012 10:30 am

AlecM,
Your reflection is quantitative, I do not have the means to speak about it. I made ​​a qualitative critique. When I say that GHG reduce the amount of energy evacuated directly from the ground, I do not know how much it is, it could be zero, I don’t know.

Stephen Wilde
December 15, 2012 10:42 am

“If the ocean cannot absorb energy from DLR,
then where is the approximately 340 W/m2 coming from
to keep the ocean from freezing?”
Descending air converting potential energy back to kinetic energy with the maximum effect at the surface.
http://climaterealists.com/index.php?id=10775
” The Ignoring Of Adiabatic Processes – Big Mistake”

December 15, 2012 11:50 am

As I said, nobody has been able to answer that question to date, nor do I expect an answer.
Except that you ignore when people do answer.

AlecM
December 15, 2012 12:21 pm

Willis Eschenbach: ‘Radiation, ~ 400 W/m2
Conduction/convection of sensible heat, ~ 20 W/m2
Evaporative loss of latent heat, ~ 80 W/m2
TOTAL LOSSES, ~ 500 W/m2’

Oh where did you get your physics? 400 W.m^2 is the S-B prediction for an isolated black body in a vacuum at ~ 19.62 deg C. Assuming the adjacent ~100 m atmosphere is at the same temperature – 1 K, and its emissivity is 0.83, it radiates ~341 W/m^2 back meaning net IR = ~59 W.m^2..
This is split 2:1 into atmospheric window IR which does no work and water vapour sidebands which end up causing more convection.
Total losses ~ 159 W.m^2, about the same as the DSR. There is no DLR – it’s the artefact of the shielding around a pyrgeometer sensor. The DLR and the ULR are combined as net IR, the only way that IR can do work. There can be no two streams each heating the atmosphere.
For further info see Poynting’s Theorem 1884.

Curious George
December 15, 2012 1:27 pm

Errors large and small: Global warming enthusiasts predict a temperature rise of about 3 degrees Celsius, which is almost exactly 1% rise in the absolute temperature. The error from considering the earth as a blackbody in the IR is 3%-4% – not “quite small” in the circumstances.

Mack
December 15, 2012 2:26 pm

Scroll up to my comments to Tim the Toolman to see what I think of Trenberth and his Earth’s Energy budget figures Willis.

Mack
December 15, 2012 3:01 pm

Willis,
The only things that are measured is an incoming solar radiation of about 1360w/sq.m.( a “solar constant”) which is a yearly global average which cannot be divided but remains simply as that, and that simply attenuates to a yearly global average at the Earth’s surface of about 340 w/sq.m. which corresponds to measured land based readings .

jae
December 15, 2012 4:36 pm

Willis: you ask:
“PPS—Why on earth would you expect Atlanta and Phoenix to have the same temperatures, or for Atlanta to be warmer?”
One of the common ways to study physical phenomena is to try to keep all variables constant, except for the one you are investigatging. The Phoenix/Atlanta comparison is an attempt to hold (almost) everything constant EXCEPT the amount of GHG. Same latitude, same elevation; therefore, same solar insolation at TOA on a given day. But the amount of GHG in Atlanta is somewhere around 4 times as much as in Phoenix. Now, would not one wonder, if the amount of downwelling radiation is proportional to the amount of GHG and if downwelling radiation has ANYTHING to do with temperature, why it is not hotter in Atlanta on some clear day in July. I suspect that it is because there is so much more water available for evaporation in Atlanta, causing a cooling effect. HOWEVER, doesn’t this also show that the temperature at any location on Earth is not very dependant upon the amount of GHGs? I think it’s probably dependant only upon insolation and the amount of water available. Which is maybe why your thermostat hypothesis works, also.
Relative to your kind advice:
“A word to the wise. Capital letters are the equivalent of shouting, and in the amount you used them, they are the infallible mark of someone with lots of passion and not much science. You should avoid that kind of shouting if you wish to be taken seriously.”
I’m sorry that you read the caps as shouting. I mean them to be only for emphasis. I’m just too lazy to underline, italicize, or bold.

richard verney
December 15, 2012 5:51 pm

Willis
Sorry for the late response but I have been without internet access.
I recall your article (radiating the oceans); it was thought provoking (as is usually the case with your articles) and led to some interesting comments. I always enjoy your articles, because they are often insightful (sometimes very) and almost always give rise to lively debate.
I am not seeking to create a new form of radiation. Any electrician will know that some measurements do not always tell the full story. A common example is a flat car battery. If you strap a voltmeter across a flat battery, it will often read 12 volts suggesting that everything is fine. It is only when you load it (say with the starter motor) that the voltage drops to say less than half. Sometimes, one can measure things but that does not necessarily mean that they have the ability to do work. I am not saying my example of a voltmeter and a flat car battery is completely analogous to the measurement of W/m2
I am suggesting that perhaps (note I use the word perhaps) the 255K DWLWIR which is being measured is incapable of imparting sensible energy/work on 277K (or 288K) water. Why do I say this? Simples, because it does not appear to cause water to evaporate in the way that one would expect given the absorption characteristics of LWIR in water (60% of all LWIR being absorbed within the first 4 microns). This leads me to consider that there is a potential problem with the DWLWIR theory that requires an explanation.
May be 255K DWLWIR can do all sorts of wonderful things on a 200K rock, or another gas molecule (such as an oxygen or nitrogen molecule) at 244K, but it appears that 255K DWLWIR has little (if any) impact on water. I have previously pointed out this problem with oceans, but one can also see the problem on land with dew and frost.
It may well be the case that UWLWIR emitted at TOA into space may have a significant effect on the Earth facing side of orbiting satellites, but, if so, that merely demonstrates that one has to look at the environment in which the radiation is operating to see whether it possess sensible energy capable of performing working in that environ. I emphasise that I am only commenting upon DWLWIR in relation to water (in its many phases on planet Earth).
Consider more carefully my example of dew (or a light film of frost) in a shady hollow. Let’s just do some maths. Let’s assume that an entire 1sqm piece of ground in a shady hollow is entirely covered by dew to a thickness of 1mm (of course, it will be less thick than this in many places). 1m by 1m by 1mm is 1 litre of water, weighing 1kg. DWLWIR, global ave is 333 W/m2. Latent Heat of Evaporation is 2260 KJ/Kg. So theoretically, DWLWIR has enough energy to evaporate water over a 1 sqm area at the rate of 0.14 gram/sec. Accordingly all the dew (1kg worth) in the shadow area would theoretically be evaporated by DWLWIR in less than 2 hours if DWLWIR can perform sensible work.
In addition one has to bear in mind that once the sun went down the night before and temperatures began to drop, the dew as it was forming was receiving the DWLWIR all night long. So why is dew around at the break of daylight? After all, it has already had the benefit of more than 7 hours of DWLWIR if the dew formed sometime before midnight. If DWLWIR could do sensible work, as the dew is forming the DWLWIR would be burning it off such that one might reasonably not expect to see morning dew at all, certainly one would not expect to see dew hang around all day (or most of the day). Dew in the shady hollow that disappears in the early afternoon is probably burnt off by the rise in ambient air temperature taking place from sun up, not form DWLWIR.
In contrast, we know that sunlight can do sensible work. Once the sun gets up, the dew (or frost) on the sunny side of the hollow can be burnt off quite quickly, maybe within 30 mins to 1 hour of sun up. This is notwithstanding that morning sun is very weak. Global ave is about 170 W/m2 (ie., only about half that of the DWLWIR) and due to the low incidence of morning sunlight, we could be talking of no more than 20% of that figure and yet this is sufficient to burn off the dew (or melt the frost unless the ambient air temp is very cold). You do not need many W/m2 because the film is very thin and there is little mass of water to be burnt off.
So you need to explain why on a given day (with its then prevailing ambient temperature and humidity), low incident sunlight (which on any basis is weak) can do something within less than an hour that DWLWIR (which is claimed to be about twice as strong and even more in relation to low incident sunlight) cannot achieve in the whole day.
Given the importance of the water cycle, the latent heat content of water (and its phase changes), the thermal capacity of the oceans in comparison to that of air, the different absorption characteristics of LWIR and Solar Radiance in water etc, if climate science was a real science there would be thousands of experiments conducted on the effect of DWLWIR revealing proper empirical data, and yet there is none!
Willis, I am not saying that I am right and you are wrong. Like most readers of this blog, I have an open (but sceptical mind) and I am here to learn. I am merely pointing out that dew is a potential problem for the concept of DWLWIR (in the figures claimed for it) and I have yet to see any reasonable explanation why low incident sunlight can do something measurable in matter of minutes which DWLWIR cannot achieve in hours.
Willis, before responding with generalities such as the oceans would freeze but for DWLWIR so we know that DWLWIR must be real and must be doing something, just take a little time and do the maths. Please address my specific point with the dew in a shady hollow which can be burnt off by direct low incident morning sun light, but cannot be burnt off in the shadow area by the DWLWIR.

richard verney
December 15, 2012 6:09 pm

Mack says:
December 15, 2012 at 3:37 am
//////////////////////////////////////////////////////////////////
You are correct to point this out. This is one of the many problems caused by the AGW proponents love for using averages. The use of averages prevents one from seeing what is truly happening in the real world, since the devil is in the detail.
The figure of 170 W/m2 is a calculated global average for solar. However, we know that the sun is imputting more than that figure into the oceans since the oceans are not equally distributed over the globe.
Once you take out the Arctic, Antartic, the land masses of the Northern Hemishpere, it is quite apparent that the bulk of the oceans occupy a more equitorial position and therefore receive more than the global average..
It is of course, the equitorial ocean that acts as the heat pump for the planet. There is plenty of solar to keep this ocean warm (and not freeze). Indeed, Willis wrote an article on ARGO and the cap of temperature of 30degC. In that article, he accepted that theoretically, the tropical oceans would be far warmer if one considers the available energy from the strong equitorial sun.
The null hypothesis is that the oceans on average radiate only about 70 W/m2 and therefore they would not freeze.

george e. smith
December 15, 2012 6:52 pm

“””””…..denniswingo says:
December 14, 2012 at 6:22 am
Wills
I find all of this quite interesting and am throwing my 0.5 cents in.
………………………….
I could blast a terawatt per square meter at 0.7 microns but since nothing at this wavelength absorbs that energy it is as if it does not exist……”””””
Dennis, why don’t you try blasting your TW/m^2 laser of 0.7 micron wavelength radiation into a bucket of water. Don’t worry about standing close to it; since nothing absorbs at that wavelength then of course nothing will happen.
For the legal disclaimer, I have not personally tried this, so I recommend that nobody try it, until they have given you their signed release from liability, or alternatively, you have given them a written gaurantee of liability for any untoward result.

richard verney
December 15, 2012 7:19 pm

Willis
You seem to consider that the oceans would freeze but for DWLWIR. Perhaps you should ask yourself, why the oceans freeze if DWLWIR is keeping them warm.
My case is simple, the tropical ocean receives plenty of solar such that it will not freeze (even without DWLWIR). Theoretically, the tropical ocean could get up to a temperature of well over 40degC (may be even about 50degC). It does not reach this theoretical temperature for a variety of reasons, one important one being the conveyor belt currents which take away heat and distribute that heat elsewhere. It is this heat pump that that prevents the oceans nearer the poles from being frozen all year round.
As I have mentioned to you many times before, there is a problem with DWLWIR and the oceans. The problem is this.
• The absorption characteristics of LWIR in water is such that 60% of all LWIR is absorbed within 4 microns. LWIR (unlike solar) does not penetrate to any significant depth. All the energy is concentrated in the first 12 or so microns, and 60% of it within the first 4 microns.
• The very top surface skin layer of the ocean is cooler than the bulk ocean below. The heat flow at this layer of the ocean is therefore upwards. Because of the direction of heat flow, any energy absorbed in the top 4 microns (as would be the bulk of the DWLWIR) cannot be conducted downwards to the bulk ocean below. Willis I have shown you this profile before, (but unfortunately, I do not have the link to hand) look it up if you disagree.
• The top micron layer is being heated very quickly. Photons travel at the speed of light, and the energy being inputted is in watts per second. Any heat in the top micron layer cannot therefore be effectively overturned as part of the ocean overturning because that process is a mechanical process that takes hours (not seconds).
• LWDWIR (global average) is said to be about 333 W/m2. Latent Heat of Evaporation is 2260 KJ/Kg. So theoretically, DWLWIR has enough energy to evaporate water (over a 1 sqm area) at the rate of 0.14 gram/sec.
• We know that the water ‘must’ be evaporating at the 0.14 gram/sec rate, because the heat that resulted from the DWLWIR being absorbed in the top micron layer cannot be conducted downwards (which if it could be would slow down the rate of evaporation) because the temperature profile in this layer of the ocean is upward directing (not downward directing and heat cannot travel against the flow). We know that the 0.14 gram/sec evaporation cannot be nullified by ocean overturning because that is a mechanical process that takes about ½ day, not seconds.
So herein lies the problem. If DWLWIR has sensible energy and has the power claimed, it would give rise to water being evaporated at the rate of 0.14grams per second. Annually, this is equivalent to rainfall of about 4.5metres. But we know that annual rainfall of this magnitude is not occurring. The oceans are clearly not evaporating at this rate (which they would be if DWLWIR ‘heats’ the oceans since due to its absorption characteristics, if it does anything it can only heat the top microns of the ocean leading to quick evaporation). The fact that we are not observing this magnitude of rainfall suggests that DWLWIR lacks sensible energy.
You need to explain why the oceans are not evaporating at the rate equivalent to 4.5 metres of rainfall annually. Of course, if you can put forward a physical mechanism explaining how heat concentrated in the top microns layer can be conducted downwards even though this is against the temperature profile of the top microns/millimetres of the ocean, and/or how ocean over turning can turn over the top micron layer faster than the rate of evaporation (0.14 grams per second), then I would be interested in considering this physical process.
PS. We are fortunate that the absorption characteristics of solar is so very different, because if it was the same as LWIR, the oceans would have boiled off eons ago.
PPS. If you read the comments to your earlier post on radiating the oceans, you will note the many commentators who consider that the oceans are only losing heat via radiating at the rate of 70 W/m2. The oceans are in perfect balance without the need for the double accounting that you apply.

Mack
December 15, 2012 7:28 pm

Richard Verney says…
“The figure of 170w/sq.m is a calculated global average for solar” The key word you used here is “calculated” . It is not measured but “calculated” . I only take heed of true measurements Richard, because the moment you start calculating you move away from reality and start drawing pictures of the sun and earth on the blackboard, so to speak .

Michael Moon
December 15, 2012 9:39 pm

Excuse me. Willis Eschenback is a massage therapist with a degree in psychology. Transport of Heat and Mass does not appear in his curriculum. The Second Law also does not appear. Just to clarify, the Sun is at over 5500 degrees Kelvin. The atmosphere is virtually always slightly cooler than the surface of the Earth. Heat goes from hot to cold, not the other way. Pyrometers and so-called pyrgeometers measure the temperature of the atmosphere.
News flash, if the atmosphere is cooler than the Earth surface, the atmosphere CANNOT warm the Earth’s surface. Carnot, anyone, Bueller? No that was not a miss-spelling, the guy’s name is Carnot, as in Carnot Cycle.
Really, massage? Dude, you belong here why? Carnot is rather important in the engneering world, the engineers who built those power plants that keep the lights on, the heat on in your home, the roads, the cars, the trains, the Internet, the CPU, in your computer, the power grid, and on and on.
Willis, the thing is, radiation can only TRANSFER heat from WARM things to COLDER things. And, by the way, when HEAT is TRANSFERRED, things get HOTTER. Accept it, consider it, make it an important part of your life, and never try to contradict it again. Yes cooler things still radiate. If the warmer things to which they radiate got HOTTER and began to radiate MORE, what stops this runaway????????
Good Lord, no one one here new that???

Michael Moon
December 15, 2012 9:43 pm

Knew that, the CPU in your computer, sorry sorry

December 15, 2012 9:49 pm

Dennis, why don’t you try blasting your TW/m^2 laser of 0.7 micron wavelength radiation into a bucket of water. Don’t worry about standing close to it; since nothing absorbs at that wavelength then of course nothing will happen.
Lets go there. Since water does not absorb radiation at 0.7 microns it is unlikely that anything would happen. Beyond that I did not say that nothing would absorb I said the atmosphere would not. If it did then the planet would have fried long ago as the insolation at that wavelength is in the several hundred watts per square meter. Solar power satellite designs use lasers at wavelengths near this to transfer megawatts per square meter with little atmospheric effects. Green lasers at about 550 nanometers are proposed for laser weapons with gigawatts/m2 for this same purpose. Thanks for helping me prove my point. Unless you understand the couplings between radiation sources and sinks you can’t understand this subject and damn few people do understand it. Here is a nice website on the absorption spectrum for water.
http://www.lsbu.ac.uk/water/vibrat.html

Michael Moon
December 15, 2012 9:51 pm

Or the First Law, or the Third Law, or the Zero’th Law. Really? Do we all remember all four??? Can we recite them? I can….

Robert Clemenzi
December 15, 2012 10:36 pm

richard verney says:
December 15, 2012 at 5:51 pm
Your arguments about DWLWIR and dew are spot on except for one thing – the ground is also emitting radiation. Assuming that the ground and the atmosphere are at the exact same temperature, then the upwelling radiation will always be greater than the downwelling radiation because the ground emits with an almost continuous blackbody spectrum and the atmosphere has a significant range of frequencies where it is transparent (the spectral window). As a result, at night the ground cools until the atmosphere is warmer and the downwelling and upwelling radiation are equal.
When the humidity is high enough, dew, fog, or frost will form. Each of these also emits upwelling radiation. Limiting the discussion to dew, the downwelling radiation keeps it from freezing. In the case of the poles, during the 6 month winter, without the downwelling radiation, the nitrogen in the atmosphere would condense into a liquid.
In the morning, before sun up, the upwelling and downwelling radiation are about equal. In this case, the additional heat from the Sun makes the difference and burns off the dew.
Since you understand electronics, think of W/m2 as current and T^4 as voltage. The surface and atmosphere are capacitors, the Sun is a current source, and deep space is ground. Add resistors as necessary to produce the correct currents.
By the way, solar radiation is UV plus visible plus SWIR, and about 50% of the energy is in the SWIR spectrum. Most of that is absorbed more than 2cm below the surface of the ocean.
http://omlc.ogi.edu/spectra/water/gif/hale73.gif
http://omlc.ogi.edu/spectra/water/data/hale73.dat

December 15, 2012 11:42 pm

Black body Energy in -> Energy out-> a blanket <-Energy back in Energy out -> SPACE

george e. smith
December 16, 2012 1:31 am

“””””…..Willis Eschenbach says:
December 13, 2012 at 12:00 am
Geoff Sherrington says:
December 12, 2012 at 10:28 pm
My son, a surveyor, heard me mention watt per sq km at the troposphere and immediately interjected “At what altitude?”
Unless light passing through a horizonal sq m is parallel, like some laser light, then because there are more sq m of conceptual surface at a higher altitude quasi-sphere around the globe, the flux through each is less when there is a steady source. How does one define the flux when the “number of sq m” on a surface is forever changing rapidly with altitude, but by definition requires measurement to be taken when steady state is approached?
I have no idea about what models do about the variation in numbers of sq m shared by a watt as altitude is changed. Can someone assure me that the effect is built into the math?
If it is not, then a watt per sq m is ‘more powerful’ at low altitude than one at high and it would be hard to construct a Willis fig 1.
Geoff, your son is right, but it is a difference that doesn’t make a difference. Here’s why……”””””
Geoff, does your son the surveyor, know that the radius of the earth orbit (mean) is 93 million miles. Compared to that, the 15 or so miles of atmosphere thickness, makes no difference to W/m^2 coming from the sun.

Mack
December 16, 2012 1:31 am

Willis,
Consider the fact that in equatorial regions a sq.m metal plate (inches thick) will get hot enough to fry eggs on. Think of the equivalent electrical wattage required to achieve this and you are more likely to agree that the Earth’s surface receives a solar incoming radiation yearly average of 340odd w/sq.m. instead of the IPCC’s 161w/sq.m.
Unless you say that some of this is caused by DLR? In which I would have to say to you in the nicest possible fashion…It’s the sun stupid. 😉

richard verney
December 16, 2012 3:53 am

Mack says:
December 15, 2012 at 7:28 pm
//////////////////////////////////////////////////////////
I agree with your comment. I have many issues with the K&T energy budget, but I am merely trying to point out one issue that arises IF that budget were correct. IF DWLWIR possesses sensible energy (in the Earth environ) and IF it is as powerful as claimed, why is it not causing dew to evaporate, or the oceans to evaporate?
The optical absorption of LWIR in water would suggest that there should be significant evaporation since the ‘theoretical’ energy being imparted cannot be spread throughout the bulk volume of the liquid. With Solar the energy penetrates to great depth such that Solar results in the very gradual and relatively slight heating of the bulk liquid. DWLWIR by comparison would concentrate all the energy that it possesses in just a few microns, rapidly raising the temperature of those few microns sufficiently to cause rapid evaporation.
We are not seeing this rapid evaporation of dew, or, for that matter, of the oceans. So why not?
All I am asking is what is the explanation as to why we are not witnessing this in the real natural world? That is what I want Willis to address.

December 16, 2012 4:36 am

I’d like to add to Mr. Clemenzi’s electrical analogy…
Since you understand electronics, think of W/m2 as current and T^4 as voltage. The surface and atmosphere are capacitors, the Sun is a current source, and deep space is ground. Add resistors as necessary to produce the correct currents.
Sea water and dense things like concrete and bricks are big integrating capacitors. Comparatively, the atmosphere represents a tiny, irrelevant capacitor…comparatively to the day/night cycle and thermal capacity of water, of course.
I wonder about Willis’ analytic skill if he really can’t figure out how tropical surface water, directly heated to an average temperature of greater than 20C by the sun, can be liquid without atmospheric backradiation. As usual, if you let your enemy frame the argument, you will lose. When integrating, it’s important that the dt be appropriate. Water is not heated by average insolation. It’s heated by peak insolation (1KW/m^2) and has thermal mass, so it’s able to store the peak energy it’s exposed to. Of course, it’s possible for a small thermal mass to heat a large thermal mass, by being a lot frigging hotter. To extend the electrical analogy, a small capacitor can add charge to a large capacitor, but only if the the voltage is huge. 1/2 C V^2, you know? Otherwise, the big capacitor charges the small capacitor, right? There’s no need to resort to nonphysical gyrations (ha, epicycles) of radiation to figure this out.

richard verney
December 16, 2012 5:01 am

Robert Clemenzi says:
December 15, 2012 at 10:36 pm
////////////////////////////////////////////////////////////////////////
My analogy with the car battery, was not intended as a direct analogy (as I mentioned). It was intended to illustrate a conceptual point, namely there are times when we measure something and not fully comprehend/appreciate the significance (or lack thereof) of what we have measured.
I am well aware that Solar has a wide spectrum and because of it’s spectrum, and the absorption characteristics of water, it penetrates deep into water. Much more than the couple of centremetres mentioned by you. That is one of my points. That is why Solar does not burn of the oceans. It results in the very gradual heating of the bulk, which is to be contrasted with the position that arises with DWLWIR which if it does anything must rapidly heat just the first few microns of the ocean thereby leading to rapid evaporation of the top micron layer.
You omit conduction. The ground also conducts heat to the atmosphere above. In my example of morning dew, for all practical purposes, Solar does not heat the dew at all. If the thin film of dew is 1mm thick, due to the absorption characteristics of water (to the various wavelengths of Solar), for all practical purposes, non of it is absorbed by the dew and it all goes to heat the ground below. The dew is burnt off by conduction brought about by Solar heating the ground below the dew.
To evaporate water you need to consider energy over time. Just before sun up, whatever amount of energy is being conducted and radiated from the ground in the hollow, is for practical purposes the same all over the hollow. Likewise the ambient temperature and humidity of the air above the hollow, At sun up, the only additional factor is the input of Solar on the sunny side of the hollow. This additional energy could be measured, but we know it is slight since there is not much energy in low incident sunlight especially in the late autumn, winter, early spring. This slight additional energy is sufficient to burn off dew within 30 mins (may be a bit more or a bit less – depends upon the prevailing ambient conditions which prevailing ambient conditions are the same in the shady side of the hollow).
As the day warms, the sunny side of the hollow conducts (and you would say radiates) more heat and the atmosphere gradually warms. At this stage of the day DWWIR is increasing. It is now more than was the case when the sun was first up. DWLWIR is omnidirectional so we know that the shady side of the hollow is now receiving some of this increased DWLWIR. The shady side of the hollow may experience this increase in the DWLWIR for hours and yet the dew is not burnt off from the shady side of the hollow. Given the length of time that the dew in the shady side of the hollow is exposed to this increase in DWLWIR, we know that the total energy received from this increased package must be more than the additional energy that was inputted to the sunny side of the hollow by Solar (alone) within the first 30 mins (or so) of daybreak.
One can visualise what is going on without knowing the precise figures. But if the relation between Solar and DWLWIR is in the order suggested by the K&T energy budget, it would not take long for DWLWIR to burn off a thin film of dew especially given the absorption characteristics of water to DWLWIR. When considering this, one must bear in mind the thermal inertia of the ground. The ground is being heated by Solar but heating of the ground could be slow, whereas water theortectically absorbs LWIR almost instantaneously and nearly all the energy from DWLWIR is contrated in the first few microns such that one would expect rapid evaporation to occur. .

mkelly
December 16, 2012 8:29 am

Mr. Verney and Mr. Clemenzi your analogy to electrical circuits is quite proper. My heat transfer book uses circuit diagrams as analogy for resistance to heat flow.

Pamela Gray
December 16, 2012 10:18 am

A case in point. Decades ago, most psychiatrists were firm in the scientific conclusion that autism was caused by distant and cold mothering. Mothers called that notion bunk and balderdash. Many used foul language in their description of the consensus and worse in their description of psychiatry.
So no, it does not take a card-carrying scientist to refute a consensus and eventually be proven correct. If you were one of those people who discredited people who’s job description was to raise their children, you were eventually labeled a fool.
My advice to AGWers, do NOT walk in those shoes.

Robert Clemenzi
December 16, 2012 10:26 am

richard verney says:
December 16, 2012 at 5:01 am
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At night, the ground radiates energy and cools much faster than the air above it. Via conduction, this pulls energy (heat) from the air immediately above it. At this point, “hollows” are a bit more complicated than the Great Plains. In a hollow, this cold air runs downhill and collects in the valley. As a result, the cold layer above a plain, or calm ocean, may be only a few centimeters thick, but in a hollow it will be several meters thick. This is easy to see in the Smokey’s (an Eastern US mountain chain) where the hollows contain fog long after the peaks are clear.
I agree, the Sun can not heat the dew directly. It heats the leaf, grain of sand, etc. and that heats the dew. Because the conduction from the leaf to the water is much greater than from the leaf to the ground (direct contact verses an air gap), the thermal mass of the bulk surface has almost no effect until after the dew is evaporated.
When evaluating a transient response, the Solar radiation suggested by the K&T energy budget is misleading. 170 W/m2 is averaged over a full day and a full year. The peak radiation to burn off the dew is more like 1,361 W/m2, the TOA value. As for heating the leaf, remember that dew drops are small lenses and that means that the “temperature” can be much higher than predicted by SB. However, for the DWLWIR the dew is opaque, meaning – no lens effect. So comparing an increase of 40 W/m2 to an increase of 1,361 W/m2 plus a lens might explain the difference.
The angle of the sun light is an interesting problem. In general, the amount of energy available is related to the cosine of the angle. However, because leaves are at all angles, and because of the lensing effect, a much higher than expected amount of energy is available to evaporate dew. As long as the sun shines, the amount of energy available is very significant. Passive (non-focusing) collectors are able to boil water in Canada during the winter. The cold air has more to do with where it comes from and the number of daylight hours than with the angle of the Sun.
With respect to dew, in the very early morning, refraction is also an issue. This, plus the extra thickness of the atmosphere (6 miles tropical noon, 154 miles in the morning), is why the morning and evening suns feel so cool.

Pamela Gray
December 16, 2012 10:43 am

Ocean heating: The Coriolis affect combined with surface obstacles and currents define the major components of Earth’s unique oceanic fluid dynamics. Ocean layers are heated, especially around the equatorial belt, with IR heating (not long wave) which itself is filtered to various levels at the surface because of cloud variation. The heat absorbed by the oceans is then mixed and sloshed around or allowed to sit and layer calmly via trade wind variations.
Oceanic heat loss: My educated guess is that in various ways, heat is largely released back into the atmosphere, sometimes because of storms, and usually because of its eventual trip to the poles where it evaporates and eventually escapes the confines of our planet. Some years and decades (long and short) the oceans lose a lot of heat, and some years and decades that cooling system doesn’t work so well. This gain/loss see-saw is anything but predictable.
I think longwave radiation oceanic heating (the kind that is re-emitted from the ocean and re-emitted back into the ocean from the atmosphere) is a tiny player, buried in the noise, in the above description, and certainly not at all in measurable trends.

Frank
December 16, 2012 11:11 am

Phi replied to me: “Well, science is not the stock market! Either radiative forcing has a meaning, or it does not.”
The IPCC’s definition of radiative forcing provides a very clear meaning. Radiative forcing is the result of a calculation that determines how the net flux of radiation through the atmosphere is expected to change as the atmosphere and/or the radiation entering the atmosphere changes. The IPCC has specified where (the tropopause) and how (allow stratosphere to reach radiative equilibrium) that calculation is to be done, and their choices make sense. To do the calculations, one needs carefully measured parameters from many reproducible laboratory experiments, atmospheric conditions and composition, and how much radiation enters the atmosphere from the sun, the surface and cloud tops. (All but the parameters change from location to location, so there is no single correct answer.) At least some of the software for doing these calculations is available on the Internet. If anything in climate science desires to be called “settled science”, it might be radiative forcing.
Unfortunately, “radiative forcing” reported in W/m2 doesn’t mean anything to non-specialists. The IPCC chapters that I have read about GHG forcing don’t tell the public how to interpret radiative forcing in terms of a temperature increase. Instead they wait until they have amplified radiative forcing with poorly understood feedbacks to produce climate sensitivity or processed through climate models containing parameters that have been tuned so that 20th century warming is attributed to increasing GHGs.
Here’s a reasonable interpretation of radiative forcing in terms of warming, generally called the no-feedbacks climate sensitivity. W = eoT^4, so W+dW = eo(T+dT)^4 where dW and dT are small changes, deltaW and deltaT. With a little algebra and ignoring terms with higher powers of dT, one can derive an equation I rarely encounter: dW/W = 4*(dT/T). It says that the percent change in radiation will be four times bigger than the associated percent equilibrium change in temperature (in degK). To compensate for a radiative forcing of 3.7 W/m2 (1.54% change) we need a 0.38% increase in temperature. Since the surface and lower troposphere receive about 240 W/m2 of solar radiation (after account for albedo), the equivalent blackbody temperature (255 degK) appears to be the most relevant temperature to use, affording a 1.0 degK warming. Where will this warming take place? Calculations can’t tell us, but 90% of the photons escaping to space are emitted from the upper troposphere, so a reasonable answer is that upper troposphere (where the temperature is near 255 degK) will warm 1.0 degK. If the lapse rate remains unchanged, the surface will warm by 1.0 degC too.

Mack
December 16, 2012 11:56 am

Willis says……..Dec15th 201 7.28pm
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Everything you say is IF there Willis. Did Micheal Moon say you worked as a massage therapist? Probably explains why you you were asking everybody for a bit of traction further up .

phi
December 16, 2012 12:37 pm

Frank,
“If anything in climate science desires to be called “settled science”, it might be radiative forcing.”
How is computed the CO2 forcing? By adding with a thought experiment some CO2 created ex nihilo. What is the temperature of this CO2? Obviously not defined. We uses the temperature profile of the atmosphere before the add but it’s a perfectly arbitrary operation. It is absolutely necessary to leave the new CO2 to adapt to its environment before making any calculation. Unfortunately, the principle of radiative forcing requires an instantaneous balance. This calculation is actually under-determined and the value obtained arbitrary.
We can being aware of that also considering the schematic equation of heat transfer through the atmosphere: Ts = L * P + k * P ^ 0.25 where Ts is the surface temperature, P the heating power, L and K synthetics factors. The addition of CO2 occurs in L and can not be modeled by an alteration of P. There is an exception with the energy evacuated directly from the ground and now intercepted by the added CO2 but it is a small amount.
“If the lapse rate remains unchanged, the surface will warm by 1.0 degC too.”
1 ° C may be correct as an average, by cons, we know that it’s not valid for surface because the main effect of the added CO2 is to change the lapse rate by reducing the potential radiative losses along the column.
This warming being not valid at the surface is critical because it is the surface temperature that initiates the feedback on water vapor.

Greg House
December 16, 2012 1:40 pm

Willis Eschenbach says, December 16, 2012 at 9:55 am: “While NET heat flow always goes from hot to cold as you claim, with radiation there are always two flows of energy going on. There is a flow of energy from the warmer to the cooler area, and there is also a flow of energy from the cooler to the warmer area. It is only the NET flow that is constrained to always go in one direction. Get a college-level beginner’s physics book, they’ll cover it in there. Here’s an example illustration, from the University of Sydney.”
============================================================
Willis, your illustration does illustrate an IDEA, but like any other illustration it does not prove the idea to be right or to be a scientific fact. You illustration only demonstrates that there are other people who share your idea about “NET”.
Second, and this is important, this “NET” thing makes physical sense with regard to temperature only, if “a flow of energy from the cooler to the warmer area”, as you put it, has an effect on the temperature of the warmer area. If it does not, your “NET” does not make physical sense at all.
Even if there is some effect on the temperature of the warmer area, but this effect is disproportionately or even vanishingly small, you can not just add or subtract energy and then derive temperature from the result of this arithmetic operation. This would be no science.
Which means, until this “NET” thing is not proven experimentally, it remains a pure fiction.
Please, Willis, present a clear reference to and a description of a REAL scientific experiment (not just a “thought experiment”) proving your idea about this “NET” effect.

TimTheToolMan
December 16, 2012 5:35 pm

Willis writes “For TimTheToolMan and the others who think that downwelling longwave radiation can’t warm the ocean, not one single one of you has answered my simple question.”
I do think DLR slows the rate of cooling, Willis. I’ve told you this a number of times in the other thread. I’ve not said it doesn’t in this or any other thread so I dont know where that accusation keeps coming from.
I am pointing out they so far I believe your understanding on how the ocean warms as a result is lacking. You’ve stated that there is no difference between warming and slower cooling but then you’ve made statements that clearly indicate you believe the DLR increases the SST. Directly!. It does not. It cant because the ocean is cooling Willis. The sun warms it at depth, not at the very surface when, for example, clouds come over.
There IS a difference between slower cooling and warming because when you think “warming”, when you think more DLR means +100 W/m2 you can and do get it wrong.

jae
December 16, 2012 5:58 pm

Willis notes:
“Greg, I doubt you can find what you are asking for. Such basic statements are rarely accompanied by experiments. The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.’
BUT, sir, I think I just gave you THE experiment (Phoenix vs. Atlanta). The amount of radiation from greenhouse gases seems to matter very little, if at all. Water and water vapor can probably explain it all.
I noticed that you have not responded to my last comment on this subject, and I am wondering why…

December 16, 2012 5:58 pm

I don’t know anyone who disputes the fact that things with mass and temperature radiate (except perhaps Myrhh, I’m not sure what he or she thinks). The Kent drawing shows two independent sources radiating toward each other. No problem. Going further, I don’t know anyone disagreeing with the idea of a passive receptor heating up and radiating. The problem is trying to do useful work with re-radiated energy…and in the presence of convection? Good luck. You can’t get an average temperature increase of 33C from backward atmospheric re-radiation. Something this blatant would be easy to instrument in a lab environment.

JazzyT
December 16, 2012 6:58 pm

There seems to be some confusion about the idea of GHGs emitting IR which is absorbed at the surface, and heating it. It might be better to say, “transferring heat to the surface” rather than “heating the surface” because, at night, in the absence of other effects, the surface will inevitably cool, but will cool more slowly than without the GHGs. During the day, the surface is warmed by the sun.
The whole climate change issue involves questions of whether the Earth might warm, on the average, by 2 degrees C over the next century, or by 3 degrees, perhaps 5 degrees, or by 1 degree or less, with various consequences or lack thereof.. This projected temperature change, over a century, is superimposed on daily and seasonal changes. Daily changes of 10 degrees is quite ordinary, and seasonal changes of 30 degrees or more are quite common, as compared to a much smaller AVERAGE change over 100 years.
There is NO need to worry about where the heat would come from for global warming; it comes in every day. If the Earth were to heat up, on the average, by 10 degrees C over the next century, most climate scientists think that truly cataclysmic results would follow–but NOBODY thinks that this is a remotely serious possibiltiy. The balance between incoming and outgoing energy is fine enough that one day’s sunlight changes the temperature much more than any average change that anyone seriously considers for the next century. (Heating the top layers of the oceans would take more than a day, but still much shorter than a century.)
GHGs do indeed radiate IR to the surface and transport heat thereby. But the surface cannot actually warm from this unless the atmosphere is actually warmer (which could happen if, say, a warm front moves through). But the effect of GHGs is to slow the rate at which the surface cools through IR emission. The actual mechanism is complicated, but that’s the end result.
We could try to calculate the temperature at the surface in terms of how much the atmosphere radiates to the ground, various heat capacities, etc. but that’s difficult to conceive of, and difficult to solve. There’s an easier way: despite the differences of night and day, seasons, weather, etc. the incoming and outgoing energy is always very close to balanced. Were this not so, the average temperature would go up, not over a century, but over a matter of weeks or days. It would probably be very exciting. But the main regulatory mechanism, the sharp dependence of IR emission on temperature, puts the brakes on quickly. As Frank mentioned above, emission goes as the fourth power of temperature. For every 1% increase in absolute temperature, which is about 3 degrees C, there’s about a 4% increase in IR emission. So, if nothing else has changed, that cools the surface down pretty quickly. (For a demo, wait until sunset.)
So, to solve it the easy (or easier) way, we look at the percentage of IR that gets through (at each wavelength, it’s not quite simple) and then ask: what temperature would the surface have to be so that the heat that escapes through IR in 24 hours is equal to the heat received from the sun in one day? The more the GHGs “block” the IR the higher that temperature is. The word “block” is a gross oversimplirication, but in the end, that’s why the temperature goes up–more solar heat is retained, until the surface is warm enough to shove through an amount of energy that equals, on the average, what the sun puts in.
Of course, after that, you look at the IR absorption spectra of the greenhouse gases, blackbody emission spectra, heat transport through air, currents, ocean currents, evaporation, precipitation, thermals caused by politicians emitting large amounts of hot air…lots of complications. But one thing you DON”T have to worry about is how the energy gets to the ground to heat it up. The sun takes care of that, every day.

Greg House
December 16, 2012 10:22 pm

Willis Eschenbach says:, December 16, 2012 at 5:23 pm: “Greg, […] The fact that solid objects radiate depending on their temperature is such basic science that it is presented as a statement of fact rather than with an experiment.
Here is some information from MIT on exactly how to calculate the radiative heat exchange between two surfaces. They have no doubt that there are two separate radiative flows.
Now, if you don’t believe what the folks at MIT say about radiation, I’m afraid you’ll have to take it up with them.”

=====================================================
Willis, I am not questioning the fact that solid objects radiate depending on their temperature, or that there are separate radiative flows. You have probably misunderstood the point. I am questioning the alleged effect of radiation from colder bodies on TEMPERATURE of warmer bodies, and logically the calculations of what you call “NET” transfer. I hope you have understood the point now.
Your references to “folks at MIT” are not a scientific argument unfortunately, anyway not when the very essence of your point is questioned. It is a wide spread logical fallacy called “appeal to authority”. I am very sure that “folks at MIT” can give references to other “folks” and so on. Such a circle of references is, however, not a scientific argumentation.
You are apparently a knowledgeable man, so I guess, if you can not refer to a real scientific experiment confirming your (and other “folks’) assertion about that “NET” thing, you probably are capable of understanding that this assertion can not be considered a scientific fact, it is just a fiction.

gymnosperm
December 16, 2012 10:55 pm

So here’s a stab at explaining why the oceans don’t freeze for lack of DLR:
Take a water heater and remove the dip tube. Install a big pump that recycles water between the holes at the top of the tank. What you are doing is installing a flowmeter and thermometer on the outflow side and saying, “Wow, there is so much energy leaving this thing would freeze if it weren’t warmed by water coming back in.”…

TimTheToolMan
December 16, 2012 11:27 pm

Willis writes “Here’s the thing. There’s no difference in result between the blanket and the torches.”
You’ve made another strawman argument without even realising it… The thing is that DLR (and a blanket) redirects the object’s energy back at the object. Back radiation. And its just not possible for a blanket to make the block of iron warmer than when you started. Just because you can set a propane torch up to account for the same energy loss means nothing and doesn’t mean warming is the same as cooling.
Do you stand by this statement that you made to tallbloke?
“The very surface warms … until it’s slightly warmer than the bulk … which then heats up slightly, until the surface is slightly cooler than the bulk, and the previous condition (cooler surface) is restored.”
Because this is very explicitly saying that DLR can warm the ocean warmer than it was before you started. This is like saying I can put a blanket on the iron or put coffee in a thermos and it will get warmer than when I started.

AlecM
December 16, 2012 11:34 pm

Willis says this: ‘Regarding heat flow, you are missing the boat entirely. While NET heat flow always goes from hot to cold as you claim, with radiation there are always two flows of energy going on. There is a flow of energy from the warmer to the cooler area, and there is also a flow of energy from the cooler to the warmer area. It is only the NET flow that is constrained to always go in one direction. Get a college-level beginner’s physics book, they’ll cover it in there. Here’s an example illustration, from the University of Sydney.’
This is plain wrong. Unfortunately most people are taught it as if it’s correct. The reason is they are taught the S-B equation and the difference, net energy flux is the difference between the two S-B predictions. This is incredibly misleading though and you along with most have fallen in to the trap.
To prove this you go back two steps in the physics. The first is the Planck Irradiation Function which when integrated over the wavelengths or the wave numbers is the S-B equation. The second is to understand the the PIF at any wavelength/wave number is for a collimated beam identical to the Poynting Vector. This transfers to physics to Maxwell’s Equations.
The wonder of this is Poynting’s Theorem which is that the net PV at any point in space is the vector sum of all the arriving PVs. In other words there can never be two streams doing work. Only the vector sum at a point can do work.
This is why the climate models are total, absolute bunkum. You can’t compute absorption of energy from the two streams because it exaggerates warming by a factor of ~7. What’s more as there is zero net CO2 IR emitted from the surface [basic radiation physics at thermal equilibrium means net PV is zero where the two PVs are near black body iamplitude] there can never be any CO2-AGW or positive feedback.
Please Willis, learn the real physics and tell these IPCC charlatans they have got it wrong. It goes back to Houghton using the two-stream approximation to calculate energy absorption. If he had had the scientific wit to work out that the DOWN warming is negative to offsets most of the UP warming we might have avoided this incredible mess of fake science.

AlecM
December 17, 2012 12:17 am

I shall also add a point I haw made in the past which is that the belief in DLR is entirely baseless because a pyrgeometer measurement is the artefact of the instrument’s shielding. The proof it to have two pyrgeometers back to back in zero temperature gradient – net signal = 0. Take one away and the signal jumps to a measure of the S-B flux for that body at its particular temperature and emissivity.
However, until that electromagnetic wave combines destructively with the wave from the opposite direction, it cannot do any thermodynamic work. ‘Back radiation’ is an imaginary energy flow all of which is destroyed at the Earth’s surface. The only work that can be done from the Earth’s surface is the 63 W/m^2 net IR flux, the 17 W/m^2 convection and the 80 W/m^2 evapo-transpiration.
Anyone who claims differently has to justify it from Maxwell’s Equations and they can’t.

AlecM
December 17, 2012 12:47 am

Willis: The citation you reference, about radiative heat transfer calculations, is exactly what I was taught as a Metallurgical Engineer. Until 3 years’ ago when I set out to study exactly what the IPCC claimed, I have never thought beyond the S-B analysis. However, once I realised the climate models have in them a perpetual motion machine, creating much more IR energy to be absorbed than real net IR, I decided to find out why.
The two-stream approximation cannot be used to calculate heat absorption unless one stream is defined as negative, the cooler to the warmer. You cannot have a perpetual motion machine. Unfortunately, the grant money would disappear because the CAGW scare vanishes, and we can’t have real science interfering with capitalism, can we?

Mack
December 17, 2012 12:51 am

Dec 14th @10.34 am “So what is providing…….. energy that is keeping the ocean from freezing? I keep waiting and waiting for you folks to say it is not DLR to explain why the oceans are not frozen” (then later)… “where is the energy coming from to keep them unfrozen?”
So then he ,Willis (hopefully) reads what I said to him about Trenberth’s unreal geometrically derived IPPC crap of 160w/sq.m. reaching the Earths surface. But no he persists..2 days later..
Dec16th 2012 @6.38pm “So the problem remains- the sun alone is not enough to keep it liquid”.
You know how you might be reading something right in front of you but cannot believe what your eyes are seeing. This is exactly my case here. How can this guy Willis who calls himself a scientist (maybe that’s his problem) say such stuff.
Willis…
If you were to take the earth and for hypothesis sake freeze the seas. Then introduce a sun which will apply a continuous and timeless input of energy to the surface of 340w/sq.m. I’m sure that before too long we would approach what we have today in reality. I’m going to say this again Willis without any smilies or winkies . It’s the sun stupid.

AlecM
December 17, 2012 1:00 am

Willis: ‘If DLR is not keeping the ocean from freezing, what is?’
The answer is very simple. The oceans do not emit 400 W/m^2 ULR. The only IR they emit is the net 60 W/m^2. This is an inescapable fact – only net energy flows are real.

TimTheToolMan
December 17, 2012 1:28 am

Willis writes “But until someone can come up with an alternate source of 340 W/m2 of energy to keep the ocean liquid, I’m calling BS on the claim that the ocean is not warmer with DLR than it would be without it.”
Its not about the warming, its about the process. And mostly about the “warming is the same as reduced cooling” At least try to understand this. I’ve done you the courtesy of understanding your posts.
This….
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-721248
And then this…
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-723255
And I think the only missing bit between those posts is that when the DLR increases and if there is still DSR to keep warming, then the slowing conduction in the skin will cause the skin’s temperature gradient to reduce (less is being lost from below now) and then the hook along with the SST will increase. At no point does the very surface warm beyond the temperature below due to the DLR.

D Böehm
December 17, 2012 2:05 am

I have to agree with Willis. The DLR is the reason the oceans are not frozen. If that is wrong, please provide a credible alternate explanation.

AlecM
December 17, 2012 2:39 am

D Boehm: ‘DLR’ is an artefact, Poynting Vectors representing the temperature of the emitter(s) in the viewing angle of the pyrgeometer sensor. It’s not real otherwise at twice average solar flux, you’d be able to feel it on the back of your hand by the heat sensors!
It cannot do any thermodynamic work until it combines vectorially with information from the hotter body, This is why to measure real energy flux you need two pyrgeometers back to back and take the difference signal.
The Trenberth ‘Energy Budget’ is total bunkum unless you give negative warming power for PVs from the cooler to warmer bodies. It’s far better to work with net energy flows!

Edim
December 17, 2012 3:28 am

“Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss (global 24/7 average ≈ 500 W/m2). The calculation is out of balance by about 340 W/m2. I (and most every scientist I know of) say that this 340 W/m2 is the energy that the ocean is absorbing from the DLR.”
There’s no missing energy – simple Earth’s energy budget show that the surface absorbs 51% of the incoming solar energy and loses 23% by evaporation, 21% by IR radiation (6% radiated diractly to space) and 7% by convection.
51 = 23 + 21 + 7

Edim
December 17, 2012 3:28 am
AlecM
December 17, 2012 3:56 am

Willis:I’m sorry to state this so bluntly, but pyrometers do not measure energy flux, they measure temperature. OK, a pyrgeometer is calibrated against a cavity black body in Power Units but that is the S-B power that would be emitted from the isolated emitter in a vacuum. The artefact is because the shield at the back of the detector stops EM information coming the other way.
The only real energy flow is the difference signal between back-to back instruments. This is what the manufacturers state, see the bottom of this page: http://www.kippzonen.com/?product/16132/CGR+3.aspx
These people have sold 1000s of pyrgeometers to climate research knowing full well that they are being misused because Meteorologists and Climate Scientists are taught ‘back radiation’ must be real because the single pyrgeometer is calibrated in W/m^2. This is the most blatant and longest [50 years] misunderstanding in science I have ever come across. and it must be stopped by forcing these academics to teach correct physics. [The other big mistakes are wrong IR physics and wrong cloud physics].
So, there is no opposing 400 W/m*2 and 340 W/m^2 power fluxes. These are instrumental artefacts, a temperature signal expressed in the EM continuum. There is no energy transfer until the information in those two streams combines vectorially as the net energy flux and even then other emitters are involved, e.g. the cosmic microwave background for the atmospheric window radiation.
‘Back Radiation’ does not exist. For normal temperature gradients, it all disappears at the Earth’ surface.

Ryan
December 17, 2012 3:58 am

Willis: “Simple energy calculations show that the incoming sunlight at the surface (global 24/7 average ≈ 160 W/m2) is not enough to balance the known losses from radiation, conduction/convection, and latent heat loss”
Well, I for one don’t see why I need to justify myself any further. The incoming heat at the equator is not 160W/sqm. It is 1000W/sqm. The water at the equator is forced to flow E-W and then has nowhere to go but northwards until it rotates around the poles and back down south like some enormous central heating system. That is why the oceans are not frozen. The specific heat capacity of the oceans is such that they can easily abosrb enough energy at the equator to remain unfrozen across the rest of the globe. This also explains why the Arctic ocean is not frozen all the way to the sea bed in winter despite having 6 months with no solar radiation let alone no DLR and with an opaque layer of ice on top. It also explains why the seas around Britain can be 10Celsius always in the winter even though the air temperature can persist below 0Celsius for days on end. You can’t use an average because the areas with less daytime heating will also have less cooling – i.e. the maths is different for the radiators of a central heating system than for the heat exchanger.
If you want to prove to yourself that Trenberth’s figures must be nonsense, go and get a big mirror and put it in the garden in a cold night over your head facing down – see how much warmer it is? No? But surely it must be – after all now you are getting 100% DLR at close range, you don’t have to rely on the weak effect of unsatured CO2 spread throughout the atmosphere. Sit in a shed however and you might well feel a bit warmer because now you have reduced the conduction/convection effect.
Radiation balances are only relevant when seen from space. Below the tropopause only the conduction/convection/evaporation is relevant. If it wasn’t for the cooling effect of the troposphere the oceans would be a lot WARMER, because vibrating atoms conduct heat between them in a far more effective manner than the emission of photons. Bear that in mind – if there was no air above the oceans they would only be able to lose heat by re-radiating photons – a much slower process than by radiating photons. They would be much hotter and would boil into space.

AlecM
December 17, 2012 4:18 am

Willis; Firstly your warm desk. Put two pyrgeometers back to back between the desk and the sofa. The net signal will be near zero. This is a fact. A single pyrgeometer measures temperature expressed in power units. It is not a real energy flow.
That only takes place when the two information streams combine. A rider to this is that the common idea that such streams are strings of photons is also very wrong. This is because the photons only exist at the instant energy quanta are transferred, and that can only happen once the net energy flux is made.
As for the two lamps, solving that problem mathematically is very complex. The 40 W will rise in temperature more than the 100 W because its cooler glass will be a sink for the warmer 100 W envelope.

AlecM
December 17, 2012 4:45 am

Right, a better explanation of the two lamps.
7% of the energy is emitted as SW light. The mechanism by which it does work is that EM vectors from the cooler body combine vectorially with the vectors from the hotter body and the net energy then passes from the hotter body to the cooler. Because the cooler body is at much lower temperature, the net flux is the same as the emitted flux.
93% of the energy is IR. Some is absorbed by the glass bulb and this then radiates and convects. Concerning the former, what happens is that radiation from cooler bodies combines vectorially and because we are dealing with a greater proportion of the energy from the cooler body, the radiative heat loss is a smaller proportion than for the SW case. As a result, convection increases.
Now consider the oceans. All the GHG thermal energy received from the atmosphere disappears, removing the same from the 400 W/m^2 you would get for an isolated ocean in a vacuum! So, the energy flow is one way, 60 W/m^2.
Thinks of the opposing Poynting Vectors as an impedance to real IR energy flux. Higher amplitude at any given wavelength reduces that flux from the surface, and it does so unevenly depending on the spectral nature of the radiation from the other direction.

TimTheToolMan
December 17, 2012 5:11 am

Willis writes “My question is, if your process is correct and the ocean is NOT absorbing 340 W/m2 from the DLR … then where is it absorbing that amount of energy from?”
The surface of the ocean IS absorbing the energy in the top 10um or so…and radiating it back up at about the same rate (some may be used in evaporation) for no net positive effect in the bulk. That is the vital bit. There is no energy gain in the bulk or in the top 1 mm or anywhere due to the DLR.
The ocean is cooling at the surface and the DLR is not warming it. Not even a little bit. Warming comes from sunlight absorbed below and the SST is driven by the energy from below, not from the DLR although the DLR is a factor in determining where the balance is.
Mathematically it accounts for some of the energy the ocean must radiate according to S-B.
Do you understand why I think your statement to tallbloke is wrong?

Mack
December 17, 2012 5:35 am

Willis,
It’s quite difficult for me (layman) to explain, but here goes…Trenberth and academics and AGW people seem not to be content with just accepting that the TOA receives a real measured ( since 1902 by waterflow and now by sattilite) reading of an incoming 1360w/sq.m. but they take this figure and like all true teachers get some idea that this is the amount emitted by the sun and immediatly go to the blackboard and draw a picture of two little circles of the sun and earth. At this point they have stepped out of reality. They then draw rays from the sun to the earth and through a process of maths and geometry work out what THEY think is the incoming TOA solar radiation. A sort of buggering up of a real measurement .Sort of one foot in reality and the other in an AGW fantasy. As I’ve told you before Willis this 1360w/sq.m. is a yearly global average which cannot be divided. It is the real deal. This is what we get. However you and the academics arrive at this CALCULATED figure of 340 w/sq.m incoming at the TOA, which then transforms into figures tearing around in your hysterical imaginations. The neat and paradoxical thing for you AGW deceiving liars is that what the earth receives in reality at the surface ie 340w/sq.m ,happens to coincide with the 340 w/sq.m. you lot work out for the TOA. In this way it easily confuses students because they are unaware of which surface you are referring to …. the actual earths surface or the “surface” of the TOA. When drawing your little pictures of the sun and earth on the blackboard what student is going to pick up on the little distance of the width of the atmosphere eh Willis. This is probably one reason your crap science has done so well. So get real Willis, The reality is that this earth receives 340 oddw/sq.m. at its surface (the sq,m metal plate in equatorial regions getting hot enough to fry eggs on ,remember?) get used to it and you can take Trenberths and your figures a shove em where there ain’t no w/sq.m.

AlecM
December 17, 2012 5:55 am

TimTheRToolMan, the operational IR emissivity of the oceans is negatively dependent on temperature so highest IR emission is from temperate seas. specifically off New England and Japan.

December 17, 2012 6:29 am

I have no problem with modeling and simulation and superposition…these things are commonly used in engineering. However, I do have a problem with creating nonphysical influences and using them in the model. If you want to spread insolation across a disk, that’s fine if there is a physical mechanism that actually does that. I’m not going to spread insolation across a disk when I’m modeling a rotating sphere. If you want to average incoming energy into an average representing a daily dose, that’s fine if you can point to the physical mechanism that does that. I’m not going to average daily insolation across a 24-hour period if there is no physical mechanism. If you want to store and integrate thermal energy, then store it in something with thermal mass like water and dense masses and use those storage mechanisms in your integrating storage/delay/feedback model.
I like the example of an independent 40W lamp radiating into a 100W lamp. Now replace the 40W lamp with a snowglobe and imagine how how much you must increase the 100W to get 40W back from the passive snowglobe…with unconstrained convection. This will give you a sense of proportion for the work done via “back radiation”.
I like the idea of replacing an insulating blanket with a torch to create an equivalent (slower) cooling profile (that’s clever and wise). Now imagine what the torch must do to increase the average temperature of the warm block by 33C. Good luck, my friends.

AlecM
December 17, 2012 6:42 am

kencoffman: as an engineer, what do you think of the modelling when it uses the two stream approximation yet only net IR can do thermodynamic work!
Including a TOA error, the result is to increase the IR energy absorbed in the atmosphere from 23 W/m^2 to 134.5 W/m^2, a factor of 5.84. this is the cause of the imaginary positive feedback. They offset higher temperature by exaggerating cloud albedo.

Ryan
December 17, 2012 7:02 am

The light bulbs. You are both wrong. If 7% of the energy from a bulb is visible light, it is not true that the remaining 93% is infra-red. Furthermore even if it was it is irrelevant because the laws of entropy make it clear that if photons are obsorbed it doesn’t matter what the frequency they will both decay to heat in the end – i.e. visible light photons cause just as much heating as infra-red.
The reason the hot bulb will not measurably heat the coller bulb is because the bulb has a filament which is probably at 3000Celsius. It is so hot that it is actually boiling single electrons from its surface. The gas in the bulb is inert, so it doesn’t capture these free electrons, but the gas does get seriously hot. The envelope of the bulb does get extremely hot but radiation has almost nothing to do with this because most of the photons are by definition going straight through. Because the envelope of the bulb gets so hot (but not as hot as the filament because the area is much bigger) there will be considerable heating of the air around the bulb and this will take away almost all the heat from the hotter bulb and hence most of the energy. With all this going on around both bulbs the relatively small amount of radiation transfer would be difficult to measure reliably. However, if you ramped down the current in the cooler bulb so it was no longer emitting in the visible spectrum but was still a little warm, you of course would see it was illuminated by the hotter bulb so clearly it is receiving (photon) radiation albeit at a difference frequency. But clearly it isn’t getting much energy through that route because if you put your hand in between the two you won’t feel the heat very much (you will if you put it just above the bulb).
If you wanted to do the experiment properly you would want the bulbs to be in a vacuum like the old bulbs because then there is little conduction as such – but then the tungsten gets so hot it actually starts to boil off whole atoms onto the inside of the glass (so there is conduction in reality and the bulbs can’t run so hot anyway).
What is happening on the microscopic scale in the conduction case is that whole atoms are violently vibrating in a solid and impacting the atoms in the less dense and free-flowing gas around it. In the case of radiation massless photons are being absorbed by electrons causing them to move to a slightly higher orbit around the atom – you can imagine that this latter mechanism really isn’t as effective as conduction in taking heat away from a cooling object.

Ryan
December 17, 2012 7:16 am

Willis: The reason you are getting confused is that black-body radiation is measured relative to absolute zero. So, in your example the desk is radiating at 400W. Does this mean putting another desk in front of it will cause 800W of radiation? No. The two desks are at the same temperature so there can be no heat transfer between them. Both are radiating at 400W which seems mighty energetic, but since they are both at that temperature there really isn’t any energy transfer between them (oh, and you won’t easily be able to distinguish between them using a thermal imaging camera either). They would be receiving photons from each other for sure (i.e. the 1st desk can see the photons of the brown second desk just as you can) but the net energy transfer is zero (the 2nd desk is emitting just as much visible light to the first desk)
Bearing in mind what I said about conduction being a much stronger effect than radiation, what happens to happens to still air above a large body of water? The conduction of the heat from the water causes the air to eventually reach the same temperature as the water (note this cannot happens the other way around as heat in free-flowing gases and liquids rises and the water is much desner than the air). If they are at the same temperature due to conduction, how much energy transfer can there be by radiation? The answer is surely zero.
That, Willis, is why the AGW crowd have got it all wrong. Because conduction occurs much more strongly at ground level than radiation, the air is much the same temperature as the land and sea due to conduction. If the air is much the same temperature as land and sea, there can be no net radiative transfer from land or sea to the air. If there is no net radiative transfer from land or sea to air near ground level, then CO2 in the air cannot make a difference to the radiative transfer. Radiation only exceeds conduction above the tropopause when the low density of the air means conduction cannot readily occur and then radiation becomes the only viable way of losing the heat to the dead cold of space.

gymnosperm
December 17, 2012 7:33 am

Well of course DLR warms the skin, but that’s about all it warms. If it didn’t the ocean WOULD freeze. What we have is an enormous eddy of energy (amounting to nearly 60% of the total incoming) cycling between the skin and water vapor in the lower atmosphere. This is precisely the part of the atmosphere where a few extra molecules of CO2 will be lost in the vapor equivalent of 13000 cubic kilometers of water the atmosphere contains on average.

AlecM
December 17, 2012 7:45 am

Ryan: what’s so incredibly wrong about the two-stream approximation is that the IPCC consensus claims there is 333 W/m^2 real energy flow from the lower atmosphere to the surface and 396 W.m^2 heading upwards, heating the air both ways as the IR is absorbed by GHGs.
These data are obtained by pyrgeometers and the meteorologists have always believed them to be real energy flows. However, only 63 W/m^2 real energy is radiated and these people can’t imagine that a body at 16°C, the surface, does not emit 396 W/m^2.
Yet this is only the case for an isolated body in a vacuum. in reality, as you show, the lower atmosphere is the same temperature and for the specific case of the 15 micron CO2 band, there is zero net radiation from the surface to be absorbed by the CO2 in the air.
In other words there can never be any CO2-AGW. Yet these people go on to claim fantastic warming by AGW oblivious to the realities of heat transfer, as this article shows.
What’s worse., they get the IR wrong: There is no possibility of any thermalisation even if there were CO2 IR to be absorbed. The IR that is absorbed is indirectly thermalised at clouds mainly, and much of that goes into grey body radiation, 22% of which from the top escapes directly to space.
In short they have made 7 mistakes in basic physics, 3 of which are embarrassing to professionals like me because they have had to have been really dumb, or fraudulent. Everything that could go wrong has gone wrong, not least 1000s of pyrgeometers measuring the temperature of the atmosphere which is then put into the Energy Budget as real energy. You could not make up this level of incompetence nor the blind persistence in the face of real science in insisting there are no errors.

Phil.
December 17, 2012 7:52 am

Regarding light bulbs there are now higher efficiency incandescent bulbs where the glass envelope is coated by a dichroic layer which reflects IR back towards the filament. The dichroic passes visible however. This back reflected light heats the filament thus producing more light emission. The result of this is to be able to run the light bulb on lower current to produce the same visible output. It’s not practical to heat the filament to a higher temperature because it will melt but by recycling the IR output you get more visible out in lumens/watt.

Ryan
December 17, 2012 9:25 am

Willis: Sorry you are not right. When you are measuring the IR you are not measuring the radiated energy from the desk, say, to the thermal imaging camera. You are measuring the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius (i.e. absolute zero). So no, it is not an energy you can sense. There is no net energy transferred from the desk to you, for instance, even if the measurement tell you that it is emitting 400W/sqm. There is a small amount of net energy transferred from you to the desk because you are ever so slightly warmer, but it would really be negligible. Similarly there can be no net transfer of photonic energy from the ocean to the air if the air is already at the same temperature, otherwise this would violate the laws of entropy.

gymnosperm
December 17, 2012 9:33 am

The 340 comes from DLW but it doesn’t go any deeper than the skin. It is cycling so fast that in a quantum sense it is the same energy. Like cycling money between your checking to savings accounts and wondering why you’re still not broke. The small (@ 10%) losses from the eddy have been more than compensated for during the satellite era by incoming SW.

Ryan
December 17, 2012 9:50 am

1. The ocean absorbs DLR.
Correct.
2. The ocean ends up warmer with the DLR than it would be without it.
Wrong. Because the ocean is emitting at the same rate, being as it is at the same temperature as the air from which it is receiving radiated heat. There is no net flow of radiated heat from ocean to air at the boundary between the two. They are at the same temperature due to conduction. Conduction drives the process to the ocean and air until they are at the same temperature, and then the net radiation in both directions cancels out. In entropy terms they are at the same levels of disorder already due to conduction, so further changes in disorder due to radiation cannot occur.

Ryan
December 17, 2012 9:54 am

“To see that, just remove the sun. The rock cools.”
The sun is much hotter than the rock, so it can heat the rock. The air is not much hotter than the rock, so it cannot heat the rock. In any case, the hot rock is busily heating the air by conduction until the two are the same temperature. If the rock is colder than the air it will remove heat from the nearby air until they are at the same temperature. Once this happens radiation cannot play a part at all.

Ryan
December 17, 2012 10:02 am

Willis: Last comment for the evening. Do you appreciate that just because the atmosphere might be emitting 340W/sqm this only is relevant for a black body at -273Celsius? I.e. only a black body at -273Celsius can absorb this power? The Earth is not black body at -273Celsius, therefore it cannot absorb 340W/sqm. The figure of 340W/sqm is almost completely irrelevant. Deep space is at -273Celsius so losing the heat to space it is relevant, but not for the Earth.
Watts are a rate of heat transfer. If I have two bodies at the same temperature I don’t get a heat transfer so I don’t get any Watts to measure. Period. Doesn’t matter if it is conduction or radiation.

jae
December 17, 2012 10:18 am

Willis:
At 6:47 on 12/16 you wrote:
“Well, yeah … except that one of them is in the desert and one is not. As a result, they have very different responses to downwelling IR. There is an interesting discussion of these issues here.”
Well, as you say, “well, yeah…” The link doesn’t work so I don’t know what you are getting at. My point is exactly that one place is a desert and the other one is not!! Why is the desert so much hotter than the humid area, despite having only 1/4 the amount of the magic GHGs??
Please check link as I would certainly like to go there. I’ve been bringing this same issue up for about 5 years now, and still have not had any success in reconciling this very simple empirical observation with the postulated radiative GHE.
Thanks!
[Link fixed, thanks. -w]

AlecM
December 17, 2012 10:21 am

Willis: ‘You measure another physically real and measurable signal in the same band. They are about equal. As you would imagine, the difference between them is about zero.
Your conclusion?
Your conclusion from the fact that they are equal is that the signals that you measured are imaginary. They are not real. They do not contain energy.
You realize how crazy you are sounding?’

I am not mad, just a rather better physicist than most..Climate science is full of people who look at Stefan-Boltzmann and imagine they know everything. So let me explain things in a different way.
Your desk and you sofa are in radiative equilibrium. If you were to measure the energy arriving at a pyrgeometer sensor between them over 360°, it would be zero.
Now put a metal shield between the sensor and the desk. The pyrgeometer signal immediately jumps to the temperature signal for the sofa, the S-B level in a vacuum. But there is still no radiative transfer between the desk and the sofa so there is no energy flow from the pyrgeometer sensor to the sofa despite the output meter for a pyrgeometer being in units of power.
Do you get it now? Just because the pyrgeometer displays W/m^2, that signal is only true for transmission to absolute zero. The only reason you detect it is because the device shields the signal from the other direction. <It is an artefact which does not exist without the shield in place.
This most basic failure to understand what a pyrgeometer measures has meant climate science is based on an entirely false premise and has been so for 50 years because the Meteorologists and now climate scientists, even mainstream physics in poor institutions are taught false science.
Most of the IR energy in Trenberth’s Energy budget does not exist.
If anyone else cares to comment on this please do because I must ram this message home; climate science has fallen at the first hurdle and the alchemists in it are beavering away repeating the same mistake over and over again.. That is the real definition of madness – at least I vary my teaching approach!

gymnosperm
December 17, 2012 10:37 am

One must also account for the observation that the oceans continued to warm after 1997 without wayming the atmosphere. DLR by definition is absorbed by the atmosphere and would warm it equally. Further, one must account for the crazy quilt distribution of ocean surface warming and cooling Bob Tisdale has shown.

December 17, 2012 11:14 am

AlecM, I am perfectly comfortable with superposition where you isolate the effect of multiple independent sources and sum them. It’s a little more complex when the sources are dependent, but we have a strategy for that too…small signal analysis with preset initial conditions. Weather and climate are state machines where the next state is determined by input signals and the recorded history. I wish, like an academic, I could simply invent forcings and transfer functions to suit my pet theory, but sadly, as an engineer, I’m bound to physical reality. Greenhouse gases increase the Earth’s surface temperature by an average of 33C. Don’t worry about how that’s thermo-mechanically possible, just take their word for it.

AlecM
December 17, 2012 11:52 am

kencoffman: greenhouse gases do not cause the Earths temperature to rise by 33K. That was the first big scam of Hansenkoism/Houghtonism!
This is because adiabatic dry LR, Cp/g, is independent of the gas and because there is no mechanism for CO2 to increase water vapour concentration, there is no link with moist LR.
The real GHE is ~ 9 K, the remaining 24 K from gravitational potential energy. The 9 K is the heating of the surface because emissivity falls as GHG thermal emission switches of specific wavelength bands!

JazzyT
December 17, 2012 12:38 pm

AlecM,
You’ve really managed to confuse yourself on the whole issue of Poynting vectors and Poynting’s theorem. It’s a perfectly good way of analyzing electromagnetic waves to find the net flow of energy through a location, and the work done on charges in that location, and the change in energy density at that location. Energy can be radiated by an object while it simultaneously absorbs radiant energy from elsewhere; add these up and you’ll get the net energy gained or lost. If you work with energy, that’s a scalar. If you want to work with Poynting vectors, you can do so; it’s not that useful. But, recognize that for for IR, the emitted and absorbed radiation do NOT cancel each other to leave only their vector sum. The vector sum will give the amount and direction of the NET energy flow at some point, including any work done on electric charges. But that doesn’t negate the components. By the same token, if you deposit your paycheck and pay your rent, and then balance your checkbook, that does not mean that your rent is suddenly unpaid and your paycheck was cut; it just tells you how much is left over for other things.
Waves can add up and cancel; that’s destructive interference. They can also add up and reinforce, with constructive interference. Vector sums of electric fields are useful for this; but if you take the cross product of electric and magnetic fields to get the Poynting vector, you lose polarization information, and can’t analyze interference. Also, you’d need phase information, given by path length differences. Besides that, the Earth’s surface and atmosphere are not going to demonstrate interference anyway; as extended, thermal sources, they don’t exhibit the necessary spatial and temporal coherence.
Poynting vectors are great for anyzing the interactions of EM waves and conductors. But for heat transfer by IR, you really need to start with an outlook based on quantum physics. When IR is emitted or absorbed, this is the interaction between molecules and photons. Only rarely is it helpful to consider the IR as an EM wave interacting with wires or other configurations of conductors. So, the Poynting vector approach isn’t all that useful for heat transfer by IR. It’s still valid, as long as you don’t go tripping yourself up by confusing the vector sum of Poyning vectors at a point with the actual propigation of waves beyond that point, or confusing time-dependent vs. time-independent Poynting vectors, or similar confusions.
Here’s somthing to sum it up:
AlecM says:
December 17, 2012 at 1:00 am

Willis: ‘If DLR is not keeping the ocean from freezing, what is?’
The answer is very simple. The oceans do not emit 400 W/m^2 ULR. The only IR they emit is the net 60 W/m^2. This is an inescapable fact – only net energy flows are real.

No, both energy flows are real. IR from the surface can transfer energy to the atmosphere; IR from the atmosphere can transfer energy to the surface. Note, I said “transfer energy,” not “warm.” It’s only a warming process if the net heat transfer into an object is positive. But any object that radiates without absorbing at least an equal amount of radiation, will cool.
All energy flows can transfer heat, but only NET energy flows can give NET heat transfer. So, in the above, the net energy flow represents the net heat loss by the ocean, which is given by the difference between IR emitted upward and downward IR absorbed.
If you say that “the problem with this whole analysis is that they physicists don’t understand the physics,” then you’d better be a world-class physicist yourself. Otherwise, you probably missed something.

AlecM
December 17, 2012 12:54 pm

JazzyT: for a collimated beam, the Poynting Vector is identical to the Planck Irradiance Function at that wavelength.
The vector sum of the PVs UP and Down is simply a wavelength dependent version of the S-B1 – S-B2 calculation………..

D.I.
December 17, 2012 1:25 pm

JazzyT says:
“If you say that “the problem with this whole analysis is that they physicists don’t understand the physics,” then you’d better be a world-class physicist yourself. Otherwise, you probably missed something”.
So what is a ‘World Class Physicist’ ? Is It some-one who has discovered a ‘New Law’ or some-one who is Popular for reciting ‘Old Laws’?
What is your definition of ‘World Class’?.
Just curious.

December 17, 2012 1:57 pm

I also note that you don’t seem to understand that the temperature of the camera doesn’t change the flow of IR from the desk. The desk doesn’t know what the temperature of the camera is, it radiates the same amount of IR regardless of the temperature of the camera.
In other words, what you call “the infra-red energy that WOULD flow if the thermal imaging camera was at -273Celsius” is exactly the same as the amount of energy that WOULD flow from the desk if the thermal imaging camera was at room temperature.

If the camera was at a temperature lower than the desk energy would flow in the direction of the camera. This would be an extremely small amount of energy as the desk is radiating in other directions proportional to the area that the desk “sees”.
If the camera were hotter than the desk then energy would flow in the direction of the desk, proportional to the area that the camera sees, also providing that all other energy sources are at a lower temperature than the camera.
This is just basic thermodynamics.
Thuswise, if the ocean is warmer than the sky/atmosphere it is radiating toward, it will radiate energy. It does not matter how much downwelling IR there is, from CO2 or a cloud, if the temperature of the ocean is greater than what it is looking at it will radiate energy and thus cool. It will also conduct energy away through molecular interaction with the atmosphere. There will also be convection that will speed the process along. However, to answer your question is that in order for the ocean to freeze, enough energy at the surface has to radiate away so as to bring the temperature at the surface down to freezing at -2c. Even then 80 more calories per cubic cm of energy has to radiate away in order to remove enough energy to support the phase change from water to ice.
The ocean, if its temperature is less than the sky/atmosphere, will absorb energy from the atmosphere/sky. This is just basic thermodynamics.
In the Arctic and Antarctic there is a phenomenon called flash freezing whereby the rough water is continually radiating energy away due to the greater surface area of the water. The temperature can go well below -2c at the surface of the water. Then when the wind dies, tens of square kilometers of water will freeze (phase change to ice) in seconds. Google this, there are videos of it out there some where.

Greg House
December 17, 2012 3:11 pm

Willis Eschenbach says, December 17, 2012 at 12:24 am:
Greg, […] I see that in fact you agree that everything radiates.
It appears that you believe, however, that when radiation from something cooler is absorbed by something warmer than the spot where the radiation originated, no energy of any kind is transferred to the warmer object.
Is this in fact your belief? And if so …
[…]Put a 100-watt incandescent bulb in a fixture, […] Me, I can do it as a thought experiment, I don’t need to build the apparatus to be clear about what will happen.

========================================================
No, Willis, there was absolutely nothing in my comments that suggested what you said about my “belief”. But nevertheless thank you for this nice example of a straw-man argument.
You “thought experiment” is only an illustration of how you think or (want to make people to believe) how things work. It is not real, it is a product of fantasy and not a scientific argument at all. Your assertion about “NET” you made previously is not science, Willis, it is a fiction, because unproven experimentally. At best you can call it an “experimentally unproven hypothesis”, but not a scientific fact.
The fact that a colder body radiates in a warmer body’s direction ALONE does not prove any particular effect on this warmer body including effect on the warmer body’s TEMPERATURE.
Apparently, neither you nor warmists I talked to on this and other blogs can present a REAL experiment confirming that notion about “NET”. Which inevitably leads to the conclusion that there has never been one and the whole notion is just a fiction.